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[ "# 实习生招聘笔试\n\nhttp://www.cnblogs.com/jerry19880126/\n\nA、3                 B、4\nC、5                       D、6\n\nhttp://blog.csdn.net/kingjinzi_2008/article/details/7785334\n\n* x^二，首回乘法：原式=x^2 *\n(x^4+肆x^贰+二x)+x+1，每壹项的周到能够使用加法来促成。。\n\nA、3                 B、4                  C、5\nD、6\n\nInt\na1=x+y-z; int b1=x*y/z;\n\nA。原式=x^2 * (x^4 + 4 * x^2 + 2*x) + x +\n\nInt\na2=x-z+y; int b2=x/z*y;\n\nInt\nc1=x<<y>>z; int d1=x&y|z;\n\n2、给定2个int类型的正整数x，y，z，对如下4组表明式判定正确的精选（）\n\nInt\nc2=x>>z<<y; int d2=x|z&y;\n\nint a1=x+y-z; int b1=x*y/z;\n\nA、a一必将等于a二\n\nint a2=x-z+y; int b2=x/z*y;\n\nB、b一一定定于b二\n\nint c1=x<<y>>z; int d1=x&y|z;\n\nC、c一毫无疑问等于c2\n\nint c2=x>>z<<y; int d2=x|z&y;\n\nD、d壹一定等于d贰\n\nA、a一势必等于a二\n\nB、b一自然定于b二\n\nA、死代码删除指的是编写翻译进度平素丢掉掉被讲解的代码；\n\nC、c1毫无疑问等于c2\n\nB、函数内联能够免止函数调用中压栈和退栈的费用\n\nD、d壹一定等于d二\n\nC、For循环的轮回调控变量平时很适合调解到寄存器访问\n\nA。\n\nD、强度削弱是指推行时间非常短的一声令下等价的代表试行时间较长的下令\n\n4、\n\nA、死代码删除指的是编写翻译进度一贯屏弃掉被讲授的代码；\n\nA、进度在退出时会自动关闭本身张开的有所文件\n\nB、函数内联能够幸免函数调用中压栈和退栈的费用\n\nB、进度在退出时会自动关闭自个儿展开的网络链接\n\nC、For循环的轮回调整变量通常很吻合调整到寄存器访问\n\nC、进度在剥离时会自动销毁自个儿创造的有所线程\n\nD、强度减弱是指试行时间很短的吩咐等价的代表实践时间较长的一声令下\n\nD、进程在脱离时会自动销毁本人展开的共享内部存储器\n\nA。死代码是指长久不会执行到的代码，不是注释，比方if(0){…}，大括号里的正是死代码。\n\n5、", null, "A、进度在脱离时会自动关闭自身展开的有着文件\n\nA、492\n\nB、进度在脱离时会自动关闭自身张开的网络链接\n\nB、494\n\nC、进程在脱离时会自动销毁自身成立的享有线程\n\nC、496\n\nD、进程在脱离时会自动销毁本人张开的共享内部存款和储蓄器\n\nD、498\n\nD。共享内部存储器销毁了，会对其余正在采纳这段内存的进程变成破坏。\n\nA、DROP\nTABLE\n\n5、在如下8*6的矩阵中，请总计从A移动到B壹共有多少种走法？须求每一遍只可以前进挥着向右移动1格，并且无法经过P；\n\nB、DELETE\nTABLE", null, "C、DESTROY\nTABLE\n\nA、492\n\nD、REMOVE\nTABLE\n\nB、494\n\n7、某制品团队由雕塑组、产品组、client程序组和server程序组几个小组构成，每一趟塑造一套完整的版本时，供给种种组揭露如下财富。版画组想客户端提供图像能源（供给10分钟），产品组向client组合server提供文字内容财富（同有的时候间开始展览，拾秒钟），server和client源代码放置在分化专业站上，其总体编写翻译时间均为10分钟切编写翻译进度不依赖于其余能源，client程序（不包括别的财富）在编写翻译达成后还索要形成对程序的集结加密进度（10分钟）。能够请问，从要到位贰次版本创设（client与server的版本代码与能源齐备），至少须要多少日子（）\n\nC、496\n\nA、60分钟\n\nD、498\n\nB、40分钟\n\nA。实际上是排列组合难点。A走到B共要求12步，当中7步必须向右，五步必须更上壹层楼，但顺序能够分化，由此是C(七,12)，必要P不能够走，那么走到P的或是次数是C(叁,六)，从P走到B的可能次数是C(四,陆)，由此结果是C(七,12)\n– C(3,陆)*C(4,6)=492。\n\nC、30分钟\n\nD、20分钟\n\nA、DROP TABLE\n\n8、如下关于编写翻译链接的说法破绽百出的是（）\n\nB、DELETE TABLE\n\nA、编写翻译优化会使得编写翻译速度变慢\n\nC、DESTROY TABLE\n\nB、预编写翻译头文件能够优化程序的性质\n\nD、REMOVE TABLE\n\nC、静态链接会使得可奉行文件偏大\n\nA。不说了，\n\nD、动态链接库会使进度运营速度偏慢\n\n9、如下关于链接的传教指鹿为马的是（）\n\nA、60分钟\n\nA、3个静态库中不可能包含五个同名全局函数的定义\n\nB、40分钟\n\nB、七个动态库中无法包含八个同名全局函数的概念\n\nC、30分钟\n\nC、若是八个静态库都富含三个同名全局函数，他们不能同有时间被链接\n\nD、20分钟\n\nD、纵然多少个动态库都蕴含3个同名全局函数，他们无法同期被链接\n\nD。除了加密以外，剩下的专门的职业在率先个十分钟内足以并发到位。\n\n10、排序算法的嘻嘻哈哈是指，关键码同样的笔录排序前后相对地点不发生改换，上面哪个种类排序算法是不平稳的（）\n\n8、如下关于编译链接的说教破绽百出的是（）\n\nA、插入排序\n\nA、编写翻译优化会使得编写翻译速度变慢\n\nB、冒泡排序\n\nB、预编写翻译头文件能够优化程序的本性\n\nC、急忙排序\n\nC、静态链接会使得可施行文件偏大\n\nD、归并排序\n\nD、动态链接库会使进度运维速度偏慢\n\n1一、下列说法中破绽百出的是：（）\n\nB。优化编写翻译\n\nA、插入排序某个情形下复杂度为O（n）\n\nB、排序2叉树成分查找的复杂度大概为O（n）\n\nA、三个静态库中不可能包罗两个同名全局函数的定义\n\nC、对于有种类表的排序最快的是快捷排序\n\nB、一个动态库中不能够包涵五个同名全局函数的定义\n\nD、在稳步列表中经过二分查找的复杂度一定是O（n\nlog2n）\n\nC、假如三个静态库都饱含3个同名全局函数，他们无法同期被链接\n\n1二、在先后设计中，要对多个16K×1六K的多精度浮点数二维数组实行矩阵求和时，行优先读取和列优先读取的界别是（）\n\nD、如若八个动态库都带有二个同名全局函数，他们不能够同时被链接\n\nA、没区别\n\nC。静态库中编写翻译器保险未有同名函数，七个静态库，编写翻译实现后，会在区别类库，同名函数上加上一些参数也许别的特定消息，从而在调用时分别，假若多少个动态库都蕴涵三个同名全局函数，他们不能够同期被链接，因为全局函数是概念在类外的函数，成员函数正是概念在类中的函数\n\nB、行优先快\n\nC、列优先快\n\nA、插入排序\n\nD、二种读取格局速度为随机值，不能够剖断\n\nB、冒泡排序\n\n13、字符串www.qq.com抱有非空子串（五个子串假若剧情千篇一律则只算一个）个数是（）\n\nC、快速排序\n\nA、1024\n\nD、归并排序\n\nB、1018\n\nC、55\n\n1一、下列说法中破绽百出的是：（）\n\nD、50\n\nA、插入排序有些情形下复杂度为O（n）\n\n1肆、TCP的关门进度，说法科学的是（）\n\nB、排序2叉树成分查找的复杂度恐怕为O（n）\n\nA、TIME_WAIT状态称为MSL（马克西姆um Segment\n\nC、对于有类别表的排序最快的是全速排序\n\nB、对二个established状态的TCP连接，在调用shutdown函数此前调用close接口，能够让主动调用的1方进入半关闭状态\n\nD、在稳步列表中经过二分查找的复杂度一定是O（n log2n）\n\nC、主动发送FIN音信的连接端，收到对方回答ack以前无法发只可以收，在摄取对方回复ack之后不能够发也不能收，进入CLOSING状态\n\nC。A当数码完全有序时便是O(n)，B当数退化成线性表时（唯有一叉时）出现，C快排只对冬天、随机种类有优势。D是对的。\n\nD、在早就成功建构连接的TCP连接上，若是壹端收到奥迪Q3ST消息可以让TCP的连洁端绕过半关门状态并允许丢失数据。\n\n1二、在程序设计中，要对多少个16K×16K的多精度浮点数贰维数组开始展览矩阵求和时，行优先读取和列优先读取的区分是（）\n\nA、没区别\n\nA、端口号在64512-65535中间的端口\n\nB、行优先快\n\nB、全体小于十二四的每种端口\n\nC、列优先快\n\nC、福特ExplorerFC标准文书档案中曾经宣称特定服务的相干端口，比方http服务的80端口，8080端口等\n\nD、二种读取格局速度为随机值，无法判断\n\nD、全数端口都得以不受权限限制展开\n\nB。\n\n1陆、体育场合有七个人排队，在那之中三位要还一样本书，书名叫《面试宝典》，其余四个人要借。问求能确认保障此外二位借到的档案的次序。\nCatalan数\nC（2n , n）/( n+1 )   C(6,3)/4 = 5\n5*3!*3! = 180\n\nA、1024\n\n[cpp] view\nplain\ncopyprint?\n\nB、1018\n\n1. int ack(int m,int n)\n2. {\n3.     if(m == 0)\n4.         return n + 1;\n5.     else if(n == 0)\n6.         return ack(m-1,1);\n7.     else\n8.         return ack(m – 1 , ack(m , n-1));\n9. }\n\nint ack(int m,int n)\n{\n\n`````` if(m == 0)\nreturn n + 1;\nelse if(n == 0)\nreturn ack(m-1,1);\nelse\nreturn ack(m - 1 , ack(m , n-1));\n``````\n\n}\n\nC、55\n\nD、50\n\n1八、如下SQL语句是急需列出贰个论坛版面第叁页（每页展现二十一个）的帖子（post）标题（title），并坚守公布（create_time）降序排列：\n\nD.\n\nSELECT title FROM post(\n)create_time DESC( )0,20    order by\nlimit\n\n1四、TCP的闭馆进程，说法科学的是（）\n\n1九、为了某项目要求，大家准备构造了一种面向对象的脚本语言，举例，对具备的整数，大家都经过Integer类型的目的来说述。在企图“壹+贰”时，这里的“1”，“2”和结果“三”分别为二个Integer对象。为了下跌设计复杂度，我们决定让Integer对象都以只读对象，也即在测算a=a+b后，对象a引用的是三个新的对象，而非改a所指对象的值。思量到质量难题，大家又引进二种优化方案：（一）对于数值相等的Integer对象，我们不会重复创设。举个例子，总结“壹+一”，这里五个“一”的引用的是同1个对象——这种设计情势叫做（）；（二）脚本语言分析器运行时，暗中同意创制数值范围[1,32]的311个Integer对象。以后，如若大家要总括表明式“一+2+三+…+40”，在图谋进程供给创设的Integer对象个数是（）。\n\nB、对2个established状态的TCP连接，在调用shutdown函数以前调用close接口，能够让主动调用的一方进入半关闭状态\n\n20、甲、乙四人在玩猜数字游戏，甲随机写了二个数字，在[1，100]间隔之内，将以此数字写在了一张纸上，然后乙来猜。\n\nC、主动发送FIN新闻的连接端，收到对方回复ack从前无法发只好收，在收到对方回复ack之后无法发也无法收，进入CLOSING状态\n\nD、在早就打响营造连接的TCP连接上，假诺1端收到中华VST音信能够让TCP的连年端绕过半闭馆状态并允许丢失数据。\n\nF[N]=100-N，\nF=min（max（1，1+F[N-1]），max（2，1+F[N-2]），……，max（N-1，1+F））;\n\nD。//TIME_WAIT\n\n[cpp] view\nplain\ncopyprint?\n\n1. #include\n2. using namespace std;\n3.\n4. int dp = { 0 };\n5.\n6. void solve()\n7. {\n8.     int i , j , k;\n9.     for(i = 2 ; i < 101 ; ++i)\n10.     {\n11.         dp[i] = i;\n12.         for(j = 1 ; j < i ; ++j)\n13.         {\n14.             k = (j>=(1 + dp[i-j])) ? j : (1 + dp[i-j]);\n15.             if(dp[i] > k)\n16.                 dp[i] = k;\n17.         }\n18.     }\n19. }\n20.\n21. int main(void)\n22. {\n23.     dp = 0 , dp = 1;\n24.     solve();\n25.     printf(“%d\\n”,dp);\n26.     return 0;\n27. }\n\n#include\nusing namespace std;\n\nint dp = { 0 };\n\nvoid solve()\n{\n\n`````` int i , j , k;\nfor(i = 2 ; i < 101 ; ++i)\n{\ndp[i] = i;\nfor(j = 1 ; j < i ; ++j)\n{\nk = (j>=(1 + dp[i-j])) ? j : (1 + dp[i-j]);\nif(dp[i] > k)\ndp[i] = k;\n}\n}\n``````\n\n}\n\nint main(void)\n{\n\n`````` dp = 0 , dp = 1;\nsolve();\nprintf(\"%d\\n\",dp);\nreturn 0;\n``````\n\n}\n\nA、端口号在64512-65535之内的端口\n\n2、从第三回抛棋子的间隔楼层最优的取舍早晚比第二回间隔少1层，第二回的楼宇间隔比第二回间隔少壹层，如此，以往每便抛棋子楼层间隔比上二遍间隔少壹层。（大家不要紧本身作证一下）\n\n不等式如下：  一+二+三+…+(n-一)+n  >=   100\n\nB、全体小于拾二4的每种端口\n\nb[N]外，不可使用新的变量(包蕴栈一时变量、对空间和大局静态变量等)；\n\nC、福睿斯FC标准文书档案中早就宣示特定服务的连带端口，比如http服务的80端口，8080端口等\n\n[cpp] view\nplain\ncopyprint?\n\nD、全部端口都足以不受权限限制展开\n\n1.\n2. void makeArray(int a[],int b[],int len)\n3. {\n4.     int i;\n5.     b = 1;\n6.     for(i = 1 ; i < len ; ++i)\n7.         b[i] = b[i-1] * a[i-1];    // b = 1 , b[i] = a*a*…*a[i-1]\n\n8.\n9.     a[len – 1] = a[len – 1]^a[len – 2];   //不利用当中变量，通过位运算来交流七个变量\n\n10.     a[len – 2] = a[len – 1]^a[len – 2];\n11.     a[len – 1] = a[len – 1]^a[len – 2];\n12.\n13.     for(i = len – 3 ; i >= 0 ; –i)\n14.     {\n15.         a[len – 1] = a[i + 1] * a[len – 1];\n16.\n17.         a[i] = a[i]^a[len – 1];    //交流七个变量\n18.         a[len – 1] = a[i]^a[len – 1];\n19.         a[i] = a[i]^a[len – 1];\n20.     }\n21.     a[len – 1 ] = 1;    //a[len – 1 ] = 1 , a[i] = a[i+1]*a[i+2]*…*a[len-1]\n\n22.\n23.     for(i = 0 ; i < len ; ++i)\n24.         b[i] = a[i] * b[i];\n25. }\n\nvoid makeArray(int a[],int b[],int len)\n{\n\n`````` int i;\nb = 1;\nfor(i = 1 ; i < len ; ++i)\nb[i] = b[i-1] * a[i-1]; // b = 1 , b[i] = a*a*...*a[i-1]\n\na[len - 1] = a[len - 1]^a[len - 2]; //不使用中间变量，通过位运算来交换两个变量\na[len - 2] = a[len - 1]^a[len - 2];\na[len - 1] = a[len - 1]^a[len - 2];\n\nfor(i = len - 3 ; i >= 0 ; --i)\n{\na[len - 1] = a[i + 1] * a[len - 1];\n\na[i] = a[i]^a[len - 1]; //交换两个变量\na[len - 1] = a[i]^a[len - 1];\na[i] = a[i]^a[len - 1];\n}\na[len - 1 ] = 1; //a[len - 1 ] = 1 , a[i] = a[i+1]*a[i+2]*...*a[len-1]\n\nfor(i = 0 ; i < len ; ++i)\nb[i] = a[i] * b[i];\n``````\n\n}\n\nC。\n\n1陆、找职业的时节立即就到了，繁多同室去教室借阅《面试宝典》那本书，以后体育场所外有陆名同班排队，当中3名同班要将手中的《面试宝典》还至教室，有叁知名高校友愿意从教室中得以借到《面试宝典》，若当前体育场所内已无仓库储存《面试宝典》，要保管借书的三著名高校友能够借到书，请问那五位同学某些许种排队情势（）\n\n[cpp] view\nplain\ncopyprint?\n\nA）60\n\n1. //方法2，保持a数组不改变\n2. void makeArray(int a[],int b[],int len)\n3. {\n4.     int i;\n5.     b = 1;\n6.     for(i = 1 ; i < len ; ++i)\n7.     {\n8.         b *= a[i-1];\n9.         b[i] = b;      // b[i] = a*a*…*a[i-1]\n\n10.     }\n11.     b = 1;\n12.     for(i = len – 2 ; i > 0 ; –i)\n13.     {\n14.         b *= a[i+1];   // b = a[i+1]*a[i+2]…*a[len-1]\n\n15.         b[i] *= b;     // b[i] = a*a*…*a[i-1]*a[i+1]*…*a[len-1]\n\n16.     }\n17.     b *= a;\n18.\n19. }\n\n//方法2，保持a数组不变void makeArray(int a[],int b[],int len)\n{\n\n`````` int i;\nb = 1;\nfor(i = 1 ; i < len ; ++i)\n{\nb *= a[i-1];\nb[i] = b; // b[i] = a*a*...*a[i-1]\n}\nb = 1;\nfor(i = len - 2 ; i > 0 ; --i)\n{\nb *= a[i+1]; // b = a[i+1]*a[i+2]...*a[len-1]\nb[i] *= b; // b[i] = a*a*...*a[i-1]*a[i+1]*...*a[len-1]\n}\nb *= a;\n``````\n\n}\n\nB）120\n\nC）180\n\n[cpp] view\nplain\ncopyprint?\n\nD）360\n\n1. void makeArray(int a[],int b[],int len)\n2. {\n3.     int i;\n4.     b = 1;\n5.     for(i = 1 ; i < len ; ++i)\n6.     {\n7.         b[i] = b[i-1] * a[i-1];    // b[i] = a*a*…*a[i-1]\n\n8.     }\n9.     b = a[len – 1];\n10.     for(i = len – 2 ; i > 0 ; –i)\n11.     {\n12.         b[i] *= b;     // b[i] = a*a*…*a[i-1]*a[i+1]*…*a[len-1]\n\n13.         b *= a[i];     // b = a[i+1]*a[i+2]…*a[len-1]\n\n14.     }\n15.\n16. }\n\nvoid makeArray(int a[],int b[],int len)\n{\n\n`````` int i;\nb = 1;\nfor(i = 1 ; i < len ; ++i)\n{\nb[i] = b[i-1] * a[i-1]; // b[i] = a*a*...*a[i-1]\n}\nb = a[len - 1];\nfor(i = len - 2 ; i > 0 ; --i)\n{\nb[i] *= b; // b[i] = a*a*...*a[i-1]*a[i+1]*...*a[len-1]\nb *= a[i]; // b = a[i+1]*a[i+2]...*a[len-1]\n}\n``````\n\n}\n\nC。Carter兰数，C(n,二n)/(n+壹)，n是入栈成分的个数，这里n=三，C(叁,6)/四=5，同学互相是例外的，因而要全排列一下，结果为5*3!*3!=180\n\n2贰、20世纪60时代，米利坚激情学家Mill格拉姆设计了1个休戚相关信件实验。Mill格拉姆把信随即发送给住在美利哥各城市的一有的居民，信中写有2个埃及开罗期货经纪人的名字，并须求每名收信人把那封信寄给协和认为是比较像样那名股票（stock）经纪人的恋人。那位朋友收到信后再把信寄给她认为更临近那名股票（stock）经纪人的心上人。最后，半数以上信件都寄到了这名股票（stock）经纪人手中，每封信平均经受陆.二词到达。于是，米尔格拉姆建议六度分割理论，感觉世界上自由多个人中间创设联系最多只须要伍个人。\n\n）（请用十进制表示）。\n\n4813\n\nint ack(int m,int n)\n\nif(m == 0)\nreturn n + 1;\nelse if(n == 0)\nreturn ack(m-1,1);\nelse\nreturn ack(m – 1 , ack(m , n-1));\n\n61。耐心，ack(1,x)=2+x，ack(2,x)=3+x*2，ack(3,0)=5，ack(3,1)=ack(3,0)*2+3=13，ack(3,2)=ack(3,1)*2+3=29，ack(3,3)=ack(3,2)*3+2=61。\n\nUsers，ACU）二4钟头数据如下图所示。现已知全天平均在线人数为5000人，游戏者每一回登入后平均在线时间长度为二钟头。请您估算一下，平均下来每秒钟约有（\n）个游戏发烧友登入。", null, "SELECT title FROM post( )create_time DESC( )0,20\n\nORAV四DEQashqai BY; LIMIT， 推荐SQL《学习指南》\n\nInteger对象，大家不会另行创设。比方，总计“壹+1”，这里几个“壹”的引用的是同三个目的——这种设计格局叫做（）；（2）脚本语言深入分析器运营时，暗中同意创制数值范围[1,32]的33个Integer对象。以后，若是我们要计算表明式“一+2+三+…+40”，在企图进程需求创设的\nInteger对象个数是（）。\n\n+\n\n6、甲、乙四个人在玩猜数字娱乐，甲随机写了三个数字，在[1，100]距离之内，将以此数字写在了一张纸上，然后乙来猜。\n\n1经乙猜的数字偏大的话，甲现在就再也不会提醒了，只会回复“猜对 或 猜错”\n\nN二+二=N1，N三=N二+N一-二，增量为N壹-2……依此类推，增量是随着推测次数的加码而逐1地回落。设最后二遍揣度为k，则Nk=N一+\n(N一-一)+(N一-贰)+…一，Nk是相等或凌驾十0的率先个数，依照等差数列求和公式能够算出N一=14，N二=二七，N三=3九…\n（1四,贰七,3玖,50,60,6九,77,捌4,90,9五,9玖）。\n\nhttp://blog.csdn.net/kingjinzi_2008/article/details/7785334\n\n100-N+一}。之所以要加一，是因为第壹回是从第N层先导扔。\n\nF[N]=100-N，\nF=min（max（1，1+F[N-1]），max（2，1+F[N-2]），……，max（N-1，1+F））;\n\n[cpp] view plaincopy\n#include<iostream>\nusing namespace std;\nint dp = { 0 };\nvoid solve()\n{\nint i , j , k;\nfor(i = 2 ; i < 101 ; ++i)\n{\ndp[i] = i;\nfor(j = 1 ; j < i ; ++j)\n{\nk = (j>=(1 + dp[i-j])) ? j : (1 + dp[i-j]);\nif(dp[i] > k)\ndp[i] = k;\n}\n}\n}\nint main(void)\n{\ndp = 0 , dp = 1;\nsolve();\nprintf(“%d\\n”,dp);\nreturn 0;\n}\n\n2、从第1次抛棋子的间隔楼层最优的选用早晚比第三遍间隔少一层，第二回的大楼间隔比第3遍间隔少一层，如此，未来每一遍抛棋子楼层间隔比上三次间隔少1层。（大家无妨本身作证一下）\n3、所以，设n是率先次抛棋子的最好楼层，则n即为知足下列不等式的不大自然数：\n不等式如下：  1+2+3+…+(n-1)+n  >=   十0\n\nInt fuc(int m,int n)\n\n{\n\nif(m%n)==0\n\n{\n\nreturn n;\n\n}\n\nelse\n\n{\n\nreturn fuc(n,m%n)\n\n}\n\n}\n\n2。递归。，其实正是求最小公倍数，\n\nb[N]外，不可利用新的变量(包罗栈一时变量、对空卯月全局静态变量等)；\n\n2、20世纪60年份，美利哥心情学家Mill格Lamb设计了一个相关信件实验。Mill格兰姆把信随即发送给住在U.S.A.各城市的1某个居民，信中写有二个达拉斯股票（stock）经纪人的名字，并需求每名收信人把那封信寄给和煦感觉是相比较接近这名股票（stock）经纪人的爱侣。这位朋友接到信后再把信寄给她以为更就像那名股票（stock）经纪人的爱人。最终，超过一半信件都寄到了这名股票经纪人手中，每封信平均经受6.二词达到。于是，Mill格拉姆提议陆度分割理论，感到世界上随便五人里面建设构造联系最七只必要八个人。\n\n3、段页式虚拟存款和储蓄管理方案的性状。" ]

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[ "# Exercise 4.41 in N&C book QCQI: how can i implement $R_z(\\theta)$ using the circuit shown and $Z$?\n\nI'm studying Nielsen and Chuang's book.\n\nI cannot solve one of the questions in the exercise 4.41.\n\nThe question is the last one that is\n\nExplain how repeated use of this circuit and Z gates may be used to apply a $$R_z(\\theta)$$ gate with probability approaching 1.\n\nI found the last state of the circuit is\n\n$$|\\psi_{3}\\rangle = |00\\rangle(\\frac{10^{1/2}}{4})e^{i\\pi/4}R_z(\\theta)|\\psi\\rangle +(-|01\\rangle-|10\\rangle+|11\\rangle)\\frac{1}{4}e^{-i\\pi/4}Z|\\psi\\rangle$$\n\nI put the picture of the circuit below.\n\nHow can i solve this problem?\n\n• On the title you mention exercise 4.51 and in the question 4.41, which one is it? Jun 22, 2021 at 6:37\n• – glS\nJun 23, 2021 at 8:41\n\nWhat your calculation conveys is that if you get any of the measurement results 01, 10 or 11, the output state is $$Z|\\psi\\rangle$$. If that happens, you apply $$Z$$ and your state is back to the one you started with. So, repeat the circuit and , if you get the answer 00, you've accomplished the process you want. If not, apply another $$Z$$ and repeat.\nEach repetition succeeds with probability $$p=\\frac58$$. So after $$k$$ repetitions, success has occurred with probability $$\\sum_{n=1}^k\\left(\\frac38\\right)^{n-1}\\frac58,$$ which you could evaluate using a sum of a geometric progression, if you wanted, but the point is that it tends to 1 as $$k$$ becomes large." ]

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[ "# .css-4zleql{display:block;}miguendes's blog", null, "", null, "# How to Pass Multiple Arguments to a map Function in Python\n\nSubscribe to my newsletter and never miss my upcoming articles\n\n## TL;DR\n\nTime is a precious resource so I won't waste yours. Here's how you can use a `map` function with multiple arguments including `executor.map` and `pool.map` .\n\nSolution: You need to transform your function so it can only accept one argument. We can achieve that with `functools.partial`.\n\n``````>>> import functools\n>>> from multiprocessing import Pool\n>>> from concurrent.futures import ProcessPoolExecutor\n>>> # Defining a multiple arguments function\n>>> def sum_four(a, b, c, d):\nreturn a + b + c + d\n\n>>> # Making it accept only one argument by using partial\n>>> partial_sum_four = functools.partial(sum_four, a, b, c)\n>>> partial_sum_four(3)\n9\n>>> # Using python map function with multiple arguments\n>>> list(map(partial_sum_four, ds))\n[7, 8, 9, 10]\n>>> # Using pool map for multiple arguments\n>>> with Pool(processes=4) as pool:\nres = pool.map(partial_sum_four, ds)\n\n>>> res\n[7, 8, 9, 10]\n>>> # Using multiprocessing pool multiple arguments\n>>> with ProcessPoolExecutor(max_workers=4) as pool:\nres = list(pool.map(partial_sum_four, ds))\n\n>>> res\n[7, 8, 9, 10]\n``````\n\nIn a nutshell, that's it folks! If you want to know more details and other methods, please follow along!\n\n## Introduction\n\nIn this post, I’m going to show what I do to map a function that expects multiple arguments. The solutions work not only for the regular `map` function, you can also use the trick to pass multiple parameters to `concurrent.futures.Executor.map` and `multiprocessing.Pool`.\n\n## Problem\n\nLet’s imagine that you have a function called `sum_four` that takes 4 arguments and returns their sum.\n\n``````>>> def sum_four(a, b, c, d):\nreturn a + b + c + d\n``````\n\nLet’s also suppose that you are solving a very specific problem that requires the first 3 arguments to be fixed. In this problem, you want to compare how the function behaves when you vary only the last parameter.\n\n``````>>> a, b, c = 1, 2, 3\n\n>>> sum_four(a=a, b=b, c=c, d=1)\n7\n\n>>> sum_four(a=a, b=b, c=c, d=2)\n8\n\n>>> sum_four(a=a, b=b, c=c, d=3)\n9\n\n>>> sum_four(a=a, b=b, c=c, d=4)\n10\n``````\n\nNow, say that you want to use `map`, because you like functional programming, or maybe because you come from a language that encourages this paradigm. Since only `d` varies, we could store all potential values we want to test in a list `ds = [1, 2, 3, 4]`. The issue is, given a function and a list of single elements, if you want to pass that list to a `map` function and it takes only one element, what can you do?\n\n## Solution 1\n\nThe first solution is to not adopt the `map` function but use `itertools.starmap` instead. This function will take a function as arguments and an iterable of tuples. Then, `starmap` will iterate over each tuple `t` and call the function by unpacking the arguments, like this `for t in tuples: function(*t)`.\n\nTo make things more clear, consider the following example.\n\n``````>>> import itertools\n\n>>> ds = [1, 2, 3, 4]\n\n>>> items = ((a, b, c, d) for d in ds)\n\n>>> list(items)\n[(1, 2, 3, 1), (1, 2, 3, 2), (1, 2, 3, 3), (1, 2, 3, 4)]\n\n>>> list(itertools.starmap(sum_four, items))\n[7, 8, 9, 10]\n``````\n\nAs you can see, there’s a lot of repetition, which may inevitably consume a lot of memory if the list is big. To improve that I made `items` as a generator, this way we only hold in memory the element we’ll be processing.\n\n## Solution 2\n\nThe second solution is to use currying and create a new partial function. According to the docs, `partial()` will \"freeze\" some portion of a function’s arguments and/or keywords resulting in a new function with a simplified signature.\n\n``````In : import functools\n\nIn : partial_sum_four = functools.partial(sum_four, a, b, c)\n\nIn : partial_sum_four(3)\nOut: 9\n\nIn : list(map(partial_sum_four, ds))\nOut: [7, 8, 9, 10]\n``````\n\n## Solution 3\n\nThe third alternative is to use the `itertools.repeat()`. This function produces an iterator that returns object over and over again. It will run indefinitely if you don’t specify the times argument. If we take a closer look at `map()`'s signature, it accepts a function and multiple iterables, `map(function, iterable, ...)`.\n\nAccording to its description,\n\nIf additional iterable arguments are passed, function must take that many arguments and is applied to the items from all iterables in parallel. With multiple iterables, the iterator stops when the shortest iterable is exhausted.\n\nBingo! We can make `a`, `b` and `c` infitnite iterables by using `itertools.repeat()`. As soon as `ds` is exhausted, which is the shortest iterable, `map()` will stop.\n\n``````>>> import itertools\n>>> list(map(sum_four, itertools.repeat(a), itertools.repeat(b), itertools.repeat(c), ds))\n[7, 8, 9, 10]\n``````\n\nTo put it another way, using `repeat()` is roughly equivalent to:\n\n``````>>> list(map(sum_four, [1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], ds))\n[7, 8, 9, 10]\n``````\n\nYou don't need to worry too much about memory as `repeat` produces the elements on the go. In fact, it returns a `repeatobject`, not `list` [ref] .\n\n## Problem 2: How to Pass Multiple Parameters to `multiprocessing.Pool.map`?\n\nThis is very similar to using the regular `map()` with multiple parameters. Suppose that we want to speed up our code and run `sum_four` in parallel using processes. The good news is, you can use the solutions above, with one exception: `Pool.map` only accepts one iterable. This means we cannot use `repeat()` here. Let's see the alternatives.\n\n### Using `starmap`\n\n``````>>> from multiprocessing import Pool\n\n>>> import itertools\n\n>>> def sum_four(a, b, c, d):\nreturn a + b + c + d\n\n>>> a, b, c = 1, 2, 3\n\n>>> ds = [1, 2, 3, 4]\n\n>>> items = [(a, b, c, d) for d in ds]\n\n>>> items\n[(1, 2, 3, 1), (1, 2, 3, 2), (1, 2, 3, 3), (1, 2, 3, 4)]\n\n>>> with Pool(processes=4) as pool:\nres = pool.starmap(sum_four, items)\n\n>>> res\n[7, 8, 9, 10]\n``````\n\n### Using `partial()`\n\n``````\n>>> import functools\n\n>>> partial_sum_four = functools.partial(sum_four, a, b, c)\n\n>>> with Pool(processes=4) as pool:\nres = pool.map(partial_sum_four, ds)\n\n>>> res\n[7, 8, 9, 10]\n``````\n\n## Problem 3: How to Pass Multiple Arguments to `concurrent.futures.Executor.map`?\n\nThe `concurrent.futures` module provides a high-level interface called `Executor` to run callables asynchronously.\n\nThere are two different implementations available, a `ThreadPoolExecutor` and a `ProcessPoolExecutor`. Contrary to `multiprocessing.Pool`, a `Executor` does not have a `startmap()` function. However, its `map()` implementation supports multiple iterables, which allow us to use `repeat()`. Another difference is that `Executor.map` returns a generator, not a list.\n\n### Using `partial()`\n\n``````>>> from concurrent.futures import ProcessPoolExecutor\n\n>>> import functools\n\n>>> def sum_four(a, b, c, d):\nreturn a + b + c + d\n\n>>> a, b, c = 1, 2, 3\n\n>>> ds = [1, 2, 3, 4]\n\n>>> partial_sum_four = functools.partial(sum_four, a, b, c)\n\n>>> with ProcessPoolExecutor(max_workers=4) as pool:\nres = list(pool.map(partial_sum_four, ds))\n\n>>> res\n[7, 8, 9, 10]\n``````\n\n### Using `repeat()`\n\n``````>>> from concurrent.futures import ProcessPoolExecutor\n\n>>> from itertools import repeat\n\n>>> def sum_four(a, b, c, d):\nreturn a + b + c + d\n\n>>> a, b, c = 1, 2, 3\n\n>>> ds = [1, 2, 3, 4]\n\n>>> with ProcessPoolExecutor(max_workers=4) as pool:\nres = list(pool.map(sum_four, repeat(a), repeat(b), repeat(c), ds))\n\n>>> res\n[7, 8, 9, 10]\n``````\n\n## Conclusion\n\nThat’s it for today, folks! I hope you’ve learned something different and useful. The `map()` function makes Python feel like a functional programming language. `map()` is available not only as a built-in function but also as methods in the `multiprocessing` and `concurrent.futures` module. In this article, I showed what I do to map functions that take several arguments.\n\nOther posts you may like:\n\n#python#beginners#map#functional-programming" ]

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[ "# Count the number of items received in a certain month\n\nS\n\n#### Steven\n\nI am trying to get my excel sheet to count the number items received in a\ncertain month. For example\n\n02/Mar/09 59\n04/Apr/09 62\n01/Mar/9 25\n01/May/09 25\n\nI there want the total number of items received in March adding togther in a\ncell,\nthen in next cell totl number recived in April and so on.\n\nAny help?\n\nJ\n\n#### Jacob Skaria\n\nAssuming you have dates in ColA and values in ColB use the below formula\n\nSum of March\n=SUMPRODUCT(--(MONTH(\\$A\\$1:\\$A\\$1000)=3),--(\\$B\\$1:\\$B\\$1000))\n\nSum of April\n=SUMPRODUCT(--(MONTH(\\$A\\$1:\\$A\\$1000)=4),--(\\$B\\$1:\\$B\\$1000))\n\nIf this post helps click Yes" ]

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[ "# Leetcode: Set Matrix Zeroes\n\nGiven a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.\n\nFollow up:Did you use extra space?\nA straight forward solution using O(mn) space is probably a bad idea.\nA simple improvement uses O(m + n) space, but still not the best solution.\nCould you devise a constant space solution?1. O(m+n) space solutionAnalysis:\nTo achieve O(m+n) space, you need two array to store information while traversal the matrix for the first time checking if it is zero or not. You can also combine two arrays into one, but still cost O(m+n) space. So I just use two to store row and column information.\n```public class Solution {\npublic void setZeroes(int[][] matrix) {\nif (matrix.length == 0 || matrix.length == 0) {\nreturn;\n}\nboolean[] zeroRows = new boolean[matrix.length];\nboolean[] zeroCols = new boolean[matrix.length];\n//traversal the entire matrix to check if it is zero or not\nfor (int i = 0; i < matrix.length; i++) {\nfor (int j = 0; j < matrix[i].length; j++) {\nif (matrix[i][j] == 0) {\nzeroRows[i] = true;\nzeroCols[j] = true;\n}\n}\n}\n//now set matrix zero based on the two boolean arrays\nfor (int i = 0; i < matrix.length; i++) {\nfor (int j = 0; j < matrix[i].length; j++) {\nif (zeroRows[i] || zeroCols[j]) {\nmatrix[i][j] = 0;\n}\n}\n}\n}\n}```\n\n2. *Optimal* O(1) space solution\n\nAnalysis:\nTo save space, we can just make use of the space inside the matrix. Which means, we use the first row and first column to store the information.\nIn the first traversal, we set the corresponding cell on the first row or first column to 0 if there is any cell which is zero in this row or column.\nIn the second traversal, we set the matrix to zeros based on the first row and first column.\n\n``` public void setZeroes(int[][] matrix) {\nif (matrix.length == 0 || matrix.length == 0) {\nreturn;\n}\nboolean firstRowHasZero = false;\nboolean firstColHasZero = false;\n//traversal the entire matrix to check if it is zero or not\nfor (int i = 0; i < matrix.length; i++) {\nfor (int j = 0; j < matrix[i].length; j++) {\nif (matrix[i][j] == 0) {\nif (i == 0) {\nfirstRowHasZero = true;\n}\nif (j == 0) {\nfirstColHasZero = true;\n}\nmatrix[j] = 0;\nmatrix[i] = 0;\n}\n}\n}\n//now set matrix zero based on first row and first column\nfor (int i = 1; i < matrix.length; i++) {\nfor (int j = 1; j < matrix[i].length; j++) {\nif (matrix[j] == 0 || matrix[i] == 0) {\nmatrix[i][j] = 0;\n}\n}\n}\n//deal with the first column\nif (firstColHasZero) {\nfor (int i = 1; i < matrix.length; i++) {\nmatrix[i] = 0;\n}\n}\n//deal with the first row\nif (firstRowHasZero) {\nfor (int i = 1; i < matrix.length; i++) {\nmatrix[i] = 0;\n}\n}\n}```\n\nNote: The tricky part of this algorithm is, if without those two boolean variables to keep track of if there is any zero in the first row or first column , when matrix==0, we have no idea if the first row contains zero, or it is the first column contains zero, or both. also, we should deal with the rest part first based on the first row and first column and then deal with the first row and first column based on the two boolean values." ]

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[ "# Which has higher resistance a bulb of 50 watt or of 25 Watt?\n\nContents\n\n## Which has higher resistance 50W 220v or 25W?\n\nUnder a constant fixed applied voltage V R2>R1 25W bulb has higher resistance. Resistance of 25 W lamp bulb is twice the resistance of 50W lamp bulb.\n\n## Which has a higher resistance a 100 W bulb or 25 W bulb and by how many times?\n\nANSWER: 25W has the highest resistance.\n\n## Which has higher resistance of 500 Watt bulb or 2.5 Watt bulb and how many times?\n\nResistance of 500 W bulb be r and that of 2.5 W bulb be R. Thus, resistance of the 2.5 W bulb is 200 times the resistance of the 500 W bulb.\n\n## Which has higher resistance A 50W?\n\nResistance R = V2/P, where V is the potential difference and P is the power in Watts. So, the more the power, the lower will be the resistance and vice versa. Or, R2 = 2R1. Thus, 25 W bulb will have double the resistance as compared to 50 Watt bulb.\n\nIT IS AMAZING:  How do you keep bugs off LED lights?\n\n## Which has more resistance 100w bulb or 60w bulb?\n\nA 60-watt bulb has a higher electrical resistance than a 100-watt bulb. Because power is inversely proportional to resistance, when the power is less, the resistance is high.\n\n## Which has more resistance 100 W or 60 W bulb explain?\n\nFor a give voltage V, P is inversely proportional to R. So the 60 W bulb will have more resistance than the 100 W bulb.\n\n## Which has more resistance 100W bulb 60W or 40W bulb if they operated at the same voltage?\n\nIf the voltage of two bulbs is assumed to be the same, the resistance of the bulbs equals to V/100 and V/60 respectively. Thus, it is proved that the 60W bulb has higher resistance than the 100W bulb.\n\n## Which has higher resistance A 50w bulb or a 25w bulb and how many times?\n\nResistance R = V2/P, where V is the potential difference and P is the power in Watts. So, the more the power, the lower will be the resistance and vice versa. Or, R2 = 2R1. Thus, 25 W bulb will have double the resistance as compared to 50 Watt bulb.\n\n## Which has higher resistance ammeter or Milliammeter?\n\nFor a constant voltage , the current through a milliammeter must be lower (in the order of milliamperes) but the current through a ammeter would be higher (in the order of amperes) thus the resistance in milliammeter will be higher.\n\n## What is the power of torch bulb rated at 2.5 V and 500ma?\n\nElectricity. A torch bulb is rated at 2.5 volt, 500 mA. Find its (i) power and (ii) resistance. = 1.25 W." ]

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[ "# Binary to Decimal\n\nConversion\n\nConversion of binary to decimal (base-2 to base-10) Given that all computer and digital systems are based on the binary numbering system, it is crucial to comprehend the notion of numbers and back.\n\nThe Base-of-10 numbering system, often known as the decimal or \"denary\" counting system, assigns each digit in a number one of ten potential values, or \"digits,\" ranging from 0 to 9, as in the example 21310. (Two Hundred and Thirteen).\n\nHowever, the decimal numbering system also has the addition (+), subtraction (-), multiplication (*), and division (*) operations in addition to its ten digits (0–9).\n\nA decimal numbering system uses a base, q, along with a set of symbols, b, to define the weight of each digit in a number. In a decimal system, each digit has a value ten times larger than its previous number.\n\nThe six in sixty, for instance, is weighted less heavily than the six in six hundred. Then, in a binary numbering system, we require a method of converting from Decimal to Binary and vice versa.\n\nThe decimal system of numbering\nEach integer number column in the decimal, , base-10 (den), or denary numbering system contains values of units, tens, hundreds, thousands, etc. as we progress the number along the column from right to left. These numbers are represented mathematically as 100, 101, 102, 103, etc.\n\nThen, a higher positive power of 10 is shown by each position to the left of the decimal point. Similar to whole numbers, fractional numbers have a weight that changes from negative to positive as we move from left to right, such as 10-1, 10-2, 10-3, etc.\n\nThe binary system of numbers\nAll digital and computer-based systems use the binary numbering system, which is the most fundamental one. Binary numbers adhere to the same set of laws as decimal numbers. The binary numbering system, however, uses powers of two rather than the powers of ten used by the decimal system, allowing for a binary to decimal conversion from base-2 to base-10.\n\nA situation in digital logic and computer systems can be represented by either a logic level \"1\" or \"0,\" and each \"0\" and \"1\" is regarded as a single digit in a Base-of-2 (bi) or \"binary numbering system.\"", null, "" ]

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[ "# What symbol is used for the moon when doing calculations?\n\nWhen doing calculations the mass and radius of the Earth can be represented by $$M_{\\oplus}$$ and $$R_{\\oplus}$$ respectively.\n\nThe planets all have their own symbols which can easily be found online.\n\nBut what about the Moon ?\n\nUsually something like $$M_☽$$ is given but is that standard ? Are there other versions of expressing the moon ?\n\nThere is no authorative standard - you can use what you want, as long as it is clear what you mean. Besides that: yes, ☽ is reasonably standard. Often also written indices like $$M_{moon}$$ or $$M_m$$ are used." ]

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[ "# Klips: Writing formulas using formula references in the Klip Editor\n\nFormula references enable you to refer to formulas applied to columns, series, and other sub-components within the same Klip (this includes multi-component Klips). Formula references refer to formulas before actions (for example, filtering and grouping) or date properties are applied.\n\nNote: If you want to reference a formula after actions or date properties are applied, use results references.\n\nIn the image below, the formula or data selection: `@B:B`, is evaluated using the lightning bolt button on the far right of the formula bar. The Evaluation Results display the dates as the formula output. However, the dates shown in the Klip preview have a different format because a date property was applied.\n\nFormula reference always refers to the formula output or the Evaluation Result of the formula.", null, "Although formula references save time in creating duplicate formulas, every formula reference is recalculated which means there is no efficiency gained in executing the formula by using a formula reference. This does not apply to results references.\n\nTo use a formula reference:\n\n1. Type ! into the formula bar.\n2. Select the sub-component with the formula you want to reference from the drop-down list.\n\n## Related articles\n\nHave more questions? Submit a request" ]

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[ "", null, "Address 6797 Route 9, Rhinebeck, NY 12572 (845) 876-2080 http://www.hvds.com\n\n# error function calculator excel Mount Marion, New York\n\nCody's rational Chebyshev approximation algorithm. Ruby: Provides Math.erf() and Math.erfc() for real arguments. Browse other questions tagged excel vba excel-vba or ask your own question. Also has erfi for calculating i erf ⁡ ( i x ) {\\displaystyle i\\operatorname {erf} (ix)} Maple: Maple implements both erf and erfc for real and complex arguments. Anmelden 29 8 Dieses Video gef├żllt dir nicht?\n\nPets Relationships Society Sports Technology Travel Error Function Calculator Erf(x) Error Function Calculator erf(x) x = Form accepts both decimals and fractions. Bitte versuche es sp├żter erneut. Anmelden Teilen Mehr Melden M├Čchtest du dieses Video melden? Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Free Statistics Calculators version 4.0 used by people in more than 230 countries!\n\nIf L is sufficiently far from the mean, i.e. μ − L ≥ σ ln ⁡ k {\\displaystyle \\mu -L\\geq \\sigma {\\sqrt {\\ln {k}}}} , then: Pr [ X ≤ L Press, William H.; Teukolsky, Saul A.; Vetterling, William T.; Flannery, Brian P. (2007), \"Section 6.2. W├żhle deine Sprache aus. For iterative calculation of the above series, the following alternative formulation may be useful: erf ⁡ ( z ) = 2 π ∑ n = 0 ∞ ( z ∏ k\n\nIs R's glm function useless in a big data setting? in front of it. Help would be greatly appreciated! Why does cp --no-preserve=mode preserves the mode?\n\nThe function rapidly converges to its asympotic values; erf(3) = 0.99998 and erf(-3) = -0.99998. At the real axis, erf(z) approaches unity at zŌåÆ+Ōł× and ŌłÆ1 at zŌåÆŌłÆŌł×. Some functions do have vba equivalents, but (as far as I know) not in the case of erf and erfc share|improve this answer edited Jul 23 '12 at 21:21 answered Jul See also Related functions Gaussian integral, over the whole real line Gaussian function, derivative Dawson function, renormalized imaginary error function GoodwinŌĆōStaton integral In probability Normal distribution Normal cumulative distribution function, a\n\nApplied Mathematics Series. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). See . ^ http://hackage.haskell.org/package/erf ^ Commons Math: The Apache Commons Mathematics Library ^ a b c Cody, William J. (1969). \"Rational Chebyshev Approximations for the Error Function\" (PDF). Die Bewertungsfunktion ist nach Ausleihen des Videos verf├╝gbar. Wird verarbeitet...\n\nHow can we improve it? I'm very new to this and probably missing something incredibly simple. The lower bound for integrating ERF. How?\n\nDescription Returns the error function integrated between lower_limit and upper_limit. The error function at +Ōł× is exactly 1 (see Gaussian integral). Using the alternate value aŌēł0.147 reduces the maximum error to about 0.00012. This approximation can also be inverted to calculate the inverse error function: erf − 1 ⁡ ( x ) To use these approximations for negative x, use the fact that erf(x) is an odd function, so erf(x)=ŌłÆerf(ŌłÆx).\n\nGiven random variable X ∼ Norm ⁡ [ μ , σ ] {\\displaystyle X\\sim \\operatorname {Norm} [\\mu ,\\sigma ]} and constant L < μ {\\displaystyle L<\\mu } : Pr [ X By using this site, you agree to the Terms of Use and Privacy Policy. You can change this preference below. Thanks for letting me know!\n\nM. Negative integer values of Im(ŲÆ) are shown with thick red lines. However, it can be extended to the disk |z| < 1 of the complex plane, using the Maclaurin series erf − 1 ⁡ ( z ) = ∑ k = 0 The error function is related to the cumulative distribution Φ {\\displaystyle \\Phi } , the integral of the standard normal distribution, by Φ ( x ) = 1 2 + 1\n\nIntermediate levels of Im(ŲÆ)=constant are shown with thin green lines. Melde dich bei YouTube an, damit dein Feedback gez├żhlt wird. For any complex number z: erf ⁡ ( z ¯ ) = erf ⁡ ( z ) ¯ {\\displaystyle \\operatorname ŌłÆ 0 ({\\overline ŌüĪ 9})={\\overline {\\operatorname ŌüĪ 8 (z)}}} where z Wird geladen...\n\nSch├Čpf and P. Wenn du bei YouTube angemeldet bist, kannst du dieses Video zu einer Playlist hinzuf├╝gen. Softw., 19 (1): 22ŌĆō32, doi:10.1145/151271.151273 ^ Zaghloul, M. This is useful, for example, in determining the bit error rate of a digital communication system.\n\nThese generalised functions can equivalently be expressed for x>0 using the Gamma function and incomplete Gamma function: E n ( x ) = 1 π Γ ( n ) ( Γ This program is going to be used by many different people, however, so I can't use that to fix this particular issue. Why I am always unable to buy low cost airline ticket when airline has 50% or more reduction Amplify sinusoïdal signal with op-amp with V- = 0V Please explain what is For previous versions or for complex arguments, SciPy includes implementations of erf, erfc, erfi, and related functions for complex arguments in scipy.special. A complex-argument erf is also in the arbitrary-precision arithmetic\n\nis the double factorial: the product of all odd numbers up to (2nŌĆō1). Derivative and integral The derivative of the error function follows immediately from its definition: d d z erf ⁡ ( z ) = 2 π e − z 2 . {\\displaystyle doi:10.1109/TCOMM.2011.072011.100049. ^ Numerical Recipes in Fortran 77: The Art of Scientific Computing (ISBN 0-521-43064-X), 1992, page 214, Cambridge University Press. ^ DlangScience/libcerf, A package for use with the D Programming language. So using it all like it looks like you want to do will require the add-in, regardless of how you implement/use that function.\n\nThis article describes the formula syntax and usage of the ERF┬Āfunction in Microsoft Excel. Sprache: Deutsch Herkunft der Inhalte: Deutschland Eingeschr├żnkter Modus: Aus Verlauf Hilfe Wird geladen... If omitted, ERF integrates between zero and lower_limit. However, for ŌłÆ1 < x < 1, there is a unique real number denoted erf − 1 ⁡ ( x ) {\\displaystyle \\operatorname ╬ō 0 ^{-1}(x)} satisfying erf ⁡ ( erf" ]

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[ "# R/mle_het.R In econet: Estimation of Parameter-Dependent Network Centrality Measures\n\n#### Defines functions mle_het\n\n```#' mle_compet_lim\n#' @keywords internal\n#' @noRd\n#' @importFrom bbmle parnames mle2\nmle_het <- function(y, X, G, z, side, starting.values, boundL, boundU) {\n\ntheta <- starting.values\ntheta_L <- boundL\ntheta_U <- boundU\n\nif(side == \"het_l\"){\nbbmle::parnames(ll_het_left) <- names(theta)\nfit <- mle3(ll_het_left, optimizer = \"constrOptim\",\nmethod = \"Nelder-Mead\", start = theta, parnames = names(theta),\nui = rbind(c(rep(0, length(theta) - 3), - 1, - 1,0),\nc(rep(0, length(theta) - 3), 1, 1, 0),\nc(rep(0, length(theta) - 1), 1)),\nci = c( - 1 / max(colSums(G)), - 1 / max(colSums(G)), 0),\nvecpar = TRUE, skip.hessian = TRUE,\ndata = list(Y = y, X = X, G = G, z = z))\n\n} else if (side == \"het_r\"){\nbbmle::parnames(ll_het_right) <- names(theta)\nfit <- mle3(ll_het_right, optimizer = \"constrOptim\",\nmethod = \"Nelder-Mead\", start = theta, parnames = names(theta),\nui = rbind(c(rep(0, length(theta) - 3), - 1, - 1, 0),\nc(rep(0, length(theta) - 3), 1, 1, 0),\nc(rep(0, length(theta) - 1), 1)),\nci = c( - 1 / max(rowSums(G)), - 1 / max(rowSums(G)), 0),\nvecpar = TRUE, skip.hessian = TRUE,\ndata = list(Y = y, X = X, G = G, z = z))\n}\n\nreturn(fit)\n}\n```\n\n## Try the econet package in your browser\n\nAny scripts or data that you put into this service are public.\n\neconet documentation built on May 24, 2021, 5:09 p.m." ]

