Q. 12 - 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$F\left(x\right)={\int }_{x_0}^{x}f\left(t\right)dt$

12. $\sum _{k=0}^{\infty }\frac{5^k}{k!}(x-3{)}^k$

Step-by-Step Solution

Answer

(a). The interval of convergence of the power series  is $\mathrm{ℝ}$

(b). The power series in $x-x_0$ for $f^{\text{'}}$ is $f^{\text{'}}\left(x\right)=\sum _{k=0}^{\infty }\frac{5^{k+1}}{k!}(x-3{)}^k$

(c). The power series in $x-x_0$ for $F\left(x\right)={\int }_{x_0}^{x}f\left(t\right)dt$ is $F\left(x\right)=\sum _{k=1}^{\infty }\frac{5^{k-1}}{k!}(x-3{)}^k$

Part(a) Step 1 : Given information

Given function : $\sum _{k=0}^{\infty }\frac{5^k}{k!}(x-3{)}^k$

Part (a): Step 2 : Simplification

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

Let us first assume 

$$\begin{gathered} b_k=\frac{5^k}{k!}(x-3{)}^k \\ \Rightarrow b_{k+1}=\frac{5^{k+1}}{(k+1)!}(x-3{)}^{k+1} \end{gathered}$$

Therefore,

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{\frac{5^{k+1}(x-3{)}^{k+1}}{(k+1)!}}{\frac{5^k(x-3{)}^k}{k!}}\right| \\ =\underset{k\to \infty }{\lim }\frac{5}{k+1}|x-3| \end{gathered}$$

Now, for $k\to \infty , \underset{k\to \infty }{\lim }\frac{5}{k+1}|x-3|=0,$ 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 $x$.

Therefore, the interval of convergence of the power series is $\mathbf{ℝ}$

Part(b) Step 1 : Given information

Given function : $\sum _{k=0}^{\infty }\frac{5^k}{k!}(x-3{)}^k$

Part (b): Step 2 : Simplification

Since $f\left(x\right)=\sum _{k=0}^{\infty }\frac{5^k}{k!}(x-3{)}^k$ , 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,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\sum _{k=0}^{\infty }\frac{5^k}{k!}(x-3{)}^k\right] \\ =\sum _{k=0}^{\infty }\frac{5^k}{k!}\frac{d}{dx}(x-3{)}^k \\ =\sum _{k=0}^{\infty }\frac{5^k}{k!}k(x-3{)}^{k-1} \\ =\sum _{k=0}^{\infty }\frac{5^k}{(k-1)!}(x-3{)}^{k-1} \end{gathered}$$

Now, we change the index in the final step So, the power series in $x-x_0$ for $f\text{'}$ is $f^{\text{'}}\left(x\right)=\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{5}}^{\mathbf{k}\mathbf{+}\mathbf{1}}}{\mathbf{k}\mathbf{!}}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{3}{\mathbf{)}}^{\mathbf{k}}$

Part(c) Step 1 : Given information

Given function : $\sum _{k=0}^{\infty }\frac{5^k}{k!}(x-3{)}^k$

Part (c): Step 2 : Simplification

Also, to find the power series in $x-x_0$ for $F$, let us integrate the function from $x_0 \mathrm{to} x$ 

Therefore,

$$\begin{gathered} F\left(x\right)={\int }_{x_0}^x\left[\sum _{k=0}^{\infty }\frac{5^i}{k!}(t-3{)}^k\right]dt \\ =\sum _{k=0}^{\infty }\frac{5^k}{k!}{\int }_{x_0}^x\left[(t-3{)}^k\right]dt \\ =\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{5}}^{\mathbf{k}}}{\mathbf{k}\mathbf{!}}{\left[\frac{(t-3{)}^{k+1}}{k+1}\right]}_{{\mathbf{x}}_{\mathbf{0}}}^{\mathbf{x}} \end{gathered}$$

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

$\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{5}}^{\mathbf{k}}}{\mathbf{k}\mathbf{!}}{\mathbf{\left[}\frac{\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{3}{\mathbf{)}}^{\mathbf{k}\mathbf{+}\mathbf{1}}}{\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{\right]}}_{{\mathbf{x}}_{\mathbf{0}}}^{\mathbf{x}}$