% file: grad/iw/analysis.tex \documentclass[12 pt]{article} \usepackage[dvips]{graphicx} \usepackage{amssymb} \newcommand \qed{\vrule height5pt width5pt} \newcommand \no{\noindent} \newcommand \implies{\Rightarrow} \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand{\bea}{\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}} \newcommand{\beann}{\begin{eqnarray*}} \newcommand{\eeann}{\end{eqnarray*}} \def\reff#1{(\ref{#1})} \newcommand \field{\mathcal{F}} \newcommand{\reals}{\mathbb{R}} \newcommand{\complex}{\mathbb{C}} \newcommand{\integers}{\mathbb{Z}} \newcommand{\half}{\mathbb{H}} \newcommand{\disc}{\mathbb{D}} \newcommand{\lerwdisc}{\mathbb{U}} \newcommand{\compose}{\circ} \newcommand{\boundary}{\partial} \newcommand{\ra}{\rightarrow} \newcommand{\calS}{\mathcal{S}} \newcommand{\calA}{\mathcal{A}} %%%%%%%%%%%%%%%%%%% macros %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \setlength{\textheight}{21cm} \setlength{\textwidth}{16cm} \oddsidemargin 0.0in \evensidemargin 0.0in \topmargin 0.0in \pagestyle{plain} \newtheorem{definition}{Definition} \newtheorem{proposition}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{theorem}{Theorem} \newtheorem{exercise}{Exercise} \begin{document} \bibstyle{ams} \centerline{\Large \bf Integration Workshop 2011} \bigskip \centerline{\Large \bf Calculus/Analysis notes} \section{Calculus} \subsection{Multivariate differential calculus} \begin{definition} Let $O \subset \reals^n$ be open and $f:O \ra \reals^m$. Let $c \in O$ and let $\epsilon$ be small enough that $B(c,\epsilon) \subset O$. The function $f$ is said to be {\it differentiable} at $c$ if there is a linear function, called the {\it total derivative}, $T^f_c : \reals^n \ra \reals^m$, such that \bea f(c+v) = f(c) + T^f_c(v) + E_c(v) \eea for $|v|<\epsilon$, where the error term satisfies \bea \lim_{v \ra 0} {E_c(v) \over |v|} =0 \eea We say $f$ is differentiable on $O$ if it is differentiable at every point in $O$. We say that $f$ is continuously differentiable if $T^f_c$ is a continuous function of $c$. \end{definition} If the total derivative exists, then for all directions the directional derivative exists \beann D_v f(c) = \lim_{\epsilon \ra 0} {f(c+\epsilon v)-f(c) \over \epsilon} \eeann and equals $T^f_c(v)$. Let $e_1,e_2,\cdots,e_n$ be the standard basis for $\reals^n$. Then the partial derivatives are the directional derivative in the directions of the standard basis: \beann {\partial f \over \partial x_k} = D_{e_k} f \eeann Note that both sides of this equation are vectors in $\reals^m$. The components are \beann {\partial f_j \over \partial x_k} = D_{e_k} f_j \eeann The matrix representation of $T$ in this basis is called the {\it Jacobian matrix} \beann Df(c) = \pmatrix{ {\partial f_1 \over \partial x_1}(c) & {\partial f_1 \over \partial x_2}(c) & \cdots & {\partial f_1 \over \partial x_n}(c) \\ {\partial f_2 \over \partial x_1}(c) & {\partial f_2 \over \partial x_2}(c) & \cdots & {\partial f_2 \over \partial x_n}(c) \\ . & . & & . \\ . & . & & . \\ . & . & & . \\ {\partial f_m \over \partial x_1}(c) & {\partial f_m \over \partial x_2}(c) & \cdots & {\partial f_m \over \partial x_n}(c) } \eeann \begin{theorem}[Chain rule] Suppose that $g: U \subset \reals^k \ra O$ and $f: O \subset \reals^n \ra \reals^m$ are differentiable at $p \in U$ and $g(p) \in O$ respectively. Then $h=f \compose g$ is differentable at $p$ and \beann T^h_p = T^f_{g(p)} \compose T^g_p \eeann In matrix form \beann Dh(p) = Df(g(p)) \, Dg(p) \eeann \end{theorem} Let $L(x_1, x_2) = \{\lambda x_1 + (1 - \lambda) x_2 : 0 \le \lambda \le 1\}$ be the line segment connecting $x_1$ and $x_2$ in $\reals^n$. \begin{theorem} [Mean Value Theorem] Let $O$ be an open subset of $\reals^n$ and assume that $f : O \ra \reals^m$ is continuously differentiable on $O$. Choose $x_1$ and $x_2$ so that $L(x_1, x_2) \subset O$. Then for every vector $a \in \reals^m$, there is a point $c \in L(x_1, x_2)$ such that \beann a \cdot (f(x_2) - f(x_1)) = a \cdot T^f_c (x_2 - x_1) \eeann \end{theorem} We now consider higher order derivatives. \begin{theorem} Let $f : \reals^n \ra \reals^m$. Then the following conditions are sufficient for the equality of the mixed partial derivatives \beann {\partial^2 f \over \partial x_i \partial x_j}(c)= {\partial^2 f \over \partial x_j \partial x_i}(c) \eeann 1. Both ${\partial f / \partial x_i}\ and {\partial f / \partial x_j}$\ exist in an $n$-ball $B(c,\delta)$ and are differentiable at $c$. 2. Both ${\partial f /\partial x_i}\ and\ {\partial f / \partial x_j}$\ exist in an $n$-ball $B(c,\delta)$\ and\ ${\partial^2 f / \partial x_i \partial x_j}$\ and\ ${\partial^2 f / \partial x_j \partial x_i}$ are both continuous at c. \end{theorem} Call $\alpha = (\alpha_1, \cdots , \alpha_n)$ a multi-index if each of its entries are non-negative integers. Write $|\alpha| = \alpha_1 + \cdots + \alpha_n$. This allows for the notational abbreviations \beann x^\alpha = x^{\alpha_1} \cdots x^{\alpha_n}, \quad D^\alpha = {\partial^{\alpha_1} \over \partial x^{\alpha_1}} \cdots {\partial^{\alpha_n} \over \partial x^{\alpha_n}} \eeann and provides for a compact notation for Taylor's formula for functions $f$ from $\reals^n$ to $\reals$. Write for $a,x \in \reals^n$ \beann f^{(k)}(a; x) = \sum_{\alpha: |\alpha|=k} \, D_\alpha f(a) x^\alpha \eeann and assume that $f$ and all of its partial derivatives of order up to $m-1$ are differentiable at each point of an open set $S \subset \reals^n$. Choose $x$ and $a$ so that $L(a, x) \subset S$. Then for some $c \in L(a, x)$, \beann f(x) = f(a) + \sum_{k=1}^{m-1} \, {1 \over k!} f^{(k)}(a; x-a) + {1 \over m!} f^{(m)}(c; x-a) \eeann \subsection{Implicit functions} Let $A$ be an $n \times n$ matrix. Then, for $y \in \reals^n$, $Ax = y$ has a unique solution $x$ whenever $A$ has nonzero determinant. This suggests that in looking for a unique solution to $f(x) = y$, we consider the Jacobian determinant, the determinant of the Jacobian matrix, \beann J_f(x) = \det Df(x) = {\partial{(f_1,\cdots,f_n)} \over \partial{(x_1,\cdots,x_n)}} \eeann \begin{theorem}[Inverse function theorem] Let $f : S \ra \reals^n$ be continuously differentiable on $S \subset \reals^n$. If the Jacobian determinant $J_f(a) \ne 0$ for some point $a \in S$, then there exists two open sets $X \subset S$ and $Y \subset f(S)$ and a uniquely determined function $g$ defined on $Y$ such that 1. $a \in X$ and $f(a) \in Y$ 2. $Y = f(X)$ 3. $f$ is one-to-one on $X$ 4. $g(Y ) = X$ 5. $g(f(x)) = x$ for every $x \in X$ 6. $g$ is continuously differentiable on $Y$ . \end{theorem} Note that for $y = f(x)$, $Dg(y) \, Df(x)$ is the identity linear transformation. \begin{theorem}[Implicit function