Question

Use a pocket calculator to check the accuracy of Stirling's approximation for$N=50$ . Also check the accuracy of equation $2.16$for$\ln N!$ .

Short answer

By using pocket calculator as,

$50!=3.0414\times {10}^{64};\ln (50!)=148.4778$and

By using Stirling's approximation as,

$50!\approx 3.0363\times {10}^{64,};\ln (50!)\approx 145.6012.$

Step-by-step solution

Step: 1 Using pocket calculator:

By using pocket calculator $n=50$as

$\begin{array}{l}N!=50!=3.0414\times {10}^{64}\\ \ln (N!)=\ln (50!)=148.4778\end{array}$

We have,

$\begin{array}{l}\left.N!=50!\approx {50}^{(}50\right)e^{-50}\sqrt{2\pi 50}\\ N!=3.0363\times {10}^{64}\\ 50!\approx 3.0363\times {10}^{64}.\end{array}$

Step: 2 Using Stirling's approximation:

By using Stirling's approximation as,

$\begin{array}{l}N!\approx N^N{e}^{-N}\sqrt{2\pi N}\\ \ln (N!)\approx Nln\left(N\right)-N\end{array}$

We have,

$\begin{array}{l}\ln (N!)=\ln (50!)\approx 50\ln \left(50\right)-50\\ ln(N!)=145.6012\\ \ln (50!)\approx 145.6012\end{array}$

The Stirling's approximation is applicable for even$n=50.$