Chapter 9, Problem 38RE

To determine

To find: The Taylor series of the function $x^3-2x^2+5\text{ at }a=-1$ .

Answer to Problem 38RE

  $2+7\left(x+1\right)-5{\left(x+1\right)}^2+{\left(x+1\right)}^3$

Explanation of Solution

Given information:

  $x^3-2x^2+5\text{ at }a=-1$

Concept Used:

Taylor Series of $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(a\right)}{n!}{\left(x-a\right)}^n}$

  ${\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(a\right)}{n!}{\left(x-a\right)}^n}=f\left(a\right)+\frac{f\text{'}\left(a\right)}{1!}\left(x-a\right)+\frac{f\text{'}\text{'}\left(a\right)}{2!}{\left(x-a\right)}^2+\mathrm{...}$

Calculation:

Consider the function, $f\left(x\right)=x^3-2x^2+5$ .

Now find the derivative and their values at $a=-1$ .

  $\begin{array}{cccc}f\left(x\right)=x^3-2x^2+5& & & f\left(-1\right)=2\\ f\text{'}\left(x\right)=3x^2-4x& & & f\text{'}\left(-1\right)=7\\ f\text{'}\text{'}\left(x\right)=6x-4& & & f\text{'}\text{'}\left(-1\right)=-10\\ f\text{'}\text{'}\text{'}\left(x\right)=6& & & f\text{'}\text{'}\text{'}\left(-1\right)=6\\ f^4\left(x\right)=0& & & f^4\left(-1\right)=0\end{array}$

Substitute the values into the Taylor series and yields,

  $\begin{array}{c}{x}^3-2x^2+5=\frac{2}{0!}{\left(x-\left(-1\right)\right)}^0+\frac{7}{1!}{\left(x-\left(-1\right)\right)}^1+\frac{-10}{2!}{\left(x-\left(-1\right)\right)}^2+\frac{6}{3!}{\left(x-\left(-1\right)\right)}^3+\mathrm{...}\\ =2+7\left(x+1\right)-5{\left(x+1\right)}^2+{\left(x+1\right)}^3+\mathrm{...}\end{array}$

Thus, the first four non zero terms of the series is $2+7\left(x+1\right)-5{\left(x+1\right)}^2+{\left(x+1\right)}^3$ .