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.

Things to Remember

• The geometric series is a series where each term is a multiple of the previous term.

  <math>a_1 = a,\ a_n = a_{n-1}r = ar^n</math>

• The partial sum of the first n terms of the geometric series is:

  <math>S_n = {a(1-r^n) \over 1-r}</math>.

• The sum of any series is:

  <math>S = \lim_{n \to \infty}S_n</math>.

  • Series without sums are divergent.
  • Series with sums are convergent.
• The sum of the geometric series is:

  <math>S = {a \over 1-r}, |r| < 1</math>

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.

#12

Amount of impurity removed after an infinite number of steps:

<math>\begin{align}

S &= {1 \over n} + {1 \over n^2} + {1 \over n^3} + ... \\

 &= {a \over 1 - r},\ a = r = {1 \over n} \\
 &= {{1 \over n} \over {n-1 \over n}} \\
 &= {1 \over n-1} \\
 &= \begin{cases}
      1 & \text{if }n = 2 \\
      {1 \over 2} & \text{if }n = 3 \\
    \end{cases} \\

\end{align} </math>

Since the complete sum is 1 when <math>n=2</math>, you can stop when you have removed enough impurities. Unfortunately, you can't any exact amount, since removing 1/3 of the impurity isn't a partial sum of the series. (The first step removes half!)

#13

This one is only complicated if you don't take everything into account and aren't precise.

We'll start with the assumption that this is geometric series. But <math>r</math> is not necessarily the interest rate: we're depositing money each month.

Let <math>d = 10</math> be the monthly deposit.

Let <math>i = 1.005</math> be the monthly interest rate.

<math> \begin{align} d &= 10 \text{ (Monthly deposit)} \\ i &= 1.005 \text{ (Monthly interest factor)} \\ S_0 &= 0 \text{ (Initial balance)} \\ S_n &= d + (d + S_{n-1})i \text{ (Each month, you earn your deposit plus interest on the deposit and the previous month.)} \\ S_1 &= d + di \text{ (Balance after the first month)} \\

   &= a \\

S_2 &= d + (d + S_1))i \text{ (Balance after the second month)} \\

   &= d + di + (d + di)i \\
   &= a + ar \\

r &= i \\ S_{120} &= {a(1-r^{120}) \over 1-r} \text{ Balance after 120 months)}\\

     &= {(d+di)(1-i^{120}) \over 1-i} \\
     &= {(20.05)(-8.1939673403) \over -0.005} \\
     &= 3,285.78 \\

\end{align} </math>

#14

So we have <math>D</math> in the bank, it earns interest at <math>1.005D</math> per month. Next month, we want to have <math>D+10</math>, so we can withdraw 10 and still have <math>D</math> left.

This is rather trivial. It's simple algebra.

<math> \begin{align}

  D+10 &= 1.005 D \\

0.005 D &= 10 \\

     D &= {10 \over 0.005} \\
       &= 2000 \\

\end{align} </math>

#15

<math> \begin{align} S &= 8 + 8\left({3\over 4}\right) + 8\left({3\over4}\right)^2 + ... \\

 &= {8 \over 1 - {3 \over 4}} \\
 &= 32 in\\

\end{align} </math>

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