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

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  • #1
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Homework Statement


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


Homework Equations





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?
 

Answers and Replies

  • #2
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Homework Statement


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


Homework Equations





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?
 
  • #3
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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
Dick
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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?
 
  • #5
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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
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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!
 
  • #7
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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
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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?
 
  • #9
Dick
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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.
 
  • #10
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Why would that be wrong?
 
  • #11
Dick
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Why would that be wrong?

You forgot to use the chain rule. Multiply by the derivative of 1-x.
 
  • #12
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But that's the thing, I didn't forget to use the chain rule (to the best of my knowledge)!
 
  • #13
Dick
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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?
 
  • #14
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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
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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
 
  • #16
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That explains it :-)
 

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