# 6.3 Taylor and maclaurin series  (Page 7/13)

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## Key concepts

• Taylor polynomials are used to approximate functions near a value $x=a.$ Maclaurin polynomials are Taylor polynomials at $x=0.$
• The n th degree Taylor polynomials for a function $f$ are the partial sums of the Taylor series for $f.$
• If a function $f$ has a power series representation at $x=a,$ then it is given by its Taylor series at $x=a.$
• A Taylor series for $f$ converges to $f$ if and only if $\underset{n\to \infty }{\text{lim}}{R}_{n}\left(x\right)=0$ where ${R}_{n}\left(x\right)=f\left(x\right)-{p}_{n}\left(x\right).$
• The Taylor series for e x , $\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x$ converge to the respective functions for all real x .

## Key equations

• Taylor series for the function $f$ at the point $x=a$
$\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\text{⋯}+\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}+\text{⋯}$

In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.

$f\left(x\right)=1+x+{x}^{2}$ at $a=1$

$f\left(x\right)=1+x+{x}^{2}$ at $a=-1$

$f\left(-1\right)=1;{f}^{\prime }\left(-1\right)=-1;f\text{″}\left(-1\right)=2;f\left(x\right)=1-\left(x+1\right)+{\left(x+1\right)}^{2}$

$f\left(x\right)=\text{cos}\left(2x\right)$ at $a=\pi$

$f\left(x\right)=\text{sin}\left(2x\right)$ at $a=\frac{\pi }{2}$

${f}^{\prime }\left(x\right)=2\phantom{\rule{0.1em}{0ex}}\text{cos}\left(2x\right);f\text{″}\left(x\right)=-4\phantom{\rule{0.1em}{0ex}}\text{sin}\left(2x\right);{p}_{2}\left(x\right)=-2\left(x-\frac{\pi }{2}\right)$

$f\left(x\right)=\sqrt{x}$ at $a=4$

$f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ at $a=1$

${f}^{\prime }\left(x\right)=\frac{1}{x};f\text{″}\left(x\right)=-\frac{1}{{x}^{2}};{p}_{2}\left(x\right)=0+\left(x-1\right)-\frac{1}{2}{\left(x-1\right)}^{2}$

$f\left(x\right)=\frac{1}{x}$ at $a=1$

$f\left(x\right)={e}^{x}$ at $a=1$

${p}_{2}\left(x\right)=e+e\left(x-1\right)+\frac{e}{2}{\left(x-1\right)}^{2}$

In the following exercises, verify that the given choice of n in the remainder estimate $|{R}_{n}|\le \frac{M}{\left(n+1\right)\text{!}}{\left(x-a\right)}^{n+1},$ where M is the maximum value of $|{f}^{\left(n+1\right)}\left(z\right)|$ on the interval between a and the indicated point, yields $|{R}_{n}|\le \frac{1}{1000}.$ Find the value of the Taylor polynomial p n of $f$ at the indicated point.

[T] $\sqrt{10};a=9,n=3$

[T] ${\left(28\right)}^{1\text{/}3};a=27,n=1$

$\frac{{d}^{2}}{d{x}^{2}}{x}^{1\text{/}3}=-\frac{2}{9{x}^{5\text{/}3}}\ge -0.00092\text{…}$ when $x\ge 28$ so the remainder estimate applies to the linear approximation ${x}^{1\text{/}3}\approx {p}_{1}\left(27\right)=3+\frac{x-27}{27},$ which gives ${\left(28\right)}^{1\text{/}3}\approx 3+\frac{1}{27}=3.\overline{037},$ while ${\left(28\right)}^{1\text{/}3}\approx 3.03658.$

[T] $\text{sin}\left(6\right);a=2\pi ,n=5$

[T] e 2 ; $a=0,n=9$

Using the estimate $\frac{{2}^{10}}{10\text{!}}<0.000283$ we can use the Taylor expansion of order 9 to estimate e x at $x=2.$ as ${e}^{2}\approx {p}_{9}\left(2\right)=1+2+\frac{{2}^{2}}{2}+\frac{{2}^{3}}{6}+\text{⋯}+\frac{{2}^{9}}{9\text{!}}=7.3887\text{…}$ whereas ${e}^{2}\approx 7.3891.$

