# Zwiebach page 221

1. Oct 31, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
On this page Zwiebach says that Riemann zeta(s) turns out to be finite for all values of s except s = 1?

I have no idea what he means since

$$\zeta(0.5) = \sum_{n=1}^{\infty} ( 1/n^{1/2})$$

is clearly not finite.

2. Relevant equations

3. The attempt at a solution

2. Oct 31, 2007

### Avodyne

The definition only applies for Re(s)>1, where the sum converges. For Re(s)<1, the zeta function is defined by analytic continuation.

For an explanation of how this works, I recommend Riemann's original paper. You can find a free English translation here:

http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta

3. Nov 1, 2007

### ehrenfest

I guess I just do not understand what the statement "zeta(s) turns out to be finite for all values of s except s = 1" means exactly.

zeta(s) is not even defined at s = 1 as you said...

4. Nov 1, 2007

### Jimmy Snyder

Can you rephrase this as a question. I doubt that the quoted sentence can be made any clearer.

He didn't say that.

Do you understand that the sum defines zeta(s) only when Re(s) > 1? This means that zeta(.5) is not defined by the sum.
Do you understand that analytic continuation defines zeta(s) for Re(s) < 1?

5. Nov 1, 2007

### ehrenfest

He said that zeta is only defined when Re(s) > 1. So, zeta(s) is not defined when s =1, right?

6. Nov 1, 2007

### Jimmy Snyder

You are misreading what he wrote. Once again, zeta(s) is only defined by the sum when Re(s) > 1. This does not mean that zeta(s) is not defined for Re(s) < 1, only that it is not defined by the sum. zeta(s) is defined by analytic continuation when Re(s) < 1. However, your final conclusion is correct, zeta(s) is not defined when s = 1.

Last edited: Nov 1, 2007
7. Nov 1, 2007

### ehrenfest

I see your point now. But "zeta(s) turns out to be finite for all values of s except s = 1" still seems like an incorrect or at least vacuous statement to me since as you just said zeta(s) is not defined when s= 1. So, since zeta(1) is not finite because it is not defined, is that what Zwiebach is saying?

Also, that statement implies that zeta(1+i) is finite somehow, even though it is not in either of the two catagories Re(s) < 1 or > 1?

8. Nov 1, 2007

### Avodyne

By a series of very clever manipulations, Riemann was able to show that the sum over n (which converges only for Re(s)>1) could be written in the form 1/(s-1) + terms which are obviously finite for any finite value of s (inlcuding complex values). We then use this new expression (which makes sense for all complex values of s other than s=1, where it has a simple pole) to define the zeta function.

Physicists would say zeta is infinite at s=1, mathematicians would say zeta is not defined at s=1.

Last edited: Nov 1, 2007
9. Nov 1, 2007

### Jimmy Snyder

Certainly not vacuous, it speaks volumes about zeta(s) when s is not 1. The only issue is what it says about zeta(s) when zeta is 1. Aristotle comes to Zwiebach's defense here. Is zeta(1) finite?

In my opinion, you are splitting hairs. Be pleased to read Zwiebach as:
"zeta(s) turns out to be finite for all values of s other than s = 1". Stated this way, it says absolutely nothing about zeta(1).

I have been waiting for this to come up. I am not familiar with the zeta function and am not inclined to look too deeply into it. My guess though is that analytic continuation works for all values of s except 1. Information on this is surely available on the net. You can grep for 'Riemann zeta function' or follow the link that was provided by Avodyne.

10. Nov 1, 2007

### ehrenfest

OK. I see what Zwiebach meant, and maybe I am splitting hairs, but I think it would have been clearer if Zwiebach had said instead:

"The function $$\zeta(s) = \sum_{n=1}^{\infty} ( 1/n^{s})$$ turns out to be finite for all values of s with Re(s) > 1 and can be analytically extended so that its value for s with Re(s) < 1 are finite as well. That leaves only numbers with Re(s) = 1 undefined by the zeta function"

or something like that.

11. Nov 1, 2007

### Jimmy Snyder

As you have written it, it is both confusing and incorrect. I would write it as he did because it is correct and clear.

12. Nov 1, 2007

### Avodyne

I looked up Zweibach, this is what he actually says:

"We consider the zeta function $\zeta(s)$, which is defined by the infinite sum
$$\zeta(s)=\sum_{n=1}^\infty {1\over n^s}, \;\;\; \Re(s) > 1.$$
The argument $s$ of the zeta function is assumed to be a complex number, but as indicated, the above sum only converges if the real part of the argument is greater than one. We can use analytic continuation to define the zeta function for all possible values of the argument. $\zeta(s)$ turns out to be finite for all values of $s$ except $s=1$."

13. Nov 1, 2007

### ehrenfest

OK. Zwiebach's right.