Is Zeta regularization real?

[zetafunction]

Is "Zeta regularization" real??

in many pages of the web i have found the intringuing result

[tex] \sum _{n=0}^{\infty} n^{s}= \zeta (-s) [/tex]

but the first series on the left is divergent ¡¡¡ for s >0 at least

other webpages use even more weird results


[tex] \sum _{n=0}^{\infty} h^{s+1}(a/h + n)^{s}\approx \int_{0}^{\infty}dx (a+x)^{s} [/tex] (h is an step)

but can we rely on these results for divergent series and integrals ? ,if so why there are still unsolved divergencies.

 

[alxm]

What's so 'intriguing' about that? It's just took the definition of the Riemann zeta function with a change of sign. And yes, the Riemann zeta function is divergent in a lot of cases.

In the latter case, I suspect you're missing a limit for h somewhere, but yes, you can rely on those results. If a series is approximated by an integral (such as in the integral test of convergence, which you can easily find a proof of) then the integral will be divergent if the series is divergent.

This has nothing to do with quantum physics?!

 

[zetafunction]

i put here becuase the realtion of Casimir effect and the series

[tex] \sum_{n=0}^{\infty} n^{3} = -1/120 [/tex]

this -1/120 gives correct results in the calculation of Casimir effect although the series of the sum of cubes diverges

 

[alxm]

It does not diverge for an odd negative integer, then the zeta function is:
[tex]\zeta(-n)=-\frac{B_{n+1}}{n+1}[/tex]

The fourth Bernoulli number is -1/30.

 

[Hurkyl]

(1) It may be the case that if you pay more careful attention to your calculation, it turns out that you were really trying to calculate [itex]\zeta(-3)[/itex] the entire time. And since you didn't use the fact that sum was a sum in any essential way, you can reinterpret your calculation as if the sum was just a stand-in for [itex]\zeta[/itex].


(2) Generally speaking, you should use the tool appropriate for the job. The infinite summation operation you learned in elementary calculus is a sledgehammer -- it's simple, well-behaved, and mysteriously manages to be applicable in a lot of situations unrelated to calculus.

Given how important the Riemann zeta function (and other similar functions) are in number theory, it shouldn't be that surprising that sophisticated zeta-related techniques are more widely applicable to number theoretic problems than infinite sums -- and so when infinite sums fail to solve a problem, you might get results by turning things into zeta-like functions.

(Of course, you would still have to prove that the zeta-like function actually tells you what you wanted to know)

I don't know why the technique turned out to be applicable to physics -- maybe point (1) I mentioned turns out to happen a lot? *shrug*


It's important to understand that zeta regularization doesn't say "the infinite sum of 1 + 8 + 27 + ... is equal to -1/120" -- it says "the zeta regularization of 1 + 8 + 27 + ... is equal to -1/120". Zeta regularization is simply a different operation than the infinite sum... and there are lots of different operations that you can apply to infinite sequences of numbers. Wikipedia mentions some examples. And different operations can give different values for the same sum.

 

[Hurkyl]

It does not diverge for an odd negative integer, then the zeta function is:
[tex]\zeta(-n)=-\frac{B_{n+1}}{n+1}[/tex]

The fourth Bernoulli number is -1/30.

The sum diverges. Fortunately, zeta isn't defined to be that sum.

 

[alxm]

The sum diverges. Fortunately, zeta isn't defined to be that sum.

Well I think you can define it that way, but only for positive reals > 1 (which I forgot in my first post..)

Two sloppy posts in a row, it seems.

 

[zetafunction]

but this kind of zeta regularization wouldn't be just cheating ?? in fact you get a negative answer -1/120 summing only positive terms 1+8+27+125+...

perhaps it works becuase the analytic continuation or somehow similar of the function f(x) to negative values f(-n) in a similar way to the Gamma function for factorials and so on

 

[lbrits]

but this kind of zeta regularization wouldn't be just cheating ?? in fact you get a negative answer -1/120 summing only positive terms 1+8+27+125+...

perhaps it works becuase the analytic continuation or somehow similar of the function f(x) to negative values f(-n) in a similar way to the Gamma function for factorials and so on

The point is that you were never performing an infinite sum to begin with (the theory isn't defined to arbitrarily short wavelengths). If you had a theory defined to arbitrarily small wavelengths, then the actual expression would look like your sum at low orders, and look like something else at very high orders. What no-one told you, though, is that in your theory (QED, presumably), *some* questions are such that short distance physics is irrelevant to long-distance physics, and simply have the effect of rescaling things like coupling constants. This property is known as "renormalizability". With that knowledge, you can use whatever convenient function to "regularize" your sum, since your regulator, and some other regulator, and the true theory, only differ at high energies, which is irrelevant to your problem anyway.

If zeta function regularization bothers you, wait 'til you hit dimensional regularization (although the paper by 't Hooft and Veltman might clear things up for you: Nucl. Phys. B44: 189-213)