Stirling's Approximation
$$ x! \approx x^{x}e^{-x}\sqrt{ 2\pi x } \iff \ln x! \approx x\ln x - x + \frac{1}{2} \ln(2\pi x) $$This proof (more like an interesting justification). is taken from page 2 of (MacKay 2003). $$ P(r \mid \lambda) = \frac{e^{-\lambda}\lambda^{r}}{r!} $$$$ e^{-\lambda} \frac{\lambda^{\lambda}}{\lambda!} \approx \frac{1}{\sqrt{ 2\pi \lambda }} \implies \lambda! \approx \lambda^{\lambda}e^{-\lambda}\sqrt{ 2\pi \lambda } $$ Which finishes the derivation of the approximation. Approximation of the binomial A quick derivation with the Stirling’s approximation gives a nice approximation for log of the binomials ...