# Exponential Function

## Introduction

Exponential function $f(x) = \exp{x}$ has different form of definitions but all of the definitions are mathematically equivalent. When I was an undergraduate, I did not spend too much time thinking why the definitions are equivalent. Many years later, in this blog post, I would like to discuss the equivalence of the exponential function definitions.

## Prerequisites

### Series Absolute Convergence

A real or complex series $\sum_{n=0}^{\infty} a_n$ is said to converge absolutely if

\begin{align} \sum_{n=0}^{\infty} \lvert a_n \rvert &= \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \lvert a_k \rvert \\ &= L \\ \end{align}

for some real number $L$.

If a real or complex series $\sum_{n=0}^{\infty} a_n$ converges absolutely, we have

\begin{align*} \lim_{n \rightarrow \infty} \lvert a_n \rvert &= \lim_{n \rightarrow \infty} \Big( \sum_{k=0}^{n} \lvert a_k \rvert - \sum_{k=0}^{n-1} \lvert a_k \rvert \Big) \\ &= \lim_{n \rightarrow \infty} \Big( \sum_{k=0}^{n} \lvert a_k \rvert \Big) - \lim_{n \rightarrow \infty} \Big( \sum_{k=0}^{n-1} \lvert a_k \rvert \Big) \\ &= \lim_{n \rightarrow \infty} \Big( \sum_{k=0}^{n} \lvert a_k \rvert \Big) - \lim_{n - 1 \rightarrow \infty} \Big( \sum_{k=0}^{n-1} \lvert a_k \rvert \Big) \\ &= L - L \\ & = 0 \end{align*}

Because $\lim_{n \rightarrow \infty} \lvert a_n \rvert = 0$, we also have $\lim_{n \rightarrow \infty} a_n = 0$. This can be proved by contradiction and we will skip the proof here.

### Ratio Test

Given a real or complex series $\sum_{n=0}^{\infty} a_n$ where $a_n$ is non-zero when $n$ is large, with the limit

$$L = \lim_{n \rightarrow \infty} \Big\lvert \frac{a_{n+1}}{a_n} \Big\rvert$$

the ratio test informs us that

• if $L < 1$, the series converges absolutely;
• if $L > 1$, the series is divergence;
• if $L = 1$ or the limit does not exist, we don’t have a conclusion about convergence.

The ratio could be proved mainly using geometric series and we will skip the proof here.

## Exponential Definitions

There are two common definitions for exponential function, including the series definition and the limit definition. Our goal is to prove the equivalence between the two definitions.

### Series Definition

For a real or complex number $x$, the exponential function $\exp(x) = e^x$ is defined as follows.

\begin{align} \exp{x} &= \sum_{n=0}^{\infty} \frac{x^n}{n!} \\ &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \\ \end{align}

### Limit Definition

For a real or complex number $x$, the exponential function $\exp(x) = e^x$ is defined as follows.

\begin{align} \exp{x} &= \lim_{n \rightarrow \infty} \Big( 1 + \frac{x}{n} \Big)^n \end{align}

## Equivalence of Definitions

Proof

Given real or complex number $x$, we could apply binomial expansion.

\begin{align} \Big( 1 + \frac{x}{n} \Big)^n &= \sum_{k=0}^{n} {n \choose k} \frac{x^k}{n^k} \\ &= \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \frac{x^k}{n^k} \\ &= \sum_{k=0}^{n} \frac{n (n-1) \cdots (n - k + 1)}{k!} \frac{x^k}{n^k} \\ &= \sum_{k=0}^{n} \frac{x^k}{k!} \frac{n (n-1) \cdots (n - k + 1)}{n^k} \\ &= \sum_{k=0}^{n} \frac{x^k}{k!} \Big(1 - \frac{1}{n}\Big) \Big(1 - \frac{2}{n}\Big) \cdots \Big(1 - \frac{k - 1}{n}\Big) \\ \end{align}

We define

$$u_k(n) = \begin{cases} \frac{x^k}{k!} \Big(1 - \frac{1}{n}\Big) \Big(1 - \frac{2}{n}\Big) \cdots \Big(1 - \frac{k - 1}{n}\Big) & \text{if k \leq n} \\ 0 & \text{if k > n} \\ \end{cases}$$

Let’s see if $\lim_{n \rightarrow \infty} u_k(n)$ exists for every possible $k$.

When $k > n$, it is obvious that $\lim_{n \rightarrow \infty} u_k(n) = 0$.

When $k \leq n$,

\begin{align} \lim_{n \rightarrow \infty} u_k(n) &= \lim_{n \rightarrow \infty} \bigg[ \frac{x^k}{k!} \Big(1 - \frac{1}{n}\Big) \Big(1 - \frac{2}{n}\Big) \cdots \Big(1 - \frac{k - 1}{n}\Big) \bigg] \\ &= \bigg(\lim_{n \rightarrow \infty} \frac{x^k}{k!} \bigg) \bigg(\lim_{n \rightarrow \infty} \Big(1 - \frac{1}{n}\Big) \bigg) \bigg(\lim_{n \rightarrow \infty} \Big(1 - \frac{2}{n}\Big) \bigg) \cdots \bigg(\lim_{n \rightarrow \infty} \Big(1 - \frac{k-1}{n}\Big) \bigg)\\ &= \lim_{n \rightarrow \infty} \frac{x^k}{k!} \end{align}

So the question becomes if $\lim_{n \rightarrow \infty} \frac{x^k}{k!}$ exists for every possible $k \leq n$. For example, does $\lim_{n \rightarrow \infty} \frac{x^k}{k!}$ exist when $k = n, n - 1, n - 2, \cdots$?

By ratio test, it is not difficult to find that the series $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ converges absolutely, and $\lim_{n \rightarrow \infty} \frac{x^n}{n!} = 0$, $\lim_{n \rightarrow \infty} \frac{x^{n - 1}}{(n - 1)!} = 0$, etc. The limit $\lim_{n \rightarrow \infty} \frac{x^k}{k!}$ must exist for every possible $k \leq n$.

\begin{align} \lim_{n \rightarrow \infty} u_k(n) &= \lim_{n \rightarrow \infty} \frac{x^k}{k!} \\ &= \frac{x^k}{k!} \\ \end{align}

Therefore,

\begin{align} \lim_{n \rightarrow \infty}\Big( 1 + \frac{x}{n} \Big)^n &= \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{x^k}{k!} \Big(1 - \frac{1}{n}\Big) \Big(1 - \frac{2}{n}\Big) \cdots \Big(1 - \frac{k - 1}{n}\Big) \\ &= \lim_{n \rightarrow \infty} \sum_{k=0}^{\infty} u_k(n) \\ &= \sum_{k=0}^{\infty} \lim_{n \rightarrow \infty} u_k(n) \\ &= \sum_{k=0}^{\infty} \frac{x^k}{k!} \\ &= \sum_{n=0}^{\infty} \frac{x^n}{n!} \\ \end{align}

This concludes the proof.

Lei Mao

12-27-2021

12-27-2021