Using Stirling’s approximation, find the leading behavior of:
\(\binom { 2 N } { N }\) for large N.
\(S = \log Z ( N , m )\) with \(\begin{array} { r } { Z = \frac { N ! } { ( N - m ) ! m ! } } \end{array}\) for large N.
P-TRIV-CA30
Using Stirling’s approximation, find the leading behavior of:
\(\binom { 2 N } { N }\) for large N.
\(S = \log Z ( N , m )\) with \(\begin{array} { r } { Z = \frac { N ! } { ( N - m ) ! m ! } } \end{array}\) for large N.