Question

Show as in Problem 11that the Maclaurin series for${\mathit{e}}^{\mathbf{x}}$converges to${\mathit{e}}^{\mathbf{x}}$.

Short answer

The Maclaurin series for $e^x$ converges to $e^x$.

Step-by-step solution

Given Information

The function is $e^x$.

Definition of the convergence of the Series

The Series is said to be convergent if the terms of the Series are moving toward zero.

Verify the statement

To prove the Maclaurin series for $e^x$converges to $e^x$, we need to show that

$f\left(X\right)={\lim }_{\mathrm{n}\to \infty }{\mathrm{T}}_{\mathrm{n}}\left(\mathrm{X}\right)$ which is equivalent to showing that the remainder

$R_n\left(X\right)=f\left(X\right)-T_n\left(X\right)\to 0$. The function is $f\left(X\right)=e^x$.

Plugging this into the inequality, and taking the limit, we have:

$$\begin{gathered} \underset{n\to \infty }{\lim }\left|R_n\left(X\right)\right|\le \underset{n\to \infty }{\lim }\frac{x^{n+1}{e}^x}{\left(n+1\right)!} \\ =e^x\underset{n\to \infty }{\lim }\frac{x^{n+1}}{\left(n+1\right)!} \\ =0 \end{gathered}$$

Hence, $\underset{n\to \infty }{\lim }\frac{x^{n+1}{e}^x}{\left(n+1\right)!}=0\mathrm{and}{\mathrm{T}}_{\mathrm{n}}\left(\mathrm{X}\right)\to \mathrm{f}\left(\mathrm{X}\right).$and .

Thus, the Maclaurin series for $e^x$converges to $e^x$.