Maths Analysis II - Homework #23 20220519

Author： Bi_Gu
发布时间：May 26, 2022
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# Maths Analysis II

## Homework #23

### 12.1.3

$(1)$

$$
\begin{align}
g(x)&=\begin{cases}f(x)&0<x<\pi\\\pi^2&x=0\\f(x+\pi)&-\pi<x<0\end{cases}=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)
\\&=\frac{2}{3}\pi^2+\frac{1}{\pi}\sum_{n=1}^{\infty}\bigg(\frac{4}{n^2}(\cos n\pi+1)\pi\cos nx-\frac{2}{n}(\cos n\pi+1)\pi^2\sin nx\bigg)
\\&=\frac{2}{3}\pi^2+\sum_{n=1}^{\infty}\frac{2}{n}\bigg(2\cos 2nx-\frac{\pi}{n}\sin 2nx\bigg).
\end{align}
$$

$(3)$

$$
\begin{align}
g(x)&=\begin{cases}f(x)&0<x<\pi\\f(x+\pi)&-\pi<x<0\end{cases}
=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)
\\&=\frac{\sin 2nh}{2n\pi}+\frac{1}{\pi}\sum_{n=1}^{\infty}\bigg(\frac{1-\cos2nh}{2n^2h\pi}\cos nx+\big(\frac{1}{n\pi}-\frac{\sin 2nh}{2n^2h\pi}\big)\sin nx\bigg).
\end{align}
$$

### 12.1.4

$(1)$ Since that $a_0=0$, we have

$$
\frac{1}{\pi}\int_{-\pi}^{\pi}a(2a-|x|)\mathrm{d}x=0\Rightarrow a(4a-\pi)=0\Rightarrow a=\frac{\pi}{4}.
$$

$(2)$ Since that $b_1=1$, we have

$$
\frac{1}{\pi}\int_{-\pi}^{\pi}ax\sin x\mathrm{d}x=1\Rightarrow 2a=1\Rightarrow a=\frac{1}{2}.
$$

### 12.1.5

$(1)$ Define $\displaystyle \langle f,g\rangle_d=\frac{1}{d}\int_{-d}^{d}f(x)g(x)\mathrm{d}x$, then $\langle f(d\cdot),g(d\cdot)\rangle_1=\langle f,g\rangle_d$.

$$
\begin{align}
g(x)&=\begin{cases}\displaystyle f(\frac{x}{\pi})&0\leq x<\pi\\g(x+\pi)&-\pi\leq x<0\\g(x\pm 2\pi)&x\leq -\pi\text{ or }x\geq \pi\end{cases}\\&=\frac{\langle g,1\rangle_{\pi}}{2\pi}+\frac{1}{\pi}\sum_{n=1}^{\infty}(\langle g,\cos n\cdot\rangle_{\pi}\cos nx+\langle g,\sin n\cdot\rangle_{\pi}\sin nx)\\&=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx\\&=S(\frac{x}{\pi})
\end{align}
$$

Therefore $S(x)=g(\pi x)$. Based on such result, we have

$$
S(\frac{9}{4})=f(\frac{1}{4})=\frac{1}{4};\quad S(-\frac{5}{2})=f(\frac{1}{2})=1.
$$

$(2)$

$$
\begin{align}
S(3\pi)&=\frac{1}{2}\bigg(S(-\pi+0)+S(\pi-0)\bigg)=\frac{\pi^2}{2};
\\S(4\pi)&=\frac{1}{2}\bigg(S(0+0)+S(0-0)\bigg)=0.
\end{align}
$$

### 12.1.9

First we have that

$$
\begin{align}
g(x)&=\begin{cases}f(x)&0\leq x<\pi\\f(-x)&-\pi\leq x<0\end{cases}=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\cos nx)
\\&=1+\frac{\pi}{2}-\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1-(-1)^n}{n^2}\cos nx\\&=1+\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos(2n+1)x}{(2n+1)^2}
\end{align}
$$

Based on such results, we have

$(1)$

$$
\sum_{n=1}^{\infty}\frac{\cos(2n-1)}{(2n-1)^2}=\frac{\pi}{4}\big(1+\frac{\pi}{2}-g(1)\big)=\frac{\pi}{4}(\frac{\pi}{2}-1)=\frac{\pi^2}{8}-\frac{\pi}{4}.
$$

