# 6.3 Taylor and maclaurin series  (Page 2/13)

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## Uniqueness of taylor series

If a function $f$ has a power series at a that converges to $f$ on some open interval containing a , then that power series is the Taylor series for $f$ at a .

The proof follows directly from [link] .

To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as Taylor polynomials    .

Visit the MacTutor History of Mathematics archive to read brief biographies of Brook Taylor and Colin Maclaurin and how they developed the concepts named after them.

## Taylor polynomials

The n th partial sum of the Taylor series for a function $f$ at $a$ is known as the n th Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by

$\begin{array}{c}{p}_{0}\left(x\right)=f\left(a\right),\hfill \\ {p}_{1}\left(x\right)=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right),\hfill \\ {p}_{2}\left(x\right)=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2},\hfill \\ {p}_{3}\left(x\right)=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\frac{f\text{‴}\left(a\right)}{3\text{!}}{\left(x-a\right)}^{3},\hfill \end{array}$

respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of $f$ at $a,$ respectively. If $x=a,$ then these polynomials are known as Maclaurin polynomials for $f.$ We now provide a formal definition of Taylor and Maclaurin polynomials for a function $f.$

## Definition

If $f$ has n derivatives at $x=a,$ then the n th Taylor polynomial for $f$ at $a$ is

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

The n th Taylor polynomial for $f$ at 0 is known as the n th Maclaurin polynomial for $f.$

We now show how to use this definition to find several Taylor polynomials for $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ at $x=1.$

## Finding taylor polynomials

Find the Taylor polynomials ${p}_{0},{p}_{1},{p}_{2}$ and ${p}_{3}$ for $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ at $x=1.$ Use a graphing utility to compare the graph of $f$ with the graphs of ${p}_{0},{p}_{1},{p}_{2}$ and ${p}_{3}.$

To find these Taylor polynomials, we need to evaluate $f$ and its first three derivatives at $x=1.$

$\begin{array}{cccccccc}\hfill f\left(x\right)& =\hfill & \text{ln}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill f\left(1\right)& =\hfill & 0\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & \frac{1}{x}\hfill & & & \hfill {f}^{\prime }\left(1\right)& =\hfill & 1\hfill \\ \hfill f\text{″}\left(x\right)& =\hfill & -\frac{1}{{x}^{2}}\hfill & & & \hfill f\text{″}\left(1\right)& =\hfill & -1\hfill \\ \hfill f\text{‴}\left(x\right)& =\hfill & \frac{2}{{x}^{3}}\hfill & & & \hfill f\text{‴}\left(1\right)& =\hfill & 2\hfill \end{array}$

Therefore,

$\begin{array}{ccc}\hfill {p}_{0}\left(x\right)& =\hfill & f\left(1\right)=0,\hfill \\ \hfill {p}_{1}\left(x\right)& =\hfill & f\left(1\right)+{f}^{\prime }\left(1\right)\left(x-1\right)=x-1,\hfill \\ \hfill {p}_{2}\left(x\right)& =\hfill & f\left(1\right)+{f}^{\prime }\left(1\right)\left(x-1\right)+\frac{f\text{″}\left(1\right)}{2}{\left(x-1\right)}^{2}=\left(x-1\right)-\frac{1}{2}{\left(x-1\right)}^{2},\hfill \\ \hfill {p}_{3}\left(x\right)& =\hfill & f\left(1\right)+{f}^{\prime }\left(1\right)\left(x-1\right)+\frac{f\text{″}\left(1\right)}{2}{\left(x-1\right)}^{2}+\frac{f\text{‴}\left(1\right)}{3\text{!}}{\left(x-1\right)}^{3}\hfill \\ & =\hfill & \left(x-1\right)-\frac{1}{2}{\left(x-1\right)}^{2}+\frac{1}{3}{\left(x-1\right)}^{3}.\hfill \end{array}$

The graphs of $y=f\left(x\right)$ and the first three Taylor polynomials are shown in [link] .

