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[ "", null, "", null, "Home", null, "Puzzle Contest", null, "Single Shift Puzzle", null, "Flip Flat Rubik", null, "Double Shift Puzzle", null, "Rotate 3x3 Puzzle", null, "Rotate 5x5 Puzzle", null, "Rotate 7x7 Puzzle", null, "Nine Faces Rubik", null, "Six Faces Rubik", null, "24 Faces Magic Cube", null, "4D Rubik Cube", null, "Sudoku", null, "Quotes By Author", null, "Quotes By Subject", null, "Quote Of The Day", null, "Detective Mystery", null, "Reversi", null, "Logic Test", null, "Math Test", null, "Word Of The Day", null, "Puzzles", null, "Unsolved Problems", null, "Hall of Fame", null, "Contact Us", null, "Login", null, "Sign Up", null, "Sign Out", null, "My Account", null, "Change Password", null, "About Us", null, "Contributors", null, "Link To Us", null, "Privacy Policy", null, "Terms of Service", null, "Site Map\n\n## Flat Rubik - 4D Rubik Cube Puzzle Rules", null, "Flat 4D Rubik Cube Puzzle is a flat equivalent of a four-dimensional variation of the Rubik Cube puzzle. A four-dimensional space (4D) is an abstract mathematical generalization of a three-dimensional space.\n4D Rubik Cube consists of 8 cells. Each cell represents the original three-dimensional (3D) Rubik Cube. Cell #1 is presented on Figure 1. As the original 3D Rubik Cube, it consists of 6 faces. Each face has a different color.", null, "Figure 1\nA vector with four elements (xyzv) can be used to represent a position of a 4D cube in four-dimensional space. See more details in Appendix 1.\n\n4D Cube has 24 faces. Each face belongs to two cells (3D cubes). For example, the face 1.2 belongs to the cell #1. It has coordinates X=0, Y=0 (See Appendix 1). Face 2.1 belongs to the cell #2 and has coordinates X=0, Y=0. Since the face 2.1 has the same coordinates as the face 1.2, it is just a mirror of the same face.\n\nThe central square of each face displays two digits separated by dot. The first digit is the cell number. The second digit shows the cell number for the mirror face. For example, face 1.2 belongs to the cell #1, and its mirror (2.1) face belongs to the cell #2. In this example the mirror face 2.1 belongs to the cell #2, and its mirror face (1.2) belongs to the cell #1. All squares except the central squares show only the cell number.\n\nYou can rotate the cell by clicking on any of the central squares. There are four rotating options: Rotate One Layer Clockwise, Rotate One Layer Counterclockwise, Rotate 3D Cube (the whole cell) Clockwise, and Rotate 3D Cube (the whole cell) Counterclockwise. For the puzzle to be solved, all squares should return to their original places.\n\nRotate One Layer example\nFor example, if you select the ‘Rotate One Layer Clockwise' option and click on “1.2” square, the following elements of the 4D Cube will be moved (See Appendix 2 for details):\n\n• One layer attached to the face 1.2 in the cell #1 rotates clockwise. One layer rotation is similar to the rotation of the Six Faces Flat Rubik Puzzle (See Appendix 3)\n• One layer attached to the face 2.1 in the cell #2 rotates counterclockwise. One layer rotation is similar to the rotation of the Six Faces Flat Rubik Puzzle (See Appendix 3)\n• The edge AB in the cell #8 moves to the cell #3 and replaces the edge DA.\n• The edge BC in the cell #7 moves to the cell #8 and replaces the edge AB.\n• The edge CD in the cell #4 moves to the cell #7 and replaces the edge BC.\n• The edge DA in the cell #3 moves to the cell #4 and replaces the edge CD.\n\nThe results of the rotation are shown on Figure 2.\nFigure 2\n Cell #1(X=0)", null, "", null, "Cell #2(Y=0)", null, "", null, "Cell #3(Z=0)", null, "", null, "Cell #4(V=0)", null, "", null, "Cell #7(Z=1)", null, "", null, "Cell #8(V=1)", null, "", null, "Rotate 3D Cube (the whole cell) example\nIf you select the ‘Rotate Cell Clockwise' option and click on “1.2” square, the following elements of the 4D Cube will be moved:\n\n• The whole cell #1 rotates clockwise including the following elements: the face 1.2 rotates clockwise; the face 1.6 rotates counterclockwise; the face 1.3 moves to 1.4; face 1.4 moves to 1.7; the face 1.7 moves to 1.8; and the face 1.8 moves to 1.3.\n• One layer attached to the face 2.1 in cell #2 rotates counterclockwise\n• One layer attached to the face 6.1 in cell #6 rotates clockwise\n• One layer attached to the face 3.1 in cell #3 moves to the layer 4.1 in cell #4\n• One layer attached to the face 4.1 in cell #4 moves to the layer 7.1 in cell #7\n• One layer attached to the face 7.1 in cell #7 moves to the layer 8.1 in cell #8\n• One layer attached to the face 8.1 in cell #8 moves to the layer 3.1 in cell #3\n\nThe result of the rotation is shown on Figure 3.\nFigure 3", null, "", null, "", null, "Cell #1 Cell #2 Cell #3", null, "", null, "", null, "Cell #4 Cell #7 Cell #8\nAppendix 1\n\nThe Figure 4 below shows the relationship between 4D Rubik’s Cube faces, Flat Rubik faces, and four (xyzv) coordinates for each cell.\nFigure 4\n Cell #1: X = 0 Y=0 =>1.2 Y=1=>1.6 Z=0=>1.3 Z=1=>1.7 V=0=>1.4 V=1=>1.8", null, "", null, "Cell #2: Y = 0 X=0=>2.1 X=1=>2.5 Z=0=>2.3 Z=1=>2.7 V=0=>2.4 V=1=>2.8", null, "", null, "Cell #3: Z = 0 X=0=>3.1 X=1=>3.5 Y=0=>3.2 Y=1=>3.6 V=0=>3.4 V=1=>3.8", null, "", null, "Cell #4: V = 0 X=0=>4.1 X=1=>4.5 Y=0=>4.2 Y=1=>4.6 Z=0=>4.3 Z=1=>4.7", null, "", null, "Cell #5: X = 1 Y=0 =>5.2 Y=1=>5.6 Z=0=>5.3 Z=1=>5.7 V=0=>5.4 V=1=>5.8", null, "", null, "Cell #6: Y = 1 X=0=>6.1 X=1=>6.5 Z=0=>6.3 Z=1=>6.7 V=0=>6.4 V=1=>6.8", null, "", null, "Cell #7: Z = 1 X=0=>7.1 X=1=>7.5 Y=0=>7.2 Y=1=>7.6 V=0=>7.4 V=1=>7.8", null, "", null, "Cell #8: V = 1 X=0=>8.1 X=1=>8.5 Y=0=>8.2 Y=1=>8.6 Z=0=>8.3 Z=1=>8.7", null, "", null, "Appendix 2\n\nEach face has four edges and eight vertices. The vertices on the face 1.2 have the following coordinates: A(0,0,0,1), B(0,0,1,1), C(0,0,1,0), D(0,0,0,0). The vertices A, B, C, and D besides cell #1 belong to the cells #2, 3, 4, 7, and 8 (See the Figure 5).\n\nIf you select the ‘Rotate One Layer Clockwise' option and click on “1.2” square, the vertices will have the following coordinates: B(0,0,0,1), C(0,0,1,1), D(0,0,1,0), A(0,0,0,0) (See Figure 5 below).\nFigure 5", null, "", null, "", null, "Cell #1 (X=0) Cell #2 (Y=0) Cell #3 (Z=0)", null, "", null, "", null, "Cell #4 (V=0) Cell #7 (Z=1) Cell #8 (V=1)\nAppendix 3\n\nRotating one layer is similar to the rotation of the Six Faces Flat Rubik Puzzle. Figure 6 displays the one layer (attached to the face 1.2 in cell #1) rotation for the original Rubik Cube.", null, "Figure 6\nFigure 7 displays the equivalent of one layer rotation for the Flat Rubik Puzzle. It shifts and rotates clockwise 12 elements around the selected face 1.2.", null, "Figure 7" ]

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[ "# On the global structure of the Gromov-Hausdorff metric space\n\nThis is a purely idle question, which emerged during a conversation with a friend about what is (not) known about the space of compact metric spaces. I originally asked this question at math.stackexchange (https://math.stackexchange.com/questions/1356066/global-structure-of-the-gromov-hausdorff-space), but received no answers even after bountying.\n\nBackground. If $$A, B$$ are compact subsets of a metric space $$M$$, the Hausdorff distance between $$A$$ and $$B$$ is the worst worst best distance between their points: $$d_H(A, B)=\\max\\{\\sup\\{\\inf\\{d(a, b): b\\in B\\}: a\\in A\\}, \\sup\\{\\inf\\{d(a, b): a\\in A\\}: b\\in B\\}\\}.$$ For two compact metric spcaes $$X, Y$$, their Gromov-Hausdorff distance is the infimum (in fact, minimum) over all isometric embeddings of $$X, Y$$ into $$Z$$ via $$f, g$$ of $$d_H(f(X), g(Y))$$. The Gromov-Hausdorff space $$\\mathcal{GH}$$ is then the \"set\" of isometry classes of compact metric spaces, with the metric $$d_{GH}$$.)\n\nQuestion. How homogeneous is $$\\mathcal{GH}$$? For example: while distinct points in $$\\mathcal{GH}$$ are in fact distinguishable in a broad sense, it's not clear that distinct points can always be distinguished just by the metric structure of $$\\mathcal{GH}$$, so it's reasonable to ask:\n\nAre there any nontrivial self-isometries of $$\\mathcal{GH}$$?\n\nThere are of course lots of related questions one can ask (e.g., autohomeomorphisms instead of isometries); I'm happy for any information about the homogeneity of $$\\mathcal{GH}$$.\n\n• For any $r>0$, there is an autohomeomorphism of $\\mathcal{GH}$ that takes a compact metric space and scales its metric by $r$. Jul 26, 2015 at 19:46\n• On the first place: does $d_{GH}$ behaves as a distance? does it satisfy the triangle inequality? is $d(X,Y)>0$ whenever $X,Y$ are non-isometric? (Btw, the Hausdorff distance is for non-empty compact subsets, and hence $d_{GH}$ is also for non-empty compact metric spaces only.)\n– YCor\nJul 26, 2015 at 20:50\n• @EricWofsey yes, for autohomeomorphisms the question should be about (anything approaching a) classification rather than existence. Jul 26, 2015 at 22:55\n• @YCor I assume all metric spaces are nonempty as a rule :). Yes, $d_{GH}$ (restricted to the compact spaces!) is a metric, and in particular satisfies the triangle inequality; see e.g. helsinki.fi/~cristina/pdfs/gromovHausdorff.pdf Jul 26, 2015 at 22:57\n• @YCor: That the space GH is geodesic is proved in arxiv.org/abs/1504.03830. The question of whether GH is homeomorphic to $\\ell^2$ is Problem 1062 in [T.Banakh, R.Cauty, M.Zarichnyi, Open problems in infinite-dimensional topology. in: Open Problems in Topology, II (E.Pearl ed.), Elsevier, 2007, p.601–624] electronically available at franko.lviv.ua/faculty/mechmat/Departments/Topology/…. Jul 30, 2015 at 10:57\n\nTo answer the main question -- there are no nontrivial self-isometries of $\\mathcal{GH}$.\n\nI can give a proof of this, but as it is getting rather long, I will state some facts in $\\mathcal{GH}$ without proof for now, and will come back and provide provide proofs or references.\n\nFirst, a bit of notation. Let ${\\rm diam}(A)=\\sup\\{d(a,b)\\colon a,b\\in A\\}$ denote the diameter of a metric space $A$. For any $\\lambda > 0$ use $\\lambda A$ to denote the space with the same points as $A$, but with scaled metric given by $d_\\lambda(a,b)=\\lambda d(a,b)$. Let $\\Delta_n$ denote the (n-1)-simplex - a space consisting of $n$ points all at unit distance from each other. So, $\\Delta_1$ is the space consisting of a single point, and $\\lambda\\Delta_n$ is a space consisting of $n$ points all at distance $\\lambda$ from each other.\n\nNow, the following are standard, \\begin{align} &d_{GH}(\\Delta_1,A)=\\frac12{\\rm diam}(A),\\\\ &d_{GH}(A,B)\\le\\frac12\\max({\\rm diam}(A),{\\rm diam}(B)),\\\\ &d_{GH}(\\lambda A,\\mu A)=\\frac12\\lvert\\lambda-\\mu\\rvert{\\rm diam}(A). \\end{align} For reference, I am using Some Properties of Gromov–Hausdorff Distances by Fecundo Mémoli for the standard properties of $\\mathcal{GH}$. Then, the one-point space $\\Delta_1$ is distinguished just in terms of the metric on $\\mathcal{GH}$ as follows,\n\n1. $A$ is isometric to $\\Delta_1$ if and only if $d_{GH}(B,C)\\le\\max(d_{GH}(A,B),d_{GH}(A,C))$ for all $B,C\\in\\mathcal{GH}$.\n\nThat this inequality holds for $A=\\Delta_1$ follows from the first two standard properties above. On the other hand, if $A$ contains more than one point, so ${\\rm diam}(A)>0$, we can take $B=\\Delta_1$ and $C=\\lambda A$ for any $\\lambda > 1$, $$\\begin{eqnarray} d_{GH}(B,C)&=&d_{GH}(\\Delta_1,\\lambda A)=\\frac\\lambda2{\\rm diam}(A) \\\\ &>&\\frac12\\max(1,\\lambda-1){\\rm diam}(A)=\\max(d_{GH}(A,B),d_{GH}(A,C)) \\end{eqnarray}.$$ So, statement 1 holds in both directions. Therefore, any isometry fixes $\\Delta_1$ and, by the first standard property of $\\mathcal{GH}$ above, it preserves the diameter of spaces.\n\nNext, the simplexes $\\Delta_n$ are distinguished in terms of the metric as follows,\n\n2. The following are equivalent for any $A\\in\\mathcal{GH}$.\n\n• ${\\rm diam}(A)=1$ and $d_{GH}(B,C)\\le\\max(d_{GH}(A,B),d_{GH}(A,C))$ for all $B,C\\in\\mathcal{GH}$ with diameter less than or equal to $1$.\n• $A=\\Delta_n$ for some $n\\ge2$.\n\nThe proof of this is given below. So, an $\\iota$ is an isometry on $\\mathcal{GH}$ maps the set of simplifies $\\{\\Delta_n\\colon n\\ge2\\}$ to itself. Now, the finite spaces in $\\mathcal{GH}$ can be identified.\n\n3. For any $n\\ge2$ and $A\\in\\mathcal{GH}$, the following are equivalent.\n\n• $d_{GH}(A,B)=d_{GH}(\\Delta_n,B)$ for all $B\\in\\mathcal{GH}$ with $d_{GH}(\\Delta_n,B)={\\rm diag}(B)\\ge\\max({\\rm diag}(A),1)$.\n• $A$ is a finite space $n$ or fewer points.\n\nTherefore, if $\\mathcal{GH}_n$ denotes the finite metric spaces with $n$ or fewer points, (2) and (3) imply that $\\iota$ permutes $\\{\\mathcal{GH}_n\\colon n\\ge2\\}$. As it must preserve the inclusions $\\mathcal{GH}_n\\subset\\mathcal{GH}_{n+1}$, it follows that $\\iota$ maps $\\mathcal{GH}_n$ to itself for each $n\\ge2$.\n\nNow, fix $n\\ge2$, let $N=n(n-1)/2$, and $S$ be the subset of $\\mathbb{R}^N$ consisting of points $\\mathbf x=(x_{ij})_{1\\le i < j\\le n}$ with $x_{ij}>0$ and $x_{ik}\\le x_{ij}+x_{jk}$ for all distinct $i,j$ (here, I am using $x_{ij}\\equiv x_{ji}$ whenever $i > j$. Let $G$ be the group of linear transformations of $\\mathbb{R}^N$ mapping $\\mathbf x$ to $g_ \\sigma(\\mathbf x)=(x_{\\sigma(i)\\sigma(j)})$ for permutations $\\sigma\\in S_n$. Then, $S$ is a region in $\\mathbb{R}^N$ with nonempty interior bounded by a finite set of hyperplanes, and maps in $G$ take the interior of $S$ into itself. We can define $\\theta\\colon S\\to\\mathcal{GH}$ by letting $\\theta(\\mathbf x)$ be the space with $n$ points $a_1,\\ldots,a_n$ and $d(a_i,a_j)=x_{ij}$, and define a (continuous multivalued) function $f\\colon S\\to S$ by $\\theta\\circ f=\\iota\\circ\\theta$. Then, $\\theta$ maps $S$ onto the metric spaces with $n$ points, the values of $f(\\mathbf x)$ are orbits of $G$, and the fact that $\\iota$ is an isometry means that if $\\mathbf y=f(\\mathbf x)$ and $\\mathbf y^\\prime=f(\\mathbf x^\\prime)$, then $\\min_g\\lVert g(\\mathbf y)-\\mathbf y^\\prime\\rVert=\\min_g\\lVert g(\\mathbf x)-\\mathbf x^\\prime\\rVert$, with the minimum taken over $g\\in G$ and using the $\\ell_\\infty$ norm on $\\mathbb{R}^N$.\n\nNow, let $X\\subset\\mathbb{R}^N$ consist of the fixed points of elements of $G$, which is a finite union of hyperplanes, and $S^\\prime=S\\setminus\\ X$. Note that $f$ maps $S^\\prime$ into itself -- suppose that $\\mathbf y = f(\\mathbf x)\\in X$ for some $\\mathbf x\\in S^\\prime$. Then, $g(\\mathbf y)=\\mathbf y$ for some $g\\in G$. Choosing $f(\\mathbf x^\\prime)=\\mathbf y^\\prime$ arbitrarily close to $\\mathbf y$ with $h(\\mathbf y^\\prime)\\not=\\mathbf y^\\prime$ (all $h\\in G$), we have $\\mathbf x^\\prime$ and and $g(\\mathbf x^\\prime)$ arbitrarily close to $\\mathbf x$, so $g(\\mathbf x)=\\mathbf x$, contradicting the assumption. So, in the neighbourhood of any point in $S^\\prime$, we can take a continuous branch of $f$, in which case $f$ is locally an isometry under the $\\ell_\\infty$ norm, which is only the case if $f(\\mathbf x)=P\\mathbf x+\\mathbf b$ (where $P$ permutes and possibly flips the signs of elements of $\\mathbf x$, and $\\mathbf b\\in\\mathbb{R}^N$), with $P$ and $\\mathbf b$ constant over each component of $S^\\prime$. As $\\mathbf x\\to 0$, $\\theta(\\mathbf x)$ tends to $\\Delta_1$, from which we see that $\\mathbf b=0$.\n\nSo, we have $f(\\mathbf x)=P\\mathbf x$, and as the components of $f(\\mathbf x)$ are positive, $P$ is a permutation matrix. In order that $f$ is continuous across the hyperplanes in $X$, we see that $P$ is constant over all of $S^\\prime$ (choosing continuous branches of $f$ across each hyperplane). Then, $P^{-1}gP\\in G$, for all $g\\in G$, as $f$ is invariant under the action of $G$. So, $P$ is in the normalizer of $G$. Now, it can be seen that centraliser of $G$ in the group of permutations (acting on $\\mathbb{R}^N$ by permuting the elements) is trivial, which implies that its normaliser is itself (Permutation Groups, see the comment preceding Theorem 4.2B). Hence $P\\in G$, so $\\iota$ acts trivially on the spaces with $n$ points. As the finite spaces are dense in $\\mathcal{GH}$, $\\iota$ is trivial.\n\nI'll now give a proof of statement (2) above, for which the following alternative formulation of the Gromov-Hausdorff distance will be useful. A correspondence, $R$, between two sets $A$ and $B$ is a subset of $A\\times B$ such that, for each $a\\in A$ there exists $b\\in B$ such that $(a,b)\\in R$ and, for each $b\\in B$, there is an $a\\in A$ with $(a,b)\\in R$. The set of correspondences between $A$ and $B$ is denoted by $\\mathcal R(A,B)$. If $A,B$ are metric spaces then the discrepancy of $R$ is, $${\\rm dis}(R)=\\sup\\left\\{\\lvert d(a_1,a_2)-d(b_1,b_2)\\rvert\\colon (a_1,b_1),(a_2,b_2)\\in R\\right\\}.$$ The Gromov-Hausdorff distance is the infimum of ${\\rm dis}(R)/2$ taken over $R\\in\\mathcal R(A,B)$.\n\nNow, lets prove (2), starting with the case where $A=\\Delta_n$, some $n\\ge2$, for which we need to prove $$d_{GH}(B,C)\\le\\max(d_{GH}(\\Delta_n,B),d_{GH}(\\Delta_n,C))$$ whenever $B,C$ have diameter bounded by $1$. As we have, $d_{GH}(B,C)\\le1/2$, the required inequality is trivial unless $d_{GH}(\\Delta_n,B)$ and $d_{GH}(\\Delta_n,C)$ are strictly less than $1/2$, so we suppose that this is the case. Denote the points of $\\Delta_n$ by $p_1,p_2,\\ldots,p_n$. If $R\\in\\mathcal R(\\Delta_n,B)$ is such that ${\\rm dis}(R)/2 < 1/2$, then letting $B_i$ consist of the points $b\\in B$ with $(p_i,b)\\in R$, the sets $B_1,\\ldots,B_n$ cover $B$. For any $b,b^\\prime\\in B_i$ then $d(b,b^\\prime)=\\lvert d(p_i,p_i)-d(b,b^\\prime)\\rvert\\le{\\rm dis}(R)$, so the $B_i$ have diameters bounded by ${\\rm dis}(R)$. Also, for any $i\\not=j$, if $b\\in B_i\\cap B_j$ then ${\\rm dis}(R)\\ge\\lvert d(p_i,p_j)-d(b,b)\\rvert=1$, giving a contradiction. So, the $B_i$ are disjoint sets covering $B$. Similarly, if $S\\in\\mathcal R(\\Delta_n,C)$ has ${\\rm dis}(S)/2 < 1/2$ then we can partition $C$ into $n$ sets, $C_i$, of diameter bounded by ${\\rm dis}(S)$. Defining $T=\\bigcup_{i=1}^n(B_i\\times C_i)\\in\\mathcal R(B,C)$, it can be seen that ${\\rm dis}(T)\\le\\max({\\rm dis}(R),{\\rm dis}(S))$, from which the required inequality follows.\n\nNow, we prove the converse - if $A$ has diameter $1$ and is not isometric to $\\Delta_n$ for any $n$, then we can find spaces $B,C$ of diameter $1$ with $d_{GH}(B,C)=1/2$ and with $d_{GH}(A,B)$, $d_{GH}(A,C)$ strictly less than $1/2$. I'll consider first the case where $A$ is finite with $m\\ge2$ points, so $A=\\{a_1,\\ldots,a_m\\}$. As $A$ is not isometric to $\\Delta_m$, there must exist two points separated by less than unit distance. Wlog, take $d(a_{m-1},a_m)=x < 1$. Then, $m > 2$, otherwise $A$ would have diameter $x < 1$. We can define $R\\in\\mathcal R(A,\\Delta_{m-1})$ to be $\\{(a_i,p_i)\\colon i=1,\\ldots,m-1\\}\\cup\\{(a_m,p_{m-1})\\}$, which has discrepancy bounded by the max of $\\lvert d(a_i,a_j)-1\\rvert$ over $i\\not=j$ and $d(a_{m-1},a)m)=x$. So, $d_{GH}(A,\\Delta_{m-1}) < 1/2$. Next, we can define $R\\in\\mathcal R(A,\\Delta_m)$ to be the collection of pairs $(a_i,p_i)$ for $i=1,\\ldots,m$. Its discrepancy is the max of $d(a_i,a_j)-1$ over $i\\not=j$, which is strictly less than $1$, so $d_{GH}(A,\\Delta_m) < 1/2$. However, $d_{GH}(\\Delta_{m-1},\\Delta_m)=1/2$, so $B=\\Delta_{m-1}$ and $C=\\Delta_m$ satisfies the desired properties.\n\nNow, suppose that $A$ is not a finite space. For any $m\\ge2$ there exists a collection $a_1,a_2,\\ldots,a_m$ of $m$ distinct points in $A$. Then, by compactness, we can cover $A$ with a finite collection of nonempty sets $A_1,\\ldots,A_r$ of diameter bounded by $1/2$. Set $A_{r+i}=\\{a_i\\}$ for $i=1,\\ldots,m$. Let $S=\\{s_1,\\ldots,s_{m+r}\\}$ be the finite space with $d(s_i,s_j)=1$ for $i,j > r$ and with distance $1/2$ between all other pairs of points. The correspondence $R=\\bigcup_{i=1}^{m+r}(A_i\\times\\{s_i\\})$ has discrepancy $${\\rm dis}(R)=\\max\\left\\{1/2,1-d(a_i,a_j)\\colon 1\\le i < j\\le m\\right\\} < 1.$$ So, $S$ is finite with diameter $1$, contains a subset isometric to $\\Delta_m$, and $d_{GH}(A,S) < 1/2$.\n\nNow, we can let $B$ be any finite set with diameter $1$ and $d_{GH}(A,B) < 1/2$. If $m$ greater than the number of points in $B$, let $C$ be a finite set with diameter $1$, a subset isometric to $\\Delta_m$, and $d_{GH}(A,C) < 1/2$. Then, $d_{GH}(B,C)=1/2$, and satisfies the required properties, proving (2).\n\n• I'll come back and add the proofs of (2) and (3) when I have some time. Aug 5, 2015 at 1:03\n• Lets consider the map $\\phi$ on $\\mathcal{GH}$ with $\\phi(X)=$The hyperspace of $X$. However it is not surjective but I wonder if this is an isometry? Jun 13, 2018 at 10:37\n\nWe have written a text with more details on the proof given by George Lowther. Here one can see the result: https://arxiv.org/pdf/1806.02100.pdf However, locally there are many self-isometries, see https://arxiv.org/pdf/1611.04484.pdf" ]

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[ "", null, "### Popular Searches\n\n• ##### How To Calculate Belt Conveyor Speed\n\nWe have how to calculate belt conveyor speed,The standard speed for most unit handling conveyors is 65 FPM feet per minute which works out to the average speed a person walks when carrying a 50pound box This pace is ideal for manybut not allorder picking and assembly operations...\n\nKnow More\n• ##### How To Calculate Belt Pull and Power\n\nApr 27, 2018· Mike Gawinski explains how to calculate required belt pull and required conveyor drive power on a package handling belt conveyor , Speed, Torque & Horsepower - Duration: ....\n\nKnow More\n• ##### How To Calculate Belt Conveyor Speed\n\nHow To Calculate Belt Conveyor Speed The standard speed for most unit handling conveyors is 65 FPM feet per minute which works out to the average speed a person walks when carrying a 50pound box This pace is ideal for manybut not allorder picking and assembly operations...\n\nKnow More\n• ##### Line Speed Calculator\n\nWhile both companies are running the same number of containers per minute, they are running at different conveyor line speeds Calculating conveyor speed in FPM is actually quite easy Divide the number of containers per minute by the number of containers per foot to get line speed ,...\n\nKnow More\n• ##### Belt Velocity Calculator\n\ngetcalc's Belt Velocity Calculator is an online mechanical engineering tool to calculate the rotational speed of the belt, in both US customary & metric (SI) units Definition & Formula Belt Velocity is a measure of rotational speed of the belt at which the rotational force is being transferred from one pulley to another...\n\nKnow More\n• ##### How to calculate conveyor belt speed\n\nHow to calculate conveyor belt speed A conveyor usually consists of a rubber belt, serving as a platform, which is free to rotate around a series of separated rollers Objects and materials placed on top of the conveyor belt will be moved from one edge to the other as the rollers spin...\n\nKnow More\n• ##### How to Measure Conveyor Belt Speed with Encoders\n\nHow to Measure Conveyor Belt Speed with Encoders Rotary encoders are commonly used to measure belt speed in conveyor applications including synchronizing conveyors to pick-and-place operations and synchronizing the operation of multiple conveyor belts to one another...\n\nKnow More\n• ##### Conveyor Calc\n\nSorry - there is a problem with the database We're in the process of rectifying the problem....\n\nKnow More\n• ##### Calculation for conveyor belt speed\n\nCalculation for conveyor belt speed GOOD VALIDATION PRACTICE (cGVP) Manoj_Singh 2017-10-15 06:19:53 UTC #1 How to calculate conveyor belt speed in vial depyrogenated tunnelif we have following data-vial diameter ;machine output;conveyor width;length of hot zone Home ....\n\nKnow More\n• ##### How to Calculate Belt Length\n\nHow to Calculate Belt Length If the centre distance (C) is known, belt length or pitch length can be determined as follows: So if you know the distance between the center points of the pulleys, and their diameters, you can calculate the length of belt you need using this formula...\n\nKnow More\n• ##### Belt Speed\n\nNov 02, 2019· The Speed of a Belt calculator computes the speed at which a linear length of belt travels around a pulley based on the diameter (D) of the pulley and the rotation rate (R) Belt INSTRUCTIONS: Choose units and enter the following: (D) Diameter of the pulley(RPM) Rotation Rate of the pulleyBelt Speed (BS): The calculator computes the belt speed (BS) in meters per second...\n\nKnow More\n• ##### Belt Transmissions\n\nBelt Gearing The relationship between the rotational speed of the motor and the fan and the disc diameter can be expressed as d f n f = d m n m (3) Horsepower If belt tension and belt velocity are known - 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Dec 14, 2018 - The motor speed divided by the speed reducer ratio is the output speed of the reducer, that is, the pulley speed; It is also the speed of the drum, multiplied by the diameter of the drum multiplied by 314 is the speed of the belt per minute;...\n\nKnow More\n• ##### Conveyor Capacity\n\nConveyor capacity is determined by the belt speed, width and the angle of the belt - and can be expressed as Q = ρ A v (1) where Q = conveyor capacity (kg/s, lb/s) ρ = density of transported material (kg/m 3, lb/ft 3) A = cross-sectional area of the bulk solid on the belt (m 2, ft 2)...\n\nKnow More" ]

[ null, "https://www.postsv-husum.de/themes/images/banner-list.jpg", null ]

[ "Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.\n\n# Trapped air metamaterial concept for ultrasonic sub-wavelength imaging in water\n\n## Abstract\n\nAcoustic metamaterials constructed from conventional base materials can exhibit exotic phenomena not commonly found in nature, achieved by combining geometrical and resonance effects. However, the use of polymer-based metamaterials that could operate in water is difficult, due to the low acoustic impedance mismatch between water and polymers. Here we introduce the concept of “trapped air” metamaterial, fabricated via vat photopolymerization, which makes ultrasonic sub-wavelength imaging in water using polymeric metamaterials highly effective. This concept is demonstrated for a holey-structured acoustic metamaterial in water at 200–300 kHz, via both finite element modelling and experimental measurements, but it can be extended to other types of metamaterials. The new approach, which outperforms the usual designs of these structures, indicates a way forward for exploiting additive-manufacturing for realising polymer-based acoustic metamaterials in water at ultrasonic frequencies.\n\n## Introduction\n\nThe properties of a given material are typically characterized by a number of parameters such as electrical permittivity $$\\epsilon$$ and magnetic permeability $$\\mu$$ for electromagnetic waves, or by the density $$\\rho$$ and bulk modulus $$K$$ for acoustics. These parameters normally assume positive values for natural materials. In 1968, Veselago1 showed theoretically that phenomena such as negative refractive index could be obtained when these parameters are negative. Acoustic metamaterials (AMs) can exhibit exotic properties by using sub-wavelength features (referred as meta-atoms) in the form of resonators and scatterers. These are typically arranged in a periodic way so that they behave like a bulk continuous material but with ‘on-demand’ effective properties2,3,4,5,6,7,8,9,10,11,12,13,14,15,16. For example, Zhang et al.3 and Yang et al.4 describe materials exhibiting negative effective density $$\\rho_{eff}$$ and bulk modulus $$K_{eff} ,$$ resulting in a negative acoustic refractive index ($${\\upeta }_{eff} < 0$$).\n\nThe manipulation of $$\\rho_{eff}$$ and $$K_{eff}$$ for a given frequency $$\\left( f \\right)$$ and wavelength ($$\\lambda )$$ is of interest for ultrasonic imaging, in applications such as nondestructive evaluation17,18,19,20 and diagnostic biomedical imaging21,22,23. This arises because of the so-called “diffraction limit”, a consequence of the inability to capture the evanescent field which carries the finer sub-wavelength details of an image. The minimum feature scale $$w$$ that can be distinguished during an ultrasonic test is given via the relation1\n\n$$w \\cong c/\\left( {2 \\cdot f} \\right) \\cong \\lambda /2$$\n(1)\n\nwhere $$c$$ is the speed of sound of the medium. AMs can be used to overcome this diffraction limit, because the resolution of the image will be dictated by their sub-wavelength internal structure and the resulting exotic effective properties. For instance, Holey-Structured Acoustic Metamaterials (HSAMs) have been described which produce this enhanced sub-wavelength imaging resolution in air at audible and low ultrasound frequencies. This is achieved by the coupling of evanescent waves via Fabry–Pérot Resonance (FPR) mechanisms10,11,12. Sub-wavelength imaging is very attractive for many imaging methods, one example being near-field optical microscopy. In ultrasound, sub-wavelength imaging could lead to significant improvements. For example, one consequence for biomedical imaging is that lower frequencies could be used with an AM to obtain the same imaging resolution as that from a conventional measurement at higher frequencies, and this would allow an increased penetration into the body.\n\nIn their pioneering work24, Christensen et al. showed that FPRs exist within the holes of a 2D array of square-shaped apertures fabricated within a bulk material. Consider a HSAM having a thickness $$h$$, through-thickness channels of width $$a$$ (hole width), and a distance between hole centres, i.e. the lattice constant, $$\\Lambda .$$ Usually $$h > \\Lambda > a$$. The transmission process of acoustic waves through such HSAM is regulated by the resonant mode within each hole, and the transmission coefficient $$T^{00}$$ is unity when resonances occur within the through-thickness channels, so that\n\n$$\\lambda_{{FPR_{m} }} = \\frac{2 \\cdot h}{m} \\to \\left| {T^{00} } \\right| = 1,$$\n(2)\n\nwhere $$\\lambda_{{FPR_{m} }}$$ is the wavelength at resonance and $$m = 1,2, \\ldots ,N$$ with $$N \\in {\\mathbb{N}}$$.\n\nEquation (2) means that all the acoustic information, including that from evanescent waves, is transferred from one side of the metamaterial to the other whenever the perpendicular component of the incident field wavevector $$k_{ \\bot }$$ is such that $$\\frac{2\\pi }{{k_{ \\bot } }} = \\lambda_{{FPR_{m} }}$$, as the transmission coefficient has a modulus of unity11.The resolution is then not dictated by the global material properties but by the lattice constant $$\\Lambda$$, i.e. the final resolution can go well-beyond the limit set by Eq. (1); in addition, $$\\Lambda /a$$ also determines the $$\\rho_{eff}$$ of the HSAM11,12,25,26,27,28,29. Examples of imaging in air at resolutions of up to $$w$$/50 at audible frequencies have been reported11,12, where typically the substrate material of the HSAMs has a large acoustic impedance ($$Z$$) compared to the medium inside the hole, confining the acoustic energy within each channel. For the case of a polymer immersed in water, the difference in acoustic impedance is much less, resulting in an exchange of acoustic energy between the holes. This is likely to lead to a loss in performance, as the effect relies on independence of the channels to preserve the conditions for evanescent wave interaction. It is thus essential that there is a sufficient difference in $$Z$$ between the bulk material ($$Z_{b}$$) of the AM and the water-filled holes ($$Z_{water}$$) within which the acoustic signal is contained. Metallic substrates could be used; however, there is an attractiveness to using polymers as they are readily translated to additive manufacturing technologies, where the use of metallic substrates is limited due to the high aspect ratio $$h/a$$ of the holes and their relatively-high cost.\n\nAlthough some work has been reported showing polymer-based HSAMs operating in water30,31, the effects of acoustic coupling from the water-filled holes into the polymer substrate would be expected to degrade their performance. As an example, Amireddy et al.31 demonstrated operation in water at $$f = 250$$ kHz, but significant unwanted acoustic energy within the solid was seen, and an amplitude enhancement due to the additional evanescent wave contribution to the image was not observed, as would be expected for efficient operation of an AM11. Note that Estrada et al.13 reported that conventional polymer substrates would not be expected to be efficient for producing such devices for water-coupled applications.\n\nIn this work, we overcome the above-mentioned limitations and demonstrate that polymer-based HSAM structures can in fact be constructed for efficient use in water. This uses a new “trapped air metamaterial” (TAM) concept, where the acoustic impedance mismatch between a polymer and water is strongly enhanced if air is trapped within the bulk material in a particular way. This uses the fact that the acoustic impedance of air ($$Z_{air}$$) is very low (so that $$Z_{air}$$ << $$Z_{b}$$, $$Z_{water}$$), enhancing acoustic isolation. To gain a quantitative insight into the impedance mismatch problem, consider both nickel ($$\\rho_{nickel} = 8,900 \\,{\\text{kg}}\\,{\\text{m}}^{ - 3}$$, $$c_{nickel} = 5,600 \\,{\\text{m}}\\,{\\text{s}}^{ - 1}$$) and a typical polymer ($$\\rho_{polymer} = 1,400\\, {\\text{kg}}\\,{\\text{m}}^{ - 3}$$, $$c_{polymer} = 1,500 \\,{\\text{m}}\\,{\\text{s}}^{ - 1}$$), which are the types of materials to be compared in this paper. When used in air at room temperature ($$\\rho_{air} = 1.2\\, {\\text{kg}}\\,{\\text{m}}^{ - 3}$$, $$c_{air} = 343 \\,{\\text{m}}\\,{\\text{s}}^{ - 1}$$), the acoustic impedance mismatch of the two materials with respect to that of the air $$Z_{air}$$ is large, having values of:\n\n\\begin{aligned} & \\frac{{Z_{nickel} }}{{Z_{air} }} = 1.2 \\times 10^{5} ; \\\\ & \\frac{{Z_{polymer} }}{{Z_{air} }} = 5.1 \\times 10^{3} . \\\\ \\end{aligned}\n(3)\n\nThis results in efficient confinement of acoustic energy within each single channel in both cases9,10,11,12,13,32,33. However, if the same polymer was used with water-filled channels ($$\\rho_{water} = 1,000\\, {\\text{kg}}\\,{\\text{m}}^{ - 3}$$, $$c_{water} = 1,480\\, {\\text{m}}\\,{\\text{s}}^{ - 1} )$$, then we obtain\n\n$$\\frac{{Z_{polymer} }}{{Z_{water} }} = 1.4,$$\n(4)\n\nso that now energy will pass much more easily from water into the polymer substrate. However, if a layer of air is trapped between the channels and the polymer substrate, we obtain\n\n$$\\frac{{Z_{water} }}{{Z_{air} }} = 3.6 \\times 10^{3} ,$$\n(5)\n\ni.e. of the same order of magnitude as in Eq. (3), meaning that the novel strategy can be used successfully in water. This can be investigated in more detail via Finite Element Modelling, as showed in the next section. Note that the resultant TAM is expected to outperform the use of metal (nickel) for constructing the metamaterial; this is because:\n\n$$\\frac{{Z_{nickel} }}{{Z_{water} }} = 33.6 \\ll \\frac{{Z_{water} }}{{Z_{air} }}.$$\n(6)\n\nFor a last overview of the impedance mismatch issue, the theoretical transmission coefficients values $$t_{1 - 2} = 1 - \\left[ {\\left( {Z_{2} - Z_{1} )/(Z_{2} + Z_{1} } \\right)} \\right]^{2}$$ for the above mentioned combinations are:\n\n\\begin{aligned} & t_{polymer - water} = { }0.97; \\\\ & t_{nickel - water} = 0.11; \\\\ & t_{air - water} = 0.0011. \\\\ \\end{aligned}\n(7)\n\nThe very low value of $$t_{air - water}$$ is thus likely to improve the performance of a TAM when compared to the use of simple metal or polymer substrate. Finite element modeling (FEM) simulations and experimental measurements demonstrate both the imaging capabilities of the Trapped Air design and the inability of standard polymer HSAMs to operate well in water. An additional comparison to a nickel structure indicates that the proposed new TAM design is more effective.\n\n### Trapped air metamaterial design\n\nThe new TAM design was developed to allow a layer of air to completely surround each water-filled channel, in such a way that metamaterials could be constructed in polymer using vat photopolymerization (see \"Methods\" section). Figure 1a shows the construction of the two layers. This was designed with $$\\Lambda$$ = 1.2 mm, $$a = 0.8$$ mm, and $$h = 6$$ mm, producing FPRs in the $$f = 100 {-} 300$$ kHz range so as to match the available ultrasonic transducer operating bandwidth (see \"Methods\" section). The bottom layer to the left contains an array of hollow polymer rods, in this case with an air spacing between them. The second layer, shown to the right of Fig. 1a, contains an array of holes to match the channels in each of the hollow rods. This is inverted and placed on top of the first layer and the seal between the two layers made watertight. When immersed in water, the channels would then fill with water, but each would be separated from the other by the air trapped within the sealed device.\n\nA schematic diagram of an assembled TAM is shown in Fig. 1b, with the dimensions used for the FEM simulations also shown. It can be seen how the air layers act to insulate the water-filled channels acoustically from each other.\n\nA TAM was fabricated for experimental testing using the vat photopolymerization process, and a conventional HSAM was also printed using the same polymer for comparison. These were compared to a nickel HSAM fabricated using powder bed fusion techniques. The FEM simulations shown below used values of $$a_{I} =0.1$$  mm and $$a_{II} =0.2$$  mm respectively, these values being present in the fabricated polymer and metallic HSAMs. Due to limitations imposed during manufacture of the more complicated TAM structure, (see \"Methods\" section and Supplementary Information), the walls and air gap had to be thicker in this case ($$a_{I} = 0.2 {\\text{mm}} \\,\\,{\\text{and}} \\,\\,a_{II} = 0.6 {\\text{mm}})$$, and the overall device had fewer holes across its structure (16 × 16 compared to 24 × 24).\n\nThe devices were placed in water at room temperature together with an E-shaped aperture machined into a ~ 0.9 mm thick brass plate, as shown in Fig. 2. The arms of the “E” aperture had a width of 1 mm. At the expected simulated FPR of 246 kHz, λ in water is ≈ 6 mm. Hence, for operation at or close to this frequency the aperture was of λ/6 in size, which is beyond the conventional diffraction limit. An ultrasonic piezoelectric transducer was placed at ~ 150 mm from the brass slab and fed with a frequency modulated “chirp” signal over the 100–400 kHz frequency range. A miniature hydrophone (Precision Acoustics) of 0.2 mm active diameter collected the ultrasonic pressure field transmitted through the metamaterial at various locations across the xy plane.\n\nThe received signals were processed with Chirp Z-Transform algorithm33 and images obtained for comparison to FEM predictions.\n\n## Results\n\n### Finite element simulations\n\nA 2-D FEM was built to obtain the transmission coefficient $$\\left| {T^{00} } \\right|$$ of three designs — a TAM in polymer containing trapped air, and HSAMs with simply holes within either a polymer or nickel substrate, each having the same values of $$\\Lambda$$, $$a$$ and $$h$$ as shown in Fig. 1b. A plane wave acoustic field propagating along the z-axis direction was employed in the modelling together with a perfectly- matched layer domain. Predictions for $$\\left| {T^{00} } \\right|$$ were obtained over the $$f = 0 {-} 300$$ kHz range and are plotted in Fig. 3. The reader is referred to \"Methods\" section for more details of the FEM model. It is clear that both the new trapped air design and its bulky nickel counterpart lead to the successful establishment of FPRs, whilst a flattened behaviour is obtained for a standard polymer design. This indicates that, without this trapped air acoustic isolation, a severe disruption to the operation of the polymer metamaterial is likely to occur—this is in agreement with the predictions of Eqs. (3)–(5). Moreover, it can be noted that more resonant peaks are seen for the trapped air design with respect to its metal counterpart and to the theoretical prediction of Eq. (2). Thus, the TAM is likely to behave as a medium with an effective speed of sound $$c_{eff - TAM} = 972 \\,{\\text{m}}\\,{\\text{s}}^{ - 1}$$, being estimated from substitution of the value of 81 kHz in Eq. (2) for the first observed FPR mode (m = 1).\n\nMore insight into the impedance mismatch problem, and thus about the validity of the proposed trapped air design and the establishment of FPR within the holes, is found if $$x - z$$ sections of the structures are investigated. Figure 4a–c shows the absolute value of the pressure field at peak frequencies for the two HSAMs and the TAM respectively. While FPRs are well-established and isolated from each other in the TAM, the same it is not true for the polymer substrate HSAM, where FPRs are not well-established. This is due to widespread exchange of acoustic energy between individual channels. Finally note that the TAM design show more distinct standing waves/FPRs than those of a nickel HSAM, which agrees with the estimated value of $$c_{eff - TAM}$$.\n\n### Sub-wavelength imaging\n\n3-D FEM simulations were conducted to check whether sub-wavelength details could be resolved using the three designs, see Figs. 5 and 6. Figure 5 shows a series of two-dimensional sections of the total pressure field magnitude across $$x {-} y$$ planes for the polymer HSAM design at a fixed location but at different frequencies. In Fig. 5a a wide frequency range is analyzed: sub-wavelength details are lost over the whole range due to the global resonance of the whole channeled structure, but an enhanced transmission occurs around the predicted resonance frequency values. By zooming around the peak, see Fig. 5b, it is also confirmed that sub-wavelength details cannot be appreciated at frequencies where enhanced transmission occur.\n\nConversely, sub-wavelength imaging capabilities are exhibited at frequencies around a resonance by both nickel HSAM and polymer TAM designs, see Fig. 6a,b respectively. In particular, the TAM shows much better results than the nickel one as all the “E” edges, including the vertical one, are well-imaged.\n\n### Experimental results\n\nExperiments were performed in a water tank for all three designs, and the experimental images that were obtained are shown in Fig. 7a–c. Despite the larger resultant hole size and separation in the TAM, the results were excellent, showing the TAM design to be very effective compared to HSAM designs in either polymer or metal. The “E” details are not imaged by the polymer HSAM (Fig. 7a), as expected from the simulations of Fig. 5 and the impedance mismatch prediction of Eq. (4). Conversely, the finer details of the “E” shaped aperture are seen at frequencies close to the FPR frequency values for both the nickel HSAM (Fig. 7b), and TAM (Fig. 7c) designs. Good results are obtained for the TAM, noting that a shift toward the lower frequencies for the best image was obtained experimentally due to its slightly greater thickness ($$h$$ ~ 6.4 mm, see \"Methods\" section and Supplementary Information).\n\nThe earlier FEM simulations of Fig. 6 predicted that the TAM would outperform the metal HSAM in imaging performance for the same basic unit cell parameters. It should be noted that the HSAMs were simulated and fabricated with a 24 × 24 matrix of holes, based on a unit cell with solid walls (0.4 mm) between each 0.8 mm hole, but that the TAM measured experimentally had a 16 × 16 matrix over the same area. This is because the unit cell spacing had to be increased to allow the polymer walls trapping the air between each hole to be structurally sound. Despite this, the TAM still outperformed the two HSAMs experimentally. Note also that it is the dimensions of the water-filled holes that predominantly control the resonances and frequency response, and so it was essential to retain these dimensions (a and h) across all samples (both HSAMs and the TAM) for the comparison. The reduction in the density of the holes in a given area of the TAM would have adversely affected both the imaging resolution and the ρeff (which is a function of Λ/a11,12,25,26,27,28,29); even so, the TAM was still the best performing structure in our studies.\n\nFigure 8a shows the best image experimentally-obtained using the TAM, together with the results obtained from the 3-D FEM simulation (Fig. 8b), and a cross-section of the amplitude obtained at $$x = 10$$ mm for both the experiment and the simulations (Fig. 8c). A good agreement between the two is found, as the “E” profile is well-reconstructed. This further demonstrates the capability and potential of the trapped air design strategy. Note that the greater separation of the holes of the realised TAM device with respect to the FEM design results in a lower spatial resolution (see Fig. 8c). In fact, by considering the TAM as a periodic structure along the surface plane, $$\\pi /\\Lambda$$ determines the limit of the Brillouin zone, so the smaller is $$\\Lambda$$, the larger is the Brillouin zone11,12,25,26,27,28,29 . However, it is still evident that the TAM is still an excellent structure for imaging purposes.\n\n## Conclusions\n\nIt has been shown that the use of trapped air within an acoustic metamaterial fabricated from polymers in a standard additive manufacturing process can enable sub-wavelength imaging in water. The performance of the resulting TAMs has been modelled and shown to be similar to that obtained experimentally at ultrasonic frequencies in water. As expected, the air layers have greatly reduced acoustic cross-coupling from the water-filled channels into the polymer substrate, thus enabling a new class of acoustic metamaterial to be considered for ultrasonic sub-wavelength imaging, sound shielding/extraordinary transmission, and in other acoustic metamaterial designs where high-impedance mismatches between channels are strongly desired3,4,33.\n\n## Methods\n\nA detailed description of the materials and methods employed in the present research is reported below.\n\n### 2-D finite element modelling simulations\n\nA 2-D pressure acoustics frequency domain finite element model was realised in COMSOL Multiphysics to characterise the frequency behaviour of the tested HSAMs. Two different geometric arrangements were tested: one for simulating either polymer or nickel HSAMs, while the latter employed a volume of trapped air enclosed in a polymer shell. The geometry employed for the polymer and nickel cases is illustrated in Fig. 9.\n\nA control volume of 20 mm was used both before and after the metamaterial in the z-direction, and the whole geometry was surrounded by a 10 mm perfectly matched layer (PML) to simulate an infinite water tank, i.e. anechoic condition. The geometry was meshed so to have at least 10 elements to represent each channel width. A 1 Pa (peak-to-peak) plane wave travelling along the z direction was incident onto the AM at different frequencies in the 100–300 kHz range (at 100 Hz increments). The transmission coefficient $$\\left| {T^{00} } \\right|$$ of each structure was obtained as the ratio of the outlet pressure energy to that at the inlet. A different geometry was employed to demonstrate the effect of the trapped air inside the polymer, represented in white in Fig. 10.\n\n## 3D finite element modeling simulations for sub-wavelength imaging\n\nThe 3D model consisted of a plane wave travelling through an “E” shaped aperture having sub-wavelength thickness, carved out of a 1 mm thick brass plate. A plane wave radiation condition was applied to the first boundary in z direction and a soft boundary condition was applied to the last boundary in z direction to avoid reflections. Data were analysed by imaging the pressure field amplitude at 0.1 mm from the HSAMs outlet, see Fig. 11.\n\nThe TAM was manufactured using vat polymerisation containing a 16 × 16 array of holes with $$a = 0.8$$ mm, $$h = 6$$ mm, $$\\Lambda = 1.8$$ mm, $$a_{I} = 0.2$$ mm and $$a_{II} = 0.6$$ mm (see Fig. 1a for a diagram of dimensions). These differ from the FEM intend due to the manufacturing tolerances on wall thickness ($$a_{I}$$) and air gap ($$a_{II}$$) to allow reliable production. Vat polymerisation utilised Daylight precision hard white resin (25 µm layer height) and a Liquid Crystal Precision 1.5 3-D printer both supplied by the Photocentric Group. All additive manufacturing preparation required the controlled use of support material to prevent any support being applied to the holes which would occlude them. Vat polymerisation was also used for the polymer HSAM. Metallic HSAM was produced through powder bed fusion of Inconel 625 (a Nickel based alloy) with an array of 24 × 24 holes with $$a = 0.8$$ mm, $$h = 6$$ mm, and $$\\Lambda = 1.2$$ mm. This used a Renishaw AM250 SLM Powder Bed Fusion printer with a laser power of 200 W and a laser spot size of 150 µm. This conventional design was also manufactured for comparison in vat polymerisation to allow material comparison without trapped air gaps. This used the same materials and equipment settings as those used for the TAM.\n\n### Experimental setup\n\nThe experiments were conducted in a 400 mm long, 350 mm wide and 200 mm deep volume of water enclosed in a custom-made acrylic tank. The input voltage signal was generated by a NI PXI-5421 Arbitrary Waveform Generator, amplified by a NCA1000-2E amplifier and emitted by a custom made 25.4 mm diameter piezocomposite transducer. An ultrasonic chirp signal of 50 µs duration and with frequencies sweeping from 100 to 400 kHz was used. The radiated ultrasonic signal passed through a ⁓ 0.9 mm thick plate containing the sub-wavelength aperture in the form of the letter “E”. The through-transmitted energy was then collected by the tested metamaterials (one at a time), and transferred to its far side to be then acquired at a distance of ⁓ 0.15 mm from the outlet surface using a 0.2 mm diameter Precision Acoustics needle hydrophone. Signals were then captured using a National Instruments PXI-5122 14 Bit Digitizer sampling at a rate of 100 MS·s−1. Both the Waveform Generator and the Digitizer were enclosed in a NI PXI-1042Q chassis. A grid of 131 × 131 measurement points was acquired by means of a 3D motorized stage with steps of 0.14 mm for an area of 18.34 × 18.34 mm2. In order to approximate a flat far field wave front, the distance between source and hydrophone was chosen to be 150 mm. Note that the setup employed is similar to that described in34.\n\nFor each scanned point, the DC component was removed from the time domain waveform and the region of interest, selected via a rectangular time window, was converted into frequency domain data via a Chirp Z-Transform algorithm33.\n\n## Data availability\n\nThe datasets generated during and/or analysed during the current study are available from the corresponding author on reasonable request.\n\n## Code availability\n\nThe FEM simulations were performed using COMSOL 5.4 (licence no. 7077568). The code is available from ([email protected]) on reasonable request.\n\n## References\n\n1. 1.\n\nVeselago, V. G. The electrodynamics of substances with simultaneously negative values of ε and μ. Phys. Usp. 10(4), 509–514 (1968).\n\n2. 2.\n\nCummer, S. A., Christensen, J. & Alù, A. Controlling sound with acoustic metamaterials. Nat. Rev. Mater. 1(3), 16001 (2016).\n\n3. 3.\n\nZhang, X. & Liu, Z. Negative refraction of acoustic waves in two-dimensional phononic crystals. Appl. Phys. Lett. 85, 341–343 (2004).\n\n4. 4.\n\nYang, S. et al. Focusing of sound in a 3D phononic crystal. Phys. Rev. Lett. 93(2), 24301 (2004).\n\n5. 5.\n\nLi, J. & Chan, C. T. Double-negative acoustic metamaterial. Phys. Rev. E 70(5), 055602 (2004).\n\n6. 6.\n\nLiu, Z. et al. Locally resonant sonic materials. Science 289, 1734–1736 (2000).\n\n7. 7.\n\nFleury, R. & Alù, A. Extraordinary sound transmission through density-near-zero ultranarrow channels. Phys. Rev. Lett. 111, 055501 (2013).\n\n8. 8.\n\nLanoy, M. et al. Subwavelength focusing in bubbly media using broadband time reversal. Phys. Rev. B 91, 224202 (2015).\n\n9. 9.\n\nMolerón, M. & Daraio, C. Acoustic metamaterial for subwavelength edge detection. Nat. Commun. 6, 8037 (2015).\n\n10. 10.\n\nChristensen, J. & García de Abajo, F. J. Acoustic field enhancement and subwavelength imaging by coupling to slab waveguide modes. Appl. Phys. Lett. 97, 164103 (2010).\n\n11. 11.\n\nZhu, J. et al. A holey-structured metamaterial for acoustic deep-subwavelength imaging. Nat. Phys. 7, 52–55 (2011).\n\n12. 12.\n\nLaureti, S. et al. High-resolution acoustic imaging at low frequencies using 3D-printed metamaterials. AIP Adv. 6(12), 121701 (2016).\n\n13. 13.\n\nEstrada, H. et al. Extraordinary sound screening in perforated plates. Phys. Rev. Lett. 101(8), 084302 (2008).\n\n14. 14.\n\nGliozzi, A. S. et al. Proof of concept of a frequency-preserving and time-invariant metamaterial-based nonlinear acoustic diode. Sci. Rep. 9(1), 1–9 (2019).\n\n15. 15.\n\nYablonovitch, E. Inhibited spontaneous emission in solid-state physics and electronics. Phys. Rev. Lett. 58, 2059 (1987).\n\n16. 16.\n\nPendry, J. B. Negative refraction makes a perfect lens. Phys. Rev. Lett. 85(18), 3966 (2000).\n\n17. 17.\n\nRizwan, M. K. et al. Contextual application of pulse-compression and multi-frequency distance-gain size analysis in ultrasonic inspection of forging. J. Nondestr. Eval. 38(3), 72 (2019).\n\n18. 18.\n\nLaureti, S. et al. Detection of rebars in concrete using advanced ultrasonic pulse compression techniques. Ultrasonics 85, 31–38 (2018).\n\n19. 19.\n\nMason, T. J., Paniwnyk, L. & Lorimer, J. P. The uses of ultrasound in food technology. Ultrason. Sonochem. 3(3), S253–S260 (1996).\n\n20. 20.\n\nChemat, F. & Khan, M. K. Applications of ultrasound in food technology: processing, preservation and extraction. Ultrason. Sonochem. 18(4), 813–835 (2011).\n\n21. 21.\n\nGanpathi, I. S., So, J. Y. & Ho, K. Y. Endoscopic ultrasonography for gastric cancer. Surg. Endosc. Interv. Tech. 20(4), 559–562 (2006).\n\n22. 22.\n\nKane, D., Balint, P. V. & Sturrock, R. D. Ultrasonography is superior to clinical examination in the detection and localization of knee joint effusion in rheumatoid arthritis. J. Rheumatol. 30(5), 966–971 (2003).\n\n23. 23.\n\nGoddard, P. J. Veterinary ultrasonography (Acribia, Zaragoza, 2000).\n\n24. 24.\n\nChristensen, J., Martin-Moreno, L. & Garcia-Vidal, F. J. Theory of resonant acoustic transmission through subwavelength apertures. Phys. Rev. Lett. 101(1), 14301 (2008).\n\n25. 25.\n\nWiltshire, M. C. K., Hajnal, J., Pendry, J. B., Edwards, D. & Stevens, C. Metamaterial endoscope for magnetic field transfer: Near field imaging with magnetic wires. Opt. Expr. 11, 709–715 (2003).\n\n26. 26.\n\nBelov, P. A. & Silveirinha, M. G. Resolution of subwavelength transmission devices formed by a wire medium. Phys. Rev. E 73, 056607 (2006).\n\n27. 27.\n\nSilveirinha, M. G., Belov, P. A. & Simovski, C. R. Ultimate limit of resolution of subwavelength imaging devices formed by metallic rods. Opt. Lett. 33, 1726–1728 (2008).\n\n28. 28.\n\nShvets, G., Trendafilov, S., Pendry, J. B. & Sarychev, A. Guiding, focusing, and sensing on the subwavelength scale using metallic wire arrays. Phys. Rev. Lett. 99, 053903 (2007).\n\n29. 29.\n\nKawata, S., Ono, A. & Verma, P. Subwavelength colour imaging with a metallic nanolens. Nat. Photon 2, 438–442 (2008).\n\n30. 30.\n\nAmireddy, K. K., Balasubramaniam, K. & Rajagopal, P. Holey-structured metamaterial lens for subwavelength resolution in ultrasonic characterization of metallic components. Appl. Phys. Lett. 108(22), 224101 (2016).\n\n31. 31.\n\nAmireddy, K., Balasubramaniam, K. & Rajagopal, P. Deep subwavelength ultrasonic imaging using optimized holey structured metamaterials. Sci. Rep. 7, 7777 (2017).\n\n32. 32.\n\nXie, Y., Konneker, A., Popa, B. I. & Cummer, S. A. Tapered labyrinthine acoustic metamaterials for broadband impedance matching. Appl. Phys. Lett. 103(20), 201906 (2013).\n\n33. 33.\n\nMolerón, M., Serra-Garcia, M. & Daraio, C. Acoustic fresnel lenses with extraordinary transmission. Appl. Phys. Lett. 105(11), 114109 (2014).\n\n34. 34.\n\nRabiner, L., Schafer, R. W. & Rader, C. The chirp z-transform algorithm. IEEE Trans. Audio Electroacoust. 17(2), 86–92 (1969).\n\n35. 35.\n\nAstolfi, L. et al. Negative refraction in conventional and additively manufactured phononic crystals. In IEEE International Ultrasonics Symposium, 2529–2532 (2019).\n\n## Acknowledgements\n\nThe Authors are thankful to Mr. Frank Courtney (Senior Technician, School of Engineering, University of Warwick, UK) for his invaluable help in the 3D printing of the polymer and Trapped Air metamaterial. Funding for this work was provided through the UK Engineering and Physical Sciences Research Council (EPSRC), Grant Numbers EP/N034163/1, EP/N034201/1 and EP/N034813/1.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nDr. S.L. conceived the idea of trapping air inside a Holey-Structured Acoustic Metamaterial, designed and performed experiments and finite element modelling, interpreted and compared the experimental data and the model, wrote and revised the manuscript and guided the project. Prof. D.A.H. and Prof. M.R. supervised, guided, founded the project and revised the paper. Prof. M.R. has also developed the post-processing scripts for experimental data interpretation. PhD student Mr. L.A. designed the metamaterials, performed experiments and finite element modelling, wrote and revised the paper. Dr. R.L.W. supervised the design and fabrication of the metamaterials and revised the manuscript. Prof. P.J.T. gave conceptual advice and revised the paper. Prof. P.B. reviewed the work. Dr. L.N., Prof. S.F., Dr. M.A. and Prof. A.T.C. were project collaborators, the last two also being in charge of the design and fabrication of the metal sample.\n\n### Corresponding author\n\nCorrespondence to Stefano Laureti.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare no competing interests.\n\n### Publisher's note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Rights and permissions\n\nReprints and Permissions", null, "", null, "" ]

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[ "# How to determine which bulb will be the brightest in the series and parallel connection?\n\nI stumbled upon these sentences while studying and this seems to be horribly confusing.\n\nresistance is proportional to the inverse of the power of the bulb in series connection and so the bulb with the lowest wattage(power) will have maximum resistance and it will glow the brightest.\n\nBut what i don't seem to understand is that the brightness of the bulb should depend on the power or heat dissipated(in case of an incandescent light bulb) then why is it that in this case the bulb with the lowest power or the highest resistance glows the brightest.\n\nMoreover in the next paragraph, which is for bulbs in parallel connection, it is said that\n\nAs the resistance of the highest wattage (power) bulb is minimum, it will glow the brightest.\n\nAnother question that arises at this point is that why are the two paragraphs contradicting?\n\nFrom both the paragraphs it is clear that the relationship used is\n\nresistance is proportional to the inverse of the power $$P ∝ \\frac{1}{R}$$\n\nbut then why wasn't the other two formulas used\n\n$$P= VI = {I^2}R$$\n\naccording to\n\n$$P= {I^2}R$$\n\nresistance is directly proportional to the electric power so shouldn't the power increase with increase in resistance or vice-versa.\n\nOne thing that is to be kept in mind is that the bulbs were manufactured for working of the same voltage.\n\n• Can you tell us where you got these claims from? Was it a book, a lecture or a website? – user137661 Feb 12 '19 at 18:03\n• It is from a book. I think there is something wrong. – ayush paul Feb 12 '19 at 18:04\n• Which book is it? Which page? We need more context in order to understand the question you're posing and the possible answers. Books can be wrong, but also students can misunderstand paragraphs, that is why I am asking. – user137661 Feb 12 '19 at 18:13\n• Well i am not sure if it would actually help because this book is only available in India. But the book is new simplified physics class 12 by S.L. Arora Dhanpatrao publications, page no. 3.62 chapter current electricity , topic: power consumption in a combination of appliances. – ayush paul Feb 12 '19 at 18:24\n• Re, \"...resistance is proportional to the inverse of power...\" The power dissipated by a resistor can be computed as $I^2R$. If several resistors are connected in series, then the current must be the same in each one. Therefore, the power dissipated by each one will be directly proportional to its $R$, not inversely proportional. (NOTE: I say \"resistor\" instead of \"bulb\" because I am ignoring the fact that the resistance of an incandescent light bulb filament depends on its temperature.) – Solomon Slow Feb 12 '19 at 18:24\n\n. . . . the bulb with the lowest wattage(power) will have maximum resistance and it will glow the brightest.\n\n$$R= \\dfrac {V^2}{P}$$ so for a given supply voltage $$V$$ the bulb with the higher power rating will have a lower resistance.\n\nWhen two bulbs are connected in series to a power supply, the current $$I$$ through both bulbs is the same.\nAs the power dissipated in a bulb is $$I^2\\,R$$ the bulb with the higher resistance (lower power when connected directly across the supply) will dissipate the greater power.\n\nThe analysis above assumes that the resistance of the bulbs does not depend on the voltage across them.\nThis is in fact not true and if the voltage does not change by too much, the variation of the resistance of a tungsten filament bulb is approximately $$R \\propto V^{0.5}$$ and the power dissipated $$P \\propto V^{1.5}$$.\n\nThere is more about the properties of such a bulb in the answer to the question entitled Car headlight voltage.\n\nWhen the book says:\n\nAs the resistance of the highest wattage (power) bulb is minimum, it will glow the brightest.\n\nWhat it is referring to is the power rating of the bulb. This power rating assumes the bulb has a potential difference across it of 120 V (in North America at least - maybe 220 V in India). So, if you compare a bulb rated at 60 watts and a bulb rated at 100 watts the 100 watt bulb will have a lower resistance and draw more current when they are connected in parallel (they way they are designed to used).\n\nIf, on the other hand, you connect the bulbs in series, then what is constant is the current not the voltage. The bulb with the highest power rating, which has the lowest resistance, will thus have the smaller voltage drop across it and thus will dissipate the smaller amount of power.\n\n• I really like this answer! I was confused about this for some time, but now it's all clear. Thanks! – Apekshik Panigrahi Aug 24 '19 at 10:13" ]

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[ "### Difference between foo(void) and foo()\n\nSource: http://stackoverflow.com/\n\nProblem:\nConsider these two function definitions\n``````void foo(){ ... }\nvoid foo(void){ ..... }``````\nWhat is the difference between these two functions?\n\nHint: The answer depends whether this is C code or C++ code.\n\n1.", null, "The difference is this,\nfoo(void) is the same in C/C++ a function with no arguments.\nfoo() in c++ also means the same as above.\nfoo() in c means that we there is not specified about the arguments of function foo() and about the type.\n\n2.", null, "Taken from the book :The Complete Reference C++ - Page 626 - Edition #4\n\n---\nint f();\nIn C this means that the function f may or Might not have any parameters !!\n--\n---\nIn C++ \"Super Set Of C\"\nint f() is same as int f(void)\nbecause void parameter list is optional in C++\nBut programmers put void inside () making it clear that no parameters are accepted by function !!\n\nExample ::\n\nC Code :\n#include\nvoid f()\n{\nputs(\"hello there\");\n}\nint main()\n{\nf(66);\n}\n----\n\nThe above code will compile without error and print \"hello there\"\n\nC++ example:\n\n#include\nusing namespace std;\nvoid f( )\n{\ncout<<\"from f\"<<endl;\n}\nint main()\n{\nf(3);\nreturn 0;\n}\n\nThe above code will say that(compiler)\ntoo many arguments supplied !!\n---" ]

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[ "# Download PDF by J. W. S. Cassels, A. Frohlich: Algebraic Number Theory: Proceedings of an Instructional", null, "By J. W. S. Cassels, A. Frohlich\n\nThis publication offers a brisk, thorough remedy of the principles of algebraic quantity conception on which it builds to introduce extra complicated themes. all through, the authors emphasize the systematic improvement of innovations for the specific calculation of the elemental invariants corresponding to earrings of integers, type teams, and devices, combining at each one level thought with particular computations.\n\nRead or Download Algebraic Number Theory: Proceedings of an Instructional Conference Organized by the London Mathematical Society (A Nato Advanced Study Institute W) PDF\n\nSimilar number theory books\n\nNew PDF release: The Prime Numbers and Their Distribution (Student\n\nWe now have been eager about numbers--and top numbers--since antiquity. One extraordinary new course this century within the research of primes has been the inflow of principles from chance. The target of this booklet is to supply insights into the major numbers and to explain how a series so tautly decided can contain one of these impressive volume of randomness.\n\nMathematical Modeling for the Life Sciences by Jacques Istas PDF\n\nProviding quite a lot of mathematical versions which are at the moment utilized in existence sciences can be considered as a problem, and that's exactly the problem that this e-book takes up. in fact this panoramic examine doesn't declare to supply an in depth and exhaustive view of the numerous interactions among mathematical types and lifestyles sciences.\n\nGet The Theory of Algebraic Number Fields PDF\n\nThis publication is a translation into English of Hilbert's \"Theorie der algebraischen Zahlkrper\" most sensible often called the \"Zahlbericht\", first released in 1897, within which he supplied an elegantly built-in review of the improvement of algebraic quantity thought as much as the tip of the 19th century. The Zahlbericht supplied additionally an organization beginning for extra learn within the topic.\n\nAdditional resources for Algebraic Number Theory: Proceedings of an Instructional Conference Organized by the London Mathematical Society (A Nato Advanced Study Institute W)\n\nExample text\n\nAs kL 3 kL, =3 k,, and as (kL : kLO) is a power of p it follows that f(L,/L,) = 1. Also e(L1/LO) is prime to p. Thus Li is tamely ramified over L,. In the sequel we shall use repeatedly the fact that, for any tower E 3 F 13 K of fields, EJK is tamely ramified if and only if E/F and F/K are tamely ramified. This follows from the definition of tame ramification. As a first consequence we conclude the L, and its subfields are tamely ramified over K. Let L’ be a subfield of L, tamely ramified over K.\n\nNow write a H a for the residue class map S + S/‘Q LEMMA 2. t&a) Proof. Remark: (4) = et,,,,(a). By Lemma 1, with A = S/pS, N = \\$3/p% Analogously one can show that N&a) = Nk,&V. PROPOSITION 4. uL(z)) 2 e-l. Proof. Write again A = SJpS, N = p/pS and denote by I the image in A of an element x of S, Choose a k-basis (al) of A, so that for 1 5 i I (e- 1)f the a, form a k-basis of N. We can lift (ai} back to an R-basis {xl} of S, so that jY, = a,. e. e. 4 2 (e - l>f, LOCAL 21 FIELDS L is said to be non-ramified over K, if (i) e(L/K) = 1.\n\nUK(bi) > 0 for all i. COROLLARY. K has a totally ramified extension of prescribed degree e. Proof. Let uK(c) = 1 and E(X) = xp--cx-c. It is worth mentioning some results, which follow on from Proposition 1, but which we shall not be able to give in this course. (See Serre, Ch. e. F z k). Here the characteristics of k and K coincide. Conversely if this is the case then K is (to within value isomorphism) the field of formal power series over k. There remains the case when x = p # 0 but the characteristic of K is zero." ]

