Question

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

30.$f\left(x\right)=x^{-4}{x}_0=1$

Short answer

The required expression is .$\sum _{n-0}^{\infty }{(-1)}^n(x-1)$

Step-by-step solution

Taylor series

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

Derivatives of function at x0.

We have to calculate the Taylor series expansion for, f (x) = x^-1 at x₀=1.

Calculating the derivatives of function at x₀.

$f\left(x\right)=x^{-1}thenf\left(x_0\right)=1$

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

$f\text{'}\text{'}\left(x\right)=2x^{-3}thenf\text{'}\text{'}\left(x_0\right)=2$

$f\text{'}\text{'}\text{'}\left(x\right)=6x^{-4}thenf\text{'}\text{'}\text{'}\left(x_0\right)=-6$

$f\text{'}\text{'}\text{'}\text{'}\left(x\right)=24x^{-5}thenf\text{'}\text{'}\text{'}\text{'}\left(x_0\right)=24$

Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at x₀=1, then,

$$\begin{gathered} {(x-1)}^{-1}=1-1(x-1)+2.\frac{{(x-1)}^2}{2!}-6\frac{{(x-1)}^3}{3!}+24.\frac{{(x-1)}^4}{4!}+.... \\ =1-(x-1)+{(x-1)}^2-{(x-1)}^3+{(x-1)}^4+.... \\ =\sum _{n-0}^{\infty }{(-1)}^n.{(x-1)}^n \end{gathered}$$

Hence, the required expression is$=\sum _{n-0}^{\infty }{(-1)}^n.{(x-1)}^n$