Exponential of Product/Proof 3
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Theorem
Let $x, y \in \R$ be real numbers.
Let $\exp x$ be the exponential of $x$.
Then:
- $\map \exp {x y} = \paren {\exp y}^x$
Proof
For $n \in \Z_{\ge 0}$:
| \(\ds \map \exp {n y}\) | \(=\) | \(\ds \map \exp {\sum_{k \mathop = 1}^n y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \prod_{k \mathop = 1}^n \exp y\) | Exponential of Sum of Real Numbers | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \paren {\exp y}^n\) |
That is:
- $\forall n \in \Z_{\ge 0}: \map \exp {n y} = \paren {\exp y}^n$
Now let $n \in \Z_{<0}$.
It follows that $-n \in \Z_{>0}$, so:
| \(\ds \exp n\) | \(=\) | \(\ds \map \exp {-\paren {-n y} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {\map \exp {-n y} }\) | Reciprocal of Real Exponential | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {\paren {\exp y}^{-n} }\) | from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\exp y}^n\) | Real Number to Negative Power: Positive Integer |
Thus:
- $(2): \quad \forall m \in \Z: \map \exp {m y} = \paren {\exp y}^m$
Next, for $n \in \Z_{>0}$:
| \(\ds \exp y\) | \(=\) | \(\ds \map \exp {n \frac y n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\map \exp {\frac y n} }^n\) | from $(1)$ | |||||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \sqrt [n] {\exp y}\) | \(=\) | \(\ds \map \exp {\frac y n}\) |
So fix $r \in \Q$.
Let $r = \dfrac m n$, where $m \in \Z$ is an integer and $n \in \Z_{>0}$ is a strictly positive integer.
From the above:
| \(\ds \map \exp {r y}\) | \(=\) | \(\ds \map \exp {\frac m n y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt [n] {\map \exp {m y} }\) | from $(3$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt [n] {\paren {\exp y}^m}\) | from $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\exp y}^r\) |
Thus, from the definition of $\paren {\exp y}^x$ as the unique continuous extension of $r \mapsto \paren {\exp y}^r$ from $\Q$ to $\R$:
- $\map \exp {x y} = \paren {\exp y}^x$
$\blacksquare$