# Zeta (2k) formula derivation!

1. Oct 21, 2007

### yasiru89

Using the not so legitimate sum,
$$\sum_{n = 0}^{ + {\infty}} \exp {in{\theta}} = \frac {1}{2} + \frac {1}{2} i\cot {\frac {{\theta}}{2}}$$

I was able to obtain (by differentiation) valid expressions for the analytic continuation of the $${\zeta}$$ series for negative integers(and 0)

Putting $${\theta} = {\pi} + {\theta}$$
in the real part of the above equation I have,
$$\sum_{n = 1}^{ + {\infty}} ( - 1)^{n - 1} \cos {n{\theta}} = \frac {1}{2}$$

I proceeded to integrate this equation 0 to $${\theta}$$ 2k times in the hopes of finding the general formula for $${\zeta}(2k)$$

However I am stuck at,
$$\frac {( - 1)^{k}{\theta}^{2k}}{2{(2k)}!} = \sum_{n = 1}^{ + {\infty}} ( - 1)^{n - 1} n^{ - 2k} \cos {n{\theta}} + \sum_{n = 1}^{ + {\infty}} ( - 1)^{n - 1} n^{ - 2k} \sum_{r = 0}^{k - 1} ( - 1)^{r - 1} \frac {n^{2r}}{{(2r)}!} {\theta}^{2r}$$

No idea what to make of that troublesome finite series or how to introduce the Bernoulli numbers to the mix!

Any assistance would be greatly appreciated! Thanks!

PS- people who'll feed me the convergence issues shouldn't even bother replying! I need this derivation only, I know many other proofs but none so interesting. I think the Bernoulli numbers might come into the mix somewhere between the calculation!

2. Oct 21, 2007

### Count Iblis

Why not take the Laplace transformation w.r.t.$$\theta$$of both sides of the first equation and then expand both sides in powers of 1/s?

3. Oct 21, 2007

### Hurkyl

Staff Emeritus
Rather than pre-emptively attack people for pointing out that none of this makes sense given the usual meaning of the symbols you're using -- why not state what meaning you are giving them? :grumpy:

4. Oct 21, 2007

### Hurkyl

Staff Emeritus
As long as we're throwing caution to the wind, did you notice that interchanging the order of that summation yields one of the usual sums for the Zeta function?

5. Oct 22, 2007

### yasiru89

Well aren't we grumpy today? I'm only trying to show with this particular instance that the analytic continuation of the geometric series when applied to the zeta function is consistent. While that's a mouthful, people are too stuck up to even consider this and harp on about convergence issues, I only meant to save them the trouble of replying (I suppose that's lost on you)

And duh, I got to the point where we can say that,

$${\zeta}(2k) = \sum_{r = 0}^{k} (-1)^{r-1} \frac{{\pi}^{2r}}{{(2r)}!} (1 - 2^{1-2(k-r)}) {\zeta}(2(k-r))$$

The relation is recursive and works(try k=1 for the classic Basel problem), I just need a general form for the zeta s with the Bernoulli numbers in it!!!

And I don't want to bring in integral transforms at all, I'm convinced a very simple operation should do the trick!!

6. Oct 22, 2007

### Gib Z

$$\sum_{n = 0}^{ + {\infty}} \exp {in{\theta}} = \frac {1}{2} + \frac {1}{2} i\cot {\frac {{\theta}}{2}}$$

And that series is obviously incorrect because comparing only real parts, the function being summed corresponds to cos ntheta, and $$\sum_{n=0}^{\infty} \cos (n\theta) = \frac{1}{2}$$ for any theta..which does not seem to be the case.

7. Oct 22, 2007

### Count Iblis

$$\sum_{n=0}^{\infty}\exp\left(i n\theta\right)=\frac{i}{2}\frac{\exp\left(-i\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}$$

8. Oct 22, 2007

### Gib Z

Yes I don't see how that changes my argument; when you take only the Real or Imaginary co efficients it implies certain results about the summations of the cosines and sines (from Euler's Formula). It leads to the cosine summation in post 6 to be equal to a certain constant, for all values of theta, which is definitely not true.

