Question

${\mathit{e}}^{\mathbf{cos}\mathit{x}\mathbf{-}\mathbf{1}}$. Hint:${\mathit{e}}^{\mathbf{c}}\mathit{o}\mathit{s}\mathit{x}\mathbf{=}\mathit{e}\mathbf{\cdot }{\mathit{e}}^{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}\mathbf{-}\mathbf{1}}$

Short answer

$e^{\cos \left(x\right)}=e\cdot e^{cos\left(x\right)-1}$

$e^{\cos \left(x\right)}=e·\left[1-\frac{x^2}{3}+\frac{x^4}{6}+\dots \right]$

Step-by-step solution

Given information.

The Maclaurin expansion of the function$e^{\cos x}$ is to be evaluated.

Definition of Maclaurin series.

The definition of the Maclaurin series is defined.

$$\begin{gathered} \mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{720}}\mathbf{+}\mathbf{\dots } \\ \mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{720}}\mathbf{+}\mathbf{\dots } \\ {\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{+}\mathbf{\dots } \end{gathered}$$

Begin by stating the Maclaurin series.

State the Maclaurin series

$$\begin{gathered} \cos \left(x\right)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\dots \\ \cos \left(x\right)-1=-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\dots \\ {e}^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \end{gathered}$$

Make appropriate substitution in the Maclaurin series and evaluate.

Replace x with the expression$\left[-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\dots \right]$in the third defined Maclaurin series.

role="math" localid="1657382741919" $$\begin{gathered} e^{\cos \left(x\right)-1}=1+\left[-\frac{x^2}{2}+\frac{x^4}{24}+\dots \right]+\frac{{\left[-\frac{x^2}{2}+\cdots \right]}^2}{2}+\dots \\ {e}^{\cos \left(\omega \right)-1}=1+\frac{\left(-3x^2+x^4\right)}{6}+\dots \\ {e}^{\cos \left(x\right)}=e\cdot e^{cos\left(x\right)-1} \\ {e}^{\cos \left(x\right)}=e·\left[1-\frac{x^2}{3}+\frac{x^4}{6}+\dots \right] \end{gathered}$$

Thus, the final answer is $e^{\cos \left(x\right)}=e·\left[1-\frac{x^2}{3}+\frac{x^4}{6}+\dots \right]$