Zeta Function for dummies.


by hawaiifiver
Tags: dummies, function, zeta
hawaiifiver
hawaiifiver is offline
#1
Sep26-11, 05:24 AM
P: 57
Hello to all.

This could be quite long. Apologies. I am a physics student trying to understand the Zeta function and the Riemann hypothesis. Its not on my coursework, but I am interested in pure mathematics. I have a few questions. Perhaps you can help me out. Thank you.

My questions are concerned with the Wikipedia page on the Riemann Hypothesis.

(Q1) In the formula for the Zeta function i.e. [tex] \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} [/tex] is the [tex] s [/tex] a complex number like [tex] s = a + bi [/tex] Can [tex] a [/tex] and [tex] b [/tex] take any value?

(Q2)

On the Wikipedia page for the Riemann Hypothesis, there is a diagram in the top right hand corner. They state that the diagram is a plot of [tex] s = \frac{1}{2} + i \ x [/tex] . Does that mean that I would have to compute the sum of the Zeta function for each value of x, in order to plot that diagram?


(Q3) So how do I calculate the sum of an infinite series if that is the case? For instance, how would I calculate [tex] \zeta(\frac{1}{2} + i \ 14.135) [/tex]. I ask this because I want to see how they arrive at the zero of the Zeta function in that Wikipedia diagram.

(Q4) Does the value of [tex]a[/tex] and [tex]b[/tex] determine whether I can calculate the sum of the infinite series of the zeta function?

(Q5) Could you recommend a good introductory book on the Zeta function. As you can see I am pretty much flying solo on figuring this out.

Thanks for your help.
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chiro
chiro is offline
#2
Sep26-11, 07:43 AM
P: 4,570
Hey hawaiifiver.

In terms of functions, functions only make sense when the output is something that is finite, and in this same spirit, the Zeta function only makes sense for values that converge.

Wikipedia has some specific details about the convergence of the Zeta function:

http://en.wikipedia.org/wiki/Riemann_zeta_function

If you want to study things like convergence, you should probably look at some areas of math like analysis and topology. Calculus I and II have preliminary ideas about how to test whether certain sequences converge or not, if you need some background information before going to the higher level material.

So to answer your question, if s is a complex number, then if s = a + bi and Zeta(s) converges or can be shown to converge, then s is a valid part of the domain of the Zeta function.
micromass
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#3
Sep26-11, 09:21 AM
Mentor
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P: 16,690
Hi hawaiifiver!

You seem to be missing something. That is, the definition

[tex]\zeta(s)=\sum_{n=0}^{+\infty}{\frac{1}{n^s}}[/tex]

does not hold for each s. That definition is only for the s with Re(s)>1. That is, for the s=a+bi with a>1.

For all other s, the Riemann zeta function is defined as the analytic continuation of the above sum. To know the values of the Riemann zeta function is points like [itex]\frac{1}{2}+ix[/itex], we can not use the above sum. We will have to use another form.

For more information about the Riemann zeta function, I suggest you read a good book on complex analysis first. I like the book by Freitag & Busam. It includes some information on the zeta and gamma functions.

RamaWolf
RamaWolf is offline
#4
Sep26-11, 10:33 AM
P: 96

Zeta Function for dummies.


Best bid for an introduction to the Rieman Zeta (and very nice to read!) is:

John Derbyshire: Prime Obsession, Bernhard Riemann and the Greatest Unsolved
Problem in Mathematics
RamaWolf
RamaWolf is offline
#5
Sep26-11, 10:54 AM
P: 96
If you just want to know the value of [tex] \zeta(\frac{1}{2} + i \ 14.135) [/tex]

got to WolframAlpha and enter

http://www.wolframalpha.com/input/?i=RiemannZeta[1%2F2%2B14.135+i

If you want to write your own computer function which calculates zeroes
of the Riemann zeta, I could assist you with my experience in writing a
(PowerBasic) function for that purpose.
hawaiifiver
hawaiifiver is offline
#6
Oct1-11, 09:17 AM
P: 57
Quote Quote by micromass View Post
Hi hawaiifiver!

You seem to be missing something. That is, the definition

[tex]\zeta(s)=\sum_{n=0}^{+\infty}{\frac{1}{n^s}}[/tex]

does not hold for each s. That definition is only for the s with Re(s)>1. That is, for the s=a+bi with a>1.

For all other s, the Riemann zeta function is defined as the analytic continuation of the above sum. To know the values of the Riemann zeta function is points like [itex]\frac{1}{2}+ix[/itex], we can not use the above sum. We will have to use another form.

For more information about the Riemann zeta function, I suggest you read a good book on complex analysis first. I like the book by Freitag & Busam. It includes some information on the zeta and gamma functions.
I'll check out that book. Thank you.



I found this formula from Konrad Knopp on wikipedia which is a globally convergent series.

[tex]

\zeta(s)=\frac{1}{1-2^{1-s}}
\sum_{n=0}^\infty \frac {1}{2^{n+1}}
\sum_{k=0}^n (-1)^k {n \choose k} (k+1)^{-s} [/tex]

Does this converge then, when substituting in s = 1/2 + 14.1347 i
lostcauses10x
lostcauses10x is offline
#7
Oct1-11, 09:48 AM
P: 94
All good advice. Also find a course or book on Complex analysis if you have not already had such.
RamaWolf
RamaWolf is offline
#8
Oct1-11, 11:32 PM
P: 96
I am preparing a post:

'Programming details on the computation of the Riemann zeta function using Aribas'

There you can find all details to compute ζ([itex]\frac{1}{2}[/itex] + 14.1347 i)
hawaiifiver
hawaiifiver is offline
#9
Oct3-11, 06:08 AM
P: 57
Thanks.


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