Einstein is supposed to have said that compound interest is the eighth wonder of the world. This quote is apocryphal, and there is no authoritative source for this. Nevertheless, it is a good place to appreciate compound interest, which is the most common "real-world" phenomenon relating to the exponential function. This is just a collection of derivations of rough approximations related to the compound interest.
There's nothing of great significance here. I intend to collect over the years, a bunch of heuristics related to compound interest. The rules are all at high-school level, even though I found that online sources do not give derivations.
The rule of 72 says that if the compound interest rate is N% per annum, where N is between 6 and 10, then the number of years to double the initial capital is 72/N.
So for example, if your interest rate is 6 percent, then it will take approximately 72/6 = 12 years to double.
The origins of this rule go back to Luca Pacioli, the giant of medieval mathematics.
The derivation is "straight"forward: Suppose the initial capital is 1, the rate is r, and the number of years is N, then for doubling, we must have
N
(1+r) = 2.
Taking logarithms on both sides, we have
N = 1/ (log (1+r))
2
So N follows an inverse logarithmic curve in 1+r. It just so
happens that the convenient approximation for reasonable rates
(between 6 and 10) happens to be 72/100r. [See comparison at
Wolfram Alpha.] Outside this range,
the approximation is very inaccurate.
This phenomenon, approximating log(1+x) with constant multiples of x, is a basic inequality often used in information theory.
What about higher interest rates? Are there simple ways to modify the above rule to get good precision for higher interest rates? Simple calculations show the values of K/100r (note that r is a number between 0 and 1 - so 33 percent compound interest would be represented by r=0.33) to be approximately correct at the following points. (For in-between rates, we can use intermediate values.)
| Interest rate | K for "rule of K" |
| .1 | 72 |
| .15 | 74 |
| .2 | 76 |
| .25 | 77 |
| .33 | 80 |
| .4 | 82 |
| .5 | 85 |
Note on a possible point of confusion: when the interest rate increases, time to double should decrease. So it may seem like K should decrease when interest rate increases. The actual time to double decreases. However, the question here is different - we are asking which K/(100r) best approximates 1/log2(1+r) around the given value of r. For this question, K mildly increases when r increases, as above.
Suppose you get a loan of 1 lakh from a bank at r percent monthly interest. Then how do we calculate the EMI if the amount has to be paid back over, say 20 months? This is a nice application of the geometric progression. Suppose the EMI we finally determine is called M.
Month 1: The bank gives 1 lakh.
Month 2: the bank would have earned interest if it had not
given it to you. This amount is 1 (1+r). You pay M. So the
debt to the bank is now
1(1+r) - M.
2
Month 3: the bank would have had 1(1+r) . You gave
M in month 1, it would have earned interest, and is
equivalent to M(1+r). You pay another M this month, so the
debt to the bank is now
2
1(1+r) - M(1+r) - M
Month 4:
3 2
1(1+r) - M(1+r) - M(1+r) - M
...
Month N:
N N-1 N-2
1(1+r) - M(1+r) - M(1+r) - ... - M
Suppose at this stage, the payment is over. Then using the formula for the sum of a finite geometric series, we get
N
M (1+r) - 1 N
---------- = 1(1+r)
r
from which we get
N
M = loan x r x (1+r)
------------
N
[(1+r) - 1]
How do we make sense of this to decide our repayment schedule? Let's use some realistic values, from the perspective of loans in India as of this writing. For annual interest rate of 10 percent, the monthly rate is about 0.0083. Supposing that it is to be paid back over 10 years = 120 months, the multiplying factor (1+r)N/((1+r)N-1) is about 1.6. For the same r, but with 240 month payment schedule, the mutliplication factor is about 1.15. This shows that as the repayment schedule gets longer, the EMI reduces, but the total amount paid over the entire period is often much more. (EMI decreases at a lower rate, whereas the total amount increases linearly with N).