8
$\begingroup$

At least according to the answer to this question, $\zeta(1) = \gamma $ (once reqularized, of course).

Let me rephrase that by stating that:

$$ \sigma(\zeta(1)) = \gamma $$ Here, $\sigma(x)$ is the 'summation-function'. It's a function that assigns a value to any $x$, using Borel, Abel, Ramanujan, Euler, Cesaro or any other summation method that works (e.g. It makes a divergent series summable). The $\sigma$-function 'chooses' a summation method that suits $x$ best (to assign a (finite) constant to it). We assume that the different summation methods dont have different 'working' values for the same $x$ (I now call upon this question).

Furthermore, we denote $C$ as a converging series and $D$ as a diverging one.

What would $\sigma(C + D) $ be? Is it $\sigma(C) + \sigma(D)$ ? Or what would, for example, $\sigma(\zeta(1)^3 + \zeta(2))$ be?

So, to summarize my question: Could you please explain the properties of the $\sigma$-function to me, with relation to $C$ and $D$ ?

Thanks a lot in advance.

P.S. A bonus question: What do you think of the 'summation-function'? is it useful or just mathematical bogus? Or has it been defined (even more) properly already?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Have you read terrytao.wordpress.com/2010/04/10/… ? $\endgroup$
    – Qiaochu Yuan
    Commented Jun 11, 2010 at 20:40
  • $\begingroup$ No, I haven't, but it looks very interesting (and relevant)! Thanks, mister Yuan. $\endgroup$
    – Max Lonysa Muller
    Commented Jun 11, 2010 at 20:55
  • 6
    $\begingroup$ You should read the first chapter of Hardy's book "Divergent Series". I think you're going to find that defining what it means for $\sigma$ to "choose a summation method that fits $x$ best" will be very slippery... $\endgroup$
    – David Hansen
    Commented Jun 11, 2010 at 21:22
  • $\begingroup$ Hm I guess so, mister Hansen, I still have a lot to learn ;). $\endgroup$
    – Max Lonysa Muller
    Commented Jun 12, 2010 at 15:18

4 Answers 4

Reset to default
7
$\begingroup$

Making sense of "picks a summation method that works" is very difficult, because for many series there are different reasonable choices. A standard method of summing bad series is "zeta-function regularization" --- for example, the method is popular in physics, because S. Hawking uses it to compute QFT on curved backgrounds. In its easiest form, let $\sum a_n$ be the series you want to sum: then you can consider the function $\zeta_a(s) = \sum a_n^{-s}$. When the sequence $a_n$ is positive and grows at least as $n^\epsilon$ for some $\epsilon>0$, then $\zeta_a$ will converge in the far-right part of the complex plane. Now you can hope that it has a singly-valued analytic continuation to $s = -1$.

However, this summation method will not satisfy the linearity that you want. One example: you can look up values for zeta functions of the form $\sum (an+b)^{-s}$ and see directly the failure of additivity.

More generally, you should look at Hardy, Divergent Series. Among other statements in there are some no-go theorems, of the form: there is no function $\{\text{series}\} \to \{\text{numbers}\}$ that agrees with the Cauchy convergence on convergent series and that satisfies some natural requirements. (Unfortunately, I don't have the book with me, and I don't remember any exact versions of such a theorem.)

Improve this answer
$\endgroup$
11
  • $\begingroup$ Ok, thanks a lot, mister Johnson-Freyd! You're not the first to recommend me Hardy's book on divergent series. I'm not sure if I'm able enough to understand it already, though... (I'm still a high-school student). I was also wondering what would happen if we plunge in some converging series in the 'summation-function'. Would the value of these series as an argument of this function be different from their actual evaluation? $\endgroup$
    – Max Lonysa Muller
    Commented Jun 12, 2010 at 21:08
  • $\begingroup$ I generally think that a "summation function" should be required to agree with Cauchy's summation function (classical convergence). But for something like zeta-function regularization, it does not: indeed, if $\sum a_n$ converges, then in particular $a_n \to 0$, and so $a_n^{-s} \to \infty$ for large $s$, and in particular $\sum a_n^{-s}$ diverges for large $s$. You can compare zeta-function summation with Abel summation in the regime that $n^\epsilon < a_n < n^\delta$ eventually, and in general they do not agree. Abel summation is an industry standard, and agrees with Cauchy. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jun 14, 2010 at 17:36
  • $\begingroup$ More on comparing these different methods: Another form of "zeta regularization" is for infinite products $\prod a_n$. The idea is that, formally, $\prod a_n = \exp\bigl(-\frac{d}{ds}\zeta_a(s)\bigr|_{s=0}\bigr)$, and so if $\zeta_a$ is regular near $s=0$, you can define the product. (It does not agree with zeta-regularized $\exp( \sum \log a_n)$.) If $a_n \sim n^\epsilon$ for $\epsilon > 1$, then $\prod a_n^{-1}$ converges in the Cauchy sense, and I don't remember if in fact $\prod a_n^{-1} = \bigl(\prod a_n\bigr)^{-1}$ where the RHS is zeta-regularized. I think not? $\endgroup$
    – Theo Johnson-Freyd
    Commented Jun 14, 2010 at 17:41
  • $\begingroup$ Wait, actually, I'm talking crap. You still can't compare Abel and Zeta a priori. The rules for Abel summation are that $\sum a_n x^n$ should converge for $|x|<1$, and then take the limit as $x\to 1$. This is necessarily infinite if $a_n \to \infty$ are all positive. What you can try is an "Eulerian" summation (he had many) where you ask that $\sum a_n x^n$ converge for $|x|$ small, and ask that it have an analytic continuation to $x = 1$. Alternately you can try to get a grip on signed zeta-regularized sums. This is required for various "index" theorems. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jun 14, 2010 at 17:48
  • $\begingroup$ Mister Johnson-Freyd, thanks a lot. I realize now that still have to learn a lot in order to produce any valuable research on this. Although I understand most of what you've written (and I very much respect the fact that you took the time to think about and answer my question 2 days after answering the original question), I want to understand everything in detail. After looking at the first couple of pages of Hardy's book on Divergent Series, I realized that I don't posess the required prerequisite knowledge to understand it. Most of my knowledge on 'higher' mathematics I (see new comment) $\endgroup$
    – Max Lonysa Muller
    Commented Jun 14, 2010 at 18:52
5
$\begingroup$

