Maths Analysis II - Homework #24 20220524

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# Maths Analysis II

## Homework #24 20220524

### 12.2.1

First we have that

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

and

$$
b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx=0
$$

From which we can get

$$
f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx=\frac{a_0}{2}+\sum_{n=1}^{\infty}\frac{2\sin an}{\pi n}\cos nx;
$$

Thus we have

$(1)$

$$
\begin{align}
&\quad\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)\mathrm{d}x=\frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)
\\&\Rightarrow \frac{2a}{\pi}=\frac{1}{2}\bigg(\frac{2a}{\pi}\bigg)^2+\bigg(\frac{2}{\pi}\bigg)^2\sum_{n=1}^{\infty}\frac{\sin^2an}{n^2}
\\&\Rightarrow \sum_{n=1}^{\infty}\frac{\sin^2an}{n^2}=\frac{a(\pi-a)}{2}.
\end{align}
$$

$(2)$

$$
\sum_{n=1}^{\infty}\frac{\cos^2 an}{n^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{\sin^2 an}{n^2}=\frac{\pi^2-3a\pi+3a^2}{6}.
$$

### 12.2.2

Since that $f(x)$ is both integrable and square integrable, both $\displaystyle \int_{-\pi}^{\pi}f^2(x)\mathrm{d}x$ and $\displaystyle\int_{-\pi}^{\pi}f(x)\mathrm{d}x$ exist.

Thus we have

$$
\begin{align}
\sum_{n=1}^{\infty}\bigg(\frac{\left|a_n\right|}{n}+\frac{\left|b_n\right|}{n}\bigg)&\leq\sum_{n=1}^{\infty}\frac{1}{2}\bigg(a_n^2+\frac{1}{n^2}\bigg)+\frac{1}{2}\bigg(b_n^2+\frac{1}{n^2}\bigg)
\\&=\frac{1}{2}\sum_{n=1}^{\infty}\bigg(a_n^2+b_n^2\bigg)+\frac{\pi^2}{6}
\\&=\frac{1}{2\pi}\int_{-\pi}^{\pi}f^2(x)\mathrm{d}x-\frac{1}{4}\bigg(\int_{-\pi}^{\pi}f(x)\mathrm{d}x\bigg)^2
\end{align}
$$

which exists, which means that both $\displaystyle\sum_{n=1}^{\infty}\frac{\left|a_n\right|}{n}$ and $\displaystyle\sum_{n=1}^{\infty}\frac{\left|b_n\right|}{n}$ converges.

Which is equivalent to both $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{n}$ and $\displaystyle\sum_{n=1}^{\infty}\frac{b_n}{n}$ converges.

**The proof is complete.**

### 12.2.3

First we have that

$$
a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\mathrm{d}x=0
$$

and

$$
b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin x\mathrm{d}x=\frac{2}{\pi}\cdot\frac{1-\cos n\pi}{n}
$$

From which we can get

$$
f(x)=\sum_{n=1}^{\infty}b_n\sin nx=\sum_{n=1}^{\infty}\frac{2}{\pi}\cdot\frac{1-(-1)^2}{n}\sin nx=\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\sin(2n-1)\pi}{2n-1};
$$

Thus we have

$$
\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\mathrm{d}x=\sum_{n=1}^{\infty}b_n^2\Rightarrow 2=\bigg(\frac{4}{\pi}\bigg)^2\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}\Rightarrow \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}.
$$

$(2)$

$$
\begin{align}
&\quad\int_{0}^{\pi}f(x)\mathrm{d}x=\frac{\pi}{2}+\sum_{n=1}^{\infty}-\frac{b_n}{n}\cos nx\\&\Rightarrow\left|x\right|=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos(2n-1)\pi}{(2n-1)^2}
\\&\Rightarrow \sum_{n=1}^{\infty}\frac{\cos(2n-1)x}{(2n-1)^2}=\frac{\pi}{4}(x-\frac{\pi}{2}).
\end{align}
$$

