## 张祖锦2023年数学专业真题分类70天之第23天
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507、 11、 已知 $\displaystyle f(x)=\sum\_\{n=0\}^\infty \frac\{1\}\{2^n+x\}$. 证明: (1)、 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上可导, 并一致连续; (2)、 反常积分 $\displaystyle \int\_0^\infty f(x)\mathrm\{ d\} x$ 发散. (华南理工大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle \frac\{1\}\{2^n+x\}\leq\frac\{1\}\{2^n\}$ 及 Weierstrass 判别法知 $\displaystyle f(x)$ 关于 $\displaystyle x\geq 0$ 一致收敛. 又由
\begin\{aligned\} \left|\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} x\}\left(\frac\{1\}\{2^n+x\}\right)\right|=\frac\{1\}\{(2^n+x)^2\}\leq \frac\{1\}\{4^n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Weierstrass 判别法知 $\displaystyle f$ 可导, 且
\begin\{aligned\} |f'(x)|=\left|\sum\_\{n=0\}^\infty \frac\{-1\}\{(2^n+x)^2\}\right| \leq \sum\_\{n=0\}^\infty \frac\{1\}\{4^n\}=\frac\{4\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle f$ 一致连续 (对 $\displaystyle \varepsilon > 0$, 可取 $\displaystyle \delta=\frac\{3\varepsilon\}\{4\}$ 即可). 最后由
\begin\{aligned\} f(x)\geq&\sum\_\{n=0\}^\infty \int\_n^\{n+1\}\frac\{1\}\{2^t+x\}\mathrm\{ d\} t =\int\_0^\infty \frac\{\mathrm\{ d\} t\}\{2^t+x\} \stackrel\{2^t=s\}\{=\}\frac\{1\}\{\ln 2\}\int\_1^\infty \frac\{1\}\{s+x\}\cdot\frac\{\mathrm\{ d\} s\}\{s\}\\\\ =&\frac\{\ln (1+x)\}\{x\ln 2\} \geq \frac\{1\}\{x\}, x\geq 1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \int\_0^\infty f(x)\mathrm\{ d\} x$ 发散.跟锦数学微信公众号. [在线资料](
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508、 7、 (10 分) 设函数列 $\displaystyle \left\\{f\_n(x)\right\\}$ 在集合 $\displaystyle I$ 上逐点收敛于极限函数 $\displaystyle f(x)$. (1)、 叙述函数列 $\displaystyle f\_n(x)$ 在集合 $\displaystyle I$ 上一致收敛于极限函数 $\displaystyle f(x)$ 的定义; (2)、 利用一致收敛的定义, 判断函数列 $\displaystyle f\_n(x)=\sqrt\{x^2+\frac\{1\}\{n^2\}\}$ 在 $\displaystyle D=(-1,1)$ 上是否一致收敛. (华南师范大学2023年数学分析考研试题) [函数项级数 ]
[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)、 $\displaystyle \forall\ \varepsilon > 0, \exists\ N,\mathrm\{ s.t.\} \forall\ n\geq N, \forall\ x\in I, |f\_n(x)-f(x)| < \varepsilon$. (2)、
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ N=\left\[\frac\{1\}\{\varepsilon\}\right\]+1,\mathrm\{ s.t.\} \forall\ n\geq N, \forall\ x\in (-1,1),\\\\ & \left|\sqrt\{x^2+\frac\{1\}\{n^2\}\}-|x|\right| =\left|\sqrt\{x^2+\frac\{1\}\{n^2\}\}-\sqrt\{x^2\}\right| =\frac\{\frac\{1\}\{n^2\}\}\{\sqrt\{x^2+\frac\{1\}\{n^2\}\}+\sqrt\{x^2\}\}\\\\ &\leq \frac\{\frac\{1\}\{n^2\}\}\{\frac\{1\}\{n\}\}=\frac\{1\}\{n\} < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\_n\rightrightarrows |x|, x\in (-1,1)$.跟锦数学微信公众号. [在线资料](
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509、 5、 (15 分) 设函数 $\displaystyle f\_0$ 在有界区间 $\displaystyle [0,a]$ 上可积, 定义函数列
\begin\{aligned\} f\_n(x)=\int\_0^x f\_\{n-1\}(t)\mathrm\{ d\} t, x\in [0,a]\left(n=1,2,\cdots\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 函数列 $\displaystyle \left\\{f\_n\right\\}$ 在 $\displaystyle [0,a]$ 上一致收敛. (华中师范大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle M=\max\_\{[0,a]\}|f\_0|$, 则由
\begin\{aligned\} |f\_k(x)|\leq \frac\{Mx^k\}\{k!\}\Rightarrow |f\_\{k+1\}(x)|=\left|\int\_0^x f\_k(t)\mathrm\{ d\} t\right| \leq \int\_0^x \frac\{Mt^k\}\{k!\}\mathrm\{ d\} t=\frac\{Mx^\{k+1\}\}\{(k+1)!\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及数学归纳法知
\begin\{aligned\} |f\_n(x)|\leq\frac\{Mx^n\}\{n!\}\Rightarrow \max\_\{[a,b]\}|f\_n|\leq \frac\{Ma^n\}\{n!\}\xrightarrow\{n\to\infty\}0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\_n\rightrightarrows 0$ 于 $\displaystyle [0,a]$.跟锦数学微信公众号. [在线资料](
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510、 6、 (15 分) 求级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{\sin(nx)\}\{\mathrm\{e\}^\{nx\}\}$ 的收敛域. (华中师范大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle x > 0$ 时, 由 $\displaystyle \left|\frac\{\sin(nx)\}\{\mathrm\{e\}^\{nx\}\}\right|\leq (\mathrm\{e\}^\{-x\})^n$ 及 Weierstrass 判别法知原级数收敛. (2)、 当 $\displaystyle x\leq 0$ 时, (2-1)、 若 $\displaystyle x=k\pi, k\in -\mathbb\{N\}$, 则通项为 $\displaystyle 0$, 而级数收敛. (2-2)、 若 $\displaystyle x\neq k\pi , k\in-\mathbb\{N\}$, 往证 $\displaystyle \lim\_\{n\to\infty\}\frac\{\sin(nx)\}\{\mathrm\{e\}^\{nx\}\}=0$ 不成立, 而原级数发散. 用反证法. 若 $\displaystyle \lim\_\{n\to\infty\}\frac\{\sin(nx)\}\{\mathrm\{e\}^\{nx\}\}=0$, 则
