中册 6.5 傅里叶级数 第11题

\section*{解题过程：} （1）与题9（5）的余弦级数展开相同． 傅里叶级数为 $\displaystyle x^{2}=\frac{\pi^{2}}{3}+4 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos n x, x \in(-\pi, \pi)$ ． 由此得 $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos n x=\frac{1}{4}\left(x^{2}-\frac{\pi^{2}}{3}\right), x \in(-\pi, \pi)$ ． 取 $x=\pi$ 得 $\displaystyle \pi^{2}=\frac{\pi^{2}}{3}+4 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos n \pi=\frac{\pi^{2}}{3}+4 \sum_{n=1}^{\infty} \frac{1}{n^{2}}$ ．由此得 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ ． 由帕塞瓦尔等式 $\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2}(x) \mathrm{d} x=\frac{a_{0}^{2}}{2}+\sum_{n=1}^{\infty}\left(a_{n}^{2}+b_{n}^{2}\right)$ ，有 $$ \frac{1}{\pi} \int_{-\pi}^{\pi}\left(x^{2}\right)^{2} \mathrm{~d} x=\frac{1}{2} \cdot\left(\frac{2}{3} \pi^{2}\right)+\sum_{n=1}^{\infty}\left(\frac{4(-1)^{n}}{n^{2}}\right)^{2} . $$ 故 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{90}$ ． 最后证明 $\displaystyle \int_{0}^{1} \frac{\ln x}{1-x} \mathrm{~d} x=-\frac{\pi^{2}}{6}$ ． 令 $t=1-x$ ，则 $\displaystyle \int_{0}^{1} \frac{\ln x}{1-x} \mathrm{~d} x=\int_{0}^{1} \frac{\ln (1-t)}{t} \mathrm{~d} t$ ．由 $\displaystyle \ln (1-t)=-\sum_{n=1}^{\infty} \frac{t^{n}}{n}$ 得 $$ \int_{0}^{1} \frac{\ln x}{1-x} \mathrm{~d} x=-\int_{0}^{1} \sum_{n=1}^{\infty} \frac{t^{n-1}}{n} \mathrm{~d} t=-\sum_{n=1}^{\infty} \int_{0}^{1} \frac{t^{n-1}}{n} \mathrm{~d} t=-\sum_{n=1}^{\infty} \frac{1}{n^{2}}=-\frac{\pi^{2}}{6} . $$ （2）如图 6．24，$f(x)$ 按段光滑，由收玫定理知，它可以展成傅立叶级数。 $$ \begin{aligned} & a_{0}=\frac{1}{\pi} \int_{0}^{2 \pi} x^{2} \mathrm{~d} x=\frac{8}{3} \pi^{2} \\ & a_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} x^{2} \cos n x \mathrm{~d} x=\frac{4}{n^{2}}, n=1,2, \cdots \\ & b_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} x^{2} \sin n x \mathrm{~d} x=-\frac{4 \pi}{n}, n=1,2, \cdots \end{aligned} $$ 所以在 $[0,2 \pi]$ 上 $\displaystyle f(x)=\frac{4 \pi^{2}}{3}+4 \sum_{n=1}^{\infty}\left(\frac{\cos n x}{n^{2}}-\frac{\pi \sin n x}{n}\right)$ ． 令 $x=0$ 得 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ ． \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/f12c872f-2d5e-49ed-83c0-1c7e59b03c5d-399.jpg?height=782&width=1880&top_left_y=2693&top_left_x=690} \captionsetup{labelformat=empty} \caption{图 6.24} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/f12c872f-2d5e-49ed-83c0-1c7e59b03c5d-399.jpg?height=782&width=1762&top_left_y=2693&top_left_x=3453} \captionsetup{labelformat=empty} \caption{图 6.25} \end{figure} （3）如图 6．25，因为 $f(x)=\pi^{2}-x^{2}$ 为偶函数，所以 $$ \begin{aligned} b_{n} & =0, n=1,2, \cdots \\ a_{0} & =\frac{2}{\pi} \int_{0}^{\pi}\left(x^{2}-\pi^{2}\right) \mathrm{d} x=-\frac{4}{3} \pi^{2} \\ a_{n} & =\frac{2}{\pi} \int_{0}^{\pi}\left(x^{2}-\pi^{2}\right) \cos n x \mathrm{~d} x=\frac{2}{n \pi}\left[\left.\left(x^{2}-\pi^{2}\right) \sin n x\right|_{0} ^{\pi}-2 \int_{0}^{\pi} x \sin n x \mathrm{~d} x\right] \\ & =\left.