没有数字的得数

王守恩

王守恩 发表于 2019-1-17 16:21
对 9 楼的公式作简化。

我们的目标很明确：

左边用电脑也不好算，试试用10楼的算式。

     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{1}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+0)!}=\frac{1441729e}{40320}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+1)!}=\frac{4596553e}{362880}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}\cdot n_{2}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+2)!}=\frac{1441729e}{403200}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}\cdot n_{2}\cdot n_{3}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+3)!}=\frac{32989969e}{39916800}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}\cdot n_{2}\cdot n_{3}\cdot n_{4}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+4)!}=\frac{76751233e}{479001600}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}\cdot n_{2}\cdot n_{3}\cdot n_{4}\cdot n_{5}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+5)!}=\frac{2628983e}{98841600}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}\cdot n_{2}\cdot n_{3}\cdot n_{4}\cdot n_{5}\cdot n_{6}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+6)!}=\frac{335262313e}{87178291200}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}\cdot n_{2}\cdot n_{3}\cdot n_{4}\cdot n_{5}\cdot n_{6}\cdot n_{7}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+7)!}=\frac{642451441e}{1307674368000}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}\cdot n_{2}\cdot n_{3}\cdot n_{4}\cdot n_{5}\cdot n_{6}\cdot n_{7}\cdot n_{8}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+8)!}=\frac{130469561e}{2324754432000}\)
     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\sum_{n_{9}=0}^{\infty}\frac{n_{1}\cdot n_{2}\cdot n_{3}\cdot n_{4}\cdot n_{5}\cdot n_{6}\cdot n_{7}\cdot n_{8}\cdot n_{9}}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8}+n_{9})!}=e\sum_{m=0}^{8}\frac{8!}{m!(8-m)!(m+9)!}=\frac{158433901e}{27360571392000}\)

王守恩

王守恩 发表于 2019-1-19 08:04
左边用电脑也不好算，试试用10楼的算式。

     \(\D\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\inf ...

    求助：这有限怎么就替代了无限，能点拨一下吗？

A,B=0, 1, 2, 3, 4, 5,......


\[\D\frac{\D\sum_{m=0}^{A}\D \frac{A!}{m!\ (A-m)!\ (m+B)!}}{\D\sum_{n=0}^{\infty}\ \ \frac{(n+A)!}{n!\ (A+B)!\ \ (n-B)!}}=\D\frac{1}{\D\ e\ }\]

王守恩

mathematica 发表于 2018-12-30 14:13
这样的问题不要发论坛上了，没人会证的，
真的太难了，又是拉马努金一样的等式，
看到就头疼

右边用电脑也不好算，试试用左边的算式。

\(\D\sum_{m=0}^{0}\frac{0!× e}{m!\ (0-m)!\ (0-m)!\ }=\sum_{n_{1}=0}^{\infty}\frac{1}{n_{1}!\ }\)
\(\D\sum_{m=0}^{1}\frac{1!× e}{m!\ (1-m)!\ (1-m)!\ }=\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\frac{1}{(n_{1}+n_{2})!\ }\)
\(\D\sum_{m=0}^{2}\frac{2!× e}{m!\ (2-m)!\ (2-m)!\ }=\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\frac{1}{(n_{1}+n_{2}+n_{3})!\ }\)
\(\D\sum_{m=0}^{3}\frac{3!× e}{m!\ (3-m)!\ (3-m)!\ }=\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\frac{1}{(n_{1}+n_{2}+n_{3}+n_{4})!\ }\)
\(\D\sum_{m=0}^{4}\frac{4!× e}{m!\ (4-m)!\ (4-m)!\ }=\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\frac{1}{(n_{1}+n_{2}+n_{3}+n_{4}++n_{5})!\ }\)
\(\D\sum_{m=0}^{5}\frac{5!× e}{m!\ (5-m)!\ (5-m)!\ }=\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\frac{1}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6})!\ }\)
\(\D\sum_{m=0}^{6}\frac{6!× e}{m!\ (6-m)!\ (6-m)!\ }=\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\frac{1}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7})!\ }\)
\(\D\sum_{m=0}^{7}\frac{7!× e}{m!\ (7-m)!\ (7-m)!\ }=\sum_{n_{1}=0}^{\infty}\sum_{n_{2}=0}^{\infty}\sum_{n_{3}=0}^{\infty}\sum_{n_{4}=0}^{\infty}\sum_{n_{5}=0}^{\infty}\sum_{n_{6}=0}^{\infty}\sum_{n_{7}=0}^{\infty}\sum_{n_{8}=0}^{\infty}\frac{1}{(n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}+n_{8})!\ }\)


