Euler's constant e

Euler's constant, denoted by $e$ , is the number

e=2.71828182845904523536028747135266249775724709...

This number is the basis of some functions with very nice properties, some of which we will get to know later in greater detail. It is also part of arguably one of the most beautiful equations we know:

e^{i\cdot \pi}=-1

This equation is regarded as beautiful because it combines three very important and seemingly disparate numbers ( $\pi, e$ , and the imaginary number $i$ ) in an unexpected way, and the result is ... $-1$ .

Like $\pi$ , the number $e$ is an irrational number and as such has a never ending, non-repeating decimal representation. We can give geometrical meaning to $\pi$ - it is the circumference of a circle (of diameter $1$ ). In this section, we discuss where the number $e$ occurs.

The constant $e$ occurs naturally in a financial context. Assume you have a bank account with $1\text{ EUR}$ in it. Let's assume the bank is a generous one, and you receive $100\%$ interest per year. That is, after one year you get $100\%$ of the amount in your account, so $1\text{ EUR}$ . After one year, you therefore have

2\text{ EUR}

in your account.

Now, here is the interesting part. You asked the bank if you could split the interest in two, and get the interest paid every half year as follows: after half a year you get $50\%$ of the amount in your account, which is $0.50\text{ EUR}$ , so you have now $1.50 \text{ EUR}$ in your account. After another half a year, you get the other $50\%$ from the current amount (which is $1.50\text{ EUR}$ ), so you get another $0.75\text{ EUR}$ . Thus, the total amount in your account after one year is now:

2.25\text{ EUR}

Hmm, so get get more money out of your $100\%$ interest. Splitting the interest in this way is called compound interest, which makes sense, as one of the meanings of the word "compound" means "das Gemisch". Here you get interest from your money (the $1\text{ EUR}$ ) but also from the interest of $0.50\text{ EUR}$ . We also say that the $100\%$ interest is compounded twice per year.

The question is, of course, if we compound more often per year, can we get even more money? Simply by compounding every day, for example, can we become millionaires? Here is an exercise that answers this question.

Exercise 1

You have $1\text{ EUR}$ in your account and the bank pays $100\%$ interest per year. How much money do you have in your account after a year if

compounded three times per year?
$365$ times per year (that is, every day)?
$n$ times per year, where $n$ is an arbitrary natural number?

If you keep increasing $n$ , will you get an unlimited amount of money after a year, or is there a limit of what you can get? If so, what is this limit?

Solution

Denote by $y$ the amount in your account after one year.

Every third part of the year, you get
$\frac{100\%}{3}=33.\overline{3}\%$
of the current amount. This is exponential growth with a growth factor of
$1+\frac{33.\overline{3}}{100}=1+\frac{1}{3}$ $\begin{array}{rllll} \text{amount}:& 1 &\xrightarrow[]{\cdot (1+1/3)} & \frac{4}{3} &\xrightarrow[]{\cdot (1+1/3)} & ... & \xrightarrow[]{\cdot (1+1/3)} & y\\ \text{year}:& 0 & \xrightarrow[]{+1/3} & \frac{1}{3} & \xrightarrow[]{+1/3} & \frac{2}{3} & \xrightarrow[]{+1/3} & 1\\ \end{array}$
Thus, after one year the amount is
$y=1\cdot \left(1+\frac{1}{3}\right)^3 =\underline{2.370...}\text{ EUR}$
Percent increase every $365^\text{th}$ part of the year is
$\frac{100\%}{365}=\frac{100}{365}\%$
of the current amount. This is exponential growth with a growth factor of
$1+\frac{\frac{100}{365}}{100}=1+\frac{1}{365}$ $\begin{array}{rllll} \text{amount}:& 1 &\xrightarrow[]{\cdot (1+1/365)} & \frac{366}{365} &\xrightarrow[]{\cdot (1+1/365)} & ... & \xrightarrow[]{\cdot (1+1/365)} & ...& \xrightarrow[]{+1/365} & y\\ \text{year}:& 0 & \xrightarrow[]{+1/365} & \frac{1}{365} & \xrightarrow[]{+1/365} & \frac{2}{365} & \xrightarrow[]{+1/365} & ... & \xrightarrow[]{+1/365} & 1\\ \end{array}$
Thus, after one year the amount is
$y=1\cdot \left(1+\frac{1}{365}\right)^{365} =\underline{2.714...}\text{ EUR}$
Similar to above, we get $y=1\cdot \left(1+\frac{1}{n}\right)^n$ Euros.

Inserting a couple of $n$ into the formula above, we see that for increasing $n$ the amount $y$ approaches the value $e$ !
$\begin{array}{r| l | l} n & y=\left(1+\frac{1}{n}\right)^n\\\hline 1\,000 & \mathbf{2.71}69239... \\ 10\,000 & \mathbf{2.718}1459...\\ 100\,000 & \mathbf{2.7182}682...\\ 1\,000\,000 & \mathbf{2.71828}04...\\ 10\,000\,000 & \mathbf{2.718281}6...\\ \end{array}$
The digits in bold are similar to the ones in the number $e$ .

To summarise the example above:

Theorem 1

For $n$ approaching $\infty$ , the expression

\left(1+\frac{1}{n}\right)^n

approaches $e$ (and actually my serve as a definition of the number $e$ ). In short, we can write

\left(1+\frac{1}{n}\right)^n \xrightarrow[]{n\rightarrow \infty} e

Can you generalise compound interest to arbitrary initial amounts and arbitrary percent interest? See the next exercise.

Exercise 2

You have $120\text{ EUR}$ in your account. The bank gives you $5\%$ interest per year. If compounded every month, how much money do you have after one year?
You have $A$ Euros in a bank account, and the interest per year is $p\%$ . If you compound $n$ times per year, what is the amount $y$ after one year?

Solution

Every month the amount increases by $\frac{5\%}{12}=\frac{5}{12}\%$ . Thus, the growth factor is
$1+\frac{\frac{5}{12}}{100}=1+\frac{5}{12\cdot 100}$
and thus
$y=120\cdot \left(1+\frac{5}{12\cdot 100}\right)^{12}=126.139\text{ EUR}$
Similar to above, we get
$y=A\cdot \left(1+\frac{p}{n\cdot 100}\right)^n$

Exercise 3

What values does the term below approach for $n\rightarrow \infty$ ?

$\left(1+\frac{x}{n}\right)^n$
$A\cdot \left(1+\frac{p}{n\cdot 100}\right)^n$

Solution

$\left(1+\frac{x}{n}\right)^n=\left(1+\frac{1}{n/x}\right)^n = \left(\left(1+\frac{1}{n/x}\right)^{n/x}\right)^x =\left(\left(1+\frac{1}{m}\right)^{m}\right)^x \rightarrow e^x$
$A\cdot \left(1+\frac{p}{n\cdot 100}\right)^n=A\cdot \left(1+\frac{p/100}{n}\right)^n = Ae^{p/100}$