Mathematical Methods in the Physical Sciences/Chapter 1/Section 1

Overview

I remember that the professor who taught this course told us that we could never be physicists unless we could be careful enough to get all the answers right.

If you're going to make a mistake here, it's likely due to sloppiness.

Problems

#1

Read the question very, very carefully. Read it about 10 times.

Height of the tenth rebound: First, I have to think carefully about what a "rebound" is. Whenever someone talks about the nth of anything, I always think about what the first, second, and third items look like, then I can find a pattern.

The first bounce is 2/3 yd. The second, 4/9 yd. The third is 8/27 yd.

So, building a table:

Bounce	n	height !
First	1	2/3 yd
Second	2	4/9 yd
Third	3	8/27 yd
nth	n	<math>(2/3)^n</math> yd

So the height of the 10th rebound is now obvious: <math>(2/3)^10 yd= 0.0173 yd</math>.

Distance traveled after the tenth rebound: Again, parse this carefully.

The ball travels up and down for each bounce, an equal distance. The bounces after the 10th are: <math>2(2/3)^11 + 2(2/3)^12 + 2(2/3)^13 + ...</math>. (Units are yards.) In this geometric series, <math>a = a_0 = 2(2/3)^11 = 0.0231 yd</math>, and <math>r = 2/3</math>.

The sum of the infinite series is simply:

<math>S = {a \over 1-r} = 3a = 0.0694 yd</math>

(The solutions say <math>0.104 yd</math>. I can't understand why I'm wrong. Please enlighten me.)

The total distance traveled is:

<math> \begin{align} T &= 1 + 2((2/3) + (2/3)^2 + (2/3)^3 + ...) \\

 &= 1 + 2 \left ( {2/3 \over 1 - 2/3} \right ) \text{ (Substituting in the infinite series sum)}\\
 &= 1 + 2 ( 2 ) \\
 &= 5\,yd

\end{align} </math>

#2

Deriving the formula is indeed simple.

<math> \begin{align} S_n &= a + ar + ar^2 + ... + ar^{n-1} \\ r S_n &= ar + ar^2 + ar^3 + ... + ar^n \\ S_n - r S_n &= a -ar^n \\ S_n (1-r) &= a(1-r^n) \\ S_n &= {a(1-r^n) \over 1-r} \\ \end{align} </math>

To show that it converges, all we have to do is show that it has a sum. That's rather easy.

<math> \begin{align} S &= \lim_{n\to\infty}S_n \\

 &= \lim_{n\to\infty}{a(1-r^n) \over 1-r} \\
 &= {a \over 1-r} \left (1 - \lim_{n\to\infty}r^n \right ) \\
 &= \begin{cases}
      {a \over 1-r}    & |r| < 1 \\
      \text{undefined} & |r| \ge 1 \\
    \end{cases} \\

\end{align}

</math>

#3

<math> \begin{align} 0.555... &= 5/10 + 5/100 + 5/1000 + ..., \ a = {5 \over 10},\ r= {1 \over 10} \\

        &= {a \over 1 - r} \\
        &= {{5 \over 10} \over 1-{1 \over 10}} \\
        &= {5 \over 9} \\

\end{align} </math>

#4 - #11

I'll let you do these on your own. They're trivial and quite fun.