   
   
   
   
   

   

王守恩

右边很好算，左边用电脑也不好算。
2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226,
257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785,
842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601,
1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501, .................

\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-0)!\times e}=1^2+1=2\)
\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-1)!\times e}=2^2+1=5\)
\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-2)!\times e}=3^2+1=10\)
\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-3)!\times e}=4^2+1=17\)
\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-4)!\times e}=5^2+1=26\)
\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-5)!\times e}=6^2+1=37\)
\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-6)!\times e}=7^2+1=50\)
\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-7)!\times e}=8^2+1=65\)
\(\D\sum_{k=0}^{\infty}\frac{k^2}{(k-8)!\times e}=9^2+1=82\)
         

王守恩

漂亮的公式，活跃的气氛。

\[\D\sum_{n=0}^{\infty}\frac{\pi^2}{n^2+\pi^2}=\frac{e^{\pi*\pi}+\pi*\pi*e^{\pi*\pi}+\pi*\pi*e^{-\pi*\pi}-e^{-\pi*\pi}}{e^{\pi*\pi}+e^{\pi*\pi}-e^{-\pi*\pi}-e^{-\pi*\pi}}\]

\[\D\sum_{n=1}^{\infty}\frac{\pi^2}{n^2+\pi^2}=\frac{e^{\pi*\pi}-\pi*\pi*e^{\pi*\pi}-\pi*\pi*e^{-\pi*\pi}-e^{-\pi*\pi}}{e^{-\pi*\pi}-e^{\pi*\pi}-e^{\pi*\pi}+e^{-\pi*\pi}}\]

王守恩

王守恩 发表于 2019-2-12 10:07
漂亮的公式，活跃的气氛。

\[\D\sum_{n=0}^{\infty}\frac{\pi^2}{n^2+\pi^2}=\frac{e^{\pi*\pi}+\pi*\pi ...

\[\D\prod_{n=2}^{\infty}\frac{n^2-1}{n^2+1}=\frac{\pi+\pi}{e^{\pi}-e^{-\pi}}\]

\[\D\prod_{n=3}^{\infty}\frac{n^2-1}{n^2+1}=\frac{\pi+\pi+\pi+\pi+\pi+\pi+\pi+\pi+\pi+\pi}{e^{\pi}-e^{-\pi}+e^{\pi}-e^{-\pi}+e^{\pi}-e^{-\pi}}\]

dlpg070

[quote]王守恩 发表于 2019-3-9 21:00
\(\prod _{n=m}^{\infty } \frac{n^2-1}{n^2+1}\)


m=2 ,3,---,10化简:
\(\left\{\left(
\begin{array}{c}
\pi  \text{csch}(\pi ) \\
\frac{5}{3} \pi  \text{csch}(\pi ) \\
\frac{25}{12} \pi  \text{csch}(\pi ) \\
\frac{85}{36} \pi  \text{csch}(\pi ) \\
\frac{1105}{432} \pi  \text{csch}(\pi ) \\
\frac{8177 \pi  \text{csch}(\pi )}{3024} \\
\frac{204425 \pi  \text{csch}(\pi )}{72576} \\
\frac{13287625 \pi  \text{csch}(\pi )}{4572288} \\
\frac{108958525 \pi  \text{csch}(\pi )}{36578304} \\
\end{array}
\right)\right.\)

