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

## Homework Statement

I am trying to represent the function 2x/(1+x2) as a series and determine its radius of convergence.

## 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).
Could someone please confirm that this is indeed the correct answer?

Mark44
Mentor

## Homework Statement

I am trying to represent the function 2x/(1+x2) as a series and determine its radius of convergence.

## 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).
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.
Could someone please confirm that this is indeed the correct answer?

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)?

Dick
Science Advisor
Homework Helper
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)?

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?

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?

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?

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!

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.

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

Dick
Science Advisor
Homework Helper
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.

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

Why would that be wrong?

Dick
Science Advisor
Homework Helper
Why would that be wrong?

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

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

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

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

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?

I initially had (at my disposal) the series for ln(1-x) = Ʃ(n=1,infinity) xn/n

Series for ln(1-x) is -Ʃ(n=1, ∞)xn/n

That explains it :-)