Q. 27 - Calculus

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. 

27. $\left(\mathrm{a}\right) \frac{1}{1 + 8x^3} \left(\mathrm{b}\right) {\frac{x^2}{ {(1 + 8x^3)}^2}}^{}$

Step-by-Step Solution

Answer

a) The interval of convergence for the series is $\boxed{\mathbf{(}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}}$

b) The Maclaurin series for the given function is $\boxed{\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{3}\mathbf{(}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{)}}\mathit{k}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{3}\mathbf{k}\mathbf{-}\mathbf{1}}}$

Part (a) Step 1 : Given Information

The equation is $\frac{1}{1 + 8x^3}$.

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 in the Maclaurin series.

Therefore,

  $$\begin{gathered} \frac{1}{1 + 8x^3}=\sum _{k=0}^{\infty }{(-8x^3)}^k \\ \Rightarrow \frac{1}{1 + 8x^3}=\boxed{\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{k}}\mathbf{ }{\mathbf{8}}^{\mathbf{k}}\mathbf{ }{\mathit{x}}^{\mathbf{3}\mathbf{k}}} \end{gathered}$$

Part (a) Step 3: Simplification

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

Take

 $$\begin{gathered} b_k={(-1)}^k{8}^k{x}^{3k} \\ \Rightarrow b_{k+1}={(-1)}^{k+1}{8}^{k+1}{x}^{3k+3} \end{gathered}$$

As a result,

 $$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{(-1)}^{k+1}{8}^{k+1}{x}^{3k+3}}{{(-1)}^k{8}^k{x}^{3k+1}}\right| \\ =\underset{k\to \infty }{\lim }\left|(-1)8x^3\right| \end{gathered}$$

Because we're evaluating the limit $k\to \infty$, its value is $\left|\mathbf{-}\mathbf{ }\mathbf{8}{\mathit{x}}^{\mathbf{3}}\right|$.

The series will converge according to the ratio test of absolute convergence when $-\frac{1}{2}<x<\frac{1}{2}$

As a result, the convergence interval will include the value $(-\frac{1}{2},\frac{1}{2})$

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

As a result, for $x =-\frac{1}{2}$, the series becomes $$\begin{gathered} \frac{1}{1 + 8x^3}=\sum _{k=0}^{\infty }{(-1)}^k 8^k {(-\frac{1}{2})}^{3k} \\ =\sum _{k=0}^{\infty }\left(1\right) \end{gathered}$$

This means that the series will diverge.

Similarly, the series will diverge for $x=\frac{1}{2}$

So, the interval convergence is $\boxed{\mathbf{(}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}}$

Part (b) Step 1: Given information

The equation is$\frac{x^2}{ {(1 + 8x^3)}^2}$

Part (b) Step 2: Simplification

 The Maclaurin series for the function $f\left(x\right)$ is:

$\frac{1}{1+8x^3}=\sum _{k=0}^{\infty }(-1{)}^k{8}^k{x}^{3k}$

Now, the derivate of the function $f\left(x\right)$ is

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{1}{1+8x^3}\right) \\ =\frac{\left(1+8x^3\right)0-1\left(24x^2\right)}{{\left(1+8x^3\right)}^2} \\ =\frac{-24x^2}{{\left(1+8x^3\right)}^2} \end{gathered}$$

Thus, the Maclaurin series for $\frac{3x^2}{{\left(1-x^3\right)}^2}$ is:

$\begin{array}{rcl}\frac{-24x^2}{{\left(1+8x^3\right)}^2}& =& \frac{d}{dx}\left(\sum _{k=0}^{\infty }(-2{)}^{3k}{x}^{3k}\right)\\ & =& \sum _{k=0}^{\infty }(-2{)}^{3k}\frac{d}{dx}\left(x^{3k}\right)\\ & =& \sum _{k=0}^{\infty }(-2{)}^{3k}3k\left(x^{3k-1}\right)\end{array}$

This implies that the Maclaurin series for$\frac{x^2}{{\left(1+8x^3\right)}^2}$ is:

$\begin{array}{rcl}\frac{x^2}{{\left(1+8x^3\right)}^2}& =& -\frac{1}{24}\sum _{k=0}^{\infty }(-2{)}^{3k}3k\left(x^{3k-1}\right)\\ & =& \sum _{k=0}^{\infty }(-2{)}^{3(k-1)}k\left(x^{3k-1}\right)\end{array}$