# Equivalence of Definitions of Euler's Number

## Theorem

The following definitions of the concept of Euler's Number are equivalent:

### Limit of Series

The series $\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ converges to a limit.

This limit is Euler's number $e$.

### Limit of Sequence

The sequence $\sequence {x_n}$ defined as $x_n = \paren {1 + \dfrac 1 n}^n$ converges to a limit as $n$ increases without bound.

That limit is called Euler's Number and is denoted $e$.

### Base of Logarithm

The number $e$ can be defined as the number satisfied by:

$\ln e = 1$

where $\ln e$ denotes the natural logarithm of $e$.

That $e$ is unique follows from Logarithm is Strictly Increasing.

### Exponential Function

The number $e$ can be defined as the number satisfied by:

$e := \exp 1 = e^1$

where $\exp 1$ denotes the exponential function of $1$.

## Proof 1

See Equivalence of Definitions of Real Exponential Function: Inverse of Natural Logarithm implies Limit of Sequence for how $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n = e$ follows from the definition of $e$ as the number satisfied by $\ln e = 1$.

See Euler's Number: Limit of Sequence implies Limit of Series for how $\displaystyle e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ follows from $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n = e$.

Now suppose $e$ is defined as $\displaystyle e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$.

Let us consider the series $\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$.

From Series of Power over Factorial Converges, this is convergent for all $x$.

We differentiate $f \left({x}\right)$ with respect to $x$ term by term (justified by Power Series is Differentiable on Interval of Convergence), and get:

 $\displaystyle D_x \left({f \left({x}\right)}\right)$ $=$ $\displaystyle D_x \left({1}\right) + D_x \left({\frac {x} {1!} }\right) + D_x \left({\frac {x^2} {2!} }\right) + D_x \left({\frac {x^3} {3!} }\right) + \cdots + D_x \left({\frac {x^n} {n!} }\right) + D_x \left({\frac {x^{n+1} } {\left({n+1}\right)!} }\right) + \cdots$ $\displaystyle$ $=$ $\displaystyle 0 + 1 + \frac {2 x} {2!} + \frac {3 x^2} {3!} + \cdots + \frac {n x^{n-1} } {n!} + \frac {\left({n+1}\right) x^n} {\left({n+1}\right)!} + \cdots$ $\displaystyle$ $=$ $\displaystyle 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n-1} } {\left({n-1}\right)!} + \frac {x^n} {n!} \cdots$ $\displaystyle$ $=$ $\displaystyle f \left({x}\right)$

Thus we have:

$D_x \left({f \left({x}\right)}\right) = f \left({x}\right)$
$f \left({x}\right) = e^x$
$D_x \left({f^{-1} \left({x}\right)}\right) = \dfrac 1 {f^{-1} \left({x}\right)}$

Hence from Derivative of Natural Logarithm Function:

$f^{-1} \left({x}\right) = \ln x$

It follows that $e$ can be defined as that number such that $\ln e = 1$.

Hence all the definitions of $e$ as given here are equivalent.

$\blacksquare$

## Proof 2

### 1 implies 2

$\Box$

### 2 implies 3

$\Box$

### 3 implies 4

Let $e$ be the unique solution to the equation $\ln x = 1$.

We want to show that $\exp 1 = e$, where $\exp$ is the exponential function.

 $\displaystyle \exp 1 = 0$ $\leadstoandfrom$ $\displaystyle \map \ln {\exp 1} = \ln e$ Logarithm is Injective $\displaystyle$ $\leadstoandfrom$ $\displaystyle 1 = \ln e$ Exponential is Inverse of Logarithm and Inverse of Inverse

where the final equation holds by hypothesis.

Hence the result.

$\Box$

### 4 implies 1

Let $e = \exp 1$, where $\exp$ denotes the exponential function.

We want to show that:

$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!} = e$
$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!} = \exp 1$

And $\exp 1 = e$ by hypothesis.

Hence the result.

$\blacksquare$