13.3 习题课讲解

解 (1) 计算可得$$\begin{array}{}\text{(1)}& f(x)\sim \sum _{n=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{n-1}}{n}\sin nx=:S(x)\end{array}$$ 由Dirichlet判别法知，$\sum _{n=1}^{+\mathrm{\infty }}\frac{(-{\mathrm{e}}^{\mathrm{i}x}{)}^n}{n}$在区间$[-\pi +\delta ,\pi -\delta ]$上一致收敛，故原Fourier级数在区间$(-\pi ,\pi )$上内闭一致收敛到$f$。故和函数$S$在$(-\pi ,\pi )$上连续且有$S(x)=f(x)$，另有$S(-\pi )=S(\pi )=0$。

(2) 由Parseval等式可得$$\begin{array}{}\text{(2)}& \frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^2\mathrm{d}x=\frac{A_0^2}{2}+\sum _{n=1}^{+\mathrm{\infty }}(A_n^2+B_n^2)=\sum _{n=1}^{+\mathrm{\infty }}\frac{4}{n^2}\Rightarrow \sum _{n=1}^{+\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}\end{array}$$

(3) 令$x=\frac{\pi }{2}$可得$$\begin{array}{}\text{(3)}& \frac{\pi }{2}=\sum _{n=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{n-1}}{n}\sin \frac{n\pi }{2}=\sum _{k=1}^{+\mathrm{\infty }}\frac{2(-1{)}^{k-1}}{2k-1}\Rightarrow \sum _{k=1}^{+\mathrm{\infty }}\frac{(-1{)}^{k-1}}{2k-1}=\frac{\pi }{4}\end{array}$$ $◻$

解 (1) 计算可得$$\begin{array}{}\text{(4)}& f(x)\sim \frac{\pi }{2}-\sum _{n=1}^{+\mathrm{\infty }}\frac{4}{(2n-1{)}^2\pi }\cos (2n-1)x\end{array}$$ 其一致收敛到$f$自身。

(2) 令$x=0$可得$$\begin{array}{}\text{(5)}& 0=\frac{\pi }{2}-\sum _{n=1}^{+\mathrm{\infty }}\frac{4}{(2n-1{)}^2\pi }\Rightarrow \sum _{n=1}^{+\mathrm{\infty }}\frac{1}{(2n-1{)}^2}=\frac{{\pi }^2}{8}\end{array}$$

(3) 由Parseval等式可得$$\begin{array}{}\text{(6)}& \frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^2\mathrm{d}x=\frac{A_0^2}{2}+\sum _{n=1}^{+\mathrm{\infty }}(A_n^2+B_n^2)=\frac{{\pi }^2}{2}+\sum _{n=1}^{+\mathrm{\infty }}\frac{16}{(2n-1{)}^4{\pi }^2}\Rightarrow \sum _{n=1}^{+\mathrm{\infty }}\frac{1}{(2n-1{)}^4}=\frac{{\pi }^4}{96}\end{array}$$

(4) 设级数收敛于$S$，则有$$\begin{array}{}\text{(7)}& S=\frac{S}{2^4}+\frac{{\pi }^4}{96}\Rightarrow S=\frac{{\pi }^4}{90}\end{array}$$ $◻$

解 计算可得$$\begin{array}{}\text{(8)}& f(x)=x^2\mathrm{sgn}x\sim \sum _{n=1}^{+\mathrm{\infty }}[\frac{(-1{)}^{n-1}}{n\pi }-\frac{4(1-(-1{)}^n)}{n^3\pi }]\sin nx,x\in (-\pi ,\pi )\end{array}$$ $◻$