Chapter 1: Infinite Series, Power Series

Repeat Problem 14b for the following points and directions.

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\hat{\mathbf{i}}\mathbf{+}\hat{\mathbf{j}} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{4}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\hat{\mathbf{j}} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{-}\mathbf{3}\hat{\mathbf{i}}\mathbf{+}\hat{\mathbf{j}} \\ \mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{4}\hat{\mathbf{i}}\mathbf{-}\mathbf{3}\hat{\mathbf{j}} \end{gathered}$$

Generalize Problem $\mathbf{14}$to any mass $\mathbf{M}$of circular cross-section and moment of inertia $\mathbf{I}$. Consider a hoop, a disk, a spherical shell, a solid spherical ball; order them as to which would first reach the bottom of the inclined plane. (For moments of inertia, see Chapter $\mathbf{5}$, Section $\mathbf{4}$.)

Find the geodesics on a plane using polar coordinates.

Use the series you know to show that:

$\mathit{l}\mathit{n}\mathbf{3}\mathbf{+}\frac{{\mathbf{\left(}\mathbf{l}\mathbf{n}\mathbf{3}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{\left(}\mathbf{l}\mathbf{n}\mathbf{3}\mathbf{\right)}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{+}\mathbf{\dots }\mathbf{=}\mathbf{2}$

Find the Maclaurin series $\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\left(x\right)}}$.

Find the sum of each of the following series by recognizing it as the Maclaurin series for a function evaluated at a point$\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{2}}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}$.

As in Problem 14, let the displacements be ${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathit{s}\mathit{i}\mathit{n}\left(\frac{t}{\sqrt{2}}\right)$ and${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{t}$ . The pendulums start together at t = 0. Make computer plots to estimate when they will be together again and then, by computer, solve the equation ${\mathbf{y}}_{\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{2}}$ for the root near your estimate.

Suppose a large number of particles are bouncing back and forth between $\mathit{x}\mathbf{=}\mathbf{0}\mathbf{\text{and}}\mathit{x}\mathbf{=}\mathbf{1}$, except that at each endpoint some escape. Let r be the fraction reflected each time; then (1 - r) is the fraction escaping. Suppose the particles start at $\mathit{x}\mathbf{=}\mathbf{0}$ heading toward $\mathit{x}\mathbf{=}\mathbf{1}$; eventually all particles will escape. Write an infinite series for the fraction which escape at $\mathit{x}\mathbf{=}\mathbf{1}$and similarly for the fraction which escape at $\mathit{x}\mathbf{=}\mathbf{0}$. Sum both the series. What is the largest fraction of the particles which can escape at $\mathit{x}\mathbf{=}\mathbf{0}$? (Remember that r must be between 0 and 1.)

In testing$\mathbf{\sum }\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}}$ for convergence, a student evaluates$\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathit{n}}^{\mathbf{-}\mathbf{2}}\mathbf{ }\mathit{d}\mathit{n}\mathbf{=}{-n^{-1}|}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{=}\mathbf{0}\mathbf{+}\mathbf{\infty }\mathbf{=}\mathbf{\infty }$ and concludes (erroneously) that the series diverges. What is wrong?

Find the Maclaurin series of the following functions.

${\mathbf{e}}^{\mathbf{-}\mathbf{1}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}$