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[ "# 1. It is desired to have the first joint of a six-axis robot to move from the initial, ? 0 = 15°, to\n\n1.       It is desired to have the first joint of a six-axis robot to move from the initial, θ0 = 15°, to a final position, θf = 75°, in 3 seconds using a cubic polynomial.   (a) Determine the trajectory.   (b) Calculate joint angle at 2 seconds.   (c) Comment on its end point velocities and accelerations.   (d) Compare the present result with those obtained in Example 12.1.\n\n2.       Repeat the problem in Exercise 12.1 but this time initial acceleration and final deceleration are specified as 5°/s2 .\n\n3.       A two-degree-of-freedom planar robot is to follow a straight line between the start (3,10) cm and the end (8,15) cm points of the motion segment. Find the joint variable for the robot if the path is divided into 5 segments. Assume each link is 10 cm long.\n\n4.       Find the coefficients of two cubic polynomials which are connected in a two-segment spline with continuous acceleration at the intermediate via points.\n\n5.       A single cubic trajectory is given by", null, "which is used between t = 0 and 1 s. What are the starting and final positions, velocities, and accelerations?", null, "## Plagiarism Checker\n\nSubmit your documents and get free Plagiarism report\n\nFree Plagiarism Checker" ]

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[ "", null, "", null, "### Home > PC > Chapter 8 > Lesson 8.1.4 > Problem8-61\n\n8-61.", null, "", null, "", null, "Find the slope of the line and the y-intercept. Write the equation in y = mx + b form.", null, "", null, "Find the area under the curve after 1 minute and then after 5 minutes.\n\n198 feet", null, "$A = \\frac{1}{2}(3 + (3 + 0.01T ))T = 3T + 0.005T ^{2}$" ]

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[ "", null, "Succeed with maths – Part 1\n\nStart this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.\n\nFree course\n\n# 6 Order of operations\n\nSince many everyday problems that you encounter require the use of more than one operation, you need to make sure you know how to correctly proceed, write and carry out the calculation.\n\nThe five operations you have looked at so far are joined by brackets when considering order of operation.\n\nBrackets indicate the highest priority, followed by any exponents (powers). Then carry out the multiplication and division, and finally any addition and subtraction. If part of the calculation involves only multiplication and division or only addition and subtraction, work through from left to right.\n\nThe correct order to carry out the operations can be summarised by using the mnemonic BEDMAS, where the letters stand for Brackets, Exponents, Division, Multiplication, Addition and Subtraction.\n\nThe following everyday problem involving more than one operation demonstrates the importance of BEDMAS.\n\nLet’s suppose you purchased four very large pepperoni pizzas that cost £15.99 each and you want to split the total cost among six people. One way to do this is by using addition and division.\n\nYou might think that you can enter this into your calculator as:\n\n15.99 + 15.99 + 15.99 + 15.99 ÷ 6.\n\nThis will give you £50.635 as the answer! That is clearly not correct as each pizza only costs £15.99.\n\nThe calculator has actually worked out:\n\n15.99 + 15.99 + 15.99 + (15.99 ÷ 6) = 47.97 + 2.665 = 50.635. This is because it follows BEDMAS, dividing only the last £15.99, not the total, by six. To obtain the correct answer you must use brackets:\n\n(15.99 + 15.99 + 15.99 + 15.99) ÷ 6 = 10.66\n\nNow the total cost of the pizzas is divided by 6 (remember, B comes before D in BEDMAS), to give a much more sensible answer of £10.66.\n\nYou can now see how important it is to get the correct order for calculations.\n\nNow, it is your turn to have a go. Try applying this rule in the next activity.\n\nBefore you start this activity you need to ensure that your calculator does know the BEDMAS rules. So, check this by working out 3 + 5 × 2 , without pressing the equals sign until after the final 2. If your calculator knows the rules the answer will be 13, if it doesn’t it will give you 16.\n\nBut don’t worry! Just be aware of the rules and work through the calculations carefully yourself.\n\n## Activity _unit3.6.1 Activity 7 BEDMAS on paper versus the calculator\n\nTiming: Allow approximately 10 minutes\n\nFirst of all, try these examples on paper without using a calculator. Then, check your answer using a calculator.\n\n(a) (3 + 4) × 2\n\n### Comment\n\nRemember BEDMAS. Are there brackets? If so, then do the calculation inside those first. Next, look for exponents, then multiplication, division, addition, and subtraction.\n\n(a) (3 + 4) × 2\n\nCarry out the calculation in brackets first: 3 + 4 = 7. Now multiply by two and the answer is 14.\n\n(b) 2 + 32\n\n(b) 2 + 32\n\nNo brackets, this time, so start with the exponent: 32 = 9. Add the two and you get 11.\n\n(c) (2 + 3)2\n\n(c) (2 + 3)2\n\nThis looks like part (b), but this time there are brackets, so you must do the calculation inside the brackets first: 2 + 3 = 5. Now square the result, and the answer is 25.\n\n(d) 32 + 42\n\n(d) 32 + 42\n\nWork out the exponents first: 32 = 9 and 42 = 16. Finally add 9 + 16 = 25.\n\n(e) 2 + 3 × 4 + 5. Take care with this one!\n\nIt can help to make calculations like this easier to read if you put brackets around the part that you need to do first.\n\n(e) 2 + 3 × 4 + 5\n\nThis time, you have addition and multiplication, so you must do the multiplication first: 3 ×4 = 12.\n\nSo now, the calculation is 2 + 12 + 5 = 19.\n\nUsing brackets you would write 2 + (3 × 4) + 5.\n\nThis will give you the same answer but just might make your job easier!\n\nThe last few sections have been activities that were not related to any real world problems so could seem a little abstract in nature. Now, you will apply this maths to the real world again and also continue your work on problem solving skills from Week 1." ]

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[ "# Codeforces Round #412 (rated Archive\n\n## Codeforces Round #412 (rated, Div. 1, based on VK Cup 2017 Round 3), problem: (F) Test Data Generation, Accepted Solution In C/C++\n\n#include<bits/stdc++.h> #define ll long long ll multmod(ll a, ll b, ll moder){ a = (a % moder + moder) % moder, b = (b % moder + moder) % moder; ll ret = 0; for …\n\n## Codeforces Round #412 (rated, Div. 1, based on VK Cup 2017 Round 3), problem: (E) Blog Post Rating Solution In JAVA\n\nimport static java.lang.Double.parseDouble; import static java.lang.Integer.parseInt; import static java.lang.Long.parseLong; import static java.lang.Math.min; import static java.lang.System.exit; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.Random; import java.util.StringTokenizer; public class E { static BufferedReader …\n\n## Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3), problem: (F) Perishable Roads Solution In JAVA\n\nimport java.io.BufferedReader; import java.io.BufferedWriter; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.StringTokenizer; public class Div2_412F { public static void main(String args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); PrintWriter printer = …\n\n## Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3), problem: (E) Prairie Partition Solution In C/C++\n\n#include<bits/stdc++.h> using namespace std; int n; long long a; int check(int x){ int i,j,res,ress,t,sum; long long A,AA; i=0;ress=0; for(j=0;j<60;j++){ A=(1LL<<j); res=0; for(;i<n;i++){ if(a==A){ res++; } else if(a<A){ ress++; } else break; } if(ress>x) return 0; …\n\n## Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3), problem: (D) Dynamic Problem Scoring Solution In C/C++\n\n#include<bits/stdc++.h> using namespace std; int a; int cnt; int n; int getsc(int id, bool f, int s) { if(a==-1) return 0; int sum; if(a == -1 || a == -1 || a<a) sum = cnt; …\n\n## Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3), problem: (B) T-Shirt Hunt Solution In C/C++\n\n#include <bits/stdc++.h> using namespace std; int p,x,y,i; int luc(int s){ s = (s/50)%475; for (int i=0;i<25;i++) if((s=(s*96+42)%475)+26==p)return 1; return 0; } int main(){ cin>>p>>x>>y; for(i=x%50;!luc(i)||i<y;i+=50){} return cout<<(i<=x?0:(i-x+99)/100),0; }\n\n## Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3), problem: (A) Is it rated? Solution In C/C++\n\n#include <iostream> using namespace std; int n,i,a,a1,b,f; string s={“maybe”,”unrated”,”rated”}; int main(){ cin>>n; for(;i<n;i++){ cin>>a>>b; if(a!=b){f=2;break;} if(i>0&&a>a1)f=1; a1=a; } cout<<s; }" ]

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[ "Theory and Modern Applications\n\n# Estimation on Certain Nonlinear Discrete Inequality and Applications to Boundary Value Problem\n\n## Abstract\n\nWe investigate certain sum-difference inequalities in two variables which provide explicit bounds on unknown functions. Our result enables us to solve those discrete inequalities considered by Sheng and Li (2008). Furthermore, we apply our result to a boundary value problem of a partial difference equation for estimation.\n\n## 1. Introduction\n\nVarious generalizations of the Gronwall inequality [1, 2] are fundamental tools in the study of existence, uniqueness, boundedness, stability, invariant manifolds, and other qualitative properties of solutions of differential equations and integral equation. There are a lot of papers investigating them (such as ). Along with the development of the theory of integral inequalities and the theory of difference equations, more attentions are paid to some discrete versions of Gronwall-Bellman-type inequalities (such as ). Some recent works can be found, for example, in and some references therein.\n\nWe first introduce two lemmas which are useful in our main result.\n\nLemma 1.1 (the Bernoulli inequality ).\n\nLet", null, "and", null, ", then", null, ".\n\nLemma 1.2 ().\n\nAssume that", null, "are nonnegative functions and", null, "is nonincreasing for all natural numbers, if for all natural numbers,", null, "(1.1)\n\nthen for all natural numbers,", null, "(1.2)\n\nSheng and Li considered the inequalities", null, "(1.3)\n\nwhere", null, "for", null, ".\n\nIn this paper, we investigate certain new nonlinear discrete inequalities in two variables:", null, "(1.4)", null, "(1.5)", null, "(1.6)\n\nwhere", null, "for", null, ".\n\nFurthermore, we apply our result to a boundary value problem of a partial difference equation for estimation. Our paper gives, in some sense, an extension of a result of .\n\n## 2. Main Result\n\nThroughout this paper, let", null, "denote the set of all real numbers, let", null, "be the given subset of", null, ", and", null, "denote the set of nonnegative integers. For functions", null, ", their first-order differences are defined by", null, ",", null, ", and", null, ". We use the usual conventions that empty sums and products are taken to be 0 and 1, respectively. In what follows, we assume all functions which appear in the inequalities to be real-value,", null, "and", null, "are constants, and", null, ".\n\nLemma 2.1.\n\nAssume that", null, "are nonnegative functions defined for", null, ", and", null, "is nonincreasing in each variable, if", null, "(2.1)\n\nthen", null, "(2.2)\n\nProof.\n\nDefine a function", null, "by", null, "(2.3)\n\nThe function", null, "is nonincreasing in each variable, so is", null, ", we have", null, "(2.4)\n\nUsing Lemma 1.2, the desired inequality (2.2) is obtained from (2.1), (2.3), and (2.4). This completes the proof of Lemma 2.1.\n\nTheorem 2.2.\n\nSuppose that", null, "and", null, "are nonnegative functions defined for", null, ",", null, "satisfies the inequality (1.4). Then", null, "(2.5)\n\nwhere", null, "(2.6)\n\nProof.\n\nDefine a function", null, "by", null, "(2.7)\n\nFrom (1.4), we have", null, "(2.8)\n\nBy applying Lemma 1.1, from (2.8), we obtain", null, "(2.9)", null, "(2.10)\n\nIt follows from (2.9) and (2.10) that", null, "(2.11)\n\nwhere we note the definitions of", null, "and", null, "in (2.6). From (2.6), we see", null, "is nonnegative and nonincreasing in each variable. By applying Lemma 2.1, the desired inequality (3.3) is obtained from (2.9) and (2.11). This completes the proof of Theorem 2.2.\n\nTheorem 2.3.\n\nSuppose that", null, "and", null, "are nonnegative functions defined for", null, ",", null, "satisfies", null, "(2.12)\n\nwhere", null, ", and", null, "satisfies the inequality (1.5). Then", null, "(2.13)\n\nwhere", null, "(2.14)", null, "(2.15)\n\nProof.\n\nDefine a function", null, "by", null, "(2.16)\n\nThen, as in the proof of Theorem 2.2, we have (2.8), (2.9), and (2.10). By (2.12),", null, "(2.17)\n\nIt follows from (2.8), (2.9), (2.10), and (2.17) that", null, "(2.18)\n\nwhere we note the definitions of", null, "and", null, "in (2.14) and (2.15). From (2.14) we see", null, "is nonnegative and nonincreasing in each variable. By applying Lemma 2.1, the desired inequality (2.19) is obtained from (2.9) and (2.18). This completes the proof of Theorem 2.3.\n\nTheorem 2.4.\n\nSuppose that", null, "are the same as in Theorem 2.3,", null, "satisfies the inequality (1.6). Then", null, "(2.19)\n\nwhere", null, "(2.20)", null, "(2.21)\n\nProof.\n\nDefine a function", null, "by", null, "(2.22)\n\nThen, as in the proof of Theorem 2.2, we have (2.8), (2.9), and (2.10). By (2.12),", null, "(2.23)\n\nIt follows from (2.8), (2.9), (2.10), and (2.23) that", null, "(2.24)\n\nwhere", null, "and", null, "are defined by (2.20) and (2.21), respectively. From (2.20), we see", null, "is nonnegative and nonincreasing in each variable. By applying Lemma 2.1, the desired inequality (2.19) is obtained from (2.9) and (2.24). This completes the proof of Theorem 2.4.\n\n## 3. Applications to Boundary Value Problem\n\nIn this section, we apply our result to the following boundary value problem (simply called BVP) for the partial difference equation:", null, "(3.1)", null, "satisfies", null, "(3.2)\n\nwhere", null, "and", null, "are constants,", null, ", functions", null, "are given, and functions", null, "are nonincreasing. In what follows, we apply our main result to give an estimation of solutions of (3.1).\n\nCorollary 3.1.\n\nAll solutions", null, "of BVP (3.1) have the estimate", null, "(3.3)\n\nwhere", null, "(3.4)\n\nProof.\n\nClearly, the difference equation of BVP (3.1) is equivalent to", null, "(3.5)\n\nIt follows from (3.2) and (3.5) that", null, "(3.6)\n\nLet", null, ". Equation (3.6) is of the form (1.4), here", null, ". Applying our Theorem 2.2 to inequality (3.6), we obtain the estimate of", null, "as given in Corollary 3.1.\n\n## References\n\n1. Bellman R: The stability of solutions of linear differential equations. Duke Mathematical Journal 1943,10(4):643–647. 10.1215/S0012-7094-43-01059-2\n\n2. Gronwall TH: Note on the derivatives with respect to a parameter of the solutions of a system of differential equations. Annals of Mathematics 1919,20(4):292–296. 10.2307/1967124\n\n3. Agarwal RP, Deng S, Zhang W: Generalization of a retarded Gronwall-like inequality and its applications. 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Tamkang Journal of Mathematics 2001,32(3):217–223.\n\n## Acknowledgments\n\nThis work is supported by Scientific Research Foundation of the Education Department Guangxi Province of China (200707MS112) and by Foundation of Natural Science and Key Discipline of Applied Mathematics of Hechi University of China.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Wu-Sheng Wang.\n\n## Rights and permissions\n\nReprints and Permissions\n\nWang, WS. Estimation on Certain Nonlinear Discrete Inequality and Applications to Boundary Value Problem. Adv Differ Equ 2009, 708587 (2009). https://doi.org/10.1155/2009/708587\n\n• Accepted:\n\n• Published:\n\n• DOI: https://doi.org/10.1155/2009/708587\n\n### Keywords\n\n• Differential Equation\n• Integral Equation\n• Partial Differential Equation\n• Ordinary Differential Equation\n• Functional Analysis", null, "" ]

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[ "## Mathematics Paper 1 Questions and Answers - Bungoma Diocese Mock Exams 2021/2022\n\nINSTRUCTIONS TO CANDIDATES\n\n• The paper contains TWO sections: Section A and B\n• Answer ALL questions in section I and STRICTLY ANY FIVE questions from section II.\n• All working and answers must be written on the question paper in the spaces provided below each question.\n• Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.\n• Marks may be awarded for correct working even if the answer is wrong.\n• Non-programmable silent electronic calculators and KNEC Mathematical tables may be used except where stated otherwise.\n\nSECTION A (50MKS)\nANSWER ALL THE QUESTIONS IN THIS SECTION\n\n1. Simplify  (1³/₇ − ⁵/₈) + ²/₃ of 1¹/₅     (4mks)\n¾ + 1⁵/₇ ÷ ⁴/₇ of 2¹/₃\n2. A straight line ax + by = 16 passes through A (2, 5) and B (3, 7). Find the values of a and b (3mks)\n3. Simplify     2−x−x     (3mks)\n3x2−2x−1\n4. Solve for X where 0 ≤ x ≤ = 90°   (2mks)\nsin 2x − cos (x−30) = 0\n5. Solve for X in                    (3mks)\n2x − 4  ≤ 3x + 2 < 10 − x\nHence represent your solution on a number line\n6. Two similar cylindrical solids have heights of 18 cm and 24 cm. The volume of the larger cylinder is 320cm3, find the volume of the smaller cylinder (4mks)\n7. Solve for X      (3mks)\n83x − 2 x 16½x  = ¼\n8. A quantity P varies jointly as Q and inversely as on the square root of R. If Q is increased by 10% and R is reduced by 19%, find the percentage change in P (3mks)\n9. Okedi sold goods whose marked price is sh. 340,000 at a discount of 2%. He was paid sh. 16660 as commission for the total sales. Calculate the percentage rate of commission (3mks)\n10. The interior angle of a regular polygon is three and a half times the exterior angle. Determine the sides of the polygon (3mks)\n11. Give that A =", null, ", B =", null, ";find matrix C where AC = B (3mks)\n12. Amoit bought 2 pens and 5 exercise books at a cost of sh. 275. Allan bought 4 such pens and exercise books from the same shop at a cost of sh. 415 by letting sh. X and y to be the costs of a pen and a book respectively, find the cost of each item (4mks )\n13. Okech left some money in his will to be shared amongst his wife, son and daughter in the ratio 4:3:2 respectively. If the daughter received sh. 120,000 less than the mother’s share, find the total amount of money Okech left in his will. (2mks)\n14. Use tables of reciprocals to find the reciprocal of 0.3758. Hence find the value 3√0.125 correct to 4.S.f (4mks)\n0.3758\n15. A major sector of a circle subtends an angle of 150 at the centre. The radius of the circle is 7cm and the centre is at O as shown", null, "If the sector is folded into a conical shape, calculate the radius of the cone correct to 1 d.p (3mks)\n16. A Kenyan bank buys and sells currencies at the exchange rates below\n Currency Buying (ksh) Selling (ksh) 1 euro 1 us dollar 147.87    74.22 148.00   74.50\nAn American tourist arrived in Kenya with 24,000 Euros. He converted all the euros to Kenya shillings at the bank. He spent a total sh. 200,000 while in Kenya and converted the rest into US dollars at the bank. Find the amount in dollars that he received. (3mks)\n\nSECTION II (50MKS)\nANSWER ANY FIVE QUESTIONS IN THIS SECTION\n\n1. The diagonals of a rectangle P, Q, R, S intersect at (5, 3). Given that the equation of line PQ is 4y − 9x =13 and that of line PS is y − 4x =5\n1. The co-ordinators of P (3mks)\n2. The co-ordinates of R (2mks)\n3. The equation of line RQ (2mks)\n4. The equation of a perpendicular line drawn to meet PR at (5,3) (3mks)\n2. A bus left Malaba town at 6.00am and travelled at an average speed of 80km/h towards Nairobi which is 510km away. At 6.30am a salon car left Nairobi the same day following the same route and travelled at average speed of 100km/h towards Malaba. After 1 hour, the car had a puncture which took 15minutes to repair before proceeding with the journey;\nDetermine\n1. The distance covered by the bus in 30minutes (1mks)\n2. The time of the day when car met the bus. (6mks)\n3. The distance from Nairobi to the point where the car met with the bus (2mks)\n4. The time of the day to the nearest minute when the bus got to Nairobi (1mk)\n3. Points P, Q and R are a straight line on a level ground. An electricity pole is erected at P with a point X and Y on it. From point X, the angle of depression of point Q is 48° while the angle of depression of R from Y which is 3m above X is 60°\n1. Illustrate the position of X, Y, P and R by sketching. (1mk)\n2. Hence calculate to 1 d.p.\n1. The length XP (3mks)\n2. The distance YQ (2mks)\n3. The distance PQ (2mks)\n4. The angle of elevation of Y from R given that PR = 8cm (2mks)\n4.\n1. The figure shows a velocity- time graph of a car", null, "1. Find the total distance covered by the car in metres (3mks)\n2. Calculate the deceleration of the car (3mks)\n2. A lorry left kisumu at 8.00am and travelled towards the Nakuru at an average speed of 72km/h. At 8.30am a matatu left kisumu and followed the lorry at an average speed of 96km/h.\nDetermine the time of the day when the matatu caught up with the lorry (4mks)\n5. The table below shows marks scored by 48 students in a geography exam.\n Marks % 30-39 40-49 50-59 60-69 70-79 80-89 Students 6 10 x 9 12 2\n1. Determine the value of x (2mks)\n2. State the modal class (1mk)\n3. Calculate the (3mks)\n1. Mean mark\n2. Median mark (4mks)\n6.\n1. Complete the table below for the equation Y = x2 + 3x − 6 where −7 ≤ × ≤ 4\n x −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 y 4 −6 −2\n(3mks)\n2. Using the scale 1 cm to represent 1 unit on the X- axis and 1cm to represent 2 units on the Y - axis, draw the graph of y = x2 + 3x − 6 for −7≤ ×≤4 (4mks)\n3. Use your graph to solve for x in\nx2 + 3x − 6 =0 (2mks)\n4. State the;\n1. Turning point of the curved (1mk)\n2. Equation of the line symmetry (1mk)\n7. the figure shows triangle ABC inscribed in a circle where AC = 10cm, BC = 7cm and AB =11cm", null, "Calculate correct 1 d p ( use π = 22/7 )\n1. The size of the angle CAB (4mks)\n2. The radius of the circle (2mks)\n3. Hence, find the area of the shaded region (4mks)\n8. ABCDEFGA is a belt tied around two wheels whose centres are O and Q forming a pulley system. Given that Q =36cm, AO = 5cm BQ = 7cm. calculate correct 1 d.p (Take π = 22/7 )", null, "1. Angle AOQ (3mks)\n2. The length of the belt in contact with\n1. The wheel whose centre is O (2mks)\n2. The wheel whose centre is Q (2mks)\n3. The length of AB, hence the total length of the belt (3mks)", null, "## MARKING SCHEME\n\n1. Num (¹⁰/₇ − ⁵/₈) + ²/₃ of ⁶/₅\n= 80 − 35 + 4\n56       5\n45/564/5\n= 225 + 224\n280\n449\n280\nDen  ¾ + ¹²/₇ ÷ ⁴/₇ of ⁷/₃\n= ¾ + ¹²/₇ × ⁷/₄ of ¾\n= ¾ + ¹²/₇ × ¾\n= 21 + 36\n28\n57/28\n449/280 × 28/57449/570\n2. 2a + 5b = 16\n3a + 7b = 16\n6a + 15b = 48\n6a + 14b = 32\nb = 16\na = −32\n3. 2  − x  − x2 = 2  − 2x + x  − x2\n= 2(1  −  x) + x ( 1  − x)\n= (2 + x)(1 − x)\n3x2  − 3x + x  − 1\n= 3x(x  − 1) +1(x  − 1)\n=(3x+1)(x −1)\n∴  =(2 + x) (1 − x2)\n(3x + 1)(x  − 1)\n=  − 2 + x\n3x + 1\n4. 2x + x − 30 = 90\n3x = 120\nx = 40\n5. 2x − 4 ≤ 3x + 2\n−6 ≤ x\n3x + 2 < 10 − x\n4x < 8\nx < 2", null, "6. LSF = 18/24 = ¾\nVSF = (¾)327/64\n27/64x/320\nx = 135cm3\n7.  23(3x − 2) x 24(½x)  = 4−1\n23(3x − 2) x 24(½x)  = 2−2\n9x − 6 + 2x = −2\n11x = 4\nx = 4/11\n8.  P ∝\n√R\nP = KQ\n√R\nNew P =    1.1 P\n√0.81\n= 1.1 P = 1.2222P\n0.9\n%Increase = 0.2222 × 100%\n= 22.22%\n9.  Sales = 100 − 2 × sh 340,000\n100\n= 98/100 × sh 340,000\n= Sh 333,200\n%Commission =  16660  × 100%\n333200\n=5%\n10. Let ext angle = x\n3½x + x = 180°\n4.5x = 180\nx = 40°\n11. AC = B\nC = A−1 = B", null, "12. 2x + 5y = 275\n4x + 7y = 415\n8x + 20y = 1100\n8x + 14y = 830\n6y = 270\ny = 45\nx = 25\n13. W : S : D\n4 : 3 : 2\nLet total amount = x\n(4/9 − 2/9)x = sh 120,000\n2/9x = 120,000\nx = Sh 540,000\n14. 0.3758 =    1\n0.3758\n=          1\n1 × 3.758 × 10−1\n= 1 × 0.2661 × 101\n= 2.661\n3√0.125 = 0.05\n∴ = 2.661 ×  0.5\n= 1.3305\n= 1.331\n15.  Area = 150/360 ×  22/7 ×  7 ×  7\n= 64..17cm2\n∴ 64.17 = 22/7 × R × 7\n2.917 = R\n2.9 = R\n16.  Amount on arrival = 24000  × 147.87\n= Sh. 3,548,880\nBalance = Sh. 3,548,880 − Sh 200,000\n= Sh 3,348,880\n74.50\n= 44951dollars\n17.", null, "1. y − 4x = 5 ......(i)\n4y − 9x = 13 .....(ii)\n4y − 16x = 20\n− 4y − 9x = 13\n−7x = 7\nx = −1\ny = 1\nHence P (−1,1)\n2. x = 2(5) − (−1) = 11\ny = 2(3) −1      = 5\nHence R(11,5)\n3. SP // RQ\nHence M1 = M2 = 4\n∴ 4 = y − 11\nx − 5\ny − 11 = 4(x−5)\ny = 4x − 20 + 11\ny = 4x − 9\n\n4. PR ∴ P(−1,1) and (5,3)\nGrad = M1 = 3−1 = 2 = 1\n5+1    5    3\nM2  = −3\nHence −3 = y − 3\nx −5\ny−3 = −3(x − 5)\ny = −3x + 15 + 3\ny = − 3x + 18\n18.", null, "1. Distance covered = 80 × ½h             ½h + 1h + ¼h = 1¾h\n= 40km\n2. Relative distance = 510 − (80 × 1¾ + 100 × 1)\n= 510 − (80 × 7/4 + 100)\n= 510 − (140 + 100)\n= 510 − 240\n= 270km\nTime taken =    270km           = 270\n(100 + 80)km/h    180\n= 1½hrs\nTime of day = 6.00am + ½hr + 1hr + ¼hr + 1½hr\n= 6.00am + 3hr 15mins\n= 9.15a.m\n3. Distance from NRB = 100 × 1 + 100 × 3/2\n= 100 + 150\n= 250km\n4.  Time taken = 510km  = 6.375hr\n80km/h\n= 6hr 23mins\nTime of day = 6.00am + 6hrs 23 mins\n= 12.23pm\n19.\n\n1.", null, "2.\n1. Tan 42 = h/t\nh1 = t tan 42°\nTan 30 =   h\nt + 3\nh2 = tan 30(t+3)\nHence h1 = h2\nt tan 42 = tan 30(t+3)\n0.9004t = 0.5774(t+3)\n0.9004t = 0.5774t + 1.7322\n0.9004t − 0.5774t = 1.7322\n0.323t = 1.7322\nt = 1.7322\n0.323\n= 5.363\n≅ 5.4m\n2. Cos 30° = 8.362\nYR\nYR = 8.362\ncos 30\n= 9.656 = 9.7m\n3. PQ = (3 + 5.362) tan 30°\n= 8.362 tan 30\n= 4.828 = 4.8m\n4. Tan x = 8.362\n8\nx = tan−1 (8.362)\n8\n= 46.27°\n= 46.3°\n20.\n21.\n22.\n1.\n x −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 y 22 12 4 −2 −6 −8 −8 −6 −2 4 12 22\n\n2.", null, "23.\n1. Area of √14(14−11)(14−10)(14−7)                     Alternative method\nΔ ABC = √(14 × 3 × 4 ×7                                  Cos A = 11² + 10² − 7²\n=√1156                                                               2 × 11 × 10\n= 34.292                                                           = 172/220 = 0.7818\nA = ½ab sin θ                                                            A = Cos−1 (0.7818)\n34.292 = ½ × 11 × 10 sin A                                          = 38.57\nSin A = 34.292 × 2                                                    = 38.6°\n110\n= 0.6235\nA = sin−1 0.6235\n= 38.57\n= 38.6°\n2 sin A\n=         7\n2 sin 38.57\n= 5.614\n= 5.6\n3. Shaded Area = πr2 − ½ab sinθ                              Area of circle and evidence of subtraction\n= 22/7 × 5.614² − 34.292\n= 99.05 − 34.292\n= 64.758\n= 64.8\n24.", null, "Sin x = 2/36 = 0.05556\nx = sin−1 (2/36)\n= 3.185°\n1. < AOQ = 90 + 3.185\n= 93.185 = 93.2\n<AOE = 360 − 2 × 93.185\n= 360 − 186.37\n= 173.63   ≅ 173.6\n2.\n1. Arc length = θ/360 × 2πr\n= 173.63 × 2 × 22 × 5\n360              7\n= 15.16      ≅ 15.2\n2. Arc length =  186.37 × 2 × 22 × 5\n360              7\n= 22.78   ≅ 22.8\n3. AB = √(362 − 22)\n= √1292\n= 35.94    ≅ 35.9\nTotal length = 15.16 + 22.78 + 2 × 35.94\n= 109.82\n\n• ✔ To read offline at any time.\n• ✔ To Print at your convenience\n• ✔ Share Easily with Friends / Students" ]

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[ "", null, "", null, "", null, "Main Page | See live article | Alphabetical index\n\nWaring's Problem, proposed in 1770 by Edward Waring, asks whether for every natural number k there exists an associated positive integer s such that every natural number is the sum of at most s kth powers of natural numbers. The affirmative answer was provided by David Hilbert in 1909. Sometimes this topic is described as Hilbert-Waring's theorem.\n\nFor every k, we denote the least such s by g(k). Note we have g(1) = 1. Some simple computations show that 7 requires 4 squares, 23 requires 9 cubes, and 79 requires 19 fourth-powers. Waring conjected that these values were in fact the best possible.\n\nLagrange's Four Square Theorem from 1770 states that every natural number is the sum of at most four squares; since three squares are not enough, this theorem establishes g(2) = 4. Lagrange's Four Square Theorem was conjectured by Fermat in 1640 and was first stated in 1621.\n\nOver the years various bounds were established, using increasingly sophisticated and complex proof techniques. For example, Liouville showed that g(4) is at most 53. Hardy and Littlewood showed that all sufficiently large numbers are the sum of at most 19 fourth powers.\n\nThat g(3) = 9 was established from 1909 to 1912 by Wieferich and A. J. Kempner, g(4) = 19 in 1986 by R. Balasubramanian, F. Dress, and J.-M. Deshouillers, g(5) = 37 in 1964 by Jing-run Chen and g(6) = 73 in 1940 by Pillai.\n\nAll the other values of g are now also known, as a result of work by Dickson, Pillai, Rubugunday and Niven. Their formula contains two cases, and it is conjectured that the second case never occurs; in the first case, the formula reads\n\ng(k) = floor((3/2)k) + 2k - 2     for k ≥ 6.", null, "", null, "" ]

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[ "Main content\n\n# Adding & subtracting rational numbers: 79% - 79.1 - 58 1/10\n\nCCSS Math: 7.NS.A.1d\n\n## Video transcript\n\n79% minus 79.1 minus 58 and 1/10. And I encourage you to pause this video and try to compute this expression on your own. Well, the thing that jumps out at you is that these are in different formats. This is a percentage. These are different representations. There's a percentage. This is a decimal. This is a mixed number. And so to make sense of it, it's probably a good idea to get them all in the same format. And it seems like we could get all of these into a decimal format pretty easily. So let's go that way. So 79%, that literally means 79 per 100. If you wanted to write it as a fraction, it would be 79/100. But if you wanted to write it as a decimal, it's 0.79, which could be 0-- or we would write it down a 0.79. Now, 79.1 is already written as a decimal, so we'll just write it again. So minus 79.1. And then, 58 and 1/10. Well, 1/10 is the same thing as 0.1. So you could view this as 58-- well, and literally as 1/10. So it's minus 58 and 1/10. Or, you could view this as 58.1. So now they're all in the same format, let's actually do the computation. Now, the first thing that jumps out at you is you have a fairly small number here. Small positive number. It's less than 1. And you're subtracting fairly large numbers over here. So your whole answer is going to be negative. And to make sense of this a little bit, what I'm going to do is I'm going to factor out a negative sign. And that'll make the computation-- at least in my brain, it's going to make it a little bit easier. So if we factor out a negative sign, this becomes-- so we're going to factor it out. Actually, let me just do it this way. So if we factor out a negative sign, then this will become negative. This would be positive. And this would be positive. And just to verify this, imagine distributing this negative sign, or if this was a negative 1. Negative 1 times this is positive. Negative 1 times this is negative. Negative 1 times this is negative. So these two expressions are the exact same thing. And the reason why I did that is now we'll do the more natural thing of we will add these two numbers. We'll get a positive number, a larger positive number than what we're going to subtract from it right over here. So we can use our traditional method. Although, we can't forget about this negative out here. So let's first do that. Let's add 79.1 plus 58.1. So 79.1 plus 58.1. So 0.1 plus a 0.1 is 0.2. 9 plus 8 is 17. So that's seven 1's and one 10. So one 10 plus seven 10's is going to get us to 0.8. Plus 0.5 gets us to thirteen 10's, or 130. So we have 137.2 is this part right over here. So 100. Let me write this down. So we have 137.2. And then from that, adding a negative 0.79 is equivalent of subtracting 0.79. So let's do that. Let's subtract 0.79, making a point to align our decimal points so that we're subtracting the right place from the right place. And now let's do our subtraction. So right now we're subtracting 9 from nothing. We could write a 0 right over here, but we still face an issue in the hundredths place. We're also subtracting a 0.7 from 0.2. So we're going to have to regroup a little bit in the numerator in order to subtract. Or at least, in order to subtract using the most traditional technique. So let's take a tenth from the 2, so it's only one tenth now, and give it to the hundredths. So one tenth is ten hundredths. So we could subtract that ten hundredths minus nine hundredths is one hundredth. Now in the tenths place. We don't have enough up here, so let's take 1 from the one's place. So that becomes a 6. 1 is ten tenths. So now we have 11 tenths. 11 minus 7 is 4. Add our decimal place. 6 minus 0 is 6. And then we got our 13 just like that. So outside the parentheses, I still have the negative sign. When I computed all of this inside the parentheses, I got 136.41. And then we can't forget about the negative sign out here. So this whole thing computes to negative 136.41." ]

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[ "# Angr hooking - de/recompiling - chainbreaker\n\nThe executable that I am going to analyze here is interesting in several ways. First, no attempt at code obfuscation is made and the decompilers from both `Ghidra` and `IDA` produce code that can be recompiled with little effort. Also, the program is very short, making it a good example from an academic standpoint. The input is passed via a command line argument and so `angr` or `PinCTF` are natural first approaches to find the flag. However, there is just enough complication to prevent this from working and that is what makes this such a nice example to show progressive improvements to the `angr` input file using hooks. Let's see what the program does:\n\n``\\$ ./chainbreaker``\n``usage: ./chainbreaker SEED``\n\n``\\$ ./chainbreaker a``\n``Starting chainbreaker!``\n``ERROR``\n``Seed must be an integer!``\n\n``\\$ ./chainbreaker 0``\n``Starting chainbreaker!``\n``Seed 0 requires 96 links``\n\n``LINK 1 0 -> ERROR``\n``Invalid chain produced!``\n\n``\\$ ./chainbreaker 2``\n``Starting chainbreaker!``\n``Seed 2 requires 96 links``\n\n``LINK 1 2 -> 5 Sleeping for 5ms``\n``LINK 2 5 -> 20 Sleeping for 20ms``\n``LINK 3 20 -> 119 Sleeping for 119ms``\n``LINK 4 119 -> 426 Sleeping for 426ms``\n``LINK 5 426 -> 1529 Sleeping for 1529ms``\n``LINK 6 1529 -> 8160 Sleeping for 8160ms``\n``[ ...]``\n``LINK 94 2049 -> 16472 Sleeping for 16472ms``\n``LINK 95 16472 -> -445 Sleeping for 0ms``\n``LINK 96 -445 -> -1570 Sleeping for 0ms``\n``ERROR``\n``Starting seed doesnt match final output``\n\nOk. The program wants an integer seed, from which it produces a chain link number and then iterates through a series of computations to check if the seed is correct. The usage of the sleep timer precludes `PinCTF`. It is easy to patch the program to always sleep for 0 time, but different seeds use different link sizes and that alone causes any run time differential due to the final check to drown in the signal. Another natural thing to try is `angr`. Lets look at the decompiled code:\n\n`int __cdecl main(int argc, const char **argv, const char **envp)``{`` unsigned int v4; // [rsp+1Ch] [rbp-64h]`` char v5; // [rsp+20h] [rbp-60h]`` char v6; // [rsp+4Fh] [rbp-31h]`` char v7; // [rsp+50h] [rbp-30h]`` int inputasint; // [rsp+5Ch] [rbp-24h]`` int v9; // [rsp+60h] [rbp-20h]`` int i; // [rsp+64h] [rbp-1Ch]`` int v11; // [rsp+68h] [rbp-18h]`` int InputasInt2; // [rsp+6Ch] [rbp-14h]`\n` v9 = 100;`` if ( argc == 2 )`` {`` puts(\"Starting chainbreaker!\");`` std::allocator<char>::allocator(&v6);`` std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::basic_string(&v5, argv, &v6);`` inputasint = std::__cxx11::stoi((__int64)&v5, 0LL, 0xAu);`` std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::~basic_string(&v5);`` std::allocator<char>::~allocator(&v6);`` InputasInt2 = inputasint;`` v11 = 21309 * ((inputasint ^ 0x7B) + (inputasint ^ 0x141)) % 100;`` if ( v11 >= 0 )`` {`` if ( !v11 )`` v11 = 10; // if V11 is zero, make it 10`` }`` else`` {`` v11 = -v11; // else if v11 is negative, make it positive`` }`` printf(\"Seed %s requires %d links\\n\\n\", argv, (unsigned int)v11);`` for ( i = 0; i <= 99 && i < v11; ++i )`` {`` printf(\"LINK %d\\t\\t%d\\t->\\t\", (unsigned int)(i + 1), (unsigned int)InputasInt2);`` InputasInt2 = parse(inputasint, InputasInt2, i);`` v4 = InputasInt2;`` if ( InputasInt2 < 0 )`` v4 = 0;`` printf(\"%d\\t Sleeping for %dms\\n\", (unsigned int)InputasInt2, v4);`` std::chrono::duration<long,std::ratio<1l,1000l>>::duration<int,void>(&v7, &v4);`` std::this_thread::sleep_for<long,std::ratio<1l,1000l>>(&v7);`` }`` if ( i == 99 )`` {`` puts(\"ERROR\\nMaximum allowed iterations reached\");`` exit(1);`` }`` if ( inputasint == InputasInt2 && i == v11 )`` {`` puts(\"You have broken the chain!\");`` exit(0);`` }`` puts(\"ERROR\\nStarting seed doesnt match final output\");`` }`` else`` {`` puts(\"usage: ./chainbreaker SEED\");`` }`` return 0;``}`\n\nThe 'parse' function looks like this:\n\n`__int64 __fastcall parse(int a1, int a2, int a3)``{`` int v4; // [rsp+8h] [rbp-18h]`` int i; // [rsp+18h] [rbp-8h]`` int v6; // [rsp+1Ch] [rbp-4h]`\n` v4 = a2;`` if ( !a2 )`` {`` puts(\"ERROR\\nInvalid chain produced!\");`` exit(0);`` }`` if ( a2 < -4096 || a2 > 4096 )`` v4 = -(a2 % 4096);`` v6 = 1;`` for ( i = 0; i <= 2; ++i )`` v6 ^= v4 << i;`` return v6 + (v4 ^ (unsigned int)(a1 + a3 - 1)) + a1 - 15;``}`\n\nIt's fairly simple code. Many of the 'std::' calls are compiler generated from higher level string functions the original author used, as described here.\n\nNote that the input gets parsed into a 32 bit integer value. If we can get the computation to run fast enough, this challenge could be solved brute force. Since it's simple to do, first I want to try to recompile the code with some changes to optimize run time. I used Visual Studio for this case:\n\n`#include \"stdafx.h\"`\n`int invalid_chain = 0;``int ii =0;`\n`__int64 __fastcall parse(int a1, int a2, int a3)``{`` int v4; // [rsp+8h] [rbp-18h]`` int i; // [rsp+18h] [rbp-8h]`` int v6; // [rsp+1Ch] [rbp-4h]`\n` v4 = a2;`` if ( !a2 )`` {`` invalid_chain = 1; // Invalid chain produced`` return 0;`` }`` if ( a2 < -4096 || a2 > 4096 )`` v4 = -(a2 % 4096);`` v6 = 1;`` for ( i = 0; i <= 2; ++i )`` v6 ^= v4 << i;`` return v6 + (v4 ^ (unsigned int)(a1 + a3 - 1)) + a1 - 15;``}`\n`int _tmain(int argc, _TCHAR* argv[])``//int __cdecl main(int argc, const char **argv, const char **envp)``{`` unsigned int v4; // [rsp+1Ch] [rbp-64h]`` char v5; // [rsp+20h] [rbp-60h]`` char v6; // [rsp+4Fh] [rbp-31h]`` char v7; // [rsp+50h] [rbp-30h]`` int v8; // [rsp+5Ch] [rbp-24h]`` int v9; // [rsp+60h] [rbp-20h]`` int i; // [rsp+64h] [rbp-1Ch]`` int v11; // [rsp+68h] [rbp-18h]`` int v12; // [rsp+6Ch] [rbp-14h]`\n` v9 = 100;`\n` for (int counter=0x00000000;counter<=0xffffffff;counter++)`` {`` v8 = counter;`` v12 = v8;`` v11 = 0x533D * ((v8 ^ 0x7B) + (v8 ^ 0x141)) % 100;`` if ( v11 >= 0 )`` {`` if ( !v11 )`` v11 = 10;`` }`` else`` {`` v11 = -v11;`` }`` if (0 == (counter % 65536)) // limit the amount of information going to console`` printf(\"Seed %8.8x requires %d links\\n\", counter, (unsigned int)v11);`` for ( i = 0; i <= 99 && i < v11; ++i )`` {`` v12 = parse(v8, v12, i);`` if (invalid_chain== 1) continue; // if the parse() function determines an invalid chain, try next loop value`` v4 = v12;`` if ( v12 < 0 )`` v4 = 0;`` }`` if ( i == 99 )`` {`` continue; // Maximum allowed iterations reached - go try next loop value`` }`` if ((i <99)&& v8 == v12 && i == v11 &&(0==invalid_chain)) `` {`` puts(\"You have broken the chain!\");`` printf(\" %d \\n\", counter);`` return 1;`` }`` invalid_chain = 0;`` }`` return 0;``}`\n\nOn a LG Gram 17 laptop, this program can try the whole 32-bit keyspace in about 15 minutes. Modern computers are awesome!\n\n``Seed ffffd32a requires 72 links``\n``You have broken the chain!``\n`` -11478``\n\nBut this article is supposed to be about `angr` hooks, so lets continue there. A first crack at this should avoid code sections that 'puts' error messages and look for code paths that lead to `puts(\"You have broken the chain!\")`.\n\n``import time``\n``import claripy``\n``import angr``\n\n``# specify good and bad endpoints``\n``bad1 = (0x40149D) # ERROR\\nStarting seed doesnt match final``\n``bad2 = (0x4014AB) # usage: ./chainbreaker SEED``\n``bad3 = (0x401461) # ERROR\\nMaximum allowed iterations reached``\n``bad4 = (0x40122C) # ERROR\\nInvalid chain produced!``\n``good = (0x401487) # You have broken the chain!``\n\n``project = angr.Project(\"./chainbreaker\", auto_load_libs=True)``\n``#create an initial state with a symbolic bit vector as argv1``\n``sym_arg_size = 1 #Length in Bytes because we will multiply with 8 later``\n``sym_arg = claripy.BVS('sym_arg', 8*sym_arg_size)``\n``argv = [project.filename]``\n``argv.append(sym_arg)``\n\n``initial_state = project.factory.full_init_state(args=argv)``\n``for byte in sym_arg.chop(8):``\n`` initial_state.add_constraints(byte >= '\\x30') # '0'``\n`` initial_state.add_constraints(byte <= '\\x39') # '9'``\n``sm = project.factory.simulation_manager(initial_state)``\n\n``t0 = time.time()``\n``sm.explore(find=good,avoid=[bad1,bad2,bad3,bad4])``\n``t1 = time.time()``\n``print (t1-t0)``\n``found = sm.found``\n``result = found.solver.eval(sym_arg, cast_to=bytes)``\n``print(result)``\n\nThis code limits the input seed to a single digit, and yet angr has not finished hours later. And we would only get an error message at that point any way because the correct input has 6 digits and starts with a '-'. Since we already have the key, we can investigate what goes wrong. With the input modified like this:\n\n``data = '-1147' # this is the start of the correct password``\n``stringAsList = list(data); # turn string into list``\n``stringAsList.append(sym_arg) # declare how long the ``\n``bytestring = claripy.Concat(*stringAsList) # turn list into bytestring``\n``argv.append(bytestring)``\n``print(argv)``\n\nIf we ignore the fact that the answer is a negative number, we can see that the original angr input file does, in fact, find the right solution eventually - it's just very slow.\n\n``(angr) \\$ python angrfile ``\n\n``[...]``\n\n``<BV48 0x2d31313437 .. sym_arg_47_8>``\n``['./chainbreaker', <BV48 0x2d31313437 .. sym_arg_47_8>]``\n\n``[...]``\n\n``813.7370481491089``\n``b'8'``\n\nAbout 800 seconds for a single symbol; about 1600 for two, but after many hours 3 had not yet finished.\n\nNext, lets patch out the delay so that angr does not symbolize it. From Ghidra, we get the addresses to hook (after rebasing the position-independent executable to 0x400000)\n\n``00401441 e8 e8 02 CALL std::chrono::duration<long,std--ratio<1l,1000l undefined duration<int,void>(dur``\n`` 00 00``\n``00401446 48 8d 45 LEA RAX => local_38 ,[RBP + -0x30 ]``\n`` d0``\n``0040144a 48 89 c7 MOV param_1 ,RAX``\n``0040144d e8 cb 04 CALL std::this_thread::sleep_for<long,std--ratio<1l void sleep_for<long,std--ratio<1``\n`` 00 00``\n``00401452 83 45 e4 ADD dword ptr [RBP + local_24 ],0x1``\n`` 01``\n\nAnd it's pretty easy to hook these calls:\n\n``def durationhook(state):``\n`` print(\"0x401441 hooked\")``\n``project.hook(0x401441, durationhook, length=5) # hook this address, execute durationhook() and then skip 5 bytes, skipping the 'call' at this address``\n``def sleep_forhook(state):``\n`` print(\"0x40144d hooked\")``\n``project.hook(0x40144d, sleep_forhook, length=5) # hook this address, execute sleep_forhook() and then skip 5 bytes, skipping the 'call' at this address``\n\nAn interesting fact in this case is that while the angr documentation states:\n\n\"The CFG analysis does not distinguish between code from different binary objects. This means that by default, it will try to analyze control flow through loaded shared libraries. This is almost never intended behavior, since this will extend the analysis time to several days, probably. To load a binary without shared libraries, add the following keyword argument to the `Project` constructor: `load_options={'auto_load_libs': False}` \"\n\nthe reality is that `auto_load_libs` has the opposite effect here.\n\nThe two hooks yield a computation time (for 1-symbolic-digit integers) of about 30 minutes. Can anything else be done for the current case? There are some messages during the angr execution that point to some potential targets:\n\n`WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | The program is accessing memory or registers with an unspecified value. This could indicate unwanted behavior.``WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | angr will cope with this by generating an unconstrained symbolic variable and continuing. You can resolve this by:``WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | 1) setting a value to the initial state``WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | 2) adding the state option ZERO_FILL_UNCONSTRAINED_{MEMORY,REGISTERS}, to make unknown regions hold null``WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | 3) adding the state option SYMBOL_FILL_UNCONSTRAINED_{MEMORY_REGISTERS}, to suppress these messages.``WARNING | 2020-01-12 16:13:03,411 | angr.state_plugins.symbolic_memory | Filling memory at 0x7ffffffffff0000 with 207 unconstrained bytes referenced from 0x409d2a0 (strlen+0x0 in libc.so.6 (0x9d2a0))``WARNING | 2020-01-12 16:13:03,566 | angr.state_plugins.symbolic_memory | Filling memory at 0x7fffffffffeff00 with 6 unconstrained bytes referenced from 0x40b9ff0 (memcpy+0x0 in libc.so.6 (0xb9ff0))``WARNING | 2020-01-12 16:13:03,690 | angr.state_plugins.symbolic_memory | Filling memory at 0x7fffffffffefe68 with 8 unconstrained bytes referenced from 0x4049340 (strtoq+0x0 in libc.so.6 (0x49340))`\n\nso `angr` does not know that the `sym_arg_size = 1` constrains these values. I have seen these warnings many times for programs that use string functions, so it's valuable to see how much of a problem this really is.\n\n`printf` and puts seem obvious candidates to hook, as we don't need their output in this analysis:\n\n`def puts_hook(state):`` print(\"puts hooked\")``project.hook_symbol('puts', puts_hook)`\n`def printfhook(state):`` pass`` #print(\"printf hooked\")``project.hook(0x4013b1, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address``project.hook(0x4013ec, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address``project.hook(0x40142e, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address`\n\nWhich results in a 25% run time reduction, but the messages don't all go away.\n\n`WARNING | 2020-01-12 20:03:19,308 | angr.project | Address is already hooked, during hook(0x4083cc0, <function puts_hook at 0x7f535df95620>). Re-hooking.``puts hooked``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | The program is accessing memory or registers with an unspecified value. This could indicate unwanted behavior.``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | angr will cope with this by generating an unconstrained symbolic variable and continuing. You can resolve this by:``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | 1) setting a value to the initial state``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | 2) adding the state option ZERO_FILL_UNCONSTRAINED_{MEMORY,REGISTERS}, to make unknown regions hold null``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | 3) adding the state option SYMBOL_FILL_UNCONSTRAINED_{MEMORY_REGISTERS}, to suppress these messages.``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | Filling memory at 0x7ffffffffff0000 with 207 unconstrained bytes referenced from 0x409d2a0 (strlen+0x0 in libc.so.6 (0x9d2a0))``WARNING | 2020-01-12 20:03:21,816 | angr.state_plugins.symbolic_memory | Filling memory at 0x7fffffffffeff00 with 6 unconstrained bytes referenced from 0x40b9ff0 (memcpy+0x0 in libc.so.6 (0xb9ff0))`\n\nAnd it looks like the `puts` function was already hooked by the library simulation code. Re-hooking removes the call to `stroq`. The other calls are due to `stoi` and its' support calls:\n\n``std::allocator<char>::allocator(&v6);``\n``std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::basic_string(&v5, argv, &v6);``\n``inputasint = std::__cxx11::stoi((__int64)&v5, 0LL, 0xAu);``\n``std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::~basic_string(&v5);``\n``std::allocator<char>::~allocator(&v6);``\n\nThe hooking of `stoi` requires quite a bit of changes, so here the complete script:\n\n`import time``import claripy``import angr`\n`# specify good and bad endpoints``bad1 = (0x40149D) # ERROR\\nStarting seed doesnt match final``bad2 = (0x4014AB) # usage: ./chainbreaker SEED``bad3 = (0x401461) # ERROR\\nMaximum allowed iterations reached``bad4 = (0x40122C) # ERROR\\nInvalid chain produced!``good = (0x401487) # You have broken the chain!`\n`project = angr.Project(\"./chainbreaker\", auto_load_libs=True)`\n`argv = [project.filename]``data = '-1147' # dummy data, since we are hooking stoi``argv.append(data)``print(argv)``initial_state = project.factory.full_init_state(args=argv)``sm = project.factory.simulation_manager(initial_state)`\n`aux_arg = claripy.BVS('sym_arg', 8*4) # declase a 32 bit symbolic variable``initial_state.add_constraints(aux_arg <= -11000) # constrain variable to range we know the answer to be in``initial_state.add_constraints(aux_arg >= -12000) # constrain variable to range we know the answer to be in`\n`def durationhook(state):`` print(\"0x401441 hooked\")``project.hook(0x401441, durationhook, length=5) # hook this address, execute durationhook() and then skip 5 bytes, skipping the 'call' at this address``def sleep_forhook(state):`` print(\"0x40144d hooked\")``project.hook(0x40144d, sleep_forhook, length=5) # hook this address, execute sleep_forhook() and then skip 5 bytes, skipping the 'call' at this address``def allocatorhook(state):`` print(\"allocator hooked\")``project.hook(0x4012e5, allocatorhook, length=5) # hook this address, execute allocatorhook() and then skip 5 bytes, skipping the 'call' at this address``def allocatordestructorhook(state):`` print(\"allocatordestructor hooked\")``project.hook(0x401334, allocatordestructorhook, length=5) # hook this address, execute allocatordestructorhook() and then skip 5 bytes, skipping the 'call' at this address``def basicstringhook(state):`` print(\"basicstring hooked\")``project.hook(0x401303, basicstringhook, length=5) # hook this address, execute basicstringhook() and then skip 5 bytes, skipping the 'call' at this address``def basicstringdestructorhook(state):`` print(\"basicstringdestructor hooked\")``project.hook(0x401328, basicstringdestructorhook, length=5) # hook this address, execute basicstringdestructorhook() and then skip 5 bytes, skipping the 'call' at this address``def stoihook(state):`` state.regs.rax = aux_arg`` print(\"stoi hooked\")`` print(state.regs.rax)``project.hook(0x401319, stoihook, length=5) # hook this address, execute stoihook() and then skip 5 bytes, skipping the 'call' at this address``def puts_hook(state):`` print(\"puts hooked\")``project.hook_symbol('puts', puts_hook)``def printfhook(state):`` pass`` #print(\"printf hooked\")``project.hook(0x4013b1, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address``project.hook(0x4013ec, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address``project.hook(0x40142e, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address`\n`t0 = time.time()``sm.explore(find=good,avoid=[bad1,bad2,bad3,bad4])``t1 = time.time()``print (t1-t0)``found = sm.found``print(aux_arg)``result = found.solver.eval(aux_arg)``print(result)``result = result - 4294967296``print(result)`\n\nAnd the warnings about the unconstrained bytes disappeared:\n\n`WARNING | 2020-01-13 01:26:35,067 | cle.loader | The main binary is a position-independent executable. It is being loaded with a base address of 0x400000.``['./chainbreaker', '-1147']``WARNING | 2020-01-13 01:26:36,018 | angr.project | Address is already hooked, during hook(0x4083cc0, <function puts_hook at 0x7f2cf0c88ae8>). Re-hooking.``puts hooked``allocator hooked``basicstring hooked``stoi hooked``<SAO <BV64 0x0 .. sym_arg_47_32>>``basicstringdestructor hooked``allocatordestructor hooked`\n\nNow an interesting thing happens, for both\n\n``initial_state.add_constraints(aux_arg <= -11470) # constrain variable to range we know the answer to be in``\n``initial_state.add_constraints(aux_arg >= -11480) # constrain variable to range we know the answer to be in``\n\nand\n\n``initial_state.add_constraints(aux_arg <= -11400) # constrain variable to range we know the answer to be in``\n``initial_state.add_constraints(aux_arg >= -11500) # constrain variable to range we know the answer to be in``\n\nthe execution time is about 900 seconds, but for\n\n``initial_state.add_constraints(aux_arg <= -11000) # constrain variable to range we know the answer to be in``\n``initial_state.add_constraints(aux_arg >= -12000) # constrain variable to range we know the answer to be in``\n\nAfter setting all these hooks, all that is left of the program is symbolic execution of the user space, but in this case angr is incapable of solving it. The hooking however does show its value. Instead of not being able to solve for even the last digit, `angr` can now solve 2. We also learned that the common warnings about the `printf` symbolization do not necessarily have to be too much of a concern." ]

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[ "# Set-builder notation\n\nSet-builder notation\n\nIn set theory and its applications to logic, mathematics, and computer science, set-builder notation (sometimes simply \"set notation\") is a mathematical notation for describing a set by stating the properties that its members must satisfy. Forming sets in this manner is also known as set comprehension, set abstraction or as defining a set's intension.\n\nBuilding sets\n\nLet &Phi;(\"x\") be a schematic formula in which \"x\" appears free. Set builder notation has the form {\"x\" : &Phi;(\"x\")} (some write {\"x\" | &Phi;(\"x\")}, using the vertical bar instead of the colon), denoting the set of all individuals in the universe of discourse satisfying the predicate &Phi;(\"x\"), that is, the set whose members are every individual \"x\" such that &Phi;(\"x\") is true: formally, the extension of the predicate. Set builder notation binds the variable \"x\" and must be used with the same care applied to variables bound by quantifiers.\n\nExamples (the universe of discourse can be taken to be, for example, all complex numbers):\n* $\\left\\{x : x = x^2 \\right\\} ,!$ is the set $\\left\\{0, 1\\right\\}$,\n* $\\left\\{x : x in mathbb\\left\\{R\\right\\} and x > 0\\right\\}$ is the set of all positive real numbers. This, like all sets involving real numbers, is an example of a set that cannot be given by enumeration.\n* $\\left\\{k : n in mathbb\\left\\{N\\right\\} and k = 2n\\right\\}$ is the set of all even natural numbers,\n* $\\left\\{a : exists p, q in mathbb\\left\\{Z\\right\\} \\left(q ot = 0 land aq=p\\right) \\right\\}$ is the set of rational numbers, or numbers that can be written as the ratio of two integers.\n* $mathbb\\left\\{N\\right\\}_m = \\left\\{x in mathbb\\left\\{Z\\right\\} : x ge m \\right\\} = \\left\\{ m, m + 1, m + 2, ...\\right\\}.$ Thus, e.g., $x_1 = m, x_2 = m + 1$, etc. (n.b.: in the case of sets, the order is not important, so that one could call $x_1 = m + 2$ and so forth). As an example, $mathbb\\left\\{N\\right\\}_3 = \\left\\{x in mathbb\\left\\{Z\\right\\} : x ge 3 \\right\\} = \\left\\{ 3, 4, 5, ...\\right\\}.$ This shows how convenient set-builder notation is.\n\nThe $and$ sign stands for \"and\", requiring both conditions be fulfilled simultaneously. It is often replaced by a comma (,) semicolon (;) or written out as \"and\". Alternatively, sets of the form $\\left\\{ x : x in X and phi\\left(x\\right) \\right\\}$ can be written as $\\left\\{ x in X : phi\\left(x\\right) \\right\\}$. The set of positive real numbers would then be notated $\\left\\{ x in mathbb\\left\\{R\\right\\} : x > 0 \\right\\}$.\n\nLogical equivalence\n\nAn important fact is the logical equivalence:\n\n.\n\nThis means that sets two sets are equal if and only if their \"membership requirements\" are logically equivalent.\n\nFor example, $\\left\\{ x in mathbb\\left\\{R\\right\\} : x^2 = 1 \\right\\} = \\left\\{ x in mathbb\\left\\{R\\right\\} : |x| = 1 \\right\\}$ because, for any real number \"x\", \"x\"2 = 1 if and only if |\"x\"| = 1.\n\nRussell's paradox\n\nLet {\"S\" : \"S\" is a set and \"S\" does not belong to \"S\"} denote the set of all sets that do not belong to themselves. This set cannot exist; Russell's paradox explains why.\n\nSolutions to the paradox restrict set-builder notation in certain ways. Let \"X\" = {\"x\" ∈ \"A\" : \"P\"(\"x\")} denote the set of every element of \"A\" satisfying the predicate \"P\"(\"x\"). The canonical restriction on set builder notation asserts that \"X\" is a set only if \"A\" is already known to be a set. This restriction is codified in the axiom schema of separation present in standard axiomatic set theory. Note that this axiom schema excludes {\"S\" : \"S\" is a set and \"S\" does not belong to \"S\"} from sethood.\n\nOther problems\n\nThe notation can be complicated, especially as in the previous example, and abbreviations are often employed when context indicates the nature of a variable. For example:\n* {\"x\" : \"x\" > 0}, in a context where the variable \"x\" is used only for real numbers, indicates the set of all positive real numbers;\n* {\"p\"/\"q\" : \"q\" is not zero}, in a context where the variables \"p\" and \"q\" are used only for integers, indicates the set of all rational numbers; and\n* {\"S\" : \"S\" does not belong to \"S\"}, in a context where the variable \"S\" is used only for sets, indicates the set of all sets that don't belong to themselves.As the last example shows, such an abbreviated notation again might not denote an actual nonparadoxical set, unless there is in fact a set of all objects that might be described by the variable in question.\n\nVariations\n\nDefining sets in terms of other sets\n\nAnother variation on set-builder notation describes the members of the set in terms of members of some other set.Specifically, {\"F\"(\"x\") : \"x\" in \"A\"}, where \"F\" is a function symbol and \"A\" is a previously defined set, indicates the set of all values of members of \"A\" under \"F\".For example:\n* {2\"n\" : \"n\" in N}, where N is the set of all natural numbers, is the set of all even natural numbers.In axiomatic set theory, this set is guaranteed to exist by the axiom schema of replacement.\n\nThese notations can be combined in the form {\"F\"(\"x\") : \"x\" in \"A\", \"P\"(\"x\")}, which indicates the set of all values under \"F\" of those members of \"A\" that satisfy \"P\".For example:\n* {\"p\"/\"q\" : \"p\" in Z, \"q\" in Z, \"q\" is not zero}, where Z is the set of all integers, is the set of all rational numbers (Q).This example also shows how multiple variables can be used (both \"p\" and \"q\" in this case). This notation is acceptable even though e.g. 2/3 and 4/6 are both included in this definition, and a set can not contain multiple copies of an element; the case \"p\"=4, \"q\"=6 says with harmless redundancy that 2/3 is in the set.\n\nParallels in programming languages\n\nSet-builder notation is closely related to a construct in some programming languages, most notably Python and Haskell, called list comprehension.\n\nIn Python, list comprehensions are denoted by square brackets, and have a different syntax to set-builder, but are fundamentally the same. Consider these examples, given in both set-builder notation and Python list comprehension.\n* Set-builder:\n** {\"l\": \"l\" ∈ \"L\"}\n** \"k\", \"x\"}: \"k\" ∈ \"K\" ⋀ \"x\" ∈ \"X\" ⋀ \"P(x)\" }\n*List comprehension:\n** Python:\n*** ` [l for l in L] `\n*** ` [(k, x) for k in K for x in X if P(x)] `\n** Haskell\n*** (Haskell does not allow uppercase names so ks = K, xs = X and p = P)\n*** ` [l | l <- L] `\n*** ` [(k,x) | k <- ks, x <- xs, p(x)] `\n\nNote that in Python ` [l for l in L] ` equals to just `list(L)`.\n\nWhile Python's list comprehension works similarly to set-builder notation, it does not denote a set but rather creates a mathematical tuple (as opposed to Python's native tuple datatype; the actual returned value's type is list) based on existing tuples. It is possible to use true sets in Python with the set keyword and set class, but this causes additional deviations from set-builder notation:\n* `set(l for l in L)`\n* `set((k, x) for k in K for x in X if P(x))`The first can be written just as:\n* `set(L)`.\n\nNote that the upcoming Python 3.0 supports proper set comprehensions using a hybrid of mathematical set-builder and Python list comprehension notation: [ [http://docs.python.org/dev/3.0/reference/expressions.html#set-displays Set displays - Python 3.0 Language Reference] ]\n* `{(k, x) for k in K for x in X if P(x)}`\n\nReferences\n\nee also\n\n*Set notation\n*SQL - Structured Query Language, used to implement set operations upon a relational database\n\nWikimedia Foundation. 2010.\n\n### Look at other dictionaries:\n\n• set-builder notation — noun a mathematical notation for describing a set by stating the properties that its members must satisfy …   Wiktionary\n\n• Set notation — Sets are fundamental objects in mathematics. Intuitively, a set is merely a collection of elements or members . There are various conventions for textually denoting sets. In any particular situation, an author typically chooses from among these… …   Wikipedia\n\n• Set (mathematics) — This article gives an introduction to what mathematicians call intuitive or naive set theory; for a more detailed account see Naive set theory. For a rigorous modern axiomatic treatment of sets, see Set theory. The intersection of two sets is… …   Wikipedia\n\n• Notation — The term notation can refer to: Contents 1 Written communication 1.1 Biology and Medicine 1.2 Chemistry 1.3 Dance and movement …   Wikipedia\n\n• Naive set theory — This article is about the mathematical topic. For the book of the same name, see Naive Set Theory (book). Naive set theory is one of several theories of sets used in the discussion of the foundations of mathematics. The informal content of… …   Wikipedia\n\n• Implementation of mathematics in set theory — This article examines the implementation of mathematical concepts in set theory. The implementation of a number of basic mathematical concepts is carried out in parallel in ZFC (the dominant set theory) and in NFU, the version of Quine s New… …   Wikipedia\n\n• Intersection (set theory) — Intersections of the Greek, Latin and Russian alphabet (upper case graphemes) (The intersection of Greek and Latin letters is used for the Greek licence plates.) …   Wikipedia\n\n• List comprehension — A list comprehension is a syntactic construct available in some programming languages for creating a list based on existing lists. It follows the form of the mathematical set builder notation (set comprehension) as distinct from the use of map… …   Wikipedia\n\n• Interval (mathematics) — This article is about intervals of real numbers. For intervals in general mathematics, see Partially ordered set. For other uses, see Interval. In mathematics, a (real) interval is a set of real numbers with the property that any number that lies …   Wikipedia\n\n• Natural number — Natural numbers can be used for counting (one apple, two apples, three apples, ...) from top to bottom. In mathematics, the natural numbers are the ordinary whole numbers used for counting ( there are 6 coins on the table ) and ordering ( this is …   Wikipedia\n\n### Share the article and excerpts\n\n##### Direct link\nDo a right-click on the link above\nand select “Copy Link”" ]

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[ "Skip to main content\n\n# How to generate a log-random permeability field\n\nI was looking for a simple method to generate a log-random permeability field with specified Dykstra-Parsons coefficient and correlation length. I found this webpage with some well-written Matlab codes for the generation of rough surfaces. I'm literally illiterate when it comes to signal processing. All I know is that the Dykstra-Parsons coefficient is defined as $$V_{DP}=\\frac{\\text{Standard deviation of }\\log(k)}{\\text{Average of }\\log(k)}=\\frac{\\log(k)_{P50}-\\log(k)_{P84.1}}{\\log(k)_{P50}}$$ We can rearrange the above equation to obtain the standard deviation of the permeability field: $$\\sigma = -\\log(1-V_{DP})$$ I used this standard deviation and the average permeability value, $k_{ave}$, to generate a random distribution, and pass it through the filter that I found in the mentioned Matlab codes to generate a log-random permeability field. I showed the result to a Geologist colleague of mine, and he confirmed that it can indeed be considered as an acceptable permeability field. Here's one sample code for a 2D field generation, in Julia. You can easily convert it to Matlab.\n\nIn :\nfunction permfieldlogrndg(Nx,Ny,k_avrg,V_dp,clx,cly)\n# 2D random field generator:\n# hdf: Gaussian\n# acf: Gaussian\n# The surface has a Gaussian height distribution function\n# and Gaussian autocovariance functions\nLx=1.0\nX = linspace(-Lx/2,Lx/2,Nx)\nY = linspace(-Lx/2,Lx/2,Ny)'\n\ns = -log(1-V_dp)\nmu = log(k_avrg)-s^2/2\nZ = s*randn(Nx,Ny)\n# Gaussian filter\nF = exp(-(X.^2/(clx^2/2.0).+Y.^2/(cly^2/2.0)))\n# correlated surface generation including convolution (faltning) and inverse\n# Fourier transform and normalizing prefactors\nf = 2.0/sqrt(pi)*Lx/sqrt(Nx*Ny)/sqrt(clx)/sqrt(cly)*ifft(fft(Z).*fft(F))\nperm = exp(mu+real(f))\nend;\n\n\nNow we can test the code to generate a field on a Nx x Ny grid, an average permeability of k_avrg, a Dykstra-Parsons coefficient V_dp, and correlation length of clx in x direction and cly in y direction.\n\nIn :\nusing PyPlot;\n\nIn :\nNx=100\nNy=50\nclx=1.0\ncly=0.05\nk_avg= 1e-12 # [m^2]\nV_dp=0.9\nk= permfieldlogrndg(Nx,Ny,k_avg,V_dp,clx,cly)\nimshow(k')\ncolorbar()\n#tight_layout();", null, "The above figure shows a permeability field, for a layered reservoir (correlation length of 1.0 in the x direction). Play a bit with the code, and see what happens. I'll really appreciate your comments about the flaws in this simple code.\n\n## Comments\n\nComments powered by Disqus" ]