theorem] Let $S \subset \reals^n \times \reals^k$ and suppose that $f : S \ra \reals^n$ is continuously differentiable. Assume that $f(x_0, y_0) = 0$ and that the determinant of the $n \times n$ matrix $\partial f_j / \partial x_i (x_0,y_0)$ where $i,j$ both range from $1$ to $n$ is not zero. Then there exists an open set $Y_0 \subset \reals^k$ containing $y_0$ and a unique function $ g : Y_0 \ra \reals^n$ such that 1. $g$ is continuously differentiable 2. $g(y_0) = x_0$ 3. $f(g(y), y) = 0$ for every $y \in Y_0$. \end{theorem} \subsection{Multivariable Riemann integrals} Let $A=[a_1,b_1]\times\cdots\times[a_n,b_n]\subset\reals^n$ and let ${\cal P}_k$ be a partition of $[a_k,b_k]$ into $m_k$ intervals. Then \beann {\cal P} = {\cal P}_1\times\cdots\times{\cal P}_n \eeann divides $A$ into $m_1\times\cdots\times m_n$ $n$ dimensional intervals. A partition ${\cal Q}$ is called a refinement of ${\cal P}$ if ${\cal P}\subset{\cal Q}$. For $I\in {\cal P}$ let $\hbox{vol}(I)$ be the product of the lengths of the one dimensional intervals that determine $I$. Let $f$ be a real valued function defined on $A$. For any choice of sample points $t_I \in I$, define the Riemann sum \beann S(f,\mathbf{t}, {\cal P})= \sum_I f(t_I)\hbox{vol}(I) \eeann \begin{theorem} The function $f$ is Riemann integrable on $A$ if there exist a number, $r$, having the following property: For every $\epsilon > 0$, there exists a partition ${\cal P}_\epsilon$ such that for any refinement $\cal Q$ of ${\cal P}_\epsilon$ and any choice of sample points $t_I \in I$. \beann |S(f,\mathbf{t}, {\cal P}) - r| < \epsilon. \eeann \end{theorem} We typically write the integral \beann \int_A f(x_1,\ldots,x_n)\ d(x_1,\ldots,x_n)= \int_A f(x)\ dx. \eeann \subsection{Change of variable formulas for integrals} Let $\tau$ be a one-to-one continuously differentiable mapping of an open set $V \subset \reals^k$ into $\reals^k$ such that the Jacobian determinant $J_\tau(x) \ne 0$ for all $x \in V$. Let $f$ be a continuous function on $\reals^k$ whose support is compact and lies in $\tau(V)$. Then \beann \int_{\reals^k} \, f(y) \, dy = \int_{\reals^k} \, f(\tau(x)) \, |J_\tau(x)| \, dx \eeann \subsection{Differential forms and Stokes theorem} Let $K \subset \reals^k$ be compact and let $V \subset \reals^n$ be open. A $k$-surface is a continuously differentiable mapping $\Phi : K \ra V$. For example, each component of a 1-surface is called a curve. A {\it differential form of order $k$}, or briefly, a $k$-form, is a function $\omega$, represented symbolically by \beann \omega= \sum a_{i_1 \cdots i_k}(x) \, dx_{i_1} \wedge \cdots \wedge dx_{i_k} \eeann that assigns to each $k$-surface $\Psi$ in $V$ a number \beann \int_\Phi \omega = \int_K \, \sum a_{i_1 \cdots i_k}(\Phi(u)) \, {\partial(x_{i_1},\cdots,x_{i_k}) \over \partial(u_1,\cdots,u_k)} du \eeann A 0-form is defined to be a continuous function of $V$ . Integrals of 1-forms are called line integrals. Let $c \in \reals$ and let $\omega, \omega_1, \omega_2$ be $k$-forms on $V$. Then \beann \int_\Phi \, c \omega = c \int_\Phi \, \omega \eeann \beann \int_\Phi \, (\omega_1 + \omega_2) = \int_\Phi \, \omega_1 + \int_\Phi \, \omega_2 \eeann For $\omega=a_{{i_1} \cdots i_k}(x) \, dx_{i_1} \wedge \cdots \wedge dx_{i_k}$ and for $\bar{\omega}$ obtained from $\omega$ by interchanging some pair of