## 张祖锦2023年数学专业真题分类70天之第25天
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553、 (8)、 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{(2n+1)2^\{2n+1\}\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\left.\sum\_\{n=1\}^\infty \frac\{t^\{2n+1\}\}\{2n+1\}\right|\_\{t=\frac\{1\}\{2\}\} =\sum\_\{n=1\}^\infty \int\_0^\frac\{1\}\{2\}s^\{2n\}\mathrm\{ d\} s =\int\_0^\frac\{1\}\{2\}\sum\_\{n=1\}^\infty s^\{2n\}\mathrm\{ d\} s\\\\ =&\int\_0^\frac\{1\}\{2\}\frac\{s^2\}\{1-s^2\}\mathrm\{ d\} s =\int\_0^\frac\{1\}\{2\}\left(-1+\frac\{1\}\{1-s^2\}\right)\mathrm\{ d\} s\\\\ =&-\frac\{1\}\{2\}+\frac\{1\}\{2\}\int\_0^\frac\{1\}\{2\}\left(\frac\{1\}\{1-s\}+\frac\{1\}\{1+s\}\right)\mathrm\{ d\} s =-\frac\{1\}\{2\}+\left.\frac\{1\}\{2\}\ln\frac\{1+s\}\{1-s\}\right|\_\{s=0\}^\{s=\frac\{1\}\{2\}\}\\\\ =&\frac\{1\}\{2\}\ln 3-\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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554、 8、 (15 分) 求函数 $\displaystyle f(x)=\arctan \frac\{1-2x\}\{1+2x\}$ 在 $\displaystyle x=0$ 处的幂级数展开式, 并求 $\displaystyle \sum\_\{n=0\}^\infty \frac\{(-1)^n\}\{2n+1\}$ 的和. (西北大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} f'(x)=-\frac\{2\}\{1+4x^2\} =-2\sum\_\{n=0\}^\infty (-1)^n (4x^2)^n, |x| < \frac\{1\}\{2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} f(x)&=f(0)+\int\_0^xf'(t)\mathrm\{ d\} t\\\\ &=\frac\{\pi\}\{4\}-2\sum\_\{n=0\}^\infty \frac\{(-1)^n 4^nx^\{2n+1\}\}\{2n+1\}, |x|\leq \frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x=\frac\{1\}\{2\}$ 得
\begin\{aligned\} 0=\frac\{\pi\}\{4\}-2\sum\_\{n=0\}^\infty \frac\{(-1)^n\frac\{1\}\{2\}\}\{2n+1\} \Rightarrow \sum\_\{n=0\}^\infty \frac\{(-1)^n\}\{2n+1\}=\frac\{\pi\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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555、 2、 (20 分) 求级数 $\displaystyle \sum\_\{n=0\}^\infty \frac\{n^2+1\}\{\left(\frac\{1\}\{2\}\right)^n n!\}x^n$ 的和函数. (西南财经大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\sum\_\{n=0\}^\infty \frac\{n(n-1)+n+1\}\{n!\}(2x)^n\\\\ =&4x^2\sum\_\{n=2\}^\infty \frac\{(2x)^\{n-2\}\}\{(n-2)!\} +2x\sum\_\{n=1\}^\infty \frac\{(2x)^\{n-1\}\}\{(n-1)!\} +\sum\_\{n=0\}^\infty \frac\{(2x)^n\}\{n!\}\\\\ =&4x^2\mathrm\{e\}^\{2x\}+2x\mathrm\{e\}^\{2x\}+\mathrm\{e\}^\{2x\} =(4x^2+2x+1)\mathrm\{e\}^\{2x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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556、 5、 (15 分) 求幂级数 $\displaystyle \sum\_\{n=0\}^\infty \frac\{x^\{2n+1\}\}\{2n+1\}$ 的收敛域及和函数. (新疆大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 已知级数的收敛域为 $\displaystyle (-1,1)$. 当 $\displaystyle |x| < 1$ 时,