王守恩

1，十个数码：0,1,2,3,4,5,6,7,8,9可以有无穷变化！
2，交换分子分母位置，得到的就是分母是 “1” 的数。

      \(\D\prod_{n=2}^{\infty}\frac{(n-1)^1\ (n+1)}{(n-0)^2}=\frac{1}{2!}\)
      \(\D\prod_{n=3}^{\infty}\frac{(n-2)^2\ (n+1)}{(n-1)^3}=\frac{1}{3!}\)
      \(\D\prod_{n=3}^{\infty}\frac{(n-2)^3\ (n+2)}{(n-1)^4}=\frac{1}{4!}\)
      \(\D\prod_{n=4}^{\infty}\frac{(n-3)^4\ (n+2)}{(n-2)^5}=\frac{1}{5!}\)
      \(\D\prod_{n=4}^{\infty}\frac{(n-3)^5\ (n+3)}{(n-2)^6}=\frac{1}{6!}\)
      \(\D\prod_{n=5}^{\infty}\frac{(n-4)^6\ (n+3)}{(n-3)^7}=\frac{1}{7!}\)
      \(\D\prod_{n=5}^{\infty}\frac{(n-4)^7\ (n+4)}{(n-3)^8}=\frac{1}{8!}\)
      \(\D\prod_{n=6}^{\infty}\frac{(n-5)^8\ (n+4)}{(n-4)^9}=\frac{1}{9!}\)

dlpg070

王守恩 发表于 2019-2-12 10:07
漂亮的公式，活跃的气氛。


\[\D\sum_{n=0}^{\infty}\frac{\pi^2}{n^2+\pi^2}=\frac{e^{\pi*\pi}+\pi*\pi ...

化简 m=0,1,---,8如下:

\(\left(
\begin{array}{c}
\frac{1}{2} \left(1+\pi ^2 \coth \left(\pi ^2\right)\right) \\
\frac{1}{2} \left(\pi ^2 \coth \left(\pi ^2\right)-1\right) \\
\frac{\pi ^2 \left(\left(1+\pi ^2\right) \coth \left(\pi ^2\right)-3\right)-1}{2 \left(1+\pi ^2\right)} \\
-\frac{5}{2}+\frac{1}{1+\pi ^2}+\frac{4}{4+\pi ^2}+\frac{1}{2} \pi ^2 \coth \left(\pi ^2\right) \\
-\frac{7}{2}+\frac{1}{1+\pi ^2}+\frac{4}{4+\pi ^2}+\frac{9}{9+\pi ^2}+\frac{1}{2} \pi ^2 \coth \left(\pi ^2\right) \\
-\frac{9}{2}+\frac{1}{1+\pi ^2}+\frac{4}{4+\pi ^2}+\frac{9}{9+\pi ^2}+\frac{16}{16+\pi ^2}+\frac{1}{2} \pi ^2 \coth \left(\pi ^2\right) \\
-\frac{11}{2}+\frac{1}{1+\pi ^2}+\frac{4}{4+\pi ^2}+\frac{9}{9+\pi ^2}+\frac{16}{16+\pi ^2}+\frac{25}{25+\pi ^2}+\frac{1}{2} \pi ^2 \coth \left(\pi ^2\right) \\
-\frac{13}{2}+\frac{1}{1+\pi ^2}+\frac{4}{4+\pi ^2}+\frac{9}{9+\pi ^2}+\frac{16}{16+\pi ^2}+\frac{25}{25+\pi ^2}+\frac{36}{36+\pi ^2}+\frac{1}{2} \pi ^2 \coth \left(\pi ^2\right) \\
-\frac{15}{2}+\frac{1}{1+\pi ^2}+\frac{4}{4+\pi ^2}+\frac{9}{9+\pi ^2}+\frac{16}{16+\pi ^2}+\frac{25}{25+\pi ^2}+\frac{36}{36+\pi ^2}+\frac{49}{49+\pi ^2}+\frac{1}{2} \pi ^2 \coth \left(\pi ^2\right) \\
\end{array}
\right) \text{Table}\left[\left\{\text{FullSimplify}\left[\sum _{n=m}^{\infty } \frac{\pi ^2}{n^2+\pi ^2}\right]\right\},\{m,0,8\}\right]\)