proof of values of the Riemann zeta function in terms of Bernoulli numbers
This article proves part of the theorem given in the article.
Theorem 1.
Proof.
The method is as follows. Using Fourier series together with induction on n, we derive a formula
for the Bernoulli periodic function B2n(x) involving an infinite sum. On setting x to 0, this sum reduces to a constant times the appropriate zeta function
, and the result follows.
We first compute the Fourier series for B2(x). B2(x) is periodic with period 1, so
cn=∫10B2(x)e-2πinx𝑑x=∫10x2e-2πinx𝑑x-∫10xe-2πinx𝑑x+16∫10e-2πinx𝑑x |
We have
∫10e-2πinx𝑑x=0 | ||
∫10xe-2πinxdx=-12πnxe-2πinx|10+12πin∫10e-2πinxdx=i2πn | ||
∫10x2e-2πinxdx=-12πinx2e-2πinx|10+22πin∫10xe-2πinxdx=12π2n2+i2πn |
so that
cn=12π2n2 |
But then bn=cn-c-n=0 for all n, a0=0, and for n>0, an=cn+c-n=1π2n2 (where an are the coefficients of cos and bn the coefficients of sin in the Fourier series). Thus
B2(x)=∞∑k=11π2k2cos(2πkx)=1π2∞∑k=11k2cos(2πkx) |
Using this case as an inductive hypothesis, assume that for some n≥2
B2(n-1)(x)=(-1)n2⋅(2(n-1))!(2π)2(n-1)∞∑k=11k2(n-1)cos(2πkx) |
Then on (0,1)
B2n′′ |
and thus
Since , the sum converges absolutely, so we can move the sum outside the integrals, and we get
Thus we have established this formula for all . Setting , then, we get
or, trivially rewriting,
But clearly for , so it must be that the alternate in sign, and thus
Note that as a effect of this proof, we see that the even-index Bernoulli numbers alternate in sign!
Title | proof of values of the Riemann zeta function![]() ![]() ![]() |
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Canonical name | ProofOfValuesOfTheRiemannZetaFunctionInTermsOfBernoulliNumbers |
Date of creation | 2013-03-22 17:46:37 |
Last modified on | 2013-03-22 17:46:37 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 5 |
Author | rm50 (10146) |
Entry type | Proof |
Classification | msc 11M99 |