Q.11 - 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 $f$

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

11. $\sum _{k=0}^{\infty }{2}^k{x}^k$

Step-by-Step Solution

Answer

(a). The interval of convergence of the power series$f\left(x\right)=\sum _{k=0}^{\infty }\frac{1}{k+2}{x}^{3+1}$

(b). The power series in$x-x_0$ for $f^{\text{'}}$is $f^{\text{'}}\left(x\right)=\sum _{k=1}^{\infty }\frac{3k+4}{k+3}{x}^{3+1}$

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

Part(a) Step 1 : Given information

Given function :$f\left(x\right)=\sum _{k=0}^x{2}^k{x}^k$

Part (a): Step 2 : Simplification

Consider the power series $f\left(x\right)=\sum _{k=0}^x{2}^k{x}^k$

The power series contains the factors of the terms of the series which involves the kth power. So, to find the interval of convergence of the power series, let us use the modified root test on the series.

Since $b_k=2^k{x}^k$ so we evaluate $\underset{k\to \infty }{\lim }\sqrt[4]{\left|b_k\right|}$

Therefore,

$$\begin{gathered} \underset{k\to \infty }{\lim }\sqrt[k]{\left|b_k\right|}=\underset{k\to \infty }{\lim }\sqrt[4]{2^k{x}^k\mid } \\ =\underset{k\to \infty }{\lim } \end{gathered}$$

Now, for $k\to \infty ,$the value of the limit is$2\left|x\right|.$  Hence, by the ratio test for absolute convergence, we know that the series will converge when $2\left|x\right|<1$

That is,

$\left|x\right|<\frac{1}{2}$

So, the interval of convergence of the series

$\sum _{k=0}^{\infty }{2}^k{x}^k \text{ is }\left(-\frac{1}{2},\frac{1}{2}\right)$

Part(b) Step 1 : Given information

Given function :$f\left(x\right)=\sum _{k=0}^x{2}^k{x}^k$

Part (b): Step 2 : Simplification

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

Therefore,

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

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

$$\begin{gathered} f^{\text{'}}\left(x\right)=\sum _{k=0}^{\infty }(k+1)2^{k+1}{x}^k \\ =\mathit{\sum }_{k\mathit{=}\mathit{0}}^{\mathit{\infty }}{2}^{k}k{x}^{k\mathit{-}\mathit{1}} \\ \Rightarrow f^{\text{'}}\left(x\right)=\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{(}\mathit{k}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{2}}^{\mathbf{k}\mathbf{+}\mathbf{1}}{\mathit{x}}^{\mathbf{k}} \end{gathered}$$

Part(c) Step 1 : Given information

Given function :$f\left(x\right)=\sum _{k=0}^x{2}^k{x}^k$

Part (c): Step 2 : Simplification

To find the power series, let us Integrate the function from $x_0 \mathrm{to} x$

Therefore,

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

here, $x_0=0$. So

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

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

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