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[T] The Fresnel integrals are used in design applications for roadways and railways and other applications because of the curvature properties of the curve with coordinates $\left(C\left(t\right),S\left(t\right)\right).$ Plot the curve $\left({C}_{50},{S}_{50}\right)$ for $0\le t\le 2\pi ,$ the coordinates of which were computed in the previous exercise.
Estimate ${\int}_{0}^{1\text{/}4}\sqrt{x-{x}^{2}}}dx$ by approximating $\sqrt{1-x}$ using the binomial approximation $1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{2128}-\frac{7{x}^{5}}{256}.$
${\int}_{0}^{1\text{/}4}\sqrt{x}}\left(1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{128}-\frac{7{x}^{5}}{256}\right)\phantom{\rule{0.1em}{0ex}}dx$
$=\frac{2}{3}{2}^{\mathrm{-3}}-\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{2}{5}{2}^{\mathrm{-5}}-\frac{1}{8}\phantom{\rule{0.2em}{0ex}}\frac{2}{7}{2}^{\mathrm{-7}}-\frac{1}{16}\phantom{\rule{0.2em}{0ex}}\frac{2}{9}{2}^{\mathrm{-9}}-\frac{5}{128}\phantom{\rule{0.2em}{0ex}}\frac{2}{11}{2}^{\mathrm{-11}}-\frac{7}{256}\phantom{\rule{0.2em}{0ex}}\frac{2}{13}{2}^{\mathrm{-13}}=0.0767732...$
whereas ${\int}_{0}^{1\text{/}4}\sqrt{x-{x}^{2}}}dx=0.076773.$
[T] Use Newton’s approximation of the binomial $\sqrt{1-{x}^{2}}$ to approximate $\pi $ as follows. The circle centered at $\left(\frac{1}{2},0\right)$ with radius $\frac{1}{2}$ has upper semicircle $y=\sqrt{x}\sqrt{1-x}.$ The sector of this circle bounded by the $x$ -axis between $x=0$ and $x=\frac{1}{2}$ and by the line joining $\left(\frac{1}{4},\frac{\sqrt{3}}{4}\right)$ corresponds to $\frac{1}{6}$ of the circle and has area $\frac{\pi}{24}.$ This sector is the union of a right triangle with height $\frac{\sqrt{3}}{4}$ and base $\frac{1}{4}$ and the region below the graph between $x=0$ and $x=\frac{1}{4}.$ To find the area of this region you can write $y=\sqrt{x}\sqrt{1-x}=\sqrt{x}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\left(\text{binomial expansion of}\sqrt{1-x}\right)$ and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate $\pi .$
Use the approximation $T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)$ to approximate the period of a pendulum having length $10$ meters and maximum angle ${\theta}_{\text{max}}=\frac{\pi}{6}$ where $k=\text{sin}\left(\frac{{\theta}_{\text{max}}}{2}\right).$ Compare this with the small angle estimate $T\approx 2\pi \sqrt{\frac{L}{g}}.$
$T\approx 2\pi \sqrt{\frac{10}{9.8}}\left(1+\frac{{\text{sin}}^{2}\left(\theta \text{/}12\right)}{4}\right)\approx 6.453$ seconds. The small angle estimate is $T\approx 2\pi \sqrt{\frac{10}{9.8}\approx 6.347}.$ The relative error is around $2$ percent.
Suppose that a pendulum is to have a period of $2$ seconds and a maximum angle of ${\theta}_{\text{max}}=\frac{\pi}{6}.$ Use $T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)$ to approximate the desired length of the pendulum. What length is predicted by the small angle estimate $T\approx 2\pi \sqrt{\frac{L}{g}}?$
Evaluate ${\int}_{0}^{\pi \text{/}2}{\text{sin}}^{4}\theta d\theta$ in the approximation $T=4\sqrt{\frac{L}{g}}{\displaystyle {\int}_{0}^{\pi \text{/}2}\left(1+\frac{1}{2}{k}^{2}{\text{sin}}^{2}\theta +\frac{3}{8}{k}^{4}{\text{sin}}^{4}\theta +\text{\cdots}\right)\phantom{\rule{0.1em}{0ex}}d\theta}$ to obtain an improved estimate for $T.$
${\int}_{0}^{\pi \text{/}2}{\text{sin}}^{4}\theta d\theta}=\frac{3\pi}{16}.$ Hence $T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}+\frac{9}{256}{k}^{4}\right).$
[T] An equivalent formula for the period of a pendulum with amplitude ${\theta}_{\text{max}}$ is $T\left({\theta}_{\text{max}}\right)=2\sqrt{2}\sqrt{\frac{L}{g}}{\displaystyle {\int}_{0}^{{\theta}_{\text{max}}}\frac{d\theta}{\sqrt{\text{cos}\phantom{\rule{0.1em}{0ex}}\theta}-\text{cos}\left({\theta}_{\text{max}}\right)}}$ where $L$ is the pendulum length and $g$ is the gravitational acceleration constant. When ${\theta}_{\text{max}}=\frac{\pi}{3}$ we get $\frac{1}{\sqrt{\text{cos}\phantom{\rule{0.1em}{0ex}}t-1\text{/}2}}\approx \sqrt{2}\left(1+\frac{{t}^{2}}{2}+\frac{{t}^{4}}{3}+\frac{181{t}^{6}}{720}\right).$ Integrate this approximation to estimate $T\left(\frac{\pi}{3}\right)$ in terms of $L$ and $g.$ Assuming $g=9.806$ meters per second squared, find an approximate length $L$ such that $T\left(\frac{\pi}{3}\right)=2$ seconds.
