Anslytic Functions

Exercise4.1

1. Find the radius of convergence.
(a) $\sum \frac{nx^{n}}{3^{n}} \htmlref{(b)}{enshu:4-1-1-b} \sum \frac{n^{n}x^{n}}{n!}$
2. Show the followings.
(a) $\sum_{n=0}^{\infty}x^{n} = \frac{1}{1 - x}, \vert x\vert < 1$
(a) $\log{(1+x)} = \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}, \vert x\vert \leq 1$
3. For $f(x) = \sum_{n=1}^{\infty}\frac{\sin{nx}}{n^3}$，Show the followings.
(a) $f(x)$ is uniformly convergent on $-\infty < x < \infty$.
(b) $f^{\prime}(x) = \sum_{n=1}^{\infty}\frac{\cos{nx}}{n^2}$
(c) $\int_{0}^{\pi}f(x)dx = \sum_{n=1}^{\infty}\frac{2}{(2n-1)^{4}}$

Answer
1.
(a)

$$\rho = \lim_{n \rightarrow \infty}\vert\frac{a_{n}}{a_{n+1}}\vert = \lim_{n \rightarrow \infty}\vert\frac{n/3^{n}}{(n+1)/3^{n+1}}\vert$$

$$= \lim_{n \rightarrow \infty}\vert\frac{3n}{n+1}\vert = 3.$$

(b)

$$\rho = \lim_{n \rightarrow \infty}\vert\frac{a_{n}}{a_{n+1}}\vert = \lim_{n \rightarrow \infty}\vert\frac{n^n/n!}{(n+1)^{n+1}/(n+1)!}\vert$$

$$= \lim_{n \rightarrow \infty}\vert(\frac{n}{n+1})^{n}\vert = \lim_{n \rightarrow \infty}\vert(\frac{1}{1 + 1/n})^{n}\vert = \frac{1}{e}.$$

2.
(a) Let $S_{n} = \sum_{k=0}^{n}x^{k}$. Then

$$S_{n}-xS_{n} = \sum_{k=0}^{n}x^{k} - \sum_{k=0}^{n}x^{k+1} = 1 - x^{n+1}$$

Thus, $S_{n} = \frac{1 - x^{n+1}}{1 - x}$ . Note that

$$\lim_{n \rightarrow \infty}x^{n} = \left\{\begin{array}{cl} \infty & x > 1\\ 0 & \vert x\vert < 1\\ vibrate & x < -1 \end{array} \right .$$

$$S = \lim S_{n} = \frac{1}{1-x}, \vert x\vert < 1.$$

(b)

$$\log{(1+x)} = \int_{0}^{x}\frac{1}{1+t}dt = \int_{0}^{x}\sum_{n=0}^{\infty}(-1)^{n}t^{n}dt (term by term integration)$$

$$= \sum_{n=0}^{\infty}\int_{0}^{x}(-1)^{n}t^{n}dt = \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}, \vert x\vert < 1$$

For $x = 1$, $\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}$ converges.

$$\log{(1+x)} =\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}, -1 < x \leq 1$$

3.
(a) Let $\vert\frac{\sin{nx}}{n^3}\vert \leq \frac{1}{n^3} = M_{n}$.

$$\sum M_{n} = \sum \frac{1}{n^3} < \infty$$

Then by Weierstrass'M-test it converges uniformly.
(b) Let $f(x) = \sum_{n=1}^{\infty}\frac{\sin{nx}}{n^3}$. Then

$$f^{\prime}(x) = (\sum_{n=1}^{\infty}\frac{\sin{nx}}{n^3})^{\prime}$$

By (a), $f(x)$ is uniformly convergent. Thus term by term differentiation is possible. Thus,

$$f^{\prime}(x) = \sum_{n=1}^{\infty}(\frac{\sin{nx}}{n^3})^{\prime} = \sum_{n=1}^{\infty}\frac{\cos{nx}}{n^2}$$

(c) Let $f(x) = \sum_{n=1}^{\infty}\frac{\sin{nx}}{n^3}$. Then

$$\int_{0}^{\pi}f(x)dx = \int_{0}^{\pi}\sum_{n=1}^{\infty}\frac{\sin{nx}}{n^3}$$

By (a), $f(x)$ is uniformly convergent. Thus, term by term integration is possible. Therefore,

$$\sum_{n=1}^{\infty}\int_{0}^{\pi}\frac{\sin{nx}}{n^3}dx = \sum_{n=1}^{\infty}\frac{2}{(2n-1)^{4}}$$