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\begin{center}{\bf Math 214 Advanced Calculus II \\
Final Exam \\ SOLUTIONS}\end{center}
June 1, 2001 \hfill Ali Sinan Sert\"{o}z \atla
{\bf 1) }
\parbox[t]{\olcu}{
Construct a function $f:\Rr^2\goes\Rr$ such that all directional
derivatives of $f$ at the origin exist but $f$ is not differentiable
at the origin. Is it possible to construct this $f$ such that it is
differentiable at the origin but some directional derivatives at the
origin do not exist? Explain.
}\atla
{\bf Solution: }
The first part is Exercise 3-b on page 295 and solved in detail in class.
The second part is the content of Theorem~6.7 on page 288.
\\ \mbox{}\hrulefill\mbox{}
\atla
{\bf 2) } \parbox[t]{\olcu}{
Consider the functions
\begin{eqnarray*}
F(t) &=(5t^3+7t^2+12t+13, \lambda t-9) \\
G(x,y) &=x^2+y^2+3xy+2x-18y+1.
\end{eqnarray*}
Find those $\lambda$ for which \\
{\bf (i) } $F\circ G$ is invertible around the origin. \\
{\bf (ii) } $G\circ F$ is invertible around the origin.
}\atla
{\bf Solution: }
\[ DF(t)=\left( \begin{array}{c}15t^2+14t+12 \\ \lambda \end{array}
\right), \;\;\;
DG(x,y)=\left( 2x+3y+2, 2y+3x-18 \right). \]
{\bf (i) } $\dis G(0,0)=1$, so $\dis DG(0,0)=(2,-18)$ and $\dis
DF(1)=\left( \begin{array}{c}41\\ \lambda \end{array} \right)$. Hence
$\dis D(F\circ G)(0,0)=DF(1)\cdot DG(0,0)=\left( \begin{array}{cc} 82
& -738 \\ 2\lambda & -18\lambda \end{array}\right)$. Since $det
D(F\circ G)(0,0)=0$, the function $F\circ G$ is not invertible around
the origin for any value of $\lambda$. \\
{\bf (ii) } $\dis F(0)=(13,-9)$ and $\dis DF(0)=\left(
\begin{array}{c}12 \\ \lambda \end{array}\right), \;\; DG(13, -9)=(1,
3)$. Hence $\dis D(G\circ F)(0)=DG(13,-9)\cdot
DF(0)=12+3\lambda$. Thus the function $G\circ F$ is invertible around
the origin for all values of $\lambda\neq -4$.
\\ \mbox{}\hrulefill\mbox{}
\atla
{\bf 3) } \parbox[t]{\olcu}{
Let $\mathbf{v}_j=(v_{j1},\dots,v_{jn})\in \Rr^n$, $j=1,\dots,n$, be
fixed. The parallelpiped determined by the vectors $\mathbf{v}_j$'s is
the set
\[ \mathcal{P}:=\{ t_1\mathbf{v}_1+\cdot+t_n\mathbf{v}_n \; | \; t_j\in
[0,1]\; \} .
\]
Prove that
\[ Vol (\mathcal{P})=|\det (v_{ij})_{n\times n}|. \]
}\atla
{\bf Solution: }
Consider the map $\phi :\Rr^n\goes \Rr^n$ defined by $\phi
(x_1,\dots,x_n)= x_1\mathbf{v}_1+\cdots+x_n\mathbf{v}_n$. Let $C=\{
(x_1,\dots,x_n)\in\Rr^n \; | \; 0\leq x_1,\dots,x_n\leq 1\; \}$ be the
unit n-cube in $\Rr^n$. Then $\phi (C)=\mathcal{P}$ and
\[ Vol(\mathcal{P})=\int_{\phi (C)}du=\int_C |\Delta_\phi (x)|dx. \]
Since $\Delta_\phi (x)= \det (v_{ij})_{n\times n}$, the result follows.
