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.