Question

In Exercises 25–30, find Maclaurin series for the given pairs of functions, using these steps:

(a) Use substitution and/or multiplication and the Maclaurin series for $\frac{1}{1-x}$to find the Maclaurin series for the given function. Also, provide the interval of convergence for the series you found.

(b) Use Theorem 8.11 and your answer from part (a) to find the Maclaurin series for the given function. Also, provide the interval of convergence for the series you found.

28.$\left(\mathrm{a}\right)\frac{1}{3-x}\left(\mathrm{b}\right)\frac{1}{{(3-x)}^2}$

Short answer

a) The interval of convergence for the series is$\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}$

b) The Maclaurin series for the given function is $\boxed{\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{k}\mathbf{-}\mathbf{1}}}$

Step-by-step solution

Part (a) Step 1 : Given Information

Given equation :

$\left(\mathrm{a}\right)\frac{1}{3-x}$

Part (a) Step 2: Simplification

The Maclaurin series for the function $g\left(x\right)=\frac{1}{1-x}\mathrm{is}\frac{1}{1-x}=\underset{k=0}{\overset{\infty }{\sum x^k}}$

To determine the Maclaurin series for the function $f\left(x\right)$using substitution and/or multiplication and the Maclaurin series for $\frac{1}{1-x}$, we substitute $x$by$x-2$in the Maclaurin series.

Therefore,

$$\begin{gathered} \frac{1}{1-(x-2)}=\sum _{k=0}^{\infty }{(x-2)}^k \\ \Rightarrow \frac{1}{3-x}=\boxed{\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathit{k}}} \end{gathered}$$

Part (a) Step 3: Simplification

To get the interval convergence, we must first find the absolute convergence using the ratio test.

Take $$\begin{gathered} b_k={(x-2)}^k \\ \Rightarrow b_{k+1}={(x-2)}^{k+1} \end{gathered}$$

As a result,

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

Because we're evaluating the limit $k\to \infty$, its value is $\left|x-2\right|$.

The series will converge according to the ratio test of absolute convergence when $1<x<3$

As a result, the convergence interval will include the value $(1,3)$

We examine the behavior of the series at the interval's endpoints because it is a finite interval.

As a result, for $x=1$, the series becomes

$$\begin{gathered} \frac{1}{3-x}=\sum _{k=0}^{\infty }{(1-2)}^k \\ =\sum _{k=0}^{\infty }{(-1)}^k \end{gathered}$$

This means that the series will converge.

Similarly, the series will diverge for $x=3$

So, the interval convergence is $\boxed{\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}}$

Part (b) Step 1: Given information

Given equation :$\left(\mathrm{b}\right)\frac{1}{{(3-x)}^2}$

Part (b) Step 2: Given information

Finding the derivative of the function, we have :

$f\text{'}\left(x\right)=\frac{1}{{(3-x)}^2}$

Hence, the Maclaurin series for the function is

$\boxed{\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{k}\mathbf{-}\mathbf{1}}}$