In fact you could have asked for more. Let AC the set of absolutely convergent series, and $S:AC\rightarrow \mathbb{C}$ the $\mathbb{C}$-algebra homomorphism that associates to a convergent series its sum.

Then we may ask for an extension $\sigma$ of $S$, defined on some subalgebra $D_1$ of the set D of all series, that satisfies the following rules.

  • regularity: if $s\in D_1$ is converging, then $\sigma(s)=S(s)$,

  • invariance by translation: $\sigma(\sum_0^\infty a_n)=a_0+\sigma(\sum_1^\infty a_n)$,

  • linearity: $\sigma$ is $\mathbb{C}$-linear,

  • product: $\sigma$ is an homomorphism for multiplication.

Abelian summations methods satisfy these four rules. These methods associate to a divergent sum $\sum a_n$ a function, say $\sum a_n x^n$, and try to take its value at $x=1$ by some process related to analytic continuation. If you can read French, the first chapter of the book "Series divergentes et theories asymptotiques", by J.P. Ramis, is a nice introduction to these questions. The author surveys resummation methods for divergent series, from Leibniz to Ecalle.

Improve this answer
$\endgroup$
1
  • $\begingroup$ coudy, do you have some good references where this type of multiplicative functionals have been studied? Thanks! $\endgroup$
    – M.G.
    Commented Jun 11, 2010 at 23:02
3
$\begingroup$

Define $\tau(C+D):=\sigma(C)+\sigma(D)$. Then $\tau$ is a summation method for $C+D$, and by your assumption of uniqueness it follows that $\sigma(C+D)=\tau(C+D)=\sigma(C)+\sigma(D)$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Ok.. Thank you. Would this also imply that $\sigma(\zeta(1)^3)=\gamma^3$? $\endgroup$
    – Max Lonysa Muller
    Commented Jun 12, 2010 at 14:49
  • $\begingroup$ My impression is that such summation methods don´t need to be consistent wrt. multiplication, i.e. that $\sigma(AB)=\sigma(A)\sigma(B)$ is always valid. I am not even sure if they are consistent wrt. addition... In fact, the Laurent-series of the Riemann zeta function have the first 3 coefficients $c_{-1}=1, c_0=\gamma_0=\gamma, c_1=\gamma_1$, thus the constant term of $\zeta(s)^2$ is $\gamma + \gamma_1\neq\gamma^2$. $\endgroup$
    – M.G.
    Commented Jun 12, 2010 at 16:05
  • $\begingroup$ It is in itself an interesting question if regularization/summation methods can be throughout consistent wrt. to basic arithmetic operations, or even more general wrt. continuous functions in several variables (for several dirvegent series to plug in). $\endgroup$
    – M.G.
    Commented Jun 12, 2010 at 16:05
  • $\begingroup$ Yes I think that's a very interesting question, and a good answer would have some research applications, I believe! By the way, $\gamma_1=\gamma$? So the constant term of $\zeta(s)^2$ is $2\gamma$? We could extend this question by asking what the constant term of $\zeta(s)^n$ would be.. $\endgroup$
    – Max Lonysa Muller
    Commented Jun 12, 2010 at 16:57
  • $\begingroup$ No, $\gamma_1\neq\gamma_0=\gamma$, see Stieltjes constants -> en.wikipedia.org/wiki/Stieltjes_constants $\endgroup$
    – M.G.
    Commented Jun 12, 2010 at 17:44
0
$\begingroup$

I am currently working on a theory that assigns to divergent sums the values from a set of "extended" numbers. Each extended number has a transfinite and standard(or regular) part. The standard part corresponds to the regularized value of the series or integral.

In this theory the regular part function is linear: for extended numbers $w$, $w_1$ and $w_2$ and a regular number $a$ the following holds:

$$\operatorname{reg} (w_1+w_2)=\operatorname{reg}w_1+\operatorname{reg}w_2$$

and

$$\operatorname{reg} (a w)=a\operatorname{reg}w$$

On the other hand, the regular part of product of two transfinite numbers is not usually a product of their regular parts.

For instance,

$$\operatorname{reg} \int_0^\infty 1\, dx=0$$

but

$$\operatorname{reg} \left(\int_0^\infty 1\, dx\right)^2=-\frac1{12}$$

$$\operatorname{reg} \sum_{k=1}^\infty 1=-\frac12$$

but

$$\operatorname{reg} \left(\sum_{k=1}^\infty 1\right)^2=\frac16$$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .