A Ramanujan-like summation: is it correct? Is it extensible?

Solution 1:

A non rigorous approach, just simple "intuitive methode". Because the power is $x^2$, the period is 2. Make sure to include the zero point in it's polynomal notation. I will later try to write it for all power's of $x$.

$$ \sum_{x=1}^{\infty} (c)^{-x^2}$$ $$ \sum_{x=1}^{\infty} \sum_{n \in \mathbb{Z}}\ (\ln(c)x^2)^n(-1)^n/n!=\sum_{x=1}^{\infty} \sum_{n \in \mathbb{Z}}\ \frac{(x)^n (\ln(c))^{n/2}}{(n/2)!} (e^{i\pi n/2})\sum_{k=0}^1 \frac{e^{\frac{2i \pi*kn}{2}}}{2}$$

$$= \sum_{n \in \mathbb{Z}}\ \frac{\zeta(-n) (\ln(c))^{n/2}}{(n/2)!} \frac{(e^{i\pi n/2}+e^{3i\pi n/2})}{2}$$

First take the limit as n goes to -1. Which gives us the value $\frac{ \sqrt{\pi}}{2 \sqrt{\ln(c)}}$

There's another value at the "normal" order, mainly as $n$ goes to 0. At $n=0$ the value is $-\frac{1}{2}$.

Notice that both answers give a very very good approaximation for the sum, but are a very little off, (orders like $10^-{6}$, really small). Mathematically I can't fully show why (yet). More on intuition based, let $h$ be a very small value, and you have rewrite the function you are trying to sum as $g(n)f(n)$ with $g(n)$ being a periodically zero, with an order $h$ near the zero so $g(n+h)$ and $f(n)$ being a function that grows more/equal then a constant value, the regularised (as in the upper or lower limit is infinity) the sum is not that well defined and you just have to be careful. Lots of horrible talk and explainations I soon will be ashamed of so I am hoping for a better explaination. So here's what I found to be the solution.

If $n$ is even, the zero's don't have any influence for the sum and can be regarded as 0.

If n is uneven this isn't the case: $$= \sum_{n \in \mathbb{Z}}\ \frac{\zeta(-n) (\ln(c))^{n/2}}{(n/2)!} e^{i\pi n/2} \frac{(1+(-1)^n)}{2}$$

use that $$\zeta(-n)=\frac{\zeta(n+1) n! (i^{n+1}+(-i)^{n+1})}{(2\pi)^{n+1}} $$ $$\frac{n!}{(n/2)!2^n}=\frac{((n-1)/2)!}{\sqrt{\pi}}$$

$$ \sum_{n \in \mathbb{Z}}\ \frac{\zeta(n+1)(\ln(c))^{n/2}\big(\frac{(n-1)}{2}\big)! }{\pi^{n+3/2}} \frac{(e^{i\pi n/2}+e^{3i\pi n/2}))((e^{i\pi (n+1)/2}+e^{3i\pi (n+1)/2}))}{4}=$$

We only had to look at when $n=2m-1$.

$$ \sum_{m \in \mathbb{Z}}\ \frac{\zeta(2m)(m-1)! (ln(c))^{m-1/2} }{2\pi^{2m+1/2}} i(e^{3i\pi m}-e^{i\pi (m)})(-1)^{m}$$ Because it's summed over all integers including the negative ones, we subsitute m=-m.

$$ \sum_{m \in \mathbb{Z}} \frac{\zeta(-2m)(-m-1)! \pi^{2m-1/2}}{2(\ln(c))^{m+1/2}} i(e^{-3i\pi m}-e^{-i\pi (m)}) (-1)^{m}$$

I've had the alternating/peridiocally zero's kinda horrible defined. But approach the limit of m=m+h at the zero's/infinities to look at the (uneven) zero's. So that $e^{-3i\pi (m+h)}-e^{-i\pi (m+h)}=-2hi\pi$ I know you should work this better out with it's period, and notice that everything is defined in the same "complex" period, maybe use trigonometry functions, but those functions are less intunitive. Again it's non rigorous, only to show that there's some way to a good solution with this methode. And know that $(-n-1)!=h^{-1}/n!$ if n is an integer $$ \sum_{m \in \mathbb{Z}} \frac{\zeta(-2m)(-m-1)! \pi^{2m+1/2}}{(\ln(c))^{m+1/2}} (h) (-1)^{m}$$ $$ \frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{m \in \mathbb{Z}} \frac{\zeta(-2m) \pi^{2m}}{m!(\ln(c))^{m}} (-1)^{m}$$ Officially I'd say you'd had to write it again out also as (and maybe you couldn't ignore the other zero's, I just don't know yet, it just works, I've to polish it obviously)

$$ \frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{m \in \mathbb{Z}} \frac{\zeta(-m) \pi^{m}}{(m/2)!(\ln(c))^{m/2}} (-1)^{m/2}(1+(-1)^m)$$ $$ \frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}\sum_{m \in \mathbb{Z}} \frac{x^{m} \pi^{m}}{(m/2)!(\ln(c))^{m/2}} (-1)^{m/2}(1+(-1)^m)$$ Which is the regularised sum of $$\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{\frac{-\pi^2 x^2}{ln(c)}}$$

$$\sum_{x=1}^{\infty} (c)^{-x^2}=\frac{\sqrt{\pi}}{2 \sqrt{ln(c)}}-1/2+\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{\frac{-\pi^2 x^2}{\ln(c)}}.$$

The first two value's being your I(c^(-1),2) and Z(c^(-1),2) and the sum being your systematically error. And i can only agree that the systematically error is something counter intunitive at first, but also arise when you apply this summation methode to other series as i recall well known easier examples such as $\sum_{x=1}^{\infty} \frac{1}{2x^2-1}$.

