Odd Bernoulli Numbers Must Be 0

Date: 02/16/2003 at 17:44:02
From: Jennifer
Subject: Bernoulli Numbers

I have read about an idea whereby you take the second derivative of 
t/exp(t) -1 and find that it is an even function. Then you show that 
an even function, when expressed as a Taylor Series, has only even 
powers of t. And from there we draw the conclusion that the odd 
Bernoulli numbers have to be 0.

Could you possibly explain how this proof would work?  First of all, 
I don't understand how a function that involves exp(t) can be even 
since exp(t) is neither even nor odd. And second, I would need to 
prove that it is true that the Taylor series expansion would have 
only even powers of t; I could not just state that fact.

Date: 02/17/2003 at 12:34:40
From: Doctor Douglas
Subject: Re: Bernoulli Numbers

Hi, Jennifer.

There are a number of different ways to prove the desired result. I 
think the idea of taking the second derivative and showing that it is 
even is an excellent one - it does all of the Bernoulli numbers in one 
swoop, and you only have to take two derivatives. Admittedly, the two 
derivatives are a little messy, and require you to apply the quotient 
rule twice to

             t
  f(t) = ----------
         exp(t) - 1

But I can tell you (from having just worked it out) that it's a lot
less tedium than doing the polynomial long division.

It is true that the function f"(t) will involve exp(t), and exp(t) is 
neither even nor odd. Note that the function  g(t) = 2*sinh(t) = 
exp(t) - exp(-t) is odd, even though it is the combination of two
things that are neither even nor odd - what is important is that
g(t) = -g(-t). Similarly, you will need to show that f"(t) is even by 
considering what happens to f(-t).  

Here's a hint that will save you some work. After you take both
derivatives, in the denominator you will obtain something like
[exp(t)-1]^4. If you multiply both numerator and denominator by
exp(-t/2)^4 = exp(-2t), the denominator takes the convenient form
[exp(t-t/2) - exp(-2t)]^4 = [exp(t/2) - exp(-t/2)]^4. Note that this 
denominator does not change sign if you go from t to -t, because of 
the exponent 4.

To answer your second question, to prove that the Taylor series
expansion has only even powers of x, you can do the following:

  Let f(x) even by assumption
  thus f(x) = f(-x)   by definition of "even"

Now differentiate this equation k times to get

  f^k(x) = (-1)^k f^k(x)

Now consider what happens for odd k and at x=0: Clearly the sign of 
f(0) doesn't change sign when we evaluate f(-0), so f^k(0) = 0 for all 
odd k. 

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/