Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

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

By using pocket calculator as,

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

By using Stirling's approximation as,

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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,

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

Step: 2 Using Stirling's approximation:

By using Stirling's approximation as,

$\begin{array}{l} N! \approx N^{N} e^{- N} \sqrt{2 π N} \\ \ln (N!) \approx N l n (N) - N \end{array}$

We have,

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

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

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Short Answer

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

Step by Step Solution

Step: 1 Using pocket calculator:

Step: 2 Using Stirling's approximation:

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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! .

Step by Step Solution

Step: 1 Using pocket calculator:

Step: 2 Using Stirling's approximation:

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