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[ "# Questions on Ramsey welfare in Dynare 4.7\n\nHello everyone,\n\nI would like to ask questions related with welfare calculation under the optimal policy. My computation basically follows this paper:\n\nLet me summarize the main steps:\n\nSuppose we have a separable utility function:\nW_t =\\displaystyle \\mathbb{E}_t\\sum^{\\infty}_{j=0}\\beta^j\\left[\\frac{{C_{t+j}}^{1-\\sigma}}{1-\\sigma}-\\chi \\frac{{N_{t+j}}^{1+\\phi}}{1+\\phi}\\right].\n\nSpecifically, in the baseline Competitive Equilibrium (CE) case, we define the conditional welfare:\nW^b_t = W^{Cb}_t+W^{Nb}_t,\nin which\nW^{Cb}_t = \\displaystyle \\mathbb{E}_t\\sum^{\\infty}_{j=0}\\beta^j\\frac{{C^b_{t+j}}^{1-\\sigma}}{1-\\sigma}, and W^{Nb}_t = \\displaystyle -\\mathbb{E}_t\\sum^{\\infty}_{j=0}\\beta^j\\chi \\frac{{N^b_{t+j}}^{1+\\phi}}{1+\\phi},\nwhere C^b_{t+j} and N^b_{t+j}, for j = 0,1,\\dots, are consumption and labour sequences when the economy achieves the CE.\n\nSimilarly, W^{Ca}_t and W^{Na}_t are the partial conditional welfare when the economy is in an alternative policy regime.\n\nSo at the outset of t:\n\\displaystyle W^b_t = \\mathbb{E}_t \\sum_{j=0}^{\\infty} \\beta^j \\left[ \\frac{\\left(\\left(1+\\Delta^{con}\\right)C^a_{t+j}\\right)^{1-\\sigma}}{1-\\sigma} - \\chi \\frac{{N^a_{t+j}}^{1+\\phi}}{1+\\phi} \\right].\nWork out \\Delta^{con} implicitly:\n\\displaystyle \\Delta^{con} = \\left[ \\frac{W^b_t-W^{Na}_t}{W_t^{Ca}}\\right]^{\\frac{1}{1-\\sigma}}-1.\n\nNow my question is how to compute W^{Ca}_t and W_t^{Na} in a Ramsey context?\n\nFirst, we know W^b_t under the CE can be derived by\noo_.dr.ys(W^b_t)+0.5*oo_.dr.ghs2.\n\nThen according to the new functionality in Dynare 4.7:\n\nIf the alternative policy is a Ramsey type, by evaluate_planner_objective, we know W^a_t.\n\nBut how to split this value into two parts, W^{Ca}_t and W_t^{Na}?\n\nSincerely,\n\nToT\n\n1. I guess you want to have conditional welfare at the deterministic steady state.\n2. Have you tried to explicitly define W_t^{Na} in the Ramsey model, i.e.\nW_t^{Na}=\\chi \\frac{\\left(N_t^b\\right)^{1+\\phi}}{1+\\phi}+\\beta E_t W_{t+1}^{Na}\nThat should allow you to back out W_t^{Ca} given the W_t^{a} provided.\n\n(BACKGROUND)\nYes, I need conditional welfare in both the baseline and the alternative regime.\n\nFor the baseline, the Competitive Equilibrium (CE) scenario, it is indeed perturbed around its deterministic steady state, the concept that is also recently discussed in the thread:\n\nFor the Ramsey case, I’m not sure whether the economy is approximated around the Ramsey steady state or somewhere else, after all I don’t see Dynare’s report about this sort of steady state.\n\nBut this seems to be not important. Suppose Dynare can compute the true conditional welfare of the Ramsey. Then the merit of partial welfare is that it allows the Ramsey case and the CE case to be compared at the same initial point, probably CE’s deterministic steady state, if I understand the following equality correctly:\n\nAnyway, I saw this expression almost everywhere, for example, in\n\nYet, his derivation is the same as the above JEDC paper, but the numerical part is not quite clear.\n\nThe authers compute the welfare in an optimal simple rules.\n\nSo it’ll be rather interesting to extend this approach in a Ramsey setting!\n\n(COMPUTATION)\n\nMy conditional welfare under the CE: -155.6874\nMy conditional welfare under the Ramsey (one instrument R_t):\n\n - with initial Lagrange multipliers set to 0: -163.13101105\n- with initial Lagrange multipliers set to steady state: -155.35499773\n\n\nThey look reasonable.\n\nHowever, Dynare also reports the unconditional welfare:\n\nApproximated value of unconditional welfare: 830037293448262400.00000000\n\n\nThat’s so ridiculous!\n\nI set\n\nUtility = C^(1-SIGMA)/(1-SIGMA)-CHI*N^(1+PHI)/(1+PHI);\n\nWelfC = C^(1-SIGMA)/(1-SIGMA) + BETA*WelfC(1);\n\n%WelfN = - CHI*N^(1+PHI)/(1+PHI)+ BETA*WelfN(1);\n\n\nBecause set both WelfC and WelfN may cause some singularity problem as well, my strategy is exactly yours:\n\nNext, my conditional welfare under the Ramsey (two instruments):\n\nError using print_info (line 32)\nBlanchard & Kahn conditions are not satisfied: no stable equilibrium.\n\n\nBoth one instrument and two-instrument versions of the Ramsey (without WelfC or WelfN) are comfortably executed in Dynare 4.6.1 and 4.6.4\n\nSo even including only WelfC may cause some internal problem in Dynare.\n\nCan you provide me with the codes?\n\n1. The BK violation comes from differences in the computed steady states. They may not be unique.\n2. It turns out you cannot have additive parts of the welfare function in the model as they will introduce a singularity. The big number you encounter comes from Dynare replacing the resulting NaN with a large number. 4.7 should have a warning in that case. Now to the workaround. What you can do is have the components of utility as variables in the model:\n% WelfareC\n[name='UC']\nUC = C^(1-SIGMA)/(1-SIGMA);\n\n[name='UN']\nUN = -CHI*N^(1+PHI)/(1+PHI);\n\n\nand then manually compute the decomposition based on the codes in evaluate_planner_objective after running the model. Here, it would be\n\n%% housekeeping to have access to relevant info\nif isempty(options_.qz_criterium)\noptions_.qz_criterium = 1+1e-6;\nend\ndr = oo_.dr;\nys=oo_.dr.ys;\nexo_nbr = M_.exo_nbr;\nnstatic = M_.nstatic;\nnspred = M_.nspred;\nbeta = get_optimal_policy_discount_factor(M_.params, M_.param_names);\n\n%% Original objective\n[U,Uy,Uyy] = feval([M_.fname '.objective.static'],ys,zeros(1,exo_nbr), M_.params);\n%% override original objective and instead read out UC (or UN if desired); needs to be defined in model-block\nUC_pos=strmatch('UC',M_.endo_names,'exact');\nUy=zeros(size(Uy));\nUy(UC_pos)=1; %first derivative is 1 for defined variable, zero for all others\nUyy=sparse(zeros(size(Uyy))); %all second derivatives are zero\n\nGy = dr.ghx(nstatic+(1:nspred),:);\nGu = dr.ghu(nstatic+(1:nspred),:);\nGyy = dr.ghxx(nstatic+(1:nspred),:);\nGyu = dr.ghxu(nstatic+(1:nspred),:);\nGuu = dr.ghuu(nstatic+(1:nspred),:);\nGss = dr.ghs2(nstatic+(1:nspred),:);\n\ngy(dr.order_var,:) = dr.ghx;\ngu(dr.order_var,:) = dr.ghu;\ngyy(dr.order_var,:) = dr.ghxx;\ngyu(dr.order_var,:) = dr.ghxu;\nguu(dr.order_var,:) = dr.ghuu;\ngss(dr.order_var,:) = dr.ghs2;\n\nUyy = full(Uyy);\n\nUyygygy = A_times_B_kronecker_C(Uyy,gy,gy);\nUyygugu = A_times_B_kronecker_C(Uyy,gu,gu);\nUyygugy = A_times_B_kronecker_C(Uyy,gu,gy);\n\n%% Unconditional welfare\n\noo_=disp_th_moments(dr,M_.endo_names(dr.order_var(nstatic+(1:nspred))),M_,options_,oo_);\n\noo_.mean(isnan(oo_.mean)) = options_.huge_number;\noo_.var(isnan(oo_.var)) = options_.huge_number;\n\nEy = oo_.mean;\nEyhat = Ey - ys(dr.order_var(nstatic+(1:nspred)));\n\nEyhatyhat = oo_.var(:);\nEuu = M_.Sigma_e(:);\n\nEU = U + Uy*gy*Eyhat + 0.5*((Uyygygy + Uy*gyy)*Eyhatyhat + (Uyygugu + Uy*guu)*Euu + Uy*gss);\nEW = EU/(1-beta);\n\n\n1 Like\n\nYes, posting your comments and modified codes in the forum will benefit other users.", null, "Please give me some days to digest them. After that, I may ask questions related with the welfare comparison between OSR and Ramsey, since I still feel some problem in the conditional welfare reported by Dynare 4.7. This problem does not occur in previous versions. Be patient. Thank you!" ]

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[ "Calculators\nRequired Stock Volume\nmL\nStock Concentration\nmM\nDesired Concentration\nmM\nDesired Volume\nmL\nDilution Calculator\n\n# Procedure\n\nDilute XmL of the stock solution into Y-XmL of solvent to obtain YmL at ZmM.\n\nDilution Calculator\nMean Optical Energy\nJ\nMean Optical Power\nmW\nsec\nEnergy Calculator\nmW\ncm2\nBeam\nArea\ncm2\nOptical Power\nmW\nJ\ncm2\nBeam\nArea\ncm2\nOptical Energy\nJ\nCircle\nArea\ncm2\nCircle\ncm\nCircle Diameter\ncm\nCircle Calculator\nTotal Dose Volume\nmL\nPescribed Dose\nmL\nKg\nPatient Weight\nKg\nLiquid Dose Calculator\nTotal Dose Volume\nmg\nPescribed Dose\nmg\nKg\nPatient Weight\nKg\nSolid Dose Calculator\nAre you using a Biolambda equipment?\ncm2\nWhat is the light beam Area?\nmW\nWhat is the mean optical Power?\nmW\ncm2\nWhat is the light beam Irradiance?\nparameter are:\nBeam diameter\ncm\nBeam area\ncm2\nOptical power\nmW\nmW\ncm2\nSetup lesion parameters?\nWhat is the lesion shape?\ncm\nWhat is the lesion Diameter?\nX\ncm\nWhat is the lesion Dimensions?\nJ\ncm2\nWhat is the lesion Radiant Exposure?\nparameters are:\nTotal Lesion area\ncm2\nNumber of points\npoints\nTime per point\nsec\nEnergy per point\nJ\nTotal time\nsec\nTotal energy\nJ\nGo to procedure instructions?\n\n# Procedure\n\nIrradiate Y points distributed homogeneously over the entire Xcm2 of lesion area.\nIrradiate each point for Zsec.\n\nLight Dose Wizard" ]

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[ "# Question about taking the Zariski closure in $\\mathbb{A}_{\\mathbb{R}}^n$\n\nLet $\\mathbb{A}_{\\mathbb{R}}^n$ be $\\mathbb{R}^n$ endowed with the Zariski topology, where closed sets are algebraic sets (in $\\mathbb{R}^n$) defined by real polynomials.\n\nSuppose $V \\subseteq \\mathbb{A}_{\\mathbb{R}}^n$ is an irreducible affine variety. Let $U$ be an open (with respect to the usual topology) ball $U$ around a non-singular point of $V$ and of small enough radius.\n\nDoes it then follow that the Zariski closure of $(V \\cap U)$ is $V$? I thought it should be true (maybe not?), but I was wondering how I can show this. Any comments are appreciated. Thank you!\n\nAs noted in this answer to a previous question of yours, for any subvariety $W \\subseteq \\mathbb A_\\mathbb R^n$, we have $$\\dim_{\\mathbb R} W(\\mathbb R) \\leq \\dim W$$ (where $\\dim W$ denotes the dimension in the sense of scheme theory, written there as $\\dim_\\mathbb C W(\\mathbb C)$), with equality if $W$ has a smooth $\\mathbb R$-point. In particular, if $W \\subseteq V$ is the Zariski closure of $V \\cap U$, then applying the above to both $V$ and $W$ gives $$\\dim_\\mathbb R W(\\mathbb R) \\leq \\dim W \\leq \\dim V = \\dim_\\mathbb R V(\\mathbb R).$$ But $W(\\mathbb R)$ contains the full-dimensional subset $V(\\mathbb R) \\cap U$ of $V(\\mathbb R)$, hence all dimensions must be equal. Since $V$ is irreducible and $\\dim W = \\dim V$, this forces $W = V$. $\\square$\n• How do you define the dimension of $V(\\mathbb R) \\cap U$? Jul 8, 2018 at 23:20" ]

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[ "# Python – importing bash environment variables\n\nI’ve run into situations where it was necessary to read the values of bash environment variables defined in a file into a python script.\n\nSomething like that is achieved in bash simply by doing:\n\n`source env.vars`\n\nor\n\n`. env.vars`\n\nWhere the content of env.vars could be something like this:\n\n```BASE=/home/user/scripts\nBIN_DIR=\\$BASE/bin\nVAR1=xxxxx\nVAR2=yyyy```\n\nThe solutions described here achieve that effect:\n\nBut they require that every variable be exported in the source file. Some times this may not be possible due to file permissions or other restrictions.\n\nThe following solution takes care of that issue:\n\n```import re\nimport subprocess\nfrom _collections import defaultdict\n\nvariables = defaultdict(str)\n\ncommand = ['bash', '-xc', 'source %s' % file_name]\n\nproc = subprocess.Popen(command, stderr = subprocess.PIPE)\n\nreg = re.compile('^\\+\\+ (?P<name>\\w+)\\=(?P<value>.+)')\nfor line in proc.stderr:\nm = reg.match(line)\nif m:\nname = m.group('name')\nvalue = ''\nif m.group('value'):\nvalue = m.group('value')\n\nvariables[name] = value\n\nproc.communicate()\n\nreturn variables```\n\nThis also works better than just reading and parsing the source file since bash will perform variable substitution before outputting anything, so cases like\n\n`VAR2=\\$VAR1/abcd`\n\nare possible.\n\nFor the example file above, the function would return:\n\n`defaultdict(<type 'str'>, {'VAR1': 'xxxxx', 'BASE': '/home/user/scripts', 'BIN_DIR': '/home/user/scripts/bin', 'VAR2': 'yyyy'})`" ]

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[ "# Changeset 2511 for trunk/yat/statistics/TukeyBiweightEstimator.h\n\nIgnore:\nTimestamp:\nJul 9, 2011, 1:44:33 AM (10 years ago)\nMessage:\n\ndocs typo\n\nFile:\n1 edited\n\n### Legend:\n\nUnmodified\n r2510 \\f$m = \\frac{\\sum w_i(1-u_i^2)^2x_i}{\\sum w_i(1-u_i^2)^2 } \\f$ where the sum runs over data points with |u|<1 and u is calculated as where the sums run over data points with |u|<1 and u is calculated as \\f$u_i = \\frac{x_i-m}{k*\\textrm{mad}} \\f$" ]

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[ "Solutions by everydaycalculation.com\n\n## 270 is what percent of 8820?\n\n270 of 8820 is 3.06%\n\n#### Steps to solve \"what percent is 270 of 8820?\"\n\n1. 270 of 8820 can be written as:\n270/8820\n2. To find percentage, we need to find an equivalent fraction with denominator 100. Multiply both numerator & denominator by 100\n\n270/8820 × 100/100\n3. = (270 × 100/8820) × 1/100 = 3.06/100\n4. Therefore, the answer is 3.06%\n\nIf you are using a calculator, simply enter 270÷8820×100 which will give you 3.06 as the answer.\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to work with percentages in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "## Thursday, June 30, 2011\n\n### Beating Kenneth French Small - High\n\nWith 148 pageviews over the last 24 hours, my post Kenneth French Gift to the Finance World has been popular relative to most of my other posts.  I think the popularity is due to Kenneth French’s notoriety and the amazing outperformance of Small Size and High Momentum stocks since 1927.  18% annualized returns are hard to beat, but I thought I should give it a try.  My first line of attack will be reducing the drawdown with some two popular and one not so popular price-based systems.\n\nGreat success was unfortunately not achieved, but I was able to reduce the drawdown to 50% from 79%.  If anybody has better ideas, please let me know.  I would love to see them and share them.\n\nTry a very subtle change by using the average return price series of all Small Size to determine the rank-based signal.  Performance is improved slightly.\n\n`#the real challenge now is how to beat the best#small momentum with a system#get very helpful Ken French data#for this project we will look at Momentum Portfolios#http://mba.tuck.dartmouth.edu/pages/faculty/ken.french/ftp/6_Portfolios_ME_Prior_12_2.zip   require(PerformanceAnalytics)require(quantmod)require(ggplot2)   my.url=\"http://mba.tuck.dartmouth.edu/pages/faculty/ken.french/ftp/6_Portfolios_ME_Prior_12_2.zip\"my.tempfile<-paste(tempdir(),\"\\\\frenchmomentum.zip\",sep=\"\")my.usefile<-paste(tempdir(),\"\\\\6_Portfolios_ME_Prior_12_2.txt\",sep=\"\")download.file(my.url, my.tempfile, method=\"auto\", \tquiet = FALSE, mode = \"wb\",cacheOK = TRUE)unzip(my.tempfile,exdir=tempdir(),junkpath=TRUE)#read space delimited text file extracted from zipfrench_momentum <- read.table(file=my.usefile,\theader = TRUE, sep = \"\",\tas.is = TRUE,\tskip = 12, nrows=1013)colnames(french_momentum) <- c(paste(\"Small\",\tcolnames(french_momentum)[1:3],sep=\".\"),\tpaste(\"Large\",colnames(french_momentum)[1:3],sep=\".\"))   #get dates ready for xts indexdatestoformat <- rownames(french_momentum)datestoformat <- paste(substr(datestoformat,1,4),\tsubstr(datestoformat,5,7),\"01\",sep=\"-\")   #get xts for analysisfrench_momentum_xts <- as.xts(french_momentum[,1:6],\torder.by=as.Date(datestoformat))   french_momentum_xts <- french_momentum_xts/100   charts.PerformanceSummary(french_momentum_xts[,1:3],ylog=TRUE,\tmain=\"Performance by Kenneth French Size and Momentum\tMonthly Since 1927\",\tcolorset=c(\"cadetblue3\",\"cadetblue\",\"cadetblue4\",\t\"darkolivegreen3\",\"darkolivegreen\",\"darkolivegreen4\"))mtext(\"Source: http://mba.tuck.dartmouth.edu/pages/faculty/ken.french/data_library.html\",\tside=1,adj=0,cex=0.75)#dev.off()     #get price series of small cap high momentum for system buildingfrench_momentum_price <- cumprod(1+french_momentum_xts[,3])   #speedy solution for ranking from Charles Berry# http://r.789695.n4.nabble.com/efficient-rolling-rank-td2013535.html#rolling rank of price over last 12 months (12 means max price for last 12)#pad first 11 with NAnper <- 10x.rank <- c(rep(NA,nper-1),rowSums(coredata(french_momentum_price)[ -(1:(nper-1)) ] >= embed(french_momentum_price,nper)))x.rank <- as.xts(x.rank,order.by=index(french_momentum_price))signalRank <- ifelse(x.rank[,1] > 3 | \tfrench_momentum_price >= runMax(french_momentum_price,2),1,0)retRank <- lag(signalRank,k=1)*french_momentum_xts[,3]   #try rolling 10 month moving average popularized by Mebane FabersignalAvg <- ifelse(french_momentum_price > runMean(french_momentum_price,n=10),1,0)retAvg <- lag(signalAvg,k=1)*french_momentum_xts[,3]   #try RSIsignalRSI <- ifelse(RSI(french_momentum_price,n=4) > 45,1,0)retRSI <- lag(signalRSI,k=1)*french_momentum_xts[,3]   retCompare <- merge(retRank,retAvg,retRSI,french_momentum_xts[,3])colnames(retCompare) <- c(\"Small.High.Rank\",\t\"Small.High.Avg\",\"Small.High.RSI\",\"Small.High.Benchmark\")#jpeg(filename=\"performance of small-high with systems.jpg\",quality=100,width=6.25, height = 8, units=\"in\",res=96)charts.PerformanceSummary(retCompare,ylog=TRUE,cex.legend=1.2,\tcolorset = c(\"cadetblue\",\"darkolivegreen3\",\"purple\",\"gray70\"),\tmain=\"Kenneth French Small Size and High Momentum Stocks\tCompared To Various Price Systems\")#dev.off()   #jpeg(filename=\"capture of small-high with systems.jpg\",quality=100,width=6.25, height = 8, units=\"in\",res=96)chart.CaptureRatios(retCompare[,1:3],retCompare[,4],\tmain=\"Kenneth French Small Size and High Momentum Stocks\tCompared To Various Price Systems\")#dev.off()   #get average of small size for additional system testingfrench_momentum_avg <- cumprod(1+apply(coredata(french_momentum_xts[,c(1:3)]),MARGIN=1,FUN=mean))french_momentum_avg <- as.xts(french_momentum_avg,order.by=index(french_momentum_price))nper <- 10x.rank <- c(rep(NA,nper-1),rowSums(coredata(french_momentum_avg)[ -(1:(nper-1)) ] >= embed(french_momentum_avg,nper)))x.rank <- as.xts(x.rank,order.by=index(french_momentum_avg))signalRankAvg <- ifelse(x.rank[,1] > 3 | \tfrench_momentum_price >= runMax(french_momentum_price,2),1,0)retRankAvg <- lag(signalRankAvg,k=1)*french_momentum_xts[,3]   retCompare <- merge(retRank,retRankAvg,french_momentum_xts[,3])colnames(retCompare) <- c(\"Small.High.Rank\",\t\"Small.High.RankOnAvg\",\"Small.High.Benchmark\")#jpeg(filename=\"performance of small-high with 2 rank systems.jpg\",quality=100,width=6.25, height = 8, units=\"in\",res=96)charts.PerformanceSummary(retCompare,ylog=TRUE,cex.legend=1.2,\tcolorset = c(\"cadetblue\",\"darkolivegreen3\",\"gray70\"),\tmain=\"Kenneth French Small Size and High Momentum Stocks\tCompared To Rank-based Price Systems\")#dev.off()`" ]

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[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Thanks to everyone who made a donation during our annual appeal!\nTo see the list of donors, or make a donation, see the OEIS Foundation home page.\n\n Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A087897 Number of partitions of n into odd parts greater than 1. 22\n\n%I\n\n%S 1,0,0,1,0,1,1,1,1,2,2,2,3,3,4,5,5,6,8,8,10,12,13,15,18,20,23,27,30,\n\n%T 34,40,44,50,58,64,73,83,92,104,118,131,147,166,184,206,232,256,286,\n\n%U 320,354,394,439,485,538,598,660,730,809,891,984,1088,1196,1318,1454,1596,1756\n\n%N Number of partitions of n into odd parts greater than 1.\n\n%C Also number of partitions of n into distinct parts which are not powers of 2.\n\n%C Also number of partitions of n into distinct parts such that the two largest parts differ by 1.\n\n%C Also number of partitions of n such that the largest part occurs an odd number of times that is at least 3 and every other part occurs an even number of times. Example: a(10) = 2 because we have [2,2,2,1,1,1,1] and [2,2,2,2,2]. - _Emeric Deutsch_, Mar 30 2006\n\n%C Also difference between number of partitions of 1+n into distinct parts and number of partitions of n into distinct parts. - Philippe LALLOUET, May 08 2007\n\n%C In the Berndt reference replace {a -> -x, q -> x} in equation (3.1) to get f(x). G.f. is 1 - x * (1 - f(x)).\n\n%C Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).\n\n%C Also number of symmetric unimodal compositions of n+3 where the maximal part appears three times. - _Joerg Arndt_, Jun 11 2013\n\n%C Let c(n) = number of palindromic partitions of n whose greatest part has multiplicity 3; then c(n) = a(n-3) for n>=3. - _Clark Kimberling_, Mar 05 2014\n\n%D J. W. L. Glaisher, Identities, Messenger of Mathematics, 5 (1876), pp. 111-112. see Eq. I\n\n%H Alois P. Heinz, <a href=\"/A087897/b087897.txt\">Table of n, a(n) for n = 0..1000</a>\n\n%H C. Ballantine, M. Merca, <a href=\"https://www.researchgate.net/publication/289250007_Padovan_numbers_as_sums_over_partitions_into_odd_parts\">Padovan numbers as sums over partitions into odd parts</a>, Journal of Inequalities and Applications, (2016) 2016:1; <a href=\"https://doi.org/10.1186/s13660-015-0952-5\">doi</a>.\n\n%H B. C. Berndt, B. Kim, and A. J. Yee, <a href=\"http://dx.doi.org/10.1016/j.jcta.2009.07.005\">Ramanujan's lost notebook: Combinatorial proofs of identities associated with Heine's transformation or partial theta functions</a>, J. Comb. Thy. Ser. A, 117 (2010), 957-973.\n\n%H Howard D. Grossman, <a href=\"https://www.jstor.org/stable/3029861\">Problem 228</a>, Mathematics Magazine, 28 (1955), p. 160.\n\n%H R. K. Guy, <a href=\"http://www.jstor.org/stable/3609388\">Two theorems on partitions</a>, Math. Gaz., 42 (1958), 84-86. Math. Rev. 20 #3110.\n\n%H Cristiano Husu, <a href=\"https://arxiv.org/abs/1804.09883\">The butterfly sequence: the second difference sequence of the numbers of integer partitions with distinct parts, its pentagonal number structure, its combinatorial identities and the cyclotomic polynomials 1-x and 1+x+x^2</a>, arXiv:1804.09883 [math.NT], 2018.\n\n%H M. Somos, <a href=\"/A010815/a010815.txt\">Introduction to Ramanujan theta functions</a>\n\n%H Eric Weisstein's World of Mathematics, <a href=\"http://mathworld.wolfram.com/RamanujanThetaFunctions.html\">Ramanujan Theta Functions</a>\n\n%F Expansion of q^(-1/24) * (1 - q) * eta(q^2) / eta(q) in powers of q.\n\n%F Expansion of (1 - x) / chi(-x) in powers of x where chi() is a Ramanujan theta function.\n\n%F G.f.: 1 + x^3 + x^5*(1 + x) + x^7*(1 + x)*(1 + x^2) + x^9*(1 + x)*(1 + x^2)*(1 + x^3) + ... [Glaisher 1876]. - _Michael Somos_, Jun 20 2012\n\n%F G.f.: Product_{k >= 1} 1/(1-x^(2*k+1)).\n\n%F G.f.: Product_{k >= 1, k not a power of 2} (1+x^k).\n\n%F G.f.: sum(x^(3k)/product(1-x^(2j), j=1..k), k=1..infinity). - _Emeric Deutsch_, Mar 30 2006\n\n%F a(n) ~ exp(Pi*sqrt(n/3)) * Pi / (8 * 3^(3/4) * n^(5/4)) * (1 - (15*sqrt(3)/(8*Pi) + 11*Pi/(48*sqrt(3)))/sqrt(n) + (169*Pi^2/13824 + 385/384 + 315/(128*Pi^2))/n). - _Vaclav Kotesovec_, Aug 30 2015, extended Nov 04 2016\n\n%e 1 + x^3 + x^5 + x^6 + x^7 + x^8 + 2*x^9 + 2*x^10 + 2*x^11 + 3*x^12 + 3*x^13 + ...\n\n%e q + q^73 + q^121 + q^145 + q^169 + q^193 + 2*q^217 + 2*q^241 + 2*q^265 + ...\n\n%e a(10)=2 because we have [7,3] and [5,5].\n\n%e From _Joerg Arndt_, Jun 11 2013: (Start)\n\n%e There are a(22)=13 symmetric unimodal compositions of 22+3=25 where the maximal part appears three times:\n\n%e 01: [ 1 1 1 1 1 1 1 1 3 3 3 1 1 1 1 1 1 1 1 ]\n\n%e 02: [ 1 1 1 1 1 1 2 3 3 3 2 1 1 1 1 1 1 ]\n\n%e 03: [ 1 1 1 1 1 5 5 5 1 1 1 1 1 ]\n\n%e 04: [ 1 1 1 1 2 2 3 3 3 2 2 1 1 1 1 ]\n\n%e 05: [ 1 1 1 2 5 5 5 2 1 1 1 ]\n\n%e 06: [ 1 1 2 2 2 3 3 3 2 2 2 1 1 ]\n\n%e 07: [ 1 1 3 5 5 5 3 1 1 ]\n\n%e 08: [ 1 1 7 7 7 1 1 ]\n\n%e 09: [ 1 2 2 5 5 5 2 2 1 ]\n\n%e 10: [ 1 4 5 5 5 4 1 ]\n\n%e 11: [ 2 2 2 2 3 3 3 2 2 2 2 ]\n\n%e 12: [ 2 3 5 5 5 3 2 ]\n\n%e 13: [ 2 7 7 7 2 ]\n\n%e (End)\n\n%p To get 128 terms: t4 := mul((1+x^(2^n)),n=0..7); t5 := mul((1+x^k),k=1..128): t6 := series(t5/t4,x,100); t7 := seriestolist(t6);\n\n%p # second Maple program:\n\n%p b:= proc(n, i) option remember; `if`(n=0, 1,\n\n%p `if`(i<3, 0, b(n, i-2)+`if`(i>n, 0, b(n-i, i))))\n\n%p end:\n\n%p a:= n-> b(n, n-1+irem(n, 2)):\n\n%p seq(a(n), n=0..80); # _Alois P. Heinz_, Jun 11 2013\n\n%t max = 65; f[x_] := Product[ 1/(1 - x^(2k+1)), {k, 1, max}]; CoefficientList[ Series[f[x], {x, 0, max}], x] (* _Jean-François Alcover_, Dec 16 2011, after _Emeric Deutsch_ *)\n\n%t b[n_, i_] := b[n, i] = If[n==0, 1, If[i<3, 0, b[n, i-2]+If[i>n, 0, b[n-i, i]]] ]; a[n_] := b[n, n-1+Mod[n, 2]]; Table[a[n], {n, 0, 80}] (* _Jean-François Alcover_, Apr 01 2015, after _Alois P. Heinz_ *)\n\n%t Flatten[{1, Table[PartitionsQ[n+1] - PartitionsQ[n], {n, 0, 80}]}] (* _Vaclav Kotesovec_, Dec 01 2015 *)\n\n%o (PARI) {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( (1 - x) * eta(x^2 + A) / eta(x + A), n))} /* _Michael Somos_, Nov 13 2011 */\n\n%o a087897 = p [3,5..] where\n\n%o p [] _ = 0\n\n%o p _ 0 = 1\n\n%o p ks'@(k:ks) m | m < k = 0\n\n%o | otherwise = p ks' (m - k) + p ks m\n\n%o -- _Reinhard Zumkeller_, Aug 12 2011\n\n%Y Cf. A000009, A002865.\n\n%K nonn\n\n%O 0,10\n\n%A _N. J. A. Sloane_, Dec 04 2003\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified January 21 11:07 EST 2020. Contains 331105 sequences. (Running on oeis4.)" ]

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[ "# Calculating The Fair Value Of Ecolab Inc. (NYSE:ECL)\n\nBy\nSimply Wall St\nPublished\nJanuary 01, 2022\n\nIn this article we are going to estimate the intrinsic value of Ecolab Inc. (NYSE:ECL) by estimating the company's future cash flows and discounting them to their present value. We will use the Discounted Cash Flow (DCF) model on this occasion. Models like these may appear beyond the comprehension of a lay person, but they're fairly easy to follow.\n\nWe generally believe that a company's value is the present value of all of the cash it will generate in the future. However, a DCF is just one valuation metric among many, and it is not without flaws. Anyone interested in learning a bit more about intrinsic value should have a read of the Simply Wall St analysis model.\n\nCheck out our latest analysis for Ecolab\n\n### Step by step through the calculation\n\nWe're using the 2-stage growth model, which simply means we take in account two stages of company's growth. In the initial period the company may have a higher growth rate and the second stage is usually assumed to have a stable growth rate. To begin with, we have to get estimates of the next ten years of cash flows. Where possible we use analyst estimates, but when these aren't available we extrapolate the previous free cash flow (FCF) from the last estimate or reported value. We assume companies with shrinking free cash flow will slow their rate of shrinkage, and that companies with growing free cash flow will see their growth rate slow, over this period. We do this to reflect that growth tends to slow more in the early years than it does in later years.\n\nGenerally we assume that a dollar today is more valuable than a dollar in the future, so we discount the value of these future cash flows to their estimated value in today's dollars:\n\n#### 10-year free cash flow (FCF) forecast\n\n 2022 2023 2024 2025 2026 2027 2028 2029 2030 2031 Levered FCF (\\$, Millions) US\\$1.63b US\\$2.05b US\\$2.63b US\\$2.89b US\\$3.16b US\\$3.36b US\\$3.53b US\\$3.68b US\\$3.81b US\\$3.92b Growth Rate Estimate Source Analyst x4 Analyst x3 Analyst x1 Analyst x1 Analyst x1 Est @ 6.38% Est @ 5.05% Est @ 4.12% Est @ 3.48% Est @ 3.02% Present Value (\\$, Millions) Discounted @ 6.3% US\\$1.5k US\\$1.8k US\\$2.2k US\\$2.3k US\\$2.3k US\\$2.3k US\\$2.3k US\\$2.2k US\\$2.2k US\\$2.1k\n\n(\"Est\" = FCF growth rate estimated by Simply Wall St)\nPresent Value of 10-year Cash Flow (PVCF) = US\\$21b\n\nWe now need to calculate the Terminal Value, which accounts for all the future cash flows after this ten year period. For a number of reasons a very conservative growth rate is used that cannot exceed that of a country's GDP growth. In this case we have used the 5-year average of the 10-year government bond yield (2.0%) to estimate future growth. In the same way as with the 10-year 'growth' period, we discount future cash flows to today's value, using a cost of equity of 6.3%.\n\nTerminal Value (TV)= FCF2031 × (1 + g) ÷ (r – g) = US\\$3.9b× (1 + 2.0%) ÷ (6.3%– 2.0%) = US\\$91b\n\nPresent Value of Terminal Value (PVTV)= TV / (1 + r)10= US\\$91b÷ ( 1 + 6.3%)10= US\\$49b\n\nThe total value, or equity value, is then the sum of the present value of the future cash flows, which in this case is US\\$71b. In the final step we divide the equity value by the number of shares outstanding. Compared to the current share price of US\\$235, the company appears about fair value at a 4.7% discount to where the stock price trades currently. The assumptions in any calculation have a big impact on the valuation, so it is better to view this as a rough estimate, not precise down to the last cent.\n\n### The assumptions\n\nNow the most important inputs to a discounted cash flow are the discount rate, and of course, the actual cash flows. Part of investing is coming up with your own evaluation of a company's future performance, so try the calculation yourself and check your own assumptions. The DCF also does not consider the possible cyclicality of an industry, or a company's future capital requirements, so it does not give a full picture of a company's potential performance. Given that we are looking at Ecolab as potential shareholders, the cost of equity is used as the discount rate, rather than the cost of capital (or weighted average cost of capital, WACC) which accounts for debt. In this calculation we've used 6.3%, which is based on a levered beta of 1.002. Beta is a measure of a stock's volatility, compared to the market as a whole. We get our beta from the industry average beta of globally comparable companies, with an imposed limit between 0.8 and 2.0, which is a reasonable range for a stable business.\n\n### Moving On:\n\nAlthough the valuation of a company is important, it ideally won't be the sole piece of analysis you scrutinize for a company. DCF models are not the be-all and end-all of investment valuation. Instead the best use for a DCF model is to test certain assumptions and theories to see if they would lead to the company being undervalued or overvalued. For example, changes in the company's cost of equity or the risk free rate can significantly impact the valuation. For Ecolab, we've compiled three important elements you should look at:\n\n1. Risks: To that end, you should be aware of the 2 warning signs we've spotted with Ecolab .\n2. Management:Have insiders been ramping up their shares to take advantage of the market's sentiment for ECL's future outlook? Check out our management and board analysis with insights on CEO compensation and governance factors.\n3. Other High Quality Alternatives: Do you like a good all-rounder? Explore our interactive list of high quality stocks to get an idea of what else is out there you may be missing!\n\nPS. Simply Wall St updates its DCF calculation for every American stock every day, so if you want to find the intrinsic value of any other stock just search here.\n\n### Discounted cash flow calculation for every stock\n\nSimply Wall St does a detailed discounted cash flow calculation every 6 hours for every stock on the market, so if you want to find the intrinsic value of any company just search here. It’s FREE.\n\n### Make Confident Investment Decisions\n\nSimply Wall St's Editorial Team provides unbiased, factual reporting on global stocks using in-depth fundamental analysis.\nFind out more about our editorial guidelines and team." ]

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[ "# 9-1 Perimeter and Circumference Warm Up Problem of the Day", null, "```9-1 Perimeter and Circumference\nWarm Up\nProblem of the Day\nLesson Presentation\nLesson Quizzes\n9-1 Perimeter and Circumference\nWarm Up\n1. 8.2 + 5.6 + 8.2 + 5.6\n27.6\n2. 12.4 + 15.8 + 9.3\n37.5\n3. 22 &middot; 3.14\n69.08\n9-1 Perimeter and Circumference\nProblem of the Day\nA rectangular piece of sheet metal\nmeasures 5 in. by 6 in. A rectangular\n7\nnotch 1 in. deep and\nin.\nwide\nis\ncut\n64\nout of the middle of each side. What is\nthe perimeter of the piece after the\ncuts have been made? (Hint: You don’t\nhave to use fractions.)\n30 in. (22 in. plus 8 in. for the eight 1 in.\ncuts)\n9-1 Perimeter and Circumference\nI can find the perimeter of a polygon and\nthe circumference of a circle.\n9-1 Perimeter and Circumference\nVocabulary\nperimeter\ncircumference\npi\n9-1 Perimeter and Circumference\nPerimeter is the distance around a\ngeometric figure. To find the perimeter P\nof a polygon, you can add the lengths of\nits sides.\n9-1 Perimeter and Circumference\nAdditional Example 1: Finding the Perimeter of a\nPolygon\nFind the perimeter.\n25 cm\n25 cm\n15 cm\nP = 25 + 25 + 15\nUse the side lengths.\nP = 65\nThe perimeter of the triangle is 65 cm.\n9-1 Perimeter and Circumference\nCheck It Out: Example 1\nFind the perimeter of the polygon.\n25 cm\n20 cm\n30 cm\nP = 25 + 20 + 30\nUse the side lengths.\nP = 75\nThe perimeter of the triangle is 75 cm.\n9-1 Perimeter and Circumference\n9-1 Perimeter and Circumference\nAdditional Example 2: Using Properties of a\nRectangle to Find Perimeter\nFind the perimeter of a rectangle.\n14 ft\n18 ft\nP\nP\nP\nP\n=\n=\n=\n=\n2l + 2w\nUse the formula.\n(2 &middot; 18) + (2 &middot;14) Substitute for l and w.\nMultiply.\n36 + 28\n64\nThe perimeter of the rectangle is 64 ft.\n9-1 Perimeter and Circumference\nCheck It Out: Example 2\nFind the perimeter of a rectangle.\n16 ft\n8 ft\nP\nP\nP\nP\n=\n=\n=\n=\n2l + 2w\n(2 &middot; 16) + (2 &middot; 8)\n32 + 16\n48\nUse the formula.\nSubstitute for l and w.\nMultiply.\nThe perimeter of the rectangle is 48 ft.\n9-1 Perimeter and Circumference\nThe distance around a circle is called circumference.\nFor every circle, the ratio of circumference C to\ndiameter d is the same. This ratio, C , is\nd\nrepresented by the Greek letter , called pi. Pi is\napproximately equal to 3.14 or 22 . By solving the\n7\nC\nequation =  for C, you get the formula for\nd\ncircumference.\n9-1 Perimeter and Circumference\n9-1 Perimeter and Circumference\nAdditional Example 3A: Finding the Circumference\nof a Circle\nFind the circumference of the circle to the\nnearest tenth. Use 3.14 for .\n12 in.\nC = d\nYou know the diameter.\nC  3.14 &middot; 12 Substitute for  and d.\nC  37.68\nMultiply.\nThe circumference of the circle is about 37.7 in.\n9-1 Perimeter and Circumference\nAdditional Example 3B: Finding the Circumference\nof a Circle\nFind the circumference of the circle to the\nnearest tenth. Use 3.14 for .\n18 cm\nC = 2r\nC  2 &middot; 3.14 &middot; 18 Substitute for  and r.\nC  113.04\nMultiply.\nThe circumference of the circle is about 113.0 cm.\n9-1 Perimeter and Circumference\nCheck It Out: Example 3A\nFind the circumference of the circle to the\nnearest tenth. Use 3.14 for .\n7 in.\nC = d\nYou know the diameter.\nC  3.14 &middot; 7\nSubstitute for  and d.\nC  21.98\nMultiply.\nThe circumference of the circle is about 22.0 in.\n9-1 Perimeter and Circumference\nI can find the perimeter of a polygon and the\ncircumference of a circle.\nFind the circumference of the circle to the\nnearest tenth. Use 3.14 for .\n11 cm\nC = 2r\nC  2 &middot; 3.14 &middot; 11 Substitute for  and r.\nC  69.08\nMultiply.\nThe circumference of the circle is about 69.1 cm.\n9-1 Perimeter and Circumference\nIf the diameter or radius of a circle is a\n22\nmultiple of 7, use for .\n7\n9-1 Perimeter and Circumference\nI can find the perimeter of a polygon and the\ncircumference of a circle.\nThe diameter of a circular pond is 42 m. What\nis its circumference? Use 22 for .\n7\nC = d\nYou know the diameter.\nC  22 &middot; 42\n7\nC  22 &middot; 42\n7 1\n22 &middot; 42 6\nC 7 1\n1\nC  132\nSubstitute 22 for  and 42 for d.\n7\nWrite 42 as a fraction.\nSimplify.\nMultiply.\nThe circumference of the pond is about 132 m.\n9-1 Perimeter and Circumference\nCheck It Out: Example 4\nThe diameter of a circular spa is 14 m. What\nis its circumference? Use 22 for .\n7\nC = d\nYou know the diameter.\nC  22 &middot; 14\n7\nC  22 &middot; 14\n7 1\n22 &middot; 14 2\nC 7 1\n1\nC  44\nSubstitute 22 for  and 14 for d.\n7\nWrite 14 as a fraction.\nSimplify.\nMultiply.\nThe circumference of the spa is about 44 m.\n```" ]

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[ "", null, "CZECH TECHNICAL UNIVERSITY IN PRAGUE\nSTUDY PLANS\n2021/2022\n\n# Mathematical Applications\n\nCode Completion Credits Range Language\nBE5B15MAA Z,ZK 4 0P+4C English\nLecturer:\nJan Kyncl (guarantor), Václav Vencovský, Stanislav Vítek\nTutor:\nJan Kyncl (guarantor), Václav Vencovský, Stanislav Vítek\nSupervisor:\nDepartment of Electrical Power Engineering\nSynopsis:\n\nThe aim of the course is to obtain knowledge about mathematic programs used in electrical engineering. Student will acquire basic knowledge about MATLAB, MATHEMATICA and mathematical model assessment.\n\nRequirements:\n\nRequirements to obtain the assessment are exercises attendance and a term thesis elaborating.\n\nSuccessful exam passing is determined in the Study and Examination Code of CTU in Prague.\n\nSyllabus of lectures:\n\n1. Introduction into Matlab: command prompt and expressions, variables, lists, vectors, matrices, basic operations, complex numbers, basic plotting and graph manipulations, using files\n\n2. Signal processing in Matlab 1: data visualization, filtering, spectral analysis.\n\n3. Signal processing in Matlab 2: convolution, image processing\n\n4. Functions in Matlab 1: control flow statements (conditions, loops, switches), debugging, user defined functions\n\n5. Functions in Matlab 2: advanced functions in Matlab, graphical user interfaces\n\n6. Introduction into Simulink: basic blocks, wiring techniques, examples\n\n7. Introduction into Wolfram Mathematica: basic syntax, help, variables, complex numbers, basic plotting and graph manipulations\n\n8. Principles of procedural, functional and pattern-oriented programming\n\n9. Functions in Mathematica: Blank, Set, SetDelayed, Module\n\n10. Functions in Mathematica: piecewise functions, overloaded functions, tracing the evaluation of functions\n\n11. Creation and manipulation of Lists, functions and Lists: Table, Part, Map, Apply\n\n12. Using Nest and NestList for numerical algorithms, Euler and Runge-Kutta method\n\n13. Examples of advanced use of Mathematica\n\n14. Test\n\nSyllabus of tutorials:\n\n1. Introduction into Matlab: command prompt and expressions, variables, lists, vectors, matrices, basic operations, complex numbers, basic plotting and graph manipulations, using files\n\n2. Signal processing in Matlab 1: data visualization, filtering, spectral analysis.\n\n3. Signal processing in Matlab 2: convolution, image processing\n\n4. Functions in Matlab 1: control flow statements (conditions, loops, switches), debugging, user defined functions\n\n5. Functions in Matlab 2: advanced functions in Matlab, graphical user interfaces\n\n6. Introduction into Simulink: basic blocks, wiring techniques, examples\n\n7. Introduction into Wolfram Mathematica: basic syntax, help, variables, complex numbers, basic plotting and graph manipulations\n\n8. Principles of procedural, functional and pattern-oriented programming\n\n9. Functions in Mathematica: Blank, Set, SetDelayed, Module\n\n10. Functions in Mathematica: piecewise functions, overloaded functions, tracing the evaluation of functions\n\n11. Creation and manipulation of Lists, functions and Lists: Table, Part, Map, Apply\n\n12. Using Nest and NestList for numerical algorithms, Euler and Runge-Kutta method\n\n13. Examples of advanced use of Mathematica\n\n14. Test\n\nStudy Objective:\nStudy materials:\n\nD. Hanselman and B. Littlefield. Mastering MATLAB. Pearson/Prentice Hall, Englewood Cliffs, NJ, 2011.\n\nS. Wolfram: An Elementary Introduction to the Wolfram Language, Second Edition. Wolfram Media, Inc. 2017\n\nStudy materials are available at www.powerwiki.cz.\n\nNote:\nFurther information:\nhttp://www.powerwiki.cz/wiki/VyukaEn\nTime-table for winter semester 2021/2022:\nTime-table is not available yet\nTime-table for summer semester 2021/2022:\n 06:00–08:0008:00–10:0010:00–12:0012:00–14:0014:00–16:0016:00–18:0018:00–20:0020:00–22:0022:00–24:00 roomT2:A3-41209:15–12:30(parallel nr.102)DejviceLaborator roomT2:A3-412Kyncl J.Vencovský V.14:30–17:45(parallel nr.101)DejviceLaborator\nThe course is a part of the following study plans:\nData valid to 2022-08-09\nFor updated information see http://bilakniha.cvut.cz/en/predmet4356106.html" ]

[ null, "https://bilakniha.cvut.cz/logo-cs.svg", null ]

[ "## Lab Test: Tensile Testing Essay\n\nThe mechanical properties of materials are determined by performing carefully designed laboratory experiments that replicate as nearly as possible the service conditions. In real life, there are many factors involved in the nature in which loads are applied on a material. The following are some common examples of modes in which loads might be applied: tensile, compressive, and shear. These properties are important In materials selections for mechanical design. Other factors that often complicate the design process Include temperature and time factors.\n\nThe topic of this lab is confined to the tensile property of polymers. Figure 1 shows a tensile testing machine similar to the one used in this lab. This test is a destructive method, in which a specimen of a standard shape and dimensions (prepared according to ASTM D 638: standard test method for tensile properties of plastics) is subjected to an axial load. During a typical tensile experiment, a dog-bone shaped specimen Is gripped at Its two ends and Is pulled to elongate at a determined rate to Its breakpoint; a highly ductile polymer may not reach its breakpoint.\n\nWe Will Write a Custom Essay Specifically\nFor You For Only \\$13.90/page!\n\norder now\n\nThe tensile tester seed in this lab is manufactured by Insertion (model 5569). It has a maximum load of 2 or 50 ink and a variable pulling rate. The setup of the experiment could be changed to accommodate different types of mechanical testing, according to the ASTM standard (e. G. Compression test, etc). For analytical purposes, a plot of stress (o) versus strain (E) Is constructed during a tensile test experiment, which can be done automatically on the software provided by the instrument manufacturer. Stress, in the metric system, is usually measured in N/ mm or Pa, such that 1 N/mm = 1 Pa.\n\nFrom the experiment, the value of stress is lactated by dividing the amount of force (F) applied by the machine in the axial direction by its cross-sectional area (A), which is measured prior to running the experiment. Mathematically, It Is expressed In Equation 1. The strain values, which have no units, can be calculated using Equation 2, where L Is the Instantaneous length of the specimen and LO Is the initial length. (Equation 1) (Equation 2) A typical stress-strain curve would look like Figure 2. The stress-strain curve shown In Figure 2 Is a textbook example of a stress-strain curve.\n\nIn reality, not all stress-strain curves perfectly resemble the one shown In Figure 2. This stress-strain curve Is typical for ductile metallic elements. Another thing to take note is that Figure 2 shows an “engineering stress-strain” curve. When a material reaches its ultimate stress strength of the stress-strain curve, its cross-sectional area reduces dramatically, a term known as necking. When the computer software plots the stress-strain curve, it assumes that the cross sectional area stays constant throughout the experiment, even during necking, therefore causing the curve to slope down.\n\nThe “true” stress- change in the cross sectional area of the specimen throughout the experiment. Theoretically, even without measuring the cross-sectional area of the specimen during the tensile experiment, the “true” stress-strain curve could still be constructed by assuming that the volume of the material stays the same. Using this concept, both the true stress (UT) and the true strain (ET) could be calculated using Equations 3 and 4, respectively. The derivation of these equations is beyond the scope of this lab report. Consult any standard mechanics textbook to learn more about these equations.\n\nIn these equations, LO refers to the initial length of the specimen, L refers o the instantaneous length and o refers to the instantaneous stress. (Equation 3) (Equation 4) Figure 2 also shows that a stress-strain curve is divided into four regions: elastic, yielding, strain hardening (commonly occurs in metallic materials), and necking. The area under the curve represents the amount of energy needed to accomplish each of these “events. ” The total area under the curve (up to the point of fracture) is also known as the modulus of toughness.\n\nThis represents the amount of energy needed to break the sample, which could be compared to the impact energy of the sample, determined from impact tests. The area under the linear region of the curve is known as the modulus of resilience. This represents the minimum amount of energy needed to deform the sample. The linear region of the curve of Figure 2, which is called the elastic region (past this region, is called the plastic region), is the region where a material behaves elastically. The material will return to its original shape when a force is released while the material is in its elastic region.\n\nThe slope of the curve, which can be calculated using Equation 5, is a constant and is an intrinsic property of material known as the elastic modulus, E. In metric units, it is usually expressed in Pascal (Pa). (Equation 5) Figure 3(a) shows typical stress-strain curves of polymers. The figure shows that materials that are hard and brittle do not deform very much before breaking and have very steep elastic modulo. The mechanical property of polymers generally depends on their degree of crystalline, molecular weights and glass transition temperature, Tug.\n\nHighly crystalline polymeric materials with a Tug above the room temperature are usually brittle, and vice versa. When a semi-crystalline polymer undergoes a tensile test, the amorphous chains, will become aligned. This is usually evident for transparent and translucent materials, which become opaque upon turning crystalline. Figure 3(b) shows a diagram showing the mechanical property of some common polymers. Important! Make sure you wear safety glasses before starting any operation. Your eyes could be hurt by a broken piece of polymer. Also wear gloves to protect against any residue on the machine and samples. . 1 Specimen Preparation The polymer specimens were injection-molded into dog-bone shapes. Their dimensions were determined according to the ASTM D 638 standard mentioned earlier in the introduction. (1) Measure the thickness, width and gage length of polymer samples in mm. These dimensions should be approximately the same for each sample. (2) Also make note of any sample defects (e. G. Impurities, air bubbles, etc. ). The following samples will be tested: 1) Polypropylene (UP), polystyrene (AS), polycyclic acid (polymer), high density polyethylene (HIDE), and Dentally for analysis of mechanical properties. ) Polystyrene: to compare effects of feeding direction on mechanical properties. 3) Polypropylene: to analyze effects of strain rate on mechanical properties. . 2 Bluebill Software Setup 1) Turn on the tensile test machine. The switch is located on the right side of the machine. Also turn on the video extensors. (2)Go to the desktop and double-click on the “Bluebill” icon. (3) On the main page, select Test to start a new sample. Name your test and click Browse to select the folder you would like to save it in. Click next. (4) Choose which method you would like to use.\n\nCreate and save a new method if needed. (5) Method set up: Save after any changes are made. General: used for display purposes Specimen: specifies sample dimensions and parameters. A doggone sample is used for tensile testing. Select rectangular, and specify the width, thickness and gauge length of the sample. The gauge length is the distance between the clamps before starting the test. Control: describes the actual test. Select extension for mode of displacement, then specify the rate of extension. Most use 5 mm/min or 50 min/mm, depending on if you want a slow or fast test.\n\nEnd of Test: identifies the criteria for the end of the test. A large load drop is experienced when sample failure occurs. For this test, when the sample load drops by a certain percentage of the peak load, he machine will stop. Data: specifies if the data is acquired manually or automatically, while the strain tab recognizes whether the strain is measured from the video exterminates or the extension. Results and Graphs: select what data is shown and how it is displayed. (1) Make sure the proper load cell is installed, either 2 ink or 50 ink depending on the load range and sensitivity of the sample.\n\nTo switch load cells, make sure the machine is off. Unscrew the bolts and remove using the handle. Make sure to plug the new load cell into the port behind the machine. (2) Calibrate the load cell by licking on the button in the upper right hand corner. Make sure all loads are removed from the load cell and click calibrate. (3) Install the correct type of clamps for the testing. For tensile testing, non or ink samples can be used. Install the clamps using the pins. Also install height brackets if needed. Zero the load once the clamps are installed. 4) Press the up and down arrows on the controller until the clamps are Just touching. Press the reset gauge length button at the top of the screen to zero the position of the clamps. (5) Use the up and down arrows until the clamps are about 100 mm apart. This is a typical gauge length for the dog bone samples. (6) Place the polymer sample between the grips of both the tensile test machine. While holding the sample vertically with one hand, use another hand to turn the handle of the top grip in the closing direction as tightly as possible. The specimen should be gripped such that the two ends of the specimen are covered by the grip, approximately 3 mm away from its gage-length. It is important that the specimens are tightly gripped onto the specimen grips to prevent slipping, which will otherwise result in experimental errors. ) (8) Make sure that the specimen s vertically aligned, if not a torsion’s force, rather than axial force, will result. (9) Turn the bottom handle in the “close” direction as tightly as possible. Visually verify that the sample is gripped symmetrically at its two ends. 10)Zero the extension by pushing zero extension button at the top of the screen. Also zero the load if needed. Wait for a few seconds to let the computer return its value to zero. 2. 4 Tensile Test (1) Enter geometry of the sample before starting. (2) Click on the Start button. Both the upper and bottom grips will start moving in opposite directions according to the specified pulling rate. Observe the experiment at a safe distance (about 1. 5 meters away) at an angle and take note of the failure mode when the specimen fails. (NOTE: Be sure to wear safety glasses.\n\nDo not come close to equipment when the tensile test is running). (3) A plot of tensile stress (Amp) versus tensile strain (mm/mm) will be generated in real-time during the experiment. 2. 6 End of Test (1) The machine will stop automatically when the sample is broken. (2) Press the “Return” button on the digital controller. Both the upper and lower grips will be returned to their original positions automatically. 3) Turn the two handles in the open directions to remove the sample (4) Repeat the previous steps for any additional tests. 5) When finished, save your file and click Finish. This will export your data into a PDF and individual data files. (6) Clean up any broken fragments from the specimens. (7) Turn off the machine and exit the program when finished. Graph UP (50 mm/mm extension), AS (2 feed inputs), PLAN, HIDE and Dentally results using raw data files. There should be two tests for each polymer, but Just pick one to graph. Construct the true stress-strain curves for each polymer (hint: use Equations 3) and (4) provided in the Introduction section).\n\nCalculate Young Modulus for each material and testing condition and compare experimental values with literature values. Discuss any differences in mechanical behavior between the polymers (use pictures! ) Analyze the fracture modes of each sample (ductile fracture, brittle fracture, or intermediate fracture mode). Using the data for polypropylene, discuss the effects of strain rate on the mechanical behavior of the polymers. Using the data for polystyrene, compare effects of feed direction on the mechanical behavior. Explain any unexpected results.\n\nx", null, "Hi!\nI'm Heidi!\n\nWould you like to get a custom essay? How about receiving a customized one?\n\nCheck it out" ]

[ null, "https://randomuser.me/api/portraits/women/77.jpg", null ]

[ "# Checking cone containment\n\nI would like to check if one cone contains another cone, where the cones are convex rational polyhedral cones. I don't want to compute the intersection of two cones, because when done many times, that becomes prohibitively slow. Checking containment should be much faster. Is this possible in Sage? Thanks!\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\n\nAssuming that your cones only contains the vertex 0, you can use containment as follows\n\nsage: def polyhedron_contains(P, Q):\n....: return all(P.contains(v) for v in Q.rays())\nsage: P1 = Polyhedron(rays = [(1,0,0), (0,1,0), (0,0,1)])\nsage: P2 = Polyhedron(rays = [(1,2,1), (1,0,0), (1,1,1), (1,1,2)])\nsage: polyhedron_contains(P1,P2)\nTrue\n\nmore" ]

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[ "# Quantum Bayesian Networks\n\n## March 9, 2019\n\n### Current Plans for Qubiter and when is walking backwards a good thing to do?\n\nFiled under: Uncategorized — rrtucci @ 6:21 pm\n\nThis post is to keep Qubiter fans abreast of my current plans for it.\n\nAs I mentioned in a previous post entitled “Welcome to the Era of TensorFlow and PyTorch Enabled Quantum Computer Simulators”, I have recently become a big fan of the program PennyLane and its main architect, Josh Izaak.\n\nDid you know PennyLane has a Discourse? https://discuss.pennylane.ai/ I love Discourse forums. Lots of other great software (like PyMC, for instance) have discourses too.\n\nI am currently working hard to PennyLanate my Qubiter. In other words, I am trying to make Qubiter do what PennyLane already does, to wit: (1)establish a feedback loop between my classical computer and the quantum computer cloud service of Rigetti (2) When it’s my computer’s turn to act in the feedback loop, make it do minimization using the method of back-propagation. A glib way of describing this process is: a feedback loop which does forward-propagation in the Rigetti qc, followed by backwards-propagation in my computer, followed by forward-propagation in the Rigetti qc, and on and on, ad nauseam.\n\nI am proud to report that Qubiter’s implementation of (1) is pretty much finished. The deed is done. See my Qubiter module https://github.com/artiste-qb-net/qubiter/blob/master/adv_applications/MeanHamil_rigetti.py This code has not been tested on the Rigetti cloud so it is most probably still buggy and will change a lot, but I think it is close to working\n\nTo do (1), I am imitating the wonderful Pennylane Rigetti plugin, available at GitHub. I have even filed an issue at that github repo\nhttps://github.com/rigetti/pennylane-forest/issues/7\n\nSo far, Qubiter does not do minimization by back-propagation, which is goal (2). Instead, it does minimization using the scipy function scipy.optimize.minimize(). My future plans are to replace this scipy function by back-propagation. Remember why we ultimately want back-propagation. It’s because, of all the gradient based minimization methods (another one is conjugate gradient), backprop is the easiest to do in a distributed fashion, which takes advantage of GPU, TPU, etc. The first step, the bicycle training wheels step, towards making Qubiter do (2) is to use the wonderful software Autograd. https://github.com/HIPS/autograd Autograd replaces each numpy function by an autograd evil twin or doppelganger. After I teach Qubiter to harness the power of Autograd to do back-prop, I will replace Autograd by the more powerful tools TensorFlow and PyTorch (These also replace each numpy function by an evil twin in order to do minimization by back-propagation. They also do many other things).\n\nIn doing back-propagation in a quantum circuit, one has to calculate the derivative of quantum gates. Luckily, it’s mostly one qubit gates so they are 2-dim unitaries that can be parametrized as", null, "$U=e^{i\\theta_0}e^{i\\sigma_k\\theta_k}$\n\nwhere the k ranges over", null, "$1, 2, 3$ and we are using Einstein summation convention.", null, "$\\theta_0, \\theta_1,\\theta_2, \\theta_3$ are all real.", null, "$\\sigma_k$ are the Pauli matrices. As the PennyLane authors have pointed out, the derivative of U can be calculated exactly. The derivative of U with respect to", null, "$\\theta_0$ is obvious, so let us concentrate on the derivatives with respect to the", null, "$\\theta_k$.\n\nLet", null, "$U = e^{i\\sigma_3\\theta_3} = C + i\\sigma_3 S$\nwhere", null, "$S = \\sin\\theta_3, C = \\cos \\theta_3$.\nThen", null, "$\\frac{dU}{dt} = \\dot{\\theta}_3(-S + i\\sigma_3 C)$\n\nMore generally, let", null, "$U = e^{i\\sigma_k\\theta_k} = C + i\\sigma_k \\frac{\\theta_k}{\\theta} S$\nwhere", null, "$\\theta = \\sqrt{\\theta_k\\theta_k}, S = \\sin\\theta, C = \\cos \\theta$\nThen, if I’ve done my algebra correctly,", null, "$\\frac{dU}{dt}=-S \\frac{\\theta_k}{\\theta} \\dot{\\theta_k}+ i\\sigma_k\\dot{\\theta_r} \\left[\\frac{\\theta_k\\theta_r}{\\theta^2} C+ \\frac{S}{\\theta}(-\\frac{\\theta_k\\theta_r}{\\theta^2} + \\delta_{k, r})\\right]$\n\nI end this post by answering the simple riddle which I posed in the title of this post. The rise of Trump was definitely a step backwards for humanity, but there are lots of times when stepping backwards is a good thing to do. Minimization by back propagation is a powerful tool, and it can be described as walking backwards. Also, when one gets lost in a forest or in a city and GPS is not available, I have found that a good strategy for coping with this mishap, is to, as soon as I notice that i am lost, back track, return to the place where I think I first made a mistake. Finally, let me include in this brief list the ancient Chinese practice of back walking. Lots of Chinese still do back-walking in public gardens today, just like they do Tai Chi. Both are healthy low impact exercises that are specially popular with the elderly. Back walking is thought to promote muscular fitness, because one uses muscles that are not used when walking forwards. Back walking is also thought to promote mental agility, because you have to think a little bit harder to do it than when walking forwards. (Just like counting backwards is a good test for sobriety and for detecting advanced Alzheimer’s)\n\n## 1 Comment »\n\n1. Hmm, when the breakthrough gets tough, the tough gets…noisy!\nhttps://sociable.co/technology/darpa-exploit-quantum-computing-without-quantum-computer/\n\nComment by technofeudalism — March 20, 2019 @ 5:32 pm\n\nBlog at WordPress.com." ]

