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[T] Suppose that a set of standardized test scores is normally distributed with mean $\mu =100$ and standard deviation $\sigma =10.$ Set up an integral that represents the probability that a test score will be between $90$ and $110$ and use the integral of the degree $10$ Maclaurin polynomial of $\frac{1}{\sqrt{2\pi}}{e}^{\text{\u2212}{x}^{2}\text{/}2}$ to estimate this probability.
The probability is $p=\frac{1}{\sqrt{2\pi}}{\displaystyle {\int}_{\left(a-\mu \right)\text{/}\sigma}^{\left(b-\mu \right)\text{/}\sigma}{e}^{\text{\u2212}{x}^{2}\text{/}2}dx}$ where $a=90$ and $b=100,$ that is, $p=\frac{1}{\sqrt{2\pi}}{\displaystyle {\int}_{\mathrm{-1}}^{1}{e}^{\text{\u2212}{x}^{2}\text{/}2}dx}=\frac{1}{\sqrt{2\pi}}{\displaystyle {\int}_{\mathrm{-1}}^{1}{\displaystyle \sum _{n=0}^{5}{\left(\mathrm{-1}\right)}^{n}}\frac{{x}^{2n}}{{2}^{n}n\text{!}}dx}=\frac{2}{\sqrt{2\pi}}{\displaystyle \sum _{n=0}^{5}{\left(\mathrm{-1}\right)}^{n}}\frac{1}{\left(2n+1\right){2}^{n}n\text{!}}\approx 0.6827.$
[T] Suppose that a set of standardized test scores is normally distributed with mean $\mu =100$ and standard deviation $\sigma =10.$ Set up an integral that represents the probability that a test score will be between $70$ and $130$ and use the integral of the degree $50$ Maclaurin polynomial of $\frac{1}{\sqrt{2\pi}}{e}^{\text{\u2212}{x}^{2}\text{/}2}$ to estimate this probability.
[T] Suppose that $\sum _{n=0}^{\infty}{a}_{n}{x}^{n}$ converges to a function $f\left(x\right)$ such that $f\left(0\right)=1,{f}^{\prime}\left(0\right)=0,$ and $f\text{\u2033}\left(x\right)=\text{\u2212}f\left(x\right).$ Find a formula for ${a}_{n}$ and plot the partial sum ${S}_{N}$ for $N=20$ on $[\mathrm{-5},5].$
As in the previous problem one obtains
${a}_{n}=0$ if
$n$ is odd and
${a}_{n}=\text{\u2212}\left(n+2\right)\left(n+1\right){a}_{n+2}$ if
$n$ is even, so
${a}_{0}=1$ leads to
${a}_{2n}=\frac{{\left(\mathrm{-1}\right)}^{n}}{\left(2n\right)\text{!}}.$
[T] Suppose that $\sum _{n=0}^{\infty}{a}_{n}{x}^{n}$ converges to a function $f\left(x\right)$ such that $f\left(0\right)=0,\phantom{\rule{0.5em}{0ex}}{f}^{\prime}\left(0\right)=1,$ and $f\text{\u2033}\left(x\right)=\text{\u2212}f\left(x\right).$ Find a formula for ${a}_{n}$ and plot the partial sum ${S}_{N}$ for $N=10$ on $[\mathrm{-5},5].$
Suppose that $\sum _{n=0}^{\infty}{a}_{n}{x}^{n}$ converges to a function $y$ such that $y\text{\u2033}-{y}^{\prime}+y=0$ where $y\left(0\right)=1$ and $y\prime \left(0\right)=0.$ Find a formula that relates ${a}_{n+2},{a}_{n+1},$ and ${a}_{n}$ and compute ${a}_{0},\mathrm{...},{a}_{5}.$
$y\text{\u2033}={\displaystyle \sum _{n=0}^{\infty}\left(n+2\right)\left(n+1\right){a}_{n+2}{x}^{n}}$ and ${y}^{\prime}={\displaystyle \sum _{n=0}^{\infty}\left(n+1\right){a}_{n+1}{x}^{n}}$ so $y\text{\u2033}-{y}^{\prime}+y=0$ implies that $\left(n+2\right)\left(n+1\right){a}_{n+2}-\left(n+1\right){a}_{n+1}+{a}_{n}=0$ or ${a}_{n}=\frac{{a}_{n-1}}{n}-\frac{{a}_{n-2}}{n\left(n-1\right)}$ for all $n\xb7y\left(0\right)={a}_{0}=1$ and ${y}^{\prime}\left(0\right)={a}_{1}=0,$ so ${a}_{2}=\frac{1}{2},{a}_{3}=\frac{1}{6},{a}_{4}=0,$ and ${a}_{5}=-\frac{1}{120}.$
