# Power series

1. Dec 5, 2008

### hyper

Hello, I have a power series, and a problem with an integration with it, I dont understand why I should integrate it from zero at one point. I have attached a detalied explanation of the problem.

http://img79.imageshack.us/img79/5156/powerseriespt4.jpg [Broken]

Any help would be greatly appreciated.

Last edited by a moderator: May 3, 2017
2. Dec 5, 2008

### HallsofIvy

Staff Emeritus
No, you don't integrate anything, you simply expanded the sum incorrectly.

You are given that
$$g(x)= \sum_{n=1}^\infty \frac{x^{n+2}}{n(n+2)4^n}$$
You say, correctly, that the derivative of that is, differentiating term by term,
$$g'(x)= \sum_{n=1}^\infty \frac{x^{n+1}}{n 4^n}$$
Obviously, you can factor an "x" out of that leaving
$$g'(x)= x\sum_{n=1}^\infty \frac{x^n}{n 4^n}= x\sum{n=1}^\infty \frac{1}{n}\left(\frac{n}{4}\right)^n[/itex] [tex]g'(x)= x\sum_{n=1}^\infty \frac{1}{n}\left(\frac{x}{4}+1- 1\right)^n$$
and you should be able to recognize that last sum as the Taylor's series for ln(x) about x= 1.

Last edited: Dec 6, 2008
3. Dec 6, 2008

### hyper

Thanks for your answer! I did not recognize that it was the series for ln(x) about one.

But I would like to know how I can solve excercises like this without recognizing known series. And if you look at what I have written you'll se that I do end up with the right answer without recognizing any known series, but I would like to know why.

I think it could be nice for me learning to solve problems like this without knowing how a lot of known series look like.

Have a nice day.

4. Feb 5, 2012

### kekxi

I'm sure you're past caring by now, but actually you were doing fine. All you need to do now is evaluate the constant of integration. This can be done by plugging in any value of x within the radius of convergence. For example, x=1 or x=3 will work, but again you'd need to recognize the series for ln(1-1/4) or ln(1-3/4), which you are trying to avoid.

To get around this, you need to "plug in" x=0. Of course, this is cheating, since g'(x)/x is undefined at x=0, but we step around that problem by using the limit of g'(x)/x as x APPROACHES 0. This is perfectly well-defined and easily evaluated to be equal to 0, since your equation (3) shows that away from 0, g'(x)/x is a power series with a factor of x.

Since g'(x)/x approaches 0 as x approaches 0, your last equation implies that C= ln(4) as required.

It is often useful with term-by-term integration problems to integrate from 0 to x as you suggest. This is equivalent to plugging in x=0 as above, since the integral of any integrable function from 0 to 0 is of course 0. Because of the x in the denominator in this problem, I think it's clearer to handle the way you did it.

Incidentally, I have bypassed some mathematical niceties regarding equating the limit of a function and the value of it's series expansion. Suffice to say that there is no problem because the series in question converges and is continuous in an interval around x=0.