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", 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[ "Products", null, "Rewards\nfrom HOLOOLY\n\nWe are determined to provide the latest solutions related to all subjects FREE of charge!\n\nEnjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program\n\nHOLOOLY\n\nHOLOOLY\nTABLES\n\nAll the data tables that you may search for.\n\nHOLOOLY\nARABIA\n\nFor Arabic Users, find a teacher/tutor in your City or country in the Middle East.\n\nHOLOOLY\nTEXTBOOKS\n\nFind the Source, Textbook, Solution Manual that you are looking for in 1 click.\n\nHOLOOLY\nHELP DESK\n\nNeed Help? We got you covered.\n\n## Q. 5.26\n\nDetermine $V_{ab}, V_{cb}, and V_c$ for the network in Fig. 5.59.", null, "## Verified Solution\n\nThere are two ways to approach this problem. The first is to sketch the diagram in Fig. 5.60 and note that there is a 54 V drop across the series resistors $R_1 and R_2$. The current can then be determined using Ohm’s law and the voltage levels as follows:\n\n$I=\\frac{54V}{45\\Omega }=1.2A$\n\n$V_{ab}=IR_2=(1.2A)(25\\Omega )=30V$\n\n$V_{cb}=-IR_1=-(1.2A)(20\\Omega )=-24V$\n\n$V_c=E_1=-19V$\n\nThe other approach is to redraw the network as shown in Fig. 5.61 to clearly establish the aiding effect of $E_1 and E_2$ and then solve the resulting series circuit:\n\n$I=\\frac{E_1+E_2}{R_T}=\\frac{19V+35V}{45\\Omega } =\\frac{54V}{45\\Omega } =1.2A$\n\nand        $V_{ab}=30V\\qquad V_{cb}=-24V\\qquad V_c=-19V$", null, "", null, "" ]

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[ "# Shannon entropy in the context of machine learning and AI\n\nIn this post, I want to elaborate on the concept of Shannon entropy in the context machine learning and AI. My goal is to provide some insight into the math behind Shannon entropy, but keep the discussion relatively informal.\n\nSo let’s dive right in. It is common in machine learning to quantify the expected amount of information associated with stochastic events, and to quantify the similarity between probability distributions. In both cases, Shannon entropy is used as a measure for information content of probability distributions. It is named after the father of information theory, Claude Shannon (1916–2001).\n\n## Self-information\n\nThe fundamental concept behind Shannon entropy is the so-called self-information of an event, sometimes called surprisal. The intuition behind self-information is as follows. When an unlikely outcome of an event (random variable) is observed, we associate it with a high amount of information. Contrarily, when a more likely outcome is observed, we associate it with a smaller amount of information. It is very helpful to think of self-information as the surprise associated with an event. Consider, for example, a heavily biased coin that always lands on heads. The outcome of any coin toss is fully predictable, thus we will never be surprised about the outcome, which means that we gain zero information from such an experiment. In other words, its self-information is zero. If the coin is biased less heavily, there is some amount of surprise each time we toss the coin, although more than 50% of the time we will still see heads. Consequently, the self-information is larger than zero. If the coin is biased towards always landing on tails (instead of heads), we again end up with zero entropy, or surprise. The maximum amount of surprise is obtained in the case of an unbiased coin where the chances of landing on heads or tails are both 50%, as this is the situation where the outcome of the coin toss is least predictable.\n\nBased on the above informal requirements, we can find an appropriate function to describe self-information. For a discrete random variable X with possible values x_1, . . . , x_n and probability mass function P(X), any positive and monotonically decreasing function I(p_i) between 0 and 1 could be used as a measure for information. So far so good. One additional and crucial property, however, is additivity of independent events; the self-information of two subsequent coin tosses should be twice the self-information of a single coin toss. This makes sense for independent variables, as the amount of surprise, or unpredictability, becomes twice larger in this case. Formally, we need I(p_i·p_j) = I(p_i)+I(p_j) for independent events x_i and x_j. The function fulfilling all of these requirements is the negative logarithm,\n\nFigure 1 shows a plot of I(p).", null, "Figure 1. The self-information function I(p). Low probabilities are associated with high self-information, and vice versa.\n\nLet’s go back to our simple coin toss experiment. In the language of information theory, one bit (also called ”Shannon”) of information is required to represent the two possible outcomes of a single coin toss. Similarly, for two consecutive coin tosses, two bits of information are required to describe all four possible outcomes. In general, log_2(n) bits of information are required to describe the outcome, or equivalently the self-information, of n subsequent independent random events. Let’s walk through the simple calculations for three subsequent coin tosses. There are 2^3 = 8 possible outcomes, and the probability of any particular outcome is 0.5^3 = 0.125. Therefore, the self-information of this experiment is I(0.125) = −log_2(0.125) = 3. We need 3 bits to represent all possible outcomes, so the self-information of any particular sequence of three coin tosses equals 3.0.\n\nWe can also calculate the self-information of a continuous random variable. Figure 2 shows three different pdfs and their corresponding information content. The Dirac delta in 2a corresponds to the strong bias, zero entropy case of a biased coin that always lands on the same side. An infinitely high information content would be associated wherever p(x) = 0. However, this is hypothetical since these events will never actually occur due to zero probability. The Gaussian pdf in Figure 2B is analogous to the biased coin that tends to fall on the same side often, but not always. Lastly, Figure 2C depicts a uniform pdf, with associated uniform information content, analogous to our unbiased coin toss.", null, "Figure 2. Three different probability densities p on [−3,+3] and their self-information I(p). (A) Dirac delta function (completely deterministic) (B) Gaussian with μ = 0,σ = 0.5 (biased towards x = 0) (C.) Uniform distribution.\n\n## Entropy\n\nSo far we have only discussed self-information. For the case of a fair coin, self-information actually equals Shannon entropy, because all outcomes are equal in probability. In general, however, Shannon entropy is the average self-information (expected value) over all possible values of X,\n\nwhere b is the base of the logarithm. Above we used b = 2, other common choices are b = 10 as well as Euler’s number e, but the choice doesn’t matter much, since logarithms of varying bases are related by a constant. We will assume base 2 from here on and omit b.\n\nIf you paid close attention, you may wonder what happens to entropy when p(x_i) = 0, since in this case we have to compute 0 · log(0). It turns out that\nlim p→0 p · log(p) = 0. A mathematical proof can be established using the rule of L’Hospital.\n\nShannon entropy generalizes to the continuous domain, where it is referred to as differential entropy (some restrictions apply which shall not be discussed here). For a continuous random variable x and probability density function p(x), Shannon entropy is defined as follows,\n\nThe entropy for our three example distributions is 0 (Dirac delta), 174 (Gaussian), and 431 (uniform). The pattern that emerges from our experiment is that broad distributions have the highest entropy. It’s important for your understanding to take a closer look at Figure 2B and 2C. Despite the fact that in the Gaussian case the area under I(p) is much larger than in the uniform case, the entropy is much smaller than in the uniform case because I(p) is weighted by the probability density function p, which is close to zero on both sides of the Gaussian. The best analogy to keep in mind to remember that a broad probability density has high entropy is to imagine a gas filled in a tank. We know from physics that the entropy in a closed system increases over time, and never decreases on its own. The distribution of gas particles (number of gas particles per unit volume) converges to a uniform value across the tank, after we infuse the gas on either side of the tank. Low entropy would mean that a higher density gas particles accumulated in certain areas, which never happens on its own. Accumulation of many gas particles in a small region corresponds to our Gaussian pdf, and in an extreme case to the Dirac delta function, when all gas particles are condensed into an infinitely small region.\n\n## Cross entropy\n\nCross entropy is a mathematical tool for comparing two probability distributions p and q. It is similar to entropy, but instead of the expectation of log(p) under p, we compute the expectation of log(q) under p,\n\nIn the language of information theory, this quantity gives us the average number of bits required to code an event from q if we use the ”wrong” coding scheme q instead of p. In machine learning, it is a very useful measure for the similarity of probability distributions and serves as a loss function (more details below).\n\n# Uses in machine learning\n\nYou may wonder, at this point, how entropy is relevant in machine learning. So let’s look at a some specific areas next.\n\n## Bayesian learning\n\nFirst, the Gaussian case described above is important, as the normal distribution is a very common modeling choice in machine learning applications. The goal in machine learning is to reduce entropy. We want to make predictions, and we have to be confident about our predictions. Entropy gives us a measure to quantify this confidence. In a Bayesian setting, often a prior distribution is assumed that has a broad pdf, reflecting the uncertainty in the value of the random variable before an observation is made. When data comes in, the entropy reduces and causes the posterior to form peaks around the likely values of the parameters.\n\n## Decision tree learning\n\nIn decision tree learning, entropy is used to build the tree. Constructing a decision tree starts at the root node, by splitting the data set S into a number of subsets according to all possible values of the ”best” attribute, i.e., the one that minimizes the (combined) entropy in the resulting subsets. This procedure is repeated recursively until there are no more attributes left to split. This prodecude is called ID3 algorithm.\n\n## Classification\n\nCross entropy is the basis of the standard loss function for logistic regression and neural networks, in both binomial and multinomial classification scenarios. Usually, p is used for the true (or empirical) distribution (i.e., the distribution of the training set), and q is the distribution described by a model. Let’s look at binary logistic regression as an example. The two classes are labeled 0 and 1, and the logistic model assigns the probabilities q_(y=1) = ŷ and q_(y=0) = 1 − ŷ to each input x. This can be concisely written as q ∈ {ŷ, 1 − ŷ}. Even though the empirical labels p are always exactly 0 or 1, the same notation is used here, p ∈ {y, 1 − y}, so don’t get confused. Using this notation, the cross entropy between empirical and estimated distribution for a single sample is\n\nWhen used as a loss function, the average of all cross entropies from all N samples is used,\n\n## KL divergence\n\nClosely related to cross entropy, the KL divergence from q to p, written DKL(p||q), is another similarity measure often used in machine learning. In the language of Bayesian Inference, DKL(p||q) is a measure of the information gained when one revises one’s beliefs from the prior distribution q to the posterior distribution p, or in other words, the amount of information lost when q is used to approximate p. It is used, for example, in training the latent space representation of a variational autoencoder. KL divergence can be expressed using entropy and cross entropy,\n\nWhile cross entropy measures the average total bits needed to code an event from p when a coding scheme optimized for q is used, while the KL divergence gives the number of additional bits needed when the optimal coding scheme for q is used, instead of the optimal coding scheme for p. From this we can see that in the context of machine learning, where p is fixed, cross entropy and KL divergence are related by a constant additive term, so for the purpose of optimization they are equivalent. It may still make sense to talk about KL divergence rather than cross entropy from a theoretical perspective, and one useful property of KL divergence is that it is zero when p and q are equal.\n\n# Conclusion\n\nThe purpose of this post is to shed some light on the concept of entropy in the context of machine learning and artificial intelligence, by explaining the most essential aspects of the theory, and where it appears in applications. The Python code for generating all results presented above can be found on this Gist.\n\nWritten by" ]

[ null, "https://miro.medium.com/max/60/1*[email protected]", null, "https://miro.medium.com/max/60/1*USFoTUyFmiEtXTmLEtRG1w.png", null ]

[ "Question\n(a) On a particular day, it takes $9.60\\times 10^{3}\\textrm{ J}$ of electric energy to start a truck’s engine. Calculate the capacitance of a capacitor that could store that amount of energy at 12.0 V. (b) What is unreasonable about this result? (c) Which assumptions are responsible?\n1. $133\\textrm{ F}$\n2. A capacitor of this capacitance would be so massive and heavy that a truck would not be able to carry it. If we assumed the capacitor is a parallel plate capacitor with the highest capacity dielectric, strontium titanate, with square plates, the side length of the square would be 254 km!\n3. It's unreasonable to presume this energy could be supplied by a capacitor. Batteries are used instead.\n\n# OpenStax College Physics, Chapter 19, Problem 70 (Problems & Exercises)", null, "", null, "In order to watch this solution you need to have a subscription.\n\n•", null, "" ]

[ null, "https://collegephysicsanswers.com/sites/default/files/styles/video_thumbnail/public/solution_thumbnails/thumbs-ed1ch19pe70-00002.jpg", null, "https://collegephysicsanswers.com/themes/cpa2/images/padlock.svg", null, "https://collegephysicsanswers.com/sites/default/files/styles/calc_screenshot_mobile/public/calculator_screenshots/ch19pe70.jpeg", null ]

[ "# Concentration of Neutral ammonium fluoride\n\n## neutral ammonium fluoride: mass and molar concentration\n\n### Molar concentration per milliliter\n\n 0.01 mmol/ml 10 µmol/ml 10 000 nmol/ml 10 000 000 pmol/ml\n\n### Molar concentration per deciliter\n\n 1 mmol/dl 1 000 µmol/dl 1 000 000 nmol/dl 1 000 000 000 pmol/dl\n\n### Molar concentration per liter\n\n 10 mmol/l 10 000 µmol/l 10 000 000 nmol/l 10 000 000 000 pmol/l\n\n### Mass concentration per milliliter\n\n 0 g/ml 0.37 mg/ml 370.37 µg/ml 370 370 ng/ml 370 370 000 pg/ml\n\n### Mass concentration per deciliter\n\n 0.04 g/dl 37.04 mg/dl 37 037 µg/dl 37 037 000 ng/dl 37 037 000 000 pg/dl\n\n### Mass concentration per liter\n\n 0.37 g/l 370.37 mg/l 370 370 µg/l 370 370 000 ng/l 370 370 000 000 pg/l\n\n### Equivalent molar concentration per milliliter\n\n 0.01 meq/ml 10 µeq/ml 10 000 neq/ml 10 000 000 peq/ml\n\n### Equivalent molar concentration per deciliter\n\n 1 meq/dl 1 000 µeq/dl 1 000 000 neq/dl 1 000 000 000 peq/dl\n\n### Equivalent molar concentration per liter\n\n 10 meq/l 10 000 µeq/l 10 000 000 neq/l 10 000 000 000 peq/l\n• The units of  amount of substance (e.g. mole) per milliliter,  liter and deciliter are SI units of measurements of molar concentrations.\n• The units of molar concentration per deciliter:\n• millimole per deciliter [mm/dl],  micromole per deciliter [µm/dl],  nanomole per deciliter [nm/dl]  and  picomole per deciliter [pm/dl].\n• The units of molar concentration per milliliter:\n• millimole per milliliter [mm/ml],  micromole per milliliter [µm/ml],  nanomole per milliliter [nm/ml]  and  picomole per milliliter [pm/ml].\n• The units of molar concentration per liter:\n• millimole per liter [mm/l],  micromole per liter [µm/l],  nanomole per liter [nm/l]  and  picomole per liter [pm/l].\n• The units of  mass  per milliliter,  liter and deciliter are non-SI units of measurements of mass concentrations still used in many countries.\n• The units of mass concentration per deciliter:\n• gram per deciliter [g/dl],  milligram per deciliter [mg/dl],  microgram per deciliter [µg/dl],  nanogram per deciliter [ng/dl]  and  picogram per deciliter [pg/dl].\n• The units of mass concentration per milliliter:\n• gram per milliliter [g/ml],  milligram per milliliter [mg/ml],  microgram per milliliter [µg/ml],  nanogram per milliliter [ng/ml]  and  picogram per milliliter [pg/ml].\n• The units of mass concentration per liter:\n• gram per liter [g/l],  milligram per liter [mg/l],  microgram per liter [µg/l],  nanogram per liter [ng/l]  and  picogram per liter [pg/l].\n• The  equivalent  per milliliter,  liter and deciliter are obsolete, non-SI units of measurements of molar concentrations still used in many countries. An equivalent is the number of moles of an ion in a solution, multiplied by the valence of that ion.\n• The units of equivalent concentration per deciliter:\n• milliequivalent per deciliter [meq/dl],  microequivalent per deciliter [µeq/dl],  nanoequivalent per deciliter [neq/dl]  and  picoequivalent per deciliter [peq/dl].\n• The units of equivalent concentration per milliliter:\n• milliequivalent per milliliter [meq/ml],  microequivalent per milliliter [µeq/ml],  nanoequivalent per milliliter [neq/ml]  and  picoequivalent per milliliter [peq/ml].\n• The units of equivalent concentration per liter:\n• milliequivalent per liter [meq/l],  microequivalent per liter [µeq/l],  nanoequivalent per liter [neq/l]  and  picoequivalent per liter [peq/l].\n\n#### Foods, Nutrients and Calories\n\nTRADER JOE'S, SPARKLING WHITE CHARDONNAY GRAPE JUICE, CHARDONNAY GRAPE, UPC: 00501743 contain(s) 62 calories per 100 grams (≈3.53 ounces)  [ price ]\n\n1376 foods that contain Tocopherol, gamma.  List of these foods starting with the highest contents of Tocopherol, gamma and the lowest contents of Tocopherol, gamma\n\n#### Gravels, Substances and Oils\n\nCaribSea, Freshwater, African Cichlid Mix, White weighs 1 169.35 kg/m³ (73.00014 lb/ft³) with specific gravity of 1.16935 relative to pure water.  Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical  or in a rectangular shaped aquarium or pond  [ weight to volume | volume to weight | price ]\n\nTitanic anhydride (rutile) [TiO2  or  O2Ti] weighs 4 240 kg/m³ (264.69455 lb/ft³)  [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]\n\nVolume to weightweight to volume and cost conversions for Corn oil with temperature in the range of 10°C (50°F) to 140°C (284°F)\n\n#### Weights and Measurements\n\nA milliliter is a derived metric measurement unit of volume with sides equal to one centimeter (1cm)\n\nThe speed measurement was introduced to measure distance traveled by an object per unit of time,\n\noz t/pt to gr/mm³ conversion table, oz t/pt to gr/mm³ unit converter or convert between all units of density measurement.\n\n#### Calculators\n\nCalculate volume of a hexagonal prism and its surface area" ]

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[ "# Definition of debased\n\n## \"debased\" in the verb sense\n\n### 1. corrupt, pervert, subvert, demoralize, demoralise, debauch, debase, profane, vitiate, deprave, misdirect\n\ncorrupt morally or by intemperance or sensuality\n\n\"debauch the young people with wine and women\"\n\n\"Socrates was accused of corrupting young men\"\n\n\"Do school counselors subvert young children?\"\n\n\"corrupt the morals\"\n\n### 2. debase, alloy\n\nlower in value by increasing the base-metal content\n\ncorrupt, debase, or make impure by adding a foreign or inferior substance often by replacing valuable ingredients with inferior ones\n\n## \"debased\" in the adjective sense\n\nmixed with impurities\n\nlowered in value\n\n\"the dollar is low\"\n\n\"a debased currency\"\n\n### 3. corrupted, debased, vitiated\n\nruined in character or quality\n\nSource: WordNet® (An amazing lexical database of English)\n\nWordNet®. Princeton University. 2010.\n\n# debased in Scrabble®\n\nThe word debased is playable in Scrabble®, no blanks required.\n\nDEBASED\n(94 = 44 + 50)\n\ndebased\n\nDEBASED\n(94 = 44 + 50)\nDEBASED\n(92 = 42 + 50)\nDEBASED\n(89 = 39 + 50)\nDEBASED\n(89 = 39 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(84 = 34 + 50)\nDEBASED\n(83 = 33 + 50)\nDEBASED\n(82 = 32 + 50)\nDEBASED\n(78 = 28 + 50)\nDEBASED\n(78 = 28 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(74 = 24 + 50)\nDEBASED\n(74 = 24 + 50)\nDEBASED\n(74 = 24 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(71 = 21 + 50)\nDEBASED\n(68 = 18 + 50)\nDEBASED\n(67 = 17 + 50)\nDEBASED\n(66 = 16 + 50)\nDEBASED\n(66 = 16 + 50)\nDEBASED\n(65 = 15 + 50)\nDEBASED\n(65 = 15 + 50)\nDEBASED\n(64 = 14 + 50)\nDEBASED\n(63 = 13 + 50)\nDEBASED\n(63 = 13 + 50)\nDEBASED\n(63 = 13 + 50)\nDEBASED\n(63 = 13 + 50)\nDEBASED\n(62 = 12 + 50)\n\nDEBASED\n(94 = 44 + 50)\nDEBASED\n(92 = 42 + 50)\nDEBASED\n(89 = 39 + 50)\nDEBASED\n(89 = 39 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(86 = 36 + 50)\nDEBASED\n(84 = 34 + 50)\nDEBASED\n(83 = 33 + 50)\nDEBASED\n(82 = 32 + 50)\nDEBASED\n(78 = 28 + 50)\nDEBASED\n(78 = 28 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(76 = 26 + 50)\nDEBASED\n(74 = 24 + 50)\nDEBASED\n(74 = 24 + 50)\nDEBASED\n(74 = 24 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(72 = 22 + 50)\nDEBASED\n(71 = 21 + 50)\nDEBASED\n(68 = 18 + 50)\nDEBASED\n(67 = 17 + 50)\nDEBASED\n(66 = 16 + 50)\nDEBASED\n(66 = 16 + 50)\nDEBASED\n(65 = 15 + 50)\nDEBASED\n(65 = 15 + 50)\nDEBASED\n(64 = 14 + 50)\nDEBASED\n(63 = 13 + 50)\nDEBASED\n(63 = 13 + 50)\nDEBASED\n(63 = 13 + 50)\nDEBASED\n(63 = 13 + 50)\nDEBASED\n(62 = 12 + 50)\n(39)\n(39)\n(36)\nDEBASE\n(36)\n(36)\n(36)\n(36)\nSEABED\n(36)\n(33)\nBASED\n(33)\nDEBASE\n(33)\n(33)\n(33)\n(33)\nSEABED\n(33)\n(33)\n(33)\n(33)\n(32)\n(32)\n(30)\nSEABED\n(30)\nSEABED\n(30)\n(30)\n(30)\nSEABED\n(30)\n(30)\n(30)\n(30)\nBASED\n(30)\n(30)\nDEBASE\n(30)\nDEBASE\n(30)\nBEDS\n(30)\nDEBASE\n(30)\nSEABED\n(30)\nDEBASE\n(30)\n(28)\n(28)\n(28)\nBASED\n(28)\nBEES\n(27)\nBASE\n(27)\nBASED\n(27)\nDEBASE\n(27)\n(27)\nDABS\n(27)\nDEBASE\n(27)\n(27)\n(27)\nBASED\n(27)\n(27)\nSEABED\n(27)\nDEEDS\n(27)\nDEEDS\n(27)\n(27)\nSEABED\n(27)\n(26)\n(26)\nSEABED\n(26)\n(26)\nDEBASE\n(26)\n(26)\n(24)\n(24)\n(24)\n(24)\n(24)\n(24)\n(24)\nDABS\n(24)\n(24)\nBASED\n(24)\nBASED\n(24)\n(24)\nBASED\n(24)\n(24)\n(24)\n(24)\nBASED\n(24)\n(24)\n(24)\n(24)\nBEDS\n(24)\n(24)\n(24)\nEASED\n(24)\nDEED\n(24)\nDEEDS\n(24)\nDEEDS\n(24)\nDEED\n(24)\nDEBASE\n(22)\nDEBASE\n(22)\nBASED\n(22)\nDEEDS\n(22)\n(22)\nDEBASE\n(22)\nDEBASE\n(22)\nBASED\n(22)\n(22)\n(22)\n(22)\nSEABED\n(22)\n(22)\nDEBASE\n(22)\n(22)\nSEABED\n(22)\nSEABED\n(22)\n(22)\nSEABED\n(22)\nSEABED\n(22)\n(21)\nBEDS\n(21)\nBEDS\n(21)\nBEDS\n(21)\nBEDS\n(21)\n(21)\n(21)\n(21)\n(21)\nDEEDS\n(21)\nDEEDS\n(21)\nDEEDS\n(21)\n(21)\nDEES\n(21)\nEASED\n(21)\nEASED\n(21)\nEASED\n(21)\n(21)\nSEED\n(21)\n(21)\n(21)\nBASE\n(21)\n(21)\nDABS\n(21)\nDABS\n(21)\nDABS\n(21)\nDABS\n(21)\n(21)\n(21)\n(21)\nBEES\n(21)\n(21)\nSEABED\n(20)\n(20)\nDEBASE\n(20)\nSEABED\n(20)\n(20)\n(20)\n(20)\nSEABED\n(20)\n(20)\n(20)\n(20)\n(20)\nSEABED\n(20)\nEASED\n(20)\n(20)\n\n# debased in Words With Friends™\n\nThe word debased is playable in Words With Friends™, no blanks required.\n\nDEBASED\n(107 = 72 + 35)\n\ndebased\n\nDEBASED\n(107 = 72 + 35)\nDEBASED\n(89 = 54 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(77 = 42 + 35)\nDEBASED\n(77 = 42 + 35)\nDEBASED\n(77 = 42 + 35)\nDEBASED\n(77 = 42 + 35)\nDEBASED\n(75 = 40 + 35)\nDEBASED\n(67 = 32 + 35)\nDEBASED\n(67 = 32 + 35)\nDEBASED\n(67 = 32 + 35)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(61 = 26 + 35)\nDEBASED\n(61 = 26 + 35)\nDEBASED\n(61 = 26 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(57 = 22 + 35)\nDEBASED\n(55 = 20 + 35)\nDEBASED\n(55 = 20 + 35)\nDEBASED\n(53 = 18 + 35)\nDEBASED\n(53 = 18 + 35)\nDEBASED\n(52 = 17 + 35)\nDEBASED\n(52 = 17 + 35)\nDEBASED\n(52 = 17 + 35)\nDEBASED\n(52 = 17 + 35)\nDEBASED\n(51 = 16 + 35)\nDEBASED\n(51 = 16 + 35)\nDEBASED\n(51 = 16 + 35)\nDEBASED\n(50 = 15 + 35)\nDEBASED\n(50 = 15 + 35)\nDEBASED\n(50 = 15 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(48 = 13 + 35)\nDEBASED\n(48 = 13 + 35)\nDEBASED\n(48 = 13 + 35)\nDEBASED\n(48 = 13 + 35)\nDEBASED\n(47 = 12 + 35)\n\nDEBASED\n(107 = 72 + 35)\nDEBASED\n(89 = 54 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(83 = 48 + 35)\nDEBASED\n(77 = 42 + 35)\nDEBASED\n(77 = 42 + 35)\nDEBASED\n(77 = 42 + 35)\nDEBASED\n(77 = 42 + 35)\nDEBASED\n(75 = 40 + 35)\n(69)\nDEBASED\n(67 = 32 + 35)\nDEBASED\n(67 = 32 + 35)\nDEBASED\n(67 = 32 + 35)\nSEABED\n(66)\nDEBASE\n(66)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(63 = 28 + 35)\nDEBASED\n(63 = 28 + 35)\n(63)\nDEBASED\n(61 = 26 + 35)\nDEBASED\n(61 = 26 + 35)\nDEBASED\n(61 = 26 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(59 = 24 + 35)\nDEBASED\n(57 = 22 + 35)\n(57)\n(57)\n(57)\nDEBASED\n(55 = 20 + 35)\nDEBASED\n(55 = 20 + 35)\nSEABED\n(54)\nDEBASE\n(54)\nDEBASED\n(53 = 18 + 35)\nDEBASED\n(53 = 18 + 35)\nDEBASED\n(52 = 17 + 35)\nDEBASED\n(52 = 17 + 35)\nDEBASED\n(52 = 17 + 35)\nDEBASED\n(52 = 17 + 35)\n(51)\n(51)\nDEBASED\n(51 = 16 + 35)\nBASED\n(51)\nDEBASED\n(51 = 16 + 35)\nDEBASED\n(51 = 16 + 35)\nDEBASED\n(50 = 15 + 35)\nDEBASED\n(50 = 15 + 35)\nDEBASED\n(50 = 15 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(49 = 14 + 35)\nDEBASED\n(49 = 14 + 35)\n(48)\nDEBASED\n(48 = 13 + 35)\nDEBASED\n(48 = 13 + 35)\nDEBASED\n(48 = 13 + 35)\nDEBASED\n(48 = 13 + 35)\nBEDS\n(48)\n(48)\nDEBASED\n(47 = 12 + 35)\n(45)\nBEES\n(45)\nBASE\n(45)\n(45)\n(45)\n(45)\n(44)\n(44)\n(44)\n(44)\nDEBASE\n(42)\nDEBASE\n(42)\nSEABED\n(42)\nSEABED\n(42)\nDEBASE\n(40)\nDEBASE\n(40)\nSEABED\n(40)\nSEABED\n(40)\n(39)\n(39)\n(39)\n(39)\n(39)\n(39)\n(39)\nBASED\n(39)\n(38)\n(38)\nBASED\n(36)\nDEBASE\n(36)\nSEABED\n(36)\n(36)\nDEBASE\n(36)\nDEBASE\n(36)\nDEBASE\n(36)\nDABS\n(36)\n(36)\nSEABED\n(36)\nSEABED\n(36)\nSEABED\n(36)\nBASED\n(34)\n(34)\nBASED\n(33)\n(33)\n(33)\n(33)\n(33)\nBASED\n(33)\nDEEDS\n(33)\n(33)\n(33)\nDEEDS\n(33)\n(33)\n(33)\nDEED\n(30)\nSEABED\n(30)\nEASED\n(30)\n(30)\nSEABED\n(30)\n(30)\n(30)\nDEED\n(30)\n(30)\n(30)\nDABS\n(30)\n(30)\nDEBASE\n(30)\nDEBASE\n(30)\nBEDS\n(30)\n(30)\n(30)\nDEBASE\n(28)\nDEBASE\n(28)\nSEABED\n(28)\n(28)\nSEABED\n(28)\nDEEDS\n(28)\nBASE\n(27)\n(27)\n(27)\nDEES\n(27)\nDEEDS\n(27)\nDEEDS\n(27)\nBEES\n(27)\n(27)\n(27)\nBASED\n(27)\n(27)\nBASED\n(27)\nBASED\n(27)\nSEED\n(27)\n(26)\n(26)\nBASED\n(26)\n(26)\n(26)\n(26)\n(26)\nBASED\n(26)\n(26)\n(26)\n(26)\n(24)\nBEDS\n(24)\n(24)\nBEDS\n(24)\n(24)\n(24)\nBEDS\n(24)\nBEDS\n(24)\nEASED\n(24)\n(24)\n(24)\n(24)\n(24)\n(24)\n(24)\n(24)\nEASED\n(24)\n(24)\n(24)\n(24)\n(24)\n(24)\nEASED\n(24)\nBEDS\n(24)\nDABS\n(24)\n\nas base based\n\nas base debase\n\ndebasedness\n\nwidebased" ]

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[ "## EDMUND F ROBERTSON: ABSTRACTS\n\nBack to the list of publications .\n\n### Fibonacci semigroups\n\nAuthors: C.M. Campbell, E.F. Robertson, N. Ruskuc and R.M. Thomas\n\nJournal: J. Pure Appl. Algebra 94 (1994), 49-57\n\nAbstract: In this paper we consider semigroups S(r,n) defined by the presentations for Fibonacci groups F(r,n). We prove that S(r,n) is a union of a finite number of copies of F(r,n). We also consider semigroups S(r,n,k) defined by the presentations for generalised Fibonacci groups F(r,n,k). We show that if gcd(n,k)=1 or gcd(n,r+k-1)=1 then again S(r,n,k) is a union of a finite number of copies of F(r,n,k). Finally we show that S(2,6,3) is infinite although F(2,6,3) is finite.\n\n### Application of computational tools for finitely presented groups\n\nAuthors: G. Havas and E.F. Robertson\n\nJournal: Computational Support for Discrete Mathematics, DIMACS Series in Discrete Mathematics and Theoretical Computer Science 15 (1994), 29-39\n\nAbstract: Computer based techniques for recognizing finitely presented groups are quite powerful. Tools available for this purpose are outlined. A general computational approach for investigating finitely presented groups by way of quotients and subgroups is described and examples are presented. The techniques can provide detailed information about group structure. Under suitable circumstances a finitely presented group can be shown to be soluble and its complete derived series can be determined, using what is in effect a soluble quotient algorithm.\n\n### Defining-relations for Hurwitz groups\n\nAuthors: C.M. Campbell, M.D.E. Conder and E.F. Robertson\n\nJournal: Glasgow Math. J. 35 (1994), 363-371\n\nAbstract: This paper considers a number of questions posed by John Leech thirty years ago regarding quotients of the triangle group (2, 3, 7) which have remained unanswered, We provide answers to three of these and throw some light on a fourth one, which appears to be quite difficult. We examine a few related results. Our approach is mostly computational, using machine implementations of coset enumeration techniques.\n\n### Semigroup presentations and minimal ideals\n\nAuthors: C.M. Campbell, E.F. Robertson, N. Ruskuc and R.M. Thomas\n\nProceedings: \"The Proceedings of the ICMS Workshop on Geometric and Combinatorial Methods in Group Theory\" (A.J. Duncan, N.D. Gilbert and J. Howie (eds.), Cambridge University Press, 1994), 29-42\n\nAbstract: The purpose of this paper is first to give a survey of some recent results concerning semigroup presentations, and then to prove a new result which enables us to describe the structure of semigroups defined by certain presentations. The main theme is to relate the semigroup S defined by a presentation P to the group G defined by P. After mentioning a result of Adjan's giving a sufficient condition for S to embed in G, we consider some cases where S maps surjectively (but not necessarily injectively) onto G. In these examples, we find that S has minimal left and right ideals, and it turns out that this is a sufficient condition for S to map onto G. In this case, the kernel of S (i.e. the unique minimal two-sided ideal of S) is a disjoint union of pairwise isomorphic groups, and we describe a necessary and sufficient condition for these groups to be isomorphic to G. We then move on and expand on these results by proving a new result, which is a sort of rewriting theorem, enabling us to determine the presentations of the groups in the kernel in certain cases. We finish off by applying this new result to certain semigroup presentations and by pointing out its limitations.\n\n### Semigroup and group presentations\n\nAuthors: C.M. Campbell, E.F. Robertson N. Ruskuc and R.M. Thomas\n\nJournal: Bull. London Math. Soc. 27 (1995), 46-50\n\nAbstract: Let P be a semigroup presentation, let S be the semigroup defined by P, and let G be the group defined by P. We prove that if S has both minimal left ideals and minimal right ideal, then the natural homomorphism f:S->G; is onto. The restriction of f to a maximal subgroup H of the minimal two-sided ideal I of S is a group epimorphism. We prove that this restriction is an isomorphism if and only if the idempotents of I are closed under multiplication. Finally, we apply the obtained results to describe the structure of the semigroups defined by a semigroup variant of (l,m,n)-presentations.\n\n### Rewriting a semigroup presentation\n\nAuthors: C.M. Campbell, E.F. Robertson N. Ruskuc and R.M. Thomas\n\nJournal: Internat. J. Algebra Comput. 5 (1995), 81-103\n\nAbstract: Let S be a finitely presented semigroup having a minimal left ideal L and a minimal right ideal R. The main result gives a presentation for the group R intersection L. It is obtained by rewriting the relations of S, using the actions of S on its minimal left and minimal right ideals. This allows the structure of the minimal two-sided ideal of S to be described explicitly in terms of a Rees matrix semigroup. These results are applied to the Fibonacci semigroups, proving the conjecture that S(r,n,k) is infinite if g.c.d.(n,k)>1 and g.c.d.(n,r+k-1)>1. Two enumeration procedures, related to rewriting the presentation of S into a presentation for R intersection L, are described. The first enumerates the minimal left and minimal right ideals of S, and gives the actions of S on these ideals. The second enumerates the idempotents of the minimal two-sided ideal of S.\n\n### Reidemeister-Schreier type rewriting for semigroups\n\nAuthors: C.M. Campbell, E.F. Robertson N. Ruskuc and R.M. Thomas\n\nJournal: Semigroup Forum 51 (1995), 47-62\n\nAbstract: In this paper we develop a general method for finding presentations for subsemigroups of semigroups defined by presentations. This method is based on the idea of rewriting, akin to the Reidemeister-Schreier method for groups. We also give two applications of this method. The first gives a presentation for a two-sided ideal of a semigroup, and implies that a two-sided ideal of finite index in a finitely presented semigroup is itself finitely presented. The second gives a presentation for the Schutzenberger group of a 0-minimal two-sided ideal, and implies that this group is finitely presented if the ideal contains either finitely many 0-minimal left ideals or finitely many 0-minimal right ideals.\n\n### On semigroups defined by Coxeter type presentations\n\nAuthors: C.M. Campbell, E.F. Robertson N. Ruskuc and R.M. Thomas\n\nJournal: Proc. Royal Soc. Edinburgh , 125 (1995), 1063-1075\n\nAbstract: Presentations of Coxeter type are defined for semigroups. Minimal right ideals of a semigroup defined by such a presentation are proved to be isomorphic to the group with the same presentation. A necessary and sufficient condition for these semigroups to be finite is found. The structure of semigroups defined by Coxeter type presentations for the symmetric and alternating groups is examined in detail.\n\n### Subsemigroups of finitely presented semigroups\n\nAuthors: C.M. Campbell, E.F. Robertson N. Ruskuc and R.M. Thomas\n\nJournal: J. Algebra 180 (1996), 1-21\n\nAbstract: In this paper we investigate subsemigroups of finitely presented semigroups with respect to the properties of being finitely generated or finitely presented. We prove that a right ideal of finite index in a finitely presented semigroup is itself finitely presented. We also prove that in a free semigroup of finite rank a subsemigroup of finite index is finitely presented, and that any ideal which is finitely generated as a subsemigroup is finitely presented.\n\n### Certain one-relator products of semigroups\n\nAuthors: C.M. Campbell, E.F. Robertson, N. Ruskuc, R.M. Thomas and Y. Unlu\n\nJournal: Comm. Algebra 23 (1995), 5207-5219\n\nAbstract: We apply some recent results to investigate finiteness and structure of some (semigroup) one-relator products of two cyclic groups.\n\n### Central factors of deficiency zero groups\n\nAuthors: G. Havas and E.F. Robertson\n\nJournal: Comm. Algebra 24 (1996), 3483-3487.\n\nAbstract: We answer a question of some twenty years standing: are the central factors of nilpotent groups of deficiency zero 3-generated? We prove a negative answer by giving an explicit presentation for a 3-generator, 3-relator group of order 131072 and class 5 which has central factors which are 4-generated but not 3-generated. We outline the computational techniques which lead to this result.\n\n### On subsemigroups and ideals in free products of semigroups\n\nAuthors: C.M. Campbell, E.F. Robertson N. Ruskuc and R.M. Thomas\n\nJournal: Internat. J. Algebra Comput. , to appear\n\nAbstract: Subsemigroups and ideals of free products of semigroups are studied with respect to the properties of being finitely generated or finitely presented. It is proved that the free product of any two semigroups, at least one of which is non-trivial, contains a two-sided ideal which is not finitely generated as a semigroup, and also contains a subsemigroup which is finitely generated but not finitely presented. By way of contrast, in the free product of two trivial semigroups, every subsemigroup is finitely generated and finitely presented. Further, it is proved that an ideal of a free product of finitely presented semigroups, which is finitely generated as a semigroup, is also finitely presented. It is not known whether one-sided ideals of free products have the same property, but it is shown that they do when the free factors are free commutative.\n\n### Symmetric presentations and orthogonal groups\n\nAuthors: C.M. Campbell, G. Havas, E.F. Robertson and S.A. Linton\n\nJournal: Proc. ATLAS 10 years on, Cambridge University Press , to appear\n\nAbstract: We examine series of finite presentations which are invariant under the full symmetric group acting on the set of generators. Evidence from computational experiments reveals a remarkable tendency for the groups in these series to be closely related to the orthogonal groups. We examine cases of finite groups in such series and look in detail at an infinite group with such a presentation. We prove some theoretical results about 3-generator symmetric presentations and make a number of conjectures regarding n-generator symmetric presentations.\n\n### Generators and relations of direct products of semigroups\n\nAuthors: E.F. Robertson N. Ruskuc and J. Wiegold\n\nJournal: , submitted\n\nAbstract: Let S and T be two infinite semigroups. It is shown that S x T is finitely generated if and only if S and T are finitely generated and S^2 = S and T^2 = T. Further, necessary and sufficient conditions are given on S and T for S x T to be finitely presented. The conditions are applied to find a finite semigroup S with 11 elements and the property that given any infinite finitely presented semigroup T with T^2 = T then S x T is finitely generated but not finitely presented.\n\n### Wilsonian products in finite semigroups\n\nAuthors: P.Z. Hermann, E.F. Robertson and N. Ruskuc\n\nJournal: , submitted\n\nAbstract: Let S be a finite semigroup. Consider the set p(S) of all elements of S which can be represented as a product of all the elements of S in some order. It is shown that p(S) is contained in the minimal ideal M of S and intersects each maximal subgroup H of M in essentially the same way. It is also shown that p(S) intersects H in a union of cosets of H'.\n\n### Decidability questions in finitely presented semigroups\n\nAuthors: C.M. Campbell, E.F. Robertson N. Ruskuc and R.M. Thomas\n\nJournal: , in preparation\n\nAbstract: We examine some properties of finitely presented semigroups and investigate whether they are recursively enumerable and whether they are recursive.\n\n### Computing transformation monoids\n\nAuthors: S.A. Linton, G. Pfeiffer, E.F. Robertson and N. Ruskuc\n\nJournal: , in preparation\n\nAbstract: Efficient computational methods are available for computing with finite groups of permutations. In this paper we utilise such methods in developing an algorithm to compute the order and algebraic structure of finite transformation monoids. We start from an algorithm due to Lallement and McFadden and develop some theoretical improvements. The underlying strategy is to translate computations in the transformation monoid into computations in some permutation groups.\n\n### Groups and actions in transformation semigroups\n\nAuthors: S.A. Linton, G. Pfeiffer, E.F. Robertson and N. Ruskuc\n\nJournal: Mathematische Zeitschrift , to appear\n\nAbstract:\n\n### Presentations for subsemigroups- applications to ideals of semigroups\n\nAuthors: C.M. Campbell, E.F. Robertson N. Ruskuc and R.M. Thomas\n\nJournal: J. Pure Appl. Algebra , to appear\n\nAbstract: A method for finding presentations for subsemigroups of semigroups defined by presentations is used to investigate when ideals are finitely presented. It is shown that an ideal of a finitely presented semigroup is not necessarily finitely presented, even if it is finitely generated as a semigroup. By way of contrast, it is then proved that in a free product of two, or indeed of finitely many, finite semigroups, every right ideal which is finitely generated as a semigroup is finitely presented.\n\nBack to the list of publications ." ]

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[ "# 5178813\n\n## 5,178,813 is an odd composite number composed of three prime numbers multiplied together.\n\nWhat does the number 5178813 look like?\n\nThis visualization shows the relationship between its 3 prime factors (large circles) and 8 divisors.\n\n5178813 is an odd composite number. It is composed of three distinct prime numbers multiplied together. It has a total of eight divisors.\n\n## Prime factorization of 5178813:\n\n### 3 × 1193 × 1447\n\nSee below for interesting mathematical facts about the number 5178813 from the Numbermatics database.\n\n### Names of 5178813\n\n• Cardinal: 5178813 can be written as Five million, one hundred seventy-eight thousand, eight hundred thirteen.\n\n### Scientific notation\n\n• Scientific notation: 5.178813 × 106\n\n### Factors of 5178813\n\n• Number of distinct prime factors ω(n): 3\n• Total number of prime factors Ω(n): 3\n• Sum of prime factors: 2643\n\n### Divisors of 5178813\n\n• Number of divisors d(n): 8\n• Complete list of divisors:\n• Sum of all divisors σ(n): 6915648\n• Sum of proper divisors (its aliquot sum) s(n): 1736835\n• 5178813 is a deficient number, because the sum of its proper divisors (1736835) is less than itself. Its deficiency is 3441978\n\n### Bases of 5178813\n\n• Binary: 100111100000101101111012\n• Base-36: 32ZZX\n\n### Squares and roots of 5178813\n\n• 5178813 squared (51788132) is 26820104088969\n• 5178813 cubed (51788133) is 138896303717305813797\n• The square root of 5178813 is 2275.7005514785\n• The cube root of 5178813 is 173.0122058057\n\n### Scales and comparisons\n\nHow big is 5178813?\n• 5,178,813 seconds is equal to 8 weeks, 3 days, 22 hours, 33 minutes, 33 seconds.\n• To count from 1 to 5,178,813 would take you about twelve weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 5178813 cubic inches would be around 14.4 feet tall.\n\n### Recreational maths with 5178813\n\n• 5178813 backwards is 3188715\n• The number of decimal digits it has is: 7\n• The sum of 5178813's digits is 33\n• More coming soon!" ]

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[ "# Pyspark Dataframes as View\n\nFor a script that I am running, I have a bunch of chained views that looked at a specific set of data in sql (I am using Apache Spark SQL):\n\n``````%sql\ncreate view view_1 as\nselect column_1,column_2 from original_data_table\n``````\n\nThis logic culminates in `view_n`. However, I then need to perform logic that is difficult (or impossible) to implement in sql, specifically, the `explode` command:\n\n``````%python\ndf_1 = sqlContext.sql(\"SELECT * from view_n\")\ndf1_exploded=df_1.withColumn(\"exploded_column\", explode(split(df_1f.col_to_explode,',')))\n``````\n\n# My Questions:\n\n1. Is there a speed cost associated with switching to and from sql tables to pyspark dataframes? Or, since pyspark dataframes are lazily evaluated, is it very similair to a view?\n\n2. Is there a better way of switching from and sql table to a pyspark dataframe?", null, "On Best Solutions\n\nYou can use `explode()` and just about anything that DF has via Spark SQL (https://spark.apache.org/docs/latest/api/sql/index.html)\n\n``````print(spark.version)\n2.4.3\n\ndf = spark.createDataFrame([(1, [1,2,3]), (2, [4,5,6]), (3, [7,8,9]),],[\"id\", \"nest\"])\ndf.printSchema()\n\nroot\n|-- id: long (nullable = true)\n|-- nest: array (nullable = true)\n| |-- element: long (containsNull = true)\n\ndf.createOrReplaceTempView(\"sql_view\")\nspark.sql(\"SELECT id, explode(nest) as un_nest FROM sql_view\").show()\n\ndf.createOrReplaceTempView(\"sql_view\")\nspark.sql(\"SELECT id, explode(nest) as flatten FROM sql_view\").show()\n\n+---+-------+\n| id|flatten|\n+---+-------+\n| 1| 1|\n| 1| 2|\n| 1| 3|\n| 2| 4|\n| 2| 5|\n| 2| 6|\n| 3| 7|\n| 3| 8|\n| 3| 9|\n+---+-------+\n\n``````" ]

[ null, "https://techqa.club/v/q/pyspark-dataframes-as-view-c3RhY2tvdmVyZmxvd3w1NjExNDUyNg==", null ]

[ "# Free download audio books for android Mathematical Maturity via Discrete Mathematics\n\n## Mathematical Maturity via Discrete Mathematics. Vadim Ponomarenko", null, "Mathematical-Maturity-via.pdf\nISBN: 9780486838571 | 272 pages | 7 Mb", null, "• Mathematical Maturity via Discrete Mathematics\n• Page: 272\n• Format: pdf, ePub, fb2, mobi\n• ISBN: 9780486838571\n• Publisher: Dover Publications\n\n### Free download audio books for android Mathematical Maturity via Discrete Mathematics\n\nDesigned for a one-semester course for undergraduate majors in math, computer science, and computer engineering, this text helps students take the crucial step from consuming mathematics to producing mathematics. Author Vadim Ponomarenko employs the general concept of discrete mathematics to introduce the basic knowledge of proof techniques and their uses. Like other beginning texts on methods of proof, this treatment offers definitions, theorems, and techniques. Unlike other books, it explains how to read, interpret, and use definitions, demonstrating not only general proof strategies — like proof of induction — but also the specific methods of thought for implementing these strategies. All techniques are built from scratch to provide an intellectually consistent whole. Each chapter contains several exercises, for which the author provides hints rather than solutions to encourage creative thinking.\n\ndiscrete mathematics - Dover Books on Mathematics - Amazon.com\n1-16 of 28 results for Dover Books on Mathematics : \"discrete mathematics\" . Mathematical Maturity via Discrete Mathematics (Dover Books on Mathematics). Discrete Mathematics, 7th Edition: Richard Johnsonbaugh - Amazon\nBuy Discrete Mathematics, 7th Edition on Amazon.com ✓ FREE SHIPPING on and construct proofs - and, generally, expanding their mathematical maturity - this If you are a seller for this product, would you like to suggest updates through  Mathematical Maturity via Discrete Mathematics - Google Books Result\nVadim Ponomarenko - ‎2019 - Mathematics Is calculus necessary? - Harvard Math Department\nIt needs also mathematical maturity from the teacher. A core Calculus is inherent in every other subject, even discrete structures. Discrete Whoever calls for using stats as a replacement of calculus does not know stats. Mathematical Maturity Via Discrete Mathematics - Amazon.ca\nMathematical Maturity Via Discrete Mathematics: Vadim Ponomarenko: 9780486838571: Books - Amazon.ca. Technological and Pedagogical Innovations for Teaching\nMaturity. Math is useful for CS. Figure 2.1: A pre-requisite map for discrete math related . students are using (and it often is), the first non-programming course  Mathematical maturity via discrete mathematics - Gibert\nMathematical maturity via discrete mathematics. Livre | Date de parution : 13/11/ 2019. Soyez le premier à commenter ce produit. Offre internet. Neuf - 14,95 €. Discrete Mathematics And Its Applications 6th Ed\nmathematical maturity is needed to study discrete mathematics in a .. Instructors using this book can adjust the level of difficulty of their course by choosing. Mathematical Maturity via Discrete Mathematics - Mighty Ape\nBuy Mathematical Maturity via Discrete Mathematics by Vadim Ponomarenko for \\$34.00 at Mighty Ape NZ. Pre-order for NZ release day delivery. Designed for a  Mathematical Maturity Via Discrete Mathematics by - Dymocks\nMathematical Maturity Via Discrete Mathematics from Dymocks online bookstore. PaperBack by Vadim Ponomarenko. Spiral Notebook for Discrete Mathematics\nsee that discrete mathematics forms the foundation of many mathematical . All of these topics are crucial in the development of your mathematical maturity. Give a formula (using appropriate symbols) for each of these statements: (a) Sam   Mathematical Maturity via Discrete Mathematics - Vadim - Foyles\nMathematical Maturity via Discrete Mathematics (Paperback). Vadim Ponomarenko. £10.99. Pre-order. Despatched on publication 31/12/2019. Pre- Order.\n\n0コメント\n\n• 1000 / 1000" ]

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[ "2017年3月30日 星期四\n\n[ Intro2ML ] Ch6. Model Evaluation and Improvement - Grid Search\n\nIntroduction\nNow that we know how to evaluate how well a model generalizes, we can take the next step and improve the model’s generalization performance by tuning its parameters. We discussed the parameter settings of many of the algorithms in scikit-learn in Chapters 2 and 3, and it is important to understand what the parameters mean before trying to adjust them. Finding the values of the important parameters of a model (the ones that provide the best generalization performance) is a tricky task, but necessary for almost all models and datasets. Because it is such a common task, there are standard methods in scikit-learn to help you with it. The most commonly used method is grid search, which basically means trying all possible combinations of the parameters of interest.\n\nConsider the case of a kernel SVM with an RBF (radial basis function) kernel, as implemented in the SVC class. As we discussed in Chapter 2, there are two important parameters: the kernel bandwidth, gamma, and the regularization parameter, C. Say we want to try the values 0.001, 0.01, 0.1, 1, 10, and 100 for the parameter C, and the same for gamma. Because we have six different settings for C and gamma that we want to try, we have 36 combinations of parameters in total. Looking at all possible combinations creates a table (or grid) of parameter settings for the SVM, as shown here:\n\nSimple Grid Search\nWe can implement a simple grid search just as for loops over the two parameters, training and evaluating a classifier for each combination:\n- ch6_t09.py\n1. from sklearn.model_selection import cross_val_score\n3. from sklearn.svm import SVC\n4. from sklearn.model_selection import train_test_split\n5.\n7. X_train, X_test, y_train, y_test = train_test_split(\n8.     iris.data, iris.target, random_state=0)\n9. print(\"Size of training set: {}   size of test set: {}\".format(X_train.shape, X_test.shape))\n10.\n11. best_score = 0\n12. for gamma in [0.0010.010.1110100]:\n13.     for C in [0.0010.010.1110100]:\n14.         # for each combination of parameters, train an SVC\n15.         svm = SVC(gamma=gamma, C=C)\n16.         svm.fit(X_train, y_train)\n17.         # evaluate the SVC on the test set\n18.         score = svm.score(X_test, y_test)\n19.         # if we got a better score, store the score and parameters\n20.         if score > best_score:\n21.             best_score = score\n22.             best_parameters = {'C': C, 'gamma': gamma}\n23.\n24. print(\"Best score: {:.2f}\".format(best_score))\n25. print(\"Best parameters: {}\".format(best_parameters))\nOutput:\nSize of training set: 112 size of test set: 38\nBest score: 0.97\nBest parameters: {'C': 100, 'gamma': 0.001}\n\nThe Danger of Overfitting the Parameters and the Validation Set\nGiven this result, we might be tempted to report that we found a model that performs with 97% accuracy on our dataset. However, this claim could be overly optimistic (or just wrong), for the following reason: we tried many different parameters and selected the one with best accuracy on the test set, but this accuracy won’t necessarily carry over to new data. Because we used the test data to adjust the parameters, we can no longer use it to assess how good the model is. This is the same reason we needed to split the data into training and test sets in the first place; we need an independent dataset to evaluate, one that was not used to create the model.\n\nOne way to resolve this problem is to split the data again, so we have three sets: the training set to build the model, the validation (or developmentset to select the parameters of the model, and the test set to evaluate the performance of the selected parameters. Figure 6-5 shows what this looks like:\nFigure 6-5. A three fold split of data into training set, validation set, and test set\n\nAfter selecting the best parameters using the validation set, we can rebuild a model using the parameter settings we found, but now training on both the training data and the validation data. This way, we can use as much data as possible to build our model. This leads to the following implementation:\n- ch6_t10.py\n2. from sklearn.svm import SVC\n3. from sklearn.model_selection import train_test_split\n4.\n6. # split data into train+validation set and test set\n7. X_trainval, X_test, y_trainval, y_test = train_test_split(iris.data, iris.target, random_state=0)\n8. # split train+validation set into training and validation sets\n9. X_train, X_valid, y_train, y_valid = train_test_split(X_trainval, y_trainval, random_state=1)\n10. print(\"Size of training set: {}   size of validation set: {}   size of test set:\"\n11.       \" {}\\n\".format(X_train.shape, X_valid.shape, X_test.shape))\n12.\n13.\n14. best_score = 0\n15. for gamma in [0.0010.010.1110100]:\n16.     for C in [0.0010.010.1110100]:\n17.         # for each combination of parameters, train an SVC\n18.         svm = SVC(gamma=gamma, C=C)\n19.         svm.fit(X_train, y_train)\n20.         # evaluate the SVC on the test set\n21.         score = svm.score(X_valid, y_valid)\n22.         # if we got a better score, store the score and parameters\n23.         if score > best_score:\n24.             best_score = score\n25.             best_parameters = {'C': C, 'gamma': gamma}\n26.\n27. # rebuild a model on the combined training and validation set,\n28. # and evaluate it on the test set\n29. svm = SVC(**best_parameters)\n30. svm.fit(X_trainval, y_trainval)\n31. test_score = svm.score(X_test, y_test)\n32. print(\"Best score on validation set: {:.2f}\".format(best_score))\n33. print(\"Best parameters: \", best_parameters)\n34. print(\"Test set score with best parameters: {:.2f}\".format(test_score))\nOutput:\nSize of training set: 84 size of validation set: 28 size of test set: 38\n\nBest score on validation set: 0.96\n('Best parameters: ', {'C': 10, 'gamma': 0.001})\nTest set score with best parameters: 0.92\n\nThe best score on the validation set is 96%: slightly lower than before, probably because we used less data to train the model (X_train is smaller now because we split our dataset twice). However, the score on the test set—the score that actually tells us how well we generalize—is even lower, at 92%. So we can only claim to classify new data 92% correctly, not 97% correctly as we thought before!\n\nThe distinction between the training set, validation set, and test set is fundamentally important to applying machine learning methods in practice. Any choices made based on the test set accuracy “leak” information from the test set into the model. Therefore, it is important to keep a separate test set, which is only used for the final evaluation. It is good practice to do all exploratory analysis and model selection using the combination of a training and a validation set, and reserve the test set for a final evaluation—this is even true for exploratory visualization. Strictly speaking, evaluating more than one model on the test set and choosing the better of the two will result in an overly optimistic estimate of how accurate the model is.\n\nGrid Search with Cross-Validation\nWhile the method of splitting the data into a training, a validation, and a test set that we just saw is workable, and relatively commonly used, it is quite sensitive to how exactly the data is split. From the output of the previous code snippet we can see that GridSearchCV selects 'C': 10, 'gamma': 0.001 as the best parameters, while the output of the code in the previous section selects 'C': 100, 'gamma': 0.001 as the best parameters. For a better estimate of the generalization performance, instead of using a single split into a training and a validation set, we can use cross-validation to evaluate the performance of each parameter combination. This method can be coded up as follows:\n- ch6_t11.py\n1. import numpy as np\n3. from sklearn.svm import SVC\n4. from sklearn.model_selection import cross_val_score\n5. from sklearn.model_selection import train_test_split\n6.\n8. # split data into train+validation set and test set\n9. X_trainval, X_test, y_trainval, y_test = train_test_split(iris.data, iris.target, random_state=0)\n10. # split train+validation set into training and validation sets\n11. X_train, X_valid, y_train, y_valid = train_test_split(X_trainval, y_trainval, random_state=1)\n12. print(\"Size of training set: {}   size of validation set: {}   size of test set:\"\n13.       \" {}\\n\".format(X_train.shape, X_valid.shape, X_test.shape))\n14.\n15.\n16. best_score = 0\n17. for gamma in [0.0010.010.1110100]:\n18.     for C in [0.0010.010.1110100]:\n19.         # for each combination of parameters, train an SVC\n20.         svm = SVC(gamma=gamma, C=C)\n21.         # perform cross-validation\n22.         scores = cross_val_score(svm, X_trainval, y_trainval, cv=5)\n23.         # compute mean cross-validation accuracy\n24.         score = np.mean(scores)\n25.         # if we got a better score, store the score and parameters\n26.         if score > best_score:\n27.             best_score = score\n28.             best_parameters = {'C': C, 'gamma': gamma}\n29.\n30. # rebuild a model on the combined training and validation set,\n31. # and evaluate it on the test set\n32. svm = SVC(**best_parameters)\n33. svm.fit(X_trainval, y_trainval)\n34. test_score = svm.score(X_test, y_test)\n35. print(\"Best score on validation set: {:.2f}\".format(best_score))\n36. print(\"Best parameters: \", best_parameters)\n37. print(\"Test set score with best parameters: {:.2f}\".format(test_score))\nTo evaluate the accuracy of the SVM using a particular setting of C and gamma using five-fold cross-validation, we need to train 36 * 5 = 180 models. As you can imagine, the main downside of the use of cross-validation is the time it takes to train all these models\nWARNING\nAs we said earlier, cross-validation is a way to evaluate a given algorithm on a specific dataset. However, it is often used in conjunction with parameter search methods like grid search. For this reason, many people use the term cross-validation colloquially to refer to grid search with cross-validation.\n\nThe overall process of splitting the data, running the grid search, and evaluating the final parameters is illustrated in Figure 6-7:\nFigure 6-7. Overview of the process of parameter selection and model evaluation with GridSearchCV\n\nBecause grid search with cross-validation is such a commonly used method to adjust parameters, scikit-learn provides the GridSearchCV class, which implements it in the form of an estimator. To use the GridSearchCV class, you first need to specify the parameters you want to search over using a dictionary. GridSearchCV will then perform all the necessary model fits. The keys of the dictionary are the names of parameters we want to adjust (as given when constructing the model—in this case, C and gamma), and the values are the parameter settings we want to try out. Trying the values 0.001, 0.01, 0.1, 1, 10, and 100 for C and gamma translates to the following dictionary:\n1. param_grid = {'C': [0.0010.010.1110100],\n2.               'gamma': [0.0010.010.1110100]}\n3. print(\"Parameter grid:\\n{}\".format(param_grid))\nOutput:\nParameter grid:\n{'C': [0.001, 0.01, 0.1, 1, 10, 100], 'gamma': [0.001, 0.01, 0.1, 1, 10, 100]}\n\nWe can now instantiate the GridSearchCV class with the model (SVC), the parameter grid to search (param_grid), and the cross-validation strategy we want to use (say, five-fold stratified cross-validation):\n1. from sklearn.model_selection import GridSearchCV\n2. from sklearn.svm import SVC\n3. grid_search = GridSearchCV(SVC(), param_grid, cv=5)\nGridSearchCV will use cross-validation in place of the split into a training and validation set that we used before. However, we still need to split the data into a training and a test set, to avoid overfitting the parameters:\n1. X_train, X_test, y_train, y_test = train_test_split( iris.data, iris.target, random_state=0)\nThe grid_search object that we created behaves just like a classifier; we can call the standard methods fitpredict, and score on it. However, when we call fit, it will run cross-validation for each combination of parameters we specified in param_grid\n1. grid_search.fit(X_train, y_train)\nFitting the GridSearchCV object not only searches for the best parameters, but also automatically fits a new model on the whole training dataset with the parameters that yielded the best cross-validation performance. The GridSearchCV class provides a very convenient interface to access the retrained model using the predict and score methods. To evaluate how well the best found parameters generalize, we can call score on the test set:\n1. print(\"Test set score: {:.2f}\".format(grid_search.score(X_test, y_test)))\nOutput:\nTest set score: 0.97\n\nChoosing the parameters using cross-validation, we actually found a model that achieves 97% accuracy on the test set. The important thing here is that we did not use the test set to choose the parameters. The parameters that were found are scored in the best_params_ attribute, and the best cross-validation accuracy (the mean accuracy over the different splits for this parameter setting) is stored in best_score_\n1. print(\"Best parameters: {}\".format(grid_search.best_params_))\n2. print(\"Best cross-validation score: {:.2f}\".format(grid_search.best_score_))\nOutput:\nBest parameters: {'C': 100, 'gamma': 0.01}\nBest cross-validation score: 0.97\n\nWARNING\nAgain, be careful not to confuse best_score_ with the generalization performance of the model as computed by the score method on the test set. Using the score method (or evaluating the output of the predict method) employs a model trained on the whole training set. The best_score_ attribute stores the mean cross-validation accuracy, with cross-validation performed on the training set.\n\nSometimes it is helpful to have access to the actual model that was found—for example, to look at coefficients or feature importances. You can access the model with the best parameters trained on the whole training set using the best_estimator_ attribute:\n1. print(\"Best estimator:\\n{}\".format(grid_search.best_estimator_))\nOutput:\nBest estimator:\nSVC(C=100, cache_size=200, class_weight=None, coef0=0.0,\ndecision_function_shape=None, degree=3, gamma=0.01, kernel='rbf',\nmax_iter=-1, probability=False, random_state=None, shrinking=True,\ntol=0.001, verbose=False)\n\nANALYZING THE RESULT OF CROSS-VALIDATION\nIt is often helpful to visualize the results of cross-validation, to understand how the model generalization depends on the parameters we are searching. As grid searches are quite computationally expensive to run, often it is a good idea to start with a relatively coarse and small grid. We can then inspect the results of the cross-validated grid search, and possibly expand our search. The results of a grid search can be found in the cv_results_ attribute, which is a dictionary storing all aspects of the search. It contains a lot of details, as you can see in the following output, and is best looked at after converting it to a pandas DataFrame: (ch6_t12.py\n1. from ch6_t12 import *\n2. import pandas as pd\n3. # convert to DataFrame\n4. results = pd.DataFrame(grid_search.cv_results_)\n5. # show the first 5 rows\nOutput:\n\nEach row in results corresponds to one particular parameter setting. For each setting, the results of all cross-validation splits are recorded, as well as the mean and standard deviation over all splits. As we were searching a two-dimensional grid of parameters (C and gamma), this is best visualized as a heat map (Figure 6-8). First we extract the mean validation scores, then we reshape the scores so that the axes correspond to C and gamma\n- ch6_t13.py\n1. import numpy as np\n3. from sklearn.svm import SVC\n4. from sklearn.model_selection import GridSearchCV\n5. from sklearn.model_selection import cross_val_score\n6. from sklearn.model_selection import train_test_split\n7.\n8. from ch6_t12 import *\n9. import pandas as pd\n10. # convert to DataFrame\n11. results = pd.DataFrame(grid_search.cv_results_)\n12. # show the first 5 rows\n14.\n15. import mglearn\n16. scores = np.array(results.mean_test_score).reshape(66)\n17.\n18. # plot the mean cross-validation scores\n19. mglearn.tools.heatmap(scores, xlabel='gamma', xticklabels=param_grid['gamma'],\n20.                       ylabel='C', yticklabels=param_grid['C'], cmap=\"viridis\")\nFigure 6-8. Heat map of mean cross-validation score as a function of C and gamma\n\nEach point in the heat map corresponds to one run of cross-validation, with a particular parameter setting. The color encodes the cross-validation accuracy, with light colors meaning high accuracy and dark colors meaning low accuracy. You can see that SVC is very sensitive to the setting of the parameters. For many of the parameter settings, the accuracy is around 40%, which is quite bad; for other settings the accuracy is around 96%. We can take away from this plot several things. First, the parameters we adjusted are very important for obtaining good performance. Both parameters (C and gammamatter a lot, as adjusting them can change the accuracy from 40% to 96%. Additionally, the ranges we picked for the parameters are ranges in which we see significant changes in the outcome. It’s also important to note that the ranges for the parameters are large enough: the optimum values for each parameter are not on the edges of the plot.\n\nNow let’s look at some plots (shown in Figure 6-9) where the result is less ideal, because the search ranges were not chosen properly:\nFigure 6-9. Heat map visualizations of misspecified search grids\n\nThe first panel shows no changes at all, with a constant color over the whole parameter grid. In this case, this is caused by improper scaling and range of the parameters C and gamma. However, if no change in accuracy is visible over the different parameter settings, it could also be that a parameter is just not important at all. It is usually good to try very extreme values first, to see if there are any changes in the accuracy as a result of changing a parameter; The second panel shows a vertical stripe pattern. This indicates that only the setting of the gamma parameter makes any difference. This could mean that the gamma parameter is searching over interesting values but the C parameter is not—or it could mean the C parameter is not important.\n\nThe third panel shows changes in both C and gamma. However, we can see that in the entire bottom left of the plot, nothing interesting is happening. We can probably exclude the very small values from future grid searches. The optimum parameter setting is at the top right. As the optimum is in the border of the plot, we can expect that there might be even better values beyond this border, and we might want to change our search range to include more parameters in this region.\n\nTuning the parameter grid based on the cross-validation scores is perfectly fine, and a good way to explore the importance of different parameters. However, you should not test different parameter ranges on the final test set—as we discussed earlier, evaluation of the test set should happen only once we know exactly what model we want to use\n\nSEARCH OVER SPACES THAT ARE NOT GRIDS\nIn some cases, trying all possible combinations of all parameters as GridSearchCV usually does, is not a good idea. For example, SVC has a kernel parameter, and depending on which kernel is chosen, other parameters will be relevant. If kernel='linear', the model is linear, and only the C parameter is used. If kernel='rbf', both the C and gamma parameters are used (but not other parameters like degree). In this case, searching over all possible combinations of Cgamma, and kernel wouldn’t make sense: if kernel='linear', gamma is not used, and trying different values for gamma would be a waste of time. To deal with these kinds of “conditional” parameters, GridSearchCV allows the param_grid to be a list of dictionaries. Each dictionary in the list is expanded into an independent grid. A possible grid search involving kernel and parameters could look like this:\n1. param_grid = [{'kernel': ['rbf'],\n2.                'C': [0.0010.010.1110100],\n3.                'gamma': [0.0010.010.1110100]},\n4.               {'kernel': ['linear'],\n5.                'C': [0.0010.010.1110100]}]\n6. print(\"List of grids:\\n{}\".format(param_grid))\nIn the first grid, the kernel parameter is always set to 'rbf' (not that the entry for kernel is a list of length one), and both the C and gamma parameters are varied. In the second grid, the kernel parameter is always set to linear, and only C is varied. Now let’s apply this more complex parameter search:\n1. grid_search = GridSearchCV(SVC(), param_grid, cv=5)\n2. grid_search.fit(X_train, y_train)\n3. print(\"Best parameters: {}\".format(grid_search.best_params_))\n4. print(\"Best cross-validation score: {:.2f}\".format(grid_search.best_score_))\nOutput:\nBest parameters: {'C': 100, 'kernel': 'rbf', 'gamma': 0.01}\nBest cross-validation score: 0.97\n\nUSING DIFFERENT CROSS-VALIDATION STRATEGIES WITH GRID SEARCH\nSimilarly to cross_val_scoreGridSearchCV uses stratified k-fold cross-validation by default for classification, and k-fold cross-validation for regression. However, you can also pass any cross-validation splitter, as described in “More control over cross-validation”, as the cv parameter in GridSearchCV. In particular, to get only a single split into a training and a validation set, you can use ShuffleSplit or StratifiedShuffleSplit with n_iter=1. This might be helpful for very large datasets, or very slow models.\n\nNESTED CROSS-VALIDATION\nIn the preceding examples, we went from using a single split of the data into training, validation, and test sets to splitting the data into training and test sets and then performing cross-validation on the training set. But when using GridSearchCV as described earlier, we still have a single split of the data into training and test sets, which might make our results unstable and make us depend too much on this single split of the data. We can go a step further, and instead of splitting the original data into training and test sets once, use multiple splits of cross-validation. This will result in what is called nested cross-validation. In nested cross-validation, there is an outer loop over splits of the data into training and test sets. For each of them, a grid search is run (which might result in different best parameters for each split in the outer loop). Then, for each outer split, the test set score using the best settings is reported.\n\nThe result of this procedure is a list of scores—not a model, and not a parameter setting. The scores tell us how well a model generalizes, given the best parameters found by the grid. As it doesn’t provide a model that can be used on new data, nested cross-validation is rarely used when looking for a predictive model to apply to future data. However, it can be useful for evaluating how well a given model works on a particular dataset\n\nImplementing nested cross-validation in scikit-learn is straightforward. We call cross_val_score with an instance of GridSearchCV as the model:\n1. scores = cross_val_score(GridSearchCV(SVC(), param_grid, cv=5),\n2.                          iris.data, iris.target, cv=5)\n3. print(\"Cross-validation scores: \", scores)\n4. print(\"Mean cross-validation score: \", scores.mean())\nOutput:\nCross-validation scores: [ 0.967 1. 0.967 0.967 1. ]\nMean cross-validation score: 0.98\n\nThe result of our nested cross-validation can be summarized as “SVC can achieve 98% mean cross-validation accuracy on the iris dataset”—nothing more and nothing less.\n\nHere, we used stratified five-fold cross-validation in both the inner and the outer loop. As our param_grid contains 36 combinations of parameters, this results in a whopping 36 * 5 * 5 = 900 models being built, making nested cross-validation a very expensive procedure. Here, we used the same cross-validation splitter in the inner and the outer loop; however, this is not necessary and you can use any combination of cross-validation strategies in the inner and outer loops. It can be a bit tricky to understand what is happening in the single line given above, and it can be helpful to visualize it as for loops, as done in the following simplified implementation:\n1. def nested_cv(X, y, inner_cv, outer_cv, Classifier, parameter_grid):\n2.     outer_scores = []\n3.     # for each split of the data in the outer cross-validation\n4.     # (split method returns indices of training and test parts)\n5.     for training_samples, test_samples in outer_cv.split(X, y):\n6.         # find best parameter using inner cross-validation\n7.         best_parms = {}\n8.         best_score = -np.inf\n9.         # iterate over parameters\n10.         for parameters in parameter_grid:\n11.             # accumulate score over inner splits\n12.             cv_scores = []\n13.             # iterate over inner cross-validation\n14.             for inner_train, inner_test in inner_cv.split(\n15.                     X[training_samples], y[training_samples]):\n16.                 # build classifier given parameters and training data\n17.                 clf = Classifier(**parameters)\n18.                 clf.fit(X[inner_train], y[inner_train])\n19.                 # evaluate on inner test set\n20.                 score = clf.score(X[inner_test], y[inner_test])\n21.                 cv_scores.append(score)\n22.             # compute mean score over inner folds\n23.             mean_score = np.mean(cv_scores)\n24.             if mean_score > best_score:\n25.                 # if better than so far, remember parameters\n26.                 best_score = mean_score\n27.                 best_params = parameters\n28.         # build classifier on best parameters using outer training set\n29.         clf = Classifier(**best_params)\n30.         clf.fit(X[training_samples], y[training_samples])\n31.         # evaluate\n32.         outer_scores.append(clf.score(X[test_samples], y[test_samples]))\n33.     return np.array(outer_scores)\nNow, let’s run this function on the iris dataset:\n1. from sklearn.model_selection import ParameterGrid, StratifiedKFold\n2. scores = nested_cv(iris.data, iris.target, StratifiedKFold(5),\n3.           StratifiedKFold(5), SVC, ParameterGrid(param_grid))\n4. print(\"Cross-validation scores: {}\".format(scores))\nOutput:\nCross-validation scores: [ 0.967 1. 0.967 0.967 1. ]\n\nPARALLELIZING CROSS-VALIDATION AND GRID SEARCH\nWhile running a grid search over many parameters and on large datasets can be computationally challenging, it is also embarrassingly parallel. This means that building a model using a particular parameter setting on a particular cross-validation split can be done completely independently from the other parameter settings and models. This makes grid search and cross-validation ideal candidates for parallelization over multiple CPU cores or over a cluster. You can make use of multiple cores in GridSearchCV and cross_val_score by setting the n_jobs parameter to the number of CPU cores you want to use. You can set n_jobs=-1 to use all available cores.\n\nYou should be aware that scikit-learn does not allow nesting of parallel operations. So, if you are using the n_jobs option on your model (for example, a random forest), you cannot use it in GridSearchCV to search over this model. If your dataset and model are very large, it might be that using many cores uses up too much memory, and you should monitor your memory usage when building large models in parallel.\n\nIt is also possible to parallelize grid search and cross-validation over multiple machines in a cluster, although at the time of writing this is not supported within scikit-learn. It is, however, possible to use the IPython parallel framework for parallel grid searches, if you don’t mind writing the for loop over parameters as we did in “Simple Grid Search”. For Spark users, there is also the recently developed spark-sklearn package, which allows running a grid search over an already established Spark cluster.\n\n[ Python 常見問題 ] How to get symbolic link target in Python?\n\nSource From  Here Question How do I extract the target path from a  symbolic link ? HowTo The problem with  os .readlink()  is it will onl..." ]

