Math Analysis Homework - Sem 2 Week 3

T1

1. 求下列幂级数的收敛域: (2) $\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)^{n^2} x^n$ ;

\begin{gathered} R=1/\limsup_{n\to \infty} \sqrt[n]{(1+\dfrac1n)^{n^2}} \\ =1/\limsup_{n\to \infty} (1+\dfrac{1}{n} )^n \\ =\frac1e \\ \text{when } x=\pm\frac1e, \\ \ln [(1+\dfrac{1}{n} )^{n^2}x^n] \\ =\ln [((1+\dfrac{1}{n} )^n\frac1e)^n] \\ =n(n\ln(1+\dfrac1n)-1) \\ \to -\dfrac12 \\ \lim_{n \to \infty} |(1+\dfrac{1}{n} )^{n^2} x^n|=\dfrac{1}{e^\frac12} \ne 0, \text{not convergent} \\ \text{thus the convergent range is } (-\dfrac{1}{e} ,\dfrac{1}{e} ) \end{gathered}

T2

1. 求下列幂级数的收敛域: (6) $\sum_{n=1}^\infty \left(\frac{a^n}{n} + \frac{b^n}{n^2}\right) x^n, a>0, b>0$ ;

\begin{gathered} R=1/\limsup_n \sqrt[ n ]{ \dfrac{a^n}{n} +\dfrac{b^n}{n^2} } =\dfrac1{\max(a,b)} \\ \end{gathered}

设通项为 $c_n,f(x)=\sum_{n=1}^\infty c_nx^n$

若 $a\ge b$ : 若 $x=\dfrac1a$ 时有 $\lim_{n \to \infty} \dfrac{c_nx^n}{\dfrac1n}=1$ ,由比较判别法知发散.

$x=-\dfrac1a$ 时:

\begin{gathered} f(x)=\sum _{n = 1} ^{\infty} c_{2n}x^{2n}+c^{2n+1}x^{2n+1} \\ =\sum _{n = 1} ^{\infty} (\dfrac{1}{2n} -\dfrac{1}{2n+1}) +\dfrac{b^{2n}}{a^{2n}}(\dfrac{1}{2n} - \dfrac{a}{b(2n+1)^2} ) \end{gathered}

其中第一部分收敛,第二部分括号里有界所以也收敛,于是收敛.

若 $b>a$ , $x=\pm \dfrac1b$ :

\begin{gathered} \lim_{n \to \infty} \dfrac{|\dfrac{a^n}{n} +\dfrac{b^n}{n^2} x^n |}{\dfrac{1}{n^2} } =1 \\ \end{gathered}

由比较判别法知绝对收敛.

故收敛域为:

\begin{gathered} \begin{cases} [-\dfrac{1}{b} ,\dfrac{1}{b} ],b>a \\ [-\dfrac{1}{a} ,\dfrac{1}{a} ),a\ge b \end{cases} \end{gathered}

T3

1. 求下列幂级数的收敛域: (8) $\sum_{n=1}^\infty \frac{n!}{n^n} x^n$ ;

\begin{gathered} \lim_{n \to \infty} \dfrac{c_{n}}{c_{n-1}}=\lim_{n \to \infty} n\dfrac{(n-1)^{n-1}}{n^n} =\lim_{n \to \infty} (1-\dfrac{1}{n} )^{n-1}=\frac1e \\ \implies R=e \end{gathered}

$x\pm e$ 时:

\begin{gathered} \lim_{n \to \infty} \dfrac{n!}{n^n} e^n =\lim_{n \to \infty} \dfrac{e^n}{n^n} \sqrt{ 2n\pi } (\dfrac{n}{e} )^n=\sqrt{ 2\pi n } \ne 0 \end{gathered}

不收敛.

于是收敛域为 $(-e,e)$

T4

2. 设 $\sum_{n=0}^\infty a_n x^n$ 和 $\sum_{n=0}^\infty b_n x^n$ 的收敛半径分别为 $r_1$ 和 $r_2$ , 证明:

(1) $\sum_{n=0}^\infty (a_n + b_n) x^n$ 的收敛半径 $r \ge \min\{r_1, r_2\}$ , 且当 $r_1 \neq r_2$ 时, 有 $r = \min\{r_1, r_2\}$ ; 当 $r_1 = r_2$ 时, $r$ 可能大于 $r_1$ ;

(2) $\sum_{n=0}^\infty (a_n b_n) x^n$ 的收敛半径 $r \ge r_1 r_2$ .