\begin\{aligned\} \sum\_\{n=0\}^\infty \frac\{x^\{2n+1\}\}\{2n+1\} &=\sum\_\{n=0\}^\infty \int\_0^x t^\{2n\}\mathrm\{ d\} t =\int\_0^x \frac\{1\}\{1-t^2\}\mathrm\{ d\} t =\frac\{1\}\{2\}\ln \frac\{1+x\}\{1-x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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557、 (2)、 无穷级数 $\displaystyle \sum\_\{n=1\}^\infty n\left(\frac\{1\}\{2\}\right)^n=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (长安大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\left. \sum\_\{n=1\}^\infty nx^n \right|\_\{x=\frac\{1\}\{2\}\} =\left. x\sum\_\{n=1\}^\infty (x^n)' \right|\_\{x=\frac\{1\}\{2\}\} =\left. x\left(\sum\_\{n=1\}^\infty x^n\right)' \right|\_\{x=\frac\{1\}\{2\}\}\\\\ =&\left. x\left(\sum\_\{n=0\}^\infty x^n\right)' \right|\_\{x=\frac\{1\}\{2\}\} =\left. \frac\{x\}\{(1-x)^2\} \right|\_\{x=\frac\{1\}\{2\}\}=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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558、 (3)、 求函数 $\displaystyle f(x)=\sin x$ 在 $\displaystyle x\_0=\frac\{\pi\}\{6\}$ 处的 Taylor 级数, 并确定收敛范围. (中国海洋大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} f(x)=&\sin x=\sin \left(x-\frac\{\pi\}\{6\}+\frac\{\pi\}\{6\}\right)\\\\ =&\sin \left(x-\frac\{\pi\}\{6\}\right)\frac\{\sqrt\{3\}\}\{2\}+\cos \left(x-\frac\{\pi\}\{6\}\right)\frac\{1\}\{2\}\\\\ =&\frac\{\sqrt\{3\}\}\{2\}\sum\_\{k=0\}^\infty \frac\{(-1)^k\}\{(2k+1)!\}\left(x-\frac\{\pi\}\{6\}\right)^\{2k+1\}\\\\ &+\frac\{1\}\{2\}\sum\_\{k=0\}^\infty \frac\{(-1)^k\}\{(2k)!\}\left(x-\frac\{\pi\}\{6\}\right)^\{2k\}, |x| < \infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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559、 9、 (15 分) 若 $\displaystyle x$ 取值于 $\displaystyle \mathbb\{R\}$, 讨论级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{n\}x^n$ 的收敛集合, 并给出证明. 若 $\displaystyle x$ 取值于 $\displaystyle \mathbb\{C\}$ 重复上面的问题. (中国科学技术大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 当 $\displaystyle x\in\mathbb\{R\}$ 时, 由幂级数的收敛半径公式知 $\displaystyle R=1$, 从而收敛集合为 $\displaystyle (-1,1]$. (2)、 当 $\displaystyle x\in\mathbb\{C\}$ 时, 若 $\displaystyle |x| < 1$, 则幂级数收敛且为绝对收敛, 当 $\displaystyle x: |x|=1, x\neq -1$ 时, $\displaystyle x=\mathrm\{e\}^\{\mathrm\{ i\}\theta\}, -\pi < \theta < \pi$, 而
\begin\{aligned\} \mbox\{原式\}=\sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{n\}\mathrm\{e\}^\{\mathrm\{ i\} n\theta\} =\sum\_\{n=1\}^\infty \frac\{(-1)^n \cos n\theta\}\{n\} +\sum\_\{n=1\}^\infty \frac\{(-1)^n \sin n\theta\}\{n\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Dirichlet 判别法知即知原式收敛. 于是收敛集合为
\begin\{aligned\} \left\\{x\in\mathbb\{C\}; |x|\leq 1, x\neq -1\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
560、 7、 (15 分) 求幂级数 $\displaystyle \sum\_\{n=2\}^\infty \frac\{x^\{n-2\}\}\{n\cdot 3^n\}$ 的收敛域及和函数. (中国矿业大学(北京)2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 易知收敛域为 $\displaystyle [-3,3)$. 当 $\displaystyle x\in [-3,0)\cup (0,3)$ 时,