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[ "# Geometry worksheets - PDF\n\nGeometry math worksheets for children to practice . Suitable PDF printable geometry worksheets for children in the following grades : Pre-k, kindergarten, 1st grade, 2nd grade, 3rd grade, 4th grade, 5th grade, 6th grade and 7th grade. Worksheets cover the following geometry topics: Shapes, symmetry, angles, perimeter, area, volume, Pythagorean Theorem, radius, circumference, complex figures, kinds of triangles, tracing shapes, congruent shapes, segments, open and close curves etc" ]

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[ "Two way ANOVA - overview\n\nThis page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table\n\nTwo way ANOVA\nLogistic regression\nPearson correlation\nIndependent/grouping variablesIndependent variablesVariable 1\nTwo categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \\geqslant 2$, $J \\geqslant 2$)One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variablesOne quantitative of interval or ratio level\nDependent variableDependent variableVariable 2\nOne quantitative of interval or ratio levelOne categorical with 2 independent groupsOne quantitative of interval or ratio level\nNull hypothesisNull hypothesisNull hypothesis\nANOVA $F$ tests:\n• H0 for main and interaction effects together (model): no main effects and interaction effect\n• H0 for independent variable A: no main effect for A\n• H0 for independent variable B: no main effect for B\n• H0 for the interaction term: no interaction effect between A and B\nLike in one way ANOVA, we can also perform $t$ tests for specific contrasts and multiple comparisons. This is more advanced stuff.\nModel chi-squared test for the complete regression model:\n• H0: $\\beta_1 = \\beta_2 = \\ldots = \\beta_K = 0$\nWald test for individual regression coefficient $\\beta_k$:\n• H0: $\\beta_k = 0$\nor in terms of odds ratio:\n• H0: $e^{\\beta_k} = 1$\nLikelihood ratio chi-squared test for individual regression coefficient $\\beta_k$:\n• H0: $\\beta_k = 0$\nor in terms of odds ratio:\n• H0: $e^{\\beta_k} = 1$\nin the regression equation $\\ln \\big(\\frac{\\pi_{y = 1}}{1 - \\pi_{y = 1}} \\big) = \\beta_0 + \\beta_1 \\times x_1 + \\beta_2 \\times x_2 + \\ldots + \\beta_K \\times x_K$. Here $x_i$ represents independent variable $i$, $\\beta_i$ is the regression weight for independent variable $x_i$, and $\\pi_{y = 1}$ represents the true probability that the dependent variable $y = 1$ (or equivalently, the proportion of $y = 1$ in the population) given the scores on the independent variables.\nH0: $\\rho = \\rho_0$\n\nHere $\\rho$ is the Pearson correlation in the population, and $\\rho_0$ is the Pearson correlation in the population according to the null hypothesis (usually 0). The Pearson correlation is a measure for the strength and direction of the linear relationship between two variables of at least interval measurement level.\nAlternative hypothesisAlternative hypothesisAlternative hypothesis\nANOVA $F$ tests:\n• H1 for main and interaction effects together (model): there is a main effect for A, and/or for B, and/or an interaction effect\n• H1 for independent variable A: there is a main effect for A\n• H1 for independent variable B: there is a main effect for B\n• H1 for the interaction term: there is an interaction effect between A and B\nModel chi-squared test for the complete regression model:\n• H1: not all population regression coefficients are 0\nWald test for individual regression coefficient $\\beta_k$:\n• H1: $\\beta_k \\neq 0$\nor in terms of odds ratio:\n• H1: $e^{\\beta_k} \\neq 1$\nIf defined as Wald $= \\dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:\n• H1 right sided: $\\beta_k > 0$\n• H1 left sided: $\\beta_k < 0$\nLikelihood ratio chi-squared test for individual regression coefficient $\\beta_k$:\n• H1: $\\beta_k \\neq 0$\nor in terms of odds ratio:\n• H1: $e^{\\beta_k} \\neq 1$\nH1 two sided: $\\rho \\neq \\rho_0$\nH1 right sided: $\\rho > \\rho_0$\nH1 left sided: $\\rho < \\rho_0$\nAssumptionsAssumptionsAssumptions of test for correlation\n• Within each of the $I \\times J$ populations, the scores on the dependent variable are normally distributed\n• The standard deviation of the scores on the dependent variable is the same in each of the $I \\times J$ populations\n• For each of the $I \\times J$ groups, the sample is an independent and simple random sample from the population defined by that group. That is, within and between groups, observations are independent of one another\n• Equal sample sizes for each group make the interpretation of the ANOVA output easier (unequal sample sizes result in overlap in the sum of squares; this is advanced stuff)\n• In the population, the relationship between the independent variables and the log odds $\\ln (\\frac{\\pi_{y=1}}{1 - \\pi_{y=1}})$ is linear\n• The residuals are independent of one another\n• Variables are measured without error\nAlso pay attention to:\n• Multicollinearity\n• Outliers\n• In the population, the two variables are jointly normally distributed (this covers the normality, homoscedasticity, and linearity assumptions)\n• Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another\nNote: these assumptions are only important for the significance test and confidence interval, not for the correlation coefficient itself. The correlation coefficient just measures the strength of the linear relationship between two variables.\nTest statisticTest statisticTest statistic\nFor main and interaction effects together (model):\n• $F = \\dfrac{\\mbox{mean square model}}{\\mbox{mean square error}}$\nFor independent variable A:\n• $F = \\dfrac{\\mbox{mean square A}}{\\mbox{mean square error}}$\nFor independent variable B:\n• $F = \\dfrac{\\mbox{mean square B}}{\\mbox{mean square error}}$\nFor the interaction term:\n• $F = \\dfrac{\\mbox{mean square interaction}}{\\mbox{mean square error}}$\nNote: mean square error is also known as mean square residual or mean square within.\nModel chi-squared test for the complete regression model:\n• $X^2 = D_{null} - D_K = \\mbox{null deviance} - \\mbox{model deviance}$\n$D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.\nWald test for individual $\\beta_k$:\nThe wald statistic can be defined in two ways:\n• Wald $= \\dfrac{b_k^2}{SE^2_{b_k}}$\n• Wald $= \\dfrac{b_k}{SE_{b_k}}$\nSPSS uses the first definition.\n\nLikelihood ratio chi-squared test for individual $\\beta_k$:\n• $X^2 = D_{K-1} - D_K$\n$D_{K-1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.\nTest statistic for testing H0: $\\rho = 0$:\n• $t = \\dfrac{r \\times \\sqrt{N - 2}}{\\sqrt{1 - r^2}}$\nwhere $r$ is the sample correlation $r = \\frac{1}{N - 1} \\sum_{j}\\Big(\\frac{x_{j} - \\bar{x}}{s_x} \\Big) \\Big(\\frac{y_{j} - \\bar{y}}{s_y} \\Big)$ and $N$ is the sample size\nTest statistic for testing values for $\\rho$ other than $\\rho = 0$:\n• $z = \\dfrac{r_{Fisher} - \\rho_{0_{Fisher}}}{\\sqrt{\\dfrac{1}{N - 3}}}$\n• $r_{Fisher} = \\dfrac{1}{2} \\times \\log\\Bigg(\\dfrac{1 + r}{1 - r} \\Bigg )$, where $r$ is the sample correlation\n• $\\rho_{0_{Fisher}} = \\dfrac{1}{2} \\times \\log\\Bigg( \\dfrac{1 + \\rho_0}{1 - \\rho_0} \\Bigg )$, where $\\rho_0$ is the population correlation according to H0\nPooled standard deviationn.a.n.a.\n\\begin{aligned} s_p &= \\sqrt{\\dfrac{\\sum\\nolimits_{subjects} (\\mbox{subject's score} - \\mbox{its group mean})^2}{N - (I \\times J)}}\\\\ &= \\sqrt{\\dfrac{\\mbox{sum of squares error}}{\\mbox{degrees of freedom error}}}\\\\ &= \\sqrt{\\mbox{mean square error}} \\end{aligned} --\nSampling distribution of $F$ if H0 were trueSampling distribution of $X^2$ and of the Wald statistic if H0 were trueSampling distribution of $t$ and of $z$ if H0 were true\nFor main and interaction effects together (model):\n• $F$ distribution with $(I - 1) + (J - 1) + (I - 1) \\times (J - 1)$ (df model, numerator) and $N - (I \\times J)$ (df error, denominator) degrees of freedom\nFor independent variable A:\n• $F$ distribution with $I - 1$ (df A, numerator) and $N - (I \\times J)$ (df error, denominator) degrees of freedom\nFor independent variable B:\n• $F$ distribution with $J - 1$ (df B, numerator) and $N - (I \\times J)$ (df error, denominator) degrees of freedom\nFor the interaction term:\n• $F$ distribution with $(I - 1) \\times (J - 1)$ (df interaction, numerator) and $N - (I \\times J)$ (df error, denominator) degrees of freedom\nHere $N$ is the total sample size.\nSampling distribution of $X^2$, as computed in the model chi-squared test for the complete model:\n• chi-squared distribution with $K$ (number of independent variables) degrees of freedom\nSampling distribution of the Wald statistic:\n• If defined as Wald $= \\dfrac{b_k^2}{SE^2_{b_k}}$: approximately the chi-squared distribution with 1 degree of freedom\n• If defined as Wald $= \\dfrac{b_k}{SE_{b_k}}$: approximately the standard normal distribution\nSampling distribution of $X^2$, as computed in the likelihood ratio chi-squared test for individual $\\beta_k$:\n• chi-squared distribution with 1 degree of freedom\nSampling distribution of $t$:\n• $t$ distribution with $N - 2$ degrees of freedom\nSampling distribution of $z$:\n• Approximately the standard normal distribution\nSignificant?Significant?Significant?\n• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or\n• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\\alpha$\nFor the model chi-squared test for the complete regression model and likelihood ratio chi-squared test for individual $\\beta_k$:\n• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or\n• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\\alpha$\nFor the Wald test:\n• If defined as Wald $= \\dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chi-squared tests. Wald can be interpret as $X^2$\n• If defined as Wald $= \\dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.\n$t$ Test two sided:\n$t$ Test right sided:\n$t$ Test left sided:\n$z$ Test two sided:\n$z$ Test right sided:\n$z$ Test left sided:\nn.a.Wald-type approximate $C\\%$ confidence interval for $\\beta_k$Approximate $C$% confidence interval for $\\rho$\n-$b_k \\pm z^* \\times SE_{b_k}$\nwhere the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).\nFirst compute the approximate $C$% confidence interval for $\\rho_{Fisher}$:\n• $lower_{Fisher} = r_{Fisher} - z^* \\times \\sqrt{\\dfrac{1}{N - 3}}$\n• $upper_{Fisher} = r_{Fisher} + z^* \\times \\sqrt{\\dfrac{1}{N - 3}}$\nwhere $r_{Fisher} = \\frac{1}{2} \\times \\log\\Bigg(\\dfrac{1 + r}{1 - r} \\Bigg )$ and the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).\nThen transform back to get the approximate $C$% confidence interval for $\\rho$:\n• lower bound = $\\dfrac{e^{2 \\times lower_{Fisher}} - 1}{e^{2 \\times lower_{Fisher}} + 1}$\n• upper bound = $\\dfrac{e^{2 \\times upper_{Fisher}} - 1}{e^{2 \\times upper_{Fisher}} + 1}$\nEffect sizeGoodness of fit measure $R^2_L$Properties of the Pearson correlation coefficient\n• Proportion variance explained $R^2$:\nProportion variance of the dependent variable $y$ explained by the independent variables and the interaction effect together:\n\\begin{align} R^2 &= \\dfrac{\\mbox{sum of squares model}}{\\mbox{sum of squares total}} \\end{align} $R^2$ is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.\n\n• Proportion variance explained $\\eta^2$:\nProportion variance of the dependent variable $y$ explained by an independent variable or interaction effect:\n\\begin{align} \\eta^2_A &= \\dfrac{\\mbox{sum of squares A}}{\\mbox{sum of squares total}}\\\\ \\\\ \\eta^2_B &= \\dfrac{\\mbox{sum of squares B}}{\\mbox{sum of squares total}}\\\\ \\\\ \\eta^2_{int} &= \\dfrac{\\mbox{sum of squares int}}{\\mbox{sum of squares total}} \\end{align} $\\eta^2$ is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.\n\n• Proportion variance explained $\\omega^2$:\nCorrects for the positive bias in $\\eta^2$ and is equal to:\n\\begin{align} \\omega^2_A &= \\dfrac{\\mbox{sum of squares A} - \\mbox{degrees of freedom A} \\times \\mbox{mean square error}}{\\mbox{sum of squares total} + \\mbox{mean square error}}\\\\ \\\\ \\omega^2_B &= \\dfrac{\\mbox{sum of squares B} - \\mbox{degrees of freedom B} \\times \\mbox{mean square error}}{\\mbox{sum of squares total} + \\mbox{mean square error}}\\\\ \\\\ \\omega^2_{int} &= \\dfrac{\\mbox{sum of squares int} - \\mbox{degrees of freedom int} \\times \\mbox{mean square error}}{\\mbox{sum of squares total} + \\mbox{mean square error}}\\\\ \\end{align} $\\omega^2$ is a better estimate of the explained variance in the population than $\\eta^2$. Only for balanced designs (equal sample sizes).\n\n• Proportion variance explained $\\eta^2_{partial}$: \\begin{align} \\eta^2_{partial\\,A} &= \\frac{\\mbox{sum of squares A}}{\\mbox{sum of squares A} + \\mbox{sum of squares error}}\\\\ \\\\ \\eta^2_{partial\\,B} &= \\frac{\\mbox{sum of squares B}}{\\mbox{sum of squares B} + \\mbox{sum of squares error}}\\\\ \\\\ \\eta^2_{partial\\,int} &= \\frac{\\mbox{sum of squares int}}{\\mbox{sum of squares int} + \\mbox{sum of squares error}} \\end{align}\n$R^2_L = \\dfrac{D_{null} - D_K}{D_{null}}$\nThere are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.\n• The Pearson correlation coefficient is a measure for the linear relationship between two quantitative variables.\n• The Pearson correlation coefficient squared reflects the proportion of variance explained in one variable by the other variable.\n• The Pearson correlation coefficient can take on values between -1 (perfect negative relationship) and 1 (perfect positive relationship). A value of 0 means no linear relationship.\n• The absolute size of the Pearson correlation coefficient is not affected by any linear transformation of the variables. However, the sign of the Pearson correlation will flip when the scores on one of the two variables are multiplied by a negative number (reversing the direction of measurement of that variable).\nFor example:\n• the correlation between $x$ and $y$ is equivalent to the correlation between $3x + 5$ and $2y - 6$.\n• the absolute value of the correlation between $x$ and $y$ is equivalent to the absolute value of the correlation between $-3x + 5$ and $2y - 6$. However, the signs of the two correlation coefficients will be in opposite directions, due to the multiplication of $x$ by $-3$.\n• The Pearson correlation coefficient does not say anything about causality.\n• The Pearson correlation coefficient is sensitive to outliers.\nANOVA tablen.a.n.a.\n--\nEquivalent ton.a.Equivalent to\nOLS regression with two categorical independent variables and the interaction term, transformed into $(I - 1)$ + $(J - 1)$ + $(I - 1) \\times (J - 1)$ code variables.-OLS regression with one independent variable:\n• $b_1 = r \\times \\frac{s_y}{s_x}$\n• Results significance test ($t$ and $p$ value) testing $H_0$: $\\beta_1 = 0$ are equivalent to results significance test testing $H_0$: $\\rho = 0$\nExample contextExample contextExample context\nIs the average mental health score different between people from a low, moderate, and high economic class? And is the average mental health score different between men and women? And is there an interaction effect between economic class and gender?Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?Is there a linear relationship between physical health and mental health?\nSPSSSPSSSPSS\nAnalyze > General Linear Model > Univariate...\n• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factor(s)\nAnalyze > Regression > Binary Logistic...\n• Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)\nAnalyze > Correlate > Bivariate...\n• Put your two variables in the box below Variables\nJamoviJamoviJamovi\nANOVA > ANOVA\n• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factors\nRegression > 2 Outcomes - Binomial\n• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates\n• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well\n• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'\nRegression > Correlation Matrix\n• Put your two variables in the white box at the right\n• Under Correlation Coefficients, select Pearson (selected by default)\n• Under Hypothesis, select your alternative hypothesis\nPractice questionsPractice questionsPractice questions" ]

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[ "# HackerRank: DynamicArray\n\n## Problem Statement\n\nThere are N sequences. All of them are initially empty, and you are given a variable lastans = 0. You are given Q queries of two different types:\n\n• 1 x y” – Insert y at the end of the ((x XOR lastans) mod N)th sequence.\n• 2 x y” – Print the value of the (y mod size)th element of the ((x XOR lastans) mod N)th sequence. Here, \\$size\\$ denotes the size of the related sequence. Then, assign this integer to lastans.\n\nNote: You may assume that, for the second type of query, the related sequence will not be an empty sequence. Sequences and the elements of each sequence are indexed by zero-based numbering.\n\nYou can get more information about XOR from Wikipedia. It is defined as ^ in most of the modern programming languages.\n\nInput Format\n\nThe first line consists of \\$N\\$, number of sequences, and \\$Q\\$, number of queries, separated by a space. The following \\$Q\\$ lines contains one of the query types described above.\n\nConstraints\n1 < N,Q < 10^5\n0 < x < 10^9\n0 < y < 10^9\n\nOutput Format\n\nFor each query of type two, print the answer on a new line.\n\nSample Input\n\n``````2 5\n1 0 5\n1 1 7\n1 0 3\n2 1 0\n2 1 1\n``````\n\nSample Output\n\n``````7\n3\n``````\n\nExplanation\n\nThe first sequence is 5, 3 and the second sequence is 7.\n\n## Solution:\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Nonorthogonal Eigenvectors\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg116403] Re: Nonorthogonal Eigenvectors\n• From: Daniel Lichtblau <danl at wolfram.com>\n• Date: Sun, 13 Feb 2011 05:50:01 -0500 (EST)\n\n```----- Original Message -----\n> From: \"Kevin J. McCann\" <kjm at KevinMcCann.com>\n> To: mathgroup at smc.vnet.net\n> Sent: Saturday, February 12, 2011 4:19:39 AM\n> Subject: [mg116374] Nonorthogonal Eigenvectors\n> I have seen some threads from the past on this, but never got a\n>\n> Suppose I have an exact matrix A:\n>\n> A = {{1, 0, 0, 0, 2}, {0, 16, 0, 0, 0}, {0, 0, 9, 0, 0}, {0, 0, 0, 0,\n> 0}, {2, 0, 0, 0, 4}};\n>\n> P = Eigenvectors[A]\n>\n> produces the following\n>\n> {{0,1,0,0,0},{0,0,1,0,0},{1,0,0,0,2},{-2,0,0,0,1},{0,0,0,1,0}}\n>\n> which is not an unitary matrix, although the vectors are orthogonal,\n> just not normal, i.e.\n>\n> Transpose[P].P\n>\n> is not the identity matrix.\n>\n> However, if I make A numeric:\n>\n> nA = A//N\n>\n> then\n>\n> nP = Eigenvectors[nA]\n>\n> produces\n>\n> {{0., 1., 0., 0., 0.}, {0., 0., 1., 0., 0.}, {0.447214, 0., 0., 0.,\n> 0.894427}, {-0.894427, 0., 0., 0., 0.447214}, {0., 0., 0., -1., 0.}}\n>\n> and\n>\n> Transpose[nP].nP\n>\n> is the identity matrix.\n>\n> I do not understand why making the matrix inexact produces the result\n> that I would expect, but when the matrix is exact it doesn't. Also, I\n> don't think the inconsistency is a useful thing.\n>\n> Any ideas why someone decided to do it this way?\n>\n> Kevin\n\nNormalizing eigenvectors is not all that hard, I guess. But what would be the point of introducing radicals or Root objects where they are not needed? In my view, returning the \"simplest\" form is the one to use(not that we always know how to find it). If the vector can be cast as integer valued, then simplest would be integers with gcd equal to 1.\n\nI will mention here that exact and corresponding approximate numeric computations tend to be two very different beasts.\n\nDaniel Lichtblau\nWolfram Research\n\n```\n\n• Prev by Date: Re: changing variable in an equation\n• Next by Date: Re: Bug in Mathematica 7.0.1 multiple integration?\n• Previous by thread: NInegrate Bug\n• Next by thread: Re: solution of equation [CORRECTION]" ]

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[ "# Optimazion – Definitions and Uses\n\nOptimazion, located in Norresundby, Nordjylland, Denmark, is in the Management, Scientific, and Technical Consulting Services industry. The company has 1 employees and is estimated to generate \\$30,191 in sales annually. Optimazion is also a member of the D&B Hoovers network and has access to the company’s contact information, financial statements, and more. This listing has been curated using the D&B Hoovers subscription database.", null, "Many people confuse optimization with mathematical programming, which is the application of mathematical techniques to solve quantitative problems. The name, however, is related to a collection of principles that are used to calculate the maximum or minimum value of a function. The methods of optimization are used in various disciplines, including mathematics, engineering, computer science, and systems analysis. There are several applications for optimization, including data mining and machine learning, which will be discussed below.\n\nThe term “optimization” has many definitions and uses. It is often used in the mathematical field to solve problems that involve maximizing or minimizing a function. It is useful in many different fields, including systems analysis, linear programming, and computer science. As you might expect, the concept of optimization has grown as a result of the realization that mathematical elements are common to various problems. It can be used to help solve a variety of problems.\n\nThe most basic definition of optimization is maximizing or minimizing a function. It is a technique that applies mathematical principles and methods to solve quantitative problems in many different fields. It is widely used in engineering, business, and physics. It is an important technique for systems analysis and mathematical programming. It is an integral part of many areas of science. There are several uses for optimization. For example, optimizing a process can make the output of a system more productive.\n\nThe term “optimization” has many different meanings in different disciplines. In computer science, it is an application of a mathematical technique that seeks to maximize a function’s value. In the field of finance, optimization is also used in politics. It is a method for reducing a political candidate’s vote. This is a good example of an optimization problem. The correct definition is that of a problem and the objective.\n\nOptimazion is a mathematical process of solving problems. It involves a series of mathematical procedures used to determine the best possible solution to a problem. The concept is based on the principle of minima and maxima. It is also a technique for solving problems that can be solved by a particular algorithm. In practice, optimization is a fundamental tool for systems. It is often used in mathematics and systems analysis.\n\nOptimazion is a problem that involves maximizing a function. The best possible solution is to minimize the number of variables that are necessary to achieve the goal. The goal is to maximize the value of the objective. In other words, it is the best possible solution to a problem. For this reason, optimizazion is also known as linear programming. It is used in the field of computer science to solve optimization problems." ]

[ null, "https://i.imgur.com/SEFPbW6.jpg", null ]

[ "Time Limit : sec, Memory Limit : KB\nEnglish\n\n# Problem F: Find the Multiples\n\nYou are given a sequence a0a1...aN-1 digits and a prime number Q. For each ij with ai ≠ 0, the subsequence aiai+1...aj can be read as a decimal representation of a positive integer. Subsequences with leading zeros are not considered. Your task is to count the number of pairs (i, j) such that the corresponding subsequence is a multiple of Q.\n\n## Input\n\nThe input consists of at most 50 datasets. Each dataset is represented by a line containing four integers N, S, W, and Q, separated by spaces, where 1 ≤ N ≤ 105, 1 ≤ S ≤ 109, 1 ≤ W ≤ 109, and Q is a prime number less than 108. The sequence a0...aN-1 of length N is generated by the following code, in which ai is written as a[i].\n\n int g = S;\nfor(int i=0; i<N; i++) {\na[i] = (g/7) % 10;\nif( g%2 == 0 ) { g = (g/2); }\nelse { g = (g/2) ^ W; }\n}\n\n\nNote: the operators /, %, and ^ are the integer division, the modulo, and the bitwise exclusiveor, respectively. The above code is meant to be a random number generator. The intended solution does not rely on the way how the sequence is generated.\n\nThe end of the input is indicated by a line containing four zeros separated by spaces.\n\n## Output\n\nFor each dataset, output the answer in a line. You may assume that the answer is less than 230.\n\n## Sample Input\n\n3 32 64 7\n4 35 89 5\n5 555 442 3\n5 777 465 11\n100000 666 701622763 65537\n0 0 0 0\n\n\n## Output for the Sample Input\n\n2\n4\n6\n3\n68530\n\n\nIn the first dataset, the sequence is 421. We can find two multiples of Q = 7, namely, 42 and 21.\n\nIn the second dataset, the sequence is 5052, from which we can find 5, 50, 505, and 5 being the multiples of Q = 5. Notice that we don't count 0 or 05 since they are not a valid representation of positive integers. Also notice that we count 5 twice, because it occurs twice in different positions.\n\nIn the third and fourth datasets, the sequences are 95073 and 12221, respectively." ]

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[ "Software Products\n\nArithmetic Series: SUBTRACTION\n\nStudents learn subtraction from a variety of perspectives – counting, eliminating, graphing followed by practice, practice, practice. Once the basic subtraction concepts are learned, this math software program moves on 2-3-4-and-5-digit subtraction with borrowing – each step of the process is presented and practiced. Every screen is scored for immediate response to the student. Every four sections concludes with a comprehensive review. Extras include printable practice drills in addition and subtraction. Worksheets are printable for additional practice and reinforcement — use as classwork or homework. Problem solving worksheets help the students learn to critically discern the differences between addition and subtraction. Worksheets include critical thinking discussions and practice about temperature, vocabulary (horizontal, vertical), figural analysis and analogies, mazes, morse code, classification practice, comparison of 2-dimensional and 3-dimensional figures, and reading coins (money).\n\nIntroductions unfold…\n\nStep-by-Step Introductions\n\nPractice with HINTs\n\nConclude each section with a Quiz" ]

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[ "## May 24, 2007\n\n### Cohomology and Computation (Week 24)\n\n#### Posted by John Baez", null, "In the 1950s, Cartan, Eilenberg, Mac Lane and others systematically studied the cohomology of many different algebraic gadgets — groups, associative algebras, Lie algebras, and so on. It was later realized that underlying all these different cohomology theories there’s a marvelous unifying idea: the bar construction. In this week’s seminar on Cohomology and Computation, we began trying to understand what makes the bar construction tick:\n\n• Week 24 (May 17) - The bar construction. Why do adjoint functors give simplicial objects? First, $\\Delta_{alg}$ is the free monoidal category on a monoid object — or \"the walking monoid\", for short. Second, adjoint functors give certain monoids, called \"monads\".\n\nLast week’s notes are here; next week’s notes are here.\n\nWhy is $\\Delta_{alg}$, the free monoidal category on a monoid, called the “walking monoid”?\n\nThis is one of Jim Dolan’s jokes. Supppose some guy has really big bushy eyebrows, so that when you see him walking down the street you first notice his eyebrows and only later realize there’s a person attached. Then you might call him a \"walking pair of eyebrows\". Such a guy is basically just a minimal life support system for his eyebrows!\n\nIn math, we see the same phenomenon in things like “the free group on 2 elements”. This is a group whose sole purpose in life is to have 2 elements. To pick out 2 elements in any group $G$, we just need a homomorphism from the free group on 2 elements to $G$. Similarly, the monoidal category $\\Delta_{alg}$ is a monoidal category whose sole purpose in life is to contain a monoid object.\n\nI explained some variants of this idea back in week173 and week174 — namely, the walking equivalence, the walking adjoint equivalence, the walking adjunction, and the walking ambidextrous adjunction. Aaron Lauda discusses the walking adjunction and the eyebrows joke on page 20 of his paper Frobenius algebras and ambidextrous adjunctions. For more of the history of this joke, and how Jim sometimes feels uncomfortable when I tell his jokes in public — whoops! — read the interview by Bruce Bartlett.\n\nPosted at May 24, 2007 12:43 AM UTC\n\nTrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1285\n\n### Re: the Walking Monoid\n\nThere’s another nice fact about $\\Delta$ which fits with the picture of “the walking monoid”, but one degree up in categorical dimension.\n\nFirst, $\\Delta$, being the category of finite ordinals and order-preserving maps, is a 2-category: the 2-cells are instances $f \\leq g$ of the pointwise order between order-preserving maps. In fact, $\\Delta$ with its ordinal sum $+$ forms a monoidal 2-category.\n\nNext, taking into account the 2-category structure, there are some adjoint relationships between the 1-cells. Let $u: 0 \\to 1$ and $m: 2 \\to 1$ be the unique maps to the terminal object 1; these of course are the unit and multiplication on the monoid 1. Then, there is a string of adjunctions\n\n$(1 + u: 1 \\to 2) \\dashv (m: 2 \\to 1) \\dashv (u + 1: 1 \\to 2).$\n\nThis is very easy to see; for instance, the counit 2-cell $m(u + 1) \\leq id_1$ is an equality, and the unit 2-cell $id_2 \\leq (u + 1)m$ is also clear, because $(u + 1)m$ is the function taking both elements of 2 = {0, 1} to the maximal element 1.\n\nMonads for which the multiplication is left adjoint to the unit are called Kock-Zöberlein monads; they were first pointed out in 1972 by Anders Kock. They often arise as “cocompletion monads” of one kind of another. A typical example: consider the ‘finite coproduct completion’ of a category $C$, sometimes denoted $Fam(C)$. The objects of $Fam(C)$ are finite tuples $(c_1, ..., c_n)$ of objects of $C$, and a morphism\n\n$(c_1, ..., c_n) \\to (d_1, ..., d_m)$\n\nconsists of a function $f: [n] \\to [m]$ (where [n] = {1, …, n}) together with an n-tuple of morphisms $g_j: c_j \\to d_{f(j)}$ of $C$, composed in the obvious way. [If you know about categorical wreath products, Fam(C) can also be expressed as a wreath product $Fin \\int C$.] Finite coproducts in $Fam(C)$ are formed by concatenating tuples, and it is not hard to convince oneself that the obvious inclusion $i: C \\to Fam(C)$ has the universal property:\n\nAny functor $F: C \\to D$ to a category with finite coproducts $D$ can be extended along $i$ to a coproduct-preserving functor $Fam(C) \\to D$, uniquely so up to canonical isomorphism.\n\nNow, if $C$ already has finite coproducts, then the counit, i.e., the coproduct-preserving functor $s = sum: Fam(C) \\to C$ which extends the identity on $C$, is left adjoint to the inclusion $i: C \\to Fam(C)$, i.e., the unit! The counit of $s \\dashv i$ is an isomorphism, and the unit, applied to an object $(c_1, ..., c_n)$ in Fam(C), is just the tuple of canonical injections into the coproduct $\\sum_i c_i$ in $C$:\n\n$(c_1, ..., c_n) \\to (\\sum_i c_i).$\n\nIn cases like this, where the counit of a bi-adjunction is left adjoint to the unit in this manner, we say that the bi-adjunction is KZ or Kock-Zöberlein.\n\nHere is a more precise definition: a bi-adjunction $(F: C \\to B) \\dashv (G: B \\to C)$ between bicategories, with unit $u: 1_C \\to G F$ and counit $c: F G \\to 1_B$ and triangulators\n\n$s: (G c)(u G) \\stackrel{\\sim}{\\to} 1_G$\n\n$t: 1_F \\stackrel{\\sim}{\\to} (c F)(F u),$\n\nis said to Kock-Zöberlein if $s$ is the counit of an adjunction $G c \\dashv u G$ (equivalently, if $t$ is the unit of an adjunction $F u \\dashv c F$; proving that equivalence is a fun exercise in bicategorical algebra). Similarly, there is a notion of Kock-Zöberlein (pseudo)monad $(M, m, u)$, where there is a string of adjunctions\n\n$M u \\dashv m \\dashv u M.$\n\nIf you’ve followed me this far, you will have no trouble guessing the punch line:\n\nTheorem (Kock): The monoidal 2-category $\\Delta$ is the “walking Kock-Zöberlein monoid.”\n\nSee the penultimate paper referenced here for details.\n\nI think this result is pretty widely known within the categorical community, but a quick Google search on Kock-Zöberlein or KZ monads didn’t garner a huge number of hits, so I think the result probably bears repeating here. I haven’t thought much about what implications this result would have with respect to the bar construction.\n\nPosted by: Todd Trimble on May 26, 2007 1:04 AM | Permalink | Reply to this\n\n### Re: the Walking Monoid\n\nIn my previous post, I should have added (I think it’s pretty cool, anyway) that in $\\Delta$, if you look at the $n+1$ injections\n\n$d_0 \\leq ...\\leq d_n: [n] \\to [n+1]$\n\nand the $n$ surjections\n\n$s_1 \\leq ... \\leq s_n: [n+1] \\to [n]$\n\nthen what you get is a big long string of adjunctions\n\n$d_0 \\dashv s_1 \\dashv d_1 \\dashv s_2 \\dashv ... \\dashv s_n \\dashv d_n$\n\nas a result of the KZ property.\n\nPosted by: Todd Trimble on May 26, 2007 1:41 AM | Permalink | Reply to this\n\n### Re: the Walking Monoid\n\nThat is cool. Thanks for explaining all this stuff about Kock-Zöberlein monads. I need to lock myself in the cellar and ponder them sometime.\n\nI occasionally feel I understand them — thanks mainly to your attempts to explain them, but also to the prevalence of ‘weakly idempotent’ pseudomonads on $Cat$ having the flavor of ‘cocompletion processes’: I’ve seen lots of those in my time!\n\nBut, the relation of all this to the monoidal 2-category version of $\\Delta$ seems mysterious to me, and phrases like “$s$ is the counit of an adjunction $G c \\dashv u G$” make my mind freeze like a deer in the headlights.\n\nI don’t think there’s any more you can do to help: I just need to do some calculations to see for myself how this stuff works!\n\nif you look at the $n+1$ injections\n\n$d_0 \\leq ...\\leq d_n: [n] \\to [n+1]$\n\nand the $n$ surjections\n\n$s_1 \\leq ... \\leq s_n: [n+1] \\to [n]$\n\nthen what you get is a big long string of adjunctions\n\n$d_0 \\dashv s_1 \\dashv d_1 \\dashv s_2 \\dashv ... \\dashv s_n \\dashv d_n$\n\nas a result of the KZ property.\n\nCool! I vaguely seem to remember results that say “any sufficiently long string of adjunctions has to be sort of degenerate in some way.” I’ve looked for such results, but I don’t remember what I found. This example constrains what theorems I could possibly be imagining.\n\nDo you know any results along these lines?\n\nPosted by: John Baez on May 26, 2007 2:52 AM | Permalink | Reply to this\n\nI vaguely seem to remember results that say “any sufficiently long string of adjunctions has to be sort of degenerate in some way.” I’ve looked for such results, but I don’t remember what I found. This example constrains what theorems I could possibly be imagining. Do you know any results along these lines?\n\nThere’s an absolutely remarkable theorem of this flavor due to Richard Wood and Bob Rosebrugh; I don’t know if it’s close to what you were trying to remember. I think it goes something like this:\n\n$y: C \\to Set^{C^{op}}.$\n\nAs we all know, $y$ preserves all limits which happen to exist in $C$. But the condition that $y$ is a right adjoint: that turns out to be a very strong cocompleteness condition on $C$. In that case, people sometimes say that $C$ is total (don’t know who made up that term).\n\nPresheaf categories (and Grothendieck toposes generally) are total in this sense. In the presheaf case, the left adjoint $m$ to the Yoneda embedding\n\n$y: Set^{C^{op}} \\to Set^{(Set^{C^{op}})^{op}}$\n\nis (somewhat interestingly)\n\n$Set^{y_C^{op}}: Set^{(Set^{C^{op}})^{op}} \\to Set^{C^{op}}.$\n\n(As an aside, the condition that the Yoneda embedding on $C$ have a left exact left adjoint is actually rather close to the condition that $C$ is a Grothendieck topos!).\n\nNow, in the presheaf case $Set^{C^{op}}$, the left adjoint to Yoneda has itself a left adjoint\n\n$L_y: Set^{C^{op}} \\to Set^{(Set^{C^{op}})^{op}}$ given essentially by taking left Kan extension along a Yoneda embedding. (I’m not sure about this, but the existence of such a string of left adjoints which starts from the Yoneda embedding $y_C$ and proceeds leftwards:\n\n$L \\dashv m \\dashv y_C$\n\nmight be equivalent to the condition that $C$ is a presheaf topos.)\n\nCan the left Kan extension\n\n$L_y: Set^{C^{op}} \\to Set^{(Set^{C^{op}})^{op}}$\n\nhave a left adjoint? It can (but the noose is gradually tightening). For example, if $C$ is a small sup-lattice, then $y: C \\to Set^{C^{op}}$ has a left adjoint $\\xi: Set^{C^{op}} \\to C$, and the left adjoint to $L_y$ would be a left Kan extension $L_{\\xi}$ along $\\xi$ (more or less).\n\n“Okay, smarty-pants, can this left adjoint $L_{\\xi}$ have a left adjoint?” Well, yes it can, but now we’re really getting to the end of the rope: if the sup-lattice $C$ in the previous paragraph were the terminal category 1, then $\\xi: Set \\to 1$ has a left adjoint $i: 1 \\to Set$, namely the initial object (and the left adjoint to $L_{\\xi}$ would then be a left Kan extension $L_i$). That’s the absolute limit to how far we can go.\n\nSo, then, the Rosebrugh-Wood results are:\n\n• Theorem: There exists a string of adjoints $j \\dashv k \\dashv l \\dashv m \\dashv y_C,$ where $y_C$ is the Yoneda embedding of $C$, if and only if $C$ is equivalent to $Set$.\n• Corollary: There is no such adjoint string $i \\dashv j \\dashv k \\dashv l \\dashv m \\dashv y_C.$\n\nI see this result is downloadable here. (I saw Richard Wood give a seminar on this theorem at Macquarie; it was impressive.)\n\nPosted by: Todd Trimble on May 26, 2007 4:32 AM | Permalink | Reply to this\n\n### Re: Cohomology and Computation (Week 24)\n\nThere was one other thing about the bar construction which I wanted to mention, since John told me that his course is ending soon and he might not get around to mentioning it himself.\n\nAs John said, the bar construction is a wonderful gadget which can be used to compute cohomology for general algebraic theories (groups, algebras, Lie algebras, …), as well as do other things like constructing classifying bundles. Given any monad $M$ and an $M$-algebra $X$, the bar construction is an efficient machine for constructing a ‘free resolution’ of $X$, to which one can then apply the machinery of derived functors to compute the cohomology of $X$.\n\nWhat John gave in his course was a very nice introduction, with lots of diagrams and pictures, which gave the precise conceptual details underlying the bar construction: a simplicial object whose $n$-dimensional component is a free algebra $M M^n X$ over the monad $M$, with face operators which look like:\n\n\n... mMX\n--> mX\nMmX --> a\n... MMMX --> MMX MX --> X\n--> -->\nMMa Ma\n\n\n\nWhat I want to talk about here is the conceptual sense in which the bar construction is a resolution of $X$, i.e., the acyclicity of the bar construction as a simplicial object. I also want to talk about the sense in which the bar construction on $X$ is a universal $M$-free resolution of $X$.\n\nIn order for this note to make sense, I’ll first have to recall some of what John said. A monad $M$ is a monoid in a monoidal category, and the story begins by considering the string diagram calculus for monoids in monoidal categories. As John showed in pictures, the string diagrams, modulo the equivalence relation imposed by the monoid equations, look exactly like pictures (cospans) of order-preserving functions between finite ordinals. This was summarized in the slogan, “the category $\\Delta$ is the walking monoid”, i.e., $\\Delta$ equipped with its terminal object is precisely the initial strict monoidal category equipped with a monoid.\n\nOr, we could say that $\\Delta^{op}$ is the “walking comonoid”. It’s nice to know by the way that $\\Delta^{op}$ is equivalent to the category of finite intervals (i.e., finite totally ordered sets with a top and bottom, and maps which preserve the order and top and bottom). This is something you can see very well if you stare at John’s pictures of planar 2d cobordisms, which are planar thickenings of string diagrams for monoids, and then let your attention flicker over to the complement (of different shading) and read it upside-down as a 2d cobordism. This idea and the planar 2d cobordisms also figure prominently in Aaron Lauda’s Frobenius algebras and ambidextrous adjunctions.\n\nWith that, the bar construction is easy to construct. Let $(M, m: M M \\to M, u: 1 \\to M)$ be a monad on a category $E$. One has an adjunction\n\n$(F: E \\to E^M) \\dashv (U: E^M \\to E)$\n\nwhere $E^M$ is the Eilenberg-Moore category of $M$-algebras. This in turn gives a comonad $F U$ acting on $M$-algebras, that is to say, a comonoid in a monoidal category of endofunctors. Because $\\Delta^{op}$ is the walking comonoid, there is a unique monoidal functor (the bar construction)\n\n$Bar_M: \\Delta^{op} \\to [E^M, E^M]$\n\nwhich sends the comonoid in $\\Delta^{op}$ to $F U$. By postcomposing this by the evaluation functor at an $M$-algebra $X$, one gets a simplicial $M$-algebra\n\n$B_M(X): \\Delta^{op} \\to E^M$\n\nand this is the bar construction applied to $X$. This description appears in Mac Lane’s Categories for the Working Mathematician, which is where I first learned about it (see section VII.6 in the second edition).\n\nAs I say, the important point here is that this object is acyclic, in a very strong sense of the word which I need to explain. Part of the philosophy here is that the bar construction is a piece of pure equational logic, and we will correspondingly treat its acyclicity purely algebraically, not as a property involving an existential quantifier but as a structure, called a ‘resolution’ for lack of anything better. Roughly speaking, a resolution structure on a simplicial object $Y$ with augmentation,\n\n\n... Y --> Y --> Y,\n-->\n\n\n\nis a contracting homotopy which realizes a homotopy equivalence between $Y$ and the constant simplicial object at $Y$, and moreover a contracting homotopy with some very good properties.\n\nRecall that contracting homotopy $h$ on a simplicial object $Y: \\Delta^{op} \\to E$ is given by a collection of maps\n\n$h_n: Y[n] \\to Y[n+1]$\n\nsatisfying some equations. Observe also that the map $[n] \\mapsto [n+1]$ is the object part of a comonad $ + (-): \\Delta^{op} \\to \\Delta^{op}$. Here is the key notion:\n\n• A resolution is a simplicial object $Y$ together with a right-sided coalgebra structure $h: Y \\to Y \\circ ( + (-))$ over the comonad $ + (-)$.\n\nThis will require some unpacking, but let’s first see how this works in the case of the bar resolution. Since $Bar_M$ preserves the monoidal and comonoid structures, we have a commutative diagram\n\n\nDelta^{op} -----> [E^M, E^M]\n| Bar_M |\n| |\n| +( ) | FUo( )\nV V\nDelta^{op} -----> [E^M, E^M]\nBar_M\n\n\n\nand we augment this diagram with another:\n\n\nDelta^{op} -----> [E^M, E^M] ---> [E^M, E]\n| Bar_M | U o( ) |\n| | |\n|+( ) FUo( )| UFo( )|\nV V V\nDelta^{op} -----> [E^M, E^M] ---->[E^M, E]\nBar_M U o( )\n\n\n\nThe bar resolution is the horizontal composite $UBar_M$. Notice there is is a canonical right $F U$-coalgebra structure on $U$, given by the coaction $u U: U \\to U F U = M U$. This gives a right action of the comonad $+ (-)$ on $UBar_M$, as we see from the diagram, therefore we obtain a resolution structure with components\n\n$h_n = u M^n: M^n \\to M^{n+1}.$\n\nNow, what exactly is a resolution structure? I’d like to tease that out.\n\nA first trivial observation is that a right-sided coalgebra for the comonad $+(-)$ is the same as an ordinary (left-sided) coalgebra for the comonad on simplicial objects\n\n$E^{+(-)}: E^{\\Delta^{op}} \\to E^{\\Delta^{op}}$\n\ngiven by pulling back along $ + (-)$. I’ll call this pullback comonad $P$, for short.\n\nSecond, to make things easier (or perhaps more recognizable), let’s pretend for a moment that $E = Set$. The pullback comonad $P$ then has a left adjoint, given by left Kan extension along $+ ( )$. What is that left Kan extension? On the representable objects, it takes the affine simplex $hom(-, [n])$ to $hom(-, [n+1])$, i.e, it takes the cone of the affine simplex. This cone construction extends to all simplicial sets (by left Kan extension). I’d like to call the Kan extension the ‘cone functor’ on simplicial sets, although we should be careful here: we are not coning a space or simplicial set to a point, because that operation does not preserve disjoint sums (therefore isn’t a left adjoint). What the left Kan extension actually does is cone a space to its set of path components, i.e., it takes the disjoint union of all the cones over all the path components. But as I say, I’m still going to call this left Kan extension the ‘cone functor’, and denote it $C$.\n\nNow, we are in a general situation where we have a comonad $P$ with a left adjoint $C$. There is a general piece of abstract nonsense here, dating back to the original 1965 paper by Eilenberg and Moore, which says that the comonad structure on $P$ corresponds precisely to a monad structure on $C$, in such a way that the category of $P$-coalgebras is naturally equivalent to the category of $C$-algebras. So, now we want to know: what is a $C$-algebra?\n\nPersonally, I find it more intuitive to think about it topologically. There is an analogous cone monad $C: Top \\to Top$ acting on topological spaces, where $C X$ is defined to be the pushout of the diagram\n\n$[0, 1] \\times X \\stackrel{inj_0}{\\leftarrow} X \\stackrel{q}{\\to} \\pi_0(X)$\n\nwhere $inj_0(x) := (0, x)$ and $q$ is the canonical projection. A map $a: C X \\to X$ is thus given by a pair of maps\n\n$h: [0, 1] \\times X \\to X$ $i: \\pi_0(X) \\to X$\n\nwhere $i$ picks out a basepoint for each path component, and (by the pushout commutativity) $h$ contracts each path component to its basepoint. This already means the pair $(h, i)$ is explicit witness to the fact that $X$ is acyclic (is homotopy equivalent to its discrete space of path components), but there’s more since we have not yet accounted for the monad structure on $C$.\n\nThe monad structure on $C$ is induced from a monoid structure on $[0, 1]$, and we will take the monoid multiplication on $[0, 1]$ to be the ‘tropical’ multiplication $(x, y) \\mapsto min(x, y)$. An algebra with respect to the monad $C$ is given by a pair $(h, i)$ such that the homotopy $h$ behaves as a ‘flow’ with respect to tropical multiplication: the following equations are satisfied:\n\n$h(s, h(t, x)) = h(min(s, t), x)$ $h(1, x) = x$ $h(0, x) = iq(x).$\n\n[A note to experts: the topos of simplicial sets classifies the theory of intervals; in particular, geometric realization $Set^{\\Delta^{op}} \\to Top$ is the model which takes the generic interval to the interval [0, 1]. The generic interval carries an intrinsic monoid operation ‘min’, and geometric realization thus maps it over to the monoid $([0, 1], min)$; this is why we took the multiplication on the unit interval to be ‘tropical’.]\n\nThe cone monad on topological spaces is again left adjoint to a comonad $P$ on topological spaces, where $P X$ is the pullback of the diagram\n\n$X^{[0, 1]} \\stackrel{eval_0}{\\to} X \\stackrel{id}{\\leftarrow} |X|$\n\nwhere $|X|$ is the underlying set of $X$ equipped with the discrete topology, included in $X$ by the identity function, and a $C$-algebra structure is again equivalent to a $P$-coalgebra structure.\n\n• In summary: a resolution = a $P$-coalgebra = a $C$-algebra = a simplicial object $Y$ equipped with a ‘homotopy flow’ which contracts $Y$ to the discrete space of its path components $Y$. In other words, a resolution is a particularly nice kind of contracting homotopy witnessing acyclicity of $Y$.\n\nNow for the main theorem:\n\n• The bar resolution $UBar_M$ is a universal $M$-algebra resolution for the monad $M$.\n\nThis may require some amplification. As above, let $B_M(X)$ denote the result of applying the bar resolution functor to a particular $M$-algebra $X$, i.e., let $B_M(X)$ be the composite\n\n$\\Delta^{op} \\stackrel{Bar_M}{\\to} [E^M, E^M] \\stackrel{eval_X}{\\to} E^M \\stackrel{U}{\\to} X.$\n\nLet $Y$ be any $M$-algebra resolution, i.e., a simplicial $M$-algebra $\\Delta^{op} \\to E^M$ equipped with a $P$-coalgebra structure on the underlying simplicial object\n\n$U Y: \\Delta^{op} \\to E.$\n\nThere is a forgetful functor from the category of $M$-algebra resolutions to $M$-algebras, which remembers only the augmentation part $Y$.\n\nThe universal property of $B_M(X)$ is that given an $M$-algebra resolution $Y$ and an $M$-algebra map $X \\to Y$, there is a unique extension to a map of $M$-algebra resolutions, $B_M(X) \\to Y$. This is the familiar ‘acyclic models theorem’:\n\n• Theorem (Acyclic Models): The bar resolution $B_M(-)$ is left adjoint to the forgetful functor from $M$-algebra resolutions to $M$-algebras.\n\nSketch of proof: The unique simplicial extension $f: B_M(X) \\to Y$ of $f_0: X \\to Y$ is constructed by induction on dimension. The map $f_{n+1}: M^{n+1} X \\to Y$ is the unique $M$-algebra map which extends the composite\n\n$M^n X \\stackrel{f_n}{\\to} Y[n] \\stackrel{h_n}{\\to} Y[n+1].$\n\nIt is immediate that $f$ preserves the contracting homotopies; all that remains to be checked is that $f$ is a simplicial map. This is based on a simple inductive argument; details may be found in my notes.\n\nPosted by: Todd Trimble on May 31, 2007 8:55 PM | Permalink | Reply to this\n\n### Re: Cohomology and Computation (Week 24)\n\nShall we promote this elaboration to a guest post?\n\n(By the way, it seems you forgot to add the hyperlink to your notes. If you send me or John or David an email with the address, we can insert it into your comment.)\n\nPosted by: urs on May 31, 2007 9:02 PM | Permalink | Reply to this\n\n### Re: Cohomology and Computation (Week 24)\n\nI’d be honored – thanks very much, Urs.\n\nI’ll send you the link in a moment.\n\nPosted by: Todd Trimble on May 31, 2007 9:20 PM | Permalink | Reply to this\n\n### Re: Cohomology and Computation (Week 24)\n\nPosted by: urs on May 31, 2007 9:32 PM | Permalink | Reply to this\n\n### Re: Cohomology and Computation (Week 24)\n\nEarlier I wrote:\n\nThis already means the pair (h,i) is explicit witness to the fact that X is acyclic (is homotopy equivalent to its discrete space of path components)\n\nwhich is not quite right since we don’t yet have the equation $h(1, x) = x$. But in the scheme of things, this is a minor error.\nPosted by: Todd Trimble on May 31, 2007 9:18 PM | Permalink | Reply to this\nRead the post Cohomology and Computation (Week 25)\nWeblog: The n-Category Café" ]

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[ "# Clustering\n\n## Exercises\n\nThis chapter uses the `gbm` dataset\n\nThe method implemented in Seurat first constructs a SNN graph based on the euclidean distance in PCA space, and refine the edge weights between any two cells based on the shared overlap in their local neighborhoods (Jaccard similarity). This step is performed using the `FindNeighbors()` function, and takes as input the previously defined dimensionality of the dataset.\n\n``````gbm <- Seurat::FindNeighbors(gbm, dims = 1:25)\n``````\n\nTo cluster the cells, Seurat next implements modularity optimization techniques such as the Louvain algorithm (default) or SLM [SLM, Blondel et al., Journal of Statistical Mechanics], to iteratively group cells together, with the goal of optimizing the standard modularity function. The `FindClusters()` function implements this procedure, and contains a resolution parameter that sets the ‘granularity’ of the downstream clustering, with increased values leading to a greater number of clusters.\n\n``````gbm <- Seurat::FindClusters(gbm, resolution = seq(0.1, 0.8, by=0.1))\n``````\n\nCluster id of each cell is added to the metadata object, as a new column for each resolution tested:\n\n``````head([email protected])\n``````\n\nTo view how clusters sub-divide at increasing resolution:\n\n``````library(clustree)\nclustree::clustree([email protected][,grep(\"RNA_snn_res\", colnames([email protected]))],\nprefix = \"RNA_snn_res.\")\n``````\n\nYou can view the UMAP coloring each cell according to a cluster id like this:\n\n``````Seurat::DimPlot(gbm, group.by = \"RNA_snn_res.0.1\")\n``````\n\nExercise: Visualise clustering based on a few more resolutions. Taking the clustering and the UMAP plots into account what do you consider as a good resolution to perform the clustering?\n\nOf course, there is no ‘optimal’ resolution, but based on resolution of 0.2, it seems that clustering fits the UMAP well:\n\n``````Seurat::DimPlot(gbm, group.by = \"RNA_snn_res.0.2\")\n``````\n\n### Save the dataset and clear environment\n\nNow, save the dataset so you can use it later today:\n\n``````saveRDS(gbm, \"gbm_day2_part1.rds\")\n``````\n\n``````rm(list = ls())" ]

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[ "# Direct product of S4 and Z4\n\nView a complete list of particular groups (this is a very huge list!)[SHOW MORE]\n\n## Definition\n\nThis group is defined as the external direct product of symmetric group:S4 and cyclic group:Z4.\n\n## Arithmetic functions\n\nFunction Value Similar groups Explanation\norder (number of elements, equivalently, cardinality or size of underlying set) 96 groups with same order order of direct product is product of orders, so the order is", null, "$24 \\times 4$ where", null, "$24 =4!$ is the order of symmetric group:S4 and", null, "$4$ is the order of cyclic group:Z4.\nexponent of a group 12 groups with same order and exponent of a group | groups with same exponent of a group exponent of direct product is lcm of exponents, so the exponent is", null, "$\\operatorname{lcm}\\{12,4\\} = 12$.\nminimum size of generating set 2 groups with same order and minimum size of generating set | groups with same minimum size of generating set\n\n## GAP implementation\n\n### Group ID\n\nThis finite group has order 96 and has ID 186 among the groups of order 96 in GAP's SmallGroup library. For context, there are 231 groups of order 96. It can thus be defined using GAP's SmallGroup function as:\n\nSmallGroup(96,186)\n\nFor instance, we can use the following assignment in GAP to create the group and name it", null, "$G$:\n\ngap> G := SmallGroup(96,186);\n\nConversely, to check whether a given group", null, "$G$ is in fact the group we want, we can use GAP's IdGroup function:\n\nIdGroup(G) = [96,186]\n\nor just do:\n\nIdGroup(G)\n\nto have GAP output the group ID, that we can then compare to what we want.\n\n### Other descriptions\n\nDescription Functions used\nDirectProduct(SymmetricGroup(4),CyclicGroup(4)) DirectProduct, SymmetricGroup, CyclicGroup" ]

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[ "# A Method for Gradient Enhancement of Continuum Damage Models\n\n• B. J. Dimitrijevic\n• K. Hackl\n\n### Abstract\n\nA method for the regularization of continuum damage material models based on gradient-type enhancement of the free-energy functional is presented. Direct introduction of the gradient of the damage variable would require C1 interpolation of the displacements, which is a complicated task to achieve with quadrilateral elements. Therefore a new variable field is introduced, which makes the model non-local in nature, while preserving C0 interpolation order of the variables at the same time. The strategy is formulated as a pure minimization problem, therefore the LBB-condition does not apply in this case. However, we still take the interpolation of the displacement field one order higher than the interpolation of the field of additional (non-local) variables. That leads to increased accuracy and removes the post-processing step necessary to obtain consistent results in the case of equal interpolation order. Several numerical examples which show the performance of the proposed gradient enhancement are presented. The pathological mesh dependence of the damage model is efficiently removed, together with the difficulties of numerical calculations in the softening range. Calculations predict a development of the damage variable which is mesh-objective for fixed internal material length.", null, "Published\n2019-07-30\nIssue\nSection\nArticle" ]

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[ "# Rotation Matrix to VR Rotation\n\nConvert rotation matrix to axis/angle rotation\n\n• Library:\n• Simulink 3D Animation / Utilities\n\n•", null, "## Description\n\nThe Rotation Matrix to VR Rotation converts Rotation Matrix (defined columnwise as 3-by-3 matrix or as a 9-element column vector) into the Axis / Angle rotation representation used for defining rotations in VR.\n\n## Ports\n\n### Input\n\nexpand all\n\n3D rotation, specified as a 3-by-3 columnwise-defined matrix, also known as a direction cosine matrix.\n\nA representation of a three-dimensional spherical rotation as a 3-by-3 real, orthogonal matrix R: RTR = RRT = I, where I is the 3-by-3 identity and RT is the transpose of R. This matrix is also known as the direction cosine matrix (DCM). The DCM is the orientation of the object in space, relative to its parent node.\n\n`$R=\\left(\\begin{array}{ccc}{R}_{11}& {R}_{12}& {R}_{13}\\\\ {R}_{21}& {R}_{22}& {R}_{23}\\\\ {R}_{31}& {R}_{32}& {R}_{33}\\end{array}\\right)=\\left(\\begin{array}{ccc}{R}_{xx}& {R}_{xy}& {R}_{xz}\\\\ {R}_{yx}& {R}_{yy}& {R}_{yz}\\\\ {R}_{zx}& {R}_{zy}& {R}_{zz}\\end{array}\\right)$`\n\nData Types: `single` | `double`\n\n### Output\n\nexpand all\n\nOutput rotation, returned as a 4-element vector in axis/angle notation,. The first three elements specify the axis of rotation and the fourth element specifies the angle.\n\n## Parameters\n\nexpand all\n\nInput signal value is considered to be zero if it is equal to or lower than the value set in this parameter. By default, the parameter is set to ε = 1e-12.", null, "" ]

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[ "Home » 15 Millimeters Equals How Many Centimeters? Update\n\n# 15 Millimeters Equals How Many Centimeters? Update\n\nLet’s discuss the question: 15 millimeters equals how many centimeters. We summarize all relevant answers in section Q&A of website Domainedevilotte.com in category: Blog Technology. See more related questions in the comments below.\n\n## Is 15mm smaller than 1 cm?\n\nFirst, note that mm is the same as millimeters and cm is the same as centimeters. Thus, when you are asking to convert 15 mm to cm, you are asking to convert 15 millimeters to centimeters. A millimeter is smaller than a centimeter.\n\n## What is the answer if we convert 15 millimeter to centimeters?\n\nNow that we know what the conversion factor is, we can easily calculate the conversion of 15 mm to cm by multiplying 0.1 by the number of millimeters we have, which is 15. So, the answer to the question “what is 15 millimeters in centimeters?” is 1.5 cm.\n\n### ✅ Convert Milimeter to Centimeter (mm to cm) – Formula, Example, Convertion Factor\n\n✅ Convert Milimeter to Centimeter (mm to cm) – Formula, Example, Convertion Factor\n✅ Convert Milimeter to Centimeter (mm to cm) – Formula, Example, Convertion Factor\n\n### Images related to the topic✅ Convert Milimeter to Centimeter (mm to cm) – Formula, Example, Convertion Factor", null, "✅ Convert Milimeter To Centimeter (Mm To Cm) – Formula, Example, Convertion Factor\n\n## What size is 15mm?\n\nMM Approximate Size In Inches Exact Size In Inches\n12mm Just short of 1/2 Inch 0.47244 Inches\n13mm Little over 1/2 Inch 0.51181 Inches\n14mm 9/16 Inch 0.55118 Inches\n15mm Just short of 5/8 Inch 0.59055 Inches\n\n## How many millimeters Makes 1 cm?\n\nOne centimeter is equal to 10 millimeters.\n\n## How big is 16mm round?\n\n16mm = 5/8 inch. 17mm = almost 11/16 inch. 18mm = just over 11/16 inch. 19mm = 3/4 inch.\n\n## How do you convert mm to cm examples?\n\nOne mm is one “millimeter” or one one-thousandth of a meter (1 mm = 1/1000 m). One cm is one “centimeter” or one one-hundredth of a meter (1 cm = 1/100 m). Therefore, 1 cm = 10 mm. To convert mm to cm, divide the number of mm by 10 to get the number of cm.\n\n## What size is 18mm in SAE?\n\nSAE to Metric Conversion Chart\nSAE Metric Inch\n18mm 0.709\n23/32″ 0.719\n19mm 0.748\n3/4″ 0.75\n29 thg 4, 2020\n\n## How wide is 15 cm in inches?\n\n15 cm = 5.90551185 inches.\n\n## Is 15mm half inch?\n\nYou can join modern 15mm pipe to 1/2 inch imperial and 28mm to 1 inch using standard metric compression fittings, but to connect 22mm pipe to 3/4 inch you’ll need a 22mm compression fitting with a special oversize olive.\n\n### How to convert millimeter to centimeter | Conversion of millimeter into centometer | mm into cm\n\nHow to convert millimeter to centimeter | Conversion of millimeter into centometer | mm into cm\nHow to convert millimeter to centimeter | Conversion of millimeter into centometer | mm into cm\n\n### Images related to the topicHow to convert millimeter to centimeter | Conversion of millimeter into centometer | mm into cm", null, "How To Convert Millimeter To Centimeter | Conversion Of Millimeter Into Centometer | Mm Into Cm\n\n## What is a 1 cm?\n\n1 centimeter is equal to 0.3937 inches, or 1 inch is equal to 2.54 centimeters. In other words, 1 centimeter is less than half as big as an inch, so you need about two-and-a-half centimeters to make one inch.\n\n## Which is more cm or mm?\n\nThe centimeter (cm) is also a unit of length which is ten times larger than a millimeter and is equal to one hundredth of a meter; therefore, there are 100 centimeters in a meter. There are 2.54 centimeters in one inch.\n\n## How many mm means 1 inch?\n\nHow many millimeters in an inch? 1 inch is equal to 25.4 millimeters, which is the conversion factor from inches to millimeters.\n\n## How CM is an inch?\n\nThe value of 1 inch is approximately equal to 2.54 centimeters. To convert inches to the centimeter values, multiply the given inch value by 2.54 cm. 1 cm = 0.393701 inches.\n\n## How many mm is a dime?\n\nCoin Specifications\nDenomination Cent Dime\nWeight 2.500 g 2.268 g\nDiameter 0.750 in. 19.05 mm 0.705 in. 17.91 mm\nThickness 1.52 mm 1.35 mm\nEdge Plain Reeded\n24 thg 9, 2019\n\n## How do you add cm and mm?\n\nNext, you explain that you can add centimeters and millimeters, by first converting centimeters to millimeters. To do this you take one jump to the right, so you multiply by 10.\n\n## Is a ruler cm or mm?\n\nA metric ruler is the standard instrument for measurement in the scientific laboratory. On a metric ruler, each individual line represents a millimeter (mm). The numbers on the ruler represent centimeters (cm). There are 10 millimeters for each centimeter.\n\n### Understanding mm, cm, m, and km\n\nUnderstanding mm, cm, m, and km\nUnderstanding mm, cm, m, and km\n\n## Is there a 10mm socket?\n\nIn fact, the 10mm socket is one of the most commonly used metric size sockets of all. It’s no secret that the majority of parts under the hood and in the car require a 10mm. The applications may vary from car to car, but the 10mm is absolutely vital to complete a project.\n\n## What is a 10mm socket in standard?\n\nSo what is a 10mm socket equivalent? The equivalent imperial size of a 10mm socket is a 3/8 socket. Imperial sizes are known as SAE (Society of Automotive Engineers) or Standard.\n\nRelated searches\n\n• 30 mm to cm\n• how many centimeters are in 40 millimeters\n• 15 mm to feet\n• how many centimeters are in 15 millimeters\n• how many centimeters is 500 millimeters\n• 15 mm to cm to inches\n• 17mm to cm\n• 150 millimeters equals how many centimeters\n• 15.5 centimeters equals how many millimeters\n• 10 mm to cm\n• 20 mm to cm\n• 15 centimeters equals how many millimeters\n• how many millimeters is 1 cm\n• 15 mm to inch\n• 15 mm to cm\n\n## Information related to the topic 15 millimeters equals how many centimeters\n\nHere are the search results of the thread 15 millimeters equals how many centimeters from Bing. You can read more if you want.\n\nYou have just come across an article on the topic 15 millimeters equals how many centimeters. If you found this article useful, please share it. Thank you very much." ]

