\documentclass[12pt]{article}
\usepackage{latexsym}
\usepackage{amssymb,amsmath}
\usepackage[pdftex]{graphicx}
\usepackage{listings}
\usepackage{courier}
\usepackage{color}
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage{enumerate}
\usepackage{endnotes}
\usepackage{extpfeil}
\usepackage{stackrel}
\usepackage{bbm}
\usepackage{tikz}
\usepackage[margin=2cm]{geometry}
\usepackage{hyperref}
\hypersetup{colorlinks=true,urlcolor=MidnightBlue,citecolor=PineGreen,linkcolor=BrickRed}
\lstset{
basicstyle=\small\ttfamily,
keywordstyle=\color{blue},
language=python,
xleftmargin=16pt,
}
\newtheorem{thm}{Theorem}[section]
\newtheorem{ithm}{Theorem}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{defi}[thm]{Definition}
\newtheorem{example}[thm]{Example}
\newtheorem{exercise}[thm]{Exercise}
\newtheorem{rem}[thm]{Remark}
\def\B{{\mathbb B}}
\def\C{{\mathbb C}}
\def\D{{\mathbb D}}
\def\Fp{{\mathbb F}_p}
\def\F{{\mathbb F}}
\def\H{{\mathbb H}}
\def\M{{\mathbb M}}
\def\N{{\mathbb N}}
\def\O{{\mathcal O}}
\def\0{{\mathbb 0}}
\def\P{{{\mathbb P}}}
\def\Q{{\mathbb Q}}
\def\R{{\mathbb R}}
\def\T{{\mathbb T}}
\def\Z{{\mathbb Z}}
\newcommand{\sol}{_{a^p,b^p,c^p}}
\newcommand{\bound}{\partial}
\newcommand{\la}[1]{\mathfrak{#1}}
\newcommand{\im}{\text{Im} \hspace{0.1em} }
\newcommand{\ann}{\text{Ann} \hspace{0.1em} }
\newcommand{\rank}{\text{rank} \hspace{0.1em} }
\newcommand{\coker}[1]{\text{coker}\hspace{0.1em}{#1}}
\newcommand{\sgn}{\text{sgn}}
\newcommand{\lcm}{\text{lcm}}
\newcommand{\re}{\text{Re} \hspace{0.1em} }
\newcommand{\ext}[1]{\text{Ext}(#1)}
\newcommand{\Hom}[1]{\text{Hom}(#1)}
\newcommand{\End}[1]{\text{End(#1)}}
\newcommand{\bs}{\setminus}
\newcommand{\rpp}[1]{\mathbb{R}\text{P}^{#1}}
\newcommand{\cpp}[1]{\mathbb{C}\text{P}^{#1}}
\newcommand{\tr}{\text{tr}\hspace{0.1em} }
\newcommand{\inner}[1]{\langle {#1}\rangle}
\newcommand{\tensor}{\otimes}
\newcommand{\Cl}{\text{Cl}}
\renewcommand{\sp}[1]{\text{Sp}_{#1}}
\newcommand{\gl}[1]{\text{GL}_{#1}}
\newcommand{\pgl}[1]{\text{PGL}_{#1}}
\renewcommand{\sl}[1]{\text{SL}_{#1}}
\newcommand{\so}[1]{\text{SO}_{#1}}
\newcommand{\SO}{\text{SO}}
\newcommand{\pso}[1]{\text{PSO}_{#1}}
\renewcommand{\o}[1]{\text{O}_{#1}}
\renewcommand{\sp}[1]{\text{Sp}_{#1}}
\newcommand{\psp}[1]{\text{PSp}_{#1}}
\newcommand{\Span}{\rm Span}
\newcommand{\kron}[2]{\bigl(\frac{#1}{#2}\bigr)}
\newcommand{\leg}[2]{\Biggl(\frac{#1}{#2}\Biggr)}
\DeclareSymbolFont{bbold}{U}{bbold}{m}{n}
\DeclareSymbolFontAlphabet{\mathbbold}{bbold}
\begin{document}
\begin{center}
{\bf {\large{Some Function Problems}}}\\
\smallskip
{ \bf {\large{SOLUTIONS}}} \\
Isabel Vogt\\
Last Edited: \today \\
\end{center}
Most of these problems were written for my students in Math 23a/b at Harvard in 2011/2012 and 2012/2013. They cover basic function theory, countability, differential calculus, and point-set topology.
