Editing Laplace's method (section)

==Example: Stirling's approximation==
Laplace's method can be used to derive [[Stirling's approximation]]

:<math>N!\approx \sqrt{2\pi N} \left(\frac{N}{e}\right)^N \,</math>

for a large [[integer]] ''N''. From the definition of the [[Gamma function]], we have

:<math>N! = \Gamma(N+1)=\int_0^\infty e^{-x} x^N \, dx.</math>

Now we change variables, letting <math>x=Nz</math> so that <math>dx = Ndz.</math> Plug these values back in to obtain

:<math>\begin{align}
N! &= \int_0^\infty e^{-Nz} (Nz)^N N \, dz \\
   &= N^{N+1} \int_0^\infty e^{-Nz} z^N \, dz \\
   &= N^{N+1} \int_0^\infty e^{-Nz} e^{N\ln z} \, dz \\
   &= N^{N+1} \int_0^\infty e^{N(\ln z-z)} \, dz.
\end{align}</math>

This integral has the form necessary for Laplace's method with

:<math>f(z) = \ln{z}-z</math>

which is twice-differentiable:

:<math>f'(z) = \frac{1}{z}-1,</math>
:<math>f''(z) = -\frac{1}{z^2}.</math>

The maximum of <math>f(z)</math> lies at ''z''<sub>0</sub> = 1, and the second derivative of <math>f(z)</math> has the value −1 at this point. Therefore, we obtain

:<math>N! \approx N^{N+1}\sqrt{\frac{2\pi}{N}} e^{-N}=\sqrt{2\pi N} N^N e^{-N}.</math>