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[ "# barcode reader using c#.net LINEAR in Software Drawing Code-128 in Software LINEAR\n\nLINEAR\nScan Code 128 Code Set B In None\nUsing Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.\nCode 128 Drawer In None\nUsing Barcode drawer for Software Control to generate, create Code 128 Code Set B image in Software applications.\nCLOSED-LOOP\nRecognizing Code128 In None\nCreate Code128 In Visual C#\nUsing Barcode generation for Visual Studio .NET Control to generate, create Code 128B image in Visual Studio .NET applications.\nSYSTEMS\nEncoding Code 128 Code Set C In Visual Studio .NET\nUsing Barcode creator for ASP.NET Control to generate, create Code 128 Code Set A image in ASP.NET applications.\nCode 128C Maker In .NET Framework\nUsing Barcode encoder for .NET Control to generate, create Code128 image in VS .NET applications.\nShifting the trial point horizontally to the left will increase the sum of the angles. As a second trial point, sy = - 0.95 + 1.5 j gives for the sum of the angles 88 + 56 + 37 = 181 z TT This result is sufficiently close to V, which is (2; + 1)~ with i = 0, and we accept the point as one on the locus. In this manner, more points on the locus can be found and a curve drawn through them. 8. Gain. To determine the gain at various points along the loci, the magnitude criterion [Eq. (15.13)] is used. For example, if the gain at s = -0.95 + jl.5 (labeled si in Fig. 15.56), is wanted, we measure the distances directly with a ruler; thus Is -p1 1 = 1.50\nGenerate Code 128 Code Set A In VB.NET\nUsing Barcode printer for VS .NET Control to generate, create ANSI/AIM Code 128 image in VS .NET applications.\nUSS Code 39 Drawer In None\nUsing Barcode generation for Software Control to generate, create Code 39 Full ASCII image in Software applications.\n1 s -p2 1 =\nGTIN - 128 Generation In None\nUsing Barcode maker for Software Control to generate, create EAN128 image in Software applications.\nData Matrix Maker In None\nUsing Barcode encoder for Software Control to generate, create Data Matrix 2d barcode image in Software applications.\nIs-p31\nPaint Barcode In None\nUsing Barcode creation for Software Control to generate, create barcode image in Software applications.\nPainting Bar Code In None\nUsing Barcode generation for Software Control to generate, create bar code image in Software applications.\n= 2.52\nMake ISSN In None\nUsing Barcode maker for Software Control to generate, create ISSN - 13 image in Software applications.\nCode 39 Extended Maker In VB.NET\nUsing Barcode creator for .NET framework Control to generate, create Code 39 Full ASCII image in VS .NET applications.\n(It is important to measure the vector lengths in units that are consistent with those used on the axes of the graph.) Substituting these values into Eq. (15.13) gives\nEncode EAN-13 In Objective-C\nUsing Barcode drawer for iPhone Control to generate, create EAN-13 image in iPhone applications.\nGS1 DataBar Expanded Maker In VS .NET\nUsing Barcode generation for .NET framework Control to generate, create GS1 DataBar image in VS .NET applications.\n(1.50)(1.82)(2.52) = or K = (1.50)(1.82)(2.52) = 6.8. To find the point corresponding to K = 6.8 on the branch along the real axis to the left of p3 requires a trial-and-error solution if the graphical approach is used. For example, if s = -4.5 is tried, we obtain\nBarcode Generator In Java\nUsing Barcode drawer for Java Control to generate, create barcode image in Java applications.\nPaint Linear In Java\nUsing Barcode encoder for Java Control to generate, create 1D image in Java applications.\nIs-plI=3.5 Is-p21 = 2.5\nCode39 Encoder In None\nUsing Barcode encoder for Online Control to generate, create Code 39 image in Online applications.\nUSS-128 Maker In Java\nUsing Barcode generation for Android Control to generate, create GS1-128 image in Android applications.\n1 s -p3 I=\nfrom which we get\nK = (1.5)(2.5)(3.5) = 13.1\nWe see that s = -4.5 does not correspond to a gain of 6.8. It is therefore necessary to try other values of s greater than -4.5 until the desired value of K = 6.8 is obtained. Although this procedure may seem very tedious, the actual calculations go quite quickly as the reader will discover while working out this example. We also may find the root on the real axis more directly by applying the following theorem from algebra: The sum of the roots (rl + r2 + *. *+ rn) of the nth-order polynomial equation\naox + alx -l + ..* + a, = 0\nis given by\n(rl +\n12 + ..+r,)\n= -z\nRooTLmus\nIn this case, we have just found the complex roots for\nr2, r3\nK = 6.8 to be\n= -0.95 -c j 1.5\nThe polynomial equation is (s + l)(s + Z)(s + 3) + K which can be expanded into s3 + 6s2 + 11s + (K + 6) = 0 According to the theorem r1 + (-0.95 + j1.5) + (-0.95 - j1/5) = -4 or 6 = -[rl - 2(0.95)] or rl = -4.10 All the detailed steps needed to plot the root locus for this problem have been discussed. The complete locus is shown in Fig. 15.5~. This same plot is also shown in mom detail in Fig. 15.2. Example 15.2. Consider the block diagram for the control system shown in Fig. 15.6. This system may represent a two-tank, liquid-level system having a PID controller and a first-order measuring lag. The open-loop transfer function is 1 + 2~13 + 113s G = K (20s + l)(lOs + 1)(0.5s + 1) Rearranging this into the standatd form,\nKNID, gives\nG= where K = Kc/150\nZl = - 0 . 5 .Q = - 1 p1 = -0.05 p2 = -0.1 p3 = -2\nK(s - ZI)(S -\nes - PlXS - PZ)(S - P3)\nLINEAR txusmm~ SYSTEMS\nFIGURE 15-7 Root-locus diagram for Example 15.2.\nln this case, them am four poles at 0, -0.05, -0.1, and -2 and two zems at -0.5 and -1. These am plotted in Fig. 15.7. Note that the three-action controller\ncontributes the pole at the origin and the zeros, -0.5 and - 1. The steps for plotting the root-locus diagram are as follows: 1. Since them am four poles, there are four branches emerging from them.\n2. Three portions of the root locus are on the real axis between 0 and -0.05,\nbetween -0.10 and -0.5, and between -1 and -2." ]

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[ "# 如何运用枢轴点（Pivot Point）找到有效地支撑位、阻力位？\n\nR2 = P + (H – L) = P + (R1 – S1)\n\nR1 = (P x 2) – L\n\nP = (H + L + C) / 3\n\nS1 = (P x 2) – H\n\nS2 = P – (H – L) = P – (R1 – S1)\n\nH 代表高价位，L 代表低价位，C 代表收盘价，S 代表支撑点，R 代表阻力点，P 代表枢轴点 pivot point。", null, "" ]

[ null, "http://jlhengxinguolu.qiye.gd.cn/uploads/image/20160605165458604.png", null ]

[ "# The Struct datatype\n\nPolars `Struct`s are the idiomatic way of working with multiple columns. It is also a free operation i.e. moving columns into `Struct`s does not copy any data!\n\nFor this section, let's start with a `DataFrame` that captures the average rating of a few movies across some states in the U.S.:\n\n``````ratings = pl.DataFrame(\n{\n\"Movie\": [\"Cars\", \"IT\", \"ET\", \"Cars\", \"Up\", \"IT\", \"Cars\", \"ET\", \"Up\", \"ET\"],\n\"Theatre\": [\"NE\", \"ME\", \"IL\", \"ND\", \"NE\", \"SD\", \"NE\", \"IL\", \"IL\", \"SD\"],\n\"Avg_Rating\": [4.5, 4.4, 4.6, 4.3, 4.8, 4.7, 4.7, 4.9, 4.7, 4.6],\n\"Count\": [30, 27, 26, 29, 31, 28, 28, 26, 33, 26],\n}\n)\nprint(ratings)\n``````\n\n``````let ratings = df!(\n\"Movie\"=> &[\"Cars\", \"IT\", \"ET\", \"Cars\", \"Up\", \"IT\", \"Cars\", \"ET\", \"Up\", \"ET\"],\n\"Theatre\"=> &[\"NE\", \"ME\", \"IL\", \"ND\", \"NE\", \"SD\", \"NE\", \"IL\", \"IL\", \"SD\"],\n\"Avg_Rating\"=> &[4.5, 4.4, 4.6, 4.3, 4.8, 4.7, 4.7, 4.9, 4.7, 4.6],\n\"Count\"=> &[30, 27, 26, 29, 31, 28, 28, 26, 33, 26],\n\n)?;\nprintln!(\"{}\", &ratings);\n``````\n\n``````shape: (10, 4)\n┌───────┬─────────┬────────────┬───────┐\n│ Movie ┆ Theatre ┆ Avg_Rating ┆ Count │\n│ --- ┆ --- ┆ --- ┆ --- │\n│ str ┆ str ┆ f64 ┆ i64 │\n╞═══════╪═════════╪════════════╪═══════╡\n│ Cars ┆ NE ┆ 4.5 ┆ 30 │\n│ IT ┆ ME ┆ 4.4 ┆ 27 │\n│ ET ┆ IL ┆ 4.6 ┆ 26 │\n│ Cars ┆ ND ┆ 4.3 ┆ 29 │\n│ … ┆ … ┆ … ┆ … │\n│ Cars ┆ NE ┆ 4.7 ┆ 28 │\n│ ET ┆ IL ┆ 4.9 ┆ 26 │\n│ Up ┆ IL ┆ 4.7 ┆ 33 │\n│ ET ┆ SD ┆ 4.6 ┆ 26 │\n└───────┴─────────┴────────────┴───────┘\n``````\n\n## Encountering the `Struct` type\n\nA common operation that will lead to a `Struct` column is the ever so popular `value_counts` function that is commonly used in exploratory data analysis. Checking the number of times a state appears the data will be done as so:\n\n``````out = ratings.select(pl.col(\"Theatre\").value_counts(sort=True))\nprint(out)\n``````\n\n``````let out = ratings\n.clone()\n.lazy()\n.select([col(\"Theatre\").value_counts(true, true)])\n.collect()?;\nprintln!(\"{}\", &out);\n``````\n``````shape: (5, 1)\n┌───────────┐\n│ Theatre │\n│ --- │\n│ struct │\n╞═══════════╡\n│ {\"NE\",3} │\n│ {\"IL\",3} │\n│ {\"SD\",2} │\n│ {\"ME\",1} │\n│ {\"ND\",1} │\n└───────────┘\n``````\n\nQuite unexpected an output, especially if coming from tools that do not have such a data type. We're not in peril though, to get back to a more familiar output, all we need to do is `unnest` the `Struct` column into its constituent columns:\n\n``````out = ratings.select(pl.col(\"Theatre\").value_counts(sort=True)).unnest(\"Theatre\")\nprint(out)\n``````\n\n``````let out = ratings\n.clone()\n.lazy()\n.select([col(\"Theatre\").value_counts(true, true)])\n.unnest([\"Theatre\"])\n.collect()?;\nprintln!(\"{}\", &out);\n``````\n``````shape: (5, 2)\n┌─────────┬────────┐\n│ Theatre ┆ counts │\n│ --- ┆ --- │\n│ str ┆ u32 │\n╞═════════╪════════╡\n│ NE ┆ 3 │\n│ IL ┆ 3 │\n│ SD ┆ 2 │\n│ ME ┆ 1 │\n│ ND ┆ 1 │\n└─────────┴────────┘\n``````\n\nWhy `value_counts` returns a `Struct`\n\nPolars expressions always have a `Fn(Series) -> Series` signature and `Struct` is thus the data type that allows us to provide multiple columns as input/ouput of an expression. In other words, all expressions have to return a `Series` object, and `Struct` allows us to stay consistent with that requirement.\n\n## Structs as `dict`s\n\nPolars will interpret a `dict` sent to the `Series` constructor as a `Struct`:\n\n``````rating_Series = pl.Series(\n\"ratings\",\n[\n{\"Movie\": \"Cars\", \"Theatre\": \"NE\", \"Avg_Rating\": 4.5},\n{\"Movie\": \"Toy Story\", \"Theatre\": \"ME\", \"Avg_Rating\": 4.9},\n],\n)\nprint(rating_Series)\n``````\n\n``````// Don't think we can make it the same way in rust, but this works\nlet rating_series = df!(\n\"Movie\" => &[\"Cars\", \"Toy Story\"],\n\"Theatre\" => &[\"NE\", \"ME\"],\n\"Avg_Rating\" => &[4.5, 4.9],\n)?\n.into_struct(\"ratings\")\n.into_series();\nprintln!(\"{}\", &rating_series);\n``````\n\n``````shape: (2,)\nSeries: 'ratings' [struct]\n[\n{\"Cars\",\"NE\",4.5}\n{\"Toy Story\",\"ME\",4.9}\n]\n``````\n\nConstructing `Series` objects\n\nNote that `Series` here was constructed with the `name` of the series in the begninng, followed by the `values`. Providing the latter first is considered an anti-pattern in Polars, and must be avoided.\n\n### Extracting individual values of a `Struct`\n\nLet's say that we needed to obtain just the `movie` value in the `Series` that we created above. We can use the `field` method to do so:\n\n``````out = rating_Series.struct.field(\"Movie\")\nprint(out)\n``````\n\n``````let out = rating_series.struct_()?.field_by_name(\"Movie\")?;\nprintln!(\"{}\", &out);\n``````\n``````shape: (2,)\nSeries: 'Movie' [str]\n[\n\"Cars\"\n\"Toy Story\"\n]\n``````\n\n### Renaming individual keys of a `Struct`\n\nWhat if we need to rename individual `field`s of a `Struct` column? We first convert the `rating_Series` object to a `DataFrame` so that we can view the changes easily, and then use the `rename_fields` method:\n\n``````out = (\nrating_Series.to_frame()\n.select(pl.col(\"ratings\").struct.rename_fields([\"Film\", \"State\", \"Value\"]))\n.unnest(\"ratings\")\n)\nprint(out)\n``````\n\n``````let out = DataFrame::new([rating_series].into())?\n.lazy()\n.select([col(\"ratings\")\n.struct_()\n.rename_fields([\"Film\".into(), \"State\".into(), \"Value\".into()].to_vec())])\n.unnest([\"ratings\"])\n.collect()?;\n\nprintln!(\"{}\", &out);\n``````\n``````shape: (2, 3)\n┌───────────┬───────┬───────┐\n│ Film ┆ State ┆ Value │\n│ --- ┆ --- ┆ --- │\n│ str ┆ str ┆ f64 │\n╞═══════════╪═══════╪═══════╡\n│ Cars ┆ NE ┆ 4.5 │\n│ Toy Story ┆ ME ┆ 4.9 │\n└───────────┴───────┴───────┘\n``````\n\n## Practical use-cases of `Struct` columns\n\n### Identifying duplicate rows\n\nLet's get back to the `ratings` data. We want to identify cases where there are duplicates at a `Movie` and `Theatre` level. This is where the `Struct` datatype shines:\n\n``````out = ratings.filter(pl.struct(\"Movie\", \"Theatre\").is_duplicated())\nprint(out)\n``````\n\n``````let out = ratings\n.clone()\n.lazy()\n// .filter(as_struct(&[col(\"Movie\"), col(\"Theatre\")]).is_duplicated())\n// Error: .is_duplicated() not available if you try that\n// https://github.com/pola-rs/polars/issues/3803\n.filter(count().over([col(\"Movie\"), col(\"Theatre\")]).gt(lit(1)))\n.collect()?;\nprintln!(\"{}\", &out);\n``````\n\n``````shape: (4, 4)\n┌───────┬─────────┬────────────┬───────┐\n│ Movie ┆ Theatre ┆ Avg_Rating ┆ Count │\n│ --- ┆ --- ┆ --- ┆ --- │\n│ str ┆ str ┆ f64 ┆ i64 │\n╞═══════╪═════════╪════════════╪═══════╡\n│ Cars ┆ NE ┆ 4.5 ┆ 30 │\n│ ET ┆ IL ┆ 4.6 ┆ 26 │\n│ Cars ┆ NE ┆ 4.7 ┆ 28 │\n│ ET ┆ IL ┆ 4.9 ┆ 26 │\n└───────┴─────────┴────────────┴───────┘\n``````\n\nWe can identify the unique cases at this level also with `is_unique`!\n\n### Multi-column ranking\n\nSuppose, given that we know there are duplicates, we want to choose which rank gets a higher priority. We define Count of ratings to be more important than the actual `Avg_Rating` themselves, and only use it to break a tie. We can then do:\n\n``````out = ratings.with_columns(\npl.struct(\"Count\", \"Avg_Rating\")\n.rank(\"dense\", descending=True)\n.over(\"Movie\", \"Theatre\")\n.alias(\"Rank\")\n).filter(pl.struct(\"Movie\", \"Theatre\").is_duplicated())\nprint(out)\n``````\n\n``````let out = ratings\n.clone()\n.lazy()\n.with_columns([as_struct(&[col(\"Count\"), col(\"Avg_Rating\")])\n.rank(\nRankOptions {\nmethod: RankMethod::Dense,\ndescending: false,\n},\nNone,\n)\n.over([col(\"Movie\"), col(\"Theatre\")])\n.alias(\"Rank\")])\n// .filter(as_struct(&[col(\"Movie\"), col(\"Theatre\")]).is_duplicated())\n// Error: .is_duplicated() not available if you try that\n// https://github.com/pola-rs/polars/issues/3803\n.filter(count().over([col(\"Movie\"), col(\"Theatre\")]).gt(lit(1)))\n.collect()?;\nprintln!(\"{}\", &out);\n``````\n\n``````shape: (4, 5)\n┌───────┬─────────┬────────────┬───────┬──────┐\n│ Movie ┆ Theatre ┆ Avg_Rating ┆ Count ┆ Rank │\n│ --- ┆ --- ┆ --- ┆ --- ┆ --- │\n│ str ┆ str ┆ f64 ┆ i64 ┆ u32 │\n╞═══════╪═════════╪════════════╪═══════╪══════╡\n│ Cars ┆ NE ┆ 4.5 ┆ 30 ┆ 1 │\n│ ET ┆ IL ┆ 4.6 ┆ 26 ┆ 2 │\n│ Cars ┆ NE ┆ 4.7 ┆ 28 ┆ 2 │\n│ ET ┆ IL ┆ 4.9 ┆ 26 ┆ 1 │\n└───────┴─────────┴────────────┴───────┴──────┘\n``````\n\nThat's a pretty complex set of requirements done very elegantly in Polars!\n\n### Using multi-column apply\n\nThis was discussed in the previous section on User Defined Functions." ]

[ null ]

[ "g\n\nRotary Balancing Calculator(discs, shafts, gears, wheels, etc.)\n\nIf a rotating body is totally uniform in material and construction for 360°, all opposing centrifugal forces (Fc) anywhere within the body will be equal.\n\nIf a mass (m₁) is located at a fixed radius (r₁) anywhere on this body, a different mass (m₂) must also be added to the body, 180° to m₁, at radius r₂ according to the relationship; r₂ = r₁.m₂/m₁\n\nA problem arises, however, if the body features multiple mass anomalies (m₁, m₂, m₃, m₄, … mᵢ) in which case it would be necessary to find an equivalent single mass, along with its position, that will counterbalance their combined effect.\n\nRotating Disc", null, "Fig 1. Unbalanced Disc\n\nFor example, the model described in Fig 1 shows a disc with three positive masses and one negative mass at known but random locations. It should be noted that the positive masses (m₁, m₂ & m₃) are shown as insets within the disc structure and as such must be calculated as mass differentials; i.e. the difference between the mass of the inset and the mass of the disc material removed to allow fitting. Whereas m₄ is a mass of material removed from the disc and therefore must be treated as a negative value. The radial location of each mass is from the axis of rotation to each CofG and all angles are measured from a common datum. Unless all these masses are deliberately and accurately located to ensure balance, the disc will vibrate and if it is allowed to resonate⁽¹⁾ this vibration could cause the disc to shatter prematurely due to excessive localised centrifugal force differentials and internal stresses.\n\nThe combination of all these masses can be replaced with a single equivalent force vector (i.e. a directional centrifugal force; Fcₑ)\n\nThe solution to this problem is to identify a single additional mass and its position that will counterbalance Fcₑ. The magnitude of this mass will depend upon the radial location according to the relationship r = m.v²/Fc and its angular orientation must be 180° to Fcₑ\n\nIt doesn’t matter how irregular its shape, if all the opposing centrifugal forces in a rotating body are equal it will not vibrate.\n\nRotating Shaft", null, "Fig 2. Unbalanced Shaft\n\nThe solution to an unbalanced shaft is similar to that of a disc, however, because these unbalancing masses can be located anywhere along the length of a shaft, the problem becomes one in 3-D. That is, masses placed at various positions along the length of the shaft will induce a longitudinal moment couple from the various unequal centrifugal forces.\n\nIt therefore becomes necessary, in such a case, to incorporate two balancing masses at different positions along the shaft that will generate an equal and opposite moment couple. The further apart these masses are located the smaller they will be.\n\nFor example, the shaft in Fig 2 shows two unbalancing masses (m₁ & m₂) located at distances a₁ & a₂ from plane A and on which the balancing masses (mᴬ & mᴮ) will be fixed at both ends of the shaft (Planes A & B).\n\nIf the balancing masses (mᴬ & mᴮ) are correctly sized and located the shaft will be stabilised centrifugally and longitudinally; i.e. no vibration.\n\nBalancing Calculator – Technical Help\n\nUnits", null, "Fig 3. Balancing a Rotary Disc\n\nYou may use any units you like but you will get out whatever you put in; i.e. you can enter radii in metres and masses in lbs but your output units will comprise the same mixture of units.\n\nExample Calculation for a Disc (1-Plane)\n\nA 0.012m thick aluminium disc of diameter 0.5m (Fig 3) is fitted with two tungsten inserts (m₁ & m₂) flush to both faces of the disc, and one cut-out (m₃)\nm₁: Ø₁ = 0.05m; R₁ = 0.179m; θ₁ = 70°\nm₂: Ø₂ = 0.035m; R₂ = 0.214m; θ₂ = 240°\nm₃: A₃ = 0.03m x 0.2m; R₃ = 0.24m; θ₃ = 270°\nNote: δm = π.ز.t.(ρᵂ - ρᴬˡ)/4\nm₁ = π x 0.05² x 0.012 x (19293 - 2713) / 4 = 0.390657 kg\nm₂ = π x 0.035² x 0.012 x (19293 - 2713) / 4 = 0.191422 kg\nm₃ = -0.03 x 0.2 x 0.012 x 2713 = -0.19534 kg\n\nThe balancing mass is to be placed at radius (R) = 0.245m;\n\nThe balancing mass (m) = 0.315071 kg (Fig 4)\nand must be located (θ) = 267.449909° from the common datum\n\nCalculation for a Shaft (2-Planes)", null, "Fig 4. Calculation Example\n\nA shaft is calculated in exactly the same manner a disc with the following additional inputs:\n\nW: The distance between the planes (A & B) in which the two balancing masses (mᴬ & mᴮ) must be located.\nThese planes can be anywhere along the shaft you like as long as they are normal to the axis of rotation.\n\na₁ to a₁₀: Are the distances of each unbalancing mass (m₁ to m₁₀) from Plane A (Fig 2)\n\nRᴬ and Rᴮ: Are the radial locations of balancing masses (mᴬ & mᴮ) within Planes A & B respectively\n\nθᴬ and θᴮ: Are the angular locations of balancing masses (mᴬ & mᴮ) within Planes A & B respectively\n\nAssembly Calculations\n\nDisc and shaft assemblies should be balanced separately. I.e. you balance the disc as a separate entity and then balance the shaft, as described above.\n\nApplicability\n\nBalancing applies to any rotating shape, including; wheels discs, all shafts (cam, crank, plain, drive, etc.).\n\nAccuracy\n\nThere is no error margin in these calculations; i.e. the output data is as accurate as the input data\n\nNotes\n\n1. 1) A perfectly balanced disc has no natural frequency and therefore cannot resonate (see Vibration Damping)" ]

[ null, "http://www.calqlata.com/prodimages/Balance%202-1.png", null, "http://www.calqlata.com/prodimages/Balance%202-2.png", null, "http://www.calqlata.com/prodimages/Balance%202-3.png", null, "http://www.calqlata.com/prodimages/Balance%202-4.png", null ]

[ "Volume 12, Issue 3, February 2000\n\nISSN: 1434-6044 (Print) 1434-6052 (Online)\n\nIn this issue (15 articles)\n\n1.", null, "Experimental physics\n\nStudy of the QED processes $$e^+e^-\\to e^+e^-\\gamma, e^+e^-\\gamma\\gamma$$ with the SND detector at VEPP-2M\n\nPages 369-374\n2.", null, "Experimental physics\n\nGlobal QCD analysis of parton structure of the nucleon: CTEQ5 parton distributions\n\nPages 375-392\n3.", null, "Experimental physics\n\nMeasurement of the spin-density matrix elements in exclusive electroproduction of $$\\rho^0$$ mesons at HERA\n\nPages 393-410\n4.", null, "Experimental physics\n\nMeasurement of high- $$Q^2$$ charged-current $$e^+p$$ deep inelastic scattering cross sections at HERA\n\nPages 411-428\n5.", null, "Experimental physics\n\n$$\\bar{\\rm p}{\\rm p}$$ -annihilation into $$\\omega\\pi^0$$ , $$\\omega\\eta$$ and $$\\omega\\eta^\\prime$$ at 600, 1200 and 1940 MeV/ $$c$$\n\nPages 429-439\n6.", null, "Theoretical physics\n\nRadiation zeros in $$W^+W^-\\gamma$$ production at high-energy colliders\n\nPages 441-450\n7.", null, "Theoretical physics\n\nYukawa coupling corrections to the decay $$H^+ \\to W^+ A^0$$\n\nPages 451-460\n8.", null, "Theoretical physics\n\nAsymptotic structure of perturbative series for $$\\tau$$ lepton observables\n\nPages 461-467\n9.", null, "Theoretical physics\n\nChiral perturbation theory with virtual photons and leptons\n\nPages 469-478\n10.", null, "Theoretical physics\n\nAnalytic approach to confinement and monopoles in lattice SU (2)\n\nPages 479-488\n11.", null, "Theoretical physics\n\nTensor glueball–meson mixing phenomenology\n\nPages 489-498\n12.", null, "Theoretical physics\n\nNon-extensive statistics, fluctuations and correlations in high-energy nuclear collisions\n\nPages 499-506\n13.", null, "Theoretical physics\n\nNeutrino masses, mixing angles and the unification of couplings in the MSSM\n\nPages 507-520\n14.", null, "Theoretical physics\n\nRunge–Kutta methods and renormalization\n\nPages 521-534\n15.", null, "Theoretical physics\n\nOn the trees of quantum fields\n\nPages 535-549" ]

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[ "# FOURIER SERIES AND FOURIER TRANSFORM , Definition , Complex Exponential , Inverse\n\nBy   November 14, 2019\n\nBRIEF REVIEW OF FOURIER SERIES AND FOURIER TRANSFORM\nIn the field of communication engineering, we need to analyze a given signal. To do so, we have to express the signal in its frequency domain. The translation of a signal from time domain to frequency domain is obtained by using the tools such as Fourier series and Fourier transform.\n1.18.1. Fourier Series: Definition and Basic Concept\nSine waves and cosine waves are the basic building functions for any periodic signal. This means that any periodic signal basically consists of sine waves having different amplitudes, of different frequencies and having different relative phase shifts.\nFourier series represents a periodic waveform in the form of sum of infinite number of sine and cosine terms. It is a representation of the signal in a time domain series form.\nFourier series is a tool used to analyze any periodic signal. After the analysis, we obtain the following information about the signal :\n(i) What all frequency components are present in the signal ?\n(ii) Their amplitudes and\n(iii) The relative phase difference between these frequency components.\nAll the frequency components are nothing else but sine waves at those frequencies.\n1.18.2. Exponential Fourier Series (i.e. Complex Exponential Fourier Series)\nSubstituting the sine and cosine functions in terms of exponential function in the expression for the quadrature Fourier series, we can obtain another type of Fourier series called the exponential Fourier series.\nA periodic signal x(t) is expressed in the exponential Fourier series form as under :\nEquation\nAmplitude and phase spectrums\nThe amplitude spectrum of the signal x(t) is denoted by,\nEquation\nThe phase spectrum of x(t) is given by,\nEquation\nThe amplitude spectrum is a symmertic or even function. This means that  Cn = C-n. But the phase spectrum is an asymmetric or odd function. This means that arg (Cn) = -arg (C-n)\n1.18.3 Fourier Transform: Mathematical Representation\nA Fourier transform is the limiting case of Fourier series. It is used for the analysis of non-periodic signals. The Fourier transform of a signal x(t) is defined as follows:\nFourier transform\nEquation\nor                                                   Equation\nThese equations are known as analysis equations.\n1.18.4 Inverse Fourier Transform : Mathematical Representation\nThe signal x(t) can be obtained back from Fourier transform X(t) by using the inverse Fourier transform. The inverse Fourier transform (IFT) is defined as under:\nInverse Fourier transform\nEquation\nOr                                                  Equation\nAmplitude and phase spectrum\n(i)      The amplitude and phase spectrums are continuous rather than being discrete in nature. Out of them, the amplitude spectrum of a real valued function x(t) exhibits an even symmetry.\nTherefore, we have\nX(t) = X(-f)\n(ii)     Also, the phase spectrum has an odd symmetry. This means that\n1.18.5 Various Properties of Fourier Transform\nTable 1.4    Various properties of Fourier Transform\n\n Sr. No. Name of Property Mathematical expression 1. linearity or superposition Equation 2. Time scaling Equation 3. Daulity or symmetry Equation 4. Time shifting Equation 5. Area under x(t) Equation 6. Area under X(f) Equation 7. Frequency shifting Equation 8. Differentiation in time domain 9. Integration in time domain 10. Conjugate functions 11. Multiplication in time domain 12. Convolution in time domain\n\n1.18.6 Fourier Transforms of Few Standard Signals\nThe Fourier transform of few standard signals is given in table 1.5.\nTable 1.5    Fourier transform of few standard signals\n\n Sr. No. Name of signal Mathematical representation Fourier transfrom 1. Rectangular pulse of amplitude A and duration T 2. Sinc pulse 3. Decaying exponential signal for t > 0 4. Rising exponential pulse for t < 0 5. Double exponential pulse 6. Unit impulse 7. DC signal 8. Cosine signal 9. Sine Signal 10. Signum function 11. Unit step" ]

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[ "carlosgaldino · home\n\n# Consistent Hashing\n\n## Introduction\n\nTraditional hash tables map keys to an array index using the following process:\n\n``````hash = hashFunc(key)\nindex = hash % arraySize\n``````\n\nWhen the `arraySize` changes, all keys need to be remapped because the index is calculated by a modular operation.\n\nThe same technique can be used to partition the data from some application into a number of databases1 by calculating the hash of a key for the data modulo the number of databases, and you could have a situation like the following:", null, "If a new database is added to the cluster, or one existing db is removed (or fails), all keys would need to be remapped just like in the hash table example. Now, if you are dealing with a lot of data you can imagine that remapping all keys would take quite some time which is not very attractive.", null, "## Consistent Hashing\n\nThat’s when consistent hashing comes as an alternative. First, let’s consider the output range of a function `f`:", null, "If we connect both ends we end up with a ring:", null, "Using the same function `f`, we can map each node to a point in the ring:", null, "The interval between two nodes in the ring form a partition. If we use the same function `f`2 over the key we will end up with a projection of that key in the ring:", null, "With this notion we can define the server responsible for our key as the first node in a clockwise direction after the key projection3.", null, "So in this case, `\"Mars\"` would be stored in the server `10.9.5.1`. The same process is applied for any other key:", null, "Note that `f(\"Venus\")` mapped to a point after the last node and before the maximum value of function `f`. Since both ends are connected, there is no problem in walking clockwise to find the responsible node which in this case is `10.1.2.3`.\n\nAdding a new node to the ring does not mean that all keys will need to be remapped. Only a fraction4 of keys will need to move to a different node:", null, "A fraction of the keys are also remapped when a node leaves the ring:", null, "And that’s the essence of Consistent Hashing. The idea was presented in a paper by Karger et al. in 1997 . Nodes are mapped to a ring, forming partitions, which are then used to find the node responsible for a key by mapping the key to the same ring, and finding the first node in a clockwise direction after the key position.\n\nSome examples of systems that use consistent hashing are: Amazon’s Dynamo , Riak , Akka , and Chord .\n\nWith consistent hashing is easier to avoid hotspots by using a function `f` that mixes well, so even if the keys are very similar they end up projected in different and distant points in the ring, causing them to be stored at different nodes. Another benefit is the smoothness for moving keys when nodes join or leave the ring, only the immediate neighbors of a node are impacted and other nodes remain unaffected.\n\nA system using consistent hashing can apply other techniques to reduce even more the impact of changes in the ring structure. If nodes are data stores, like the initial example in this post, the system could replicate the data in the next `N` nodes after the original node, `N1`, that is responsible for that data. This gives the advantage that if `N1` leaves the ring, its immediate neighbors will already have the data that was stored at `N1`, preventing an increase of network traffic after a node (in this case, `N1`) departs. The same technique helps avoiding hotspots even more since requests for the data can be handled by `N1` or any of its next `N` neighbors in the ring.\n\nUsually, systems using consistent hashing construct their rings as the output range of a hash function like SHA-1, or SHA-2, for example. The range for SHA-1 goes from 0 to 2160, and SHA-2 has different output ranges, SHA-256 is 0 to 2256, SHA-512 is 0 to 2512, etc. Using any of these functions to map the nodes in the ring will have the effect of placing the nodes in random positions. The partitions will very likely have different sizes which means nodes will be responsible for different amounts of data. It may not be attractive to have this characteristic and since the range is well defined, the ring could be split into partitions of equal size and then each node could claim ownership of a number of these partitions, guaranteeing that each node handles more or less the same amount of data5.\n\nAnother important technique is the usage of virtual nodes, which we will see next.\n\n### Virtual Nodes\n\nThe ring shown above had a one-to-one mapping between physical nodes and nodes in the ring. This approach presents a challenge that randomly placing the nodes in the ring might lead to a non-uniform distribution of data between nodes.", null, "This problem becomes more evident when a node leaves the ring which requires that all the data handled by that node need to be moved entirely to a single other node.", null, "To avoid overloading a single node when another one leaves the ring, and to distribute the keys more evenly, the system can create a different mapping between physical nodes and nodes in the ring. Instead of having a one-to-one mapping, the system creates virtual nodes, creating a M-to-N mapping between physical nodes and virtual nodes in the ring.", null, "With virtual nodes, each physical node becomes responsible for multiple partitions in the ring. Then if a node leaves the ring, the load handled by this node is evenly dispersed across the remaining nodes in the ring.", null, "And similarly when a node joins the ring, it receives a roughly equivalent amount of data from the other nodes in the ring. The virtual nodes scheme also helps when the system is comprised of heterogeneous nodes in terms of resources such as CPU cores, RAM, disk space, etc. With an heterogeneous cluster the number of virtual nodes for each physical node can be chosen considering the characteristics of each physical node.\n\n## `concha`: A consistent hashing library in Erlang\n\n`concha` is a consistent hashing library in Erlang that I built. The ring represents the output range of SHA-256, and the same function is used for mapping nodes and keys to the ring. It provides lookup of nodes based on the given keys, creating the ring with virtual nodes, adding and removing nodes, etc. For more information and examples of usage you can visit its repository on GitHub.\n\n1. And also to a bunch of other situations, for example, to spread requests to different servers.\n\n2. Or any other function with the same output range.\n\n3. If the projection is equal to a node’s projection, then that node is responsible for the key.\n\n4. Usually `K/n`, where `K` is the number of keys, and `n` is the number of nodes in the ring.\n\n5. Depending on the key distribution in the ring." ]

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[ "", null, "OpenCV  3.4.8 Open Source Computer Vision\nsamples/cpp/tutorial_code/ImgProc/Morphology_2.cpp", null, "Sample screenshot\n\nCheck the corresponding tutorial for more details\n\n#include <iostream>\nusing namespace cv;\nMat src, dst;\nint morph_elem = 0;\nint morph_size = 0;\nint morph_operator = 0;\nint const max_operator = 4;\nint const max_elem = 2;\nint const max_kernel_size = 21;\nconst char* window_name = \"Morphology Transformations Demo\";\nvoid Morphology_Operations( int, void* );\nint main( int argc, char** argv )\n{\nCommandLineParser parser( argc, argv, \"{@input | baboon.jpg | input image}\" );\nif (src.empty())\n{\nstd::cout << \"Could not open or find the image!\\n\" << std::endl;\nstd::cout << \"Usage: \" << argv << \" <Input image>\" << std::endl;\nreturn EXIT_FAILURE;\n}\nnamedWindow( window_name, WINDOW_AUTOSIZE ); // Create window\ncreateTrackbar(\"Operator:\\n 0: Opening - 1: Closing \\n 2: Gradient - 3: Top Hat \\n 4: Black Hat\", window_name, &morph_operator, max_operator, Morphology_Operations );\ncreateTrackbar( \"Element:\\n 0: Rect - 1: Cross - 2: Ellipse\", window_name,\n&morph_elem, max_elem,\nMorphology_Operations );\ncreateTrackbar( \"Kernel size:\\n 2n +1\", window_name,\n&morph_size, max_kernel_size,\nMorphology_Operations );\nMorphology_Operations( 0, 0 );\nwaitKey(0);\nreturn 0;\n}\nvoid Morphology_Operations( int, void* )\n{\n// Since MORPH_X : 2,3,4,5 and 6\nint operation = morph_operator + 2;\nMat element = getStructuringElement( morph_elem, Size( 2*morph_size + 1, 2*morph_size+1 ), Point( morph_size, morph_size ) );\nmorphologyEx( src, dst, operation, element );\nimshow( window_name, dst );\n}" ]

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[ "### 概念 #\n\n$$dx(t)=p(x(t),t)dt + \\sqrt{\\alpha} dW_t.\\tag{48}$$\n\n### 线性随机微分方程} #\n\n$$dx(t)=x(t)dt + \\sqrt{\\alpha} dW_t,\\tag{49}$$\n\n$$P [x(t)]\\mathscr{D}x(t)= P [W(t)]\\mathscr{D}W(t).\\tag{50}$$\n\n$$P [W(t)]=\\exp\\left(-\\int_{t_a}^{t_b} \\frac{1}{2}\\dot{W}^2(t)dt\\right).\\tag{51}$$\n\n\\begin{aligned} P [x(t)]\\mathscr{D}x(t)=& P [W(t)]\\mathscr{D}W(t)\\\\ =&\\exp\\left(-\\int_{t_a}^{t_b} \\frac{1}{2}\\dot{W}^2(t)dt\\right)\\mathscr{D}W(t)\\\\ =&\\exp\\left(-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}(t)-x(t)\\right]^2 dt\\right)\\mathscr{D}W(t)\\end{aligned},\\tag{52}\n\n$$\\mathscr{D}W(t)=\\mathcal{J}[x(t)]\\mathscr{D}x(t),\\tag{53}$$\n\n\\begin{aligned} P [x(t)]\\mathscr{D}x(t)=&\\exp\\left(-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}(t)-x(t)\\right]^2 dt\\right)\\mathscr{D}W(t)\\\\ =&\\exp\\left(-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}(t)-x(t)\\right]^2 dt\\right)\\mathscr{D}x(t)\\end{aligned},\\tag{54}\n\n$$P [x(t)]=\\exp\\left(-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}(t)-x(t)\\right]^2 dt\\right).\\tag{55}$$\n\n\\begin{aligned}\\int P [x(t)]\\mathscr{D}x(t)=&\\int\\exp\\left(-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}^2(t)+x^2(t)-2\\dot{x}(t) x(t)\\right] dt\\right)\\mathscr{D}x(t)\\\\ =&\\exp\\left(\\frac{1}{2\\alpha}\\left(x_b^2-x_a^2\\right)\\right)\\times\\\\ &\\qquad\\int\\exp\\left(-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}^2(t)+x^2(t)\\right] dt\\right)\\mathscr{D}x(t)\\end{aligned}.\\tag{56}\n\n$$K(x_b,t_b;x_a,t_a)=\\int\\exp\\left(-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}^2(t)+x^2(t)\\right] dt\\right)\\mathscr{D}x(t)\\tag{57}$$\n\n### 计算雅可比行列式 #\n\n$$dx(t)=p(x(t), t)dt + \\sqrt{\\alpha} dW_t ,\\tag{58}$$\n\n\\begin{aligned} P [x(t)]\\mathscr{D}x(t)=& P [W(t)]\\mathscr{D}W(t)\\\\ =&\\exp\\left(-\\int_{t_a}^{t_b} \\frac{1}{2}\\dot{W}^2 dt\\right)\\mathscr{D}W(t)\\\\ =&\\exp\\left(-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}-p(x,t)\\right]^2 dt\\right)\\mathscr{D}W(t)\\end{aligned}.\\tag{59}\n\n$$(x_k - x_{k-1}) - \\frac{1}{2}\\epsilon\\left[p(x_k,t_k)+p(x_{k-1},t_{k-1})\\right]=\\sqrt{\\alpha} (W_k-W_{k-1}),\\tag{60}$$\n\n\\left\\{\\begin{aligned}&\\mathscr{D}x(t)\\approx dx_1 dx_2\\dots dx_{k-1}\\\\ &\\mathscr{D}W(t)\\approx dW_1 dW_2\\dots dW_{k-1}\\end{aligned}\\right. .\\tag{61}\n\n$$\\frac{\\partial W_k}{\\partial x_k}=\\frac{1}{\\sqrt{\\alpha}}\\left(1-\\frac{1}{2}\\epsilon\\frac{\\partial p(x_k,t_k)}{\\partial x_k}\\right)\\approx \\frac{1}{\\sqrt{\\alpha}}\\exp\\left(-\\frac{1}{2}\\epsilon\\frac{\\partial p(x_k,t_k)}{\\partial x_k}\\right),\\tag{62}$$\n\n$$\\mathcal{J}[x(t)]\\approx \\prod_{k=1}^{n-1} \\left(1-\\frac{1}{2}\\epsilon\\frac{\\partial p(x_k,t_k)}{\\partial x_k}\\right)\\approx \\exp\\left[-\\frac{1}{2}\\left(\\sum_{k=1}^{n-1}\\frac{\\partial p(x_k,t_k)}{\\partial x_k}\\right)\\epsilon\\right],\\tag{63}$$\n\n$$\\mathcal{J}[x(t)]=\\exp\\left(-\\frac{1}{2}\\int_{t_a}^{t_b}\\frac{\\partial p(x)}{\\partial x}dt\\right),\\tag{64}$$\n\n\\begin{aligned} P [x(t)]\\mathscr{D}x(t)=&\\exp\\left\\{-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\dot{x}-p(x,t)\\right]^2 dt\\right\\}\\\\ &\\qquad\\qquad\\exp\\left(-\\frac{1}{2}\\int_{t_a}^{t_b}\\frac{\\partial p(x,t)}{\\partial x}dt\\right)\\mathscr{D}x(t)\\\\ =&\\exp\\left\\{-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\left(\\dot{x}-p(x,t)\\right)^2 + \\alpha\\frac{\\partial p(x,t)}{\\partial x} \\right]dt\\right\\}\\mathscr{D}x(t) \\end{aligned},\\tag{65}\n\n\\begin{aligned} P [x(t)]=&\\exp\\left\\{-\\frac{1}{2\\alpha}\\int_{t_a}^{t_b} \\left[\\left(\\dot{x}-p(x,t)\\right)^2 + \\alpha\\frac{\\partial p(x,t)}{\\partial x} \\right]dt\\right\\}\\\\ =&\\exp\\left\\{-\\frac{1}{\\alpha}\\int_{t_a}^{t_b}\\left[\\frac{1}{2}\\dot{x}^2-\\dot{x}p(x,t)+\\frac{1}{2}p^2(x,t)+\\frac{1}{2}\\alpha\\frac{\\partial p(x,t)}{\\partial x}\\right]dt\\right\\} \\end{aligned}.\\tag{66}\n\n\\begin{aligned}\\frac{d}{dt}\\int p(x,t)dx =& \\dot{x}\\frac{\\partial}{\\partial x}\\int p(x,t)dx+\\frac{\\partial}{\\partial t}\\int p(x,t)dx\\\\ =&\\dot{x}p(x,t)+\\int \\frac{\\partial p(x,t)}{\\partial t}dx \\end{aligned},\\tag{67}\n\n$$P [x(t)]=\\exp\\left[-\\frac{1}{\\alpha}\\int_{t_a}^{t_b}\\left(\\frac{1}{2}\\dot{x}^2-V(x,t)\\right)dt\\right]\\exp\\left(\\frac{1}{\\alpha}\\int_{x_a}^{x_b}p(x,t)dx\\right).\\tag{68}$$\n\n$$V(x,t)=-\\frac{1}{2}\\left(\\alpha\\frac{\\partial p}{\\partial x}+p^2\\right)-\\int \\frac{\\partial p}{\\partial t}dx.\\tag{69}$$\n\n### 路径积分方法 #\n\n@online{kexuefm-3762,\ntitle={路径积分系列：4.随机微分方程},\nauthor={苏剑林},\nyear={2016},\nmonth={Jun},\nurl={\\url{https://spaces.ac.cn/archives/3762}},\n}" ]

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[ "NumWords.com\n\n# How to write Four hundred nine in numbers in English?", null, "We can write Four hundred nine equal to 409 in numbers in English\n\n< Four hundred eight :||: Four hundred ten >\n\nEight hundred eighteen = 818 = 409 × 2\nOne thousand two hundred twenty-seven = 1227 = 409 × 3\nOne thousand six hundred thirty-six = 1636 = 409 × 4\nTwo thousand forty-five = 2045 = 409 × 5\nTwo thousand four hundred fifty-four = 2454 = 409 × 6\nTwo thousand eight hundred sixty-three = 2863 = 409 × 7\nThree thousand two hundred seventy-two = 3272 = 409 × 8\nThree thousand six hundred eighty-one = 3681 = 409 × 9\nFour thousand ninety = 4090 = 409 × 10\nFour thousand four hundred ninety-nine = 4499 = 409 × 11\nFour thousand nine hundred eight = 4908 = 409 × 12\nFive thousand three hundred seventeen = 5317 = 409 × 13\nFive thousand seven hundred twenty-six = 5726 = 409 × 14\nSix thousand one hundred thirty-five = 6135 = 409 × 15\nSix thousand five hundred forty-four = 6544 = 409 × 16\nSix thousand nine hundred fifty-three = 6953 = 409 × 17\nSeven thousand three hundred sixty-two = 7362 = 409 × 18\nSeven thousand seven hundred seventy-one = 7771 = 409 × 19\nEight thousand one hundred eighty = 8180 = 409 × 20\n\nSitemap" ]

[ null, "https://www.numwords.com/images/409-words-to-number-en.jpg", null ]

[ "## Data Visualization And Data Analytics\n\n#### GET ASSIGNMENT HELP", null, "# Physics Homework Help With Rotational Motion\n\n## Rotational Motion\n\nConsider a rigid body rotation about a fixed axis. A body is said to be rigid, if distance between any two particles remain same OR in other words, it can not be deformed.\n\nAs the given body is in pure rotational motion every point on it are moving in a circular motion about fixed axis having different radii. Consider two point P and Q located on a body, moving in a circle of radii r1 and r2 respectively. Now, the angle (q) they have turned will be same, but the distances along arc are unequal as their radii are different.", null, "### Equations defining the rotation of a rigid body about a fixed axis\n\nThe following physical quantities are used to describe a rotational motion.\n\n### (A) Angular displacement:\n\nIt is defined as the change in angular position (from 1 to 2) of the position vector with respect to any reference line.\n\nHence the angular displacement Dq is given by Dq = q2 - q1\n\nIf angular displacement is large, it is scalar. If it is very small, then it is treated as vectors.", null, "### (B) Angular velocity:\n\nThe average angular velocity of the body in the time interval t (t2 – t1) is given by", null, "The instantaneous angular velocity", null, "is", null, "The direction of", null, "is always be perpendicular to the plane containing rotation.\n\n### (C) Angular Acceleration:\n\nThe average angular acceleration of the rotating body in the interval. (t2 – t1) ,\n\nis given by", null, "The instantaneous angular acceleration,", null, "", null, "From figure we have S = qrs ® arc length traveled by the particle\n\n Differentiating,", null, "… (i) Þ vT = wr Where vT is tangential speed. Differentiating, equation (i) once again", null, "", null, "Þ aT = ar … (ii)\n\nas the point P is moving in a circular path a centripetal acceleration is associated with a which is given as\n\naC = w2r = vT2 / r … (iii)\n\nThe resultant acceleration of quantity is", null, "### Homework Help For Rotational Motion\n\nassignmenthelp.net provides best Online assignment Help service in Physics for all standards. Our Tutor provide their high quality and optimized Tutorial help to fulfill all kind of need of Students.\n\nTo Schedule a Rotational Motion tutoring session\n\nPhysics tutors | Online tutors | Physics Homework help | Online tutoring term paper help | Assignment help | Physics projects | Physics tutorial | Physics questions | physics problems | Physics online help | Science physics | Physics tutor | high school physics | Homework tutor | science help | physics experiments | Online Tutoring | Physics tutors | Homework tutoring\n\nAssignment Help Features\nAssignment Help Services\n• Assignment Help\n• Homework Help\n• Writing Help", null, "", null, "" ]

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[ "# MCMC in Python: sim and fit with same model\n\nHere is a github issue and solution that I saw the other day. I think it’s a nice pattern.\n\n```def generate_model(values={'mu': true_param, 'm': None}):\n\n#prior\nmu = pymc.Uniform(\"mu\", lower=-10, upper=10, value=values['mu'],\nobserved=(values['mu'] is not None))\n\n# likelihood function\nm = pymc.Normal(\"m\", mu=mu, tau=tau, value=values['m'],\nobserved=(values['m'] is not None))\n\nreturn locals()\n```\n\nComments Off on MCMC in Python: sim and fit with same model\n\nFiled under statistics" ]

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[ "# Transfer Function Calculation Matlab with Example\n\nWant create site? Find Free WordPress Themes and plugins.\n\n## Transfer Functions Representation\n\nConsider a fixed single-input/single-output linear system with input u(t) and output y(t) given by the differential equation\n\n$\\begin{matrix} \\overset{..}{\\mathop{y}}\\,+6\\overset{.}{\\mathop{y}}\\,+5y=4\\overset{.}{\\mathop{u}}\\,+3u & \\cdots & (1) \\\\\\end{matrix}$\n\nApplying the Laplace transform to both sides of (1) with zero initial conditions, we obtain the transfer function of the system from the input U (s) to the output Y(s) in TF form as the ratio of polynomials:\n\n$\\begin{matrix} G(s)=\\frac{4s+3}{{{s}^{2}}+6s+5} & \\cdots & (2) \\\\\\end{matrix}$\n\nAlternatively, this transfer function can be expressed in terms of its zeros zi, poles pj, and gain K in the factored, or Zero-Pole (ZP) form, as\n\n$\\begin{matrix} G(s)=\\frac{4(s+0.75)}{(s+1)(s+5)} & \\cdots & (3) \\\\\\end{matrix}$\n\nWhich shows that G(s) has a single zero at s = -0.75, two poles at s = -1 and -5, and a gain of 4.\n\n## Transfer Function Using Numerator and Denominator Coefficients\n\nIf we know the numerator and denominator polynomials of G(s), we can represent the model in MATLAB by creating a pair of row vectors containing the coefficients of the powers of s, in descending order, of the numerator and denominator polynomials. For the transfer function in (2), we would enter numG = [4 3] and denG = [1 6 5].\n\nIf the transfer function is known in terms of its zeros, poles, and gain, we can create the model by entering column vectors for the zeros and poles, and enter the gain as a scalar. Then the model is represented by these three quantities, which can be used as arguments in commands for performing calculations. To create the system described in (3), we would enter the commands zG = -0.75, pG = [-1; -5],and kG = 4.\n\nWhen a model has been described in MATLAB in either one of these forms, the Control System Toolbox provides several functions to obtain the description in the other form.\n\n## Commands to Create Transfer Functions\n\nFor example, if the numerator and denominator polynomials are known as the vectors numG and denG, we merely enter the MATLAB command [zz, pp, kk] = tf2zp (numG, denG). The result will be the three-tuple [zz, pp, kk] , which consists of the values of the zeros, poles, and gain of G(s), respectively. Alternatively, if the zeros, poles, and gain are known as the column vectors zG and pG and the scalar kG, we would enter the command [num, den] = zp2tf (zG, pG, kG) .\n\n## Transfer Function Graphical Representation\n\nOnce we have the system model in either the TF or the ZP form, we can obtain a graphical representation of the transfer functions poles and zeros by using the pzmap command from the Control System Toolbox. The two forms of the command that are applicable to transfer-function models are pzmap(numG,denG) and pzmap(pG, zG), where in the first case the arguments are row vectors and in the latter case, the arguments are column vectors. In either case, a region of the s-plane is shown with the zeros indicated by “O” and the poles by “X”.\n\n## Transfer Function Matlab Example\n\nConvert G(s) to Factored Form\n\nFind the transfer function in both the polynomial (TF) and factored (ZP) forms for the fourth-order system whose differential equation is\n\n$\\overset{….}{\\mathop{y}}\\,+3\\overset{…}{\\mathop{y}}\\,+4\\overset{..}{\\mathop{y}}\\,+2.5\\overset{.}{\\mathop{y}}\\,+0.8y=3\\overset{..}{\\mathop{u}}\\,+5\\overset{.}{\\mathop{u}}\\,+7u$\n\nSolution\n\nTaking the Laplace transform of the differential equation with zero initial conditions, we obtain the polynomial form of the transfer function as\n\n$G(s)=\\frac{3{{s}^{2}}+5s+3}{{{s}^{4}}+3{{s}^{3}}+4{{s}^{2}}+2.5s+0.8}$\n\nThe MATLAB commands in Script 1 will create the polynomial form of G(s) and then use it to determine the factored form. From the numerical output we find that G(s) has zeros at s = -0.8333± j1.2802, poles at s == -0.4548±j0.5420 and s = -1.0452±j0.7110, and a gain of 3.\n\n## Matlab Code to Find the Transfer Function\n\n\n% Script 1: Matlab Code to Find the transfer function\n\n%in both the polynomial (Transfer-Function) and factored (Zero-Pole) forms\n\nnumG = [3 5 7] % enter transfer function numerator\n\ndenG = [1 3 4 2.5 0.8] % ...and denominator polynomials\n\n% convert to factored (ZP) form\n\n[zz,pp,kk] = tf2zp(numG,denG)\n\npzmap(numG,denG) % pzmap from TF form\n\npzmap(pp,zz) % pzmap from ZP form", null, "Did you find apk for android? You can find new Free Android Games and apps." ]

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[ "Home | | Strength of Materials for Mechanical Engineers | Solved Answers: Stress, Strain and Deformation of Solids\n\n# Solved Answers: Stress, Strain and Deformation of Solids\n\nMechanical - Strength of Materials - Stress, Strain and Deformation of Solids\n\nProblem - 1:\n\nA mild steel rod 2m long &3cm diameter is subjected to an axial pull of 10kN. E for Youngs modulus for steel is 2 x 105 N/mm2. Find the stress, strain.\n\nSol:", null, "Problem - 2:\n\nA steel bar 900mm long its 2ends are 40mm &30mm in diameters &the length of each rod is 200mm. The middle portion of the bar is 15mm in diameters &500mm in long if the bar is subjected an axial tensile load of 15KN. Determine the stress in each section &total extension. Take E=200 103N/mm2.\n\nSol:", null, "Problem –3:\n\nThe following are the results of a tensile on test on a molested rod is 16 mm gaugh length is 50 mm, load at proportionality limit is 48.5 kN. Extension at the proportionally limit is 0.05mm. Load at yield point is 50.3 KN ultimate load is 90KN. Final length between gauge points is 64mm, dia of the neck at   fracture   is   13.7   mm.   Determine   the e you Stress, % of elangation & % of reduction in area.\n\nGiven data:\n\nDia of rod d          =       16mm\n\nDia of Fracture of rod fd         =       13.7 mm\n\nGauge length l      =       50mm\n\nFinal length  \\$  L  =       645 mm\n\nLoad at yield point                  =       50.3KN\n\nLoad at   limit       =       48.5 KN\n\nExtension of ( inf ) limit \\$   l   =       0.05.mm", null, "", null, "Problem - 4:\n\nA Cylindrical pipe of diameter 1.5mm & thickness 1.5cm is subjected to an internal fluid pressure of 1.2 N/mm2. Find the longitudinal stress & hoop stress developed in the pipe.", null, "Problem - 5:\n\nA Cyliner of internal dia 0.5m contains air at apressure of 7 N/mm2.If the maximum permissible stress induced in material is 80 N/mm2. Find the thickness of the cylinder .", null, "Problem - 6:\n\nCalculate change in length, change in diameters, change in volume of the thin cylinder 100cm dia & 1cm thickness, 5m long E= 2 105 N/mm2 poisson’s ratio=0.3& internal(1/m)fluid (or pressure 3 N/mm2.", null, "", null, "", null, "Problem - 7:\n\nA bar of 20mm diameter is tested in tension it is observed that when a lead of 40KN is applied the extension measured over a gauge length of 200mm us 0.12mm&contraction in diameter is 0.0036mm. Find    poisson’s    ratio,    young’s modulas.", null, "", null, "Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\nMechanical : Strength of Materials : Stress, Strain and Deformation of Solids : Solved Answers: Stress, Strain and Deformation of Solids |" ]

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[ "top of page", null, "# Groupe de #TwinsProd\n\nPublic·54 membres\n\n# Learn from the Expert: Mechanics of Aircraft Structures Second Edition by C. T. Sun, a Renowned Professor and Researcher\n\n## Mechanics Of Aircraft Structures Second Edition C T Sun Solution Manual\n\nIf you are looking for a comprehensive textbook that covers the fundamentals of structural mechanics and aerospace engineering with an emphasis on new materials and recent advances, then you might want to check out Mechanics Of Aircraft Structures Second Edition by C. T. Sun. This book is designed to help students get a solid background in structural mechanics as well as to help professionals keep up to date on recent developments in the field. In this article, we will give you an overview of what this book is about, who is the author, what are the main features and updates of the second edition, and how you can use it to learn or review the topics covered in each chapter. We will also provide some example problems and solutions from each chapter to give you a taste of what you can expect from this book.\n\n## Introduction\n\nMechanics Of Aircraft Structures Second Edition is a textbook that covers the basic theory and analysis methods of structural mechanics with applications to aircraft structures. It also introduces the mechanics of composite materials and laminated structures, which are increasingly used in modern aircraft design. The book is divided into eight chapters, each with its own objectives, key concepts, terms, examples, problems, and solutions. The book also includes appendices that provide useful formulas and tables for reference.\n\nThe author of this book is C. T. Sun, who is the Neil A. Armstrong Distinguished Professor in the School of Aeronautics and Astronautics at Purdue University in West Lafayette, Indiana. He is also the recipient of the 2004 Purdue University Research Award. He has over 40 years of experience in teaching and research in the areas of structural mechanics, composite materials, fracture mechanics, and nanomechanics. He has published over 300 journal papers and 10 books on these topics. He is also a Fellow of the American Society of Mechanical Engineers (ASME), the American Institute of Aeronautics and Astronautics (AIAA), the American Society for Composites (ASC), and the American Academy of Mechanics (AAM).\n\nThe second edition of this book was published in 2006 by John Wiley & Sons. It has been extensively updated and expanded to reflect the latest developments and advances in the field of structural mechanics and aerospace engineering. Some of the new features and updates of the second edition are:\n\n• A new chapter on fracture mechanics that introduces the basic concepts and tools for studying crack growth and damage tolerance in aircraft structures.\n\n• A new section on analysis of composite laminates that covers the stiffness and compliance matrices, failure criteria, and delamination effects.\n\n• More examples and problems that illustrate the applications of the theory and methods to realistic aircraft structures.\n\n• More details and explanations on some of the derivations and solutions to enhance the clarity and understanding of the material.\n\n• More references and suggestions for further reading or practice at the end of each chapter.\n\n## Chapter 1: Characteristics of Aircraft Structures and Materials\n\nThe first chapter of this book provides an introduction to the characteristics of aircraft structures and materials. It covers the following topics:\n\n• The types and functions of aircraft structures, such as wings, fuselage, tail, landing gear, etc.\n\n• The concepts and definitions of stress, strain, stiffness, strength, and failure modes in structural mechanics.\n\n• The properties and classifications of materials used in aircraft structures, such as metals, alloys, composites, ceramics, etc.\n\nSome of the key concepts and terms introduced in this chapter are:\n\n• Aircraft structures: The components or parts that support or transmit loads in an aircraft.\n\n• Loads: The forces or moments that act on a structure or a material.\n\n• Stress: The intensity or distribution of internal forces in a material or a cross-section.\n\n• Strain: The measure or degree of deformation or change in shape or size of a material or a structure due to external loads.\n\n• Stiffness: The resistance or ability of a material or a structure to deform under external loads.\n\n• Strength: The maximum stress or load that a material or a structure can withstand without failure.\n\n• Failure modes: The ways or mechanisms by which a material or a structure fails under external loads, such as yielding, fracture, buckling, fatigue, creep, etc.\n\n• Materials: The substances or matter that make up a structure or a component.\n\nHere are some example problems and solutions from this chapter:\n\n### Example 1.1\n\nA thin-walled cylindrical pressure vessel with an inner radius of 0.5 m and a wall thickness of 10 mm is subjected to an internal pressure of 2 MPa. Determine the hoop stress and the longitudinal stress in the wall.\n\n#### Solution\n\nThe hoop stress $\\sigma_h$ is given by:\n\n$$\\sigma_h = \\fracprt$$ where $p$ is the internal pressure, $r$ is the inner radius, and $t$ is the wall thickness. Substituting the given values, we get:\n\n$$\\sigma_h = \\frac(2\\times10^6)(0.5)0.01 = 100\\times10^6 \\text Pa = 100 \\text MPa$$ The longitudinal stress $\\sigma_l$ is given by: $$\\sigma_l = \\fracpr2t$$ where $p$ is the internal pressure, $r$ is the inner radius, and $t$ is the wall thickness. Substituting the given values, we get: $$\\sigma_l = \\frac(2\\times10^6)(0.5)2(0.01) = 50\\times10^6 \\text Pa = 50 \\text MPa$$ The longitudinal stress is usually less than the hoop stress in a thin-walled cylindrical pressure vessel. This is because the internal pressure acts on the ends of the cylinder and stretches the length of the cylinder, as shown in Figure 1.\n\nFigure 1: Longitudinal stress in a thin-walled cylindrical pressure vessel\n\n### Example 1.2\n\nA thin-walled spherical pressure vessel with an inner radius of 0.4 m and a wall thickness of 5 mm is subjected to an internal pressure of 1.5 MPa. Determine the tangential stress and the radial stress in the wall.\n\n#### Solution\n\nThe tangential stress $\\sigma_t$ is given by:\n\n$$\\sigma_t = \\fracprt$$ where $p$ is the internal pressure, $r$ is the inner radius, and $t$ is the wall thickness. Substituting the given values, we get:\n\n$$\\sigma_t = \\frac(1.5\\times10^6)(0.4)0.005 = 120\\times10^6 \\text Pa = 120 \\text MPa$$ The tangential stress accounts for the stress in the plane of the surface of the sphere. The stress normal to the walls of the sphere is called the radial stress, $\\sigma_r$. The radial stress is zero on the outer wall since that is a free surface. On the inner wall, the normal stress is $\\sigma_r = -p$, as shown in Figure 2. Since $t/r \\ll 1$, $p \\ll \\sigma_t$, and it is reasonable to take $\\sigma_r = 0$ not only on the outer wall, but on the inner wall also. The stress state in the spherical wall is then one of plane stress.\n\nFigure 2: Radial stress in a thin-walled spherical pressure vessel\n\n71b2f0854b\n\n## À propos\n\nBienvenue dans le groupe ! Vous pouvez communiquer avec d'au...\n\nbottom of page" ]

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[ "#### Solution for What is 25.622 percent of 10:\n\n25.622 percent *10 =\n\n(25.622:100)*10 =\n\n(25.622*10):100 =\n\n256.22:100 = 2.5622\n\nNow we have: 25.622 percent of 10 = 2.5622\n\nQuestion: What is 25.622 percent of 10?\n\nPercentage solution with steps:\n\nStep 1: Our output value is 10.\n\nStep 2: We represent the unknown value with {x}.\n\nStep 3: From step 1 above,{10}={100\\%}.\n\nStep 4: Similarly, {x}={25.622\\%}.\n\nStep 5: This results in a pair of simple equations:\n\n{10}={100\\%}(1).\n\n{x}={25.622\\%}(2).\n\nStep 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both\nequations have the same unit (%); we have\n\n\\frac{10}{x}=\\frac{100\\%}{25.622\\%}\n\nStep 7: Again, the reciprocal of both sides gives\n\n\\frac{x}{10}=\\frac{25.622}{100}\n\n\\Rightarrow{x} = {2.5622}\n\nTherefore, {25.622\\%} of {10} is {2.5622}\n\n#### Solution for What is 10 percent of 25.622:\n\n10 percent *25.622 =\n\n(10:100)*25.622 =\n\n(10*25.622):100 =\n\n256.22:100 = 2.5622\n\nNow we have: 10 percent of 25.622 = 2.5622\n\nQuestion: What is 10 percent of 25.622?\n\nPercentage solution with steps:\n\nStep 1: Our output value is 25.622.\n\nStep 2: We represent the unknown value with {x}.\n\nStep 3: From step 1 above,{25.622}={100\\%}.\n\nStep 4: Similarly, {x}={10\\%}.\n\nStep 5: This results in a pair of simple equations:\n\n{25.622}={100\\%}(1).\n\n{x}={10\\%}(2).\n\nStep 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both\nequations have the same unit (%); we have\n\n\\frac{25.622}{x}=\\frac{100\\%}{10\\%}\n\nStep 7: Again, the reciprocal of both sides gives\n\n\\frac{x}{25.622}=\\frac{10}{100}\n\n\\Rightarrow{x} = {2.5622}\n\nTherefore, {10\\%} of {25.622} is {2.5622}\n\nCalculation Samples" ]

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[ "# How to compute Dedekind eta function efficiently?\n\nAccording to wiki: https://en.wikipedia.org/wiki/Dedekind_eta_function, Dedekind eta function is defined in many equivalent forms. But none of them is an explicit description (say in algorithmic format) on how to computing it. Where to find such one? Thanks!\n\nEuler's formula\n\n$$\\sum\\limits_{n \\in \\mathbb{Z}} {( - 1)^n q^{\\frac{{(3n^2 - n)}} {2}} } = \\prod\\limits_{n = 1}^\\infty {(1 - q^n ),}$$\n\n(which can be proven from Jacobi’s triple product identity by using the fact that $\\prod\\limits_{n = 1}^\\infty {(1 - q^{3n} )(1 - q^{3n - 2} )} (1 - q^{3n - 1} ) = \\prod\\limits_{n = 1}^\\infty {(1 - q^n )}$) provides a good way of numerically computing\n\n$$\\eta (\\tau )=e^{\\frac {\\pi {\\rm {{i}\\tau }}}{12}}\\prod _{n=1}^{\\infty }(1-e^{2n\\pi {\\rm {{i}\\tau }}})=q^{\\frac {1}{24}}\\prod _{n=1}^{\\infty }(1-q^{n}).$$\n\nMy maple code of Gatteschi-Sokal algorithm for computing $R(t,x)=\\prod_{n=1}^{\\infty}(1-tx^n)$:\n\nGS:=proc(t,x,prec)\n\nlocal R0, a0, b0, Rn, an, bn, d, c, i, N, r, Rd;\n\nN := 100;\n\nd := 1/2; if d = t*x then d := (1/2)*d end if;\n\nr := evalf$[prec]$(1+d/(1-x));\n\na0 := 1; b0 := evalf$[prec]$(d/(d-t*x));\n\nR0 := evalf$[prec]$(r*a0+(1-r)*b0);\n\ni := 0;\n\nwhile i < N do\n\nc := evalf$[prec]$(a0*(d*a0+(1.0-d)*b0));\n\nan := evalf$[prec]$(c/b0);\n\nbn := evalf$[prec]$(c/(x*a0+(1.0-x)*b0));\n\nRn := evalf$[prec]$(r*an+(1.0-r)*bn);\n\nRd := evalf$[prec]$(abs(Rn-R0));\n\nif Rd < 10^(-prec) then i := N else\n\na0 := an; b0 := bn; R0 := Rn; i := i+1 end if;\n\nend do;\n\nreturn Rn;\n\nend proc;\n\neta:=(t,prec)->evalf$[prec]$(GS(1,exp($2*Pi*I*t$),prec));\n\nDedekind_eta := (t, prec)-> evalf$[prec]$(exp($1/12*Pi*I*t$)*eta(t, prec));\n\nTest:\n\nt:=0.3*I:\n\neta(t,40); Dedekind_eta(t,40);\n\n0.8251926470787677741036466781518992636742\n\n0.7628619270903183863013294748250092216042\n\nAlmost same with PARI/GP output:\n\neta(0.3*I,0)=0.82519264707876777410364667815189926367\n\neta(0.3*I,1)=0.76286192709031838630132947482500922160" ]