Find the Taylor polynomials ${p}_{0},{p}_{1},{p}_{2}$ and ${p}_{3}$ for $f\left(x\right)=\frac{1}{{x}^{2}}$ at $x=1.$

${p}_{0}\left(x\right)=1;{p}_{1}\left(x\right)=1-2\left(x-1\right);{p}_{2}\left(x\right)=1-2\left(x-1\right)+3{\left(x-1\right)}^{2};{p}_{3}\left(x\right)=1-2\left(x-1\right)+3{\left(x-1\right)}^{2}-4{\left(x-1\right)}^{3}$

We now show how to find Maclaurin polynomials for e x , $\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x.$ As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.

## Finding maclaurin polynomials

For each of the following functions, find formulas for the Maclaurin polynomials ${p}_{0},{p}_{1},{p}_{2}$ and ${p}_{3}.$ Find a formula for the n th Maclaurin polynomial and write it using sigma notation. Use a graphing utilty to compare the graphs of ${p}_{0},{p}_{1},{p}_{2}$ and ${p}_{3}$ with $f.$

1. $f\left(x\right)={e}^{x}$
2. $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$
3. $f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$
1. Since $f\left(x\right)={e}^{x},$ we know that $f\left(x\right)={f}^{\prime }\left(x\right)=f\text{″}\left(x\right)=\text{⋯}={f}^{\left(n\right)}\left(x\right)={e}^{x}$ for all positive integers n . Therefore,
$f\left(0\right)={f}^{\prime }\left(0\right)=f\text{″}\left(0\right)=\text{⋯}={f}^{\left(n\right)}\left(0\right)=1$

for all positive integers n . Therefore, we have
$\begin{array}{ccc}\hfill {p}_{0}\left(x\right)& =\hfill & f\left(0\right)=1,\hfill \\ \hfill {p}_{1}\left(x\right)& =\hfill & f\left(0\right)+{f}^{\prime }\left(0\right)x=1+x,\hfill \\ \hfill {p}_{2}\left(x\right)& =\hfill & f\left(0\right)+{f}^{\prime }\left(0\right)x+\frac{f\text{″}\left(0\right)}{2\text{!}}{x}^{2}=1+x+\frac{1}{2}{x}^{2},\hfill \\ \hfill {p}_{3}\left(x\right)& =\hfill & f\left(0\right)+{f}^{\prime }\left(0\right)x+\frac{f\text{″}\left(0\right)}{2}{x}^{2}+\frac{f\text{‴}\left(0\right)}{3\text{!}}{x}^{3}\hfill \\ & =\hfill & 1+x+\frac{1}{2}{x}^{2}+\frac{1}{3\text{!}}{x}^{3},\hfill \\ \hfill {p}_{n}\left(x\right)& =\hfill & f\left(0\right)+{f}^{\prime }\left(0\right)x+\frac{f\text{″}\left(0\right)}{2}{x}^{2}+\frac{f\text{‴}\left(0\right)}{3\text{!}}{x}^{3}+\text{⋯}+\frac{{f}^{\left(n\right)}\left(0\right)}{n\text{!}}{x}^{n}\hfill \\ & =\hfill & 1+x+\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{3}}{3\text{!}}+\text{⋯}+\frac{{x}^{n}}{n\text{!}}\hfill \\ & =\hfill & \sum _{k=0}^{n}\frac{{x}^{k}}{k\text{!}}.\hfill \end{array}$

The function and the first three Maclaurin polynomials are shown in [link] .
2. For $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ the values of the function and its first four derivatives at $x=0$ are given as follows:
$\begin{array}{cccccccc}\hfill f\left(x\right)& =\hfill & \text{sin}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill f\left(0\right)& =\hfill & 0\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & \text{cos}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill {f}^{\prime }\left(0\right)& =\hfill & 1\hfill \\ \hfill f\text{″}\left(x\right)& =\hfill & \text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill f\text{″}\left(0\right)& =\hfill & 0\hfill \\ \hfill f\text{‴}\left(x\right)& =\hfill & \text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill f\text{‴}\left(0\right)& =\hfill & -1\hfill \\ \hfill {f}^{\left(4\right)}\left(x\right)& =\hfill & \text{sin}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill {f}^{\left(4\right)}\left(0\right)& =\hfill & 0.\hfill \end{array}$

