I want to find the inverse function of the following expression.(range is a positive real number) I tried to find an approximate expression using series expansion, but it was not easy. How do I get a nice approximate inverse function?
$$\frac{Γ(1 +\frac{1}{k})^2}{Γ(1 +\frac{2}{k})}$$
On
I prefer to add a second answer.
Let $x=\frac 1k$ and $z=\log(y)$ and expand again $$z=\sum_{n=2}^\infty (-1)^{n-1} a_n\,x^n$$ the first $a_n$ being $$\left\{\frac{\pi ^2}{6},2 \zeta (3),\frac{7 \pi ^4}{180},6 \zeta (5),\frac{31 \pi ^6}{2835},18 \zeta (7),\frac{127 \pi ^8}{37800},\frac{170 }{3}\zeta (9),\frac{73 \pi ^{10}}{66825},186 \zeta (11),\cdots\right\}$$
Using series reversion $$\color{red}{\large\frac 1 k=x=t+\frac{a_3 }{2 a_2}t^2+\frac{5 a_3^2-4 a_2 a_4 }{8 a_2^2}t^3+\frac{2 a_3^3-3 a_2 a_3 a_4+a_2^2 a_5 }{2 a_2^3}t^4+O(t^5)}$$ where $\color{red}{t=\sqrt{-\frac{z}{a_2}}=\frac 1 \pi \sqrt{-6\log(y)}}$
Repeating the same calculations,as shown below, the results are significantly much better (notice that still better could be done adding the next powers of $t$; not done here because the next coefficient is too long to be typed here).
$$\left( \begin{array}{ccc} y & \text{estimate} & \text{solution} \\ 0.1000 & 0.40978 & 0.41134 \\ 0.2000 & 0.54120 & 0.54269 \\ 0.3000 & 0.67282 & 0.67416 \\ 0.4000 & 0.82056 & 0.82171 \\ 0.5000 & 0.99904 & 1.00000 \\ 0.6000 & 1.23071 & 1.23147 \\ 0.7000 & 1.56014 & 1.56069 \\ 0.8000 & 2.10101 & 2.10135 \\ 0.9000 & 3.30338 & 3.30352 \\ & & \\ 0.9100 & 3.52428 & 3.52441 \\ 0.9200 & 3.78515 & 3.78526 \\ 0.9300 & 4.09981 & 4.09991 \\ 0.9400 & 4.48984 & 4.48992 \\ 0.9500 & 4.99128 & 4.99134 \\ 0.9600 & 5.67008 & 5.67012 \\ 0.9700 & 6.66405 & 6.66408 \\ 0.9800 & 8.33037 & 8.33038 \\ 0.9900 & 12.0894 & 12.0894 \\ & & \\ 0.9910 & 12.7835 & 12.7835 \\ 0.9920 & 13.6039 & 13.6039 \\ 0.9930 & 14.5943 & 14.5943 \\ 0.9940 & 15.8229 & 15.8229 \\ 0.9950 & 17.4036 & 17.4036 \\ 0.9960 & 19.5449 & 19.5449 \\ 0.9970 & 22.6825 & 22.6825 \\ 0.9980 & 27.9457 & 27.9457 \\ 0.9990 & 39.8255 & 39.8255 \\ & & \\ 0.9991 & 42.0194 & 42.0194 \\ 0.9992 & 44.6128 & 44.6128 \\ 0.9993 & 47.7438 & 47.7438 \\ 0.9994 & 51.6279 & 51.6279 \\ 0.9995 & 56.6255 & 56.6255 \\ 0.9996 & 63.3958 & 63.3958 \\ 0.9997 & 73.3164 & 73.3164 \\ 0.9998 & 89.9585 & 89.9585 \\ 0.9999 & 127.524 & 127.524 \end{array} \right)$$
