Equivalence of Definitions of Real Exponential Function/Extension of Rational Exponential implies Differential Equation

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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:

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

That is, let $f \left({ x }\right)$ 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 $\R$.

$\exp \left({ x }\right)$ 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:

\(\displaystyle \map {D_x} {\exp x}\) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {\map \exp {x + h} - \exp x} h\) Definition of Derivative
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h\) Exponential of Sum
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {\exp x \paren {\exp h - 1} } h\)
\(\displaystyle \) \(=\) \(\displaystyle \exp x \paren {\lim_{h \mathop \to 0} \frac {\exp h - 1} h}\) Multiple Rule for Limits of Functions, as $\exp x$ is constant
\(\displaystyle \) \(=\) \(\displaystyle \exp x\) 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 $D_x \exp x = \exp x$.

$\Box$

\(\displaystyle \exp 0\) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac 0 n}^n\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$