Interest
Let's look at the compound interest one gets from a savings account. To see how much a given amount M of money would yield after time t, we have to consider two factors: (a) the yearly interest rate r (usually given as a percentage), and (b) how often the interest is computed and added. (In parts 1 and 2 use a TI-108 calculator; after that use either a TI-30 IIS or a TI-34 II.)
1. Let's consider first the "old" way of computing interest.
Let's say that you left $1000 in a bank, the interest rate was 5%, and the interest was computed once a year. So,
After: you had:
1 year $1000*1.05 = $1050
2 years $1050*1.05 = $1102.50
3 years $1102*1.05 = $1157.63 (.625 was rounded)
Now assume that you withdraw the money after 3 months (1/4 of a year) of the 4th year. You got only 1/4*5% = 1.25% interest for that period. So,
3 years 3 mo. $1157.63*1.0125 = $1172.10 (.0953 rounded)
2. Let's compute the interest more often.
Now assume that interest was added every 6 months. So every 6 months, 1/2*5% = 2.5% of the current sum was added. The situation would be a little different.
After: you would have:
.5 year $1000*1.025 = $1025
1 year $1025*1.025 = $1050.63
1.5 years $1050.63*1.025 = $1076.90
2 years $1076.90*1.025 = $1103.82
Thus after 2 years you would get $1.32 more than when the interest was computed yearly.
What happens if we compute the interest monthly, weekly, daily, every hour, every minute, every second, or all the time?
3. "Continuous" computation of interest.
If we compute the interest n times a year, each time we add 1/n*p% of the current sum. In other terms, each time we multiply the current amount by 1 + i/n, where i = p/100. So after a period of t years (t doesn't have to be a whole number), our initial investment M would increase to,
M*(1 + i/n)^(n*t) = M*((1 + i/n)^n)^t
Thus, if we compute interest rate n times a year we get as much as if the actual rate were (1 + i/n)^n instead of i.
3. Comparing (1 + i/n)^n to i.
Let's compute (1 + i/n)^n, for i = .05 (use a scientific calculator).
A program for the TI-34 II is, (1+.05/Ans)^Ans.
| n = | (1 + 0.05/n)^n |
| 1 | 1.05 |
| 2 | 1.050625 |
| 4 | 1.050945337 |
| 12 | 1.051161898 |
| 52 | 1.051245842 |
| 364 | 1.051267486 |
| 1000 | 1.051269782 |
| 10000 | 1.051270965 |
Study these numbers carefully. It seems that when n increases, they converge to a specific value. It can be proven to be so, and the value is e^i.
Here,
e^.05 = 1.051271096376
Remark.
Enter e^(.05) =, getting 1.0551271096. Then compute additional digits that were not displayed, by entering, Ans-1.0551271096 =.
Add the answer 3.76x10^-10 to the previous value.
The fact that e^(.05) is smaller than the last value in the table is an artifact of the calculation, which is due to rounding errors.
4. The modern way of computing interest.
Most of the time, the formula, M*e^(it) is used to compute the amount left for time t with interest rate i.
5. About the constant e.
This constant is a very important one, and it pops up in many problems involving computations. There are many very efficient method of computing approximate values of e with any required accuracy. The number e is an irrational number.