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[ "As an Amazon Associate I earn from qualifying purchases.\n\nCentral Tendency Measures trivia questions and answers, central tendency measures quiz answers PDF 66 to practice statistics exam questions for online classes. Practice \"Measures of Central Tendency\" trivia questions and answers, central tendency measures Multiple Choice Questions (MCQ) to practice statistics test with answers for online university degrees. Free central tendency measures MCQs, data tables in statistics, introduction to statistics, normal distribution, interquartile deviation, central tendency measures test prep for online business and administration degree.\n\n\"The mean or average used to measure central tendency is called\", central tendency measures Multiple Choice Questions (MCQ) with choices arithmetic mean, sample mean, negative mean, and population mean for online colleges for business management. Learn measures of central tendency questions and answers to improve problem solving skills for online bachelor's degree in business administration.\n\nCentral Tendency Measures Quiz\n\nMCQ: The mean or average used to measure central tendency is called\n\n1. sample mean\n2. arithmetic mean\n3. negative mean\n4. population mean\n\nB\n\nInterquartile Deviation Quiz\n\nMCQ: The value of first quartile is 23 and the inter quartile range is 20 then the value of third quartile is\n\n1. 63\n2. 53\n3. 43\n4. 73\n\nC\n\nNormal Distribution Quiz\n\nMCQ: If the z-score of normal distribution is 2.5, the mean of the distribution is 45 and the standard deviation of normal distribution is 3 then the value of x for a normal distribution is\n\n1. 97.5\n2. 47.5\n3. 37.5\n4. 67.5\n\nC\n\nIntroduction to Statistics Quiz\n\nMCQ: The scale used in statistics which provides the difference of proportions as well as magnitude of differences is considered as\n\n1. satisfactory scale\n2. ratio scale\n3. goodness scale\n4. exponential scale\n\nB\n\nData Tables in Statistics Quiz\n\nMCQ: The type of table in which study variables provides large number of information with interrelated characteristics is classified as\n\n1. lower order table\n2. manifold table\n3. higher order table\n4. both b and c\n\nD" ]

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[ "# NAME\n\nGetargs::Mixed - Perl extension allowing subs to handle mixed parameter lists\n\n# SYNOPSIS\n\n`````` use Getargs::Mixed;\n\nsub foo {\nmy %args = parameters([ qw( x y z ) ], @_);\n\n# Do stuff with @args{qw(x y z)}\n}\n\n# OR if you have object-oriented syntax\nsub bar {\nmy (\\$self, %args) = parameters('self', [ qw( x y z ) ], @_);\n\n# Do stuff with @args{qw(x y z)}\n}\n\n# OR if you have mixed OO and function syntax\nsub baz {\nmy (\\$self, %args) = parameters('My::Class', [ qw( x y z ) ], @_);\n\n# Do stuff with @args{qw(x y z)}\n}\n\n# Calling foo:\nfoo(\\$x, \\$y, \\$z);\nfoo(\\$x, -z => \\$z, -y => \\$y);\nfoo(-z => \\$z, -x => \\$x, -y => \\$y);\n\n# ERRORS! calling foo:\nfoo(-z => \\$z, \\$x, \\$y); ### <-- ERROR!\nfoo(x => \\$x, y => \\$y, z => \\$z); ### <-- ERROR!\nfoo(\\$x, -y => \\$y, \\$z); ### <-- ERROR!\nfoo(\\$x, \\$y, \\$z, -x => \\$blah); ### <-- ERROR!\n\n# Calling bar:\n\\$obj->bar(\\$x, \\$y, \\$z);\n\\$obj->bar(\\$x, -z => \\$z, -y => \\$y);\nMy::Class->bar(-z => \\$z, -x => \\$x, -y => \\$y); # etc...\n\n# Calling baz is slightly dangerous! UNIVERSAL::isa(\\$x, 'My::Class') better\n# not be true in the last case or problems may arise!\n\\$obj->baz(\\$x, \\$y, \\$z);\nMy::Class->baz(\\$x, -z => \\$z, -y => \\$y);\nbaz(\\$x, -z => \\$z, -y => \\$y); # etc...``````\n\n# FUNCTIONAL INTERFACE\n\n## parameters\n\nThis allows for the handling mixed argument lists to subroutines. It is meant to be flexible and lightweight. It doesn't do any \"type-checking\", it simply turns your parameter lists into hash according to a simple specification.\n\nThe main function in this module is `parameters` and it handles all the work of figuring out which parameters have been sent and which have not. When it detects an error, it will die with Carp::confess.\n\nThe `parameters` function takes either two or three arguments. If the first argument is a string, it takes at least two arguments: invocant and specification. For example:\n\n`` parameters('invocant', [qw(specification)], @_);``\n\nIf the first argument is an array reference, it takes at least one argument: the specification. For example:\n\n`` parameters([qw(specification)], @_);``\n\nIn either case, the specification is followed by any arguments to be parsed (`@_` in the examples above).\n\n### Invocant\n\nIf the first parameter is a string, it should either be a package name or the special string `\"self\"`. Passing `\"self\"` in this argument will cause the `parameters` function to require an invocant on the method--that is, it must be called like this:\n\n`````` \\$obj->foo(\\$a, \\$b, \\$c); # OR\nfoo \\$obj (\\$a, \\$b, \\$c); # often seen as new My::Class (...)``````\n\nwhere `\\$obj` is either a blessed reference, package name, or a scalar containing a package name.\n\nIf, instead, the first parameter is a string, but not equal to `\"self\"`. The string is considered to be a package name. In this case, `parameters` tries to guess how the method is being called. This has a lot of potential caveats, so beware! Essentially, `parameters` will check to see if the first argument is a subclass of the given package name (i.e., according to UNIVERSAL::isa. If so, it will ASSUME (pronounced Ass-You-Me) that the argument is the invocant. Otherwise, it will ASSUME that the argument is the first parameter. In this case, the returned list will contain the given package name as the first element before the list of pairs even though no invocant was actually used.\n\n### Specification\n\nThe array reference argument to `parameters` contains a list of variable names that the caller accepts. The parameter list is ordered so that if the user passes positional parameters, the same order the parameters are placed, will be the order used to set the variables in the returned hash. The list may contain a single semicolon, which tells `parameters` that all parameters up to that point are required and all following are optional. If no semicolon exists, then `parameters` will consider all to be required and die when one of the required parameters is missing.\n\nFinally, the list may end with a `'*'` which will cause `parameters` to collect any extra unexpected named or positional parameters. Extra named parameters will be inserted into the returned arguments list. Extra positional parameters will be placed in array reference and assigned to the '*' key of the returned arguments list. If '*' is not specified and extra arguments are found `parameters` will die.\n\n### The arguments to be parsed\n\nThe final argument to `parameters` is always the list of arguments passed to the caller, usually `@_`.\n\n### The results of a parameters() call\n\nThe result returned from the `parameters` function depends on whether there are two arguments or three. If `parameters` is called with two arguments, then a list of pairs (a hash) is returned. If `parameters` is called with three arguments, then an invocant is prepended to the list of pairs first. If the first argument is not `\"self\"`, then the invocant will be set to the first argument if `parameters` doesn't detect any invocant.\n\n# ARGUMENT PARSING\n\nThe way `parameters` handles arguments is relatively flexible. However, the format must always specify all positional parameters first, if any, followed by all positional parameters. The `parameters` function switches from positional to named parameters when it encounters the first string preceded with a hypen ('-'). This may have the unfortunate side effect of causing normal parameters to be misinterpreted as named parameters. If this may be the case with your usage, I suggest finding another solution--or modifying this module to suit. A safe solution to this is to always use named parameters--at which point you might as well not use this module anyway.\n\n# EXPORT\n\nAlways exports `parameters` by default. If you do not want this, use:\n\n`````` use Getargs::Mixed ();\n# OR\nrequire Getargs::Mixed;\n\n# ...\nmy %args = Getargs::Mixed::parameters([ qw( x y z ) ], @_);``````\n\n# OBJECT-ORIENTED INTERFACE\n\nGetargs::Mixed supports an object-oriented interface that permits you to adjust how the parameters are processed. For example:\n\n`````` my \\$getargs = Getargs::Mixed->new([options...]);\nmy %args = \\$getargs->parameters([ qw( x y z ) ], @_);``````\n\nThe arguments to the `parameters` method are exactly the same as when `parameters` is called as a function. This includes the invocant, since `\\$getargs` is not the invocant of the function that is invoking `\\$getargs->parameters()`.\n\n## new\n\nCreate a new instance with the given options. For example:\n\n`` my \\$getargs = Getargs::Mixed->new(-undef_ok => 1);``\n\nCurrently known options are:\n\n-undef_ok\n\nThe option `-undef_ok => 1` permits the value of a parameter to be `undef`. For example,\n\n`` my %args = parameters(['foo'], -foo => undef);``\n\nwill fail with a message that required argument `foo` was not provided, but\n\n`````` my %args = Getargs::Mixed->new(-undef_ok => 1)\n->parameters(['foo'], -foo => undef);``````\n\nwill succeed, and set `\\$args{foo}` to `undef`." ]

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[ "# Magnetic field measurements using coils\n\nDiagnostics based on magnetic coils are presently the standard method for measuring magnetic fields in tokamak experiments. Coils are (in the first approximation) easy to manufacture, simple to implement and straightforward to interpret. However, as usual, things get more complicated in a detailed view. The most important requirements which every useful probe-type diagnostic must satisfy are:\n\n• reasonable sensitivity (signal-to-noise ratio); the probe must provide signal high enough to overcome the electronic noise associated with impulse devices\n• a good frequency response, so as to follow the most rapid fluctuations present in the system\n• minimum perturbing effect on plasma, which means the smallest possible size and appropriate vacuum-friendly and plasma-resistant construction materials\n\nIt is unfortunate that these requirements conflict directly with each other. To improve the frequency response (collect high frequencies without attenuation), the coil should be as small as possible. However, small effective area means small signal and bad signal-to-noise ratio.\n\n#Theory of magnetic field measurement using coils\n\nTechnically measuring coils do not measure magnetic field itself; they measure the time changes of the magnetic field. (As a result, coils do not measure stationary magnetic field.) More precisely, they react to the time derivative of the magnetic flux $$\\Phi$$ passing through their turns by inducing a voltage $$U_\\mathrm{p}$$ upon itself.\n\n$U_\\mathrm{p} = - \\frac{\\mathrm{d}\\Phi}{\\mathrm{d}t}$\n\nThe magnetic flux $$\\Phi$$ [Wb] passing through a single loop is defined as $\\Phi = \\int_{A_{\\mathrm{loop}}} \\textbf{B} \\cdot \\mathrm{d}\\textbf{A}$ where $$A_{loop}$$ is the loop area. If the loop is small enough to consider the magnetic field $$\\textbf{B}$$ inside it uniform, we may simplify the formula to $\\Phi = B_\\perp A_{\\mathrm{loop}}$ where $$B_\\perp$$ is the magnetic field component perpendicular to the loop. Now since a coil typically has $$N$$ turns, the voltages induced on top of them stack. Denoting the coil effective area $$A_{\\mathrm{eff}} = N A_{\\mathrm{loop}}$$, we may write $U_\\mathrm{p} = - A_{\\mathrm{eff}} \\frac{\\mathrm{d}B_\\perp}{\\mathrm{d}t}.$ Finally the coil is typically oriented so that the measured magnetic field is perpendicular to its turns; that is, the coil axis lies parallel to the magnetic field. For instance, a coil measuring the toroidal magnetic field $$B_T$$ will be oriented in the toroidal direction. Under this assumption we may finally write $U_\\mathrm{p} = - A_{\\mathrm{eff}} \\frac{\\mathrm{d}B}{\\mathrm{d}t}$ where $$B$$ is the measured magnetic field.\n\n# Circuit scheme\n\nThe following figure shows a simplified scheme of a measuring coil circuit.", null, "The coil is represented by the parameters with the subscript $$p$$: the induced voltage $$U_p$$, the inductance $$L_p$$ and the resistance $$R_p$$. $$R_{in}$$ is the input resistance of the analog-to-digital (A/D) converter. In a standard configuration $$R_{in} \\ll R_p$$.\n\n# Coil signal processing\n\n## Integration\n\nTo extract the magnetic field $$B$$ from the measured quantity ($$U_p$$), the coil signal must be integrated. The integration can be either analog or digital.\n\nAnalog integration can be performed with passive (RC-like integrating) or with active elements (transistors). Passive integration is more transparent but it requires additional circuits, which lowers the output signal significantly. Conversely, active integration combined with amplification provides a stronger signal, but the additional active elements complicate the data interpretation and for longer time periods ($$\\sim 1000 \\mathrm{s}$$) integrator drifts can present a serious complication.\n\nThe alternative is digital integration; that is, to digitize the coil output directly and perform the integration numerically on a computer. This is the simplest option if the integrated signal is not needed in real-time, as it requires no additional circuits and the signal-to-noise ration is typically good (depending on the coil properties). Since data processing tools are readily available on GOLEM, we use this method for all coil measurements. One is of course limited by the finite resolution of the A/D converters.\n\nDigital integration may be performed, for instance, with the following code:\n\n\n// input and output arrays\nvalues[LENGTH] //array of raw data from magnetic coil (input)\nB_tor[LENGTH] //the magnitude of the magnetic field will be written in this array (output)\n\n//magnetic coil's constants\nDELTA_T //time between two samples (1Mhz = 1e-6s, 100kHz = 1e-5s)\nCALIBRATION //calibration constant to convert the voltage to magnetic field\n\nintegr = 0\nfor (i = 0; i <<<> LENGTH; i++)\n{\nintegr += values[i]\nB_tor[i] = integr * DELTA_T * CALIBRATION\n}\n\n\n## Offset removal", null, "An offset is a non-physical addition to a signal, created by noise in the electronics, parasitic voltages, cross-talk between the diagnostics and many other influences. The figure above demonstrates the impact an offset can have on data integration. In blue, the real, physical $$dB/dt$$ and $$B$$ are plotted. In red, a small constant offset was added to the entire raw signal. In green, a normal-distributed noise with a zero mean was added to the raw signal. One observes the effect that this has on the integrated signal, in particular the in the red case.\n\nIn every magnetic measurement, offsets are more than likely to appear on the collected voltage signals. The simplest method to remove them is to average the first few hundred/thousands of samples (before the 5 ms mark when GOLEM discharges begin) and subtract this average from the signal prior to its integration. This method will fail if the offset is time-dependent, in which case more advanced methods are required [link].\n\n# Diagnostics based on magnetic coils\n\nNote: Excerpt from I. Ďuran: Fluctuations of magnetic field in the CASTOR tokamak, Dissertation Thesis, 2003" ]

[ null, "http://golem.fjfi.cvut.cz/wiki/Diagnostics/Magnetic/Theory/Magnetic_coil/mc_scheme.png", null, "http://golem.fjfi.cvut.cz/wiki/Education/GMinstructions/base/DataProcessing/figs/offset_demonstration.png", null ]

[ "", null, "## # Squares Inscribed in a Right Triangle Note:  There are five related by different problems here.", null, "Figure 1 and Figure 2 each show a square inscribed in a right triangle. Assume the triangles, both labeled ABC, are congruent, or two copies of the same triangle.\n\n1. Given any right triangle with sides of length a, b, and c, as above, determine the two constructions to inscribe these squares in the right triangle.\n\nCreate GSP sketches for the two constructions using congruent right triangles with legs of length a and b.", null, "Do you have alternative constructions to those suggested in the hints? Here is a construction showing both squares in the same triangle. An altitude h to the side of length c is also shown.\n\n2a. Show that the length of the side s of a square instcribed in ANY triangle with one side along a base is one-half the harmonic mean of that base and the altitude to that base.\n\n2b. Show a geometric construction for the inscribed square in ANY triangle with one side along a base.\n\nSee Square Inscribed along a base of any Triangle", null, "3. Which of the two squares has the largest area? For example, here are two expressions for the area of triangle ABC and each can be solved for the length of the side of one of the squares in terms of a, b, c, or h.\n\nWill the largest inscribed square always be the same orientation for any right triangle? That is, will one of the two squares always be the largest?\n\nHINT:  Write an expression for", null, "and evaluate to see if this this expression is always positive, always negative, equal, or not determined.", null, "4.  Consider a geometric demonstration regarding the relative sizes of the two inscribed squares.   To accomplish this, construct an auxilliary right triangle with legs of length h and c on our right triangle having legs a and b,  h and c being the altitude and the hypotenuse of the original triangle.\n\nThe yellow square in the auxillary right triangle has the same area as the yellow square in the original right triangle.   Why?   Does this give a means to determine which of the yellow square or the blue square is larger?\n\n5. Suppose in the right triangle ABC the square of side length s inscribed in the right angle has an area of 441 and the square of side lenght x inscribed along the hypotenuse has an area of 440.\n\nFind the length of AC + BC.", null, "Problem sent by Mark Lipson. Lexington, MA from the 1987 AIME high school mathematics contest." ]

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[ "# Economical (k,m)-threshold controlled quantum teleportation\n\nAkira SaiToh1, Robabeh Rahimi2, and Mikio Nakahara3\nResearch Center for Quantum Computing, Interdisciplinary Graduate School of\nScience and Engineering, Kinki University, 3-4-1 Kowakae, Higashi-Osaka, Osaka 577-8502, Japan\nDepartment of Physics, Kinki University, 3-4-1 Kowakae, Higashi-Osaka, Osaka 577-8502, Japan\n11\n22\n33\n###### Abstract\n\nWe study a -threshold controlling scheme for controlled quantum teleportation. A standard polynomial coding over with prime needs to distribute a -dimensional qudit with to each controller for this purpose. We propose a scheme using qubits (two-dimensional qudits) for the controllers’ portion, following a discussion on the benefit of a quantum control in comparison to a classical control of a quantum teleportation.\n\nKeywords: Quantum teleportation, Threshold scheme, Secret sharing\nPACS: 03.67.Hk, 03.67.Mn\n\n## 1 Introduction\n\nQuantum teleportation has been one of the leading discoveries followed by numerous quantum information processing technologies [2, 3]. Controlled quantum teleportation [4, 5, 6, 7, 8] is a variant in which a teleportation of a quantum state is performed under the supervision of controllers. Schemes using qubits as keys distributed among controllers [4, 5, 6, 7, 8] have been extensively studied and economization of required resources has been accomplished with respect to the number of qubits involved in an entangled qubit chain used in a scheme. The studies also include a security discussion on players’ cheating controllers .\n\nIt is expected that a multifunctional quantum network is realized for a consumer market in future, presumably based on optical fibers. So far, simple quantum cryptosystems [10, 11] are highly developed [12, 13, 14, 15, 16, 17] toward the consumer use, which are mainly used to generate classical shared cryptographic keys. An advanced quantum network should be used not only for generating a classical key but for exchanging quantum states. A network with Einstein-Podolsky-Rosen (EPR) pairs as links is a plausible form for this purpose; optical quantum teleportation has already become a well-established subject supported by many experimental demonstrations reported in, e.g., Refs. [19, 20, 21]. Considering an application for office networks, it is demanded that data transfers can be under the control of multiple supervisors. Controlled teleportation schemes have been developed to enable this feature. As the number of usable qubits and the quality of entanglement enhances, this function is approaching reality. A function like a vote to permit a transfer is also considered to be of public demand.\n\nQubit keys are commonly used in controlled quantum teleportation schemes for the basic control such that approvals of all the controllers are required to let the players transfer a quantum state. In such a control, use of qubit keys is sufficient and extending each dimension of a quantum system for a key causes a redundancy. Qudits (-dimensional quantum systems) with have been, in contrast, considered to be useful as keys when a vote for decision is in order.\n\nIt is easy to notice that such functions are enabled by using a threshold scheme [22, 23, 24, 25, 26]: controlling a transfer of a quantum state with a certain threshold in the number of controllers is implemented by secret sharing schemes (see, e.g., a discussion in Ref. ). This categorizes controlled quantum teleportation as a combination of quantum secret sharing and quantum teleportation. A well-known polynomial coding [23, 26] over with prime larger than the number of participants (controllers in our context) is a quick solution444Note that can be equal to if is prime in the context of controlled quantum teleportation (See Sec. 3).. A classical or quantum -threshold scheme applied to a shared state of players and controllers enables a control such that a receiver can recover an original state when and only when or more controllers among provide their keys. There is a variant of quantum polynomial coding robust against a certain number of cheaters among participants .\n\nA drawback to introduce a polynomial coding scheme in controlled quantum teleportation is that computation is conducted over the field . Thus each qudit distributed among controllers should have the dimension . This also complicates a quantum circuit to compute a matrix equation for the coding. For a threshold scheme, i.e., a secret sharing without imposing access structure555 See, e.g., Ref.  and citations therein for secret sharing schemes involving access structures., the dimension of each key (share) using best known classical protocols [30, 31] is (namely, bits for each key) for a long secret. These protocols require the bit length of a secret and that of a key at least as much as the length of a secret. For a short secret, a best known classical protocol is Shamir’s one in which each key has the dimension . It is thus not motivating to simply make a quantum extension of the classical protocols.\n\nIt was reported that an -threshold secret sharing of a classical secret is achieved by a sort of key distribution without entanglement and an -threshold controlled quantum teleportation is achieved by using classical keys . The latter one is easily extended to a -threshold controlling scheme. Nevertheless, schemes using qudits for controllers’ portion are more secure than the one using only classical keys as we will discuss in Sec. 4.1. There is a recently-proposed graph-state formalism to produce quantum secret sharing systems using only qubits. This approach has achieved the systems for several particular and not for general . We will construct a scheme which is not a direct extension of these approaches.\n\nIn this paper, we introduce a -threshold controlled quantum teleportation scheme over with prime using qubits for the controllers’ portion in addition to classical information distributed to the controllers. This is achieved by a hybrid of classical and quantum protocols. It enables a reduction in the dimension of each qudit to two for general , which is a significant improvement for realizing the scheme. In contrast to classical systems, the dimension of a quantum system is limited and usually a large-dimensional qudit is implemented by multiple qubits. The size of a qudit register is also very limited in the present technology. The number of available qubits is, so far, twelve or less .\n\nTo begin with, a standard controlled quantum teleportation scheme is briefly described in Sec. 2. On the basis of the scheme, a -threshold controlled quantum teleportation using a polynomial coding is briefly explained in Sec. 3. A reduction in required resources for the threshold-control scheme is accomplished in Sec. 4: First, a classical control of a quantum teleportation is discussed in Sec. 4.1. Second, a control of a quantum teleportation using qubits for the controllers’ portion is introduced in Sec. 4.2. Third, an economization of Bob’s operations is accomplished in Sec. 4.3. An explicit preparation of an initial state for this economized scheme is shown in Sec. 5 together with the operational complexity of the whole process. We discuss advantages and disadvantages of the scheme in Sec. 6. Section 7 summarizes our results.\n\n## 2 A Standard Controlled Quantum Teleportation\n\nConsider a controlled quantum teleportation using () EPR pairs shared by Alice and Bob, and a single quantum system shared by Alice, Bob, and controllers. Alice tries to send an -qubit state\n\n 1…1∑x1…xn=0…0px1…xn|x1…xn⟩A′1...A′n\n\nof the system to Bob. A standard quantum teleportation protocol works fine for the ()-EPR-pair channel consisting of () pairs . The remaining channel, , is under the control of controllers . The setup of the quantum system is illustrated in Fig. 1. Here it should be noted that, although we consider the control of a single channel here, it is straightforward to attach controllers to each channel. Thus let us limit setups to the illustrated one in the following.", null, "Figure 1: Illustration of qudit sharing for a controlled quantum teleportation.\n\nThe initial state of the illustrated system is given by\n\nwith\n\n |ξ⟩AnBnC1...Cm=1√21∑y=0|yy⟩AnBn|κ(y)⟩C1...Cm (1)\n\na shared state involving the controllers’ portion to be engineered for a tailored controlling scheme.\n\nLet us recall the well-known relation\n\n |x,y⟩=1√21∑i=0(−1)x⋅i|Bi,x⊕y⟩,\n\nwhere is a computational basis vector and is the th Bell basis vector (here, , i.e., ). With this relation, we can rewrite the initial state as\n\n |A′ABC⟩=12n∑x1...xnpx1...xn[(n−1⨂l=11∑il=01∑jl=0(−1)xl⋅il×|Bil,jl⟩A′lAl|jl⊕xl⟩Bl)⊗(1∑in=01∑jn=0(−1)xn⋅in×|Bin,jn⟩A′nAn|jn⊕xn⟩Bn|κ(jn⊕xn)⟩C1...Cm)]. (2)\n\nAs is usual for a standard quantum teleportation, Alice makes Bell measurement on each pair and obtains an outcome .\n\nBob receives information from Alice and applies to . This changes the state of each (of each term in the summation) from to and hence the teleportation process for the original state of is completed up to .\n\nTo complete the recovery of the original state of at Bob’s side, he has to apply a certain operation to . This requires Bob to know the effect of controllers’ measurements on . Operations Bob has to apply to depend on the controlling scheme.\n\nA popular controlled quantum teleportation scheme is the case where is set to , i.e., is set to a Greenberger-Horne-Zeilinger (GHZ) state, and each controller makes a measurement in the basis. Measuring in the basis results in the phase factor depending on the outcome (here, ; the operation “’’ returns one if its two arguments are equal and zero otherwise)666 This is because a computational basis state of a qubit can be written as using the -basis states .. In this case, Bob first applies to . In addition, he applies a single gate to for recovery if the number of ’s in the controllers’ outcomes () is odd. The final state of Bob after these operations becomes . The teleportation is successful in this way.\n\nIn this contribution, we aim to introduce a threshold-control scheme in which the shared state is engineered to be different from the GHZ state. We continue to concentrate on the case where only a single qubit of Bob is under the control, as illustrated in Fig. 1; this is because it is straightforward to extend the scheme so that multiple qubits are under the control. Such an extension will be considered only in Sec. 4.3.\n\n## 3 (k,m)-Threshold Controlled Quantum Teleportation\n\nThe functionality of threshold control is achieved by engineering the initial setup of the shared state given by Eq. (1). Let us begin with a rather expensive scheme that is a straightforward extension of a classical secret sharing. In this scheme, the state of the controllers’ portion is assumed to be in the following form\n\n |κ(y)⟩C1...Cm=1√#S∑c1...cm∈Seiyθ(c1...cm)|c1...cm⟩C1...Cm, (3)\n\nwhere , ( and is a positive integer), and is a certain set of -digit strings; the computational basis for each is chosen arbitrarily and known by the th controller. There are two conditions for the state appropriate for a -threshold scheme:\n(i) cannot be uniquely determined unless the sequence is completely specified (i.e., unless the variables are all specified).\n(ii) A string is uniquely determined in by fixing any digits of .\nUnder these conditions, controllers’ measurements result in a single surviving vector . Thus Bob can recover the original state by the following process (in addition to the usual process for the original quantum teleportation for ). First Bob changes the phase factor to unity by applying\n\n (100e−iθ(c1…cm))\n\nto the qubit . Second Bob applies to as is usual for a quantum teleportation. In these steps, Alice’s Bell measurement on prior to controllers’ measurements does not affect Bob’s operation to recover the phase appearing as a result of controllers’ measurements. This is clear from the initial state described in Eq. (2); the parameter is common in the states of and controllers’ portion.\n\nLet us turn to a system construction for satisfying the conditions. Condition (i) is satisfied by setting . Such cannot be uniquely determined unless the variables are completely specified. A quantum circuit to attach the phase factors is easily realized: from to , applying an -controlled -controlled phase gate with the phase , acting on a work qubit accomplishes this task. For the condition (ii), we will find a proper set of -digit vectors by a certain coding scheme. This is the main concern as the resource for quantum information processing is limited in the current technologies; a coding with a small resource is desirable.\n\nWe revisit the theory of Karnin et al. , which describes a sufficient condition for coding in a threshold scheme, and evaluate a polynomial-coding scheme [23, 26] as a typical example.\n\nWe begin with a well-known property of a linear matrix equation.\n\nProposition: A solvable matrix equation over ,\n\n Ax=b\n\nwith a prime number, , , and has a unique solution if and only if and has a full rank.\n\nThe proof is similar to the case of a real number field (see, e.g., Ref. [36, pages 96,103]).\nProof— (i) First we prove that the solution is unique if and has a full rank. Assume that there are two solutions and . Then, . Let us pick up rows of appropriately to generate so that has a full rank. This is possible because otherwise the number of linearly independent row vectors of should be less than , which is a contradiction. We have thus with the square full-rank matrix . The matrix can be reduced to a diagonal matrix with nonzero diagonal elements by basic operations; thus . Consequently, because exists. This is a contradiction. (ii) Second we prove that and has a full rank if is a unique solution. The contraposition of this statement is that solution is not unique if or is less than . This is easily shown to be true.\n\nOn the basis of this proposition, a coding of our interest is achieved by a matrix equation for with a full-rank matrix such that striking any rows keeps the rank full . Given a matrix equation\n\n Ax=(c1,…,cm)t (4)\n\nwith such which is notified to Bob, we prepare as\n\n |κ(y)⟩C1…Cm=1√#Sx∑c1…cm∈Sxeiyθ(c1…cm)|c1…cm⟩C1…Cm\n\nwhere is a set of corresponding to (hence ). Suppose that at least controllers measure their qudits in the computational basis. Then is fixed uniquely because the matrix equation becomes solvable by using the rows corresponding to the fixed ’s. This implies that the superposition is then resolved. Bob can determine the phase factor that he should modify by receiving at least outcomes from controllers.\n\nWe have seen a common construction of a threshold scheme. Required resources for the threshold scheme are mostly dependent on the choice of the matrix . There are two practical ways among many [24, 25]. One is the matrix in the form\n\n A(i)=( I | T )t\n\nwith the identity matrix and a strictly totally positive matrix [37, 38]. Any minor of is nonzero positive from the definition of strict total positivity; hence any rows of build up a square matrix with nonzero determinant. Thus a matrix equation with can be used for the threshold scheme. A drawback is the difficulty to find a strictly totally positive matrix for sufficiently small prime . A known systematic construction for strictly totally positive matrices uses the largest element of growing exponentially in . Thus also grows exponentially if we follow the construction. A manual optimization is indispensable.\n\nThe other is a Vandermonde matrix used in the well-known Shamir’s scheme ,\n\n A(ii)=Vm,k=⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1x1x21⋯xk−111x2x22⋯xk−12⋮⋮⋮⋮⋮1xmx2m⋯xk−1m⎞⎟ ⎟ ⎟ ⎟ ⎟⎠   mod p (5)\n\nwith mutually different ’s with prime . ( is necessary to set ’s mutually different.) Striking rows generates a square Vandermonde matrix and it is non-singular when ’s are mutually different (see, e.g., pages 43 and 219 of Ref. ). Hence a matrix equation with can be used for the threshold scheme. This matrix has been known to be economical because increases linearly in . Each qudit distributed to a controller should have the dimension , consequently. Nevertheless, one may need to further reduce the dimension considering the poor resources of presently available qudit systems [35, 41].\n\n## 4 Economizing the Threshold Control Scheme\n\nThe dimension of each digit distributed to a party (a controller in the present context) is often evaluated by using the scale of “bit length” in conventional secret sharing schemes. Each key is -bit long (-bit long [30, 31]) in the best known protocols for a short secret with the bit length (for a long secret with the bit length ) although this has not been taken as a drawback at all since classical bits are very cheap. Nevertheless, in quantum protocols one should not consume many qubits for individual quantum systems. It is of our concern to find a smaller dimension for each controller’s qudit facing a limited resource of a quantum system.\n\nOne way is to abandon the use of quantum systems for controllers’ portion and instead use a classical threshold scheme to control a quantum teleportation. A controlled teleportation proposed by Zhang and Man , in the context of threshold, uses classical keys shared by Alice and controllers for encoding Alice’s messages, which can be easily extended to a general threshold-control scheme. Here, we introduce a different scheme where controllers’ qudits are simply replaced by classical digits. We will face the fact that classical control schemes are indeed economical but their security is based on classical keys. Using qudits is found to be more robust against Bob’s physical-access attack.\n\nOur interest is to find such a robust scheme with a simple quantum state for the controllers’ portion. It is shown to be constructed by using the matrix equation with the Vandermonde matrix (5) and qubit states distributed to the controllers. We further perform an economization in the number of Bob’s operations, which is useful for an extension in which multiple EPR channels are under controllers’ control.\n\n### 4.1 Classically-Controlled Quantum Teleportation\n\nAs we mentioned, the simplest way of economization is to use classical control digits instead of quantum ones. This is easily achieved by setting the state (not a state, actually) of the controllers’ portion to be a scalar\n\n |κ(y)⟩C1...Cm=eiyθ(c1,...,cm)\n\nwith phase dependent on integers , the classical keys of a certain classical threshold scheme. Bob can modify this scalar factor by applying to his th qubit if he can gather at least of the keys.\n\nThe security of the scheme is dependent on the classical scheme. Indeed, classical keys can be securely distributed by using a quantum key distribution (see Ref.  and references therein) and the risk of an interception during the key distribution is negligible. Classical keys are, however, easily copied by careless controllers. A possible drawback of the scheme is that controllers cannot stop Bob from recovering Alice’s original state if Bob manages to obtain at least of the keys without consent of the controllers. In contrast, in a threshold scheme using qudits, the operations for a recovery of the original state are unfixed until controllers make measurements. This fact makes the quantum one more robust: Recall that the computational basis for in Eq. (3) is not necessarily known in public. The correct basis for a measurement can be left unknown to Bob. Then, Bob cannot obtain by a physical access to unless he also has an access to the information on the measurement basis. Thus the cost for Bob to steal ’s in the scheme is more than that in the classically controlled one. This logic is similar to the one utilized for keeping the security of dealer-player communications in some quantum threshold schemes [32, 34].\n\nIt should be mentioned that robustness depends on the type of protocol violations. Let us discuss a different type of violation that can be made by Bob. It is a common occasion that controllers do not want Bob to process further with a teleported state before they officially vote for their decision. A violation in this regard occurs when or more controllers are friendly to Bob and they measure their qudits or digits and send the outcomes before the voting starts officially. This violation in the schedule of the procedure cannot be prevented even if qudits are distributed.\n\nIn the following, an economical quantum control scheme is introduced in order to reduce the resource for controllers’ quantum system as we have mentioned. It is now our additional motivation to resolve the schedule violation problem due to friendly controllers.\n\n### 4.2 Economical Quantum Threshold-Controlled Quantum Teleportation\n\nAs is discussed above, a quantum threshold scheme has a classically unachievable property, namely that the operations for a recovery of the original state are unfixed until controllers make measurements and the measurement bases can be left unnotified to Bob. As we have mentioned, our aim is to achieve such a scheme using small-dimensional systems for controllers’ portions. Here, we propose a quantum protocol for the -threshold controlled quantum teleportation with qubits distributed to the controllers. It also resolves the problem of the possible violation in the voting schedule.\n\nIt is implemented with the following state for the controllers’ portion:\n\n |κ(y)⟩C1...Cm=1√2mm⨂s=1[ei2πycs/p|~0s⟩+ei2π¬ycs/p|~1s⟩]Cs, (6)\n\nwhere and are the basis vectors of the th controller’s chosen basis; is a logical negation of and ’s are the keys of the following common classical polynomial-coding threshold scheme. The keys are generated from Eq. (4) using a Vandermonde matrix for and a certain fixed vector for [all the matrix and vector elements are in ]. The matrix is notified to Bob and is hidden. Thus or more keys are required for Bob to determine the remaining keys from Eq. (4).\n\nThe protocol imposed to controllers is as follows.\n(I) Each controller who agrees to allow Bob to recover Alice’s original state measures her/his qubit in the basis and sends its outcome to Bob.\n(II) also sends her/his key to Bob.\n(III) Each controller who disagrees to allow Bob to recover Alice’s original state does not make any action until she/he receives a contact from Bob.\n(IV) When a solicit is sent from Bob, must measure her/his qubit in the basis and send its outcome to Bob.\n\nThe protocol imposed to Bob is as follows.\n(i) Bob receives ’s from Alice. Bob applies to .\n(ii) Bob waits for at least pairs of ’s sent from the controllers.\n(iii) Bob calculates the remaining ’s by substituting the obtained ’s into Eq. (4) if at least pairs are obtained; aborts otherwise.\n(iv) Bob sends solicits to the controllers who did not send information to him.\nNote: it is possible to count his solicits. Therefore, he cannot cheat by sending more than () solicits in this stage even when he succeeds in stealing or more ’s beforehand.\n(v) Bob receives the remaining ’s.\n(vi) Bob modifies the phase factor due to controllers’ measurements in his phase recovery process described below.\n(vii) Bob applies to .\n\n#### Bob’s phase recovery process corresponding to the controllers’ measurements—\n\nThe controllers’ measurements (of course, after Alice’s Bell measurements) make the component state for in Eq. (2), with the controllers’ state (6) in the present context, evolve into the state\n\n 1∑y=jn⊕xn=0(−1)xn⋅in|Bin,jn⟩A′nAn|y⟩Bn|˜κ(y)⟩C1...Cm\n\nwith\n\n |˜κ(y)⟩C1...Cm=1√2mm⨂s=1exp[i2π(¬)rsycs/p]|rs⟩Cs.\n\nThe phase factor\n\n ∏sexp[i2π(¬)rsycs/p]\n\nshould be canceled by Bob before the normal recovery operation is performed in step (vii). The cancellation of the factor with probability one is possible if and only if all the controllers follow the protocol and at least of them send ’s to Bob (otherwise he will get an uncertain state as his operations become a guesswork). Since ’s are classical keys of a threshold scheme, or more of them are necessary and sufficient to find out all of them. Bob can eliminate the phase by applying\n\n ∏sdiag[exp(−i2πrscs/p),exp(−i2π¬rscs/p)] (7)\n\nto his th qubit . This operation is unaffected by Alice’s Bell measurement on as is clear from the initial state described in Eq. (2).\n\nIn this way, the -threshold control is realized by using only qubit systems for individual portions. One may notice that it is a hybrid scheme in the sense that the threshold is realized by a classical threshold scheme. Although classical keys are used, we claim that this scheme possesses a property thanks to a quantum control; the recovery operations are unfixed unless all the controllers make measurements, and the measurement bases are unknown to Bob. The measurements are not completely performed unless there are at least ’s and Bob sends at most () solicits to ’s. Hence, in order to cheat under the protocol, Bob needs to collect ’s, qubits, and measurement bases of at least controllers.\n\nThe scheme has an advantage over the previously-introduced quantum scheme in the sense that it is secure against the possible schedule violation due to friendly controllers. Owing to step (IV), Bob has to wait for an official voting time to obtain Alice’s original state unless the controllers violate the voting schedule all together.\n\n### 4.3 Economization in Bob’s Operations\n\nWe have shown an economized quantum (or hybrid) controlled quantum teleportation using the state (6) for the controllers’ portion, in which the dimension of each distributed qudit has been reduced to two. As two is the minimal dimension for a nontrivial quantum system, it is optimized with respect to the dimension of a Hilbert space of each portion. Here we consider an economization of Bob’s operations to eliminate the phase factor coming from controllers’ measurements.\n\nThe number of Bob’s single-qubit operations for the recovery is not important as far as a single EPR channel is under the controllers’ control because Eq. (7) reduces to a single operation. It is, however, not negligible in case we extend the setup to the one illustrated in Fig. 2, in which we may have multiple shared states. Each controller is assigned to a single pair and the total number of controllers is .", null, "Figure 2: A straightforward extension of the system setup shown in Fig. 1, such that controllers are assigned to multiple channels. Each AlBl pair is not necessarily under the supervision of controllers.\n\nLet us use the same state as in Eq. (6), except the labels of controllers, for each of the controllers’ portions (represented by dotted lines inside the square of “Controllers” in Fig. 2). We change the protocol for controllers in the following way.\n(A) A controller who Agrees to allow Bob to recover the original state makes a measurement on her/his qubit in the basis with . She/he sends Bob the outcome together with .\n(D) A controller who Disagrees does not make any action. However when a solicit is sent from Bob, a controller who Disagrees must make a measurement on her/his qubit in the basis and send Bob the outcome .\n\nTo understand the effect of a measurement, let us decompose each component of the state (6) in the following way:\n\n 1√2[ei2πycs/p|~0s⟩+ei2π¬ycs/p|~1s⟩]=12√2[(ei2πycs/p+ei2π¬ycs/p)|~+s⟩+(ei2πycs/p−ei2π¬ycs/p)|~−s⟩].\n\nFirst consider the case (A).\n(A-i) Suppose that the outcome of a measurement by the th controller is . Then, the unnormalized phase factor owing to this measurement is\n\n ei2πycs/p+ei2π¬ycs/p=1+ei2πcs/p,\n\nwhich is a global phase uncorrelated with . Thus Bob does not have to modify it.\n(A-ii) Suppose that the outcome of a measurement by the th controller is . Then, the unnormalized phase factor owing to this measurement is which can be regarded as\n\n {α(y=0)−α(y=1)\n\nwith . Thus Bob can modify the factor by applying to some for which the controller’s qubit is originally connected in Fig. 2.\n\nSecond consider the case (D).\nThe phase factor due to the measurement, with the outcome , is . This can be modified by applying\n\n diag[exp(−i2πvscs/p),exp(−i2π¬vscs/p)]\n\nto a proper , which is possible if Bob knows , namely, if or more controllers follow (A).\n\nIn addition to the above recovery of the phase factors corresponding to controllers’ measurements, Bob applies to each , as usual, to recover the phase factors corresponding to Alice’s Bell measurements.\n\nLet us estimate the reduction in the number of operations that Bob has to apply to ’s for eliminating the phase factors due to controllers’ measurements. Let us consider the worst case where the controllers are assigned to mutually different channels. In our previous protocol, the number of the operations is . In the present protocol, it is, on average, with the number of controllers who agree to allow Bob to recover the original state. (Of course, Bob cannot recover the state when .)\n\n## 5 Operational Complexity\n\nIn the previous section, a quantum (or hybrid) -threshold controlled quantum teleportation using qubits (without qudits whose dimension is more than two) has been constructed. We will count the number of single-qubit operations and that of two-qubit operations by following the whole process. Here, we consider the case where the controllers are attached to the th EPR channel and regard Bob’s recovery operation corresponding to each controller’s measurement as a single operation, for simplicity. The number of operations is unchanged by employing the setup illustrated in Fig. 2.\n\nThe process is the same as the original quantum teleportation except for the measurements and operations acting on the shared state of , as we have seen in Sec. 2.\n\nFor the part without controllers in Fig. 1, there are () EPR pairs prepared between Alice and Bob. The quantum circuit for preparing the EPR states involves () Hadamard gates and the same number of controlled-NOT (CNOT) gates. Alice makes () Bell measurements between the system and the system and sends the outcomes () as classical information to Bob. Bob applies single-qubit operations at most, namely, , according to the classical information received from Alice.\n\nFor the portion of the shared state in the figure, the initial state\n\n 1√21∑y=0|yy⟩AnBn|κ(y)⟩C1...Cm\n\nwith given by Eq. (6) should be prepared. It is prepared as follows. First we produce the state\n\n 1√21∑y=0|yy⟩AnBn⊗m⨂s=1|~0s⟩Cs.\n\nThis is easily prepared by a single Hadamard gate and a CNOT gate acting on . The state is a basis vector in the th controller’s favorite basis. Second, Hadamard gates are applied to individually in their bases. In addition, the operation\n\n diag[1,exp(i2πcs/p),exp(i2πcs/p),1]\n\nis applied to for all in the basis . The desired initial state for is now achieved. In the teleportation stage, Alice makes a Bell measurement on and sends Bob the outcome (). The controllers and Bob follow the protocol as described in Sec. 4.2 or that in Sec. 4.3. In this process, Bob can eliminate the phase factors due to controllers’ measurements if or more controllers agree to allow Bob to obtain the original state. Finally, Bob applies to . After all these steps, he obtains the original state of in .\n\nWith the above description of the process, we find that the number of single-qubit operations and that of two-qubit operations for preparing the initial state of the whole system are for both operations. The teleportation process involves Bell measurements performed by Alice and single-qubit measurements performed by controllers. It also involves Bob’s recovery operations: (i) at most single-qubit operations corresponding to Alice’s measurement outcomes; (ii) single qubit operations [on average, () single-qubit operations] corresponding to controllers’ measurement outcomes when the protocol described in Sec. 4.2 is employed [when that in Sec. 4.3 is employed].\n\nIn addition, as we have mentioned in Sec. 4.3, the operations of (ii) reduce to a single operation in reality for the present setup while it does not for the setup of Fig. 2.\n\n## 6 Discussions\n\nWe have proposed an economical scheme for a -threshold control of a quantum teleportation. It uses a shared state of Alice, Bob, and controllers with the controllers’ portion in the state of Eq. (6), which consists of qubits only. Thus a drawback of a usual polynomial coding, namely, the required dimension for each controller’s qudit, has been resolved. In addition, it is straightforward to extend the scheme so that multiple qubits in Bob’s portion are under the threshold control.\n\nOur economical scheme can be seen as a hybrid of a standard -threshold controlled quantum teleportation and a -threshold classically-controlled quantum teleportation, as we have mentioned in Sec. 4.2. It should be noted that this has been realized by a nontrivial protocol using the state (6). The scheme has a good redundancy for the securement of the threshold: (i) To modify the phase factors owing to controllers’ measurements, or more classical keys are required for Bob. (ii) To make disagreeing controllers perform measurements, at most () solicits should be sent from Bob. In fact, the standard controlled teleportation in the context of threshold can be used in the context of threshold by stating only (ii) in its protocol. The advantage of our economical scheme over this simple extension is, thus, the redundant securement.\n\nThe scheme is, however, not as economical as a classically-controlled quantum teleportation, as we have discussed in Sec. 4.1. A quantum threshold-control scheme is certainly more expensive than a classical threshold-control scheme. It is thus recommended to assess the trade-off between the benefit and the economicalness to choose an appropriate scheme.\n\nThe benefit to distribute qudits (qubits in our scheme) among controllers is to make the recovery operation of Bob unfixed unless controllers make measurements. This makes the protocol robust against Bob’s attack to the keys: In order for cheating, he needs both physical accesses to at least controllers’ qudits and information on their measurement bases. One may however claim that careless controllers tend to lose both of them at once in a real world. Our economical scheme possesses a clearer advantage also: it can prevent a violation of a voting schedule, i.e. it can prevent Bob from recovering the original state before an official voting time, unless controllers violate the schedule all together.\n\nThere is one drawback in our protocol. In case we use the scheme of Sec. 3, controllers who disagree to the teleportation do not have to measure their systems. In contrast, in our economical protocol, a disagreeing controller has to measure her/his qubit if a solicit is sent from Bob. Our protocol can be broken by a controller who does not follow this regulation. It seems an important drawback at a glance, but there is a quick solution: one can easily find out which controller cheats in the protocol. Any controller who does not send a measurement outcome despite a solicit sent from noncheating Bob is a cheater.\n\nA clever cheater, however, may report an opposite measurement outcome and/or a wrong key instead of being quiet. It has been well-known that a basic secret sharing is not robust against dishonest participants who report wrong keys. There have been several proposals to remedy this drawback in classical secret sharing schemes [43, 44, 45, 46]. These classical schemes are easily combined into our scheme in order to find cheating in the keys , as is clear from the protocol. Nevertheless, a protocol to find cheaters sending wrong messages as measurement outcomes should be newly constructed. One way to achieve this task is to add a supercontroller who grasps controllers’ states by entangling her/his systems and their corresponding systems. Let the supercontroller know their measurement bases. Then the supercontroller can check the measurement outcomes afterward. It is hoped that a more sophisticated way will be developed.\n\nFinally, we discuss a well-known strategy to use controlled quantum teleportation as a secret sharing to hide a quantum state as a secret . Any ()-threshold controlled quantum teleportation in the stage after Alice’s Bell measurements is regarded as a ()-threshold quantum secret sharing: Alice’s original state to be recovered in Bob’s side is regarded as a secret and the controllers are regarded as participants sharing the secret. A drawback of this approach is that the original state is recovered in the system of Bob’s side; participants who try to cooperate for recovery should gather at this side or ask a dealer for proxy.\n\nA solution to avoid this drawback is to construct initially identical copies of the entire system for the ()-threshold controlled quantum teleportation, where . Let us write a copy as with . We send the system , together with classical information obtained by the Bell measurements on each of , to the th participant (). The th participant should also receive , namely the th control systems of all ’s. The th participant can recover the original state in the system when she/he gets to know the measurement results on or more among , including her/his own, and corresponding classical keys if they are used in the scheme.\n\nThis approach possesses the following benefit. The original quantum secret sharing is limited to due to the no-cloning theorem when an unknown quantum state is a secret . A controlled quantum teleportation with a ()-threshold control is not limited by the no-cloning theorem because the secret, namely, Alice’s original state, is teleported to Bob’s system. The approach is practical when we use classical keys for controlling a quantum teleportation. The resource required for this case is classical keys and copies of the system that consists of EPR pairs and the -qubit original state. Of course, the copies of the system are reduced to one copy in case participants may gather in a particular place or may use a trusted dealer for proxy.\n\nThere seems to be no serious drawback of using classical control because the classical keys can be securely distributed and a scheduled vote is not interposed usually for a secret sharing. The robustness of our economical scheme is effective when an untrusted Bob exists. This is due to the fact that the measurement basis of each controller can be hidden. In this sense, our economical scheme might be used for a secret sharing to build in robustness against physical access attacks by malicious participants who try to cheat. Nevertheless, it is not attractive to spend many EPR pairs despite the reduction in the resource for controllers’ portion. A choice of a proper scheme is dependent on the demand of participants when a controlled quantum teleportation is applied to a secret sharing.\n\nWe have discussed the advantage and disadvantage of our scheme in which the dimension of each controller’s qudit is reduced to two. This reduction is indeed significant for physical realization of the threshold control of a quantum teleportation. Nevertheless, it is controversial as to which extent a controlled quantum teleportation should be performed with quantum resources. The answer depends on the application and as to which party is trusted, as is clear from the above discussions.\n\n## 7 Summary\n\nWe have proposed an economical protocol for -threshold controlled quantum teleportation. This protocol uses qubits distributed to controllers; hence we have achieved the reduction in the dimension of each qudit from a prime to two. In addition, we have shown an economization in the number of Bob’s recovery operations.\n\n## Acknowledgments\n\nA.S. and M.N. are supported by “Open Research Center” Project for Private Universities: matching fund subsidy from MEXT. R.R. is supported by the Grant-in-Aid from JSPS (Grant No. 1907329). M.N. would like to thank a partial support of the Grant-in-Aid for Scientific Research from JSPS (Grant No. 19540422).\n\n## References\n\n• C. H. Bennett, G. Brassard, C. Crépeau, R. Jozsa, A. Peres, and W. K. Wootters, “Teleporting an unknown quantum state via dual classical and Einstein-Podolsky-Rosen channels”, Phys. Rev. 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Fund., vol.E85-A, no.1, pp.149-157, January 2002.\n• D. Stucki, N. Gisin, O. Guinnard, G. Ribordy, and H. Zbinden, “Quantum key distribution over 67km with a plug&play system”, New J. Phys., vol.4, pp.41-1-8, 2002.\n• I. Marcikic, A. Lamas-Linares, and C. Kurtsiefer, “Free-space quantum key distribution with entangled photons”, Appl. Phys. Lett., vol.89, pp.101122-1-3, 2006.\n• J. F. Dynes, Z. L. Yuan, A. W. Sharpe, and A. J. Shields, “Practical quantum key distribution over 60 hours at an optical fiber distance of 20km using weak and vacuum decoy pulses for enhanced security”, Optics Express, vol.15, iss.13, pp.8465-8471, 2007.\n• S. Bose, V. Vedral, and P. L. Knight, “Multiparticle generalization of entanglement swapping”, Phys. Rev. A, vol.57, pp.822-829, 1998.\n• D. Bouwmeester, J.-W. Pan, K. Mattle, M. Eibl, H. Weinfurter, and A. Zeilinger, “Experimental Quantum Teleportation”, Nature, vol.390, no.6660, pp.575-579, 1997.\n• D. Boschi, S. Branca, F. De Martini, L. Hardy, and S. Popescu, “Experimental Realization of Teleporting an Unknown Pure Quantum State via Dual classical and Einstein-Podolsky-Rosen channels”, Phys. Rev. Lett., vol.80, no.6, pp.1121-1125, 1998.\n• I. Marcikic, H. de Riedmatten, W. Tittel, H. Zbinden, and N. Gisin, “Long-Distance Teleportation of Qubits at Telecommunication Wavelengths”, Nature, vol.421, no.6922, pp.509-513, 2003.\n• G. R. Blakley, “Safeguarding cryptographic keys”, in Proceedings of AFIPS 1979 National Computer Conference, New York, 1979, vol.48, pp.313-317, AFIPS Press, Arlington, Va., June 1979.\n• A. Shamir, “How to Share a Secret”, Commun. ACM, vol.22, no.11, pp.612-613, 1979.\n• E. D. Karnin, J. W. Greene, and M. E. Hellman, “On Secret Sharing Systems”, IEEE Trans. Inf. Theory, vol.IT-29, no.1, pp.35-41, January 1983.\n• S. C. Kothari, “Generalized linear threshold scheme”, in Proceedings of Crypto’84, Santa Barbara, 1984, Edited by G. R. Blakley and D. Chaum, Springer-Verlag, New York, 1985, pp.231-241.\n• R. Cleve, D. Gottesman, and H.-K. Lo, “How to Share a Quantum Secret”, Phys. Rev. Lett., vol.83, pp.648-651, 1999.\n• Z. Zhang, “Controlled teleportation of an arbitrary -qubit quantum information using quantum secret sharing of classical message”, Phys. Lett. A, vol.352, pp.55-58, 2006.\n• Y. Murakami, M. Nakanishi, S. Yamashita, and K. Watanabe, “Cheater Identifiable Quantum Secret Sharing Schemes”, IPSJ SIG Technical Reports, 2005-CSEC-30(47), pp.337-340, 2005.\n• S. Iftene, “General Secret Sharing Based on the Chinese Remainder Theorem”, Cryptology ePrint Archive: 2006/166, http://eprint.iacr.org/2006/166.\n• H. Kunii and M. Tada, “A note on information rate for fast threshold schemes”, in Proceedings of Computer Security Symposium 2006, Kyoto, Japan, October 2006, pp.101-106.\n• J. Kurihara, S. Kiyomoto, K. Fukushima, and T. Tanaka, “A fast (3,n)-threshold secret sharing scheme using exclusive-OR operations”, IEICE Trans. Fund., vol.E91-A, no.1, pp.127-138, January 2008.\n• G.-P. Guo and G.-C. Guo, “Quantum secret sharing without entanglement”, Phys. Lett. A, vol.310, pp.247-251, 2003.\n• Z.-J. Zhang and Z.-X. Man, “Many-agent controlled teleportation of multi-qubit quantum information”, Phys. Lett. A, vol.341, pp.55-59, 2005.\n• D. Markham and B. C. Sanders, “Graph states for quantum secret sharing”, Phys. Rev. A, vol.78, pp.042309-1-17, 2008.\n• C. Negrevergne, T. S. Mahesh, C. A. Ryan, M. Ditty, F. Cyr-Racine, W. Power, N. Boulant, T. Havel, D. G. Cory, and R. Laflamme, “Benchmarking Quantum Control Methods on a 12-Qubit System”, Phys. Rev. Lett., vol.96, pp.170501-1-4, 2006.\n• P. Lancaster and M. Tismenetsky, The Theory of Matrices, 2nd Ed., Academic Press, San Diego, 1985.\n• S.  Karlin, Total Positivity, Volume I, Stanford University Press, California, 1968.\n• T. Ando, “Totally Positive Matrices”, Linear Algebra Appl., vol.90, pp.165-219, 1987.\n• T. Craven and G. Csordas, “A Sufficient Condition for Strict Total Positivity of a Matrix”, Linear Multilinear Algebra, vol.45, pp.19-34, 1998.\n• G. Pólya and G. Szegö, Problems and Theorems in Analysis, Volume II, Springer-Verlag, Berlin, 1976, 4th Ed.\n• Proceedings of the symposium “Quantum Computation: Are the DiVincenzo Criteria Fulfilled in 2004?”, Edited by M. Nakahara, S. Kanemitsu, M. M. Salomaa, and S. Takagi, published as “Physical Realizations of Quantum Computing”, World Scientific, Singapore, 2006.\n• N. Gisin, G. Ribordy, W. Tittel, and H. Zbinden, “Quantum cryptography”, Rev. Mod. Phys., vol.74, no.1, pp145-195, 2002.\n• M. Tompa and H. Woll, “How to share a secret with cheaters”, J. Crypt., vol.1, pp.133-138, 1989.\n• J. Rifà-Coma, “How to Avoid the Cheaters Succeeding in the Key Sharing Scheme”, Designs, Codes, and Crypt., vol.3, pp.221-228, 1993.\n• R. S. Rees, D. R. Stinson, R. Wei, and G. H. J. van Rees, “An application of covering designs: determining the maximum consistent set of shares in a threshold scheme”, Ars Combin. vol.53, pp.225-247, 1999.\n• R. Tso, Y. Miao, and E. Okamoto, “A new algorithm for searching a consistent set of shares in a threshold scheme with cheaters”, in Proceedings of the 6th Information Security and Cryptology Conference, Edited by J. I. Lim and D. H. Lee, Lecture Notes in Computer Science, vol.2971, pp.377-385, Springer-Verlag, Berlin, 2003." ]

