Exponential Function is Well-Defined/Real

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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 $\exp$.

From Series of Power over Factorial Converges:

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

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


Proof 2

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

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

$f_n \left({ x }\right) = \left({ 1 + \dfrac x n }\right)^n$

Fix $x \in \R$.


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

Thus, $\left\langle{ f_n \left({ x }\right) }\right\rangle$ is bounded above.

From Exponential Sequence is Eventually Increasing:

$\exists N \in \N : \left\langle{ f_{N + n} \left({ x }\right) }\right\rangle$ is increasing

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

From Tail of Convergent Sequence, $\left\langle{ f_{n} \left({ x }\right) }\right\rangle$ converges to $z$.

Hence the result, from Limit of Function Unique.


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.


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.


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 $(1)$.

From Exponential of Zero: Proof 3:

$\map \phi 0 = 1$

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.