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Q. 2.15

Expert-verifiedFound in: Page 63

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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

By using pocket calculator as,

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

By using Stirling's approximation as,

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

By using pocket calculator $n=50$ as

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

We have,

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

By using Stirling's approximation as,

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

We have,

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

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

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