Does this double product equal an exponential function?

Question

I was looking at the graph of $$\prod_{n=1}^\infty\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\Gamma\left(n-x+1\right)}=\prod_{n=1}^\infty\prod_{k=1}^\infty\left(1-\frac{x^2}{\left(n+k\right)^2}\right)$$

I noticed that it looks much like a normal curve of height $1$.

So is this equal to the form $e^{\frac{-x^2}{v}}$ for some $v$?

Or essentially the same question: does $$\sum_{n=1}^\infty\sum_{k=1}^\infty\ln\left(1-\frac{x}{\left(n+k\right)^2}\right)=-\frac{x}{v}$$ for some $v$?

Edit Related Question on the partial product.

Answer 1

A problem is that the sum diverges.

$\begin{array}\\ s(x) &=-\sum_{n=1}^\infty\sum_{k=1}^\infty\ln\left(1-\frac{x}{\left(n+k\right)^2}\right)\\ &\ge\sum_{n=1}^\infty\sum_{k=1}^\infty \dfrac{x}{(n+k)^{2}} \quad\text{since }-\ln(1-z) \ge z \text{ if } z \ge 0\\ &=x\sum_{n=1}^\infty\sum_{k=n+1}^\infty \dfrac{1}{k^{2}}\\ &\ge x\sum_{n=1}^\infty\sum_{k=n+1}^\infty \dfrac{1}{k(k+1)}\\ &= x\sum_{n=1}^\infty\sum_{k=n+1}^\infty (\dfrac1{k}-\dfrac1{k+1})\\ &= x\sum_{n=1}^\infty\dfrac1{n+1}\\ \end{array} $

and this diverges.

Answer 2

The observation is interesting per se.

In fact, if we consider $$f_p(x)=\prod_{n=1}^p\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}$$ we can find (using a CAS) that $$f_p(x)=\frac{ G(p+2)^2 \, G(2-x)\, G(2+x) }{G(p+2-x)\, G(p+2+x)}\left(-\frac{\sin(\pi x)}{\pi x(x^2-1)}\frac{G(3-x)\, G(3+x) }{ G(2-x)\, G(2+x)}\right)^p$$ where appear the Barnes G-function.

For the simplest $p=1$, $$f_1(x)=-\frac{\sin (\pi x)}{\pi x \left(x^2-1\right)}$$ does not show such a shape but, increasing $p$ more and more, we can effectively notice what you observed.

The problem, as @marty cohen answered, is that there is no limit to this function.

What is numerically interesting is the computation of $x$ such that $f_p(x)=\frac 12$. Here are some results ( the table has been updated after the edit). $$\left( \begin{array}{ccccc}\ p & \text{exact} & \text{using } (1) &\text{using } (2) &\text{using } (3)\\ 10 & 0.543223 & 0.486469 & 0.567259 & 0.545324 \\ 20 & 0.481383 & 0.438933 & 0.497577 & 0.482529 \\ 30 & 0.452924 & 0.416538 & 0.466221 & 0.453767 \\ 40 & 0.435333 & 0.402499 & 0.447040 & 0.436024 \\ 50 & 0.422935 & 0.392505 & 0.433608 & 0.423532 \\ 60 & 0.413516 & 0.384854 & 0.423448 & 0.414049 \\ 70 & 0.406002 & 0.378714 & 0.415372 & 0.406488 \\ 80 & 0.399801 & 0.373622 & 0.408723 & 0.400251 \\ 90 & 0.394552 & 0.369293 & 0.403108 & 0.394973 \\ 100 & 0.390022 & 0.365543 & 0.398270 & 0.390419 \\ 200 & 0.363578 & 0.343397 & 0.370187 & 0.363856 \\ 300 & 0.350356 & 0.332155 & 0.356240 & 0.350588 \\ 400 & 0.341793 & 0.324812 & 0.347238 & 0.341998 \\ 500 & 0.335563 & 0.319439 & 0.340703 & 0.335749 \\ 600 & 0.330715 & 0.315240 & 0.335627 & 0.330888 \\ 700 & 0.326775 & 0.311816 & 0.331507 & 0.326938 \\ 800 & 0.323472 & 0.308938 & 0.328057 & 0.323627 \\ 900 & 0.320640 & 0.306464 & 0.325101 & 0.320788 \\ 1000 & 0.318169 & 0.304301 & 0.322524 & 0.318311 \end{array} \right)$$

