While playing on pruduct series, I noticed interesting function that it has similar behavior to $C\cdot\sin^2(\pi x)$ function . I need help about some results below.
$$f(x)=\prod\limits_{n=-\infty}^{ \infty }(1-e^{-a{(x+n)^2}}) $$ where $a>0$
The function is a periodic function because
$$f(x+1)=\prod\limits_{n=-\infty}^{ \infty }(1-e^{-a{(x+1+n)^2}}) =\prod\limits_{n=-\infty}^{ \infty }(1-e^{-a{(x+n)^2}}) $$
Some properties of the function:
- $f(x+1)=f(x)$ Period is $1$
- It is obvious that $f(k)=0$ where $k$ is integer.
- It is even function. $f(-x)=f(x)$
These properties above are same as $C.\sin^2(\pi x)$
My ideas and my aim to find out about the function are
- Can we express it as known functions?
- How closed form of $f(\frac{1}{2})$ can be found? I believe that the function behaves the same as $C\sin^2(\pi x)$ has and $f'(\frac{1}{2})=0$ but I could not prove it.
- If it is a periodic function, what is the Fourier series expansion of the function?
Any reference links about the function and help will be appreciated very much .
Thanks
UPDATE: I proved my conjecture above , $f'(\frac{1}{2})=0$
I added some extra information too
$$\frac{f'(x)}{f(x)}=2a\sum\limits_{n=-\infty}^{ \infty } \frac{x+n}{e^{a{(x+n)^2}}-1} $$
$$\frac{f'(\frac{1}{2})}{f(\frac{1}{2})}=2a\sum\limits_{n=-\infty}^{ \infty } \frac{\frac{1}{2}+n}{e^{a{(\frac{1}{2}+n)^2}}-1} $$
$$\frac{f'(\frac{1}{2})}{f(\frac{1}{2})}=a\sum\limits_{n=-\infty}^{ \infty } \frac{2n+1}{e^{a{(\frac{1}{2}+n)^2}}-1}$$
$$f'(\frac{1}{2})=af(\frac{1}{2})\sum\limits_{n=-\infty}^{ \infty } \frac{2n+1}{e^{a{(\frac{1}{2}+n)^2}}-1}$$
$$f'(\frac{1}{2})=af(\frac{1}{2})\sum\limits_{n=-\infty}^{ -1 } \frac{2n+1}{e^{a{(\frac{1}{2}+n)^2}}-1}+af(\frac{1}{2})\sum\limits_{n=0}^{ \infty } \frac{2n+1}{e^{a{(\frac{1}{2}+n)^2}}-1}$$
$$f'(\frac{1}{2})=af(\frac{1}{2})\sum\limits_{k=0}^{ \infty } \frac{-(2k+1)}{e^{a{(-1)^2(\frac{1}{2}+k)^2}}-1}+af(\frac{1}{2})\sum\limits_{n=0}^{ \infty } \frac{2n+1}{e^{a{(\frac{1}{2}+n)^2}}-1}$$
$$f'(\frac{1}{2})=-af(\frac{1}{2})\sum\limits_{k=0}^{ \infty } \frac{2k+1}{e^{a{(\frac{1}{2}+k)^2}}-1}+af(\frac{1}{2})\sum\limits_{n=0}^{ \infty } \frac{2n+1}{e^{a{(\frac{1}{2}+n)^2}}-1}$$
$$f'(\frac{1}{2})=0$$
There can be proved in same way that
$f'(\frac{2k+1}{2})=0$ where $k$ is integer.
UPDATE: Nov 13th
I tried to used roots of $1-e^{-a{(x+n)^2}}$ . I would like to share my results. Maybe it can be helpful to find out an expression with known functions.
$$1-e^{-a{(x+n)^2}}=0$$ roots are:
$$a{(x+n)^2}=2k\pi i$$ Where $k$ integer.
If $k=0$
Roots $x=-n$ where $n$ is integer as known. for $k=0$ and They are double roots ,Thus
$$\prod\limits_{n=-\infty}^{ \infty }1-e^{-a{(x+n)^2}}=x^2\prod\limits_{n=1}^{ \infty }(1-\frac{x^2}{n^2})^2\prod\limits_{n=-\infty}^{ \infty }G(x,a)$$
$$f(x)=\frac{\sin^2 (\pi x)}{\pi^2}\prod\limits_{n=-\infty}^{ \infty }G(x,a)$$
$\prod\limits_{n=-\infty}^{ \infty }G(x,a)$ will be expressed by other roots for $k\neq 0$
$$\prod\limits_{n=-\infty}^{ \infty }G(x,a)=\prod\limits_{n=-\infty}^{ \infty }\prod\limits_{k=1}^{ \infty }(1-\frac{x}{-n+\sqrt{\frac{2k \pi i }{a}}})(1-\frac{x}{-n-\sqrt{\frac{2k \pi i }{a}}})(1-\frac{x}{-n+\sqrt{\frac{-2k \pi i }{a}}})(1-\frac{x}{-n-\sqrt{\frac{-2k \pi i }{a}}})$$
$$\prod\limits_{n=-\infty}^{ \infty }G(x,a)=\prod\limits_{n=-\infty}^{ \infty }\prod\limits_{k=1}^{ \infty }(1+\frac{2nx}{n^2-\frac{2k \pi i }{a}}+\frac{x^2}{n^2-\frac{2k \pi i }{a}})(1+\frac{2nx}{n^2+\frac{2k \pi i }{a}}+\frac{x^2}{n^2+\frac{2k \pi i }{a}})$$
$$\prod\limits_{n=-\infty}^{ \infty }G(x,a)=\prod\limits_{n=-\infty}^{ \infty }\prod\limits_{k=1}^{ \infty }(\frac{(x+n)^2-\frac{2k \pi i }{a}}{n^2-\frac{2k \pi i }{a}})(\frac{(x+n)^2+\frac{2k \pi i }{a}}{n^2+\frac{2k \pi i }{a}})$$
So finally we can write ;
$$f(x)=\frac{\sin^2 (\pi x)}{\pi^2}\prod\limits_{n=-\infty}^{ \infty }\prod\limits_{k=1}^{ \infty }(\frac{(x+n)^4+\frac{4\pi^2 k^2 }{a^2}}{n^4+\frac{4 \pi^2 k^2 }{a^2}})$$
I thought to change product order can help us, so to get
$$f(x)=\frac{\sin^2 (\pi x)}{\pi^2}\prod\limits_{k=1}^{ \infty }\prod\limits_{n=-\infty}^{ \infty }(\frac{(x+n)^4+\frac{4\pi^2 k^2 }{a^2}}{n^4+\frac{4 \pi^2 k^2 }{a^2}})$$
I am till here for now, But I do not see how to go from here to get a nice expression with known functions.
