Question

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{,}\mathbf{}\mathit{a}\mathbf{=}\mathbf{3}\mathbf{.}$

Short answer

The first few terms of Taylor’s series expansion are:

$e^x=e^3·\left[1+(x-3)+\frac{(x-3{)}^2}{2}+\frac{(x-3{)}^3}{6}+\dots \right]$.

Step-by-step solution

Given Information

The given function.

$f\left(x\right)=e^x.$

Definition of Taylor’s series

The Taylor’s series is a power series that yields the expansion of a function in the vicinity of a point if the function is continuous, all of its derivatives exist, and the series converges.

Find the Taylor’s series for ex

Use the Maclaurin series of and replace $x$by $x-3$,

localid="1657388907873" $$\begin{gathered} e^x={e^{3+(x-3}}^{)} \\ {e}^x=e^3·e^{(x-3)} \\ {e}^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \\ {e}^{(x-3)}=1+(x-3)+\frac{(x-3{)}^2}{2}+\frac{(x-3{)}^3}{6}+\dots \end{gathered}$$

Write first few terms of Taylor’s series about localid="1657388920739" $a=3$

localid="1657388930350" $$\begin{gathered} e^x=e^3·e^{(x-3)} \\ =e^3·\left[1+(x-3)+\frac{(x-3{)}^2}{2}+\frac{(x-3{)}^3}{6}+\dots \right] \end{gathered}$$

Hence, first few terms of Taylor’s series expansion are:

localid="1657388838541" $e^x=e^3·\left[1+(x-3)+\frac{(x-3{)}^2}{2}+\frac{(x-3{)}^3}{6}+\dots \right]$