Compute $\Gamma(2.7)$

Question

All I know of the Gamma function is $$\Gamma(\alpha) = \int\limits_{0}^{\infty}x^{\alpha - 1}e^{-x}\text{ d}x$$ and the recursive formula $$\Gamma(\alpha) = (\alpha-1)\Gamma(\alpha - 1)\text{.}$$ A friend asked me to try to find $\Gamma(2.7)$. I assume this involves complex analysis, which I have absolutely no background on (except for the basic Euler formula). Using Wolfram, I found that $$\Gamma(2.7) = e^{0.434821}$$ which suggests to me perhaps writing out $e^{-x}$ as a Taylor series: $$\Gamma(\alpha) = \int\limits_{0}^{\infty}x^{\alpha - 1}\sum\limits_{i=1}^{\infty}\dfrac{(-x)^{i}}{i!}\text{ d}x$$ but alas, this doesn't look helpful.

How does one derive that $\Gamma(2.7) = e^{0.434821}$?

Answer 1

May be slightly better than Stirling, you could use Gosper approximation $$n! \approx \sqrt{\pi(2n+\frac{1}{3})} \Big(\frac{n}{e}\Big)^n$$ For $n=1.7$, this gives an approximate value of $1.54201$ while Stirling leads to $1.47156$ while the "exact" value is $1.54469$.

If you continue using Stirling series, that is to say $$n!=\sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}(1+\frac{1}{12 n}+\frac{1}{288 n^2}-\frac{139}{51840 n^3}-\frac{571}{2488320 n^4}+\cdots+O\left(\left(\frac{1}{n}\right)^6\right))$$ you successively find $1.47156$, $1.54369$, $1.54546$,$1.54466$, $1.54462$ ,$\cdots$.

Answer 2

Hint: Use Stirling's approximation in conjunction with the fact that $e\simeq2.7$.

Answer 3

You could use a Riemann sum to approximate $\Gamma(2.7)$

$\Gamma(2.7)$ $\approx$ $\sum_{x=1}^\infty e^{-x}x^{1.7}$

This comes out to about 1.52485

You can use Riemann sums to evaluate the gamma function of complex numbers too.

Answer 4

As stated in the comments, the Legendre duplication formula for the $\Gamma$ function gives: $$ \color{blue}{\Gamma\left(\frac{27}{10}\right)}=\frac{119}{100}\cdot 2^{3/5}\cdot\left(\color{red}{\frac{\Gamma\left(\frac{1}{5}\right)}{\Gamma\left(\frac{2}{5}\right)}}\right)^{-1}\cdot\sqrt{\pi} \tag{1}$$ hence it is enough to compute the red ratio with reasonable precision.

Due to the Euler product for the $\Gamma$ function we have: $$ \color{red}{\frac{\Gamma\left(\frac{1}{5}\right)}{\Gamma\left(\frac{2}{5}\right)}} = \lim_{n\to +\infty}n^{-1/5}\cdot\frac{\frac{2}{5}\cdot\left(\frac{2}{5}+1\right)\cdot\ldots\cdot\left(\frac{2}{5}+n\right)}{\frac{1}{5}\cdot\left(\frac{1}{5}+1\right)\cdot\ldots\cdot\left(\frac{1}{5}+n\right)}\tag{2}$$ and the RHS can be explicitly evaluated at $n=32$ to get an approximation with two accurate digits, or at $n=2^{20}$ to get an approximation with six accurate figures. Since $\frac{1}{5}$ and $\frac{2}{5}$ are quite close to zero and the $\Gamma$ function has a simple pole at the origin, it is reasonable to expect that the red ratio is close to $\color{red}{2}$, and indeed it is: $$ \color{red}{\frac{\Gamma\left(\frac{1}{5}\right)}{\Gamma\left(\frac{2}{5}\right)}}\approx \color{red}{2}.06966344. \tag{3}$$