Chapter 8, Problem 11RCC

a.

To determine

To write: the Maclaurin Series and the interval of convergence for each of the following functions.

Answer to Problem 11RCC

  $\frac{1}{1-x}={\displaystyle \sum _{n=0}^{\infty }{x}^n},R=1.$

Explanation of Solution

Given information: Given functions is $\frac{1}{1-x}$ .

Calculation:

  $\begin{array}{c}f(x)=\frac{1}{1-x}\Rightarrow f(0)=1.\\ f\text{'}(x)=\frac{1}{{(1-x)}^2}\Rightarrow f\text{'}(0)=1.\\ f\text{'}\text{'}(x)=\frac{2}{{(1-x)}^3}\Rightarrow f\text{'}\text{'}(0)=2=2!.\\ f\text{'}\text{'}\text{'}(x)=\frac{2.3}{{(1-x)}^4}\Rightarrow f\text{'}\text{'}\text{'}(0)=6=3!.\\ \mathrm{.......}\\ \mathrm{.......}\\ f^{(n)}(0)=n!.\\ \text{Therefore}\text{,}\text{the}\text{Maclaurin}\text{}\text{Series}\text{is}:\\ {\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}}{x}^n={\displaystyle \sum _{n=0}^{\infty }\frac{n!}{n!}}{x}^n={\displaystyle \sum _{n=0}^{\infty }{x}^n=1+x+x^2}+\mathrm{.....}\\ So,a_n=x^n\text{then},\\ a_{n+1}=x^{n+1}\\ \left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{x^{n+1}}{x^n}\right|=\left|x\right|\end{array}$

By the Ratio Test, the series convergence for all x such as $\left|x\right|$<1 or -1< x < 1 so ,the radius of convergence is R =1.

b.

To determine

To write: the Maclaurin Series and the interval of convergence for each of the following functions.

Answer to Problem 11RCC

  $e^x={\displaystyle \sum _{n=0}^{\infty }\frac{x^n}{n!}},R=\infty$

Explanation of Solution

Given information: Given functions is $e^x$ .

Calculation:

  $\begin{array}{c}f(x)=e^x\Rightarrow f(0)=1.\\ f\text{'}(x)=e^x\Rightarrow f\text{'}(0)=1.\\ f\text{'}\text{'}(x)=e^x\Rightarrow f\text{'}\text{'}(0)=1\\ f\text{'}\text{'}\text{'}(x)=e^x\Rightarrow f\text{'}\text{'}\text{'}(0)=1.\\ \mathrm{.......}\\ \mathrm{.......}\\ f^{(n)}(0)=1.\\ \text{Therefore}\text{,}\text{the}\text{Maclaurin}\text{}\text{Series}\text{is}:\\ {\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}}{x}^n={\displaystyle \sum _{n=0}^{\infty }\frac{1}{n!}}{x}^n={\displaystyle \sum _{n=0}^{\infty }{x}^n=1+\frac{x}{1!}+\frac{x^2}{2!}}+\mathrm{.....}\\ So,a_n=\frac{x^n}{n!}\text{then},\\ a_{n+1}=\frac{x^{n+1}}{(n+1)!}\\ \left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}\right|=\frac{\left|x\right|}{(n+1)}\to 0<1\end{array}$

By the Ratio Test, the series convergence for all x so ,the radius of convergence is R = $\infty$ .

c.

To determine

To write: the Maclaurin Series and the interval of convergence for each of the following functions.

Answer to Problem 11RCC

  $\sin x={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}},R=\infty$

Explanation of Solution

Given information: Given functions is $f(x)=\sin x$ .

Calculation:

  $\begin{array}{c}f(x)=\sin x\Rightarrow f(0)=0.\\ f\text{'}(x)=\cos x\Rightarrow f\text{'}(0)=1.\\ f\text{'}\text{'}(x)=-\sin x\Rightarrow f\text{'}\text{'}(0)=0\\ f\text{'}\text{'}\text{'}(x)=-\cos x\Rightarrow f\text{'}\text{'}\text{'}(0)=-1.\\ \mathrm{.......}\\ \mathrm{.......}\\ \text{Since}\text{the}\text{derivatives}\text{repeat}\text{in}\text{a}\text{cycle}\text{of}\text{four,}\\ f(0)+\frac{f\text{'}(0)}{1!}+\frac{f\text{'}\text{'}(0)}{2!}+\mathrm{....}={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{2n+1}}{(2n+1)!}\\ \text{Therefore}\text{,}\text{the}\text{Maclaurin}\text{}\text{Series}\text{is}:\\ {\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}}{x}^n={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{2n+1}}{(2n+1)!}.\\ $ So,a_n={(-1)}^n\frac{x^{2n+1}}{(2n+1)!}\text{then},$$ a_{n+1}={(-1)}^{n+1}\frac{x^{2n+3}}{(2n+3)!}$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{{(-1)}^{n+1}\frac{x^{2n+3}}{(2n+3)!}}{{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}}\right|=\frac{1\times 1}{(2n+1)(2n+3)}\to 0<1\end{array}$

By the Ratio Test, the series convergence for all x so ,the radius of convergence is R = $\infty$ .

d.

