# 6.3 Taylor and maclaurin series

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• Describe the procedure for finding a Taylor polynomial of a given order for a function.
• Explain the meaning and significance of Taylor’s theorem with remainder.
• Estimate the remainder for a Taylor series approximation of a given function.

In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. Here we discuss power series representations for other types of functions. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function $f$ and the series converges on some interval, how do we prove that the series actually converges to $f?$

## Overview of taylor/maclaurin series

Consider a function $f$ that has a power series representation at $x=a.$ Then the series has the form

$\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}={c}_{0}+{c}_{1}\left(x-a\right)+{c}_{2}{\left(x-a\right)}^{2}+\text{⋯}.$

What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series [link] is a representation for $f$ at $x=a,$ we certainly want the series to equal $f\left(a\right)$ at $x=a.$ Evaluating the series at $x=a,$ we see that

$\begin{array}{cc}\hfill \sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}& ={c}_{0}+{c}_{1}\left(a-a\right)+{c}_{2}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & ={c}_{0}.\hfill \end{array}$

Thus, the series equals $f\left(a\right)$ if the coefficient ${c}_{0}=f\left(a\right).$ In addition, we would like the first derivative of the power series to equal ${f}^{\prime }\left(a\right)$ at $x=a.$ Differentiating [link] term-by-term, we see that

$\frac{d}{dx}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)={c}_{1}+2{c}_{2}\left(x-a\right)+3{c}_{3}{\left(x-a\right)}^{2}+\text{⋯}.$

Therefore, at $x=a,$ the derivative is

$\begin{array}{}\\ \\ \hfill \frac{d}{dx}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& ={c}_{1}+2{c}_{2}\left(a-a\right)+3{c}_{3}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & ={c}_{1}.\hfill \end{array}$

Therefore, the derivative of the series equals ${f}^{\prime }\left(a\right)$ if the coefficient ${c}_{1}={f}^{\prime }\left(a\right).$ Continuing in this way, we look for coefficients c n such that all the derivatives of the power series [link] will agree with all the corresponding derivatives of $f$ at $x=a.$ The second and third derivatives of [link] are given by

$\frac{{d}^{2}}{d{x}^{2}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)=2{c}_{2}+3·2{c}_{3}\left(x-a\right)+4·3{c}_{4}{\left(x-a\right)}^{2}+\text{⋯}$

and

$\frac{{d}^{3}}{d{x}^{3}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)=3·2{c}_{3}+4·3·2{c}_{4}\left(x-a\right)+5·4·3{c}_{5}{\left(x-a\right)}^{2}+\text{⋯}.$

Therefore, at $x=a,$ the second and third derivatives

$\begin{array}{cc}\hfill \frac{{d}^{2}}{d{x}^{2}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& =2{c}_{2}+3·2{c}_{3}\left(a-a\right)+4·3{c}_{4}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & =2{c}_{2}\hfill \end{array}$

and

$\begin{array}{cc}\hfill \frac{{d}^{3}}{d{x}^{3}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& =3·2{c}_{3}+4·3·2{c}_{4}\left(a-a\right)+5·4·3{c}_{5}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & =3·2{c}_{3}\hfill \end{array}$

equal $f\text{″}\left(a\right)$ and $f\text{‴}\left(a\right),$ respectively, if ${c}_{2}=\frac{f\text{″}\left(a\right)}{2}$ and ${c}_{3}=\frac{f\text{‴}\left(a\right)}{3}·2.$ More generally, we see that if $f$ has a power series representation at $x=a,$ then the coefficients should be given by ${c}_{n}=\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}.$ That is, the series should be

$\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\frac{f\text{‴}\left(a\right)}{3\text{!}}{\left(x-a\right)}^{3}+\text{⋯}.$

This power series for $f$ is known as the Taylor series for $f$ at $a.$ If $x=0,$ then this series is known as the Maclaurin series for $f.$

## Definition

If $f$ has derivatives of all orders at $x=a,$ then the Taylor series    for the function $f$ at $a$ is

$\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\text{⋯}+\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}+\text{⋯}.$

The Taylor series for $f$ at 0 is known as the Maclaurin series    for $f.$

Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall from [link] that power series representations are unique. Therefore, if a function $f$ has a power series at $a,$ then it must be the Taylor series for $f$ at $a.$

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