# College Math Teaching

## August 1, 2015

### Interest Theory: discounting

Filed under: applied mathematics, elementary mathematics — Tags: , — collegemathteaching @ 10:29 pm

Some time ago, I served in the U. S. Navy. The world “Navy” was said to be an acronym for Never Again Volunteer Yourself. But I forgot that and volunteered to teach a class on Mathematical interest theory. That means, of course, I have to learn some of this, and so I am going over a classic text and doing the homework.

The math itself is pretty simple, but some of the concepts seem strange to me at this time. So, I’ll be using this as “self study” prior to the start of the semester, and perhaps I’ll put more notes up as I go along.

By the way, if you are interested in the notes for my undergraduate topology class, you can find them here.

Discounting: concepts, etc. (from this text) (Kellison)

Initial concept:

Suppose you borrow 100 dollars for one year at 8 percent interest. So at time 0 you have 100 dollars and at time 1, you pay back 100 + (100)(.08) = 108.
Now let’s do something similar via “discounting”. The contract is for 100 dollars and the rate is an 8 percent discount. The bank takes their 8 percent AT THE START and you end up with 92 dollars at time zero and pay back 100 at time 1.

So the difference is: in interest, the interest is paid upon pay back, and so the amount function is: $A(t) = (1+it)A(0)$. In the discount situation we have $A(1)(1-d(1)) = A(0)$ where $d$ is the discount rate. So the amount function is $A(t) = \frac{A(0)}{1-dt}$ where $t \in [0, \frac{1}{d})$

If we used compound interest, we’d have $A(t) = (1+i)^tA(0)$ and in compound discount we’d have $A(t) = \frac{A(0)}{(1-d)^t}$

This leads to some interesting concepts.

First of all, there is the “equivalence concept”. Think about the above example: if getting 92 dollars now lead to 100 dollars after one period, what interest rate would that be? Of course it would be $\frac{8}{92} = .087$. So what we’d have is this: $i = \frac{d}{1-d}$ or $d = \frac{i}{1+i}$.

Effective rates: this is only of interest in the “simple interest” or “simple discount” situation.

Let’s start with simple interest. The amount function is of the form $A(t) = (1 +it)A(0)$. The idea is that if you invest, say, 100 dollars earning, say, 5 percent simple interest (NO compounding), then in one year you get 5 dollars of interest, 2 years, 10 dollars of interest, 3 years 15 dollars of interest, etc. You can see the problem here; say at the end of year one your account was worth 105 dollars and at the end of year 2, it was worth 110 dollars. So, in effect, your 105 dollars earned 5 dollars interest in the second year. Effectively, you earned a lower rate in year 2. It got worse in year 3 (110 earned only 5 dollars).

So the EFFECTIVE INTEREST in period $n$ is $\frac{A(n) - A(n-1)}{A(n-1)} = \frac{1 + ni)-(1+(n-1)i)}{1+(n-1)i}=\frac{i}{1+(n-1)i}$ which you can see goes to zero as $n$ goes to infinity.

Effective discount works in a similar manner, though we divide by the amount at the end of the period, rather than the beginning of it: $\frac{A(n)-A(n-1)}{A(n)} = \frac{\frac{1}{1-nd} - \frac{1}{1-(n-1)d}}{\frac{1}{1-nd}} = \frac{d}{1-(n-1)d}$