Department of Mathematics, Statistics
and Computer Science
Wim Ruitenburg's Fall 2011 MATH 1300-101
Last updated: September 2011
Comments and suggestions: Email wimr@mscs.mu.edu
Money from chapter 10
Despite all the money talk, chapter 10 is really about exponential growth, with
examples from money management.
From the chapter we can learn:
- Increasing an amount like 1000 by 5 percent is essentially the same as
multiplying 1000 by (1 + 5/100) or by 105/100 or by 1.05, and the result equals
1050.
Decreasing an amount like 1000 by 5 percent is essentially the same as
multiplying 1000 by (1 - 5/100) or by 95/100 or by 0.95, and the result equals
950.
- Compound interest is very powerful.
When we start with just 1000 (dollars, say), and increase the amount by 5
percent each year for 10 years, the new amount equals 1000 * 1.05 * 1.05 * 1.05
* 1.05 * 1.05 * 1.05 * 1.05 * 1.05 * 1.05 * 1.05, also written 1000 * 1.05^10,
and which equals approximately 1628.8946 (a bank may round this off to
1628.89).
- We give one example of using logarithms to compute big exponents.
Suppose we need to find a number n such that 2^n = 10^25.
We take two steps.
- We search for a number x such that 2 = 10^x.
This number x equals log(2).
The calculator gives that x = log(2) is about 0.301.
So 2 = 10^0.301 approximately.
Substitute in the original equation: (10^0.301)^n = 10^25 approximately.
- Now we have 10^(0.301 * n) = 10^25 approximately.
So 0.301 * n = 25 approximately.
With the calculator we get n = 25/ 0.301 = 83 approximately.
- Suppose inflation neutral interest rates are 5 percent a year.
What is a current fair price P so that we have 1500 in 10 years?
We must solve P * 1.05^10 = 1500.
The formula is P = 1500 / 1.05^10, which equals 920.86988 approximately.
- Increasing an amount like 1000 by 5 percent and then reducing the new
amount by 5 percent is NOT neutral.
For first we multiply 1000 by 1.05 to get 1050, and then we multiply 1050 by
0.95 and get as final result only 997.5.
We spent some time on continuously compounded interest.
We approximate this through the following example.
- We have 1000 dollars in a bank account.
Suppose we get 1 percent interest each day for a year.
After 1 year we get 1000 * (1 + 1/100)^365 dollars, which is approximately
37783 dollars (checked with a calculator).
- A common question is the reverse.
For example, what interest x should we get every day so that after 1 year we
get 10 percent interest?
In other words, find x such that 1000 * (1 + x)^365 = 1000 * (1 + 1/10) = 1100.
So find x such that (1 + x)^365 = 1100 / 1000 = 1.1.
Here is a way to solve this efficiently.
- There is a so-called Euler number e = 2.71828182845999988... with
special properties described below.
- For big values of n we have (1 + 1/n)^n is approximately equal to e.
For example, (1 + 1/365)^365 = e approximately.
- Better still, for big n and for values y we have (1 + y/n)^n is
approximately equal to e^y.
For example (1 + 0.2/365)^365 = e^0.2 approximately (and equals about 1.2214).
- Back to the question about x, rewrite x = y/ 365 and first find y
such that (1 + y/ 365)^365 = 1.1.
So find y such that e^y = 1.1 approximately.
The calculator has a special logarithm function for that called ln.
So y = ln(1.1) which equals 0.09531 approximately.
Then x = y /365 = 0.09531 /365 = 0.000261 approximately.
And thus x = 0.000261 which is 0.0261 percent interest each day.