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  Does this asymptotic series appear anywhere in physics?

+ 2 like - 0 dislike

I recently discovered the following relation for arbitrary $d_r$:

$$     \lim_{k \to \infty}  \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left(  f(\frac{k}{n}r)\frac{k}{n} \right) \sim  \underbrace{\frac{1}{\zeta(s)}   \sum_{r=1}^\infty  \frac{d_r}{r^s}}_{\text{removable singularity}}  \times \int_0^\infty f(x) \, dx  $$

As $s \nearrow 1$ and $\int f(x) dx$ absolutely converges and $f(\infty)= f(0) = 0$

Some obvious choices are $d_r= 1$ , $d_r = s^r r^s$ or $d_r = \frac{\delta_{1,r}}{s-1}$ where the R.H.S $\neq 0$. Is there any possible use of this? (I was thinking along the lines of divergent series in QFT). I know this relation looks very bizzare so I've added a proof:


Consider an integral such that $$ \int_0^\infty f(x) \, dx = C,$$where, $f(x)$ is a smooth and continuous function and absolutely converges.

Now we raise both sides to the power s:

$$\left(\int_0^\infty f(x) \, dx\right)^s = C^s $$

We substitute $x$ with $rx$ to get:

$$\left(\int_0^\infty f(rx) \, dx\right)^s = (C/r)^s $$

Multiplying both sides by an arbitrary coefficient:

$$ (b_r)\left(\int_0^\infty f(rx) \, dx\right)^s = (b_r)( C/r)^s $$

Taking their sum:

$$ \sum_{r=1}^\infty b_r \left(\int_0^\infty f(rx) \, dx\right)^s = C^s \underbrace{\sum_{r=1}^\infty \frac{b_r}{r^s}}_{\text{dirichlet series}}   $$

We write the integral as a limit of a Riemann sum:

$$ \sum_{r=1}^\infty  \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s  = C^s \sum_{r=1}^\infty \frac{b_r}{r^s} $$

Using the mobius inversion formula:

$$ \sum_{r=1}^\infty  \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s  = C^s \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{d_r}{r^s} $$

We define $ d_r = \sum_{e|r} b_e $


$$ (\frac{b_1}{1^s} + \frac{b_2}{2^s} + \frac{b_3}{3^s} + \frac{b_4}{4^s} + \dots) \times (\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots) = \frac{b_1}{1^s}  + \frac{b_1 + b_2}{2^s} + \frac{b_1 + b_3}{3^s} + \frac{b_1 + b_2 + b_4}{4^s} + \dots $$

Now focusing on the L.H.S ($s \nearrow  1 $):

$$ \sum_{r=1}^\infty  \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s \sim \sum_{r=1}^\infty  \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right) $$

Focusing on the R.H.S of the asymptote:

$$ \lim_{n \to \infty} b_1 ((f(\frac{k}{n}) + f(2 \frac{k}{n}) + f(3 \frac{k}{n}) +f(4 \frac{k}{n}) + \cdots)\frac{k}{n} $$
$$ \lim_{n \to \infty} b_2 (0.f(\frac{k}{n}) + f(2 \frac{k}{n}) + 0.f(3 \frac{k}{n}) +f(4 \frac{k}{n}) +\cdots) \frac{k}{n}$$
$$ \vdots $$
$$ = \lim_{n \to \infty} (\underbrace{b_1}_{d_1} (f(\frac{k}{n}) + \underbrace{(b_1 + b_2)}_{d_2}f(2 \frac{k}{n}) + \underbrace{(b_1 + b_3)}_{d_3}f(3 \frac{k}{n}) +\underbrace{(b_1 + b_2 + b_4)}_{d_4}f(4 \frac{k}{n}) + \cdots)\frac{k}{n} $$

Hence, for special $d_r$ the R.H.S converges:

$$     \lim_{k \to \infty}  \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left(  f(\frac{k}{n}r)\frac{k}{n} \right) \sim \lim_{s \to 1} \underbrace{\frac{1}{\zeta(s)}   \sum_{r=1}^\infty  \frac{d_r}{r^s}}_{\text{removable singularity}}  \times \int_0^\infty f(x) \, dx  $$

asked Jan 15, 2017 in Mathematics by Asaint (90 points) [ revision history ]
edited Jan 15, 2017 by Arnold Neumaier

 Is there any possible use of this? - Most likely the way you arrived at the consideration of this realtion hints at where possible uses are. How did you discover the formula?

Long answer: I arrived at this via consideration of a certain basis and realized that I could create this formula. Short answer: pure mathematics (which I was hoping could be applied to something). Also one of my friends said it might be useful for some path integral as it is essentially a limit of a sum of some kind.

1 Answer

+ 1 like - 0 dislike

Let me quickly brainstorm at what this might be useful for in physics. Let us first take the n limit of the expression

$$     \lim_{k \to \infty}  \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left(  f(\frac{k}{n}r)\frac{k}{n} \right)   $$

This is simply a Riemann integral which reads

$$\lim_{k \to \infty} \int_0^\infty \mathcal{D}(x) f(k x) k \mathrm{d}x$$

where $\mathcal{D}(x)$ will be some kind of horrible random distribution of the sort of white noise. What I mean by horrible is that it can be obtained as an infinitely tight packing of delta functions which have the value $d_r$ at the respective limiting points. By a simple substitution we can rewrite this integral as

$$\lim_{k \to \infty} \int_0^\infty \mathcal{D}(\frac{x}{k}) f(x) \mathrm{d}x \equiv \int_0^\infty \tilde{\mathcal{D}}(x) f(x) \mathrm{d}x$$

which only increases the tight-packing of the peaks of $\mathcal{D}$ to a new formal distribution $\tilde{\mathcal{D}}$.

Hence, in principle, I guess this kind of formula might be useful to consider in the following type of situation. We have a stochastic-differential process characterized by $\tilde{\mathcal{D}}(x) $, and we are able to step-by-step derive the properties of the sequence $d_r$; we can then use your formula e.g. to compute some $f(x)$-momenta or mean values of some observables.

answered Jan 16, 2017 by Void (1,645 points) [ revision history ]
edited Jan 16, 2017 by Void

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