\begin\{aligned\} &\lim\_\{n\to\infty\}\sin nx=\lim\_\{n\to\infty\}\frac\{\sin(nx)\}\{\mathrm\{e\}^\{nx\}\}\cdot (\mathrm\{e\}^\{-|x|\})^n\xlongequal[\tiny\mbox\{无穷小\}]\{\tiny\mbox\{有界 $\displaystyle \cdot$\}\} 0\\\\ \Rightarrow&0=\lim\_\{n\to\infty\}\sin(n+1)x =\lim\_\{n\to\infty\}\left\[\sin nx\cos x+\cos nx\sin x\right\]\\\\ &\xlongequal[\tiny\mbox\{无穷小\}]\{\tiny\mbox\{有界 $\displaystyle \cdot$\}\} \sin x\lim\_\{n\to\infty\}\cos nx\Rightarrow \lim\_\{n\to\infty\}\cos nx=0\\\\ \Rightarrow&0=\left\[\lim\_\{n\to\infty\}\sin nx\right\]^2 +\left\[\lim\_\{n\to\infty\}\cos nx\right\]^2=\lim\_\{n\to\infty\}\left\[\sin^2nx+\cos^2nx\right\]\\\\ &=\lim\_\{n\to\infty\}1=1,\mbox\{这是一个矛盾. 故有结论.\} \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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511、 7、 (15 分) 记
\begin\{aligned\} C[a,b]=\left\\{f; f\mbox\{为有界区间 $\displaystyle [a,b]$ 上的连续函数\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对任意的 $\displaystyle f\in C[a,b]$. 令 $\displaystyle T(f)=\max\_\{x\in [a,b]\}|f(x)|$. (1)、 证明: 对任意的 $\displaystyle f,g\in C[a,b]$, 有 $\displaystyle T(f+g)\leq T(f)+T(g)$. (2)、 设 $\displaystyle C[a,b]$ 中的函数列 $\displaystyle \left\\{f\_n\right\\}$ 满足: 对任意的 $\displaystyle \varepsilon > 0$, 存在 $\displaystyle N > 0$ 使得 $\displaystyle m,n > N$ 时, 有 $\displaystyle T(f\_m-f\_n) < \varepsilon$. 证明: 存在 $\displaystyle f\in C[a,b]$, 使得
\begin\{aligned\} \lim\_\{n\to\infty\}T(f-f\_n)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华中师范大学2023年数学分析考研试题) [函数项级数 ]
[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)、 由定义,
\begin\{aligned\} \forall\ x\in [a,b], |f(x)+g(x)|\leq |f(x)|+|g(x)|\leq T(f)+T(g). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取上确界即知 $\displaystyle T(f+g)\leq T(f)+T(g)$. (2)、 由题设,
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ N,\mathrm\{ s.t.\} \forall\ m,n > N, \forall\ x\in [a,b], |f\_m(x)-f\_n(x)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
固定 $\displaystyle x\in [a,b]$ 知 $\displaystyle \left\\{f\_n(x)\right\\}$ 是 $\displaystyle \mathbb\{R\}$ 中的 Cauchy 列. 由 Cauchy 收敛准则知 $\displaystyle \lim\_\{n\to\infty\}f\_n(x)=f(x)$ 存在. 在上式中令 $\displaystyle m\to\infty$ 得
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ N,\mathrm\{ s.t.\} \forall\ n > N, &\boxed\{\forall\ x\in [a,b], |f(x)-f\_n(x)|\leq\varepsilon\}\\\\ \Leftrightarrow& T(f-f\_n)\leq \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \lim\_\{n\to\infty\}T(f-f\_n)=0$, 即 $\displaystyle f\_n\rightrightarrows f$ 于 $\displaystyle [a,b]$. 由 $\displaystyle f\_n$ 的连续性即知 $\displaystyle f\in C[a,b]$.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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512、 3、 解答题. (1)、 记 $\displaystyle f\_n(x)=xn^k\mathrm\{e\}^\{-nx\}, x\in (0,+\infty)$. 试问 $\displaystyle k$ 在哪个区间时, $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle (0,+\infty)$ 上一致收敛. (暨南大学2023年数学分析考研试题) [函数项级数 ]
[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 L'Hospital 法则知 $\displaystyle \lim\_\{n\to\infty\}f\_n(x)=0$. 由 $\displaystyle f\_n'(x)=\mathrm\{e\}^\{-nx\}n^k(1-nx)$ 知
\begin\{aligned\} \sup\_\{x > 0\}f\_n(x)=f\_n\left(\frac\{1\}\{n\}\right)=\frac\{1\}\{\mathrm\{e\} n^\{1-k\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当且仅当 $\displaystyle k < 1$ 时, $\displaystyle \sup\_\{x > 0\}f\_n(x)\to 0\Leftrightarrow f\_n\rightrightarrows 0$ 于 $\displaystyle (0,+\infty)$.跟锦数学微信公众号. [在线资料](
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513、 9、 设 $\displaystyle f(x)=\sum\_\{n=1\}^\infty \frac\{x^n\}\{n^2\ln(1+n)\}$. 证明: (1)、 $\displaystyle f(x)$ 在 $\displaystyle [-1,1]$ 上连续; (2)、 $\displaystyle f(x)$ 在 $\displaystyle x=1$ 处不可导. (南昌大学2023年数学分析考研试题) [函数项级数 ]
