Question

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{,}\mathbf{}\mathit{a}\mathbf{=}\mathbf{1}$

Short answer

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

Step-by-step solution

Given information

The Maclaurin expansion of the function $f\left(x\right)=\frac{1}{x},a=1$is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

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

Begin by stating the Maclaurin series

State the Maclaurin series.

$$\begin{gathered} f\left(x\right)=\frac{1}{x} \\ =\frac{1}{1+(x-1)} \\ \frac{1}{1+x}=1-x+x^2-x^3+\dots \end{gathered}$$

Make appropriate substitution in the Maclaurin series and evaluate

Replace $\mathit{x}$with the expression $(x-1)$in the third defined Maclaurin series.

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

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

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