\begin\{aligned\} \mbox\{原式\}=&\frac\{1\}\{x^2\}\sum\_\{n=2\}^\infty \frac\{1\}\{n\}\left(\frac\{x\}\{3\}\right)^n =\frac\{1\}\{x^2\}\sum\_\{n=2\}^\infty \int\_0^\frac\{x\}\{3\}t^\{n-1\}\mathrm\{ d\} t =\frac\{1\}\{x^2\}\int\_0^x \frac\{t\}\{1-t\}\mathrm\{ d\} t =\frac\{-x-3\ln\frac\{3-x\}\{3\}\}\{3x^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原式 $\displaystyle =\left\\{\begin\{array\}\{llllllllllll\}\frac\{-x-3\ln\frac\{3-x\}\{3\}\}\{3x^2\},&x\in [-3,0)\cup (0,3),\\\\ 0,&x=0.\end\{array\}\right.$跟锦数学微信公众号. [在线资料](
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561、 3、 (15 分) 求 $\displaystyle f(x)=\frac\{x^2\}\{(1+x^2)^2\}$ 的幂级数展开式. (中山大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} f(x)=&\left. \frac\{t\}\{(1+t)^2\} \right|\_\{t=x^2\}=\left. t\left(\sum\_\{n=0\}^\infty (-1)^\{n-1\}t^n\right)' \right|\_\{t=x^2\}\\\\ =&\left. t\sum\_\{n=1\}^\infty (-1)^\{n-1\} nt^\{n-1\} \right|\_\{t=x^2\} =\sum\_\{n=1\}^\infty (-1)^\{n-1\} nx^\{2n\}, |x| < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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562、 8、 (10 分) 求幂级数 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n \frac\{x^n\}\{n\}$ 的收敛半径, 收敛域以及和函数, 并求 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^\{n-1\}\}\{n\}$. (重庆大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 易知收敛半径为 $\displaystyle 1$, 收敛域为 $\displaystyle (-1,1]$. 当
\begin\{aligned\} \mbox\{原式\}=&\sum\_\{n=1\}^\infty \frac\{(-x)^n\}\{n\} =\sum\_\{n=1\}^\infty \int\_0^\{-x\}t^\{n-1\}\mathrm\{ d\} t =\int\_0^\{-x\}\frac\{\mathrm\{ d\} t\}\{1-t\} =-\ln (1+x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^\{n-1\}\}\{n\}=\ln 2$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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563、 (5)、 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{4n-2\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. [题目有问题, 跟锦数学微信公众号没法做哦.] (重庆师范大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [题目有问题, 跟锦数学微信公众号没法做哦.]跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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564、 (3)、 求 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{\sqrt\{n\}\}x^n$ 的收敛区间和收敛域. (重庆师范大学2023年数学分析考研试题) [幂级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle a\_n=\frac\{1\}\{\sqrt\{n\}\}$, 则 $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_\{n+1\}\}\{a\_n\}=1$, 而原幂级数的收敛半径为 $\displaystyle R=1$, 收敛区间为 $\displaystyle (-1,1)$. 由 Leibniz 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{\sqrt\{n\}\}$ 收敛, 而原幂级数的收敛域为 $\displaystyle [-1,1)$.跟锦数学微信公众号. [在线资料](
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565、 5、 (15 分) (1)、 求函数 $\displaystyle f(x)=|x|, x\in [-\pi,\pi]$ 的傅里叶级数. (2)、 证明 $\displaystyle \sum\_\{k=1\}^\infty \frac\{1\}\{k^2\}=\frac\{\pi^2\}\{6\}$. (北京科技大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设