\begin{enumerate}
\item {\bf T/F: }The function $q:\mathbb{R} \rightarrow \mathbb{R}$ given by
$$ q(x)= \left\{
\begin{array}{lr}
x & : x \in \mathbb{Q}\\
0 & : x \notin \mathbb{Q}
\end{array}
\right.$$
is nowhere continuous. \\
-----------------
{\bf Solution:}
{\bf FALSE}\\
This interesting function is actually continuous at 0! Let's see why: \\
In order for $q$ to be continuous at $0$, $$\forall \epsilon >0, \exists \delta >0 \ s.t. \forall y \in \mathbb{R}, |y-0|<\delta \rightarrow |q(y)-q(0)| < \epsilon$$
Well, $q(0) = 0$, so we just need an appropriate $\delta$ such that for all $|y| < \delta, |q(y)| < \epsilon$. \\
Claim: for all $\epsilon$ let $\delta = \frac{\epsilon}{2}$. Let's check the requirements. There are two possibilities, $y \in \mathbb{Q}$ and $y \notin \mathbb{Q}$
For $y \notin \mathbb{Q}$, $q(y) = 0 \stackrel{\checkmark}{<} \epsilon$ \\
For $y \in \mathbb{Q}$, $|q(y)| = |y| < \frac{\epsilon}{2} \stackrel{\checkmark}{<} \epsilon$ \\
So $q$ is continuous at $0$! \\
\item {\bf T/F:} The space of results of ``infinite coin flips" (i.e. vectors where the $i^{\text{th}}$ position is H/T depending on the result of the flip) is countably infinite. \\
-----------------
{\bf Solution:}
{\bf FALSE}\\
Imagine a new coin flip vector $f'$ whose $i^{\text{th}}$ position is H if the $i^{\text{th}}$ coin flip of the $i^{\text{th}}$ flip vector was T and conversely T if it was H. This differs in at least one position from every ``counted" flip vector. Thus the set is not countable - i.e. it is uncountable infinite.
\item Let $f$ be a function $f:D \mapsto C$.
\begin{enumerate}
\item We can think of the function as merely a set of ordered pairs $(x,y)$. What set must this be a subset of?
-----------------
{\bf Solution:}
A function is merely an ordered pair $(x,f(x))$ where $x \in D$ and $f(x) \in C$. So $S_f$ (the function $f$ as a set) is a subset of the Cartesian product $D \times C$.
\item Let $D=C=\mathbb{Q}$. And let $f$ be any function $f:D \mapsto C$. Thinking of $f$ as a set, what is its cardinality? Prove your answer.
-----------------
{\bf Solution:}
By definition of being a function, $f$ is \emph{well-defined} for all $x \in D$. So this means that $\forall x \in D$, $(x,f(x)) \in S_f$, and if $x=x'$, then $f(x)=f(x')$ -uniqueness. Thus we can create a bijection between the set $S_f$ and $D$ which is merely $(x,f(x)) \mapsto x$. This is injective and surjective by the above conditions on $f$. Thus $S_f$ must be of the same cardinality as $D$ - in this case countably infinite.
\item Let $D=C=\mathbb{Q}$. What is the cardinality of the set of functions $f:D \mapsto C$. Justify your answer (it might help to think of what we call a similar set).