Edit

Taking logarithms of the product and using Stirling approximation, a very crude approximation would be $$\color{blue}{f_p(x) \sim \exp\left({-x^2 H_p}\right)}\tag 1$$ A better one would be $$\color{blue}{f_p(x) \sim \exp\left({-x^2 \left(H_p-\frac{1}{2}H_p^{(2)}\right)}\right)}\tag 2$$ Just computing the second derivative at $x=0$ would give as a much much better approximation (this is the exact second order Taylor expansion of $f_p(x)$ buil at $x=0$) $$\color{red}{f_p(x) \sim \exp\left(-x^2 \left(\psi ^{(0)}(p+2)+(p+1)\, \psi ^{(1)}(p+2)+\gamma-\frac{\pi ^2}{6} \right)\right)}\tag 3$$

Considering the asymptotics of the constant $k_{i,p}$ which appears in equation $(i)$, we can see that they are very closely related to each other $$k_{1,p}=\log(p)+\gamma +\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$

$$k_{2,p}= \log(p)+\left(\gamma-\frac{\pi ^2}{12}\right) +\frac{1}{p}+O\left(\frac{1}{p^2}\right) $$

$$k_{3,p}=\log(p)+\left(\gamma-\frac{\pi ^2}{6} +1\right)+\frac{1}{p}+O\left(\frac{1}{p^2}\right) $$

Moreover, computing, we can notice that $k_{3,p}\approx k_{1,p}-\frac 23$.

The advantage of these appromations is that, if we need to solve for $x$, $f_p(x)=a$, we have a very good estimate to start Newton method. For illustration, considering $p=100$ and $a=\frac 14$, we would get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.55213548373373596669 \\ 1 & 0.55099521268758128733 \\ 2 & 0.55099729839656297058 \\ 3 & 0.55099729840353755347 \end{array} \right)$$ which is the solution for twenty significant figures.

Answer 3

I prefer to add another answer for an extension of the work.

Considering $$f_p(x)=\prod_{n=1}^p\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}$$

$$\log\left(f_p(x) \right)=\sum_{n=1}^p \log\left(\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)} \right)$$

Using Taylor expansion $$\log\left(\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}\right)=-2\sum_{k=1}^\infty \frac{\psi ^{(2 k-1)}(n+1)}{(2 k)!}x^{2k}$$ and what remain is to compute the sums over $n$.

Writing $$\log\left(f_p(x) \right)=c_1 x^2+c_2 x^4+c_3 x^6 + c_4 x^8+\cdots$$ that is to say $$f_p(x)=\exp(c_1 x^2+c_2 x^4+c_3 x^6 + c_4 x^8+\cdots)$$ we should have $$c_1=-\psi ^{(0)}(p+2)-(p+1) \psi ^{(1)}(p+2)-\gamma+\frac{\pi^2}6$$ $$c_2=\frac{-45 \psi ^{(2)}(p+2)-15 (p+1) \psi ^{(3)}(p+2)-90 \zeta (3)+\pi ^4}{180} $$ $$c_3=\frac{-315 \psi ^{(4)}(p+2)-63 (p+1) \psi ^{(5)}(p+2)+8 \left(\pi ^6-945 \zeta (5)\right)}{22680}$$ $$c_4=\frac{-105 \psi ^{(6)}(p+2)-15 (p+1) \psi ^{(7)}(p+2)+8 \left(\pi ^8-9450 \zeta (7)\right)}{302400}$$ For any value of $p$, all coefficients are negative and they are smaller and smaller; this justifies the approximation given by $(3)$ in the previous answer.

For infinitely large values of $p$, we have $$\frac {c_2}{c_1} \sim \frac{90 \zeta (3)-\pi^4}{30 \left(6 \log (p)-\pi ^2+6 \gamma +6\right)}$$ $$\frac {c_3}{c_2} \sim \frac{4 \left(\pi ^6-945 \zeta (5)\right)}{63 \left(\pi ^4-90 \zeta (3)\right)}\approx 0.109046$$ $$\frac {c_4}{c_3} \sim \frac{3 \left(\pi ^8-9450 \zeta (7)\right)}{40 \left(\pi ^6-945 \zeta (5)\right)}\approx 0.163594$$

Warning

Better material in my answer to this question.