Q. 39 - Calculus

Question

In Exercises 35–40, 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.

\((a)\frac{1}{1-3x},x_{0}=3\)

\((b)\frac{1}{\left ( 1-3x \right )^{2}}\) 

Step-by-Step Solution

Answer

Part (a) The Taylor series for the given function is \(f(x)=-\frac{1}{8}\sum_{k=0}^{\infty}\left ( -\frac{3}{8} \right )^{k}\left ( x-3 \right )^{k}\) and the interval of convergence of the series is \(\left ( \frac{1}{3},\frac{17}{3} \right )\). 

Part (b) The Taylor series for the given function is \(\begin{aligned}\frac{1}{(1-3 x)^2}&=-\frac{1}{24} \sum_{k=0}^{\infty}\left(-\frac{3}{8}\right)^{k+1} (k+1)(x-3)^{k}\end{aligned}\) and the interval of convergence of the series is \(\left ( \frac{1}{3},\frac{17}{3} \right )\). 

Part (a) Step 1. Given Information

We have to find the Taylor series for the given function at the value of \(x_{0}=3\)  and determine the interval of convergence for the series.

Part (a) Step 2. Find the Taylor series

To find the Taylor series we will use the formula \(f(x)=f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(x-x_0\right)+f^{\prime \prime}\left(x_0\right) \frac{\left(x-x_0\right)^2}{2 !}+f^{\prime \prime}\left(x_0\right) \frac{\left(x-x_0\right)^3}{3 !}+\ldots\).

So,

\(f\left(x_0\right)=\frac{1}{1-3(3)}=-\frac{1}{8}\)

And

\(\begin{aligned}f^{\prime}(x) &=\frac{d}{d x}[f(x)] \\&=\frac{d}{d x}\left(\frac{1}{1-3 x}\right) \\&=\frac{(1-3 x) 0-1(-3)}{(1-3 x)^2} \\&=\frac{3}{(1-3 x)^2}\end{aligned}\)

At \(x=3\)

\(\begin{aligned}f^{\prime}(3) &=\frac{3}{(1-3 \left ( 3 \right ))^2}\\f^{\prime}(3) &=\frac{3}{64}\end{aligned}\)

Part (a) Step 3. Find the Taylor series

Now, the second derivative

\(\begin{aligned}f^{\prime \prime}(x) &=\frac{d}{d x}\left[f^{\prime}(x)\right] \\&=\frac{d}{d x}\left[\frac{3}{(1-3 x)^2}\right] \\&=\frac{18}{(1-3 x)^3}\end{aligned}\)

At \(x=3\)

\(\begin{aligned}f^{\prime \prime}(3)&=\frac{18}{(1-3 \left ( 3 \right ))^3}\\f^{\prime \prime}(3)&=\frac{18}{512}\\f^{\prime \prime}(3)&=\frac{9}{256}\\\end{aligned}\)

Now, the third derivative

\(\begin{aligned}f^{\prime \prime \prime}(x) &=\frac{d}{d x}\left[f^{\prime \prime}(x)\right] \\&=\frac{d}{d x}\left[\frac{18}{(1-3 x)^3}\right] \\&=\frac{162}{(1-3 x)^4}\end{aligned}\)

At \(x=3\)

\(\begin{aligned}f^{\prime \prime \prime}(3)&=\frac{162}{(1-3 (3))^4}\\f^{\prime \prime \prime}(3)&=\frac{162}{(8)^4}\\f^{\prime \prime \prime}(3)&=\frac{81}{(2048}\\\end{aligned}\)

Therefore, the function is

\(\begin{aligned}f(x) &=-\frac{1}{8}+\frac{3}{64}(x-3)+\frac{9}{256}\left[\frac{(x-3)^2}{2 !}\right]+\frac{81}{2048}\left[\frac{(x-3)^3}{3 !}\right]+\ldots \\&=-\frac{1}{8}+\frac{3}{64}(x-3)+\frac{9}{512}(x-3)^2+\frac{27}{4096}(x-3)^3+\ldots\end{aligned}\)

We can write the function as

\(f(x)=-\frac{1}{8}\sum_{k=0}^{\infty}\left ( -\frac{3}{8} \right )^{k}\left ( x-3 \right )^{k}\).

