Added function seqs

chapters/multivar_calc.tex

@@ -6,4 +6,5 @@
 
     \subfile{sections/partial_total_diff.tex}
     \subfile{sections/higher_derivs.tex}
+    \subfile{sections/functionseq_and_diff.tex}
 \end{document}

chapters/sections/functionseq_and_diff.tex

@@ -0,0 +1,252 @@
+% !TeX root = ../../script.tex
+\documentclass[../../script.tex]{subfiles}
+
+\begin{document}
+\section{Function Sequences and Differentiability}
+
+\begin{eg}
+    Consider $\seq{f}$:
+    \begin{align*}
+        f_n: \realn &\longrightarrow \cmpln \\
+        x &\longmapsto \frac{1}{n} e^{inx}
+    \end{align*}
+    Then 
+    \begin{gather*}
+        \supnorm{f_n} = \frac{1}{n} \conv{} 0 \\
+        \iff\\
+        \seq{f} \text{ converges uniformly to the zero function}
+    \end{gather*}
+    But 
+    \[
+        f_n'(x) = ie^{inx} = i(e^{ix})^n
+    \]
+    converges (pointwise even) only for $x = 2k\pi, ~~k \in \intn$.
+\end{eg}
+
+\begin{rem}
+    Let $f: X \rightarrow V$ where $V$ is a normed vector space. Define 
+    \[
+        \supnorm{f} = \sup\set[x \in X]{\norm{f(x)}}
+    \]
+    the supermum norm. Also define 
+    \begin{itemize}
+        \item $B(X, V)$ the space of bounded functions from $X \rightarrow V$
+        \item $C_B(X, V)$ the space of continuous, bounded functions from $X \rightarrow V$
+    \end{itemize}
+\end{rem}
+
+\begin{thm}
+    Let $U \subset \realn^n$ be open and $f_n: U \rightarrow \realn^m$ continuously differentiable $\forall n \in \natn$.
+    If $\seq{f}$ and $(Df_n)$ converge uniformly to $f: U \rightarrow \realn^m$ and $g:U \rightarrow \realn^{m \times m}$,
+    then $f$ is differentiable and $Df = g$.
+\end{thm}
+\begin{proof}
+    First consider $m = 1$. We use the operator norm on $\realn^{m \times m}$.
+    First, let $Df_n$ be continuous $\forall n$ and thus $g$ is continuous.
+    Choose $x \in U$ and $\epsilon > 0$, then 
+    \begin{equation}
+        \exists \delta > 0: ~~\norm{g(y) - g(x)} < \frac{\epsilon}{3} ~~\text{ if }~~ \norm{y - x} < \delta
+    \end{equation}
+    Furthermore 
+    \begin{equation}
+        \exists N \in \natn: ~~\supnorm{Df_n - g} < \frac{\epsilon}{3} ~~\forall n > N
+    \end{equation}
+    Let $y \in \oball[\delta](x)$. Then according to the intermediate value theorem,
+    \begin{equation}
+        \forall n \in \natn ~\exists \xi_n \in \oline = \set[{t \in [0, 1]}]{x + t(y - x)}
+    \end{equation}
+    such that 
+    \begin{equation}
+        f_n(y) - f_n(x) = Df_n(\xi_n)(y - x)
+    \end{equation}
+    We have $\xi_m \in \oball[\delta](x)$. Then 
+    \begin{equation}
+        \begin{split}
+              &\frac{1}{\norm{y - x}} \left| f_n(y) - f_n(x) - Df_n(x)(y - x) \right| \\ 
+            = &\frac{1}{\norm{y - x}} \underbrace{\abs{(Df_n(\xi_n) - Df_n(x))(y - x)}}_{\norm{Df_n(\xi_n)} - Df_n(x)\norm{y - x}} \\
+            \le &\norm{Df_n(\xi_n) - Df_n(x)} \\
+            \le &\norm{Df_n(\xi_n) - g(\xi_n)} + \norm{g(\xi_n) - g(x)} + \norm{g(x) - Df_n(x)} \\
+            \le &\supnorm{Df_n - g} + \norm{g(\xi_n) - g(x)} + \supnorm{g - Df_n} \\
+            = &2\supnorm{Df_n - g} + \norm{g(\xi_n) - g(x)} < \epsilon
+        \end{split}
+    \end{equation}
+    For $n \rightarrow \infty$ we have 
+    \begin{equation}
+        \frac{1}{\norm{y - x}} \abs{f(y) - f(x) - g(x)(y - x)} < \epsilon ~~\forall y \in \oball[\delta](x)
+    \end{equation}
+    Since $\epsilon > 0$ is arbitrary, we get 
+    \begin{equation}
+        \limes{y}{x} \frac{1}{\norm{y - x}} \abs{f(y) - f(x) - g(x)(y - x)} = 0
+    \end{equation}
+    This means that $f$ is differentiable in $x$ with $Df(x) = g(x)$.
+\end{proof}
+
+\begin{rem}
+    On $C_B^1(U, \realn^m)$ (the space of continuous, differentiable and bounded functions with bounded derivative) we can define a norm:
+    \[
+        \cnorm{f} := \supnorm{f} + \supnorm{Df}
+    \]
+    Then the above theorem is equivalent to the statement that $C_B^1(U, \realn^m)$ with $\cnorm{f}$ is complete.
+\end{rem}
+
+\begin{thm}
+    Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ be a power series with positive convergence radius $\rho$.
+    Then $f$ is differentiable on $\oball[\rho](0)$ and 
+    \[
+        f'(x) = \sum_{n=0}^{\infty} n a_n x^{n-1}
+    \]
+\end{thm}
+\begin{proof}
+    We need to inspect the convergence radius $R$ of 
+    \begin{equation}
+        \sum_{n=0}^{\infty} n a_n x^{n-1} = \frac{1}{x} \sum_{n=0}^{\infty} n a_n x^{n}
+    \end{equation}
+    $(\sqrt[n]{n})$ converges to $1$, so $\exists \epsilon > 0$ such that for sufficiently big $n$ we have 
+    \begin{equation}
+        (1 - \epsilon)\sqrt[n]{a_n} \le \sqrt{n a_n} \le (1 + \epsilon)^n \sqrt{a_n}
+    \end{equation}
+    and thus 
+    \begin{equation}
+        \frac{1 - \epsilon}{\rho} = (1 - \epsilon) \cdot \limsupn \sqrt[n]{\abs{a_n}} \le \limsupn \sqrt[n]{\abs{n a_n}} = \frac{1}{R} \le \frac{1 + \epsilon}{\rho}
+    \end{equation}
+    So 
+    \begin{equation}
+        \implies \frac{1 - \epsilon}{\rho} \le \frac{1}{R} \le \frac{1 + \epsilon}{\rho}
+    \end{equation}
+    Since this holds for every $\epsilon$, this implies $\rho = R$. 
+    Now for $x \in \oball[\rho](0)$ set
+    \begin{equation}
+        g(x) := \series{k} n a_n x^{n - 1}
+    \end{equation}
+    Let $x \in \oball[\rho](0)$ be fixed and choose $a > 0$ such that $\abs{x} < a < \rho$.
