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将 latex 术语方程式转换为适合 wolfram alpha 的版本

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将 latex 术语方程式转换为适合 wolfram alpha 的版本

我在数学堆栈交换上问了一个组合数学问题，得到了以下答复LaTeX 风格的答案。我尝试将该方程式转换为 wolfram alpha 可以理解的内容；但是，似乎行不通。关于如何编辑方程式的语法，您有什么想法吗？我已在下面发布了所使用的语法。

\[
n!+\sum_{i=1}^{n-1}(-1)^i\sum_{k=1}^i\binom{i-1}{i-k}\binom{n-i}{k}2^k(n-i)!
\]

答案1

目录：

  • 初始答案使用xintexpr和 OP 的公式。实现是可扩展的……

  • 使用递归公式给出的第二个答案http://oeis.org/A002464。第二种方法一次性构造给定表达式的值数组。它使用包数值表达式用于大整数计算。

您可以使用 TeX 或 LaTeX 来计算。

代码注释中详细说明了一些有点痛苦的细节：主要的是，如果 y>x，二项式（x，y）函数当前会发出错误，而不是默默返回零。

\documentclass{article}

\usepackage{xintexpr}[2016/03/12]% 1.2f or more recent needed for binomial

\begin{document}

% unfortunately the default binomial(x,y) function raises an error
% if y>x. Hence define wrapper to intercept the case.
\xintdefiifunc bbinomial(x,y):=if(y>x,0,binomial(x,y));

% unfortunately we can not use \xintdefiifunc F(n):=....; syntax because the
% function variable "n" appears in the summation range. This is current
% limitation, documented in §10.10.3 of xint.pdf.

% Hence we use a macro interface, which will need braces: \myF{10}.

% We employ parentheses around #1 in case
% it is itself some math expression.

% we choose \xintiiexpr rather than \xinttheiiexpr, for efficiency
% if used in other expressions.

\newcommand\myF [1]{% n = #1
   \xintiiexpr (#1)!+
   add((-1)^i % probably faster: ifodd(i, -1, +1)
       *add(binomial(i-1,i-k)*bbinomial((#1)-i,k)*2^k, 
       k=1..i)*((#1)-i)!, 
   i=1..(#1)-1)\relax }

% unfortunately in xintiiexpr, currently 1..0 does not evaluate
% to empty range, but proceeds by negative steps, hence evaluate to
% 1, 0 which we don't want.

% 1..[1]..0 would create such an empty range.
% but further problem is that add(<expression>, i = <range>) syntax
% currently does not work with range being empty.

% Consequently the above expression requires n > 1.

% test
% \xinttheiiexpr seq(\myF{n}, n=2..10)\relax

\medskip
\begin{tabular}{c|r}
  $n$&$F(n)$\\
\hline
\xintFor* #1 in {\xintSeq{2}{20}}
\do
{$#1$&$\xintthe\myF{#1}$\\}
\end{tabular}

\end{document}

在此处输入图片描述

更现实的方法是一次性定义扩展到组合值的宏。基于递归公式。

\documentclass{article}

\usepackage{bnumexpr}

% http://oeis.org/A002464

% If n = 0 or 1 then a(n) = 1; if n = 2 or 3 then a(n) = 0; otherwise a(n) =
% (n+1)*a(n-1) - (n-2)*a(n-2) - (n-5)*a(n-3) + (n-3)*a(n-4)

% 1, 1, 0, 0, 2, 14, 90, 646, 5242, 47622, 479306, 5296790, 63779034

\begin{document}

\begingroup
\makeatletter

\@namedef{F<0>}{1}
\@namedef{F<1>}{1}
\@namedef{F<2>}{0}
\@namedef{F<3>}{0}

\count@=4

\loop
% no \@nameedef in latex
  \expandafter\edef\csname F<\the\count@>\endcsname
       {\thebnumexpr (\count@+1)*\@nameuse{F<\the\numexpr\count@-1>}
                    -(\count@-2)*\@nameuse{F<\the\numexpr\count@-2>}
                    -(\count@-5)*\@nameuse{F<\the\numexpr\count@-3>}
                    +(\count@-3)*\@nameuse{F<\the\numexpr\count@-4>}\relax}%
\ifnum\count@<30
\advance\count@\@ne
\repeat

\count@=0

\ttfamily

http://oeis.org/A002464

\loop
\the\count@: \@nameuse{F<\the\count@>}\endgraf
\ifnum\count@<30
\advance\count@\@ne
\repeat

\endgroup

\end{document}

在此处输入图片描述

答案2

F(n)这是基于LuaLaTeX 的计算方法n。相对于您提供的版本，公式略有改写，以 (a) 添加括号和方括号以提供一些视觉分组和 (b) 添加一些解释性文字以突出显示外部（“i”）和内部（“k”） for 循环。

该表以n=2since开头F(1)=1。(F(1) 是顺便说一下，计算正确。）对于表，我设置n_{\max}为 15；事实上，只要 ，计算方法就不会产生溢出n<20。

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath} % for "\binom" macro
\usepackage{luacode} % for "luacode" env. and "\luaexec" macro
\begin{luacode}
-- First, define two helper functions: "factorial" and "mchoose"
function factorial ( n )
   local k
   if n==0 or n==1 then
      return 1
   else
      return n * factorial(n-1)
   end
end

-- 'mchoose' is patterned after the posting in http://stackoverflow.com/a/15302448. 
-- Thanks, @egreg, for pointing me to this posting!
function mchoose( n, k )
     if ( k == 0 or k == n ) then
        return 1
     else
        return ( n * mchoose(n - 1, k - 1)) / k 
     end
end

-- Second, set up the function "F"
function F ( n )
   local i, k, result, kterm
   result = factorial ( n ) 
   for i=1,n-1 do           -- outer loop is entered only if n>1
      kterm=0  -- (re)set "kterm" to 0
      for k=1,i do
         kterm = kterm + mchoose(i-1,i-k) * mchoose(n-i,k) * 2^k 
      end
      result = result + ((-1)^i) * factorial(n-i) * kterm
   end
   return result
end

\end{luacode}

\begin{document}
\[
F(n)=n!+\biggl\{\,
     \underbrace{ \sum_{i=1}^{n-1} (-1)^i (n-i)! \biggl[\,
         \underbrace{\sum_{k=1}^i \binom{i-1}{i-k} \binom{n-i}{k} 2^k}_%
         {\text{inner or $k$ loop}}\biggr]}_%
     {\text{outer or $i$ loop}}\biggr\}
\]

\bigskip
% print values of n and F(n) for n=2,...,15
\[
\begin{array}{@{}rr@{}}
\hline
n & F(n) \\
\hline
\luaexec{for n=2,15,1 do
           tex.sprint(n .. "&" .. math.floor(F(n)) .. "\\\\")
         end}
\hline
\end{array}
\]
\end{document}

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