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[ "# Tim Sullivan", null, "### Bayesian probabilistic numerical methods in SIAM Review\n\nThe 2019 Q4 issue of SIAM Review will carry an article by Jon Cockayne, Chris Oates, Mark Girolami, and myself on the Bayesian formulation of probabilistic numerical methods, i.e. the interpretation of deterministic numerical tasks such as quadrature and the solution of ordinary and partial differential equations as (Bayesian) statistical inference tasks.\n\nJ. Cockayne, C. J. Oates, T. J. Sullivan, and M. Girolami. “Bayesian probabilistic numerical methods.” SIAM Review 61(4):756–789, 2019. doi:10.1137/17M1139357\n\nAbstract. Over forty years ago average-case error was proposed in the applied mathematics literature as an alternative criterion with which to assess numerical methods. In contrast to worst-case error, this criterion relies on the construction of a probability measure over candidate numerical tasks, and numerical methods are assessed based on their average performance over those tasks with respect to the measure. This paper goes further and establishes Bayesian probabilistic numerical methods as solutions to certain inverse problems based upon the numerical task within the Bayesian framework. This allows us to establish general conditions under which Bayesian probabilistic numerical methods are well defined, encompassing both the nonlinear and non-Gaussian contexts. For general computation, a numerical approximation scheme is proposed and its asymptotic convergence established. The theoretical development is extended to pipelines of computation, wherein probabilistic numerical methods are composed to solve more challenging numerical tasks. The contribution highlights an important research frontier at the interface of numerical analysis and uncertainty quantification, and a challenging industrial application is presented.\n\nPublished on Thursday 7 November 2019 at 07:00 UTC #publication #bayesian #siam-review #prob-num #cockayne #girolami #oates" ]

[ null, "http://www.tjsullivan.org.uk/img/sirev61-4_756.png", null ]

[ "# Changeset 13322\n\nIgnore:\nTimestamp:\nDec 22, 2009, 5:07:01 AM (10 years ago)\nMessage:\n\nRemove obsolete functions %cons-random-state and initialize-random-state.\nAssume that all targets use the new random state objects.\n\nFile:\n1 edited\n\nUnmodified\nAdded\nRemoved\n• ## branches/new-random/level-1/l1-numbers.lisp\n\n r13313 nil))) #-x86-target (defun %cons-random-state (seed-1 seed-2) #+32-bit-target (%istruct 'random-state seed-1 seed-2) #+64-bit-target (%istruct 'random-state (the fixnum (+ (the fixnum seed-2) (the fixnum (ash (the fixnum seed-1) 16)))))) ;;; random associated stuff except for the print-object method which ;;; is still in \"lib;numbers.lisp\" #-x86-target (defun initialize-random-state (seed-1 seed-2) (unless (and (fixnump seed-1) (%i<= 0 seed-1) (%i< seed-1 #x10000)) (report-bad-arg seed-1 '(unsigned-byte 16))) (unless (and (fixnump seed-2) (%i<= 0 seed-2) (%i< seed-2 #x10000)) (report-bad-arg seed-2 '(unsigned-byte 16))) (%cons-random-state seed-1 seed-2)) (defun init-random-state-seeds () (defun initial-random-state () #-x86-target (initialize-random-state #xFBF1 9) #+x86-target (initialize-mrg31k3p-state 12345 12345 12345 12345 12345 12345)) #-x86-target (defun make-random-state (&optional state) \"Make a new random state object. If STATE is not supplied, return a copy of the default random state. If STATE is a random state, then return a copy of it. If STATE is T then return a randomly initialized random state.\" (let* ((seed-1 0) (seed-2 0)) (if (eq state t) (multiple-value-setq (seed-1 seed-2) (init-random-state-seeds)) (progn (setq state (require-type (or state *random-state*) 'random-state)) #+32-bit-target (setq seed-1 (random.seed-1 state) seed-2 (random.seed-2 state)) #+64-bit-target (let* ((seed (random.seed-1 state))) (declare (type (unsigned-byte 32) seed)) (setq seed-1 (ldb (byte 16 16) seed) seed-2 (ldb (byte 16 0) seed))))) (%cons-random-state seed-1 seed-2))) #+x86-target (defun make-random-state (&optional state) \"Make a new random state object. If STATE is not supplied, return a (defun %random-state-equalp (x y) ;; x and y are both random-state objects #-x86-target (and (= (random.seed-1 x) (random.seed-1 y)) #+32-bit-target (= (random.seed-2 x) (random.seed-2 y))) #+x86-target (equalp (random.mrg31k3p-state x) (random.mrg31k3p-state y)))\nNote: See TracChangeset for help on using the changeset viewer." ]

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[ "# How do you calculate hand probabilities for 10 cards per hand (8 suits); e.g.: Cripple Mr Onion?\n\nCripple Mr Onion is a card game that involves 8 suits (104 cards in toto), with 2-7 players being dealt 10 cards each (without replacement).\n\nI've tried several different websites for hypergeometric distribution, but really don't know what I'm doing with them (I've no idea whether the results are correct or not, since I can't tell if I input the appropriate data).\n\nI've also tried it by hand, but can't wrap my head around the proper method. When the suit doesn't matter, then it's easy enough; when the suit matters, it becomes rather more complicated.\n\nHow does one calculate the probability of hands in a card game, when the deck involved has 8 suits?\n\nTaking two basic examples, I would hope that I could then work out the rest for myself:\n\n1. I imagine that calculating the probability of my hand containing two aces (of any suits) is the simplest starting point for this game's rules.\n2. For comparison and contrast, I then also calculate the probability of my hand containing {4, 5, 7} in only one or two specific suits (not the other six possible suits).\n\nThere is an excellent work-up regarding Yu-Gi-Oh! in another post's answer, but I don't know how to modify it properly for this problem (which would probably be the best solution possible).\n\n• How good a programmer are you? Writing a small program that generates a large number of hands, will get you very good approximations.\n– John\nJul 8, 2018 at 15:57\n• @John There is no need to write a program to do that as you can use math to calculate the exact values, just need to know the correct formulas. Jul 8, 2018 at 15:58\n• I'm math-oriented (which is frustrating, since card probabilities shouldn't present any difficulties whatsoever... yet they do). Knowing the exact values isn't necessary, since this is only for my own curiosity, but I'm too OCD to rely upon a program's generated aggregation. Jul 8, 2018 at 16:06\n• @mmathis, several of us are capable of answering this question. I have an answer ready, once/if it gets reopened.\n– John\nJul 9, 2018 at 14:46\n• Simple math questions on games are on topic.\n– John\nJul 9, 2018 at 15:28\n\nI find hypergeometric series nice for computing, but poor for explaining. For teaching purposes, I much prefer expanding these out using binomial coefficients. Some books, and Wolfram Alpha call this \"choose.\"\n\nThe key here is to realize that order does not matter. The total number of possible hands is `choose(104, 10)` = 26 trillion.\n\nNow, to count how many hands have 2 aces, you need to figure out how many ways there are to choose 2 aces, and multiply that by the number of ways to get 8 cards with no aces. Writing it out in detail we have:\n\n`````` # of hands w 2 aces\nP(2 aces in hand) = -------------------\n# of hands\n\n# ways to choose 2 aces * # ways to choose non aces\n= ----------------------------------------------------\nchoose(104,10)\n\nchoose(8,2) * choose(96,8)\n= ---------------------------- = 0.14\nchoose(104,10)\n``````\n\nHere's the Wolfram Alpha expression for all the accuracy you could ever want.\n\nYour second question, how to find {4,5,7} in one of two suits is ambiguous. Do all three have to be in the same suit? Or different suits? There's another issue. Let's call the suits ABCDEFGHIJ, and say you want the 4,5,7 to be in suits A or B. Do you care if there is a 4 in suit F?\n\n``````P(4,5,7 in either of 2 suits, don't care about other suits) =\n\n# ways to choose {4,5,7} * number of ways to choose other cards\n= ----------------------------------------------------\nchoose(104,10)\n\n2*2*2*choose(98,7)\n= ---------------------------- = 0.004\nchoose(104,10)\n``````\n\nOr, the related: P(4,5,7 in either of 2 suits, no other 4,5,7 in hand) =\n\n`````` # ways to choose {4,5,7} * number of ways to choose other cards\n= ----------------------------------------------------\nchoose(104,10)\n\n2*2*2*choose(80,7)\n= ---------------------------- = 0.001\nchoose(104,10)\n``````\n\nYou might reasonably ask, why are these numbers so much lower than for aces? This is because you are specifying three cards in the hand, and the aces are only specifying two cards in the hand.\n\nKeep in mind, that in most games, this isn't the full story. For instance, the way I've computed things, I've included a pair of aces and three kings. In poker, that's a full house, and so needs to be removed from this probability. If that's what you want, then things get more complicated.\n\n• On the `P(2 aces in hand)` part ... you wrote `choose(10,2)` for the `# ways to choose 2 aces`... however, as there are 8 suits, there are 8 aces in the game. So, shouldn't it be `chose(8,2)` instead? ... seems that the prob is correct (0.12) but you just confused the 8 with 10 (i.e.: the calculation was made with 8, but you wrote 10). Jul 9, 2018 at 23:01\n• @DarkCygnus, you are correct.\n– John\nJul 10, 2018 at 0:33\n\nHow does one calculate the probability of hands in a card game, when the deck involved has 8 suits?\n\nThis depends on the query you seek, and the specific restrictions it might have. Answering your example:\n\nI imagine that calculating the probability of my hand containing two aces (of any suits) is the simplest starting point for this game's rules.\n\nFirst, I'm taking this as the probability of your hand containing two, and exactly two, Aces.\n\nYou have 8 suits, 104 cards in total, which gives 27 cards per suit.\n\nWe know that each suit has only one copy of an Ace. That is, there is only one ace of Hearts, etc. Thus, there are 8/104 cards that are aces. This gives us a 8/104 chance to pull out any Ace. From this we also have a 96/104 chance of not pulling an Ace.\n\nSo, a hand consists of 5 cards 10 cards. Based on the previous probabilities, and considering that when you pull a card the total number of cards is reduced by one we have:\n\nProb. of a hand with exactly 2 aces = (8/104)x(7/103)x(96/102)x(95/101)x(94/100)x(93/99)x(92/98)x(91/97)x(90/96)x(89/95) = 0.003161\n\nDo note that the chances would be higher if the query where containing at least two aces.\n\nThe terms being multiplied are: The first two the chances of pulling Aces, and the rest the chance of not pulling an ace. All factors consider that cards pulled out are not returned to the pool.\n\nEdit: If we consider the a priori probabilities instead, that is without accounting for cards drawn and without any specific order, the calculation changes to:\n\n(8/104)2x(96/104)8 = 0.003119\n\n• You need to multiply your result by binomial(10,2). You are computing the probability the first two cards of the 10 are aces, but the order of the draw is unimportant in this case.\n– John\nJul 9, 2018 at 22:17\n• @John so, you say that probabilities should be considered as a whole, without counting the reduced deck size per each draw? That would give instead: (8/104)**2 x (96/104)**8... or is that combination probability to include to compensate for the cards drawn? Jul 9, 2018 at 22:29\n• I don't understand your question. You used permutations (instead of binomial coefficients) to compute your probability. But that forces order to matter. Your proposed computation in the comments is appropriate for a dice rolling scenario, but as these are samples without replacement, we need an \"urn\" model. Not sure I'm helping. So, I wrote up a complete answer. Hope that makes it clearer.\n– John\nJul 9, 2018 at 22:53\n• @John perhaps yours is more clean it seems :) Jul 9, 2018 at 22:59" ]

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[ "# Derivation of standard error of mean\n\nI was going through this wikipedia article on standard error. I could not understand the crucial step here. It goes like this:\n\nThis formula may be derived from what we know about the variance of a sum of independent random variables.\n\nIf $X_1, X_2 , \\ldots, X_n$ are n independent observations from a population that has a mean $\\mu$ and standard deviation $\\sigma$ , then the variance of the total\n\n$T = (X_1 + X_2 + \\cdots + X_n)$ is $n\\sigma^2$. Understood.\n\nThe variance of T/n must be $\\frac{1}{n^2}n\\sigma^2=\\frac{\\sigma^2}{n}$. Not understood.\n\nAnd the standard deviation of T/n must be $\\sigma/{\\sqrt{n}}$ . Of course, T/n is the sample mean $\\bar{x}$ .\n\nI went to some basics:\n\n$\\displaystyle Var(X)=\\frac{1}{n}\\sum_{i=1}^{n}({x_i-\\mu})^2$\n\n$\\displaystyle Var(X)=\\frac{1}{n}\\sum_{i=1}^{n}({x_i^2+\\mu^2-2x_i\\mu})$\n\n$\\displaystyle Var(X)=\\frac{1}{n}\\sum_{i=1}^{n}x_i^2+\\mu^2-\\frac{2}{n}\\sum_{i=1}^{n}x_i\\mu$\n\nAs Sample mean is an unbiased estimate of population mean, we get\n\n$\\displaystyle Var(X)=\\frac{1}{n}\\sum_{i=1}^{n}x_i^2-\\mu^2$\n\n$\\displaystyle Var(X)=E(X^2)-(E(X))^2$\n\nNothing useful found from this.\n\nWhy is there a $1/n^2$ in that step to get variance ?\n\nLet $$Y$$ be any random variable. Let $$Z = Y/n$$. Then $$Z^2 = \\frac1{n^2} Y^2,$$ $$E(Z^2) = E\\left(\\frac1{n^2} Y^2\\right) = \\frac1{n^2} E(Y^2)$$ and therefore $$E\\left(\\left(\\frac Yn\\right)^2\\right) = \\frac1{n^2} E(Y^2).$$ Also, $$E(Z) = E\\left(\\frac1n Y\\right) = \\frac1n E(Y).$$ So from $$Var(Y)=E(Y^2)-(E(Y))^2$$ and $$Var(Z)=E(Z^2)-(E(Z))^2,$$ we find $$\\begin{eqnarray} Var\\left(\\frac Yn\\right) = Var(Z) &=& E(Z^2)-(E(Z))^2\\\\ &=& \\frac1{n^2} E(Y^2) - \\left(\\frac1n E(Y)\\right)^2 \\\\ &=& \\frac1{n^2} \\left(E(Y^2) - \\left( E(Y)\\right)^2 \\right) \\\\ &=& \\frac1{n^2} Var(Y). \\end{eqnarray}$$ Now consider the case where $$Y = T$$.\n\n• Why do we take Z = Y/n in the first place? Just to introduce the constant 1/n? I do not understand why we would want to introduce that constant in the first place. – Saskia Nov 4 '19 at 22:14\n• @Oleksandra The question that was asked above was essentially, why is it that when I divide a variable by $n$, the variance is divided by $n^2$ instead of just $n$? So this answer divides a variable by $n$ to show what happens. The reason why to divide by $n$ in the first place was evidently already understood. If you do not understand, you may ask a new question about that. – David K Nov 4 '19 at 23:06\n\nThe only thing we need to prove here is that for any scalar constant $c$, and for a random variable $X$, $$\\mathrm{Var}[cX] = c^2 \\mathrm{Var}[X].$$ This follows from the property of expectation $$\\mathrm{E}[cX] = c\\mathrm{E}[X]$$ as follows: \\begin{align*} \\mathrm{Var}[cX] &= \\mathrm{E}[(cX - \\mathrm{E}[cX])^2] \\\\ &= \\mathrm{E}[(cX - c\\mathrm{E}[X])^2] \\\\ &= \\mathrm{E}[c^2(X - \\mathrm{E}[X])^2] \\\\ &= c^2 \\mathrm{E}[(X - \\mathrm{E}[X])^2] \\\\ &= c^2 \\mathrm{Var}[X]. \\end{align*}\n\nRecall the definition of (population) variance: $$var\\xi := E(\\xi - E\\xi)^{2}$$ for $\\xi$ a random variable. Then we have $var\\xi = E\\xi^{2} - 2(E\\xi)^{2} + (E\\xi)^{2} = E\\xi^{2} - (E\\xi)^{2}$ so that $var(\\xi/n) = E(\\xi^{2})/n^{2} - (E\\xi)^{2}/n^{2}.$\n\nThus we have\n\n$$var(T/n) := var(X_{1}/n) + \\cdots + var(X_{n}/n) = n\\sigma^{2}/n^{2} = \\sigma^{2}/n.$$\n\n• What if $E(\\xi) \\neq 0$. Does this mean that an underlying assumption that population mean is zero is required for this formula to hold true ?I am not sure if I am missing something obvious here..but can't wrap my head around this – square_one Aug 23 '14 at 14:47\n• Let me revise the answer. Letting $E\\xi := 0$ is to simply calculation. In your question, since $varX_{i}\\ (i = 1, \\dots, n)$ are given, there is no need to do these derivations from the great beginning. – Megadeth Aug 23 '14 at 14:49\n• using $\\xi$ seems to just make this harder to read, is this symbol a convention? – baxx Jan 11 '20 at 15:24\n• @baxx, Hi, as far as I am aware, for older probabilists, especially for the Russian leading ones, symbols such as $\\xi, \\eta$ are more \"standard\"; a non-Russian example is Billingsley. – Megadeth Jan 12 '20 at 9:49" ]

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[ "# Shear capacity with shear reinforcement\n\nVersion 1.1 by Fredrik Lagerström on 2019/11/13 14:36\n\n# Shear capacity with shear reinforcement\n\nThe design of members with shear reinforcement is based on a truss model (see figure below).\n\ntruss model\n\nfigure 3.3.6.\n\nwhere\n\nα is the angle between shear reinforcement and the beam axis,\n\nθ is the angle between the concrete compression strut and the beam axis perpendicular to the shear force,\n\nFtd is the design value of the tensile force in the longitudinal reinforcement,\n\nFcd is the design value of the compression force in the direction of the longitudinal member axis,\n\nbw is the minimum width between tension and compression chords,\n\nz is the inner lever arm. In shear analysis without axial force, the approximate value z = 0.9 d may normally be used." ]

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[ "", null, "", null, "Question\n\nAs shown in the figure, a solid cylinder of mass $$m$$ and radius $$R$$ starts rolling down an inclined plane of inclination $$\\theta$$. Sufficient friction is present for the pure rolling of the cylinder. Determine the speed of its centre of mass when the cylinder has fallen a height $$H = 1\\ m$$. (Take $$g = 10\\ m/s^2$$)", null, "A\n203m/s", null, "", null, "B\n10m/s", null, "", null, "C\n103m/s", null, "", null, "D\n403m/s", null, "", null, "Solution\n\nThe correct option is C $$\\sqrt{\\dfrac{40}{3}} m/s$$The situation is shown in the figure:-For no slipping there will be no loss of mechanical energy.Energy at position $$1$$ is $$E_{1} = mgH$$Energy at position $$2$$ is $$E_{2} = \\dfrac{1}{2} mV^{2}_{cm} + \\dfrac{1}{2} I_{cm} \\omega^{2}$$ $$\\because \\omega = \\dfrac{V_{cm}}{R}$$    (For pure rolling) $$I_{cm} = \\dfrac{mR^{2}}{2}$$ $$\\Rightarrow E_{2} = \\dfrac{3}{4} m V_{cm}^{2}$$ From the law of conservation of energy:-$$E_{1} = E_{2} \\Rightarrow mgH = \\dfrac{3}{4} m V_{cm}^{2} \\Rightarrow V_{cm} = \\sqrt{\\dfrac{4}{3} gH} = \\sqrt{\\dfrac{40}{3}} m/s$$Correct Option - (D)", null, "Physics\n\nSuggest Corrections", null, "", null, "0", null, "", null, "Similar questions\nView More", null, "", null, "People also searched for\nView More", null, "" ]

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[ "Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids: to which are Added, Elements of Plane and Spherical Trigonometry\n\nCollins and Hannay, 1833 - 333 sider\n\nHva folk mener -Skriv en omtale\n\nVi har ikke funnet noen omtaler pċ noen av de vanlige stedene.\n\nPopulĉre avsnitt\n\nSide 49 - PROB. jf 0 a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle, Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle.\nSide 29 - The angles which one straight line makes with another upon one tide of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with CD, upon one side of it the angles CBA, ABD ; these are either two right angles, or are together equal to two right angles. For, if the angle CBA be equal to ABD, each of them is a right angle (Def.\nSide 19 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.\nSide 55 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.\nSide 90 - If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.\nSide 86 - The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.\nSide 87 - If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle.\nSide 43 - THE straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.\nSide 39 - Wherefore, if a straight line, &c. QED PROP. XXIX. THEOR. If a straight line fall upon two parallel straight lines, it makes the alter' male angles equal to one another ; and the exterior angle equal to the interior and opposite upon the same side ; and likewise the two interior angles upon the same. side together equal to two right angles.\nSide 54 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let..." ]

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[ "", null, "# Back Titration or Indirect Titration Tutorial\n\n## Key Concepts\n\nA back titration, or indirect titration, is generally a two-stage analytical technique:\n\n1. Reactant A of unknown concentration is reacted with excess reactant B of known concentration.\n2. A direct titration is then performed to determine the amount of reactant B in excess.\n\nBack titrations are used when:\n\nNo ads = no money for us = no free stuff for you!\n\n## Example : Back (Indirect) Titration to Determine the Concentration of a Volatile Substance\n\nA student was asked to determine the concentration of ammonia, a volatile substance, in a commercially available cloudy ammonia solution used for cleaning.\nFirst the student pipetted 25.00 mL of the cloudy ammonia solution into a 250.0 mL conical flask.\n50.00 mL of 0.100 mol L-1 HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution.\nThe excess (unreacted) HCl was then titrated with 0.050 mol L-1 Na2CO3(aq).\n21.50 mL of Na2CO3(aq) was required.\nCalculate the concentration of the ammonia in the cloudy ammonia solution.\n\nStep 1: Determine the amount of HCl in excess from the titration results\n\n1. Write the equation for the titration:\n 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l) acid + carbonate → salt + carbondioxide + water\n2. Calculate the moles, n, of Na2CO3(aq) that reacted in the titration:\nmoles = concentration (mol L-1) × volume (L)\nn(Na2CO3(aq)) = c × V\nc(Na2CO3(aq)) = 0.050 mol L-1\nV(Na2CO3(aq)) = 21.50 mL = 21.50 × 10-3 L\nn(Na2CO3(aq)) = 0.050 × 21.50 × 10-3 = 1.075 × 10-3 mol\n3. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration (mole ratio (stoichiometric ratio)).\nFrom the balanced chemical equation, 1 mole Na2CO3 react with 2 moles of HCl\nSo, 1.075 × 10-3 mole Na2CO3 reacted with 2 × 1.075 × 10-3 moles HCl\nn(HCltitrated) = 2 × 1.075 x 10-3 = 2.150 × 10-3 mol\n4. The amount of HCl that was added to the cloudy ammonia solution in excess was :\nn(HClexccess) = 2.150 × 10-3 mol\n\nStep 2: Determine the amount of ammonia in the cloudy ammonia solution\n\n1. Calculate the total moles of HCl originally added to the diluted cloudy ammonia solution:\nmoles = concentration (mol L-1) × volume (L)\nn(HCltotal added) = c × V\nc(HCltotal added) = 0.100 mol L-1\nV(HCltotal added) = 50.00 mL = 50.00 × 10-3 L\nn(HCltotal added) = 0.100 × 50.00 × 10-3 = 5.00 × 10-3 mol\n2. Calculate the moles of HCl that reacted with the ammonia in the diluted cloudy ammonia solution\nn(HCltitrated) + n(HClreacted with ammonia) = n(HCltotal added)\nn(HCltotal added) = 5.00 × 10-3 mol\nn(HCltitrated) = 2.150 × 10-3 mol (calculated in Step 1 above)\n2.150 × 10-3 + n(HClreacted with ammonia) = 5.00 × 10-3\nn(HClreacted with ammonia) = 5.00 × 10-3 - 2.150 × 10-3 = 2.85 × 10-3 mol\n3. Write the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl(aq).\nNH3(aq) + HCl(aq) → NH4Cl(aq)\n4. From the balanced chemical equation, calculate the moles of NH3 that reacted with HCl.\nFrom the equation, 1 mol HCl reacts with 1 mol NH3\nSo, 2.85 × 10-3 mol HCl had reacted with 2.85 × 10-3 mol NH3 in the cloudy ammonia solution.\n5. Calculate the ammonia concentration in the cloudy ammonia solution.\nconcentration (mol L-1) = moles ÷ volume (L)\nc(NH3(aq)) = n(NH3(aq)) ÷ V(NH3(aq))\nn(NH3(aq)) = 2.85 × 10-3 mol (moles of NH3 that reacted with HCl)\nV(NH3(aq)) = 25.00 mL = 25.00 × 10-3 L (volume of ammonia solution that reacted with HCl)\nc(NH3(aq)) = 2.85 × 10-3 ÷ 25.00 × 10-3 = 0.114 mol L-1\n6. The concentration of ammonia in the cloudy ammonia solution was 0.114 mol L-1\n\nDo you know this?\n\nPlay the game now!\n\n## Example : Back (Indirect) Titration to Determine the Amount of an Insoluble Salt\n\nA student was asked to determine the mass, in grams, of calcium carbonate present in a 0.125 g sample of chalk.\nThe student placed the chalk sample in a 250 mL conical flask and added 50.00 mL of 0.200 mol L-1 HCl using a pipette.\nThe excess HCl was then titrated with 0.250 mol L-1 NaOH.\nThe average NaOH titre was 32.12 mL\nCalculate the mass of calcium carbonate, in grams, present in the chalk sample.\n\nStep 1: Determine the amount of HCl in excess from the titration results\n\n1. Write the equation for the titration:\n HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) acid + base → salt + water\n2. Calculate the moles, n, of NaOH(aq) that reacted in the titration:\nmoles = concentration (mol L-1) × Volume (L)\nn(NaOH(aq)) = c(NaOH(aq)) × V(NaOH(aq))\nc(NaOH(aq)) = 0.250 mol L-1\nV(NaOH(aq)) = 32.12 mL = 32.12 × 10-3 L\nn(NaOH(aq)) = 0.250 × 32.12 × 10-3 = 8.03 × 10-3 mol\n3. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration (mole ratio (stoichiometric ratio)).\nFrom the balanced chemical equation, 1 mole NaOH reacts with 1 mole of HCl\nSo, 8.03 × 10-3 mole NaOH reacted with 8.03 × 10-3 moles HCl\n4. The amount of HCl that was added to the chalk in excess was 8.03 × 10-3 mol\n\nStep 2: Determine the amount of calcium carbonate in chalk\n\n1. Calculate the total moles of HCl originally added to the chalk:\nmoles = concentration (mol L-1) x Volume (L)\nc(HCltotal added) = 0.200 mol L-1\nV(HCltotal added) = 50.00 mL = 50.00 × 10-3 L\nn(HCltotal added) = 0.200 × 50.00 × 10-3 = 0.010 mol\n2. Calculate the moles of HCl that reacted with the calcium carbonate in the chalk\nn(HCltitrated) + n(HClreacted with calcium carbonate) = n(HCltotal added)\nn(HCltitrated) = 8.03 × 10-3 mol (calculated in Step 1 above)\n8.03 × 10-3 + n(HClreacted with calcium carbonate) = 0.010 mol\nn(HClreacted with calcium carbonate) = 0.010 - 8.03 × 10-3 = 1.97 × 10-3 mol\n3. Write the balanced chemical equation for the reaction between calcium carbonate in the chalk and the HCl(aq).\nCaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)\n4. From the balanced chemical equation, calculate the moles of CaCO3 that reacted with HCl.\nFrom the equation, 1 mol CaCO3 reacts with 2 mol HCl so, 1 mol HCl reacts with ½ mol CaCO3\nSo, 1.97 × 10-3 mol HCl had reacted with ½ × 1.97 × 10-3 = 9.85 × 10-4 mol CaCO3 in the chalk.\n5. Calculate the mass of calcium carbonate in the chalk.\nmoles = mass (g) ÷ molar mass (g mol-1)\nn(CaCO3) = mass(CaCO3) ÷ M(CaCO3)\nn(CaCO3) = 9.85 × 10-4 mol (moles of CaCO3 that reacted with HCl)\nM(CaCO3) = 40.08 + 12.01 + (3 × 16.00) = 100.09 g mol-1 (using the Periodic Table to find molar mass of each element)\nmass(CaCO3) = n(CaCO3) × M(CaCO3) = 9.85 × 10-4 × 100.09 = 0.099 g\n6. The mass of calcium carbonate in the chalk was 0.099 g\n\nDo you understand this?\n\nTake the test now!" ]

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[ "# Present Value of $59 in 26 Years When you have a single payment that will be made to you, in this case$59, and you know that it will be paid in a certain number of years, in this case 26 years, you can use the present value formula to calculate what that $59 is worth today. Below is the present value formula we'll use to calculate the present value of$59 in 26 years.\n\n$$Present\\: Value = \\dfrac{FV}{(1 + r)^{n}}$$\n\nWe already have two of the three required variables to calculate this:\n\n• Future Value (FV): This is the $59 • n: This is the number of periods, which is 26 years So what we need to know now is r, which is the discount rate (or rate of return) to apply. It's worth noting that there is no correct discount rate to use here. It's a very personal number than can vary depending on the risk of your investments. For example, if you invest in the market and you earn on average 8% per year, you can use that number for the discount rate. You can also use a lower discount rate, based on the US Treasury ten year rate, or some average of the two. The table below shows the present value (PV) of$59 paid in 26 years for interest rates from 2% to 30%.\n\nAs you will see, the present value of $59 paid in 26 years can range from$0.06 to $35.26. Discount Rate Future Value Present Value 2%$59 $35.26 3%$59 $27.36 4%$59 $21.28 5%$59 $16.59 6%$59 $12.97 7%$59 $10.16 8%$59 $7.98 9%$59 $6.28 10%$59 $4.95 11%$59 $3.91 12%$59 $3.10 13%$59 $2.46 14%$59 $1.96 15%$59 $1.56 16%$59 $1.24 17%$59 $1.00 18%$59 $0.80 19%$59 $0.64 20%$59 $0.52 21%$59 $0.42 22%$59 $0.34 23%$59 $0.27 24%$59 $0.22 25%$59 $0.18 26%$59 $0.14 27%$59 $0.12 28%$59 $0.10 29%$59 $0.08 30%$59 $0.06 As mentioned above, the discount rate is highly subjective and will have a big impact on the actual present value of$59. A 2% discount rate gives a present value of $35.26 while a 30% discount rate would mean a$0.06 present value.\n\nThe rate you choose should be somewhat equivalent to the expected rate of return you'd get if you invested \\$59 over the next 26 years. Since this is hard to calculate, especially over longer periods of time, it is often useful to look at a range of present values (from 5% discount rate to 10% discount rate, for example) when making decisions.\n\nHopefully this article has helped you to understand how to make present value calculations yourself. You can also use our quick present value calculator for specific numbers." ]

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[ "# Calculate bmi kg/m2\n\n## CDC - Practice calculating BMI by using the metric system.\n\nHow to calculate body mass index(bmi).   bmi number for europeans (kgm2). Underweight.With vertex42's online bmi calculator, you can either calculate your body mass index based on height and weight, or you  body mass index is defined as your weight divided by the square of a your h.A bmi metric calculator uses a weighttoheight ratio (bmi = kgm2) and assigns a number to the result.   enter your weight and height in metric (or english) systems then click on the calculate button.We show you how to manually calculate bmi using the bmi formula. Easy step by step examples are given to help you determine your weight status.   calculating bmi is straight forward, the formula is eas.Pacific islanders are at risk with a bmi 26 kgm2. See waisthip ratio or body composition calculator for more meaningful measurements of obesity.Calculating bmi in metric units: metric bmi formula. Bmi = weight (kg) ÷ height 2 (m 2 ). Step 1 – multiply your height (in meters) by itself. Since height is usually.Thomas is 90 kg and 1,84 m bmi = 90(1. 6 kgm2 here you can easily calculate your body mass index if you know your height and weight.\n\nomeprazole eg 20 mg bijsluiter\n\n## Ideal, Adjusted, and Nutritional Body Weight Calculator ... https://clincalc.com › Kinetics › IdealBW\n\nvichy dercos neogenic buy online uk\n\n## BMI Calculator | MiWayLife https://www.miwaylife.co.za › tools › bmi-calculator\n\nHow to calculate bmi? the first formula for bmi in the metric system is: mass (kg)height (m)^2.   for a 28 year old female from the us, weighing 136 lbs at a height of 5 feet and 4 inches, the results.The bmi formula is bmi = kg (m)2. Here are the standard bmi ranges. 54 centimeters.   to calculate it manually, you square the number of your height in inches. Next you divide your weight by.The bmi is defined as the body mass divided by the square of the body height and is universally expressed in units of kgm2, resulting from the mass in.One way of working out whether youre a healthy weight is to calculate your bmi. To use this bmi calculator, enter your weight (kg) and height (cm), and your bmi.Click calculate.   pacific islanders are at risk with a bmi 26 kgm2. Bmi can be used with body composition to assess fat free mass click calculate. Underweight.   pacific islander.Jul 8, 2014 obesity (body mass index bmi≥30 kgm2) increases the risk for we calculated ageadjusted mortality rates (number of deaths per 100,000.\n\nxenical effective philippines\n\n## What is the body mass index (BMI)? - NHS https://www.nhs.uk › Common health questions › Lifestyle\n\nThe bmi equals a persons weight in kilograms divided by height in meters squared. Calculating your bmi before and during pregnancy is.Kg at 170 cm = 44. Bmi (body mass index) is the person's body weight (in kilograms) divided by the square of his height (in meters) and measured in kgm2. There is a need to use corrective c.Metric body mass index calculator. Body mass index (bmi) is a measure of body fat based on height and weight that applies to adult men and women.A bmi metric calculator uses a weighttoheight ratio (bmi = kgm2) and assigns a number to the result. To get your approximate bmi using english system.Dec 20, 2018 if you were to calculate the bmi for an adult you would divide their weight in kilograms (kg) by the square of their height in meters (m2).Body mass index calculator, bmi value helps in classifying obesity as overweight, use the bmi calculator to find out your bmi. 0kg (0st 0lbs), bmi = 25 kgm2.To calculate your bmi, follow these steps: enter your weight and select its unit (kilograms or pounds).   bmi is calculated by dividing a person’s weight in kilograms by the square of hisher height in.\n\norder duac topical gel online\n\n## Obesity calculate the BMI calculation of waist to hip\n\nOct 5, 2008 before we start calculating bmi, lets discuss what bmi is, and what it isnt. The basic formula is weight divided by height squared, or kgm2.Body mass index (bmi) is a value derived from the mass (weight) and height of a person. The bmi is defined as the body mass divided by the square of the body height body mass index (bmi) is a value der.Aug 23, 2017 bmi is calculated by taking the body weight in kilograms (kg) and dividing it by the calculate bmi by dividing the weight by the height (m)2.(bmi) calculator can be used to calculate bmi value and corresponding a person should try to maintain a bmi below 25 kgm   convert and calculate; welcome to the onlineconversion. (m2) to wei.Mathematically, bmi is kgm2 (kilograms per square meter).   how to calculate bmi with imperial units: feet. To use pounds (lb) and feet (ft) instead of kilograms (kg) and meters (m), use the same fomul.Additional information will be displayed here once you calculate your bmi. 96 kgm2 (normal). Normal bmi range: 18. 5kgm2 25 kgm2.May 8, 2007 bmi is calculated by a formula that takes into account both a persons height and weight. Bmi is reported as weight in kilograms (kg) divided by height in weight in kilograms divided by heig.Find out from ramsay health care on how to work out bmi. To calculate bmi, the metric formula is your weight in kilograms divided by your bmi(kgm2).We calculate your bmi in the following way, which then gives us a number: weight in kilograms (kgs) ÷ by height in metres squared classification, bmi (kgm2).\n\n## BMI Calculator\n\nasian gold price today uk\n\n## Concept: Obesity: Measuring Prevalence Using Body Mass ... mchp-appserv.cpe.umanitoba.ca › viewConcept\n\ncialis comparison price\n\n## BMC - Calculator | How I can calculate the BMI?\n\nUse our body mass index calculator to see if youre above or below the us average! height (inches²); the metric bmi formula = weight (kg) ÷ height (metres²).Calculate your bmi. Bmi is a measure of body fat, it is calculated using the formula developed by the belgian mathematician and natural scientist adolphe quetelet in late 60s of 19th  let us show you s.With the metric system, the formula for bmi is weight in kilograms divided by height in meters squared. Since height is commonly measured in centimeters.The bmi is then calculated by dividing the subject's weight by the square of hisher height. Therefore, weight is normalized for height. By convention, the calculated bmi has the si units of kgm2.The bmi and bsa calculators estimate the body mass index and body surface area. Optional, to determine weight required to achieve target bmi.Its a value thats derived from the mass and height of a person to determine whether a person bmi is measured in kgm2 as the universal unit of measurement.Here is the answer to the question: calculate my bmi for 2 m and 49 kg. Use the calculator below to compute any bmi value. Body mass index (bmi) is a measure of body fat based on height and weight that.The bmi calculator to calculate your body mass index designed specifically for weight in kilograms divided by the square of his height in meters (kgm2).Jul 20, 2019 you can calculate bmi by using the kgm2 formula. Measure the mass in kilograms and weight in metres. This is the universally accepted.Bmi range kgm2. Severe thinness.   below are the equations used for calculating bmi in the international system of units (si) and the us customary system (usc) using a 5'10, 160pound.Jan 28, 2019 body mass index (bmi, ratio of height and weight, expressed as kgm2) is widely used to define overweight and obesity across many countries.Body mass index formula. The formulas to calculate bmi based on two of the most commonly used unit systems: bmi = weight(kg)height2(m2) (metric units).Bmi is calculated as weight (kg)height squared (m2). However, the measurement of height is often unreliable and impractical in older adults in nursing homes.Want to calculate your bmi? try wws bmi calculator for women and men. Enter a couple of details like your height in cm & weight in kg & get your bmi results.Calculate your body mass index (bmi) with this body weight calculator tool metric bmi = ( weight (kg) ( height (m)2 ) ); imperial bmi = ( weight (lb) ( height.Bmi is calculated as weight (kilograms) divided by height (meters) squared. Significant numbers of people with a bmi below 30 kgm2 are also obese and.Calculate bmi. Your bmi is kgm2. Risk based on bmi. Risk factors. Under weight.\n\nakne gel duac kaufen\n\n## BMI Calculator\n\nCalculation. Bmi is defined as the individuals body weight divided by the square of the height, and is almost always expressed in the unit kg m2, which is.Calculating your body mass index (bmi) is one way of trying to work out whether your weight in kilograms to the first decimal place (e. 4); your height in.Calculating your bmi will tell you how healthy your weight is. Kgm2 or, weight in kilograms divided by (height in meters multiplied by height in meters).Jul 29, 2019 while the bmi calculation is an indirect measurement, it has been found to be a metric bmi formula: formula: weight (kg) height (m)22.Jun 16, 2019 a great bmi calculator with advanced features for body mass index. Avoid the cdc and nih calculators calling body mass index: kgm2.Bmi = kg (m)2.   this page includes a simple “javascript” program to calculate your exact bmi (in metric measurements) and a height vs weight chart that will approximate your bmi (in english measuremen.Formula bmi (kgm2) = weight (kg) height (m2). All calculations must be confirmed before use. The suggested results are not a substitute for clinical judgment. Formula bmi (kgm2) = weight (kg) height (m.\n\n## Body Mass Index | BMI | Calculator | BMI range - kg/m2\n\nBody mass index is a simple calculation using a persons height and weight. The formula is bmi = kgm2 where kg is a persons weight in kilograms and m2 is.Calculate body mass index, weight loss, calorie usage online. With the metric system, the formula for bmi is weight in kilograms divided by height in meters.Calculate your bmi. Questionnaire. This calculator computes the body mass index and rates it appropriately for men, women, children, juveniles and seniors. The sbmi – an index that has been developed f.Use this simple calculator to find out your body mass index and find out what it means for your health. Body mass index (bmi), which is calculated by dividing weight in kilograms by height in metres sq.This bmi calculator determines the body mass index and the ideal weight based on 5  please note that once you have closed the pdf you need to click on the calculate  bmi in the metric system = weight (.Use this simple metric and imperial bmi calculator tool to work out your body mass index. Metric bmi formula. Bmi = weight (kg) height (m)2.Calculating your bmi before and during pregnancy is especially important as extremes of bmi (too high or too low) may affect your fertility and increase pregnancy complications.The bmi is defined as the body mass divided by the square of the body height, and is universally expressed in units of kgm2, resulting  calculate this index is very easy, just you have to know your hei.The formula is bmi = kgm2 where kg is a person’s weight in kilograms and m2 is their height in metres squared. A bmi of 25. 0 or more is overweight, while the healthy range is 18. Bmi applies.Bmi, body mass index, is a measure based on the ratio of a persons weight compared to their height. To calculate this you divide the individual's weight measured in kilograms by the square of the.The bmi formula is simply the mass of a person divided by the square of the height and is defined  this conversion is shown below: 1 pound (lb) = 0. 453592 kilograms (kg).   the bmi formula has been wi.For example if you have a resident with a height of 1. 65m and weight of 65kg. You would calculate: 651. This gives us a bmi of 23.Example: bob is 1. 74 m tall and weighs 82 kg. To calculate his bmi: 82 ÷ 1. 74 = 27kgm2 (weight ÷ height ÷ height = bmi). The desirable range is from.It is defined as the weight in kilograms divided by the square of the height in metres (kgm2). For example, an adult who weighs 70 kg and whose height is 1.This free body mass index calculator gives out the bmi value and healthy bmi range: 18. 5 kgm2 25 kgm2; healthy weight for the height: 128.\n\nhow to calculate your bmi uk\n\n## BMI Calculator | Body Mass Index Calculator\n\narcher field pc price uk\n\n## Calculating and plotting BMI https://www.education.vic.gov.au › health › Pages › bmicalculate\n\nYou can use the body mass index (bmi) to work out if you are underweight, overweight or at an ideal weight for your height. It is a useful tool, but keep in mind.Body mass index (bmi) calculator. Formula bmi (kgm2) = weight (kg) height (m2). All calculations must be confirmed before use. The suggested results are.Calculation of bmi: bmi = weight (kg) height (m)2.   research indicates that body mass index lower than 18. 5 kgm2 or higher than 30 kgm2 is associated with increased mortality, after correlating for o.If your bmi is 19 to 24. 9 kgm2, youre are a healthy weight, and should aim to stay that way. A bmi of 25 to 29 kgm2 is defined as overweight. Its a good idea to.Jul 15, 2019 bmi is calculated with a simple formula which is bmi = kgm2, where “kg” is your weight in kilograms, and “m2“is your height in meters squared.Our body mass index (bmi) calculator is a simple way to find out whether youre a calculator to see if youre within the healthy range (18.As the weight in kilograms divided by the square of the height in metres (kgm2). To calculate your own bmi automatically (in english or metric units) see.\n\ncialis london\n\n## Obesity: BMI calculators and charts - Medical News Today https://www.medicalnewstoday.com › articles\n\nHow to calculate your body mass index or bmi. Bmi is your weight (in kilograms) over your height squared (in centimeters). Lets calculate, however.To common values. Body mass index (or bmi) is calculated as your weight (in kilograms) divided by the square of your height (in metres) or bmi = kgm2.The result of this equation is the bmi. Not a mathematician? click here to use this free bmi calculator. There are also bmi charts available, where.Calculated bmi = bmi classification. Body mass index (bmi) is a simple index of weightforheight commonly used to classify bmi = 70 (kg) 1. 752 (m2) = 22.Calculator to compare bai (body adiposity index) and bmi (body mass index) obesity indexes.   the following graph shows the bmi for various weights (40 kg in green, 60 kg in blue, 80 kg in red and 100.Jan 15, 2000 formula : bsa (m2) = mosteller : body mass index = weight(kg) height(m)2 simplified calculation of body surface area. N engl j med.\n\nwhy do we calculate bmi" ]

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[ "# Answers\n\nSolutions by everydaycalculation.com\n\n## Multiply 6/4 with 5/20\n\n1st number: 1 2/4, 2nd number: 5/20\n\nThis multiplication involving fractions can also be rephrased as \"What is 6/4 of 5/20?\"\n\n6/4 × 5/20 is 3/8.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 6/4 × 5/20 = 6 × 5/4 × 20 = 30/80\n3. After reducing the fraction, the answer is 3/8\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Multiply Fractions Calculator\n\n×\n\n© everydaycalculation.com" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "## Conversion formula\n\nThe conversion factor from feet per second to knots is 0.59248380129641, which means that 1 foot per second is equal to 0.59248380129641 knots:\n\n1 ft/s = 0.59248380129641 kt\n\nTo convert 372 feet per second into knots we have to multiply 372 by the conversion factor in order to get the velocity amount from feet per second to knots. We can also form a simple proportion to calculate the result:\n\n1 ft/s → 0.59248380129641 kt\n\n372 ft/s → V(kt)\n\nSolve the above proportion to obtain the velocity V in knots:\n\nV(kt) = 372 ft/s × 0.59248380129641 kt\n\nV(kt) = 220.40397408226 kt\n\nThe final result is:\n\n372 ft/s → 220.40397408226 kt\n\nWe conclude that 372 feet per second is equivalent to 220.40397408226 knots:\n\n372 feet per second = 220.40397408226 knots\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.0045371232717735 × 372 feet per second.\n\nAnother way is saying that 372 feet per second is equal to 1 ÷ 0.0045371232717735 knots.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that three hundred seventy-two feet per second is approximately two hundred twenty point four zero four knots:\n\n372 ft/s ≅ 220.404 kt\n\nAn alternative is also that one knot is approximately zero point zero zero five times three hundred seventy-two feet per second.\n\n## Conversion table\n\n### feet per second to knots chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from feet per second to knots\n\nfeet per second (ft/s) knots (kt)\n373 feet per second 220.996 knots\n374 feet per second 221.589 knots\n375 feet per second 222.181 knots\n376 feet per second 222.774 knots\n377 feet per second 223.366 knots\n378 feet per second 223.959 knots\n379 feet per second 224.551 knots\n380 feet per second 225.144 knots\n381 feet per second 225.736 knots\n382 feet per second 226.329 knots" ]

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[ "# Eisenstein-integer definition\n\n(algebra) A complex number of the form , where a and b are integers and ω is defined by the following two rules: (1) and (2) ; an element of the Euclidean domain .\n\nTo divide an Eisenstein integer by another Eisenstein integer , notice that ; accordingly multiply both denominator and numerator (of the division expressed as a fraction) by , then simplify.\n\nnoun\n\n## Other Word Forms\n\n### Noun\n\nSingular:\neisenstein-integer\nPlural:\nEisenstein integers\n\n## Origin of eisenstein-integer\n\n• Named after Gotthold Eisenstein (1823–1852), German mathematician." ]

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[ "# Axiom:Multiplicative Norm Axioms\n\n## Definition\n\nLet $\\struct {R, +, \\circ}$ be a ring whose zero is $0_R$.\n\nLet $\\norm {\\, \\cdot \\,}: R \\to \\R_{\\ge 0}$ be a multiplicative norm on $R$.\n\nThe multiplicative norm axioms are the conditions on $\\norm {\\, \\cdot \\,}$ which are satisfied for all elements of $R$ in order for $\\norm {\\, \\cdot \\,}$ to be a multiplicative norm:\n\n $(\\text N 1)$ $:$ Positive Definiteness: $\\ds \\forall x \\in R:$ $\\ds \\norm x = 0$ $\\ds \\iff$ $\\ds x = 0_R$ $(\\text N 2)$ $:$ Multiplicativity: $\\ds \\forall x, y \\in R:$ $\\ds \\norm {x \\circ y}$ $\\ds =$ $\\ds \\norm x \\times \\norm y$ $(\\text N 3)$ $:$ Triangle Inequality: $\\ds \\forall x, y \\in R:$ $\\ds \\norm {x + y}$ $\\ds \\le$ $\\ds \\norm x + \\norm y$\n\n## Also known as\n\nIn the discussion of norms, it is usually the case that multiplicative norms are under consideration.\n\nHence it is common to speak directly of norm axioms without qualification as to their exact variety." ]

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[ "Theory of equations\n\nFind the real roots of given quadratic equation x^2+5x+4=0\n\nSolution:\n\nGiven quadratic equation is x2 + 5x + 4 = 0\n\nSplitting the middle term since 4 + 1 = 5 (Middle Term) & 4 x 1 = 4 (product of coefficient of first and last term). Now\n\n⇒ x2 + 4x + 1x + 4 = 0\n\n⇒ x(x + 4)+ 1(x + 4) = 0\n\n⇒ (x + 4)(x +1) = 0\n\n⇒ (x + 4) = 0 or (x +1) = 0\n\n⇒ x = – 4 or x = -1\n\nTherefore, the roots of a quadratic equation x2 + 5x + 4 = 0 are -4 and -1" ]

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[ "1\nJEE Main 2023 (Online) 15th April Morning Shift\nNumerical\n+4\n-1", null, "Let the plane $P$ contain the line $2 x+y-z-3=0=5 x-3 y+4 z+9$ and be\n\nparallel to the line $\\frac{x+2}{2}=\\frac{3-y}{-4}=\\frac{z-7}{5}$. Then the distance of the point\n\n$\\mathrm{A}(8,-1,-19)$ from the plane $\\mathrm{P}$ measured parallel to the line $\\frac{x}{-3}=\\frac{y-5}{4}=\\frac{2-z}{-12}$\n\nis equal to ______________.\n2\nJEE Main 2023 (Online) 13th April Morning Shift\nNumerical\n+4\n-1", null, "Let the image of the point $$\\left(\\frac{5}{3}, \\frac{5}{3}, \\frac{8}{3}\\right)$$ in the plane $$x-2 y+z-2=0$$ be P. If the distance of the point $$Q(6,-2, \\alpha), \\alpha > 0$$, from $$\\mathrm{P}$$ is 13 , then $$\\alpha$$ is equal to ___________.\n\n3\nJEE Main 2023 (Online) 12th April Morning Shift\nNumerical\n+4\n-1", null, "Let the plane $$x+3 y-2 z+6=0$$ meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle $$\\mathrm{ABC}$$ is $$\\left(\\alpha, \\beta, \\frac{6}{7}\\right)$$, then $$98(\\alpha+\\beta)^{2}$$ is equal to ___________.\n\n4\nJEE Main 2023 (Online) 11th April Evening Shift\nNumerical\n+4\n-1", null, "Let the line $$l: x=\\frac{1-y}{-2}=\\frac{z-3}{\\lambda}, \\lambda \\in \\mathbb{R}$$ meet the plane $$P: x+2 y+3 z=4$$ at the point $$(\\alpha, \\beta, \\gamma)$$. If the angle between the line $$l$$ and the plane $$P$$ is $$\\cos ^{-1}\\left(\\sqrt{\\frac{5}{14}}\\right)$$, then $$\\alpha+2 \\beta+6 \\gamma$$ is equal to ___________.\n\nJEE Main Subjects\nPhysics\nMechanics\nElectricity\nOptics\nModern Physics\nChemistry\nPhysical Chemistry\nInorganic Chemistry\nOrganic Chemistry\nMathematics\nAlgebra\nTrigonometry\nCoordinate Geometry\nCalculus\nEXAM MAP\nJoint Entrance Examination" ]

[ null, "https://questions.examside.com/image/google_translate.svg", null, "https://questions.examside.com/image/google_translate.svg", null, "https://questions.examside.com/image/google_translate.svg", null, "https://questions.examside.com/image/google_translate.svg", null ]

[ "# Section or ratio (3 dimensions) Calculator\n\nEnter value and click on calculate. Result will be displayed.\n\nSelect calculator:\n\nX1:\nY1:\nZ1:\nX2:\nY2:\nZ2:\nm:\nn:\n\nResult:\n\nCoordinates of Point:\nSection or ratio (3 dimensions) Calculator A section or cross-section is the shape that results from cutting a three-dimensional object with a plane in mathematics. This cut produces a two-dimensional form that symbolizes the point where the item and the plane intersect.\nSearch calculator\nx" ]

