This is a thread meant to help those who are having trouble with power series. Also I will cover some useful uses for these power series.

Part 1. What is a power series?

A power series bascially works of this observation

"Hey at x=0 $\displaystyle e^{x}$ and $\displaystyle x+1$ are equal, and also for numbers close to x=0 $\displaystyle x+1$ is a good approximation for $\displaystyle e^{x}$. I wonder if I can find a way to find a good approximation for numbers like x=7?"

Well you do this by first taking one of these cases and saying since $\displaystyle e^{x}=x+1,x=0$. Next you can say "well if they increase at a similar rate($\displaystyle f'(x)$ the approximation will be better...and hey if the rate of the rate($\displaystyle f''(x)$) increases similarly then the approximation will be even better..and if the rate of the rate of the rate....ad infinitum"

This is how The classic $\displaystyle f(c)=\sum_{x=a}^{\infty}\frac{f^{(n)}(c)(x-c)^n}{n!}$ arises....

This series will ensure that for all the values at which the series converges the sereis will follow the pattern of the series's rate being comparable to the rate of the function being approximated.

Now I am writing this assuming you knew this. So That is all I am going to go into regarding that

Part 2. Why power series converge only for a certain values?

To show this make a show of good faith and assume that the following is correct

$\displaystyle \ln(1-x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1},\forall{x}\ in[-1,1)$

I will show later how this was gotten...You may ask why there has to be a limiting interval?

This is called the interval of convergence. This is exactly what it sounds like, the series only converges for x values that are an element of the interval. To show this I will take one of the values contained x=-1.

If we imput that into our series we have

$\displaystyle -\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1}$

Now since $\displaystyle a_{n+1}<a_n$ and $\displaystyle \lim_{n\to\infty}a_n=0$ we see this converges by the alternating series test

So we see that this converges for x=-1

Next let's show why x=1 is excluded. When x=1 the series becomes

$\displaystyle -\sum_{n=0}^{\infty}\frac{1}{n+1}$

which diverges by the integral test

$\displaystyle \int_0^{\infty}\frac{dx}{x+1}=\ln(x+1)\bigg|_0^{\i nfty}=\infty$

Now lets test another random value...say $\displaystyle x=e^{\pi}$ Gelfond's constant

Imputting that we get $\displaystyle -\sum_{n=0}^{\infty}\frac{e^{n\pi}}{n+1}$

Which obviously diverges due to the n-th term test

$\displaystyle \lim_{n\to\infty}\frac{e^{n\pi}}{n+1}\ne{0}$

This is also true $\displaystyle \forall{x}\notin[-1,1)$

Part 3. How do I find power series for unfamilar functions?(I will do this for Maclaurin only)

First off let me be blunt. You are going to have much difficulty in this topic if you don't memorize the basic power series

To learn them look here Maclaurin Series -- from Wolfram MathWorld

The must knows are

$\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n},\forall{x}\in[-1,1]$

$\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^{n},\forall{x}\in[-1,1]$

$\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!},\forall{x}\in\mathbb{R}$

$\displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!},\forall{ x}\in\mathbb{R}$

From these you can use tricks to find all the others you'll most likely encounter

First lets look at the previously stated power series for $\displaystyle \ln(1-x)$

Here is how this was derived since

$\displaystyle -\int\frac{dx}{1-x}=\ln(1-x)$

and we know that $\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$

we can make the jump that

$\displaystyle \ln(1-x)=-\int\frac{dx}{1-x}=-\int\sum_{n=0}^{\infty}x^{n}dx=-\sum_{n=0}^{\infty}\int{x^{n}}dx=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$

Note that although integrating a power series does not change its radius of convergence it can change its endpoint behavior

I will go over three more

THe first is this: Find the power series for $\displaystyle \sin(\sqrt{x})$

Well first we need to find the power series for $\displaystyle \sin(x)$ before we can tackle this...first we make this observation

$\displaystyle \int\cos(x)dx=\sin(x)\therefore\sin(x)=\int\sum_{n =0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$

You can verify the details missed yourself

So if

$\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\Rightarrow{\sin(\sqrt{x}) =\sum_{n=0}^{\infty}\frac{(-1)^{n}(\sqrt{x})^{2n+1}}{(2n+1)!}}$

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The second one is $\displaystyle \frac{6x}{5-2x}$

The key for these type is to get it into one of our normal forms $\displaystyle \frac{1}{1-x}$

To do this we first need to have a 1-u(x) on the bottom so we rewrite it as this $\displaystyle \frac{6x}{5}\frac{}{1-\frac{2}{5}x}$

The second part of which looks like our known formula so we apply the power series replacing

$\displaystyle \frac{1}{1-\frac{2}{5}x}$ with $\displaystyle \sum_{n=0}^{\infty}\bigg(\frac{2}{5}x\bigg)^{n}$

To get $\displaystyle \frac{6x}{5-2x}=\frac{6x}{5}\sum_{n=0}^{\infty}\bigg(\frac{2}{ 5}\bigg)^{n}x^{n}=\frac{6}{5}\sum_{n=0}^{\infty}\b igg(\frac{2}{5}\bigg)^{n}x^{n+1}$

Check interval of convergence by using ratio test

I will be back later for third example and uses.