The constant known as e

What is $e$?

To grasp the concept of $e$, imagine this scenario: You've invested $100$ units into a financial scheme promising a $100$ return over a year. While such a scheme is definitely a scam in real life, let's overlook that for this explanation.

A year later, your investment doubles to $200$ units, marking a growth factor of $2$. But what happens if you opt for a semester-long investment instead, achieving a $50$ return, then reinvest the total (now $150$ units) for another semester? The growth factor of your investment becomes:

$$(1,5)^2 = 2,25$$

Notice how it increases. Let's break down the investment further into quarters, each with a 25% return. The growth now becomes to:

$$(1,25)^4 = 2,441$$

What if the investment were even more frequent, say monthly? This is the total growth factor at the end of the year:

$$(1 + \frac{1}{12})^{12} = 2,613$$

The formula for calculating the growth factor is thus:

$$(1 + \frac{1}{n})^n$$

Does this mean the growth factor increases infinitely? Not quite. Eventually, it approaches a constant value, known as $e$. Hence, $e$ is defined by the term:

$$\lim\limits_{n \to \infty} (1 + \frac{1}{n})^n$$

But, how do we calculate this? Well, this is where binomial expansion comes to place. Remember the formula for binomial expansion:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

Where $\binom{n}{k}$ is defined as:

$$\frac{n!}{(k!)(n-k)!}$$

Now let's plug our definition for $e$ to the binomial expansion:

$$\begin{split} e &= \lim\limits_{n \to \infty} \sum_{k=0}^{n} (\frac{n!}{(k!)(n-k)!}) 1^{n-k} (\frac{1}{n})^k \\ &= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{n!}{(k!)(n-k)!} \cdot \frac{1}{n^k}\\ &= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{1}{k!} \cdot \frac{n!}{(n^k)(n-k)!} \end{split}$$

Now, we can focus on this part of the equation:

$$\frac{n!}{(n^k)(n-k)!} = \frac{n!}{(n-k)!} \cdot \frac{1}{n^k}$$

The fraction $\frac{n!}{(n-k)!}$ means $(n)(n-1)(n-2)\cdots(n-k+1)$. Since $n-k$ becomes closer to $n$ as $n$ reaches infinity, then:

$$\begin{split} \lim\limits_{n \to \infty} \frac{n!}{(n-k)!} \cdot \frac{1}{n^k} &= \lim\limits_{n \to \infty} \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-k+1}{n} \\ &= 1 \cdot 1 \cdots 1 \\ &= 1 \end{split}$$

From here we can incorporate it to our binomial expansion:

$$\begin{split} e &= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{n!}{(n^k)(n-k)!} \cdot \frac{1}{k!} \\ e &= \sum_{k=0}^{\infty} \frac{1}{k!} \end{split}$$

Getting the value of $e$ using binomial expansion is basically implementing a specific case of $e^x$ where $x=1$

Let us try to get the approximation of $e^x$ then, by expanding $\lim\limits_{n \to \infty} (1 + \frac{1}{n})^{nx}$:

$$\begin{split} e^x &= \lim\limits_{n \to \infty} \sum_{k=0}^{n} \binom{nx}{k} 1^{nx-k}(\frac{1}{n})^k \\ &= \lim\limits_{n \to \infty} \sum_{k=0}^{n} (\frac{nx!}{(k!)(nx-k)!})(\frac{1}{n})^k \\ &= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{nx!}{(k!)(nx-k)!} \cdot \frac{1}{n^k}\\ &= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{1}{k!} \cdot \frac{nx!}{(n^k)(nx-k)!} \end{split}$$

Just like with $\frac{n!}{(n-k)!}$ we can expand $\frac{nx!}{(nx-k)!}$ as $n$ approaches infinity, $nx-k$ becomes closer to $nx$:

$$\begin{split} \lim\limits_{n \to \infty} \frac{nx!}{(nx-k)!} \cdot \frac{1}{n^k} &= \lim\limits_{n \to \infty} \frac{nx}{n} \cdot \frac{nx-1}{n} \cdots \frac{nx-k+1}{n} \\ &= x \cdot x \cdots x \\ &= x^k \end{split}$$

Plugging it back to the sum, we get:

$$\begin{split} e^x &= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{nx!}{(n^k)(nx-k)!} \cdot \frac{1}{k!} \\ &= \sum_{k=0}^{\infty} \frac{x^k}{k!} \end{split}$$

To derive $e^x$ we can do the following:

$$\begin{split} e^x &= \sum_{k=0}^{\infty} \frac{x^k}{k!}\\ &= \frac{x^0}{0!} + \sum_{k=1}^{\infty} \frac{x^k}{k!}\\ &= 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}\\ \frac{de^x}{dx} &= 0 + \sum_{k=1}^{\infty} \frac{k \cdot x^{k-1}}{k!} \\ &= \sum_{k=1}^{\infty} \frac{x^{k-1}}{(k-1)!}\\ &= \sum_{k=0}^{\infty} \frac{x^{k}}{k!}\\ &= e^x \end{split}$$

TODO