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\textbf{Notebook:A set of questions and solutions.}

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1. Use mathematical analysis to show that:

e and $\pi$ are irrational numbers;

Proof.

For a), For any positive integer $N$ we have
$$\begin{array}{rl}e& =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\\ e_N& =\sum _{n=0}^N\frac{1}{n!}=\frac{p_N}{q_N}<e\end{array}$$$$e-e_N<\frac{e}{(N+1)!},q_N⩽\frac{1}{N!}$$
Assume that $e$ is rational, then we can write $e=\frac{p}{q}$ for some $p,q\in \mathbb{N}$ with $q>0.$ Choose $N$ sufficiently large so that $N>p,$ then we get $e<\frac{N+1}{q}$ thus $\frac{1}{q}>\frac{e}{N+1}.$ We have
$$e-e_N=\frac{p}{q}-\frac{p_N}{q_N}=\frac{p{q}_N-q{p}_N}{q{q}_N}⩾\frac{1}{q{q}_N}⩾\frac{1}{qN!}⩾\frac{e}{(N+1)!}$$
which is a contradiction. So $e$ must be irrational.

2.$$\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n\sum _{j=1}^n\sum _{m=1}^n\frac{\arctan \frac{j}{n}\arctan \frac{k^2}{n^2}}{k{n}^5+k{j}^2{n}^3}{m}^3(\ln m-\ln n{)}^3(\ln (n+j)-\ln n).$$

Solution

$$\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n\sum _{j=1}^n\sum _{m=1}^n\frac{\arctan \frac{j}{n}\arctan \frac{k^2}{n^2}{m}^3(\ln m-\ln n{)}^3[\ln (j+n)-\ln n]}{k{n}^5+k{j}^2{n}^3}$$

$$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n^3}\sum _{k=1}^n\sum _{j=1}^n\sum _{m=1}^n\frac{\arctan \frac{j}{n}\arctan \frac{k^2}{n^2}{\left(\frac{m}{n}\right)}^3{(\ln \frac{m}{n})}^3[\ln (1+\frac{j}{n})]}{\frac{k}{n}(1+\frac{j^2}{n^2})}$$

$$={\int }_0^1\frac{\arctan z^2}{z}dz{\int }_0^1(y\ln y{)}^{3}dy{\int }_0^1\frac{\arctan x\ln (1+x)}{1+x^2}dx$$

$$=\frac{G}{2}{\int }_0^1(y\ln y{)}^{3}dy{\int }_0^1\frac{\arctan x\ln (1+x)}{1+x^2}dx$$

$$=\frac{G}{2}\cdot \frac{-3}{128}{\int }_0^1\frac{\arctan x\ln (1+x)}{1+x^2}dx$$

let
$$I_3={\int }_0^1\frac{\arctan x\ln (1+x)}{1+x^2}dx$$

$$I_3={\int }_0^{\frac{\pi }{4}}x\ln (1+\tan x)dx={\int }_0^{\frac{\pi }{4}}x\ln (\cos x+\sin x)dx-{\int }_0^{\frac{\pi }{4}}x\ln (\cos x)dx$$

$$={\int }_0^{\frac{\pi }{4}}x\ln (\sqrt{2}\sin (x+\frac{\pi }{4}))dx-{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}(\frac{\pi }{2}-x)\ln (\sin x)dx$$

$$=\frac{{\pi }^2\ln 2}{64}x-\frac{3\pi }{4}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\ln \sin xdx$$

$$=\frac{{\pi }^2\ln 2}{64}+2{\psi }_1-\frac{3\pi }{4}{\psi }_2.$$
Try to compute ${\psi }_1,{\psi }_2$,

$$\ln \sin x=-\ln 2-\sum _{n=1}^{\mathrm{\infty }}\frac{\cos \left(2nx\right)}{n},0<x<\pi$$

$${\psi }_1={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}x\ln \sin xdx={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}x[-\ln 2-\sum _{n=1}^{\mathrm{\infty }}\frac{\cos \left(2nx\right)}{n}]dx$$

$$=-\frac{3{\pi }^2\ln 2}{32}-\sum _{n=1}^{\mathrm{\infty }}\frac{n\pi \sin \left(n\pi \right)+\cos \left(n\pi \right)-\left(\frac{n\pi }{2}\right)\sin \left(\frac{n\pi }{2}\right)-\cos \left(\frac{n\pi }{2}\right)}{4n^3}$$
$$=-\frac{3{\pi }^2\ln 2}{32}-\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^3}+\frac{\pi }{8}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{(2n-1{)}^2}-\frac{1}{32}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^3}$$

$$=\frac{21}{128}\zeta (3)-\frac{\pi }{8}-\frac{{\pi }^2\ln 2}{64}.$$

$${\psi }_2=\frac{G}{2}-\frac{\pi \ln 2}{4}$$ the answer is
$$-\frac{3G}{256}(\frac{21}{64}\zeta (3)-\frac{\pi }{8}G+\frac{{\pi }^2\ln 2}{64}).$$

3.
Find an explicit conformal transformation of an open set $U=\{|z|>$
1}$\mathrm{\setminus }(-\mathrm{\infty },-1]$ to the unit disc.

Proof.
We will construct the mapping in several steps(*≧ω≦)

1. The mapping $w=-\frac{1}{z}$ maps $U$ onto $D_1\mathrm{\setminus }[0,1]$

2. $w^{\prime }=\sqrt{w}$ maps $D_1\mathrm{\setminus }[0,1]$ onto $\{z||z\mid <1,\mathrm{Im}(z)>0\}$

3. The mapping $w^{\prime \prime }=\frac{w^{\prime }-1}{w^{\prime }+1}$ maps $\{z||z\mid <1,\mathrm{Im}(z)>0\}$ onto
   $\{z\mid \mathrm{Re}(z)>0,\mathrm{Im}(z)>0\}$

4. $u=w^{\prime \prime 2}\mathrm{maps}\{z\mid \mathrm{Re}(z)>0,\mathrm{Im}(z)>0\}$ onto $\{z\mid \mathrm{Im}(z)>0\}$

5. $v=\frac{u-i}{u+i}$ maps the upper half plane onto the unit disc.

6. Composition of the mappings 1)-4) is $u={\left(\frac{\sqrt{-\frac{1}{2}}-1}{\sqrt{-\frac{1}{z}}+1}\right)}^2\cdot$ Its composition with 5 ) gives the desired mapping.