+    This means that 
+    \begin{align*}
+        f_N(x) := \sum_{n=0}^{N} a_n x^n && \text{and} && g_N(x) := \sum_{n=0}^{N} a_nx^{n-1}
+    \end{align*}
+    converge uniformly on $\oball[a](0)$ to $f$ and $g$.
+    Obviously, $f_N' = g_N$, so $f$ is differentiable and $f' = g$.
+    Since differentiabiility is a local property, the desired statement follows $\forall x \in \oball[\rho](0)$.
+\end{proof}
+
+\begin{cor}
+    Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ be a power series with convergence radius $\rho > 0$. Then $f \in C^{\infty}(\oball[r](0))$, and 
+    \[
+        a_k = f^{(k)}(0) \cdot \inv(k!)
+    \]
+    Furthermore, the series representation (if it exists) is unique.
+\end{cor}
+\begin{proof}
+    The infinite Differentiability follows inductively from the previous theorem. Also inductively we have 
+    \begin{equation}
+        f^{(k)}(x) = \sum_{n=0}^{\infty} n(n-1) \cdots (n - k + 1) a_nx^{n-k}
+    \end{equation}
+    Choose $x = 0$ and receive 
+    \begin{equation}
+        f^{(k)}(0) = n (n-1) \cdots (n - k + 1) a_n
+    \end{equation}
+\end{proof}
+
+\begin{eg}[Derivative of the exponential function]
+    \[
+        (e^x)' = \sum_{n=0}^{\infty} \left(\frac{x^n}{n!}\right)' = \sum_{n=1}^{\infty} \frac{n x^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x
+    \]
+\end{eg}
+
+\begin{rem}[Taylor Series]
+    We can define the Taylor series for $f: \field \rightarrow \field$
+    \[
+        \sum_{n=0}^{\infty} \frac{f^(n)(0)}{n!} x^n = f(x)
+    \]
+    \begin{itemize}
+        \item In general, this doesn't hold true for all $x$, not even for $f \in C^{\infty}$.
+        \item The convergence radius could be $0$
+        \item There are examples of convergent Taylor series that don't converge to the initial function, e.g.
+        \[
+            f: x \mapsto \begin{cases}
+                \exp\left(-\frac{1}{x}\right), & x > 0 \\
+                0, & \text{else}
+            \end{cases}
+        \]
+        $f$ is infinitely continuously differentiable in $0$, but the Taylor series would converge to $0$.
+    \end{itemize}
+\end{rem}
+
+\begin{defi}
+    Let $a_{\eta} \in \field$ (Multiindex notation) be coefficients $\forall \eta \in \natn_0^d$. Then 
+    \[
+        \sum_{\eta \in \natn_0^d} a_{\eta} x^{\eta}
+    \]
+    is said to be a (formal) power series with $d$ variables.
+    
+    A function $f: U \rightarrow \field$ with $U$ neighbourhood around $0$ is said to be analytic in $0$, if and only if 
+    \[
+        \exists \epsilon > 0, a_{\eta} \in \field: ~~f(x) = \sum_{\eta \in \natn_0^d} a_{\eta} x^{\eta} ~~\forall x \in \oball(0)
+    \]
+\end{defi}
+
+\begin{rem}
+    \begin{enumerate}[(i)]
+        \item The convergence of the series to $S(x)$ can be defined as follows: 
+        $\forall \epsilon > 0 ~\exists A \subset \natn_0^d \text{ finite}$ such that $\forall B \supset A$ finite we have 
+        \[
+            \abs{\sum_{\eta \in B} a_{\eta} x^{\eta} - S(x)} < \epsilon
+        \]
+
+        \item If the series converges in $(y_1, \cdots, y_n)$, then it also absolutely converges in the open cuboid 
+        \[
+            \set[\abs{x_i} < \abs{y_i} ~~\forall i \in \set{1, \cdots, d}]{x \in \realn^d}
+        \]
+        which means 
+        \[
+            \sum_{\eta \in \natn_0^d} \abs{a_{\eta}} (\abs{x_1}, \cdots, \abs{x_d})^{\eta} < \infty
+        \]
+
+        \item If the power series converges on a neighbourhood $U$ around $0$, then it is infinitely differentiable and 
+        \[
+            a_{\eta} = \frac{\partial^{\eta} f(0)}{\eta!}
+        \]
+        with 
+        \begin{align*}
+            \partial^{\eta} := \partial_1^{\eta_1} \partial_2^{\eta_2} \cdots \partial_d^{\eta_d} && \eta! := \eta_1! \eta_2! \cdots \eta_d!
+        \end{align*}
+
+        \item The formula above is only rarely useful to calculate the Taylor series. By inverting it we can calculate the 
+        derivative of a known series representation. E.g.
+        \[
+            f(x) = xe^{x^2} = x \cdot \sum_{k = 0}^{\infty} \frac{(x^2)^k}{k!} = \series_{k=0}^{\infty} \frac{x^{2k + 1}}{k!} ~~\forall x \in \field
+        \]
+        $f^{(k)}(0) = 0$ is $k$ is even, and it is something else if $k$ is odd.
+
+        \item $C^{\omega}(U)$ is the space of all analytic functions.
+        \[
+            C(U) \supset C^1(U) \supset C^2(U) \supset \cdots \supset C^k(U) \supset \cdots \supset C^{\infty}(U) \supset C^{\omega}(U)
+        \]
+
+        \item The analytic functions are closed among sums, products and concatinations. 
+        A power series is analytic within its converges radius.
+    \end{enumerate}
+\end{rem}
+
+\begin{eg}
+    Consider the power series 
+    \[
+        \sum_{n=0}^{\infty} (xy)^n = \sum_{\eta \in \natn_0^2} (x y)^{\eta} \cdot a_{\eta}
+    \]
+    with 
+    \begin{align*}
+        &a_{\eta} = 1 \text{ if } \eta_1 = \eta_2 \\
+        &a_{\eta} = 0 \text{ else}
+    \end{align*}
+    This series converges on 
+    \[
+        \set[\abs{xy} < 1]{(x, y)}
+    \]
+    to $\frac{1}{1 - xy}$.
+
+    \begin{center}
+        \includegraphics[scale=0.5]{../../assets/conv_rad.pdf}
+    \end{center}
+
+    So the convergence area must not necessarily be a sphere. The limit function is also defined outside of the convergence area.
+\end{eg}
+\end{document}

