\documentclass[12pt]{article}
\usepackage{fullpage,amsmath,amsfonts,graphicx,mathtools}

\def\d{{\mathrm d}}
\def\e{{\mathrm e}}
%\def\a{{\boldsymbol a}}
%\def\b{{\boldsymbol b}}
%\def\x{{\boldsymbol x}}
%\def\y{{\boldsymbol y}}
\def\R{{\mathbb R}}
%\def\v{\boldsymbol{v}}

\begin{document}

\title{Analysis II}
\author{Homework 11}
\date{Due in class Monday, May 6, 2019}
\maketitle

\begin{enumerate}

\item Recall the Lagrange multiplier theorem: Let $E \subset \R^d$ be
open, $h \in C^1(E,\R)$, and $f \in C^1(E, \R^n)$ with $n < d$.
Suppose that $h$ has a local extremum subject to the constraint
$f(x) = 0$ at a point $a \in E$, and suppose that $f'(a)$ has full
rank.  Then there exists a unique vector $\lambda \in \R^n$, the
\emph{Lagrange multipliers}, such that
\[
  \nabla h(a) =  \lambda^T \, \nabla f(a) \,.
\]
(Here, we think of $\nabla f$ as a row vector and $\lambda$ as a  column
vector.)

\begin{enumerate}
\item Show that, under the assumptions of the Lagrange multiplier
theorem, that if $h$ has a local extremum subject to the constraint
$f(x) = 0$ at a point $a \in E$, then the function
$H \colon E \times \R^n \to \R$,
\[
  H(x,\mu) = h(x) - \mu^T f(x) \,,
\]
has a critical point $(a,\lambda)$, where $\lambda$ is the vector of
Lagrange multipliers.

\item Give an example which shows that the critical point of $H$ in
part (a) may not correspond to a local extremum of $H$.

\emph{Hint:} Choose simple functions for $d=2$ and $n=1$, show that
$\operatorname{Hess}H$ is indefinite, and appeal to the second
derivative test.
\end{enumerate}


\item Find the minimum and maximum values of
\[
  h(x,y) = x \, y
\]
subject to the constraint
\[
  \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \,.
\]

\item
\begin{enumerate}
\item Calculate the iterated integral
\[
  \int_0^1 \int_0^1 \frac{x}{(1+x^2)(1+xy)} \, \d x \, \d y
\]
in two different ways, and prove thereby that
\[
  \int_0^1 \frac{\ln (1+x)}{1+x^2} \, \d x
  = \frac{\pi \ln 2}{8} \,.
\]
\item Conclude that
\[
  \int_0^1 \frac{\arctan x}{1+x} \, \d x
  = \frac{\pi \ln 2}{8} \,.
\]
\end{enumerate}

\item Consider the function
\[
  f(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2}
\]
on the rectangle $I = [0,1] \times [\alpha, 1]$ for $\alpha \in
(0,1)$.

\begin{enumerate}
\item Show that
\[
  \int_0^1 f(x,y) \, \d x = - \frac1{1+y^2}
\]
for every fixed $y \in [\alpha,1]$.

\emph{Hint:} Note that
\[
  \frac{x^2-y^2}{(x^2+y^2)^2}
  = \frac{1}{x^2+y^2} + y \, \frac{\partial}{\partial y}
    \frac{1}{x^2+y^2} \,.
\]

\item Note that the result from (a) continuously extends to the unit
square $I = [0,1]^2$ and conclude that
\[
  \int_0^1 \int_0^1 f(x,y) \, \d x \, \d y = - \frac\pi4
\]
while 
\[
  \int_0^1 \int_0^1 f(x,y) \, \d y \, \d x = \frac\pi4 \,.
\]

\item Does this contradict the theorem on the exchange of partial
integration proved in class?  Explain!

\end{enumerate}


\end{enumerate}

\end{document}
