# On Riemann Hypothesis

1. Sep 3, 2007

### kezman

Can somebody explain me about the trivial zeros?

Why $$\zeta(-2) = \zeta(-4) = \zeta(-6) = 0 = \zeta(k)$$

So $$\zeta(k) \sum_{n=1}^{ \infty} \frac{1}{n^k} = 0$$?

2. Sep 3, 2007

### Hurkyl

Staff Emeritus
That series representation is only valid when the real part of k is bigger than 1. The functional equation is one way to see the trivial zeroes.

3. Sep 3, 2007

### Kummer

The idea here is analytic continuation. To extend the domain of the Riemann zeta function analytically. Now the important concept here is the complex gamma function. It can be show as Hurkyl says that the Riemann zeta function and complex gamma function satisfy the functional equation. Now we can extend the definition of the Riemann function by this property since the gamma function is still defined here. This immediately leads to the "trivial zeros".

4. Sep 4, 2007

### kezman

How do you "Extend the domain?"

5. Sep 5, 2007

### LorenzoMath

two holomorphic functions that coincide in a sub-domain which has an accumulation point coincide on the intersection of their domains. So you have a unique function defined on the union of their domains. we call this analytic continuation.

6. Sep 5, 2007

### LorenzoMath

this property depends on the fact that a holomorphic function has a unique tayler series expansion.

7. Sep 5, 2007

### Kummer

By using the functional equation. You see the Riemann zeta function agree on some region with the functional equation involving the gamma function. Since they agree on this region they must agree everywhere (as Lozerno say) if you want to extend the Riemann function analytically (i.e. analytic continuation) that is how it must be done.

I recommend to learn the complex gamma function extremely well before attempting to do this in detail. Have you ever learn Complex Analysis? It would be impossible to understand otherwise (even if you are Grothendieck)?

8. Sep 5, 2007

### kezman

I was going to ask you if i had to wait for complex analysis. Im on my way.

"You see the Riemann zeta function agree on some region with the functional equation involving the gamma function. Since they agree on this region they must agree everywhere (as Lozerno say) if you want to extend the Riemann function analytically (i.e. analytic continuation) that is how it must be done."

Thats interesting. So the functional equation evaluated in for example 2, gives the same as in the series form.

9. Sep 6, 2007

### Kummer

No! The series form of the Zeta function only works for real parts strictly greater than 1. It can be proved that functional equation is the same as the infinite series when the real part is greater than 1. However, what happens if the real part is not greater than 1? Then the series diverges. However, the functional equation still has sense. So we define the Zeta function to be equal to this value. It is a perfectly reasonable definition. That is what Bernhard Riemann did. The Zeta function was orginally discovered in the 18th Century by the great Euler. However, it only worked for real part greater than 1 domains. Riemann found this functional equation and got the idea that we can extend this function in a natural way.

So even though -2,-4,... make the series diverge we rather look at the functional equation for this to make sense. And we can easily see that these are the "trivial zeros".

Let me try to explain paragraph #2. {-2,-4,...} are not trivial zeros of Euler's Zeta function (because it diverges) (they are not even zeros!). However, {-2,-4,...} are trivial zeros of Riemann's Zeta function because this function is more than just infinite series and it is defined and zero at these values.

Last edited: Sep 6, 2007
10. Sep 7, 2007

### kezman

So the functional equation in 2 is $$\pi^2/6$$?