In similair rough fashion for $$ \sum_{x=1}^{\infty} (c)^{-x^d}$$

$$ \sum_{m \in \mathbb{Z}}\ \frac{(-1)^{1/d}\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \sum_{k=0}^{d-1} \frac{e^{\frac{2ipi*km}{d}}}{d}$$

$$\sum_{m \in \mathbb{Z}}\ \frac{\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{e^{i\pi m/d}(e^{2 i \pi m}-1)}{d(e^{ 2i \pi m/d}-1)}$$

Now the value at m=0, will always be -1/2. at limit m goes to -1 gives:

$$\frac{-\pi}{\sin(\pi/d) (\ln(c))^{1/d}(-\frac{1+d}{d})!}=\frac{(1/d)!}{\ln(c)^{1/d}}$$

Which is similair to the integeral you stated, if d is not even you got to indeed sum is over the other value's and don't forget the error.

A try of mine to find the error:

$$\sum_{m \in \mathbb{Z}} \frac{\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{e^{i\pi m/d}(e^{2 i \pi m}-1)}{d(e^{ 2i \pi m/d}-1)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{-\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{h \pi}{d \sin(m \pi /d)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{-\zeta(m+1) m!(ln(c))^{m/d}}{(2 \pi)^{m+1} (m/d)!} \frac{h \pi (i^{m+1}+(-i)^{m+1})}{d \sin(m \pi /d)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(m+1) m!(ln(c))^{m/d}}{(2 \pi)^{m} (m/d)!} \frac{h \pi (sin(m \pi/2)}{d \sin(m \pi /d)}$$

Use that:

$$\frac{m!}{(m/d)!}= d^{m+1/2} (2 \pi)^{(1-d)/2} \prod_{j=1}^{d-1} \big(\frac{m-j}{d}\big)!$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(m+1) (ln(c))^{m/d}}{} \frac{h (sin(m \pi/2)}{\sin(m \pi /d)}\frac{d^{m-1/2} (2 )^{(3-d)/2-m} (\pi)^{(1-d)/2-m}\prod_{j=1}^{d-1} \big(\frac{m-j}{d}\big)!}{}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1) (ln(c))^{-m/d}}{} \frac{h (sin(-m \pi/2)}{\sin(-m \pi /d)}\frac{d^{-m-1/2} (2 )^{(3-d)/2+m} (\pi)^{(1-d)/2+m}\prod_{j=1}^{d-1} \big(\frac{-m-j}{d}\big)!}{}$$

Split up for the zero's in $\prod_{j=1}^{d-1} $

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1) (ln(c))^{-m/d}}{} \frac{h (sin(-m \pi/2)}{\sin(-m \pi /d)}\frac{d^{-m-1/2} (2 )^{(3-d)/2+m} (\pi)^{(3-d)/2+m}}{ \prod_{j=1}^{d-1}\sin(\frac{\pi (m+j)}{d}) \big(\frac{m+j}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{} \frac{h (sin((-dm-k) \pi/2)}{\sin((-dm-k) \pi /d)}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(3-d)/2+dm+k}}{ \prod_{j=1}^{d-1}\sin(\frac{\pi (dm+j+k)}{d}) \big(\frac{ \pi (dm+j+k)}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{(sin((-dm-k) \pi/2)}{\sin((-dm-k) \pi /d)}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d}) \big(\frac{dm+j+k}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=0)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d})}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}$$

$$\prod_{(j=0)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d})=d2^{1-d}(-1)^{?}$$ I don't have the minus sign for now. Will come back at that later, assuming it's not there cause i've already enough in one equation.

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{}\frac{d^{(-dm-k)-3/2} (2 )^{(1+d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}$$

to make it readable $$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}\frac{(2 )^{(1+d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{d^{(dm+k)+3/2} }$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!} \frac{(\frac{2\pi}{d})^{dm+k+1/2} (\frac{2}{\pi})^{d/2}}{d}$$

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(dm+k)}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!} $$

I am already suprised the formula above seem to work for d=2. If only the m didn't sneaked into the product life would be good. Kinda used this is a note pad, i will clean it up one day, but results count.

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(dm+k)}}{} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!} $$

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(m)}}{} \frac{\sin((m) \pi/2)}{\prod_{(j=1)}^{d-1} \big(\frac{m+j}{d}-1\big)!} $$

It shows how I think you can proceed (I let it rest, as it got messy, and i probably made some errors, and made things unnessecary hard), with a doable solution of d=2, as was the original question. Maybe you can or should add $m \in d\mathbb{Z}$, to make it smooth, Again i do not know yet.