(1):

\begin{gathered} R_a=\limsup_{n\to \infty} \sqrt[n]{|a_n|} =\dfrac1{r_1} \\ R_b=\limsup_{n\to \infty} \sqrt[n]{|b_n|} =\dfrac1{r_2} \\ R=\limsup_{n\to \infty} \sqrt[n]{|a_n+b_n|} \le \sqrt[ n ]{ 2\max(|a_n|,|b_n|) } \\ \implies R \le \max(R_a,R_b) \\ \implies r \ge \min(r_1,r_2) \\ \text{when } r_1 \ne r_2, \\ \text{assume } r_1 < r_2, \\ R=\limsup_{n\to \infty} \sqrt[n]{|a_n+b_n|} \\ \ge \limsup_{n\to \infty} \sqrt[n]{|a_n|} =R_a \\ \implies r \le r_1 \implies r=r_1 \\ \text{when } r_1=r_2, \\ \text{let } a_n=1,b_n=-1,r=\infty>r_1=r_2=1 \\ \end{gathered}

(2):

\begin{gathered} \limsup _{n\to \infty} \sqrt[n]{|a_nb_n|} \le \limsup_{n\to \infty} \sqrt[n]{|a_n|}\cdot \limsup_{n\to \infty} \sqrt[n]{|b_n|} =\dfrac{1}{r_1r_2} \\ \implies r \ge r_1r_2 \end{gathered}

对上极限的不等号,考虑取 $a_nb_n$ 取最大值的子列为 $a_{n_k}b_{n_k}$ ,则显然 $\lim_{k\to \infty} \sqrt[n_k]{|a_{n_k}b_{n_k}|} = \lim_{k\to \infty} \sqrt[n_k]{|a_{n_k}|}\cdot \sqrt[n_k]{|b_{n_k}|} \le \limsup \sqrt[n]{|a_n|}\cdot \limsup \sqrt[n]{|b_n|}$ .

T5

3. 设 $f(x) = \sum_{n=0}^\infty a_n x^n$ 在 $(-r, r)$ 中收敛, 且 $\sum_{n=0}^\infty \frac{a_n}{n+1} r^{n+1}$ 收敛. 证明: 无论 $\sum_{n=0}^\infty a_n x^n$ 在 $x=r$ 处是否收敛, 都有 $\int_0^r f(x)dx = \sum_{n=0}^\infty \frac{a_n}{n+1} r^{n+1}$ 并由此证明 $\int_0^1 \frac{\ln \frac{1}{1-x}}{x} dx = \sum_{n=1}^\infty \frac{1}{n^2}$

(1):

由优级数判别法, $\forall x\in [-l,l]\subset (-r,r)$ , 有

\begin{gathered} \sum _{n = 0} ^{\infty} a_n|x|^n \le (\sum _{n = 0} ^{\infty} a_nr^n(\dfrac{l}{r} )^n) \\ \end{gathered}

其中因为 $a_nr^n$ 收敛到 $0$ 所以有界,优级数绝对收敛,所以幂级数 $f$ 内闭一致收敛.于是可以逐项积分,得到 $\forall r'\in (-r,r)$ :

\begin{gathered} \lim_{r' \to r} \int_0^{r'} f(x)dx = \lim_{r' \to r} \sum_{n=0}^\infty \int_0^{r'} a_n x^n dx = \lim_{r' \to r} \sum_{n=0}^\infty \frac{a_n}{n+1} (r')^{n+1} \end{gathered}

此时已知右边的级数在闭区间上收敛所以一致收敛所以连续,于是即

\begin{gathered} \int_0^r f(x)dx = \sum_{n=0}^\infty \frac{a_n}{n+1} r^{n+1} \end{gathered}

(2):

\begin{gathered} \int_0^1 \dfrac{\ln \frac{1}{1-x}}{x} dx \\ =\int_0^1 \sum _{n = 0} ^{\infty} \dfrac{x^n}{n+1} dx \\ \end{gathered}

注意到该级数的收敛于为 $(-1,1)$ ,而我们知道 $\sum_n \dfrac1{n^2}$ 收敛,于是

\begin{gathered} =\sum _{n = 0} ^{\infty} \dfrac{1}{n+1} \int_0^1 x^n dx \\ =\sum _{n = 0} ^{\infty} \dfrac{1}{(n+1)^2} \\ =\sum _{n = 1} ^{\infty} \dfrac{1}{n^2} \end{gathered}

