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\noindent Solutions for Homework 11
\bigskip
 
\noindent{\bf Page 358/ 1b: Solution.}
Note that $f(u,v) = (0,1)$ if $(u,v) = (0,0)$ (other points are possible).
Hence we check,
$$
Df_{(0,0)} = 
\left.\left(\begin{array}{cc}
1 & 1 \\
\cos u & -\sin v
\end{array}
\right)\right|_{(u,v)=(0,0)} =
\left(\begin{array}{cc}
1 & 1 \\
1 & 0
\end{array}
\right).
$$
So clearly, $f$ is $C^1$ near $(0,0)$, and $Df_{(0,0)}$ is invertible.
By the inverse function theorem, we conclude that there is a local
inverse $g$ defined near $f(0,0) = (0,1)$ and that
$$
Dg_{(0,1)} = (Df_{(0,0)})^{-1} = 
\left(\begin{array}{cc}
0 & 1 \\
1 & -1
\end{array}\right).
$$

\bigskip
\noindent{\bf Page 358/ 6: Solution.}
\bigskip

\begin{description}
\item[a]
To check the $1-1$ assertion, suppose that $f(x_1,y_1) = f(x_2,y_2)$
for two points in $(x_1,y_1),(x_2,y_2)\in E$.  Then 
$x_1 + x_2 = y_1 + y_2$ and $x_1y_1 = x_2y_2$.  Solving the first
equation for $x_1$ gives $x_1 = y_1+y_2-x_2$.  Substituting this
in the second equation and rearranging gives $(y_2-x_2)(y_2 - y_1) = 0$
So either $y_2 = x_2$ or $y_1 = y_2$.  The first case is impossible
since $x_2 > y_2$ (by definition of $E$).  The second case implies that
$x_1 = x_2$, as well.

To check the onto assertion, let $(s,t)\in\R^2$ be a point satisfying
$s > 2\sqrt{t}$ and $t> 0$.  We find $(x,y)\in E$ such that 
$f(x,y) = (s,t)$ by solving 
\begin{eqnarray*}
x+y & = & s \\
xy & = & t
\end{eqnarray*}
for $x$ and $y$.  This can be done by solving the first equation for
$x$, substituting the result in the second equation and solving for
$y$.  The result is
\begin{eqnarray*}
x & = & \frac{s + \sqrt{s^2 - 4t}}{2} \\
y & = & \frac{s - \sqrt{s^2 - 4t}}{2},
\end{eqnarray*}
where our choice of signs is governed by the requirement that $x>y$.
We note that since $s\geq 2\sqrt{t}$, then $s, s^2 - 4t>0$.  Since $t > 0$,
we have $s^2 - 4t < s^2$.  Hence $x > y > 0$ as desired.  Thus
$$
f^{-1}(s,t) = (\frac{s + \sqrt{s^2 - 4t}}{2},\frac{s - \sqrt{s^2 - 4t}}{2}).
$$
\medskip

\item[b]
$$
Df^{-1}_{f(x,y)} = (Df_{(x,y)})^{-1} = 
\left(\begin{array}{cc}
1 & 1 \\
y & x 
\end{array}\right)^{-1} 
=
\frac{1}{x-y}
\left(\begin{array}{cc}
x & -1 \\
-y & 1 
\end{array}\right).
$$

\item[c]
Using the formula from part (a) gives
$$
Df^{-1}_{(s,t)} = 
\left(\begin{array}{cc}
\frac{1}{2} + \frac{s}{2\sqrt{s^2 - 4t}} & 
\frac{-1}{\sqrt{s^2 - 4t}} \\
\frac{1}{2} - \frac{s}{2\sqrt{s^2 - 4t}} &  
\frac{1}{\sqrt{s^2 - 4t}}
\end{array}\right).
$$
Now we substitute $(s,t) = (x+y,xy)$ and obtain
\begin{eqnarray*}
Df^{-1}_{f(x,y)} & = &
\left(\begin{array}{cc}
\frac{1}{2} + \frac{x+y}{2\sqrt{(x+y)^2 - 4xy}} & 
\frac{-1}{\sqrt{(x+y)^2 - 4xy}} \\
\frac{1}{2} - \frac{x+y}{2\sqrt{(x+y)^2 - 4xy}} &  
\frac{2}{\sqrt{(x+y)^2 - 4xy}}
\end{array}\right)
=
\left(\begin{array}{cc}
\frac{1}{2} + \frac{x+y}{2\sqrt{(x-y)^2}} & 
\frac{-1}{\sqrt{(x-y)^2}} \\
\frac{1}{2} - \frac{x+y}{2\sqrt{(x-y)^2}} &  
\frac{1}{\sqrt{(x-y)^2}}
\end{array}\right) \\
& = & 
\left(\begin{array}{cc}
\frac{x}{x-y} & 
\frac{-1}{x-y} \\
\frac{-y}{x-y} &  
\frac{1}{x-y}
\end{array}\right)
\end{eqnarray*}
which agrees with part (b).


