4

This may be a stupid question but is there a way to calculate Riemann's Zeta Function by hand exactly or can you only estimate it?

MJD
  • 65,394
  • 39
  • 298
  • 580
The Diamante
  • 195
  • 2
    You can certainly compute the values at the even integers "by hand." Which values do you want? – Potato Apr 10 '13 at 01:45
  • As simple as I can get. Just trying it out. – The Diamante Apr 10 '13 at 01:48
  • Well, then it was no joke: Potato already told you that the values of zeta at the even ( negative, I'm guessing he meant) integers are pretty easy to check by hand...even sleeping, in the dark and dancing rock'n'roll – DonAntonio Apr 10 '13 at 01:52
  • Thank you! How would you go about doing this though? – The Diamante Apr 10 '13 at 01:54
  • This question seems to be relevant http://math.stackexchange.com/questions/3271/how-to-evaluate-riemann-zeta-function?rq=1 – Baby Dragon Apr 10 '13 at 01:55
  • From the parts I understood it looks like the right answer. All that's left to do is stare at the formulas for a few hours and then I will have a "EUREKA!" moment. – The Diamante Apr 10 '13 at 02:04
  • I wonder what you mean by "exactly." If you mean symbolically, then sure. Indeed, if it weren't for the fact that humans found the explicit values by hand in the first place long ago, our computers today likely wouldn't "know" how to compute explicit values either. If you mean a finite decimal expansion that is not simply an approximation, then no. (It's not like we can write out an infinite decimal expansion.) – anon Apr 10 '13 at 02:13
  • By exactly I meant symbolically. Thanks for pointing that out. – The Diamante Apr 10 '13 at 02:16

3 Answers3

3

The functional equation of the zeta function gives you the trivial zeros at once:

$$\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$$

if you input now $\,s=-2k\;,\;\;k\in\Bbb N\,$ , you get

$$\zeta(-2k)=2^{-2k}\pi^{-2k-1}\sin\left(\frac{\pi (-2k)}{2}\right)\Gamma(1+2k)\zeta(1+2k)=0\ldots$$

Question: Why in the negative even integers and not in the positive ones? Hint: check stuff about the gamma function.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • I have found many questions and pages in the Gamma Function and understand the basics of it but can not find what the individual variables represent/their values. – The Diamante Apr 10 '13 at 02:34
  • Ok @TheDiamante, then you probably know the Gamma function has poles on the negative integers, and that's why in the above relation we had to take even negative integers... – DonAntonio Apr 10 '13 at 02:53
  • Yes, I understand this but I do not understand how to find all the variable of the Gamma Function. How would this be achieved? – The Diamante Apr 10 '13 at 03:05
  • I don't understand what you mean by "all the variable of the Gamma Function"... – DonAntonio Apr 10 '13 at 03:06
  • There are variables within the function that I can't find how to find their values. As here http://en.wikipedia.org/wiki/Gamma_function, I don't know the values of the variables d, e and t in the second formula. – The Diamante Apr 10 '13 at 03:11
  • In that formula $,d,$ is not a vriable but part of $,dt,$ in the integral. For the other variables I really advice you to read in a book: it's way too long to do it here. – DonAntonio Apr 10 '13 at 03:16
  • Thank you for clearing that up. Sad they only give so much room to explain. – The Diamante Apr 10 '13 at 03:18
  • @TheDiamante Do you know calculus? You will surely need it to tackle anything significant about the zeta function on your own terms. – anon Apr 10 '13 at 04:12
1

Assuming you can calculate $n^s$ for natural $n$ and whatever $s$ you are interested in, you could try using:

$$\zeta(s)=\frac{\eta(s)}{1-2^{1-s}}$$

where

$$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s},\quad\Re(s)>0$$

Define the partial sums of this to be

$$\eta(s,k)=\sum_{n=1}^k\frac{(-1)^{n+1}}{n^s}$$

While it is true that we have

$$\eta(s)=\lim_{k\to\infty}\underbrace{\eta(s,k)}_{\eta(s,k,0)}$$

we actually have the faster converging limit:

$$\eta(s)=\lim_{k\to\infty}\underbrace{\frac{\eta(s,k)+\eta(s,k+1)}2}_{\eta(s,k,1)}$$

Or even faster:

$$\eta(s)=\lim_{k\to\infty}\frac{\eta(s,k,1)+\eta(s,k+1,1)}2$$

Generally, you can define

$$\eta(s,k,n)=\begin{cases}\eta(s,k),&n=0\\\frac{\eta(s,k,n-1)+\eta(s,k+1,n-1)}2&,n>0\end{cases}$$

And then

$$\eta(s)=\lim_{k\to\infty}\eta(s,k,k)$$

converges fairly rapidly. The advantage here being that this approach converges for any complex $s$ and there is no difficult arithmetic involved.

After a few minutes of computation by hand, I approximated $\zeta(2)$ using only the first 6 terms of $\eta(2)$:

k | η(2,0,k)   | η(2,1,k)   | η(2,2,k)   | η(2,3,k)   | η(2,4,k)   | η(2,5,k)
------------------------------------------------------------------------------------
1 | 1.00000000 | 0.87500000 | 0.84027778 | 0.82899306 | 0.82498264 | 0.82347222
2 | 0.75000000 | 0.80555555 | 0.81770833 | 0.82097222 | 0.82196181
3 | 0.86111111 | 0.82986111 | 0.82423611 | 0.82295139
4 | 0.79861111 | 0.81861111 | 0.82166667
5 | 0.83861111 | 0.82472222
6 | 0.81083333

Compare this to the known value of $\eta(2)=\pi^2/12\approx0.82246703$

Simply Beautiful Art
  • 74,685
1

Riemann did it - that's how he came up with his famous hypotheses. Of course, it was very arduous and he had to derive some pioneering results to do it, but that's why he is famous and I am not.

Do a search for "computing the Riemann zeta function" and you will get many hits. A good reference is Edwards' book "Riemann's Zeta Function".

marty cohen
  • 107,799