Since the fourth derivative is $\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ the pattern repeats. That is, ${f}^{\left(2m\right)}\left(0\right)=0$ and ${f}^{\left(2m+1\right)}\left(0\right)={\left(-1\right)}^{m}$ for $m\ge 0.$ Thus, we have
$\begin{array}{c}{p}_{0}\left(x\right)=0,\hfill \\ {p}_{1}\left(x\right)=0+x=x,\hfill \\ {p}_{2}\left(x\right)=0+x+0=x,\hfill \\ {p}_{3}\left(x\right)=0+x+0-\frac{1}{3\text{!}}{x}^{3}=x-\frac{{x}^{3}}{3\text{!}},\hfill \\ {p}_{4}\left(x\right)=0+x+0-\frac{1}{3\text{!}}{x}^{3}+0=x-\frac{{x}^{3}}{3\text{!}},\hfill \\ {p}_{5}\left(x\right)=0+x+0-\frac{1}{3\text{!}}{x}^{3}+0+\frac{1}{5\text{!}}{x}^{5}=x-\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}},\hfill \end{array}$

and for $m\ge 0,$
$\begin{array}{cc}\hfill {p}_{2m+1}\left(x\right)& ={p}_{2m+2}\left(x\right)\hfill \\ & =x-\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}}-\text{⋯}+{\left(-1\right)}^{m}\frac{{x}^{2m+1}}{\left(2m+1\right)\text{!}}\hfill \\ & =\sum _{k=0}^{m}{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{\left(2k+1\right)\text{!}}.\hfill \end{array}$

Graphs of the function and its Maclaurin polynomials are shown in [link] .
3. For $f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x,$ the values of the function and its first four derivatives at $x=0$ are given as follows:
$\begin{array}{cccccccc}\hfill f\left(x\right)& =\hfill & \text{cos}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill f\left(0\right)& =\hfill & 1\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & \text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill {f}^{\prime }\left(0\right)& =\hfill & 0\hfill \\ \hfill f\text{″}\left(x\right)& =\hfill & \text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill f\text{″}\left(0\right)& =\hfill & -1\hfill \\ \hfill f\text{‴}\left(x\right)& =\hfill & \text{sin}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill f\text{‴}\left(0\right)& =\hfill & 0\hfill \\ \hfill {f}^{\left(4\right)}\left(x\right)& =\hfill & \text{cos}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \hfill {f}^{\left(4\right)}\left(0\right)& =\hfill & 1.\hfill \end{array}$

Since the fourth derivative is $\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ the pattern repeats. In other words, ${f}^{\left(2m\right)}\left(0\right)={\left(-1\right)}^{m}$ and ${f}^{\left(2m+1\right)}=0$ for $m\ge 0.$ Therefore,
$\begin{array}{c}{p}_{0}\left(x\right)=1,\hfill \\ {p}_{1}\left(x\right)=1+0=1,\hfill \\ {p}_{2}\left(x\right)=1+0-\frac{1}{2\text{!}}{x}^{2}=1-\frac{{x}^{2}}{2\text{!}},\hfill \\ {p}_{3}\left(x\right)=1+0-\frac{1}{2\text{!}}{x}^{2}+0=1-\frac{{x}^{2}}{2\text{!}},\hfill \\ {p}_{4}\left(x\right)=1+0-\frac{1}{2\text{!}}{x}^{2}+0+\frac{1}{4\text{!}}{x}^{4}=1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}},\hfill \\ {p}_{5}\left(x\right)=1+0-\frac{1}{2\text{!}}{x}^{2}+0+\frac{1}{4\text{!}}{x}^{4}+0=1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}},\hfill \end{array}$

and for $n\ge 0,$
$\begin{array}{cc}\hfill {p}_{2m}\left(x\right)& ={p}_{2m+1}\left(x\right)\hfill \\ & =1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}}-\text{⋯}+{\left(-1\right)}^{m}\frac{{x}^{2m}}{\left(2m\right)\text{!}}\hfill \\ & =\sum _{k=0}^{m}{\left(-1\right)}^{k}\frac{{x}^{2k}}{\left(2k\right)\text{!}}.\hfill \end{array}$

Graphs of the function and the Maclaurin polynomials appear in [link] .

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