Suppose that $\sum _{n=0}^{\infty}{a}_{n}{x}^{n}$ converges to a function $y$ such that $y\text{\u2033}-{y}^{\prime}+y=0$ where $y\left(0\right)=0$ and ${y}^{\prime}\left(0\right)=1.$ Find a formula that relates ${a}_{n+2},{a}_{n+1},$ and ${a}_{n}$ and compute ${a}_{1},\mathrm{...},{a}_{5}.$
The error in approximating the integral ${\int}_{a}^{b}f\left(t\right)\phantom{\rule{0.1em}{0ex}}dt$ by that of a Taylor approximation ${\int}_{a}^{b}{P}_{n}\left(t\right)\phantom{\rule{0.1em}{0ex}}dt$ is at most ${\int}_{a}^{b}{R}_{n}\left(t\right)\phantom{\rule{0.1em}{0ex}}dt}.$ In the following exercises, the Taylor remainder estimate ${R}_{n}\le \frac{M}{\left(n+1\right)\text{!}}{\left|x-a\right|}^{n+1}$ guarantees that the integral of the Taylor polynomial of the given order approximates the integral of $f$ with an error less than $\frac{1}{10}.$
[T] $\int}_{0}^{\pi}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}dt};{P}_{s}=1-\frac{{x}^{2}}{3\text{!}}+\frac{{x}^{4}}{5\text{!}}-\frac{{x}^{6}}{7\text{!}}+\frac{{x}^{8}}{9\text{!$ (You may assume that the absolute value of the ninth derivative of $\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}$ is bounded by $0.1.)$
a. (Proof) b. We have ${R}_{s}\le \frac{0.1}{\left(9\right)\text{!}}{\pi}^{9}\approx 0.0082<0.01.$ We have ${\int}_{0}^{\pi}\left(1-\frac{{x}^{2}}{3\text{!}}+\frac{{x}^{4}}{5\text{!}}-\frac{{x}^{6}}{7\text{!}}+\frac{{x}^{8}}{9\text{!}}\right)\phantom{\rule{0.1em}{0ex}}dx}=\pi -\frac{{\pi}^{3}}{3\xb73\text{!}}+\frac{{\pi}^{5}}{5\xb75\text{!}}-\frac{{\pi}^{7}}{7\xb77\text{!}}+\frac{{\pi}^{9}}{9\xb79\text{!}}=\mathrm{1.852...},$ whereas ${\int}_{0}^{\pi}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}dt}=\mathrm{1.85194...},$ so the actual error is approximately $0.00006.$
[T] $\int}_{0}^{2}{e}^{\text{\u2212}{x}^{2}}}dx;{p}_{11}=1-{x}^{2}+\frac{{x}^{4}}{2}-\frac{{x}^{6}}{3\text{!}}+\text{\cdots}-\frac{{x}^{22}}{11\text{!$ (You may assume that the absolute value of the $23\text{rd}$ derivative of ${e}^{\text{\u2212}{x}^{2}}$ is less than $2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{14}.)$
The following exercises deal with Fresnel integrals .
The Fresnel integrals are defined by $C\left(x\right)={\displaystyle {\int}_{0}^{x}\text{cos}\left({t}^{2}\right)\phantom{\rule{0.1em}{0ex}}dt}$ and $S\left(x\right)={\displaystyle {\int}_{0}^{x}\text{sin}\left({t}^{2}\right)\phantom{\rule{0.1em}{0ex}}dt}.$ Compute the power series of $C\left(x\right)$ and $S\left(x\right)$ and plot the sums ${C}_{N}\left(x\right)$ and ${S}_{N}\left(x\right)$ of the first $N=50$ nonzero terms on $[0,2\pi ].$
Since
$\text{cos}\left({t}^{2}\right)={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{t}^{4n}}{\left(2n\right)\text{!}}}$ and
$\text{sin}\left({t}^{2}\right)={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{t}^{4n+2}}{\left(2n+1\right)\text{!}}},$ one has
$S\left(x\right)={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{x}^{4n+3}}{\left(4n+3\right)\left(2n+1\right)\text{!}}}$ and
$C\left(x\right)={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{x}^{4n+1}}{\left(4n+1\right)\left(2n\right)\text{!}}}.$ The sums of the first
$50$ nonzero terms are plotted below with
${C}_{50}\left(x\right)$ the solid curve and
${S}_{50}\left(x\right)$ the dashed curve.
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