# Exponential Function Series Definition

## An alternative definition of the exponential function through infinite series

### Motivation

My previous post about the definition of the exponential function has provided no connection between a well-known characterization (or an alternative definition) of the exponential function:

$$\exp(s) = \lim_{n\to\infty} \sum_{k=1}^n \frac{s^k}{k!}.$$

The term to be summed is simpler than the one in the binomial expansion of $(1 + s / n)^n$.

### Solution

We want this sum to be as small as possible as $n \to \infty$.

$$\sum_{k=2}^n \left( 1 - \frac{n \cdot \dots \cdot (n + 1 - k)} {\underbrace{n \cdot \dots \cdot n}_{n^k}} \right) \, \frac{x^k}{k!}$$

Observe that the fraction is a product

$$\left(1 - \frac1n\right)\left(1 - \frac2n\right) \dots \left(1 - \frac{k-1}{n}\right).$$

The reminds me a first-order approximation inequality

$$1 - \sum_{k=1}^n x_k \le \prod_{k=1}^n (1 - x_k)$$

for each $x_k \in [0,1], k = 1,\dots,n$.

I call this “first-order approximation” because if you put $x_k = c_k \varepsilon$ for each $k = 1,\dots,n$, the LHS would become the first-order approximation of the product on the RHS.

This inequality is a nice induction exercise. The base case for $n = 1$ is trivial. The inductive step is a bit tricky. We can’t apply the case for $n = 2$ first because we aren’t sure what the range of $x_1 + \dots + x_n$ is. Instead we apply the induction hypothesis on $1 - (x_1 + \dots + x_n)$.

\begin{align} &\quad 1 - \sum_{k=1}^{n+1} x_k \\ &\le \left(\prod_{k=1}^n (1-x_k)\right) - x_{n+1} \\ &= \left(\prod_{k=1}^n (1-x_k)\right) (1 {\color{blue}-x_{n+1}}) \underbrace{{\color{blue}+\left(\prod_{k=1}^n (1-x_k)\right) x_{n+1}} - x_{n+1}}_{\le 0} \\ &\le \prod_{k=1}^{n+1} (1-x_k) \end{align}

Put $x_i = \frac{i}{n}$ for each $i = 1,\dots,k$

\begin{align} 1 - \sum_{i=1}^k \frac{i}{n} &\le \prod_{i=1}^k \left(1-\frac{i}{n} \right) <1\\ 0 &< 1 - \prod_{i=1}^k \left(1-\frac{i}{n}\right) \le \sum_{i=1}^k \frac{i}{n} = \frac{k(k+1)}{2n} \end{align}

Put this back into the sum that we want to minimize.

\begin{align} &\quad \sum_{k=2}^n \left( 1 - \frac{n \cdot \dots \cdot (n + 1 - k)} {\underbrace{n \cdot \dots \cdot n}_{n^k}} \right) \, \frac{x^k}{k!} \\ &\le \sum_{k=2}^n \frac{(k-1)k}{2n} \, \frac{x^k}{k!} \\ &\le \frac{x^2}{2n} \sum_{k=2}^n \frac{{\lvert x \rvert}^{k-2}}{(k-2)!} \\ &\le \frac{x^2}{2n} \, M_{\lvert x \rvert} \end{align}

In the last line, we used the fact that the series on the right is absolutely convergent, which I have already shown in equations (4), (5) and (10) in my previous post. To end this post, take $n \to \infty$.

### 1 comment #### AnonymousNov 18, 2023 23:15:51

Amazing blog. Thanks mate!