Q . 13 - 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 $x-x_0 \mathrm{for} f^{\text{'}}.$

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

13. $\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^k$

Step-by-Step Solution

Answer

(a). The interval of convergence of the power series is  every value of $x$

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

$f^{\text{'}}\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^{k+1}(k+1)}{(2k+2)!}{x}^k$

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

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

Part(a) Step 1 : Given information

Given function : $\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^k$

Part (a): Step 2 : Simplification

Let us first assume 

$$\begin{gathered} b_k=\frac{(-1{)}^k}{\left(2k\right)!}{x}^k \\ \Rightarrow b_{k+1}=\frac{(-1{)}^{k+1}}{\left[2\right(k+1\left)\right]!}{x}^{k+1} \end{gathered}$$

Therefore,

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

Now, for $k\to \infty , \underset{k\to \infty }{\lim }\frac{-1}{(2k+2)(2k+1)}\left|x\right|=0$, that is the value of limit will be zero no matter what value the variable$x$ takes.

Hence, by the ratio test the series converges absolutely for every value of $x$.

Part(b) Step 1 : Given information

Given function : $\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^k$

Part (b): Step 2 : Simplification

Since $f\left(x\right)=\sum _{k=0}^5\frac{(-1{)}^k}{\left(2k\right)!}{x}^4$ , so to find the power series in $x-x_0$ for $f\text{'}$, let us take the derivative of the function $f\left(x\right)$

Therefore,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^k\right] \\ =\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}\frac{d}{dx}{x}^k \\ =\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}k{x}^{k-1} \\ \mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}}\mathbf{k}}{\mathbf{\left(}\mathbf{2}\mathbf{k}\mathbf{\right)}\mathbf{!}}{\mathit{x}}^{\mathbf{k}\mathbf{-}\mathbf{1}} \end{gathered}$$

Part(c) Step 1 : Given information

Given function : $\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^k$

Part (c): Step 2 : Simplification

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

Therefore,

$$\begin{gathered} F\left(x\right)={\int }_{x_0}^x\left[\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{t}^k\right]dt \\ =\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\int }_{k_k}\left[t^k\right]dt \\ =\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\left[\frac{t^{k+1}}{k+1}\right]}_{k_0}^x \end{gathered}$$

here, $x_0=0$

Thus,

$F\left(x\right)=\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}}}{\mathbf{\left(}\mathbf{2}\mathbf{k}\mathbf{\right)}\mathbf{!}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathit{x}}^{\mathbf{k}\mathbf{+}\mathbf{1}}$