True or False? In the following exercises, justify your answer with a proof or a counterexample.
If the radius of convergence for a power series $\sum _{n=0}^{\infty}{a}_{n}{x}^{n}$ is $5,$ then the radius of convergence for the series $\sum _{n=1}^{\infty}n{a}_{n}{x}^{n-1}$ is also $5.$
True
Power series can be used to show that the derivative of ${e}^{x}\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}{e}^{x}.$ ( Hint: Recall that ${e}^{x}={\displaystyle \sum _{n=0}^{\infty}\frac{1}{n\text{!}}}{x}^{n}.)$
For small values of $x,\text{sin}\phantom{\rule{0.1em}{0ex}}x\approx x.$
True
The radius of convergence for the Maclaurin series of $f\left(x\right)={3}^{x}$ is $3.$
In the following exercises, find the radius of convergence and the interval of convergence for the given series.
$\sum _{n=0}^{\infty}{n}^{2}}{\left(x-1\right)}^{n$
ROC: $1;$ IOC: $\left(0,2\right)$
$\sum _{n=0}^{\infty}\frac{{x}^{n}}{{n}^{n}}$
$\sum _{n=0}^{\infty}\frac{3n{x}^{n}}{{12}^{n}}$
ROC: $12;$ IOC: $\left(\mathrm{-16},8\right)$
$\sum _{n=0}^{\infty}\frac{{2}^{n}}{{e}^{n}}}{\left(x-e\right)}^{n$
In the following exercises, find the power series representation for the given function. Determine the radius of convergence and the interval of convergence for that series.
$f\left(x\right)=\frac{{x}^{2}}{x+3}$
$\sum _{n=0}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n}}{{3}^{n+1}}}{x}^{n};$ ROC: $3;$ IOC: $\left(\mathrm{-3},3\right)$
$f\left(x\right)=\frac{8x+2}{2{x}^{2}-3x+1}$
In the following exercises, find the power series for the given function using term-by-term differentiation or integration.
$f\left(x\right)={\text{tan}}^{\mathrm{-1}}\left(2x\right)$
integration: $\sum _{n=0}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n}}{2n+1}}{\left(2x\right)}^{2n+1$
$f\left(x\right)=\frac{x}{{\left(2+{x}^{2}\right)}^{2}}$
In the following exercises, evaluate the Taylor series expansion of degree four for the given function at the specified point. What is the error in the approximation?
$f\left(x\right)={x}^{3}-2{x}^{2}+4,a=\mathrm{-3}$
${p}_{4}\left(x\right)={\left(x+3\right)}^{3}-11{\left(x+3\right)}^{2}+39\left(x+3\right)-41;$ exact
$f\left(x\right)={e}^{1\text{/}\left(4x\right)},a=4$
In the following exercises, find the Maclaurin series for the given function.
$f\left(x\right)=\text{cos}\left(3x\right)$
$\sum _{n=0}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n}{\left(3x\right)}^{2n}}{2n\text{!}}$
$f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(x+1\right)$
In the following exercises, find the Taylor series at the given value.
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x,a=\frac{\pi}{2}$
$\sum _{n=0}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n}}{\left(2n\right)\text{!}}}{\left(x-\frac{\pi}{2}\right)}^{2n$
$f\left(x\right)=\frac{3}{x},a=1$
In the following exercises, find the Maclaurin series for the given function.
$f\left(x\right)={e}^{\text{\u2212}{x}^{2}}-1$
$\sum _{n=1}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n}}{n\text{!}}}{x}^{2n$
$f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x-x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x$
In the following exercises, find the Maclaurin series for $F\left(x\right)={\displaystyle {\int}_{0}^{x}f\left(t\right)}dt$ by integrating the Maclaurin series of $f\left(x\right)$ term by term.
$f\left(x\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}$
$F\left(x\right)={\displaystyle \sum _{n=0}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n}}{\left(2n+1\right)\left(2n+1\right)\text{!}}}{x}^{2n+1}$
$f\left(x\right)=1-{e}^{x}$
Use power series to prove Euler’s formula : ${e}^{ix}=\text{cos}\phantom{\rule{0.1em}{0ex}}x+i\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x$
Answers may vary.
The following exercises consider problems of annuity payments .
For annuities with a present value of $\text{\$}1$ million, calculate the annual payouts given over $25$ years assuming interest rates of $1\text{\%},5\text{\%},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}10\text{\%}.$
A lottery winner has an annuity that has a present value of $\text{\$}10$ million. What interest rate would they need to live on perpetual annual payments of $\text{\$}\mathrm{250,000}?$
$2.5\text{\%}$
Calculate the necessary present value of an annuity in order to support annual payouts of $\text{\$}\mathrm{15,000}$ given over $25$ years assuming interest rates of $1\text{\%},5\text{\%},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}10\text{\%}.$
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