0
$\begingroup$

Let $n$ postive integer,Assmue that $\Gamma(x)$ is Gamma function $$\dfrac{\Gamma\left(\dfrac{1}{2}+\dfrac{1}{2n}\right)}{\Gamma\left(1+\dfrac{1}{2n}\right)}=A+\dfrac{B}{n}+\dfrac{C}{n^2}+\dfrac{D}{n^3}+\cdots$$

use $$\dfrac{d}{dz}\left(\ln{\Gamma\left(\dfrac{1}{2}+z\right)}-\ln{\Gamma\left(1+z\right)}\right)=\varepsilon(z+1/2)-\varepsilon(z+1)$$ where $\varepsilon(z)$ is Digamma function Now I have prove $$A=\sqrt{\pi},B=-\sqrt{\pi}\log{2}$$ But can't find $C,D$

$\endgroup$
1
$\begingroup$

Considering $$A(z)=\log(Y(z))=\log \left(\Gamma \left(\frac{1}{2}+z\right)\right)-\log (\Gamma (z+1))$$ you are looking for the Taylor expansion built around $z=0$. This means $$A(z)= A(0)+\frac{A'(0)} {1!}z+\frac{A''(0)} {2!}z^2+\frac{A'''(0)} {3!}z^3+\cdots$$

You are given for the derivatives $$A'(z)=\psi ^{(0)}\left(z+\frac{1}{2}\right)-\psi ^{(0)}(z+1)$$ which generalizes to $$A''(z)=\psi ^{(1)}\left(z+\frac{1}{2}\right)-\psi ^{(1)}(z+1)$$ $$A'''(z)=\psi ^{(2)}\left(z+\frac{1}{2}\right)-\psi ^{(2)}(z+1)$$

which, now, need to be simplified using $z=0$.

So $$A(0)=\frac{\log (\pi )}{2}\qquad A'(0)=-\log (4)\qquad A''(0)=\frac{\pi ^2}{3}\qquad A'''(0)=-12 \zeta (3)$$ This makes $$A(z)=\frac{\log (\pi )}{2}-\log (4)z+\frac{\pi ^2 }{6}z^2-2 \zeta (3)z^3+O\left(z^4\right)$$ Now, Taylor again $$Y(z)=e^{A(z)}=\sqrt{\pi }-\sqrt{\pi } \log (4)z+\frac{1}{6} \sqrt{\pi } \left(\pi ^2+3 \log ^2(4)\right)z^2-\frac{1}{6} \left(\sqrt{\pi } \left(12 \zeta (3)+\log ^3(4)+\pi ^2 \log (4)\right)\right)z^3 +O\left(z^4\right)$$ Now, to finish, replace $z$ by $\frac 1 {2n}$.

Just to see how accurate is the expansion, $$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation}\\ 5 & 1.565345082 & 1.563836840 \\ 10 & 1.660110105 & 1.660007284 \\ 15 & 1.695365291 & 1.695344347 \\ 20 & 1.713776688 & 1.713769957 \end{array} \right)$$

$\endgroup$
1
$\begingroup$

Doing a Google search for "gamma function asymptotics" leads to a number of relevant links including this:

http://dlmf.nist.gov/5.11

This, in turn, has a number of references. This one is free:

T. Burić and N. Elezović (2011) Bernoulli polynomials and asymptotic expansions of the quotient of gamma functions. J. Comput. Appl. Math. 235 (11), pp. 3315–3331.

It is available at

http://www.sciencedirect.com/science/article/pii/S0377042711000562

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.