# 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 .

#### Questions & Answers

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research.net
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That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
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