Question

In$\left(2-e^{-x}\right)$

Short answer

$\ln \left(1+\left[1-e^{-x}\right]\right)=x-x^2+x^3-\frac{13x^4}{12}+\dots$

Step-by-step solution

Given information.

The Maclaurin expansion of the function$\ln \left(2-e^{-x}\right)$is to be evaluated.

Definition of Maclaurin series.

The definition of the Maclaurin series is defined.

$\mathbf{1}\mathbf{-}{\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{\dots }$

$\mathit{l}\mathit{n}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{\dots }$

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

State the Maclaurin series.

$\ln \left(2-e^{-x}\right)=\ln \left(1+\left[1-e^{-x}\right]\right)$

$e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}\cdots$

$1-e^x=x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}\dots$

$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$

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

Replace $x$with $\left[x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}\dots \right]$and evaluate the Maclaurin series.

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

$\ln \left(1+\left[1-e^{-x}\right]\right)=\frac{\left(36x-36x^2+36x^3-39x^4\right)}{36}+\dots$

$\ln \left(1+\left[1-e^{-x}\right]\right)=x-x^2+x^3-\frac{13x^4}{12}+\dots$

Thus, the final answer is$\ln \left(1+\left[1-e^{-x}\right]\right)=x-x^2+x^3-\frac{13x^4}{12}+\dots$