Question: In Problems 29–34, determine the Taylor series about the point X₀for the given functions and values of X₀.

31. $f\left(x\right)=\frac{1+x}{1-x}.$x₀ = 0 ,

For a function ^{$f\left(x\right)$} the Taylor series expansion about a point^$x_0$is given by,$f(x-x_0)=f\left(x_0\right)+f\text{'}(x_{0)}.(x-x_0)+f\text{'}\text{'}(x_0).\frac{{(x-x_0)}^2}{2!}+f\text{'}\text{'}\text{'}(x_0)\frac{{(x-x_0)}^3}{3!}+....$

We have to calculate the Taylor series expansion for, f(x) = $\frac{1+x}{1-x}$ at x₀=0.

The function f(x) can be further simplified for easier calculations,

$$\begin{gathered} \frac{1+x}{1-x}=\frac{-(1+x)}{x-1} \\ =\frac{-(2+x-1)}{x-1} \\ =\frac{-2}{x-1}-\frac{x-1}{x-1} \\ =\frac{2}{1-x}-1 \end{gathered}$$

Calculating the derivatives of function at x₀.

$f\left(x\right)=\frac{2}{1-x}-1$then^{$f\left(x_0\right)=1$}

$f\text{'}\left(x\right)=\frac{2}{{(1-x)}^2}$then$f\text{'}\left(x_0\right)=2$

$f\text{'}\text{'}\left(x_{}\right)=\frac{4}{{(1-x)}^2}$then $f\text{'}\text{'}\left(x_0\right)=4$

$f\text{'}\text{'}\text{'}\left(x\right)=\frac{12}{{(1-x)}^4}$then$f\text{'}\text{'}\text{'}\left(x_0\right)=12$

$f\text{'}\text{'}\text{'}\text{'}\left(x_0\right)=\frac{48}{{(1-x)}^5}$then$f\text{'}\text{'}\text{'}\text{'}\left(x_0\right)=48$

Substituting the above derivatives in Taylor series expansion for the function at x₀=0, then,

$\frac{1+x}{1-x}=1-2.(x-0)+4\frac{{\left(x-0\right)}^2}{2!}-12.\frac{{\left(x-0\right)}^3}{3!}+48.\frac{{\left(x-0\right)}^4}{4!}+....$

= $1+2x+2x^2+2x^3+2x^4+....$

= $1+\sum _{n-1}^{\infty }2{\left(x\right)}^n$

Hence, the required expression is