[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
\begin\{aligned\} \sup\_\{x\in [-1,1]\}\left|\frac\{x^n\}\{n^2\ln (1+n)\}\right|\leq\frac\{1\}\{n^2\ln 2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Weierstrass 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{x^n\}\{n^2\ln(1+n)\}$ 在 $\displaystyle [-1,1]$ 上一致收敛, 而和函数 $\displaystyle f(x)$ 在 $\displaystyle [-1,1]$ 上连续. (2)、 对 $\displaystyle \forall\ n\geq 1$, 我们有
\begin\{aligned\} 0 < x < 1\Rightarrow f'(x)\geq \sum\_\{k=1\}^n \frac\{x^\{k-1\}\}\{k\ln(1+k)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x\to 1^-$ 得
\begin\{aligned\} \varliminf\_\{x\to 1^-\}f'(x)\geq \lim\_\{x\to 1^-\}\sum\_\{k=1\}^n \frac\{x^\{k-1\}\}\{k\ln(1+k)\} =\sum\_\{k=1\}^n \frac\{1\}\{k\ln(1+k)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle n\to\infty$ 得
\begin\{aligned\} \varliminf\_\{x\to 1^-\}f'(x)\geq \sum\_\{k=1\}^\infty \frac\{1\}\{k\ln(1+k)\}=+\infty\Rightarrow \lim\_\{x\to 1^-\}f'(x)=+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} \lim\_\{x\to 1^-\}\frac\{f(x)-f(1)\}\{x-1\}\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to 1^-\}f'(\xi\_x) \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\} \lim\_\{x\to 1^-\}f'(x)=+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle f$ 在 $\displaystyle x=1$ 处不可导.跟锦数学微信公众号. [在线资料](
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514、 4、 (25 分) 设 $\displaystyle f(x)=\sum\_\{n=1\}^\infty \frac\{\mathrm\{e\}^\{-nx\}\}\{n^2+1\}$, $\displaystyle x\geq 0$. 证明: $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 可导, 且 $\displaystyle f'\_+(0)$ 不存在. (南开大学2023年数学分析考研试题) [函数项级数 ]
[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) / 对 $\displaystyle \forall\ k\geq 1, \forall\ \delta > 0$ , 由
\begin\{aligned\} \sup\_\{x\geq \delta\}\left|\left(\frac\{\mathrm\{e\}^\{-nx\}\}\{n^2+1\}\right)^\{(k)\}\right| &=\sup\_\{x\geq \delta\} \frac\{n^k \mathrm\{e\}^\{-nx\}\}\{n^2+1\} \leq n^\{k-2\}\mathrm\{e\}^\{-n\delta\} =n^\{k-2\}\mathrm\{e\}^\{-n\frac\{\delta\}\{2\}\}\mathrm\{e\}^\{-n\frac\{\delta\}\{2\}\}\\\\ & \leq \frac\{n^\{k-2\}\}\{\frac\{1\}\{k!\}\left(\frac\{n\delta\}\{2\}\right)^k\}\mathrm\{e\}^\{-n\frac\{\delta\}\{2\}\} \leq \frac\{2^kk!\}\{\delta^k\}\mathrm\{e\}^\{-n\frac\{\delta\}\{2\}\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle \sum\_\{n=1\}^\infty \frac\{2^kk!\}\{\delta^k\}\mathrm\{e\}^\{-n\frac\{\delta\}\{2\}\}$ 收敛, 及 Weierstrass 判别法知 $\displaystyle \left(\frac\{\mathrm\{e\}^\{-nx\}\}\{n^2+1\}\right)^\{(k)\}$ 在 $\displaystyle x\geq \delta$ 上一致收敛, 而 $\displaystyle f(x)$ 可任意次逐项求导,
\begin\{aligned\} f^\{(k)\}(x)=\sum\_\{k=0\}^\infty \left(\frac\{\mathrm\{e\}^\{-nx\}\}\{n^2+1\}\right)^\{(k)\}, \forall\ x\geq \delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \delta > 0$ 的任意性即知 $\displaystyle f$ 在 $\displaystyle x > 0$ 上有任意阶导数. 进一步,
\begin\{aligned\} &\left|\frac\{f(x)-f(0)\}\{x\}\right|=\left|\sum\_\{n=1\}^\infty \frac\{\mathrm\{e\}^\{-nx\}-1\}\{(n^2+1)x\}\right|\\\\ &=\sum\_\{n=1\}^\infty \frac\{1-\mathrm\{e\}^\{-nx\}\}\{(n^2+1)x\} \geq \sum\_\{n=1\}^N \frac\{1-\mathrm\{e\}^\{-nx\}\}\{(n^2+1)x\},\quad \forall\ N\\\\ \stackrel\{x\to 0^+\}\{\Rightarrow\}&\varliminf\_\{x\to0^+\}\left|\frac\{f(x)-f(0)\}\{x\}\right|\geq \sum\_\{n=1\}^N \frac\{n\}\{n^2+1\}, \quad \forall\ N\\\\ \stackrel\{N\to\infty\}\{\Rightarrow\}&\varliminf\_\{x\to0^+\}\left|\frac\{f(x)-f(0)\}\{x\}\right|\geq +\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f'\_+(0)$ 不存在.跟锦数学微信公众号. [在线资料](
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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
515、 3、 设 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛, 证明: $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{n^x\}$ 关于 $\displaystyle x\in [0,+\infty)$ 一致收敛. (厦门大学2023年数学分析考研试题) [函数项级数 ]