$(2)$

$$
\sum_{n=1}^{\infty}\frac{\cos4(2n-1)}{(2n-1)^2}=\frac{\pi}{4}\big(1+\frac{\pi}{2}-g(4)\big)=\frac{\pi}{4}(\frac{\pi}{2}-(2\pi-4))=\pi-\frac{3}{8}\pi^2.
$$

### 12.1.10

Since that

$$
F_{+n}=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(x)e^{\mp\mathrm{i}n\omega x}\mathrm{d}x=\frac{H}{T}\int_{-\frac{r}{2}}^{\frac{r}{2}}e^{\mp\mathrm{i}n\omega x}\mathrm{d}x=\frac{H\sin\frac{n\pi r}{T}}{n\pi}
$$

We have

$$
f(x)=\sum_{n=-\infty}^{\infty}F_ne^{\mathrm{i}n\omega x}=H\sum_{n=-\infty}^{\infty}\frac{\sin\frac{n\pi r}{T}}{n\pi}e^{\mp\mathrm{i}n\frac{2\pi}{T}x}.
$$

### 12.3.2

$(2)$ Since $a_n=0$ and

$$
b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin ax\sin nx\mathrm{d}x=\frac{1}{\pi}\bigg(\frac{\sin(a-n)\pi}{a-n}-\frac{\sin(a+n)\pi}{a+n}\bigg)
$$

We have that

$$
\begin{align}
\sin ax&=\frac{1}{\pi}\sum_{n=1}^{\infty}\bigg(\big(\frac{(-1)^n\sin a\pi}{a-n}-\frac{(-1)^n\sin a\pi}{a+n}\big)\sin nx\bigg)
\\&=\frac{2\sin ax}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n}{a^2-n^2}\sin nx.
\end{align}
$$

$(3)$ Since $b_n=0$ and

$$
a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x\sin x\cos nx\mathrm{d}x=\frac{2\pi\cos n\pi}{(n-1)(n+1)}-\frac{2\pi\sin n\pi}{(n-1)(n+1)}
$$

We have that

$$
x\sin x=1-2\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2-1}\cos nx.
$$

### 12.3.4

First we have that

$$
\begin{align}
\cos ax&=\frac{\sin a\pi}{a\pi}+\frac{1}{\pi}\sum_{n=1}^{\infty}\bigg(\frac{\sin(a-n\pi}{a-n}+\frac{\sin(a+n)\pi}{a+n}\bigg)\cos nx\\&=
\frac{\sin a\pi}{a\pi}+\frac{1}{\pi}\sum_{n=1}^{\infty}\bigg(\frac{1}{a-n}+\frac{1}{a+n}\bigg)(-1)^n\sin a\pi\cos nx
\\&=\frac{\sin a\pi}{\pi}\bigg(\frac{1}{a}+\sum_{n=1}^{\infty}\frac{2a(-1)^n}{a^2-n^2}\cos nx\bigg)
\end{align}
$$

$(1)$ Let $x=\pi$, then $\cos a\pi=\displaystyle \frac{\sin a\pi}{\pi}\bigg(\frac{1}{a}+\sum_{n=1}^{\infty}\frac{2a}{a^2-n^2}\bigg)$; then let $a=\displaystyle\frac{x}{\pi}$ and we have

$$
\cot x=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^2-n^2\pi^2}.
$$

$(2)$ Let $x=0$, then $1=\displaystyle\frac{\sin a\pi}{\pi}\bigg(\frac{1}{a}+\sum_{n=1}^{\infty}\frac{2a(-1)^n}{a^2-x^2}\bigg)$; then let $\displaystyle a=\frac{x}{\pi}$ and we have

$$
\frac{1}{\sin x}=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2(-1)^nx}{x^2-n^2\pi^2}.
$$

### 12.3.5

When $x\in(-\pi,\pi)$, $\cos\displaystyle\frac{x}{2}>0$; then we have

$$
\left|\cos x\right|=\cos\bigg(\frac{1}{2}(2x)\bigg)=\frac{1}{\pi}\bigg(2+\sum_{n=1}^{\infty}\frac{(-1)^n}{\frac{1}{4}-n^2}\cos 2nx\bigg)=\frac{2}{\pi}+\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4n^2-1}\cos 2nx
$$

Similarly we have

$$
\begin{align}
\left|\sin x\right|&=\left|\cos(x-\frac{\pi}{2})\right|\\&=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4n^2-1}\cos (2nx-n\pi)\\&=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{4n^2-1}\cos 2nx.
\end{align}
$$

Last modification：May 26, 2022

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Maths Analysis II - Homework #23 20220519