\begin\{aligned\} f(x)\sim \frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty (a\_n\cos nx+b\_n\sin nx), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由 $\displaystyle f(x)$ 为偶函数知 $\displaystyle b\_n=0$,
\begin\{aligned\} a\_0=&\frac\{1\}\{\pi\}\int\_\{-\pi\}^\pi f(x)\mathrm\{ d\} x=\pi,\\\\ a\_n=&\frac\{1\}\{\pi\}\int\_\{-\pi\}^\pi f(x)\cos nx\mathrm\{ d\} x =\frac\{2[-1+(-1)^n]\}\{\pi n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} f(x)\sim \frac\{\pi\}\{2\}+\sum\_\{k=1\}^\infty \frac\{-4\}\{\pi(2k-1)^2\}\cos(2k-1)x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x=0$, 即得 $\displaystyle \sum\_\{k=1\}^\infty \frac\{1\}\{(2k-1)^2\}=\frac\{\pi^2\}\{8\}$,
\begin\{aligned\} S\equiv \sum\_\{k=1\}^\infty \frac\{1\}\{k^2\}=&\sum\_\{k=1\}^\infty \frac\{1\}\{(2k-1)^2\}-\sum\_\{k=1\}^\infty \frac\{1\}\{(2k)^2\} =\frac\{\pi^2\}\{8\}-\frac\{1\}\{4\}S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
解得 $\displaystyle S=\frac\{\pi^2\}\{6\}$.跟锦数学微信公众号. [在线资料](
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---
566、 10、 (15 分, 其中第一问 9 分, 第二问 6 分) 解答下列问题, 其中 $\displaystyle a\in (0,1)$. (1)、 求 $\displaystyle f(x)=\cos ax$ 在 $\displaystyle [-\pi,\pi]$ 上傅里叶级数; (2)、 证明:
\begin\{aligned\} \pi=\int\_0^\pi \frac\{\sin x\}\{x\}\mathrm\{ d\} x+\sum\_\{n=1\}^\infty (-1)^n\int\_0^\pi \frac\{2x\sin x\}\{x^2-n^2\pi^2\}\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(北京师范大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 $\displaystyle f$ 是偶函数知 $\displaystyle f(x)\sim \frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty a\_n\cos nx$, 则
\begin\{aligned\} a\_0&=\frac\{2\}\{\pi\}\int\_0^\pi f(x)\mathrm\{ d\} x=\frac\{2\sin \pi a\}\{\pi a\},\\\\ n\geq 1\Rightarrow a\_n&=\frac\{2\}\{\pi\}\int\_0^\pi f(x)\cos nx\mathrm\{ d\} x =\frac\{2(-1)^n a \sin \pi a\}\{\pi(a^2-n^2)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f(x)\sim \frac\{\sin \pi a\}\{\pi a\} +\frac\{2a \sin \pi a\}\{\pi\}\sum\_\{n=1\}^\infty \frac\{(-1)^n \}\{a^2-n^2\} \cos nx$. (2)、 令 $\displaystyle x=0$ 得
\begin\{aligned\} 1=&\frac\{\sin \pi a\}\{\pi a\}+\frac\{2a\sin \pi a\}\{\pi\}\sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{a^2-n^2\}\\\\ =&\frac\{\sin \pi a\}\{\pi\}\left\[\frac\{1\}\{a\}+\sum\_\{n=1\}^\infty \frac\{(-1)^n 2a\}\{a^2-n^2\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle a=\frac\{x\}\{\pi\}$ 得
\begin\{aligned\} 1=&\frac\{\sin x\}\{\pi\}\left\[\frac\{\pi\}\{x\}+\sum\_\{n=1\}^\infty \frac\{(-1)^n \cdot 2\frac\{x\}\{\pi\}\}\{\frac\{x^2\}\{\pi^2\}-n^2\}\right\]\\\\ =&\frac\{\sin x\}\{x\}+\sum\_\{n=1\}^\infty \frac\{(-1)^n 2x\sin x\}\{x^2-n^2\pi^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle x$ 在 $\displaystyle [0,\pi]$ 上积分, 并将求和与积分交换次序 (利用实变函数非负可测函数的逐项积分) 即知结论成立.跟锦数学微信公众号. [在线资料](
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---