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[ "# Constant of integration\n\n\nConstant of integration\n\nIn calculus, the indefinite integral of a given function (i.e., the set of all antiderivatives of the function) is only defined up to an additive constant, the constant of integration. This constant expresses an ambiguity inherent in the construction of antiderivatives. If a function f(x) is defined on an interval and F(x) is an antiderivative of f(x), then the set of all antiderivatives of f(x) is given by the functions F(x) + C, where C is an arbitrary constant. The constant of integration is sometimes omitted in lists of integrals for simplicity.\n\n## Origin of the constant\n\nThe derivative of any constant function is zero. Once one has found one antiderivative F(x), adding or subtracting a constant C will give us another antiderivative, because", null, "$(F(x) + C)' = F\\,'(x) + C\\,' = F\\,'(x)$. The constant is a way of expressing that every function has an infinite number of different antiderivatives.\n\nFor example, suppose one wants to find antiderivatives of cos(x). One such antiderivative is sin(x). Another one is sin(x) + 1. A third is sin(x) − π. Each of these has derivative cos(x), so they are all antiderivatives of cos(x).\n\nIt turns out that adding and subtracting constants is the only flexibility we have in finding different antiderivatives of the same function. That is, all antiderivatives are the same up to a constant. To express this fact for cos(x), we write:", null, "$\\int \\cos(x)\\,dx = \\sin(x) + C.$\n\nReplacing C by a number will produce an antiderivative. By writing C instead of a number, however, a compact description of all the possible antiderivatives of cos(x) is obtained. C is called the constant of integration. It is easily determined that all of these functions are indeed antiderivatives of cos(x):", null, "\\begin{align} \\frac{d}{dx}[\\sin(x) + C] &= \\frac{d}{dx}[\\sin(x)] + \\frac{d}{dx}[C] \\\\ &= \\cos(x) + 0 \\\\ &= \\cos(x) \\end{align}\n\n## Necessity of the constant\n\nAt first glance it may seem that the constant is unnecessary, since it can be set to zero. Furthermore, when evaluating definite integrals using the fundamental theorem of calculus, the constant will always cancel with itself.\n\nHowever, trying to set the constant equal to zero doesn't always make sense. For example, 2sin(x)cos(x) can be integrated in two different ways:", null, "\\begin{align} \\int 2\\sin(x)\\cos(x)\\,dx &=& \\sin^2(x) + C &=& -\\cos^2(x) + 1 + C \\\\ \\int 2\\sin(x)\\cos(x)\\,dx &=& -\\cos^2(x) + C &=& \\sin^2(x) - 1 + C \\end{align}\n\nSo setting C to zero can still leave a constant. This means that, for a given function, there is no \"simplest antiderivative\".\n\nAnother problem with setting C equal to zero is that sometimes we want to find an antiderivative that has a given value at a given point (as in an initial value problem). For example, to obtain the antiderivative of cos(x) that has the value 100 at x = π, then only one value of C will work (in this case C = 100).\n\nThis restriction can be rephrased in the language of differential equations. Finding an indefinite integral of a function f(x) is the same as solving the differential equation", null, "$\\frac{dy}{dx} = f(x)$. Any differential equation will have many solutions, and each constant represents the unique solution of a well-posed initial value problem. Imposing the condition that our antiderivative takes the value 100 at x = π is an initial condition. Each initial condition corresponds to one and only one value of C, so without C it would be impossible to solve the problem.\n\nThere is another justification, coming from abstract algebra. The space of all (suitable) real-valued functions on the real numbers is a vector space, and the differential operator", null, "$\\frac{d}{dx}$ is a linear operator. The operator", null, "$\\frac{d}{dx}$ maps a function to zero if and only if that function is constant. Consequently, the kernel of", null, "$\\frac{d}{dx}$ is the space of all constant functions. The process of indefinite integration amounts to finding a preimage of a given function. There is no canonical preimage for a given function, but the set of all such preimages forms a coset. Choosing a constant is the same as choosing an element of the coset. In this context, solving an initial value problem is interpreted as lying in the hyperplane given by the initial conditions.\n\n## Reason for a constant difference between antiderivatives\n\nThis result can be formally stated in this manner: Let", null, "$F:\\mathbb{R}\\rightarrow\\mathbb{R}$ and", null, "$G:\\mathbb{R}\\rightarrow\\mathbb{R}$ be two everywhere differentiable functions. Suppose that", null, "$F\\,'(x) = G\\,'(x)$ for every real number x. Then there exists a real number C such that F(x) − G(x) = C for every real number x.\n\nTo prove this, notice that [F(x) − G(x)]' = 0. So F can be replaced by F-G and G by the constant function 0, making the goal to prove that an everywhere differentiable function whose derivative is always zero must be constant:\n\nChoose a real number a, and let C = F(a). For any x, the fundamental theorem of calculus says that", null, "\\begin{align} \\int_a^x 0\\,dt &= F(x)-F(a)\\\\ &= F(x)-C, \\end{align}\n\nwhich implies that F(x) = C. So F is a constant function.\n\nTwo facts are crucial in this proof. First, the real line is connected. If the real line were not connected, we would not always be able to integrate from our fixed a to any given x. For example, if we were to ask for functions defined on the union of intervals [0,1] and [2,3], and if a were 0, then it would not be possible to integrate from 0 to 3, because the function is not defined between 1 and 2. Here there will be two constants, one for each connected component of the domain. In general, by replacing constants with locally constant functions, we can extend this theorem to disconnected domains. For example, there are two constants of integration for", null, "$\\textstyle\\int dx/x$ and infinitely many for", null, "$\\textstyle\\int \\tan x\\,dx.$\n\nSecond, F and G were assumed to be everywhere differentiable. If F and G are not differentiable at even one point, the theorem fails. As an example, let F(x) be the Heaviside step function, which is zero for negative values of x and one for non-negative values of x, and let G(x) = 0. Then the derivative of F is zero where it is defined, and the derivative of G is always zero. Yet it's clear that F and G do not differ by a constant. Even if it is assumed that F and G are everywhere continuous and almost everywhere differentiable the theorem still fails. As an example, take F to be the Cantor function and again let G = 0.\n\nWikimedia Foundation. 2010.\n\n### Look at other dictionaries:\n\n• Constant of integration — Constant Con stant, n. 1. That which is not subject to change; that which is invariable. [1913 Webster] 2. (Math.) A quantity that does not change its value; used in countradistinction to {variable}. [1913 Webster] 3. (Astron.) A number whose… …   The Collaborative International Dictionary of English\n\n• constant of integration — integravimo pastovioji statusas T sritis automatika atitikmenys: angl. constant of integration vok. Integrationskonstante, f rus. постоянная интегрирования, f pranc. constante d intégration, f …   Automatikos terminų žodynas\n\n• constant of integration — integravimo konstanta statusas T sritis fizika atitikmenys: angl. constant of integration vok. Integrationskonstante, f rus. константа интегрирования, f; постоянная интегрирования, f pranc. constante d’intégration, f …   Fizikos terminų žodynas\n\n• constant of integration — Math. a constant that is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant. * * * …   Universalium\n\n• constant of integration — Math. a constant that is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant …   Useful english dictionary\n\n• constant of integration — noun An unspecified constant term added to a particular antiderivative to make it represent its whole family of antiderivatives …   Wiktionary\n\n• Constant — or The Constant may refer to: Contents 1 In Mathematics 2 Other concepts 3 People 4 Organization 5 …   Wikipedia\n\n• Constant — Con stant, n. 1. That which is not subject to change; that which is invariable. [1913 Webster] 2. (Math.) A quantity that does not change its value; used in countradistinction to {variable}. [1913 Webster] 3. (Astron.) A number whose value, when… …   The Collaborative International Dictionary of English\n\n• Constant of aberration — Constant Con stant, n. 1. That which is not subject to change; that which is invariable. [1913 Webster] 2. (Math.) A quantity that does not change its value; used in countradistinction to {variable}. [1913 Webster] 3. (Astron.) A number whose… …   The Collaborative International Dictionary of English\n\n• Integration by parts — Topics in Calculus Fundamental theorem Limits of functions Continuity Mean value theorem Differential calculus  Derivative Change of variables Implicit differentiation Taylor s theorem Related rates …   Wikipedia" ]

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[ "", null, "# Roll Call 54 on H.R. 3248\n\n## On Motion to Suspend the Rules and Pass\n\nTo designate the facility of the United States Postal Service located at 112 South 5th Street in Saint Charles, Missoui, as the “Lance Corporal Drew W. Weaver Post Office Building”\n\n(Committee on Oversight and Reform)\nCongress: 2/14/2012 – 112th Congress, 2nd Session\n\n## Result\n\nPassed\nYay Nay Pres NV\nGOP 233 0 0 8\nDEM 179 0 0 13\nIND 0 0 0 0\nTOTAL 412 0 0 21\n\nYay (412)\n\nAckerman (D, NY)\nAkin (R, MO)\nAlexander (R, LA)\nAltmire (D, PA)\nAmash (R, MI)\nAmodei (R, NV)\nAndrews (D, NJ)\nAustria (R, OH)\nBaca (D, CA)\nBachmann (R, MN)\nBachus (R, AL)\nBaldwin (D, WI)\nBarletta (R, PA)\nBarrow (D, GA)\nBartlett (R, MD)\nBarton (TX) (R, TX)\nBass (CA) (D, CA)\nBass (NH) (R, NH)\nBenishek (R, MI)\nBerg (R, ND)\nBerkley (D, NV)\nBerman (D, CA)\nBiggert (R, IL)\nBilbray (R, CA)\nBilirakis (R, FL)\nBishop (GA) (D, GA)\nBishop (NY) (D, NY)\nBishop (UT) (R, UT)\nBlack (R, TN)\nBlackburn (R, TN)\nBlumenauer (D, OR)\nBonamici (D, OR)\nBonner (R, AL)\nBono Mack (R, CA)\nBoren (D, OK)\nBoswell (D, IA)\nBoustany (R, LA)\nBrady (PA) (D, PA)\nBrady (TX) (R, TX)\nBraley (IA) (D, IA)\nBrooks (R, AL)\nBroun (GA) (R, GA)\nBrown (FL) (D, FL)\nBuchanan (R, FL)\nBucshon (R, IN)\nBuerkle (R, NY)\nBurgess (R, TX)\nBurton (IN) (R, IN)\nButterfield (D, NC)\nCalvert (R, CA)\nCamp (R, MI)\nCanseco (R, TX)\nCantor (R, VA)\nCapito (R, WV)\nCapps (D, CA)\nCapuano (D, MA)\nCardoza (D, CA)\nCarnahan (D, MO)\nCarney (D, DE)\nCarson (IN) (D, IN)\nCarter (R, TX)\nCassidy (R, LA)\nCastor (FL) (D, FL)\nChabot (R, OH)\nChaffetz (R, UT)\nChandler (D, KY)\nChu (D, CA)\nCicilline (D, RI)\nClarke (MI) (D, MI)\nClarke (NY) (D, NY)\nClay (D, MO)\nCleaver (D, MO)\nClyburn (D, SC)\nCoble (R, NC)\nCoffman (CO) (R, CO)\nCole (R, OK)\nConaway (R, TX)\nConnolly (VA) (D, VA)\nConyers (D, MI)\nCooper (D, TN)\nCosta (D, CA)\nCostello (D, IL)\nCourtney (D, CT)\nCravaack (R, MN)\nCrawford (R, AR)\nCrenshaw (R, FL)\nCritz (D, PA)\nCrowley (D, NY)\nCuellar (D, TX)\nCulberson (R, TX)\nCummings (D, MD)\nDavis (IL) (D, IL)\nDavis (KY) (R, KY)\nDeFazio (D, OR)\nDeGette (D, CO)\nDeLauro (D, CT)\nDenham (R, CA)\nDent (R, PA)\nDesJarlais (R, TN)\nDeutch (D, FL)\nDiaz-Balart (R, FL)\nDicks (D, WA)\nDingell (D, MI)\nDold (R, IL)\nDonnelly (IN) (D, IN)\nDoyle (D, PA)\nDreier (R, CA)\nDuffy (R, WI)\nDuncan (SC) (R, SC)\nDuncan (TN) (R, TN)\nEdwards (D, MD)\nEllison (D, MN)\nEllmers (R, NC)\nEmerson (R, MO)\nEngel (D, NY)\nEshoo (D, CA)\nFarenthold (R, TX)\nFarr (D, CA)\nFattah (D, PA)\nFilner (D, CA)\nFincher (R, TN)\nFitzpatrick (R, PA)\nFlake (R, AZ)\nFleischmann (R, TN)\nFleming (R, LA)\nFlores (R, TX)\nForbes (R, VA)\nFortenberry (R, NE)\nFoxx (R, NC)\nFrank (MA) (D, MA)\nFranks (AZ) (R, AZ)\nFrelinghuysen (R, NJ)\nFudge (D, OH)\nGallegly (R, CA)\nGaramendi (D, CA)\nGardner (R, CO)\nGarrett (R, NJ)\nGerlach (R, PA)\nGibbs (R, OH)\nGibson (R, NY)\nGingrey (GA) (R, GA)\nGohmert (R, TX)\nGonzalez (D, TX)\nGoodlatte (R, VA)\nGosar (R, AZ)\nGowdy (R, SC)\nGranger (R, TX)\nGraves (GA) (R, GA)\nGraves (MO) (R, MO)\nGreen, Al (D, TX)\nGriffin (AR) (R, AR)\nGriffith (VA) (R, VA)\nGrijalva (D, AZ)\nGrimm (R, NY)\nGuthrie (R, KY)\nGutierrez (D, IL)\nHahn (D, CA)\nHall (R, TX)\nHanabusa (D, HI)\nHanna (R, NY)\nHarper (R, MS)\nHarris (R, MD)\nHartzler (R, MO)\nHastings (FL) (D, FL)\nHastings (WA) (R, WA)\nHayworth (R, NY)\nHeck (R, NV)\nHeinrich (D, NM)\nHensarling (R, TX)\nHerrera Beutler (R, WA)\nHiggins (D, NY)\nHimes (D, CT)\nHinchey (D, NY)\nHinojosa (D, TX)\nHochul (D, NY)\nHolden (D, PA)\nHolt (D, NJ)\nHonda (D, CA)\nHoyer (D, MD)\nHuelskamp (R, KS)\nHuizenga (MI) (R, MI)\nHultgren (R, IL)\nHunter (R, CA)\nHurt (R, VA)\nInslee (D, WA)\nIsrael (D, NY)\nIssa (R, CA)\nJackson (IL) (D, IL)\nJackson Lee (TX) (D, TX)\nJenkins (R, KS)\nJohnson (GA) (D, GA)\nJohnson (OH) (R, OH)\nJohnson, E. B. (D, TX)\nJohnson, Sam (R, TX)\nJones (R, NC)\nJordan (R, OH)\nKaptur (D, OH)\nKeating (D, MA)\nKelly (R, PA)\nKildee (D, MI)\nKind (D, WI)\nKing (IA) (R, IA)\nKing (NY) (R, NY)\nKingston (R, GA)\nKinzinger (IL) (R, IL)\nKissell (D, NC)\nKline (R, MN)\nKucinich (D, OH)\nLamborn (R, CO)\nLance (R, NJ)\nLandry (R, LA)\nLangevin (D, RI)\nLankford (R, OK)\nLarsen (WA) (D, WA)\nLarson (CT) (D, CT)\nLatham (R, IA)\nLaTourette (R, OH)\nLatta (R, OH)\nLee (CA) (D, CA)\nLevin (D, MI)\nLewis (CA) (R, CA)\nLewis (GA) (D, GA)\nLipinski (D, IL)\nLoBiondo (R, NJ)\nLoebsack (D, IA)\nLofgren, Zoe (D, CA)\nLong (R, MO)\nLowey (D, NY)\nLucas (R, OK)\nLuetkemeyer (R, MO)\nLuján (D, NM)\nLummis (R, WY)\nLungren, Daniel E. (R, CA)\nLynch (D, MA)\nMack (R, FL)\nMaloney (D, NY)\nManzullo (R, IL)\nMarchant (R, TX)\nMarino (R, PA)\nMarkey (D, MA)\nMatheson (D, UT)\nMatsui (D, CA)\nMcCarthy (CA) (R, CA)\nMcCarthy (NY) (D, NY)\nMcCaul (R, TX)\nMcClintock (R, CA)\nMcCollum (D, MN)\nMcCotter (R, MI)\nMcDermott (D, WA)\nMcGovern (D, MA)\nMcHenry (R, NC)\nMcIntyre (D, NC)\nMcKeon (R, CA)\nMcKinley (R, WV)\nMcMorris Rodgers (R, WA)\nMcNerney (D, CA)\nMeehan (R, PA)\nMeeks (D, NY)\nMica (R, FL)\nMichaud (D, ME)\nMiller (FL) (R, FL)\nMiller (MI) (R, MI)\nMiller (NC) (D, NC)\nMiller, Gary (R, CA)\nMiller, George (D, CA)\nMoran (D, VA)\nMurphy (CT) (D, CT)\nMurphy (PA) (R, PA)\nMyrick (R, NC)\nNapolitano (D, CA)\nNeugebauer (R, TX)\nNoem (R, SD)\nNugent (R, FL)\nNunes (R, CA)\nNunnelee (R, MS)\nOlson (R, TX)\nOlver (D, MA)\nOwens (D, NY)\nPalazzo (R, MS)\nPallone (D, NJ)\nPascrell (D, NJ)\nPastor (AZ) (D, AZ)\nPaulsen (R, MN)\nPearce (R, NM)\nPelosi (D, CA)\nPence (R, IN)\nPerlmutter (D, CO)\nPeters (D, MI)\nPeterson (D, MN)\nPetri (R, WI)\nPingree (ME) (D, ME)\nPitts (R, PA)\nPlatts (R, PA)\nPoe (TX) (R, TX)\nPolis (D, CO)\nPompeo (R, KS)\nPosey (R, FL)\nPrice (GA) (R, GA)\nPrice (NC) (D, NC)\nQuayle (R, AZ)\nQuigley (D, IL)\nRahall (D, WV)\nReed (R, NY)\nRehberg (R, MT)\nReichert (R, WA)\nRenacci (R, OH)\nReyes (D, TX)\nRibble (R, WI)\nRichardson (D, CA)\nRichmond (D, LA)\nRigell (R, VA)\nRivera (R, FL)\nRoby (R, AL)\nRoe (TN) (R, TN)\nRogers (AL) (R, AL)\nRogers (KY) (R, KY)\nRogers (MI) (R, MI)\nRohrabacher (R, CA)\nRokita (R, IN)\nRooney (R, FL)\nRos-Lehtinen (R, FL)\nRoskam (R, IL)\nRoss (AR) (D, AR)\nRoss (FL) (R, FL)\nRothman (NJ) (D, NJ)\nRoybal-Allard (D, CA)\nRoyce (R, CA)\nRunyan (R, NJ)\nRush (D, IL)\nRyan (OH) (D, OH)\nRyan (WI) (R, WI)\nSánchez, Linda T. (D, CA)\nSanchez, Loretta (D, CA)\nSarbanes (D, MD)\nScalise (R, LA)\nSchakowsky (D, IL)\nSchiff (D, CA)\nSchilling (R, IL)\nSchmidt (R, OH)\nSchwartz (D, PA)\nSchweikert (R, AZ)\nScott (SC) (R, SC)\nScott (VA) (D, VA)\nScott, Austin (R, GA)\nScott, David (D, GA)\nSensenbrenner (R, WI)\nSessions (R, TX)\nSewell (D, AL)\nSherman (D, CA)\nShimkus (R, IL)\nShuler (D, NC)\nShuster (R, PA)\nSimpson (R, ID)\nSires (D, NJ)\nSlaughter (D, NY)\nSmith (NE) (R, NE)\nSmith (NJ) (R, NJ)\nSmith (TX) (R, TX)\nSmith (WA) (D, WA)\nSoutherland (R, FL)\nSpeier (D, CA)\nStark (D, CA)\nStearns (R, FL)\nStivers (R, OH)\nStutzman (R, IN)\nSullivan (R, OK)\nSutton (D, OH)\nTerry (R, NE)\nThompson (CA) (D, CA)\nThompson (MS) (D, MS)\nThompson (PA) (R, PA)\nThornberry (R, TX)\nTiberi (R, OH)\nTierney (D, MA)\nTipton (R, CO)\nTonko (D, NY)\nTowns (D, NY)\nTsongas (D, MA)\nTurner (NY) (R, NY)\nTurner (OH) (R, OH)\nUpton (R, MI)\nVan Hollen (D, MD)\nVelázquez (D, NY)\nVisclosky (D, IN)\nWalberg (R, MI)\nWalden (R, OR)\nWalsh (IL) (R, IL)\nWalz (MN) (D, MN)\nWasserman Schultz (D, FL)\nWaters (D, CA)\nWaxman (D, CA)\nWebster (R, FL)\nWelch (D, VT)\nWest (R, FL)\nWestmoreland (R, GA)\nWhitfield (R, KY)\nWilson (FL) (D, FL)\nWilson (SC) (R, SC)\nWittman (R, VA)\nWolf (R, VA)\nWomack (R, AR)\nWoodall (R, GA)\nWoolsey (D, CA)\nYarmuth (D, KY)\nYoder (R, KS)\nYoung (AK) (R, AK)\nYoung (IN) (R, IN)\n\nNay (0)\n\nPresent (0)\n\nNot Voting (21)\n\nBecerra (D, CA)\nCampbell (R, CA)\nCohen (D, TN)\nDavis (CA) (D, CA)\nDoggett (D, TX)\nGreen, Gene (D, TX)\nGuinta (R, NH)\nHerger (R, CA)\nHirono (D, HI)\nJohnson (IL) (R, IL)\nMoore (D, WI)\nMulvaney (R, SC)\nNeal (D, MA)\nPaul (R, TX)\nPayne (D, NJ)\nRangel (D, NY)\nRuppersberger (D, MD)\nSchock (R, IL)\nSerrano (D, NY)\nWatt (D, NC)\nYoung (FL) (R, FL)" ]

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[ "# Charged Particle Traps: Physics and Techniques of Charged by Fouad G. Major, Viorica N. Gheorghe, Günther Werth", null, "By Fouad G. Major, Viorica N. Gheorghe, Günther Werth\n\nThis publication offers an creation and advisor to fashionable advances in charged particle (and antiparticle) confinement by way of electromagnetic fields. Confinement in several capture geometries, the impression of catch imperfections, classical and quantum mechanical description of the trapped particle movement, various equipment of ion cooling to low temperatures, and non-neutral plasma homes (including Coulomb crystals) are the most topics. They shape the root of such functions of charged particle traps as high-resolution optical and microwave spectroscopy, mass spectrometry, atomic clocks, and, possibly, quantum computing.\n\nRead Online or Download Charged Particle Traps: Physics and Techniques of Charged Particle Field Confinement PDF\n\nBest atomic & nuclear physics books\n\nAnomalous X-Ray Scattering for Materials Characterization (Springer Tracts in Modern Physics)\n\nThe evolution of our realizing of so much houses of latest useful fabrics is said to our wisdom in their atomic-scale constitution. To additional this, numerous X-ray and neutron strategies are hired. The anomalous X-ray scattering (AXS) technique, exploiting the so-called anomalous dispersion influence close to the absorption fringe of the constituent point, is among the strongest tools for picking the actual partial constitution services of person pairs of parts or the environmental services round particular parts in multicomponent platforms.\n\nPrinciples of Laser Spectroscopy and Quantum Optics\n\nIdeas of Laser Spectroscopy and Quantum Optics is an important textbook for graduate scholars learning the interplay of optical fields with atoms. It additionally serves as an awesome reference textual content for researchers operating within the fields of laser spectroscopy and quantum optics. The e-book presents a rigorous creation to the prototypical difficulties of radiation fields interacting with - and three-level atomic structures.\n\nTheoretical Nuclear Physics\n\nThe final two decades have witnessed a tremendous improvement of nuclear physics. a number of facts have amassed and plenty of experimental evidence are identified. because the experimental options have accomplished higher and bigger perfection, the theoretical research and interpretation of those facts became correspondingly extra exact and specific.\n\nExtra resources for Charged Particle Traps: Physics and Techniques of Charged Particle Field Confinement\n\nSample text\n\n18) into the radial variable r = x2 + y 2 , we obtain 2 2 + R− + 2R+ R− cos(ω1 t + ϕ+ − ϕ− )]1/2 . 19) is bounded by the cylinder |R+ − R− | ≤ r ≤ R+ + R− , |z| < Rz . 18) is a figure which can be described as an epitrochoid (see also Appendix B). The x − y projection of the motion is periodic if ω+ /ω− is rational; otherwise if ω+ /ω− is irrational, the radial motion is quasiperiodic. Some trajectories with the initial conditions x(0) = x0 = 0, y(0) = 0, and vx (0) = 0 are illustrated in Figs. 3.\n\n54 3 The Penning Trap Fig. 2. 5R+ . Periodic orbits for (a) ω+ /ω √− = 2; (b) ω+ /ω− = 8; (c) ω+ /ω− = 9/2; (d) quasiperiodic orbit for ω+ /ω− = 2 17 Fig. 3. 2R+ . (a) Periodic orbit for ω+ /ωz = 6; (b) quasiperiodic orbit for ω+ /ωz = 35 The energy content of the different motions can be derived using the Hamiltonian formalism. We choose the symmetric gauge with the vector potential A = B0 (−y/2, x/2, 0). 1 Theory of the Ideal Penning Trap 55 where p = M v is the momentum vector and the axial component of the canonical angular momentum is Lz = xpy − ypx .\n\n32) ωs = 2QV0 /M Ω(r02 + 2z02 ) , and Φsc (r) is the space charge potential that is itself determined by the density through the Poisson equation. 33) where R is the value of r at which the cloud density is reduced to 1/e of its center value, and can be roughly considered as the cloud radius. The ion oscillation frequency, shifted by space charge, now depends on the distance from the trap center. The numerical results agree with those previously 32 2 The Paul Trap Fig. 15. 2 . Copyright (2003) by the American Physical Society obtained by a statistical model using similar density distributions and graphically represented in Fig." ]

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[ "This page is a companion for the paper \"Optimal sequential contests\" by Toomas Hinnosaar. For the notation and the computation algorithm, please refer to the paper. Download the Matlab codes here.\n\nThere are two relevant inputs:\n g(X) = = -h(X)/h'(X) characterizes the payoff function. Click here to specify h instead. n = (n1,...,nT) = (), where n1 is the number of players before the first announcement, n2 the second, and so on.\nNote: Condition 1 is satisfied => the characterization result holds.\n\nvalue period description\nX* 0.7968   Total effort", null, "x1* 0.3603 1 Effort by player 1", null, "x2* 0.1091 2 Effort by player 2", null, "x3* 0.1091 2 Effort by player 3", null, "x4* 0.1091 2 Effort by player 4", null, "x5* 0.1091 2 Effort by player 5", null, "g(X) = (2^(-X)-0.5)/(log(2))", null, "g,\n0, yscale=[-0.001,0.7223]\nh(X) = exp[-∫0X1/g(t) dt]", null, "h,\n0, yscale=[-0.001,0.9872]", null, "f2,\n0, yscale=[-0.001,1.001]", null, "f1,\n0, yscale=[-2.8864,1.001]", null, "f0,\n0, yscale=[-6.4931,1.001]\ng1(X)=-2^(-X)", null, "g1,\n0, yscale=[-0.001,0.7223]\ng2(X)=0.5 2^(-2 X) (2^X - 4.)", null, "g2,\n0, yscale=[-0.001,0.7223]", null, "X h(X),\n0, yscale=[-0.001,0.1828]" ]

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[ "Opposite $$\\angle$$'s of a parallelogram are equal: $$\\hat{X} = \\hat{V}$$ and $$\\hat{W} = \\hat{U}$$. Q\\hat{T}R = T\\hat{R}S & \\text{alt } \\angle \\text{s } QT \\parallel RS \\\\ Both pairs of opposite angles of $$MNOP$$ are equal. $$\\therefore \\hat{x} = 180^{\\circ} - 36^{\\circ} - 102^{\\circ} = 42^{\\circ}$$. Polygons. This video shows how to prove that the the diagonals of a rhombus are perpendicular. Prove that the quadrilateral $$MNOP$$ is a parallelogram. Quadrilateral $$XWVU$$ with sides $$XW \\parallel UV$$ and $$XU \\parallel WV$$ is given. The adjective “Euclidean” is supposed to conjure up an attitude or outlook rather than anything more specific: the course is not a course on the Elements but a wide-ranging and (we hope) interesting introduction to a selection of topics in synthetic plane geometry… Euclidean geometry deals with space and shape using a system of logical deductions. Quadrilateral $$XWST$$ is a parallelogram and $$TV$$ and $$XW$$ have lengths $$b$$ and $$2b$$, respectively, as shown. His ideas seemed so logical and obvious, yet I had not been using them! You need to prove that $$NPTS$$ is a parallelogram. A ratio describes the relationship between two quantities … PNQ is a tangent to the circle at N. Calculate, giving reasons, the size of: L̂1 Ô M̂ 2 N̂2 N̂1 51 17 3 Q P 2 1 2 2 2 1 1 1 1 N O M K L JENN: LEARNER MANUAL EUCLIDEAN GEOMETRY GRADE … $$PQ=TQ$$. Study content slides on the topic (1 – 2 hours in total). \\text{Steps} & \\text{Reasons} \\\\ 1 tangent s e c a n t d i a m e t e r c h or d arc r a d i u s sector.. seg ment CHAPTER 8 EUCLIDEAN GEOMETRY … $$AD \\parallel BC (AE \\parallel CF, ~ AECF$$ is a parallelogram), $$CF = AE$$ ($$AECF$$ is a parallelogram), $$ABCD$$ is a parallelogram (two sides are parallel and equal). 10.1.2 10.1.1 10.1 QUESTION 10 2 1 In the diagram below, O is the centre of circle KLNM. Geometry can be split into Euclidean geometry and analytical geometry. Terminology. This lesson also traces the history of geometry… Provide learner with additional knowledge and understanding of the topic, Enable learner to gain confidence to study for and write tests and exams on the topic, Provide additional materials for daily work and use on the topic. Chapter 11: Euclidean geometry. First show $$\\triangle ADW\\equiv \\triangle CBY$$. Corollary 1. QR = TS \\text{ and } RS = QT & \\text{congruent triangles (AAS)} \\\\ In parallelogram $$ABCD$$, the bisectors of the angles ($$AW$$, $$BX$$, $$CY$$ and $$DZ$$) have been constructed. Corollary 2. Euclidean Geometry for Grade 12 Maths – Free Example. Quadrilateral $$QRST$$ with sides $$QR \\parallel TS$$ and $$QT \\parallel RS$$ is given. 2. \\hat{P} &= \\hat{T_1} \\quad \\text{(}\\angle \\text{s opp equal sides)} \\\\ Is this correct? 12.7 Topic Euclidean Geometry … $$\\angle$$'s in a $$\\triangle = 180 ^{\\circ}$$, $$\\therefore \\hat{X} + X\\hat{W}U + X\\hat{U}W = 180 ^{\\circ}$$. Then show $$\\triangle PDW\\equiv \\triangle NBY$$. Download the Show Notes: http://www.mindset.co.za/learn/sites/files/LXL2013/LXL_Gr10Mathematics_26_Euclidean%20Geometry_26Aug.pdf In this live Grade 10 … 1.3. Grade: 12. You are also given $$AD = CB$$, $$DB = AC$$, $$AD \\parallel CB$$, $$DB \\parallel AC$$, $$\\hat{A} = \\hat{B}$$ and $$\\hat{D} = \\hat{C}$$. To do 19 min read. \\therefore XW = UV and XU = WV & \\text{congruent triangles (AAS)} \\\\ by this license. \\text{Steps} & \\text{Reasons} \\\\ Maths and Science Lessons > Courses > Grade 10 – Euclidean Geometry. Grade 10 – Euclidean Geometry. Complete the interactive assignment (30 min in total). Now we know that $$\\hat{X} = \\hat{V} = 36^{\\circ}$$ and that $$X\\hat{U}W = 42^{\\circ}$$. This lesson introduces the concept of Euclidean geometry and how it is used in the real world today. Prove that $$QRST$$ is a parallelogram. Aims and outcomes of tutorial: Improve marks and help you achieve 70% or more! X\\hat{U}W = U\\hat{W}V & \\text{alt } \\angle \\text{s; } XU \\parallel WV \\\\ \\text{In} \\triangle XWU \\text{ and } \\triangle WVU \\text{ side } WU = WU &\\text{common side} \\\\ Euclidean geometry also allows the method of superposition, in which a figure is transferred to another point in space. Chapter 11: Euclidean geometry. AD &= BC \\text{ (opp sides of } \\parallel \\text{m)}\\\\ EUCLIDEAN GEOMETRY: (±50 marks) EUCLIDEAN GEOMETRY: (±50 marks) Grade … \\hat{Q} = \\hat{S} & \\text{congruent triangles (AAS)} \\\\ \\end{array}\\]. \\begin{align*} EUCLIDEAN GEOMETRY TEXTBOOK GRADE 11 (Chapter 8) Presented by: Jurg Basson MIND ACTION SERIES Attending this Workshop = 10 SACE Points. A perpendicular bisector is a perpendicular line that passes through the midpoint Euclidean Geometry (Revision of Gr 11 Circle Geometry). Euclidean Geometry.The golden ratio | Introduction to Euclidean geometry | Geometry | Khan Academy.Drawing line segments example | Introduction to Euclidean geometry | Geometry | Khan Academy.Geometry - Proofs for Triangles.Quadrilateral overview | Perimeter, area, and volume | Geometry | Khan Academy.Euclid as the father of geometry | Introduction to Euclidean geometry | Geometry … \\hat{P} &= \\hat{R} ~(\\text{ opp} \\angle\\text{s of } \\parallel\\text{m)} \\\\ Calc presentation … 27 Jul. Parallelogram $$ABCD$$ and $$BEFC$$ are shown below. You are also given $$AB=CD$$, $$AD=BC$$, $$AB\\parallel CD$$, $$AD\\parallel BC$$, $$\\hat{A}=\\hat{C}$$, $$\\hat{B}=\\hat{D}$$. In $$\\triangle CDZ$$ and $$\\triangle ABX$$, In $$\\triangle XAM$$ and $$\\triangle ZCO$$. We use this information to present the correct curriculum and Mathematics » Euclidean Geometry » Circle Geometry. If all the sides of a polygon of n sides are … A theorem is a hypothesis (proposition) that can be shown to be true by accepted mathematical operations and … Earn a badge for having successfully completed the tutorial and assignment. … Two triangles in the figure are congruent: $$\\triangle QRS \\equiv \\triangle QPT$$. One of the authors of the Mind Action Series mathematics textbooks had a workshop that I attended. GRADE 10_CAPS Curriculum 10.7 Euclidean Geometry10.7 Euclidean Geometry ---- Angles Angles Angles 1.1 Complete the following geometric facts.1.1 Complete the following geometric facts. In parallelogram $$ADBC$$, the bisectors of the angles $$(A, D, B, C)$$ have been constructed, indicated with the red lines below. Additionally, $$SN = SR$$. We will now apply what we have learnt about geometry and the properties of polygons (in particular triangles and quadrilaterals) to prove some of these properties. / 07. Prove that $$XWVU$$ is a parallelogram. Prove that $$MNOP$$ is a parallelogram. We will also look at how we can prove a particular quadrilateral is one of the special quadrilaterals. We can solve this problem in two ways: using the sum of angles in a triangle or using the sum of interior angles in a quadrilateral. Let us help you to study smarter to achieve your goals. The sum of the interior $$\\angle$$'s in a quadrilateral is $$360^{\\circ}$$. \\hat{T_1} &= \\hat{Q_1}\\quad \\text{ (alt } \\angle \\text{s; } (PS \\parallel QR)\\text{)} \\\\ \\hline a) All parallelograms are … V\\hat{U}W = X\\hat{W}U & \\text{alt } \\angle \\text{s; } XW \\parallel UV \\\\ Option 2: sum of interior angles in a quadrilateral. For example to prove a quadrilateral is a parallelogram it is not enough to show that both pairs of sides are parallel, learners will also need to show that either the opposite angles are equal or both pairs of opposite sides are equal in length. 10 | Page The following investigation is about the perpendicular bisector of a chord. Grade 11 Euclidean Geometry 2014 10 OR Theorem 1 The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord. Siyavula Practice guides you at your own pace when you do questions online. The following terms are regularly used when referring to circles: Arc — a portion of the circumference of a circle. It is basically introduced for flat surfaces. Fill in the missing reasons and steps to prove that the quadrilateral $$ABCD$$ is a parallelogram. \\text{In} \\triangle QRT \\text{ and } \\triangle RST \\text{ side } RT = RT &\\text{common side} \\\\ Grade 11 Euclidean Geometry 2014 11 . This video shows how to prove that the opposite angles of a parallelogram are equal. Once you have learned the basic postulates and the properties of all the shapes and lines, you can begin to use this information to solve geometry … All Siyavula textbook content made available on this site is released under the terms of a The sum of the interior $$\\angle$$'s in a quadrilateral is $$360 ^{\\circ}$$. $$PQRS$$ is a parallelogram. Mathematics Grade 12; Euclidean geometry; Ratio and proportion; Previous. We know that $$\\hat{Q} = \\hat{S} = 34^{\\circ}$$ and that $$R\\hat{T}S = 41^{\\circ}$$. Posted on July 27, 2015 January 19, 2018 by Maths @ SHARP. BC &= EF \\text{ (opp sides of } \\parallel \\text{m)}\\\\ Everything Maths, Grade 10. On this page you can read or download euclidean geometry grade 10 pdf in PDF format. Therefore $$MNOP$$ is a parallelogram. There is a lot of work that must be done in the beginning to learn the language of geometry. \\therefore XWVU \\text{is a parallelogram } & \\text{opp sides of quad are } = This chapter focuses on solving problems in Euclidean geometry and proving riders. You are also given that: $$\\hat{Q} = y$$ and $$\\hat{S} = 34^{\\circ}$$; $$Q\\hat{T}R = x$$ and $$R\\hat{T}S = 41^{\\circ}$$. You need to prove that $$\\triangle TVU \\equiv \\triangle SVW$$. \\therefore \\hat{Q_1} &= \\hat{R} \\therefore \\triangle QRT \\equiv \\triangle STR &\\text{congruent (AAS)} \\\\ Section 11 1-notes_2 kerrynix. Redraw the diagram and mark all given and known information: Study the diagram below; it is not necessarily drawn to scale. Arc — a straight line joining the euclidean geometry grade 10 notes of an … Everything Maths, Grade 8 Maths, Grade –! Referring to circles: Arc — a portion of the interior \\ ( XWVU\\ ) with sides \\ ( {! Whether the following terms are regularly euclidean geometry grade 10 notes when referring to circles: Arc — a straight line the... … Everything Maths, Grades Euclidean geometry and proving riders PROBLEMS in Euclidean geometry and it! A lot of work that must be done in the missing reasons and steps to prove that \\ ( )... Of … Euclidean geometry … on this page you can read or notes! On the topic ( 1 – 2 hours in total ) tutorial and.. Circumference of a euclidean geometry grade 10 notes is less than two right angles EF\\ ) bisect other..., by completing a digital, interactive assignment ( 30 min in ). 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[ "## Swift Division Assignment Operator\n\nIn Swift, Division Assignment Operator is used to compute the division of given value in left operand by the value in right operand, and assign the quotient in this division operation to the variable given as left operand.\n\nIn this tutorial, we will learn about Division Assignment Operator in Swift language, its syntax, and how to use this operator in programs, with examples.\n\n### Syntax\n\nThe syntax of Division Assignment Operator is\n\n`x /= value`\n\nwhere\n\nThe expression `x += value` computes x + value and assigns this computed value to variable x.\n\nTherefore the expression `x += value` is same as `x = x + value`.\n\n## Examples\n\n### 1. Division Assignment of integer value to x\n\nIn the following program, we shall division assign an integer value of 25 to variable x using Division Assignment Operator.\n\nmain.swift\n\n```var x = 110\nx /= 25\nprint(\"x is \\(x)\")```\n\nOutput\n\n`x is 4`\n\nExplanation\n\n```x /= 25\n\nx = x / 25\n= 110 / 25\n= 4 (integer quotient)```\n\n### 2. Division Assign with float values\n\nIn the following program, we shall do division assignment with floating point values.\n\nmain.swift\n\n```var x: Float = 3.14\nvar y: Float = 1.2\nx /= y\nprint(\"x is \\(x)\")```\n\nOutput\n\n`x is 2.6166666`\n\nExplanation\n\n```x /= y\n\nx = x / y\n= 3.14 / 1.2\n= 2.6166666```\n\n## Summary\n\nIn this Swift Operators tutorial, we learned about Division Assignment Operator, its syntax, and usage, with examples." ]

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[ "# Milliliters to Cubic Inches Converter\n\nEnter the volume in milliliters below to get the value converted to cubic inches.\n\nResults in Cubic Inches:", null, "1 mL = 0.061024 in³\n\nDo you want to convert cubic inches to milliliters?\n\n## How to Convert Milliliters to Cubic Inches\n\nTo convert a measurement in milliliters to a measurement in cubic inches, multiply the volume by the following conversion ratio: 0.061024 cubic inches/milliliter.\n\nSince one milliliter is equal to 0.061024 cubic inches, you can use this simple formula to convert:\n\ncubic inches = milliliters × 0.061024\n\nThe volume in cubic inches is equal to the volume in milliliters multiplied by 0.061024.\n\nFor example, here's how to convert 5 milliliters to cubic inches using the formula above.\ncubic inches = (5 mL × 0.061024) = 0.305119 in³\n\n## What is a Milliliter?\n\nThe milliliter is a unit of volume equal to 1 cubic centimeter, 1/1,000 of a liter, or about 0.061 cubic inches.\n\nThe milliliter is an SI unit of volume in the metric system. In the metric system, \"milli\" is the prefix for thousandths, or 10-3. A milliliter is sometimes also referred to as a millilitre. Milliliters can be abbreviated as mL, and are also sometimes abbreviated as ml or mℓ. For example, 1 milliliter can be written as 1 mL, 1 ml, or 1 mℓ.\n\nSince one milliliter is equivalent to one cubic centimeter, milliliters are sometimes expressed using the abbreviation for a cubic centimeter (cc) for things such as medical dosages or engine displacements.\n\n## What is a Cubic Inch?\n\nA cubic inch is a unit of volume equal to the space consumed by a cube with sides that are one inch in all directions. One cubic inch is equivalent to about 16.387 cubic centimeters or 0.554 fluid ounces.\n\nThe cubic inch is a US customary and imperial unit of volume. A cubic inch is sometimes also referred to as a cubic in. Cubic inches can be abbreviated as in³, and are also sometimes abbreviated as cu inch, cu in, or CI. For example, 1 cubic inch can be written as 1 in³, 1 cu inch, 1 cu in, or 1 CI.\n\n## Milliliter to Cubic Inch Conversion Table\n\nTable showing various milliliter measurements converted to cubic inches.\nMilliliters Cubic Inches\n1 mL 0.061024 in³\n2 mL 0.122047 in³\n3 mL 0.183071 in³\n4 mL 0.244095 in³\n5 mL 0.305119 in³\n6 mL 0.366142 in³\n7 mL 0.427166 in³\n8 mL 0.48819 in³\n9 mL 0.549214 in³\n10 mL 0.610237 in³\n11 mL 0.671261 in³\n12 mL 0.732285 in³\n13 mL 0.793309 in³\n14 mL 0.854332 in³\n15 mL 0.915356 in³\n16 mL 0.97638 in³\n17 mL 1.0374 in³\n18 mL 1.0984 in³\n19 mL 1.1595 in³\n20 mL 1.2205 in³\n21 mL 1.2815 in³\n22 mL 1.3425 in³\n23 mL 1.4035 in³\n24 mL 1.4646 in³\n25 mL 1.5256 in³\n26 mL 1.5866 in³\n27 mL 1.6476 in³\n28 mL 1.7087 in³\n29 mL 1.7697 in³\n30 mL 1.8307 in³\n31 mL 1.8917 in³\n32 mL 1.9528 in³\n33 mL 2.0138 in³\n34 mL 2.0748 in³\n35 mL 2.1358 in³\n36 mL 2.1969 in³\n37 mL 2.2579 in³\n38 mL 2.3189 in³\n39 mL 2.3799 in³\n40 mL 2.441 in³\n\n## References\n\n1. National Institute of Standards and Technology, Specifications, Tolerances, and Other Technical Requirements for Weighing and Measuring Devices, Handbook 44 - 2019 Edition, https://nvlpubs.nist.gov/nistpubs/hb/2019/NIST.HB.44-2019.pdf\n2. National Institute of Standards and Technology, Specifications, Tolerances, and Other Technical Requirements for Weighing and Measuring Devices, Handbook 44 - 2019 Edition, https://nvlpubs.nist.gov/nistpubs/hb/2019/NIST.HB.44-2019.pdf" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Don't want to calculate scalar product\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg19743] Re: Don't want to calculate scalar product\n• From: Martin Kraus <Martin.Kraus at informatik.uni-stuttgart.de>\n• Date: Wed, 15 Sep 1999 03:52:59 -0400\n• Organization: Institut fuer Informatik, Universitaet Stuttgart\n• References: <7ree45\\[email protected]>\n• Sender: owner-wri-mathgroup at wolfram.com\n\nToshiyuki (Toshi) Meshii wrote:\n>\n> Hi,\n>\n> I understand that one of the unique point of Mathematica is that you don't\n> have to be conscious about whether a vector is a \"row\" or a \"column\" vector.\n\nActually I always think of Mathematica lists as row vectors.\n(Though some functions treat them as if they were column vectors. :)\n\n> However, I think there is a case in which you have to specify whether you\n> want Mathematica to recognize a vector as a row vector or a column vector.\n> Let me explain my case.\n>\n> a={1,2,3}; b={4,5,6}\n>\n> I want to define \"a\" as a column vector (3x1 matrix) and \"b\" as a row vector\n> (1x3 matrix).\n\nWell, just do it:\n\na={{1},{2},{3}}; b={{4,5,6}}; a.b\n\nreturns\n\n{{4, 5, 6}, {8, 10, 12}, {12, 15, 18}}\n\n> I expect the product\n> a.b\n> as a 3x3 matrix, but the answer Mathematica gives is a scalar product \"32\".\n\nproduct.\nThe inner product is what \".\" calculates. For the outer product\nuse Outer[Times, a, b] with the vectors a and b:\n\na = {1, 2, 3}; b = {4, 5, 6}; Outer[Times, a, b]\n\n{{4, 5, 6}, {8, 10, 12}, {12, 15, 18}}\n\n> Please give me an advice to get the 3x3 matrix as a product.\n>\n> -Toshi\n\nHope that helps\n\nMartin Kraus\n\n• Prev by Date: Subject: Re: RE: Finding a relative prime\n• Next by Date: Re: RealTime3D in v4.0: Capabilities and compatibilities\n• Previous by thread: Re: Don't want to calculate scalar product\n• Next by thread: Re: Finding a relative prime (corrected)" ]

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[ "# Extensions of dynamic programming for multi-stage combinatorial optimization\n\nMichal Mankowski*, Mikhail Moshkov\n\n*Corresponding author for this work\n\n7 Citations (Scopus)\n\n## Abstract\n\nWe propose a dynamic programming framework for an exact multi-stage (lexicographic) combinatorial optimization. Unlike conventional algorithms of dynamic programming that return one optimal solution, two dynamic programming algorithms proposed in this paper are coping with the whole set of optimal solutions or with its essential part. We describe the set of elements for optimization by a labeled directed acyclic graph, which in some sense, is similar to the structure of subproblems of the considered problem. For a given cost function (objective), the first algorithm constructs a subgraph of this graph that describes the whole set of optimal elements or its essential part. This algorithm can be used for multi-stage optimization of elements relative to a sequence of cost functions. The second algorithm counts elements before the optimization and after each optimization step. The considered labeled directed acyclic graph is a kind of circuit. This circuit builds the set of elements for optimization from one-element sets attached to input nodes. It uses the operation of union of sets attached to unifying nodes and functional operations attached to functional nodes. The algorithms for optimization and counting elements are defined for an arbitrary circuit without repetitions in which each element is generated exactly one time. For a considered problem with a known conventional dynamic programming solution algorithm, usually, it is easy to construct a corresponding circuit without repetitions. Once the circuit and cost functions are defined, our framework provides the correctness proofs and detailed time complexity analysis for the proposed algorithms. To make this approach more intuitive, we consider an illustrative example related to the maximum subarray problem. We tested our approach on the following nine combinatorial optimization problems: matrix chain multiplication, global sequence alignment, optimal paths in directed graphs, binary search trees, optimal bitonic tour, segmented least squares, convex polygon triangulation, one-dimensional clustering, and line breaking (text justification). We consider the last three problems in detail: construct a circuit without repetitions, describe at least two cost functions, evaluate a number of operations and time required by the algorithms, discuss an example, and consider the results of computer experiments with randomly generated instances of the problem.\n\nOriginal language English 106-132 27 Theoretical Computer Science 844 https://doi.org/10.1016/j.tcs.2020.08.009 Published - 6 Dec 2020\n\n## Fingerprint\n\nDive into the research topics of 'Extensions of dynamic programming for multi-stage combinatorial optimization'. Together they form a unique fingerprint." ]

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[ "# Power of Number in C\n\nPower of Number in C. The power of a number determines how many times it should be multiplied. The product of multiplying a number by itself is called power. What number is being multiplied is indicated by the base number. The exponent, which is a little number printed above and to the right of the base number, indicates the number of times the base number is multiplied.\n\n`#include <stdio.h>int main(){int base, exp;long double value = 1.0;printf(\"Enter a base number\");scanf(\"%d\", &base);printf(\"Enter an exponent\");scanf(\"%d\", &exp);while (exp != 0){value *= base;--exp;}printf(\"Answer = %.0Lf\", value);return 0;}`\n\nOutput:\n\nEnter a base number4\nEnter an exponent5\nSolution is 1024\n\n## Power of Number in C to Calculate Decimal Number\n\nA number base is the amount of digits or combinations of digits used to represent numbers in a counting system. Any whole number greater than 0 can be used as a basis. The decimal system, sometimes known as base 10, is the most widely used number system.\n\n## What is Exponent Number in C Program\n\nAn exponent is a number or letter that is written above and to the right of the base in a mathematical statement. Power of Number in C this program will produce the result for exponent value in c program.\n\n## Predecrement Operator in C Program\n\nBefore utilising a variable in an expression, a pre-decrement operator is used to decrement its value. The value of the variable is first decremented and then utilised in an expression with the pre-decrement operator.\n\n## What is Format Specifier in C\n\nDuring input and output, the format specifier is utilised. It’s a technique to notify the compiler what kind of data is in a variable when using scanf() or printf() to take input (). Percentages such as c, d, and f are some instances.\n\nLong integer values are represented using the percent ld format specifier. For printing the lengthy integer value contained in the variable, the printf() function is used." ]

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[ "# Tangent – Explanation & Examples\n\nIn the context of a right triangle, we can simply define the tangent function or any other trigonometric function using the terms hypotenuse, opposite, and adjacent in a right-angled triangle. Sounds interesting? Yes, it is. But, how can we define the tangent function using a right-angled triangle?\n\nThe tangent function is defined by determining the ratio of the length of the side opposite a reference angle (acute angle) of a right triangle to the length of the adjacent side of a right triangle.\n\nAfter studying this lesson, we are expected to learn the concepts driven by these questions and be qualified to address accurate, specific, and consistent answers to these questions.\n\n• What is a tangent function?\n• How can we determine the formula for tangent function from a right-angled triangle?\n• How can we solve actual problems using trigonometric functions?\n\nThis lesson aims to clear up any confusion you might have about the concepts involving the tangent function.\n\n## What is tangent?\n\nIn the context of a triangle, the tangent function is the ratio of the opposite side to the adjacent side. For an angle $\\alpha$, the tangent function is denoted by $\\tan \\alpha$. In other words, the tangent is a trigonometric function of any given angle.\n\nThe following figure 5-1 represents a typical right triangle. The lengths of the three legs (sides) of the right triangle are named $a$, $b$, and $c$. The angles opposite the legs of lengths $a$, $b$, and $c$ are named $\\alpha$, $\\beta$, and $\\gamma$. The tiny square with the angle $\\gamma$ shows that it is a right angle.\n\nUse the diagram in Figure 5-1 to determine the tangent function from the perspective of the angle $\\alpha$.\n\nLooking at Figure 5-1, we can determine the tangent function from the right-angled triangle if we divide the length of the side opposite the reference angle $\\alpha$ (acute angle) by the length of the adjacent side.\n\nThe following figure 5-2 represents a tangent function.\n\nLooking at Figure 5-2, we can identify that the side of length $a$ is the opposite side that lies exactly opposite the reference angle $\\alpha$, and the side of length $b$ is the adjacent side that lies right next to the reference angle $\\alpha$. Thus,\n\nOpposite = $a$\n\nAdjacent = $b$\n\nTherefore, the tangent of an angle $\\alpha$ is\n\n$\\tan \\alpha ={\\frac {a}{b}}$\n\nTherefore, we conclude that the tangent function is the ratio of the opposite side to the adjacent side.\n\nTangent function from the perspective of the angle $\\beta$\n\nWe should be cautious when we apply the terms opposite and adjacent because the meaning of these terms is dependent on the reference angle we are using.\n\nThe following figure 5-3 represents a typical right triangle from the perspective of the angle $\\beta$.\n\nYou can observe that now the roles of the sides have been shifted.\n\nLooking at Figure 5-3, it is clear now the length of the side $a$ is right next to the reference angle $\\beta$, and the length of the side $b$  lies exactly opposite the reference angle $\\beta$. Thus, in relation to the angle measuring $\\beta$, now we have\n\nAdjacent = $a$\n\nOpposite = $b$\n\nWhile the hypotenuse $c$ remains the same. This is why the hypotenuse is very special in a right-angled triangle.\n\nThe following figure 5-4 represents a tangent function from the perspective of the angle $\\beta$.\n\nLooking at Figure 5-4, we can identify that the side of length $b$ is the opposite side that lies exactly opposite the reference angle $\\beta$, and the length of the side $a$  lies right next to the reference angle $\\beta$. Thus,\n\nOpposite = $b$\n\nAdjacent = $a$\n\nWe know that the tangent function is the ratio of the opposite side to the adjacent side.\n\nTherefore, the tangent of an angle $\\beta$ is\n\n$\\tan \\beta ={\\frac {b}{a}}$\n\n## What is the formula for the tangent?\n\nThe following figure 5-5 illustrates a clear comparison of how we determined the ratios of tangent function from the perspective of both the angles$\\alpha$ and $\\beta$." ]

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[ "• Accueil\n• Math\n• 1. find the derivative of f(x) = 2x2 – 16x + 35​...\n\n# 1. find the derivative of f(x) = 2x2 – 16x + 35​\n\n• Réponse publiée par: nila93", null, "solution\n\nf(x) = 2x2 – 16x + 35", null, "", null, "35 = 0\n\ntherefore", null, "• Réponse publiée par: snow01\n\nIf you meant f(x) = 1/x - 1\n\nBut if you meant f(x) = 1/(x-1)\n\n• Réponse publiée par: candace08\n\nnot sure sa sagot ko, double check mo nalang", null, "• Réponse publiée par: homersoncanceranguiu\n\nThe domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.\n\nInterval Notation:\n\n(−∞,∞)\n\nSet-Builder Notation:\n\n{x|x∈R}y=\n\nis a straight line perpendicular to the y-axis at point\n\n(0,7),\n\nwhich means that the range is a set of one value\n\n{7}.\n\nSet-Builder Notation:\n\n{7}\n\nDetermine the domain and range.\n\nDomain:\n\n(−∞,∞),\n\n{x|x∈R}\n\nRange: {7}\n\n• Réponse publiée par: calmaaprilgrace\nANSWER:", null, "^_^\n• Réponse publiée par: sherelyn0013\nThe derivative is 6x-3  , multiply the exponent to its coefficient to get 6x , the derivative of 3x is 3\n• Réponse publiée par: Rosalesdhan\nThe derivative of the given function is 9x²-4x+1\nConnaissez-vous la bonne réponse?\n1. find the derivative of f(x) = 2x2 – 16x + 35​..." ]

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[ "# User Input and Screen Output – 2 MCQ’s\n\nThis set of MATLAB Multiple Choice Questions & Answers (MCQs) focuses on “User Input and Screen Output – 2″.\n\n1. What is the output of the following code?\n\n`sprintf(‘%E’,912)`\n\na) 9.120000E+02\nb) 9.120E+02\nc) 9.1200E+02\nd) 9.12000E+02\n\n2. What is the output of the following code?\n\n`sprintf(‘%d %d %d’,.1, .2.3)`\n\na) Error\nb) ‘1.000000e-01 2.000000e-01 3.000000e-01’\nc) 1.000000e-01 2.000000e-01 3.000000e-01\nd) []\n\n3. What is the output of the following code?\n\n`sprint(‘%i’,91293)`\n\na) 91293\nb) Syntactical Error\nc) The format specifier does not exist\nd) ‘91293’\n\n4. What is the output of the following code?\n\n`disp({12 })`\n\na) \nb) 12\nc) [12 ]\nd) Error\n\n5. What is the output of the following code?\n\n`sprintf('%04.2f',2.9121)`\n\na) ‘2.91210000000000000000’\nb) ‘2.91’\nc) 2.91\nd) Error\n\n6. What is the output of the following code?\n\n`size(‘’)`\n\na) 0\nb) Error\nc) 0 0\nd) 1 1\n\n7. What is the output of the following code?\n\n`sprintf(); disp();`\n\na) Error due to disp\nb) Error due to sprintf()\nc) Both gives an error\nd) No output is displayed\n\n8. How can we influence the disp command with format specifiers?\na) Not possible\nb) Via the sprintf() command\nc) Use format specifiers as string input\nd) Give format specifiers as input only\n\n9. What is the output of the following code?\n\n`P=disp(234)`\n\na) Error\nb) P=234\nc) 234\nd)\n\n``` P=\n234```\n\n10. What is the output of the following code?\n\n`fprintf(“%f”,.123)`\n\na) .123000\nb) %f\nc) Error\nd) .123\n\n11. What is the output of the following code?\n\n`disp(“12”)`\n\na) ‘12’\nb) “12”\nc) 12\nd) Error\n\n12. What is the size of ans variable after the following code?\n\n```sprintf('23 23')\nans=\n’23 23’```\n\na) 14 bytes\nb) 10 bytes\nc) Error\nd) 1*5", null, "5 Star\n0%\n4 Star\n0%\n3 Star\n0%\n2 Star\n0%\n1 Star\n0%" ]

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[ "# Specializations of nonsymmetric Macdonald–Koornwinder polynomials\n\n## Abstract\n\nThe purpose of this article is to work out the details of the Ram–Yip formula for nonsymmetric Macdonald–Koornwinder polynomials for the double affine Hecke algebras of not-necessarily reduced affine root systems. It is shown that the $$t\\rightarrow 0$$ equal-parameter specialization of nonsymmetric Macdonald polynomials admits an explicit combinatorial formula in terms of quantum alcove paths, generalizing the formula of Lenart in the untwisted case. In particular, our formula yields a definition of quantum Bruhat graph for all affine root systems. For mixed type, the proof requires the Ram–Yip formula for the nonsymmetric Koornwinder polynomials. A quantum alcove path formula is also given at $$t\\rightarrow \\infty$$. As a consequence, we establish the positivity of the coefficients of nonsymmetric Macdonald polynomials under this limit, as conjectured by Cherednik and the first author. Finally, an explicit formula is given at $$q\\rightarrow \\infty$$, which yields the p-adic Iwahori–Whittaker functions of Brubaker, Bump, and Licata.\n\nThis is a preview of subscription content, access via your institution.\n\n1. 1.\n\nIn , the authors use $$E_\\lambda (X;q^{-1};t^{-1})$$ and correspondingly send $$q\\rightarrow 0$$; also uses this convention for q and t.\n\n## References\n\n1. 1.\n\nBraverman, A., Finkelberg, M.: Weyl modules and $$q$$-Whittaker functions. Math. Ann. 359(1–2), 45–59 (2014)\n\n2. 2.\n\nBraverman, A., Finkelberg, M.: Twisted zastava and $$q$$-Whittaker functions. Preprint. arXiv:1410.2365\n\n3. 3.\n\nBrubaker, B., Bump, D., Licata, A.: Whittaker functions and Demazure operators. J. Number Theory 146, 41–68 (2015)\n\n4. 4.\n\nBrenti, F., Fomin, S., Postnikov, A.: Mixed Bruhat operators and Yang–Baxter equations for Weyl groups. Int. Math. Res. Not. (8), 419–441 (1999)\n\n5. 5.\n\nChari, V., Ion, B.: BGG reciprocity for current algebras. Compos. Math. 151(7), 1265–1287 (2015)\n\n6. 6.\n\nCherednik, I.: Double affine Hecke algebras, Knizhnik–Zamolodchikov equations, and Macdonald’s operators. Int. Math. Res. Not. (9), 171–180 (1992)\n\n7. 7.\n\nCherednik, I.: Nonsymmetric Macdonald polynomials. Int. Math. Res. Not. 1995(10), 483–515 (1995)\n\n8. 8.\n\nCherednik, I.: Intertwining operators of double affine Hecke algebras. Sel. Math. (N.S.) 3(4), 459–495 (1997)\n\n9. 9.\n\nCherednik, I.: Double Affine Hecke Algebras. London Mathematical Society Lecture Note Series, 319. Cambridge University Press, Cambridge (2005)\n\n10. 10.\n\nCherednik, I., Feigin, E.: Extremal part of the PBW-filtration and E-polynomials. Adv. Math. 282, 220–264 (2015)\n\n11. 11.\n\nCherednik, I., Orr, D.: Nonsymmetric difference Whittaker functions. Math. Z. 279(3–4), 879–938 (2015)\n\n12. 12.\n\nFourier, G., Littelmann, P.: Weyl modules, Demazure modules, KR-modules, crystals, fusion products and limit constructions. Adv. Math. 211(2), 566–593 (2007)\n\n13. 13.\n\nGaussent, S., Littelmann, P.: LS galleries, the path model, and MV cycles. Duke Math. J. 127(1), 35–88 (2005)\n\n14. 14.\n\nGörtz, U.: Alcove walks and nearby cycles on affine flag manifolds. J. Algebraic Combin. 26(4), 415–430 (2007)\n\n15. 15.\n\nHaglund, J., Haiman, M., Loehr, N.: A combinatorial formula for nonsymmetric Macdonald polynomials. Am. J. Math. 130(2), 359–383 (2008)\n\n16. 16.\n\nHaiman, M.: Cherednik algebras, Macdonald polynomials and combinatorics. Int. Congr. Math. III, 843–872, Eur. Math. Soc., Zürich (2006)\n\n17. 17.\n\nIon, B.: Nonsymmetric Macdonald polynomials and Demazure characters. Duke Math. J. 116(2), 299–318 (2003)\n\n18. 18.\n\nIon, B.: Involutions of double affine Hecke algebras. Compos. Math. 139(1), 67–84 (2003)\n\n19. 19.\n\nIon, B.: Standard bases for affine parabolic modules and nonsymmetric Macdonald polynomials. J. Algebra 319, 3480–3517 (2008)\n\n20. 20.\n\nKac, V.: Infinite dimensional Lie algebras, 3rd edn. Cambridge University Press, Cambridge (1990)\n\n21. 21.\n\nKnop, F.: Integrality of two variable Kostka functions. J. Reine Angew. Math. 482, 177–189 (1997)\n\n22. 22.\n\nLam, T., Shimozono, M.: Quantum cohomology of $$G/P$$ and homology of affine Grassmannian. Acta Math. 204, 49–90 (2010)\n\n23. 23.\n\nLenart, C.: From Macdonald polynomials to a charge statistic beyond type A. J. Combin. Theory Ser. A 119(3), 683–712 (2012)\n\n24. 24.\n\nLenart, C., Schilling, A.: Crystal energy functions via the charge in types A and C. Math. Z. 273(1–2), 401–426 (2013)\n\n25. 25.\n\nLenart, C., Naito, S., Sagaki, D., Schilling, A., Shimozono, M.: A uniform model for Kirillov–Reshetikhin crystals I: lifting the parabolic quantum Bruhat graph. Int. Math. Res. Not. IMRN (7), 1848–1901 (2015)\n\n26. 26.\n\nLenart, C., Naito, S., Sagaki, D., Schilling, A., Shimozono, M. (in preparation)\n\n27. 27.\n\nMacdonald, I.G.: Affine Hecke algebras and orthogonal polynomials. Séminaire Bourbaki, vol. 1994/1995, Astérisque 237, 189–207, Exp. No. 797, 4 (1996)\n\n28. 28.\n\nMacdonald, I.G.: Affine Hecke Algebras and Orthogonal Polynomials. Cambridge Tracts in Mathematics, vol. 157. Cambridge University Press, Cambridge (2003)\n\n29. 29.\n\nOpdam, E.: Harmonic analysis for certain representations of graded Hecke algebras. Acta Math. 175, 75–121 (1995)\n\n30. 30.\n\nRam, A., Yip, M.: A combinatorial formula for Macdonald polynomials. Adv. Math. 226(1), 309–331 (2011)\n\n31. 31.\n\nThe Sage-Combinat Community, Sage-Combinat: Enhancing Sage as a Toolbox for Computer Exploration in Algebraic Combinatorics. http://combinat.sagemath.org (2008)\n\n32. 32.\n\nSahi, S.: Nonsymmetric Koornwinder polynomials and duality. Ann. Math. (2) 150(1), 267–282 (1999)\n\n33. 33.\n\nSanderson, Y.: On the connection between Macdonald polynomials and Demazure characters. J. Algebr. Combin. 11(3), 269–275 (2000)\n\n34. 34.\n\nSchwer, C.: Galleries, Hall–Littlewood polynomials, and structure constants of the spherical Hecke algebra. Int. Math. Res. Not. (2006). Art. ID 75395, 31 pp\n\n35. 35.\n\nStokman, J.: Macdonald–Koornwinder Polynomials. Preprint. arXiv:1111.6112\n\n## Acknowledgements\n\nThanks to Bogdan Ion, Arun Ram, and Siddartha Sahi, for patient explanations about the double affine Hecke algebra. Thanks to Anne Schilling and Nicolas Thiéry for implementing nonsymmetric Macdonald polynomials in sage ; we used their program extensively. Thanks to ICERM, which provided the venue for the above activities. Thanks to Cristian Lenart, Satoshi Naito, Daisuke Sagaki, and Anne Schilling for related collaborations. Thanks to Ivan Cherednik for helpful discussions. The second author thanks the NSF for the support from Grant NSF DMS-1200804, and both authors thank the NSF for partial support from Grant NSF DMS-1600653.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Daniel Orr.\n\n## Rights and permissions\n\nReprints and Permissions" ]

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[ "Community Profile", null, "# Jesús Gutiérrez\n\nLast seen: 6 months ago Active since 2020\n\n#### Statistics\n\nAll\n•", null, "•", null, "•", null, "#### Content Feed\n\nView by\n\nQuestion\n\nHow do I make the legend color rectangles smaller?\nIn barplots legends I find too wide the boxes that show each color. I want they to be more like squares so I can have a narrower...\n\n7 months ago | 1 answer | 0\n\n### 1\n\nHow to export high quality plot in eps format ?\nI use these lines to get high quality eps and import it to LaTex documents: set(gcf,'units','centimeters','position',[xini,yini...\n\n1 year ago | 3\n\nQuestion\n\nSurface fit doesnt cover all datapoints\nI am triying to plot a surface response with some input points from a parametric study with the following code: [xData,yData,zD...\n\n1 year ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nUsing remote matlab on a cluster: Lost connection but job keeps calculating. How to reopen matlab gui?\nJust lost connection while a script on matlab was running remotely on a linux cluster. I was running it from the remote matlab G...\n\n1 year ago | 1 answer | 0\n\n### 1\n\nSolved\n\nFind the sum of all the numbers of the input vector\nFind the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...\n\n1 year ago\n\nSolved\n\nMake the vector [1 2 3 4 5 6 7 8 9 10]\nIn MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...\n\n1 year ago\n\nSolved\n\nFind all elements less than 0 or greater than 10 and replace them with NaN\nGiven an input vector x, find all elements of x less than 0 or greater than 10 and replace them with NaN. Example: Input ...\n\n1 year ago\n\nSolved\n\nTimes 2 - START HERE\nTry out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...\n\n1 year ago" ]

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[ "# Category Archive for: Investment Analysis Portfolio Management\n\nYield to Maturity Assignment Help Introduction Yield to maturity is the overall return that will be made by somebody who available a bond and holds it up until its maturity date. The yield to maturity may likewise be described as yield, internal rate of return, or the marketplace rate of interest at the time that…\n\nReal Rate Assignment Help Introduction Real rate is a rate of interest that has actually been adapted to eliminate the impacts of inflation to show the real expense of funds to the debtor and the real yield to the loan provider. The real rate of interestof a financial investment is computed as the quantity by…\n\nReturn on Assets Assignment Help Introduction Return on assets is the ratio of yearly earnings to typical overall assets of a company throughout a fiscal year. It determineseffectiveness of business in utilizing its assets to produce earnings. It is an earnings ratio. Return on assets shows the number of cents made on each dollar of…\n\nRatio Analysis Assignment Help Introduction A ratio analysis is a quantitative analysis of info consisted of in a business’s monetary declarations. Ratio analysis is based on line products in monetary declarations like the balance sheet, earnings declaration and money circulation declaration, the ratios of one product or a mix of products to another product or…\n\nRate of Return Assignment Help Introduction The rate of return (ROR), in some cases called (ROI), is the ratio of the annual earningsfrom a financial investment to the original financial investment. The preliminary quantitygotten (or payment), the portion of succeeding invoices (or payments) and any last invoice (or payment), all play a consider identifying the…\n\nPrice of Bond Assignmet Help Introduction A bond is a financial obligation security that pays a set quantity of interesttill maturity. When a bond grows, the primary quantity of the bond is returned to the shareholder. Lots of financiers determine the present value of a bond. The price of a bond is the amount of…\n\nNominal Rate Assignment Help Introduction The nominal rate of interest (or cash rate of interest) is the portion boost in cash you pay the loan provider for using the cash you obtained. Think of that you obtained \\$100 from your bank one year earlier at 8 % interest on your loan. You have to pay…\n\nFixed-Income and Credit-Sensitive Instruments Assignment Help Introduction A financial investment that supplies a return through fixed routine payments and the ultimate return of principal at maturity. Unlike a variable-income security, where payments are based upon some hidden step such as short-term rate of interest, the payments of a fixed-income security are understood ahead of time.…" ]

