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2026-03-26

Math Analysis Homework - Sem 2 Week 4

2026-03-26_posts/homework/sem2
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Math Analysis Homework - Sem 2 Week 4

T1

1. 将下列函数在指定点展开成泰勒级数： (3) 1x2,x=1\frac{1}{x^2}, x = 1;

f(x)=x−2f(n)=(−1)n(n+1)!x−2−n  ⟹  f(x)=∑i=0∞f(i)(1)i!(x−1)i=1+∑i=1∞(−1)i(i+1)(x−1)i\begin{gathered} f(x)=x^{-2} \\ f^{(n)}= (-1)^n (n+1)! x^{-2-n} \implies \\ f(x)=\sum _{i = 0} ^{\infty} \dfrac{f^{(i)}(1)}{i!} (x-1)^i \\ =1+\sum _{i = 1} ^{\infty}(-1)^i(i+1)(x-1)^i \end{gathered}

余项为

Rn(x)=(−1)n(n+1)(θx+1−θ)−2−n(x−1)n=(−1)n(n+1)(x−1θx+1−θ)n1(θx+1−θ)2\begin{gathered} R_n(x)=(-1)^{n}(n+1)(\theta x+1-\theta)^{-2-n} (x-1)^n \\ =(-1)^n (n+1) (\dfrac{x-1}{\theta x+1-\theta} )^n \dfrac{1}{(\theta x+1-\theta)^2} \end{gathered}

则∣x−1∣<1|x-1|<1时收敛.x=2,0x=2,0通项极限不为00显然发散.

T2

1. 将下列函数在指定点展开成泰勒级数： (5) ln⁡(1+x+x2+x3),x=0\ln (1 + x + x^2 + x^3), x = 0;

f(x)=ln⁡(1−x41−x)=ln⁡(1−x4)−ln⁡(1−x)=∑i=1∞xii−∑i=1∞x4ii=∑i=1∞aixi,ai={1i−14i,i=4k,k∈{Z}1i,otherwise\begin{gathered} f(x)=\ln(\dfrac{1-x^4}{1-x} )=\ln(1-x^4)-\ln(1-x) \\ =\sum _{i = 1} ^{\infty} \dfrac{x^i}{i} -\sum _{i = 1} ^{\infty} \dfrac{x^{4i}}{i} \\ =\sum _{i = 1} ^{\infty} a_ix^i,a_i=\begin{cases} \dfrac{1}{i} -\dfrac{1}{4i} ,i=4k,k\in \{ Z \} \\ \dfrac{1}{i} ,\text{otherwise} \end{cases} \end{gathered}

两个ln⁡\ln的级数收敛域均为(−1,1)(-1,1),故收敛于为(−1,1)(-1,1)

T3

3. 求函数 f(x)=arctan⁡2x1−x2f(x) = \arctan \frac{2x}{1-x^2} 在 x=0x=0 处的幂级数展开式，并求 ∑n=0∞(−1)n2n+1\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} 的值.

f′(x)=21+x2=∑i=0∞2⋅(−1)ix2i  ⟹  f(x)=∫f′(x)=∑i=0∞2⋅(−1)ix2i+12i+1,x∈(−1,1)\begin{gathered} f'(x)=\dfrac{2}{1+x^2} \\ =\sum _{i = 0} ^{\infty} 2\cdot (-1)^ix^{2i} \\ \implies f(x)=\int f'(x)=\sum _{i = 0} ^{\infty} \dfrac{2\cdot (-1)^i x^{2i+1}}{2i+1} ,x\in (-1,1) \end{gathered}

由莱布尼茨判别法知级数在x=1x=1时收敛,由阿贝尔定理知级数的和函数s(x)s(x)连续,故所求即 s(1)=lim⁡x→1s(x)=lim⁡x→1f(x)=π2s(1)=\lim_{x \to 1} s(x)=\lim_{x \to 1} f(x)=\dfrac\pi2

T4

4. 设函数 f(x)=11−x−x2f(x) = \frac{1}{1-x-x^2}，记 an=f(n)(0)n!a_n = \frac{f^{(n)}(0)}{n!}. 证明：

  • (1) 级数 ∑n=0∞1an\sum_{n=0}^{\infty} \frac{1}{a_n} 收敛；
  • (2) a0=a1=1,an+2=an+1+an,n=0,1,2,⋯a_0 = a_1 = 1, a_{n+2} = a_{n+1} + a_n, n=0, 1, 2, \cdots；
  • (3) 级数 ∑n=0∞an+1anan+2\sum_{n=0}^{\infty} \frac{a_{n+1}}{a_n a_{n+2}} 收敛，并求其和.