567、 (6)、 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}x,&x\in [-\pi,0),\\\\ 0,&x\in [0,\pi)\end\{array\}\right.$ 的 Fourier 级数的和函数 $\displaystyle S(x)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (电子科技大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}x,&-\pi < x < 0,\\\\ 0,&0 < x < \pi,\\\\ -\frac\{\pi\}\{2\},&x=\pm \pi.\end\{array\}\right.$跟锦数学微信公众号. [在线资料](
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---
568、 (8)、 将 $\displaystyle f(x)=\arcsin(\sin x)$ 展成傅里叶级数. (广西大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f$ 是奇函数, 且
\begin\{aligned\} 0\leq x\leq\frac\{\pi\}\{2\}\Rightarrow& f(x)=x,\\\\ \frac\{\pi\}\{2\}\leq x\leq \pi\Rightarrow& 0\leq \pi-x\leq \frac\{\pi\}\{2\}\Rightarrow f(x)=\arcsin \left(\sin (\pi-x)\right)=\pi-x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}x,&0\leq x\leq \frac\{\pi\}\{2\},\\\\ \pi-x,&\frac\{\pi\}\{2\}\leq x\leq \pi.\end\{array\}\right.$ 设 $\displaystyle f(x)\sim\sum\_\{n=1\}^\infty b\_n\sin nx$, 其中
\begin\{aligned\} b\_n=\frac\{2\}\{\pi\}\int\_0^\pi f(x)\sin nx\mathrm\{ d\} x =\left\\{\begin\{array\}\{llllllllllll\}0,&n=2k,\\\\ \frac\{4(-1)^k\}\{\pi(2k+1)^2\},&n=2k+1.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle f(x)\sim\frac\{4\}\{\pi\}\sum\_\{k=0\}^\infty \frac\{(-1)^k\}\{(2k+1)^2\}\sin (2k+1)x$.跟锦数学微信公众号. [在线资料](
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---
569、 8、 设 $\displaystyle f(x)$ 在 $\displaystyle (-\infty,+\infty)$ 上可导, 若
\begin\{aligned\} f(x)=f(x+2k)=f(x+b), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle k$ 为正整数, $\displaystyle b$ 为正无理数, 则 $\displaystyle f(x)$ 为一常数 (用傅里叶级数证明). (华中科技大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 $\displaystyle f(x)=f(x+2k)$ 知 $\displaystyle f$ 是 $\displaystyle 2k$-周期函数. 又由 $\displaystyle f$ 可导知可设
\begin\{aligned\} f(x)=\frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty \left(a\_n\cos \frac\{n\pi\}\{k\} x+b\_n\sin \frac\{n\pi\}\{k\}x\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} a\_n=\int\_\{-k\}^k f(x)\cos \frac\{n\pi\}\{k\}x\mathrm\{ d\} x, b\_n=\int\_\{-k\}^k f(x)\sin \frac\{n\pi\}\{k\} x\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle f(x)=f(x+b)$ 知
\begin\{aligned\} a\_n&=\int\_\{-k\}^k f(x+b) \cos \frac\{n\pi\}\{k\} x\mathrm\{ d\} x \stackrel\{x+b=t\}\{=\}\int\_\{-k+b\}^\{k+b\} f(t)\cos \frac\{n\pi\}\{k\} (t-b)\mathrm\{ d\} t\\\\ &=\int\_\{-k+b\}^\{k+b\}f(t)\left\[\cos \frac\{n\pi\}\{k\} t\cos b\frac\{n\pi\}\{k\} +\sin \frac\{n\pi\}\{k\} t\sin b\frac\{n\pi\}\{k\}\right\]\mathrm\{ d\} t\\\\ &=\cos b\frac\{n\pi\}\{k\} \int\_\{-k+b\}^\{k+b\}f(t) \cos \frac\{n\pi\}\{k\} t\mathrm\{ d\} t +\sin b\frac\{n\pi\}\{k\} \int\_\{-k+b\}^\{k+b\}f(t)\sin \frac\{n\pi\}\{k\} t\mathrm\{ d\} t\\\\ &=\cos b\frac\{n\pi\}\{k\} \int\_\{-k\}^k f(t)\cos \frac\{n\pi\}\{k\} t\mathrm\{ d\} t +\sin b\frac\{n\pi\}\{k\} \int\_\{-k\}^k f(t)\sin \frac\{n\pi\}\{k\} t\mathrm\{ d\} t\\\\ &=a\_n\cos b\frac\{n\pi\}\{k\} +b\_n\sin b\frac\{n\pi\}\{k\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理,
