# Finding a power series representation

1. Apr 10, 2007

### superdave

1. The problem statement, all variables and given/known data

Start with the power series representation 1/(1-x) = sum from n=0 to inf. of x^n for abs(x) < 1 to find a power series representation for f(x) and determine the radius of convergence.

f(x)=ln(5+x^2)

2. Relevant equations

3. The attempt at a solution

Okay, so I get

1/(1-q)= sum q^n

Now, the problem lies in the fact that x is to a power of 2.

so if q= (-(x^2)-4) I get 1 / (1-(-(x^2)-4)) but if I want to integrate that, now that x is squared, I don't get the ln.

Can I do it the other way?

integrate first to get:

ln (1-q) = sum q^(n+1) / (n+1) and then replace q? I feel like that shouldn't work.

2. Apr 10, 2007

### Mystic998

My thoughts: f'(x) = 2x/(5 + x^2), right? So, keeping in mind that x is a constant term with respect to summation and all the work you've done, how would you write that as an infinite series?

Of course, I haven't done this in a while, so I could be completely wrong. Heh.

3. Apr 10, 2007

### HallsofIvy

Staff Emeritus
Yes. The derivative of ln(5+ x2 is 2x/(5+ x2). Now, you want to make that look like the sum of a geometric series a/(1- r). First, divide both numerator and denominator by 5: (2/5)x/(1+ x^2/5)= (2/5)x/(1- (-x^2/5)). Do see what a and r must be? Once you have the geometric series, integrate term by term to go back to the logarithm.