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Chapter 11: Q11.4P (page 554)

Use Stirling鈥檚 formula to evaluate $\underset{n 鈫� 鈭�}{l i m} \frac{(2 n)! \sqrt{n}}{2^{2 n} {(n!)}^{2}}$ .

Short Answer

Expert verified

The value of the function is $\frac{1}{\sqrt{蟺}}$ .

Step by step solution

01

Given Information

The function is $\lim_{n 鈫� 鈭�} \frac{(2 n)! \sqrt{n}}{2^{2 n} {(n!)}^{2}}$ .

02

Definition of the sterling’s formula.

Sterling鈥檚 formula is used to simplify formulas involving factorial.

$n! 鈻� n^{n} e^{- n} \sqrt{2 蟺苍}$ .

03

Find the value of the function.

The function is $\lim_{n 鈫� 鈭�} \frac{(2 n)! \sqrt{n}}{2^{2 n} {(n!)}^{2}}$ .

The sterling鈥檚 formula is $n! 鈻� n^{n} e^{- n} \sqrt{2 蟺苍}$ .

$\begin{array}{rcl} (2 n)! & 鈮� & {(2 n)}^{n} e^{- 2 n} \sqrt{4 蟺苍} \\ {(n!)}^{2} & 鈮� & (n) \end{array}$ ⁿe^-n2蟺苍2

Substitute above valuesin the equation mentioned below.

$\begin{array}{rcl} \underset{n 鈫� 鈭�}{l i m} \frac{(2 n)! \sqrt{n}}{2^{2 n} {(n!)}^{2}} & = & \frac{{(2 n)}^{n} e^{- 2 n} \sqrt{4 蟺苍}}{{(n^{n} e^{- n} \sqrt{2 蟺苍})}^{2}} \\ = & \frac{1}{\sqrt{蟺}} \end{array}$

Hence, the value of the function is $\begin{array}{rcl} \frac{1}{\sqrt{蟺}} \end{array}$ .

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Most popular questions from this chapter

The logarithmic integralis $l i (x) = {鈭�}_{0}^{x} \frac{dt}{In t}$ . Express as exponential integrals

$l i (x)$
$l i (e^{x})$
$l i (x) = {鈭�}_{0}^{x} \frac{dt}{In (\frac{1}{t})}$

The integral ${鈭�}_{0}^{鈭�} t^{p - 1} e^{- t} d t = 螕' (p, x)$ is called an incomplete $螕'$ function. [Note that if x = 0, this integral is $螕' (p)$ .] By repeated integration by parts, find several terms of the asymptotic series for $螕' (p, x)$ .

${鈭�}_{0}^{\frac{蟺}{2}} \sqrt{1 - \frac{s i n^{2} 胃}{9} d 胃}$

In Chapter 1, equations (13.5) and (13.6), we defined the binomial coefficients $(p n)$ whereis a non-negative integer butmay be negative or fractional. Show that $(p n)$ can be written in terms of $螕$ functions as

role="math" localid="1664340642097" $(\frac{p}{n}) = \frac{螕 (p + 1)}{n! 螕 (p - n + 1)}$

Write the integral in equation (12.7) as an elliptic integral and show that (12.8)gives its value. Hints: Write $c o s 胃 = 1 - 2 s i n^{2} (\frac{胃}{2})$ and a similar equation for $c o s 伪$ . Then make the change of variable $x = \frac{\sin (\frac{胃}{2})}{\sin (\frac{伪}{2})}$ .

See all solutions

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