# Riemann Zeta Function?

1. Mar 24, 2006

### pivoxa15

They claim that the trivial 0s (zeta(z)=0) occur when z is a negative odd integer (with no imaginary component). But it seems obviously wrong.

Take z=-2

zeta(-2)=1+1/(2^-2)+1/(3^-2)....
=1+4+9....

Obviously this series will not equal 0.

Where have I gone wrong?
Have I misunderstood the meaning of 'zeros of the zeta function'?

Last edited: Mar 24, 2006
2. Mar 24, 2006

### matt grime

That is not the taylor series of the zeta function for the domain in which you're looking. Surely you noticed that the sum doesn't actually exist, never mind not being zero.

You need to use analytic continuation to define the function for any z (or s usually) with real part less than or equal to 1.

3. Mar 24, 2006

### HallsofIvy

Staff Emeritus
Perhaps you misunderstood the meaning of "odd"!

4. Mar 24, 2006

### shmoe

5. Mar 24, 2006

### waht

This formula work if the exponent is greater than 1. There is another integral formula that works for all domain, and can be simplified to the series sum which you descibe, also, it simplifies to bernoulli numbers resulting 0's of negative even integers.

6. Mar 24, 2006

### pivoxa15

So the formula I quoted $$\zeta(z)=\sum_{n=1}^{\infty}1/n^{z}$$
and the most popular form to define the zeta function is undefined (or divergent) for Re(z)<=1.

The second most popular form
$\zeta(z)=\prod_{p\text{ prime}}(1-p^{-z})^{-1}$ is also undefined for Re(z)<=1.

The form that is defined for all Re(z) is
$$\zeta(z)=2^{z}\pi^{z-1}\sin(\pi z/2)\Gamma(1-z)\zeta(1-z)$$

I realise that this form is iterative in that the zeta function appears on both sides so this form of $$\zeta(z)$$ is just like an infinite series or product.

Since the Riemann Hypothesis is concerned with z when Re(z)=1/2 and the first two forms of the zeta function is undefined when Re(z)=1/2, why qupte them in popular texts that describe the Riemann Hypothesis?

Even my lecturer when describing the Riemann Hypothesis quoted $\zeta(z)=\sum_{n=1}^{\infty}1/n^{z}$ and said that the goal is to find non trivial 0s of this function. If the function is undefined for Re(z)<=1 and non zero for Re(z)>1 than there are no zeros of this function in that form.

Last edited: Mar 24, 2006
7. Mar 24, 2006

### waht

8. Mar 24, 2006

### shmoe

The Dirichlet series and the Euler product are the 'natural' way that the zeta function came to be. With just these guys alone and restricted to real variables s>1, Euler managed an alternate proof that there are infinitely many primes by considering what happens as s->1.

Riemann then considered complex values of s and showed many crucial things. Firstly, where the Dirichlet series converges, that is the half plane with real part of s>1, this defines an analytic function. Next, this analytic function can be extended to the entire complex plane, with a simple pole at s=1. Like all analytic continuations, this extension is unique- we aren't arbitrarily defining it in the rest of the plane, there is exactly one way of doing this. In otherwords the Dirichlet series definition is enough to uniquely determine the zeta function (in the entire complex plane), so this is the standard starting point.

Last edited: Mar 24, 2006
9. Apr 3, 2006

### eljose

i have another "little" question..we know that Riemann zeta function can be extended to negative values of "s" and is related to Dirichlet Eta function..then my question is about the Prime zeta function:

$$P(s)= \sum_{p}p^{-s}$$

a)can P(s) be extended to argument s<1 and s negative?
b)would be a relationship between P(s) and a "so called" alternating prime zeta function?..thanks.

EDIT: Pivoxa made a "funny" arguemtn using the negative values of Riemann zeta function, in fact this "negative" values in the form:

$$1+2^{m}+3^{m}+................ \rightarrow{\zeta(-m)}$$

are used by physicist in wich they call "Zeta regularization" and was an idea i had for my PhD to use this "regularization" for giving a finite meaning to integrals in the form:

$$\int_{0}^{\infty}dpp^{m}$$ believe it or not the 2regularization" for divergent series appears in G.H Hardy book "divergent series" and it,s widely used in physics for the case m=-2n n=1,2,3,4,.... the sum has a "zero" value

Last edited: Apr 3, 2006
10. Sep 25, 2008

### yasiru89

elijose-

a) Yes. Since the prime zeta function can be expressed in terms of the Riemann zeta with a Mobius inversion (see the mathworld article for prime zeta).

b) Since there's no easy relationship between an 'alternating' prime zeta function and the prime zeta function (since, trivially- the sum is taken over primes), I would use the relation I mentioned in a) for the analytic continuation.

Regarding the 'problem' of extending the series definition, we know that the analytic continuation has a uniqueness property, hence the definitions (in terms of the contour integral or the globally convergent series vs the 'standard' series definition) are compatible. Given extended summation senses, all the series results (even the one that reads 'the sum of every positive integer to the power of a positive even integer is zero') can be made rigorous independent of the analytic continuation and the resulting reflective functional equation (this was indeed foreseen by Euler arguably- see Hardy's Divergent Series for more).

11. Feb 24, 2009

### yasiru89

See http://mathrants.blogspot.com if anyone's interested in summation methods by which we may recover the analytic continuation of Riemann's zeta function.