\begin\{aligned\} b\_n=b\_n\cos b\frac\{n\pi\}\{k\}-a\_n\sin b\frac\{n\pi\}\{k\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
联立即可求得 $\displaystyle a\_n=b\_n=0\ (n\geq 1)$. 故 $\displaystyle f(x)=\frac\{a\_0\}\{2\}$ 是常数.跟锦数学微信公众号. [在线资料](
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---
570、 (2)、 设 $\displaystyle f(x)=\pi^2-x^2, x\in (-\pi,\pi]$, 求 $\displaystyle f$ 的 Fourier 级数. (华中师范大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} a\_0=\frac\{2\}\{\pi\} \int\_0^\pi x^2\mathrm\{ d\} x=\frac\{2\pi^2\}\{3\},\quad n\geq 1\Rightarrow a\_n=\frac\{2\}\{\pi\} \int\_0^\pi x^2\cos nx\mathrm\{ d\} x =\frac\{4(-1)^n\}\{n^2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle x^2\sim\frac\{\pi^2\}\{3\}+4\sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{n^2\}\cos nx.$ 故
\begin\{aligned\} f(x)\sim\frac\{2\pi^2\}\{3\}+4\sum\_\{n=1\}^\infty \frac\{(-1)^\{n-1\}\}\{n^2\}\cos nx. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
571、 7、 (15 分) 幂级数 $\displaystyle \sum\_\{n=1\}^\infty c\_nx^n$ 的收敛半径为 $\displaystyle +\infty$, 其和函数 $\displaystyle f(x)$ 是周期为 $\displaystyle 2\pi$ 的奇函数, $\displaystyle f(x)$ 的傅里叶级数为 $\displaystyle \sum\_\{n=1\}^\infty b\_n\sin nx$. 求证: (1)、 (5 分) $\displaystyle c\_\{2k\}=0, \forall\ k\in\mathbb\{N\}$. (2)、 (10 分) $\displaystyle \forall\ k\in\mathbb\{N\}, \sum\_\{n=1\}^\infty n^k |b\_n|$ 收敛, 并求 $\displaystyle \sum\_\{n=1\}^\infty n^\{2k-1\}b\_n$. (用幂级数系数表示) (南京大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由题设, $\displaystyle f(x)=\sum\_\{n=1\}^\infty c\_nx^n$ 是奇函数, 而
\begin\{aligned\} &\sum\_\{n=1\}^\infty c\_n(-x)^n=f(-x)=-f(x)=-\sum\_\{n=1\}^\infty c\_nx^n\\\\ \Rightarrow&\sum\_\{k=1\}^\infty c\_\{2k\}x^\{2k\}=-\sum\_\{k=1\}^\infty c\_\{2k\}x^\{2k\} \Rightarrow 2\sum\_\{k=1\}^\infty c\_\{2k\}x^\{2k\}=0\Rightarrow c\_\{2k\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由第 1 步知
\begin\{aligned\} &\sum\_\{k=0\}^\infty c\_\{2k+1\}x^\{2k+1\} =f(x)=\sum\_\{n=1\}^\infty b\_n \sin nx\\\\ =&\sum\_\{n=1\}^\infty b\_n \sum\_\{k=0\}^\infty \frac\{(-1)^k\}\{(2k+1)!\}(nx)^\{2k+1\} =\sum\_\{k=0\}^\infty \left\[\sum\_\{n=1\}^\infty n^\{2k+1\}b\_n \frac\{(-1)^k\}\{(2k+1)!\}\right\] x^\{2k+1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} &c\_\{2k+1\}=\sum\_\{n=1\}^\infty n^\{2k+1\}b\_n \frac\{(-1)^k\}\{(2k+1)!\}\\\\ \Rightarrow& \sum\_\{n=1\}^\infty n^\{2k+1\}b\_n=(-1)^k (2k+1)!c\_\{2k+1\}\\\\ \Rightarrow&\sum\_\{n=1\}^\infty n^\{2k-1\}b\_n=(-1)^\{k-1\} (2k-1)!c\_\{2k-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \lim\_\{n\to\infty\}n^\{2k+3\}sb\_n=0\Rightarrow \lim\_\{n\to\infty\}\frac\{n^kb\_n\}\{\frac\{1\}\{n^\{k+3\}\}\}=0 \stackrel\{\mbox\{比较判别法\}\}\{\Rightarrow\}\sum\_\{n=1\}^\infty n^k |b\_n| < \infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
