Evaluating the Riemann Zeta Function: Step-by-Step Guide for \zeta(c + xi)

[epkid08]

Can someone show me the steps to evaluating \zeta(c + xi), where 0 \leq c<1?

 

[Count Iblis]

Can someone show me the steps to evaluating \zeta(c + xi), where 0 \leq c<1?

There are many ways. In this case, as long as x is nonzero, you can simply use the definition of the zeta function

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

If s has a nonzero imaginary part, the summation will converge. You can then analytically continue this to the positive real axis by adding and subtracting the pole at s = 1.

 

[epkid08]

There are many ways. In this case, as long as x is nonzero, you can simply use the definition of the zeta function

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

If s has a nonzero imaginary part, the summation will converge. You can then analytically continue this to the positive real axis by adding and subtracting the pole at s = 1.

I have some questions.

The definition of the zeta function you posted, doesn't that only work for where re(s)>1?

I don't really understand the correct usage of s. Isn't s defined as n + xi? So then how can s just equal 1? Or does s refer only to the real part of it?

The non-trivial zeros of the zeta function happen when re(\zeta(s)) + im(\zeta(s))=0, and are said to be equal to \zeta(1/2+xi), right?

Hopefully you can see through my confusion

 

[CRGreathouse]

I don't really understand the correct usage of s. Isn't s defined as n + xi? So then how can s just equal 1? Or does s refer only to the real part of it?

Re(s) is the real part of s, which is complex. I don't know what you mean by "how can s just equal 1", since no one's claimed that to be the case. Were you talking about the pole at s = 1? That's the thing that stops you from evaluating zeta(1) -- which is obvious since you're trying to sum the harmonic sequence there.

The non-trivial zeros of the zeta function happen when re(\zeta(s)) + im(\zeta(s))=0, and are said to be equal to \zeta(1/2+xi), right?

Yes, \zeta(s)=0 at s=1/2+xi for some real values of x.

 

[epkid08]

Re(s) is the real part of s, which is complex. I don't know what you mean by "how can s just equal 1", since no one's claimed that to be the case. Were you talking about the pole at s = 1? That's the thing that stops you from evaluating zeta(1) -- which is obvious since you're trying to sum the harmonic sequence there.



Yes, \zeta(s)=0 at s=1/2+xi for some real values of x.

I'm not quite sure how to evaluate the function where s equals say, 1/2 + i, mainly the i part.

 

[Santa1]

To evaluate for 0<Re(s)<1 you can use the Dirichlet eta function;

http://mathworld.wolfram.com/DirichletEtaFunction.html

 

[Kurret]

What is the definition of the zeta function anyway? I only found this:\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}
on wikipedia, but then I read that zeros occur at even negative integers, but I can't see how that fits with this definition...

 

[Santa1]

That is the definition for Re(s)>1 only. You have to use the analytic continuation to evaluate for the negative even integers.

 

[Count Iblis]

What is the definition of the zeta function anyway? I only found this:\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}
on wikipedia, but then I read that zeros occur at even negative integers, but I can't see how that fits with this definition...

That summation defines an analytical function for Re(s)>1. You can then analytically continue that function to the whole complex plane. You then get a (unique) meromorphic function on the complex plane that is identical to the given summation for Re(s) > 1.

 

[Count Iblis]

If Im(s) is different from zero, then doesn't the summation converge for 0<Re(s)<1? You'll have an oscillatory summand with decreasing terms...

 

[CRGreathouse]

If Im(s) is different from zero, then doesn't the summation converge for 0<Re(s)<1? You'll have an oscillatory summand with decreasing terms...

It doesn't converge nicely, that's for sure... it's wildly oscillating up to 10^7, at the least, in the OP's example.

 

[Kurret]

That is the definition for Re(s)>1 only. You have to use the analytic continuation to evaluate for the negative even integers.

Which means?

 

[Santa1]

Let's recap here,

If you want to evaluate the zeta function With Re(s) > 1 you can use the definition.

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

If you want to evaluate for Re(s) > 0 you can use the Dirichlet eta function for an analytic continuation.

\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}

Which satisfies the relationship:

\frac{\eta(s)}{1-2^{1-s}}=\zeta(s)

To evaluate for Re(s) < 0 you can use another analytic continuation.

\zeta(s)=2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s)