Q. 15 - Calculus

Question

Perform the following steps for the power series in $x-x_0$ in Exercises 11-16:

(a) Find the interval of convergence, $I$, for the series.

(b) Let $f$ be the function to which the series converges on $I$. Find the power series in $x-x_0$ for $f^{\text{'}}$.

(c) Find the power series in $x-x_0$ for $F\left(x\right)={\int }_{x_0}^{x}f\left(t\right)dt$.

15. $f\left(x\right)=\sum _{i=1}^{\infty }\frac{(-1{)}^d}{k}(x+5{)}^k$

Step-by-Step Solution

Answer

(a). The interval of convergence of the power series is $\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{]}$

(b). The power series in $x-x_0$ for $f^{\text{'}}$ is

${\mathit{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{5}{\mathbf{)}}^{\mathbf{k}}$

(c). The power series in $x-x_0$ for $F\left(x\right)={\int }_{x_0}^{x}f\left(t\right)dt$  is $\mathit{F}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{2}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}\mathbf{-}\mathbf{1}}}{\mathbf{k}\mathbf{(}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{5}{\mathbf{)}}^{\mathbf{k}}$

Part(a) Step 1 : Given information

Given function : $f\left(x\right)=\sum _{i=1}^{\infty }\frac{(-1{)}^d}{k}(x+5{)}^k$

Part (a): Step 2 : Simplification

Let us first assume

$$\begin{gathered} b_k=\frac{(-1{)}^k}{k}(x+5{)}^k \\ \Rightarrow b_{k+1}=\frac{(-1{)}^{k+1}}{k+1}(x+5{)}^{k+1} \end{gathered}$$

Therefore,

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\mid \frac{\frac{(-1{)}^{k+1}}{k+1}(x+5{)}^{k+1}}{(-1{)}^k}(x+5{)}^k \\ =\underset{k\to \infty }{\lim }\frac{-k}{(k+1)}|x+5| \end{gathered}$$

Now, by the ratio test for absolute convergence, the series will converge only when $|x+5|<1$

Therefore, $x\in (-6,-4)$

Now, we check the series at the end points

So, when $x=-6$

$$\begin{gathered} \sum _{k=1}^{\infty }\frac{(-1{)}^k}{k}(-6+5{)}^k=\sum _{k=1}^{\infty }\frac{(-1{)}^{2k}}{k} \\ =\sum _{k=1}^{\infty }\frac{1}{k} \end{gathered}$$

This series will diverge.

When $x=4$

$\sum _{k=1}^{\infty }\frac{(-1{)}^k}{k}(4+5{)}^k=\sum _{k=1}^{\infty }\frac{(-1{)}^k{9}^k}{k}$

This series will converge.

Therefore, the interval of convergence of the power series is $\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{]}$

Part(b) Step 1 : Given information

Given function : $f\left(x\right)=\sum _{i=1}^{\infty }\frac{(-1{)}^d}{k}(x+5{)}^k$

Part (b): Step 2 : Simplification

Derivative of the function $f\left(x\right)$

Therefore,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\sum _{i=1}^{\infty }\frac{(-1{)}^k}{k}(x+5{)}^k\right] \\ =\sum _{i=1}^{\infty }\frac{(-1{)}^d}{k}\frac{d}{dx}(x+5{)}^k \\ =\sum _{i=1}^{\infty }\frac{(-1{)}^k}{k}k(x+5{)}^{k-1} \\ =\sum _{i=1}^{\infty }(-1{)}^k(x+5{)}^{k-1} \end{gathered}$$

Now, we change the index in the final step. So, the power series in $x-x_0$ for $f\text{'}\left(x\right)$  is 

$f^{\text{'}}\left(x\right)=\mathbf{\sum }_{\mathbf{t}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{5}{\mathbf{)}}^{\mathbf{k}}$

Part(c) Step 1 : Given information

Given function : $f\left(x\right)=\sum _{i=1}^{\infty }\frac{(-1{)}^d}{k}(x+5{)}^k$

Part (c): Step 2 : Simplification

Also, to find the power series in $x-x_0$ for $F$ , let us integrate the function$f\left(x\right)$ from $x_0$ to $x$

Therefore,

$$\begin{gathered} F\left(x\right)={\int }_k^x\left[\sum _{i=1}^k\frac{(-1{)}^k}{k}(t+5{)}^k\right]dt \\ =\sum _{k=1}^5\frac{(-1{)}^k}{k}{\int }_{5_0}^k\left[(t+5{)}^k\right]dt \\ =\sum _{k=1}^k\frac{(-1{)}^k}{k}{\left[\frac{(t+5{)}^{k+1}}{k+1}\right]}_k^k \end{gathered}$$

Thus,

$F\left(x\right)=\sum _{k=1}^{\infty }\frac{(-1{)}^2}{k(k+1)}(x+5{)}^{k+1}$

Now, we change the index in the final step

So, the power series in $x-x_0$ for $F\left(x\right)={\int }_{x_0}^{x}f\left(t\right)dt$ is

$F\left(x\right)=\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{2}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}\mathbf{-}\mathbf{1}}}{\mathbf{k}\mathbf{(}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{5}{\mathbf{)}}^{\mathbf{k}}$