572、 8、 将 $\displaystyle f(x)=1-x^2\ (0\leq x\leq \pi)$ 展开为余弦级数, 并求 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{n^2\}$. (南京航空航天大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle g(x)=x^2$,
\begin\{aligned\} g(x)\sim \frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty a\_n\cos nx, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} a\_0&=\frac\{2\}\{\pi\}\int\_0^\pi g(x)\mathrm\{ d\} x=\frac\{2\pi^2\}\{3\},\\\\ n\geq 1\Rightarrow a\_n&=\frac\{2\}\{\pi\}\int\_0^\pi g(x)\cos nx\mathrm\{ d\} x=\frac\{4(-1)^n\}\{n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} g(x)&\sim \frac\{\pi^2\}\{3\}+4\sum\_\{n=1\}^\infty \frac\{(-1)^n\cos nx\}\{n^2\},\\\\ f(x)&\sim 1-\frac\{\pi^2\}\{3\}-4\sum\_\{n=1\}^\infty \frac\{(-1)^n\cos nx\}\{n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x=0$ 得
\begin\{aligned\} 1=1-\frac\{\pi^2\}\{3\}-4\sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{n^2\} \Rightarrow \sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{n^2\}=-\frac\{\pi^2\}\{12\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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573、 12、 把函数 $\displaystyle f(x)=x\ (0\leq x\leq \pi)$ 展开为余弦级数, 并指出收敛性, 再利用该级数求 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}$. (上海大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设
\begin\{aligned\} f(x)\sim \frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty a\_n\cos nx, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} a\_0&=\frac\{2\}\{\pi\}\int\_0^\pi f(x)\mathrm\{ d\} x=\pi,\\\\ n\geq 1\Rightarrow a\_n&=\frac\{2\}\{\pi\}\int\_0^\pi f(x)\cos nx\mathrm\{ d\} x=\left\\{\begin\{array\}\{llllllllllll\}0,&n=2k,\\\\ -\frac\{4\}\{(2k-1)^2\pi\},&n=2k-1.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f(x)\sim\frac\{\pi\}\{2\}-\frac\{4\}\{\pi\}\sum\_\{k=1\}^\infty \frac\{\cos (2k-1)x\}\{(2k-1)^2\}$. 令 $\displaystyle x=0$ 得
\begin\{aligned\} \sum\_\{k=1\}^\infty \frac\{1\}\{(2k-1)^2\}=\frac\{\pi^2\}\{8\}\Rightarrow& S\equiv \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}=\sum\_\{k=1\}^\infty \frac\{1\}\{(2k)^2\}+\sum\_\{k=1\}^\infty \frac\{1\}\{(2k-1)^2\} =\frac\{1\}\{4\}S+\frac\{\pi^2\}\{8\}\\\\ \Rightarrow& \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}=S=\frac\{\pi^2\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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574、 4、 设 $\displaystyle f(x)=|x|, x\in [-\pi,\pi]$. (1)、 求 $\displaystyle f(x)$ 的 Fourier 展开式, 并证明 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}=\frac\{\pi^2\}\{6\}$; (2)、 求 $\displaystyle \int\_0^\frac\{1\}\{2\}\frac\{\ln(1+2x)\}\{x\}\mathrm\{ d\} x$. (天津大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 可设
