好题

Eufisky posted @ 2017年10月20日 01:55 in 数学分析 with tags 积分 , 727 阅读

设$a,b>0$,而$\displaystyle \int_0^{\infty}{\frac{tdt}{\sqrt[3]{\left( a^3+t^3 \right) \left( b^3+t^3 \right) ^2}}}$.证明:

(1) $\displaystyle I\left( a,b \right) =I\left( \frac{a+2b}{3},\sqrt[3]{b\frac{a^2+ab+b^2}{3}} \right)$.

(2) 数列$\{a_n\},\{b_n\}$满足$a_0=a,b_0=b$,且满足$a_{n+1}=\frac{a+2b}{3},b_{n+1}=\sqrt[3]{b_n\frac{a_n^2+a_nb_n+b_n^2}{3}}$,求证$\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\frac{I(1,1)}{I(a,b)}$.

线性系统理论代数基础，韩京清：http://book.ixueshu.com/book/15b7a59545f6afe2.html

http://book.ixueshu.com/book/b9f03de828640dda.html

Ramanujan's golden ratio equation

\begin{align*}R\left( e^{-2\pi} \right) &=\frac{e^{-\frac{2\pi}{5}}}{1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\ddots}}}=\sqrt{\frac{5+\sqrt{5}}{2}}-\phi ,\\R\left( e^{-2\sqrt{5}\pi} \right) &=\frac{e^{-\frac{2\pi}{\sqrt{5}}}}{1+\frac{e^{-2\pi \sqrt{5}}}{1+\frac{e^{-4\pi \sqrt{5}}}{1+\ddots}}}=\frac{\sqrt{5}}{1+\left( 5^{3/4}\left( \phi -1 \right) ^{5/2}-1 \right) ^{1/5}}-\phi ,\end{align*}

其中$\phi=\frac{1+\sqrt5}{2}$为黄金分割数(golden ratio).

这里利用了Rogers–Ramanujan continued fraction

\begin{align*}R(q) & = q^{\frac{1}{5}}\prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})} \\ &= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}.\end{align*}

Ramanujan–Sato series,https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series

\[\frac{1}{\pi} = \frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{26390k+1103}{396^{4k}}.\]

Chudnovsky algorithm，https://en.wikipedia.org/wiki/Chudnovsky_algorithm

\[\frac{1}{\pi} = 12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (545140134k + 13591409)}{(3k)!(k!)^3 \left(640320\right)^{3k + 3/2}}.\]此公式对$\pi$有非常好的计算性能.

Ramanujan's Hypergeometric Identity,http://mathworld.wolfram.com/RamanujansHypergeometricIdentity.html

$$\sum_{k=-\infty}^\infty 2^k = 0.$$

Plot the graphs of the functions $$f(x)=\dfrac{2(x^2+|x|-6)}{3(x^2+|x|+2)}+\sqrt{16-x^2}$$ and $$g(x)=\dfrac{2(x^2+|x|-6)}{3(x^2+|x|+2)}-\sqrt{16-x^2}$$ in $x\in[-4,4]$ on the same plane.

![enter image description here][1]

[1]: http://i.stack.imgur.com/9PdtB.jpg

\[\sum_{n=1}^{\infty} \frac{n^{13}}{e^{2\pi n} - 1} = \frac{1}{24}.\]

美丽有两种，一种是全新的数学工具；一种是时刻浮现在脑海里你甜美的笑容。

sinx小于x小于tanx，Young不等式，等周问题
张筑生，谢惠民