[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) / $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛, $\displaystyle \frac\{1\}\{n^x\}$ 关于 $\displaystyle n$ 递减, 且一致有界 (为 $\displaystyle 1$). 由 Abel 判别法即知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{n^x\}$ 关于 $\displaystyle x\in [0,+\infty)$ 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)/
---
516、 (2)、 求级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{2^n\}\tan \frac\{x\}\{2^n\}$ 的和函数. (上海大学2023年数学分析考研试题) [函数项级数 ]
[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\} &\frac\{1\}\{2^n\}\tan\frac\{x\}\{2^n\} =\frac\{\sin \frac\{x\}\{2^n\}\}\{2^n\cos \frac\{x\}\{2^n\}\} =\frac\{2\sin^2\frac\{x\}\{2^n\}\}\{2^\{n+1\}\sin \frac\{x\}\{2^n\}\cos\frac\{x\}\{2^n\}\} =\frac\{1-\cos \frac\{x\}\{2^\{n-1\}\}\}\{2^n\sin \frac\{x\}\{2^\{n-1\}\}\}\\\\ =&\frac\{\left(2\cos^2\frac\{x\}\{2^n\}-\cos\frac\{x\}\{2^\{n-1\}\}\right)-\cos \frac\{x\}\{2^\{n-1\}\}\}\{2^n\sin\frac\{x\}\{2^n\}\} =\frac\{2\cos^2\frac\{x\}\{2^n\}-2\cos \frac\{x\}\{2^\{n-1\}\}\}\{2^n\sin \frac\{x\}\{2^\{n-1\}\}\}\\\\ =&\frac\{\cos^2\frac\{x\}\{2^n\}\}\{2^n\cdot 2\sin \frac\{x\}\{2^n\}\cos\frac\{x\}\{2^n\}\} -\frac\{1\}\{2^\{n-1\}\}\cot \frac\{x\}\{2^\{n-1\}\} =\frac\{1\}\{2^n\}\cot \frac\{x\}\{2^n\}-\frac\{1\}\{2^\{n-1\}\}\cot\frac\{x\}\{2^\{n-1\}\} \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\} \sum\_\{n=1\}^N \frac\{1\}\{2^n\}\tan \frac\{x\}\{2^n\} =&-\sum\_\{n=1\}^N \left\[\frac\{1\}\{2^\{n-1\}\}\cot \frac\{x\}\{2^\{n-1\}\}-\frac\{1\}\{2^n\}\cot \frac\{x\}\{2^n\}\right\]\\\\ =&-\left\[\cot x-\frac\{1\}\{2^N\}\cdot \cot \frac\{x\}\{2^N\}\right\]. \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\} \sum\_\{n=1\}^\infty \frac\{1\}\{2^n\}\tan \frac\{x\}\{2^n\} =&\lim\_\{N\to\infty\}\frac\{1\}\{2^N\}\cdot \frac\{x\}\{2^N\}-\cot x =\lim\_\{N\to\infty\} \cos \frac\{x\}\{2^N\}\cdot \frac\{\frac\{x\}\{2^N\}\}\{\sin \frac\{x\}\{2^N\}\}\cdot \frac\{1\}\{x\}-\cot x\\\\ =&\frac\{1\}\{x\}-\cot 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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
517、 5、 (20 分) 设可微函数列 $\displaystyle \left\\{f\_n\right\\}$ 在 $\displaystyle [a,b]$ 上收敛, $\displaystyle \left\\{f\_n'\right\\}$ 在 $\displaystyle [a,b]$ 上一致有界, 证明: $\displaystyle \left\\{f\_n\right\\}$ 在 $\displaystyle [a,b]$ 上一致收敛. (上海交通大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle \left\\{f\_n'\right\\}$ 在 $\displaystyle [a,b]$ 上一致有界知
\begin\{aligned\} \exists\ M > 0,\mathrm\{ s.t.\} \forall\ x\in [a,b], |f\_n'(x)|\leq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle K\in\mathbb\{Z\}\_+$, 使得 $\displaystyle \frac\{b-a\}\{K\} < \frac\{\varepsilon\}\{2M\}$, 并将 $\displaystyle [a,b]$ $\displaystyle K$ 等分, 记分点为 $\displaystyle a=x\_0 < \cdots < x\_K=b$, 则由
\begin\{aligned\} \lim\_\{n\to\infty\}f\_n(x\_k)=f(x\_k), k=0,1,\cdots,K \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Cauchy 收敛准则知
\begin\{aligned\} \exists\ N,\mathrm\{ s.t.\} \forall\ m > n\geq N, \forall\ k=0,1,\cdots,K, |f\_m(x\_k)-f\_n(x\_k)| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle m > n\geq N$ 时, 对 $\displaystyle \forall\ a\leq x\leq b$,
\begin\{aligned\} \exists\ 0\leq k\leq K,\mathrm\{ s.t.\} |x-x\_k| < \frac\{b-a\}\{K\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} |f\_m(x)-f\_n(x)|&\leq|f\_m(x)-f\_m(x\_k)| +|f\_m(x\_k)-f\_n(x\_k)| +|f\_n(x\_k)-f\_n(x)|\\\\ &=|f\_m'(\xi\_k)(x-x\_k)| +|f\_m(x\_k)-f\_n(x\_k)| +|f\_n'(\eta\_k)(x-x\_k)|\\\\ &\leq 2\cdot M\frac\{b-a\}\{K\}+\frac\{\varepsilon\}\{2\}=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了$\left\\{f\_n(x)\right\\}$ 在 $\displaystyle [a,b]$ 上一致收敛.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)/
---
518、 4、 设 $\displaystyle f(x,y)$ 在 $\displaystyle a\leq x\leq A, b\leq y\leq B$ 上连续, $\displaystyle \left\\{\varphi\_n(x)\right\\}$ 在 $\displaystyle [a,A]$ 上一致收敛并 $\displaystyle b\leq \varphi\_n(x)\leq B\ (n=1,2,\cdots)$. 证明
\begin\{aligned\} F\_n(x)=f\left(x,\varphi\_n(x)\right)\ (n=1,2,\cdots,n) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle [a,A]$ 上一致收敛. (首都师范大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle f(x,y)$ 在 $\displaystyle a\leq x\leq A, b\leq y\leq B$ 上连续知一致连续, 而