subscripts, $\bar{\omega} = - \omega$. Write the basic $k$-form $dx_I = dx_{i_1} \wedge \cdots \wedge dx_{i_k}$, for $1 \le i_1 < \cdots < i_k$, giving the standard presentation \beann \omega = \sum_i a_I(x) dx_I \eeann \subsection{Differentiation of forms} \smallskip The operator $d$ is a mapping from $k$-forms to $(k + 1)$-forms defined as follows: 1. For a class $C^1$ $0$-form $f$ , \beann df = \sum_{i=1}^n {\partial f \over \partial x_i} dx_i \eeann 2. For the class $C^1$ $k$-form $\omega$ above in the standard presentation, \beann d \omega = \sum_I \, (da_I) \wedge dx_I \eeann For i = 1, 2, let $\omega_i$ be class $C^1$ $k_i$-forms. Then \beann d(\omega_1 \wedge \omega_2) = (d \omega_1) \wedge \omega_2 + (-1)^{k_1} \omega_1 \wedge (d \omega_2) \eeann If $\omega$ is of class $C^2$, $d(d\omega) = 0$. \begin{definition} A $k$-form $\omega$ is called exact if $\omega = d \eta$ for some $(k - 1)$-form $\eta$. A class $C^1$ $k$-form is called closed if $d \omega = 0$. \end{definition} Every exact class $C^1$ form is closed. If the domain is a convex set, then the Poincare lemma states that the converse is true. \begin{theorem}[General Stokes' theorem] If $\Psi$ is a $k$-chain of class $C^2$ in an open set $V \subset \reals^m$ and if $\omega$ is a $(k - 1)$-form of class $C^1$ in $V$ , then \beann \int_\Psi \, d \omega = \int_{\boundary \Psi} \, \omega \eeann \end{theorem} Various theorems from calculus and vector calculus are special cases of this general theorem as indicated in the following table. \begin{table} \begin{center} \begin{tabular}{|r|r|l|} \hline $k$ & $m$ & theorem \\ \hline 1 & 1 & fundamental theorem \\ 2 & 2 & Green's theorem \\ 3 & 3 & divergence theorem \\ 2 & 3 & classical Stokes theorem \\ \hline \end{tabular} \end{center} \end{table} \begin{theorem} [Green's Theorem] Let $C$ be a simple closed curve in the $xy$-plane Let $M(x,y)$ and $N(x,y)$ be continuously differentiable on an open set containing $C$ and $R$, the region it encloses. Then \beann \int_C (M dx + N dy) = \int_R ( {\partial N \over \partial x} - {\partial M \over \partial y}) dx dy \eeann \end{theorem} \begin{theorem} [Divergence Theorem] Let $F$ be a continuously differentiable vector field on an open set $V \subset \reals^3$, and let $C \subset V$ be closed with positively oriented boundary $\boundary C$. Then \beann \int_C (\nabla \cdot F) dV = \int_{\boundary C} (F \cdot \mathbf{n} ) \, dA \eeann where $\mathbf{n}$ is a unit normal vector, pointing outwards. \end{theorem} \begin{theorem} [classical Stokes' Theorem] Let $F$ be a continuously differentiable vector field on an open set $V \subset \reals^3$, and let $S \subset V$ be a 2-surface of class $C^2$. Then \beann \int_S (\nabla \times F) \cdot \mathbf{n} \, dV = \int_{\boundary S} (F \cdot \mathbf{t}) \, ds \eeann where $\mathbf{t}$ is a oriented unit tangent vector. \end{theorem} \section{Real Analysis} \subsection{Sequences and series} \begin{definition} Let $a_n$ be a sequence of real or complex numbers. We say that $a_n$ {\it converges} to $a$ if for every $\epsilon > 0$ there is an $N$ such that $n \ge N$ implies $|a_n-a|<\epsilon$. We say that $a_n$ is a Cauchy sequence if for every $\epsilon > 0$ there is an $N$ such that $n,m \ge N$ implies $|a_n-a_m|<\epsilon$. \end{definition} By the completeness axiom of the real numbers, a monotone