\documentclass{math23a}
\problemset{9}
\year1999\month12\day11
\author{Nils~R. Barth}

\begin{document}
\item[1]
 \begin{enumerate}

   \item % a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-
    First, $f(2,2)=0$, so $(2,2)\in C_0$.
    Second,
 \begin{equation}\label{eqn:Df}
[Df](x,y) =\begin{bmatrix}3x^2+6x-12& 2y-6\end{bmatrix},
 \end{equation} 
    so $Df(2,2)=\begin{bsmallmatrix}12& -2 \end{bsmallmatrix}$, which
    is not equal to zero, so $(2,2)$ is a smooth point.
    
   \item % b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-
    Since $D_2f(2,2)\neq 0, D_1f(2,2)\neq 0$, we conclude from the
    inverse function theorem that (respectively), there exist
    $\epsilon_f, \epsilon_g$ such that $y=f(x), x=g(y)$ on
    $(2-\epsilon_f,2+\epsilon_f$ and $(2-\epsilon_g,2+\epsilon_g)$,
    respectively. Taking $\epsilon=\min(\epsilon_f,\epsilon_g)$ yields
    the desired result.

   \item % c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-c-
    The tangent line of the curve $f(x,y)=0$ at $(x_0,y_0)$ is given
    by $[D_1f](x_0,y_0)\cdot(x-x_0)+[D_2f](x_0,y_0)\cdot
    (y-y_0)=0$. In our case, this yields $12(x-2)-2(y-2)=0$.

    The tangent line is a kind of an affine copy of the tangent space,
    which is given by $[D_1f](x_0,y_0)\dot x+[D_2f](x_0,y_0)\dot y=0$,
    or in our case $12\dot x-2\dot y =0$.

   \item % d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-d-
    It is natural to solve this and the next part simultaneously.
    $C_c$ will be singular at \emph{some} point iff
    $Df=\begin{bsmallmatrix}0& 0\end{bsmallmatrix}$ at some
    point of $C_c$. Now $Df=0$ for $(x,y)=(-1\pm \sqrt{5},3)$ (by
    solving the linear and quadratic polynomials in \eqnref{Df}), so
    $C_c$ is singular iff $(-1\pm \sqrt{5},3)\in C_c$, \ie iff
    $c=f(-1\pm \sqrt{5},3)$.

   \item % e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-e-
    See above; the singular points of $C_{f(-1\pm\sqrt{5},3)}$ are
    $(-1\pm \sqrt{5},3)$, respectively.

   \item % f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-f-
    By reasoning as above, $Df(x,y)=\begin{bsmallmatrix}p'(x) &
    -q'(y)\end{bsmallmatrix}$. This is zero iff $p'(x)=q'(y)=0$. Now
    $p', q'$ are polynomials of degrees $m-1,n-1$, respectively, so
    they have roots (possibly repeated)
    $x_1,\dots,x_{m-1},y_1,\dots,y_{n-1}$.
    So $(x,y)$ is a singular point iff $Df(x,y)=0$, that is, for
    $(x_i,y_j)$. Now these points lie on the curves $C_{f(x_i,y_j)}$,
    and thus there are at most $(m-1)(n-1)<mn$ singular curves.

  \end{enumerate}

\item[2]
 \begin{enumerate}

   \item % a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-a-
    Note that squaring is locally $C^1$ invertible at a point iff the
    Jacobian at that point is invertible (and the partials are $C^1$,
    which they are).
    Via the attached Mathematica output, we see that
 \begin{equation}
  DS
  \begin{bmatrix}
   a & 0\\
   0 & b
  \end{bmatrix}
   =
  \begin{bmatrix}
   2a\\
   & a+b\\
   && a+b\\
   &&&2b
  \end{bmatrix}.
 \end{equation} 
    Further, note that a diagonal matrix is invertible iff all its
    diagonal entries are non-zero, and we see that the desired $g$
    exists iff $2a\neq 0$, $2b\neq 0$, and $a+b\neq 0$. A slick way of
    saying all of this is: $ab(a+b)\neq 0$.

   \item % b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-
    As above, but now we get
 \begin{equation}
  DS
  \begin{bmatrix}
   a & 0\\
   0 & b
  \end{bmatrix}
   =
  \begin{bmatrix}
   5a^4\\
   & a^4+a^3b+a^2b^2+ab^3+b^4\\
   && a^4+a^3b+a^2b^2+ab^3+b^4\\
   &&&5b^4
  \end{bmatrix},
 \end{equation} 
   so $g$ exists iff $ab(a^4+a^3b+a^2b^2+ab^3+b^4)\neq 0$.

   Incidentally, you may have noticed that the big middle term is also
   $\frac{a^5 - b^5}{a-b}$, so the middle term factors as
   $\prod_{i=1}^4 (a-\zeta_5^i b)$, where $\zeta=e^{2\pi i/5}$ (we
   call $\zeta_5$ a \emph{primitive fifth root of unity}).
   This is because $a^n-b^n=\prod_{i=1}^n (a-\zeta_n^i b)$, and we've
   just gotten rid of the $(a-b)$ term.

   Looking at this, one could conjecture that for diagonal matrices,
   the map $A\mapsto A^n$ is locally $C^1$ invertible iff
   $ab\frac{a^n-b^n}{a-b}\neq 0$ (where one actually divides out the
   fraction). Can you prove this? Hint: it follows almost immediately
   from the expansion $Df(A)(H)=HA^{n-1}+AHA^{n-2}+\dots+A^{n-1}H$ for
   $f(A)=A^n$.
  \end{enumerate}

\end{document}
