几个重要定理

Eufisky posted @ 2017年8月07日 01:50 in 数学分析 with tags 定理 Ramanujan , 1282 阅读

1.Mittag-Leffler's theorem.

设 $\Omega$ 是平面内的开集, $A\subset \Omega$ , $A$ 在 $\Omega$ 内没有极限点,且对每个 $a\in A$ ,对应有一个正整数 $m(\alpha)$ 和一个有理函数 $P_\alpha (z)=\sum_{j=1}^{m(\alpha)}c_{j,\alpha}(z-\alpha)^{-j},$ 则在 $\Omega$ 内存在一个亚纯函数 $f$ ,它在每个 $\alpha\in A$ 处的主要部分是 $P_\alpha$ 且在 $\Omega$ 内没有其它极点.详见Rudin实分析与复分析P216.

这里有亚纯函数极展开的一些例子.

$\begin{align*}\frac{1}{\sin \left( z \right)}&=\sum_{n\in \mathbb{Z}}{\frac{\left( -1 \right) ^n}{z-n\pi}}=\frac{1}{z}+2z\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{z^2-\left( n\,\pi \right) ^2}},\\\cot \left( z \right) &\equiv \frac{\cos \left( z \right)}{\sin \left( z \right)}=\sum_{n\in \mathbb{Z}}{\frac{1}{z-n\pi}}=\frac{1}{z}+2z\sum_{k=1}^{\infty}{\frac{1}{z^2-\left( k\,\pi \right) ^2}},\\\frac{1}{\sin ^2\left( z \right)}&=\sum_{n\in \mathbb{Z}}{\frac{1}{\left( z-n\,\pi \right) ^2}},\\\frac{1}{z\sin \left( z \right)}&=\frac{1}{z^2}+\sum_{n\ne 0}{\frac{\left( -1 \right) ^n}{\pi n\left( z-\pi n \right)}}=\frac{1}{z^2}+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n\,\pi}}\frac{2z}{z^2-\left( n\,\pi \right) ^2}.\end{align*}$

2.Ramanujan's Master Theorem

假设

$x=0$ 的一些邻域中有

$F(x)=\sum_{k=0}^\infty\frac{\phi (k)(-x)^k}{k!}$ 对某些函数（称为解析或可积的）

$\phi (k)$ 成立.那么

$\int_0^\infty x^{n-1}F(x)dx=\Gamma (n)\phi (-n).$

参考：这里以及

Ramanujan's Proof

This proof was given by none other than Ramanujan.

Recall Euler's integral representation of the Gamma Function -

$\int_0^\infty e^{-mx}x^{n-1}dx = m^{-n}\Gamma(n).$

where

$m,n>0$ . Let

$m=r^k$ with

$r>0$ , multiply both sides by

$\frac{f^{(k)}h^k}{k!}$ and sum on

$k, 0 \leq k <\infty$ , to obtain

$\sum_{k=0}^\infty \frac{f^{(k)}(a)h^k}{k!}\int_0^\infty e^{-r^k x}x^{n-1}dx=\Gamma(n) \sum_{k=0}^\infty \frac{f^{(k)}(a)(hr^{-n})^k}{k!}.$

Next, expand

$e^{-r^k x}, 0\leq k<\infty$ , in its Maclaurin Series, invert the order of summation and integration, and apply Taylor's Theorem to deduce that

$\int_0^\infty x^{n-1}\sum_{j=0}^\infty \frac{f(h r^j+a)}{j!}(-x)^j dx = \Gamma(n)f(hr^{-n}+a).$

Now define

$f(hr^m+a)=\varphi(m)$ , where

$m$ is real and

$a,h$ and

$r$ are regarded as constants. Then

$\int_0^\infty x^{n-1}\sum_{j=0}^\infty \frac{\varphi(j) (-x)^j}{j!}dx= \Gamma(n) \varphi(-n).$

This completes Ramanujan's proof. Ramanujan was very fond of this clever, original technique and he used it many contexts.

