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\vspace*{-.7in}

\noindent
\textbf{Math 5616H} \hfill \textbf{Homework 6}

\noindent
\textbf{Posted: 3/8; Updated: 3/20; Due: Monday, 3/23/2015}\\
The problem set is due at the beginning of the class on
Monday. \smallskip

\noindent
\textbf{Reading}: 9.15-18, 20-21, 39, 41. \smallskip

\begin{prob}
  Let $U \subset \nr^n$ be an open set. Show that if $f_k : U \to
  \nr$, $k = 1, \dots, m$ are differentiable, then $f = (f_1, \dots,
  f_m)$ is differentiable on $U$ and $Df = (Df_1, \dots, Df_m)$.
\end{prob}

\begin{prob}
  Show that the maximum value of the directional derivative of $f: U
  \to \nr$, $U \subset \nr^n$ open, at a point $x$ is the direction of
  the gradient $\nabla f(x)$ and the value of this derivative is
  $\abs{\abs{\nabla f(x)}}$. \emph{Hint}: You can use the relation
  between the dot product of two vectors, their norms, and the angle
  between them.
\end{prob}

\begin{prob}
Let 
\[
f(x,y) = \begin{cases} 0 & \text{when $(x,y) = (0,0)$},\\
\frac{xy}{x^2+y^2} & \text{otherwise}.
\end{cases}
\]
Show that $D_1f(0,0) = D_2f(0,0) = 0$, but nonetheless $f$ is not
continuous at the origin, and hence not differentiable there.
\end{prob}

\begin{prob}
Let 
\[
f(x,y) = \begin{cases} 0 & \text{when $(x,y) = (0,0)$},\\
\frac{x^3}{x^2+y^2} & \text{otherwise}.
\end{cases}
\]
Show that $f$ has a directional derivative in every direction at the
origin, but $f$ is not differentiable at the origin. \emph{Hint}: Show
that the formula $D_v f (x) = \nabla f(x) \cdot v$ for $\abs{\abs{v}}
= 1$ (Eqn.\ (40) in Chapter 9) fails.
\end{prob}

\begin{prob}
Let 
\[
f(x,y) = \sqrt{\abs{x} + \abs{y}}. 
\]
Find those points in $\nr^2$ at which $f$ is
differentiable. \emph{Hint}: If near a point $(x,y)$, say, in the
second quadrant, the function is equal to $\sqrt{-x+y}$, then it will
be the composition of two differentiable functions, thereby,
differentiable by the ``Chain Rule.''
\end{prob}

\begin{prob}
Let $f: \nr^n \times \nr^n \to \nr$ be defined by
\[
f(x,y) = x \cdot y. 
\]
Show that $f$ is differentiable on $\nr^n \times \nr^n$ and that
$Df(a,b) (x,y) = b \cdot x + a \cdot y$.
\end{prob}

\begin{prob}[Cauchy-Riemann Equations]
  Let $U \subset \nc$ be an open set. A function $f: U \to \nc$ is
  called \emph{complex differentiable} at $z_0 \in U$ if
\[
\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z-z_0}
\]
exists. A function $f$ is called \emph{analytic} on $U$ if $f$ is
complex differentiable at each point of $U$. Let $f (z) = u(x,y) +
iv(x,y)$, where $u$ and $v$ are functions $U \to \nr$ and $z$ is
written in the form $z = x+iy$ with $x,y \in \nr$. Show that if $f$ is
analytic on $U$, then
\[
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{
  and } \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial
  x}
\]
at all points of $U$.
\end{prob}

\begin{prob}[Mean Value Theorem]
  Let $U \subset \nr^n$ be an open set and $f: U \to \nr$ be
  differentiable on $U$. Let $x, y$ be two distinct points in $U$ such
  that the line segment joining them lies entirely in $U$. Show that
  there exists $\xi \in (0,1)$ such that
\[
f(y) - f(x) = Df(z) (y-x),
\]
where $z = (1-\xi) x + \xi y$.
\end{prob}

\begin{prob}
  Deduce from the previous problem that if $Df(x) = 0$ for all $x \in
  U$, then $f$ is constant on $U$, provided $U$ is
  connected. \emph{Hint}: This problem does not obviously follow from
  the previous one, because a connected set is not necessarily
  convex. To use connectedness, pick a point $x \in U$ and show that
  the set of points $y \in U$ at which $f(y) = f(x)$ is open.
\end{prob}

\begin{prob}
Let 
\[
f(x,y) = \begin{cases} 0 & \text{when $(x,y) = (0,0)$},\\
\frac{x^3y - xy^3}{x^2+y^2} & \text{otherwise}.
\end{cases}
\]
Show that $f$ is differentiable everywhere. Show that $D_{12}f(0,0)$
and $D_{21}f(0,0)$ exist, but $D_{12}f(0,0) \ne D_{21}f(0,0)$.
\end{prob}

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