mathematica

Mathematica 讲座笔记本 [ 下载 ] 第一章 Mathematica 简介 [ 观看 ] [ 下载 ] 第二章 Mathematica 界面和编程语言 [ 观看 ] [ 下载 ] 第三章 符号运算 [ 观看 ] [ 下载 ] 第四章 数值运算 [ 观看 ] [ 下载 ] 第五章 图形运算 [ 观看 ] [ 下载 ] 第六章 案例分析一 [ 观看 ] [ 下载 ] 第七章 案例分析二 [ 观看 ] [ 下载 ] 第八章 案例分析三 [ 观看 ] [ 下载 ] 第九章 与其它程序的交流 [ 观看 ] [ 下载 ] 第十章 Mathematica 讲座总结 [ 观看 ] [ 下载 ] 来源： https://www.cnblogs.com/liangliangdetianxia/p/4042135.html

【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 我上次面试时遇到的一个问题： 设计一个函数 f ，使得： f(f(n)) == -n 其中 n 是一个32位有 符号整数 ; 您不能使用复数算法。 如果您不能为整个数字范围设计这样的函数，请为最大范围设计它。 有任何想法吗？ #1楼 x86 asm（AT＆T风格）： ; input %edi ; output %eax ; clobbered regs: %ecx, %edx f: testl %edi, %edi je .zero movl %edi, %eax movl $1, %ecx movl %edi, %edx andl $1, %eax addl %eax, %eax subl %eax, %ecx xorl %eax, %eax testl %edi, %edi setg %al shrl $31, %edx subl %edx, %eax imull %ecx, %eax subl %eax, %edi movl %edi, %eax imull %ecx, %eax .zero: xorl %eax, %eax ret 检查代码，传递所有可能的32位整数，错误-2147483647（下溢）。 #2楼 该Perl解决方案 适用于整数，浮点数和字符串 。 sub f { my $n =

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: How do you set up manipulate so that you can control a variable with a vertical slider instead of a horizontal slider in Mathematica? 回答1: From the help .... Manipulate[u, {u, 0, 1, ImageSize -> Small}, ControlType -> VerticalSlider, ControlPlacement -> Left] 文章来源: Putting a VerticalSlider into Mathematica's Manipulate?

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: Can I set the default value for a function argument to be something that's not constant? Example: tod := Mod [ AbsoluteTime [], 86400 ] f [ x_ : tod ] := x In the above, 'tod' changes every time I evaluate it, but "f[]" does not. "?f" yields: f [ x_ : 42054.435657 `11.376386798562935] := x showing the default value was hardcoded when I created the function. Is there a workaround here? 回答1: It seems to work if the function holds its arguments: tod := Mod [ AbsoluteTime [], 86400 ] SetAttributes [ f , HoldAll ]; f [ x_ : tod ] := x

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I find this a bit annoying. I make a 3D plot, initially it come up in the default orientation. Then, using the mouse, I rotate it some way. Now I run the command again, expecting to obtain the original shape (i.e the original orientation before I rotated it by mouse), but instead, it just gives me the same plot I have on the screen, i.e. it seems to have kept/remembered the last viewpoint in that output cell. I wanted it to go back to the original viewpoint. So, I delete the output cell to make it happen. Do you think this is how

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Does anybody know if there is a built-in function in Mathematica for getting the lhs of downvalue rules (without any holding)? I know how to write the code to do it, but it seems basic enough for a built-in For example: a[1]=2; a[2]=3; BuiltInIDoNotKnowOf[a] returns {1,2} 回答1: This seems to work; not sure how useful it is, though: a[1] = 2 a[2] = 3 a[3] = 5 a[6] = 8 Part[DownValues[a], All, 1, 1, 1] 回答2: This is like keys() in Perl and Python and other languages that have built in support for hashes (aka dictionaries). As your example

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: At mathkb.com, I found a interesting post "Another review of Mathematica's debugger" (by berniethejet) talking about debugging in wolfram workbench. http://www.mathkb.com/Uwe/Threads/List.aspx/mathematica/20986 I think this is a good question worth discussing and I would like hear some experiences of using workbench, even though I've never touched workbench. Is workbench a real debugger but a watcher? what's its advantage over mathematica? How do you debug when you writting big or small codes? mabye workbench is for debugging small codes and

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: How do you plot legends for functions without using the PlotLegends package? 回答1: I, too, was disappointed by the difficulty of getting PlotLegend to work correctly. I wrote my own brief function to make my own custom figure legends: makePlotLegend[names_, markers_, origin_, markerSize_, fontSize_, font_] := Join @@ Table[{ Text[ Style[names[[i]], FontSize -> fontSize, font], Offset[ {1.5*markerSize, -(i - 0.5) * Max[markerSize,fontSize] * 1.25}, Scaled[origin] ], {-1, 0} ], Inset[ Show[markers[[i]], ImageSize -> markerSize], Offset[ {0.5

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I've been working through problems on Project Euler , and some of the solutions that other people have posted use a triple-at-sign, i.e. '@@@'. In the help browser for v7, I find an entry for @@ (which says it's the infix version of 'Apply') but none for @@@. What does it mean? EDIT: Here's an example, which I think I can post without violating the spirit of Project Euler: bloc[n_, f_][t_] := {f @@@ #, #~Tr~f} & /@ Join @@ Partition[t, {n, n}, 1]; 回答1: As others have noted, @@@ is, technically, shorthand for Apply with an optional third

mathematica学习笔记 代数运算 Expand[]函数 Factor[]用于因式分解 Together[]用于合并分式 Apart[] Collect[]合并指定符号项 Simplify[]和FullSimplify[] 方程求解 Solve[] Reduce[] FindRoot[] 代数运算 mathematica遇到没有预先定义的字符串会将其处理为变量，比如 软件自动将表达式进行了化简处理 Expand[]函数 Expand[]函数用于展开多项式： Factor[]用于因式分解 Factor[]与Expand[]恰好相反 Together[]用于合并分式 Apart[] Collect[]合并指定符号项 Simplify[]和FullSimplify[] 当Simplify[]不能满足要求时，可使用FullSimplify进行完全化简，FullSimplify将会尝试更多的变换形式，通常 首先使用Simplify，如果得到的结果不满意再考虑使用FullSimplify 方程求解 Solve[] Solve[] 需要两个参数，第一个参数是方程组，第二个参数是自变量： 1.当输入多个方程和多个变量是需要用｛｝构成列表 2.此时需要使用两个等于号 Reduce[] Solve非常擅长解方程，但有时候需要特定的指令来求解。 此时已经出现提示：可能没有给出所有解