\\ \mbox{}\hrulefill\mbox{}
\atla
{\bf 4) } \parbox[t]{\olcu}{
Let
\[ \dis F(x,y,z)=\left( \frac{x}{(3x^2+5y^2+7z^2)^{3/2}},
\frac{y}{(3x^2+5y^2+7z^2)^{3/2}} , \frac{z}{(3x^2+5y^2+7z^2)^{3/2}}
\right) . \]
Show that $div F=0$. Show that there is no vector field $G(x,y,z)$
such that $curl G=F$.
}\atla
{\bf Solution: }
This is almost the same as Remark 2 on page 451. That $div F=0$ is
straightforward. The rest of the solution follows exactly the same
lines of proof on page 452 except that you have to observe that for
this problem $\dis \int_S F\cdot \mathbf{n}d\sigma$ is not easy to
evaluate. However you can easily observe that it must be
positive. This then supplies the required contradiction.
\\ \mbox{}\hrulefill\mbox{}
\atla
{\bf 5) } \parbox[t]{\olcu}{
Let $\omega=(y+z)dxdy+(x+z)dydz+(x+y)dzdx$ be a 2-form on $\Rr^3$, and
let $S$ be the unit sphere centered at the origin in $\Rr^3$. Evaluate
\[ \int_S \omega . \]
In the above integral there is a difference between $dxdy$ and
$dydx$. However Fubini's theorem tells us that we can integrate with
respect to any order. How do you explain this {\it inconsistency}?
}\atla
{\bf Solution: }
Let $B$ be the unit ball centered at the origin in $\Rr^3$. \\
$d\omega=(dy+dz)dxdy+(dx+dz)dydz+(dx+dy)dzdx=3dxdydz$. \\
\[ \int_S\omega=\int_Bd\omega=\int_B 3dxdydz = 3 Vol(B)=4\pi.\]
In the Stokes theorem there is a difference between $dxdy$ and
$dydx$. This tells us which integral is equal to which integral. But
Fubini's theorem is about evaluating the integral. Once you decide
which integral to evaluate, say using Stokes theorem, then Fubini's
theorem says that you can integrate it as an iterated integral in any
order you like, under certain mild conditions. \\
Stokes Theorem: page 515. Fubini's Theorem: page 358.
\\ \mbox{}\hrulefill\mbox{} \\
\end{document}
\atla
{\bf Notes to the novice:} You must write your answers in a readable
and self explanatory manner. I do not try to read what you mean. I
simply read what you write. If what you have in mind and what you have
on paper are different, beware that I mark only what is on the
paper. Intentions don't count! Moreover, if your method is correct,
whatever that means, you do not automatically get partial
credits. Your partials are determined on how close you get to the
correct result, and let me be the judge of it!! \\
Good luck...
{\bf 6) } \parbox[t]{\olcu}{
Find $G(\mathbb{C},\mathbb{R})$, the group of automorphisms of the
field $\mathbb{C}$ keeping the elements of the field $\mathbb{R}$
pointwise fixed. Is $\mathbb{C}/\mathbb{R}$ a normal extension? \\
{\tiny (See page 239, and also 245.)}}
\atla
{\bf 7) } \parbox[t]{\olcu}{
Let $K/F$ be some field extension and $a\in K$.
Let $f(x)\in F[x]$ be a monic polynomial
of smallest positive degree which has $a$ as a root. Let $g(x)\in
F[x]$ be another monic polynomial which has $a$ as a root. Show that
if $\deg f=\deg g$ then $f=g$. \\
{\tiny (See page 211.)}
}
\atla
{\bf 8) } \parbox[t]{\olcu}{
If $G$ is a group, $Z$ its center, and if $G/Z$ is cyclic,
prove that $G$ must be abelian. {\tiny (Problem 16 on page 91.)}
} \atla
\setlength{\olcu}{\textwidth}
\settowidth{\eksi}{\bf Bonus: }
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{\bf Bonus: } \parbox[t]{\olcu}{
If $G$ is a group and $H$ is a subgroup of index $p$ in
$G$, where $p$ is the smallest prime dividing the order of $G$, then
prove that $H$ is a normal subgroup of $G$
} \atla
{\bf All questions are 7.5 points except the Bonus which is 15 points.}
\end{document}
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