# Symmetry of Zeta Function

• Jun 20th 2008, 10:02 AM
rman144
Symmetry of Zeta Function
I know that the functional equation for the Zeta function has a symmetrical property such that f(s)=f(1-s), but I was curious as to whether or not this applies to the more simplistic version of the equation where it is basically:

sigma[1/(n^(a+bi))], n=1, n=infinity.

Basically, does anyone know if:

sigma[1/(n^(1/2+R+bi))]=sigma[1/(n^(1/2-R+bi))]

Would be true for a zero off the line, should one occur.
• Jun 20th 2008, 10:14 AM
ThePerfectHacker
I think you are assuming that the Zeta function is expressed in terms of a power series, it is not. It is only expressiable as a power series when the real part of $s$ is greater than $1$, otherwise one cannot use power series.

If we define $f(s) = \pi ^{-s/2} \Gamma (s/2) \zeta (s)$ then $f(1-s) = f(s)$. But this only works for the analytically extended Zeta function. Not the power series.
• Jun 20th 2008, 10:38 AM
rman144
Why would the power series not extend to those values?
• Jun 20th 2008, 11:55 AM
Mathstud28
Quote:

Originally Posted by rman144
Why would the power series not extend to those values?

I assume it would be because

$\zeta(s<1)=\sum_{n=0}^{\infty}\frac{1}{n^{s}}$

Which would be a divergent power series.
• Jun 20th 2008, 12:43 PM
ThePerfectHacker
Quote:

Originally Posted by rman144
Why would the power series not extend to those values?

Quote:

Originally Posted by rman144
What if "s" is complex? Then the equation seperates into the real and imaginary parts such that:

0=sigma[cos(b ln(n))/n^a], 0=sigma[sin(b ln(n))/n^a], both of which do not diverge for 0<a<1.

It makes no difference whether $s$ is complex or not. In order to have convergence in the power series we need to have $\Re (s) > 1$. The way Riemann defined the Zeta function for divergent values if by using the functional equation.
• Jun 20th 2008, 12:58 PM
rman144
So if that is correct, then how does Riemann's functional equation account for the divergence? Basically, what does the functional equation provide that the power series is incapable of doing, and more importantly, how?
• Jun 20th 2008, 01:38 PM
ThePerfectHacker
Quote:

Originally Posted by rman144
So if that is correct, then how does Riemann's functional equation account for the divergence? Basically, what does the functional equation provide that the power series is incapable of doing, and more importantly, how?

Have you ever studied complex analysis? Here is an example, consider, $f(z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}z^n}{n}$. This function is only defined on $|z-1|<1$. However, there is a way to extend the domain of this function. Eventhough power series are not going to work if we let $g(z) = \log z$ then $g(z)$ would analytically extend $f(z)$ to the whole complex plane minus the non-positive axis. The similar situation with the Zeta function, but it is a lot more difficult.
• Jun 21st 2008, 12:03 AM
Opalg
There seems to be some confusion here. The series $\zeta(s)=\sum_{n=0}^{\infty}\frac{1}{n^{s}}$ is not a power series. A power series is a series of the form $\sum a_ns^n$, where s is the variable, and the power to which it is raised is the summation index. In the series for ζ(s) it is the other way round. The summation index n is raised to the power -s.

The Riemann zeta function is analytic everywhere except for a pole at s=1. It must therefore have a power series representation in any disk or annulus that does not contain the point 1. But as far as I know there is no convenient way of calculating an explicit power series representation for it. Power series do not seem to be a useful tool in the analysis of the Riemann function.