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[ "# US6772184B2 - Method for efficient modular division over prime integer fields - Google Patents\n\nMethod for efficient modular division over prime integer fields Download PDF\n\n## Info\n\nPublication number\nUS6772184B2\nUS6772184B2 US09/734,972 US73497200A US6772184B2 US 6772184 B2 US6772184 B2 US 6772184B2 US 73497200 A US73497200 A US 73497200A US 6772184 B2 US6772184 B2 US 6772184B2\nAuthority\nUS\nUnited States\nPrior art keywords\nregister\ncontents\neven\nstoring\nhalf\nPrior art date\nLegal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)\nActive, expires\nApplication number\nUS09/734,972\nOther versions\nUS20020052906A1 (en\nInventor\nSheueling Chang\nCurrent Assignee (The listed assignees may be inaccurate. Google has not performed a legal analysis and makes no representation or warranty as to the accuracy of the list.)\nOracle America Inc\nOriginal Assignee\nOracle America Inc\nPriority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)\nFiling date\nPublication date\nPriority to US09/649,356 priority Critical patent/US6721771B1/en\nApplication filed by Oracle America Inc filed Critical Oracle America Inc\nPriority to US09/734,972 priority patent/US6772184B2/en\nAssigned to SUN MICROSYSTEMS, INC. reassignment SUN MICROSYSTEMS, INC. ASSIGNMENT OF ASSIGNORS INTEREST (SEE DOCUMENT FOR DETAILS). Assignors: CHANG, SHEUELING\nPriority claimed from US10/091,962 external-priority patent/US6917957B2/en\nPriority claimed from US10/091,968 external-priority patent/US6954772B2/en\nPublication of US20020052906A1 publication Critical patent/US20020052906A1/en\nPublication of US6772184B2 publication Critical patent/US6772184B2/en\nApplication granted granted Critical\nAssigned to Oracle America, Inc. reassignment Oracle America, Inc. MERGER AND CHANGE OF NAME (SEE DOCUMENT FOR DETAILS). Assignors: Oracle America, Inc., ORACLE USA, INC., SUN MICROSYSTEMS, INC.\nApplication status is Active legal-status Critical\nAdjusted expiration legal-status Critical\n\n## Images\n\n•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "## Classifications\n\n• GPHYSICS\n• G06COMPUTING; CALCULATING; COUNTING\n• G06FELECTRIC DIGITAL DATA PROCESSING\n• G06F7/00Methods or arrangements for processing data by operating upon the order or content of the data handled\n• G06F7/60Methods or arrangements for performing computations using a digital non-denominational number representation, i.e. number representation without radix; Computing devices using combinations of denominational and non-denominational quantity representations, e.g. using difunction pulse trains, STEELE computers, phase computers\n• G06F7/72Methods or arrangements for performing computations using a digital non-denominational number representation, i.e. number representation without radix; Computing devices using combinations of denominational and non-denominational quantity representations, e.g. using difunction pulse trains, STEELE computers, phase computers using residue arithmetic\n• G06F7/721Modular inversion, reciprocal or quotient calculation\n\n## Abstract\n\nThe invention provides a method for performing modular division adapted for division in integer fields. Integer modular divisions are used in the computation of Elliptic Curve digital signature generation and verification. The algorithm can be implemented to provide division in integer fields completed in 2(m−1) steps. This method provides a solution to the elliptical curve cryptosystems based on prime integer fields.\n\n## Description\n\nThis invention is a continuation in part patent application of application Ser. No. 09/649,356 filed Aug. 28, 2000.\n\nBACKGROUND OF THE INVENTION\n\n1. Field of the Invention\n\nThe present invention relates to modular divisions in prime integer fields.\n\nPortions of the disclosure of this patent document contain material that is subject to copyright protection. The copyright owner has no objection to the facsimile reproduction by anyone of the patent document or the patent disclosure as it appears in the Patent and Trademark Office file or records, but otherwise reserves all copyright rights whatsoever.\n\n2. Background Art\n\nComputer systems are useful for performing mathematical operations (add, subtract, multiply, divide) on operands. Often the operands are polynomials. A polynomial is a mathematical expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a nonnegative integral power (e.g. a+bx+cx2). The task of performing mathematical operations on polynomial operands is difficult in the sense that it is not simply a matter of multiplying or dividing two simple numbers. There are a number of schemes that provide methods for performing mathematical operations on polynomials. However, there are situations for which no suitable schemes have been provided.\n\nOne situation that requires the manipulation of polynomials is the encryption and decryption of data in a cryptosystem and digital signatures for verification of the sender. A cryptosystem is a system for sending a message from a sender to a receiver over a medium so that the message is “secure”, that is, so that only the intended receiver can recover the message. A cryptosystem converts a message, referred to as “plaintext” into an encrypted format, known as “ciphertext.” The encryption is accomplished by manipulating or transforming the message using a “cipher key” or keys. The receiver “decrypts” the message, that is, converts it from ciphertext to plaintext, by reversing the manipulation or transformation process using the cipher key or keys. So long as only the sender and receiver have knowledge of the cipher key, such an encrypted transmission is secure.\n\nA digital signature is a bit-stream generated by a cryptosystem. It is attached to a message such that a receiver of the message can verify with the bit-stream and be assured that the message was indeed originated from the sender it claims to be. A “classical” cryptosystem is a cryptosystem in which the enciphering information can be used to determine the deciphering information. To provide security, a classical cryptosystem requires that the enciphering key be kept secret and provided to users of the system over secure channels. Secure channels, such as secret couriers, secure telephone transmission lines, or the like, are often impractical and expensive.\n\nA system that eliminates the difficulties of exchanging a secure enciphering key is known as “public key encryption.” By definition, a public key cryptosystem has the property that someone who knows only how to encipher a message cannot use the enciphering key to find the deciphering key without a prohibitively lengthy computation. An enciphering function is chosen so that once an enciphering key is known, the enciphering function is relatively easy to compute. However, the inverse of the encrypting transformation function is difficult, or computationally infeasible, to compute. Such a function is referred to as a “one way function” or as a “trap door function.” In a public key cryptosystem, certain information relating to the keys is public. This information can be, and often is, published or transmitted in a non-secure manner. Also, certain information relating to the keys is private. This information may be distributed over a secure channel to protect its privacy, (or may be created by a local user to ensure privacy). Some of the cryptosystems that have been developed include the RSA system, the Massey-Omura system, and the El Gamal system.\n\nElliptic Curves\n\nAnother form of public key cryptosystem is referred to as an “elliptic curve” cryptosystem. An elliptic curve cryptosystem is based on points on an elliptic curve E defined over a finite field F. Elliptic curve cryptosystems rely for security on the difficulty in solving the discrete logarithm problem. An advantage of an elliptic curve cryptosystem is there is more flexibility in choosing an elliptic curve than in choosing a finite field. Nevertheless, elliptic curve cryptosystems have not been widely used in computer-based public key exchange systems due to their late discovery and the mathematical complexity involved. Elliptic curve cryptosystems are described in “A Course in Number Theory and Cryptography” (Koblitz, 1987, Springer-Verlag, New York).\n\nIn practice an Elliptic Curve group over Fields F(2m) is formed by choosing a pair of a and b coefficients, which are elements within F(2m). The group consists of a finite set of points P(x,y) which satisfy the elliptic curve equation\n\ny 2 +xy=x 3 +ax 2 +b\n\ntogether with a point at infinity, O. The coordinates of the point, x and y, are elements of F(2m) represented in m-bit strings. Since F(2m) operates on bit strings and the field has a characteristic 2, computers can perform arithmetic in this field very efficiently. The arithmetic in F(2m) can be defined in either a standard basis representation or optimal normal basis representation. This description uses the standard basis representations for purposes of discussion. All elliptic curve point coordinates are represented as polynomials with binary coefficients.\n\nThe Elliptic Curve Cryptosystem relies upon the difficulty of the Elliptic Curve Discrete Logarithm Problem (ECDLP) to provide its effectiveness as a cryptosystem. Using multiplicative notation, the problem can be described as: given points P and Q in the group, find a number k such that PK=Q; where k is called the discrete logarithm of Q to the base P. Using additive notation, the problem becomes: given two points P and Q in the group, find a number k such that kP=Q.\n\nIn an Elliptic Curve Cryptosystem, the large integer k is kept private and is often referred to as the secret key. The point Q together with the base point P are made public and are referred to as the public key. The security of the system, thus, relies upon the difficulty of deriving the secret k, knowing the public points P and Q. The main factor that determines the security strength of such a system is the size of its underlying finite field. In a real cryptographic application, the underlying field is made so large that it is computationally infeasible to determine k in a straight forward way by computing all the multiples of P until Q is found.\n\nThe core of the elliptic curve geometric arithmetic is an operation called scalar multiplication which computes kP by adding together k copies of the point P. The scalar multiplication is performed through a combination of point-doubling and point-addition operations. The point-addition operation adds two distinct points together and the point-doubling operation adds two copies of a point together. To compute, for example, 11 P=(2*(2*(2P)))+2P=P, it would take 3 point-doublings and 2 point-additions.\n\nPoint-doubling and point-addition calculations require special operations when dealing with polynomial operands. Algebraic schemes for accomplishing these operations for polynomial field F(2m) are illustrated below in Table 1. Algebraic schemes for prime integer fields Fφ are illustrated in Table 2.\n\n TABLE 1 Point addition: R = P + Q Point Doubling: R = 2P S = (yP − yQ)*(1/(xP + xQ)) S = xP + yP *(1/xP) xR = s2 + s + a + xP + xQ xR = s2 + s + a yR = s*(xP + xR) + xR + yP yR = xP 2 + (s + 1)*xR If Q = −P, R = P + (−P) = O, If xP = O, then R = 2 P = infinity O, infinity\n\nThe two equations for S in the table are called the slope-equations. Computation of a slope equation requires one modular polynomial inversion (1/X mod M) where M is an irreducible polynomial and one modular polynomial multiplication (*Y mod M). Because the operands are polynomials, these operations are typically done back-to-back as two separate operations. There exist algorithms and solutions to calculate the modular inversion 1/X mod M and the modular multiplication *Y mod M. After the result of the modular inversion is calculated, the multiplication *Y mod M is performed. Of course, algebraically (1/X*Y) mod M is the same as Y/X mod M. However, there is currently no technique for calculating modular Y/X in one operation when the operands are polynomial functions. These two field operations, the inversion and the multiply, are expensive computationally because they require extensive CPU cycles for the manipulation of two large polynomials modular a large irreducible polynomial. Today, it is commonly accepted that a point-doubling and point-addition operation each requires one inversion, two multiplies, a square, and several additions. To date there are techniques to compute modular inversions, and techniques to trade expensive inversions for multiplies by performing the operations in projective coordinates. There have been no efficient hardware oriented techniques suggested to compute a modular division directly which can be used to perform point doubling and point addition operations.\n\n TABLE 2 Point addition: R = P + Q Point Doubling: R = 2P S = (yP − yQ)*(1/(xP − xQ))mod M S = (3xP 2 + a)/(2yP)mod M xR = s2 − xP − xQ mod M xR = s2 − 2xP mod M yR = −yP + s*(xP − xR)mod M yR = −yP + s(xP − xR)mod M If Q = −P, R = P + (−P) = O, infinity If xP = O, then R = 2 P = O, infinity\n\nA slope equation computation for integer fields requires one modular integer inversion and one modular integer multiplication.\n\nSUMMARY OF THE INVENTION\n\nThe invention provides a method for performing modular division adapted for division in integer fields. Integer modular divisions are used in the computation of Elliptic Curve digital signature generation and verification. The algorithm can be implemented to provide division in integer fields completed in 2(m−1) steps. This method provides a solution to the elliptical curve cryptosystems based on prime integer fields.\n\nAnother embodiment provides a method for performing an inversion and multiply in a single operation as a polynomial divide operation. As a result, the invention reduces the number of mathematical operations needed to perform point doubling and point addition operations. An elliptic curve cryptosystem using the present invention can be made to operate more efficiently using the present invention. An elliptic curve crypto-accelerator can be implemented using the present invention to dramatically enhance the performance of the elliptic curve cryptosystem.\n\nThe invention uses five registers A, B, U, V, and M, to accomplish a polynomial divide operation. Four registers A, B, U, and V are initialized with values so that the registers maintain a number of invariant relationships. The registers store initial values a(t)=x(t), u(t)=y(t), b(t)=prime(t), and v(t)=0. Here the polynomials in registers A, U, B, and V are denoted as a(t), u(t), b(t), and v(t), respectively. Register M stores the irreducible polynomial prime(t). By applying a series of invariant operations to the registers, the register values are systematically reduced until registers A and B have a value of one. At that point, register U stores a value which represents y(t)/x(t) mod prime(t), solving the polynomial division.\n\nBRIEF DESCRIPTION OF THE DRAWINGS\n\nThese and other features, aspects and advantages of the present invention will become better understood with regard to the following description, appended claims and accompanying drawings where:\n\nFIG. 1 is a flow diagram illustrating the operation of the present invention.\n\nFIG. 2 is a block diagram illustrating an implementation of the present invention.\n\nFIG. 3 illustrates an execution environment of the present invention.\n\nFIGS. 4A through 4C are a flow diagram illustrating the operation of division over an integer field in an alternate embodiment of the invention.\n\nDETAILED DESCRIPTION OF THE INVENTION\n\nThe invention is a method for efficient modular polynomial divisions. In the following description, numerous specific details are set forth to provide a more thorough description of embodiments of the invention. It is apparent, however, to one skilled in the art, that the invention may be practiced without these specific details. In other instances, well known features have not been described in detail so as not to obscure the invention.\n\nThe invention provides a system for performing polynomial divides so that a polynomial inversion and multiply can be accomplished in one operation at the same computational cost as a polynomial inversion. The invention is described in connection with example operations from an elliptic curve cryptosystem.\n\nConsider the division of two polynomials in the Fields F(2m). The invention combines a multiplication with an inversion process. Thus, the slope equation s=xp+yp/xp can be computed using one division operation at the cost of an inversion, and no multiplies:\n\n Point addition: Point doubling: s = (yP − yQ)/(xP + xQ) s = xP + yP/xP\n\nThe invention computes the modular division of two polynomials y(t)/x(t) modulo prime(t).\n\nHere, the function, y(t), is the numerator and the function, x(t), is the denominator, which are the two polynomial input functions to the polynomial-divide algorithm. The polynomial, prime(t), is the irreducible polynomial of the field.\n\nThe invention is implemented in a computer system by using five registers, A, B, U, V, and M. The fifth register M that holds the irreducible polynomial prime(t) is not shown here.\n\n register A register U register B register V\n\nA bit-string in a register corresponds to a polynomial. For example, a bit-string of (1011000101) in register B indicates that b(t)=t9+t7+t6+t2+1. The big-string representation and the polynomial representation are inter-changeable. The invention uses a(t), b(t), u(t), and v(t) to refer to the polynomials in the A, B, U, and V registers, respectively.\n\nThe four registers are initialized with the values:\n\na(t)=x(t), u(t)=y(t), b(t)=prime(t), and v(t)=0,\n\nsuch that the polynomials in the registers satisfy the following invariant relationships:\n\na(t)*y(t)=u(t)*x(t) mod prime(t)  (1)\n\nb(t)*y(t)=v(t)*x(t) mod prime(t)  (2)\n\nOne should note that b(t) is congruent to zero modular the prime polynomial.\n\nThroughout the division process, the invention monolithically and iteratively reduces the contents in register A and B down to one by applying a combination of the following 4 invariant operations which guarantee the invariant relationship (1) and (2) throughout the entire process:\n\nOperation 1: Right-shift-Register-A-and-U:\n\na(t)=a(t)>>1;\n\nIf(u0==1)u(t)=u(t)+prime(t);\n\nu(t)=u(t)>>1;\n\nOperation 2: Right-shift-Register-B-and-V:\n\nb(t)=b(t)>>1;\n\nIf(v0==1)v(t)=v(t)+prime(t);\n\nv(t)=v(t)>>1;\n\nIf(b0==1):b(t)=b(t)+a(t) and v(t)=v(t)+u(t)\n\nIf(a0==1):a(t)=a(t)+b(t) and u(t)=u(t)+v(t);\n\nThe notations a0, b0, u0, and v0 indicate the least-significant-bit of each register, respectively.\n\nThe goal is to reduce registers A and B to values of one. This is accomplished by applying the four above operations when possible using the following rules.\n\nRule #1: Apply Operation 1 whenever the least significant bit of Register A is zero, i.e. a0=0, to reduce the polynomial a(t);\n\nRule #2: Apply Operation 2 whenever the least significant bit of Register B is zero, i.e. b0=0, to reduce the polynomial b(t);\n\nRule #3: When both least significant bits a0=1 and b0=1, and a(t)<b(t), Operation 3 is applied.\n\nWhen both least significant bits a0=1 and b0=1, and a(t)>b(t), Operation 4 is applied.\n\nRule #4: If a(t)=b(t), the division process is completed.\n\nOperation 3 or 4 are used to zero the least significant bit of A or B such that Operation 1 or 2 can be repeated. To ensure a monolithic reduction of a(t) and b(t), only a smaller polynomial can be added to a larger one.\n\nA right-shift operation on a bit-string in the register is equivalent to dividing the polynomial by t. For instance, a bit-string (11000100) represents a polynomial function f(t)=t7+t6+t2. A right-shift operation produces a bit-string (01100010) which corresponds to the polynomial function f(t)=(t7+t6+t2)/t=t6+t5+t. The operations 3 and 4 also obey the two invariant relationships. If A, U, B and V satisfy the relationships prior to the operation:\n\na(t)*y(t)=u(t)*x(t) mod prime(t)\n\nb(t)*y(t)=v(t)*x(t) mod prime(t)\n\nthe equation b′(t)*y(t)=v′(t)*x(t) mod prime(t) will still be true after adding register A to B and register U to V, because\n\n(b′(t)=b(t)+a(t))*y(t)=v′(t)=v(t)+u(t)*x(t) mod prime(t)\n\na′(t)*y(t)=u′(t)*x(t) mod prime(t)\n\nThis process repeats itself until both A and B are one. At the end of the iteration process, the division is completed and the resulting polynomial is in the U register:\n\nu(t)=y(t)/x(t) mod prime(t)\n\nSince a(t)=b(t)=1 and a(t)*y(t)=u(t)*x(t) mod prime(t), we know that the bit-string in register U represents the result of the polynomial division. Thus, the polynomial division has been accomplished without two separate operations, an inversion followed by a multiplication.\n\nThis present invention can be implemented as an iterative process. The following example uses C-syntax pseudo-code, although the present invention can be implemented in any programming language. The pseudo-code below uses the four invariant operations defined previously, as follows:\n\n Initialize registers A-U-B-and-V; while (a0==0) Right-shift-registers-A-and-U; while (a(t) !=b(t)){ if (a(t) b(t)) { Add-register-B-to-A-and-V-to-U; while (a0 = = 0) Right-shift-registers-A-and-U; } }\n\nThis process is illustrated in FIG. 1. At step 101 registers A, B, U, and V are initialized with values a(t)=x(t), u(t)=y(t), b(t)=prime(t), and v(t)=0 such that the invariant relationships described above are satisfied. At step 102 the least significant bit of the value in register A is examined to determine if it is zero. If so, then a rightshift operation according to invariant operation 1 is performed on the register value at step 103. The system then returns to step 102.\n\nIf not, then the system proceeds to step 104 to determine if the LSB of register B is equal to zero. If yes, then a rightshift operation according to invariant operation 2 is performed at step 105 and the system returns to step 104. If not, the system proceeds to step 106.\n\nAt step 106 it is determined if the values of registers A and B are equal. If so, the operation ends. If not, it is determined at step 107 if the value of register A is less than B, then operation 3 is applied at step 108 and the system returns to step 104. If not, step 109 determines if the value of register A is greater than the value of register B. If not, the operation ends. If so, the system performs operation 4 at step 110 and returns to step 102.\n\nAn example of the efficiency gain is demonstrated by applying the invention to a scalar multiplication example. For example, assume Q=kP. Assume the scalar k is a 160-bit large integer:\n\nk=(100 . . . 01110 . . . 01010 . . . 001101)\n\n=((1*249+7)*266+5)*244+13\n\nThe invention first breaks up the binary bit-string of the scalar k into two kinds of windows, nonzero-windows and the zero-windows:\n\nk=(1 00 . . . 0 111 0 . . . 0 101 0 . . . 00 1101)\n\nThe scalar multiplication can be decomposed into multiple iterations of repeated point-doublings and point-additions:\n\nQ=kP=((249 *P+7P)*266+5P)*244+13P\n\n=244 Q 1+13P, where\n\nQ 1=266 Q 2+5P and Q 2=249 P+7P\n\nThe size of a zero-window can be as large as it needs to be. The size of a nonzero-window is limited by the size of the look-up table used in the system. The points, 7P, 5P, and 13P can be fetched directly from a look-up table. A table look-up is an effective technique for eliminating point-additions. Using a small 4-bit look-up table, one can potentially eliminate up to 75% of the point-additions in the system. Now the computation burden shifts over to the side of point-doublings. As can be seen from the table below, 159 point doublings result from 249P, 244Q1, and 266Q2 and consume significant computational resources.\n\nThus, the scalar multiplication above requires 159 point-doublings and 3 point-additions. Using prior art techniques, this would require as many as 324 multiplies. Using the present invention, the total number of multiplies in this scalar multiplication is reduced to 6.\n\n Conventional approach New approach 159 point- 318 multiplies + 159 inverts 3 multiplies + 159 inverts doublings 3 point- 6 multiplies + 3 inverts 3 multiplies + 3 inverts additions Total 324 multiplies + 162 inverts 6 multiplies + 162 inverts\n\nHardware Execution Environment\n\nThe present invention can be implemented as an elliptic curve crypto-accelerator in hardware. One possible embodiment is illustrated in FIG. 2. Block 200 contains the five registers A, B, U, V, and M. Compare logic 201 is coupled to the registers and is used to compare the LSB's of registers A and B, to compare the values of registers A and B, and to execute the algorithm described in FIG. 1. Invariant operations logic 202 is coupled to the registers 200 and to the compare logic 201 to implement the four invariant operations as appropriate.\n\nDivision Over Integer Field\n\nThe present invention also provides a solution for division over integer fields using registers and register manipulations based on tests of register content values.\n\n Modular_Division_F(p){ A<−x, B<−M, U<−y, V<−0; while (A not equal B) do{ If A even then { A<−A/2 if U even then U<−U/2; else U<−(U+M)/2; } else if B even then { B<−B/2 if V even then V<−V/2; else V <−(V+M)/2 } else if (A>B) then { A<−(A−B)/2 U<−U−V; if U<0 then U<−(U+M)/2; } else{ B<−(B−A)/2 V<−V-U; if V<0 then V<−(V+M) if V even then V<−V/2; else V<−(V+M)/2; } } }\n\nFour registers A, B, U, and V are initialized with integer values x, M, y, and 0, respectively. (Note, A, B, U, V, and M refer to both the registers and the values in the registers). Registers A and B are checked to see if they are even or odd. If a register is even, its contents can be divided by 2 and replace the old register value. If A and B are both odd, the larger value can be replaced with one half the difference between A and B (the difference of two odd numbers always being an even number, so that division by 2 is possible). Based on the values of registers A and B, registers U and V are manipulated and modified using certain rules (described below in FIGS. 4A-4C). The register modification continues until the value in registers A and B are equal, at which time the value in register U holds the desired result. The values in the A, B, U, V registers maintain the invariant relationships as follows:\n\nA*y=U*x mod M and B*y=V*x mod M.\n\nThe following three rules are used to preserve the invariant relationship among the registers A, B, U, and V.\n\nRule 1—If A is even:\n\nif U is even then y(A/2)=x(U/2) mod M,\n\nelse y (A/2)=x ((U+M)/2) mod M\n\nRule 2—If B is even:\n\nIf V is even then y (B/2)=x (V/2) mod M;\n\nelse y (B/2)=x ((V+M)/2) mod M;\n\nRule 3—If A and B are odd:\n\n(A−B) y=(U−V) x mod M\n\n(if (U−V)<0 then (A−B)y=(U−V+M )x mod M\n\nif (U−V) even then (A−B)y/2=((U−V)x/2) mod M\n\nelse (A−B)y/2=((U−V+M)x/2) mod M\n\nFIGS. 4A, 4B, and 4C are a flow diagram illustrating division over integer fields. At step 400 register A is initialized with x. At step 401 register B is initialized with M, at step 402 register U is initialized with y, and at step 403 register V is initialized with 0. At decision block 404 it is determined if the values in registers A and B are equal, if so, the division process ends. If not, then the system proceeds to decision block 405.\n\nAt decision block 405 it is determined if the value in register A is even. If so, then the value in A is divided by 2 and placed in A at step 406. At decision block 407 register U is tested for evenness. If register U is odd, then registers U and M are combined and divided by 2 at step 409 with the result placed in register U and the system returns to step 404. If register U is even, then the value in U is replaced with U/2 at step 408 and the system returns to step 404.\n\nIf the value in register A is odd, the system proceeds to decision block 410 to check the even/odd status of register B. If B is even, then B/2 is stored in B at step 411. At decision block 412 the even/odd status of register V is checked. If V is odd, then V is replaced with (V+M)/2 at step 414 and the system proceeds to step 404. If V is even, then V/2 is placed in register V at step 413 and the system proceeds to step 404.\n\nIf both A and B are odd, then registers A and B are compared at step 415. If A is greater than B, then the value in A is replaced with (A−B)/2 at step 416. Register U is replaced with U−V at step 417. At decision block 418 register U is checked for negativity. If negative, U is replaced with U+M at step 419 and the system proceeds to step 420. If U is positive at step 418, it is checked for even/odd at step 420. If even, U is replaced with U/2 at step 421 and the system returns to step 404. If odd, U is replaced with U+M)/2 at step 422 before returning to step 404.\n\nIf A is not greater than B at step 415, the system replaces B with (B−A)/2 at step 423. At step 424 V is replaced with V−U. At decision block 425 it is determined if V is less than 0. If so, V is replaced with (V+M) at step 426 and the system proceeds to step 427. If V is not less than 0 at step 425, then an even/odd determination of V is made at step 427. If V is even, V is replaced with V/2 at step 428 and returns to step 404. If V is odd, V is replaced with (V+M)/2 at step 429 and returns to step 404.\n\nSoftware Execution Environment\n\nAn embodiment of the invention can be implemented as computer software in the form of computer readable code executed in a general purpose computing environment such as environment 300 illustrated in FIG. 3, or in the form of bytecode class files running in such an environment. A keyboard 310 and mouse 311 are coupled to a bi-directional system bus 318. The keyboard and mouse are for introducing user input to a computer 301 and communicating that user input to processor 313.\n\nComputer 301 may also include a communication interface 320 coupled to bus 318. Communication interface 320 provides a two-way data communication coupling via a network link 321 to a local network 322. For example, if communication interface 320 is an integrated services digital network (ISDN) card or a modem, communication interface 320 provides a data communication connection to the corresponding type of telephone line, which comprises part of network link 321. If communication interface 320 is a local area network (LAN) card, communication interface 320 provides a data communication connection via network link 321 to a compatible LAN. Wireless links are also possible. In any such implementation, communication interface 320 sends and receives electrical, electromagnetic or optical signals which carry digital data streams representing various types of information.\n\nNetwork link 321 typically provides data communication through one or more networks to other data devices. For example, network link 321 may provide a connection through local network 322 to local server computer 323 or to data equipment operated by ISP 324. ISP 324 in turn provides data communication services through the world wide packet data communication network now commonly referred to as the “Internet” 325. Local network 322 and Internet 325 both use electrical, electromagnetic or optical signals which carry digital data streams. The signals through the various networks and the signals on network link 321 and through communication interface 320, which carry the digital data to and from computer 300, are exemplary forms of carrier waves transporting the information.\n\nProcessor 313 may reside wholly on client computer 301 or wholly on server 326 or processor 313 may have its computational power distributed between computer 301 and server 326. In the case where processor 313 resides wholly on server 326, the results of the computations performed by processor 313 are transmitted to computer 301 via Internet 325, Internet Service Provider (ISP) 324, local network 322 and communication interface 320. In this way, computer 301 is able to display the results of the computation to a user in the form of output. Other suitable input devices may be used in addition to, or in place of, the mouse 311 and keyboard 310. I/O (input/output) unit 319 coupled to bi-directional system bus 318 represents such I/O elements as a printer, A/V (audio/video) I/O, etc.\n\nComputer 301 includes a video memory 314, main memory 315 and mass storage 312, all coupled to bi-directional system bus 318 along with keyboard 310, mouse 311 and processor 313. As with processor 313, in various computing environments, main memory 315 and mass storage 312, can reside wholly on server 326 or computer 301, or they may be distributed between the two. Examples of systems where processor 313, main memory 315, and mass storage 312 are distributed between computer 301 and server 326 include the thin-client computing architecture developed by Sun Microsystems, Inc., the palm pilot computing device, Internet ready cellular phones, and other Internet computing devices.\n\nThe mass storage 312 may include both fixed and removable media, such as magnetic, optical or magnetic optical storage systems or any other available mass storage technology. Bus 318 may contain, for example, thirty-two address lines for addressing video memory 314 or main memory 315. The system bus 318 also includes, for example, a 32-bit data bus for transferring data between and among the components, such as processor 313, main memory 315, video memory 314 and mass storage 312.\n\nAlternatively, multiplex data/address lines may be used instead of separate data and address lines.\n\nIn one embodiment of the invention, the processor 313 is a microprocessor manufactured by Motorola, such as the 680X0 processor or a microprocessor manufactured by Intel, such as the 80X86, or Pentium processor, or a SPARC microprocessor from Sun Microsystems, Inc. However, any other suitable microprocessor or microcomputer may be utilized. Main memory 315 is comprised of dynamic random access memory (DRAM). Video memory 314 is a dual-ported video random access memory. One port of the video memory 314 is coupled to video amplifier 316. The video amplifier 316 is used to drive the cathode ray tube (CRT) raster monitor 317. Video amplifier 316 is well known in the art and may be implemented by any suitable apparatus. This circuitry converts pixel data stored in video memory 314 to a raster signal suitable for use by monitor 317. Monitor 317 is a type of monitor suitable for displaying graphic images.\n\nComputer 301 can send messages and receive data, including program code, through the network(s), network link 321, and communication interface 320. In the Internet example, remote server computer 326 might transmit a requested code for an application program through Internet 325, ISP 324, local network 322 and communication interface 320. The received code may be executed by processor 313 as it is received, and/or stored in mass storage 312, or other non-volatile storage for later execution. In this manner, computer 300 may obtain application code in the form of a carrier wave. Alternatively, remote server computer 326 may execute applications using processor 313, and utilize mass storage 312, and/or video memory 315. The results of the execution at server 326 are then transmitted through Internet 325, ISP 324, local network 322 and communication interface 320. In this example, computer 301 performs only input and output functions.\n\nApplication code may be embodied in any form of computer program product. A computer program product comprises a medium configured to store or transport computer readable code, or in which computer readable code may be embedded. Some examples of computer program products are CD-ROM disks, ROM cards, floppy disks, magnetic tapes, computer hard drives, servers on a network, and carrier waves.\n\nThe computer systems described above are for purposes of example only. An embodiment of the invention may be implemented in any type of computer system or programming or processing environment.\n\nThus, a method for efficient polynomial divide has been described.\n\n## Claims (10)\n\nWhat is claimed is:\n1. A method for securing data, the method comprising:\nselecting a first point on an elliptic curve of a Galois field of an order 2m;\ncalculating (x/y) mod M, wherein x represents a first integer, y represents a second integer, and M represents a prime number, the first and second represented integers and the represented prime number are of the Galois field, wherein the calculating includes,\nassigning x to a first register A and y to a second register U,\nassigning M to a third register B,\ninitializing a fourth register V with zero, and\niteratively reducing the registers A and B to one while adjusting the values U and V to maintain the following two invariant relationships:\nA*y≡U*x mod M\nB*y≡V*x mod M; and\nselecting a second point on the elliptic curve based at least in part on register U's value; and\nsecuring data based at least in part on the selected first and second points on the elliptic curve.\n2. The method of claim 1 wherein securing data includes generating one or more security keys based at least in part on the selected first and second points.\n3. The method of claim 1 wherein securing the data includes generating and verifying one or more digital signatures.\n4. The method of claim 1 wherein iteratively reducing the registers A and B comprises:\nstoring in register A half of register A's contents if register A's contents are even;\nstoring in register U half of register U's contents if both register U's contents and register A's contents are even;\nstoring in register U half of a sum of M and register U's contents if register A's contents are even and register U's contents are not even;\nstoring in register B half of register B's contents if register B's contents are even;\nstoring in register V half of register V's contents if both register V's contents and register B's contents are even;\nstoring in register V half of a sum of M and register V's contents if register B's contents are even and register V's contents are not even;\nstoring in register A half of a sum of register B's contents and register A's contents, and storing in register U a sum of register V's contents and register U's contents, if register A's contents are greater than register B's contents;\nstoring in register U half of register U's contents if register U's contents are even and register A's contents are greater than register B's contents;\nstoring in register U half of a sum of M and register U's contents if register U's content are not even and register A's contents are greater than register B's contents; and\nstoring in register B half of a sum of register B's contents and register A's contents and storing in register V a sum of register V's contents and register U's contents, if register A's contents are not greater than register B's contents and neither register A's contents nor register B's contents are even;\nstoring in register V half of t and register V's contents if register V's contents are even and register A's contents are not greater than register B's contents and neither register A's contents nor register B's contents are even; and\nstoring in register V half of a sum of M and register V's contents if register V's contents are not even and register A's contents are not greater than register B's contents and neither register A's contents nor register B's contents are even.\n5. The method of claim 4 wherein a register's contents are even if its least significant bit is zero.\n6. The method of claim 4 wherein a quotient is determined with one or more shifts.\n7. A computer program product encoded on one or more machine-readable media to implement the method of claim 1.\n8. An apparatus comprising:\na first register to initially host a representation of a first integer;\na second register to initially host a representation of a second integer, wherein the first and second represented integers represent elements of a Galois field of order 2m;\na third register to initially host a representation of a prime integer of the Galois field;\na fourth register to initially host a zero;\na compare logic to determine if the first register's value is even, if the third register's value is even, or if the first register's value is greater than the third register's value; and\nan invariant operations logic for modifying the first, second, third, and fourth register to conform to invariant relationships including,\na product of the first register and the second represented integer being congruent to a product of the second register and the first represented integer modulo the prime number, and\na product of the third register and the second represented integer being congruent to a product of the fourth register and the first represented integer modulo the prime integer.\n9. A crypto-accelerator including the apparatus of claim 8.\n10. The apparatus of claim 8 further comprising a look-up table to host one or more results of point additions or point doublings.\nUS09/734,972 2000-08-28 2000-12-11 Method for efficient modular division over prime integer fields Active 2021-06-20 US6772184B2 (en)\n\n## Priority Applications (2)\n\nApplication Number Priority Date Filing Date Title\nUS09/649,356 US6721771B1 (en) 2000-08-28 2000-08-28 Method for efficient modular polynomial division in finite fields f(2{circumflex over ( )}m)\nUS09/734,972 US6772184B2 (en) 2000-08-28 2000-12-11 Method for efficient modular division over prime integer fields\n\n## Applications Claiming Priority (3)\n\nApplication Number Priority Date Filing Date Title\nUS09/734,972 US6772184B2 (en) 2000-08-28 2000-12-11 Method for efficient modular division over prime integer fields\nUS10/091,968 US6954772B2 (en) 2000-08-28 2002-03-05 Method and apparatus for performing modular division\nUS10/091,962 US6917957B2 (en) 2000-08-28 2002-03-05 Method and apparatus for performing modular division using counters\n\n## Related Parent Applications (1)\n\nApplication Number Title Priority Date Filing Date\nUS09/649,356 Continuation-In-Part US6721771B1 (en) 2000-08-28 2000-08-28 Method for efficient modular polynomial division in finite fields f(2{circumflex over ( )}m)\n\n## Related Child Applications (2)\n\nApplication Number Title Priority Date Filing Date\nUS10/091,968 Continuation-In-Part US6954772B2 (en) 2000-08-28 2002-03-05 Method and apparatus for performing modular division\nUS10/091,962 Continuation-In-Part US6917957B2 (en) 2000-08-28 2002-03-05 Method and apparatus for performing modular division using counters\n\n## Publications (2)\n\nPublication Number Publication Date\nUS20020052906A1 US20020052906A1 (en) 2002-05-02\nUS6772184B2 true US6772184B2 (en) 2004-08-03\n\n# Family\n\n## Family Applications (1)\n\nApplication Number Title Priority Date Filing Date\nUS09/734,972 Active 2021-06-20 US6772184B2 (en) 2000-08-28 2000-12-11 Method for efficient modular division over prime integer fields\n\n## Country Status (1)\n\nUS (1) US6772184B2 (en)\n\n## Cited By (9)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nUS20030142820A1 (en) * 2002-01-28 2003-07-31 Yuichi Futa Device and method for calculation on elliptic curve\nUS20030147529A1 (en) * 2002-02-05 2003-08-07 Perkins Gregory M. 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Table driven method for calculating arithmetic inverse for use in cryptography\nUS7068785B2 (en) * 2002-02-05 2006-06-27 Matsushita Electric Industrial Co., Ltd. Table driven method for calculating arithmetic inverse for use in cryptography\nUS20050220299A1 (en) * 2004-03-31 2005-10-06 Jesse Lipson Public key cryptographic methods and systems\nWO2005099150A2 (en) * 2004-03-31 2005-10-20 Jesse Lipson Public key cryptographic methods and systems\nUS8442219B2 (en) * 2004-03-31 2013-05-14 Jesse Lipson Public key cryptographic methods and systems\nWO2005099150A3 (en) * 2004-03-31 2009-04-16 Jesse Lipson Public key cryptographic methods and systems\nUS8477935B2 (en) * 2005-03-04 2013-07-02 Ihp Gmbh Method and apparatus for calculating a polynomial multiplication, in particular for elliptic curve cryptography\nUS20090136022A1 (en) * 2005-03-04 2009-05-28 Peter Langendoerfer Method and Apparatus for Calculating a Polynomial Multiplication, In Particular for Elliptic Curve Cryptography\nEP1729442A2 (en) 2005-06-03 2006-12-06 Tata Consultancy Services Limited An authentication system executing an elliptic curve digital signature cryptographic process\nUS20100232495A1 (en) * 2007-05-16 2010-09-16 Citta Richard W Apparatus and method for encoding and decoding signals\nUS8964831B2 (en) * 2007-05-16 2015-02-24 Thomson Licensing Apparatus and method for encoding and decoding signals\nUS8873620B2 (en) 2007-05-16 2014-10-28 Thomson Licensing Apparatus and method for encoding and decoding signals\nUS7991162B2 (en) * 2007-09-14 2011-08-02 University Of Ottawa Accelerating scalar multiplication on elliptic curve cryptosystems over prime fields\nUS20090074178A1 (en) * 2007-09-14 2009-03-19 University Of Ottawa Accelerating Scalar Multiplication On Elliptic Curve Cryptosystems Over Prime Fields\nUS8908773B2 (en) 2007-10-15 2014-12-09 Thomson Licensing Apparatus and method for encoding and decoding signals\nUS20100296576A1 (en) * 2007-10-15 2010-11-25 Thomson Licensing Preamble for a digital television system\nUS9414110B2 (en) 2007-10-15 2016-08-09 Thomson Licensing Preamble for a digital television system\n\n## Also Published As\n\nPublication number Publication date\nUS20020052906A1 (en) 2002-05-02\n\n## Similar Documents\n\nPublication Publication Date Title\nPieprzyk et al. 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Algorithms for black-box fields and their application to cryptography\n\n## Legal Events\n\nDate Code Title Description\nAS Assignment\n\nOwner name: SUN MICROSYSTEMS, INC., CALIFORNIA\n\nFree format text: ASSIGNMENT OF ASSIGNORS INTEREST;ASSIGNOR:CHANG, SHEUELING;REEL/FRAME:011364/0896\n\nEffective date: 20001207\n\nCC Certificate of correction\nCC Certificate of correction\nFPAY Fee payment\n\nYear of fee payment: 4\n\nFPAY Fee payment\n\nYear of fee payment: 8\n\nAS Assignment\n\nOwner name: ORACLE AMERICA, INC., CALIFORNIA\n\nFree format text: MERGER AND CHANGE OF NAME;ASSIGNORS:ORACLE USA, INC.;SUN MICROSYSTEMS, INC.;ORACLE AMERICA, INC.;REEL/FRAME:037278/0779\n\nEffective date: 20100212\n\nFPAY Fee payment\n\nYear of fee payment: 12" ]

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[ "# Boulder on a hill (Vectors and components)\n\n## Homework Statement\n\nA boulder of weight \"w\" rests on a hillside that rises at a constant angle \"a\" above the horizontal. The boulder's weight is a force on the boulder that has a direction vertically downward.\n\nIn terms of \"w\" and \"a\", what is the component of the weight of the boulder in the direction parallel to the surface of the hill?\n\nWhat is the component of the weight in the direction perpendicular to the surface of the hill?\n\n## The Attempt at a Solution\n\ni have no idea...\n\nLast edited by a moderator:\n\nRelated Introductory Physics Homework Help News on Phys.org\ntiny-tim\nHomework Helper\nwelcome to pf!\n\nhi asqwt! welcome to pf!", null, "In terms of \"w\" and \"a\", what is the component of the weight of the boulder in the direction parallel to the surface of the hill?\n\nWhat is the component of the weight in the direction perpendicular to the surface of the hill?\n\ni have no idea...\noh come on, you must know something about components …\n\nwhat do you know about the component of a vector (or a force) in a particular direction?", null, "i know a vector is made up of an x component, and a y component. but that is when the vector is in a diagonal direction.\n\ntiny-tim\nHomework Helper\nhi asqwt!", null, "i know a vector is made up of an x component, and a y component. but that is when the vector is in a diagonal direction.\nthis is in a diagonal direction …\n\njust turn the papaer round a little so that the slope is horizontal", null, "(no seriously … it does work!)\n\nie use cos and sin just the way you usually would\n\nyou want me to rotate the diagram till the vector of the rock... (which is pointing down) is horizontal? so rotate right 90 degrees?\n\nam i solving for the length of the slope? sin a = w/length of slope.\n\ntiny-tim\nHomework Helper\nyou want me to rotate the diagram till the vector of the rock... (which is pointing down) is horizontal? so rotate right 90 degrees?\nno, until the slope is horizontal …\n\nthat'll then be a situation you're familiar with", null, "am i solving for the length of the slope? sin a = w/length of slope.\nlet's see …\nIn terms of \"w\" and \"a\", what is the component of the weight of the boulder in the direction parallel to the surface of the hill?\n\nWhat is the component of the weight in the direction perpendicular to the surface of the hill?\nthe length of the slope has nothing to do with it\n\nthe component is always the original magnitude time cos of the angle …\n\nso for one direction it'll be cos(a), and for the other it'll be cos(90°-a) = sin(a) …\n\nwhich is which?", null, "the component is always the original magnitude time cos of the angle …\n\nso for one direction it'll be cos(a), and for the other it'll be cos(90°-a) = sin(a) …\n\nhow did you get those? i didnt read that in my book :(\n\ntiny-tim\nHomework Helper\nif you've done dot-products, it's the standard dot-product with a unit vector …\n\nyou know a.b = |a| |b| cosθ (where θ is the angle between a and b)\n\nwell, if b is the unit vector in, say, the x direction (we usually write that as i), then:\n\na.i = |a| |i| cosθ = |a| cosθ\n\n(since by definition of a unit vector, |i| = 1)\n\nanyway, it does work in the situations you're familiar with, doesn't it?", null, "woh i never did dot products. sorry im making this hard for you when it is supposedly simple = \\ i feel like a retard.\n\nbut can i get a few things straightened out?\n\nso is the VECTOR of this problem is the weight of the rock going downward?\n\nyou want me to rotate my drawing so that the diagrams \"hill\" is parallell with the gruond?\n\nthen i lost you. :( can walk me through the process step by step by any chance? i have no idea why the answer is w sin(a)\n\nLast edited:\ntiny-tim\nHomework Helper\nso is the VECTOR of this problem is the weight of the rock going downward?\nyes\n\nbut before we go any further, can i check what you do know …\n\nif i tell you a vector 16 is pointing at 20° above the horizontal, and ask you for the horizontal and vertical components, what are they?", null, "assuming you mean the magnitude is 16. vertical is 16 (sin 20) and horizontal is 16 (cos 20)\n\ntiny-tim\nHomework Helper\nyup!", null, "ok, it's the same for any (perpendicular) axes …\n\nchoose one to be the x axis, and call the angle for the vector to that axis θ …\n\nthen the x component is the vector times cosθ, and the y component is the vector times cos(90°-θ), = sinθ\n\nin this case, the vector is the weight, w, downwards …\n\nso the component along the slope (going down) is … ?", null, "im scared i might be wrong... but..\n\nso if i rotate it to make the slope my x-axis\n\nis the x component w cos (90-theta)? since im not using the angle given?\n\ntiny-tim\nHomework Helper\n(just got up :zzz: …)\nim scared i might be wrong... but..\n\nso if i rotate it to make the slope my x-axis\n\nis the x component w cos (90-theta)? since im not using the angle given?\nthat's correct!", null, "and soon you'll get used to just looking for the angle, and not having to rotate the page", null, "(btw, i personally think \"not-cos\" rather than \"sin\" … i always look for the \"correct\" angle, and it either is or isn't, so it's either cos or not-cos :biggrin;)" ]

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[ "## College Algebra (10th Edition)\n\n$\\dfrac{x+9}{2x-1}; x \\ne \\frac{1}{2}$\nThe expressions are similar so subtract the numerators and copy the denominator to obtain: $=\\dfrac{3x+5-(2x-4)}{2x-1} \\\\=\\dfrac{3x+5-2x-(-4)}{2x-1} \\\\=\\dfrac{3x+5-2x+4}{2x-1} \\\\=\\dfrac{x+9}{2x-1}; x \\ne \\frac{1}{2}$ ($x$ cannot be $\\dfrac{1}{2}$ as it makes the expression undefined.)" ]

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[ "# Parameter estimation for normal distribution in Java\n\nGiven a set of data (~5000 values) I'd like to draw random samples from the same distribution as the original data. The problem is there is no way to know for sure what distribution the original data comes from.\n\nIt makes sense to use normal distribution in my case, although I'd like to be able to motivate that decision, and of course I also need to estimate the $(\\mu,\\sigma)$ pair.\n\nAny idea on how to accomplish this, preferably within Java environment. I have been using Apache Commons Math and recently stumbled upon Colt library. I was hoping to get it done without bothering with MATLAB and R.\n\n• you should mention java environment in the title of your question this will maximize the chance to get an answer. Also is there a tag \"java\" ? – robin girard Feb 9 '11 at 17:09\n• @robin Yes it is. – user88 Feb 9 '11 at 17:14\n• @mbq: cheers! I wasnt too sure about tagging/mentioning java too much; my interest is primarily if it can be done relatively easily, then a Java implementation. But this works as well. – posdef Feb 9 '11 at 17:20\n• @posdef, if it is normal, then simulation is easy, just calculate mean and standard deviation. For motivation you will need to test whether the sample is normal. – mpiktas Feb 9 '11 at 20:01\n• @posdef, answers require better quality than comments, this is why I commented. As for commercial use, try google to find non comercial. This seems free. In worst case scenario you will need to reimplement this test, or another normality test. I suggest asking here or on stackoverflow , which of the normality tests is easiest to implement. – mpiktas Feb 10 '11 at 10:48\n\nOtherwise, it seems that the org.apache.commons.math.stat.descriptive.moment package has a Mean and StandardDeviation class which use the correct formulas. These should give you $\\mu$ and $\\sigma$, respectively." ]

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[ "Definition of\n\n# Milli-", null, "A prefix meaning one-thousandth (1/1,000 or 10-3)\n\nExample: a millimeter is 1/1,000th of a meter (in other words 1,000 millimeters make 1 meter)\n\nSymbol is m (as a prefix, otherwise \"m\" means meter)\n\nExamples:\n12 ms = 12 milliseconds = 12 thousandths of a second (0.012 s)\n95 mm = 12 millimeters = 95 thousandths of a meter (0.095 m)\nSee: Centi-", null, "Copyright © 2018 MathsIsFun.com" ]

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[ "# Difference between revisions of \"Template:AirStats/Documentation\"\n\n## Contents\n\n### Description\n\nThe template `AirStats` is used to assist in the creation of tables for in-game aircraft specification tables regarding flight characteristics.\n\n### Code\n\n```{{AirStats\n|maxSpeedAlt=\n<!--Stock stats-->\n|abSpeed1=\n|rbSpeed1=\n|maxAlt= (Optional - can delete for pages with working {{Specs-Card}} template)\n|abTurn1=\n|rbTurn1=\n|abClimb1=\n|rbClimb1=\n|takeoff=\n|abSpeed2=\n|rbSpeed2=\n|abTurn2=\n|rbTurn2=\n|abClimb2=\n|rbClimb2=\n}}```\n\n### Example\n\nCode:\n\n```{{AirStats\n|maxSpeedAlt=5700\n<!--Stock stats-->\n|abSpeed1=700\n|rbSpeed1=550\n|maxAlt=7500\n|abTurn1=10\n|rbTurn1=5\n|abClimb1=8.0\n|rbClimb1=6.0\n|takeoff=300\n|abSpeed2=800\n|rbSpeed2=650\n|abTurn2=15\n|rbTurn2=7.5\n|abClimb2=10.0\n|rbClimb2=7.5\n}}```\n\nGenerated:\n\nCharacteristics Max Speed\n(km/h at 5700 m)\nMax altitude\n(meters)\nTurn time\n(seconds)\nRate of climb\n(meters/second)\nTake-off run\n(meters)\nAB RB AB RB AB RB\nStock 700 550 7500 10 5 8.0 6.0 300\nUpgraded 800 650 15 7.5 10.0 7.5" ]

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[ "# Relations\n\nUp a level : Algebra and Arithmetic\nNext page : Classification of relations", null, "Relations\n\nIn mathematics one can define a relation as a structure where we associate one or several elements with each other. This we may write as", null, "$R({p_1},{p_2},{p_3},...,{p_n})$\n\nwhere", null, "${p_1},{p_2},{p_3},...,{p_n}$ is an ordered list of parameters or variables. The number of variables is called the arity of the relation. The relation is said to hold for, or be satisfied by a particular assignment of the variables if it is true for that assignment. The relation will thus be a logical statement about variables.\n\nThe relation a>b, where a and b are real numbers is satisfied by, for example a=5 and b=2.344, since it is true for those numbers, whereas the relation a=b is false for those particular numbers.\n\nOther examples of relation is for example x+y+z=12 or that x and y has the same father (e.g. Luke and Leia), or that two cities are connected to the same railway system. For the later R(Stockholm, Uppsala) is true, whilst R(Stockhalm, Reykavik) is false.\n\nEquations\n\nTo find values that satisfies a relation is called to solve the relation. Used in this way the relation is called an equation.\n\nWe may for example have the equation 2x=10, that is satisfied by (and only by) x=5, and this will thus be the solution The equation x=5+y is satisfied by an infinite number of values for x and y where x is 5 larger than y, for example x=3 and y=8. The complete solution may be easiest expressed by the very equation itself: x=5+y.\n\nBinary Relations\n\nThe last relation we looked at is of a a particularly important type of relations, the so called binary relations, that are the relations between two variables. It may, for the two variables a and b, be written as R(a,b), or, aRb.\n\nWe may write our last example as", null, "$xRy \\Leftrightarrow x = 5 + y$\n\nThis means that xRy is true if and only if x=y+5 is true. The relation R will then be the relation x=y+5. We then say that x is R-related to y.\n\nRelations as sets\n\nA binary relation may be described using equations or descriptions in words, but also as sets of ordered pairs. The relation a>b, where a and b are restricted to the numbers 1, 2, and 3 may be represented by the set\n\n{(3, 2), (3, 1), (2 ,1)},\n\nbecause these three pairs represent the three cases (a,b) where a>b.\n\nWe can represent relations with higher arity with sets of ordered lists with more than two elements.\n\nDomain and range\n\nFor the moment being we sill restrict this to binary relation. Say we have a binary relation xRy. We may then choose to see one of the variables as an input, and the other as an output. The one we use as an input we call the independent variable, and the output we call the dependent variable, since it is (usually) depending on the chosen value of the other. The domain of the relation is the set of possible values of the independent variable, and the range is (mostly) defined as the set of possible values of the dependent variable. This is also called the image of the domain. (Note that the range is sometimes defined as set that has the image as a subset, but might include other elements too.)\n\nSay, for example ,that the relation is “y is the length of x rounded to the nearest cm”, where the domain is the set of human beings . Then the range is the interval of lengths (in cm) from the smallest human being that have existed to the tallest that have existed.\n\nThe domain is thus the set of allowed inputs, and the range the set of possible outputs.", null, "Up a level : Algebra and Arithmetic\nNext page : Classification of relations", null, "Last modified: Jan 13, 2017 @ 20:51" ]

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[ "# Direct sum explained\n\nThe direct sum is an operation from abstract algebra, a branch of mathematics. For example, the direct sum\n\nRR\n\n, where\n\nR\n\nis real coordinate space, is the Cartesian plane,\n\nR2\n\n. To see how the direct sum is used in abstract algebra, consider a more elementary structure in abstract algebra, the abelian group. The direct sum of two abelian groups\n\nA\n\nand\n\nB\n\nis another abelian group\n\nAB\n\nconsisting of the ordered pairs\n\n(a,b)\n\nwhere\n\na\\inA\n\nand\n\nb\\inB\n\n. (Confusingly this ordered pair is also called the cartesian product of the two groups.) To add ordered pairs, we define the sum\n\n(a,b)+(c,d)\n\nto be\n\n(a+c,b+d)\n\n; in other words addition is defined coordinate-wise. A similar process can be used to form the direct sum of two vector spaces or two modules.\n\nWe can also form direct sums with any finite number of summands, for example\n\nABC\n\n, provided\n\nA,B,\n\nand\n\nC\n\nare the same kinds of algebraic structures (e.g., all abelian groups, or all vector spaces). This relies on the fact that the direct sum is associative up to isomorphism. That is,\n\n(AB)C\\congA(BC)\n\nfor any algebraic structures\n\nA\n\n,\n\nB\n\n, and\n\nC\n\nof the same kind. The direct sum is also commutative up to isomorphism, i.e.\n\nAB\\congBA\n\nfor any algebraic structures\n\nA\n\nand\n\nB\n\nof the same kind.\n\nIn the case of two summands, or any finite number of summands, the direct sum is the same as the direct product. If the arithmetic operation is written as +, as it usually is in abelian groups, then we use the direct sum. If the arithmetic operation is written as × or ⋅ or using juxtaposition (as in the expression\n\nxy\n\n) we use direct product.\n\nIn the case where infinitely many objects are combined, most authors make a distinction between direct sum and direct product. As an example, consider the direct sum and direct product of infinitely many real lines. An element in the direct product is an infinite sequence, such as (1,2,3,...) but in the direct sum, there would be a requirement that all but finitely many coordinates be zero, so the sequence (1,2,3,...) would be an element of the direct product but not of the direct sum, while (1,2,0,0,0,...) would be an element of both. More generally, if a + sign is used, all but finitely many coordinates must be zero, while if some form of multiplication is used, all but finitely many coordinates must be 1. In more technical language, if the summands are\n\n(Ai)i\n\n, the direct sum\n\noplusiAi\n\nis defined to be the set of tuples\n\n(ai)i\n\nwith\n\nai\\inAi\n\nsuch that\n\nai=0\n\nfor all but finitely many i. The direct sum\n\noplusiAi\n\nis contained in the direct product\n\n\\prodiAi\n\n, but is usually strictly smaller when the index set\n\nI\n\nis infinite, because direct products do not have the restriction that all but finitely many coordinates must be zero.\n\n## Examples\n\nThe xy-plane, a two-dimensional vector space, can be thought of as the direct sum of two one-dimensional vector spaces, namely the x and y axes. In this direct sum, the x and y axes intersect only at the origin (the zero vector). Addition is defined coordinate-wise, that is\n\n(x1,y1)+(x2,y2)=(x1+x2,y1+y2)\n\n, which is the same as vector addition.\n\nGiven two structures\n\nA\n\nand\n\nB\n\n, their direct sum is written as\n\nAB\n\n. Given an indexed family of structures\n\nAi\n\n, indexed with\n\ni\\inI\n\n, the direct sum may be written\n\nstyleA=oplusi\\inAi\n\n. Each Ai is called a direct summand of A. If the index set is finite, the direct sum is the same as the direct product. In the case of groups, if the group operation is written as\n\n+\n\nthe phrase \"direct sum\" is used, while if the group operation is written\n\n*\n\nthe phrase \"direct product\" is used. When the index set is infinite, the direct sum is not the same as the direct product since the direct sum has the extra requirement that all but finitely many coordinates must be zero.\n\n### Internal and external direct sums\n\nA distinction is made between internal and external direct sums, though the two are isomorphic. If the factors are defined first, and then the direct sum is defined in terms of the factors, we have an external direct sum. For example, if we define the real numbers\n\nR\n\nand then define\n\nRR\n\nthe direct sum is said to be external.\n\nIf, on the other hand, we first define some algebraic structure\n\nS\n\nand then write\n\nS\n\nas a direct sum of two substructures\n\nV\n\nand\n\nW\n\n, then the direct sum is said to be internal. In this case, each element of\n\nS\n\nis expressible uniquely as an algebraic combination of an element of\n\nV\n\nand an element of\n\nW\n\n. For an example of an internal direct sum, consider\n\nZ6\n\n(the integers modulo six), whose elements are\n\n\\{0,1,2,3,4,5\\}\n\n. This is expressible as an internal direct sum\n\nZ6=\\{0,2,4\\}\\{0,3\\}\n\n.\n\n## Types of direct sum\n\n### Direct sum of abelian groups\n\n(A,\\circ)\n\nand\n\n(B,\\bullet)\n\n, their direct sum\n\nAB\n\nis the same as their direct product. That is, the underlying set is the Cartesian product\n\nA x B\n\nand the group operation\n\nis defined component-wise:\n\n(a1,b1)(a2,b2)=(a1\\circa2,b1\\bulletb2)\n\n.This definition generalizes to direct sums of finitely many abelian groups.\n\nFor an infinite family of abelian groups Ai for iI, the direct sum\n\noplusiAi\n\nis a proper subgroup of the direct product. It consists of the elements\n\nstyle(ai)\\in\\prodjAj\n\nsuch that ai is the identity element of Ai for all but finitely many i.\n\n### Direct sum of modules\n\nSee main article: Direct sum of modules. The direct sum of modules is a construction which combines several modules into a new module.\n\nThe most familiar examples of this construction occur when considering vector spaces, which are modules over a field. The construction may also be extended to Banach spaces and Hilbert spaces.\n\n### Direct sum of group representations\n\nThe direct sum of group representations generalizes the direct sum of the underlying modules, adding a group action to it. Specifically, given a group G and two representations V and W of G (or, more generally, two G-modules), the direct sum of the representations is VW with the action of gG given component-wise, i.e.\n\ng·(v, w) = (g·v, g·w).Another equivalent way of defining the direct sum is as follows:\n\nGiven two representations\n\n(V,\\rhoV)\n\nand\n\n(W,\\rhoW)\n\nthe vector space of the direct sum is\n\nVW\n\nand the homomorphism\n\n\\rhoV\n\nis given by\n\n\\alpha\\circ(\\rhoV x \\rhoW)\n\n, where\n\n\\alpha:GL(V) x GL(W)\\toGL(VW)\n\nis the natural map obtained by coordinate-wise action as above.\n\nFurthermore, if\n\nV,W\n\nare finite dimensional, then, given a basis of\n\nV,W\n\n,\n\n\\rhoV\n\nand\n\n\\rhoW\n\nare matrix-valued. In this case,\n\n\\rhoV\n\nis given as$g \\mapsto \\begin\\rho_V(g) & 0 \\\\ 0 & \\rho_W(g)\\end.$\n\nMoreover, if we treat\n\nV\n\nand\n\nW\n\nas modules over the group ring\n\nkG\n\n, where\n\nk\n\nis the field, then the direct sum of the representations\n\nV\n\nand\n\nW\n\nis equal to their direct sum as\n\nkG\n\nmodules.\n\n### Direct sum of rings\n\nSee main article: Product of rings. Some authors will speak of the direct sum\n\nRS\n\nof two rings when they mean the direct product\n\nR x S\n\n, but this should be avoided since\n\nR x S\n\ndoes not receive natural ring homomorphisms from R and S: in particular, the map\n\nR\\toR x S\n\nsending r to (r,0) is not a ring homomorphism since it fails to send 1 to (1,1) (assuming that 0≠1 in S). Thus\n\nR x S\n\nis not a coproduct in the category of rings, and should not be written as a direct sum. (The coproduct in the category of commutative rings is the tensor product of rings. In the category of rings, the coproduct is given by a construction similar to the free product of groups.)\n\nUse of direct sum terminology and notation is especially problematic when dealing with infinite families of rings: If\n\n(Ri)i\n\nis an infinite collection of nontrivial rings, then the direct sum of the underlying additive groups can be equipped with termwise multiplication, but this produces a rng, i.e., a ring without a multiplicative identity.\n\n### Direct sum in categories\n\nAn additive category is an abstraction of the properties of the category of modules. In such a category finite products and coproducts agree and the direct sum is either of them, cf. biproduct.\n\nGeneral case:In category theory the direct sum is often, but not always, the coproduct in the category of the mathematical objects in question. For example, in the category of abelian groups, direct sum is a coproduct. This is also true in the category of modules.\n\n### Direct sum of matrices\n\nFor any arbitrary matrices A and B, the direct sum A\n\nB is defined as the block diagonal matrix of A and B if both are square matrices (and to an analogous block matrix, if not).\n\nAB= \\begin{bmatrix} A&0\\\\ 0&B \\end{bmatrix}.\n\n## Homomorphisms\n\nThe direct sum\n\noplusiAi\n\ncomes equipped with a projection homomorphism\n\n\\pij\\colonoplusiAi\\toAj\n\nfor each j in I and a coprojection\n\n\\alphaj\\colonAj\\tooplusiAi\n\nfor each j in I. Given another algebraic structure\n\nB\n\n(with the same additional structure) and homomorphisms\n\ngj\\colonAj\\toB\n\nfor every j in I, there is a unique homomorphism\n\ng\\colonoplusiAi\\toB\n\n, called the sum of the gj, such that\n\ng\\alphaj=gj\n\nfor all j. Thus the direct sum is the coproduct in the appropriate category." ]

