Power series

1. Apr 10, 2005

tandoorichicken

Two parts to this problem. On the first part I need someone to check my work, and I need help on solving the second part.

(a) Find a power series representation for f(x) = ln(1+x).
$$\frac{df}{dx} = \frac{1}{1+x} = \frac{1}{1-(-x)} = 1-x+x^2-x^3+x^4-...$$
$$\int_{n} \frac{1}{1+x} = \int_{n} (1-x+x^2-x^3+...)\dx = x-\frac{x^2}{2}+\frac{x^3}{3}-...+C = \sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1} +C$$
sub x=0 in original equation: ln(1+0) = ln(1) = 0 = C.
$$\ln(1+x) = \sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1}$$
with radius of convergence = 1.

(b) Find a power series representation for f(x) = x*ln(1+x).
If this function is differentiated, you get
$$\ln(1+x) + \frac{x}{x+1}$$
which is the same as a sum of power series
$$\sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1} + \sum^{\infty}_{n=0} (-x)^{n+1}$$
have I gone too far? or where do I go from here?
I know I will eventually have to integrate back to get the series for the original function.

2. Apr 10, 2005

StatusX

The first series looks right except for a sign, but why don't you just replace n+1 by n and run the index from 1 to infinity? For the second one, you can just multiply each term from the expansion for ln(1+x) by x.

Last edited: Apr 10, 2005
3. Apr 10, 2005

Hurkyl

Staff Emeritus
(a) looks fine (except for the sign), though it's usually good form to reindex the power series, and rearrange the terms, so that it looks like

$$\sum_{n = ?}^{\infty} (\mathrm{something}) x^n$$

(b) You don't have to do any differentiation at all... but you could do it that way if you really want to, by combining the two sums.