Chapter 8.7, Problem 15E

To determine

To find: The Taylor series for $f\left(x\right)$ cantered at $a$ .

Answer to Problem 15E

Theratio test shows that the series converges for all the value of $x$ therefore the radius of the convergence sis $R=\infty$ and the required series is ${\displaystyle \sum _{n\to 0}^{\infty }\frac{e^3}{n!}{\left(x-3\right)}^n}$ .

Explanation of Solution

Given:

Thegiven function is $f\left(x\right)=\cos x$ .

The series is centred at $a=\pi$ .

Calculation:

Consider the series that is centred at $a$ is of the form.

  $\begin{array}{c}f\left(x\right)=f\left(a\right)+\frac{f^{\prime }\left(a\right)}{1!}+\frac{f^{\prime }\left(a\right)}{2!}{\left(x-a\right)}^2+\mathrm{....}\\ ={\displaystyle \sum _{n!}^{\infty }\frac{f^{\left(n\right)}\left(a\right)}{n!}}{\left(x-a\right)}^n\end{array}$

Then, for the given function as,

  $\begin{array}{l}f\left(x\right)=\cos x\\ f\left(\pi \right)=\cos \left(\pi \right)\\ f\left(\pi \right)=-1\end{array}$

Differentiate again,

  $\begin{array}{l}{f}^{\prime }\left(\pi \right)=-\sin \left(\pi \right)\\ f\left(\pi \right)=0\end{array}$

Differentiate again,

  $\begin{array}{l}{f}^{″}\left(\pi \right)=-\cos \left(\pi \right)\\ f\left(\pi \right)=1\end{array}$

Differentiate again,

  $\begin{array}{l}{f}^{‴}\left(\pi \right)=\sin \pi \\ f^{‴}\left(\pi \right)=0\end{array}$

The differentiation continues in this way.

Then, the series is,

  $\begin{array}{c}f\left(\pi \right)+\frac{f^{\prime }\left(\pi \right){\left(x-\pi \right)}^2}{1!}+\frac{f^{″}\left(\pi \right){\left(x-\pi \right)}^2}{2!}+\frac{f^{‴}\left(\pi \right){\left(x-\pi \right)}^3}{3!}+\frac{f^4\left(\pi \right){\left(x-\pi \right)}^3}{4!}\mathrm{....}=\left[\begin{array}{l}\left(-1\right)+\frac{{\left(x-\pi \right)}^2}{2!}\\ +\frac{{\left(x-\pi \right)}^4}{4!}+\mathrm{...}\end{array}\right]\\ ={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^{n+1}}\frac{{\left(x-\pi \right)}^{2\pi }}{\left(2n\right)!}\end{array}$

Then substitute $a=3$ in the series as,

  $\begin{array}{c}f\left(x\right)={\displaystyle \sum _{n\to 0}^{\infty }\frac{f^n\left(a\right)}{n!}}{\left(x-a\right)}^n\\ ={\displaystyle \sum _{n\to 0}^{\infty }\frac{f^n\left(3\right)}{n!}}{\left(x-3\right)}^n\\ ={\displaystyle \sum _{n\to 0}^{\infty }\frac{e^3}{n!}{\left(x-3\right)}^n}\end{array}$

Consider $a_n=\frac{{\left(-1\right)}^{n+1}{\left(x-\pi \right)}^{2n}}{\left(2n\right)!}$ then,

  $\begin{array}{c}\underset{n\to \infty }{\text{lim}}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\text{lim}}\left|\frac{{\left(-1\right)}^{n+1}{\left(x-\pi \right)}^{2n}}{\left(2n+2\right)!}\times \frac{\left(2n\right)!}{{\left(-1\right)}^{n+1}{\left(x-\pi \right)}^{2\pi }}\right|\\ =\underset{n\to \infty }{\text{lim}}\frac{{\left|x-\pi \right|}^2}{\left(2n+2\right)\left(2n+\right)}\\ =0<1\text{for}\text{all}x\end{array}$

Thus, the ratio test shows that the series converges for all the value of $x$ therefore the radius of the convergence is $R=\infty$ .