\end{description}






\noindent{\bf Other Problems}
\bigskip

\noindent{\bf A.}  
          In problem 1b, use the method described in class to approximate
          a solution $(x,y)$ of $f(x,y) = (.1,1.2)$.  Then use your
          first approximation to generate a second approximation.

\bigskip 
\noindent{\bf Solution: }
Taking $\mathbf{x}^0 = (0,0)$ and $\mathbf{y} = (.1,1.2)$, we use the formula
$$
\mathbf{x}^1 = \mathbf{x}^0 + (Df_{\mathbf{x}^0})^{-1}\cdot
               (\mathbf{y}-f(\mathbf{x}^0)) 
             = \left(\begin{array}{c} 0 \\ 0
               \end{array}\right) + 
\left(\begin{array}{cc}
0 & 1 \\
1 & -1
\end{array}\right)\cdot
\left(\left(\begin{array}{c}
.1 \\
1.2 
\end{array}\right)
-
\left(\begin{array}{c}
0 \\
1 
\end{array}\right)\right) 
=
\left(\begin{array}{c}
.2 \\
-.1 
\end{array}\right)
$$
to generate a new guess.  Applying the same formula one more time gives
$$
\mathbf{x}^2 = \left(\begin{array}{c} .2 \\ -.1
               \end{array}\right) + 
\left(\begin{array}{cc}
1 & 1 \\
\cos .2 & \sin(-.1)
\end{array}\right)^{-1}\cdot
\left(\left(\begin{array}{c}
.1 \\
1.2 
\end{array}\right)
-
\left(\begin{array}{c}
.1 \\
1.19367 
\end{array}\right)\right) 
=
\left(\begin{array}{c}
.207187 \\
-.107187 
\end{array}\right)
$$



\bigskip
\noindent{\bf B.} This problem is designed to show what can go wrong when 
the hypotheses of the inverse function theorem fail.
\begin{enumerate} 
\item Define $f:\R^2\to\R^2$ by $f(x,y) = (x^2 + y^2, x-2y)$.  

\bigskip
\item At which points $(x,y)\in\R^2$ are the hypotheses of the inverse
      function theorem satisfied?

\bigskip
\noindent{\bf Solution:}
Since $\det Df_{(x,y)} = -4x - 2y$, the hypotheses of the inverse function
theorem are satisfied (i.e $Df$ is invertible) if and only if $y\neq -2x$.


\bigskip
\item Pick any point (your choice---I want you to use actual numbers here) 
      $(x_0,y_0)\in\R^2$ where the hypthoses 
      of the inverse function theorem fail, and let 
      $(r_0,s_0) = f(x_0,y_0)$.  Pick a number $r$ slightly larger than
      $r_0$.  How many real solutions $(x,y)$ of $f(x,y) = (r,s_0)$ are there
      near $(x_0,y_0)$?  What if $r$ is slightly less than $r_0$?  

\bigskip
\noindent{\bf Solution (example):} Take $(x_0,y_0) = (1,-2)$.  Then
$f(x_0,y_0) = (5,5)$.  Set $r = 5.1$.  Then $f(x,y) = (5.1,5)$
implies that $x = 2y + 5$ and $(2y+5)^2 + y^2 = x^2 + y^2 = 5.1$.
Simplifying this last equation and solving for $y$ using the quadratic
formula gives
$$
y = \frac{-20\pm\sqrt{400 - 20\cdot 19.9}}{10} = -2 \pm .141421
$$
Hence, we obtain {\it two} nearby points 
$$
(x,y) = (1.28284,-1.85858) \qquad\text{and}\qquad 
(x,y) = (0.717157,-2.14142)
$$
satisfying $f(x,y) = (5.1,5)$.  If, on the other hand, we set $r=4.9$,
we obtain
$$
y = \frac{-20\pm\sqrt{400 - 20\cdot 20.1}}{10} = -2 \pm \frac{sqrt{-2}}{10},
$$
so there are {\it no} points $(x,y)$ satisfyin $f(x,y) = (4.9,5)$.

\bigskip
\item Explain why your answers to the last part rule out the existence of
      a local inverse for $f$.  Explain what's going on here in geometric 
      terms.

\bigskip
\noindent{\bf Solution:} 
If there were a local inverse $g$ for $f$ defined near $(5,5)$, we
would have that $f(g(\mathbf{y})) = \mathbf{y}$ for every 
$\mathbf{y}$ near $(5,5)$---in particular, $(x,y) = g(4.9,5)$ would
satisfy $f(x,y) = (4.9,5)$.  Since there is no such point, $g$ cannot
exist.

In geometric terms, solving $f(x,y) = (a,b)$ amounts to finding the 
intersection
of a circle of radius $\sqrt{a}$ and center $(0,0)$ with the line
$x-2y = b$.  The hypotheses of the inverse function theorem fail precisely
when $(x_0,y_0)$ is chosen so that the circle and the line defined
by equating first and second coordinates of $f(x_0,y_0) = f(x,y)$ are
tangent to each other.  When the circle is expanded or contracted
slightly (i.e. $r$ is chosen slightly larger or slightly smaller), the 
single intersection either splits into two intersections
or disappears altogether.


\end{enumerate}




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