Growth formulas:

We started with the wheat and chessboard problem where one grain of wheat is placed on the first square of a chessboard, two on the next, four on the next ... etc. until we reach the 64th.

This situation was modeled by: 1) A=1*2t

We then convinced ourselves that a more general formula would be:

2) A=A0*Bt

where our present (not original) amount was multiplied by B each time unit.

From there , we concluded that increasing our present amount by r each year was the same as multiplying it by 1+ r, so we got the following formula:

  1. A=A0*(1+r)t

We then looked at Moore’s law of computer transistors doubling every 18 months and came up with:

4) A=2500*2t/1.5

This led us to the general formula::

  1. A=A0*Bt/k where k is the number of time units that it takes to multiply our original amount by B. The division by k adjusts for the fact we are not multiplying by B each time unit but at some (proper or improper) fraction of a time unit.

One of the students in our class, Lauren, asked an interesting question concerning formula 3) where interest problems are concerned: What would happen if you divided the r and multiplied the t by the same number? Would this just change the time unit and give us the same answer? The answer is no. The resulting formula:

6) A=A0*(1+r/m)mt

This is the formula for compounding your interest m times a time unit (usually years in this case) This formula will yield a higher amount when m>1 since you are getting interest four times during the year. A chart of interest compounded yearly vs. quarterly is helpful in showing this fact.

 

Let A0=1000, and r = 5%

0ms 3ms 6ms 9ms 1 yr

yearly: 1000 1000 1000 1000 1050

quarterly: 1000 1012.50 1025.16 1037.97 1050.95

Question: What would happen if your compounded every day? Minute? Second? Microsecond? Continuously?

This means that in the formula:

A=A0*(1+r/m)mt you let m go to infinity. Since r is fixed, this will yield the same result as letting m/r go to infinity.

A=A0*(1+r/m)mt

A=A0*[(1+r/m)m/r]rt

Let m/r= some number n, then let n go to infinity:

A=A0*[(1+1/n)n]rt

As n goes to infinity (1+1/n)n goes to the number e which is approximately 2.72.

Which gives us the new formula for continuous compounding:

7) A=A0*ert

Let’s look at a problem involving three of our formulas:

Find the amount after five years if $1000 is invested at 5% compounded:

  1. Yearly: A = A0*(1+r)t -> A = 1000(1.05)^5 =1276.28

b) Quarterly: A = A0*(1+r/m)mt -> A = 1000(1.0125)^20 =1282.04

c) Continuously: A = A0ert -> A = 1000e^.25 =1284.03