\frac{4}{n^{2} \pi}\left(x \cos n x-\frac{\sin n x}{n}\right)\right|_{0} ^{\pi}=\frac{4(-1)^{n}}{n^{2}}, n=1,2, \cdots \end{aligned} $$ 由于 $f(x)$ 在 $(-\infty,+\infty)$ 内连续，所以 $\displaystyle x^{2}-\pi^{2}=-\frac{2}{3} \pi^{2}+\sum_{n=1}^{\infty} \frac{4(-1)^{n}}{n^{2}} \cos n x, x \in[-\pi, \pi]$ ． 于是 $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos n x=\frac{1}{4}\left(x^{2}-\frac{\pi^{2}}{3}\right), x \in[-\pi, \pi]$ ． 在上式中分别令 $x=\pi, x=0$ ，得 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}, \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}=-\frac{\pi^{2}}{12}$ ． 由帕塞瓦尔等式 $\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2}(x) \mathrm{d} x=\frac{a_{0}^{2}}{2}+\sum_{n=1}^{\infty}\left(a_{n}^{2}+b_{n}^{2}\right)$ 有 $$ \frac{1}{\pi} \int_{-\pi}^{\pi}\left(x^{2}-\pi^{2}\right)^{2} \mathrm{~d} x=\frac{1}{2} \cdot \frac{16 \pi^{4}}{9}+16 \sum_{n=1}^{\infty} \frac{1}{n^{4}} . $$ 故 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{90}$ ． （4）如图 6．26，$f(x)$ 按段光滑，由收玫定理知它可以展成傅里叶级数。 $$ a_{0}=\frac{1}{\pi} \int_{0}^{2 \pi} 4 x(2 \pi-x) \mathrm{d} x=\frac{16}{3} \pi^{2} $$ $$ \begin{aligned} a_{n} & =\frac{1}{\pi} \int_{0}^{2 \pi} 4 x(2 \pi-x) \cos n x \mathrm{~d} x=\frac{1}{n \pi}\left[\left.4 x(2 \pi-x) \sin n x\right|_{0} ^{2 \pi}-8 \int_{0}^{2 \pi}(\pi-x) \sin n x \mathrm{~d} x\right] \\ & =\left.\frac{8}{n^{2} \pi}\left[(\pi-x) \cos n x+\frac{\sin n x}{n}\right]\right|_{0} ^{2 \pi}=-\frac{16}{n^{2}}, n=1,2, \cdots . \\ b_{n} & =\frac{1}{\pi} \int_{0}^{2 \pi} 4 x(2 \pi-x) \sin n x \mathrm{~d} x=-\frac{1}{n \pi}\left[\left.4 x(2 \pi-x) \cos n x\right|_{0} ^{2 \pi}-8 \int_{0}^{2 \pi}(\pi-x) \cos n x \mathrm{~d} x\right] \\ & =\left.\frac{8}{n^{2} \pi}\left[(\pi-x) \sin n x-\frac{\cos n x}{n}\right]\right|_{0} ^{2 \pi}=0, n=1,2, \cdots . \end{aligned} $$ 所以 $\displaystyle 4 x(2 \pi-x)=\frac{8}{3} \pi^{2}-16 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \cos n x, x \in[0,2 \pi]$ ． 由帕塞瓦尔等式 $\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2}(x) \mathrm{d} x=\frac{a_{0}^{2}}{2}+\sum_{n=1}^{\infty}\left(a_{n}^{2}+b_{n}^{2}\right)$ 得 $$ \frac{1}{\pi} \int_{-\pi}^{\pi} 16 x^{2}(2 \pi-x)^{2} \mathrm{~d} x=\frac{1}{2}\left(\frac{16 \pi^{2}}{3}\right)^{2}+16^{2} \sum_{n=1}^{\infty} \frac{1}{n^{4}} . $$ 故 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{90}$ ． \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/f12c872f-2d5e-49ed-83c0-1c7e59b03c5d-400.jpg?height=781&width=1879&top_left_y=4068&top_left_x=870} \captionsetup{labelformat=empty} \caption{图 6.26} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/f12c872f-2d5e-49ed-83c0-1c7e59b03c5d-400.jpg?height=719&width=1638&top_left_y=4130&top_left_x=3432} \captionsetup{labelformat=empty} \caption{图 6.27} \end{figure} （5）如图 6．