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Reversing Taylor Series to find the original function

  1. Jul 4, 2012 #1
    1. The problem statement, all variables and given/known data


    I need to find the convergence a unknown function. Now I know the Taylor series of it which is 1/3+2/(3^2)+3/(3^3+4/(4^4+...+k/(3^k). Which mean I can just take the Riemann sum of k/(3^k) from say 0 to 50 and that would give me 3/4.

    However this is not enough I need to confirm the answer by finding the unknown function. Normally this isnt too difficult as I would manipulate the series to get it into some sort of form that is familiar like e^x or cos(x), but this one is throwing me for a loop.

    The bottom part (3^k) fits x/(1-x) when x=3 but Im not sure about the top

    any help or hints would be great!
     
  2. jcsd
  3. Jul 4, 2012 #2
    I don't know if i am write or wrong , as i have never dealed with them but cant we write them as [summation of (K-1)/3^K]
     
  4. Jul 4, 2012 #3

    I like Serena

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    Welcome to PF, delta59! :smile:

    Perhaps you could start with substituting for instance x=1/3?
    What would you get?

    Afterward, try to think of defining a suitable function f(x) of which you know what the series sum is if you integrate f(x).
    (I'll explain later.)
     
  5. Jul 4, 2012 #4
    x/(1-x) when x=1/3 gives the correct bottom but then I just need to square the bottom function to get the top
    so the f(x)=x/((1-x))^2 gives the expansion of 1/3+2/(3^2)+3/(3^3+4/(4^4+...+k/(3^k)

    Thanks bouncing ideas I got now.
     
  6. Jul 4, 2012 #5

    I like Serena

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    Yep. That works too! :)
     
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