T6

4. 求下列幂级数的和函数, 并指出等式成立的范围: (6) $f(x)=\sum_{n=1}^\infty \frac{2n+1}{n!} x^{2n}$ ;

收敛半径为

\begin{gathered} R=\dfrac{1}{\limsup_{n\to \infty} \sqrt[ n ]{ \dfrac{2n+1}{n!} } } =\infty \end{gathered}

由T5,因为 $\forall X\in R$ ,级数:

\begin{gathered} \sum _{n = 1} ^{\infty} \dfrac{\dfrac{2n+1}{n!} }{2n+1} x^{2n+1}=xe^{x^2}-x \end{gathered}

收敛,所以

\begin{gathered} \int_0^X f(x)dx \\ =\int_0^X \sum_{n=1}^\infty \frac{2n+1}{n!} x^{2n} dx \\ =\sum_{n=1}^\infty \frac{2n+1}{n!} \int_0^X x^{2n} dx \\ =\sum _{n = 1} ^{\infty} \dfrac{X^{2n+1}}{n!} \\ =Xe^{X^2}-X \end{gathered}

于是

\begin{gathered} f(x)=\dfrac{d}{dx} \int_0^X f(x)dx \\ =\dfrac{d}{dx} (Xe^{X^2}-X) \\ =e^{X^2}+2X^2e^{X^2}-1 \\ \end{gathered}

对任意 $x$ 成立.

T7

4. 求下列幂级数的和函数, 并指出等式成立的范围: (9) $f(x)=\sum_{n=1}^\infty (-1)^{n-1} n^2 x^n$ ;

\begin{gathered} R=1/\limsup_{n\to \infty} \sqrt[n]{n^2} =1 \\ \forall X\in (-1,1), \int_0^X \dfrac1x fdx=\sum _{n = 1} ^{\infty} \int_0^x (-1)^{n-1} n^2 x^{n-1} dx \\ =\sum _{n = 1} ^{\infty} (-1)^{n-1} n X^n \\ \int_0^Y \dfrac1X(\int_0^X \dfrac{1}{x} fdx) dX \\ =\sum _{n = 1} ^{\infty} (-1)^{n-1}Y^{n} \\ =\dfrac{Y}{1+Y}\\ \end{gathered}

所以

\begin{gathered} \int \dfrac1x(\int \dfrac1x f)=\dfrac{x}{1+x}\\ \implies f(x)=\dfrac{x(1-x)}{(1+x)^3} \end{gathered}

当 $x\in (-1,1)$ 时成立.

T8

5. 利用幂级数求下列级数的和: (2) $\sum_{n=2}^\infty \frac{1}{(n^2-1)2^n}$ ;

设

\begin{gathered} f(x)=\sum _{n = 2} ^{\infty} \dfrac{1}{(n+1)(n-1 )} x^{n-1} \\ =\sum _{n = 2} ^{\infty} \dfrac{1}{2} (\dfrac{1}{n-1} -\dfrac{1}{n+1} ) x^{n-1} \\ =\dfrac{1}{2} \sum _{n = 1} ^{\infty} \dfrac{x^n}{n}-\dfrac{1}{2} x^{-2}\sum _{n = 3} ^{\infty} \dfrac{x^n}{n} \\ =-\dfrac{1}{2} \ln (1-x)+\dfrac{1}{2} x^{-2}(\ln (1-x)+\dfrac{x^2}{2} +\dfrac{x}{1} ) \end{gathered}

其中 $\ln (1-x)$ 的收敛域是 $(-1,1)$ .

于是

\begin{gathered} Ans=\dfrac12f(\dfrac{1}{2} )=\dfrac{5}{8} -\dfrac{3}{4} \ln 2 \end{gathered}

T9

5. 利用幂级数求下列级数的和: (3) $\sum_{n=1}^\infty \frac{n(n+2)}{2^{2(n+1)}}$ ;

\begin{gathered} \text{let } f(x)=\sum _{n = 2} ^{\infty} (n-1)(n+1)x^n \\ \end{gathered}

由比值法易知收敛半径为 $1$ ,当 $x<\dfrac12$ 时一致收敛可逐项积分:

\begin{gathered} \int \dfrac1{x^3} \int f(x) =\sum _{n = 2} ^{\infty} x^{n-1}=\dfrac{1}{1-x}-1 \\ \implies f(x)=\dfrac{3-x}{(1-x)^3}x^2 \\ Ans=f(\dfrac{1}{4} )=\dfrac{11}{27} \end{gathered}

T10

6. 证明: (1) $y = \sum_{n=0}^\infty \frac{x^{4n}}{(4n)!}$ 满足 $y^{(4)} = y$ ;

(2) $y = \sum_{n=0}^\infty \frac{x^n}{(n!)^2}$ 满足 $xy'' + y' - y = 0$ .