script.tex

@@ -9,7 +9,10 @@
 \usepackage{tikz, pgfplots}
 \usepackage{kbordermatrix}
 \usepackage{fancyhdr}
-\usetikzlibrary{calc,trees,positioning,arrows,fit,shapes,angles}
+\usepackage{pdfpages}
+\usetikzlibrary{calc,trees,positioning,arrows,fit,shapes,angles,patterns}
+
+\graphicspath{assets}
 
 \usepackage{color}
 \usepackage{hyperref}
@@ -66,6 +69,7 @@
 \newcommand{\dnorm}{\norm{\cdot}}
 \newcommand{\normed}[1][V]{(#1, \dnorm)}
 \newcommand{\supnorm}[1]{\norm{#1}_{\infty}}
+\newcommand{\cnorm}[2][1]{\norm{#2}_{C_{#1}}}
 
 \newcommand{\oball}[1][\epsilon]{B_{#1}}
 \newcommand{\cball}[1][\epsilon]{K_{#1}}
@@ -141,6 +145,9 @@
 
 \pgfplotsset{compat=1.17}
 
+\newcommand{\drawge}{-- (rel axis cs:1,0) -- (rel axis cs:1,1) -- (rel axis cs:0,1) \closedcycle}
+\newcommand{\drawle}{-- (rel axis cs:1,1) -- (rel axis cs:1,0) -- (rel axis cs:0,0) \closedcycle}
+
 \begin{document}
 \begin{titlepage}
   \begin{center}