Chapter 11, Problem 1CCE

To determine

The $n$ th Taylor polynomial of $f(x)$ at $x=a$.

Answer to Problem 1CCE

Solution:

The $n$ th Taylor polynomial of $f\left(x\right)$ at $x=a$ is the polynomial $p_n\left(x\right)$ is defined by

$p_n\left(x\right)=f\left(a\right)+\frac{f\text{'}\left(a\right)}{1!}\left(x-a\right)+\frac{f"\left(a\right)}{2!}{\left(x-a\right)}^2+\mathrm{........}+\frac{f^n\left(a\right)}{n!}{\left(x-a\right)}^n$.

Explanation of Solution

Given information:

Instruction to describe the $n$ th Taylor polynomial of $f(x)$ at $x=a$.

Explanation:

To approximate given function $f(x)$ by a polynomial for values of $x$ near some number $a$. Since the behavior of $f(x)$ near $x=a$ is determined by the values of $f(x)$ and its derivative at

$x=a$.

To approximate $f(x)$ by a polynomial $p(x)$ for which the values of $p(x)$ and its derivative at $x=a$ are the same as those of $f(x)$. It can be done if the polynomial is of the form

$p(x)=a_0+a_1\left(x-a\right)+a_1{\left(x-a\right)}^2+...+a_n{\left(x-a\right)}^n$, which is the polynomial in $x-a$.

Where $a_0,a_1,a_2...a_n$ are given numbers and $a_n\ne 0$.

It is easy to compute, and so on, because setting $x=a$ in $p(x)$ or one of its derivative makes most of the terms equal to zero.

The following result yield,

The $n$ th Taylor polynomial of $f\left(x\right)$ at $x=a$ is the polynomial $p_n\left(x\right)$ is defined by

$p_n\left(x\right)=f\left(a\right)+\frac{f\text{'}\left(a\right)}{1!}\left(x-a\right)+\frac{f"\left(a\right)}{2!}{\left(x-a\right)}^2+\mathrm{........}+\frac{f^n\left(a\right)}{n!}{\left(x-a\right)}^n$.