\begin\{aligned\} &\forall\ \varepsilon > 0, \exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ a\leq x,x'\leq A, b\leq y, y'\leq B,\\\\ &|f(x,y)-f(x',y')| < \varepsilon.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \left\\{\varphi\_n(x)\right\\}$ 在 $\displaystyle [a,A]$ 上一致收敛及 Cauchy 收敛准则知
\begin\{aligned\} \exists\ N,\mathrm\{ s.t.\} \forall\ m > n\geq N, \forall\ a\leq x\leq A, |\varphi\_m(x)-\varphi\_n(x)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle m > n\geq N$ 时, 对 $\displaystyle \forall\ a\leq x\leq A$,
\begin\{aligned\} &|x-x|=0 < \delta, |\varphi\_m(x)-\varphi\_n(x)| < \delta\\\\ \Rightarrow&|F\_m(x)-F\_n(x)| =|f\left(x,\varphi\_m(x)\right)-f\left(x,\varphi\_n(x)\right)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \left\\{F\_n(x)\right\\}$ 在 $\displaystyle [a,A]$ 上一致收敛.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)/
---
519、 5、 设 $\displaystyle \sum\_\{n=1\}^\infty u\_n(x)$ 在 $\displaystyle [a-c,a+c]\ (c > 0)$ 上一致收敛, 对任意 $\displaystyle n\in\mathbb\{N\}\_+$, 存在 $\displaystyle c\_n$, 使得 $\displaystyle \lim\_\{x\to a\}u\_n(x)=c\_n$. 证明: (1)、 $\displaystyle \sum\_\{n=1\}^\infty c\_n$ 收敛; (2)、 $\displaystyle \lim\_\{x\to a\}\sum\_\{n=1\}^\infty u\_n(x)=\sum\_\{n=1\}^\infty c\_n$. (首都师范大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle \sum\_\{n=1\}^\infty u\_n(x)$ 在 $\displaystyle [a-c,a+c]\ (c > 0)$ 上一致收敛及 Cauchy 收敛准则知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ N,\mathrm\{ s.t.\} \forall\ m > n\geq N, \forall\ x\in [a-c,a+c], \left|\sum\_\{k=n\}^m u\_k(x)\right| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x\to a$ 得 $\displaystyle \left|\sum\_\{k=n\}^m c\_k\right|\leq\varepsilon$. 由数项级数收敛的 Cauchy 收敛准则知 $\displaystyle \sum\_\{n=1\}^\infty c\_n$ 收敛. (2)、 写出
\begin\{aligned\} &\left|\sum\_\{n=1\}^\infty u\_n(x)-\sum\_\{n=1\}^\infty c\_n\right| \leq\left|\sum\_\{n=N+1\}^\infty u\_n(x)\right| +\left|\sum\_\{n=N+1\}^\infty c\_n\right|+\left|\sum\_\{n=1\}^N[u\_n(x)-c\_n]\right|\\\\ \leq&\sup\_\{t\in [a-c,a+c]\}\left|\sum\_\{n=N+1\}^\infty u\_n(t)\right| +\left|\sum\_\{n=N+1\}^\infty c\_n\right|+\left|\sum\_\{n=1\}^N[u\_n(x)-c\_n]\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x\to a$ 得
\begin\{aligned\} \varlimsup\_\{x\to a\}\left|\sum\_\{n=1\}^\infty u\_n(x)-\sum\_\{n=1\}^\infty c\_n\right| \leq\sup\_\{t\in [a-c,a+c]\}\left|\sum\_\{n=N+1\}^\infty u\_n(t)\right| +\left|\sum\_\{n=N+1\}^\infty c\_n\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle N\to\infty$ 得
\begin\{aligned\} 0\leq\varlimsup\_\{x\to a\}\left|\sum\_\{n=1\}^\infty u\_n(x)-\sum\_\{n=1\}^\infty c\_n\right|\leq \varlimsup\_\{x\to a\}\left|\sum\_\{n=1\}^\infty u\_n(x)-\sum\_\{n=1\}^\infty c\_n\right|\leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lim\_\{x\to a\}\left|\sum\_\{n=1\}^\infty u\_n(x)-\sum\_\{n=1\}^\infty c\_n\right|=0\Rightarrow \lim\_\{x\to a\}\sum\_\{n=1\}^\infty u\_n(x)=\sum\_\{n=1\}^\infty c\_n$.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)/
---
520、 5、 (本题 10 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续, $\displaystyle f(x) > 0$. 证明: $\displaystyle \left\\{\sqrt[n]\{f(x)\}\right\\}$ 在 $\displaystyle [a,b]$ 上一致收敛于 $\displaystyle 1$. (四川大学2023年数学分析考研试题) [函数项级数 ]
[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) / 由连续函数的最值定理知
\begin\{aligned\} 0 < m=\min\_\{[a,b]\}f\leq M=\max\_\{[a,b]\}f < \infty. \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\} &\sqrt[n]\{m\}-1\leq \sqrt[n]\{f(x)\}-1\leq \sqrt[n]\{M\}-1, \forall\ x\in[a,b]\\\\ \Rightarrow&\max\_\{x\in[a,b]\}|\sqrt[n]\{f(x)\}-1|\leq \max\left\\{|\sqrt[n]\{m\}-1|,|\sqrt[n]\{M\}-1|\right\\}\\\\ &=\frac\{|\sqrt[n]\{m\}-1|+|\sqrt[n]\{M\}-1|-\left||\sqrt[n]\{m\}-1|-|\sqrt[n]\{M\}-1|\right|\}\{2\}\xrightarrow\{n\to\infty\}0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \left\\{\sqrt[n]\{f(x)\}\right\\}$ 在 $\displaystyle [a,b]$ 上一致收敛于 $\displaystyle 1$.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)/
