Question

Find the first few terms of the Taylor series of the following function about the given points.

${\mathit{e}}^{\mathbf{x}}\mathbf{,}\mathit{a}\mathbf{=}\mathbf{1}$

Short answer

The first few terms of Taylor's series of

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

Step-by-step solution

Taylor series and the given function.

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

Calculation by the Taylor series.

Differentiate the function;

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

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:

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

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

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