Equivalence of Definitions of Real Exponential Function/Extension of Rational Exponential implies Differential Equation
Theorem
The following definition of the concept of the real exponential function:
As an Extension of the Rational Exponential
Let $e$ denote Euler's number.
Let $f: \Q \to \R$ denote the real-valued function defined as:
- $\map f x = e^x$
That is, let $\map f x$ denote $e$ to the power of $x$, for rational $x$.
Then $\exp : \R \to \R$ is defined to be the unique continuous extension of $f$ to $\R$.
$\map \exp x$ is called the exponential of $x$.
implies the following definition:
As the Solution of a Differential Equation
The exponential function can be defined as the unique solution $y = \map f x$ to the first order ODE:
- $\dfrac {\d y} {\d x} = y$
satisfying the initial condition $\map f 0 = 1$.
Proof
Let $\exp x$ be the real function defined as the extension of rational exponential.
Then we have:
\(\ds \map {D_x} {\exp x}\) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\map \exp {x + h} - \exp x} h\) | Definition of Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h\) | Exponential of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\exp x \paren {\exp h - 1} } h\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \exp x \paren {\lim_{h \mathop \to 0} \frac {\exp h - 1} h}\) | Multiple Rule for Limits of Real Functions, as $\exp x$ is constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \exp x\) | Derivative of Exponential at Zero: Proof 2 |
The application of Derivative of Exponential at Zero is not circular as the referenced proof does not depend on $D_x \exp x = \exp x$.
$\Box$
\(\ds \exp 0\) | \(=\) | \(\ds \lim_{n \mathop \to +\infty} \paren {1 + \frac 0 n}^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$