---
521、 5、 讨论 $\displaystyle f\_n(x)=\frac\{x+2\}\{x+1\}\cos(x^n)$ 在 $\displaystyle \left\[0,1\right\]$ 上的一致收敛性. (天津大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle \lim\_\{n\to\infty\}f\_n(x)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{x+2\}\{x+1\},&0\leq x < 1,\\\\ \frac\{3\}\{2\}\cos 1,&x=1\end\{array\}\right.$ 不连续知 $\displaystyle f\_n$ 在 $\displaystyle [0,1]$ 上不一致收敛.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)/
---
522、 4、 对每个正整数 $\displaystyle n$, $\displaystyle f\_n(x)$ 在 $\displaystyle [a,b]$ 上单调, 若 $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle [a,b]$ 上收敛于连续函数 $\displaystyle f(x)$, 证明: $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle [a,b]$ 上一致收敛. (同济大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上连续知其在 $\displaystyle [a,b]$ 上一致连续,
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ a\leq x,x'\leq b, |x-x'| < \delta, |f(x)-f(x')| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle K\in\mathbb\{Z\}\_+$ 使得 $\displaystyle \frac\{b-a\}\{K\} < \delta$, 并将 $\displaystyle [a,b]$ $\displaystyle K$ 等分, 记分点为 $\displaystyle x\_0,\cdots,x\_K$, 则由 $\displaystyle \lim\_\{n\to\infty\}f\_n(x\_k)=f(x\_k), 0\leq k\leq K$ 知
\begin\{aligned\} \exists\ N,\mathrm\{ s.t.\} \forall\ n\geq N, \forall\ 0\leq k\leq K, |f\_n(x\_k)-f(x\_k)| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle n\geq N$ 时, 对 $\displaystyle \forall\ x\in [a,b], \exists\ 1\leq k\leq K,\mathrm\{ s.t.\} x\_\{k-1\}\leq x\leq x\_k$. 若 $\displaystyle f\_n$ 递增, 则
\begin\{aligned\} f\_n(x)-f(x)=&\left\[f\_n(x)-f\_n(x\_k)\right\]+\left\[f\_n(x\_k)-f(x\_k)\right\]+\left\[f(x\_k)-f(x)\right\]\\\\ \leq&0+\frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\varepsilon,\\\\ f\_n(x)-f(x)=&\left\[f\_n(x)-f\_n(x\_\{k-1\})\right\]+\left\[f\_n(x\_\{k-1\})-f(x\_\{k-1\})\right\]+\left\[f(x\_\{k-1\})-f(x)\right\]\\\\ \geq&0-\frac\{\varepsilon\}\{2\}-\frac\{\varepsilon\}\{2\}=-\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle f\_n$ 递减, 则只要将上述 $\displaystyle x\_k,x\_\{k-1\}$ 的位置对调. 故而总有 $\displaystyle |f\_n(x)-f(x)| < \varepsilon$. 这就证明了 $\displaystyle f\_n\rightrightarrows f$, 于 $\displaystyle [a,b]$.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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---
523、 (3)、 已知 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, 且 $\displaystyle f(1)=0$, 则 $\displaystyle \left\\{x^n\right\\}$ 与 $\displaystyle \left\\{x^nf(x)\right\\}$ 在 $\displaystyle [0,1]$ 上是否一致收敛, 说明你的理由. (武汉理工大学2023年数学分析考研试题) [函数项级数 ]
[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) / (3-1)、 由 $\displaystyle \lim\_\{n\to\infty\}x^n=\left\\{\begin\{array\}\{llllllllllll\}0,&0\leq x < 1,\\\\ 1,&x=1\end\{array\}\right.$ 知极限函数不连续, 而 $\displaystyle \left\\{x^n\right\\}$ 在 $\displaystyle [0,1]$ 上不一致收敛. (3-2)、 由 $\displaystyle f\in C[0,1]$ 及 $\displaystyle f(1)=0$ 知
\begin\{aligned\} &\exists\ M > 0,\mathrm\{ s.t.\} \forall\ x\in [0,1], |f(x)|\leq M;\\\\ &\forall\ \varepsilon > 0,\exists\ \delta\in (0,1),\mathrm\{ s.t.\} \forall\ 1-\delta < x\leq 1, |f(x)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \lim\_\{n\to\infty\}(1-\delta)^n M=0$ 知
\begin\{aligned\} \exists\ N,\mathrm\{ s.t.\} \forall\ n\geq N, (1-\delta)^nM < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle n\geq N$ 时,
\begin\{aligned\} 0\leq x\leq 1-\delta&\Rightarrow |x^nf(x)|\leq (1-\delta)^nM < \varepsilon,\\\\ 1-\delta < x\leq 1&\Rightarrow |x^nf(x)|\leq 1\cdot |f(x)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle x^nf(x)\rightrightarrows 0$.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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---
524、 (3)、 已知 $\displaystyle \alpha > 0$, 回答以下问题: (3-1)、 证明: $\displaystyle \sum\_\{n=1\}^\infty \frac\{\sin nx\}\{n^\alpha\}$ 关于 $\displaystyle x$ 在 $\displaystyle (0,2\pi)$ 上内闭一致收敛; (3-2)、 证明: $\displaystyle \int\_0^x \sum\_\{n=1\}^\infty \frac\{\sin nt\}\{n^\alpha\}\mathrm\{ d\} t=\sum\_\{n=1\}^\infty \frac\{1-\cos nx\}\{n^\{\alpha+1\}\}$. (西安交通大学2023年数学分析考研试题) [函数项级数 ]
