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

  1. Jan 1, 2013 #1
    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).
    Could someone please confirm that this is indeed the correct answer?
     
  2. jcsd
  3. Jan 1, 2013 #2

    Mark44

    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.
     
  4. Jan 1, 2013 #3
    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)?
     
  5. Jan 1, 2013 #4

    Dick

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    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?
     
  6. Jan 2, 2013 #5
    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?
     
  7. Jan 2, 2013 #6
    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!
     
  8. Jan 2, 2013 #7
    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.
     
  9. Jan 2, 2013 #8
    You must have made an algebraic mistake. Why don't you post what you did?
     
  10. Jan 2, 2013 #9

    Dick

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    Did you maybe calculate ##\frac{d}{dx}(ln(1-x)) = \frac{1}{1-x}##? That would be wrong.
     
  11. Jan 2, 2013 #10
    Why would that be wrong?
     
  12. Jan 2, 2013 #11

    Dick

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    You forgot to use the chain rule. Multiply by the derivative of 1-x.
     
  13. Jan 2, 2013 #12
    But that's the thing, I didn't forget to use the chain rule (to the best of my knowledge)!
     
  14. Jan 2, 2013 #13

    Dick

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    I give up trying to guess how you got the -1 factor. Can you show us how you did?
     
  15. Jan 2, 2013 #14
    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?
     
  16. Jan 2, 2013 #15
    Series for ln(1-x) is -Ʃ(n=1, ∞)xn/n
     
  17. Jan 2, 2013 #16
    That explains it :-)
     
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