91Ó°ÊÓ

Function is $e^x\text{and}a=1$

The Taylor series of $\mathbf{\text{f(x)}}$about is expressed as follows:

role="math" localid="1664257910058" $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{1}\mathbf{!}}\mathbf{(}\mathit{x}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{2}\mathbf{!}}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{3}\mathbf{!}}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}{\mathbf{)}}^{\mathbf{3}}\mathbf{+}\mathbf{…}\mathbf{…}\mathbf{.}\mathbf{.}$

Differentiate the function;

$$\begin{gathered} e^x=e+\frac{e}{1!}(\mathrm{π}-1)+\frac{e}{2!}(\mathrm{π}-1{)}^2+\frac{e}{3!}(\mathrm{π}-1{)}^3+\frac{e}{4!}(\mathrm{π}-1{)}^4+\frac{e}{5!}(\mathrm{π}-1{)}^5……. \\ {e}^x=e+e(\mathrm{π}-1)+\frac{e}{2}(\mathrm{π}-1{)}^2+\frac{e}{6}(\mathrm{π}-1{)}^3+\frac{e}{24}(\mathrm{π}-1{)}^4+\frac{e}{120}(\mathrm{π}-1{)}^5……. \end{gathered}$$

Substitute $1$ for $x$as follows:

$\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& \frac{d{e}^x}{dx}\\ f^{\text{'}}\left(x\right)& =& e^x\\ f^{\text{'}}\left(1\right)& =& e^1\\ f^{\text{'}}\left(1\right)& =& e\\ & & \end{array}$

Differentiate the function $e^x$ two times and substitute $1$ for$ x$ as follows:

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d^2{e}^x}{d{x}^2}\backslash \mathrm{hfill} \\ {f}^{\text{'}\text{'}}\left(x\right)=e^x \\ {f}^{\text{'}\text{'}}\left(1\right)=e^1 \\ {f}^{\text{'}\text{'}}\left(1\right)=e \end{gathered}$$

Similarly, all the values of derivatives of the function $e^x$at $1$ are.

Substitute all the values of derivatives and $1$ for $x$in Taylor’s equation as follows:

$$\begin{gathered} e^x=e+\frac{e}{1!}(\mathrm{π}-1)+\frac{e}{2!}(\mathrm{π}-1{)}^2+\frac{e}{3!}(\mathrm{π}-1{)}^3+\frac{e}{4!}(\mathrm{π}-1{)}^4+\frac{e}{5!}(\mathrm{π}-1{)}^5…….\backslash \mathrm{hfill} \\ {e}^x=e+e(\mathrm{π}-1)+\frac{e}{2}(\mathrm{π}-1{)}^2+\frac{e}{6}(\mathrm{π}-1{)}^3+\frac{e}{24}(\mathrm{π}-1{)}^4+\frac{e}{120}(\mathrm{π}-1{)}^5…….\backslash \mathrm{hfill} \end{gathered}$$

Thus, the first few terms of the Taylor series of $e^x\text{at}a=1$are:

$e+e(\mathrm{π}-1)+\frac{e}{2}(\mathrm{π}-1{)}^2+\frac{e}{6}(\mathrm{π}-1{)}^3+\frac{e}{24}(\mathrm{π}-1{)}^4+\frac{e}{120}(\mathrm{π}-1{)}^5……..$