Exponential Function is Well-Defined/Real/Proof 5

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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.


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.