# Series representation for 2x/(1+x[SUP]2[/SUP])

1. Jan 1, 2013

### peripatein

1. The problem statement, all variables and given/known data
I am trying to represent the function 2x/(1+x2) as a series and determine its radius of convergence.

2. Relevant equations

3. The attempt at a solution
Using the series for ln(1-x) I have come up with the following series:
Ʃ(n=1,infinity) [2*(-1)n*x2n-1], with radius of convergence 1 (for 0<x<1).

2. Jan 1, 2013

### Staff: Mentor

You'll have to fill me in on the connection between ln(1 - x) and 2x/(1 + x2).

A much simpler way to go is to use the fact that 1/(1 + t) = 1 - t + t2 - t3 +- ...

Now substitute x2 for t to get the series for 1/(1 + x2).

Finally, multiply each term in the preceding series by 2x.

It could be that this results in the same thing you got, but this is a more straightforward way to go.

3. Jan 1, 2013

### peripatein

How could it be more straightforward than using a power series formula for ln(1-x) and the fact that the expression in the sum is merely the detivative of ln(1+x^2)?

4. Jan 1, 2013

### Dick

What's easy might be a matter of opinion. I'd use geometric series myself. But you still have a problem. If you put n=1 then the first term in your series is -2x. Is that really what you want?

5. Jan 2, 2013

### peripatein

You're right. Why is it that I am having this problem? Should it not have been equivalent? It has to be, as everything I did was just and correct. Would you explain?
Supposing I wished to use the series I came up with, how may I correct it?

6. Jan 2, 2013

### MrWarlock616

This happens to me too! But you can do it using your method too. The method easiest is always preferable!

$\frac{2x}{1+x^2}=\frac{d}{dx}(ln(1+x^2))$

Then use the expansion for ln(1+x) and replace $x$ by $x^2$ in it!

7. Jan 2, 2013

### peripatein

That's exactly what I did, but then I did get the first term to be -2x, which is incorrect! I could somewhat "cheat" and replace (-1)^n with (-1)^(n+1), but I doubt that's how it should be handled.

8. Jan 2, 2013

### MrWarlock616

You must have made an algebraic mistake. Why don't you post what you did?

9. Jan 2, 2013

### Dick

Did you maybe calculate $\frac{d}{dx}(ln(1-x)) = \frac{1}{1-x}$? That would be wrong.

10. Jan 2, 2013

### peripatein

Why would that be wrong?

11. Jan 2, 2013

### Dick

You forgot to use the chain rule. Multiply by the derivative of 1-x.

12. Jan 2, 2013

### peripatein

But that's the thing, I didn't forget to use the chain rule (to the best of my knowledge)!

13. Jan 2, 2013

### Dick

I give up trying to guess how you got the -1 factor. Can you show us how you did?

14. Jan 2, 2013

### peripatein

I initially had (at my disposal) the series for ln(1-x) = Ʃ(n=1,infinity) xn/n
I then substituted (-x^2), which yielded the following series: Ʃ(n=1,infinity) (-1)n*x2n/n
I then differentiated wrt x. This yielded: Ʃ(n=1,infinity) (-1)n*2x2n-1
Would you agree?

15. Jan 2, 2013

### MrWarlock616

Series for ln(1-x) is -Ʃ(n=1, ∞)xn/n

16. Jan 2, 2013

### peripatein

That explains it :-)