Thanks for helps
EDIT: 1/9/2017
After @Yuriy S 's wonderful graphics, I realized that we may express the $f(x,a)$ in approximation.
$$f(x,a)=\prod\limits_{n=-\infty}^{ \infty }(1-e^{-a{(x+n)^2}}) $$ where $a>0$
Known values of $f(x,a)$:
- $f(x,0)=0$
- $\lim_{a\to\infty} f(x,a)=1$
- $f(1/2+k,a)$ has maximum, where $k$ is integer
- $f(k,a)=0$, where $k$ is integer
- $f(x,a)$ is periodic function over $x$
I conjecture that $f(x,a) \approx \cfrac{a^k \sin^2(\pi x)}{a^k \sin^2(\pi x)+c}$ ,where $a>0$, $k,c$ are positive real numbers,
From Graphics below, I calculated $k$ and $c$.
in sixth graph,
$$f(1/2,1)\approx 0.04$$ $$f(1/2,0.6)\approx 0.01$$
From these results, I got $k$ and $c$.
$k \approx 2.774$ and $c \approx 24$.
Maybe someone can help me to calculate $k$ and $c$ more accurately.
Thanks Best Regards
I post this answer separately, so that the analytics are not lost among the illustrative material from my first post.
It's often easier to deal with the logarithms of infinite products, since they are represented as infinite series.
$$\ln f(x)=\sum_{n=-\infty}^\infty \ln(1-e^{-a{(x+n)^2}})=$$
Since $a(x+n)^2>0$ we can always expand the logarithm into its Taylor series:
$$=-\sum_{n=-\infty}^\infty \sum_{k=1}^\infty \frac{1}{k} e^{-ka(x+n)^2}=-\sum_{k=1}^\infty \frac{1}{k} e^{-kax^2} \sum_{n=-\infty}^\infty e^{-kan(n+2x)}$$
The latter sum fits the definition of a Jacobi Theta Function $\vartheta_3$, thus:
$$\sum_{n=-\infty}^\infty e^{-kan(n+2x)}=\sqrt{\frac{\pi}{ka}} e^{kax^2} \vartheta_3 \left(\pi x, e^{-\frac{\pi^2}{ka}}\right)$$
This result is confirmed by Wolfram Alpha.
This looks complicated, but there is another series representation for $\vartheta_3$ which might help:
$$\vartheta_3 \left(\pi x, e^{-\frac{\pi^2}{ka}}\right)=1+2\sum_{l=1}^\infty e^{-\frac{\pi^2l^2}{ka}} \cos(2 \pi l x)$$
Now we obtain:
$$\ln f(x)=-\sqrt{\frac{\pi}{a}} \sum_{k=1}^\infty \frac{1}{k^{3/2}} \left(1+2\sum_{l=1}^\infty e^{-\frac{\pi^2l^2}{ka}} \cos(2 \pi l x) \right)$$
And finally:
$$\ln f(x)=-\sqrt{\frac{\pi}{a}} \zeta \left( \frac32 \right) -2\sqrt{\frac{\pi}{a}} \sum_{l=1}^\infty \sum_{k=1}^\infty \frac{1}{k^{3/2}} e^{-\pi^2l^2/(ka)} \cos(2 \pi l x)$$
Or, introducing:
The familiar form of a Fourier series emerges (with the first term being the constant term $l=0$):
The convergence is not very good, however the $l$ sum converges much faster than the $k$ sum (which should be obvious by looking at the exponential term in $C_l$). Thus, we don't need a lot of Fourier terms, but the coefficients have to be computed precisely.
Here is an illustration for $a=1$:
This Fourier series allows us excellent insight into similarities with the $C \sin^2 (\pi x)$ function.
$$\frac{(\sin^2 (\pi x))'}{\sin^2 (\pi x)}=2 \pi \cot (\pi x)$$
$$\frac{f'(x)}{f(x)}=(\ln f(x))'=4 \pi \sqrt{\frac{\pi}{a}} \sum_{l=1}^\infty l~C_l \sin(2 \pi l x)$$
It turns out, these new functions are very close for $a=1$ (if we compute $C_l$ accurately enough):
However, there is a correlation - the larger $a$, the faster $k$-sum converges, but the slower $l$-sum converges. Thus, if $a$ is small you need to take more terms in $k$-sum, but if $a$ is large you ned to take more terms in the $l$-sum.
Here are plots for the same large number of terms for $a=0.2,1,5$:
Disclaimer
The Fourier series has been introduced in this answer, I have not seen it before writing my own, even though it's linked above in the comments to the OP. I acknowledge that Semiclassical got there first.