[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) / (3-1)、 由
\begin\{aligned\} &\left|2\sin\frac\{x\}\{2\}\sum\_\{k=1\}^n \sin kx\right| =\left|\sum\_\{k=1\}^n \left\[\cos \left(k-\frac\{1\}\{2\}\right)x-\cos \left(k+\frac\{1\}\{2\}\right)x\right\]\right|\\\\ =&\left|\cos \frac\{x\}\{2\}-\cos\left(n+\frac\{1\}\{2\}\right)x\right|\leq 2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知对 $\displaystyle 0 < \delta < \pi$,
\begin\{aligned\} \sup\_\{\delta\leq x\leq 2\pi-\delta\}\left|\sum\_\{k=1\}^n \sin kx\right|\leq \sup\_\{\delta\leq x\leq 2\pi-\delta\}\frac\{1\}\{\sin\frac\{x\}\{2\}\}\leq \frac\{1\}\{\sin \frac\{\delta\}\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \frac\{1\}\{n^\alpha\}\searrow 0$ 及 Dirichlet 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{\sin nx\}\{n^\alpha\}$ 关于 $\displaystyle x$ 在 $\displaystyle [\delta,2\pi-\delta]$ 上一致收敛. 由 $\displaystyle \delta$ 的任意性知结论成立. (3-2)、 对 $\displaystyle \forall\ 0 < \delta < x < 2\pi$, 由第 1 步知
\begin\{aligned\} \int\_\delta^x \sum\_\{n=1\}^\infty \frac\{\sin nt\}\{n^\alpha\}\mathrm\{ d\} t =\sum\_\{n=1\}^\infty \int\_\delta^x \frac\{\sin nt\}\{n^\alpha\}\mathrm\{ d\} t =\sum\_\{n=1\}^\infty \frac\{\cos n\delta-\cos nx\}\{n^\{\alpha+1\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
左端被积函数在 $\displaystyle [0,x]$ 上可积, 右端函数项级数有优级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{2\}\{n^\{\alpha+1\}\}$ 而关于 $\displaystyle \delta$ 一致收敛. 故令 $\displaystyle \delta\to 0^+$ 即得结论.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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---
525、 6、 (15 分) 讨论函数 $\displaystyle f(x)=\sum\_\{n=1\}^\infty n\mathrm\{e\}^\{-nx\}$ 在 $\displaystyle (0,+\infty)$ 上的连续性, 并说明理由. (西南大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle \sup\_\{x\geq a\}n\mathrm\{e\}^\{-nx\}\leq n\mathrm\{e\}^\{-na\}$, $\displaystyle \sum\_\{n=1\}^\infty n\mathrm\{e\}^\{-na\}$ 收敛及优级数判别法知 $\displaystyle \sum\_\{n=1\}^\infty n\mathrm\{e\}^\{-nx\}$ 在 $\displaystyle [a,+\infty)$ 上一致收敛. 从而 $\displaystyle f(x)$ 在 $\displaystyle [a,+\infty)$ 上连续. 由 $\displaystyle a > 0$ 的任意性知 $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 上连续.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)/
---
526、 7、 (10 分) 证明: $\displaystyle F(x)=\sum\_\{n=1\}^\infty \mathrm\{e\}^\{-nx\}\cos nx$ 在 $\displaystyle (0,+\infty)$ 上不一致收敛, 但是连续的. (西南交通大学2023年数学分析考研试题) [函数项级数 ]
[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
\begin\{aligned\} \sup\_\{x > 0\}\sum\_\{k=n\}^n \mathrm\{e\}^\{-kx\}\cos kx \geq \lim\_\{x\to 0^+\} \mathrm\{e\}^\{-nx\}\cos nx=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Cauchy 收敛准则知 $\displaystyle \sum\_\{n=1\}^\infty \mathrm\{e\}^\{-nx\}\cos nx$ 在 $\displaystyle (0,+\infty)$ 上不一致收敛. (2)、 对 $\displaystyle \forall\ \delta > 0$, 由
\begin\{aligned\} \sup\_\{x\geq \delta\}|\mathrm\{e\}^\{-nx\}\cos nx|\leq \sup\_\{x\geq \delta\} \mathrm\{e\}^\{-nx\}=\mathrm\{e\}^\{-n\delta\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及优级数判别法知 $\displaystyle \sum\_\{n=1\}^\infty \mathrm\{e\}^\{-nx\}\cos nx$ 在 $\displaystyle [\delta,+\infty)$ 上一致收敛, 而和函数 $\displaystyle s(x)$ 在 $\displaystyle [\delta,+\infty)$ 上连续. 由 $\displaystyle \delta > 0$ 的任意性知 $\displaystyle s(x)$ 在 $\displaystyle (0,+\infty)$ 上连续.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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---
527、 5、 证明函数项级数 $\displaystyle \sum\_\{n=1\}^\infty x\mathrm\{e\}^\{-nx^2\}$ (1)、 在 $\displaystyle [\delta,+\infty)$ 上一致收敛 (其中 $\displaystyle \delta > 0$); (2)、 在 $\displaystyle [0,+\infty)$ 上非一致收敛. (湘潭大学2023年数学分析考研试题) [函数项级数 ]
[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)、 设 $\displaystyle f\_n(x)=x\mathrm\{e\}^\{-nx^2\}$, 则由 $\displaystyle f\_n'(x)=\mathrm\{e\}^\{-nx^2\}(1-2nx)$ 知
