% vim: tw=50
% 25/02/2022 11AM

\noindent
Now we restrict to $\RR$:
\begin{theorem}
    \begin{enumerate}[(i)]
        \item $e : \RR \to \RR$ is everywhere
            differentiable and $e'(x) = e(x)$
        \item $e(x + y) = e(x)e(y)$
        \item $e(x) > 0$ for all $x \in \RR$
        \item $e$ is strictly increasing
        \item $e(x) \to \infty$ as $x \to \infty$,
            $e(x) \to 0$ as $x \to -\infty$
        \item $e : \RR \to (0, \infty)$ is a
            \emph{bijection}.
    \end{enumerate}
\end{theorem}

\begin{proof}
    \begin{enumerate}[(i)]
        \item Already done.
        \item Clearly
            \[ e(x) > 0 \qquad \forall x \ge 0 \]
            and $e(0) = 1$. Also
            \[ e(0) = e(x - x) = e(x)e(x) = 1
            \implies e(-x) > 0 \]
            for all $x > 0$.
        \item Already done.
        \item $e'(x) = e(x) > 0$ so $e$ is
            strictly increasing.
        \item $e(x) > 1 + x$ for $x > 0$ so if $x
            \to \infty$, $e(x) \to \infty$. For $x
            > 0$ since
            \[ e(-x) = \frac{1}{e(x)} \]
            then $e(x) \to 0$ as $x \to -\infty$.
        \item Injectivity follows right away from
            being strictly increasing.
            Surjectivity: Take $y \in (0 \in
            \infty)$. From (v) there exist $a, b
            \in \RR$ such that
            \[ e(a) < y < e(b) \]
            so by the Intermediate Value Theorem
            there exists $x \in \RR$ such that
            $e(x) = y$.
    \end{enumerate}
\end{proof}

\begin{remark*}
    $e : (\RR, +) \to ((0, \infty), \times)$ is a
    \emph{group isomorphism}.
\end{remark*}

\noindent
Since $e$ is a bijection we have an inverse:
\[ l : (0, \infty) \to \RR \]

\begin{theorem}
    \begin{enumerate}[(i)]
        \item $l : (0, \infty) \to \RR$ is a
            bijection and $l(e(x)) = x$ for all $x
            \in \RR$ and $r(l(t)) = t$ for all $t
            \in (0, \infty)$.
        \item $l$ is differentiable and $l'(t) =
            \frac{1}{t}$.
        \item $l(xy) = l(x) + l(y)$ for all $x, y
            \in (0, \infty)$. 
    \end{enumerate}
\end{theorem}

\begin{proof}
    \begin{enumerate}[(i)]
        \item Obvious from the definition of $l$.
        \item Inverse rule (Theorem 3.6) $l$ is
            differentiable and
            \[ l'(t) = \frac{1}{e(l(t))} =
            \frac{1}{t} \]
        \item From IA Groups if $e$ is an
            isomorphism, so is its inverse.
    \end{enumerate}
\end{proof}

\myskip
Now define for $\alpha \in \RR$ and $x > 0$:
\[ r_\alpha (x) \stackrel{\text{def}}{=} e(\alpha
l(x)) \]

\begin{theorem}
    Suppose $x, y > 0$ and $\alpha, \beta \in
    \RR$. Then
    \begin{enumerate}[(i)]
        \item $r_\alpha(xy) =
            r_\alpha(x)r_\alpha(y)$
        \item $r_{\alpha + \beta}(x) =
            r_\alpha(x)r_\beta(x)$
        \item $r_\alpha(r_\beta(x)) =
            r_{\alpha\beta}(x)$
        \item $r_1(x) = x$, $r_0(x) = 1$.
    \end{enumerate}
\end{theorem}

\begin{proof}
    \begin{enumerate}[(i)]
        \item \eqnoskip
            \begin{align*}
                r_\alpha(xy)
                &= e(\alpha l(xy)) \\
                &= e(\alpha l(x) + \alpha l(y)) \\
                &= e(\alpha l(x)) e(\alpha l(y))
                \\
                &= r_\alpha (x) r_\alpha (y)
            \end{align*}
        \item \eqnoskip
            \begin{align*}
                r_{\alpha + \beta}(x)
                &= e((\alpha + \beta)l(x)) \\
                &= e(\alpha l(x)) e(\beta l(x)) \\
                &= r_\alpha(x) r_\beta(x)
            \end{align*}
        \item \eqnoskip
            \begin{align*}
                r_\alpha(r_\beta(x))
                &= r_\alpha(e(\beta l(x))) \\
                &= e(\alpha le(\beta l(x))) \\
                &= e(\alpha \beta l(x)) \\
                r_{\alpha \beta}(x)
            \end{align*}
        \item $r_1(x) = e(l(x)) = x$, $r_0(x) =
            e(0) = 1$.
    \end{enumerate}
\end{proof}

\myskip
For $n \ge 1$, $n \in \ZZ$
\[ r_n(x) = r_{1 + \cdots + 1}(x) = x \cdots x =
x^n \]
\[ r_1(x)r_{-1}(x) = r_0(x) = 1 \]
\[ \implies r_{-1}(x) = \frac{1}{x} \]
\[ r_{-n}(x) = \frac{1}{x^n} \]
\[ (r_{\frac{1}{q}}(x))^q = r_1(x) = x \]
($q \in \ZZ$, $q \ge 1$)
\[ \implies r_{\frac{1}{q}}(x) = x^{\frac{1}{q}} \]
\[ r_{\frac{p}{q}}(x) = (r_{\frac{1}{q}}(x))^p =
x^{\frac{p}{q}} \]
Thus $r_\alpha(x)$ agrees with $x^\alpha$ when $a
\in \QQ$ as previously defined. Now we give them
names:
\begin{align*}
    \exp(x)
    &= e(x)
    &&x \in \RR \\
    \log x
    &= l(x)
    &&x \in (0, \infty) \\
    x^\alpha
    &= r_\alpha(x)
    &&\alpha \in \RR, x \in (0, \infty)
\end{align*}
\[ e(x) = e(x \log e) = e_x(e) = e^x \]
where
\[ e = \stackrel{\text{def}}{=} = \sum_{n =
0}^\infty \frac{1}{n!} \]
$\exp(x)$ is also a power, which we may as well
write as $e^x$. Finally we compute
\begin{align*}
    (x^\alpha)'
    &= (e^{\alpha \log x})' \\
    &= e^{\alpha \log x} \frac{\alpha}{x} \\
    &= \alpha x^{\alpha - 1}
\end{align*}
$f(x) = a^x$, $a > 0$ then
\[ f'(x) = (e^{x \log a})' = e^{x \log a} \log a =
a^x \log a \]
