diff --git a/chapters/multivar_calc.tex b/chapters/multivar_calc.tex index ee2c7a2..de486bf 100644 --- a/chapters/multivar_calc.tex +++ b/chapters/multivar_calc.tex @@ -5,4 +5,5 @@ \chapter{Multivariable Calculus} \subfile{sections/partial_total_diff.tex} + \subfile{sections/higher_derivs.tex} \end{document} \ No newline at end of file diff --git a/chapters/sections/higher_derivs.tex b/chapters/sections/higher_derivs.tex new file mode 100644 index 0000000..945c5ff --- /dev/null +++ b/chapters/sections/higher_derivs.tex @@ -0,0 +1,271 @@ +\documentclass[../../script.tex]{subfiles} +% !TEX root = ../../script.tex + +\begin{document} +\section{Higher Derivatives} + +\begin{defi} + Let $U \subset \realn^ n$ and let $f$ be (the only) partial derivative of order $0$. + Now define recursively + \begin{enumerate}[(i)] + \item $f$ is said to be $(k + 1)$-times partially differentiable if all partial derivatives of order $k$ are partially differentiable. + \item The partial derivatives of order $(k + 1)$ are the functions $\partial_i g ~~i \in \set{1, \cdots, n}$ where $g$ is the partial derivative of order $k$ of $f$. + \end{enumerate} + The $k$-th partial derivative in terms of $i$ of $f$ is denoted as + \[ + \partial_i^k f + \] + $f$ is said to be $k$-times continuously differentiable if all partial derivatives of order $k$ are continuous. + $C^k(U, \realn^m)$ is the vector space of all $k$-times continuously differentiable functions. + + $f$ is said to be infinitely differentiable (or smooth) is it is $k$-times differentiable $\forall k \in \natn$, and the vector space + of all infinitely differentiable functions is denoted as $C^{\infty}(U, \realn^m)$. + + For total differentiability we have + \begin{align*} + f: \realn^n \longrightarrow \realn^m && Df: \realn^m \longrightarrow \realn^{m \times n} + \end{align*} +\end{defi} + +\begin{rem} + Let $f: \realn^n \rightarrow \realn^m$ be sufficiently often differentiable. Consider for $u \in \realn^n$ + \[ + x \longmapsto Df(x) u = \underbrace{\limes{k}{0} \frac{f(x + hu) - f(x)}{h}}_{\substack{\text{Directional derivative along } u}} + \] + Now consider for fixed $x$ + \begin{align*} + D^2 f(x) : \realn^n \times \realn^n &\longrightarrow \realn^m \\ + (u ,v) &\longmapsto D(Df(\cdot)u)(x)v + \end{align*} + $D^2f(x)$ is linear in $v$ and $u$, and + \begin{align*} + D^2 f(x) (u_1 + \lambda u_2, v) &= D(Df(\cdot)(u_1 + \lambda u_2))(x) v \\ + &= D(Df(\cdot)u_1 + \lambda Df(\cdot)u_2)(x) v \\ + &= D(Df(\cdot)u_1)(x) v + \lambda D(Df(\cdot)u_2)(x) v \\ + &= D^2f(x)(u_1, v) + \lambda D^2f(x)(u_2, v) + \end{align*} + $D^2f(x)$ is a bi-linear mapping. +\end{rem} + +\begin{defi} + Let $U \subset \realn^n$ and $f: U \rightarrow \realn^m$. Define recursively for $k \ge 1$: + \begin{enumerate}[(i)] + \item $f$ is said to be $(k+1)$ times (totally) differentiable on $U$, if the term $D^k(\cdot)(u_1, \cdots, u_k)$ is differentiable on $U \forall u_1, \cdots, u_k \in \realn^n$. + \item The $(k+1)$-th derivative of $f$ in $x \in U$ is the multi-linear mapping + \begin{align*} + D^{k+1} f(x): (\realn^n)^{k+1} &\longrightarrow \realn^m \\ + (u_1, \cdots, u_k, v) &\longmapsto D(D^kf(\cdot)(u_1, \cdots, u_k))(x) v + \end{align*} + \end{enumerate} +\end{defi} + +\begin{rem} + Let $f_1, \cdots, f_m: U \rightarrow \realn$, then the function + \begin{align*} + f: U &\longrightarrow \realn^m \\ + x &\longmapsto (f_1(x), \cdots, f_m(x)) + \end{align*} + is $k$-times totally differentiable if and only if the $f_1, \cdots, f_n$ are totally differentiable. + \[ + (D^kf(x)(u_1, \cdots, U_k))_j = D^k f_j(x)(u_1, \cdots, u_k) + \] +\end{rem} + +\begin{rem} + $D^k f(x)$ really is multi-linear (linear in every point) $\forall k \in \natn$. + Other multi-linear mappings are + \begin{enumerate}[(i)] + \item The scalar product on $\realn^n$ + \[ + \realn^n \times \realn^n \longrightarrow \realn + \] + \item The determinant + \[ + \realn^{n \times n} \longrightarrow \realn + \] + \end{enumerate} +\end{rem} + +\begin{rem} + A matrix $A \in \realn^{m \times n}$ is uniquely determined by its effect on the canonical basis $e_1, \cdots, e_n$. + This means if $v \in \realn$, then $\exists \alpha_1, \cdots, a_n \in \realn$ that are uniquely determined such that + \[ + v = \alpha_1, e_1 + \cdots + \alpha_n e_n + \] + Then + \[ + Av = \alpha_1 Ae_1 + \cdots + \alpha_n Ae_n + \] + $Ae_i$ is the $i$-th column of $A$. + An analogous statement for multi-linear mappings would be, that + \[ + A: \realn^{n \times k} \longrightarrow \realn^m + \] + is uniquely determined if $A(e_{i_1}, e_{i_2}, \cdots, e_{i_k})$ known $\forall i_1, \cdots, i_k \in \set{1, \cdots, n}$. +\end{rem} + +\begin{thm} + Let $U \subset \realn^n$ be open, $f: U \rightarrow \realn^m$ $k$-times differentiable in $x$ and + let $e_1, \cdots, e_n$ be the canonical basis of $\realn^n$. Then + \[ + D^k f(x) (e_{i_1}, \cdots, e_{i_k}) = \partial_{i_k} \cdots \partial_{i_1} f(x) + \] + $\forall i_i, \cdots, i_k \in \set{1, \cdots, n}$. +\end{thm} +\begin{proof} + For $k = 1$ this is already proven. So we can use proof by induction; + assume the statement holds for a $k$, i.e. $\forall i_1, \cdots, i_k \in \set{1, \cdots, k}$ + \[ + D^k f(x) (e_{i_1}, \cdots, e_{i_k}) = \partial_{i_k} \cdots \partial_{i_1} f(x) + \] + Then for $i_1, \cdots, i_k, i_{k+1} \in \set{1, \cdots, n}$ + \begin{equation} + \begin{split} + D^{k+1} f(x) (e_{i_1, \cdots, e_{i_k}}) &= D(D^k f(\cdots)(e_{i_1}, \cdots, e_{i_k}))(x) \cdot e_{i_{k+1}} \\ + &= D(\partial_{i_k}, \cdots \partial_{i_1} f(\cdot))(x) e_{i_{k+1}} \\ + &= \partial_{i_{k+1}}\partial_{i_k} \cdots \partial_{i_1} f(x) + \end{split} + \end{equation} + The order in which partial derivatives are applied is important! +\end{proof} + +\begin{eg} + Consider + \begin{align*} + f: \realn^2 &\longrightarrow \realn \\ + (x_1, x_2) &\longmapsto x_1^2 \cos(x_2) + \end{align*} + Then we can calculate + \begin{align*} + D^2 f(x) (u, v) ~~ u = u_1e_1 + u_2e_2, v = v_1e_1 + v_2e_2 + \end{align*} + As follows + \begin{align*} + D^2f(x)(u, v) &= u_1v_1D^2f(x)(e_1, e_1) + u_1v_2D^2f(x)(e_1, e_2) \\ + & ~~+ u_2v_1D^2f(x)(e^2, e^1) + u_2v_2D^2f(x)(e^2, e^2) \\ + &= u_1v_1 \cdot 2 \cdot \cos(x_2) - 2x_1\sin(x_2)u_1v_2 \\ + & ~~-2x_1\sin(x_2)v_1u_2 - x_1^2 \cos(x_2)u_2v_2 + \end{align*} +\end{eg} + +\begin{thm} + Let $U \subset \realn^n$ be open, and $f: U \rightarrow \realn^m$ $k$-times continuously differentiable. + Then $f$ is $k$-times totally differentiable. +\end{thm} +\begin{proof} + This is already proveb for $k = 1$. So we can use induction over $k$; + assume the statement is correct for $k \in \natn$. Let $u_1, \cdots, u_k \in \realn^n$, then $D^kf(\cdot)(u_1, \cdots, u_k)$ + is a linear combination of the partial derivative of $f$ of order $k$, and is thus continuously differentiable once more. + Therefore $D^2f(\cdot)(u_1, \cdots, u_k)$ is totally differentiable, and thus $f$ is $(k+1)$-times totally differentiable. +\end{proof} + +\begin{thm}[Theorem of Schwarz] + Let $U \subset \realn^n$ be open, and also $f \in C^2(U, \realn^m)$. Then + \[ + \forall x \in U ~\forall u, v \in \realn^n: ~~D^2f(x)(u, v) = D^2f(x)(v, u) + \] + and + \[ + \forall x \in U ~\forall i_1, i_2 \in \set{1, \cdots, n}: ~~\partial_{i_1}\partial_{i_2} f(x) = \partial_{i_2}partial_{i_1} f(x) + \] +\end{thm} +\begin{proof} + Let $m = 1$, $x \in U$, $\epsilon > 0$ such that $\oball(x) \subset U$. + If $u = 0$ or $v = 0$ then both sides of the equation vanish, so let $u, v \in \realn^n \setminus \set{0}$ and + \begin{equation} + 0 < t < c := \frac{\epsilon}{2 \cdot \max\set{\norm{u}, \norm{v}}} + \end{equation} + Define the helper function + \begin{equation} + \begin{split} + g_1: [0, t] &\longrightarrow \realn \\ + s &\longmapsto f(x + tv + su) - f(x + su) + \end{split} + \end{equation} + And apply the one dimensional intermediate value theorem. $\exists \xi \in (0, t)$ such that + \begin{equation} + g_1(t) - g_1(0) = g_1'(\xi) \cdot t = (Df(x + tv + \xi u) u - Df(x + \xi u)u) \cdot t + \end{equation} + Analogously, define and apply the intermediate value theorem to + \begin{equation} + \begin{split} + g_2: [0, t] &\longrightarrow \realn \\ + s &\longmapsto Df(x + sv + \xi u) u + \end{split} + \end{equation} + and get $\eta \in (0, t)$ + \begin{equation} + \begin{split} + g_2(t) - g_2(0) = g_2'(\eta) t &= D(Df(\cdot)u)(x + \eta v +\xi u)uvt \\ + &= D^2 f(x + \eta v + \xi u)(u, v)t + \end{split} + \end{equation} + using these results, we can get $\xi, \eta \in (0, t)$ for all $t \in (0, c)$ such that + \begin{equation} + \begin{split} + f(x + &tv + tu) - f(x + tv) - f(x + tu) + f(x) \\ + &= g_1(t) - g_1(0) = (Df(x + tv + \xi u)u - Df(x + \xi u)u) t \\ + &= (g_2(t) - g_2(0))t = D^2 f(x + \eta v + \xi u)(u, v) t^2 + \end{split} + \end{equation} + So we can write + \begin{equation} + \begin{split} + \limes{t}{0} &\frac{f(x + tv + tu) - f(x + tv) - f(x + tu) + f(x)}{t^2} \\ + &= \limes{t}{0} D^2 f\underbrace{(x + \eta v + \xi u)}_{\conv{} x}(u, v) \\ + &= D^2 f(x)(u, v) + \end{split} + \end{equation} + The left side is symmetric in terms of swapping $u$ and $v$, so the right side must be as well. +\end{proof} + +Note, that +\[ + D^2f(x)(e_{i_1}, e_{i_2}) = \partial_{i_2} \partial_{i_1} f(x) = \partial_{i_1} \partial_{i_2} f(x) = D^2f(x)(e_{i_2}, e_{i_1}) +\] + +\begin{rem} + Via induction: + \begin{enumerate}[(i)] + \item $D^kf(x)(u_1, \cdots, u_k)$ is independent from the order of the $u_i$, if $D^kf$ is continuous. + \item The limit of the second derivaative is useful in the numerical discussion of differential equations. + \end{enumerate} +\end{rem} + +\begin{thm}[Taylor's Theorem] + Let $U \subset \realn^n$ be open, $f: U \rightarrow \realn$ be $(l + 1)$-times differentiable and $h \in \realn^n$ such that + $x + th \in U$ $\forall t \in [0, 1]$. Then $\exists \theta \in [0, 1]$ such that + \[ + f(x + h) = \series[l]{k} \frac{1}{k!} D^kf(x)(h, \cdots, h) + \frac{1}{(l+1)!}D^{l+1}f(x + \theta h)(h, \cdots, h) + \] +\end{thm} +\begin{hproof} + Apply the one dimensional Taylor theorem with Lagrange error bound onto a helper function + \begin{equation} + \begin{split} + g: [0, 1] &\longrightarrow \realn \\ + t &\longmapsto f(x + th) + \end{split} + \end{equation} +\end{hproof} + +\begin{rem} + \begin{enumerate}[(i)] + \item Consider $h = \sum_{i=1}^n h_ie_i$. Then + \[ + D^2f(x)(h, h) = \sum_{i, j = 1}^n h_i h_j D^2f(x)(e_i, e_j) = \sum_{i, j = 1}^n \partial_i \partial_j f(x) h_i h_j + \] + + \item Analogously to one dimension, we can formulate criteria for local extrema: + \[ + Df(x) = 0, \cdots, D^{l-1}f(x) = 0 \text{ and } D^lf(x) \ne 0 + \] + \begin{itemize} + \item $x$ is a local minimum if $l$ is even and $D^lf(x)$ is positive. + \item $x$ is a local maximum if $l$ is even and $D^lf(x)$ is negative. + \item $x$ is no local extremum of $l$ is odd or if $D^lf(x)$ is undefined. + \end{itemize} + Definedness is complicated to determine for $l > 2$. + \end{enumerate} +\end{rem} +\end{document} \ No newline at end of file diff --git a/chapters/sections/partial_total_diff.tex b/chapters/sections/partial_total_diff.tex index dfbcc69..12bde9a 100644 --- a/chapters/sections/partial_total_diff.tex +++ b/chapters/sections/partial_total_diff.tex @@ -292,7 +292,7 @@ They denote the connecting line between $x$ and $y$. \end{center} \begin{thm}[Intermediate value theorem for $\realn$-valued functions] - Let $U \subset \realn^n$ be oppen, $x, y \in U$ and $\cline \subset U$. + Let $U \subset \realn^n$ be open, $x, y \in U$ and $\cline \subset U$. Now let $f: U \rightarrow \realn$ differentiable on $\oline$ and continuous in $x, y$. Then \[ \exists \xi \in \cline: ~~f(y) - f(x) = Df(\xi) (y-x) diff --git a/script.pdf b/script.pdf index dd78404..3ebb785 100644 Binary files a/script.pdf and b/script.pdf differ diff --git a/script.tex b/script.tex index c788a30..eec9227 100644 --- a/script.tex +++ b/script.tex @@ -8,6 +8,7 @@ \usepackage{multicol} \usepackage{tikz, pgfplots} \usepackage{kbordermatrix} +\usepackage{fancyhdr} \usetikzlibrary{calc,trees,positioning,arrows,fit,shapes,angles} \usepackage{color} @@ -154,6 +155,8 @@ \tableofcontents +\pagestyle{headings} + \subfile{chapters/FaN.tex} \subfile{chapters/real_analysis_1.tex} \subfile{chapters/linear_algebra.tex}