Finding the Inverse of the Product of Two Gamma Functions

Question

I'm currently doing a calculation that pertains to Pi-day this March 14$^\textrm{th}$. Without going into details as they're not relevant to point of the question. I have the function $$ y = 4 \cdot \Gamma(1+\frac{1}{x}) \cdot \Gamma(2-\frac{1}{x}) $$

I'd like to find the inverse functions for the ranges domains $[1, \infty)$. If there happens to be a property of the inverse of the product of two continuous functions, but I know from looking at this function in Mathematica that for $f_1(x) = \Gamma(1+\frac{1}{x})$ and $f_2(x) = \Gamma(2-\frac{1}{x})\:$, $\:f_1(x) \biggr{|}_{\approx 2.16623} = \: f_2(x) \biggr{|}_{\approx 1.85739} \approx 0.885603$, however this may not be useful information.

Answer 1

Using mainly the reflexion formula of the gamma function $$y = 4 \, \Gamma\left(1+\frac{1}{x}\right) \, \Gamma\left(2-\frac{1}{x}\right)=4 \, \Gamma\left(1+\frac{1}{x}\right) \,\left(1-\frac{1}{x}\right)\, \Gamma\left(1-\frac{1}{x}\right)$$ $$y=4\left(1-\frac{1}{x}\right)\frac{\pi \csc \left(\frac{\pi }{x}\right)}{x}=\frac{4 \pi (x-1) }{x^2\sin \left(\frac{\pi }{x}\right)}$$ which seems to be difficult to invert.

For a first approximation, you could use the magnificent $$\sin(t) \sim \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad (0\leq t\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician. This would give $$y \sim \frac \pi 4\left(5-\frac 4 x +\frac 4 {x^2}\right)$$ and you just need to solve a quadratic. The problem is that it will be good only for $1 \lt x \leq 10$.

For improvement, we need to take into account the fact that $y$ goes through a minimum for $x=2$ (at this point $y=\pi$).

Much better would be to build the $[2,2]$ Padé approximant which write

• For $1 \leq x \leq 2$ $$y= \frac {4+a_1 (x-1)+a_2(x-1)^2 } { 1+b_1 (x-1)+b_2(x-1)^2 }$$ $$a_1=-\frac{2 \left(7 \pi ^2-60\right)}{5 \left(\pi ^2-6\right)}\qquad a_2=\frac{\pi ^2}{5}\qquad b_1=\frac{23 \pi ^2-120}{10 \left(\pi ^2-6\right)}\qquad b_2=\frac{-360+138 \pi ^2-7 \pi ^4}{60 \left(\pi ^2-6\right)}$$and a quadratic equation to solve to obtain the inverse.
• For $2 \leq x \leq \infty$ $$y=\frac {4+\frac {a_1} {x}+\frac {a_2} {x^2} } {1+\frac {b_1} {x}+\frac {b_2} {x^2} }$$ $$a_1=\frac{2 \left(13 \pi ^2-60\right)}{5 \left(\pi ^2-6\right)}\qquad a_2=\frac{\pi ^2}{5}\qquad b_1=\frac{3 \pi ^2}{10 \left(\pi ^2-6\right)}\qquad b_2=-\frac{\pi ^2 \left(7 \pi ^2-60\right)}{60 \left(\pi ^2-6\right)}$$ and again a quadratic equation to solve to obtain the inverse.

To have approximate solution of the above, we could expand $y$ as series and use series reversion to get

• For $1 \leq x \leq 2$ $$x=1+t+\left(1+\frac{\pi ^2}{6}\right) t^2+\frac{1}{18} \left(18+3 \pi ^2+\pi ^4\right) t^3+\frac{\left(1080+180 \pi ^2+21 \pi ^4+25 \pi ^6\right) t^4}{1080}+O\left(t^5\right)\qquad \text{where} \qquad t=1-\frac y 4$$

• For $2 \leq x \leq \infty$ $$x=\frac{1}{t}-\frac{\pi ^2}{6}+\left(\frac{\pi ^2}{6}-\frac{\pi ^4}{36}\right) t+\left(\frac{23 \pi ^4}{360}-\frac{\pi ^6}{108}\right) t^2+\frac{\left(-234 \pi ^4+216 \pi ^6-25 \pi ^8\right) t^3}{6480}+O\left(t^5\right)\qquad \text{where} \qquad t=1-\frac y 4$$

For illustration purposes using $y=\frac{\pi+4}2$

• Padé approximants give as solutions $x_1=1.15350$ and $x_2=7.51476$
• Inverse series give $x_1=1.15260$ and $x_2=7.52000$
• Exact solutions are $x_1=1.15351$ and $x_2=7.51419$

Update

Looking for a unique representation of $y$ at the price of a small loss of accuracy, I obtained $$y=\frac {\pi +\pi (x-2)-\frac{\pi\left(\pi ^2-8\right) }{8 (\pi -4)}(x-2)^2 } {1+(x-2)-\frac{\pi\left(\pi ^2-8\right) }{32 (\pi -4)}(x-2)^2 }$$ which reproduces exactly the values of $y$ for $x=1$, $x=2$ and for $x \to \infty$ as well as the fact that the minimum occurs at $x=2$.

For $x=1$, this approximation gives $y'=-\frac{32 (\pi -4)^2}{\pi \left(\pi ^2-8\right)}=-4.01455$ instead of $-4$. For $x=2$, the approximation gives the exact value of $y''$.