27，函数的傅氏系数为 $$ \begin{aligned} & b_{n}=0, n=1,2, \cdots \\ & a_{0}=\frac{2}{1} \int_{0}^{1} x^{2} \mathrm{~d} x=\frac{2}{3} \\ & a_{n}=\frac{2}{1} \int_{0}^{1} x^{2} \cos n \pi x \mathrm{~d} x=(-1)^{n} \frac{4}{\pi^{2} n^{2}}, n=1,2, \cdots \end{aligned} $$ 故 $\displaystyle f(x)=\frac{1}{3}+\frac{4}{\pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos n \pi x, x \in[0,1]$ ． 令 $x=1$ ，得 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ ． （6）如图 6．28，函数的傅氏系数为 $$ \begin{aligned} & a_{0}=\int_{0}^{1} x^{2} \mathrm{~d} x=\frac{1}{3} . \\ & a_{n}=\int_{0}^{1} x^{2} \cos n \pi x \mathrm{~d} x=(-1)^{n} \frac{2}{\pi^{2} n^{2}}(n=1,2, \cdots) \\ & b_{n}=\int_{0}^{1} x^{2} \sin n \pi x \mathrm{~d} x=\frac{1}{n \pi}\left[-\left.x^{2} \cos n \pi x\right|_{0} ^{1}+2 \int_{0}^{1} x \cos \pi n x \mathrm{~d} x\right] \end{aligned} $$ $$ \begin{aligned} & =-\frac{1}{n \pi} \cos n \pi+\left.\frac{2}{n^{2} \pi^{2}}\left[x \sin n \pi x+\frac{1}{n \pi} \cos n \pi x\right]\right|_{0} ^{1} \\ & =-\frac{1}{n \pi}(-1)^{n}+\frac{2}{n^{3} \pi^{3}}\left[(-1)^{n}-1\right], n=1,2, \cdots \end{aligned} $$ 故 $\displaystyle f(x)=\frac{1}{6}+\sum_{n=1}^{\infty} \frac{2(-1)^{n}}{\pi^{2} n^{2}} \cos n \pi x+\left\{-\frac{1}{n \pi}(-1)^{n}+\frac{2}{n^{3} \pi^{3}}\left[(-1)^{n}-1\right]\right\} \sin n \pi x, x \in[0,1)$ ． 令 $x=1$ ，得 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ ． \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/f12c872f-2d5e-49ed-83c0-1c7e59b03c5d-401.jpg?height=643&width=1555&top_left_y=2569&top_left_x=1036} \captionsetup{labelformat=empty} \caption{图 6.28} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/f12c872f-2d5e-49ed-83c0-1c7e59b03c5d-401.jpg?height=726&width=1658&top_left_y=2486&top_left_x=3267} \captionsetup{labelformat=empty} \caption{图 6.29} \end{figure} （7）如图 6．29，$f(x)$ 按段光滑，由收玫定理知它可以展成傅里叶级数。 $$ \begin{aligned} & a_{0}=\frac{1}{\pi} \int_{0}^{2 \pi}\left(\frac{\pi-x}{2}\right)^{2} \mathrm{~d} x=\frac{\pi^{2}}{6} \\ & a_{n}=\frac{1}{4 \pi} \int_{0}^{2 \pi}(\pi-x)^{2} \cos n x \mathrm{~d} x=\frac{1}{n^{2}}, n=1,2, \cdots \\ & b_{n}=\frac{1}{4 \pi} \int_{0}^{2 \pi}(\pi-x)^{2} \sin n x \mathrm{~d} x=0, n=1,2, \cdots \end{aligned} $$ 所以 $\displaystyle \frac{1}{4}(\pi-x)^{2}=\frac{1}{12} \pi^{2}+\sum_{n=1}^{\infty} \frac{1}{n^{2}} \cos n x, x \in[0,2 \pi]$ ． 由此得 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} \cos n x=\frac{1}{4}(\pi-x)^{2}-\frac{1}{12} \pi^{2}, x \in[0,2 \pi]$ 令 $x=0$ 得 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ ． 令 $x=\pi$ 得 $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}=-\frac{\pi^{2}}{12}$ ． 由帕塞瓦尔等式 $\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2}(x) \mathrm{d} x=\frac{a_{0}^{2}}{2}+\sum_{n=1}^{\infty}\left(a_{n}^{2}+b_{n}^{2}\right)$ 有 $$ \frac{1}{\pi} \int_{-\pi}^{\pi}\left(\frac{x-\pi}{2}\right)^{2} \mathrm{~d} x=\frac{1}{2} \cdot \frac{\pi^{4}}{36}+\sum_{n=1}^{\infty} \frac{1}{n^{4}} $$ 故 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{90}$ ．