91Ó°ÊÓ

Given function :$f\left(x\right)=\underset{k=0}{\overset{x}{∑}}{2}^k{x}^k$

Consider the power series$f\left(x\right)=\underset{k=0}{\overset{x}{∑}}{2}^k{x}^k$

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$\underset{k→∞}{\lim }\sqrt[4]{\left|b_k\right|}$

Therefore,

$$\begin{gathered} \underset{k→∞}{\lim }\sqrt[k]{\left|b_k\right|}=\underset{k→∞}{\lim }\sqrt[4]{2^k{x}^kâˆ\pounds } \\ =\underset{k→∞}{\lim } \end{gathered}$$

Now, for $k→∞,$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

$\underset{k=0}{\overset{∞}{∑}}{2}^k{x}^k\text{is}\left(-\frac{1}{2},\frac{1}{2}\right)$

Given function :$f\left(x\right)=\underset{k=0}{\overset{x}{∑}}{2}^k{x}^k$

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

Therefore,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\underset{k=0}{\overset{∞}{∑}}{2}^k{x}^k\right] \\ =\underset{k=0}{\overset{∞}{∑}}{2}^k\frac{d}{dx}{x}^k \\ =\underset{k=0}{\overset{∞}{∑}}{2}^{k}k{x}^{k-1} \end{gathered}$$

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

$$\begin{gathered} f^{\text{'}}\left(x\right)=\underset{k=0}{\overset{∞}{∑}}(k+1)2^{k+1}{x}^k \\ =\underset{k\mathit{=}\mathit{0}}{\overset{\mathit{∞}}{\mathit{∑}}}{2}^{k}k{x}^{k\mathit{-}\mathit{1}} \\ ⇒f^{\text{'}}\left(x\right)=\underset{\mathbf{k}\mathbf{=}\mathbf{0}}{\overset{\mathbf{∞}}{\mathbf{∑}}}\mathbf{(}\mathit{k}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{2}}^{\mathbf{k}\mathbf{+}\mathbf{1}}{\mathit{x}}^{\mathbf{k}} \end{gathered}$$

Given function:$f\left(x\right)=\underset{k=0}{\overset{x}{∑}}{2}^k{x}^k$

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

Therefore,

$$\begin{gathered} F\left(x\right)={∫}_{x_0}^x\left[\underset{k=0}{\overset{∞}{∑}}{2}^k{t}^k\right]dt \\ =\underset{k=0}{\overset{∞}{∑}}{2}^k{∫}_{x_0}^x\left[t^k\right]dt \\ =\underset{k=0}{\overset{∞}{∑}}{2}^k{\left[\frac{t^{k+1}}{k+1}\right]}_k^x \end{gathered}$$

here,$x_0=0$. So

$F\left(x\right)=\underset{k=0}{\overset{∞}{∑}}\frac{2^k}{k}{x}^{k+1}$

Now, we change the index in the final step.

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

$F\left(x\right)=\underset{k=1}{\overset{∞}{∑}}\frac{2^{k-1}}{k}{x}^k$