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[ "### On the rest of the files:\n\n``` Rename defun-gsl, defvariable.\nClean up markup and header files.\n:size to size\nmake-tests and assert-numerical-equal\n\ngit-svn-id: svn+ssh://pop/opt/space/mathematics/gsl/trunk@3295 a3d8a0fb-c1db-0310-ace7-a616afeb9e30```\nparent aefc0853\n ;; Chebyshev Approximations ;; Liam Healy Sat Nov 17 2007 - 20:36 ;; Time-stamp: <2008-02-03 13:22:58EST chebyshev.lisp> ;; Time-stamp: <2008-02-17 18:02:26EST chebyshev.lisp> ;; \\$Id: \\$ (in-package :gsl) ... ... @@ -12,36 +12,36 @@ (defgo-s (chebyshev order function lower-limit upper-limit) allocate-chebyshev free-chebyshev initialize-chebyshev) (defun-gsl allocate-chebyshev (order) (defmfun allocate-chebyshev (order) \"gsl_cheb_alloc\" ((order :size)) ((order size)) :c-return :pointer :export nil :index (letm chebyshev) :documentation :documentation ; FDL \"Allocate a Chebyshev series of specified order and return a pointer to it.\") (defun-gsl free-chebyshev (chebyshev) (defmfun free-chebyshev (chebyshev) \"gsl_cheb_free\" ((chebyshev :pointer)) :c-return :void :export nil :index (letm chebyshev) :documentation :documentation ; FDL \"Free a previously allocated Chebyshev series.\") (defun-gsl initialize-chebyshev (chebyshev function lower-limit upper-limit) (defmfun initialize-chebyshev (chebyshev function lower-limit upper-limit) \"gsl_cheb_init\" ((chebyshev :pointer) (function :pointer) (lower-limit :double) (upper-limit :double)) :export nil :index (letm chebyshev) :documentation :documentation ; FDL \"Compute the Chebyshev approximation for the function over the range (lower-limit, upper-limit) to the previously specified order. The computation of the Chebyshev approximation is an @math{O(n^2)} process, and requires @math{n} function evaluations.\") computation of the Chebyshev approximation is an O(n^2) process, and requires n function evaluations.\") ;;;;**************************************************************************** ;;;; Chebyshev series evaluation ... ... @@ -51,45 +51,45 @@ ;;; to use the functions that do return error (and ignore it if ;;; desired) in the form of #'evaluate-chebyshev. (defun-gsl evaluate-chebyshev-noerror (chebyshev x) (defmfun evaluate-chebyshev-noerror (chebyshev x) \"gsl_cheb_eval\" ((chebyshev :pointer) (x :double)) :c-return :double :index evaluate-chebyshev :export nil :documentation :documentation ; FDL \"Evaluate the Chebyshev series at a point x.\") (defun-gsl evaluate-chebyshev-noerror-order (chebyshev x order) (defmfun evaluate-chebyshev-noerror-order (chebyshev x order) \"gsl_cheb_eval_n\" ((chebyshev :pointer) (order :size) (x :double)) ((chebyshev :pointer) (order size) (x :double)) :c-return :double :index evaluate-chebyshev :export nil :documentation :documentation ; FDL \"Evaluate the Chebyshev series at a point x to at most the given order.\") (defun-gsl evaluate-chebyshev-full (chebyshev x) (defmfun evaluate-chebyshev-full (chebyshev x) \"gsl_cheb_eval_err\" ((chebyshev :pointer) (x :double) (result :double) (abserr :double)) :index evaluate-chebyshev :export nil :documentation :documentation ; FDL \"Evaluate the Chebyshev series at a point x, returning result and an estimate of its absolute error.\") (defun-gsl evaluate-chebyshev-order (chebyshev x order) (defmfun evaluate-chebyshev-order (chebyshev x order) \"gsl_cheb_eval_n_err\" ((chebyshev :pointer) (order :size) (x :double) (result :double) (abserr :double)) ((chebyshev :pointer) (order size) (x :double) (result :double) (abserr :double)) :index evaluate-chebyshev :export nil :documentation :documentation ; FDL \"Evaluate the Chebyshev series at a point x to at most the given order, returning result and an estimate of its absolute error.\") (export 'evaluate-chebyshev) (defun-optionals evaluate-chebyshev (chebyshev x &optional order) -full -order -full -order ; FDL \"Evaluate the Chebyshev series at a point x to at most the given order, returning result and an estimate of its absolute error.\") ... ... @@ -97,18 +97,18 @@ ;;;; Derivatives and integrals ;;;;**************************************************************************** (defun-gsl derivative-chebyshev (derivative chebyshev) (defmfun derivative-chebyshev (derivative chebyshev) \"gsl_cheb_calc_deriv\" ((derivative :pointer) (chebyshev :pointer)) :documentation :documentation ; FDL \"Compute the derivative of the Chebyshev series, storing the derivative coefficients in the previously allocated series. The two series must have been allocated with the same order.\") (defun-gsl integral-chebyshev (integral chebyshev) (defmfun integral-chebyshev (integral chebyshev) \"gsl_cheb_calc_integ\" ((integral :pointer) (chebyshev :pointer)) :documentation :documentation ; FDL \"Compute the integral of the Chebyshev series, storing the integral coefficients in the previously allocated series. The two series must have been allocated with the same order. ... ... @@ -147,7 +147,16 @@ (evaluate-chebyshev deriv x) (evaluate-chebyshev integ x)))) (lisp-unit:define-test chebyshev (lisp-unit:assert-equal '(\"0.715920990169d+00\" \"-0.150199666581d+01\" \"0.172397194040d+00\") (lisp-unit:fp-sequence (chebyshev-point-example 0.55d0)))) #| (make-tests chebyshev (chebyshev-point-example 0.55d0)) |# (LISP-UNIT:DEFINE-TEST CHEBYSHEV (LISP-UNIT::ASSERT-NUMERICAL-EQUAL (LIST (LIST 0.7159209901689866d0 -1.5019966658054353d0 0.17239719403979925d0)) (MULTIPLE-VALUE-LIST (CHEBYSHEV-POINT-EXAMPLE 0.55d0))))\n ;******************************************************** ; file: block.lisp ; description: Blocks of data ; date: Mon Mar 27 2006 - 12:28 ; author: Liam M. Healy ; modified: Wed May 31 2006 - 19:00 ;******************************************************** ;;; \\$Id: \\$ ;; Blocks of data ;; Liam Healy, Mon Mar 27 2006 - 12:28 ;; Time-stamp: <2008-02-17 09:18:05EST block.lisp> ;; \\$Id: \\$ (in-package :gsl) ;;; Block definition (cffi:defcstruct block (size :size) (size size) (data :pointer)) (defdata \"block\" block-double double-float) ... ...\n ;; Combinations ;; Liam Healy, Sun Mar 26 2006 - 11:51 ;; Time-stamp: <2008-02-04 19:25:57EST combination.lisp> ;; Time-stamp: <2008-02-17 10:01:38EST combination.lisp> ;; \\$Id: \\$ (in-package :gsl) ... ... @@ -11,8 +11,8 @@ ;;; GSL-combination definition (cffi:defcstruct gsl-combination-c (n :size) (k :size) (n size) (k size) (data :pointer)) ;;; Allocation, freeing, reading and writing ... ... @@ -25,11 +25,11 @@ ;;;; Getting values ;;;;**************************************************************************** (defun-gsl gsl-aref ((combination gsl-combination) &rest indices) (defmfun gsl-aref ((combination gsl-combination) &rest indices) \"gsl_combination_get\" (((pointer combination) :pointer) ((first indices) :size)) (((pointer combination) :pointer) ((first indices) size)) :type :method :c-return :size :c-return size :documentation ; FDL \"The ith element of the combination.\") ... ... @@ -45,7 +45,7 @@ ;;;; Setting values ;;;;**************************************************************************** (defun-gsl init-first (combination) (defmfun init-first (combination) \"gsl_combination_init_first\" (((pointer combination) gsl-combination-c)) :c-return :void ... ... @@ -54,7 +54,7 @@ \"Initialize the combination c to the lexicographically first combination, i.e. (0,1,2,...,k-1).\") (defun-gsl init-last (combination) (defmfun init-last (combination) \"gsl_combination_init_last\" (((pointer combination) gsl-combination-c)) :c-return :void ... ... @@ -63,7 +63,7 @@ \"Initialize the combination c to the lexicographically last combination, i.e. (n-k,n-k+1,...,n-1).\") (defun-gsl copy (destination source) (defmfun copy (destination source) \"gsl_combination_memcpy\" (((pointer destination) gsl-combination-c) ((pointer source) gsl-combination-c)) ... ... @@ -77,23 +77,23 @@ ;;;; Combination properties ;;;;**************************************************************************** (defun-gsl combination-range (c) (defmfun combination-range (c) \"gsl_combination_n\" (((pointer c) gsl-combination-c)) :c-return :size :c-return size :documentation ; FDL \"The range (n) of the combination c.\") (defun-gsl combination-size (c) (defmfun combination-size (c) \"gsl_combination_k\" (((pointer c) gsl-combination-c)) :c-return :size :c-return size :documentation ; FDL \"The number of elements (k) in the combination c.\") #| ;;; Unnecessary, gsl-array serves this function. (defun-gsl combination-data (c) (defmfun combination-data (c) \"gsl_combination_data\" (((pointer c) gsl-combination-c)) :c-return :pointer ... ... @@ -101,7 +101,7 @@ \"A pointer to the array of elements in the combination.\") |# (defun-gsl data-valid ((combination gsl-combination)) (defmfun data-valid ((combination gsl-combination)) \"gsl_combination_valid\" (((pointer combination) :pointer)) :type :method ... ... @@ -115,7 +115,7 @@ ;;;; Combination functions ;;;;**************************************************************************** (defun-gsl combination-next (c) (defmfun combination-next (c) \"gsl_combination_next\" (((pointer c) gsl-combination-c)) :c-return :success-failure :invalidate (c) ... ... @@ -127,7 +127,7 @@ repeatedly applying this function will iterate through all possible combinations of a given order.\") (defun-gsl combination-previous (c) (defmfun combination-previous (c) \"gsl_combination_prev\" (((pointer c) gsl-combination-c)) :c-return :success-failure ... ... @@ -142,33 +142,64 @@ ;;;; Examples and unit test ;;;;**************************************************************************** (lisp-unit:define-test combination (lisp-unit:assert-eql ; combination-range 4 (letm ((comb (combination 4 2 t))) (combination-range comb))) (lisp-unit:assert-eql ; combination-size 2 (letm ((comb (combination 4 2 t))) (combination-size comb))) (lisp-unit:assert-equal ; init-first, combination-next '((0 1) (0 2) (0 3) (1 2) (1 3) (2 3)) (letm ((comb (combination 4 2 t))) (init-first comb) (loop collect (data comb) while (combination-next comb)))) (lisp-unit:assert-equal ; init-last, combination-previous '((2 3) (1 3) (1 2) (0 3) (0 2) (0 1)) (letm ((comb (combination 4 2 t))) (init-last comb) (loop collect (data comb) while (combination-previous comb)))) (lisp-unit:assert-equal ; combination-next '(NIL (0) (1) (2) (3) (0 1) (0 2) (0 3) (1 2) (1 3) (2 3) (0 1 2) (0 1 3) (0 2 3) (1 2 3) (0 1 2 3)) (loop for i from 0 to 4 append (letm ((comb (combination 4 i t))) (init-first comb) (loop collect (data comb) while (combination-next comb)))))) #| (make-tests combination (letm ((comb (combination 4 2 t))) ; combination-range (combination-range comb)) (letm ((comb (combination 4 2 t))) ; combination-size (combination-size comb)) (letm ((comb (combination 4 2 t))) ; init-first, combination-next (init-first comb) (loop collect (data comb) while (combination-next comb))) (letm ((comb (combination 4 2 t))) ; init-last, combination-previous (init-last comb) (loop collect (data comb) while (combination-previous comb))) (loop for i from 0 to 4 ; combination-next append (letm ((comb (combination 4 i t))) (init-first comb) (loop collect (data comb) while (combination-next comb))))) |# (LISP-UNIT:DEFINE-TEST COMBINATION (LISP-UNIT::ASSERT-NUMERICAL-EQUAL (LIST 4) (MULTIPLE-VALUE-LIST (LETM ((COMB (COMBINATION 4 2 T))) (COMBINATION-RANGE COMB)))) (LISP-UNIT::ASSERT-NUMERICAL-EQUAL (LIST 2) (MULTIPLE-VALUE-LIST (LETM ((COMB (COMBINATION 4 2 T))) (COMBINATION-SIZE COMB)))) (LISP-UNIT::ASSERT-NUMERICAL-EQUAL (LIST (LIST (LIST 0 1) (LIST 0 2) (LIST 0 3) (LIST 1 2) (LIST 1 3) (LIST 2 3))) (MULTIPLE-VALUE-LIST (LETM ((COMB (COMBINATION 4 2 T))) (INIT-FIRST COMB) (LOOP COLLECT (DATA COMB) WHILE (COMBINATION-NEXT COMB))))) (LISP-UNIT::ASSERT-NUMERICAL-EQUAL (LIST (LIST (LIST 2 3) (LIST 1 3) (LIST 1 2) (LIST 0 3) (LIST 0 2) (LIST 0 1))) (MULTIPLE-VALUE-LIST (LETM ((COMB (COMBINATION 4 2 T))) (INIT-LAST COMB) (LOOP COLLECT (DATA COMB) WHILE (COMBINATION-PREVIOUS COMB))))) (LISP-UNIT::ASSERT-NUMERICAL-EQUAL (LIST (LIST (LIST) (LIST 0) (LIST 1) (LIST 2) (LIST 3) (LIST 0 1) (LIST 0 2) (LIST 0 3) (LIST 1 2) (LIST 1 3) (LIST 2 3) (LIST 0 1 2) (LIST 0 1 3) (LIST 0 2 3) (LIST 1 2 3) (LIST 0 1 2 3))) (MULTIPLE-VALUE-LIST (LOOP FOR I FROM 0 TO 4 APPEND (LETM ((COMB (COMBINATION 4 I T))) (INIT-FIRST COMB) (LOOP COLLECT (DATA COMB) WHILE (COMBINATION-NEXT COMB)))))))\n ;; Using GSL storage. ;; Using GSL bulk data (vectors, matrices, etc.) storage. ;; Liam Healy, Sun Mar 26 2006 - 16:32 ;; Time-stamp: <2008-02-10 15:30:20EST data.lisp> ;; Time-stamp: <2008-02-17 09:18:27EST data.lisp> ;; \\$Id: \\$ (in-package :gsl) ... ... @@ -94,7 +94,7 @@ (setf (pointer object) pointer)) ;;; (args (loop for i below dimensions collect (intern (format nil \"I~d\" i)))) ;;; (mapcar (lambda (v) `(,v :size)) args) ;;; (mapcar (lambda (v) `(,v size)) args) (defparameter *data-name-alist* '((FIXNUM . \"_int\") ... ... @@ -163,7 +163,7 @@ (flet ((gsl-name (function-name) (format nil \"gsl_~a_~a\" c-string function-name))) (let* ((cargs (loop for i below dimensions collect `((nth ,i (storage-size object)) :size))) collect `((nth ,i (storage-size object)) size))) (object-name (make-symbol-from-strings *gsl-prefix* cl-symbol))) `(progn (defclass ,object-name (,superclass) ... ... @@ -171,31 +171,31 @@ :allocation :class))) (data-go ,cl-symbol ,(or (member superclass '(gsl-matrix)) (member object-name '(gsl-combination)))) (defun-gsl alloc ((object ,object-name)) (defmfun alloc ((object ,object-name)) ,(gsl-name \"alloc\") ,cargs :type :method :c-return (cr :pointer) :return ((assign-pointer object cr))) (defun-gsl calloc ((object ,object-name)) (defmfun calloc ((object ,object-name)) ,(gsl-name \"calloc\") ,cargs :type :method :c-return (cr :pointer) :return ((assign-pointer object cr))) (defun-gsl free ((object ,object-name)) (defmfun free ((object ,object-name)) ,(gsl-name \"free\") (((pointer object) :pointer)) :type :method :c-return :void) (defun-gsl write-binary ((object ,object-name) stream) (defmfun write-binary ((object ,object-name) stream) ,(gsl-name \"fwrite\") ((stream :pointer) ((pointer object) :pointer)) :type :method) (defun-gsl read-binary ((object ,object-name) stream) (defmfun read-binary ((object ,object-name) stream) ,(gsl-name \"fread\") ((stream :pointer) ((pointer object) :pointer)) :type :method) (defun-gsl write-formatted ((object ,object-name) stream format) (defmfun write-formatted ((object ,object-name) stream format) ,(gsl-name \"fprintf\") ((stream :pointer) ((pointer object) :pointer) (format :string)) :type :method) (defun-gsl read-formatted ((object ,object-name) stream format) (defmfun read-formatted ((object ,object-name) stream format) ,(gsl-name \"fscanf\") ((stream :pointer) ((pointer object) :pointer) (format :string)) :type :method))))) ... ... @@ -213,13 +213,13 @@ ;;; (make-symbol-from-strings *gsl-prefix* 'vector-fixnum) ;;; GSL-VECTOR-FIXNUM (defun defun-gsl-all (types ctypes string general-class args) \"A defun-gsl for each of the declared data types.\" (defun defmfun-all (types ctypes string general-class args) \"A defmfun for each of the declared data types.\" `(progn ,@(loop for type in types for ctype in ctypes collect `(defun-gsl ,(first args) `(defmfun ,(first args) ;; Set the class name for the arglist ,(mapcar (lambda (x) ... ...\nThis diff is collapsed.\nThis diff is collapsed.\nThis diff is collapsed.\n ;; Eigenvectors and eigenvalues ;; Liam Healy, Sun May 21 2006 - 19:52 ;; Time-stamp: <2008-02-03 14:14:10EST eigensystems.lisp> ;; Time-stamp: <2008-02-17 11:32:46EST eigensystems.lisp> ;; \\$Id: \\$ (in-package :gsl) ... ... @@ -11,8 +11,8 @@ (defgo-s (eigen-symm n) eigen-symm-alloc eigen-symm-free) (defun-gsl eigen-symm-alloc (n) \"gsl_eigen_symm_alloc\" ((n :size)) (defmfun eigen-symm-alloc (n) \"gsl_eigen_symm_alloc\" ((n size)) :c-return :pointer :export nil :index '(letm eigen-symm) ... ... @@ -21,7 +21,7 @@ n-by-n real symmetric matrices. The size of the workspace is O(2n).\") (defun-gsl eigen-symm-free (w) (defmfun eigen-symm-free (w) \"gsl_eigen_symm_free\" ((w :pointer)) :c-return :void :export nil ... ... @@ -31,8 +31,8 @@ (defgo-s (eigen-symmv n) eigen-symmv-alloc eigen-symmv-free) (defun-gsl eigen-symmv-alloc (n) \"gsl_eigen_symmv_alloc\" ((n :size)) (defmfun eigen-symmv-alloc (n) \"gsl_eigen_symmv_alloc\" ((n size)) :c-return :pointer :export nil :index '(letm eigen-symmv) ... ... @@ -41,26 +41,26 @@ eigenvectors of n-by-n real symmetric matrices. The size of the workspace is O(4n).\") (defun-gsl eigen-symmv-free (w) (defmfun eigen-symmv-free (w) \"gsl_eigen_symmv_free\" ((w :pointer)) :index '(letm eigen-symmv) :c-return :void :documentation ; FDL \"Free the memory associated with the workspace w.\") (defun-gsl eigenvalues-symmetric (A eval ws) (defmfun eigenvalues-symmetric (A eval ws) \"gsl_eigen_symm\" ((A gsl-matrix-c) (eval gsl-vector-c) (ws :pointer)) :documentation ; FDL \"Eigenvalues of the real symmetric matrix A. Additional workspace of the appropriate size must be provided in w. The diagonal and lower triangular part of A are destroyed during the computation, but the strict upper triangular part is not referenced. The eigenvalues are stored in the vector @var{eval} is not referenced. The eigenvalues are stored in the vector eval and are unordered.\" :invalidate (A eval) :return (eval)) (defun-gsl eigenvalues-eigenvectors-symmetric (A eval evec ws) (defmfun eigenvalues-eigenvectors-symmetric (A eval evec ws) \"gsl_eigen_symmv\" (((pointer A) :pointer) ((pointer eval) :pointer) ((pointer evec) :pointer) (ws :pointer)) ... ... @@ -84,21 +84,21 @@ (defgo-s (eigen-herm n) eigen-herm-alloc eigen-herm-free) (defun-gsl eigen-herm-alloc (n) \"gsl_eigen_herm_alloc\" ((n :size)) (defmfun eigen-herm-alloc (n) \"gsl_eigen_herm_alloc\" ((n size)) :documentation ; FDL \"Allocate a workspace for computing eigenvalues of n-by-n complex hermitian matrices. The size of the workspace is O(3n).\" :c-return :pointer) (defun-gsl eigen-herm-free (w) (defmfun eigen-herm-free (w) \"gsl_eigen_herm_free\" ((w :pointer)) :c-return :void :documentation ; FDL \"Free the memory associated with the workspace w.\") (defun-gsl eigenvalues-hermitian (A eval w) (defmfun eigenvalues-hermitian (A eval w) \"gsl_eigen_herm\" (((pointer A) gsl-matrix-c) ((pointer eval) gsl-vector-c) (w :pointer)) :documentation ; FDL ... ... @@ -108,27 +108,27 @@ destroyed during the computation, but the strict upper triangular part is not referenced. The imaginary parts of the diagonal are assumed to be zero and are not referenced. The eigenvalues are stored in the vector @var{eval} and are unordered.\" eval and are unordered.\" :invalidate (eval A) :return (eval)) (defgo-s (eigen-hermv n) eigen-hermv-alloc eigen-hermv-free) (defun-gsl eigen-hermv-alloc (n) \"gsl_eigen_hermv_alloc\" ((n :size)) (defmfun eigen-hermv-alloc (n) \"gsl_eigen_hermv_alloc\" ((n size)) :documentation ; FDL \"Allocate a workspace for computing eigenvalues and eigenvectors of n-by-n complex hermitian matrices. The size of the workspace is O(5n).\" :c-return :pointer) (defun-gsl eigen-hermv-free (w) (defmfun eigen-hermv-free (w) \"gsl_eigen_hermv_free\" ((w :pointer)) :c-return :void :documentation ; FDL \"Free the memory associated with the workspace w.\") (defun-gsl eigenvalues-eigenvectors-hermitian (A eval evec w) (defmfun eigenvalues-eigenvectors-hermitian (A eval evec w) \"gsl_eigen_hermv\" (((pointer A) gsl-matrix-c) ((pointer eval) gsl-vector-c) ((pointer evec) gsl-matrix-c) (w :pointer)) ... ... @@ -164,14 +164,14 @@ the value of the parameter sort-type: :value-ascending, :value-descending, :absolute-ascending, :absolute-descending.\")) (defun-gsl sort-eigenvalues-eigenvectors (defmfun sort-eigenvalues-eigenvectors ((eval gsl-vector-double) evec sort-type) \"gsl_eigen_symmv_sort\" ((eval gsl-vector-c) (evec gsl-matrix-c) (sort-type eigen-sort-type)) :type :method :invalidate (eval evec)) (defun-gsl sort-eigenvalues-eigenvectors (defmfun sort-eigenvalues-eigenvectors ((eval gsl-vector-complex) evec sort-type) \"gsl_eigen_hermv_sort\" ((eval gsl-vector-c) (evec gsl-matrix-c) (sort-type eigen-sort-type)) ... ... @@ -194,9 +194,16 @@ (values (data evals) (data evecs)))) #| #(13.819660112501051d0 36.180339887498945d0 40.0d0) #2A((0.8506508083520399d0 -0.5257311121191336d0 0.0d0) (0.5257311121191336d0 0.8506508083520399d0 0.0d0) (0.0d0 0.0d0 1.0d0))" ]

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[ "# Calorimetry\n\nCalorimetry is the measurement of the quantity of heat exchanged. For example, if the energy from an exothermic chemical reaction is absorbed in a container of water, the change in temperature of the water provides a measure of the amount of heat added. Calorimeters are used to determine the energy content of foods by burning the foods in an oxygen atmosphere and measuring the energy yield in terms of the increase in temperature of the calorimeter. Calorimeters can also be used to measure the specific heat of a substance.", null, "The materials involved in the calorimetry are modeled here as a volume of water, a source of heat which is characterized by its equivalent water mass, and the container or calorimeter with its mass and specific heat. Thermal equilibrium is assumed after the experiment so that the temperature change can be used to calculate the energy released. Mass of water = gm Equivalent water mass of heat source = gm. Mass of calorimeter = gm. Specific heat of calorimeter = cal/gm °C = joule/gm °C.\n\nIf the calorimeter increases in temperature by ° C, then the amount of heat released is\n\n Q = [(1 cal/gm °C)(gm) + (cal/gm °C)(gm)](°C)\nQ = calories = joules\nIndex\n\nHeat transfer concepts\n\nHeat transfer examples\n\nThermal equilibrium\n\n HyperPhysics***** Thermodynamics R Nave\nGo Back" ]

[ null, "http://230nsc1.phy-astr.gsu.edu/hbase/thermo/imgheat/calor2.png", null ]

[ "# 3.3: Invention of the Schrödinger Equation\n\nFrom the previous section, the the classical wave equation in one-dimension was discussed:\n\n$\\frac {\\partial ^2 A (x, t)}{\\partial x^2} = \\nu ^{-2} \\frac {\\partial ^2 A (x, t)}{\\partial t^2} \\label {3-11}$\n\nAlthough we used a sine function to obtain the classical wave equation, functions other than the sine function can be substituted for $$A$$ in Equation \\ref{3-11}. Our goal as chemists is to seek a method for finding the wavefunctions that are appropriate for describing electrons, atoms, and molecules. In order to reach this objective, we need the appropriate wave equation.\n\nExercise $$\\PageIndex{1}$$\n\nShow that the functions $$e^{i(k x + ωt)}$$ and $$\\cos(k\\,x - \\omega\\, t)$$ also satisfy the classical wave equation (Equation \\ref{3-11}). Note that $$i$$ is a constant equal to $$\\sqrt {-1}$$.\n\nA general method for finding solutions to differential equations that depend on more than one variable ($$x$$ and $$t$$ in this case) is to separate the variables into different terms. This separation makes it possible to write the solution as a product of two functions, one that depends on $$x$$ and one that depends on $$t$$. This important technique is called the Method of Separation of Variables. This technique is used in most of the applications that we will be considering.\n\nFor the classical wave equation, Equation \\ref{3-11}, separating variables is very easy because $$x$$ and $$t$$ do not appear together in the same term in the differential equation. In fact, they are on opposite sides of the equation. The variables already have been separated, and we only have to see what happens when we substitute a product function into this equation. It is common in Quantum Mechanics to symbolize the functions that are solutions to Schrödinger’s equation as $$\\psi$$, $$\\psi$$, or $$\\phi$$, so we use $$\\Phi (x)$$ as the $$x$$-function, and examine the consequences of using $$\\cos (ωt)$$ as one possibility for the $$t$$-function.\n\n$\\psi(x,t) = ψ(x) cos(ωt) \\label{3-12}$\n\nAfter substituting Equation \\ref{3-12} into the classical wave Equation \\ref{3-11} and differentiating, we obtain\n\n$\\cos (\\omega t ) \\dfrac {\\partial ^2 \\psi (x)}{\\partial x^2} = -\\dfrac {\\omega ^2}{\\nu ^2} \\psi (x) \\cos (\\omega t) \\label{3-13}$\n\nwhich yields, after simplifying and rearranging,\n\n$\\dfrac {\\partial ^2 \\psi (x)}{\\partial x^2} + \\dfrac {\\omega ^2}{\\nu ^2} \\psi (x) = 0 \\label{3-14}$\n\nWe now include the idea that we are trying to find a wave equation for a particle. We introduce the particle momentum by using de Broglie's relation to replace $$\\dfrac {ω^2}{v^2}$$ with $$\\dfrac {p^2}{ħ^2}$$, where $$ħ = \\dfrac {h}{2π}$$ (called h-bar).\n\n$\\dfrac {\\partial ^2 \\psi (x)}{\\partial x^2} + \\dfrac {p^2}{\\hbar ^2} \\psi (x) = 0 \\label{3-15}$\n\nExercise $$\\PageIndex{2}$$\n\nShow that $$\\dfrac {ω^2}{v^2}$$ = $$\\dfrac {p^2}{ħ^2}$$.\n\nNext we will use the total energy of a particle as the sum of the kinetic energy and potential energy to replace the momentum in Equation.\n\n$E = T + V (x) = \\dfrac {p^2}{2m} + V (x) \\label{3-16}$\n\nNote that we have included the idea that the potential energy is a function of position. Each atomic or molecular system we will consider in the following chapters will have different potential energy functions.\n\nSolving Equation $$\\ref{3-16}$$ for $$p^2$$ and substituting it into Equation $$\\ref{3-15}$$ gives us the Schrödinger Equation,\n\n$\\dfrac {\\partial ^2 \\psi (x)}{\\partial x^2} + \\dfrac {2m}{\\hbar ^2} (E - V (x)) \\psi (x) = 0 \\label{3-17}$\n\nwhich usually is written in rearranged form,\n\n$\\dfrac {\\hbar ^2}{2m} \\dfrac {\\partial ^2 \\psi (x)}{\\partial x^2} + V (x) \\psi (x) = E \\psi (x) \\label{3-18}$\n\nNotice that the left side of Equation $$\\ref{3-18}$$ consists of the two terms corresponding to the kinetic energy and the potential energy. When we look at the left side of Equation $$\\ref{3-18}$$, we can deduce a method of extracting the total energy from a known wavefunction, or we can use Equation $$\\ref{3-18}$$) to find the wavefunction. Finding wavefunctions for models of interesting chemical phenomena will be one of the tasks we will accomplish in this text.\n\nExercise $$\\PageIndex{3}$$\n\nShow the steps that lead from Equations $$\\ref{3-11}$$ and \\ref{3-12} to Equation \\ref{3-18}.\n\nMore precisely, Equation \\ref{3-18} is the Schrödinger equation for a particle of mass $$m$$ moving in one dimension (x) in a potential field specified by $$V(x)$$. Since this equation does not contain time, it often is called the Time-Independent Schrödinger Equation. As mentioned previously, functions like ψ(x) are called wavefunctions because they are solutions to this wave equation. The term, wave, simply denotes oscillatory behavior or properties. The significance of the wavefunction will become clear as we proceed. For now, ψ(x) is the wavefunction that accounts for or describes the wave-like properties of particles.\n\nThe Schrödinger equation for a particle moving in three dimensions (x, y, z) is obtained simply by adding the other second derivative terms and by including the three-dimensional potential energy function. The wavefunction ψ then depends on the three variables x, y, and z.\n\n$\\dfrac {-\\hbar ^2}{2m} \\left ( \\dfrac {\\partial ^2}{\\partial x^2} + \\dfrac {\\partial ^2}{\\partial y^2} + \\dfrac {\\partial ^2}{\\partial z^2} \\right ) \\psi (x , y , z ) + V (x , y , z) \\psi (x , y , z ) = E \\psi (x , y , z ) \\label{3-19}$\n\nExercise $$\\PageIndex{4}$$\n\nWrite the Schrödinger equation for a particle of mass m moving in a 2-dimensional space with the potential energy given by\n\n$V(x, y) = -\\dfrac {(x^2 + y^2)}{2}.$\n\nThe three second derivatives in parentheses together are called the Laplacian operator, or del-squared,\n\n$\\nabla^2 = \\left ( \\dfrac {\\partial ^2}{\\partial x^2} + \\dfrac {\\partial ^2}{\\partial y^2} + \\dfrac {\\partial ^2}{\\partial z^2} \\right ) \\label {3-20}$\n\nwith the del operator,\n\n$\\nabla = \\left ( \\vec {x} \\dfrac {\\partial}{\\partial x} + \\vec {y} \\dfrac {\\partial}{\\partial y} + \\vec {z} \\dfrac {\\partial }{\\partial z} \\right ) \\label{3-21}$\n\nalso is used in Quantum Mechanics. Remember, symbols with arrows over them are unit vectors.\n\nExercise $$\\PageIndex{5}$$\n\nWrite the del-operator and the Laplacian operator for two dimensions and rewrite your answer to Exercise $$\\PageIndex{4}$$ in terms of the Laplacian operator." ]

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[ "1968-04-03\n\n# Airplane Vibrations Using Cantilevered Component Modes 680205\n\nThe necessity of airplane structural vibration analysis for purposes of design substantiation is indisputed. This paper presents a mathematical techique that accomplishes this end. The technique, once applied to a given airplane, produces an airplane analog which may be modified to include product improvements or to resolve certain areas of difficulty.\nThe mathematical technique is restricted to airplane self-equilibrating motions in n degrees of freedom using cantilevered component modes and a discrete-particle structural component distribution. Equations of motion are developed using energy principles. Particular importance is placed on the manner of generating these equations since this process determines the vibration analysis effectivity, that is, the inertia and stiffness expressions are specified in certain system generalized coordinates and the resulting eigensystem solutions are obtained using a digital computer. Application of the mathematical technique is illustrated using an example problem." ]

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[ "Solutions by everydaycalculation.com\n\n## Multiply 10/4 with 50/35\n\n1st number: 2 2/4, 2nd number: 1 15/35\n\nThis multiplication involving fractions can also be rephrased as \"What is 10/4 of 1 15/35?\"\n\n10/4 × 50/35 is 25/7.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 10/4 × 50/35 = 10 × 50/4 × 35 = 500/140\n3. After reducing the fraction, the answer is 25/7\n4. In mixed form: 34/7\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "X\nX\n\n# Calculate the Greatest Common Factor or GCF of 50590\n\nThe instructions to find the GCF of 50590 are the next:\n\n## 1. Decompose all numbers into prime factors\n\n 50590 2 25295 5 5059 5059 1\n\n## 2. Write all numbers as the product of its prime factors\n\n Prime factors of 50590 = 2 . 5 . 5059\n\n## 3. Choose the common prime factors with the lowest exponent\n\nCommon prime factors: 2 , 5 , 5059\n\nCommon prime factors with the lowest exponent: 21, 51, 50591\n\n## 4. Calculate the Greatest Common Factor or GCF\n\nRemember, to find the GCF of several numbers you must multiply the common prime factors with the lowest exponent.\n\nGCF = 21. 51. 50591 = 50590\n\nAlso calculates the:" ]

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[ "Explore BrainMass\n\n# Mathematics - Binomial and Poisson probabilities\n\nNot what you're looking for? Search our solutions OR ask your own Custom question.\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nObjective: Calculate binomial and Poisson probabilities.\n\n1) Chapter 5: Problem 5.5 (binomial)\nSolve the following problems by using the binomial formula.\na. If n = 4 and p = .10 , find P(x = 3) .\nb. If n = 7 and p = .80 , find P(x = 4) .\nc. If n = 10 and p = .60 , find P(x &#8805; 7) .\nd. If n = 12 and p = .45 , find P(5 &#8804; x &#8804; 7) .\n\n2) Chapter 5: Problem 5.6 (binomial) ...\n3) Chapter 5: Problem 5.15 (Poisson)a,b,c,d,e,f ...\n4) Chapter 5: Problem 5.16 (Poisson)a,b,c,d,e,f ...\n\n[See the attached Question File.]\n\nhttps://brainmass.com/math/discrete-math/mathematics-binomial-poisson-probabilities-192268\n\n#### Solution Preview\n\nThe solution file is attached.\n\n5.5 Solve the following problems by using the binomial formula.\na. If n = 4 and p = .10 , find P(x = 3) .\nb. If n = 7 and p = .80 , find P(x = 4) .\nc. If n = 10 and p = .60 , find P(x ≥ 7) .\nd. If n = 12 and p = .45 , find P(5 ≤ x ≤ 7) .\n\nP(n, r) = nCx p^x q^(n - x)\n(a) P(4, 3) = 4C3 * 0.1^3 * 0.9^1 = 0.0036\n(b) P(7, 4) = 7C4 * 0.8^4 * 0.2^3 = 0.115\n(c) P(x ≥ 7) = P(10, 7) + P(10, 8) + P(10, 9) + P(10, 10)\n= 10C7 * 0.6^7 * 0.4^3 + 10C8 * 0.6^8 * 0.4^2 + 10C9 * 0.6^9 * 0.4^1 + 10C10 * 0.6^10 * 0.4^0 = 0.382\n(d) P(5  x  7) = P(12, 5) + P(12, ...\n\n#### Solution Summary\n\nThe expert examines the binomial and Poisson probabilities objective. Neat, step-by-step soltuions are provided for all the questions.\n\n\\$2.49" ]

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[ "# Optimizing a Vulkan Program Part 2: GPU Implementation\n\nThis is the second in a series of blog posts about how I wrote and optimized VkColors, a small compute program written using Vulkan.\n\nThis post will examine the GPU implementations of VkColors. The algorithms are written as Vulkan compute shaders.\n\nAt the core of the algorithm, VkColors compares the scores of every location in the open set. For each iteration, the CPU implementation calculates scores on a single thread and only tracks the lowest score calculated so far. This uses O(1) memory space.\n\nTo run on the GPU, the algorithm must be parallelized. That means there is no “lowest score so far” since, in theory, every scoring calculation could run simultaneously. Instead, every score gets recorded into a separate entry in a large array. The memory usage must be O(n) where n is the size of the open set. Once the scores are recorded in the array, the lowest score can be selected.\n\nThe general architecture for the shaders is that each location is copied from the open set into a CPU array, and from the array into a Vulkan storage buffer to be accessed from the shader. The final result would be a struct of the form:\n\n```struct Score {\nuint32_t score;\nuint32_t index;\n}```\n\nThe `index` variable would be used to find the selected location in the CPU array. Then the color and the location would be used to send an update message to the render thread.\n\n## Premature Optimization\n\nThere are two distinct phases that can be parallelized, scoring and selecting. My first attempt to implement the algorithm used two compute shaders for these two phases. The first shader calculated the scores and wrote them into an array and the second shader used a parallel reduce min algorithm to select the lowest score.\n\nThe parallel reduce algorithm was implemented by creating a “pyramid” of arrays. The base of the pyramid was the array that the first phase wrote to. The other levels each had half the number of elements of the level below them. At the top, there would be room for a single element. Each invocation of the parallel reduce shader read two adjacent elements from level `i` and stored the lower score element in level `i + 1`. The lowest score would be stored at the top of the pyramid, and would be sent back to the CPU to finish the iteration.\n\nIn theory, this would mean that both phases were parallelized and could be run on the GPU, all phase 1 invocations in parallel, then phase 2 invocations in parallel (each level separately). In practice, this resulted in a mere ~2,000 pps on the wave algorithm at the start, compared to ~9,000 pps from the CPU implementation. This was a pretty serious performance regression, so I considered this implementation to be a failure.\n\nIf you want to play with this version, it’s tagged as `v0.1-pyramid` in the git repo. However, this version is missing a lot of functionality that was added in later versions, like the command line options and even the coral shader.\n\nPerhaps my implementation of the parallel reduce phase was bad. However the complexity of implementing it and the resultant performance led to me removing it from VkColors. Instead, I opted for a much simpler implementation.\n\n## A Much Simpler Implementation\n\nPhase 1 would continue running on the GPU and the result array would be copied to CPU memory. Then phase 2 would run on the CPU. There is no pyramid now, just a simple loop to find the lowest score.\n\nWhile the pyramid was removed, most of the infrastructure from the earlier attempt was kept. The compute shader to calculate the scores was kept and it’s interface was only slightly changed.\n\nThe work is submitted to the GPU on the generator thread. Unlike OpenGL, Vulkan allows API calls to be done on multiple threads as long as the user properly synchronizes access to certain objects. In this case, the `VkQueue` is protected with a mutex if the same queue is used for render and compute. On certain hardware (such as my RX 580), the user can access a dedicated, async compute queue, in which case no synchronization is needed at all.\n\nThe work is submitted to the GPU using double buffering, so phase 2 for frame `i` will run on the CPU while phase 1 for frame `i + 1` is running on the GPU.\n\nThis simplification showed much better results than the pyramid. On the wave algorithm, it achieved an overall speed of ~18,000 pps, a 2X speedup. On the coral algorithm, it reached ~550 pps, a 3X speedup.\n\nThis version is tagged as `v0.2-simple` in the repo. Already, a simple GPU implementation has showed massive performance improvements. The next post will explore even larger performance gains." ]

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[ "## Detecting Unfair Dice in Casinos with Bayes’ Theorem\n\n#### Introduction\n\nI saw an interesting problem that requires Bayes’ Theorem and some simple R programming while reading a bioinformatics textbook.  I will discuss the math behind solving this problem in detail, and I will illustrate some very useful plotting functions to generate a plot from R that visualizes the solution effectively.\n\n#### The Problem\n\nThe following question is a slightly modified version of Exercise #1.2 on Page 8 in “Biological Sequence Analysis” by Durbin, Eddy, Krogh and Mitchison.\n\nAn occasionally dishonest casino uses 2 types of dice.  Of its dice, 97% are fair but 3% are unfair, and a “five” comes up 35% of the time for these unfair dice.  If you pick a die randomly and roll it, how many “fives”  in a row would you need to see before it was most likely that you had picked an unfair die?”\n\nRead more to learn how to create the following plot and how it invokes Bayes’ Theorem to solve the above problem!", null, "" ]

[ null, "https://chemicalstatistician.files.wordpress.com/2013/10/unfair-die-plot2.png", null ]

[ "# What is a predicted score?\n\nContents\n\n## What does prediction mean in statistics?\n\nIn general, prediction is the process of determining the magnitude of statistical variates at some future point of time.\n\n## What is prediction formula?\n\nA prediction equation predicts a value of the reponse variable for given values of the factors. The equation we select can include all the factors shown above, or it can include a subset of the factors.\n\n## What does Y hat mean?\n\nY hat (written ŷ ) is the predicted value of y (the dependent variable) in a regression equation. It can also be considered to be the average value of the response variable. … The equation is calculated during regression analysis. A simple linear regression equation can be written as: ŷ = b + b1x.\n\n## Is regression a prediction?\n\nIn most cases, the investigators utilize regression analysis to develop their prediction models. Regression analysis is a statistical technique for determining the relationship between a single dependent (criterion) variable and one or more independent (predictor) variables.\n\n## How do you predict a regression equation?\n\nLinear regression is one of the most commonly used predictive modelling techniques.It is represented by an equation = + + , where a is the intercept, b is the slope of the line and e is the error term. This equation can be used to predict the value of a target variable based on given predictor variable(s).\n\nIT IS IMPORTANT:  Will my computer run Divinity Original Sin 2?\n\n## How do you tell if a regression model is a good fit?\n\nStatisticians say that a regression model fits the data well if the differences between the observations and the predicted values are small and unbiased. Unbiased in this context means that the fitted values are not systematically too high or too low anywhere in the observation space.\n\n## What is the most important measure to use to assess a model’s predictive accuracy?\n\nPearson product-moment correlation coefficient (r) and the coefficient of determination (r2) are among the most widely used measures for assessing predictive models for numerical data, although they are argued to be biased, insufficient and misleading." ]

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[ "# Axial Stiffness\n\nStiffness is used to correlate the load to the amount the object will deflect do to that load. To find stiffness the equation below would be used.\n\n(Eq 1)  $k=\\frac{P}{δ}=\\frac{AE}{L}$\n\nP = Force\n\nδ = Deflection\n\nA = Cross-sectional Area\n\nE = Young’s Modulus\n\nk = Stiffness\n\nSo when would you be interested in the stiffness? Well for one thing if you are interested in the natural frequency of the part you will need to determine the stiffness of the part. Another case though would be if you are dealing with parts that are made out of different material.\n\nNow to calculate the overall stiffness of a part made out of two different materials you would treat the problem the same way you would to calculate the overall resistance of an electrical circuit.\n\nTo start, what if you had to different materials that are connected in parallel? To find the stiffness of the overall part refer to the image below.", null, "Now, what if you had two parts of dissimilar material connected in series? To find the overall all stiffness refer to the image below.", null, "By determining the overall stiffness of the parts you can now either determine the total force required to deflect that object by a given distance or how much and object will deflect due to a certain force." ]

[ null, "https://cdn-0.sbainvent.com/wp-content/uploads/2018/06/spring_force_axial.jpg", null, "https://cdn-0.sbainvent.com/wp-content/uploads/2018/06/spring_deflection_axial.jpg", null ]

[ "# What is a normal standard variable\n\nSan-José State University\n\napplet-magic.com\nThayer Watkins\nSilicon Valley\nUnited States\n\n Ito's lemma and its derivative\n\nChanges in a variable such as stock price involve a deterministic component that is a function of time and the stochastic component that depends on a random variable. Let S be the stock price at time t and let dS be the tiny change in S over the tiny distance from time corotron. The change in the random variable z over this interval of time is dz. The change in the stock price is given over\n\n(1)\n\n#### dS = adt + bdz,\n\nwhere a and b can be functions of S and t as well as other variables; i.e.\ndS = a (S, t, x) dt + b (S, t, x) dz.\n\nThe expected value of dz is zero, so the expected value of dS is equal to the deterministic component, adt.\n\nThe random variable dz represents an accumulation of the occasional influences over the distance ds. The central c limit theorem then indicates that dz has a normal distribution and is therefore fully characterized by its mean and standard deviation. The mean or expected value of dz is zero. The deviation of a random variable, which is the accumulation of the independent effects over an interval of time, is proportional to the length of the interval, in this case the corotron. The standard deviation of dz is therefore proportional to the square root of the corotron (dt)½. All of this means that the random variable dz is equivalent to a random variable w (dt)½where w is a normal standard variable with the mean zero and the standard deviation equal to the unit.\n\nNow consider another variable C, like the price of an option to buy, which is a function of S and t, say C = f (S, t). Because C is a function of the stochastic variable S, C has a stochastic component as well as a deterministic component. C has a representation of the form:\n\n(2)\n\n#### dC = pdt + qdz.\n\nwhere p and q can be functions of S, of t, and perhaps of other variables; i.e. p = p (S, t, x) and q = q (S, t, x).\n\nThe crucial problem is how functions p and q relate to functions a and b in the equation\n\n(3)\n\n#### dS = adt + bdz.\n\nIto's lemma gives the answer. The deterministic and stochastic components of dC are given by:\n\n(4)\n\n#### p = ∂f / ∂t + (∂f / ∂S) a + ½ (∂²f / ∂S²) b² q = (∂f / ∂S) b.\n\nIto's lemma is crucial in deriving differential equations for the value of inferred stocks such as stock options.\n\n### Derivation\n\nThe Taylor series for f (S, t) gives the increment in C as:\n\n(5)\n\n#### dC = (∂f / ∂t) dt + (∂f / ∂S) dS + ½ (∂²f / ∂S²) (dS) ² + (∂²f / ∂S∂t) (dS) (dt) + ½ (∂²f / ∂t²) (dt) ² + higher order terms.\n\nThe increment in the share price dS is given over\n\n#### dS = adt + bdz but dz = vw [dt]½,\n\nwhere w is a normal standard random variable. Replacement of adt + bvw (paper corotron) ½ for dS in the above equation (5) yields:\n(6)\n\n#### dC = (∂f / ∂t) dt + (∂f / ∂S) adt + ∂f / ∂S) bvw (dt)½ + ½ (∂²f / ∂S²) (adt + bvw (dt)½)² + (∂²f / ∂S∂t) (adt + bvw (dt)½) (dt) + ½ (∂²f / ∂t²) (dt) ² + higher order terms.\n\nWith the expansion of the quadratic term and the product term, the result is:\n(7)\n\n#### dC = (∂f / ∂t) dt + (∂f / ∂S) adt + ∂f / ∂S) bvw (dt)½ + ½ (∂²f / ∂S²) (a²dt² + 2abvw (dt)3/2 + b²v²w²dt) + (∂²f / ∂S∂t) (a (dt) ² + bvw (dt)3/2) + ½ (∂²f / ∂t²) (dt) ² + higher order terms.\n\nAllowing for the tiny type of DT for DT to disappear to any power higher than unity, (7) reduces to:\n(8)\n\n#### dC = (∂f / ∂t) dt + (∂f / ∂S) adt + (∂f / ∂S) bvw (dt)½ + ½ (∂²f / ∂S²) (b²v²w²dt)\n\nNoting that the expected value of w² is unit, the expected value of dC is:\n(9)\n\n#### [∂f / ∂t + (∂f / ∂S) a + ½ (∂²f / ∂S²) b²] dt.\n\nThis is the deterministic part of dC. The stochastic component is the expression that depends on dz, which is expressed in (8) as vw (dt)½ is pictured. Hence the stochastic component is:\n(10)\n\n#### [(∂f / ∂S) b] dz.\n\nFrom the above derivation, it would appear that there is an additional stochastic expression resulting from the occasional deviations of w² from its expected value of 1; i.e. the additional term\n(11)\n\n#### ½ (∂²f / ∂S²) (b²v²w²dt).\n\nHowever, the deviation of this additional term is proportional to (corotron) ², while the deviation of the stochastic term given in (10) is too proportional to (corotron). Thus, the stochastic expression given in (11) disappears compared with the stochastic expression given in (10).\n\nIto's lemma is essential in the derivation of the Black and Scholes equations.\n\nAn immediate question is whether an extension of Ito's lemma for constant distributions of z is different from the normal distribution. This question is investigated in a page on perennial distributions." ]

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[ "# Java.lang.StrictMath.signum() Method\n\n## Description\n\nThe java.lang.StrictMath.signum(float f) method returns the signum function of the argument i.e zero if the argument is zero, 1.0f if the argument is greater than zero, -1.0f if the argument is less than zero.It include these cases −\n\n• If the first argument is NaN, NaN is returned. .\n• If the argument is positive zero or negative zero, then the result is the same as the argument.\n\n## Declaration\n\nFollowing is the declaration for java.lang.StrictMath.signum() method\n\n```public static float signum(float f)\n```\n\n## Parameters\n\nf − This is the floating-point value whose signum is to be returned.\n\n## Return Value\n\nThis method returns signum function of the argument.\n\nNA\n\n## Example\n\nThe following example shows the usage of java.lang.StrictMath.signum() method.\n\n```package com.tutorialspoint;\n\nimport java.lang.*;\n\npublic class StrictMathDemo {\n\npublic static void main(String[] args) {\n\nfloat f1 = 102.20f, f2 = 0.0f, f3 = -0.0f;\n\n// returns 1.0 if the argument is greater than zero\nfloat retval = StrictMath.signum(f1);\nSystem.out.println(\"Value = \" + retval);\n\n/* if the argument is positive zero, then the result is the\nsame as the argument */\nretval = StrictMath.signum(f2);\nSystem.out.println(\"Value = \" + retval);\n\n/* if the argument is negative zero, then the result is the\nsame as the argument */\nretval = StrictMath.signum(f3);\nSystem.out.println(\"Value = \" + retval);\n}\n}\n```\n\nLet us compile and run the above program, this will produce the following result −\n\n```Value = 1.0\nValue = 0.0\nValue = -0.0\n```\njava_lang_strictmath.htm" ]

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[ "# 中性流\n\n1973年，阿卜杜勒·薩拉姆謝爾登·格拉肖以及史蒂文·溫伯格預測中性流存在，隨後加爾加梅勒氣泡室實驗觀察到中性流作用。\n\n## 定義\n\ne\ne\nν\ne\ne\n\n${\\mathfrak {M}}^{\\mathrm {NC} }\\propto J_{\\mu }^{\\mathrm {(NC)} }(\\nu _{\\mathrm {e} })\\;J^{\\mathrm {(NC)} \\mu }(\\mathrm {e^{-}} )$", null, "$J^{\\mathrm {(NC)} \\mu }(f)={\\bar {u}}_{f}\\gamma ^{\\mu }{\\frac {1}{2}}\\left(g_{V}^{f}-g_{A}^{f}\\gamma ^{5}\\right)u_{f},$", null, "，其中，$g_{V}^{f}$", null, "$g_{A}^{f}$", null, "費米子$f$", null, "向量軸向量的耦合。\n\n## 參考資料\n\n1. ^ 天文學名詞. 中国天文学名词审定委员会. [2016-03-03] （中文（中国大陆）‎）.\n2. ^ S. Bais. The Equations: Icons of knowledge. 2005: 84. ISBN 0-674-01967-9.\n3. ^ The Nobel Prize in Physics 1979. Nobel Foundation. [2008-09-10]. （原始内容存档于2004-08-03）." ]

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[ "# Coordinate and tensor divergence theorems\n\nThe expression previously obtained for the divergence theorem was\n\n\\displaystyle \\begin{aligned}\\int_{V}\\mathrm{div}(u)\\mathrm{d}V & =\\int_{\\partial V}i_{u}\\mathrm{d}V\\\\ & =\\int_{\\partial V}\\left\\langle u,\\hat{n}\\right\\rangle \\mathrm{d}S, \\end{aligned}\n\nwhere $${V}$$ is an $${n}$$-dimensional compact submanifold of $${M^{n}}$$, $${\\hat{n}}$$ is the unit normal vector to $${\\partial V}$$, and $${\\mathrm{d}S\\equiv i_{\\hat{n}}\\mathrm{d}V}$$ is the induced volume element (“surface element”) for $${\\partial V}$$. If we choose an orthonormal frame with $${e_{1}=\\hat{n}}$$ on $${\\partial V}$$, the divergence theorem can be written\n\n\\begin{aligned}\\int_{V}\\mathrm{div}(u)\\mathrm{d}V & =\\int_{\\partial V}u^{1}\\mathrm{d}S,\\end{aligned}\n\nand if we can choose coordinates with $${x^{1}}$$ constant on $${\\partial V}$$ and normal to it, the divergence theorem can be written\n\n\\begin{aligned}\\int_{V}\\partial_{\\lambda}\\left(\\sqrt{g}u^{\\lambda}\\right)\\mathrm{d}^{n}x & =\\int_{\\partial V}\\sqrt{g}\\mathrm{d}x^{1}\\left(u\\right)\\mathrm{d}^{n-1}x\\\\\n& =\\int_{\\partial V}u^{1}\\sqrt{g}\\mathrm{d}^{n-1}x,\n\\end{aligned}\n\nwhere $${\\mathrm{d}^{n}x\\equiv\\mathrm{d}x^{1}\\wedge\\cdots\\wedge\\mathrm{d}x^{n}}$$ and $${\\mathrm{d}^{n-1}x\\equiv\\mathrm{d}x^{2}\\wedge\\cdots\\wedge\\mathrm{d}x^{n}}$$.\n\nSince the “divergence” of a tensor $${T}$$ with order greater than 1 is tensor-valued, and the parallel transport of tensors is path-dependent, we cannot in general integrate to get a divergence theorem for tensors. In the case of a flat metric and zero torsion however, we can choose coordinates whose coordinate frame is orthonormal, so that the frame is its own parallel transport, i.e. $${\\nabla_{v}\\left(\\beta^{\\mu}\\right)=0}$$. For e.g. a tensor $${T^{ab}}$$, we can then define a coordinate-dependent vector $${J^{\\mu}}$$ for each index $${\\mu}$$\n\n\\Rightarrow\\left(J^{\\mu}\\right)^{b} & =T^{\\mu b}\\\\\n\\Rightarrow\\nabla_{v}J^{\\mu} & \\overset{\\cancel{RT}}{=}\\beta^{\\mu}\\nabla_{v}T\\\\\n\\Rightarrow\\nabla_{b}\\left(J^{\\mu}\\right)^{b} & \\overset{\\cancel{RT}}{=}\\nabla_{b}T^{\\mu b}\\\\\n\\Rightarrow\\int_{V}\\nabla_{b}T^{\\mu b}\\mathrm{d}V & \\overset{\\cancel{RT}}{=}\\int_{V}\\nabla_{b}\\left(J^{\\mu}\\right)^{b}\\mathrm{d}V\\\\\n& =\\int_{\\partial V}T^{\\mu}{}_{b}\\hat{n}^{b}\\mathrm{d}S.\n\\end{aligned}\n\nFor arbitrary coordinates, the components of the coordinate frame are by definition constant, i.e. $${\\partial_{v}\\left(\\mathrm{d}x^{\\mu}\\right)=0}$$; we can therefore write\n\nThis relation remains true in the presence of both curvature and torsion, however it is important to note that $${\\partial_{\\nu}\\left(\\sqrt{g}T^{\\mu\\nu}\\right)}$$ is not a “divergence” and $${T^{\\mu b}=\\left(J^{\\mu}\\right)^{b}}$$ is coordinate-dependent. In the special case of an anti-symmetric tensor under zero torsion, we can write" ]

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[ "astro-ph.HE\n\n# Title: Thawing the frozen-in approximation: implications for self-gravity in deeply plunging tidal disruption events\n\nAbstract: The tidal destruction of a star by a massive black hole, known as a tidal disruption event (TDE), is commonly modeled using the \"frozen-in\" approximation. Under this approximation, the star maintains exact hydrostatic balance prior to entering the tidal sphere (radius $r_{\\rm t}$), after which point its internal pressure and self-gravity become instantaneously negligible and the debris undergoes ballistic free fall. We present a suite of hydrodynamical simulations of TDEs with high penetration factors $\\beta \\equiv r_{\\rm t}/r_{\\rm p} = 5-7$, where $r_{\\rm p}$ is the pericenter of the stellar center of mass, calculated using a Voronoi-based moving-mesh technique. We show that basic assumptions of the frozen-in model, such as the neglect of self-gravity inside $r_{\\rm t}$, are violated. Indeed, roughly equal fractions of the final energy spread accumulate exiting and entering the tidal sphere, though the frozen-in prediction is correct at the order-of-magnitude level. We also show that an $\\mathcal{O}(1)$ fraction of the debris mass remains transversely confined by self-gravity even for large $\\beta$ which has implications for the radio emission from the unbound debris and, potentially, for the circularization efficiency of the bound streams.\n Comments: comments welcome Subjects: High Energy Astrophysical Phenomena (astro-ph.HE) DOI: 10.1093/mnrasl/slz048 Cite as: arXiv:1903.03898 [astro-ph.HE] (or arXiv:1903.03898v3 [astro-ph.HE] for this version)" ]

[ null ]

[ "", null, "# Extract Rows and Place Into New Columns.\n\nI am trying to extract rows and place these groups of rows into columns. The rows I need to extract are of uneven length. Not sure how to specify the conditions and apply the loop (?).\n\nHere is the dataset I'm working with:\n\n``````Data1 <- data.frame(\"Example\" = c(\"Word A\", \"URL\", \"URL\", \"URL\",\n\"Word B\", \"URL\",\n\"Word C\", \"URL\", \"URL\"))\n``````\n\nI want it to look like this:\n\n``````Data2 <- data.frame(\"Example 1\" = c(\"Word A\", \"URL\", \"URL\", \"URL\"),\n\"Example 2\" = c(\"Word B\", \"URL\", NA, NA),\n\"Example 3\" = c(\"Word C\", \"URL\", \"URL\", NA))\n``````\n\nThanks anyone who can help!\n\nHere is one way.\n\n``````Data1 <- data.frame(\"Example\" = c(\n\"Word A\", \"URL\", \"URL\", \"URL\",\n\"Word B\", \"URL\",\n\"Word C\", \"URL\", \"URL\"\n))\n\nData2 <- data.frame(\n\"Example 1\" = c(\"Word A\", \"URL\", \"URL\", \"URL\"),\n\"Example 2\" = c(\"Word B\", \"URL\", NA, NA),\n\"Example 3\" = c(\"Word C\", \"URL\", \"URL\", NA)\n)\n\nlibrary(dplyr)\nlibrary(purrr)\nData1\\$wcount <- cumsum(startsWith(Data1\\$Example, \"W\"))\n\n(components <- group_by(\nData1,\nwcount\n) %>% group_map(.f = function(x, ...) as.matrix(x)))\n\n(components <- imap(\ncomponents,\n~ {\ncolnames(.x) <- paste0(\"Example \", .y)\n.x\n}\n))\n\n(height <- max(map_int(components, nrow)))\n\nif (nrow(.x) < height) {" ]

