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Taylor series of e^x/(x-1)

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex] g(x) = \frac{x}{e^x - 1} = \sum_{n=0}^{\infty} \frac{B_n}{n!} x^n[/itex] be the taylor series for g about 0. Show B_0 = 1 and [itex] \sum_{k=0}^{n} \binom{n+1}{k} B_k = 0 [/itex].


    2. Relevant equations



    3. The attempt at a solution
    [itex] g(x) = \sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!} x^n [/itex], but [itex] g^{(n)}(0) [/itex] is always undefined at 0. So I don't see how any of these relations can hold.
     
  2. jcsd
  3. Apr 4, 2013 #2
    Does anyone have any suggestions?
     
  4. Apr 4, 2013 #3

    tiny-tim

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    l'Hôpital? :wink:
     
  5. Apr 4, 2013 #4

    Mark44

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    You can get the first part pretty easily by long division of x by ex - 1 (= x + x2/2! + x3/3! + ... + xn/n! + ...).
     
  6. Apr 4, 2013 #5
    That's true, but that's not very rigorous as it stands. What allows you to obtain series by dividing a polynomial by another series? Furthermore, how can you prove that the nth coefficient of the series obtained in this fashion is B_n for all n? This seems difficult to accomplish.
     
  7. Apr 4, 2013 #6

    Mark44

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    Sure it is.
    There's not just one way to obtain a series. For example, the Maclaurin series for 1/(1 - x) is frequently obtained by long division.
    The coefficient of the nth term is Bn, pretty much by definition. What you need to prove is that $$\sum_{k=0}^{n} \binom{n+1}{k} B_k = 0 $$
    Well, maybe, but what I've suggested is what I would try.
     
  8. Apr 4, 2013 #7

    vela

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    Do you know about absolute convergence and its implication for the multiplication and division of series?
     
  9. Apr 5, 2013 #8
    Yes, but here it is not a series being divided by a polynomial, but the other way around. I have not really encountered this before, so I suppose the complete answer to your question is no.

    I really don't see a good way of doing this. What we are obtaining through this division is some series for which a general term for B_n is not known (and I don't think there is some simple formula, after writing out the first few terms).
     
  10. Apr 5, 2013 #9

    vela

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    You can think of a polynomial as a series where all but a finite number of terms vanish.
     
  11. Apr 5, 2013 #10
    Ok; do you have any suggestions for proving the second query?
     
  12. Apr 5, 2013 #11

    vela

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    Start with
    $$\frac{x}{e^x-1} = \frac{1}{\frac{e^x-1}{x}}.$$ Then convert the problem into one of multiplying two series. It works out pretty easily. You don't need to solve for the ##B_k##'s explicitly.
     
  13. Apr 6, 2013 #12
    Alright its simple enough. On a side note (although it is not necessary because it is given in the problem statement), if we were not given that g was given by its taylor series in a neighborhood of zero, how to prove that it is? The series manipulations require us knowing g is given by its Taylor series.
     
  14. Apr 7, 2013 #13
    Any thoughts?
     
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