Q. 35 - Calculus

Question

Explore the Taylor series for the given pairs of functions, using these steps: (a) Find the Taylor series for the given function at the specified value of x 0 and determine the interval of convergence for the series. (b) Use Theorem 8.11 and your answer from part (a) to find the Taylor series for the given function for the same value of x 0. Also, find the interval of convergence for your series.

(a) $\frac{1}{1-x}$, $x_0=3$

(b) $\frac{1}{{\left(1-x\right)}^2}$

Step-by-Step Solution

Answer

The Taylor series for the given functions are (a) $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{1}{{\left(-2\right)}^{k+1}}{\left(x-3\right)}^k\end{array}$, $\left(1,5\right)$and (b) $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{k+1}{{\left(-2\right)}^{k+2}}{\left(x-3\right)}^k\end{array}$ $\left(1,5\right)$.

Step 1. Given information.

Consider the given functions are, (a) $\frac{1}{1-x}$, $x_0=3$ and (b) $\frac{1}{{\left(1-x\right)}^2}$.

Step 2. Use formula of Taylor series.

For a function with an nth-order derivative at the point $x_0$, the nth Taylor polynomial for function at $x_0$ will be, $P_n\left(x\right)=\sum _{k=0}^{\infty }\frac{f^{\left(k\right)}{x}_0}{k!}{\left(x-x_0\right)}^k$.

Step 3. Find Taylor polynomial for the function 1 1 - x .

The Taylor polynomial for the function $\frac{1}{\left(1-x\right)}$ is $P_n\left(x\right)=\sum _{k=0}^{\infty }\frac{1}{{\left(1-x\right)}^{k+1}}{\left(x-x_0\right)}^k$.

Find Taylor polynomial for $x_0=3$.

  $\begin{array}{rcl}{P}_n\left(x\right)& =& \sum _{k=0}^{\infty }\frac{1}{{\left(1-3\right)}^{k+1}}{\left(x-3\right)}^k\\ & =& \sum _{k=0}^{\infty }\frac{1}{{\left(-2\right)}^{k+1}}{\left(x-3\right)}^k\end{array}$ 

Step 4. Find interval of convergence.

$\begin{array}{rcl}\underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|& =& \underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{1}{{\left(-2\right)}^{k+2}}}{\left(x-3\right)}^{k+1}}{\frac{1}{{\left(-2\right)}^{k+1}}{\left(x-3\right)}^k}\right|\\ & =& \underset{k\to \infty }{\lim }\left|\frac{1}{\left(-2\right)}\left(x-3\right)\right|\\ & =& \frac{1}{2}\left|-x+3\right|\end{array}$

To find interval of convergence $\begin{array}{rcl}& & \frac{1}{2}\end{array}\begin{array}{rcl}& & \left|-x+3\right|\end{array}<1$.

$\begin{array}{rcl}-2& <& -x+3<2\\ 2& >& x-3>-2\\ 5& >& x>1\\ & & \end{array}$

Therefore, the interval of convergence will be $\left(1,5\right)$.

Step 5. Apply theorem.

Differentiation of a power series is used if $\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k$ is power series in $\left(x-x_0\right)$ that converges to a function on an interval I, then derivative of function can be written as,

 $\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& \frac{d}{dx}\left(\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k\right)\\ & =& \sum _{k=0}^{\infty }k{a}_k{\left(x-x_0\right)}^{k-1}\end{array}$.

Step 6. Find differentiation of power series.

Apply theorem into the power series.

$\begin{array}{rcl}\frac{d}{dx}\left(\sum _{k=0}^{\infty }\frac{1}{{\left(-2\right)}^{k+1}}{\left(x-3\right)}^k\right)& =& \sum _{k=0}^{\infty }\frac{d}{dx}\left(\frac{1}{{\left(-2\right)}^{k+1}}{\left(x-3\right)}^k\right)\\ & =& \sum _{k=0}^{\infty }\frac{1}{{\left(-2\right)}^{k+1}}\frac{d}{dx}\left({\left(x-3\right)}^k\right)\\ & =& \sum _{k=0}^{\infty }\frac{k}{{\left(-2\right)}^{k+1}}{\left(x-3\right)}^{k-1}\\ & =& \sum _{k=0}^{\infty }\frac{k+1}{{\left(-2\right)}^{k+2}}{\left(x-3\right)}^k\end{array}$

Step 7. Find Taylor series for 1 1 - x 2 .

It can be observed that $\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{{\left(1-x\right)}^2}$

The Taylor series for $\frac{d}{dx}\left(\frac{1}{1-x}\right)$ is $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{k+1}{{\left(-2\right)}^{k+2}}{\left(x-3\right)}^k\end{array}$.

Therefore, the Taylor series for $\frac{1}{{\left(1-x\right)}^2}$ will be $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{k+1}{{\left(-2\right)}^{k+2}}{\left(x-3\right)}^k\end{array}$ and the interval of convergence will be $\left(1,5\right)$.