Exponential of Real Number is Strictly Positive/Proof 5/Lemma

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Theorem

Let $x$ be a real number.

Let $\exp$ denote the (real) Exponential Function.


Then:

$\forall x \in \R: \exp x \ne 0$


Proof

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

That is, suppose $\exp$ satisfies:

$ (1): \quad D_x \exp x = \exp x$
$ (2): \quad \exp \left({0}\right) = 1$

on $\R$.


Aiming for a contradiction, suppose that $\exists \alpha \in \R: \exp \alpha = 0$.

Suppose that $\alpha > 0$.

Let $J = \left[{0 \,.\,.\, \alpha}\right]$.


From Exponential Function is Continuous, $\exp$ is continuous on $J$.

From Max and Min of Function on Closed Real Interval:

$\exists K \in \R: \forall x \in J: \left\vert{\exp x}\right\vert < K$


Then, $\forall n \in \N : \exists c_n \in J$ such that:

\(\displaystyle 1\) \(=\) \(\displaystyle \exp 0\) $\quad$ By hypothesis $(2)$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 0}^{n-1} \frac{\exp^{j} \left({\alpha}\right)}{j!} \left({-\alpha}\right)^{j} + \frac{\exp c_n}{n!} \left({-\alpha}\right)^{n}\) $\quad$ Taylor's Theorem for Univariate Functions $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 0}^{n-1} \frac{\exp \left({\alpha}\right)}{j!} \left({-\alpha}\right)^{j} + \frac{\exp c_n}{n!} \left({-\alpha}\right)^{n}\) $\quad$ By hypothesis $(1)$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 0}^{n-1} \frac 0 {j!} \left({-\alpha}\right)^{j} + \frac{\exp c_n}{n!} \left({-\alpha}\right)^{n}\) $\quad$ From our assumption aiming at a contradiction $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac{\exp c_n}{n!} \left({-\alpha}\right)^{n}\) $\quad$ $\quad$


So $\forall n \in \N :1 \le K \dfrac{\alpha^{n}}{n!}$

That is, dividing both sides by $K$:

$\forall n \in \N: \dfrac 1 K \le \dfrac{\alpha^{n}}{n!}$

But from Power over Factorial, $\dfrac{\alpha^{n}}{n!} \to 0$.


This contradicts our assumption.


The same argument, mutatis mutandis proves the result for $\alpha < 0$.


By hypothesis $(2)$:

\(\displaystyle \alpha = 0 \ \ \) \(\displaystyle \implies \ \ \) \(\displaystyle \exp \alpha\) \(=\) \(\displaystyle 1\) $\quad$ $\quad$
\(\displaystyle \) \(\ne\) \(\displaystyle 0\) $\quad$ $\quad$


Hence the result.

$\blacksquare$