### 12.3.6

$$
\begin{align}
&\quad e^{ax}\\&=\frac{1}{2\pi}\int_0^{2\pi}e^{ax}\mathrm{d}x+\sum_{n=1}^{\infty}\bigg(\frac{1}{\pi}\int_0^{2\pi}e^{ax}\cos nx\mathrm{d}x \cos nx+\frac{1}{\pi}\int_0^{2\pi}e^{ax}\sin nx\mathrm{d}x \sin nx\bigg)
\\&=\frac{e^{2a\pi}-1}{\pi}\bigg(\frac{1}{2a}+\sum_{k=1}^{\infty}\frac{a\cos kx-k\sin kx}{k^2+a^2}\bigg).
\end{align}
$$

**The proof is complete.**

### 12.3.7

First we calculate that

$$
\begin{align}
x^2&=2\int_0^xt\mathrm{d}t=4\sum_{n=1}^{\infty}(-1)^n\frac{1-\cos nx}{n^2}=4\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\cos nx}{n^2}+\frac{\pi^2}{3};
\\x^3&=3\int_0^xt^2\mathrm{d}t=12\sum_{n=1}^{\infty}(-1)^n\frac{\sin nx}{n^3}+\pi^2 x=2\sum_{n=1}^{\infty}(-1)^n\bigg(\frac{6}{n^3}-\frac{\pi^2}{n}\bigg)\sin nx;
\\x^4&=4\int_0^xt^3\mathrm{d}t=48\sum_{n=1}^{\infty}(-1)^n\frac{1-\cos nx}{n^4}+2\pi^2x^2\\&=8\sum_{n=1}^{\infty}(-1)^{n-1}\bigg(\frac{\pi^2}{n^2}-\frac{6}{n^4}\bigg)\cos nx+\frac{\pi^4}{5}.
\end{align}
$$

Then apply **Pasavel Equality** to each and we have

$$
\begin{cases}
\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}x^4\mathrm{d}x=\sum_{n=1}^{\infty}\bigg(\frac{4}{n^2}\bigg)^2+\frac{1}{2}\bigg(\frac{2}{3}\pi^2\bigg)^2
\\
\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}x^6\mathrm{d}x=2^2\sum_{n=1}^{\infty}\bigg(\frac{6}{n^3}-\frac{\pi^2}{n}\bigg)^2
\\
\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}x^6\mathrm{d}x=8^2\sum_{n=1}^{\infty}(-1)^{n-1}\bigg(\frac{\pi^2}{n^2}-\frac{6}{n^4}\bigg)\cos nx+\frac{1}{2}\bigg(\frac{2}{5}\pi^4\bigg)^2
\end{cases}
$$

From which we can obtain

$$
\begin{cases}
\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90};\\
\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^6}=\frac{\pi^4}{945};\\
\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^8}=\frac{\pi^6}{9450}.
\end{cases}
$$

### 12.3.8

Apply odd extension then we have $a_n=0$ and

$$
b_n=\frac{2}{\pi}\bigg(\int_0^1\frac{\pi-1}{2}x\sin x\mathrm{d}x+\int_1^{\pi}\frac{\pi-x}{2}\sin x\mathrm{d}x)=\frac{\sin n}{n^2}
$$

So that

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

**The proof is complete.**

### 12.3.9

$(1)$

$$
\sum_{n=1}^{\infty}\frac{\sin n}{n}=f'(0)=\frac{\pi-1}{2};\sum_{n=1}^{\infty}\bigg(\frac{\sin n}{n}\bigg)^2=f(1)=\frac{\pi-1}{2}.
$$

$(2)$

$$
\begin{align}
\sum_{n=1}^{\infty}\frac{\sin^2 n}{n^4}&=\frac{2}{\pi}\int_0^{\pi}f^2(x)\mathrm{d}x=\frac{2}{\pi}\bigg(\int_0^1\big(\frac{\pi-1}{2}x\big)^2\mathrm{d}x+\int_1^x\big(\frac{\pi-x}{2}x\big)^2\mathrm{d}x\bigg)
\\&=\frac{(\pi-1)^2}{6}.
\end{align}
$$

**The proof is complete.**

### 12.c.1

Since that $\displaystyle\frac{1}{\ln x}$ uniformly converge to zero and $\displaystyle\sum_{n=1}^{\infty}\sin nx$ is uniformly bounded when $n\notin 2k\pi,k\in\mathbb{Z}$,

$\displaystyle\sum_{n=2}^{\infty}\frac{\sin nx}{\ln x}$ is uniformly convergent when $n\notin 2k\pi, k\in\mathbb{Z}$.