(1):

对求导 $i$ 次后的级数, $R_i=1/\lim_{n \to \infty}\dfrac{(n-i-1)!}{(n-i)!}=\infty$ .收敛,内闭一致收敛,可以逐项求导,得到:

\begin{gathered} y^{(4)}=\sum _{n = 0} ^{\infty} \dfrac{d^4}{ {dx}^4} \dfrac{x^{4n}}{(4n)!}=y \end{gathered}

(2):

$R=1/\lim_{n \to \infty} \dfrac{((n-1)!)^2}{(n!)^2} =\infty$ ,收敛,一致收敛,可逐项求导.且

规定 $0!=1,(-n)!=0$

\begin{gathered} xy''+y'-y \\ \sum _{n = 2} ^{\infty} -\dfrac{x^n}{(n!)^2} +\dfrac{x^{n-1}}{n!(n-1)!} +x\dfrac{x^{n-2}}{n!(n-2)!} \\ =-\sum _{n = 0} ^{\infty} \dfrac{x^n}{(n!)^2} +\sum _{n = 0} ^{\infty} x^{n-1}\dfrac{1}{((n-1)!)^2} \\ =0 \end{gathered}

T11

7. 设 $f(x) = \sum_{n=0}^\infty a_n x^n$ 的收敛半径 $r = +\infty$ . 令 $f_n(x) = \sum_{k=0}^n a_k x^k$ . 证明: $\{f[f_n(x)]\}$ 在 $[a, b]$ 上一致收敛于 $f[f(x)]$ .

因为 $f_n(x)$ 每项连续, $f_n$ 一致收敛到 $f$ ,于是 $f$ 连续.所以 $f$ 在 $[a,b]$ 一致连续且有界.设界为 $[A,B]$ .

又有 $f$ 在 $[A,B]$ 上连续而一致连续.满足 $\forall \epsilon_1>0,\exists \delta>0,\forall |x-y|<\delta,|f(x)-f(y)|<\epsilon_1$ .

又因为 $f(x)$ 一致收敛到 $f_n(x)$ ,所以 $\forall \epsilon_2>0,\exists N,\forall n>N,|f(x)-f_n(x)|<\epsilon_2$ .

于是取 $\epsilon_1=\epsilon,\epsilon_2=\delta$ :

\begin{gathered} n>N \implies |f_n(x)-f(x)|<\delta \implies |f(f_n(x))-f(f(x))|<\epsilon \end{gathered}

得证.

T12

8. 设定项级数 $\sum_{n=0}^\infty a_n$ 发散, 且 $\lim_{n\to\infty} \frac{a_n}{a_0 + a_1 + \dots + a_n} = 0$ . 证明: 幂级数 $\sum_{n=0}^\infty a_n x^n$ 的收敛半径 $r=1$ .

\begin{gathered} \sum _{n = 0} ^{\infty} a_n 1^n=\sum _{n = 0} ^{\infty} a_n=\infty \end{gathered}

故由Abel收敛定理知收敛半径 $R\le 1$ .

$\forall x\in (-1,1),x<1$ :

设 $S_n$ 为 $a_n$ 部分和,则易知 $\lim_{n \to \infty} \dfrac{S_{n-1}}{S_n}=1$

\begin{gathered} \sum _{n = 0} ^{\infty} a_nx^n \\ =a_0+\sum _{n = 1} ^{\infty} (S_n-S_{n-1})x^n \\ =\sum _{n = 0} ^{\infty} S_n(x^n-x^{n+1})+S_nx^n \end{gathered}

最后一项显然极限为 $0$ ,对前面的部分

使用比值:

\begin{gathered} \lim_{n \to \infty} \dfrac{S_n}{S_{n-1}} \dfrac{x^n-x^{n+1}}{x^{n-1}-x^n} \\ =1\cdot x<1 \end{gathered}

故收敛.

故收敛域为 $(-1,1)$ ,收敛半径为 $1$ .