Chapter 12.5, Problem 50AYU

Stirling’s Formula An approximation for $n!$, when $n$ is large, is given by

$n!\approx \sqrt{2n\pi }{\left(\frac{n}{e}\right)}^n\left(1+\frac{1}{12n-1}\right)$

Calculate $12!,20!\text{and}\text{25!}$ on your calculator. Then use Stirling’s formula to approximate $12!,20!\text{and}\text{25!}$

To determine

To find: Calculate $12!$, $20!$ And $25!$ on calculator and using Stirling’s formula to approximate $12!$, $20!$ And $25!$.

Answer to Problem 50AYU

Solution:

Calculator solutions are

$12!=479001600,20!=2.4329020082\times {10}^{18},25!=1.5511210043\times {10}^{25}$

Stirling’s formula approximate solutions are

$\begin{array}{l}12!\approx 479476936.386\\ 20!\approx 2.447819394721279303\times {10}^{18}\\ 25!\approx 1.5210370352028733913312624\times {10}^{25}\end{array}$

Explanation of Solution

Given:

Stirling’s formula for approximation of $n!$ is given as

$n!\approx \sqrt{2n\pi }{\left(\frac{n}{e}\right)}^n\left(1+\frac{1}{12n-1}\right)$

Consider the number $12!$, $20!$ And $25!$ using graphic calculator.

Therefore, $12!=479001600,20!=2.4329020082\times {10}^{18},25!=1.5511210043\times {10}^{25}$

Now finding the values using the stirling’s formula, $n=12$,

$\begin{array}{l}\sqrt{2n\pi }{\left(\frac{n}{e}\right)}^n\left(1+\frac{1}{12n-1}\right)=\sqrt{2(12)\pi }{\left(\frac{12}{e}\right)}^{12}\left(1+\frac{1}{12(12)-1}\right)\\ =\sqrt{2(12)\times 3.14}{\left(\frac{12}{2.718}\right)}^{12}\left(1+\frac{1}{144-1}\right)\\ =\sqrt{75.36}{\left(4.415\right)}^{12}\left(1+\frac{1}{23}\right)\\ 12!=8.681\times 54848972.34\times 1.007=479476936.386\end{array}$

$n=20$,

$\begin{array}{l}\sqrt{2n\pi }{\left(\frac{n}{e}\right)}^n\left(1+\frac{1}{12n-1}\right)=\sqrt{2(20)\pi }{\left(\frac{20}{e}\right)}^{20}\left(1+\frac{1}{12(20)-1}\right)\\ =\sqrt{2(20)\times 3.14}{\left(\frac{20}{2.718}\right)}^{20}\left(1+\frac{1}{240-1}\right)\\ =\sqrt{125.6}{\left(7.36\right)}^{20}\left(1+\frac{1}{239}\right)\\ =11.207\times 217548596967646739.66\times 1.004\\ 20!=2447819394721279303.86\approx 2.447819394721279303\times {10}^{18}\end{array}$

$n=25$,

$\begin{array}{l}\sqrt{2n\pi }{\left(\frac{n}{e}\right)}^n\left(1+\frac{1}{12n-1}\right)=\sqrt{2(25)\pi }{\left(\frac{25}{e}\right)}^{25}\left(1+\frac{1}{12(25)-1}\right)\\ =\sqrt{2(25)\times 3.14}{\left(\frac{25}{2.718}\right)}^{25}\left(1+\frac{1}{300-1}\right)\\ =\sqrt{157}{\left(9.19\right)}^{25}\left(1+\frac{1}{299}\right)\\ =12.53\times 1210285373092910726186375\times 1.003\\ =15210370352028733913312624.586\\ 25!=1.5210370352028733913312624\times {10}^{25}\end{array}$