\begin\{aligned\} f(x)\sim \frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty a\_n\cos nx, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} a\_0&=\frac\{2\}\{\pi\}\int\_0^\pi |f(x)|\mathrm\{ d\} x=\pi,\\\\ n\geq 1\Rightarrow a\_n&=\frac\{2\}\{\pi\}\int\_0^\pi |f(x)|\cos nx\mathrm\{ d\} x=\left\\{\begin\{array\}\{llllllllllll\}0,&n=2k,\\\\ -\frac\{4\}\{(2k-1)^2\pi\},&n=2k-1.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f(x)\sim \frac\{\pi\}\{2\}-4\sum\_\{k=1\}^\infty \frac\{\cos(2k-1)x\}\{(2k-1)^2\}$. 令 $\displaystyle x=0$ 得 $\displaystyle \sum\_\{k=1\}^\infty \frac\{1\}\{(2k-1)^2\}=\frac\{\pi^2\}\{8\}$,
\begin\{aligned\} \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}=\sum\_\{k=1\}^\infty \frac\{1\}\{(2k-1)^2\}+\sum\_\{k=1\}^\infty \frac\{1\}\{(2k)^2\} =\frac\{\pi^2\}\{8\}+\frac\{1\}\{4\}\sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}\Rightarrow \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}=\frac\{\pi^2\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、
\begin\{aligned\} \mbox\{原式\}&\stackrel\{2x=t\}\{=\}\int\_0^1 \frac\{\ln (1+t)\}\{t\}\mathrm\{ d\} t =\int\_0^1 \sum\_\{n=0\}^\infty (-1)^n\frac\{t^n\}\{n+1\}\mathrm\{ d\} t\\\\ =&\sum\_\{n=0\}^\infty \frac\{(-1)^n\}\{n+1\}\int\_0^1 t^n\mathrm\{ d\} t =\sum\_\{n=0\}^\infty \frac\{(-1)^n\}\{(n+1)^2\} =\sum\_\{k=1\}^\infty \frac\{1\}\{(2k-1)^2\}-\sum\_\{k=1\}^\infty \frac\{1\}\{(2k)^2\}\\\\ =&\frac\{\pi^2\}\{8\}-\frac\{1\}\{4\}\cdot\frac\{\pi^2\}\{6\}=\frac\{\pi^2\}\{12\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里积分与级数可交换是因为由 Abel 判别法, $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{n+1\}\cdot t^n$ 关于 $\displaystyle t\in [0,1]$ 一致收敛.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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575、 6、 (15 分) 把函数 $\displaystyle f(x)=(x-\pi)^2$ 在 $\displaystyle (0,\pi)$ 上展开成余弦级数, 并进一步证明
\begin\{aligned\} \frac\{\pi^2\}\{6\}=1+\frac\{1\}\{2^2\}+\frac\{1\}\{3^2\}+\cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(新疆大学2023年数学分析考研试题) [Fourier级数 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设
\begin\{aligned\} f(x)\sim \frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty a\_n\cos nx, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} a\_0&=\frac\{2\}\{\pi\}\int\_0^\pi f(x)\mathrm\{ d\} x=\frac\{2\pi^2\}\{3\},\\\\ n\geq 1\Rightarrow a\_n&=\frac\{2\}\{\pi\}\int\_0^\pi f(x)\cos nx\mathrm\{ d\} x=\frac\{4\}\{n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f(x)\sim\frac\{\pi^2\}\{3\}+4\sum\_\{n=1\}^\infty \frac\{\cos nx\}\{n^2\}$. 令 $\displaystyle x=0$ 得
\begin\{aligned\} \pi^2=\frac\{\pi^2\}\{3\}+4\sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}\Rightarrow \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}=\frac\{\pi^2\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 09:09:36