You want to solve for $k$ the equation $$y=\frac{\Gamma \left(1+\frac{1}{k}\right)^2}{\Gamma \left(1+\frac{2}{k}\right)}\tag 1$$
There no formal series for $\Gamma (1+\epsilon )$ around $\epsilon=0$. However $$\log (\Gamma (1+\epsilon))=\sum_{n=1}^\infty \frac{\psi ^{(n-1)}(1)}{n! }\, \epsilon^n$$ Taking logarithms of both sides, expanding the series and transforming into one of the simplest Padé approximants, gives, after simplifications , the quadratic equation $$k^2+\frac{12 \zeta (3)}{\pi ^2}k+\frac{1}{6} \pi ^2 \left(\frac{1}{\log (y)}+\frac{864 \zeta (3)^2}{\pi^6}-\frac{7}{5}\right)=0$$ that is to say $$\color{red}{\large k\sim-\frac{6 \zeta (3)}{\pi ^2}+\sqrt{\frac{7 \pi ^2}{30}-\frac{\pi ^2}{6 \log (y)}-\frac{108 \zeta (3)^2}{\pi ^4}}}$$
Notice that the error associated to the used $[2,2]$ Padé approximant is $$\frac{-14 \pi ^6 \zeta (3)+4320 \zeta (3)^3+90 \pi ^4 \zeta (5)}{15 \pi ^4 k^5} \sim \frac{0.284}{k^5}$$ Below is a very detailed table when applying this formula for all the range of $y$. $$\left( \begin{array}{ccc} y & \text{estimate} & \text{solution} \\ 0.1000 & 0.45888 & 0.41134 \\ 0.2000 & 0.58184 & 0.54269 \\ 0.3000 & 0.70698 & 0.67416 \\ 0.4000 & 0.84913 & 0.82171 \\ 0.5000 & 1.02252 & 1.00000 \\ 0.6000 & 1.24939 & 1.23147 \\ 0.7000 & 1.57417 & 1.56069 \\ 0.8000 & 2.11045 & 2.10135 \\ 0.9000 & 3.30821 & 3.30352 \\ & & \\ 0.9100 & 3.52864 & 3.52441 \\ 0.9200 & 3.78904 & 3.78526 \\ 0.9300 & 4.10323 & 4.09991 \\ 0.9400 & 4.49279 & 4.48992 \\ 0.9500 & 4.99375 & 4.99134 \\ 0.9600 & 5.67206 & 5.67012 \\ 0.9700 & 6.66554 & 6.66408 \\ 0.9800 & 8.33137 & 8.33038 \\ 0.9900 & 12.0899 & 12.0894 \\ & & \\ 0.9910 & 12.7840 & 12.7835 \\ 0.9920 & 13.6043 & 13.6039 \\ 0.9930 & 14.5946 & 14.5943 \\ 0.9940 & 15.8232 & 15.8229 \\ 0.9950 & 17.4039 & 17.4036 \\ 0.9960 & 19.5451 & 19.5449 \\ 0.9970 & 22.6827 & 22.6825 \\ 0.9980 & 27.9458 & 27.9457 \\ 0.9990 & 39.8255 & 39.8255 \\ & & \\ 0.9991 & 42.0195 & 42.0194 \\ 0.9992 & 44.6129 & 44.6128 \\ 0.9993 & 47.7438 & 47.7438 \\ 0.9994 & 51.6280 & 51.6279 \\ 0.9995 & 56.6255 & 56.6255 \\ 0.9996 & 63.3958 & 63.3958 \\ 0.9997 & 73.3165 & 73.3164 \\ 0.9998 & 89.9585 & 89.9585 \\ 0.9999 & 127.524 & 127.524 \end{array} \right)$$
This gives for the asymptotics $$\color{blue}{\large k=-\frac{6 \zeta (3)}{\pi ^2}+\frac{\pi }{ \sqrt{6(1-y)}}+ \left(\frac{3 \pi }{40}-\frac{54 \zeta (3)^2}{\pi ^5}\right)\sqrt{6(1-y)}+O\left((1-y)^{3/2}\right)}$$