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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 $a=\mathrm{-1}$
$f\left(\mathrm{-1}\right)=1;{f}^{\prime}\left(\mathrm{-1}\right)=\mathrm{-1};f\text{\u2033}\left(\mathrm{-1}\right)=2;f\left(x\right)=1-\left(x+1\right)+{\left(x+1\right)}^{2}$
$f\left(x\right)=\text{cos}\left(2x\right)$ at $a=\pi $
$f\left(x\right)=\text{sin}\left(2x\right)$ at $a=\frac{\pi}{2}$
${f}^{\prime}\left(x\right)=2\phantom{\rule{0.1em}{0ex}}\text{cos}\left(2x\right);f\text{\u2033}\left(x\right)=\mathrm{-4}\phantom{\rule{0.1em}{0ex}}\text{sin}\left(2x\right);{p}_{2}\left(x\right)=\mathrm{-2}\left(x-\frac{\pi}{2}\right)$
$f\left(x\right)=\sqrt{x}$ at $a=4$
$f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ at $a=1$
${f}^{\prime}\left(x\right)=\frac{1}{x};f\text{\u2033}\left(x\right)=-\frac{1}{{x}^{2}};{p}_{2}\left(x\right)=0+\left(x-1\right)-\frac{1}{2}{\left(x-1\right)}^{2}$
$f\left(x\right)=\frac{1}{x}$ 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 $\left|{R}_{n}\right|\le \frac{M}{\left(n+1\right)\text{!}}{\left(x-a\right)}^{n+1},$ where M is the maximum value of $\left|{f}^{\left(n+1\right)}\left(z\right)\right|$ on the interval between a and the indicated point, yields $\left|{R}_{n}\right|\le \frac{1}{1000}.$ Find the value of the Taylor polynomial p _{n} of $f$ at the indicated point.
[T] $\sqrt{10};a=9,n=3$
[T] ${\left(28\right)}^{1\text{/}3};a=27,n=1$
$\frac{{d}^{2}}{d{x}^{2}}{x}^{1\text{/}3}=-\frac{2}{9{x}^{5\text{/}3}}\ge \mathrm{-0.00092}\text{\u2026}$ 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 ${e}^{2}\approx {p}_{9}\left(2\right)=1+2+\frac{{2}^{2}}{2}+\frac{{2}^{3}}{6}+\text{\cdots}+\frac{{2}^{9}}{9\text{!}}=7.3887\text{\u2026}$ 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 $\frac{{d}^{n}}{d{x}^{n}}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)={\left(\mathrm{-1}\right)}^{n-1}\frac{\left(n-1\right)\text{!}}{{x}^{n}},{R}_{1000}\approx \frac{1}{1001}.$ One has ${p}_{1000}\left(1\right)={\displaystyle \sum _{n=1}^{1000}\frac{{\left(\mathrm{-1}\right)}^{n-1}}{n}}\approx 0.6936$ whereas $\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2\right)\approx 0.6931\text{\cdots}.$
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 ${e}^{x}\approx 1+x+\frac{{x}^{2}}{2}+\text{\cdots}+\frac{{x}^{6}}{720}$ evaluated at − x ^{2} to approximate ${\int}_{0}^{1}{e}^{\text{\u2212}{x}^{2}}dx}.$
${\int}_{0}^{1}\left(1-{x}^{2}+\frac{{x}^{4}}{2}-\frac{{x}^{6}}{6}+\frac{{x}^{8}}{24}-\frac{{x}^{10}}{120}+\frac{{x}^{12}}{720}\right)\phantom{\rule{0.1em}{0ex}}dx$
$=1-\frac{{1}^{3}}{3}+\frac{{1}^{5}}{10}-\frac{{1}^{7}}{42}+\frac{{1}^{9}}{9\xb724}-\frac{{1}^{11}}{120\xb711}+\frac{{1}^{13}}{720\xb713}\approx 0.74683$ whereas ${\int}_{0}^{1}{e}^{\text{\u2212}{x}^{2}}dx}\approx 0.74682.$
In the following exercises, find the smallest value of n such that the remainder estimate $\left|{R}_{n}\right|\le \frac{M}{\left(n+1\right)\text{!}}{\left(x-a\right)}^{n+1},$ where M is the maximum value of $\left|{f}^{\left(n+1\right)}\left(z\right)\right|$ on the interval between a and the indicated point, yields $\left|{R}_{n}\right|\le \frac{1}{1000}$ on the indicated interval.
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ on $\left[\text{\u2212}\pi ,\pi \right],a=0$
$f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ 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 $\left|x-0\right|\le \frac{\pi}{2},$ 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 ${R}_{7}\le 0.00092.$
$f\left(x\right)={e}^{\mathrm{-2}x}$ on $\left[\mathrm{-1},1\right],a=0$
$f\left(x\right)={e}^{\text{\u2212}x}$ on $\left[\mathrm{-3},3\right],a=0$
Since ${f}^{\left(n+1\right)}\left(z\right)=\text{\xb1}{e}^{\text{\u2212}z}$ one has $M={e}^{3}.$ Since $\left|x-0\right|\le 3,$ 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 $\left|{R}_{1}\right|\le \frac{\text{max}\left|f\text{\u2033}\left(z\right)\right|}{2}{R}^{2}$ on $\left[a-R,a+R\right]$ occurs at a or $a\pm R.$ Estimate the maximum value of R such that $\frac{\text{max}\left|f\text{\u2033}\left(z\right)\right|}{2}{R}^{2}\le 0.1$ on $\left[a-R,a+R\right]$ by plotting this maximum as a function of R .
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