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\begin{document}



\title{ Math 3220-3 Homework 3}
\author{Due March 6 }
\maketitle

\begin{enumerate}  
\item 
Define  $f:\R^2\to\R$ by
$$
  f(x,y) = \begin{cases}
      \frac{x^3}{ x^2 + y^2} & \text{if  } (x,y)\ne (0,0), \\
      0 & \text{if } (x,y)= (0,0).
\end{cases}
$$
Prove the following  statements:
  \begin{enumerate}
  \item  $f$ is continuous on $\R^2$.
  \item $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist on all of
$\R^2$ and are bounded.
  \item At $(0,0)$ the directional derivatives $D_v f$ exist  for all unit vectors $v\in\R^2$.
  \item $f$ is not  differentiable at $(0,0)$.
  \end{enumerate} 
  \vskip10pt              
 \item Define $f:\R^2\to\R$ by
  $$
  f(x,y) = \begin{cases}
      \frac{x^4}{ x^2 + y^2} & \text{if  } (x,y)\ne (0,0), \\
      0 & \text{if } (x,y)= (0,0).
\end{cases}
$$
\begin{enumerate}
  \item Prove, directly from the definition of differentiability, that $f$ is differentiable at $(0,0)$, and find its derivative $d_{(0,0)}f$.
  \item Show that the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist and are continuous on all of $\R^2$.  Observe that this gives another proof of the differentiability of  $f$ at $(0,0)$.
  \vskip10pt
   
\end{enumerate}

\item Let $f:\R^2\to\R^2$ be defined  by
$$
f(x,y) = (x^2 - y^2, 2 x y) = (u,v)
$$
\begin{enumerate}
  \item Observe that $f(-x,-y) = f(x,y)$, so $f$ is not (globally) injective.
  \item Use the Inverse Function Theorem to prove that if $(x_0,y_0)\ne (0,0)$, then $(x_0,y_0)$ has a neighborhood $U$ with the property that $f$ maps $U$ bijectively to its image $V = f(U)$.
  \item Prove that $(0,0)$ has no such neighborhood.
  \item Find explicit formulas for a local inverse of $f|_U$ where $U$ is a neighborhood of $(1,0)$.
\end{enumerate}
\vskip10pt


\item (Rudin Chap 9, Ex 16) Let 
$$
f(t) =\begin{cases}
    t + 2\  t^2 \sin(\frac{1}{t})  & \text{ if  } t\ne 0, \\
     0 & \text{ if } t = 0.
\end{cases} 
$$
 Show
\begin{enumerate}
  \item $f$ is differentiable.
  \item $f'(0) = 1$.
  \item $f'$ is bounded on $(-1,1)$.
  \item $f$ is not one-to-one in any neighborhood of $0$.  Thus the continuity of $f'$ is needed in the inverse function theorem.
\end{enumerate}
\vskip10pt


\item Let $U\subset \R^n$ be open and let $f:U\to \R$ be of class $\cC^1$ (continuously differentiable).  Recall that if $p\in U$ and $v$ is a unit vector, the \emph{directional derivative of $f$ at $p$ in direction $v$}, $(D_vf)(p)$ is defined to be
$$
D_vf(p) =( \frac{d}{dt}f(p+tv))|_{t=0} = d_pf(v) = \nabla_pf\cdot v
$$
the second equality by the chain rule, the third the definition of the gradient.  A point $p\in U$ is called a \emph{critical point of $f$} if $d_pf = 0 \Longleftrightarrow \nabla_p f = 0$

\begin{enumerate}
  \item Prove that if $p$ is a local maximum of $f$, then it is a critical point of $f$.  Same for a local minimum.
  
  \item (This is a quick explanation of the Lagrange multiplier method. More details later in class)  
  
   If $g:U\to \R$ is also $\cC^1$, if $G = \{p\in U\ :\ g(p) = 0\}$ and $d_pg\ne 0$ for all $p\in G$,  then  the implicit function theorem can be used to rigorously define critical points of the restriction $f|_G$ and to prove that a local maximum or minimum of this restriction is a critical point.  Moreover, there is a useful criterion for $p_0\in G$ to be critical for $f|_G$, the \emph{Lagrange multiplier method}:
  $$
  p_0\in G \text{ is critical for } f|_G \Longleftrightarrow \exists \lambda\in \R \text{ s.t. } \nabla_{p_0}f   =   \lambda \nabla_{p_0} g.
  $$
  Since, by assumption, $\nabla_pg\ne 0$ for all $p\in G$, the orthogonal complement $\nabla_pg^\perp$ is the \emph{tangent space to $G$ at $p$} and the Lagrange multiplier condition is equivalent to
  $$
  \nabla_{p_0} f \text{ is perpendicular to } (\nabla_{p_0} g)^\perp
  $$
  or, briefly, $\nabla_{p_0}f$ is perpendicular to $G$ at $p_0$.
  
  Let's take all this for granted.
  
  \item Let $x_0\in \R^n$ and let $f:\R^n\to \R$ be square distance from $x_0$:
  $$
  f(x) = |x-x_0|^2 =(x-x_0)\cdot(x-x_0)
  $$
  where $u\cdot v$ is the usual dot product of vectors in $\R^n$. Find $\nabla_xf$.
  
\emph{Suggestion}: Expand $(x+h -x_0)\cdot (x + h  -x_0 )$ and compute $d_xf(h)$ directly from the definition of $d_xf$.
  

  
  \item As above, let $g:\R^n\to \R$ be $\cC^1$,  let $G=\{g=0 \}$ and suppose $\nabla_pg\ne 0$ for all $p\in G$.  Suppose $x_0\notin G$ and suppose $x_1\in G$ minimizes the distance $|x-x_0]$ for $x\in G$. Prove that $x_1-x_0$ is perpendicular to $G$.  
  
  \emph{Comment}: We have used this in the past for $g$ a linear function.
  \end{enumerate}
\end{enumerate}

 \end{document}

