Kapalci

2022-06-22

I have this problem where I need to rewrite the logarithmic form of Stirling's series
$\mathrm{ln}\left(z!\right)=\frac{1}{2}\mathrm{ln}\left(2\pi \right)+\left(z+\frac{1}{2}\right)\mathrm{ln}\left(z\right)-z+\frac{1}{12z}-\frac{1}{360{z}^{3}}+\frac{1}{1260{z}^{5}}-...$

Esteban Johnson

We have
$\mathrm{log}\left(z!\right)\sim \frac{1}{2}\mathrm{log}\left(2\pi \right)+\left(z+\frac{1}{2}\right)\mathrm{log}z-z+\frac{1}{12z}-\frac{1}{360{z}^{3}}+\dots .$
Taking the exponential of both sides gives
$\begin{array}{rl}z!& \sim \mathrm{exp}\left(\frac{1}{2}\mathrm{log}\left(2\pi \right)+\left(z+\frac{1}{2}\right)\mathrm{log}z-z+\frac{1}{12z}-\frac{1}{360{z}^{3}}+\dots \right)\\ & =\mathrm{exp}\left(\mathrm{log}\left(\sqrt{2\pi }{z}^{z+1/2}{e}^{-z}\right)\right)\mathrm{exp}\left(\frac{1}{12z}-\frac{1}{360{z}^{3}}+\dots \right)\\ & =\sqrt{2\pi }{z}^{z+1/2}{e}^{-z}\mathrm{exp}\left(\frac{1}{12z}-\frac{1}{360{z}^{3}}+\dots \right).\end{array}$
To finish the derivation, we use the Maclaurin series of the exponential:
$\begin{array}{rl}\mathrm{exp}\left(\frac{1}{12z}-\frac{1}{360{z}^{3}}+\dots \right)& =1+\left(\frac{1}{12z}-\frac{1}{360{z}^{3}}+\dots \right)+\frac{{\left(\frac{1}{12z}-\frac{1}{360{z}^{3}}+\dots \right)}^{2}}{2}+\dots \\ & =1+\frac{1}{12z}+\frac{1}{288{z}^{2}}-\frac{139}{51840{z}^{3}}-\dots .\end{array}$

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