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[ "# 56.25 inches to feet converter\n\n## How much feet in an inch?\n\nLet’s discuss ways to calculate the length of units, for example, 56.25 in to ft. How long is 56.25 in in feet?\n\nYou can only calculate 56.25 inches in ft if you know the conversion factor of one inch in ft.\n\n1 in = 1/12 feet.\n\n• What is an inch to feet?\n• An inch equals how many feet?\n• What is conversion inches to ft?\n• How to change 1 inch to ft?\n\n## Facts of Inch\n\nAn inch (symbol in) is an Anglo-American unit of length measurement.. Its symbol is in. In several different European languages, the term “inch” is identical to or comes from “thumb”. The thumb of a man is about an inch wide.\n\nUse:\n\n• Electronic components such as the dimensions of the display.\n• Dimensions of tires for cars and trucks.\n\n## Meaning of Foot\n\nFeet, also known as foot (symbol: ft) is a measure of length in the Anglo-American customary system of measuring It is equals to one third of a yard or 12 inches.\n\nApplication:\n\n• For measuring heights, shorter distances, field lengths.\n• Human foot size.\n\n## How to Transfer 56.25 Inches to Feet?\n\nEach region and country has its own system of conversion. So what is the ratio of 56.25 inches to ft?\n\nTo convert a value in inches to the corresponding value in feet, you simply multiply the amount in inches by 0.083333..\n\n56.25 inches to feet = 56.25 inches × 0.083333 = 4.68748125 feet\n\n• How many in in feet?\n\n1 One = 0.083333 feet. To calcualte more, use cminchesconverter.\n\n• connection between inches and feet?\n\n1 foot = 12 inches\n\n1 inch = 0.08333 feet\n\n• What is formula for inches to feet?\n\nThe conversion factor to convert in to feet is 0.083333. So just divide the feet by 0.083333 to get feet.\n\n• How to convert in to ft?\n\nft = inch × 0.083333\n\nFor example:\n\n56.25 inches to ft = 0.083333 × 56.25 = 4.68748125 ft\n\n## Formula for Converting Inches to Feet\n\nValue in feet = value in inches × 0.083333\n\n## Conclusion\n\nUp to now, do you know the number of 56.25 in to ft?" ]

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[ "# How to force overfiting of Deep Learning Network for Classification\n\n8 views (last 30 days)\nWojciech Czop on 13 Jan 2020\nCommented: Greg Heath on 18 Jan 2020\nHow to force overfiting neural network proposed at documentation https://www.mathworks.com/help/deeplearning/examples/create-simple-deep-learning-network-for-classification.html trained on MNIST dataset ?\n\n#### 1 Comment\n\nGreg Heath on 18 Jan 2020\nSpelling: overfitting has 2 \"t\"s\nHTH\nGreg\n\nSrivardhan Gadila on 17 Jan 2020\nAs your question is specific to overfitting the proposed network in the example \"Create Simple Deep Learning Network for Classification\" , I can suggest you the following:\nFirst one:\nWhile splitting the dataset for training & validation, do not split them randomly. Instead do it normally as follows:\n[imdsTrain,imdsValidation] = splitEachLabel(imds,numTrainFiles); %remove the input argument 'randomize'\nThen train the network until the training loss/accuracy saturates\nSecond one:\nTake only 10% of the original dataset provided in the example. Train on 75% of the new dataset & validate on the other 25%. You can then see that the network will overfit as the network is too big for the new dataset and over the epochs it will overfit.\nThe following code can help you for getting 10% of the original dataset\ndigitDatasetPath = fullfile(matlabroot,'toolbox','nnet','nndemos', ...\n'nndatasets','DigitDataset');\nimds = imageDatastore(digitDatasetPath, ...\n'IncludeSubfolders',true,'LabelSource','foldernames');\n%display count of number of samples per each label\nimds.countEachLabel\nnumFiles = 100;\n%Taking random 10% of the original dataset with equal samples for each cateogry\nimds = splitEachLabel(imds,numFiles,'randomize');\n%display count of number of samples per each label after taking 10% of samples from original set\nimds.countEachLabel\nnumTrainFiles = 75;\n[imdsTrain,imdsValidation] = splitEachLabel(imds,numTrainFiles,'randomize');\nYou can also change the default trainingOptions too like 'momentum', 'L2Regularization' etc.\nYou can also refer to Improve Shallow Neural Network Generalization and Avoid Overfitting and Questions related to overfitting in MATLAB Answers Community.\n\nGreg Heath on 18 Jan 2020\nOVERFITTING = More training unknowns (e.g., weights) than training vectors.\nOVERTRAINING1 = Training an overfit network to or past convergence (DANGEROUS)\nOVERTRAINING2 = Training any network past convergence (STUPID BUT NOT NECESSARILY DANGEROUS)\nHOPE THIS HELPS\nTHANK YOU FOR FORMALLY ACCEPTING MY ANSWER\nGREG" ]

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[ "1. Read Chapter 8 in Introductory Fisheries Analyses with R and these notes and consider the items below.\n• What are the three most common condition indices?\n• What is the formula for Fulton’s condition factor?\n• What fundamental assumption about fish “growth” does Fulton’s condition factor make? Why does this make this metric largely useless?\n• What is the formula for the relative condition factor of LeCren?\n• What is the primary limitation for using the relative condition factor?\n• What is the formula for a relative weight?\n• What possible relationship with condition should be checked for before computing an overall measure of condition for a stock of fish?\n• What R function is used to find the coefficients of the standard weight equation for a given fish species? What are the arguments to this function?\n• What statistical test can be used to compare mean relative weights among Gabelhouse length categories?\n2. Read Pope and Kruse (2007) Sections 10.1, 10.3, 10.6, & 10.7 (the complete chapter is here) and consider the items below.\n• What is “condition” a measure of?\n• Condition indices must control for what confounding effect? Put this into your own words.\n• Why is measuring fish condition important to a fisheries scientist?\n• What are the three most common condition indices?\n• What is the formula for Fulton’s condition factor?\n• What fundamental assumption about fish “growth” does Fulton’s condition factor make? Why does this make this metric largely useless?\n• What is the formula for the relative condition factor of LeCren?\n• What is the primary limitation for using the relative condition factor?\n• What is the formula for a relative weight?\n• What value of Wr represents fish in “good condition”?\n• What abiotic and biotic factors have been found to be related to fish condition?" ]

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[ "# js中函数的作用\n\n• 被赋值给一个变量\n• 被赋值为对象的属性\n• 作为参数被传入别的函数\n• 作为函数的结果被返回\n\n//声明一个函数，接受两个参数，返回其和\n```function add(x, y){\nreturn x + y;\n}\nvar a = 0;\nvar b = a(2, 3);//调用这个新的函数a\nprint(b);\n```\n\n```var obj = {\nid : \"obj1\"\n}\nobj.func(2, 3);//返回5\n```\n\n//高级打印函数的第二个版本\n```function adPrint2(str, handler){\nprint(handler(str));\n}\n```\n\n//将字符串转换为大写形式，并返回\n```function up(str){\nreturn str.toUpperCase();\n}\n```\n//将字符串转换为小写形式，并返回\n```function low(str){\nreturn str.toLowerCase();\n}\n```\n\nHELLO, WORLD\nhello, world\n\n```function currying(){\nreturn function(){\nprint(\"curring\");\n}\n}\n```\n\n```function (){\nprint(\"curring\");\n}\n```\n\ncurrying()();\n\ncurrying" ]

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[ "# 2020 AMC 10A Problems/Problem 5\n\n## Problem\n\nWhat is the sum of all real numbers", null, "$x$ for which", null, "$|x^2-12x+34|=2?$", null, "$\\textbf{(A) } 12 \\qquad \\textbf{(B) } 15 \\qquad \\textbf{(C) } 18 \\qquad \\textbf{(D) } 21 \\qquad \\textbf{(E) } 25$\n\n## Solution 1 (Casework and Factoring)\n\nSplit the equation into two cases, where the value inside the absolute value is positive and nonpositive.\n\nCase 1:\n\nThe equation yields", null, "$x^2-12x+34=2$, which is equal to", null, "$(x-4)(x-8)=0$. Therefore, the two values for the positive case is", null, "$4$ and", null, "$8$.\n\nCase 2:\n\nSimilarly, taking the nonpositive case for the value inside the absolute value notation yields", null, "$-x^2+12x-34=2$. Factoring and simplifying gives", null, "$(x-6)^2=0$, so the only value for this case is", null, "$6$.\n\nSumming all the values results in", null, "$4+8+6=\\boxed{\\textbf{(C) }18}$.\n\n## Solution 2 (Casework and Vieta)\n\nWe have the equations", null, "$x^2-12x+32=0$ and", null, "$x^2-12x+36=0$.\n\nNotice that the second is a perfect square with a double root at", null, "$x=6$, and the first has two distinct real roots. By Vieta's, the sum of the roots of the first equation is", null, "$-(-12)$ or", null, "$12$.", null, "$12+6=\\boxed{\\textbf{(C) }18}$.\n\n## Solution 3 (Casework and Graphing)\n\nCompleting the square gives", null, "\\begin{align*} \\left|(x-6)^2-2\\right|&=2 \\\\ (x-6)^2-2&=\\pm2. \\hspace{15mm}(\\bigstar) \\end{align*} Note that the graph of", null, "$y=(x-6)^2-2$ is an upward parabola with the vertex", null, "$(6,-2)$ and the axis of symmetry", null, "$x=6;$ the graphs of", null, "$y=\\pm2$ are horizontal lines.\n\nWe apply casework to", null, "$(\\bigstar):$\n\n1.", null, "$(x-6)^2-2=2$\n2. The line", null, "$y=2$ intersects the parabola", null, "$y=(x-6)^2-2$ at two points that are symmetric about the line", null, "$x=6.$\n\nIn this case, the average of the solutions is", null, "$6,$ so the sum of the solutions is", null, "$12.$\n\n3.", null, "$(x-6)^2-2=-2$\n4. The line", null, "$y=-2$ intersects the parabola", null, "$y=(x-6)^2-2$ at one point: the vertex of the parabola.\n\nIn this case, the only solution is", null, "$x=6.$\n\nFinally, the sum of all solutions is", null, "$12+6=\\boxed{\\textbf{(C) } 18}.$\n\n~MRENTHUSIASM\n\n## Video Solution 2\n\nEducation, The Study Of Everything\n\n~IceMatrix\n\n~bobthefam\n\n~savannahsolver\n\n## Video Solution 5\n\n~ pi_is_3.14\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]

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[ "#1fa834 Color Information\n\nIn a RGB color space, hex #1fa834 is composed of 12.2% red, 65.9% green and 20.4% blue. Whereas in a CMYK color space, it is composed of 81.5% cyan, 0% magenta, 69% yellow and 34.1% black. It has a hue angle of 129.2 degrees, a saturation of 68.8% and a lightness of 39%. #1fa834 color hex could be obtained by blending #3eff68 with #005100. Closest websafe color is: #339933.\n\n• R 12\n• G 66\n• B 20\nRGB color chart\n• C 82\n• M 0\n• Y 69\n• K 34\nCMYK color chart\n\n#1fa834 color description : Dark lime green.\n\n#1fa834 Color Conversion\n\nThe hexadecimal color #1fa834 has RGB values of R:31, G:168, B:52 and CMYK values of C:0.82, M:0, Y:0.69, K:0.34. Its decimal value is 2074676.\n\nHex triplet RGB Decimal 1fa834 `#1fa834` 31, 168, 52 `rgb(31,168,52)` 12.2, 65.9, 20.4 `rgb(12.2%,65.9%,20.4%)` 82, 0, 69, 34 129.2°, 68.8, 39 `hsl(129.2,68.8%,39%)` 129.2°, 81.5, 65.9 339933 `#339933`\nCIE-LAB 60.376, -57.89, 48.064 15.187, 28.543, 7.958 0.294, 0.552, 28.543 60.376, 75.242, 140.298 60.376, -53.23, 63.97 53.426, -42.755, 28.567 00011111, 10101000, 00110100\n\nColor Schemes with #1fa834\n\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #a81f93\n``#a81f93` `rgb(168,31,147)``\nComplementary Color\n• #4fa81f\n``#4fa81f` `rgb(79,168,31)``\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #1fa879\n``#1fa879` `rgb(31,168,121)``\nAnalogous Color\n• #a81f4f\n``#a81f4f` `rgb(168,31,79)``\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #791fa8\n``#791fa8` `rgb(121,31,168)``\nSplit Complementary Color\n• #a8341f\n``#a8341f` `rgb(168,52,31)``\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #341fa8\n``#341fa8` `rgb(52,31,168)``\n• #93a81f\n``#93a81f` `rgb(147,168,31)``\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #341fa8\n``#341fa8` `rgb(52,31,168)``\n• #a81f93\n``#a81f93` `rgb(168,31,147)``\n• #136720\n``#136720` `rgb(19,103,32)``\n• #177d27\n``#177d27` `rgb(23,125,39)``\n• #1b922d\n``#1b922d` `rgb(27,146,45)``\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #23be3b\n``#23be3b` `rgb(35,190,59)``\n• #27d341\n``#27d341` `rgb(39,211,65)``\n• #39da52\n``#39da52` `rgb(57,218,82)``\nMonochromatic Color\n\nAlternatives to #1fa834\n\nBelow, you can see some colors close to #1fa834. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #2ca81f\n``#2ca81f` `rgb(44,168,31)``\n• #21a81f\n``#21a81f` `rgb(33,168,31)``\n• #1fa829\n``#1fa829` `rgb(31,168,41)``\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #1fa83f\n``#1fa83f` `rgb(31,168,63)``\n• #1fa84b\n``#1fa84b` `rgb(31,168,75)``\n• #1fa856\n``#1fa856` `rgb(31,168,86)``\nSimilar Colors\n\n#1fa834 Preview\n\nThis text has a font color of #1fa834.\n\n``<span style=\"color:#1fa834;\">Text here</span>``\n#1fa834 background color\n\nThis paragraph has a background color of #1fa834.\n\n``<p style=\"background-color:#1fa834;\">Content here</p>``\n#1fa834 border color\n\nThis element has a border color of #1fa834.\n\n``<div style=\"border:1px solid #1fa834;\">Content here</div>``\nCSS codes\n``.text {color:#1fa834;}``\n``.background {background-color:#1fa834;}``\n``.border {border:1px solid #1fa834;}``\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000201 is the darkest color, while #f1fcf3 is the lightest one.\n\n• #000201\n``#000201` `rgb(0,2,1)``\n• #031306\n``#031306` `rgb(3,19,6)``\n• #07240b\n``#07240b` `rgb(7,36,11)``\n• #0a3410\n``#0a3410` `rgb(10,52,16)``\n• #0d4515\n``#0d4515` `rgb(13,69,21)``\n• #10551a\n``#10551a` `rgb(16,85,26)``\n• #13661f\n``#13661f` `rgb(19,102,31)``\n• #167625\n``#167625` `rgb(22,118,37)``\n• #19872a\n``#19872a` `rgb(25,135,42)``\n• #1c972f\n``#1c972f` `rgb(28,151,47)``\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #22b939\n``#22b939` `rgb(34,185,57)``\n• #25c93e\n``#25c93e` `rgb(37,201,62)``\n``#2ad845` `rgb(42,216,69)``\n• #3bdb53\n``#3bdb53` `rgb(59,219,83)``\n• #4bde62\n``#4bde62` `rgb(75,222,98)``\n• #5ce170\n``#5ce170` `rgb(92,225,112)``\n• #6ce47f\n``#6ce47f` `rgb(108,228,127)``\n• #7de78d\n``#7de78d` `rgb(125,231,141)``\n• #8dea9c\n``#8dea9c` `rgb(141,234,156)``\n• #9eedaa\n``#9eedaa` `rgb(158,237,170)``\n• #aff0b9\n``#aff0b9` `rgb(175,240,185)``\n• #bff3c7\n``#bff3c7` `rgb(191,243,199)``\n• #d0f6d6\n``#d0f6d6` `rgb(208,246,214)``\n• #e0f9e4\n``#e0f9e4` `rgb(224,249,228)``\n• #f1fcf3\n``#f1fcf3` `rgb(241,252,243)``\nTint Color Variation\n\nTones of #1fa834\n\nA tone is produced by adding gray to any pure hue. In this case, #5c6b5e is the less saturated color, while #00c71f is the most saturated one.\n\n• #5c6b5e\n``#5c6b5e` `rgb(92,107,94)``\n• #557259\n``#557259` `rgb(85,114,89)``\n• #4d7a54\n``#4d7a54` `rgb(77,122,84)``\n• #45824f\n``#45824f` `rgb(69,130,79)``\n• #3e8949\n``#3e8949` `rgb(62,137,73)``\n• #369144\n``#369144` `rgb(54,145,68)``\n• #2e993f\n``#2e993f` `rgb(46,153,63)``\n• #27a039\n``#27a039` `rgb(39,160,57)``\n• #1fa834\n``#1fa834` `rgb(31,168,52)``\n• #17b02f\n``#17b02f` `rgb(23,176,47)``\n• #10b729\n``#10b729` `rgb(16,183,41)``\n• #08bf24\n``#08bf24` `rgb(8,191,36)``\n• #00c71f\n``#00c71f` `rgb(0,199,31)``\nTone Color Variation\n\nColor Blindness Simulator\n\nBelow, you can see how #1fa834 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "### Numeracy\n\n Top Marks maths interactive maths for your level Cool Maths As the name suggests maths is cool! Primary Math games Oodles of maths games to enhance maths learning Numercy Numeracy activities from the BBC arranged in age group Rainforest MathsLots of maths learning at your own grade level Basic FractionsInteractive practice and challenging games about fractions Figure this Can you take the challenge? Math playgroundSomewhere everyone wants to do math! Games Using Table KnowlegdeA fun way to practice your times tables Maths 4 KidsABCya Math links Super maths WorldA new world where maths is supreme fun! Only some ganmes are free now. Skills wise numbersHeaps of maths games to test and extend your skills from the BBC Online Interactive MathsLots of fun interactive maths problems to work through 9-10Maths for 9 and 10 yr olds 11-13Maths for 11-13 year olds" ]

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[ "• 10以内的加减法口诀\n• 发布时间：2020-10-22 11:02 | 作者：admin | 来源：未知 | 浏览：\n\n• 　　0+4=4——1-1=0——9+5=14\n\n6+9=15——0-0=0——0+9=9\n\n1+6=7——8+3=11——2+0=2\n\n2+4=6——9+3=12——4-1=3\n\n10-6=4——2-0=2——1-0=1\n\n5-3=2——7+2=9——7+5=12\n\n3+9=12——7-0=7——7-7=0\n\n10-8=2——5-4=1——5-0=5\n\n1-0=1——0-0=0——6+0=6\n\n2-2=0——8+7=15——1-0=1\n\n2+5=7——4-1=3——3+2=5\n\n5+2=7——6+1=7——6-5=1\n\n0+7=7——7+2=9——3-3=0\n\n0-0=0——6+7=13——4-0=4\n\n0-0=0——7-5=2——6-6=0\n\n0-0=0——7+2=9——4+7=11\n\n8-6=2——3-0=3——8+4=12\n\n8+10=18——8-6=2——0+4=4\n\n10+0=10——4-4=0——2+1=3\n\n5+8=13——5+7=12——8-4=4\n\n1+4=5——9-8=1——10-6=4\n\n8-5=3——6+5=11——4+2=6\n\n6+2=8——7+4=11——2-2=0\n\n1+3=4——4+0=4——9-6=3\n\n2+2=4——3+3=6——7-6=1\n\n7+8=15——4+6=10——7+0=7\n\n9+3=12——6-1=5——3+7=10\n\n6-1=5——4+3=7——7+7=14\n\n7-4=3——8+8=16——4-2=2\n\n8+5=13——2-2=0——3+10=13\n\n5-1=4——1-0=1——6+3=9\n\n6+4=10——7-3=4——1-1=0\n\n0-0=0——3-3=0——4-0=4\n\n5+5=10——10+1=11——0+8=8\n\n9+6=15——7-2=5——8+3=11\n\n5-3=2——4+4=8——0-0=0\n\n1+2=3——5+4=9——0-0=0\n\n7-6=1——10+8=18——2+5=7\n\n2+10=12——2-2=0——0-0=0\n\n1-0=1——7-4=3——9+10=19\n\n9+4=13——3+4=7——9+8=17\n\n8+0=8——5+3=8——9+0=9\n\n9-3=6——2+2=4——2-0=2\n\n2-1=1——8+5=13——4+0=4\n\n1-1=0——7+2=9——5+1=6\n\n3+1=4——0+4=4——3-0=3\n\n5-2=3——2+3=5——0+4=4\n\n6+3=9——1-1=0——4+5=9\n\n8-7=1——7+0=7——0-0=0\n\n5+2=7——3-1=2——10-10=0\n\n3+8=11——8+10=18——1-0=1\n\n7-2=5——8-3=5——7+5=12\n\n0-0=0——7+9=16——3+0=3\n\n9+1=10——0+5=5——9+2=11\n\n6+9=15——1-1=0——9+8=17\n\n9+10=19——3-1=2——7+2=9\n\n6-0=6——8-6=2——2+4=6\n\n3-1=2——5+7=12——6+10=16\n\n3-3=0——5+3=8——0+1=1\n\n3+6=9——3+9=12——0+1=1\n\n9+4=13——1+3=4——7-2=5\n\n2-2=0——2-2=0——8+2=10\n\n10+7=17——4+0=4——3-1=2\n\n5-5=0——3+0=3——0+4=4\n\n9-1=8——9-5=4——3-3=0\n\n4-3=1——9+6=15——1+4=5\n\n6-1=5——1+2=3——2+8=10\n\n5-0=5——1+10=11——7-7=0\n\n4+9=13——0+6=6——1-0=1\n\n3+7=10——5+3=8——7-0=7\n\n4+8=12——5-4=1——3-3=0\n\n8-5=3——2-2=0——1-0=1\n\n8-2=6——5+10=15——8-8=0\n\n4-3=1——1-1=0——9-7=2\n\n7-1=6——3+0=3——10+5=15\n\n0-0=0——2-0=2——6-1=5\n\n10+2=12——5+5=10——7-1=6\n\n9-0=9——4-1=3——2-2=0\n\n5+9=14——5+4=9——5+7=12\n\n10+9=19——2-2=0——1-1=0\n\n9+0=9——9-7=2——10-9=1\n\n8+6=14——8+7=15——6-2=4\n\n1+3=4——4-4=0——6-1=5\n\n9+3=12——9+8=17——9+3=12\n\n5-4=1——8-4=4——3+4=7\n\n1+7=8——5+2=7——0-0=0\n\n1+0=1——1-0=1——3-0=3\n\n8-8=0——3+5=8——3-2=1\n\n1-0=1——3-1=2——9-4=5\n\n0-0=0——5-4=1——2+9=11\n\n4+10=14——10+5=15——1-0=1\n\n3-1=2——8-2=6——9+5=14\n\n4-0=4——4+10=14——9+8=17\n\n3+7=10——0-0=0——5+6=11\n\n9-8=1——2-1=1——8+7=15\n\n10-2=8——10-2=8——9+6=15\n\n9+0=9——8+10=18——2+7=9\n\n0+4=4——9+1=10——1+0=1\n\n6+1=7——10-3=7——5+10=15\n\n10-4=6——4-1=3——2+8=10\n转载请注明：http://www.51luopang.cn/a/xedu/yinianji/2020/0817/12770.html\n• 相关内容\n• 推荐排行\n• 随机浏览\n• 打赏站长\n•", null, "•", null, "• 小学教育 | 中学教育 | 作文鉴赏 | 辅导资料 | |网站地图|热门搜索\n• 投稿联系：QQ 547403730【可带品牌、联系方式、网址、超链接】" ]

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[ "Select Page\n\nA picture frame that lost a staple or other fastener can shift of out square, like the one on the right. The picture doesn’t fit and the backing shows through. With some old fashioned math and a new fastener, make this off-kilter frame perfect again.\n\nPencils in a Basket by Public Domain Pictures.Net   Circled Pencils by Stux\n\n## Square 90 Degree Corners the Easy Way\n\nMany home repair and remodeling projects require you to keep square corners square. A square corner has a 90-degree angle. A perfect square or rectangle has four corners that each measure 90 degrees. If the corner angles shift away from 90 degrees, the shape becomes a parallelogram.\n\nPre-hung interior doors or picture frames like those above are good examples and fairly simple projects that virtually any homeowner could undertake by themselves. Both require square corners.\n\nIn the out-of-square picture frame above, the upper left and lower right corners are less than 90 degrees. High school geometry taught us that a square or a rectangle has four 90 degree corners with opposite sides parallel.\n\nEqual diagonal measurements of a square or rectangle will verify that the corners are right angles or 90 degrees. In the picture to the left, the square has four equal sides and each corner is 90 degrees, or a right angle. If you draw a line from one corner to the opposite corner, the square or rectangle turns into two right triangles. A line drawn between other two corners will be exactly the same length if the corners are 90 degrees.\n\nMeasure from A to C and from B to D. If the two measurements are equal, all the corners are 90 degrees. If one diagonal is longer than the other, as in the picture frame on the left above, the frame is out of square. The corners with the longer diagonal measurement are smaller than 90 degrees, and the other two corners are larger than 90 degrees.\n\nThis makes it easy to determine if any rectangular frame is square or needs adjustment.\n\n## The Trick\n\nIf the two diagonal measurements are not equal, just move one of the corners on the longer diagonal toward one of the corners on the shorter diagonal until both measurements are equal.\n\nLook at the picture frame above on the left. The longer diagonal is from upper left to lower right. Move the lower right corner up and each frame corner becomes a 90 degree square when both diagonals have the same length.\n\n## Make it Easier\n\nYou could measure both diagonals repeatedly until both are equal. This tends to turn into a frustrating dance of trial and error that takes a lot of time, especially when you’re dealing with a door or window frame that you want as perfect as possible. We can simplify by determining what the diagonal measurement should be in the first place. We do this using the Pythagorean Theorem from geometry. The length of the diagonal side of any right triangle is the square root of the sum of the squares of the two sides. Put simply, side A squared + side B squared = side C squared or A2 + B2 = C 2. In the rectangle shown, we can easily calculate the expected length of the diagonal with virtually any calculator. Assume side A is 10 units, and side B is 9 units.", null, "A2 = 10 x 10 = 100. B2 = 9 x 9 = 81. A2 + B2 = 100 + 81 = 181. C2 = 181. C = √ 181. C = 13.45.\n\n## Use a Calculator\n\nEnter the length of Side A: 10    (Length of A)\n\nPress X2 and the display reads 100 (A2 = 10 x 10 = 100)\n\nPress +\n\nEnter the length of Side B:  9 (Length of B)\n\nPress X2 and the display reads 81 (B2 = 9 x 9 = 81)\n\nPress = and the display reads 181   (A2 + B2)\n\nPress √ to display the answer 13.45362404707371031716308546217\n\nYou don’t need all those numbers. Just round it off to two or three digits or 13.45", null, "## Now the Square Corners with 90 Degree Angles are Easy\n\nNo need to measure multiple times to get it perfect. Now you know the distance between opposite corners must be 13.45. Hold your tape measure or other measuring tool across two opposite corners. Move the sides until the measure reads 13.45. If you measure the other side, it will read 13.45 also.\n\nMany of us in the USA use good old inches and feet to measure things. That’s fine and it is what we are used to. At least, until you’re scratching your head trying to figure out what fraction on your tape measure corresponds to 13.45.\n\nThe really great thing about this math is that it works for any measurement system from inches or feet to parsecs, astronomical units, or… meters, centimeters, and millimeters.\n\n100 centimeters (cm) = 1 meter (m).\n\n10 millimeters (mm) = 1 centimeter (mm).\n\nIf you measured in centimeters or millimeters, you wouldn’t worry about that fraction at all. You’d just adjust the frame diagonal to 134.5 mm and be done. No conversion. No anything. Just measure everything millimeters and be done with it.\n\nIf you’re still stuck on your fractions, 13.45 is almost, but not quite equal to 13 29/64. Be sure to wear your reading glasses to see those 1/64 marks on your tape rule." ]

[ null, "https://willyknows.com/wp-content/uploads/2023/08/pythagoream-theorem.png", null, "https://willyknows.com/wp-content/uploads/2023/08/windows-10-calculator.jpg", null ]

[ "## Modelling fatality curves of COVID-19 and the effectiveness of intervention strategies", null, "## Introduction\n\nThe response interventions to the pandemic of the novel coronavirus disease (COVID-19) have varied from country to country. Several countries, especially those first hit by the disease, have adopted a standard progressive protocol, from containment to mitigation to supression (World Health Organization (WHO), 2020). As these strategies failed to deter the spread of the virus, government authorities introduced ever more stringent measures on their citizens’ movements in an attempt to suppress or sharply reduce the propagation of the virus. More recently, countries have adopted drastic countermeasures at the very outset of the outbreak. For example, on March 24, 2020, India announced a 3-week total ban on people ‘venturing out’ of their homes (Corera, 2020), even though there were fewer than 500 confirmed cases and only nine people had died from COVID-19 in a country with a population of 1.3 billion people. One difficulty in deciding the best approach to counter the spread of the novel coronavirus (SARS-CoV-2) is that the virus propagation dynamics is not yet well understood.\n\nIn this stark context, it becomes relevant to have simple models for the evolution of the COVID-19 epidemic, so as to be able to obtain estimates—however tentative—for the rise in the number of infected people as well as in the number of fatal cases, both in the near and in the more distant future. Such estimates are, of course, prone to high uncertainty: the less data available and the further in the future, the greater the uncertainty. Notwithstanding their inherent shortcomings, simple mathematical models provide valuable tools for quickly assessing the severity of an epidemic and help to guide the health and political authorities in defining or adjusting their national strategies to fight the disease (Crokidakis, 2020; Sameni, 2020; Castorina, Iorio & Lanteri, 2020; Dehkordi et al., 2020; Mair et al., 2016; Siegenfeld & Bar-Yam, 2020; Bastos & Cajueiro, 2020; Schulz, Coimbra-AraÃojo & Costiche, 2020; Manchein et al., 2020).\n\nIn the same vein, it would be desirable to have simple methods to assess the effectiveness of intervention measures as a function of the time at which they are adopted. As a general rule, the sooner an intervention is put in place the more effective it is expected to be. It is however difficult to model a priori how effective any given set of interventions will be. The effectiveness of interventions are often investigated through complex agent-based simulation models (Ferguson et al., 2020; Koo et al., 2020), which require a synthetic population and a host of parameters related to the human-to-human transmission, and as such they are very costly computationally and heavily dependent on the choice of values for the various model parameters.\n\nIn this article we use the Richards growth model (RGM) (Richards, 1959) to study the fatality curves, represented by the cumulative number of deaths as a function of time, of COVID-19 for the following countries: China, Brazil, France, Germany, Iran, Italy, South Korea, and Spain, which are at different stages of the epidemic. We show that the RGM describes reasonably well the fatality curves of all countries analysed in this study, except Brazil, which is in an early-to-intermediate stage of the epidemic. In the Brazilian case, we use instead the so-called generalised growth model to describe the available epidemic data. We also introduce a theoretical framework, within the context of the RGM model, to calculate the efficiency of interventions. Here an intervention strategy is modelled by assuming that its net result is captured by a change in the values of the parameters of the RGM after a given time t0. In this picture, the full epidemic dynamics is described in terms of two Richards models: one before and the other after the intervention ‘adoption time’ t0, where certain matching conditions are imposed at t0. In this way, we are able to derive an analytical formula for the efficiency of the corresponding intervention as a function of the adoption time t0. We show, in particular, that the intervention efficiency decays quickly if its adoption is delayed beyond a reasonably short period of time, thus showing that time is really of essence in containing an outbreak.\n\n## Data\n\n### Data source\n\nData used in this study were downloaded from the database made publicly available by the Johns Hopkins University (JHU, 2020), which lists in an automated fashion the number of confirmed cases and deaths per country. We have also compared the JHU data with the corresponding data from Worldometer (2020) for eventual data inconsistency and data redundancy. In all cases considered here we have used data up to May 8, 2020. In the present study we considered the mortality data of COVID-19 from the following countries: China, Brazil, France, Germany, Iran, Italy, South Korea, and Spain.\n\n### Confirmed cases vs. mortality data\n\nBecause a large proportion of COVID-19 infections go undetected (Li et al., 2020), it is difficult to estimate the actual number of infected people within a given population. As many carriers of the virus are either asymptomatic or develop only mild symptoms, they will not be detected unless they are tested. In other words, the number of confirmed cases for COVID-19 is a poor proxy for the total number of infections. Furthermore, the fraction of confirmed cases relative to the total number of infections depends heavily on the testing policy of each country, which makes it problematic to compare the evolution of confirmed cases among different countries. In contrast, the number of deaths attributed to COVID-19 is a somewhat more reliable measure of the advance of the epidemic and its severity. The official numbers of deaths attributed to COVID-19 have, of course, uncertainties of their own, as countries may differ as to the criteria and protocols for recording deaths related to the disease. For instance, some countries’ death figures do not include (or only later in the epidemic started to include) deaths outside hospitals, which naturally leads to under-reporting. There may also be delays in the reporting of deaths, also leading to uncertainties as to the number of deaths at a given time. Furthermore, other factors, such as the age structure of a population and quality of care, may affect the fraction of deaths relative to the number of confirmed cases. Nevertheless, taking all these factors into consideration, it is still reasonable to assume that the evolution of the number of confirmed deaths bears a relation to the dynamics of the number of infections (Famulare, 2020). Under these circumstances and in the absence of more reliable estimates for the number of infected cases for COVID-19, we decided here to seek an alternative approach and model the fatality curves, defined as the cumulative number of deaths as a function of time, rather than the number of confirmed cases, as is more commonly done.\n\n## Methods\n\n### Mathematical models\n\nWe model the time evolution of the number of cases in the epidemic by means of the Richards growth model (RGM), defined by the following ordinary differential equation (ODE) (Wang, Wu & Yang, 2012; Hsieh, 2009): $\\frac{dC}{dt}=rC\\left(t\\right)\\left[1-{\\left(\\frac{C\\left(t\\right)}{K}\\right)}^{\\alpha }\\right]$where C(t) is the cumulative number of cases at time t, r is the growth rate at the early stage, K is the final epidemic size, and the parameter α measures the asymmetry with respect to the S-shaped dynamics of the standard logistic model, which is recovered for α = 1.\n\nIt is worth to point out that the Richards model has an intrinsic connection to the SIR epidemic model, see, for example, Wang, Wu & Yang (2012). As a matter of fact, by identifying the variable C(t) of the Richards model with the cumulative number of deaths of a modified SIRD model, akin to the SIR model of Wang, Wu & Yang (2012), it is possible to establish a sort of ‘map’ between the parameters (α, r) of the Richards model and the parameters (β, γ1, γ2) of the SIRD model, where β is the transmission rate, γ1 is the recovery rate and γ2 is the death rate (Macêdo et al., 2020). The advantage, however, of phenomenological models, such as the Richards model, is that they allow for exact solutions (Chowell et al., 2016), which makes the data analysis much simpler. Furthermore, by avoiding ‘the description of biological mechanisms that may be difficult to identify,’ especially in an ongoing epidemic, they ‘can be utilised for efficient and rapid forecasts with quantified uncertainty’ (Bürger, Chowell & Lara-Daz, 2019). It is also worth pointing out that phenomenological growth models have been successfully applied to other epidemics, such as Zika (Chowell et al., 2016) and influenza (Bürger, Chowell & Lara-Daz, 2019), which makes these models good candidates for describing the ongoing COVID-19 epidemic, where there is still substantial uncertainty in the epidemiological parameters.\n\nAs already mentioned, in the present article we shall apply the RGM to the fatality curves of COVID-19, so that C(t) will represent the cumulative numbers of deaths in a given country at time t, where t will be counted in days from the first death. Nevertheless, in the interest of generality, in this and in the following section we shall refer to C(t) simply as the number of cases.\n\nEquation (1) must be supplemented with a boundary condition. Here it is convenient to choose $\\stackrel{¨}{C}\\left({t}_{c}\\right)=0$for some given tc, where dots denote time derivatives, that is $\\stackrel{¨}{C}\\left(t\\right)={d}^{2}C\\left(t\\right)/d{t}^{2}$. A direct integration of Eq. (1) subjected to condition Eq. (2) yields the following explicit formula: $C\\left(t;r,\\alpha ,K,{t}_{c}\\right)=\\frac{K}{{\\left\\{1+\\alpha \\mathrm{exp}\\left[-\\alpha r\\left(t-{t}_{c}\\right)\\right]\\right\\}}^{1/\\alpha }}$where in the left-hand side we have explicitly denoted the dependence of the solution of the RGM on the four parameters, namely r, α, K, and tc. In fitting Eq. (3) to empirical data it is convenient to set C(0) = C0, where C0 is the number of deaths recorded at the first day that a death was reported. Using that $C\\left(0\\right)=K/{\\left[1+\\alpha \\mathrm{exp}\\left(\\alpha r{t}_{c}\\right)\\right]}^{1/\\alpha }$, we can eliminate tc in favour of the other parameters, so that we are left with only three free-parameters, namely (r,α, K), to be numerically determined.\n\nWe note, however, that the RGM is not reliable in situations where the epidemic is in such an early stage that the available data is well below the estimated inflection point tc, that is, when the epidemic is still in the so-called exponential growth regime (Wu et al., 2020). In this case, it is preferable to use the so-called generalised growth model (Wu et al., 2020; Chowell, 2017), which is defined by the following ODE: $\\frac{dC}{dt}=r{\\left[C\\left(t\\right)\\right]}^{q}$where the parameter q provides an interpolation between the sub-exponential regime (0 < q < 1) and the exponential one (q = 1). The solution of Eq. (4) is $C\\left(t;r,q,A\\right)={\\left[A+\\left(1-q\\right)rt\\right]}^{1/\\left(1-q\\right)}={A}^{1/\\left(1-q\\right)}{e}_{q}\\left(rt/A\\right)$where the function eq(x) = [1 + (1 − q)x]1/(1 q) is known in the physics literature as the q-exponential function (Tsallis, 1988; Picoli et al., 2009). Here the parameter A is related to the initial condition, that is A = C(0)(1 q), but we shall treat A as a free parameter to be determined from the fit of Eq. (5) to a given dataset.\n\n### Intervention strategy and efficiency\n\nWe define an intervention strategy in the context of the RGM by considering that the corresponding measures, as applied to the actual population, induce at some time t0 a change in the parameters of the model. In this way, the solution for the whole epidemic curve acquires the following piecewise form: $C\\left(t\\right)=\\left\\{\\begin{array}{ll}C\\left(t;r,K,\\alpha ,{t}_{c}\\right),& \\phantom{\\rule{1em}{0ex}}t\\le {t}_{0}\\\\ C\\left(t;r\\mathrm{\\prime },K\\mathrm{\\prime },\\alpha \\mathrm{\\prime },{t}_{c}\\mathrm{\\prime }\\right),& \\phantom{\\rule{1em}{0ex}}t>{t}_{0}\\end{array}$where we obviously require that K′/K < 1. Furthermore we impose the following boundary conditions at t0: $C\\left({t}_{0};r,K,\\alpha ,{t}_{c}\\right)=C\\left({t}_{0};r\\mathrm{\\prime },K\\mathrm{\\prime },\\alpha \\mathrm{\\prime },{t}_{c}\\mathrm{\\prime }\\right)$ $\\stackrel{˙}{C}\\left({t}_{0};r,K,\\alpha ,{t}_{c}\\right)=\\stackrel{˙}{C}\\left({t}_{0};r\\mathrm{\\prime },K\\mathrm{\\prime },\\alpha \\mathrm{\\prime },{t}_{c}\\mathrm{\\prime }\\right)$\n\nNote that condition Eq. (8) takes into account, albeit indirectly, the fact that intervention measures take some time to affect the epidemic dynamics. In other words, the trend (i.e. the derivative) one sees at a given time t reflects in part the measures taken at some earlier time (or lack thereof). Thus, imposing continuity of the derivative of the epidemic curve at time t0 in our ‘intervention strategy’ seeks to capture this delay effect.\n\nIn our intervention model above, we assume that the net result of the intervention is to alter the parameters r and α of the RGM after the time t0; see Eq. (6). In other words, we assume that the parameters of the RGM can capture (albeit in an effective and simplified manner) some basic aspects of the underlying epidemic dynamics, so that changes in the mechanisms of the disease propagation—owing, say, to the introduction of intervention measures—could be described in terms of variations in the model parameters. It is in this sense that we associate actual interventions with possible changes in the model parameters, assuming of course that the growth model is still valid after the interventions. Admittedly, the difficult part is to estimate how a particular set of intervention measures (e.g. social distancing, contact tracing and quarantine, school closures, etc.) would influence these parameters. This link between actual interventions and the RGM parameters cannot, of course, be obtained within the context of the Richards model alone. It requires, for example the use more complex models, such as agent-based or compartmental models, to implement specific interventions, after which one can use the RGM to fit the resulting epidemic curves; see, for example Chowell (2017) where a similar approach was adopted albeit in the context of quantifying the uncertainty in epidemiological parameter estimates. Comparison of the RGM parameters after the intervention with those for the reference curve (i.e. without intervention) could thus shed light on how actual interventions can be mimic within the RGM approach. We are currently pursuing this strategy—namely, using the RGM formula to fit simulations from both agent-based and SIR-type models—to gain a better understanding of how interventions can be reflected in the parameters of the RGM. Such an analysis, however, is still ongoing (Macêdo et al., 2020) and is beyond the scope of the present article.\n\nNote that, as discussed above, the time t0 is not the actual time of adoption of the intervention but rather the time after which the corresponding measures start to affect the epidemic dynamics, as reflected in a change in the evolution of the number of cases. Nevertheless we shall for simplicity refer to t0 as the intervention ‘adoption time.’ We remark, furthermore, that as far as interventions go, there are basically two aims: (i) to reduce the speed of an epidemic, which is essentially to ‘flatten the curve’ of daily deaths and (ii) to reduce the epidemic size, that is the total number of deaths due to the epidemic. In this article we address the latter case, since the interventions modelled by the strategy Eq. (6) are designed to yield K′ < K. In other words, the aim here is to ‘bend’ the cumulative curve of deaths as soon as possible, so as to reach a lower plateau at the end of the epidemic (for more details, see “Discussion”).\n\nAs defined in Eq. (6), an intervention strategy adopted at time t0 can be viewed as a map (r, K, α, tc) → (r′, K′, α′, tc′) in the parameter space of the RGM, which results in the condition K′/K < 1. Let us therefore define the intervention efficiency as the relative reduction (expressed in percentage) of the total number of cases: η(t0) = (KK′)/K, where it is assumed that η(t0) > 0. Using conditions Eqs. (7) and (8) in Eq. (3), one obtains that $\\eta \\left({t}_{0}\\right)=1-\\frac{y}{{\\left[1-x\\left(1-{y}^{\\alpha }\\right)\\right]}^{1/\\alpha \\mathrm{\\prime }}}$where y = C(t0)/K and x = r/r′. Later we will exemplify the above measure of intervention efficiency, using as input the parameters r and α obtained from the fatality curve of the COVID-19 from Italy. This will allow us to investigate how the efficiency of different strategies (i.e. for different choices of r′ and α′) varies as a function of the adoption time t0.\n\n### Statistical fits\n\nAll statistical fits in the article were performed using the Levenberg–Marquardt algorithm (Moré, 1978) to solve the corresponding non-linear least square optimisation problem. In the case of the Richards model, we set C(0) = C0, where C0 is the number of deaths recorded at the first day that a death was reported, so that there remain three parameters, namely (r, K, α), to be determined. For the q-exponential growth model we also need to determine three parameters (r, q, A). The fitting procedures were implemented in the opensource software QtiPlot, which was also used to produce the corresponding plots in Figs. 13. The plots in Fig. 4 were produced with the data visualisation library Matplotlib for Python.\n\n## Results\n\n### Fatality curves\n\nIn Fig. 1 we show the official cumulative number of deaths (blue symbols) attributed to COVID-19 for China, where it is clearly visible the jump on day 84 from the first death, when the death toll was revised upward by almost 50%. Since officials in Wuhan, China, informed only that this increase ‘reflected updated reporting and deaths outside hospitals’ (BBC News, 2020), it does not seem possible to reconstruct the actual fatality curve for China; nor does it make much sense to fit any model to the data prior to this correction, owing to their unreliability. It is possible nonetheless to render the Chinese data amenable to a statistical fit, if only as a test of the model, if one somehow smooths the revised data. In this spirit, we adopted the following ad hoc procedure to obtain a smooth ‘empirical’ curve for China: we multiplied all data points prior to the revision date by a factor corresponding to the ratio between the numbers of deaths after and before the revision date, meaning that the excess deaths due to the data correction was uniformly redistributed by the same proportion for all dates before the revision. This is admittedly an arbitrary procedure (perhaps a worst case scenario in terms of change of shape of the unknown true curve), which is intended only as a numerical way of ‘welding’ the two sides of the curve at the jump. This ‘renormalised’ fatality curve (red circles) for China is shown in Fig. 1, superimposed with the corresponding statistical fit of the RGM curve (black solid line), where the fit parameters are shown in the inset of the figure. One sees from the figure that the fatality curve in this case, where the epidemic has apparently levelled off, is well adjusted by the RGM formula.\n\nThe good performance of the RGM for the renormalised data from China, exhibited in Fig. 1, encouraged us to apply the model to other countries at earlier stages of the epidemic. Here, however, care must be taken when estimating the model parameters from small time series, since it is well known that the Richards model (Wang, Wu & Yang, 2012; Hsieh, 2009) and its variants (Wu et al., 2020) are susceptible to the problem of over fitting, owing to the redundancy of the parameters. This may lead, for example, to estimation of certain parameters that are outside of biologically or otherwise reasonable ranges. For example, when applied to the number of infected cases in an epidemic, the parameter α should be constrained within the interval (0,1) (Wang, Wu & Yang, 2012). Here we apply the RGM instead to the number of deaths, but we assume that the same constraint should be observed. In other words, fits that return α outside this interval are disregarded as not reliable. Similarly, we restrict the acceptable values of r to the range (0,1), as we observed that values of r outside this interval tends to be an indication that the RGM is not quite suitable to fit the data. In other words, for our purposes here we assume that the restrictions 0 < r < 1 and 0 < α < 1 are useful empirical criteria for the validity of the RGM, which can also be verified from the map between the Richards model and the modified SIRD model (Macêdo et al., 2020). The unsuitability of the RGM is particularly evident when the available data does not encompass the inflection point tc (Chowell, 2017; Wu et al., 2020), although as more data is accumulated the model is expected to become more accurate. As an empirical criteria, we thus consider here that the RGM is only acceptable if tc is smaller than the time of the last data point; if this criteria is not fulfilled by a particular dataset, we then apply the generalised growth model as given in Eq. (5).\n\nWith these considerations in mind, we applied the RGM to the mortality data of COVID-19 from several other countries. In Fig. 2, we show the cumulative deaths for Italy, Spain, France, Germany, Iran, and South Korea, together with the respective RGM fits. Here again the RGM seems to be able to provide a reasonably good fit to the data for all the above countries. In the case of Brazil, where the epidemic curve has not yet reached its inflection point, the RGM is not justifiable. In such cases, it is more advisable to use a simpler growth model, such as the q-exponential. In Fig. 3 we show the fit of the q-exponential curve Eq. (5) to the Brazilian data, where one sees that the agreement is very good. From the fit we get q = 0.72, indicating that the fatality curve is in a sub-exponential growth regime.\n\n#### Intervention efficiency\n\nAs already mentioned, an intervention strategy in our model is defined by the two parameters r′ and α′ of the new Richards model after the adoption time t0; see Eq. (6). It is premature at this stage to establish a more direct link between actual intervention measures and a corresponding change in the parameters of effective growth models, as already discussed. As a matter of fact, we are currently pursuing this connection between our phenomenological model of interventions and concrete measures (such as social distancing, quarantine, school closures, etc.) by implementing these measures in more complex models, such as agent-based population models and SIRD-type models, and then use the RGM to fit the resulting fatality curves with and without interventions. In this way, we hope to establish how a possible combination of variations in the parameters (r, α) can mimic (at least approximately) a given real intervention. Such an endeavour, however, is beyond the scope of the present article. Our main goal here is to introduce a quantitative measure of the effectiveness of interventions (in the context of the Richard model) and highlight some of its main and important features. More specifically, we shall take a reference curve from the RGM, that is, with a given set of parameters (r, α, tc, K), consider different intervention ‘strategies’ as defined in Eq. (6), and then discuss their respective efficiencies.\n\nAs the reference RGM curve we take here that obtained by fitting the Italian data up to April 1, 2020, when Italy was somewhat in the middle of the outbreak (this corresponds to t = 40 in Fig. 2A), which gave r = 0.44 and α = 0.21. In Fig. 4A we show the efficiency curves as a function of the adoption time t0 for three different interventions, namely: (i) r′ = 0.5 and α′ = 0.21 (red dot-dashed curve); (ii) r′ = 0.6 and α = 0.21 (green dashed curve); and (iii) r′ = 0.44 and α′ = 0.9 (blue solid curve). In Fig. 4B we show the resulting fatality curves, as compared to the no-action case (black curve), after implementing the three intervention actions indicated by the black dots on the red, green, and blue curves of Fig. 4A.\n\n## Discussion\n\nWe have seen above that the RGM describes rather well the fatality curves of COVID-19 from different countries, which are at different stages of the pandemic. For example, in the case of China, whose fatality curve has pretty much levelled off indicating a near-end of the epidemic, the Richards Eq. (3) fits rather well the entire epidemic curve. Here, however, because of the upward revision of the Chinese mortality data on April 17, 2020, we had to adopt an ad hoc data-correction procedure, as explained above, so as to smooth the ‘discontinuous’ empirical curve and thus render it amenable to a mathematical description. In spite of the fact that we were therefore forced to work with rescaled data for the case of China, the good fit provided by the RGM observed in Fig. 1 may be seen as a good indication of the validity of the RGM to describe mortality data of COVID-19 for the full epidemic course. The RGM was also in good agreement with the empirical data for other countries, such as Italy, Spain, France, Germany, Iran, and South Korea. In all these cases, the last data point is sufficiently beyond the inflection point tc obtained from the fit of the RGM Eq. (3) to lend some credibility to the model predictions.\n\nAs emphasised earlier, we considered here that a statistical fit with the RGM is only acceptable if tc is smaller than the time of the last data point. In addition, we used two additional admissibility criteria, namely that the parameters r and α should be both in the interval (0,1). For countries where the epidemic is still in a relatively early stage, these criteria are usually not met. For example, when we applied the RGM to the mortality data from Brazil, we found that all three empirical criteria above were not satisfied, indicating that the fit with the RGM was ‘premature,’ as there were not yet enough data point in the Brazilian fatality curve to make reliable estimations of the model parameters. In this situation, we then fitted the Brazilian data with the q-exponential function given in Eq. (5) and found q < 1, indicating that the fatality curve is in a sub-exponential growth regime. This slower-than-exponential growth probably stems from the mitigation actions that have been put in place in Brazil since the onset of the outbreak; similar effect (i.e. sub-exponential growth) was also observed, for example, in China, although for the number of cases, and it was attributed to containment policies as well (Maier & Brockmann, 2020). From the fit parameters, see inset of Fig. 3, we predict that the current time for doubling the number of deaths in Brazil is 13 days, which is considerably higher than the value of about 5 days we obtained in the first version of this study (which used data up to April 1, 2020). However, the fact that there is no clear indication that the Brazilian fatality curve is near the inflection point is a cause of concern.\n\nAs discussed above, the RGM fits rather well the mortality data of several countries for which the epidemic is still ongoing. We emphasise, however, that our primary interest in the RGM is not so much aimed at its predictive capacity in the face of incomplete data, but rather more so as a mathematical framework in which one can obtain quantitative measures (in fact, an explicit formula) for the effectiveness of mitigation strategies. Using this framework, we were able to compute an explicit formula for the efficiency of an intervention as a function of the adoption time t0, some example of which are shown in Fig. 4A. From these illustrative examples, several important consequences of our efficiency formula can be obtained—which we believe are of general applicability, at least in a qualitative sense.\n\nFor example, one sees from Fig. 4A that in order to attain an efficiency of at least 80% the three interventions shown in the figure must be adopted up to the times t0 = 11, 21, and 26, respectively. However, if the interventions are further delayed by ten more days the respective efficiencies drop to about 50% or less, in all cases. Furthermore, a delay of additional 20 days above the time-window of 80% efficiency brings the efficiency to less than 30% in all cases exemplified in Fig. 4A. This shows that, in general, delaying interventions beyond a reasonable early period of time—the so-called window of opportunity—can have the adverse effect of reducing considerably the effectiveness of an intervention. As for the role of the parameters r and α in affecting the intervention efficiency, one more careful analysis of the model shows that a larger r (with α kept fixed) implies a smaller inflection time tc, which in turn leads to a smaller K′, and hence greater efficiency. A similar but stronger effect is obtained with increasing α (for r fixed): the larger the α, the sooner the curve ‘bends’ toward the plateau, thus yielding a lower final death toll. In fact, the control in the asymmetry of the ‘S-shaped’ curve, and thus the value of tc, was the original motivation for introducing the parameter α in the Richards model. It is thus natural to expect a stronger effect in changing α (for r fixed) than changing r (for α fixed). It is also clear from Fig. 4 that stronger interventions (e.g. with higher values of α) provide wider windows of opportunity, which makes epidemiological sense.\n\nFor the three examples of interventions shown in Fig. 4A, we illustrated in Fig. 4B the fatality curves resulting from theses interventions, when they are adopted at the particular time indicated by the black dots on the curves of Fig. 4A. It is interesting to observe that the qualitative behaviour seen in Fig. 4B for the fatality curves after the interventions is reminiscent of similar curves, albeit for the number of infections, obtained in agent-based simulations of actual intervention measures (Weng & Ni, 2015). This may be seen as further evidence that the Richards model can indeed be extended to model interventions, as proposed above.\n\n## Conclusions\n\nTo summarise, this article provides important insights into the time evolution of the accumulated number of deaths attributed to COVID-19, especially for countries that are in the middle of the outbreak or only recently have passed the inflection point in the curve of accumulated deaths. Our modelling of the fatality curves is particularly relevant for the COVID-19 epidemic because the actual number of infections in this case remains largely unknown, and so one is forced to deal with proxy measures, such as mortality data, to gain a better understanding of the actual severity of the epidemic.\n\nThe article also shows how simple and soluble mathematical models can provide a rich theoretical framework in which to investigate some basic and deep aspects of epidemic dynamics. In particular, we have successfully applied the Richards growth model to describe the fatalities curves of seven different countries at different stages of the COVID-19 outbreak, namely China, France, Germany, Italy, Iran, South Korea, and Spain. We also analysed the case of Brazil, which is in an earlier stage of the outbreak, and so we had to resort to a modified exponential growth model, known as the generalised growth model. This model also gave a good fit of the rising fatality curve of Brazil, from which we could estimate that the current time for doubling the number of fatalities from COVID-19 is 13 days.\n\nAnother important contribution of the present study is to provide an analytical formula to quantitatively assess the efficiency of intervention measures in an ongoing epidemic. Interventions strategies were defined in the context of the Richards model as a change in the model parameters at some specified time, referred to as the intervention adoption time. Our formula shows that, in general, the efficiency of an intervention strategy decays quite quickly as the adoption time is delayed, thus showing that time is really of essence in containing an outbreak. The present work can be extended in several ways. For example, a direct connection with a SIRD-type model can be explored in order to find the underlying epidemiological meaning of the parameters r and α of the RGM. Another possible direction of research consists in seeking to identify how the parameters of the RGM can be varied, perhaps even continuously in time, so as to mimic the effect of actual intervention measures. One could then apply such an extended Richards model with time-dependent parameters to improve its fitting performance whenever needed. All these interesting research avenues are currently being explored in our group.\n\n## Supplemental Information\n\n### Dataset.\n\nThe dataset is compiled from data provided by Johns Hopkins University at https://data.humdata.org/dataset/novel-coronavirus-2019-ncov-cases.\n\n### Code - python.\n\nMitigation Strategies\n\n### Code - python.\n\nIntervention efficiency\n\n### Competing Interests\n\nThe authors declare that they have no competing interests. Inês Cristina Lemos de Souza is the founder and director of 3Hippos Data Consulting (https://www.3hippos.com.br) in Brazil, which specialises in providing data analysis and training in data literacy for not-for-profit social organisations. She has no affiliations with or involvement in any organisation or entity with any financial interest, or non-financial interest in the subject matter discussed in this article. Giovani Lopes Vasconcelos and Inês Cristina Lemos de Souza are a married couple and have worked together with the rest of the team in a collaborative fashion.\n\n### Author Contributions\n\nGiovani L. Vasconcelos conceived and designed the experiments, performed the experiments, analysed the data, prepared figures and/or tables, authored or reviewed drafts of the paper, and approved the final draft.\n\nAntônio M.S. Macêdo conceived and designed the experiments, performed the experiments, analysed the data, prepared figures and/or tables, authored or reviewed drafts of the paper, and approved the final draft.\n\nRaydonal Ospina conceived and designed the experiments, performed the experiments, analysed the data, prepared figures and/or tables, authored or reviewed drafts of the paper, and approved the final draft.\n\nFrancisco A.G. Almeida conceived and designed the experiments, performed the experiments, analysed the data, prepared figures and/or tables, authored or reviewed drafts of the paper, and approved the final draft.\n\nGerson C. Duarte-Filho conceived and designed the experiments, performed the experiments, analysed the data, prepared figures and/or tables, authored or reviewed drafts of the paper, and approved the final draft.\n\nArthur A. Brum conceived and designed the experiments, performed the experiments, analysed the data, prepared figures and/or tables, authored or reviewed drafts of the paper, and approved the final draft.\n\nInês C.L. Souza conceived and designed the experiments, performed the experiments, analysed the data, prepared figures and/or tables, authored or reviewed drafts of the paper, and approved the final draft.\n\n### Data Availability\n\nThe following information was supplied regarding data availability:\n\nThe codes in Python and the dataset are available in the Supplemental Files.\n\n### Funding\n\nThis work was supported by Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq) under Grants No. 303772/2017-4 (GLV), No. 312612/2019-2 (AMSM), No. 305305/2019-0 (RO) and Coordenação de Aperfeiçoamento de Pessoal de Nível Superior (CAPES) of Brazil. The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript." ]

[ null, "https://dfzljdn9uc3pi.cloudfront.net/collections/70-thumb.jpg", null ]