计算得a0=a1=1a_0=a_1=1.

f(x)(1−x−x2)=1  ⟹  ∀n≥2,0=(f(x)(1−x−x2))(n)∣x=0=∑i=0nf(n−i)(0)((1−x−x2)(i)∣x=0)(ni)=f(n)(0)−nfn−1(0)+n(n−1)2⋅−2fn−2(0)=n!an−n!an−1−n!(an−2)=0  ⟹  an=an−1+an−2\begin{gathered} f(x)(1-x-x^2)=1 \\ \implies \forall n\ge 2,0=(f(x)(1-x-x^2))^{(n)}|_{x=0} \\ =\sum _{i = 0} ^{n} f^{(n-i)}(0)((1-x-x^2)^{(i)}|_{x=0})\binom ni \\ =f^{(n)}(0)-nf^{n-1}(0)+\dfrac{n(n-1)}{2} \cdot -2 f^{n-2}(0) \\ =n!a_n-n!a_{n-1}-n!(a_{n-2}) \\ =0 \\ \implies a_n=a_{n-1}+a_{n-2} \end{gathered}

故an=an−1+an−2>2an−2a_n=a_{n-1}+a_{n-2}>2a_{n-2},则1an<121an−2\dfrac1{a_n}<\dfrac12\dfrac1{a_{n-2}},由比较判别法知奇数项,偶数项均绝对收敛,故(1)中级数收敛.

(3):

注意到:

Sn=∑n=0Nan+1anan+2=∑n=0N1an−1an+2=2−1aN+1−1aN+2  ⟹  S=lim⁡n→∞Sn=2\begin{gathered} S_n=\sum _{n = 0} ^{N} \dfrac{a_{n+1}}{a_na_{n+2}} \\ =\sum _{n = 0} ^{N} \dfrac{1}{a_n} -\dfrac{1}{a_{n+2}} \\ =2-\dfrac{1}{a_{N+1}} -\dfrac{1}{a_{N+2}} \\ \implies S=\lim_{n \to \infty} S_n =2 \end{gathered}

T5

5. (1) 证明：函数项级数 sin⁡x+∑n=1∞(2n−1)!!sin⁡2n+1x(2n)!!(2n+1)\sin x + \sum_{n=1}^{\infty} \frac{(2n-1)!! \sin^{2n+1} x}{(2n)!! (2n+1)} 在区间 [0,π2]\left[0, \frac{\pi}{2}\right] 上一致收敛于和函数 xx.

(2) 证明 ∑n=1∞1(2n−1)2=π28\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{\pi^2}{8} 和 ∑n=1∞1n2=π26\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.

f(x)=arcsin⁡(x)  ⟹  f′(x)=11−x2=(1−x2)−12=1+∑n=1∞(2n−1)!!(2n)!!x2n,∀x∈[0,1)\begin{gathered} f(x)=\arcsin(x) \\ \implies f'(x)=\dfrac{1}{\sqrt{ 1-x^2 } } \\ =(1-x^2)^{-\frac12} \\ =1+\sum _{n = 1} ^{\infty} \dfrac{(2n-1)!!}{(2n)!!} x^{2n},\forall x\in[0,1) \end{gathered}

逐项积分得:

arcsin⁡(x)=x+∑n=1∞(2n−1)!!(2n)!!(2n+1)x2n+1,∀x∈[0,1]\begin{gathered} \arcsin(x)=x+\sum _{n = 1} ^{\infty} \dfrac{(2n-1)!!}{(2n)!!(2n+1)} x^{2n+1},\forall x\in [0,1] \end{gathered}

且为一致收敛.