However, by **Bessel Inequality**, since that

$$
\left||f\right||\geq\sum_{n=2}^{\infty}\frac{1}{\ln^2 n}\text{ which diverges},
$$

$f$ is not square-integrable.

**The proof is complete.**

### 12.c.2

$(1)$

$$
\begin{align}
\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\cos nx}{n}&=\mathrm{Re}\bigg(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{e^{\mathrm{i}nx}}{n}\bigg)=\mathrm{Re}\big(\ln(1+e^{\mathrm{i}nx})\big)
\\&=\mathrm{Re}\big(\ln(1+\cos x+\mathrm{i}\sin x)\big)
\\&=\ln\sqrt{(1+\cos x)^2+\sin ^2x}
\\&=\ln 2\cos\frac{x}{2}.
\end{align}
$$

$(2)$

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

**The proof is complete.**

### 12.c.3

First we have that

$$
\begin{align}
b_n&=\int_{0}^{2\pi}f(x)\sin nx\mathrm{d}x=\frac{1}{n}\int_{0}^{2n\pi}f(\frac{\pi}{n})\sin x\mathrm{d}x
\\&=\frac{1}{n}\sum_{i=0}^{n-1}\bigg(\int_{2i\pi}^{(2i+1)\pi}f(\frac{x}{n})\sin x\mathrm{d}x+\int_{(2i+1)\pi}^{(2i+2)\pi}f(\frac{x}{n})\sin x\mathrm{d}x)
\\&=\frac{1}{n}\sum_{i=0}^{n-1}\int_{2i\pi}^{(2i-1)\pi}\bigg(f(\frac{x}{n})-f(\frac{x+\pi}{n})\bigg)\sin x\mathrm{d}x
\end{align}
$$

So that

$(1)$ Since $f(x)$ decreases on $(0,2\pi)$, $\displaystyle f(\frac{x}{n})-f(\frac{x+\pi}{n})\leq 0$, which means that $b_n\geq 0$;

$(2)$ Since $f(x)$ increases on $(0,2\pi)$, $\displaystyle f(\frac{x}{n})-f(\frac{x+\pi}{n})\geq 0$, which means that $b_n\leq 0$.

**The proof is complete.**

### 12.c.4

Since that

$$
\begin{align}
\left|a_n\right|&=\frac{1}{\pi}\left|\int_{-\pi}^{\pi}f(x)\cos nx\mathrm{d}x\right|\\&=\frac{1}{n\pi}\left|\int_{-n\pi}^{n\pi}f(\frac{x}{n})\cos x\mathrm{d}x\right|\\&=\frac{1}{n\pi}\left|\sum_{i=0}^{n-1}\int_{(2i-n)\pi}^{(2i-n+1)\pi}f(\frac{x}{n})\cos x\mathrm{d}x\right|
\\&=\frac{1}{n\pi}\left|\sum_{i=0}^{n-1}\int_0^{2\pi}f(\frac{x}{n}+\frac{2\pi\mathrm{i}}{n}-\pi)\cos(x+2\pi\mathrm{i}-n\pi)\mathrm{d}x\right|
\\&\leq\frac{1}{n\pi}\left|\sum_{i=0}^{n-1}\int_0^{2\pi}f(\frac{2\pi\mathrm{i}}{n}-\pi)\cos x\mathrm{d}x\right|+\left|\sum_{i=0}^{n-1}\int_0^{2\pi}\triangle f_i(x)\cos x\mathrm{d}x\right|
\\&\leq0+\frac{1}{n\pi}\cdot 2\pi\cdot\sum_{i=0}^{n-1}\triangle f_i
\\&\leq\frac{2}{n}\bigg(f(\pi)-f(-\pi)\bigg)
\end{align}
$$

Therefore we have $\displaystyle a_n=O\bigg(\frac{1}{n}\bigg)$. Similarly we have $\displaystyle b_n=O\bigg(\frac{1}{n}\bigg)$.

