Odd Bernoulli Numbers Must Be 0Date: 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/ |
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