Bi_Gu • 2022 年 05 月 26 日

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# Maths Analysis II

## Homework #23

### 12.1.3

$(1)$

$(3)$

### 12.1.4

$(1)$ Since that $a_0=0$, we have

$$
\frac{1}{\pi}\int_{-\pi}^{\pi}a(2a-|x|)\mathrm{d}x=0\Rightarrow a(4a-\pi)=0\Rightarrow a=\frac{\pi}{4}.
$$

$(2)$ Since that $b_1=1$, we have

$$
\frac{1}{\pi}\int_{-\pi}^{\pi}ax\sin x\mathrm{d}x=1\Rightarrow 2a=1\Rightarrow a=\frac{1}{2}.
$$

### 12.1.5

$(1)$ Define $\displaystyle \langle f,g\rangle_d=\frac{1}{d}\int_{-d}^{d}f(x)g(x)\mathrm{d}x$, then $\langle f(d\cdot),g(d\cdot)\rangle_1=\langle f,g\rangle_d$.

Therefore $S(x)=g(\pi x)$. Based on such result, we have

$$
S(\frac{9}{4})=f(\frac{1}{4})=\frac{1}{4};\quad S(-\frac{5}{2})=f(\frac{1}{2})=1.
$$

$(2)$

$$
\begin{align}
S(3\pi)&=\frac{1}{2}\bigg(S(-\pi+0)+S(\pi-0)\bigg)=\frac{\pi^2}{2};
\\S(4\pi)&=\frac{1}{2}\bigg(S(0+0)+S(0-0)\bigg)=0.
\end{align}
$$

### 12.1.9

First we have that

Based on such results, we have

$(1)$

$$
\sum_{n=1}^{\infty}\frac{\cos(2n-1)}{(2n-1)^2}=\frac{\pi}{4}\big(1+\frac{\pi}{2}-g(1)\big)=\frac{\pi}{4}(\frac{\pi}{2}-1)=\frac{\pi^2}{8}-\frac{\pi}{4}.
$$

$(2)$

$$
\sum_{n=1}^{\infty}\frac{\cos4(2n-1)}{(2n-1)^2}=\frac{\pi}{4}\big(1+\frac{\pi}{2}-g(4)\big)=\frac{\pi}{4}(\frac{\pi}{2}-(2\pi-4))=\pi-\frac{3}{8}\pi^2.
$$

### 12.1.10

Since that

We have

$$
f(x)=\sum_{n=-\infty}^{\infty}F_ne^{\mathrm{i}n\omega x}=H\sum_{n=-\infty}^{\infty}\frac{\sin\frac{n\pi r}{T}}{n\pi}e^{\mp\mathrm{i}n\frac{2\pi}{T}x}.
$$

### 12.3.2

$(2)$ Since $a_n=0$ and

$$
b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin ax\sin nx\mathrm{d}x=\frac{1}{\pi}\bigg(\frac{\sin(a-n)\pi}{a-n}-\frac{\sin(a+n)\pi}{a+n}\bigg)
$$

We have that

$(3)$ Since $b_n=0$ and

$$
a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x\sin x\cos nx\mathrm{d}x=\frac{2\pi\cos n\pi}{(n-1)(n+1)}-\frac{2\pi\sin n\pi}{(n-1)(n+1)}
$$

We have that

$$
x\sin x=1-2\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2-1}\cos nx.
$$

### 12.3.4

First we have that

$(1)$ Let $x=\pi$, then $\cos a\pi=\displaystyle \frac{\sin a\pi}{\pi}\bigg(\frac{1}{a}+\sum_{n=1}^{\infty}\frac{2a}{a^2-n^2}\bigg)$; then let $a=\displaystyle\frac{x}{\pi}$ and we have

$$
\cot x=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^2-n^2\pi^2}.
$$

$(2)$ Let $x=0$, then $1=\displaystyle\frac{\sin a\pi}{\pi}\bigg(\frac{1}{a}+\sum_{n=1}^{\infty}\frac{2a(-1)^n}{a^2-x^2}\bigg)$; then let $\displaystyle a=\frac{x}{\pi}$ and we have

$$
\frac{1}{\sin x}=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2(-1)^nx}{x^2-n^2\pi^2}.
$$

### 12.3.5

When $x\in(-\pi,\pi)$, $\cos\displaystyle\frac{x}{2}>0$; then we have

Similarly we have

Maths Analysis II - Homework #23 20220519

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Maths Analysis II - Homework #23 20220519