例：证明

$\int_{0}^{\infty}{\left(\sum_{k=0}^{\infty}{\frac{(-1)^{k}P_{k+1}}{k!}x^{k}} \right)dx}=2,$

其中

$P_{k+1}$ 表示第

$k+1$ 个素数,记

$P_1=3$ .

由于

$\int_0^\infty x^{s-1} \left( \sum_{k=0}^\infty \frac{P_{k+1}}{k!}(-x)^k\right)dx =\Gamma(s) P_{1-s},$

令

$s=1$ 我们有

$\int_0^\infty \left( \sum_{k=0}^\infty \frac{P_{k+1}}{k!}(-x)^k\right)dx = P_{0}=2.$

3.Glasser's Master Theorem

对任意可积函数

$F(x)$ 和形如

$\phi(x)=|a|x-\sum_{n=1}^N\frac{|\alpha_n|}{x-\beta_n}$

的

$\phi(x)$ ,恒等式

$PV \int_{-\infty}^\infty F(\phi (x))dx=PV \int_{-\infty}^\infty F(x)dx$ 成立,其中

$a,\{\alpha_n\}_{n=1}^N$ 和

$\{\beta_n\}_{n=1}^N$ 为任意常数.这里,

$PV$ 表示Cauchy主值.这是从Cauchy的著名结果

$PV\int_{-\infty}^\infty F(u)dx=\int_{-\infty}^\infty F(x)dx$

归纳出来的,其中

$u=x-1/x$ .

例.求

$\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x.$

$I=\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x,$

$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x,$

$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x.$

Now, letting

$f(x)=\frac{1}{x^{2016}+1}$ , and noting that

$f(x)=f(-x)$ ,

$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\mathrm{d}x=\frac{1}{2}\int_{-\infty}^{\infty} f\left(-\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\right)\mathrm{d}x$

$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(x-\sum^{2015}_{i=1}\frac{i}{x-(-i)}\right)\mathrm{d}x \tag {1}$

Using Glasser's Master Theorem,

$I=\frac{1}{2}\int^{\infty}_{-\infty} f(x)\ \mathrm{d}x=\frac{1}{2}\int^{\infty}_{-\infty} \frac{1}{x^{2016}+1}\ \mathrm{d}x=\int^{\infty}_{0} \frac{1}{x^{2016}+1}\ \mathrm{d}x \tag {2}$

Now we know that

$B(a,b)=\int^{\infty}_0\frac{t^{a-1}}{(1+t)^{a+b}}dt$

From

$(2)$ ,after substituting

$x^{2016} =t$ ,

$I=\frac{1}{2016}\int^{\infty}_{0}\frac{t^{\frac{1}{2016}-1}}{(1+t)^{\frac{1}{2016}+\frac{2015}{2016}}}dt=\frac{1}{2016}B(\frac{1}{2016},\frac{2015}{2016})$

Therefore

$\color{red}{I=\frac{1}{2016}\frac{\Gamma(\frac{1}{2016})\Gamma(\frac{2015}{2016})}{\Gamma(1)}=\frac{\pi}{2016\sin(\frac{\pi}{2016})}\approx1.0000004047320180811575}$

这里

Rogers L-Function

Watson's Triple Integrals

[回复]

CGBSE Model Paper Cl 说:
2022年9月07日 23:53

Chhattisgarh State Department of School Education (Elementary Level Primary Education) and other private school teaching staff of the state have designed and suggested CG Board 4th Class Model Paper 2023 with sample answers along with Mock Test and Practice Questions for Term1 & Term 2 Exams of the Course to All Languages and Subjects. CGBSE Model Paper Class 4 Every 4th Standard Student of SCERT & NCERT Syllabus Studying in Government or Private schools of Hindi Medium, English Medium and Urdu Medium can download and practice the CGBSE STD-4 Question Paper 2023 Pdf for Part-A, Part-B, Part-C and Part-D exam.