故原式一致收敛到arcsin⁡(sin⁡x)=x\arcsin(\sin x)=x

(2):

注意到

∫0π2sin⁡2n+1(x)=(2n)!!(2n+1)!!\begin{gathered} \int_0^{\frac\pi2} \sin^{2n+1}(x)=\dfrac{(2n)!!}{(2n+1)!!} \end{gathered}

于是

∫0π2arcsin⁡(sin⁡x)=∑n=0∞(2n−1)!!(2n)!!(2n+1)∫0π2sin⁡2n+1dx=∑n=0∞(2n−1)!!(2n)!!(2n+1)(2n)!!(2n+1)!!=∑n=0∞1(2n+1)2=∫0π2xdx=π28\begin{gathered} \int_0^{\frac\pi2} \arcsin(\sin x)=\sum _{n = 0} ^{\infty} \dfrac{(2n-1)!!}{(2n)!!(2n+1)} \int_0^{\frac\pi2}\sin^{2n+1}dx \\ =\sum _{n = 0} ^{\infty} \dfrac{(2n-1)!!}{(2n)!!(2n+1)}\dfrac{(2n)!!}{(2n+1)!!} \\ =\sum _{n = 0} ^{\infty} \dfrac{1}{(2n+1)^2} \\ =\int_0^{\frac\pi2}xdx \\ =\dfrac{\pi^2}8 \end{gathered}

于是显然有:

∑n=1∞1n2−∑i=1∞1(2n)2=π28  ⟹  ∑n=1∞1n2=π26\begin{gathered} \sum _{n = 1} ^{\infty} \dfrac{1}{n^2}-\sum _{i = 1} ^{\infty} \dfrac{1}{(2n)^2}=\dfrac{\pi^2}{8} \\ \implies \sum _{n = 1} ^{\infty} \dfrac{1}{n^2} =\dfrac{\pi^2}{6} \end{gathered}

T6

7. 利用斯特林公式解决下列问题：

  • (1) 求极限 lim⁡n→∞n!n2\lim_{n \to \infty} \sqrt[n^2]{n!}；
  • (2) 证明：当 n→∞n \to \infty 时，ln⁡(n!)∼nln⁡n\ln(n!) \sim n \ln n；
  • (3) 判断级数 ∑n=2∞1ln⁡(n!)\sum_{n=2}^{\infty} \frac{1}{\ln(n!)} 的敛散性.

(1):

lim⁡n→∞n!n2=lim⁡n→∞exp⁡1n2ln⁡(n!)=lim⁡n→∞exp⁡1n2ln⁡(2nπ(ne)neθn12n)=lim⁡n→∞exp⁡1n(ln⁡n−1)=e\begin{gathered} \lim_{n \to \infty} \sqrt[ n^2 ]{ n! } \\ =\lim_{n \to \infty} \exp \dfrac{1}{n^2} \ln(n!) \\ =\lim_{n \to \infty} \exp \dfrac{1}{n^2} \ln (\sqrt{2n\pi}(\dfrac{n}{e})^n e^{\frac{\theta_n}{12n}} ) \\ =\lim_{n \to \infty} \exp\dfrac{1}{n} (\ln n-1) \\ =e \end{gathered}

(2):

lim⁡n→∞ln⁡n!nln⁡n=lim⁡n→∞ln⁡(2nπ(ne)neθn12n)nln⁡n=lim⁡n→∞nln⁡(n−1)+ln⁡(2nπ)+θn12nnln⁡n=1\begin{gathered} \lim_{n \to \infty} \dfrac{\ln n!}{n\ln n} \\ =\lim_{n \to \infty} \dfrac{\ln (\sqrt{2n\pi}(\dfrac{n}{e})^n e^{\frac{\theta_n}{12n}})}{n\ln n} \\ =\lim_{n \to \infty} \dfrac{n\ln(n-1)+\ln(\sqrt{ 2n\pi } )+\frac{\theta_n}{12n}}{n\ln n}=1 \end{gathered}

(3):

∑n=2∞1ln⁡(n!)\sum _{n = 2} ^{\infty} \dfrac{1}{\ln(n!)}和∑n=2∞1nln⁡n\sum _{n = 2} ^{\infty} \dfrac{1}{n\ln n}同敛散,从而与∫2∞1xln⁡xdx\int_2^\infty \dfrac1{x\ln x}dx同敛散,故发散.

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