6.4 Working with taylor series  (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 ${\displaystyle {\int }_0^{1\text{/}4}\sqrt{x-x^2}}dx$ by approximating $\sqrt{1-x}$ using the binomial approximation $1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}-\frac{5x^4}{2128}-\frac{7x^5}{256}.$

[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 $y=\sqrt{x}\sqrt{1-x}=\sqrt{x}\times \left(\text{binomial expansion of}\sqrt{1-x}\right)$ 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 $T\approx 2\pi \sqrt{\frac{L}{g}}.$

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 ${\displaystyle {\int }_0^{\pi \text{/}2}{\text{sin}}^4\theta d\theta }$ in the approximation $T=4\sqrt{\frac{L}{g}}{\displaystyle {\int }_0^{\pi \text{/}2}\left(1+\frac{1}{2}{k}^2{\text{sin}}^2\theta +\frac{3}{8}{k}^4{\text{sin}}^4\theta +\text{⋯}\right)d\theta }$ to obtain an improved estimate for $T.$

[T] An equivalent formula for the period of a pendulum with amplitude ${\theta }_{\text{max}}$ is $T\left({\theta }_{\text{max}}\right)=2\sqrt{2}\sqrt{\frac{L}{g}}{\displaystyle {\int }_0^{{\theta }_{\text{max}}}\frac{d\theta }{\sqrt{\text{cos}\theta }-\text{cos}\left({\theta }_{\text{max}}\right)}}$ where $L$ is the pendulum length and $g$ is the gravitational acceleration constant. When ${\theta }_{\text{max}}=\frac{\pi }{3}$ we get $\frac{1}{\sqrt{\text{cos}t-1\text{/}2}}\approx \sqrt{2}\left(1+\frac{t^2}{2}+\frac{t^4}{3}+\frac{181t^6}{720}\right).$ 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.

If the radius of convergence for a power series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ is $5,$ then the radius of convergence for the series ${\displaystyle \sum _{n=1}^{\infty }n{a}_n{x}^{n-1}}$ is also $5.$

Power series can be used to show that the derivative of $e^x\text{is}{e}^x.$ ( Hint: Recall that $e^x={\displaystyle \sum _{n=0}^{\infty }\frac{1}{n\text{!}}}{x}^n.)$

For small values of $x,\text{sin}x\approx x.$

The radius of convergence for the Maclaurin series of $f\left(x\right)=3^x$ is $3.$

${\displaystyle \sum _{n=0}^{\infty }{n}^2}{\left(x-1\right)}^n$

${\displaystyle \sum _{n=0}^{\infty }\frac{x^n}{n^n}}$

${\displaystyle \sum _{n=0}^{\infty }\frac{3n{x}^n}{{12}^n}}$

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

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

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

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

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

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

$f\left(x\right)=e^{1\text{/}\left(4x\right)},a=4$

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

$f\left(x\right)=\text{ln}\left(x+1\right)$

$f\left(x\right)=\text{sin}x,a=\frac{\pi }{2}$

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

$f\left(x\right)=e^{\text{−}{x}^2}-1$

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

$f\left(x\right)=\frac{\text{sin}x}{x}$

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

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

For annuities with a present value of $\text{\$}1$ million, calculate the annual payouts given over $25$ years assuming interest rates of $1\text{\%},5\text{\%},\text{and}10\text{\%}.$

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 $\text{\$}\mathrm{250,000}?$

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 $1\text{\%},5\text{\%},\text{and}10\text{\%}.$