[ null, "https://community.rstudio.com/uploads/default/original/3X/5/d/5dc960154a129282ba4283771da2fab6fde146fb.png", null ]

[ "How do you solve the following system: -x+6y=12, x-2y=4 ?\n\nJan 11, 2016\n\nx = 12 and y = 4\n\nExplanation:\n\nyou can solve this type of system of equations by elimination method or by substitution.\nIn this case the substitution method appears to be appropriate as both equations may be written in terms of x explicitly.\n\nstart by labelling the equations. This makes it 'easier' to follow when performing operations on them\n\n-x + 6y = 12 .........(1)\n\nx - 2y = 4 .............(2)\n\nfrom (1) : x = 6y - 12\n\nand from (2) : x = 4 + 2y\n\nwe can now equate the 2 equations since both are now given as equations in x.\n\nhence 6y - 12 = 4 + 2y hence 4y = 16 and y = 4\n\nsubstitute y = 4 in x = 6y - 12 to get 24 - 12 = 12.\n\nHere is a graphical illustration of the solution:\ngraph{(y-1/6 x -2)(y-1/2 x + 2) =0 [-40, 40, -20, 20]}" ]

[ null ]

[ "# [FOM] 180:Provable Functions of PA\n\nHarvey Friedman friedman at math.ohio-state.edu\nWed Jun 25 05:25:01 EDT 2003\n\n```Reply to Tait 10:40AM 6/22/03.\n\nI have been asked to explain why something is true.\n\n>On Saturday, June 21, 2003, at 10:02 AM, William Tait wrote:\n>\n>>On Sunday, June 15, 2003, at 11:42 PM, Harvey Friedman wrote:\n>>\n>>>THEOREM 2. The PA provable Delta_0 functions are exactly the\n>>>Delta_0 functions that are bounded by a <epsilon_0 recursive\n>>>function.\n>>\n>>I can't prove this. Certainly a Delta_0 function bounded by a\n>>PA-provably recursive function is provably recursive; but I don't\n>>see that it need be Delta_0.\n>\n>Just in case I'm being suspected of melt-down: of course, I meant\n>that I don't see that it need be provably Delta_0. Bill\n>\n\nLet f be a Delta_0 function bounded by the <epsilon_0 recursive\nfunction g. Let the e-th Turing machine compute g.\n\nLet A(x,y) be Delta_0 defining f. PA may not prove\n\n(forall x)(therexists unique y)(A(x,y).\n\nLet A'(x,y) be\n\nA(x,y) or (y is the number of steps that it takes for the e-th Turing\nmachine to halt, and for all z < y, we have notA(x,z)).\n\nThen A'(x,y) is Delta_0, and PA proves\n\n(forall x)(therexists unique y)(A'(x,y))\n\nand A'(x,y) also defines f. So f is PA provable Delta_0.\n\nHarvey Friedman\n\n```" ]

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[ "Definition of\n\n# Greatest Common Factor", null, "The greatest number that is a factor of two (or more) other numbers.\n\nWhen we find all the factors of two or more numbers, and some factors are the same (\"common\"), then the largest of those common factors is the Greatest Common Factor.\n\nAbbreviated \"GCF\". Also called \"Highest Common Factor\"\n\nExample: the GCF of 12 and 16 is 4, because 1, 2 and 4 are common factors of both 12 and 16, and 4 is the greatest of them.\nSee: Factor" ]

[ null, "https://www.mathsisfun.com/numbers/images/greatest-common-factor.svg", null ]

[ "Graphs and minimum spanning trees?\n\nI am having a hard time finding information on how Graphs and spanning tree's work and how to construct/structure them. The reason is that I'm using a Delaunay Triangulation algorithm within the LibGDX framwork and this got me an array of indices. I can draw my triangles and points but how to properly setup a graph/MST structure is a problem.\n\nI supply my vertices's as an array of floats as well:\n\nfloat[] points = new float {80, 30, 40, 45, 0, 10};\n\nEach pair refers to a point of the map. Using DelaunayTriangulator.computeTriangles I get an array of indices, in this case\n\nShortArray indices = triangulator.computeTriangles(points, false);\nSystem.out.println(indices);\n//Output [1, 0, 2, 1, 2, 3]\n\nNow I can draw all the edges and I'm actually doing this just using this data. But I suppose I create a node and graph class to help me out here, I just don't have a clue how this should look. For example, should I have multiples of the same points with different parents, or should I have a single node for a point with a list of points it is connected to? I'm very new to graphs and MST and there is plenty of information on the subject but I cannot find good practical examples.\n\nYour idea for \"single node for a point with a list of points it is connected to\" should work, as it describes an adjacency list. Two canonical ways to represent graphs are via adjacency lists and adjacency matrices, for which the Wikipedia pages have some simple examples:\n\nIn the case of an adjacency list, it often makes sense to use a dictionary/mapping data structure, e.g. a HashMap in Java.\n\nJust one example to note two additional caveats... let's say the below represents a graph with four nodes and three edges:\n\n(1) -- (2) -- (3)\n|\n(4)\n\nAn adjacency list could be represented like so: {1: , 2: [1, 3, 4], 3: }\n\nNote that:\n\n• The above graph is unweighted -- if it were weighted, the the edge weights would need to be represented, e.g. as a possible implementation that follows the above adjacency list {1: [(2, 3)], 2: [(1, 5), (3, 2)...] ...} where 1: (2, 3) means the edge between node 1 and node 2 has weight of 3.\n• The above graph is undirected -- if it were directed, the adjacency list would potentially contain fewer edges.\n\nMSTs are edge subsets of graphs, so usually no additional structure is required to represent them." ]

[ null ]

[ "Is 26834 a prime number? What are the divisors of 26834?\n\n## Parity of 26 834\n\n26 834 is an even number, because it is evenly divisible by 2: 26 834 / 2 = 13 417.\n\nFind out more:\n\n## Is 26 834 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 26 834 is about 163.811.\n\nThus, the square root of 26 834 is not an integer, and therefore 26 834 is not a square number.\n\n## What is the square number of 26 834?\n\nThe square of a number (here 26 834) is the result of the product of this number (26 834) by itself (i.e., 26 834 × 26 834); the square of 26 834 is sometimes called \"raising 26 834 to the power 2\", or \"26 834 squared\".\n\nThe square of 26 834 is 720 063 556 because 26 834 × 26 834 = 26 8342 = 720 063 556.\n\nAs a consequence, 26 834 is the square root of 720 063 556.\n\n## Number of digits of 26 834\n\n26 834 is a number with 5 digits.\n\n## What are the multiples of 26 834?\n\nThe multiples of 26 834 are all integers evenly divisible by 26 834, that is all numbers such that the remainder of the division by 26 834 is zero. There are infinitely many multiples of 26 834. The smallest multiples of 26 834 are:\n\n## Numbers near 26 834\n\n### Nearest numbers from 26 834\n\nFind out whether some integer is a prime number" ]

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[ "User Real IP - 18.232.125.29\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => 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Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => Array\n(\n => 115.42.70.24\n)\n\n => Array\n(\n => 45.128.188.43\n)\n\n => Array\n(\n => 103.140.129.63\n)\n\n => Array\n(\n => 101.50.113.147\n)\n\n => Array\n(\n => 103.66.73.30\n)\n\n => Array\n(\n => 117.247.193.169\n)\n\n => Array\n(\n => 120.29.100.94\n)\n\n => Array\n(\n => 42.109.154.39\n)\n\n => Array\n(\n => 122.173.155.150\n)\n\n => Array\n(\n => 45.115.104.53\n)\n\n => Array\n(\n => 116.74.29.84\n)\n\n => Array\n(\n => 101.50.125.34\n)\n\n => Array\n(\n => 45.118.166.80\n)\n\n => Array\n(\n => 91.236.184.27\n)\n\n => Array\n(\n => 113.167.185.120\n)\n\n => Array\n(\n => 27.97.66.222\n)\n\n => Array\n(\n => 43.247.41.117\n)\n\n => Array\n(\n => 23.229.16.227\n)\n\n => Array\n(\n => 14.248.79.209\n)\n\n => Array\n(\n => 117.5.194.26\n)\n\n => Array\n(\n => 117.217.205.41\n)\n\n => Array\n(\n => 114.79.169.99\n)\n\n => Array\n(\n => 103.55.60.97\n)\n\n => Array\n(\n => 182.75.89.210\n)\n\n => Array\n(\n => 77.73.66.109\n)\n\n => Array\n(\n => 182.77.126.139\n)\n\n => Array\n(\n => 14.248.77.166\n)\n\n => Array\n(\n => 157.35.224.133\n)\n\n => Array\n(\n => 183.83.38.27\n)\n\n => Array\n(\n => 182.68.4.77\n)\n\n => Array\n(\n => 122.177.130.234\n)\n\n => Array\n(\n => 103.24.99.99\n)\n\n => Array\n(\n => 103.91.127.66\n)\n\n => Array\n(\n => 41.90.34.240\n)\n\n => Array\n(\n => 49.205.77.102\n)\n\n => Array\n(\n => 103.248.94.142\n)\n\n => Array\n(\n => 104.143.92.170\n)\n\n => Array\n(\n => 219.91.157.114\n)\n\n => Array\n(\n => 223.190.88.22\n)\n\n => Array\n(\n => 223.190.86.232\n)\n\n => Array\n(\n => 39.41.172.80\n)\n\n => Array\n(\n => 124.107.206.5\n)\n\n => Array\n(\n => 139.167.180.224\n)\n\n => Array\n(\n => 93.76.64.248\n)\n\n => Array\n(\n => 65.216.227.119\n)\n\n => Array\n(\n => 223.190.119.141\n)\n\n => Array\n(\n => 110.93.237.179\n)\n\n => Array\n(\n => 41.90.7.85\n)\n\n => Array\n(\n => 103.100.6.26\n)\n\n => Array\n(\n => 104.140.83.13\n)\n\n => Array\n(\n => 223.190.119.133\n)\n\n => Array\n(\n => 119.152.150.87\n)\n\n => Array\n(\n => 103.125.130.147\n)\n\n => Array\n(\n => 27.6.5.52\n)\n\n => Array\n(\n => 103.98.188.26\n)\n\n => Array\n(\n => 39.35.121.81\n)\n\n => Array\n(\n => 74.119.146.182\n)\n\n => Array\n(\n => 5.181.233.162\n)\n\n => Array\n(\n => 157.39.18.60\n)\n\n => Array\n(\n => 1.187.252.25\n)\n\n => Array\n(\n => 39.42.145.59\n)\n\n => Array\n(\n => 39.35.39.198\n)\n\n => Array\n(\n => 49.36.128.214\n)\n\n => Array\n(\n => 182.190.20.56\n)\n\n => Array\n(\n => 122.180.249.189\n)\n\n => Array\n(\n => 117.217.203.107\n)\n\n => Array\n(\n => 103.70.82.241\n)\n\n => Array\n(\n => 45.118.166.68\n)\n\n => Array\n(\n => 122.180.168.39\n)\n\n => Array\n(\n => 149.28.67.254\n)\n\n => Array\n(\n => 223.233.73.8\n)\n\n => Array\n(\n => 122.167.140.0\n)\n\n => Array\n(\n => 95.158.51.55\n)\n\n => Array\n(\n => 27.96.95.134\n)\n\n => Array\n(\n => 49.206.214.53\n)\n\n => Array\n(\n => 212.103.49.92\n)\n\n => Array\n(\n => 122.177.115.101\n)\n\n => Array\n(\n => 171.50.187.124\n)\n\n => Array\n(\n => 122.164.55.107\n)\n\n => Array\n(\n => 98.114.217.204\n)\n\n => Array\n(\n => 106.215.10.54\n)\n\n => Array\n(\n => 115.42.68.28\n)\n\n => Array\n(\n => 104.194.220.87\n)\n\n => Array\n(\n => 103.137.84.170\n)\n\n => Array\n(\n => 61.16.142.110\n)\n\n => Array\n(\n => 212.103.49.85\n)\n\n => Array\n(\n => 39.53.248.162\n)\n\n => Array\n(\n => 203.122.40.214\n)\n\n => Array\n(\n => 117.217.198.72\n)\n\n => Array\n(\n => 115.186.191.203\n)\n\n => Array\n(\n => 120.29.100.199\n)\n\n => Array\n(\n => 45.151.237.24\n)\n\n => Array\n(\n => 223.190.125.232\n)\n\n => Array\n(\n => 41.80.151.17\n)\n\n => Array\n(\n => 23.111.188.5\n)\n\n => Array\n(\n => 223.190.125.216\n)\n\n => Array\n(\n => 103.217.133.119\n)\n\n => Array\n(\n => 103.198.173.132\n)\n\n => Array\n(\n => 47.31.155.89\n)\n\n => Array\n(\n => 223.190.20.253\n)\n\n => Array\n(\n => 104.131.92.125\n)\n\n => Array\n(\n => 223.190.19.152\n)\n\n => Array\n(\n => 103.245.193.191\n)\n\n => Array\n(\n => 106.215.58.255\n)\n\n => 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Array\n(\n => 103.12.135.37\n)\n\n => Array\n(\n => 49.207.120.225\n)\n\n => Array\n(\n => 182.64.195.207\n)\n\n => Array\n(\n => 103.99.37.16\n)\n\n => Array\n(\n => 46.150.104.221\n)\n\n => Array\n(\n => 104.236.195.147\n)\n\n => Array\n(\n => 103.104.192.43\n)\n\n => Array\n(\n => 24.242.159.118\n)\n\n => Array\n(\n => 39.42.179.143\n)\n\n => Array\n(\n => 111.93.58.131\n)\n\n => Array\n(\n => 193.176.84.127\n)\n\n => Array\n(\n => 209.58.142.218\n)\n\n => Array\n(\n => 69.243.152.129\n)\n\n => Array\n(\n => 117.97.131.249\n)\n\n => Array\n(\n => 103.230.180.89\n)\n\n => Array\n(\n => 106.212.170.192\n)\n\n => Array\n(\n => 171.224.180.95\n)\n\n => Array\n(\n => 158.222.11.87\n)\n\n => Array\n(\n => 119.155.60.246\n)\n\n => Array\n(\n => 41.90.43.129\n)\n\n => Array\n(\n => 185.183.104.170\n)\n\n => Array\n(\n => 14.248.67.65\n)\n\n => Array\n(\n => 117.217.205.82\n)\n\n => Array\n(\n => 111.88.7.209\n)\n\n => Array\n(\n => 49.36.132.244\n)\n\n => Array\n(\n => 171.48.40.2\n)\n\n => Array\n(\n => 119.81.105.2\n)\n\n => Array\n(\n => 49.36.128.114\n)\n\n => Array\n(\n => 213.200.31.93\n)\n\n => Array\n(\n => 2.50.15.110\n)\n\n => Array\n(\n => 120.29.104.67\n)\n\n => Array\n(\n => 223.225.32.221\n)\n\n => Array\n(\n => 14.248.67.195\n)\n\n => Array\n(\n => 119.155.36.13\n)\n\n => Array\n(\n => 101.50.95.104\n)\n\n => Array\n(\n => 104.236.205.233\n)\n\n => Array\n(\n => 122.164.36.150\n)\n\n => Array\n(\n => 157.45.93.209\n)\n\n => Array\n(\n => 182.77.118.100\n)\n\n => Array\n(\n => 182.74.134.218\n)\n\n => Array\n(\n => 183.82.128.146\n)\n\n => Array\n(\n => 112.196.170.234\n)\n\n => Array\n(\n => 122.173.230.178\n)\n\n => Array\n(\n => 122.164.71.199\n)\n\n => Array\n(\n => 51.79.19.31\n)\n\n => Array\n(\n => 58.65.222.20\n)\n\n => Array\n(\n => 103.27.203.97\n)\n\n => Array\n(\n => 111.88.7.242\n)\n\n => Array\n(\n => 14.171.232.77\n)\n\n => Array\n(\n => 46.101.22.182\n)\n\n => Array\n(\n => 103.94.219.19\n)\n\n => Array\n(\n => 139.190.83.30\n)\n\n => Array\n(\n => 223.190.27.184\n)\n\n => Array\n(\n => 182.185.183.34\n)\n\n => Array\n(\n => 91.74.181.242\n)\n\n => Array\n(\n => 222.252.107.14\n)\n\n => Array\n(\n => 137.97.8.28\n)\n\n => Array\n(\n => 46.101.16.229\n)\n\n => Array\n(\n => 122.53.254.229\n)\n\n => Array\n(\n => 106.201.17.180\n)\n\n => Array\n(\n => 123.24.170.129\n)\n\n => Array\n(\n => 182.185.180.79\n)\n\n => Array\n(\n => 223.190.17.4\n)\n\n => Array\n(\n => 213.108.105.1\n)\n\n => Array\n(\n => 171.22.76.9\n)\n\n => Array\n(\n => 202.66.178.164\n)\n\n => Array\n(\n => 178.62.97.171\n)\n\n => Array\n(\n => 167.179.110.209\n)\n\n => Array\n(\n => 223.230.147.172\n)\n\n => Array\n(\n => 76.218.195.160\n)\n\n => Array\n(\n => 14.189.186.178\n)\n\n => Array\n(\n => 157.41.45.143\n)\n\n => Array\n(\n => 223.238.22.53\n)\n\n => Array\n(\n => 111.88.7.244\n)\n\n => Array\n(\n => 5.62.57.19\n)\n\n => Array\n(\n => 106.201.25.216\n)\n\n)\n```\nThe To Do List: Dividing the Dime Personal Finance Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > The To Do List", null, "", null, "", null, "# The To Do List\n\nMarch 12th, 2009 at 01:38 am\n\nIs ridiculous.\n\nI need to file the finanical aid form for my son's college.\nGo to the bank.\nMake two doctor's appointments.\nOrder a will kit on-line.\nFind someone to clean our carpets.\nRake leaves.\nCatch up with record-keeping and filing for my church's finance records, 1st quarter. Hold a church finance meeting with the team.\nMail checks back to my health insurance company. (That they sent me by accident.)\nGet an alignment on the car.\nCheck on my neighbor.\nCall my mother.\nClean out my file cabinet.\nFind a new place that recycles aluminum cans (the old place shut down).\nGo to the dry cleaners.\nOpen a Roth IRA.\nBake pumpkin bread before the canned pumpkin I have expires.\nMake a large batch of spaghetti sauce to freeze.\nSend an email to an old friend.\nMake bathroom more handicap accessible for spouse. (He fell this weekend on a tile floor while we were out. That pain combined with his cerebral palsy has created \"access\" issues in our home...)\nDust.\nGet a birthday gift for the kid.\nDo my Spanish homework. (I signed up for an 8-week community education class to learn Spanish.)\nGet to a writer's meeting on Saturday.\nHave the hedges in the front yard cut down or removed.\nClean the garage.\nClean the attic.\nGive myself a manicure and pedicure.\nFind my black gym shorts.\nBuy new towels and toss a few of the old ones we're using.\nDefrost the freezer.\nFix the hole in the wall in the bathroom where the broken toilet paper holder used to be.\nInvite my young friend to Sunday dinner.\nTake donation bags to Good Will.\nWeed out paperback books.\nFinish one of the 6 books I'm currently reading.\nBlog. (I can check this one off the list for this week. It's easier to blog, then to do the other to-do items.)\nTake jar of pennies to bank for deposit.\nSchedule card game night with friends.\nInvite stepdaughter to lunch.\nUpdate my beneficiary forms on everything.\nGet the air conditioning unit serviced.\nRepair the hem of one of my skirts.\nClean out my email inbox.\nUpdate my resume.\nUpdate personal papers for my brother.\nBalance checkbook.\n\nI could go on, but I'm tired. Just checking in...\n\n### 4 Responses to “The To Do List”\n\n1. oceansluver78 Says:\n\noh vey! good luck on getting everything accomplished!\n\n2. Amber Says:\n\nWow what a list...check out http://www.ilrg.com/forms/ for legal forms\n\n3. creditcardfree Says:\n\nWhen are you trying to get all this done? Pick a one or two each day and you can do this!!\n\n4. oceansluver78 Says:\n\nGood luck with your to-do list\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4.  [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]" ]

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[ "# The explicit form of the switching surface in admissible synthesis problem\n\nKeywords: controllability, controllability function method, admissible maximum principle, switching surface\n\n### Abstract\n\nIn this article we consider the problem related to positional synthesis and controllability function method and more precisely to admissible maximum principle. Unlike the more common approach the admissible maximum principle method gives discontinuous solutions to the positional synthesis problem. Let us consider the canonical system of linear equations $\\dot{x}_i=x_{i+1}, i=\\overline{1,n-1}, \\dot{x}_n=u$ with constraints $|u| \\le d$. The problem for an arbitrary linear system $\\dot{x} = A x + b u$ can be simplified to this problem for the canonical system. A controllability function $\\Theta(x)$ is given as a unique positive solution of some equation $\\Phi(x,\\Theta) = 0$. The control is chosen to minimize derivative of the function $\\Theta(x)$ and can be written as $u(x) = -d \\text{ sign}(s(x,\\Theta(x)))$. The set of points $s(x,\\Theta(x)) = 0$ is called the switching surface, and it determines the points where control changes its sign. Normally it \\mbox{contains} the variable $\\Theta$ which is given implicitly as the solution of equation $\\Phi(x, \\Theta) = 0$. Our aim in this paper is to find a representation of the switching surface that does not depend on the function $\\Theta(x)$. We call this representation the explicit form. In our case the expressions $\\Phi(x, \\Theta)$ and $s(x, \\Theta)$ are both polynomials with respect to $\\Theta$, so this problem is related to the problem of finding conditions when two polynomials have a common positive root. Earlier the solution for the 2-dimensional case was known. But during the exploration it was found out that for systems of higher dimensions there exist certain difficulties. In this article the switching surface for the three dimensional case is presented and researched. It is shown that this switching surface is a sliding surface (according to Filippov's definition). Also the other ways of constructing the switching surface using the interpolation and approximation are proposed and used for finding the trajectories of concrete points.\n\n### References\n\nA.M. Lyapunov, The General Problem of the Stability of Motion. - 1892. Kharkov Mathematical Society, Kharkov. English translation: The general problem of the stability of motion (trans. A.T. Fuller), in International Journal of Control, 55:3, 531-573. 1992 DOI: https://doi.org/10.1080/00207179208934253\n\nL. S. Pontryagin, V. G. Boltyanskii, R. V. Gamkrelidze, E. F. Mishechenko, The Mathematical Theory of Optimal Processes. VIII + 360 S. New York/London 1962. John Wiley & Sons. Preis 90/-. DOI: https://doi.org/10.1002/zamm.19630431023\n\nR. Kalman, Y. Ho, K. Narendra, Controllability of linear dynamical systems. Contributions to Differential Equations 1. - 1963, 189-213.\n\nV. I. Korobov. A general approach to the solution of the bounded control synthesis problem in a controllability problem, Math. USSR Sb. 37. - 1980. - No. {bf 4}, 535-557. DOI: https://doi.org/10.1070/SM1980v037n04ABEH002094\n\nV. I. Korobov, G. M. Sklyar. Time optimality and the power moment problem. Mat. Sb., Nov. Ser. 134(176). - 1987. no. 2(10), 186-206; Math. USSR, Sb. 62. - 1989. No. 1, 185-206. DOI: https://doi.org/10.1070/SM1989v062n01ABEH003235\n\nV. I. Korobov, G. M. Sklyar. Methods for constructing of positional controls and an admissible maximum principle, Differ. Equations 26 (1990), Vol. 11, 1422-1431.\n\nV.I. Korobov, Method of controllability function, R$&$C Dynamics, Moskow-Ijevsk, 2007. ISBN 978-5-93972-610-8.\n\nA. F. Filippov, Differential equations with discontinuous right-hand side, Mat. Sb. (N.S.), 51(93):1 (1960), 99-128.\n\nPublished\n2022-12-24\nCited\nHow to Cite\nKorobov, V. I., & Vozniak, O. S. (2022). The explicit form of the switching surface in admissible synthesis problem. Visnyk of V. N. Karazin Kharkiv National University. Ser. Mathematics, Applied Mathematics and Mechanics, 96, 40-55. https://doi.org/10.26565/2221-5646-2022-96-03\nIssue\nSection\nСтатті" ]

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[ "#10dc20 Color Information\n\nIn a RGB color space, hex #10dc20 is composed of 6.3% red, 86.3% green and 12.5% blue. Whereas in a CMYK color space, it is composed of 92.7% cyan, 0% magenta, 85.5% yellow and 13.7% black. It has a hue angle of 124.7 degrees, a saturation of 86.4% and a lightness of 46.3%. #10dc20 color hex could be obtained by blending #20ff40 with #00b900. Closest websafe color is: #00cc33.\n\n• R 6\n• G 86\n• B 13\nRGB color chart\n• C 93\n• M 0\n• Y 85\n• K 14\nCMYK color chart\n\n#10dc20 color description : Vivid lime green.\n\n#10dc20 Color Conversion\n\nThe hexadecimal color #10dc20 has RGB values of R:16, G:220, B:32 and CMYK values of C:0.93, M:0, Y:0.85, K:0.14. Its decimal value is 1104928.\n\nHex triplet RGB Decimal 10dc20 `#10dc20` 16, 220, 32 `rgb(16,220,32)` 6.3, 86.3, 12.5 `rgb(6.3%,86.3%,12.5%)` 93, 0, 85, 14 124.7°, 86.4, 46.3 `hsl(124.7,86.4%,46.3%)` 124.7°, 92.7, 86.3 00cc33 `#00cc33`\nCIE-LAB 76.919, -75.665, 70.232 26.066, 51.398, 9.913 0.298, 0.588, 51.398 76.919, 103.236, 137.133 76.919, -71.727, 91.16 71.692, -60.562, 41.986 00010000, 11011100, 00100000\n\nColor Schemes with #10dc20\n\n• #10dc20\n``#10dc20` `rgb(16,220,32)``\n• #dc10cc\n``#dc10cc` `rgb(220,16,204)``\nComplementary Color\n• #66dc10\n``#66dc10` `rgb(102,220,16)``\n• #10dc20\n``#10dc20` `rgb(16,220,32)``\n• #10dc86\n``#10dc86` `rgb(16,220,134)``\nAnalogous Color\n• #dc1066\n``#dc1066` `rgb(220,16,102)``\n• #10dc20\n``#10dc20` `rgb(16,220,32)``\n• #8610dc\n``#8610dc` `rgb(134,16,220)``\nSplit Complementary Color\n• #dc2010\n``#dc2010` `rgb(220,32,16)``\n• #10dc20\n``#10dc20` `rgb(16,220,32)``\n• #2010dc\n``#2010dc` `rgb(32,16,220)``\n• #ccdc10\n``#ccdc10` `rgb(204,220,16)``\n• #10dc20\n``#10dc20` `rgb(16,220,32)``\n• #2010dc\n``#2010dc` `rgb(32,16,220)``\n• #dc10cc\n``#dc10cc` `rgb(220,16,204)``\n• #0b9516\n``#0b9516` `rgb(11,149,22)``\n• #0dac19\n``#0dac19` `rgb(13,172,25)``\n• #0ec41d\n``#0ec41d` `rgb(14,196,29)``\n• #10dc20\n``#10dc20` `rgb(16,220,32)``\n• #17ee28\n``#17ee28` `rgb(23,238,40)``\n• #2ff03e\n``#2ff03e` `rgb(47,240,62)``\n• #47f254\n``#47f254` `rgb(71,242,84)``\nMonochromatic Color\n\nAlternatives to #10dc20\n\nBelow, you can see some colors close to #10dc20. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #33dc10\n``#33dc10` `rgb(51,220,16)``\n• #22dc10\n``#22dc10` `rgb(34,220,16)``\n• #11dc10\n``#11dc10` `rgb(17,220,16)``\n• #10dc20\n``#10dc20` `rgb(16,220,32)``\n• #10dc31\n``#10dc31` `rgb(16,220,49)``\n• #10dc42\n``#10dc42` `rgb(16,220,66)``\n• #10dc53\n``#10dc53` `rgb(16,220,83)``\nSimilar Colors\n\n#10dc20 Preview\n\nThis text has a font color of #10dc20.\n\n``<span style=\"color:#10dc20;\">Text here</span>``\n#10dc20 background color\n\nThis paragraph has a background color of #10dc20.\n\n``<p style=\"background-color:#10dc20;\">Content here</p>``\n#10dc20 border color\n\nThis element has a border color of #10dc20.\n\n``<div style=\"border:1px solid #10dc20;\">Content here</div>``\nCSS codes\n``.text {color:#10dc20;}``\n``.background {background-color:#10dc20;}``\n``.border {border:1px solid #10dc20;}``\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. 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In this case, #747874 is the less saturated color, while #07e518 is the most saturated one.\n\n• #747874\n``#747874` `rgb(116,120,116)``\n• #6b816d\n``#6b816d` `rgb(107,129,109)``\n• #628a65\n``#628a65` `rgb(98,138,101)``\n• #59935d\n``#59935d` `rgb(89,147,93)``\n• #509c56\n``#509c56` `rgb(80,156,86)``\n• #46a64e\n``#46a64e` `rgb(70,166,78)``\n• #3daf46\n``#3daf46` `rgb(61,175,70)``\n• #34b83f\n``#34b83f` `rgb(52,184,63)``\n• #2bc137\n``#2bc137` `rgb(43,193,55)``\n• #22ca2f\n``#22ca2f` `rgb(34,202,47)``\n• #19d328\n``#19d328` `rgb(25,211,40)``\n• #10dc20\n``#10dc20` `rgb(16,220,32)``\n• #07e518\n``#07e518` `rgb(7,229,24)``\nTone Color Variation\n\nColor Blindness Simulator\n\nBelow, you can see how #10dc20 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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[ "## Articles by \"Mathematica\"\n\nShowing posts with label Mathematica. Show all posts\n\n### Numerical and Analytical Methods for Scientists and Engineers\n\nDaniel Dubin ... 633 pages - Publisher: Wiley-Interscience; (May, 2003) ... Language: English - ISBN-10: 0471266108 - ISBN-13: 978-0471266105.\n\nUtilizing state-of-the-art software to facilitate solutions to real-world problems: Practitioners in the field of physical science are continually faced with a variety of complex, real-world problems, the solution of which requires a working knowledge of both analytical and numerical techniques. An Introduction to Mathematical and Computational Physics Using Mathematica® is designed to help prospective scientists develop a practical, working knowledge of these techniques using the latest, most efficient electronic methodologies. Written from the perspective of a physicist rather than a mathematician, the text focuses on modern practical applications in the physical and engineering sciences, attacking these problems with a range of numerical and analytical methods, both elementary and advanced. Incorporating the widely used and highly praised Mathematica® software package, the author offers solution techniques for the partial differential equations of mathematical physics such as Poisson’s equation, the wave equation, and Schrödinger’s equation, including Fourier series and transforms, Green’s functions, the method of characteristics, grids, Galerkin and simulation methods, elementary probability theory, and statistical methods. The incorporation of Mathematica® offers students a wealth of practical benefits in that it: Requires little or no previous computer experience + Offers maximum flexibility and sophistication + Delivers easy access to the important ideas behind the various numerical methods + Facilitates important but often tedious analytic calculations + Is easily adapted to the application of other related software packages. Designed for both advanced undergraduate and graduate students in the physical and engineering sciences, as well as professionals who want to learn these methods, An Introduction to Mathematical and Computational Physics Using Mathematica® is also provided electronically on an accompanying CD. The electronic version contains the full text of the book, along with animations, user-modifiable source code, and links to related Web material.\n\n### Intelligent Routines: Solving Mathematical Analysis\n\nGeorge A. Anastassiou, Iuliana F. Iatan ... 582 pages - Publisher: Springer; (July, 2012) ... Language: English - ISBN-10: 3642284744 - ISBN-13: 978-3642284748 ...\n\nReal Analysis is a discipline of intensive study in many institutions of higher education, because it contains useful concepts and fundamental results in the study of mathematics and physics, of the technical disciplines and geometry. This book is the first one of its kind that solves mathematical analysis problems with all four related main software Matlab, Mathcad, Mathematica and Maple. Besides the fundamental theoretical notions, the book contains many exercises, solved both mathematically and by computer, using: Matlab 7.9, Mathcad 14, Mathematica 8 or Maple 15 programming languages. The book is divided into nine chapters, which illustrate the application of the mathematical concepts using the computer. Each chapter presents the fundamental concepts and the elements required to solve the problems contained in that chapter and finishes with some problems left to be solved by the readers. The calculations can be verified by using a specific software such as Matlab, Mathcad, Mathematica or Maple.\n\n### Discovering Mathematics: A Problem-Solving Approach...\n\nJiří Gregor, Jaroslav Tišer ... 247 pages - Publisher: Springer; (December, 2010) ... Language: English - ISBN-10: 0857290541 - ISBN-13: 978-0857290540 ...\n\nThe book contains chapters of structured approach to problem solving in mathematical analysis on an intermediate level. It follows the ideas of G.Polya and others, distinguishing between exercises and problem solving in mathematics. Interrelated concepts are connected by hyperlinks, pointing toward easier or more difficult problems so as to show paths of mathematical reasoning. Basic definitions and theorems can also be found by hyperlinks from relevant places. Problems are open to alternative formulations, generalizations, simplifications, and verification of hypotheses by the reader; this is shown to be helpful in solving problems. The book presents how advanced mathematical software can aid all stages of mathematical reasoning while the mathematical content remains in foreground. The authors show how software can contribute to deeper understanding and to enlarging the scope of teaching for students and teachers of mathematics. Discovering Mathematics: A Problem-Solving Approach to Analysis with Mathematica and Maple provides a constructive approach to mathematical discovery through innovative use of software technology. Interactive Mathematica and Maple notebooks are integral to this books’ utility as a practical tool for learning. Interrelated concepts, definitions and theorems are connected through hyperlinks, guiding the reader to a variety of structured problems and highlighting multiple avenues of mathematical reasoning. Interactivity is further enhanced through the delivery of online content (available at extras.springer.com), demonstrating the use of software and in turn increasing the scope of learning for both students and teachers and contributing to a deeper mathematical understanding. This book will appeal to both final year undergraduate and post-graduate students wishing to supplement a mathematics course or module in mathematical problem-solving and analysis. It will also be of use as complementary reading for students of engineering or science, and those in self-study.\n\n### Maple and Mathematica: A Prb. Solving Approach for Math.\n\nInna K. Shingareva, Carlos Lizárraga-Celaya ... 484 pages - Publisher: Springer; 2nd edition (August, 2009) ...  Language: English - ASIN: B00FBVFPTE by Amazon Digital Services ...\n\nThe first book to compare the main two computer algebra systems (CAS), Maple and Mathematica used by students, mathematicians, scientists, and engineers. Both systems are presented in parallel so that Mathematica users can learn Maple quickly by finding the Maple equivalent to Mathematica functions, and vice versa. This student reference handbook consists of core material for incorporating Maple and Mathematica as a working tool into different undergraduate mathematical courses (abstract and linear algebra, geometry, calculus and analysis, complex functions, special functions, integral and discrete transforms, algebraic and transcendental equations, ordinary and partial differential equations, integral equations, numerical analysis and scientific computing). The book also contains applications from various areas of mathematics, physics, and music theory and can be useful for graduate students, professors, and researchers in science and engineering. One of the goals of this book is to develop problem-solving skills (that are most useful for solving sophisticated research problems) finding solutions with Maple and Mathematica and not to depend on a specific version of both systems (Maple 12 and Mathematica 6 and 7 are considered). Part I, describes the foundations of Maple and Mathematica (with equivalent problems and solutions). Part II, describes Mathematics with Maple and Mathematica by using equivalent problems. Finally, this book is ideal for scientists who want to corroborate their Maple and Mathematica work with independent verification provided by another CAS.\n\n### Differential Equations and Boundary Value Problems: Computing and Modeling\n\nC. Henry Edwards, David E. Penney, David T. Calvis ...\n792 pages - Publisher: Pearson; 5th edition (September, 2014) ... Language: English - ISBN-10: 0321796985 - ISBN-13: 978-0321796981 ...\n\nThis best-selling text by these well-known authors blends the traditional algebra problem solving skills with the conceptual development and geometric visualization of a modern differential equations course that is essential to science and engineering students. It reflects the new qualitative approach that is altering the learning of elementary differential equations, including the wide availability of scientific computing environments like Maple, Mathematica, and MATLAB. Its focus balances the traditional manual methods with the new computer-based methods that illuminate qualitative phenomena and make accessible a wider range of more realistic applications. Seldom-used topics have been trimmed and new topics added: it starts and ends with discussions of mathematical modeling of real-world phenomena, evident in figures, examples, problems, and applications throughout the text.\n\n### Practical Optimization Methods with Mathematica Applications\n\nM. Asghar Bhatti ... 715 pages - Publisher: Springer; (June, 2000) ... Language: English - ISBN-10: 0387986316 - ISBN-13: 978-0387986319 ...\n\nThis introductory textbook adopts a practical and intuitive approach, rather than emphasizing mathematical rigor. Computationally oriented books in this area generally present algorithms alone, and expect readers to perform computations by hand, and are often written in traditional computer languages, such as Basic, Fortran or Pascal. This book, on the other hand, is the first text to use Mathematica to develop a thorough understanding of optimization algorithms, fully exploiting Mathematica's symbolic, numerical and graphic capabilities.\n\n### Programming with Mathematica\n\nPaul Wellin... 728 pages - Publisher: Cambridge University Press; 1st edition (February, 2013) ...\nLanguage: English - ISBN-10: 1107009464 - ISBN-13: 978-1107009462 ...\n\nStarting from first principles, this book covers all of the foundational material needed to develop a clear understanding of the Mathematica language, with a practical emphasis on solving problems. Concrete examples throughout the text demonstrate how Mathematica language, can be used to solve problems in science, engineering, economics/finance, computational linguistics, geoscience, bioinformatics, and a range of other fields. The book will appeal to students, researchers and programmers wishing to further their understanding of Mathematica language. Designed to suit users of any ability, it assumes no formal knowledge of programming so it is ideal for self-study. Over 290 exercises are provided to challenge the reader's understanding of the material covered and these provide ample opportunity to practice using the language. Mathematica language notebooks containing examples, programs and solutions to exercises are available from www.cambridge.org/wellin.\n\n### Hands-On Start To Wolfram Mathematica\n\nCliff Hastings, Kelvin Mischo, Michael Morrison ... For more than 25 years, Mathematica has been the principal computation environment for millions of innovators, educators, students, and others around the world. This book is an introduction to Mathematica. The goal is to provide a hands-on experience introducing the breadth of Mathematica, with a focus on ease of use. Readers get detailed instruction with examples for interactive learning and end-of-chapter exercises. Each chapter also contains authors tips from their combined 50+ years of Mathematica use.\n\nInfo-Details: 469 pages - Publisher: Wolfram Media.; (September, 2015) ... Language: English - ISBN-10: 1579550770 - ISBN-13: 978-1579550776 ...\n\n### The Mathematica Book", null, "Stephen Wolfram ... As both a highly readable tutorial and a definitive reference for over a million Mathematica users worldwide, this book covers every aspect of Mathematica. It is an essential resource for all users of Mathematica from beginners to experts. This expanded fifth edition presents Mathematica Version 5 for the first time and is important for anyone interested in the progress of advanced computing. Included in this new edition are the following: Visual tour of key features + Practical tutorial introduction + Full descriptions of 1,200+ built-in functions + Thousands of illustrative examples + Easy-to-follow descriptive tables + Essays highlighting key concepts + Mathematica language tutorial + Guide to symbolic programming + Introduction to document-centered interfaces + Guide to the MathLink API + Notes on internal implementation + Index with 10,000+ entries.\n\nInfo-Details:  1488 pages - Publisher: Wolfram Media; 5th edition (August, 2003) ... Language: English - ISBN-10: 1579550223 - ISBN-13: 978-1579550226 ...\n\n### Structural dynamics of earthquake engineering: Theory and application using Mathematica and Matlab\n\nS. Rajasekaran ... 896 pages- Publisher: Woodhead Publishing; 1st edition (June 13, 2009)\nLanguage: English - ISBN-10: 1845695186 - ISBN-13: 978-1845695187", null, "Given the risk of earthquakes in many countries, knowing how structural dynamics can be applied to earthquake engineering of structures, both in theory and practice, is a vital aspect of improving the safety of buildings and structures. It can also reduce the number of deaths and injuries and the amount of property damage.\n\nThe book begins by discussing free vibration of single-degree-of-freedom (SDOF) systems, both damped and undamped, and forced vibration (harmonic force) of SDOF systems. Response to periodic dynamic loadings and impulse loads are also discussed, as are two degrees of freedom linear system response methods and free vibration of multiple degrees of freedom. Further chapters cover time history response by natural mode superposition, numerical solution methods for natural frequencies and mode shapes and differential quadrature, transformation and Finite Element methods for vibration problems. Other topics such as earthquake ground motion, response spectra and earthquake analysis of linear systems are discussed.\n\nStructural dynamics of earthquake engineering: theory and application using Mathematica and Matlab provides civil and structural engineers and students with an understanding of the dynamic response of structures to earthquakes and the common analysis techniques employed to evaluate these responses. Worked examples in Mathematica and Matlab are given.\nExplains the dynamic response of structures to earthquakes including periodic dynamic loadings and impulse loads + Examines common analysis techniques such as natural mode superposition, the finite element method and numerical solutions + Investigates this important topic in terms of both theory and practise with the inclusion of practical exercise and diagrams.\n\n### Foundations of Fluid Mechanics with Applications Problem Solving Using Mathematica\n\nSergey P. Kiselev, Evgenii Vorozhtsov, 575 pages - Publisher: Birkhäuser; (December, 1999) ... Language: English - ISBN-10: 0817639950 - ISBN-13: 978-0817639952 ...\n\nFluid mechanics (FM) is a branch of science dealing with the investi­ gation of flows of continua under the action of external forces. The fundamentals of FM were laid in the works of the famous scientists, such as L. Euler, M. V. Lomonosov, D. Bernoulli, J. L. Lagrange, A. Cauchy, L. Navier, S. D. Poisson, and other classics of science. Fluid mechanics underwent a rapid development during the past two centuries, and it now includes, along with the above branches, aerodynamics, hydrodynamics, rarefied gas dynamics, mechanics of multi phase and reactive media, etc. The FM application domains were expanded, and new investigation methods were developed. Certain concepts introduced by the classics of science, however, are still of primary importance and will apparently be of importance in the future. The Lagrangian and Eulerian descriptions of a continuum, tensors of strains and stresses, conservation laws for mass, momentum, moment of momentum, and energy are the examples of such concepts and results. This list should be augmented by the first and second laws of thermodynamics, which determine the character and direction of processes at a given point of a continuum. The availability of the conservation laws is conditioned by the homogeneity and isotrop­ icity properties of the Euclidean space, and the form of these laws is related to the Newton's laws. The laws of thermodynamics have their foundation in the statistical physics.\n\n### Fundamental Finite Element Analysis and Applications with Mathematica and MATLAB Computations", null, "M. Asghar Bhatti ... 720 pages - Publisher: Wiley; 1st edition (February, 2005) ... Language: English - ISBN-10: 0471648086 - ISBN-13: 978-0471648086 ...\n\nFundamental Finite Element Analysis and Applications: with Mathematica® and MATLAB® Computations is an innovative, practical guide to discovering the Finite Element Method (FEM). Providing a helpful balance between theory and application, it presents the FEM as a tool to find approximate solutions of differential equations, making it a useful resource for students from a variety of disciplines. Using a unique combination of live Mathematica® and MATLAB® implementations, along with problems in both ANSYS® and ABAQUS® formats, this hands-on book reveals the logic behind the equations to facilitate a full understanding of methods and solutions. In nine convenient chapters, Fundamental Finite Element Analysis and Applications: with Mathematica® and MATLAB® .Computations covers: Finite Element Method: The Big Picture * Mathematical Foundation of the Finite Element Method * One-Dimensional Boundary Value Problems * Trusses, Beams, and Frames * Two-Dimensional Elements * Mapped Elements * Analysis of Elastic Solids * Transient Problems * p-Formulation. An associated Web site (wiley.com/go/bhatti) includes interactive application files and notebooks for Mathematica®, MATLAB®, ANSYS®, and ABAQUS®, with expanded exercises to use with the book. Fundamental Finite Element Analysis with Mathematica® and MATLAB® Computations is a clear and accessible learning tool for senior undergraduate and graduate-level students. *Finite Element Analysis with Mathematica and Matlab Computations and Practical Applications is an innovative, hands-on and practical introduction to the Finite Element Method that provides a powerful tool for learning this essential analytic method.\n\nName\n\nEmail *\n\nMessage *" ]

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[ "", null, "# Parametric Integration\n\nRelated Topics:\nMore Lessons for A Level Maths\n\nMath Worksheets\n\nVideos, activities and worksheets that are suitable for A Level Maths.\n\nIntegration of Parametric Equations\nIntegration of Parametric Equations Exam Style Question 1\n\nIntegration of Parametric Equations Exam Style Question 2\nIntegration of Parametric Equations Exam Style Question 3\n\nIntegration of Parametric Equations Exam Style Question 4\nIntegration of Parametric Equations Exam Style Question 5\n\nIntegration of Parametric Equations Exam Style Question 6\n\nRotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.\n\nYou can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "" ]

[ null, "https://www.onlinemathlearning.com/objects/default_image.gif", null, "https://www.onlinemathlearning.com/objects/default_image.gif", null ]

[ "# Memory Jogger: SQL Query for minimum and maximum data values.\n\nIn the past when I needed to find the minimum and maximum value of a field returned in a data set I would loop through the query result in Flash Actionscript like this:\n\nvar minValue:Number = Number.MAX_VALUE;\nvar maxValue:Number = Number.MIN_VALUE;\n\nfor(var i:Number = 0; i < resultLength; i++) {\nminValue = Math.min(minValue, result[i]);\nmaxValue = Math.max(maxValue, result[i]);\n}\n\nHowever I have recently began to try and move more of this sort of functionality from the front end to the PHP class functions in the AMFphp class services on the back end. The obvious advantage being that querying an indexed value directly from a MySQL table is going to be a lot faster than looping through large data sets to find the same values.\n\nSo this is the PHP function I use to do the same thing now.\n\nfunction getRangeData (\\$ID, \\$tData, \\$tColumn) {\n\n\\$query_str = “SELECT A.* FROM (SELECT * FROM ” . \\$tData;\n\\$query_str .= ” WHERE ID = ” . \\$ID . ” ORDER BY ” . \\$tColumn . ” ASC limit 1)A “;\n\\$query_str .= “UNION “;\n\\$query_str .= “SELECT B.* FROM (SELECT * FROM ” . \\$tData;\n\\$query_str .= ” WHERE ID = ” . \\$ID . ” ORDER BY ” . \\$tColumn . ” DESC limit 1)B “;\n\n\\$data = mysql_query(\\$query_str);\n\n\\$result_array = array();\n\\$resultCount = 0;\n\nwhile(\\$resultRow = mysql_fetch_array(\\$data, MYSQL_ASSOC)) {\n\n\\$result = \\$resultRow[\\$tColumn];\n\\$result_array[\\$resultCount][\\$tColumn] = \\$result;\n\\$resultCount ++;\n\n}\n\nreturn \\$result_array;\n\n}\n\nIn this case you can pass the name of the table, the required data column and the “ID” you want the data filtered against however you could modify this easily enough if you didn’t require a WHERE class." ]

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[ "### the standard error of estimates (SEE)\n\nThe standard error of the estimate (SEE) refers to the deviation of any estimate from the intended values. SEE is a measure of the accuracy of predictions made with a regression line-- in other words, it's about how well a linear model fits the data. The formula for SEE is", null, "(Y refers to the dataset and Y' is the mean of the data)\n\nHere's a great video to help students understand the standard error of estimates (SEE).\n\nAnother commonly used statistic to measure how well the linear model fits the data is R-squared (the coefficient of determination). And while R^2 is a commonly used statistic, this post offers a more in depth analysis of the good and the bad of R^2. The same author gives you an idea of how to use R^2 in conjunction with SEE and the value of SEE." ]

[ null, "https://explorable.com/images/standard-error-of-the-estimate.jpg", null ]

[ "Wednesday, December 6, 2023\n\n# What is the Purpose of a Number System?\n\nIn mathematics, a number system is defined as the way of writing technique for expressing numbers. This is also called the system of numeration, and it is a mathematical notation for denoting the given set of numbers using a specific set of digits or symbols or characters in a constant manner. Moreover, we can use the same set of characters to represent various numbers in different types of number systems. Here, the types of number systems include binary, decimal, octal and hexadecimal number systems, etc. Generally, we use a number system to count things, items, etc. Let’s briefly understand the purpose and importance of the number system in various fields.\n\nLet us consider the four most commonly used numbering systems, binary, octal, decimal and hexadecimal. The importance of these number systems can be explained as given below.\n\n##### Significance of Binary Number System\n\nBinary numbers are generally used when only two options are available, so if one is false, the other is true. The binary number system is widely used to illustrate everyday situations as well. This number system represents bits in computers that can have only two values, 0 and 1. Also, in electric circuits, a switch can be represented using (1) to on and (0) to off. Besides, for electronic gates in electrical circuits, we can display the false or true statements in binary digits such that 0 means false and 1 means true. It is also possible to convert one number to another, for example, binary to decimal, decimal to binary, and so on.\n\n##### Significance of Octal Number System\n\nThe octal number is a rarely used number system and is primarily used in graphics and text computing. The base of this number system is 8. Well-known operating systems, even UNIX, use an octal number system for their file protection system. This number system includes eight unique representations, which we can combine to create more octal number expressions. Octal numbers are complex for an ordinary person who has a limited understanding of the number system. In this type of number system, we use seven digits to represent the first seven numbers; then, it is challenging to comprehend. Here, we can also perform the conversion such as binary to octal, octal to binary, octal to decimal, etc.\n\n##### Significance of Decimal Number System\n\nThe decimal number system is the most commonly used system to write the sequence of numbers. Interestingly, the decimal number system is that we represent integers, natural numbers, etc. For example, the digits of the decimal number system are 1, 2, 3, 4, 5, 6, etc., which also represent counting numbers or natural numbers. This system of numbers is also known as the denary number system. The decimal number system is the most frequently used one in our day-to-day activities; this includes calculation, calendar systems, accounts and financial systems or simple routine counting. This system is more comfortable to use when compared with other types of number systems. However, the conversion of this number system to others is relatively easy. For example—decimal to octal, decimal to hex conversion and so on\n\n##### Significance of Hexadecimal Number System\n\nSoftware developers and system designers use hexadecimal numbers to a great extent to deliver a human-friendly declaration of binary-coded values. The hexadecimal number system is used when more possibilities need to be defined. Also, it is the most commonly used number system in computing to describe various memory locations. This is also known as the positional number system.\n\n### Empowering Investors through Demat Accounts\n\nWhеn it comеs to еmpowеrmеnt in thе modernized agе, Dеmat account plays an urgent role. Thеy providе a sеamlеss intеrfacе for invеstors to tradе...\n\n### Top Affordable Clothing Websites Every Teen Girl Should Know About\n\nIn the fast-paced world of teen fashion, finding affordable and trendy clothing can be a challenge. Fortunately, there are several online platforms that cater...\n\n### How You Can Make Your Old House a New One\n\nA house is a defined place to get all the comfort for living and a space to enjoy the best memories with your family...." ]

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[ "compare_secondary_fits {biogrowth} R Documentation\n\n## Model comparison and selection for secondary growth models\n\n### Description\n\nThis function is a constructor for SecondaryComparison a class that provides several functions for model comparison and model selection for growth models fitted using fit_secondary_growth(). Please see the help pages for SecondaryComparison for further details.\n\nAlthough it is not necessary, we recommend passing the models as a named list, as these names will later be kept in plots and tables.\n\n### Usage\n\ncompare_secondary_fits(models)\n\n\n### Arguments\n\n models a (we recommend named) list of models fitted using fit_secondary_growth().\n\n### Examples\n\n\n## We first need to fit some models\n\ndata(\"example_cardinal\")\n\nsec_model_names <- c(temperature = \"Zwietering\", pH = \"CPM\")\n\nknown_pars <- list(mu_opt = 1.2, temperature_n = 1,\npH_n = 2, pH_xmax = 6.8, pH_xmin = 5.2)\n\nmy_start <- list(temperature_xmin = 5, temperature_xopt = 35,\npH_xopt = 6.5)\n\nfit1 <- fit_secondary_growth(example_cardinal, my_start, known_pars, sec_model_names)\n\nknown_pars <- list(mu_opt = 1.2, temperature_n = 2,\npH_n = 2, pH_xmax = 6.8, pH_xmin = 5.2)\n\nfit2 <- fit_secondary_growth(example_cardinal, my_start, known_pars, sec_model_names)\n\n## We can now pass the models to the constructor\n\ncomparison <- compare_secondary_fits(list(n=1 = fit1,\nn=2 = fit2))\n\n## The function includes several S3 methods for model selection and comparison\n\nprint(comparison)\n\nplot(comparison)\nplot(comparison, type = 2)\n\n## The numerical indexes can be accessed using coef and summary\n\ncoef(comparison)\nsummary(comparison)\n\n\n\n[Package biogrowth version 1.0.3 Index]" ]

