Regulators and uniqueness

Question

Does the regularization of a divergent infinite sum yield a unique value?

I.e. do different regularization schemes acting on the same infinite sum produce the same exact value independent of the regulator?

What, exactly, do these values mean? Or what are they? My understanding is that they are not "convergent" values.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user user122066

Comments

Dangit, you beat me to it! math.stackexchange.com/questions/1854642/…

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker

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Relevant, as source/motivation of this question: physics.stackexchange.com/a/267243

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Clement C.

Answer 1

Different regularizations may lead to different regularized values. For instance $$\lim_{\lambda\to 0}\sum_{n\geq 0}n e^{-\lambda n} = +\infty$$ while the zeta regularization of $\sum_{n\geq 1} n$ gives the (in)famous value $\zeta(-1)=-\frac{1}{12}$.

If we take an hybrid between smoothed sums and the zeta regulatization we have: $$\sum_{n\geq 1}'' n = \sum_{N\geq 1}'\frac{N+1}{2} = \frac{\zeta(0)+\zeta(-1)}{2}=-\frac{7}{24}.$$

We also have a class of regularizations that depends on a positive parameter $\delta$: the Bochner-Riesz mean. There isn't a single regularization: a regularization is just a (somewhat arbitrary) way to extend the concept of convergence. About integrals, the Cauchy principal value can be interpreted as the Fourier transform of a distribution. About series, we may say that $$\sum_{n\geq 1}' a_n = L$$ à-la-Cesàro if $$\lim_{N\to +\infty}\frac{A_1+\ldots+A_N}{N}=L,$$ i.e. if the sequence of partial sums is converging on average. A convergent series is also a Cesàro-convergent series, but with such an extension $${\sum_{n\geq 0}}'(-1)^n = \frac{1}{2}=\lim_{\lambda\to 0}\sum_{n\geq 0}(-1)^n e^{-\lambda n}$$ where $\sum_{n\geq 0}(-1)^n$ is not convergent in the usual sense.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio

Comments

I would say that your example is an instance where one regulator "works" and another "fails to work." But are there any examples where two different regulators give different, finite values?

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker

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Could you explain your notation - what do the single and double primes on the sums mean? How did you go from summing over $n$ to summing over $(n+1)/2$?

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker

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@tparker: at first we perform a Cesàro regularization: the main term ($n$) is replaced by the averaged term $\frac{1}{N}\sum_{n=1}^{N}n = \frac{N+1}{2}$. A zeta regularization follows.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio

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I see. I would suggest getting rid of the primes on the $\Sigma$ symbols because they don't seem to mean anything and I find them extremely confusing.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker

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@tparker: without primes the chain of equality is simply wrong, so I prefer to leave the primes there.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio

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What do the primes actually mean? What is the general distinction between $\Sigma'$ and $\Sigma''$?

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker

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$''$ means Cesàro+Zeta regularization, a single $'$ means Zeta regularization or Cesàro regularization, depending on the context.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio

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And there is no "general dinstinction" because we are just inventing regularizations for proving a non-uniqueness theorem.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio

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I'm not convinced that this is really in the spirit of Cesaro regularization, which regularizes a series, not individual terms in a series. Can you come up with a more straightforward regularization where you find a series of analytic functions $f_n(z)$ such that $f_n(0) = n$ and whose sum converges for some open set of $z$, then analytically continue the sum to $z = 0$? I checked your original attempt in the comment you deleted, and I think you made a mistake and it doesn't converge.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker

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@tparker: all right, let me rephrase: we consider an hybrid between a particular smoothing (the main term is replaced by the average term) and the zeta regularization. Hope it is ok now. Out of curiosity, what is the purpose of this almost-non-math?

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio

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In physics, we often use the analytic-continuation regularization prescription that I just described to assign values to divergent series, and I'm curious if the prescription always yields a unique answer. Unfortunately, the question appears to remain open.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker

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@tparker: since the zeta regularition is not the only regularization, I think the question is pretty trivial, but I was interested to the problem in itself: if a series/integral is not converging in the usual sense, what is the physical meaning of taking any regularization of it? A regularization is a kind of arbitrary choice, like a choice of weights. The zeta regularization is perhaps God's weight choice? :D

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio

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That's my question as well. Physicists never use the hybrid smoothing-and-regularization process you described, but they do often use the scheme I described. But if there exists a sum such that the scheme results in two different finite answers, that seems to pose serious questions about the validity of the physical theory, as you point out. I have revised my question math.stackexchange.com/questions/1854642/… to focus on this narrower question, and I believe it is no longer a duplicate of this one.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker