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Where does Euler’s number ‘e’ come from?

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Maths has a few fun and interesting numbers – numbers that appear in nature, but yet numbers that go on forever. Often these numbers (the most famous of which is probably pi) are taught in school without giving the history behind the number, how it was discovered and where it came from.

One such number is ‘e’ – the exponential or ‘Euler’s number’. It is a transcendental number (in other words, a number that goes on forever and is not the root of any integer polynomial – see and you may have seen it in Euler’s famous equation (e^ipi = -1) – see Euler’s Identity! for more information.

The history behind ‘e’

Even though Euler’s name gets attached to this number, it was actually first discovered by Jacob Bernoulli in 1683 when he was trying to work out how money would grow if compound interest were calculated on a more regular basis than annually. It seems logical, that the more often you compound interest, the more quickly your money grows. The next question is therefore – if you compound more and more frequently, will your money keep increasing forever – and indeed, if you compound instantly, will you have an infinite amount of money in the bank? The answer is no – and the limit you reach is this constant ‘e’.

The maths behind ‘e’

Here’s the Maths behind it (please see for more information).

Suppose you put £1 in a bank. The bank pays 4% interest a year and charges no fees, and this is credited to your account at the end of a year. At end of five years you will have £$(1+0.04)^5$ (if in doubt, cast your mind back to working out ‘compound interest’).

However, if the interest (still at an annual rate of 4%) was “compounded” every quarter instead of every year, then the amount at the end of five years would be £ ($(1+0.04/4)^{4 \times 5}$) because you would be taking 4% of each new amount, every quarter. If you stick this in your calculator, you’ll see that compounding quarterly gives you more money than compounding annually (which hopefully should be fairly intuitive anyway).

Suppose your generous bank gives an interest rate of 100% annually. After one year you would have £ $(1+1)^1$ = £2 whereas if they compounded the interest quarterly, it would be £$(1+1/4)^{4 \times 1} = \pounds 2.43$. If you were even luckier and found a bank that compounded monthly, then you would have £$(1+1/12)^{12 \times 1} = \pounds 2.61$ after one year. And even better, if the bank compounded daily, you would have £ $(1+1/365)^{365 \times 1} = \pounds 2.72$.

It’s obvious that compounding more frequently results in more money in the bank. So if you keep this idea going ad infinitum i.e. you compound continuously, does this lead to an infinite amount in the bank?

A way to express this question mathematically is to look at:

 \[  \lim _{n \rightarrow \infty } (1 + 1/n)^ n. \]  

This quantity turns out again to be $e$ – the same base value with the property that the gradient of the graph is 1 at $x = 0$.

To work out $\lim _{n \rightarrow \infty } (1 + 1/n)^ n$ we do the following:

This series is convergent, and evaluating the sum far enough to give no change in the fourth decimal place (this occurs after the seventh term is added) gives an approximation for $e$ of 2.718.


And there we have it – a link between money in the bank, and a beautiful, naturally occurring number ‘e’ that goes on forever. I love that about Maths!

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