Question

$\mathit{c}\mathit{o}\mathit{s}\left(e^x-1\right)$

Short answer

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

Step-by-step solution

Given information

The Maclaurin expansion of the function$\cos \left(e^x-1\right)$ is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

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

$e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

Part-A Step 3: Begin by stating the Maclaurin series

State the Maclaurin series.

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

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

$e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

Part-B Step 3: Use the Maclaurin series to evaluate

Replace $x$with $\left[x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \right]$in the second equation.

$\cos \left(e^x-1\right)=1-\frac{{\left[x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \right]}^2}{2}+\frac{{\left[x+\frac{x^2}{2}+\dots \right]}^4}{24}+\cdots$

$\cos \left(e^x-1\right)=1+\frac{\left(-576x^2-576x^3-288x^4\right)}{1152}+\dots$

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

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