% vim: tw=50
% 18/02/2022 11AM

\myskip
To get a Taylor series for $f$ one needs to show
that $R_n \to 0$ as $n \to \infty$. This requires
``estimates'' and ``effort''.

\begin{remark*}
    Theorems 3.8 and 3.9 work equally well in an
    interval $[a + h, a]$ with $h < 0$.
\end{remark*}

% \begin{example*}
\noindent
\textbf{Example.}
    The binomial \emph{series}:
    \[ f(x) = (1 + x)^r , \qquad r \in \QQ \]
    \begin{claim*}
        If $|x| < 1$, then
        \[ (1 + x)^r = 1 + {r \choose 1}x + \cdots
        + {r \choose n} x^n + \cdots \]
        where
        \[ {r \choose n} \stackrel{\text{def}}{=}
        \frac{r(r - 1) \cdots (r - n + 1)}{n!} \]
    \end{claim*}
    \begin{proof}
        Clearly
        \[ f^{(n)} (x) = r(r - 1) \cdots (r - n +
        1) (1 + x)^{r - n} \]
        If $r \in \ZZ$, $r \ge 0$, then
        \[ f^{(r + 1)} \equiv 0 \]
        we have a polynomial of degree $r$. In
        general (Lagrange)
        \[ R_n = \frac{x^n}{n!} f^{(n)}(\theta x)
        = {r \choose n} \frac{x^n}{(1 + \theta
        x)^{n - r}} \]
        ($\theta \in (0, 1)$)
        \begin{note*}
            In principle, $\theta$ depends on both
            $x$ and $n$.
        \end{note*}
        For $0 < x < 1$,
        \[ (1 + \theta x)^{n - r} > 1 \]
        for $n > r$. Now observe that the series
        \[ \sum {r \choose n} x^n \]
        is absolutely convergent for $|x| < 1$.
        Indeed by the ratio test
        \[ a_n = {r \choose n} x^n \]
        \begin{align*}
            \implies \left| \frac{a_{n + 1}}{a_n} \right|
            &= \left| \frac{r(r - 1) \cdots (r - n
            + 1)(r - n)x^{n + 1}}{(n + 1)!}
            \right| \cdots \left| \frac{n!}{r(r -
            1) \cdots (r - n + 1)x^n} \right| \\
            &= \left| \frac{(r - n)x}{n + 1}
            \right| \to |x| < 1
        \end{align*}
        In particular $a_n \to 0$ so ${r \choose
        n}x^n \to 0$. Hence for $n > r$ and $0 < x
        < 1$, we have
        \[ |R_n| \le |{r \choose n} x^n| = |a_n|
        \to 0 \]
        as $n \to \infty$. So the claim is proved
        in the range $0 \le x < 1$. \\
        If $-1 < x < 0$ the argument above breaks,
        but Cauchy's form for $R_n$ works:
        \begin{align*}
            R_n
            &= \frac{(1 - \theta)^{n - 1} r(r -
            1) \cdots (r - n + 1)(1 + \theta x)^{r -
            n} x^n}{(n - 1)!} \\
            &= \ub{\frac{r(r - 1) \cdots (r - n +
            1)}{(n - 1)!}}_{r {r - 1 \choose n -
            1}} \frac{(1 - \theta)^{n - 1}}{(1 + \theta
            x)^{n - r}} x^n \\
            &= r{r - 1 \choose n - 1} x^n (1 +
            \theta x)^{r - 1} \left( \ub{\frac{1 -
            \theta}{1 + \theta x}}_{\substack{< 1
            \\ \forall x \in (-1, 1)}} \right)^{n
            - 1} \\
            \implies |R_n|
            &\le \left| r{r - 1 \choose n - 1} x^n
            \right| (1 + \theta x)^{r - 1}
        \end{align*}
        Check:
        \[ (1 + \theta x)^{r - 1} \le \max\{1, (1
        + x)^{r - 1}\} \]
        (do it!) Let
        \[ K_r = |r| \max\{1, (1 + x)^{r - 1}\} \]
        independent of $n$.
        \[ |R_n| \le K_r \left| {r - 1 \choose n -
        1} x^n \right| \to 0 \]
        because $a_n \to 0$, thus $R_n \to 0$.
    \end{proof}
% \end{example*}

\subsubsection*{Remarks on Complex Differentiation}

Formally for functions $f : E \subseteq \CC \to
\CC$ we have properties for sums, products, chain
rule etc. But it is \emph{much more restrictive}
than differentiability on the real line.

\begin{example*}
    $f : \CC \to \CC$, $f(z) = \ol{z}$
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/0ea9f55ea16e11ec.png}
    \end{center}
    If
    \[ z_n = z + \frac{1}{n} \to z \]
    then
    \[ \frac{f(z_n) - f(z)}{z_n - z} =
    \frac{\ol{z} + \frac{1}{n} - \ol{z}}{z +
    \frac{1}{n} - z} = 1 \]
    but on the other hand if
    \[ z_n = z + \frac{i}{n} \to z \]
    then
    \[ \frac{f(z_n) - f(z)}{z_n - z} =
    \frac{\ol{z} - \frac{i}{n} - \ol{z}}{z +
    \frac{i}{n} - z} = -1 \]
    so
    \[ \lim_{w \to z} \frac{f(w) - f(z)}{w - z} \]
    does \emph{not exist}, so it is nowhere
    $\CC$-differentiable!
\end{example*}

\begin{note*}
    IB Complex Analysis explores the consequences
    of $\CC$-differentiability.
\end{note*}
