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# Intuitive Guide on Euler's Number

Probability and Statistics
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Preliminary mathematics
schedule Oct 28, 2022
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This guide requires that you are familiar with the concept of compound interests. If not, please consult our guide on compound interests.

# Motivating example

Recall that the formula to compute compound interest:

\begin{align*} A=P\Big(1+\frac{r}{n}\Big)^{nt} \end{align*}

Where:

• $A$ is the final amount in the bank account after $t$ years.

• $P$ is the principal, that is, the starting amount in the bank account.

• $r$ is the interest rate (e.g. $0.1$).

• $n$ is the number of compounds per year.

• $t$ is the number of years elapsed.

Now, consider a simple case with the following assumptions:

• we initially start with one dollar in the bank account, that is, $P=1$.

• the interest rate is $100\%$, that is, the bank will double our money at every compound period. This would mean $r=1$.

• we are interested in our bank balance after one year, that is, $t=1$.

For this simple case, our formula for compound interest becomes:

\begin{align*} A=\Big(1+\frac{1}{n}\Big)^{n} \end{align*}

The only variable here is $n$, which is the number of compounds per year. For instance, $n=1$ for annual compounds, whereas $n=4$ for quarterly compounds. Let's now compute $A$ for different values of $n$ like so:

$n$

$A$

$n$

$A$

1 (annually)

2.0000

52 (weekly)

2.6926

2 (bi-annually)

2.2500

365 (daily)

2.7146

4 (quarterly)

2.4414

8760 (hourly)

2.7181

12 (monthly)

2.6130

525600 (per min)

2.7183

Note that the values of $A$ are all rounded to four decimal places. We can see that as $n$ increases, the value of $A$ also increases, but the amount of increase slows down. For instance, increasing $n$ from $1$ to $2$ gives us an additional $\$0.25$, but increasing$n$from$365$to$8760$only gives us around$\$0.003$ additionally.

As a result, $A$ seems to be converging to some constant of around $2.718$; this constant is the famous Euler's constant $e$. The larger the $n$, the closer we approach $e$, hence it makes sense to define $e$ as the limit of $A$ as $n$ tends to infinity.

Example.

# Formal definition of Euler's constant

Euler's constant $e$ is defined as follows:

$$e=\lim_{n\to\infty} \Big[\Big(1+\frac{1}{n}\Big)^n\Big]$$

Note that $e$ is irrational:

$$e=2.71828...$$

Let's now expand on the formal definition of Euler's constant and express $e^\lambda$ as a limit where $\lambda$ is some scalar number.

Theorem.

# Expressing the power of Euler's constant as limits

If $\lambda\in\mathbb{R}$, then:

$$e^\lambda= \lim_{n\to\infty}\Big(1+\frac{\lambda}{n}\Big)^n$$

Proof. From the basic rules of taking powers, we have that:

$$\Big(1+\frac{\lambda}{n}\Big)^{n} =\Big[\Big(1+\frac{\lambda}{n}\Big)^{n/\lambda}\Big]^\lambda$$

Now, let's take the limit as $n$ tends to infinity:

$$\begin{equation}\label{eq:yTGF12ryu589R5LxGty} \lim_{n\to\infty}\Big(1+\frac{\lambda}{n}\Big)^{n} = \lim_{n\to\infty}\Big[\Big(\Big(1+\frac{\lambda}{n}\Big) ^{n/\lambda}\Big)^\lambda\Big] \end{equation}$$

We now want to show that the right-hand side is equal to $e^\lambda$. Let's define a new variable:

$$k=\frac{n}{\lambda}$$

Notice how $k$ also tends to infinity as $n$ tends to infinity. Rewriting the right-hand side of \eqref{eq:yTGF12ryu589R5LxGty} in terms of $k$ gives:

$$\lim_{k\to\infty}\Big[\Big(\Big(1+\frac{1}{k}\Big) ^{k}\Big)^\lambda\Big]$$

Using the power rule of limits, we can take the power of $\lambda$ outside the limit:

\begin{equation}\label{eq:PxSqujAPFRRw0WW1Wlc} \begin{aligned}[b] \lim_{k\to\infty} \Big[\Big(\Big(1+\frac{1}{k}\Big)^{k}\Big)^\lambda\Big] &=\Big[ \lim_{k\to\infty} \Big(\Big(1+\frac{1}{k}\Big)^{k}\Big)\Big]^\lambda\\ &=e^\lambda\\ \end{aligned} \end{equation}

Substituting \eqref{eq:PxSqujAPFRRw0WW1Wlc} into \eqref{eq:yTGF12ryu589R5LxGty} gives us the desired identity:

$$e^\lambda= \lim_{n\to\infty}\Big(1+\frac{\lambda}{n}\Big)^n$$

This completes the proof.