# Exponential Function is Well-Defined/Real

## Theorem

Let $x \in \R$ be a real number.

Let $\exp x$ be the exponential of $x$.

Then $\exp x$ is well-defined.

## Proof 1

This proof assumes the series definition of $\exp$.

$\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges

Hence the result, from Convergent Real Sequence has Unique Limit.

$\blacksquare$

## Proof 2

This proof assumes the sequence definition of $\exp$.

Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R \to \R$ defined as:

$\map {f_n} x = \paren {1 + \dfrac x n}^n$

Fix $x \in \R$.

Then:

 $\ds \map {f_n} x$ $=$ $\ds \paren {1 + \dfrac x n}^n$ Definition of $\map {f_n} x$ $\ds$ $=$ $\ds \sum_{k \mathop = 0}^n \binom n k \frac {x^k} {n^k}$ Binomial Theorem: Integral Index $\ds$ $=$ $\ds \sum_{k \mathop = 0}^n \frac {x^k} {k!} \frac {\paren n \times \paren {n - 1} \times \paren {n - 2} \times \cdots \paren {n - k + 1} }{n \times n \times n \times \cdots n}$ Definition of Factorial $\ds$ $=$ $\ds \sum_{k \mathop = 0}^n \frac {x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n}$ $\ds$ $\le$ $\ds \size {\sum_{k \mathop = 0}^n \frac {x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n} }$ Negative of Absolute Value $\ds$ $=$ $\ds \sum_{k \mathop = 0}^n \frac {\size x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n}$ Absolute Value Function is Completely Multiplicative $\ds$ $\le$ $\ds \sum_{k \mathop = 0}^n \frac {\size x^k} {k!}$ Multiplication of Positive Number by Real Number Greater than One $\ds$ $<$ $\ds \sum_{k \mathop = 0}^\infty \frac {\size x^k} {k!}$ Sum of positive terms is increasing $\ds$ $<$ $\ds \infty$ Series of Power over Factorial Converges

Thus, $\sequence {\map {f_n} x}$ is bounded above.

$\exists N \in \N: \sequence {\map {f_{N + n} } x}$ is increasing

From Monotone Convergence Theorem (Real Analysis), $\sequence {\map {f_{N + n} } x}$ converges to some $z \in \R$.

From Tail of Convergent Sequence, $\sequence {\map {f_n} x}$ converges to $z$.

Hence the result, from Limit of Real Function is Unique.

$\blacksquare$

## Proof 3

This proof assumes the continuous extension definition of $\exp$.

Let $e$ denote Euler's number.

Let $f: \Q \to \R$ be the real-valued function defined as:

$f \left({r}\right) = e^r$

From Euler's Number to Rational Power permits Unique Continuous Extension, there exists a unique continuous extension of $f$ to $\R$.

Hence the result.

$\blacksquare$

## Proof 4

This proof assumes the definition of the exponential as the inverse of the logarithm.

From Logarithm is Strictly Increasing, $\ln$ is strictly monotone on $\R_{>0}$.

From Inverse of Strictly Monotone Function, $f$ permits an inverse mapping.

Hence the result, from Inverse Mapping is Unique.

$\blacksquare$

## Proof 5

This proof assumes the definition of $\exp$ as the solution to an initial value problem.

That is, suppose $\exp$ solves:

$(1): \quad \dfrac \d {\d x} y = \map f {x, y}$
$(2): \quad \map \exp 0 = 1$

on $\R$, where $\map f {x, y} = y$.

From Derivative of Exponential Function: Proof 4, the function $\phi : \R \to \R$ defined as:

$\displaystyle \map \phi x = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}$

satisfies $\map {\phi'} x = \map \phi x$.

So $\phi$ satisfies $(1)$.

$\map \phi 0 = 1$

So $\phi$ satisfies $(2)$.

Thus, $\phi$ is a solution to the initial value problem given.

From Exponential Function is Continuous: Proof 5 and $(1)$:

$\phi$ is continuously differentiable on $\R$.

Because $\map f {x, \phi} = \phi$:

$f$ is continuously differentiable on $\R^2$.

Thus, from Uniqueness of Continuously Differentiable Solution to Initial Value Problem, this solution is unique.

$\blacksquare$