sequence converges if and only if it is bounded. Given a sequence $a_n$, define $b_n = \sup \{ a_k : k \ge n \}$. Then $b_n$ is a nonincreasing sequence and so has a limit. Call this the limit superior or just lim sup $a_n$ and write $\limsup_{n \ra \infty} a_n$. Similarly, the limit inferior or lim inf is defined by \beann \liminf_{n \ra \infty} a_n = \lim_{n \ra \infty} \inf \{ a_k : k \ge n \} \eeann The $\liminf$ and $\limsup$ always exist although the $\liminf$ and the $\limsup$ can be $\pm\infty$. We always have $\liminf_{n\to\infty}a_n \le \limsup_{n\to\infty} a_n$. They are equal if and only if $a_n$ converges. In this case they are equal to the limit of $a_n$ and write $$\lim_{n\to\infty} a_n.$$ \subsection{Convergence tests for series and sequences:} \medskip \noindent 1. {\it Integral test}: Let $f$ be a positive decreasing function defined on $[1,\infty)$ such that $\lim_{x \ra \infty} f(x) = 0$. For $n = 1, 2, \cdots$, define \beann s_n = \sum_{k=1}^n f(k), \quad t_n = \int_1^n \, f(x) \, dx, \eeann Then $s_n$ converges if and only if $t_n$ converges. \medskip \noindent 2. {\it Ratio and root tests}: Given a series $\sum_{n=1}^\infty a_n$ of nonzero complex terms, let \beann r_- = \liminf_{n \ra \infty} \left| {a_{n+1} \over a_n } \right|, \quad r_+ = \limsup_{n \ra \infty} \left| {a_{n+1} \over a_n } \right|, \quad \rho = \limsup_{n \ra \infty} |a_n|^{1/n} \eeann (a) The series converges absolutely if either $r_+ < 1$ or $\rho< 1$. (b) The series diverges if either $r_- > 1$ or $\rho> 1$. (c) In all other cases the tests are inconclusive. \medskip \noindent 3. {\it Dirichlet's test}: Given a series $\sum_{n=1}^\infty a_n$ of nonzero complex terms such that the partial sums are a bounded sequence. Let $b_n$ be a decreasing sequence which converges to $0$. Then $\sum_{n=1}^\infty a_n b_n $ converges. \medskip \noindent 4. {\it Abel’s test:} The series $\sum_{n=1}^\infty a_n b_n$ converges if $\sum_{n=1}^\infty a_n$ converges and $b_n$ is monotone and bounded. \subsection{Infinite products} Let $u_n$ be a sequence of complex numbers. The infinite product $$\prod_{n=1}^\infty u_n$$ is said to converge if there is an $N$ such that $u_n \ne 0$ for $n \ge N$ and the sequence $p_k = \prod_{n=N+1}^k u_n$ has a nonzero limit $p$ as $k \ra \infty$. In the case of convergence, $\prod_{n=1}^\infty u_n$ is defined to be $u_1 \cdots u_N\cdot p$. There is a connection betweeen convergence of sums and of products. For $a_n>0$, the product $\prod_{n=1}^\infty (1+a_n)$ converges if and only if the series $\sum_{n=1}^\infty a_n$ converges. We say that the product $\prod_{n=1}^\infty (1+a_n)$ converges absolutely if $\prod_{n=1}^\infty (1+|a_n|)$ converges. Absolute convergence of the infinite product implies convergence of the product. \subsection{Sequences of functions} \subsubsection{Continuity and uniform continuity} A function $f$ is continuous at $a$ if for every $\epsilon>0$ there is a $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$. A function $f$ is uniformly continuous on $S$ if for every $\epsilon>0$ there is a $\delta>0$ such that for all $a \in S$ and all $x$ with $|x-a|<\delta$, we have $|f(x)-f(a)|<\epsilon$. \begin{theorem}If $S$ is compact and $f$ is continuous on $S$ then it is uniformly continuous on $S$. \end{theorem} \subsubsection{Convergence, uniform convergence and continuity} A sequence of functions is said to {\it converge pointwise} to a limit function $f$ on a set $S$ provided that for every $x \in S$, and each $\epsilon > 0$, there exists $N$, depending possibly on both $x$ and $\epsilon$ such that $n > N$ implies $|f_n(x) - f(x)| < \epsilon$. If the choice of $N$ does not depend on $x$, the sequence of functions is said to {\it converge uniformly.