# 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 , 1186 阅读

1.Mittag-Leffler's theorem.

\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

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.

3.Glasser's Master Theorem

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

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

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