[ "kaboom.js (beta) is a JavaScript library that helps you make games fast and fun!\n\nalso check out the kaboom environment on replit.com!\n\nUsage\n\nquick start\n\n``````<script src=\"https://kaboomjs.com/lib/0.1.0/kaboom.js\"></script>\n<script type=\"module\">\n\n// make kaboom functions global\nkaboom.global();\n\n// init kaboom context\ninit();\n\n// define a scene\nscene(\"main\", () => {\n\n// add a text at position (100, 100)\nadd([\ntext(\"ohhimark\", 32),\npos(100, 100),\n]);\n\n});\n\n// start the game\nstart(\"main\");\n\n</script>``````\n\nImport\n\nAll functions are under a global object 'kaboom', but you can also choose to import all functions into global namespace.\n\nkaboom.global()\n\nimport all kaboom functions into global namespace\n\n``````// 1) import everything to global\nkaboom.global();\ninit();\n\n// 2) keep them under kaboom namespace to prevent any collision\nconst k = kaboom;\nk.init();``````\n\nLifecycle\n\nApplication Lifecycle Methods\n\ninit([conf])\n\ninitialize context\n\n``````// quickly create a 640x480 canvas and get going\ninit();\n\n// options\ninit({\nwidth: 480, // width of canvas\nheight: 480, // height of canvas\ncanvas: document.getElementById(\"game\"), // use custom canvas\nscale: 2, // pixel size (for pixelated games you might want small canvas + scale)\nclearColor: rgb(0, 0, 1), // background color (default black)\nfullscreen: true, // if fullscreen\ncrisp: true, // if pixel crisp (for sharp pixelated games)\n});``````\n\nstart(scene, [...args])\n\nstart the game loop with specified scene\n\n``````scene(\"game\", () => {/* .. */});\nscene(\"menu\", () => {/* .. */});\nscene(\"lose\", () => {/* .. */});\nstart(\"game\");``````\n\nScene\n\nScenes are the different stages of a game, like different levels, menu screen, and start screen etc. Everything belongs to a scene.\n\nscene(name)\n\ndescribe a scene\n\n``````scene(\"level1\", () => {\n// all objs are bound to a scene\nadd(/* ... */)\n// all events are bound to a scene\nkeyPress(/* ... */)\n});\n\nscene(\"level2\", () => {\nadd(/* ... */)\n});\n\nscene(\"gameover\", () => {\nadd(/* ... */)\n});\n\nstart(\"level1\");``````\n\ngo(name, [...args])\n\nswitch to a scene\n\n``````// go to \"paused\" scene when pressed \"p\"\nscene(\"main\", () => {\nlet score = 0;\nkeyPress(\"p\", () => {\ngo(\"gameover\", score);\n})\n});\n\nscene(\"gameover\", (score) => {\n// display score passed by scene \"main\"\nadd([\ntext(score),\n]);\n});``````\n\nlayers(names, [default])\n\ndefine the draw layers of the scene\n\n``````// draw background on the bottom, ui on top, layer \"obj\" is default\nlayers([\n\"bg\",\n\"obj\",\n\"ui\",\n], \"obj\");\n\n// this will be added to the \"obj\" layer since it's defined as default above\nconst player = add([\nsprite(\"froggy\"),\n]);\n\n// this will be added to the \"ui\" layer cuz it's specified by the layer() component\nconst score = add([\ntext(\"0\"),\nlayer(\"ui\"),\n]);``````\n\ngravity(value)\n\nset the gravity value (defaults to 980)\n\n``````// (pixel per sec.)\ngravity(1600);``````\n\ncamPos(pos)\n\nset the camera position\n\n``````// camera position follow player\nplayer.action(() => {\ncamPos(player.pos);\n});``````\n\ncamScale(scale)\n\nset the camera scale\n\n``````if (win) {\ncamPos(player.pos);\n// get a close up shot of the player\ncamScale(3);\n}``````\n\ncamRot(angle)\n\nset the camera angle\n\n``camRot(0.1);``\n\ncamIgnore(layers)\n\nmake camera don't affect certain layers\n\n``````// make camera not affect objects on layer \"ui\" and \"bg\"\ncamIgnore([\"bg\", \"ui\"]);``````\n\nAsset Loading\n\nLoad assets into asset manager. These should be at application top.\n\nloadSprite(name, src, [conf])\n\nload a sprite\n\n``````loadSprite(\"froggy\", \"froggy.png\");\nloadSprite(\"froggy\", \"https://replit.com/public/images/mark.png\");\n\n// slice a spritesheet and add anims manually\nloadSprite(\"froggy\", \"froggy.png\", {\nsliceX: 4,\nsliceY: 1,\nanims: {\nrun: [0, 2],\njump: ,\n},\n});\n\n// load with aseprite sprite sheet\nloadSprite(\"froggy\", \"froggy.png\", {\naseSpriteSheet: \"froggy.json\", // use spritesheet exported from aseprite\n});``````\n\nloadSound(name, src, [conf])\n\nload a sound\n\n``loadSound(\"shoot\", \"shoot.ogg\");``\n\nloadFont(name, src, charWidth, charHeight, [chars])\n\nload a font\n\n``````// default character mappings: (ASCII 32 - 126)\n// const ASCII_CHARS = \" !\"#\\$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~\";\n\n// load a bitmap font called \"04b03\", with bitmap \"04b03.png\", each character on bitmap has a size of (6, 8), and contains default ASCII_CHARS\nloadFont(\"04b03\", \"04b03.png\", 6, 8);\n\n// load a font with custom characters\nloadFont(\"CP437\", \"CP437.png\", 6, 8, \"☺☻♥♦♣♠\");``````\n\nObjects\n\nGame Object is the basic unit of Kaboom, each game object uses components to compose their data and behavior.\n\nadd(comps)\n\nadd a game object to scene\n\n``````// a game object consists of a list of components\nconst player = add([\n// a 'sprite' component gives it the render ability\nsprite(\"froggy\"),\n// a 'pos' component gives it a position\npos(100, 100),\n// a 'body' component makes it fall and gives it jump()\nbody(),\n// raw strings are tags\n\"player\",\n\"killable\",\n// custom fields are assigned directly to the returned obj ref\n{\ndir: vec2(-1, 0),\ndead: false,\nspeed: 240,\n},\n]);\n\nplayer.action(() => {\nplayer.move(player.dir.scale(player.speed));\n});\n\nplayer.hidden = false; // if this obj renders\nplayer.paused = true // if this obj updates\n\n// runs every frame as long as player is not destroy() ed\nplayer.action(() => {\nplayer.move(100, 0);\n});\n\n// provided by 'sprite()'\nplayer.play(\"jump\"); // play a spritesheet animation\nconsole.log(player.frame); // get current frame\n\n// provided by 'pos()'\nplayer.move(100, 20);\nconsole.log(player.pos);\n\n// provided by 'body()'\nplayer.jump(320); // make player jump``````\n\ndestroy(obj)\n\nremove a game object from scene\n\n``````collides(\"bullet\", \"killable\", (b, k) => {\n// remove both the bullet and the thing bullet hit with tag \"killable\" from scene\ndestroy(b);\ndestroy(k);\nscore++;\n});``````\n\nobj.action(cb)\n\nupdate the object, the callback is run every frame\n\n``````player.action(() => {\nplayer.move(SPEED, 0);\n});``````\n\nobj.use(comp)\n\nadd a component to a game object\n\n``````// rarely needed since you usually specify all comps in the 'add()' step\nobj.use(scale(2, 2));``````\n\nobj.exists()\n\ncheck if obj exists in scene\n\n``````// sometimes you might keep a reference of an object that's already 'destroy()'ed, use exists() to check if they were\nif (obj.exists()) {\nchild.pos = obj.pos.clone();\n}``````\n\nobj.is(tag)\n\nif obj has certain tag(s)\n\n``````if (obj.is(\"killable\")) {\ndestroy(obj);\n}``````\n\nobj.on(event, cb)\n\nlisten to an event\n\n``````// when obj is 'destroy()'ed\nobj.on(\"destroy\", () => {\nadd([\nsprite(\"explosion\"),\n]);\n});\n\n// runs every frame when obj exists\nobj.on(\"update\", () => {\n// ...\n});\n\n// custom event from comp 'body()'\nobj.on(\"grounded\", () => {\n// ...\n});``````\n\nobj.trigger(event)\n\ntrigger an event (triggers 'on')\n\n``````obj.on(\"grounded\", () => {\nobj.jump();\n});\n\n// mainly for custom components defining custom events\nobj.trigger(\"grounded\");``````\n\nget(tag)\n\nget a list of obj reference with a certain tag\n\n``const enemies = get(\"enemy\");``\n\nevery(tag, cb)\n\nrun a callback on every obj with a certain tag\n\n``````// equivalent to destroyAll(\"enemy\")\nevery(\"enemy\", (obj) => {\ndestroy(obj);\n});``````\n\ndestroyAll(tag)\n\ndestroy every obj with a certain tag\n\nComponents\n\nBuilt-in components. Each component gives the game object certain data / behaviors.\n\npos(x, y)\n\nobject's position in the world\n\n``````const obj = add([\npos(0, 50),\n]);\n\n// get the current position in vec2\nconsole.log(obj.pos);\n\n// move an object by a speed (dt will be multiplied)\nobj.move(100, 100);``````\n\nscale(x, y)\n\nscale\n\n``````const obj = add([\nscale(2),\n]);\n\n// get the current scale in vec2\nconsole.log(obj.scale);``````\n\nrotate(angle)\n\nscale\n\n``````const obj = add([\nrotate(2),\n]);\n\nobj.action(() => {\nobj.angle += dt();\n});``````\n\ncolor(r, g, b, [a])\n\ncolor\n\n``````const obj = add([\nsprite(\"froggy\"),\n// give it a blue tint\ncolor(0, 0, 1),\n]);\n\nobj.color = rgb(1, 0, 0); // make it red instead``````\n\nsprite(id, [conf])\n\ndraw sprite\n\n``````// note: this automatically gives the obj an 'area()' component\nconst obj = add([\n// sprite is loaded by loadSprite(\"froggy\", src)\nsprite(\"froggy\"),\n]);\n\nconst obj = add([\nsprite(\"froggy\", {\nanimSpeed: 0.3, // time per frame (defaults to 0.1)\nframe: 2, // start frame (defaults to 0)\n}),\n]);\n\n// get current frame\nconsole.log(obj.frame);\n\n// play animation\nobj.play(\"jump\");\n\n// stop the anim\nobj.stop();\n\nobj.onAnimEnd(\"jump\", () => {\nobj.play(\"fall\");\n});``````\n\ntext(txt, size, [conf])\n\ndraw text\n\n``````// note: this automatically gives the obj an 'area()' component\nconst obj = add([\n// content, size\ntext(\"oh hi\", 64),\n]);\n\nconst obj = add([\ntext(\"oh hi\", 64, {\nwidth: 120, // wrap when exceeds this width (defaults to 0 no wrap)\nfont: \"proggy\", // font to use (defaults to \"unscii\")\n}),\n]);\n\n// update the content\nobj.text = \"oh hi mark\";``````\n\nrect(w, h)\n\ndraw rectangle\n\n``````// note: this automatically gives the obj an 'area()' component\nconst obj = add([\n// width, height\nrect(50, 75),\npos(25, 25),\ncolor(0, 1, 1),\n]);\n\n// update size\nobj.width = 75;\nobj.height = 75;``````\n\narea(p1, p2)\n\na rectangular area for collision checking\n\n``````// 'area()' is given automatically by 'sprite()' and 'rect()', but you can override it\nconst obj = add([\nsprite(\"froggy\"),\n// override to a smaller region\narea(vec2(6), vec2(24)),\n]);\n\n// callback when collides with a certain tag\nobj.collides(\"collectable\", (c) => {\ndestroy(c);\nscore++;\n});\n\n// similar to collides(), but doesn't pass if 2 objects are just touching each other (checks for distance < 0 instead of distance <= 0)\nobj.overlaps(\"collectable\", (c) => {\ndestroy(c);\nscore++;\n});\n\n// checks if the obj is collided with another\nif (obj.isCollided(obj2)) {\n// ...\n}\n\nif (obj.isOverlapped(obj2)) {\n// ...\n}\n\n// register an onClick callback\nobj.clicks(() => {\n// ...\n});\n\n// if the obj is clicked last frame\nif (obj.isClicked()) {\n// ...\n}\n\n// register an onHover callback\nobj.hovers(() => {\n// ...\n});\n\n// if the obj is currently hovered\nif (obj.isHovered()) {\n// ...\n}\n\n// check if a point is inside the obj area\nobj.hasPt();\n\n// resolve all collisions with objects with 'solid'\n// for now this checks against all solid objs in the scene (this is costly now)\nobj.resolve();``````\n\nbody([conf])\n\ncomponent for falling / jumping\n\n``````const player = add([\npos(0, 0),\n// now player will fall in this gravity world\nbody(),\n]);\n\nconst player = add([\npos(0, 0),\nbody({\n// force of .jump()\njumpForce: 640,\n// maximum fall velocity\nmaxVel: 2400,\n}),\n]);\n\n// body() gives obj jump() and grounded() methods\nkeyPress(\"up\", () => {\nif (player.grounded()) {\nplayer.jump(JUMP_FORCE);\n}\n});\n\n// and a \"grounded\" event\nplayer.on(\"grounded\", () => {\nconsole.log(\"horray!\");\n});``````\n\nsolid()\n\nmark the obj so other objects can't move past it if they have an area and resolve()\n\n``````const obj = add([\nsprite(\"wall\"),\nsolid(),\n]);\n\n// need to call resolve() (provided by 'area') to make sure they cannot move past solid objs\nplayer.action(() => {\nplayer.resolve();\n});``````\n\norigin(orig)\n\nthe origin to draw the object (default center)\n\n``````const obj = add([\nsprite(\"froggy\"),\n\n// defaults to \"topleft\"\norigin(\"topleft\"),\n// other options\norigin(\"top\"),\norigin(\"topright\"),\norigin(\"left\"),\norigin(\"center\"),\norigin(\"right\"),\norigin(\"botleft\"),\norigin(\"bot\"),\norigin(\"botright\"),\norigin(vec2(0, 0.25)), // custom\n]);``````\n\nlayer(name)\n\nspecify the layer to draw on\n\n``````layers([\n\"bg\",\n\"game\",\n\"ui\",\n], \"game\");\n\nadd([\nsprite(\"sky\"),\nlayer(\"bg\"),\n]);\n\n// we specified \"game\" to be default layer above, so a manual layer() comp is not needed\nconst player = add([\nsprite(\"froggy\"),\n]);\n\nconst score = add([\ntext(\"0\"),\nlayer(\"ui\"),\n]);``````\n\nEvents\n\nkaboom uses tags to group objects and describe their behaviors, functions below all accepts the tag as first arguments, following a callback\n\naction(tag, cb)\n\ncalls every frame for a certain tag\n\n``````// every frame move objs with tag \"bullet\" up with speed of 100\naction(\"bullet\", (b) => {\nb.move(vec2(0, 100));\n});\n\naction(\"flashy\", (f) => {\nf.color = rand(rgb(0, 0, 0), rgb(1, 1, 1));\n});``````\n\ncollides(tag, cb)\n\ncalls when objects collides with others\n\n``````collides(\"enemy\", \"bullet\", (e, b) => {\ndestroy(b);\ne.life--;\nif (e.life <= 0) {\ndestroy(e);\n}\n});``````\n\noverlaps(tag, cb)\n\ncalls when objects collides with others\n\n``````// similar to collides(), but doesn't pass if 2 objects are just touching each other (checks for distance < 0 instead of distance <= 0)\noverlaps(\"enemy\", \"bullet\", (e, b) => {\ndestroy(b);\ne.life--;\nif (e.life <= 0) {\ndestroy(e);\n}\n});``````\n\non(event, tag, cb)\n\nadd lifecycle events to a tag group\n\n``````// called when objs with tag \"enemy\" is added to scene\non(\"add\", \"enemy\", (e) => {\nconsole.log(\"run!!\");\n});\n\n// per frame (action() is actually an alias to this)\non(\"update\", \"bullet\", (b) => {\nb.move(100, 0);\n});\n\n// per frame but drawing phase if you want custom drawing\non(\"draw\", \"bullet\", (e) => {\ndrawSprite(...);\n});\n\n// when objs gets destroy() ed\non(\"destroy\", \"bullet\", (e) => {\nplay(\"explosion\");\n});``````\n\nInput\n\ninput events\n\nkeyDown(key, cb)\n\nruns every frame when specified key is being pressed\n\nkeyPress(key, cb)\n\nruns once when specified key is just pressed\n\nkeyRelease(key, cb)\n\nruns once when specified key is just released\n\nmouseDown(cb)\n\nruns every frame when left mouse is being pressed\n\nmouseClick(cb)\n\nruns once when left mouse is just clicked\n\nmouseRelease(cb)\n\nruns once when left mouse is just released\n\nQuery\n\ninformation about current window and input states\n\nwidth()\n\ncanvas width\n\nheight()\n\ncanvas height\n\ntime()\n\ncurrent game time\n\ndt()\n\ndelta time since last frame\n\nmousePos()\n\ncurrent mouse position\n\nTimer\n\ntimed events\n\nwait(time, cb)\n\nruns the callback after time seconds\n\n``````wait(3, () => {\ndestroy(froggy);\n});\n\n// or\nawait wait(3);\ndestroy(froggy);``````\n\nloop(time, cb)\n\nruns the callback every time seconds\n\n``````loop(0.5, () => {\nconsole.log(\"just like setInterval\");\n});``````\n\nAudio\n\nyeah\n\nplay(id, [conf])\n\nplays a sound\n\n``````on(\"destroy\", \"enemy\", (e) => {\nplay(\"explode\", {\nvolume: 2.0,\nspeed: 0.8,\ndetune: 1200,\n});\n});\n\nconst music = play(\"mysong\");\n\nmusic.pause();\nmusic.resume();\nmusic.stop();``````\n\nvolume(volume)\n\nset the master volume\n\nMath\n\nmath types & utils\n\nvec2(x, y)\n\ncreates a vector 2\n\n``````vec2() // => { x: 0, y: 0 }\nvec2(1) // => { x: 1, y: 1 }\nvec2(10, 5) // => { x: 10, y: 5 }\n\nconst p = vec2(5, 10);\n\np.x // 5\np.y // 10\np.clone(); // => vec2(5, 10)\np.add(vec2(10, 10)); // => vec2(15, 20)\np.sub(vec2(5, 5)); // => vec2(0, 5)\np.scale(2); // => vec2(10, 20)\np.dist(vec2(15, 10)); // => 10\np.len(); // => 11.58\np.unit(); // => vec2(0.43, 0.86)\np.dot(vec2(2, 1)); // => vec2(10, 10)\np.angle(); // => 1.1``````\n\nrgba(r, g, b, a)\n\ncreates a color from red, green, blue and alpha values (note: values are 0 - 1 not 0 - 255)\n\n``````const c = rgba(0, 0, 1, 1); // blue\n\np.r // 0\np.g // 0\np.b // 1\np.a // 1\n\nc.clone(); // => rgba(0, 0, 1, 1)``````\n\nrgb(r, g, b)\n\nshorthand for rgba() with a = 1\n\nrand(a, b)\n\ngenerate random value\n\n``````rand() // 0.0 - 1.0\nrand(1, 4) // 1.0 - 4.0\nrand(vec2(0), vec2(100)) // => vec2(29, 73)\nrand(rgb(0, 0, 0.5), rgb(1, 1, 1)) // => rgba(0.3, 0.6, 0.9, 1)``````\n\nrandSeed(seed)\n\nset seed for rand generator\n\n``randSeed(Date.now());``\n\nmakeRng(seed)\n\ncreate a seedable random number generator\n\n``````const rng = makeRng(Date.now());\n\nrng.gen(); // works the same as rand()``````\n\nchoose(arr)\n\nget random element from array\n\nchance(p)\n\nrand(0, 1) <= p\n\nlerp(a, b, t)\n\nlinear interpolation\n\nmap(a, b, x, y, t)\n\nmap number to another range\n\nDraw\n\nRaw immediate drawing functions (you prob won't need these)\n\nrender(cb)\n\nuse a generic draw loop for custom drawing\n\n``````scene(\"draw\", () => {\nrender(() => {\ndrawSprite(...);\ndrawRect(...);\ndrawLine(...);\n});\n});``````\n\ndrawSprite(name, [conf])\n\ndraw a sprite\n\n``````drawSprite(\"car\", {\npos: vec2(100),\nscale: 3,\nrot: time(),\nframe: 0,\n});``````\n\ndrawRect(pos, w, h, [conf])\n\ndraw a rectangle\n\n``drawRect(vec2(100), 20, 50);``\n\ndrawLine(p1, p2, [conf])\n\ndraw a rectangle\n\n``````drawLine(vec2(0), mousePos(), {\nwidth: 2,\ncolor: rgba(0, 0, 1, 1),\nz: 0.5,\n});``````\n\ndrawText(text, [conf])\n\ndraw a rectangle\n\n``````drawText(\"hi\", {\nsize: 64,\npos: mousePos(),\norigin: \"topleft\",\n});``````\n\nLevel\n\nhelpers on building tiled maps\n\naddLevel(map, ref)\n\ntakes a level drawing and turn them into game objects according to the ref map\n\n``````const characters = {\n\"a\": {\nsprite: \"ch1\",\nmsg: \"ohhi how are you\",\n},\n};\n\nconst map = addLevel([\n\" a \",\n\" ===\",\n\" ? * \",\n\" ==== ^^ \",\n\"===================\",\n], {\nwidth: 11,\nheight: 11,\npos: vec2(0, 0),\n// every \"=\" on the map above will be turned to a game object with following comps\n\"=\": [\nsprite(\"ground\"),\nsolid(),\n\"block\",\n],\n\"*\": [\nsprite(\"coin\"),\nsolid(),\n\"block\",\n],\n// use a callback for dynamic evauations per block\n\"?\": () => {\nreturn [\nsprite(\"prize\"),\ncolor(0, 1, rand(0, 1)),\n\"block\",\n];\n},\n\"^\": [\nsprite(\"spike\"),\nsolid(),\n\"spike\",\n\"block\",\n],\n// any catches anything that's not defined by the mappings above, good for more dynamic stuff like this\nany(ch) {\nif (characters[ch]) {\nreturn [\nsprite(char.sprite),\nsolid(),\n\"character\",\n{\nmsg: characters[ch],\n},\n];\n}\n},\n});\n\n// query size\nmap.width();\nmap.height();\n\n// get screen pos through map index\nmap.getPos(x, y);\n\n// destroy all\nmap.destroy();\n\n// there's no spatial hashing yet, if too many blocks causing lag, consider hard disabling collision resolution from blocks far away by turning off 'solid'\naction(\"block\", (b) => {\nb.solid = player.pos.dist(b.pos) <= 20;\n});``````\n\nDebug\n\ndebug utilities\n\nfps()\n\ncurrent frames per second\n\nobjCount()\n\ncurrent number of objects in scene\n\npause()\n\npause the game\n\nunpause()\n\nunpause the game\n\nkaboom.debug()\n\ndebug flags\n\n``````// scale the time\nkaboom.debug.timeScale = 0.5;\n\n// show the bounding box of objects with area()\nkaboom.debug.showArea = true;\n\n// hover to inspect objects (needs showArea checked)\nkaboom.debug.hoverInfo = true;``````" ]

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[ "Anonymous\nAnonymous asked in Science & MathematicsPhysics · 1 month ago\n\nA 2.0 kg wood block is launched up a wooden ramp that is inclined at a 29 ∘ angle. The block's initial speed is 9.0 m/s . The coefficient of kinetic friction of wood on wood is μk=0.200.\n\nA)What vertical height does the block reach above its starting point?\n\nB)What speed does it have when it slides back down to its starting point?\n\nRelevance\n\nA) initial KE becomes friction work and PE:\n\n½mv² = µmgdcosΘ + mgdsinΘ\n\nmass m cancels\n\n½v² = d*g*(µcosΘ + sinΘ)\n\n½(9.0m/s)² = d * 9.8m/s² * (0.200*cos29º + sin29º)\n\nsolves to\n\nd = 6.26 m\n\nand so the height is\n\nh = 6.26m*sin29º = 3.0 m\n\nB) The work done by friction for the back-and-forth route is\n\nW = 2 * 2.0kg * 9.8m/s² * 0.200 * 6.26m * cos29º = 42.9 J\n\ninitial KE = ½ * 2.0kg * (9.0m/s)² = 81 J\n\nfinal KE = initial KE - Work = 38.1 J = ½ * 2.0kg * V²\n\nso the final speed is\n\nV = 6.2 m/s\n\n•", null, "Login to reply the answers\n• A)What vertical height does the block reach above its starting point?\n\nThe initial kinetic energy will convert to potential energy and and friction work\n\n½mv² = mgh + μNd = mgh + μmgcosθ(h/sinθ)\n\n½v² = gh(1 + μcotθ)\n\nh = v²/2g(1 + μcotθ)\n\nh = 9.0² / 2(9.8)(1 + 0.200cot29) = 3.0369... ≈ 3.0 m\n\nB)What speed does it have when it slides back down to its starting point?\n\nThe initial potential energy will convert to kinetic energy and friction work\n\nmgh = ½mv² + μmgcosθ(h/sinθ)\n\ngh -  μgcosθ(h/sinθ) = ½v²\n\nv² = 2gh(1 - μcotθ) = 2(9.8)(3.0369)(1 - 0.200cot29) = 38.04678...\n\nv = 6.1682...≈ 6.2 m/s\n\n•", null, "Login to reply the answers" ]

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[ "0\n\nElectrostatic Focusing in CRT\n\n• Electrostatic Focusing System of CRT:\n\nThe focusing used in a cathode ray oscilloscope is analogous to the refraction of light beam, through a compound lens system. As light beam can be focused varying the focal length of the system, the electron beam also can be focused on to the screen, to offer a fine spot of illumination. To understand the electronic lens system it is necessary to know the nature of the electro static field lines and the equipotential lines in the CR Tube.\n\nLet us consider an electron situated at rest in an electric field. From the definition of electric field intensity we know that the force on a unit positive charge at any point in an electric field is the electric field intensity at that point.\n\nHence Ɛ = f/q V/m --------------------------- (1)\nWhere Ɛ = electric field intensity in V/m\nf = force on the charge in N\nq = charge in C\nThe charge of an electron is Ɛ = 1,602 x 10-19 C -------------------------- (2)\nThe force on the electron in an electric field is fe= -eƐN from equation (1) ------------------ (3)\n\nThe negative sign indicates that the force is acting in a direction opposite to the direction of the electric field. Figure illustrates the electron situated in the electric field. The electric field lines have been shown in figure with directions from positive plate to negative plate. The field lines experience lateral repulsion. This results in spreading of the space between the lines. Therefore the field lines will be curved at the ends of the plates. Hence the density of the field lines will be less at the ends of the plates. A line joining points of equal potentials is called an equipotential line. When such lines are drawn, they are obtained as shown in Figure.\n\nAs the force on an electron acts in a direction opposite to the direction of the field, it can also be stated that the force on an electron is in the direction normal to the equipotential surfaces or equipotential lines. The following figure illustrates the shape of the equipotential lines between two cylinders placed end to end.", null, "Shape of field and equiipotential lines for two cylinders placed end to end\nSince the density of the electric field varies in the area between the two cylinders, the equipotential lines are curved. Let us now consider regions on both sides of the equipotential surface S as shown in Figure. The potential to the left of S is negative and to the right of it is positive. An electron entering the area in the direction AB at an angle with the normal to the equipotential surface will experience a force at the surface. The velocity of the electron is taken as v1. The force acts in a direction normal to the equipotential surface. Hence the velocity of the electron increases to a new value v2; after it passes S. The tangential component, vt of the velocity on both sides of S remains the same. The normal component of the velocity vn only increases by the force at the equipotential surface to a new value v1n. From the figure we have\n\nVt = V1 sin θi = V2 sin θr,\nθi, is the angle of incidence and θr is the angle of refraction of the electron ray.\nRearranging equation (4).\nsin θi/ sin θr = V2/V1\n\nWe find that the equation (5) is identical to the expression relating the refraction of a light beam in geometrical optics. Therefore we conclude saying that the refraction or bending of an electron beam at an equipotential surface is governed by the same laws of refraction of a light beam. This is the reason for calling the electrostatic focusing system as electron lens system.\n\nHaving established the fact that the refraction or bending of the electron beam follows the laws of refraction of light, we will consider the electrostatic focusing system of the CR Tube.\n\nThe pre-accelerating electrode or anode, the focusing anode and the accelerating anode are shown along with the grid structure and the supply voltages in the following diagram.\n\nIn the above diagram we find the pre-accelerating anode, called as A1. It is hollow cylindrical in structure with metal discs placed to make the structure compartmental. There are holes in the metal discs. Electrons will be entering this structure through the hole in the left side. Immediately after this electrode we find the focusing anode. This is a hallow metal cylinder. After the focusing anode we have the accelerating anode called A2. In some constructions the focusing anode may not be there. Anode A1 and Anode A2 are arranged to give the focusing.\n\nThe pre-accelerating anode A1 and the accelerating anode A2 are connected in common to a high positive potential, supplied by the EHT. The focusing anode is connected to a lower + ve potential. As the focusing anode is negative with respect to the two accelerating anodes A1A2, the electric field lines are non-uniformly placed. The equipotential lines are only shown in the figure. The field distribution and hence the shape of the equipotential surfaces/ lines, can be altered by varying the potential difference between the focusing anode and the two accelerating anodes A1 and A2. As shown in the figure the potentiometer connected in the potential divider network that supplies voltage to the electrodes does this job of varying the voltage and hence is called the focus control. This control is made available on the front panel of the CRO as a user control. Varying this voltage changes the focal length of the electron lens. Therefore the beam of electrons can be accurately focused on the screen by proper adjustment of the voltage on the focusing anode.\n\nThe electron beam on entering the pre-accelerating anode will be under the influence of a high positive potential the electrons will be accelerated towards the screen. The discs with centrally located holes aid their travel in the form of a beam. However some electrons will be attracted by the pre-accelerating anode, that contribute the A1 current. The electrons being negative charges repel each other. Therefore we will not get a close beam of electrons. Right from the grid they travel in a diverging beam pass through the pre-accelerating anode with increased velocities and encounter the electric field between the focusing anode and the anode A1, As their paths are not normal to the equipotential lines (surfaces) the electrons undergo refraction through the electric field between the anode A1 and the focusing electrode. Later again they encounter the electric field between the focusing anode and the second anode while they travel towards the screen. The high voltage accelerates them and at the same time due to the electric field distribution they are again refracted through the field. Thus the electrons undergo refraction through the double concave lens system. The result is that they enter the region between A1 and focusing electrode with an inclination, travel parallel to the axis of the CR Tube in side focusing electrode structure latter undergo refraction and finally arrive on the screen as a fine spot. As mentioned earlier the size of the spot (beam) striking the screen is determined by the relative potential difference between A1, A2 and the focusing electrode (anode).\n\nIt must be noted that the second anode may not attract electrons due to their velocities. The A2 current is mostly by the secondary electrons from the screen." ]

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[ "1. Home\n2. PERCENT procedure\n\n# PERCENT procedure\n\nExpresses the body of a table as percentages of one of its margins (R.W. Payne).\n\n### Options\n\n`CLASSIFICATION` = factors Factors classifying the margin over which the percentages are to be calculated; if this is not set, the percentages are over the final margin (grand mean or grand total etc.) Method to use to calculate the margin if not already present (`totals`, `means`, `minima`, `maxima`, `variances`, `medians`); default `tota` Whether to put 100% values into the margin instead of the original values (`no`, `yes`); default `no`\n\n### Parameters\n\n`OLDTABLE` = tables Tables containing the original values Tables to store the percentage values; if any of these is unset, the new values replace those in the original table\n\n### Description\n\n`PERCENT` allows you to express the body of a table as percentages of the values in one of its margins. The table is specified using the `OLDTABLE` parameter. A table to store the new values can be specified using the `NEWTABLE` parameter, otherwise these replace the values of the original table. The margin is indicated by listing the factors that define it using the `CLASSIFICATION` option; the default is the final margin (the grand total, or grand mean etc). If the original table has no margins, option `METHOD` defines how these are to be calculated; the default is to form margins of totals. The values originally in the margin will be left unchanged. If you would prefer these to be replaced by values of 100%, you should set option `HUNDRED=yes`.\n\nOptions: `CLASSIFICATION`, `METHOD`, `HUNDRED`.\n\nParameters: `OLDTABLE`, `NEWTABLE`.\n\n### Method\n\nIf the `OLDTABLE` has no margins and contains no missing values, these are formed by the `MARGIN` directive. Alternatively, if there are missing values, margins other than variances can be formed using `TABULATE`. `CALCULATE` is then used to put the required margin into a table classified just by the factors that define the margin. The original table is divided by the marginal table and multiplied by 100 to give the required percentages. If option `HUNDRED=no`, the same operations are done on a dummy table that originally contains random numbers; for this table, values of 100 should occur only in the margin. Thus by using a logical test in which the values of the dummy table are compared with 100, the marginal values of the original table can be put back into the margin of the final table. The random numbers are generated using a specially written procedure `URANDOM` in case the Genstat random number generator is already in use in the program that called `PERCENT`.\n\nDirectives: `COMBINE`, `TABLE`, `TABULATE`, `MARGIN`.\n\nProcedures: `MTABULATE`, `SVSTRATIFIED`, `SVTABULATE`, `TABMODE`, `TABSORT`, `T%CONTROL`.\n\nCommands for: Calculations and manipulation.\n\n### Example\n\n```CAPTION 'PERCENT example',\\\n'Data from the Guide to Genstat, Part 1, Section 4.11.1.';\nSTYLE=meta,plain\\\nVARIATE [NVALUES=15] Quantity,Charge\nFACTOR [NVALUES=15; LABELS=!T(A,B)] Type\n& [LABELS=!T(London,Manchester,Birmingham,Bristol)] Town\nLondon 10 A Manchester 5 B Birmingham 10 B Bristol 25 A\nManchester 10 * Birmingham 100 B London 200 B Manchester 25 A\nBristol 50 A Birmingham 25 A Bristol 25 B London 25 A\nLondon 50 B Manchester 25 B London 50 A :\n& Charge\n10 20 15 15 * 60 80 30 25 15 25 15 40 * * :\nCAPTION 'Form tables Totdisp and Payment, as in Section 4.11.1 of the Manual.'\nTABULATE [PRINT=totals; CLASSIFICATION=Town,Type] Quantity; TOTALS=Totdisp\n& [CLASSIFICATION=Town] Charge; TOTALS=Payment\nPERCENT Payment; NEWTABLE=%payment\nCAPTION !T('The amount to be paid on behalf of each each town is expressed',\\\n'as a percentage of the total of all the payments.')\nPRINT %payment; FIELDWIDTH=8; DECIMALS=2; MNAME=total\nPERCENT [CLASSIFICATION=Town; HUNDRED=yes] Totdisp; NEWTABLE=%disp\nCAPTION !T('The numbers dispatched to each town are expressed',\\\n'as percentages of the total to that town; the',\\\n'Town margin contains values of 100%.')\nPRINT %disp; FIELDWIDTH=12; DECIMALS=2; MNAME=total\n```\nUpdated on June 19, 2019" ]

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[ "# Weighted Average Cost Of Capital\n\n#### \\$0.28\n\n-0.01 (-3.41%)\n###### DCFDCF Levered\nShare price \\$ 0.28 1.560 73.21 -0.36 1.07% 11.168 2.52 20.50 23.02 10.93 89.07\n\nThere are a number of methods that can be used to determine discount rates. A good approach – and the one we’ll use in this tutorial – is to use the weighted average cost of capital (WACC) – a blend of the cost of equity and after-tax cost of debt. A company has two primary sources of financing – debt and equity – and, in simple terms, WACC is the average cost of raising that money. WACC is calculated by multiplying the cost of each capital source (debt and equity) by its relevant weight and then adding the products together to determine the WACC value." ]

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[ "index a2257ee..47fe878 100644 (file)\n#ifndef VQ_H\n#define VQ_H\n\n-/* Algebraic pulse-base quantiser. The signal x is replaced by the sum of the pitch\n-   a combination of pulses such that its norm is still equal to 1 */\n-void alg_quant(float *x, int N, int K, float *p);\n+#include \"entenc.h\"\n+#include \"entdec.h\"\n+\n\n-/* Improved algebraic pulse-base quantiser. The signal x is replaced by the sum of the pitch\n+/* Algebraic pulse-base quantiser. The signal x is replaced by the sum of the pitch\na combination of pulses such that its norm is still equal to 1. The only difference with\nthe quantiser above is that the search is more complete. */\n-int alg_quant2(float *x, int N, int K, float *p);\n+void alg_quant(float *x, int N, int K, float *p, ec_enc *enc);\n\n-/* Just replace the band with noise of unit energy */\n-void noise_quant(float *x, int N, int K, float *p);\n+void alg_unquant(float *x, int N, int K, float *p, ec_dec *dec);\n\n/* Finds the right offset into Y and copy it */\n-void copy_quant(float *x, int N, int K, float *Y, int B, int N0);\n+void copy_quant(float *x, int N, int K, float *Y, int B, int N0, ec_enc *enc);\n\n#endif /* VQ_H */" ]

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[ "# Quant Boosters - Hemant Yadav - Set 1\n\n• Say, its 6n + 1, then 2k + 1 = 12n + 3, so a multiple of 3. If its 6n - 1, then 4k + 1 = 24n - 3, so a multiple of 3. Hence, k is either 2 or 3. Only 3 satisfys. So, p = 13 is the only value for which all three numbers are prime.\n\n• Q12) \"A\" and \"B\" are playing series of repeated games. A player wins if he wins 2 games before other player. Probabilities of \"A\" winning, drawing and loosing a game are 1/2, 1/3 and 1/6. What is the probability of \"A\" winning the series.\n\n[OA : 27/32]\n\n• If P(n) denotes the probability of winning of A when no of wins of B is n, then\n\nP(0) = (1/2)² + 2(1/3)(1/2)² + 3(1/3)²(1/2)² + ....\n= (1/2)²{1 + 2/3 + 3/9 + 4/27 + ....)\n\nS = 1 + 2/3 + 3/9 + 4/27 + ...\nS/3 = 1/3 + 2/9 + 3/27 + ....\n=> 2S/3 = 1 + 1/3 + 1/9 + 1/27 + ....\n=> S = 9/4\nP(0) = (1/4)(9/4) = 9/16\n\nP(1) = (2!/(1!*1!))(1/6)(1/2)² + (3!/1!*1!*1!))(1/3)(1/6)(1/2)² + (4!/(2!*1!*1!)(1/3)²(1/6)(1/2)² + ....\n= (1/6)(1/2)²{2 + 6(1/3) + 12(1/3)² + 20(1/3)³ + 30(1/3)^4 + ... )\n\nS = 2 + 6(1/3) + 12(1/3)² + 20(1/3)³ + 30(1/3)^4 + ... .....(1)\n=> S/3 = 2/3 + 6(1/3)² + 12(1/3)³ + .......... (2)\n(1) - (2) => 2S/3 = 2 + 4/3 + 6/3² + 8/3³ + .... ............(3)\n=> 2S/9 = 2/3 + 4/3² + 6/3³ + ..... ...........(4)\n(3) - (4) => 4S/9 = 2 + 2/3 + 2/3² + 2/3³ + .... = 3\n\nP(1) = (1/6)(1/2)²(27/4)\n= 9/32\n\n=> Probability of A winning = 9/16 + 9/32 = 27/32\n\n• Q13) Find the smallest n for which rightmost non-zero digit of P(n) will be odd where P(n) = (n + 5)!/(n – 3)!\n\n• p(n) = (n + 5)!/(n - 3)!\n\nIt is clear that p(n) is a product of 8 consecutive numbers and divisible by atleast 8!. That means minimum power of 2 in p(n) is 7, i.e, p(n) is divisible by 2^7.\n\nBut last non-zero digit has to be an odd number, so p(n) should be divisible by 5^7.\n\nSo 15625 (5^6) or its multiple has to be one of the consective numbers, and we have to make sure that there is no multiple of 16 among the 8 numbers.\nMultiple of 16 nearest to 15625 are 15616 and 15632.\n\nSo, least value of n = 15620.\n\np(n) = 15618 * 15619 * 15620 * 15621 * 15622 * 15623 * 15624 * 15625\nAfter removing (2^7)*(5^7), we will get\n7809 * 15619 * 781 * 15621 * 7811 * 15623 * 1953 * 1\n\nUnit digit = unit digit of 9 * 9 * 1 * 1 * 1 * 3 * 3 * 1 = 9\n\n• Q14) How many real number solutions are there for a, b, c and d such that a² + b² + c² + d² = a(b + c + d)\n\n[OA : Only one solution]\n\n• a² + b² + c² + d² = a(b + c + d)\n=> a²/4 + (a²/4 + b² - ab) + (a²/4 + c² - ac) + (a²/4 + d² - ad) = 0\n=> a²/4 + (a/2 - b)² + (a/2 - c)² + (a/2 - d)² = 0\n=> a = b = c = d = 0 is the only solution\n\n• Q15) There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner. What is the probability that they don't collide?\n\n[OA : 1/4]\n\n• Each ants can go towards two direction. so total different possibilities are 2 * 2 * 2 = 8. Out of which 2 are favorable where they all goes clockwise or anticlockwise . Hence required probability = 1/4\n\n• Q16) A motor cyclist is travelling on a straight road at 70 kmph towards his house. His pet bird travels at 80 kmph from the motorcylist to the house and back in 1.6 hours. Find the distance between the house and the place from where the bird starts its journey towards the house?\n\n[OA : 120 km]\n\n• In 1.6 hours the bird flies 128 km and in the same time the motor cyclist goes 112 km and they travel a distance twice the the distance of the house. So distance is (128 + 112)/2 = 120 km\n\n• Let x = a, y + z = b\n\nwe get (x + y + z)^2006 + (x - y - z)^2006 = k1 * a^2006 * b^0 + k2 * a^2004 * b^2 + .... + k1004 * b^2006\n\n=> number of terms = sum of the number of terms in b^0, b^2, .... , b^2006\n= 1 + 3 + 5 + ... + 2007\n= 1004² = 1008016\n\n• a) boxes and balls are all distinct and only 1 ball in one box\nIts like permuting a 5 digit number, so 5! Ways\n\nb) boxes are same and balls are distict and only 1 ball per box\nSince all the boxes are same and one ball per box, so just one way\n\nc) balls are same and boxes are different and only 1 ball per box\nHere all the balls are same and every box has one ball, so again one way\n\nd) balls and boxes are same and only 1 ball per box\nHere also only only way\n\ne) repeat all above questions with any number of balls in any box\n\ni) 5^5 ways, as all the balls and all the boxes are different.\n\nii) Here boxes are same but balls are different.\n(5, 0, 0, 0, 0) – 1 way\n(4, 1, 0, 0, 0) – C(5, 4) = 5 ways\n(3, 2, 0, 0, 0) – C(5, 3) = 10 ways\n(3, 1, 1, 0, 0) – C(5, 3) = 10 ways, we don’t have to choose one of other two as they will automatically choose any of the two boxes (all the boxes are similar)\n(2, 2, 1, 0, 0) – C(5, 2)*C(3, 2)/2 = 15, here we need to divide by 2 as as it doesn’t matter if I put the first pair in box 1 and second pair in box 2 or vice-versa\n(2, 1, 1, 1, 0) – C(5, 2) = 10 ways\n(1, 1, 1, 1, 1) – 1 way\nSo, total = 1 + 5 + 10 + 10 + 15 + 10 + 1 = 52 ways\n\niii) x1 + x2 + x3 + x4 + x5 = 5\nSo, C(9, 5) = 126 ways\n\niv) Since all balls and boxes are same, we just need to find what different combinations are possible:-\n(5, 0, 0, 0, 0)\n(4, 1, 0, 0, 0)\n(3, 2, 0, 0, 0)\n(3, 1, 1, 0, 0)\n(2, 2, 1, 0, 0)\n(2, 1, 1, 1, 0)\n(1, 1, 1, 1, 1)\nSo, 7 ways\n\n• Q17) Find the largest possible value of k for which 3^11 is expressible as the sum of k consecutive positive integers. What will be the answer if 3^11 is expressed as sum of k consecutive integers.\n\n[OA : 486, 2 * 3^11]\n\n• Let the first of the consecutive integers be a + 1, then the consecutive integers will be a + 1, a + 2, ..., a + k\nSo, (a + 1) + (a + 2) + .... + (a + k) = 3^11\nk(k + 2a + 1) = 2 * 3^11\n\nNow for the first part all the integers are positive, so both (k + 2a + 1) > k\nHence maximum value of k will be 2 * 3^5 = 486\n\nFor the second part there is no such constraint, so maximum possible value of k will be 2 * 3^11\n\n• Q18) Find the sum of 6 * 1! + 13 * 2! + 22 * 3! + 33 * 4! + 46 * 5! + ... n terms\n\n• It is (n +2)! + 2[(n + 1)!] - 4 = (n + 4)[(n + 1)!] - 4.\nKey lies in simply finding the nth term of the sequence which is (n + 1)[(n + 1)!] + 2n[n!].\n\n• Q19) How many natural numbers N < 40000 such that cube of sum of digits of N gives back the same number N.\nFor eg, 8³ = 512, sum of digits is 8, same as cube root of 512.\n\n[OA : 6]\n\n• Since N < 40000, N < 35³\n\nIt is clear from the given condition that the sum of digits and cube of sum of digits should leave same remainder when divided by 9.\n\nHence sum of digits should be of form 9k + 1 or 9k - 1 or 9k\n\n1³ = 1 (sum of digits is 1, so possible)\n8³ = 512 (sum of digits is 8, so possible)\n9³ = 729 (sum of digits is 18, so not possible)\n10³ = 1000 (not possible)\n19³ = 6859 (sum of digit is not 19, so not possible)\n18³ = 5832 (sum of digits is 18, so possible)\n17³ = 4913 (sum of digit is 17, so possible)\n26³ = 17576 (possible)\n27³ = 19683 (possible)\n28³ = 21952 (not possible)\n\nSo, 1, 512, 4913, 5932, 17576 and 19683 are the only such numbers\n\n• Q20) How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form i, j, k where i, j and k are positive integers not exceeding four?\na) 60\nb) 72\nc) 64\nd) 76\n\n[OA : 76]\n\n61\n\n51\n\n89\n\n61\n\n48\n\n61\n\n62\n\n61" ]

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[ "# Waves/Waves in One Dimension\n\n(Redirected from Waves/Waves in one Dimension)\n\nWaves : 1 Dimensional Waves\n1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13\nExamples - Problems - Solutions - Terminology\n\n## The Mathematics Of Waves\n\nWe start our discussion of waves by taking the equation for a very simple wave and describing its characteristics. The basic equation for such a wave is\n\n$y=a\\ \\sin \\left({\\frac {2\\pi x}{\\lambda }}-2\\pi ft+\\alpha \\right)$", null, "where $y$", null, "is the height of the wave at position $x$", null, "and time $t$", null, ". This equation describes a fairly simple wave, but most complex waves are just sums of simpler ones. If we freeze this equation in time at $t=0$", null, ", we get\n\n$y=a\\ \\sin \\left({\\frac {2\\pi x}{\\lambda }}+\\alpha \\right)$", null, "which looks like this: [TODO - Add a Graph]\n\nFrom the graph we can see that each of the three parameters has a meaning. $a$", null, "is the amplitude of the wave, how high it is. $\\lambda$", null, "is the wavelength, the distance from a part of the wave in one cycle to the same part of the wave in the next cycle. $\\alpha$", null, "is the phase of the wave, which shifts the wave to the left or right. The wavelength is a distance, and is usually measured in meters, millimeters or even nanometers depending on the wave. Phase is an angle, measured in radians.\n\nNow that we have mapped out the wave in space, let's instead set $x=0$", null, "and see how the wave changes over time\n\n$y=a\\ \\sin(-2\\pi ft+\\alpha )$", null, "Amplitude $a$", null, "and phase $\\alpha$", null, "remain, but the wavelength is gone and a new quantity has appeared: $f$", null, ", which is the frequency, or how rapidly the wave moves up and down. Frequency is measured in units of inverse time: in a fixed period of time, how many times does the wave move up and down? The unit usually used for this is the hertz, or inverse second.\n\nNow let's combine these two pictures and see how the wave moves. Figure 3 is a diagram of how the wave looks when you plot it in both space and time. The straight lines are the places where the simple wave reaches a maximum, minimum, or zero (where it crosses the x axis).\n\nWe can look at the zeros to determine the phase velocity of the wave. The phase velocity is how fast a part of the wave moves. We can think of it as the speed of the wave, but for more complicated waves it is only one type of speed - more on that in later sections.\n\nWe can get an equation for the zeros by setting our equation to zero.\n\n$0=a\\ \\sin \\left({\\frac {2\\pi x}{\\lambda }}-2\\pi ft+\\alpha \\right)$", null, "$0={\\frac {2\\pi x}{\\lambda }}-2\\pi ft+\\alpha$", null, "$x=f\\lambda t-{\\frac {\\alpha \\lambda }{2\\pi }}$", null, "You see here that we have the equation for a straight line, describing a point that is moving at velocity $f\\lambda$", null, ". This gives us the equation for the phase velocity of the wave, which is\n\n${\\mbox{velocity}}={\\mbox{frequency}}\\times {\\mbox{wavelength}}\\quad v=f\\lambda$", null, "Waves : 1 Dimensional Waves\n1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13\nExamples - Problems - Solutions - Terminology" ]

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[ "## GARCH Models\n\nBenjamin Poignard (Osaka University)\n\n2018年度 第2回データ科学 ミニレクチャーシリーズ\n\n### GARCH Models\n\nBenjamin Poignard (Osaka University)\n\n11/27（火） 14:40~16:10\n12/11（火） 14:40~16:10\n1/8 （火） 14:40~16:10\n\nThis tutorial focuses on volatility modelling of financial returns. The objective of the tutorial is to present the main model specifications, to derive their probabilistic properties and to analyse the relevant inference methods. The key model of interest is the generalised autoregressive conditionally heteroscedastic (GARCH) process introduced by Engle. The probabilistic and statistical properties of this time series model will be studied and confronted with the stylised facts of the financial series. We will also focus on the multivariate framework of volatility modelling, called Multivariate GARCH: both probabilistic and statistical properties will be considered.\n\nTutorial 1: GARCH processes\nRisk modelling and/or option pricing often require the modelling of the variance of the underlying asset. Hence GARCH-like processes are often used in the financial industry. We consider the univariate specification of the conditional volatility GARCH(p,q) model and motivate its use through financial applications. We also focus on the conditions to derive its stationarity property and the estimation method based on the Gaussian Quasi Maximum Likelihood method.\n1. GARCH modelling for financial series.\n2. Stationarity of GARCH processes.\n3. Estimation of GARCH processes: QML estimator.\n\nTutorial 2: GARCH and stochastic volatility model\nStarting from the pros and cons of the GARCH process, we introduce the univariate stochastic volatility model together with the inference method. Then we consider the hidden-Markov - or Markov switching (MS) - model, which is a significant parameterization among the family of dynamic models with hidden variables. This modelling is intuitive and can easily be interpreted for financial time series.\n1. General stochastic volatility model.\n2. Estimation.\n3. Markov switching models.\n\nTutorial 3: Multivariate GARCH models\nThe multivariate setting is nowadays required when performing risk analysis, where the objective is to provide a prediction of the variance/covariance matrix of a portfolio of diversified assets. To do so, we study the framework of multivariate GARCH model(MGARCH) and consider several variance covariance and correlation matrix processes. We also consider the probabilistic properties of such dynamics and the asymptotic properties of the Constant Conditional Correlation model will be derived.\n1. MGARCH processes: VEC-GARCH, BEKK, DCC.\n2. Stationarity of VEC-GARCH, BEKK and CCC models.\n3. Estimation of CCC model: analysis of the QML estimator." ]

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[ "# class Hash(K, V)\n\n## Overview\n\nA `Hash` represents a mapping of keys to values.\n\nSee the official docs for the basics.\n\nhash.cr\njson/to_json.cr\nyaml/to_yaml.cr\n\n## Constructor Detail\n\ndef self.new(block : Hash(K, V), K -> V? = nil, initial_capacity = nil) #\n\ndef self.new(ctx : YAML::ParseContext, node : YAML::Nodes::Node, &block) #\n\ndef self.new(initial_capacity = nil, &block : Hash(K, V), K -> V) #\n\ndef self.new(default_value : V, initial_capacity = nil) #\n\ndef self.new(ctx : YAML::ParseContext, node : YAML::Nodes::Node) #\n\ndef self.new(pull : JSON::PullParser) #\n\n## Class Method Detail\n\ndef self.zip(ary1 : Array(K), ary2 : Array(V)) #\n\nZips two arrays into a `Hash`, taking keys from ary1 and values from ary2.\n\n``````Hash.zip([\"key1\", \"key2\", \"key3\"], [\"value1\", \"value2\", \"value3\"])\n# => {\"key1\" => \"value1\", \"key2\" => \"value2\", \"key3\" => \"value3\"}``````\n\n## Instance Method Detail\n\ndef ==(other : Hash) #\n\nCompares with other. Returns `true` if all key-value pairs are the same.\n\ndef [](key) #\n\nSee also: `Hash#fetch`.\n\ndef []=(key : K, value : V) #\n\nSets the value of key to the given value.\n\n``````h = {} of String => String\nh[\"foo\"] = \"bar\"\nh[\"foo\"] # => \"bar\"``````\n\ndef []?(key) #\n\nReturns the value for the key given by key. If not found, returns `nil`. This ignores the default value set by `Hash.new`.\n\n``````h = {\"foo\" => \"bar\"}\nh[\"foo\"]? # => \"bar\"\nh[\"bar\"]? # => nil\n\nh = Hash(String, String).new(\"bar\")\nh[\"foo\"]? # => nil``````\n\ndef clear #\n\nEmpties a `Hash` and returns it.\n\n``````hash = {\"foo\" => \"bar\"}\nhash.clear # => {}``````\n\ndef clone #\n\nSimilar to `#dup`, but duplicates the values as well.\n\n``````hash_a = {\"foobar\" => {\"foo\" => \"bar\"}}\nhash_b = hash_a.clone\nhash_b[\"foobar\"][\"foo\"] = \"baz\"\nhash_a # => {\"foobar\" => {\"foo\" => \"bar\"}}``````\n\ndef compact #\n\nReturns new `Hash` without `nil` values.\n\n``````hash = {\"hello\" => \"world\", \"foo\" => nil}\nhash.compact # => {\"hello\" => \"world\"}``````\n\ndef compact! #\n\nRemoves all `nil` value from `self`. Returns `nil` if no changes were made.\n\n``````hash = {\"hello\" => \"world\", \"foo\" => nil}\nhash.compact! # => {\"hello\" => \"world\"}\nhash.compact! # => nil``````\n\ndef delete(key) #\n\nDeletes the key-value pair and returns the value, otherwise returns `nil`.\n\n``````h = {\"foo\" => \"bar\"}\nh.delete(\"foo\") # => \"bar\"\nh.fetch(\"foo\", nil) # => nil``````\n\ndef delete(key, &block) #\n\nDeletes the key-value pair and returns the value, else yields key with given block.\n\n``````h = {\"foo\" => \"bar\"}\nh.fetch(\"foo\", nil) # => nil\n\ndef delete_if(&block) #\n\nDeletes each key-value pair for which the given block returns `true`.\n\n``````h = {\"foo\" => \"bar\", \"fob\" => \"baz\", \"bar\" => \"qux\"}\nh.delete_if { |key, value| key.starts_with?(\"fo\") }\nh # => { \"bar\" => \"qux\" }``````\n\ndef dup #\n\nDuplicates a `Hash`.\n\n``````hash_a = {\"foo\" => \"bar\"}\nhash_b = hash_a.dup\nhash_b.merge!({\"baz\" => \"qux\"})\nhash_a # => {\"foo\" => \"bar\"}``````\n\ndef each #\n\nReturns an iterator over the hash entries. Which behaves like an `Iterator` returning a `Tuple` consisting of the key and value types.\n\n``````hsh = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\niterator = hsh.each\n\niterator.next # => {\"foo\", \"bar\"}\niterator.next # => {\"baz\", \"qux\"}``````\n\ndef each(&block) : Nil #\n\nCalls the given block for each key-value pair and passes in the key and the value.\n\n``````h = {\"foo\" => \"bar\"}\n\nh.each do |key, value|\nkey # => \"foo\"\nvalue # => \"bar\"\nend\n\nh.each do |key_and_value|\nkey_and_value # => {\"foo\", \"bar\"}\nend``````\n\ndef each_key #\n\nReturns an iterator over the hash keys. Which behaves like an `Iterator` consisting of the key's types.\n\n``````hsh = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\niterator = hsh.each_key\n\nkey = iterator.next\nkey # => \"foo\"\n\nkey = iterator.next\nkey # => \"baz\"``````\n\ndef each_key(&block) #\n\nCalls the given block for each key-value pair and passes in the key.\n\n``````h = {\"foo\" => \"bar\"}\nh.each_key do |key|\nkey # => \"foo\"\nend``````\n\ndef each_value(&block) #\n\nCalls the given block for each key-value pair and passes in the value.\n\n``````h = {\"foo\" => \"bar\"}\nh.each_value do |value|\nvalue # => \"bar\"\nend``````\n\ndef each_value #\n\nReturns an iterator over the hash values. Which behaves like an `Iterator` consisting of the value's types.\n\n``````hsh = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\niterator = hsh.each_value\n\nvalue = iterator.next\nvalue # => \"bar\"\n\nvalue = iterator.next\nvalue # => \"qux\"``````\n\ndef empty? #\n\nReturns `true` when hash contains no key-value pairs.\n\n``````h = Hash(String, String).new\nh.empty? # => true\n\nh = {\"foo\" => \"bar\"}\nh.empty? # => false``````\n\ndef fetch(key, &block) #\n\nReturns the value for the key given by key, or when not found calls the given block with the key.\n\n``````h = {\"foo\" => \"bar\"}\nh.fetch(\"foo\") { |key| key.upcase } # => \"bar\"\nh.fetch(\"bar\") { |key| key.upcase } # => \"BAR\"``````\n\ndef fetch(key, default) #\n\nReturns the value for the key given by key, or when not found the value given by default. This ignores the default value set by `Hash.new`.\n\n``````h = {\"foo\" => \"bar\"}\nh.fetch(\"foo\", \"foo\") # => \"bar\"\nh.fetch(\"bar\", \"foo\") # => \"foo\"``````\n\ndef fetch(key) #\n\nReturns the value for the key given by key. If not found, returns the default value given by `Hash.new`, otherwise raises `KeyError`.\n\n``````h = {\"foo\" => \"bar\"}\nh[\"foo\"] # => \"bar\"\n\nh = Hash(String, String).new(\"bar\")\nh[\"foo\"] # => \"bar\"\n\nh = Hash(String, String).new { \"bar\" }\nh[\"foo\"] # => \"bar\"\n\nh = Hash(String, String).new\nh[\"foo\"] # raises KeyError``````\n\ndef first_key #\n\nReturns the first key in the hash.\n\ndef first_key? #\n\nReturns the first key if it exists, or returns `nil`.\n\n``````hash = {\"foo\" => \"bar\"}\nhash.first_key? # => \"foo\"\nhash.clear\nhash.first_key? # => nil``````\n\ndef first_value #\n\nReturns the first value in the hash.\n\ndef first_value? #\n\nSimilar to `#first_key?`, but returns its value.\n\ndef has_key?(key) #\n\nReturns `true` when key given by key exists, otherwise `false`.\n\n``````h = {\"foo\" => \"bar\"}\nh.has_key?(\"foo\") # => true\nh.has_key?(\"bar\") # => false``````\n\ndef has_value?(val) #\n\nReturns `true` when value given by value exists, otherwise `false`.\n\n``````h = {\"foo\" => \"bar\"}\nh.has_value?(\"foo\") # => false\nh.has_value?(\"bar\") # => true``````\n\ndef hash(hasher) #\n\nSee `Object#hash(hasher)`\n\ndef inspect(io : IO) #\n\ndef invert #\n\nInverts keys and values. If there are duplicated values, the last key becomes the new value.\n\n``````{\"foo\" => \"bar\"}.invert # => {\"bar\" => \"foo\"}\n{\"foo\" => \"bar\", \"baz\" => \"bar\"}.invert # => {\"bar\" => \"baz\"}``````\n\ndef key(value) #\n\nReturns the first key with the given value, else raises `KeyError`.\n\n``````hash = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\nhash.key(\"bar\") # => \"foo\"\nhash.key(\"qux\") # => \"baz\"\nhash.key(\"foobar\") # raises KeyError (Missing hash key for value: foobar)``````\n\ndef key(value, &block) #\n\nReturns the first key with the given value, else yields value with the given block.\n\n``````hash = {\"foo\" => \"bar\"}\nhash.key(\"bar\") { |value| value.upcase } # => \"foo\"\nhash.key(\"qux\") { |value| value.upcase } # => \"QUX\"``````\n\ndef key?(value) #\n\nReturns the first key with the given value, else `nil`.\n\n``````hash = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\nhash.key?(\"bar\") # => \"foo\"\nhash.key?(\"qux\") # => \"baz\"\nhash.key?(\"foobar\") # => nil``````\n\ndef key_index(key) #\n\nReturns the index of the given key, or `nil` when not found. The keys are ordered based on when they were inserted.\n\n``````h = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\nh.key_index(\"foo\") # => 0\nh.key_index(\"qux\") # => nil``````\n\ndef keys #\n\nReturns a new `Array` with all the keys.\n\n``````h = {\"foo\" => \"bar\", \"baz\" => \"bar\"}\nh.keys # => [\"foo\", \"baz\"]``````\n\ndef merge(other : Hash(L, W)) forall L, W #\n\nReturns a new `Hash` with the keys and values of this hash and other combined. A value in other takes precedence over the one in this hash.\n\n``````hash = {\"foo\" => \"bar\"}\nhash.merge({\"baz\" => \"qux\"})\n# => {\"foo\" => \"bar\", \"baz\" => \"qux\"}\nhash\n# => {\"foo\" => \"bar\"}``````\n\ndef merge(other : Hash(L, W), &block : K, V, W -> V | W) forall L, W #\n\ndef merge!(other : Hash) #\n\nSimilar to `#merge`, but the receiver is modified.\n\n``````hash = {\"foo\" => \"bar\"}\nhash.merge!({\"baz\" => \"qux\"})\nhash # => {\"foo\" => \"bar\", \"baz\" => \"qux\"}``````\n\ndef merge!(other : Hash, &block) #\n\ndef pretty_print(pp) : Nil #\n\ndef rehash #\n\ndef reject(*keys) #\n\nReturns a new `Hash` without the given keys.\n\n``{\"a\" => 1, \"b\" => 2, \"c\" => 3, \"d\" => 4}.reject(\"a\", \"c\") # => {\"b\" => 2, \"d\" => 4}``\n\ndef reject(&block : K, V -> _) #\n\nReturns a new hash consisting of entries for which the block returns `false`.\n\n``````h = {\"a\" => 100, \"b\" => 200, \"c\" => 300}\nh.reject { |k, v| k > \"a\" } # => {\"a\" => 100}\nh.reject { |k, v| v < 200 } # => {\"b\" => 200, \"c\" => 300}``````\n\ndef reject!(&block : K, V -> _) #\n\nEquivalent to `Hash#reject`, but makes modification on the current object rather that returning a new one. Returns `nil` if no changes were made.\n\ndef reject!(keys : Array | Tuple) #\n\nRemoves a list of keys out of hash.\n\n``````h = {\"a\" => 1, \"b\" => 2, \"c\" => 3, \"d\" => 4}.reject!(\"a\", \"c\")\nh # => {\"b\" => 2, \"d\" => 4}``````\n\ndef reject!(*keys) #\n\ndef select(&block : K, V -> _) #\n\nReturns a new hash consisting of entries for which the block returns `true`.\n\n``````h = {\"a\" => 100, \"b\" => 200, \"c\" => 300}\nh.select { |k, v| k > \"a\" } # => {\"b\" => 200, \"c\" => 300}\nh.select { |k, v| v < 200 } # => {\"a\" => 100}``````\n\ndef select(keys : Array | Tuple) #\n\nReturns a new `Hash` with the given keys.\n\n``{\"a\" => 1, \"b\" => 2, \"c\" => 3, \"d\" => 4}.select(\"a\", \"c\") # => {\"a\" => 1, \"c\" => 3}``\n\ndef select(*keys) #\n\ndef select!(&block : K, V -> _) #\n\nEquivalent to `Hash#select` but makes modification on the current object rather that returning a new one. Returns `nil` if no changes were made\n\ndef select!(keys : Array | Tuple) #\n\nRemoves every element except the given ones.\n\n``````h = {\"a\" => 1, \"b\" => 2, \"c\" => 3, \"d\" => 4}.select!(\"a\", \"c\")\nh # => {\"a\" => 1, \"c\" => 3}``````\n\ndef select!(*keys) #\n\ndef shift(&block) #\n\nDeletes and returns the first key-value pair in the hash. Yields to the given block if the hash is empty.\n\n``````hash = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\nhash.shift { true } # => {\"foo\", \"bar\"}\nhash # => {\"baz\" => \"qux\"}\n\nhash = {} of String => String\nhash.shift { true } # => true\nhash # => {}``````\n\ndef shift #\n\nDeletes and returns the first key-value pair in the hash, or raises `IndexError` if the hash is empty.\n\n``````hash = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\nhash.shift # => {\"foo\", \"bar\"}\nhash # => {\"baz\" => \"qux\"}\n\nhash = {} of String => String\nhash.shift # raises IndexError``````\n\ndef shift? #\n\nSame as `#shift`, but returns `nil` if the hash is empty.\n\n``````hash = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\nhash.shift? # => {\"foo\", \"bar\"}\nhash # => {\"baz\" => \"qux\"}\n\nhash = {} of String => String\nhash.shift? # => nil``````\n\ndef size : Int32 #\n\ndef to_h #\n\nReturns `self`.\n\ndef to_json(json : JSON::Builder) #\n\ndef to_s(io : IO) #\n\nConverts to a `String`.\n\n``````h = {\"foo\" => \"bar\"}\nh.to_s # => \"{\\\"foo\\\" => \\\"bar\\\"}\"\nh.to_s.class # => String``````\n\ndef to_yaml(yaml : YAML::Nodes::Builder) #\n\ndef values #\n\nReturns only the values as an `Array`.\n\n``````h = {\"foo\" => \"bar\", \"baz\" => \"qux\"}\nh.values # => [\"bar\", \"qux\"]``````\n\ndef values_at(*indexes : K) #\n\nReturns a tuple populated with the elements at the given indexes. Raises if any index is invalid.\n\n``{\"a\" => 1, \"b\" => 2, \"c\" => 3, \"d\" => 4}.values_at(\"a\", \"c\") # => {1, 3}``" ]

[ null ]

[ "Gold Member\nI think I got these right, but I'd feel better if someone could tell me if I did them correctly:\n\nA projectile is launched from ground level at a 45 degree angle. How much work does gravity do on the projectile between its launch and when it hits the ground?\n\nGravity will do as much work on the object as the work performed by the\ny-component of the force that launched the projectile.\n$$W_{gravity} =\\Delta K_{gravity} \\] $\\Delta K_{gravity} =\\frac{1}{2}mv_y^2$ $\\Delta K_{gravity} =\\frac{1}{2}m\\left( {\\sin (45)v} \\right)^2$ $W_{gravity} =\\frac{1}{2}m\\left( {\\sin (45)v} \\right)^2$$ A hockey player pushes a puck of mass 0.50 kg across the ice using a constant force of 10.0 N over a distance of 0.50 m. How much work does the hockey player do? If the puck was initially stationary, what is its final speed? (Ignore friction.) $$W=Fd, W=10N*0.5m, W=5N$$ $$F=ma \\quad \\Rightarrow \\quad a=\\frac{F}{m}$ $a=\\frac{10.0N}{0.50kg} \\quad \\Rightarrow \\quad a=\\frac{10.0\\rlap{--} {k}\\rlap{--} {g}\\cdot m/s^2}{0.50\\rlap{--} {k}\\rlap{--} {g}}$ $a=20m/s^2$ $v_f^2 =v_i^2 +2a\\Delta d \\quad \\Rightarrow \\quad v=\\sqrt {v_i^2 +2a\\Delta d}$ $v=\\sqrt {\\left( {2\\cdot 20\\frac{m}{s^2}} \\right)\\cdot 0.50m}$ $v=4.5m/s$$ If a constant force F=(30.N)i, + (50.N)j acts on a particle that undergoes a displacement (4.0m)i + (1.0m)j , how much work is done on the particle? $$W=Fd$ $W=\\sqrt {\\left( {3.0N} \\right)^2+\\left( {5.0N} \\right)^2} \\cdot \\sqrt {\\left( {4.0m} \\right)^2+\\left( {1.0m} \\right)^2}$ $W=20J$$ ## Answers and Replies tony873004 said: I think I got these right, but I'd feel better if someone could tell me if I did them correctly: A projectile is launched from ground level at a 45 degree angle. How much work does gravity do on the projectile between its launch and when it hits the ground? Gravity will do as much work on the object as the work performed by the y-component of the force that launched the projectile. $$W_{gravity} =\\Delta K_{gravity}$ $\\Delta K_{gravity} =\\frac{1}{2}mv_y^2$ $\\Delta K_{gravity} =\\frac{1}{2}m\\left( {\\sin (45)v} \\right)^2$ $W_{gravity} =\\frac{1}{2}m\\left( {\\sin (45)v} \\right)^2$$ This is all right. A hockey player pushes a puck of mass 0.50 kg across the ice using a constant force of 10.0 N over a distance of 0.50 m. How much work does the hockey player do? If the puck was initially stationary, what is its final speed? (Ignore friction.) $$W=Fd, W=10N*0.5m, W=5N$$ $$F=ma \\quad \\Rightarrow \\quad a=\\frac{F}{m}$ $a=\\frac{10.0N}{0.50kg} \\quad \\Rightarrow \\quad a=\\frac{10.0\\rlap{--} {k}\\rlap{--} {g}\\cdot m/s^2}{0.50\\rlap{--} {k}\\rlap{--} {g}}$ $a=20m/s^2$ $v_f^2 =v_i^2 +2a\\Delta d \\quad \\Rightarrow \\quad v=\\sqrt {v_i^2 +2a\\Delta d}$ $v=\\sqrt {\\left( {2\\cdot 20\\frac{m}{s^2}} \\right)\\cdot 0.50m}$ $v=4.5m/s$$ Correct, slight rounding error but thats cool. If a constant force F=(30.N)i, + (50.N)j acts on a particle that undergoes a displacement (4.0m)i + (1.0m)j , how much work is done on the particle? $$W=Fd$ $W=\\sqrt {\\left( {3.0N} \\right)^2+\\left( {5.0N} \\right)^2} \\cdot \\sqrt {\\left( {4.0m} \\right)^2+\\left( {1.0m} \\right)^2}$ $W=20J$$ W is actually the dot product between F and D. Dot the two vectors. The dot product of two vectors a and b in two dimensions is: $$\\vec{a} \\bullet \\vec{b} = a_xb_x+a_yb_y$$ Also I dont know if its a typo, but in your square root you have magnitudes 5.0N and 3.0N for the force, but in the problem they are 50N and 30N. Science Advisor Gold Member Thanks, whozum.", null, "That is a typo. 3.0 should be in the question, not 30. 50 too. Isn't d a displacement, and therefore a scalar, not a vector? So the dot product wouldn't apply and it's just straight forward multiplication? The rounding error, I actually got 4.4721, but rounded t 4.5 because the inputs were all 2 significant figures. Is that what you meant by rounding error, or did I make another mistake? :yuck: tony873004 said: Thanks, whozum.", null, "That is a typo. 3.0 should be in the question, not 30. 50 too. Isn't d a displacement, and therefore a scalar, not a vector? So the dot product wouldn't apply and it's just straight forward multiplication? I just use D because thats what I did in algebra based physics back in high school, it hasnt left me since. The correct equation for work is: $$W = \\int_{x_1}^{x_2}{F}\\bullet{dx} = \\int_{x_1}^{x_2}{Fcos(\\theta){dx}$$ Only the component of force parallel to the path will contribute to the amount of work done. The dot product is necessary. Given a straight path dx which is always parallel (theta = 0) to the (assumed constant) force: $$W = \\int_{x_1}^{x_2}{F\\bullet{dx} = \\int_{x_1}^{x_2}{Fcos(0){dx} = \\int_{x_1}^{x_2}{F}{dx} = Fx]_{x_1}^{x_2}$$ The rounding error, I actually got 4.4721, but rounded t 4.5 because the inputs were all 2 significant figures. Is that what you meant by rounding error, or did I make another mistake? :yuck: Thats what I meant. Last edited: Science Advisor Gold Member Thanks a lot whozum. I'll switch that to the dot product. Our book is picky about significant figures. I've come up with answers like 14.57 m/s, and turn to the back to see the answer given as 10 m/s. Luckily, the homework grader does not care about sig figs. Doc Al Mentor tony873004 said: A projectile is launched from ground level at a 45 degree angle. How much work does gravity do on the projectile between its launch and when it hits the ground? Gravity will do as much work on the object as the work performed by the y-component of the force that launched the projectile. $$W_{gravity} =\\Delta K_{gravity}$ $\\Delta K_{gravity} =\\frac{1}{2}mv_y^2$ $\\Delta K_{gravity} =\\frac{1}{2}m\\left( {\\sin (45)v} \\right)^2$ $W_{gravity} =\\frac{1}{2}m\\left( {\\sin (45)v} \\right)^2$$ Careful with this one. The work done by gravity (which is the only force acting on the projectile) equals the change in KE. When the the projectile returns to ground, how has its KE changed? So what's the net work done by gravity? A hockey player pushes a puck of mass 0.50 kg across the ice using a constant force of 10.0 N over a distance of 0.50 m. How much work does the hockey player do? If the puck was initially stationary, what is its final speed? (Ignore friction.) $$W=Fd, W=10N*0.5m, W=5N$$ $$F=ma \\quad \\Rightarrow \\quad a=\\frac{F}{m}$ $a=\\frac{10.0N}{0.50kg} \\quad \\Rightarrow \\quad a=\\frac{10.0\\rlap{--} {k}\\rlap{--} {g}\\cdot m/s^2}{0.50\\rlap{--} {k}\\rlap{--} {g}}$ $a=20m/s^2$ $v_f^2 =v_i^2 +2a\\Delta d \\quad \\Rightarrow \\quad v=\\sqrt {v_i^2 +2a\\Delta d}$ $v=\\sqrt {\\left( {2\\cdot 20\\frac{m}{s^2}} \\right)\\cdot 0.50m}$ $v=4.5m/s$$ OK. Two comments. (1) Work is measured in Joules, not Newtons. (2) Why didn't you use $\\mbox{Work} = \\Delta \\mbox{Kinetic Energy} = 1/2 m v^2$ to solve for the speed; since you found the work, why not use it? If a constant force F=(30.N)i, + (50.N)j acts on a particle that undergoes a displacement (4.0m)i + (1.0m)j , how much work is done on the particle? $$W=Fd$ $W=\\sqrt {\\left( {3.0N} \\right)^2+\\left( {5.0N} \\right)^2} \\cdot \\sqrt {\\left( {4.0m} \\right)^2+\\left( {1.0m} \\right)^2}$ \\[ W=20J$$\nThis is incorrect, as it assumes that the force and displacement are parallel, which is not the case. Take the dot product of the two vectors. (Yes, displacement is a vector.) Since the force is constant there is no need for any integration." ]

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[ "", null, "", null, "Home", null, "A Summary of Factoring Polynomials", null, "Factoring The Difference of 2 Squares", null, "Factoring Trinomials", null, "Quadratic Expressions", null, "Factoring Trinomials", null, "The 7 Forms of Factoring", null, "Factoring Trinomials", null, "Finding The Greatest Common Factor (GCF)", null, "Factoring Trinomials", null, "Quadratic Expressions", null, "Factoring simple expressions", null, "Polynomials", null, "Factoring Polynomials", null, "Fractoring Polynomials", null, "Other Math Resources", null, "Factoring Polynomials", null, "Polynomials", null, "Finding the Greatest Common Factor (GCF)", null, "Factoring Trinomials", null, "Finding the Least Common Multiples\nTry the Free Math Solver or Scroll down to Tutorials!\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\n# show my work math?\n\nHere is a number of phrases that our users typed in recently in order to reach website.\n\nHow can this be helpful to you?\n\n• find the search keyword you are searching for (i.e. show my work math) in the leftmost column below\n\n• Click on the appropriate program demo button found in the same row  as your search phrase\n\n• If you find the software demo of help click on the buy button to buy the software at a special price extended to factoring-polynomials.com website visitors\n\n Related Search Phrase Algebrator Flash Demo Algebrator Static html Demo Buy now Free Online Algebra Solver", null, "", null, "", null, "math for dummies - calculating compound interest", null, "", null, "", null, "begining algebra worksheets", null, "", null, "", null, "algebra 2 answer", null, "", null, "", null, "math equasion solver", null, "", null, "", null, "GCF finder", null, "", null, "", null, "online fraction and percents worksheets", null, "", null, "", null, "integration calculator online free", null, "", null, "", null, "depreciation for algebra 2", null, "", null, "", null, "CLEP aLGEBRA TEST IN GEORGIA", null, "", null, "", null, "quadratic equation for TI-83", null, "", null, "", null, "solutions to dummit and foote", null, "", null, "", null, "solving system of quadratic equations with matlab", null, "", null, "", null, "log2 ti-83 plus", null, "", null, "", null, "When solving a rational equation, why it is OK to remove the denominator by multiplying both sides by the LCD", null, "", null, "", null, "multiplying different terms", null, "", null, "", null, "negative exponants", null, "", null, "", null, "ti-83 complex roots", null, "", null, "", null, "worksheets for integers adding and subtracting", null, "", null, "", null, "\"Algebra solver\"", null, "", null, "", null, "adding and subtracting fractions worksheet", null, "", null, "", null, "sample word problems for hands on algebra", null, "", null, "", null, "Prev Next" ]

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[ "## The Annals of Probability\n\n### Embedding laws in diffusions by functions of time\n\n#### Abstract\n\nWe present a constructive probabilistic proof of the fact that if $B=(B_{t})_{t\\ge0}$ is standard Brownian motion started at $0$, and $\\mu$ is a given probability measure on $\\mathbb{R}$ such that $\\mu(\\{0\\})=0$, then there exists a unique left-continuous increasing function $b:(0,\\infty)\\rightarrow\\mathbb{R}\\cup\\{+\\infty\\}$ and a unique left-continuous decreasing function $c:(0,\\infty)\\rightarrow\\mathbb{R}\\cup\\{-\\infty\\}$ such that $B$ stopped at $\\tau_{b,c}=\\inf\\{t>0\\vert B_{t}\\ge b(t)\\mbox{ or }B_{t}\\le c(t)\\}$ has the law $\\mu$. The method of proof relies upon weak convergence arguments arising from Helly’s selection theorem and makes use of the Lévy metric which appears to be novel in the context of embedding theorems. We show that $\\tau_{b,c}$ is minimal in the sense of Monroe so that the stopped process $B^{\\tau_{b,c}}=(B_{t\\wedge\\tau_{b,c}})_{t\\ge0}$ satisfies natural uniform integrability conditions expressed in terms of $\\mu$. We also show that $\\tau_{b,c}$ has the smallest truncated expectation among all stopping times that embed $\\mu$ into $B$. The main results extend from standard Brownian motion to all recurrent diffusion processes on the real line.\n\n#### Article information\n\nSource\nAnn. Probab., Volume 43, Number 5 (2015), 2481-2510.\n\nDates\nRevised: May 2014\nFirst available in Project Euclid: 9 September 2015\n\nhttps://projecteuclid.org/euclid.aop/1441792291\n\nDigital Object Identifier\ndoi:10.1214/14-AOP941\n\nMathematical Reviews number (MathSciNet)\nMR3395467\n\nZentralblatt MATH identifier\n1335.60150\n\n#### Citation\n\nCox, A. M. G.; Peskir, G. Embedding laws in diffusions by functions of time. Ann. Probab. 43 (2015), no. 5, 2481--2510. doi:10.1214/14-AOP941. https://projecteuclid.org/euclid.aop/1441792291" ]

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