**The proof is complete.**

Last modification：June 7, 2022

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Maths Analysis II - Homework #24 20220524

Bi_Gu • 2022 年 06 月 04 日

# Maths Analysis II

## Homework #24 20220524

### 12.2.1

First we have that

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

and

$$
b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx=0
$$

From which we can get

$$
f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx=\frac{a_0}{2}+\sum_{n=1}^{\infty}\frac{2\sin an}{\pi n}\cos nx;
$$

Thus we have

$(1)$

$(2)$

$$
\sum_{n=1}^{\infty}\frac{\cos^2 an}{n^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{\sin^2 an}{n^2}=\frac{\pi^2-3a\pi+3a^2}{6}.
$$

### 12.2.2

Since that $f(x)$ is both integrable and square integrable, both $\displaystyle \int_{-\pi}^{\pi}f^2(x)\mathrm{d}x$ and $\displaystyle\int_{-\pi}^{\pi}f(x)\mathrm{d}x$ exist.

Thus we have

which exists, which means that both $\displaystyle\sum_{n=1}^{\infty}\frac{\left|a_n\right|}{n}$ and $\displaystyle\sum_{n=1}^{\infty}\frac{\left|b_n\right|}{n}$ converges.

Which is equivalent to both $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{n}$ and $\displaystyle\sum_{n=1}^{\infty}\frac{b_n}{n}$ converges.

**The proof is complete.**

### 12.2.3

First we have that

$$
a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\mathrm{d}x=0
$$

and

$$
b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin x\mathrm{d}x=\frac{2}{\pi}\cdot\frac{1-\cos n\pi}{n}
$$

From which we can get

$$
f(x)=\sum_{n=1}^{\infty}b_n\sin nx=\sum_{n=1}^{\infty}\frac{2}{\pi}\cdot\frac{1-(-1)^2}{n}\sin nx=\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\sin(2n-1)\pi}{2n-1};
$$

Thus we have

$(2)$

### 12.3.6

**The proof is complete.**

### 12.3.7

First we calculate that

Then apply **Pasavel Equality** to each and we have

From which we can obtain

### 12.3.8

Apply odd extension then we have $a_n=0$ and

$$
b_n=\frac{2}{\pi}\bigg(\int_0^1\frac{\pi-1}{2}x\sin x\mathrm{d}x+\int_1^{\pi}\frac{\pi-x}{2}\sin x\mathrm{d}x)=\frac{\sin n}{n^2}
$$

So that

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

**The proof is complete.**

### 12.3.9

$(1)$

$$
\sum_{n=1}^{\infty}\frac{\sin n}{n}=f'(0)=\frac{\pi-1}{2};\sum_{n=1}^{\infty}\bigg(\frac{\sin n}{n}\bigg)^2=f(1)=\frac{\pi-1}{2}.
$$

$(2)$

**The proof is complete.**

### 12.c.1

Since that $\displaystyle\frac{1}{\ln x}$ uniformly converge to zero and $\displaystyle\sum_{n=1}^{\infty}\sin nx$ is uniformly bounded when $n\notin 2k\pi,k\in\mathbb{Z}$,

$\displaystyle\sum_{n=2}^{\infty}\frac{\sin nx}{\ln x}$ is uniformly convergent when $n\notin 2k\pi, k\in\mathbb{Z}$.

However, by **Bessel Inequality**, since that

$$
\left||f\right||\geq\sum_{n=2}^{\infty}\frac{1}{\ln^2 n}\text{ which diverges},
$$

$f$ is not square-integrable.

**The proof is complete.**

### 12.c.2

$(1)$

$(2)$

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

**The proof is complete.**

### 12.c.3

First we have that

So that

$(1)$ Since $f(x)$ decreases on $(0,2\pi)$, $\displaystyle f(\frac{x}{n})-f(\frac{x+\pi}{n})\leq 0$, which means that $b_n\geq 0$;

$(2)$ Since $f(x)$ increases on $(0,2\pi)$, $\displaystyle f(\frac{x}{n})-f(\frac{x+\pi}{n})\geq 0$, which means that $b_n\leq 0$.

**The proof is complete.**

### 12.c.4

Since that

Therefore we have $\displaystyle a_n=O\bigg(\frac{1}{n}\bigg)$. Similarly we have $\displaystyle b_n=O\bigg(\frac{1}{n}\bigg)$.

**The proof is complete.**

Maths Analysis II - Homework #24 20220524

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