Part (a) Step 4. Determine the interval of convergence for the series

We will use the ratio test for absolute convergence to determine the interval of convergence for the series.

Now, let \(a_{k}=\left ( -\frac{3}{8} \right )^{k}\left ( x-3 \right )^{k} so, a_{k+1}=\left ( -\frac{3}{8} \right )^{k+1}\left ( x-3 \right )^{k+1}\).

So,

\(\begin{aligned}\lim _{k \rightarrow \infty}\left|\frac{a_{k+1}}{a_k}\right| &=\lim _{k \rightarrow \infty}\left|\frac{\left(-\frac{3}{8}\right)^{k+1}(x-3)^{k+1}}{\left(-\frac{3}{8}\right)^k(x-3)^k}\right| \\&=\lim _{k \rightarrow \infty}\left|-\frac{3}{8}(x-3)\right|\end{aligned}\)

Now, by the ratio test of absolute convergence, the series will converge when \(\left|-\frac{3}{8}(x-3)\right|< 1\).

So, 

\(-\frac{8}{3}< x-3< \frac{8}{3}\)

\(x> \frac{1}{3} and x< \frac{17}{3}\)

Hence, the interval of convergence of the series is \(\left ( \frac{1}{3},\frac{17}{3} \right )\).

Part (b) Step 1. Find the Taylor series

From part (a) we find that the Taylor series for the function is \(f(x)=-\frac{1}{8}\sum_{k=0}^{\infty}\left ( -\frac{3}{8} \right )^{k}\left ( x-3 \right )^{k}\).

Now, the derivative of the function \(f(x)=\frac{1}{1-3x}\) is

\(\begin{aligned}f^{\prime}(x) &=\frac{d}{d x}\left(\frac{1}{1-3 x}\right) \\&=\frac{(1-3 x) 0-1(-3)}{(1-3 x)^2} \\&=\frac{3}{(1-3 x)^2}\end{aligned}\)

So, the Taylor series for \(\frac{3}{\left ( 1-3x \right )^{2}}\) is 

\(\begin{aligned}\frac{3}{(1-3 x)^2} &=\frac{d}{d x}\left[-\frac{1}{8} \sum_{k=0}^{\infty}\left(-\frac{3}{8}\right)^k(x-3)^k\right] \\&=-\frac{1}{8} \sum_{k=0}^{\infty}\left(-\frac{3}{8}\right)^k \frac{d}{d x}\left[(x-3)^k\right] \\&=-\frac{1}{8} \sum_{k=0}^{\infty}\left(-\frac{3}{8}\right)^k k(x-3)^{k-1}\end{aligned}\)

This means the Taylor series for  \(\frac{1}{\left ( 1-3x \right )^{2}}\) is \(\begin{aligned}\frac{1}{(1-3 x)^2}&=-\frac{1}{24} \sum_{k=0}^{\infty}\left(-\frac{3}{8}\right)^{k+1} (k+1)(x-3)^{k}\end{aligned}\).

Part (a) Step 2. Determine the interval of convergence for the series

We will use the ratio test for absolute convergence to determine the interval of convergence for the series.

Now, let \(a_{k}=\left ( -\frac{3}{8} \right )^{k+1}\left ( x-3 \right )^{k} so, a_{k+1}=\left ( -\frac{3}{8} \right )^{k+2}\left ( x-3 \right )^{k+1}\).

So,

\(\begin{aligned}\lim _{k \rightarrow \infty}\left|\frac{a_{k+1}}{a_k}\right| &=\lim _{k \rightarrow \infty}\left|\frac{\left(-\frac{3}{8}\right)^{k+2}(k+2)(x-3)^{k+1}}{\left(-\frac{3}{8}\right)^{k+1}(k+1)(x-3)^k}\right| \\&=\lim _{k \rightarrow \infty}\left|-\frac{3}{8}\frac{k+2}{k+1}(x-3)\right|\end{aligned}\)

Now, by the ratio test of absolute convergence, the series will converge when \(\left|-\frac{3}{8}(x-3)\right|< 1\).

So, 

\(-\frac{8}{3}< x-3< \frac{8}{3}\)

\(x> \frac{1}{3} and x< \frac{17}{3}\)

Hence, the interval of convergence of the series is \(\left ( \frac{1}{3},\frac{17}{3} \right )\).