Q. 16 - Calculus

Question

Perform the following steps for the power series in $x-x_0$ in $\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(x-7)}^{2k}$

Step-by-Step Solution

Answer

The power series in $x-x_0$ for $F\left(x\right)={\int }_x^{x_0}f\left(t\right)dt$ is $F\left(x\right)=\sum _{k=2}^{\infty }{(-1)}^{k-1}\frac{(k-1)!}{(2k-1)!}{(x-7)}^{2k-1}$ 

To find the interval of convergence of the power series, use the ratio test for absolute convergence

Let $b_k={(-1)}^k\frac{k!}{\left(2k\right)!}{(x-7)}^{2k}$

So, $b_{k+1}={(-1)}^{k+1}\frac{(k+1)!}{(2k+2)!}{(x-7)}^{2k+2}$

Therefore, $\underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\frac{-(k+1)}{(2k+2)(2k+1)}{\left|x-7\right|}^2$

Now, by the ratio test for absolute convergence, the series will converge only when ${\left|x-7\right|}^2<1$

Therefore $x\in (6,8)$

When $x=6$

$$\begin{gathered} \sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(x-7)}^{2k}=\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(6-7)}^{2k} \\ =\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(-1)}^{2k} \\ =\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!} \end{gathered}$$

This series will converge.

When 8

$$\begin{gathered} \sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(x-7)}^{2k}=\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(8-7)}^{2k} \\ =\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{\left(1\right)}^{2k} \\ =\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!} \end{gathered}$$

This series will converge.

Therefore, the interval of convergence of power series is $[6,8]$

Let us take the derivative of the function f ( x )

Therefore,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(x-7)}^{2k}\right] \\ {f}^{\text{'}}\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}\frac{d}{dx}{(x-7)}^{2k} \\ {f}^{\text{'}}\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}2k{(x-7)}^{2k-1} \\ {f}^{\text{'}}\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{(2k-1)!}{(x-7)}^{2k-1} \end{gathered}$$

Now we change the index in the final step

So, the power series in $x-x_0$ for $f^{\text{'}}$ is

$f^{\text{'}}\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^{k+1}\frac{(k+1)!}{(2k+1)!}{(x-7)}^{2k+1}$

To find the power series in x - x 0 for F , let us integrate the function f ( x ) from x 0 to x

Therefore,

$$\begin{gathered} F\left(x\right)={\int }_{x_0}^x\left[\sum _{k=0}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(t-7)}^{2k}\right]dt \\ F\left(x\right)=\sum _{k=1}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{\int }_{x_0}^x{(t-7)}^{2k}dt \end{gathered}$$

Thus,

$F\left(x\right)=\sum _{k=1}^{\infty }{(-1)}^k\frac{k!}{\left(2k\right)!}{(x-7)}^{2k+1}$

So, change the index in the final step :-

$F\left(x\right)=\sum _{k=2}^{\infty }{(-1)}^{k-1}\frac{(k-1)!}{(2k-1)!}{(x-7)}^{2k-1}$