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

  • #1

Homework Statement


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].


Homework Equations





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.
 

Answers and Replies

  • #3
tiny-tim
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l'Hôpital? :wink:
 
  • #4
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Homework Statement


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].


Homework Equations





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.
You can get the first part pretty easily by long division of x by ex - 1 (= x + x2/2! + x3/3! + ... + xn/n! + ...).
 
  • #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.
 
  • #6
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That's true, but that's not very rigorous as it stands.
Sure it is.
What allows you to obtain series by dividing a polynomial by another series?
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.
Furthermore, how can you prove that the nth coefficient of the series obtained in this fashion is B_n for all n?
The coefficient of the nth term is Bn, pretty much by definition. What you need to prove is that $$\sum_{k=0}^{n} \binom{n+1}{k} B_k = 0 $$
This seems difficult to accomplish.
Well, maybe, but what I've suggested is what I would try.
 
  • #7
vela
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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.
Do you know about absolute convergence and its implication for the multiplication and division of series?
 
  • #8
Do you know about absolute convergence and its implication for the multiplication and division of series?

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.

The coefficient of the nth term is Bn, pretty much by definition. What you need to prove is that [tex] \sum_{k=0}^{n} \binom{n+1}{k} B_k = 0 [/tex]

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).
 
  • #9
vela
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You can think of a polynomial as a series where all but a finite number of terms vanish.
 
  • #10
Ok; do you have any suggestions for proving the second query?
 
  • #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.
 
  • #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.
 
  • #13
Any thoughts?
 

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