# Thread: Symmetry of Zeta Function

1. ## 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.

2. 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.

Why would the power series not extend to those values?

4. 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.

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

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.
8. 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.