9. Oct 22, 2007

### Count Iblis

Note that for the series to converge $$\theta$$ must have a positive imaginary part. Now, Euler's formula for $$\exp\left(i z\right)$$ is, of course, also valid for complex $$z$$. However, the two terms for cannot be taken apart as they separately diverge. If you take $$\theta$$ real than we must replace $$\theta\rightarrow\theta + i\epsilon$$. So, the late terms in the summation of the cosine term diverge as $$\sim\exp\left(\epsilon n\right)$$

10. Oct 22, 2007

### Count Iblis

11. Oct 24, 2007

### yasiru89

Gib Z, see why Hurkyl went all ballistic on me! Once again allow me to stress that we are working on divergent series! Thanks for the links Count Iblis!

12. Oct 25, 2007

### Gib Z

You should have stated that instead of $\theta$ you were actually taking the limit $$\lim_{a\to 0} \theta + ia$$. There is indeed a difference...and I can not say I am familiar with this particular method of the derivation, so I will butt you and hope you know what you are doing.

13. Oct 25, 2007

### Count Iblis

Well, it's Yasiru's problem, not mine. But I'm a physicist and in physics we are usually very sloppy. We write down formulae and when we find out that it doesn't converge we plug in $$i\epsilon$$ at the right places to give the formula the right meaning.

14. Oct 25, 2007

### yasiru89

I get what you mean but here $${\theta}$$ is rather arbitrary and even if I use the incremental imaginary part it quickly disappears in the limit and $${\theta} = {\pi}$$ simply. Which reduces to the formula in my 2nd post.
I know what you mean by that it isn't a standard derivation but so far it's a beautiful one, isn't it?

15. Oct 25, 2007

### yasiru89

Further a forced convergence like that isn't necessary since the 2nd formula onward(of the 1st post) are completely valid.

16. Oct 25, 2007

### Hurkyl

Staff Emeritus
One of the key ideas about analytic functions is that knowledge of their behavior on any open set is enough to reconstruct their behavior everywhere.

e.g. if you consider the series
$$f(z) = \sum_{n = 0}^{+\infty} z^n$$
this is clearly only defined for $|z| < 1$. However, on that circle, we have the identity $f(z) = 1/(1 - z)$. $1/(1-z)$ is, of course, defined everywhere except $z=1$. However, we can still use the series to derive its properties, even though it's only defined on the circle! e.g. to find its derivative:

$$\frac{d}{dz} \frac{1}{1-z} = \frac{d}{dz} \sum_{n = 0}^{+\infty} z^n = \sum_{n = 0}^{+\infty} (n+1) z^n = \left( \sum_{m = 0}^{+\infty} z^m \right)^2 = \frac{1}{(1-z)^2}$$

Because this identity is valid on the entire disk, it must also be valid everywhere else too. (Except, of course, at $z = 1$.)

yasiru89's plan, I assume, is to use the series form of $1/2 + (1/2) i \cot (z / 2)$ to help him derive some identity, and then find a way to specialize it to z = 0. (or maybe $-\pi$, or something like that)

Last edited: Oct 25, 2007
17. Sep 25, 2008

### yasiru89

Thank you for your troubles, I managed to put [itex]\theta = \pi[/tex] and make it completely rigorous under Euler summation (the power series method Hurkyl mentioned, but without any specific reference to the analytic continuation- though this is hinted to by the formula).

The resulting recursive identity is basically equivalent to the sort of recursions you use to get the Bernoulli numbers anyway so I realised the 'closed form' isn't really any more efficient.
Along the way (to establish the base case [itex]\zeta(0) = -\frac{1}{2}[/tex]) I took a long detour and found a neat asymptotic variant of Euler's summation formula! (its already known though, see Ramanujan's Summation by Eric Delabaere at http://algo.inria.fr/seminars)
Interestingly, comparing [itex]\zeta(2k)[/tex] and [itex]\zeta(-k)[/tex] for k a positive integer (or zero) we might even be able to define Bernoulli numbers at negative values, but that might be redundant as well...