Where's my mistake?

wintersolstice · 2011-05-20 08:30:25

The Riemann Zeta function is defined as

It's trivial zeros are said be negative even numbers,

However if you put a negative whole number (even or odd) the two negatives (from the minus sign in the function and the value of "s") the result is a whole number and positive power.

So the terms of the sequence are all whole numbers, which indicates an infinite sum, so why do all sources that I've read about the Riemann Hypothisis say that the trivial zeros (values of s for which the sum is 0) are negative even numbers?

Either:

I've gone wrong somewhere (In which case I would really appreciate it if someone would tell me:D)

I don't know the true definition of a "zero" (in a function)(In which case ot would be useful to know what one is)

or something else is happening???

but this has been bothering me for a while.

thank you:D

bobbym · 2011-05-20 09:53:32

Hi;

A small part of your question can be answered with this.

From numerical analysis the zeta function for negative real values is equal to:

Where B are the Bernoulli numbers. It is easy to see that every negative even number yields an odd Bernoulli number and they are equal to 0.

bossk171 · 2011-05-21 06:49:12

My understanding (which is mediocre at best) is that there's a way to define the zeta function on complex numbers. If zeta is defined on real numbers, (as you did it) I think you're correct in saying it diverges. But when it's define on complex numbers, it converges.

I think this: http://www.proofwiki.org/wiki/Equivalence_of_Riemann_Zeta_Function_Definitions might be a good place to start.

I would love for someone with some experience to shine some light on this.

wintersolstice · 2011-05-24 23:51:04

bobbym wrote:

Hi;
A small part of your question can be answered with this.
From numerical analysis the zeta function for negative real values is equal to:
Where B are the Bernoulli numbers. It is easy to see that every negative even number yields an odd Bernoulli number and they are equal to 0.

I understand what your saying but my problem is, if you put any negative whole number into the formula it diverges, so it's sum is ∞ so how can the values be finite (in this case the Bernoulli numbers for negative whole numbers if the sum diverges?)

I was thinking that maybe I had made a mistake or maybe I may have stumbled on a mistake

sorry I'm not trying to argue it just doesn't make sense to me (although my knowledge of maths is quite high I don't know absolutly everything)

bobbym · 2011-05-25 03:27:20

Hi;

From the definition:

The real part of s has to be greater than zero. When you use s = - 3 of course you are violating that condition.

For s being a negative integer you use the definition I gave above that involves the Bernoulli numbers.

It is not uncommon for a function to have different definitions for different type inputs.

One other point, when s = 1 we have the well know Harmonic series which also diverges.

Math Is Fun Forum

#1 2011-05-20 08:30:25

Where's my mistake?

#2 2011-05-20 09:53:32

Re: Where's my mistake?

#3 2011-05-21 06:49:12

Re: Where's my mistake?

#4 2011-05-24 23:51:04

Re: Where's my mistake?

#5 2011-05-25 03:27:20

Re: Where's my mistake?

Board footer