Maths in a minute: Compound interest and e

The only good thing about debt is that it's connected to one of the most important constants in maths: the number $e$.

To see the connection, suppose you borrow £$ 100$ at an annual interest rate of $3\%$. How much will you owe after three years if you don't make any repayments? It's tempting to think that you'll owe the original £$ 100$ plus $9\%$ of £$ 100$, which is £$9$, giving a total of £$109$. Unfortunately, though, this calculation is not quite correct. After one year you will owe the original amount, £$100$, plus the interest from the first year, which is $$0.03 \times 100 = 3.$$ This gives a total debt of $$100 + 0.03 \times 100 = 100 \times (1+0.03) = 103.$$ To find out what you owe after another year, you need to add the interest from the second year, which is $$0.03 \times (100 \times (1+ 0.03)) = 3.09.$$ After the second year the interest is therefore $$100 \times (1+0.03) + 0.03 \times (100 \times (1+0.03)) = 100 \times (1 +0.03)^2 = 106.09.$$ The interest in the third year will be $$0.03 \times (100 \times (1 + 0.03)^2) = 3.18,$$ so after the third year you will owe a total of $$100 \times (1+0.03)^3 = 109.27.$$ You can see that the amount of interest accrued in each year is more than the amount accrued in the previous year. If you've borrowed more than £$ 100$ this growth, or compounding, of the interest is much more noticeable: it is the curse of debt and, equally, the blessing of saving. You may have seen a pattern in the expressions above. And indeed, there is a general expression describing how your total debt will grow: $$P (1 + r)^n$$ where $P$ is the initial amount you borrowed, $r$ is the rate of interest (where $r$ is written as a decimal number, such as $0.03$, rather than a percentage, $3\%$) and $n$ is the number of times the interest is compounded. The more often the interest is compounded, the greater the total, which is where you have to be careful. To make things simpler, suppose you borrow £$1$ ($P=1$) at an annual interest rate of $100\%$ ($r=1$). If the bank just calculates your total amount at the end of the first year, you will owe: $$1 \times (1+1)^1 = 1 \times 2 = 2.$$ But what if the bank decided to compound the interest quarterly, increasing the total every three months using an interest rate of $100/4 = 25\%$? Then after one year you would owe: $$1 \times ( 1+1/4)^4 = 2.44.$$ Compounding monthly would result in a total of $$1 \times (1 + 1/12)^{12} = 2.61.$$ Daily compounding would give $$1 \times (1+1/365)^{365} = 2.71$$ If the bank compounds $n$ times in a year you will owe: $$1 \times (1+1/n)^n$$ and the more times they compound, the greater the value for $n$, the more you will owe. If you take things to the limit, calculating this for higher and higher values of $n$, you'll find the total amount you'd owe converges to $$2.71828182845904523536028747135266249775724709369995...$$ We call this number $e$, the limit of the value for our compounding formula as $n$ tends to infinity. The number $e$ encapsulates growth, and can often be used to rewrite complicated equations describing growth in a much simpler way.

The person to first catch sight of the number $e$ in the context of compound interest was the mathematician Jacob Bernoulli in 1683. Bernoulli tried to find the limit of the expression

$$\left( 1+\frac{1}{n}\right)^n$$ as $n$ tends to infinity and managed to work out that it was between $2$ and $3$. It was the first time in history that a number was defined as a limit.