# 6.4 Working with taylor series  (Page 9/11)

 Page 9 / 11

[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 $1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{2128}-\frac{7{x}^{5}}{256}.$

${\int }_{0}^{1\text{/}4}\sqrt{x}\left(1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{128}-\frac{7{x}^{5}}{256}\right)\phantom{\rule{0.1em}{0ex}}dx$

$=\frac{2}{3}{2}^{-3}-\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{2}{5}{2}^{-5}-\frac{1}{8}\phantom{\rule{0.2em}{0ex}}\frac{2}{7}{2}^{-7}-\frac{1}{16}\phantom{\rule{0.2em}{0ex}}\frac{2}{9}{2}^{-9}-\frac{5}{128}\phantom{\rule{0.2em}{0ex}}\frac{2}{11}{2}^{-11}-\frac{7}{256}\phantom{\rule{0.2em}{0ex}}\frac{2}{13}{2}^{-13}=0.0767732...$

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 $y=\sqrt{x}\sqrt{1-x}=\sqrt{x}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{binomial expansion of}\sqrt{1-x}\right)$ 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 $T=4\sqrt{\frac{L}{g}}{\int }_{0}^{\pi \text{/}2}\left(1+\frac{1}{2}{k}^{2}{\text{sin}}^{2}\theta +\frac{3}{8}{k}^{4}{\text{sin}}^{4}\theta +\text{⋯}\right)\phantom{\rule{0.1em}{0ex}}d\theta$ 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 $T\left({\theta }_{\text{max}}\right)=2\sqrt{2}\sqrt{\frac{L}{g}}{\int }_{0}^{{\theta }_{\text{max}}}\frac{d\theta }{\sqrt{\text{cos}\phantom{\rule{0.1em}{0ex}}\theta }-\text{cos}\left({\theta }_{\text{max}}\right)}$ 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.

## Chapter review exercises

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 ${e}^{x}=\sum _{n=0}^{\infty }\frac{1}{n\text{!}}{x}^{n}.\right)$

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: $\left(-16,8\right)$

$\sum _{n=0}^{\infty }\frac{{2}^{n}}{{e}^{n}}{\left(x-e\right)}^{n}$

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.

$f\left(x\right)=\frac{{x}^{2}}{x+3}$

$\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{{3}^{n+1}}{x}^{n};$ ROC: $3;$ IOC: $\left(-3,3\right)$

$f\left(x\right)=\frac{8x+2}{2{x}^{2}-3x+1}$

In the following exercises, find the power series for the given function using term-by-term differentiation or integration.

$f\left(x\right)={\text{tan}}^{-1}\left(2x\right)$

integration: $\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{2n+1}{\left(2x\right)}^{2n+1}$

$f\left(x\right)=\frac{x}{{\left(2+{x}^{2}\right)}^{2}}$

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?

$f\left(x\right)={x}^{3}-2{x}^{2}+4,a=-3$

${p}_{4}\left(x\right)={\left(x+3\right)}^{3}-11{\left(x+3\right)}^{2}+39\left(x+3\right)-41;$ 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.

$f\left(x\right)=\text{cos}\left(3x\right)$

$\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}{\left(3x\right)}^{2n}}{2n\text{!}}$

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

In the following exercises, find the Taylor series at the given value.

$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x,a=\frac{\pi }{2}$

$\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{\left(2n\right)\text{!}}{\left(x-\frac{\pi }{2}\right)}^{2n}$

$f\left(x\right)=\frac{3}{x},a=1$

In the following exercises, find the Maclaurin series for the given function.

$f\left(x\right)={e}^{\text{−}{x}^{2}}-1$

$\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n}}{n\text{!}}{x}^{2n}$

$f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x-x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x$

In the following exercises, find the Maclaurin series for $F\left(x\right)={\int }_{0}^{x}f\left(t\right)dt$ by integrating the Maclaurin series of $f\left(x\right)$ term by term.

$f\left(x\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}$

$F\left(x\right)=\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{\left(2n+1\right)\left(2n+1\right)\text{!}}{x}^{2n+1}$

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

Use power series to prove Euler’s formula : ${e}^{ix}=\text{cos}\phantom{\rule{0.1em}{0ex}}x+i\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x$

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 $\text{}250,000?$

$2.5\text{%}$

Calculate the necessary present value of an annuity in order to support annual payouts of $\text{}15,000$ given over $25$ years assuming interest rates of $1\text{%},5\text{%},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}10\text{%}.$

#### Questions & Answers

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There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
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Bhagvanji
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I think
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scanning tunneling microscope
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biomolecules are e building blocks of every organics and inorganic materials.
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
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