} If $f_n \ra f$ uniformly on $S$ and each $f_n$ is continuous at a point $c$, then $f$ is continuous at $c$. Let $f_n$ be a sequence of functions defined on a set $S$. For each $x \in S$, set \beann s_n(x) = \sum_{k=1}^n \, f_n(x) \eeann If $s_n \rightarrow s$ uniformly on $S$, then we say that $\sum_{n=1}^\infty f_n(x)$ converges uniformly on $S$. \begin{theorem} [Weierstrass M-test] Let $M_n$ be a sequence of nonnegative numbers such that $|f_n(x)| \le M_n $ for $n = 1, 2, \cdots$ and every $x \in S$. If $\sum_{n=1}^\infty M_n$ converges, then $\sum_{n=1}^\infty f_n(x)$ converges uniformly on $S$. \end{theorem} \subsubsection{The $L^\infty$ norm } Consider the vector space $C(S)$, the real valued continuous functions on $S$, and define the infinity norm (or sup norm) by \beann ||f||_\infty = \sup_{x \in S} |f(s)| \eeann Then $|| \cdot ||_\infty$ is a norm, meaning that 1) $||f||_\infty \ge 0$ and $||f||_\infty = 0$ if and only if $f(x) = 0$ for all $x \in S$. 2) $||af||_\infty = |a| ||f||_\infty$ for every $a \in \reals$. 3) $|| f + g ||_\infty \le || f||_\infty + || g||_\infty $ This norm induces a metric $d(f, g) = ||f - g||_\infty$. The theorems above on uniform continuity show that $C(S)$ with this metric is a complete metric space. \subsubsection{Integration and differentiation} Many of the theorems on uniform convergence permit the reversal of the order of taking of limits. \begin{theorem} [Integration] Let $\alpha(x)$ have bounded variation on $[a,b]$ and assume that $f_n$ is Riemann integrable with respect to an integrator $\alpha$ having bounded variation. For $n=1,2,\cdots$. Define \beann g_n(x) = \int_a^x \, f_n(t) d \alpha(t), \quad x \in [a, b] \eeann Assume there exists $f$ so that $d(f_n,f) \ra 0$. Then (a) $f$ is Riemann integrable with respect to $\alpha$ and (b) $d(g_n,g) \ra 0$ where \beann g(x) = \int_a^x \, f(t) d \alpha(t), \quad x \in [a, b] \eeann \end{theorem} \begin{theorem}[Differentiation] Assume that $f_n$ is differentiable on $(a,b)$ and that there exist a function $g$ so that $d(f_n^\prime,g) \ra 0$ and a point $c \in (a,b)$ so that $f_n(c)$ converges. Then (a) there exists f so that $d(f_n,f) \ra 0$, and (b) $f$ is differentiable with derivative $g$. \end{theorem} \section{Complex Analysis} \subsection{Analytic functions} Let $O$ be an open subset of $\complex$. Let $f:O \ra \complex$. We say $f$ is analytic at $z_0$ if the following complex limit exists: \beann f^\prime(w) = \lim_{z \ra w} {f(z)-f(w) \over z - w} \eeann for all $w$ in a neighborhood of $z_0$. One can think of a function from $\complex$ to $\complex$ as a function from $\reals^2$ to $\reals^2$. We write $z=x+iy$ and $f(z) = u(x,y) + i v(x,y)$. Then the function is $F(x,y)=(u(x,y),v(x,y))$. It is important to understand that analyticity is a much stronger property than requiring that $F$ have a total derivative. The above limit involves complex numbers and so it includes as special cases $z$ approaching $z_0$ along any direction. These ``directional limits'' must all give the same complex number as the limit. In particular, by considering taking the limit in the coordinate directions, one obtains the Cauchy Riemann equations. \begin{theorem} $f$ is analytic at $z_0=(x_0,y_0)$ if and only if for all $(x,y)$ in a neighborhood of $(x_0,y_0)$ the total derivative of $F$ exists and \beann {\partial u \over \partial x }(x,y) = {\partial v \over \partial y }(x,y), \quad {\partial u \over \partial y }(x,y) = - {\partial v \over \partial x }(x,y). \eeann \end{theorem} \subsection{Power series} An infinite series of the form \beann f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n \eeann is called a power series centered at $z_0$. Define $r$ by $ 1/r = \limsup_{r \ra \infty} |a_n|^{1/n}$ (We make the conventions that $1/0 = \infty$ and $1/\infty = 0$.) Then by the root test, the series converges absolutely if $|z - z_0| < r$ and diverges if $|z - z_0| > r$. Furthermore: \noindent 1. The series converges uniformly on every compact subset of $B(z_0, r)$. \noindent 2. The function $f$ can be differentiated term by term for any $z \in B(z_0,r)$, \beann f^\prime(z) = \sum_{n=1}^\infty n a_n (z - z_0)^{n-1} \eeann \noindent 3. The power series for $f^\prime$ has radius of convergence $r$. \noindent 4. Repeated differentiation and evaluation of this yields $a_k = f^{(k)}(z_0)/k!$. \begin{theorem} Suppose that the power series \beann f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n \eeann has a nonzero radius $r$ of convergence. Then $f$ is analytic on $B(z_0,r)$. Conversely, if $f$ is analytic at $z_0$, then there is a power series with a nonzero radius of convergence that converges to $f$ in a neighborhood of $z_0$. \end{theorem} \subsection{Integration} \begin{definition} A domain $D$ is simply connected if the region bounded by every simple closed curve in $D$ is contained in $D$, i.e., every simple closed curve in $D$ may be continuously contracted to a point without leaving $D$. \end{definition} \begin{theorem} [one of many ``Cauchy's theorems''] If $D$ is a simply connected open set and $f$ is analytic on $D$ and $\gamma$ is a differentiable closed curve in $D$, then \be \oint_\gamma f(z) dz =0 \ee \end{theorem} \subsection{Zeroes, poles and residues} We say $f$ has a zero at $z_0$ if $f(z_0)=0$. In this case it is possible to write it in the form $f(z) = (z-z_0)^n g(z)$ in a neighborhood of $z_0$ where $g$ does not vanish on this neighborhood. The integer $n$ is unique and called the {\it order} of the zero. A neigborhood of a point $z_0$ means an open set containing $z_0$. By a deleted neighborhood of $z_0$ we will mean a neighborhood of $z_0$ with $z_0$ removed. A function $f$ has an isolated singularity at $z_0$ if it is analytic on a deleted neighborhood of $z_0$. If $f$ has an isolated singularity at $z_0$, and we can redefine it at $z_0$ so that the function is analytic at $z_0$, then we say $f$ has a removable singularity at $z_0$. Otherwise we consider $1/f$ where $1/f$ is defined to be $0$ at $z_0$. If this is anayltic at $z_0$ we say $f$ has a pole at $z_0$. The order of the pole is defined as the order of the zero of $1/f$ at $z_0$. If it does not have a pole we say it has an essential singularity. \begin{theorem} If $f$ has a pole of order $n$ at $z_0$ then \beann f(z) = {a_{-n}\over (z-z_0)^n} + {a_{-(n-1)}\over (z-z_0)^{n-1}} + \cdots + {a_{-1}\over z-z_0} + g(z) \eeann where $g$ is analytic at $z_0$. \end{theorem} The {\it principal part} of $f(z)$ (at $z_0$) is \beann {a_{-n}\over (z-z_0)^n} + {a_{-(n-1)}\over (z-z_0)^{n-1}} + \cdots + {a_{-1}\over z-z_0} \eeann The {\it residue} of $f$ at $z_0$ is $a_{-1}$ \section{Differential Equations} \subsection{First order ordinary differential equations} Let $V \subset \reals^n$ and $I = [t_0, t_f ]$ and $\phi: I \times V \ra \reals^n$. A solution to the initial value problem \beann y^\prime = \phi(t, y),\ y(t_0) = y_0, \eeann is a differentiable function $f$ on $I$ such that $f(t_0) = y_0$, $f(t) \in V $ and $f^\prime(t) = \phi(t, f(t))$. The ODE is called autonomous if $\phi$ is independent of $t$. The basic estimate used to study the dependence of solutions on initial conditions is Gronwall's inequality. \begin{theorem} [Gronwall's inequality] Let $f, g : [a, b) \ra \reals$ be continuous and nonnegative. Suppose \beann f(t) \le K + \int_a^t f(s) g(s) \, ds, \quad K \ge 0 \eeann Then \beann f(t) \le K \exp\left( \int_a^t g(s) \, ds \right) \eeann for $t \in [a, b)$. \end{theorem} \subsection{Linear ODE's} A {\it linear system} is one in which $\phi(t, y) = A(t)y + g(t)$. It is called {\it homogeneous} if $g(t) = 0$ for all $t$. For a homogeneous autonomous linear system $y^\prime = Ay$, a solution is $y(t) = \exp(t - t_0) A y_0$. By Gronwall’s inequality, this solution is unique. For nonautomonous systems having continuous $A$, the solutions of $y^\prime = A(t)y$ form a vector space of dimension $n$ over the complex numbers. An $n \times n$ matrix whose columns are linearly independent solutions is called a fundamental matrix. Once this matrix valued function has been found, we can find a solution to the non-homogeneous system by variation of constants. \begin{theorem} If $\Phi$ is a fundamental matrix of $y^\prime = A(t)y$ on $I$, then the function \beann \psi(t) = \Phi(t) \int_{t_0}^t \Phi^{-1}(s)g(s) \, ds \eeann is the unique solution of the nonhomogenous linear system above with initial condition $\psi(t_0) = 0$. \end{theorem} \subsection{Existence of solutions and iteration techniques} \begin{lemma} Let $\phi$ be Lipschitz in $y$ uniformly in $t$ and assume that $B(y_0,r) \subset V$ . Choose $M$ so that $|\phi(t, y)| \le M$ for $(t,y) \in I \times B(y_0,r)$. Set $t_0 \in I$ and $\delta = r/M$ . Then there is a unique $C^1$ function $y(t)$, $t \in (t_0 - \delta, t_0 + \delta)$ satisfying $y(t) \in B(y_0,r)$ that is a solution to the ordinary differential equation. \end{lemma} The differential equation is equivalent to \beann y(t) = y_0 + \int_{t_0}^t \phi(t,y(s)) \, ds \eeann The {\it Picard iteration technique} begins by seting $y_0(t) = y_0$ and defining inductively \beann y_{n+1}(t)=y_0 + \int_{t_0}^t \, \phi(t,y_n(s)) \, ds \eeann Let $K$ be the Lipschitz constant. By induction, we find that \beann |y_{n+1}(t) - y_n(t)| \le {M K^n |t - t_0|^{n+1} \over (n + 1)!} \eeann Thus, $y_n$ is a Cauchy sequence in the $L^\infty$ norm. Consequently, the limit $y$ is a continuous curve that is a solution. To check uniqueness, let $\hat{y}$ be another solution. Then check that \beann |y_n(t) - \hat{y}(t)| \le {M K^n |t - t_0|^{n+1} \over (n + 1)!} \eeann to see that $\hat{y}=y$. \end{document} \end