Compound Interest and Euler's number $e$

January 24, 2018

The Euler's constant $e$ appears whenever an extremely small change (increment or decrement) is applied an extremely large number of times e.g. an increment of 1% (1.01) applied a 100 times. It is easy to understand this concept of applying a small change a large number of times using compound interest.

The formula for compount interest is given by,

\begin{align} A = P\left(1 + \frac{r}{100.n}\right)^n \end{align}

where $P$ is the principal, $r$ is the yearly rate of interest in percentage, $n$ is the number of compounding periods and $A$ is the total amount at the end of 1 year. Let, $P=1$ and $r=100$.

If the interest is compounded annually, i.e. $n=1$, then

$$A = (1+1)^1 = 2$$

If the interest is compounded semiannually, i.e. $n=2$, then

$$A = \left(1+\frac{1}{2}\right)^2 = 2.25$$

If the interest is compounded quarterly, i.e. $n=4$, then

$$A = \left(1+\frac{1}{4}\right)^4 \approx 2.441$$

If the interest is compounded monthly, i.e. $n=12$, then

$$A = \left(1+\frac{1}{12}\right)^{12} \approx 2.613$$

If the interest is compounded daily, i.e. $n=365$, then

$$A = \left(1+\frac{1}{365}\right)^{365} \approx 2.714$$

If the interest is compounded hourly, i.e. $n=365\times24$, then

$$A = \left(1+\frac{1}{365\times24}\right)^{365\times24} \approx 2.71812$$

If the interest is compounded every second, i.e. $n=365\times24\times3600$, then

$$A = \left(1+\frac{1}{365\times24\times3600}\right)^{365\times24\times3600} \approx 2.71828$$

In the above situations, we observe that an extremely small change, say $(1+\frac{1}{n})$ is being applied an extremely large number of times i.e. $(1 + \frac{1}{n})^n$. In the limiting case, if the interest is compounded continuously, the maximum amount $A$ that can be obtained at the end of one year is given by

\begin{align} A &= \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n \\ &= \lim_{n\to\infty} \left(\binom n0 + \binom n1 \frac{1}{n} + \binom n2 \frac{1}{n^2} + \binom n3 \frac{1}{n^3} + \dots \right) \\ &= \lim_{n\to\infty} \left(1 + \frac{1}{1!} + \frac{n(n-1)}{2!}\frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!}\frac{1}{n^3} + \dots \right) \\ &= 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \\ &= e \\ &\approx 2.718 281 828 \end{align}

This behaviour of applying an extremely small change an extremely large number of times occurs in many natural phenomena such as charging of a capacitor, decay of radioactive isotopes etc. In such processes, the small change is generally proportional to