[T] $\text{cos}\left(\frac{\pi }{5}\right);a=0,n=4$

[T] $\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2\right);a=1,n=1000$

Since $\frac{{d}^{n}}{d{x}^{n}}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)={\left(-1\right)}^{n-1}\frac{\left(n-1\right)\text{!}}{{x}^{n}},{R}_{1000}\approx \frac{1}{1001}.$ One has ${p}_{1000}\left(1\right)=\sum _{n=1}^{1000}\frac{{\left(-1\right)}^{n-1}}{n}\approx 0.6936$ whereas $\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2\right)\approx 0.6931\text{⋯}.$

Integrate the approximation $\text{sin}\phantom{\rule{0.1em}{0ex}}t\approx t-\frac{{t}^{3}}{6}+\frac{{t}^{5}}{120}-\frac{{t}^{7}}{5040}$ evaluated at πt to approximate ${\int }_{0}^{1}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}\pi t}{\pi t}dt.$

Integrate the approximation ${e}^{x}\approx 1+x+\frac{{x}^{2}}{2}+\text{⋯}+\frac{{x}^{6}}{720}$ evaluated at − x 2 to approximate ${\int }_{0}^{1}{e}^{\text{−}{x}^{2}}dx.$

${\int }_{0}^{1}\left(1-{x}^{2}+\frac{{x}^{4}}{2}-\frac{{x}^{6}}{6}+\frac{{x}^{8}}{24}-\frac{{x}^{10}}{120}+\frac{{x}^{12}}{720}\right)\phantom{\rule{0.1em}{0ex}}dx$

$=1-\frac{{1}^{3}}{3}+\frac{{1}^{5}}{10}-\frac{{1}^{7}}{42}+\frac{{1}^{9}}{9·24}-\frac{{1}^{11}}{120·11}+\frac{{1}^{13}}{720·13}\approx 0.74683$ whereas ${\int }_{0}^{1}{e}^{\text{−}{x}^{2}}dx\approx 0.74682.$

In the following exercises, find the smallest value of n such that the remainder estimate $|{R}_{n}|\le \frac{M}{\left(n+1\right)\text{!}}{\left(x-a\right)}^{n+1},$ where M is the maximum value of $|{f}^{\left(n+1\right)}\left(z\right)|$ on the interval between a and the indicated point, yields $|{R}_{n}|\le \frac{1}{1000}$ on the indicated interval.

$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ on $\left[\text{−}\pi ,\pi \right],a=0$

$f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ on $\left[-\frac{\pi }{2},\frac{\pi }{2}\right],a=0$

Since ${f}^{\left(n+1\right)}\left(z\right)$ is $\text{sin}\phantom{\rule{0.1em}{0ex}}z$ or $\text{cos}\phantom{\rule{0.1em}{0ex}}z,$ we have $M=1.$ Since $|x-0|\le \frac{\pi }{2},$ we seek the smallest n such that $\frac{{\pi }^{n+1}}{{2}^{n+1}\left(n+1\right)\text{!}}\le 0.001.$ The smallest such value is $n=7.$ The remainder estimate is ${R}_{7}\le 0.00092.$

$f\left(x\right)={e}^{-2x}$ on $\left[-1,1\right],a=0$

$f\left(x\right)={e}^{\text{−}x}$ on $\left[-3,3\right],a=0$

Since ${f}^{\left(n+1\right)}\left(z\right)=\text{±}{e}^{\text{−}z}$ one has $M={e}^{3}.$ Since $|x-0|\le 3,$ one seeks the smallest n such that $\frac{{3}^{n+1}{e}^{3}}{\left(n+1\right)\text{!}}\le 0.001.$ The smallest such value is $n=14.$ The remainder estimate is ${R}_{14}\le 0.000220.$

In the following exercises, the maximum of the right-hand side of the remainder estimate $|{R}_{1}|\le \frac{\text{max}|f\text{″}\left(z\right)|}{2}{R}^{2}$ on $\left[a-R,a+R\right]$ occurs at a or $a±R.$ Estimate the maximum value of R such that $\frac{\text{max}|f\text{″}\left(z\right)|}{2}{R}^{2}\le 0.1$ on $\left[a-R,a+R\right]$ by plotting this maximum as a function of R .

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