\begin\{aligned\} &\sup\_\{x\geq \delta\} \left|x\sum\_\{k=n\}^\infty \mathrm\{e\}^\{-kx^2\}\right| =\sup\_\{x\geq \delta\} x\frac\{\mathrm\{e\}^\{-nx^2\}\}\{1-\mathrm\{e\}^\{-x^2\}\}\\\\ \leq& \frac\{1\}\{1-\mathrm\{e\}^\{-\delta^2\}\}\sup\_\{x\geq \delta\}f\_n(x)=\frac\{1\}\{1-\mathrm\{e\}^\{-\delta^2\}\}f\_n\left(\frac\{1\}\{\sqrt\{2n\}\}\right)=\frac\{1\}\{1-\mathrm\{e\}^\{-\delta^2\}\}\frac\{1\}\{\sqrt\{2\mathrm\{e\} n\}\}\xrightarrow\{n\to\infty\}0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle f\_n$ 在 $\displaystyle [\delta,+\infty)$ 上一致收敛. (2)、 由
\begin\{aligned\} &\sup\_\{x > 0\} \left|x\sum\_\{k=n\}^\infty \mathrm\{e\}^\{-kx^2\}\right| =\sup\_\{x > 0\} x\frac\{\mathrm\{e\}^\{-nx^2\}\}\{1-\mathrm\{e\}^\{-x^2\}\}\\\\ \geq& \frac\{1\}\{\sqrt\{n\}\} \frac\{\mathrm\{e\}^\{-1\}\}\{1-\mathrm\{e\}^\{-\frac\{1\}\{n\}\}\} \sim \frac\{1\}\{\sqrt\{n\}\} \frac\{\mathrm\{e\}^\{-1\}\}\{\frac\{1\}\{n\}\}=\frac\{\sqrt\{n\}\}\{\mathrm\{e\}\}\to+\infty\left(n\to\infty\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原函数项级数在 $\displaystyle [0,+\infty)$ 上非一致收敛.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](
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---
528、 8、 (15 分) 设 $\displaystyle f(x)=f\_1(x)=\frac\{x\}\{\sqrt\{1+x^2\}\}$, $\displaystyle f\_\{n+1\}(x)=f\left(f\_n(x)\right), n=1,2,\cdots$. (1)、 求 $\displaystyle f\_n(x)$ 的表达式; (2)、 证明函数列 $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle \mathbb\{R\}$ 上一致收敛. (郑州大学2023年数学分析考研试题) [函数项级数 ]
[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
\begin\{aligned\} f\_k(x)=\frac\{x\}\{\sqrt\{1+kx^2\}\}\Rightarrow f\_\{k+1\}(x)& =f\left(f\_k(x)\right)=\frac\{f\_k(x)\}\{\sqrt\{1+f\_k^2(x)\}\}\\\\ &=\frac\{\frac\{x\}\{\sqrt\{1+kx^2\}\}\}\{\sqrt\{1+k\frac\{x\}\{\sqrt\{1+kx^2\}\}\}\} =\frac\{x\}\{\sqrt\{1+(k+1)x^2\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及数学归纳法知 $\displaystyle f\_n(x)=\frac\{x\}\{\sqrt\{1+nx^2\}\}$. (2)、 由
\begin\{aligned\} &f\_n^2(x)=\frac\{x^2\}\{1+nx^2\}\stackrel\{g\_n(t)=\frac\{t\}\{1+nt\}\}\{=\}g\_n(x^2)\\\\ \Rightarrow&\sup\_\mathbb\{R\} f\_n^2=\sup\_\{t\geq 0\} g\_n(t)\stackrel\{g\_n'(t)=\frac\{1\}\{(1+nt)^2\} > 0\}\{=\}\lim\_\{t\to+\infty\}g\_n(t)=\frac\{1\}\{n\}\xrightarrow\{n\to\infty\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f\_n\rightrightarrows 0$ 于 $\displaystyle \mathbb\{R\}$.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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---
529、 10、 (20 分) 定义 $\displaystyle f(x)=\sum\_\{n=1\}^\infty \frac\{3^\{-nx\}\}\{1+n^2\}$. 证明: (1)、 $\displaystyle f(x)$ 的定义域为 $\displaystyle [0,+\infty)$; (2)、 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上是连续函数; (3)、 $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 上有连续的导函数. (郑州大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle \left|\frac\{3^\{-nx\}\}\{1+n^2\}\right|\leq \frac\{1\}\{n^2\}$ 及 Weierstrass 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{3^\{-nx\}\}\{1+n^2\}$ 关于 $\displaystyle x\geq 0$ 一致收敛, 而和函数 $\displaystyle f(x)$ 连续. 再者, 对 $\displaystyle \forall\ \delta > 0$,
\begin\{aligned\} \sup\_\{x\geq \delta\}\left|\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} x\}\left(\frac\{3^\{-nx\}\}\{1+n^2\}\right)\right| =\sup\_\{x\geq \delta\}\frac\{3^\{-nx\}n\ln 3\}\{1+n^2\}=\frac\{n\ln 3\}\{(1+n^2)3^\{n\delta\}\} \leq \frac\{\ln 3\}\{2\cdot 3^\{n\delta\}\}, \sum\_\{n=1\}^\infty \frac\{\ln 3\}\{2\cdot 3^\{n\delta\}\} < \infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Weierstrass 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} x\}\left(\frac\{3^\{-nx\}\}\{1+n^2\}\right)$ 在 $\displaystyle [\delta,+\infty)$ 上一致收敛, 而可逐项求导, 且导函数连续:
\begin\{aligned\} f'(x)=\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} x\}\left(\frac\{3^\{-nx\}\}\{1+n^2\}\right), x\geq \delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \delta > 0$ 任意性知 $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 上有连续的导函数.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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发表于 2023-3-5 08:52:39