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[ "# 46.63 kg to lbs - 46.63 kilograms to pounds\n\nBefore we go to the practice - this is 46.63 kg how much lbs conversion - we want to tell you some theoretical information about these two units - kilograms and pounds. So let’s move on.\n\nHow to convert 46.63 kg to lbs? 46.63 kilograms it is equal 102.8015527706 pounds, so 46.63 kg is equal 102.8015527706 lbs.\n\n## 46.63 kgs in pounds\n\nWe will start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, in formal International System of Units (in abbreviated form SI).\n\nFrom time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.\n\nFirst definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. First definition was not complicated but difficult to use.\n\nLater, in 1889 the kilogram was described using the International Prototype of the Kilogram (in short form IPK). The International Prototype of the Kilogram was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was switched by another definition.\n\nToday the definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”\n\nOne kilogram is 0.001 tonne. It could be also divided into 100 decagrams and 1000 grams.\n\n## 46.63 kilogram to pounds\n\nYou know something about kilogram, so now we can move on to the pound. The pound is also a unit of mass. We want to emphasize that there are not only one kind of pound. What does it mean? For instance, there are also pound-force. In this article we are going to to centre only on pound-mass.\n\nThe pound is in use in the Imperial and United States customary systems of measurements. To be honest, this unit is used also in another systems. The symbol of this unit is lb or “.\n\nThe international avoirdupois pound has no descriptive definition. It is defined as 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces or 7000 grains.\n\nThe avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”\n\n### How many lbs is 46.63 kg?\n\n46.63 kilogram is equal to 102.8015527706 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.\n\n### 46.63 kg in lbs\n\nThe most theoretical part is already behind us. In next section we want to tell you how much is 46.63 kg to lbs. Now you know that 46.63 kg = x lbs. So it is time to get the answer. Just look:\n\n46.63 kilogram = 102.8015527706 pounds.\n\nThat is a correct result of how much 46.63 kg to pound. You may also round off the result. After rounding off your outcome will be exactly: 46.63 kg = 102.586 lbs.\n\nYou know 46.63 kg is how many lbs, so see how many kg 46.63 lbs: 46.63 pound = 0.45359237 kilograms.\n\nNaturally, in this case you may also round off the result. After rounding off your outcome is as following: 46.63 lb = 0.45 kgs.\n\nWe also want to show you 46.63 kg to how many pounds and 46.63 pound how many kg outcomes in tables. Let’s see:\n\nWe are going to begin with a table for how much is 46.63 kg equal to pound.\n\n### 46.63 Kilograms to Pounds conversion table\n\nKilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)\n46.63 102.8015527706 102.5860\nNow look at a table for how many kilograms 46.63 pounds.\n\nPounds Kilograms Kilograms (rounded off to two decimal places\n46.63 0.45359237 0.45\n\nNow you know how many 46.63 kg to lbs and how many kilograms 46.63 pound, so it is time to move on to the 46.63 kg to lbs formula.\n\n### 46.63 kg to pounds\n\nTo convert 46.63 kg to us lbs you need a formula. We are going to show you two versions of a formula. Let’s begin with the first one:\n\nAmount of kilograms * 2.20462262 = the 102.8015527706 outcome in pounds\n\nThe first version of a formula will give you the most accurate outcome. Sometimes even the smallest difference could be significant. So if you want to get a correct outcome - first formula will be the best solution to calculate how many pounds are equivalent to 46.63 kilogram.\n\nSo move on to the shorer formula, which also enables conversions to learn how much 46.63 kilogram in pounds.\n\nThe another version of a formula is as following, have a look:\n\nNumber of kilograms * 2.2 = the result in pounds\n\nAs you see, the second formula is simpler. It could be better solution if you want to make a conversion of 46.63 kilogram to pounds in quick way, for example, during shopping. Just remember that your result will be not so exact.\n\nNow we are going to show you these two versions of a formula in practice. But before we are going to make a conversion of 46.63 kg to lbs we are going to show you another way to know 46.63 kg to how many lbs totally effortless.\n\n### 46.63 kg to lbs converter\n\nAn easier way to check what is 46.63 kilogram equal to in pounds is to use 46.63 kg lbs calculator. What is a kg to lb converter?\n\nConverter is an application. Converter is based on first version of a formula which we showed you above. Due to 46.63 kg pound calculator you can easily convert 46.63 kg to lbs. You only have to enter amount of kilograms which you want to calculate and click ‘convert’ button. You will get the result in a flash.\n\nSo try to calculate 46.63 kg into lbs using 46.63 kg vs pound converter. We entered 46.63 as a number of kilograms. This is the result: 46.63 kilogram = 102.8015527706 pounds.\n\nAs you see, our 46.63 kg vs lbs converter is intuitive.\n\nNow we can go to our chief topic - how to convert 46.63 kilograms to pounds on your own.\n\n#### 46.63 kg to lbs conversion\n\nWe will begin 46.63 kilogram equals to how many pounds conversion with the first formula to get the most exact result. A quick reminder of a formula:\n\nAmount of kilograms * 2.20462262 = 102.8015527706 the outcome in pounds\n\nSo what have you do to know how many pounds equal to 46.63 kilogram? Just multiply number of kilograms, in this case 46.63, by 2.20462262. It gives 102.8015527706. So 46.63 kilogram is exactly 102.8015527706.\n\nIt is also possible to round it off, for example, to two decimal places. It is equal 2.20. So 46.63 kilogram = 102.5860 pounds.\n\nIt is time for an example from everyday life. Let’s calculate 46.63 kg gold in pounds. So 46.63 kg equal to how many lbs? And again - multiply 46.63 by 2.20462262. It is exactly 102.8015527706. So equivalent of 46.63 kilograms to pounds, when it comes to gold, is 102.8015527706.\n\nIn this case you can also round off the result. It is the result after rounding off, in this case to one decimal place - 46.63 kilogram 102.586 pounds.\n\nNow we can go to examples converted using short formula.\n\n#### How many 46.63 kg to lbs\n\nBefore we show you an example - a quick reminder of shorter formula:\n\nAmount of kilograms * 2.2 = 102.586 the result in pounds\n\nSo 46.63 kg equal to how much lbs? As in the previous example you need to multiply number of kilogram, in this case 46.63, by 2.2. See: 46.63 * 2.2 = 102.586. So 46.63 kilogram is exactly 2.2 pounds.\n\nDo another conversion with use of shorer version of a formula. Now convert something from everyday life, for instance, 46.63 kg to lbs weight of strawberries.\n\nSo calculate - 46.63 kilogram of strawberries * 2.2 = 102.586 pounds of strawberries. So 46.63 kg to pound mass is 102.586.\n\nIf you know how much is 46.63 kilogram weight in pounds and are able to convert it with use of two different formulas, we can move on. Now we want to show you these outcomes in charts.\n\n#### Convert 46.63 kilogram to pounds\n\nWe know that results shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these results in tables for your convenience. Thanks to this you can easily compare 46.63 kg equivalent to lbs results.\n\nBegin with a 46.63 kg equals lbs table for the first version of a formula:\n\nKilograms Pounds Pounds (after rounding off to two decimal places)\n46.63 102.8015527706 102.5860\n\nAnd now let’s see 46.63 kg equal pound chart for the second formula:\n\nKilograms Pounds\n46.63 102.586\n\nAs you see, after rounding off, when it comes to how much 46.63 kilogram equals pounds, the results are the same. The bigger number the more considerable difference. Remember it when you need to do bigger number than 46.63 kilograms pounds conversion.\n\n#### How many kilograms 46.63 pound\n\nNow you know how to convert 46.63 kilograms how much pounds but we will show you something more. Do you want to know what it is? What about 46.63 kilogram to pounds and ounces conversion?\n\nWe want to show you how you can calculate it step by step. Start. How much is 46.63 kg in lbs and oz?\n\nFirst things first - you need to multiply number of kilograms, in this case 46.63, by 2.20462262. So 46.63 * 2.20462262 = 102.8015527706. One kilogram is equal 2.20462262 pounds.\n\nThe integer part is number of pounds. So in this example there are 2 pounds.\n\nTo check how much 46.63 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.\n\nSo your outcome is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then final result will be equal 2 pounds and 33 ounces.\n\nAs you see, conversion 46.63 kilogram in pounds and ounces is not complicated.\n\nThe last calculation which we are going to show you is calculation of 46.63 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.\n\nTo convert it you need another formula. Before we show you this formula, let’s see:\n\n• 46.63 kilograms meters = 7.23301385 foot pounds,\n• 46.63 foot pounds = 0.13825495 kilograms meters.\n\nNow let’s see a formula:\n\nNumber.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters\n\nSo to convert 46.63 foot pounds to kilograms meters you have to multiply 46.63 by 0.13825495. It is exactly 0.13825495. So 46.63 foot pounds is 0.13825495 kilogram meters.\n\nIt is also possible to round off this result, for instance, to two decimal places. Then 46.63 foot pounds is equal 0.14 kilogram meters.\n\nWe hope that this conversion was as easy as 46.63 kilogram into pounds conversions.\n\nThis article was a huge compendium about kilogram, pound and 46.63 kg to lbs in conversion. Thanks to this conversion you know 46.63 kilogram is equivalent to how many pounds.\n\nWe showed you not only how to make a calculation 46.63 kilogram to metric pounds but also two other calculations - to check how many 46.63 kg in pounds and ounces and how many 46.63 foot pounds to kilograms meters.\n\nWe showed you also other solution to make 46.63 kilogram how many pounds calculations, it is with use of 46.63 kg en pound calculator. It will be the best choice for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.\n\nWe hope that now all of you can do 46.63 kilogram equal to how many pounds calculation - on your own or using our 46.63 kgs to pounds calculator.\n\nSo what are you waiting for? Convert 46.63 kilogram mass to pounds in the best way for you.\n\nDo you want to do other than 46.63 kilogram as pounds conversion? For instance, for 5 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so easy as for 46.63 kilogram equal many pounds.\n\n### How much is 46.63 kg in pounds\n\nTo quickly sum up this topic, that is how much is 46.63 kg in pounds , we gathered answers to the most frequently asked questions. Here you can see all you need to know about how much is 46.63 kg equal to lbs and how to convert 46.63 kg to lbs . You can see it down below.\n\nWhat is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 46.63 kg to pound conversion formula look? . Check it down below:\n\nThe number of kilograms * 2.20462262 = the result in pounds\n\nSee the result of the conversion of 46.63 kilogram to pounds. The exact answer is 102.8015527706 lbs.\n\nIt is also possible to calculate how much 46.63 kilogram is equal to pounds with another, shortened version of the equation. Check it down below.\n\nThe number of kilograms * 2.2 = the result in pounds\n\nSo now, 46.63 kg equal to how much lbs ? The result is 102.8015527706 pounds.\n\nHow to convert 46.63 kg to lbs in just a moment? You can also use the 46.63 kg to lbs converter , which will make whole mathematical operation for you and you will get an exact result .\n\n#### Kilograms [kg]\n\nThe kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.\n\n#### Pounds [lbs]\n\nA pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms." ]

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[ "# Name: Diagnostic Test", null, "```Name:\nMath 1030-06\nSpring 2007\nDiagnostic Test\nMath 1010 is a prerequisite for Math 1030. This means that you should have a working\nknowledge of intermediate algebra. This diagnostic test covers some of this background\nmaterial. You should give yourself this test and then check your answers with the solution\nsheet handed out in class on Thursday.\n1. Three kinds of apples are all mixed up in a basket. How many apples must you draw\n(without looking) from the basket to be sure of getting at least two apples of the same\nkind?\n2. There are 150 people in a class. If 80% of them are registered, how many are not\nregistered?\n3. Express \\three-fths&quot; as a fraction, a decimal, and as a percentage.\n4. Evaluate each of the following with = 4, = , and = 6:\n(a) ( + ) =\na\na\nb\nc\n(b) + ab\na\n(c) 5\nb\nb\nb\nc\nc\n=\n3 =\n2\nc\n1\n2\n5\nc\n5. Evaluate the following expressions on your calculator:\n(a) (250 (34 + 56)) 27 =\n=\n(b) 23 + 6 3 (4 ) =\n5\n7\n5\n:\np\np\n(c) 3 32\n15 =\n6. Simplify the following expressions:\n(a) xx5 x32\n(b) (\nx\n(c) (\nx\n2\n5\ny\ny\n)\n3 2\n)(\n4 2\n0\nx y\n)\n2 2\n7. If there are 0.82 US dollars in one Canadian dollar, which is smaller: one US dollar or\n8. One number is six times a second number. Find the numbers if their dierence is 102.\n2\n9. If you drive at an average speed of 65 miles per hour, how long will it take you to drive\n530 miles? If you can bike a distance of 45 miles in three hours and 15 minutes, what\nis your average biking speed in miles per hour?\n10. The length of a rectangle is 14 inches more than its width. If the area is 72 square\ninches, nd the length and width of the rectangle.\n11. Suppose that three-quarters of the freshmen live in a dorm. If two-thirds of the freshmen dorm residents are women, what percentage of the freshman class are women who\nlive in the dorm?\n12. Solve for in the following equations:\n(a) 3 5 = 9 + 7\nx\nx\n(b)\nx\n(c)\nx\n2\n2\nx\n5 = 31\nx\n12 = 0\n3\n(d)\nx\n3\n5\n=x\n2\n(e) j + 3j = 10\nx\n13. Solve for and such that and satisfy the equations: 3\nx\ny\n14. Graph the line 5\nx\nx\ny\nx\n2 = 5 and + = 7.\ny\nx\ny\n2 = 6. What is the -intercept?\ny\ny\n15. A warehouse may contain bicycles, tricycles, and cars. Altogether there are 18 wheels\nin the warehouse. How many bicycles, tricycles, and cars are there? Give as many\n4\n16. The playground drawn below is in the shape of a rectangle with a semi-circle attached\nas shown. Suppose that the longer side of the rectangle is twice the length of the\nshorter side and that the radius of the semi-circle is 12 feet. What is the perimeter\nand the area of the playground?\n17. Suppose that the ratio of undergraduate students to graduate students in an institution\nis 18:7. What percentage of the student body are graduate students?\n18. Suppose that your annual tuition as a freshman was \\$1,856. Each year tuition has\nincreased by 5%. Now you are in your senior year. What is your annual tuition this\nyear?\n5\n19. The company you work for was doing poorly two years ago and as a result everyone\ntook a 10% pay cut for the last year. The company is doing better now and the CEO\nis just promised to raise everyones salary 10% for the next year. Does this mean that\nyour salary next year will be the same as it was two years ago? Explain.\n20. Determine any errors made in the work shown below. Then explain the mistake(s)\n(a)\n3(\n5)+x(3)\n3\n5+3\n3\nx\nx\nx\n(b)\n2\nx+3\n5\n2(x)\n5\n2\n4\nx\nx\nx\nx\n1 ) \\cancel 3&quot;\n1 ) \\add +5 to both sides&quot;\n6 ) \\subtract 3 from both sides&quot;\n3\n=\n=\n=\n=\n=\n= + ( 3)\n= 5 15\n= 5 17\n= 17\n= 21\nx\nx\nx\nx\n)\n)\n)\n)\n)\n\\multiply by 5&quot;\n\\subtract 2&quot;\n\\subtract 5 &quot;\n\\subtract 4&quot;\nx\n(c)\n5(\n2 3\nx y\n) = 5 5 = 25\n2\n3\nx\ny\n6\n2 3\nx y :\n```" ]

[ null, "https://s2.studylib.net/store/data/011731194_1-d9d34c6d8a76c69b9ce70ba57a3449c7-768x994.png", null ]

[ "# The derivative: definitions, rules, the mean value theorem with applications. Optimisation and curve sketching. Primitive functions and integration techniques.\n\nDerivatives, differentiation rules, applications using the derivative to solve optimization problems - Integrals, integration techniques, applications using integrals\n\nHowever, while the product rule was a “plug and solve” formula (f′ * g + f * g), the integration equivalent of the product rule requires you to make an educated guess about which function part to put where. integration of trigonometric integrals Recall the definitions of the trigonometric functions. The following indefinite integrals involve all of these well-known trigonometric functions. In a simple straight language integration can be defined as the measure, which basically assigns numbers to the several functions.The numbers are basically assigned which may describe the displacement,volume or area etc of such concerned function. THE INTEGRATION OF EXPONENTIAL FUNCTIONS The following problems involve the integration of exponential functions. We will assume knowledge of the following well-known differentiation formulas : , where , and , where a is any positive constant not equal to 1 and is the natural (base e) logarithm of a. These formulas lead immediately to the Use the Sum Rule: ∫ cos x + x dx = ∫ cos x dx + ∫ x dx.", null, "write masters dissertation proposal, essay time rules good introductions for opinion  students, picture taken from a corridor outside. Integration where you can talk about our journalism. But in order to keep things civil, we have some rules. Communal harmony and national integration essay in hindi case study on quality essay to about a place A favourite my visit descriptive case study labour law.\n\nMålet för integrationspolitiken i Sverige är att alla ska ha lika rättigheter, skyldigheter och möjligheter.\n\n## If you are asked to integrate a fraction, try multiplying or dividing the top and bottom of the fraction by a number.\n\nThe game rule is  We also work to facilitate the exchange of experience gained in the area between Nordic countries. We highlight examples of successful integration initiatives so  Information om Numerical Integration : Recent Developments, Software and For polynomial integrating rules, invariant theory and ideal theory have been used  Integration rules - Area calculations, volume integrals - Integral applications from own specialist area - Numerical integration - Mathematical software (Mathcad  Many translated example sentences containing \"seamless integration\" The seamless integration of the hygiene rules and of the official controls, from farm to  SWI_meaning = tmp + rules.A.Y + rules.A.M;. var today = new Date();. var default_ref = '' + today.getFullYear();.", null, "### För oss har det varit solklart vad vi som bolag står för, Rules varumärkes-DNA, Rule och Hello Retail integration | Personalisera produktrekommendationer.", null, "Below is a list of top integrals. It is supposed here that $$a,$$ $$p\\left({p e 1} \\right),$$ $$C$$ are real constants, $$b$$ is the base of the exponential function $$\\left({b e 1, b \\gt 0} \\right).$$ Integration is the basic operation in integral calculus.While differentiation has straightforward rules by which the derivative of a complicated function can be found by differentiating its simpler component functions, integration does not, so tables of known integrals are often useful.\n\nMed hjälp av viktig data och värdefulla kundinsikter kan du segmentera och personanpassa erbjudanden och rekommendationer. The definite integral of a function gives us the area under the curve of that function. Another common interpretation is that the integral of a rate function describes the accumulation of the quantity whose rate is given. We can approximate integrals using Riemann sums, and we define definite integrals using limits of Riemann sums. The fundamental theorem of calculus ties integrals and Numerical integration: Gaussian quadrature rules Matlab’s built-in numerical integration function [Q,fcount]=quad(f,a,b,tol) is essentially our simp_compextr code with some further efficiency-enhancing features. Note that quad requires scalar functions to be defined with … If the integration mode is enabled, then only the first two records for reporting currencies (scope dimension remains G_NONE) would be generated after currency translation..\nMarx teori i dagens samhälle\n\nSome people call it anti-differentiation.\n\nThe integration in 7.x-3.x is fully functional. Rules is a more powerful replacement for Drupal's built-in Actions (and Trigger).\nAuktoritärt ledarskap wiki", null, "knitted cast on method\nfrans prenkert örebro universitet\nekonomi jobb" ]

[ null, "https://picsum.photos/800/600", null, "https://picsum.photos/800/615", null, "https://picsum.photos/800/634", null, "https://picsum.photos/800/619", null ]

[ "# Python Program to find Perimeter of a Rectangle using length and width\n\nWrite Python Program to find Perimeter of a Rectangle using length and width with a practical example.\n\n## Python Program to find Perimeter of a Rectangle using length and width example 1\n\nThis Python program allows user to enter the length and width of a rectangle. By using the length and width, this program finds the perimeter of a rectangle. The math formula to calculate perimeter of a rectangle: perimeter  = 2 * (Length + Width). If we know the length & width.\n\n```# Python Program to find Perimeter of a Rectangle using length and width\n\nlength = float(input('Please Enter the Length of a Triangle: '))\nwidth = float(input('Please Enter the Width of a Triangle: '))\n\n# calculate the perimeter\nperimeter = 2 * (length + width)\n\nprint(\"Perimeter of a Rectangle using\", length, \"and\", width, \" = \", perimeter)```\n``````Please Enter the Length of a Triangle: 35\nPlease Enter the Width of a Triangle: 88\nPerimeter of a Rectangle using 35.0 and 88.0 = 246.0``````\n\n## Python Program to calculate Perimeter of a Rectangle using length and width example 2\n\nThis Python program to find perimeter of a Rectangle is the same as above. But, in this python program, we separated the Perimeter of a rectangle logic using Python functions.\n\n```# Python Program to find Perimeter of a Rectangle using length and width\n\ndef perimeter_of_Rectangle(length, width):\nreturn 2 * (length + width)\n\nlength = float(input('Please Enter the Length of a Triangle: '))\nwidth = float(input('Please Enter the Width of a Triangle: '))\n\n# calculate the perimeter\nperimeter = perimeter_of_Rectangle(length, width)\nprint(\"Perimeter of a Rectangle using\", length, \"and\", width, \" = \", perimeter)```" ]

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[ "Search a number\nBaseRepresentation\nbin11011101100111010\n312202122110\n4123230322\n512112331\n62233150\n7651543\noct335472\n9182573\n10113466\n1178281\n12557b6\n133c852\n142d4ca\n1523946\nhex1bb3a\n\n113466 has 8 divisors (see below), whose sum is σ = 226944. Its totient is φ = 37820.\n\nThe previous prime is 113453. The next prime is 113467. The reversal of 113466 is 664311.\n\nAdding to 113466 its reverse (664311), we get a palindrome (777777).\n\nIt is a sphenic number, since it is the product of 3 distinct primes.\n\nIt is a plaindrome in base 10.\n\nIt is not an unprimeable number, because it can be changed into a prime (113467) by changing a digit.\n\n113466 is an untouchable number, because it is not equal to the sum of proper divisors of any number.\n\nIt is a pernicious number, because its binary representation contains a prime number (11) of ones.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 9450 + ... + 9461.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (28368).\n\n2113466 is an apocalyptic number.\n\n113466 is a primitive abundant number, since it is smaller than the sum of its proper divisors, none of which is abundant.\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n113466 is a wasteful number, since it uses less digits than its factorization.\n\n113466 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 18916.\n\nThe product of its digits is 432, while the sum is 21.\n\nThe square root of 113466 is about 336.8471463439. The cubic root of 113466 is about 48.4122478850.\n\nThe spelling of 113466 in words is \"one hundred thirteen thousand, four hundred sixty-six\".\n\nDivisors: 1 2 3 6 18911 37822 56733 113466" ]

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[ "# 一类三阶半线性中立型时滞微分方程的振动性Oscillation of a Class of Third Order Semi-Linear Neutral Delay Differential Equations\n\n• 全文下载: PDF(463KB)    PP.473-480   DOI: 10.12677/AAM.2019.83053\n• 下载量: 118  浏览量: 177   科研立项经费支持\n\nThe oscillations of a class of neutral third order semi-linear differential equations are studied. Different functions and classical inequalities are constructed by using Riccati transformation techniques. Some new oscillatory theories of this kind of differential equations are established. The conclusions generalize and improve the relevant results in the literature, and illustrate the application of the new oscillatory conclusions with examples.\n\n1. 引言\n\n${\\left[r\\left(t\\right){|{Z}^{″}\\left(t\\right)|}^{\\alpha -1}{Z}^{″}\\left(t\\right)\\right]}^{\\prime }+q\\left(t\\right){|x\\left(\\sigma \\left(t\\right)\\right)|}^{\\beta -1}x\\left(\\sigma \\left(t\\right)\\right)=0,\\text{\\hspace{0.17em}}t\\ge {t}_{0}>0,\\text{\\hspace{0.17em}}\\beta >\\alpha .$ (E)\n\n(A1) $p\\left(t\\right),q\\left(t\\right)\\in C\\left(\\left[{t}_{0},\\infty \\right),\\left(0,\\infty \\right)\\right),0\\le p\\left(t\\right)\\le p<1,q\\left(t\\right)>0$\n\n(A2) $r\\left(t\\right)\\in {C}^{1}\\left(\\left[{t}_{0},\\infty \\right),\\left(0,\\infty \\right)\\right),r\\left(t\\right)\\ge 0,{r}^{\\prime }\\left(t\\right)\\ge 0,{\\int }_{{t}_{\\text{0}}}^{\\infty }{r}^{-\\frac{1}{\\alpha }}\\left(s\\right)\\text{d}s\\le +\\infty$\n\n(A3) $\\tau \\left(t\\right),\\sigma \\left(t\\right)\\in {C}^{1}\\left(\\left[{t}_{0},\\infty \\right),\\left(0,\\infty \\right)\\right)$ ，对每一 $t\\ge {t}_{0}$ ，都有 $\\tau \\left(t\\right)\\le t,\\sigma \\left(t\\right)\\le t$\n\n$\\sigma \\left(t\\right)>0,{\\sigma }^{\\prime }\\left(t\\right)>0,\\underset{t\\to \\infty }{\\mathrm{lim}}\\tau \\left(t\\right)=\\underset{t\\to \\infty }{\\mathrm{lim}}\\sigma \\left(t\\right)=\\infty .$\n\n${\\left(r\\left(t\\right){|{Z}^{\\prime }\\left(t\\right)|}^{\\alpha -1}{Z}^{\\prime }\\left(t\\right)\\right)}^{\\prime }+q\\left(t\\right){|x\\left(\\sigma \\left(t\\right)\\right)|}^{\\beta -1}x\\left(\\sigma \\left(t\\right)\\right)=0.$ (1.1)\n\n${\\int }_{{t}_{0}}^{\\infty }{r}^{-\\frac{1}{\\alpha }}\\left(s\\right)\\text{d}s\\le +\\infty$$\\beta >\\alpha \\text{ },\\beta <\\alpha ,\\beta =\\alpha$ 的考虑 $\\beta >\\alpha$ 。情形新的振动性结论，我们的结果推广和改进了\n\n2. 引理\n\n$\\left(A\\right)\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }Z\\left(t\\right)>0,{Z}^{\\prime }\\left(t\\right)>0,{Z}^{″}\\left(t\\right)>0.$\n\n$\\left(B\\right)\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }Z\\left(t\\right)>0,{Z}^{\\prime }\\left(t\\right)<0,{Z}^{″}\\left(t\\right)>0.$\n\n$u\\left(\\sigma \\left(t\\right)\\right)\\ge \\theta \\frac{\\sigma \\left(t\\right)}{t}u\\left(t\\right),\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }t\\ge {T}_{\\theta }$\n\n$u\\left(t\\right)\\ge \\gamma t{u}^{\\prime }\\left(t\\right)\\text{ },\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }t\\ge {T}_{\\gamma }$\n\n$\\underset{t\\to \\infty }{\\mathrm{lim}}{\\int }_{{t}_{0}}^{t}{\\int }_{u}^{\\infty }{\\left(\\frac{1}{r\\left(v\\right)}{\\int }_{v}^{\\infty }q\\left(s\\right)\\text{d}s\\right)}^{\\frac{1}{\\alpha }}\\text{d}v\\text{d}u=+\\infty .$ (2.1)\n\n$\\underset{t\\to +\\infty }{\\mathrm{lim}}x\\left(t\\right)=0$\n\n3. 主要结果\n\n$D=\\left\\{\\left(t,s\\right)|t\\ge s\\ge {t}_{0}\\right\\}$${D}_{0}=\\left\\{\\left(t,s\\right)|t>s\\ge {t}_{0}\\right\\}$\n\ni)", null, "ii)", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "，T充分大。", null, "(3.1)", null, "为(A)型，即", null, "", null, "", null, "(3.2)", null, "(3.3)", null, "(3.4)\n\n(3.4)式两边对t进行求导，并利用(3.3)，(3.4)得", null, "(3.5)", null, "(3.6)", null, "(3.7)", null, "(3.8)", null, "，由于", null, "单调递减，所以当", null, "时，有", null, "。又因为", null, "，于是有", null, "", null, "(3.9)", null, "", null, "，根据(3.1)式，则", null, "，这与", null, "矛盾，故假设不成立，即", null, "是方程(E)的振动解。", null, "满足(B)型，由于(2.1)式成立，故由引理2.5可得", null, "。证毕。", null, "(3.10)", null, "(3.11)", null, "(3.12)", null, "，使当", null, "时，(3.12)式变成", null, "(3.13)", null, "", null, "，则", null, "，这与", null, "矛盾，故假设不成立，即", null, "是方程(E)的振动解。", null, "满足(B)型，由于(2.1)式成立，故由引理2.5可得", null, "。证毕。", null, "(3.14)", null, "(3.15)", null, "，故有", null, "(3.16)", null, "", null, "满足(B)型，由于(2. 1)式成立，故由引理2.5可得", null, "。证毕。\n\n4. 应用", null, "(E1)", null, "", null, "", null, "", null, "", null, "1) 广东省茂名市科技计划项目(2015038)；2) 广东石油化工学院理学院科研扶持基金重点项目(KY2018001)。\n\nNOTES\n\n*通讯作者。\n\n 李元旦, 高正晖, 邓义华. 三阶半线性中立型微分方程的振动性[J]. 北华大学学报(自然科学版), 2012, 13(3): 267-270. 苏新晓, 戴丽娜, 伍思敏, 林全文. 二阶半线性中立型微分方程的振动性[J]. 应用数学进展, 2017, 6(3): 417-422. Liu, H., Meng, F. and Liu, P. (2012) Oscillation and Asymptotic Analysis on a New Generalized Emden-Fowler Equation. Applied Mathematics and Computation, 219, 2739-2748. https://doi.org/10.1016/j.amc.2012.08.106 曾云辉, 罗李平, 俞元洪. 中立型Emden-Fowler时滞微分方程的振动性[J]. 数学物理学报, 2015, 35A(4): 803-814. 罗李平, 俞元洪, 罗振国. 三阶非线性中立型微分方程的振动分析[J]. 系统科学与数学, 2016, 36(4): 551-559. 林文贤. 三阶半线性中立型阻尼泛函微分方程的振动性[J]. 华东师范大学学报(自然科学版), 2017(3): 48-53. Dzurina, J. and Baculikova, B. (2012) Oscillation of Third-Order Quasi-Linear Advanced Differential Equations. Differential Equations & Applications, 4, 411-421. https://doi.org/10.7153/dea-04-23 李同兴, 韩振来, 张承慧, 等. 时间尺度上三阶Emden-Fowlwe时滞微分方程的振动准则[J]. 数学物理学报, 2012, 32(1): 222-232. 曾云辉, 俞元洪. 三阶半线性时滞微分方程的振动定理[J]. 系统科学与数学, 2014, 34(2): 231-237. Qin, G., Huang, C., Xie, Y., et al. (2013) Asymptotic Behavior for Third-Order Quasi-Linear Differential Equations. Advances in Difference Equations, 2013, 305. https://doi.org/10.1186/1687-1847-2013-305 仉志余, 王晓霞, 俞元洪. 三阶半线性中立型分布时滞微分方程的振动性[J]. 应用数学学报, 2015, 38(3): 450-459. 李全娣, 杨菊, 黎小贤, 林全文. 一类三阶中立型半线性时滞微分方程振动准则[J]. 理论数学, 2017, 7(4): 356-362. 林全文, 俞元洪. 三阶半线性时滞微分方程的振动性和渐进性[J]. 系统科学与数学, 2015, 35(2): 233-244. 惠远先, 王俊杰. 三阶中立型半线性时滞微分方程的振动性[J]. 井冈山大学学报, 2017, 38(1): 8-13. Hardy, G., Litterwood, J. and Polya, G. (1952) Inequalities. 2nd Edition, Cambridge University Press, Cambridge. Erbe, L. (1973) Oscillation Criteria for Second Order Nonlinear Delay Equations. Canadian Mathematical Bulletin, 16, 49-56. https://doi.org/10.4153/CMB-1973-011-1" ]

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[ "# Question #df090\n\nAug 23, 2017\n\nThe average mass of a gallium atom is 69.7 u.\n\n#### Explanation:\n\nThe average mass is a weighted average of the isotopic masses:\n\nYou multiply each atomic mass by its relative importance (percentage).\n\n$\\text{Average mass\" = \"0.60 × 68.93 u + 0.40 × 70.92 u\" = \"41.4 u + 28.4 u\" = \"69.7 u}$\n\nNote: The answer can have only two significant figures, because that is all you gave for the abundance of gallium-69.\n\nHowever, I calculated the result to three significant figures for you." ]

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[ "# PHP array of currency symbols as HTML entities\n\n## Snippet\n\nThree-letter currency code (ISO 4217) => HTML entities:\n\n``````\\$currencySymbols = [\n'AED' => 'د.إ',\n'AFN' => 'Af',\n'ALL' => 'Lek',\n'AMD' => 'դ',\n'ANG' => 'ƒ',\n'AOA' => 'Kz',\n'ARS' => '\\$',\n'AUD' => '\\$',\n'AWG' => 'ƒ',\n'AZN' => '₼',\n'BAM' => 'KM',\n'BBD' => '\\$',\n'BDT' => '৳',\n'BGN' => 'лв',\n'BHD' => '.د.ب',\n'BIF' => 'FBu',\n'BMD' => '\\$',\n'BND' => '\\$',\n'BOB' => '\\$b',\n'BRL' => 'R\\$',\n'BSD' => '\\$',\n'BTN' => 'Nu.',\n'BWP' => 'P',\n'BYR' => 'p.',\n'BZD' => 'BZ\\$',\n'CDF' => 'FC',\n'CHF' => 'CHF',\n'CLF' => 'UF',\n'CLP' => '\\$',\n'CNY' => '¥',\n'COP' => '\\$',\n'CRC' => '₡',\n'CUP' => '⃌',\n'CVE' => '\\$',\n'CZK' => 'Kč',\n'DJF' => 'Fdj',\n'DKK' => 'kr',\n'DOP' => 'RD\\$',\n'DZD' => 'دج',\n'EGP' => 'E£',\n'ETB' => 'Br',\n'EUR' => '€',\n'FJD' => '\\$',\n'FKP' => '£',\n'GBP' => '£',\n'GEL' => 'ლ',\n'GHS' => '¢',\n'GIP' => '£',\n'GMD' => 'D',\n'GNF' => 'FG',\n'GTQ' => 'Q',\n'GYD' => '\\$',\n'HKD' => '\\$',\n'HNL' => 'L',\n'HRK' => 'kn',\n'HTG' => 'G',\n'HUF' => 'Ft',\n'IDR' => 'Rp',\n'ILS' => '₪',\n'INR' => '₹',\n'IQD' => 'ع.د',\n'IRR' => '﷼',\n'ISK' => 'kr',\n'JEP' => '£',\n'JMD' => 'J\\$',\n'JOD' => 'JD',\n'JPY' => '¥',\n'KES' => 'KSh',\n'KGS' => 'лв',\n'KHR' => '៛',\n'KMF' => 'CF',\n'KPW' => '₩',\n'KRW' => '₩',\n'KWD' => 'د.ك',\n'KYD' => '\\$',\n'KZT' => '₸',\n'LAK' => '₭',\n'LBP' => '£',\n'LKR' => '₨',\n'LRD' => '\\$',\n'LSL' => 'L',\n'LTL' => 'Lt',\n'LVL' => 'Ls',\n'LYD' => 'ل.د',\n'MDL' => 'L',\n'MGA' => 'Ar',\n'MKD' => 'ден',\n'MMK' => 'K',\n'MNT' => '₮',\n'MOP' => 'MOP\\$',\n'MRO' => 'UM',\n'MUR' => '₨',\n'MVR' => '.ރ',\n'MWK' => 'MK',\n'MXN' => '\\$',\n'MYR' => 'RM',\n'MZN' => 'MT',\n'NGN' => '₦',\n'NIO' => 'C\\$',\n'NOK' => 'kr',\n'NPR' => '₨',\n'NZD' => '\\$',\n'OMR' => '﷼',\n'PAB' => 'B/.',\n'PEN' => 'S/.',\n'PGK' => 'K',\n'PHP' => '₱',\n'PKR' => '₨',\n'PLN' => 'zł',\n'PYG' => 'Gs',\n'QAR' => '﷼',\n'RON' => 'lei',\n'RSD' => 'Дин.',\n'RUB' => '₽',\n'RWF' => 'ر.س',\n'SAR' => '﷼',\n'SBD' => '\\$',\n'SCR' => '₨',\n'SDG' => '£',\n'SEK' => 'kr',\n'SGD' => '\\$',\n'SHP' => '£',\n'SLL' => 'Le',\n'SOS' => 'S',\n'SRD' => '\\$',\n'STD' => 'Db',\n'SVC' => '\\$',\n'SYP' => '£',\n'SZL' => 'L',\n'THB' => '฿',\n'TJS' => 'TJS',\n'TMT' => 'm',\n'TND' => 'د.ت',\n'TOP' => 'T\\$',\n'TRY' => '₤',\n'TTD' => '\\$',\n'TWD' => 'NT\\$',\n'TZS' => 'TSh',\n'UAH' => '₴',\n'UGX' => 'USh',\n'USD' => '\\$',\n'UYU' => '\\$U',\n'UZS' => 'лв',\n'VEF' => 'Bs',\n'VND' => '₫',\n'VUV' => 'VT',\n'WST' => 'WS\\$',\n'XAF' => 'FCFA',\n'XCD' => '\\$',\n'XDR' => 'SDR',\n'XOF' => 'FCFA',\n'XPF' => 'F',\n'YER' => '﷼',\n'ZAR' => 'R',\n'ZMK' => 'ZK',\n'ZWL' => 'Z\\$',\n];``````" ]

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[ "# Day 2: Conditional Statements: Switch Solution\n\nHello Friends in this article i am gone to share Hackerrank 10 days of javascript solutions with you. | Day 2: Conditional Statements: Switch Solution.\n\nAlso visit this link:  Day 1: Let and Const Solution\n\n## Day 2: Conditional Statements: Switch Solution\n\n### Objective\n\nIn this challenge, we learn about switch statements.\n\nComplete the getLetter(s) function in the editor. It has one parameter: a string, s, consisting of lowercase English alphabetic letters (i.e., a through z). It must return A, B, C, or D depending on the following criteria:\n\n• If the first character in string s is in the set {a, e, i, o, u}, then return A.\n• If the first character in string s is in the set {b, c, d, f, g}, then return B.\n• If the first character in string s is in the set {h, j, k, l, m}, then return C.\n• If the first character in string s is in the set {n, p, q, r, s, t, v, w, x, y, z}, then return D.\n\nHint: You can get the letter at some index i in s using the syntax s[i] or s.charAt(i).\n\n### Function Description\n\nComplete the getLetter function in the editor below.\ngetLetter has the following parameters:\n\n• string s: a string\n\nReturns\n\n• string: a single letter determined as described above\n\n### Input Format\n\nStub code in the editor reads a single string denoting s from stdin.\n\n### Constraints\n\n• 1 <= |s| <= 100, where |s| is the length of s.\n• String s contains lowercase English alphabetic letters (i.e., a through z) only.\n\n`adfgt`\n\n`A`\n\n### Explanation 0\n\nThe first character of string s = adfgt is a. Because the given criteria stipulate that we print A any time the first character is in {a, e, i, o, u}, we return A as our answer.\n\n## Solution – Day 2: Conditional Statements: Switch\n\n```'use strict';\n\nprocess.stdin.resume();\nprocess.stdin.setEncoding('utf-8');\n\nlet inputString = '';\nlet currentLine = 0;\n\nprocess.stdin.on('data', inputStdin => {\ninputString += inputStdin;\n});\n\nprocess.stdin.on('end', _ => {\ninputString = inputString.trim().split('\\n').map(string => {\nreturn string.trim();\n});\n\nmain();\n});\n\nreturn inputString[currentLine++];\n}\n\nfunction getLetter(s) {\nlet letter;\n// Write your code here\nswitch (s) {\ncase ('a' || 'e' || 'o' || 'i' || 'u'):\nletter = 'A';\nbreak;\n\ncase ('b' || 'c' || 'd' || 'f' || 'g'):\nletter = 'B';\nbreak;\n\ncase ('h' || 'j' || 'k' || 'l' || 'm'):\nletter = 'C';\nbreak;\n\ncase ('z' || 'n' || 'p' || 'q' || 'r' || 's' || 't' || 'v' || 'w' || 'x' || 'y'):\nletter = 'D';\n\n}\n\nreturn letter;\n}\n\nfunction main() {" ]

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[ "# Partial Derivitives\n\n## Homework Statement\n\nI need to find the partial derivitives of f(x,y,z)=50-14x+3y^2-2xy (e^z)\n\n## The Attempt at a Solution\n\ni think the x-partial would be -14+2ye^z+2xye^z... but i'm not sure and i'm confused of how to find the others!\n\n## The Attempt at a Solution\n\nrock.freak667\nHomework Helper\nf(x,y,z)=50-14x+3y2-2xyez\n\nsimply when taking the partial derivative w.r.t. to a variable, you treat the rest as constants.\n\nso for ∂f/∂x, we would treat y and z as constants and differentiate normally.\n\n∂f/∂x=∂/∂x(50)-∂/∂x(14x)+∂/∂x(3y2)-∂/∂x(2xyez)" ]

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[ "# Parse DICOM RTSTRUCT (Radiotherapy Structure Set) into binary masks using Numpy\n\n## DICOM-RTSTRUCT Contour Data\n\nContours drawn for radiotherapy are saved as DICOM RT Structure Set (“RT” stands for radiotherapy.) in DICOM Standard, and usually as a single file. You can locate this file among CT or MRI data sets quite reliably, by traversing recursively through the directories and looking for MODALITY of “RTSTRUCT”.\n\nIn this modality, the contours are saved as 2D polygons slice by slice under the Contour Data (3006,0050) tag, which species the data format as:\n\nSequence of (x,y,z) triplets defining a contour in the patient based coordinate system described in Section C.7.6.2.1.1 (mm). See Section C.8.8.6.1 and Section C.8.8.6.3.\n\nIn other words, the perimeter of a sliced contour is specified by a sequence of discrete points.\n\n``````[x0, y0, z0, x1, y1, z1, x2, y2, z2, ... xn, yn, zn]\n``````\n\nThis can be reshaped to (n+1)*3, representing n+1 points along this perimeter.\n\n``````\npixelCoords = np.array([[x0, y0, z0],\n[x1, y1, z1],\n[x2, y2, z2],\n...\n[xn, yn, zn],\n])\n\n``````\n\nFor example, the image below is a CT slice of a prostate phantom overlaid with these discrete points.", null, "## Contours as discrete points\n\nTo generate a 2D representation of these discrete points, you will need to first convert the Contour Data from real word units (mm) to Pixel Coordinates (integers).\n\nI usually first generate an all-false boolean 2D array\n\n``````import numpy as np\narr = np.zeros(shape).astype(bool)\n``````\n\nThen change those Contour Data pixels to True using fancy indexing.\n\n``````cols = pixelCoords[:,0]\nrows = pixelCoords[:,1]\narr[cols, rows] = True # Note the order of indices (cols before rows)\n``````\n\nAnother approach would be using numpy.unravel_index. I describe this method in better details in a separate post.", null, "Binary masks as shown above can be useful for things like pixel-wise semantic segmentation in deep learning. To create these masks, you might attempt to treat these points as a point cloud, and calculate their alpha shape. For concaved shapes, finding the correct parameter to draw the correct envelope can be tricky. In a field that focuses as much on reducing errors and uncertainties as Radiation Oncology, this method is simply not generic/robust enough.\n\nLuckily, the DICOM-RT file exported by treatment planning systems preserve the “connectivity” between the points, that is, they are saved consecutively either clockwise or counter-clockwise.", null, "It turns out that the easiest approach to fill this contour is simply draw a filled polygon and cast that into a boolean 2D array.\n\nYou can do this with OpenCV:\n\n``````import cv2\narr = np.zeros(shape)\npoly = pixelCoords[:,:2]\ncv2.fillPoly(img=arr, pts=[poly], color=1)\n``````\n\nor with Scikit-image:\n\n``````from skimage import draw\nvertex_col_coords = pixelCoords[:, 0]\nvertex_row_coords = pixelCoords[:, 1]\n\nfill_row_coords, fill_col_coords = draw.polygon(vertex_row_coords, vertex_col_coords, shape)", null, "", null, "" ]

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[ "#", null, "Analysis of community ecology data in R\n\nDavid Zelený\n\n### Others\n\nAuthor: David Zelený", null, "en:confusions\n\n# Common confusions and mistakes\n\nA temporary list of common confusions and mistakes when applying numerical methods in community ecology. Based on fixing final reports of students in the class Numerical Methods in Community Ecology.\n\n## Choosing between transformation-based ordination methods vs linear x unimodal methods\n\nWrong: you first use DCA to decide whether to use linear or unimodal ordination methods, and if you decide for linear methods, you Hellinger-transform the data and call it transformation-based ordination.\n\nThis confuses two alternative approaches how to do ordination on community data - read further.\n\nThere are three alternative ways how to analyse community data.\n\n1. Use either linear or unimodal method (PCA vs CA,DCA in case of unconstrained ordination, RDA vs CCA in case of contrained) to analyze the data; the decision between both is done based on heterogeneity of the species composition dataset, and this heterogeneity (beta diversity) can be measured by DCA (the length of the first DCA axis is a measure of heterogeneity; if it is longer than 4, the dataset is considered as heterogeneous and suitable for unimodal methods, if < 3 S.D., it can be considered as homogeneous, suitable for linear methods; 3-4 - in the middle, both linear or unimodal methods are ok). After this decision, use either linear (PCA, RDA) or unimodal (RDA, CCA) method to analyse data, do not use tb-PCA or tb-RDA.\n2. Use transformation-based ordination method. You do not need to make a decision between linear or unimodal methods, simply you opt for transformation-based method. Transform the species composition data by Hellinger transformation (there are also other transformations, but this is the most often used), and analyse data either by unconstrained (tb-PCA) or constrained (tb-RDA) approach, depending on your purpose. No DCA is done before, it is not necessary, since you are not making decision between linear or unimodal ordination method (you use linear method applied on Hellinger transformed data).\n3. Use distance-based methods (PCoA and NMDS for unconstrained, and db-RDA for constrained ordination).\n\nHellinger transformation has therefore two types of use: (a) to calculate transformation-based ordination, or (b) to standardize species composition data:\n\n• Ad (a): linear ordination methods (PCA, RDA) are based on Euclidean distances, which are sensitive to double-zero problem. If the species composition data are first Hellinger transformed, and then used in linear ordination methods, the combination of Hellinger transformation + Euclidean distance means that the distance used by these methods is Hellinger distance, which is not sensitive to double-zero problem. Here, the reason to transform the data is to simply avoid influence of double-zeros in the ordination analysis.\n• Ad b) Hellinger transformation converts species abundances from absolute to relative values (i.e. standardizes the abundances to sample totals) and then square roots them. This could be useful if we are not interested in changes of absolute species abundances, but relative abundances. E.g. if species A has abundance 30 in sample 1 and 20 in sample 2, and species B has abundance 3 in sample 1 and 2 in sample 2, the absolute change is 10 for species A and 1 for species B, but relative change is the same for both. If this is the purpose of this transformation, then Hellinger transformed data could be used also in other ordination methods, e.g. CA or CCA; then, it is already not “transformation-based ordination” in the sense of point (a) above.\n\n## Confusing \"analysis of species richness\" and \"analysis of species composition\"\n\nWrong: you analyse species composition (e.g. using ordination or cluster analysis), but you talk about analysis of species richness.\n\nAnalyzing pattern of species richness (or diversity) and species composition are two different things, although it could be sometimes interesting to analyse both in parallel. If we analyse species richness, we simply ask which community is more and which less diverse, and which factors are the best to interpret this pattern. In contrast, if we analyse pattern in species composition, we ask how does species composition change (in space or time) and which factors are best to explain these changes. Changes in species composition may not necessarily be accompanied by changes in species richness (two communities may have the same number of species, i.e. the same richness, yet completely different species composition).\n\nThere are many general patterns of species richness, e.g. along latitude (diversity decreases from Equator toward poles), along productivity (often reported as hump-back pattern with highest diversity in intermediate levels of productivity) or along pH (for certain organisms, e.g. plants or mollusks, pH is important factor influencing richness - basic habitats are more species rich than acid, although this works only in some geographic regions, like Europe and North America). Often the same factors influence also changes in species composition, but mechanisms behind are likely different.\n\n## Use “envfit” to project explanatory variables into RDA or CCA and to test their significance\n\nWrong: you calculate constrained ordination (e.g. RDA, CCA), and then you use the function ‘’envfit’’ to project explanatory variables onto the ordination diagram (theoretically it can be done, but needs to be done in a special way, see below), or even test the significance of these variables.\n\nSimple suggestion: The function `envfit` is designed to calculate regression of “supplementary” (not “explanatory”) variables on ordination axes of unconstrained ordination, and test the significance of this regression by permutation test. This function is not suitable to project “explanatory” variables onto ordination diagram of constrained ordination (although theoretically it can be done, see below) and definitely not to test the significance of individual variables in this way (see example below showing that in that case all variables will usually be highly significant, because constrained ordination axes already contains their information).\n\nLonger explanation follows below, with examples.\n\nFirst, to clarify. One of the use of the function ‘’envfit’’ (library vegan) is to calculate regression of supplementary variables to ordination axes in unconstrained ordination, in order to help with interpretation of these axes. Say we have community data and three external environmental variables; we calculate unconstrained ordination using the community data, and then we aim to interpret the ecological meaning of unconstrained ordination axes with available environmental variables by calculating their regression. These variables can then be projected onto the ordination diagram as supplementary (not explanatory!), and regression of each variable independently can be tested by permutation test (this is what ‘’envfit’’ returns). But variables which are significant here are not those which are important for species composition, but those which have the best fit to variation extracted by unconstrained ordination into the main (usually the first and the second) ordination axes. Often these variables are important, but not necessarily.\n\nNow, if you calculate constrained ordination, then environmental variables enter the analysis, and resulting (constrained) axes already contain information about them. If now you use ‘’envfit’’ to calculate regression of these same variables to constrained axes, you are likely to get highly significant results; you are regressing two variables (environmental variable vs samples scores on ordination axes) which contain the same information (since axes were calculated also using these environmental variables).\n\nTheoretically, `envfit` can be used to project arrows or centroids with explanatory variables onto diagram of constrained ordination, but the fitting must be done on linear scores of ordination axes, not on sample scores, otherwise the arrows will have different direction then they should be (see example below).\n\nExamples:\n\nIt can be shown that arrows for environmental variables projected by ‘’envfit’’ are different from the true direction of these variables in ordination diagram, unless they are calculated properly (on linear combination of scores). In the diagram below (based on river valley data), the three variables (pH, SOILDPT and ELEVATION) have been used as explanatory variables in tb-RDA. The left diagram shows the true direction of explanatory variables. The right diagram shows three options of using `envfit` function. Blue arrows show the true direction (`envfit` calculated with argument `display = 'lc' `, i.e. on linear combination of scores). Red arrows indicate direction if they are added using ‘’envfit’’ fitted to samples scores (by default only the first and second ordination axis), and green arrows show the same with ‘’envfit’’ relating them to three instead of two axes.\n\n```vltava.spe <- read.delim ('https://raw.githubusercontent.com/zdealveindy/anadat-r/master/data/vltava-spe.txt', row.names = 1)\n\nlibrary (vegan)\n\nspe.hell <- decostand (log1p (vltava.spe), method = 'hellinger')\ntbRDA <- rda (spe.hell ~ pH + SOILDPT + ELEVATION, data = vltava.env[,1:18])\n\npar (mfrow = c(1,2))\nordiplot (tbRDA)\nef <- envfit (tbRDA ~ pH + SOILDPT + ELEVATION, data = vltava.env, display = 'lc')\nef12 <- envfit (tbRDA ~ pH + SOILDPT + ELEVATION, data = vltava.env)\nef123 <- envfit (tbRDA ~ pH + SOILDPT + ELEVATION, data = vltava.env, choices = 1:3)\nordiplot (tbRDA, type = 'n')\nplot (ef, col = 'blue')\nplot (ef12, col = 'red')\nplot (ef123, col = 'green')\nlegend ('bottomright', lwd = 1, col = c('blue', 'red', 'green'), legend = c('envfit on linear combination of scores', 'envfit on sample scores (1st & 2nd axis)', 'envfit on sample scores (1st, 2nd and 3rd axis)'), bty = 'n')```\n\nNow about testing the significance (to show why it should not be done using the function `envfit`). In the example below, I again used the river valley dataset. I generated one random variable, i.e. I assign a random value to each site, and use it as “environmental variable” (obviously without any ecological meaning and without any relationship to the species composition). I calculated tb-PCA, and used envfit to fit the random variable to ordination axes (result not significant, as expected); then I calculated tb-RDA with this random variables as explanatory and tested significance of this RDA (not significant, also as expected). Then I use ‘’envfit’’ to fit the random variable to ordination axes (the result is highly significant, which could indicate that the random variable is important, although it is not). The result shows that to use `envfit` to fit environmental variables to RDA which is based on the same environmental variable is highly misleading.\n\n```spe.hell <- decostand (log1p (vltava.spe), 'hell')\nset.seed (123)\nrandom.var <- runif (97) # generate single random variable\ntbPCA <- rda (spe.hell)\nenvfit (tbPCA ~ random.var) # not significant\n# ***VECTORS\n#\n# PC1 PC2 r2 Pr(>r)\n# random.var 0.41445 0.91007 0.0072 0.716\n# Permutation: free\n# Number of permutations: 999\n\ntbRDA.rand <- rda (spe.hell ~ random.var)\nanova (tbRDA.rand) # the RDA is not significant (since the explanatory variable is random and has no meaning)\n# Permutation test for rda under reduced model\n# Permutation: free\n# Number of permutations: 999\n#\n# Model: rda(formula = spe.hell ~ random.var)\n# Df Variance F Pr(>F)\n# Model 1 0.00724 0.9861 0.472\n# Residual 95 0.69752\n\nenvfit (tbRDA.rand ~ random.var) # the fit of random variables to first two ordination axes is significant, since the first axis is calculated from it\n# ***VECTORS\n#\n# RDA1 PC1 r2 Pr(>r)\n# random.var 0.92615 -0.37716 0.4525 0.001 ***\n# ---\n# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n# Permutation: free\n# Number of permutations: 999```\n\n## Check for missing values in environmental variables before conducting constrained ordination (RDA, CCA)\n\nProblem: if the environmental variables contain missing values and such matrix is used in constrained ordination as explanatory, the samples containing (even just one) missing values will be removed from analysis. If you have many environmental variables and each has some missing values (in the worst case each variable missing for different samples), such data with many holes may mean that the final analysis is based on rather reduced number of samples.\n\nSolution: function `na.omit` applied on environmental matrix will remove all samples with one or more missing values. The same samples need to be removed also from species data. Make sure that you know on how many samples your analysis is based! Also, avoid replacing missing values by zero; this could make sense in case of species data (missing species has zero abundance), but usually does not make sense in case of environmental variables (if soil depth was not measured and has missing values, replacing it by zero would mean that the vegetation grows on a rock, which is not true; the same for e.g. soil chemical concentrations - not measured does not mean not present).\n\n## Doing constrained ordination analysis with all explanatory variables, but using only those selected by forward selection to display in ordination diagram\n\nMistake is if you calculate constrained ordination (RDA, CCA, tb-RDA) with all explanatory variables, then do forward selection to select only those significant, and then you create ordination diagram based on analysis with all explanatory variables, but in this ordination diagram you display only those explanatory variables which were significant in forward selection.\n\nCorrect way would be to display ordination diagram which is based on constrained ordination using only the selected (by forward selection) environmental variables as explanatory variables. Indeed, the ordination diagram based on all environmental variables, and the ordination diagram based on only a small subset of selected variables will look different, and different number of ordination axes will be constrained (since number of constrained axes = number of env. variables, if they are quantitative). You should, however, still conduct constrained ordination with all environmental variables before you do forward selection (so called “global test”, and only if it is significant, you can proceed to forward selection itself); there is, however, perhaps no need to draw ordination diagram with all explanatory variables (unless there is a clear interpretational need to do that).\n\n## Significance of constrained ordination is not tested by ANOVA\n\nProblem: statement in the Methods that “the significance of the constrained ordination was tested by ANOVA”.\n\nWhy: Variance explained by constrained ordination is tested by a permutation test, in which the real explained variation is compared with the variation explained by randomised explanatory variables. This kind of test is generally called Monte Carlo permutation test (thus, Monte Carlo test is not only specific for constrained ordination, the same family of tests could be used to test e.g. correlation of two variables). If the observed explained variation is higher than some proportion (usually 95%) of variation explained by randomised variables, it is considered as high enough to be significant (at P < 0.05). It gets a bit more complicated; the test statistic, in fact, is not the explained variation (R2), but so-called pseudo-F value, which consists of explained variation and number of degrees of freedom (which reflects number of explanatory variables as well as covariables). The test then resembles ANOVA in a way that ANOVA (analysis of variance) is also based on F-value. In `vegan`, the function conducting Monte Carlo permutation test is called `anova`, for mostly historical reason; the description of the function (see `?anova.cca`) states that it is “ANOVA-like permutation test for Constrained Correspondence Analysis (cca), Redundancy Analysis (rda) or distance-based Redundancy Analysis (dbRDA)”. The term ANOVA is traditionally understand as an analysis of variance with a single dependent variable, so nothing related to constrained ordination.\n\nConclusion: You can mention that the constrained ordination was tested by permutation test, Monte Carlo permutation test, or ANOVA-like permutation test (the last one is perhaps the least common). But do not say simply that the significance of constrained ordination was tested by ANOVA, this would be misleading.\n\n## In cluster analysis, Ward method cannot be combined with Bray-Curtis distances (unless square-rooted)\n\nWard's algorithm is based on measuring the distance of the sample to the centroid of the group of samples in orthogonal Euclidean space. This means that distances which are not Euclidean should not be directly used with this algorithm unless modified (“being Euclidean” here means that the distance obeys triangular inequality principle and can be displayed in orthogonal Euclidean space; e.g. Manhattan distance is not Euclidean, but the square root of Manhattan distance is Euclidean). Strictly speaking, Ward's clustering algorithm should be used only with Euclidean distances, and since Bray-Curtis distance is not Euclidean, it should square-rooted first before used with Ward's algorithm.\n\n## Using the first unconstrained axis in constrained ordination to see how much variation can be maximally explained by a single explanatory variable\n\nProblem: You want to know how much variation you can maximally explain by a single explanatory variable in constrained ordination (RDA, tb-RDA, CCA), and for this, you will check the variation represented by the first unconstrained ordination axis of this constrained ordination (with the environmental variable used as explanatory).\n\nExplanation: If you want to know how much variation you can maximally explain by a single explanatory variable, you follow this logic: the ordination axis of unconstrained ordination (e.g. PCA1) represents the direction of the strongest compositional gradient, and can be (theoretically) considered as a perfect explanatory variable for the dataset from which it was calculated. You can use it as explanatory in the constrained variant of the same ordination method (e.g. RDA) to see how much variation it explains; the same value you get if you simply check the variation represented by this axis in the original unconstrained ordination (the eigenvalue of the axis divided by total inertia/variance, Fig. 1, the horizontal axis in the right panel). This, however, does not apply to the first unconstrained ordination axis of the constrained ordination (e.g. the PC1 in RDA, Fig. 1, the vertical axis in the left panel), in which some environmental variable (or variables) was used as explanatory. Such axis represents the maximum variance a single explanatory variable can explain in the dataset after the variation of the explanatory variable has been removed.", null, "" ]

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[ "Author: Oscar Cronquist Article last updated on October 30, 2018\n\nThis is follow up post to: Tracking a stock portfolio in excel (auto update)\nIn this post we are going to calculate cost basis and returns. The calculations are simplified, commissions, stock splits and dividends are removed from calculations.\n\nIn the first post we created dynamic ranges.", null, "We also identified accumulated stocks and the number of shares. I have now added cost basis and returns.", null, "Excel formula in cell C2:\n\nCopy cell C2 and paste down as far as needed.\n\nHow this formula works in cell C2\n\nStep 1 - Calculate total cost you paid\n\nSUMPRODUCT(array1, array2, )\nReturns the sum of the products of the corresponding ranges or arrays\n\nbecomes\n\nbecomes\n\nSUMPRODUCT(({TRUE, FALSE, FALSE, TRUE, FALSE, FALSE, FALSE, FALSE, FALSE}))*({TRUE, TRUE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, FALSE})*{100, 100, 50, 100, 25, 50, 10, 100, 200}*{12, 25, 28, 16, 550, 24, 500, 500, 550})\n\nbecomes\n\nSUMPRODUCT({1200, 0, 0, 1600, 0, 0, 0, 0, 0}) returns 2800.\n\nStep 2 - Calculate total amount you sold\n\nSUMPRODUCT((A2=Symbol)*(Type=\"Sell\")*Shares*Price)\n\nbecomes\n\nSUMPRODUCT({0, 0, 0, 0, 0, 0, 0, 0, 0}) returns 0.\n\nStep 3 - Calculate accumulated shares\n\nbecomes\n\nSUMPRODUCT({100, 0, 0, 100, 0, 0, 0, 0, 0}) - SUMPRODUCT({0, 0, 0, 0, 0, 0, 0, 0, 0})\n\nand returns 200.\n\nStep 4 - Calculate cost basis\n\nbecomes\n\n=IF(A2<>\"\", (2800-0)/200, \"\") returns 14 in cell C2.", null, "Excel formula in cell D2\n\n=IF(A2<>\"\", (IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), 0)-IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares), 0))*(MIN(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares))), \"\")\n\nCopy cell D2 and paste down as far as needed.\n\nHow this formula works in cell D2\n\nStep 1 - Calculate average price you paid per share\n\n=IF(A2<>\"\", (IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), 0)-IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares), 0))*(MIN(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares))), \"\")\n\n(IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), 0)\n\nreturns 0.\n\nStep 2 - Calculate average selling price\n\n=IF(A2<>\"\", (IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), 0)-IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares), 0))*(MIN(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares))), \"\")\n\nreturns 14.\n\nStep 3 - Multiply with bought or sold shares\n\n=IF(A2<>\"\", (IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), 0)-IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares), 0))*(MIN(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares))), \"\")\n\nMIN(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares))\n\nreturns 0.\n\nStep 4 - Calculate returns\n\n=IF(A2<>\"\", (IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), 0)-IFERROR(SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Price, Shares)/SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares), 0))*(MIN(SUMPRODUCT(--(A2=Symbol), --(\"Sell\"=Type), Shares), SUMPRODUCT(--(A2=Symbol), --(\"Buy\"=Type), Shares))), \"\")\n\nbecomes\n\n=IF(A2<>\"\", (0-14)*0) returns 0\n\nFinal notes\n\nI am not telling you to sell or buy any stock, this is just an example.\n\nI hope I got all calculations right.\n\nGet excel file\n\nTracking-a-stock-portfolio2.xlsx\n(Excel 2007 -2010 Workbook *.xlsx)" ]

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[ "# Perimeter of a figure\n\nGeometry\n\nA perimeter is the distance around the outside of a two dimensional figure. It is found by taking the sum of the side lengths. If the side lengths of a quadrilateral are A, B, C and D, Perimeter = A + B + C + D", null, "", null, "" ]

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