Using it and to make the formulae simpler, an approximation of the inverse of $y$ can write $$\frac 1 x=\frac 1 2 \pm \sqrt{2 (4-\pi )}\sqrt{\frac {(y-\pi) } {\alpha +\beta (y-\pi)}}$$ where $$\alpha=\pi(4-\pi) \left(\pi ^2-8\right)\qquad \text{and} \qquad \beta= \left(32-\pi ^3\right)$$

Below are some results for this simplistic formula

$$\left( \begin{array}{ccccc} y & \text{estimation} & \text{solution} & \text{estimation} & \text{solution}\\ 3.15 & 1.80681 & 1.80682 & 2.23944 & 2.23944 \\ 3.20 & 1.56196 & 1.56200 & 2.77949 & 2.77935 \\ 3.25 & 1.44905 & 1.44913 & 3.22695 & 3.22651 \\ 3.30 & 1.37230 & 1.37242 & 3.68603 & 3.68512 \\ 3.35 & 1.31392 & 1.31407 & 4.18557 & 4.18396 \\ 3.40 & 1.26686 & 1.26704 & 4.74732 & 4.74472 \\ 3.45 & 1.22754 & 1.22774 & 5.39491 & 5.39099 \\ 3.50 & 1.19385 & 1.19407 & 6.15851 & 6.15282 \\ 3.55 & 1.16447 & 1.16469 & 7.08001 & 7.07196 \\ 3.60 & 1.13848 & 1.13870 & 8.22116 & 8.20992 \\ 3.65 & 1.11523 & 1.11544 & 9.67817 & 9.66260 \\ 3.70 & 1.09424 & 1.09444 & 11.6109 & 11.5892 \\ 3.75 & 1.07515 & 1.07533 & 14.3063 & 14.2758 \\ 3.80 & 1.05768 & 1.05782 & 18.3383 & 18.2941 \\ 3.85 & 1.04159 & 1.04171 & 25.0454 & 24.9777 \\ 3.90 & 1.02671 & 1.02679 & 38.4421 & 38.3267 \\ 3.95 & 1.01289 & 1.01293 & 78.6014 & 78.3414 \end{array} \right)$$

$\color{blue}{\text{In this answer, there are }}\color{red}{\large{59}}$ $\color{blue}{\text{ occurences of }}\color{red}{\Large{\pi}}$

Answer 2

I prefer to add a second answer since based on a totally approach.

Considering the equation $$y=4 \pi\frac{ (x-1) }{x^2}\csc \left(\frac{\pi }{x}\right)$$ let $x=\frac{2}{2 t+1}$ to make $$y=\pi \left(1-4 t^2\right) \sec (\pi t)$$ which an even function "looking" like a parabola. The interest is that the range for $t$ is now restricted to $\left(-\frac12,\frac12\right)$

To approximate the rhs, let $$g(t)=\pi + a \,t^2+b\, t^4+c \,t^6+d \,t^8$$ The coefficients were obtained using the equations $$g''(0)=\pi \left(\pi ^2-8\right)\qquad g\left(\frac12\right)=4\qquad g'\left(\frac12\right)=4\qquad g''\left(\frac12\right)= \frac{4 \pi ^2}{3}$$ The results are $$a=\frac{1}{2} \pi \left(\pi ^2-8\right)\qquad b=332+\frac{2}{3} \pi \left(-72+\pi -9 \pi ^2\right)$$ $$c=\frac 83\left(-636+120 \pi -2 \pi ^2+9 \pi ^3\right)\qquad d=\frac {32}3\left(234-48 \pi +\pi ^2-3 \pi ^3\right)$$ The two curves almost overlap; the maximum error is $9.23\times 10^{-6}$ and the norm $$\Phi=\int_{-\frac 12}^{\frac 12}\Big[\pi \left(1-4 t^2\right) \sec (\pi t)-g(t)\Big]^2\,dt=2.51\times 10^{-11}$$

Now, looking at $g(t)$ as a series expansion to $O(t^{10})$, we can make its inverse and get $$t=Y+\alpha\, Y^3+\beta \,Y^5+O(Y^7)\qquad \text{where}\qquad Y=\sqrt{\frac{2 (y-\pi)}{\pi( \pi^2-8)}}$$ $$\alpha=\frac{2 \left(-498+72 \pi -\pi ^2+9 \pi ^3\right)}{3 \pi \left(\pi ^2-8\right)}$$ $$\beta =\frac{2 \left(1736028-563040 \pi +54780 \pi ^2-56316 \pi ^3+8503 \pi ^4-102 \pi ^5+459 \pi ^6\right)}{9 \pi ^2 \left(\pi ^2-8\right)^2}$$

Some results for the solution of $t$

$$\left( \begin{array}{ccc} y & \text{estimation} & \text{solution} \\ 3.15 & 0.053460 & 0.053460 \\ 3.20 & 0.140192 & 0.140203 \\ 3.25 & 0.190022 & 0.190068 \\ 3.30 & 0.228525 & 0.228638 \\ 3.35 & 0.260774 & 0.260992 \\ 3.40 & 0.288875 & 0.289240 \\ 3.45 & 0.313947 & 0.314505 \\ 3.50 & 0.336673 & 0.337473 \\ 3.55 & 0.357504 & 0.358596 \\ 3.60 & 0.376757 & 0.378196 \\ 3.65 & 0.394667 & 0.396508 \\ 3.70 & 0.411413 & 0.413713 \\ 3.75 & 0.427134 & 0.429951 \\ 3.80 & 0.441943 & 0.445338 \\ 3.85 & 0.455931 & 0.459964 \\ 3.90 & 0.469175 & 0.473909 \\ 3.95 & 0.481740 & 0.487235 \end{array} \right)$$