To determine

To write: the Maclaurin Series and the interval of convergence for each of the following functions.

Answer to Problem 11RCC

  $\cos x={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n}}{(2n)!}},R=\infty$

Explanation of Solution

Given information: Given functions is $f(x)=\cos x$ .

Calculation:

  $\begin{array}{c}f(x)=\cos x\Rightarrow f(0)=1.\\ f\text{'}(x)=-\sin x\Rightarrow f\text{'}(0)=0.\\ f\text{'}\text{'}(x)=-\cos x\Rightarrow f\text{'}\text{'}(0)=-1\\ f\text{'}\text{'}\text{'}(x)=\sin x\Rightarrow f\text{'}\text{'}\text{'}(0)=0.\\ \mathrm{.......}\\ \mathrm{.......}\\ \text{Since}\text{the}\text{derivatives}\text{repeat}\text{in}\text{a}\text{cycle}\text{of}\text{four,}\\ f(0)+\frac{f\text{'}(0)}{1!}+\frac{f\text{'}\text{'}(0)}{2!}+\mathrm{....}={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{2n}}{(2n)!}\\ \text{Therefore}\text{,}\text{the}\text{Maclaurin}\text{}\text{Series}\text{is}:\\ {\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}}{x}^n={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{2n}}{(2n)!}.\\ $ So,a_n={(-1)}^n\frac{x^{2n}}{(2n)!}\text{then},$$ a_{n+1}={(-1)}^{n+1}\frac{x^{2n+2}}{(2n+2)!}$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{{(-1)}^{n+1}\frac{x^{2n+2}}{(2n+2)!}}{{(-1)}^n\frac{x^{2n}}{(2n)!}}\right|=\frac{1\times 1}{(2n+1)(2n+2)}\to 0<1\end{array}$

By the Ratio Test, the series convergence for all x so, the radius of convergence is R = $\infty$ .

e.

To determine

To write: the Maclaurin Series and the interval of convergence for each of the following functions.

Answer to Problem 11RCC

  ${\tan }^{-1}x={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{2n+1}}{(2n+1)!},R=1.$

Explanation of Solution

Given information: Given functions is ${\tan }^{-1}x$ .

Calculation:

  $\begin{array}{c}\text{Therefore}\text{,}\text{the}\text{Maclaurin}\text{}\text{Series}\text{is}:\\ {\displaystyle \sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}}{x}^n={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{2n+1}}{(2n+1)!}\\ $ So,a_n={(-1)}^n\frac{x^{2n+1}}{(2n+1)!}\text{then},$$ a_{n+1}={(-1)}^{n+1}\frac{x^{2n+3}}{(2n+3)!}$\left|\frac{a_n}{a_{n+1}}\right|=\left|\frac{{(-1)}^{n+1}\frac{x^{2n+3}}{(2n+3)!}}{{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}}\right|=\frac{\left|x^3\right|}{(2n+3)(2n+2)}\to \left|x^3\right|<1\\ \left|x\right|<1\end{array}$

By the Ratio Test, the series convergence for all x such that -1< x <1 so , the radius of convergence is R =1.

f.

To determine

To write: the Maclaurin Series and the interval of convergence for each of the following functions.

Answer to Problem 11RCC

  $\ln (1+x)={\displaystyle \sum _{n=0}^{\infty }{(-1)}^{n-1}}\frac{x^n}{n!},R=1.$

Explanation of Solution

Given information: Given functions is $\ln (1+x)$ .

Calculation:

  $\begin{array}{c}\ln (1+x)={\displaystyle \sum _{n=0}^{\infty }{(-1)}^{n-1}}\frac{x^n}{n!}\\ $ So,a_n={(-1)}^{n-1}\frac{x^n}{n!}\text{then},$$ a_{n+1}={(-1)}^n\frac{x^{n+1}}{(n+1)!}$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{{(-1)}^n\frac{x^{n+1}}{(n+1)!}}{{(-1)}^{n-1}\frac{x^n}{n!}}\right|=\frac{\left|x\right|}{(n+1)}\to \left|x\right|<1\end{array}$

By the Ratio Test, the series convergence for all x such that -1 < x<1 so ,the radius of convergence is R =1.