Stirling Formula

There is a formula to approximately calculate by Stirling:

$N!≈√{2⋅π⋅N}⋅(÷{N}{e})^N$

In logarithms, we get:

$\ln N!=\ln √{2⋅π⋅N}+N⋅\ln N-N⋅\ln e$

$\ln N!=\ln √{2⋅π⋅N}+N⋅\ln N-N$

$\ln N!=÷{1}{2}⋅\ln 2⋅π⋅N+N⋅\ln N-N$

$\ln N!=÷{1}{2}⋅(\ln 2+\ln π+\ln N)+N⋅\ln N-N$

For very big N, we approximately get:

$\ln N!≈N⋅\ln N-N$

Author: Danny (remove the ".nospam" to send)

Last modification on: Sat, 04 May 2024 .