-----------------
{\bf Solution:}
What we are concerned with is the set of all possible sets $\{S_f\}_{f}$ for functions $f: \mathbb{Q} \mapsto \mathbb{Q}$. First note that we can map to any subset of $\mathbb{Q}$ with our function $f$. Thus the set $\mathcal{P}(\mathbb{Q})$ injects/imbeds into our set $\{S_f\}_f$ (note that this is not surjective because the image does not uniquely determine the map). And the cardinality of $\mathcal{P}(\mathbb{Q})$ is uncountably infinite (Cantor proved this for $\mathcal{P}(\mathbb{N})$ and the result follows for our case). Thus our set $\{S_f\}_f$ is uncountable.
\item Let $D$ be an uncountable set, $C=\{0\}$. How many different well-defined functions $f$ are there? What can you say about this/these function(s) (i.e. injectivity, surjectivity, bijectivity)?
-----------------
{\bf Solution:}
There are two (trivial) subsets of $\{0\}$ - namely $\{0\}$ and $\phi$. Thus the only possibilities for functions are the zero map $f_1:x \in \mathbb{Q} \mapsto 0$ and the trivial subset $(D\times \phi)$ which is the empty set. However, from the definition of a function, $\forall x \in D, \exists f(x) \in C$ such that $f: x \mapsto f(x)$. This is not satisfies by $\{0\}$. So the only map is the zero function $f_1$. The map $f_1$ is obviously necessarily surjective but not injective.
\item Let $D=\phi$ and $C$ be a countable set . How many different well-defined functions $f$ are there? Justify your answer based upon the your answer to part (a). If such an $f$ exists is it injective? Surjective?
-----------------
{\bf Solution:}
In this case $(\{0\},\mathbb{Q})$ is a function (vacuously) as there does not exist counter-example points in $\{0\}$ which are not mapped to the codomain. The map $\phi$ is quite interesting in how many properties are vacuously true! The definition of injective is $\forall f(x_1)=f(x_2) \Rightarrow x_1=x_2$. In order for $f$ to not be injective, there must exits $f(x_1)=f(x_2)$ with $x_1 \neq x_2$, but there doesn't exist $f(x_1)$ or $f(x_2)$ as counterexamples! But for surjectivity, the definition says $\forall y \in \mathbb{Q}, \exists x \in \{0\}$ such that $f(x)=y$. In this case we can not find $x \in \{0\}$ so this is false - it is not surjective.
\end{enumerate}
\item {\bf T/F:} Given a continuous function $f\begin{pmatrix} x \\ y \end{pmatrix}$ with $f: \R^2 \to \R$, if the Jacobian matrix at $\bold{a} \in \R^2$ exists, then $f$ is differentiable at $\bold{a}$, and its derivative is given by $[Jf(\bold{a})]$.
------------------ \\
{\bf Solution: FALSE}
Take the function $f(x,y) = \frac{x^2y}{x^2+y^2}$, with $f(0,0)=0$. This function is continuous at the origin, and both partials exist, but it is \emph{not} differentiable there.
\item {\bf T/F:} $f$ is continuous at $x_0$ if and only if $\forall \epsilon>0, \exists \delta>0$ such that $f(B_{\delta}(x_0)) \subset B_{\epsilon}(f(x_0))$.
------------------ \\
{\bf Solution: TRUE}
This is merely a restatement of the definition of continuity, i.e. $f$ is continuous if and only if $\forall \epsilon >0, \exists \delta>0$ such that $|x_0-y_0|<\delta$ implies that $|f(x_0)-f(y_0)|<\epsilon$. The first condition is that $y$ is in a $\delta$-neighborhood of $x_0$ (i.e. $B_{\delta}(x_0)$) and the second condition says that this ball is mapped to an $\epsilon$-neighborhood of $f(x_0)$.
\item {\bf T/F:} If $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is differentiable and $[Df(0)]$ is not invertible, then there is no differentiable function $g:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that
$$(g \circ f)(\vec{x})=\vec{x}$$
------------------ \\
{\bf Solution: TRUE}
By the chain rule we know that:
$$[D(g \circ f(\vec{x})]\vec{h}=[Dg(f(\vec{x}))][Df(\vec{x})]\vec{h}$$
So at $\vec{v}=\vec{0}$ we have
$$[D(g \circ f(0)]\vec{h}=[Dg(f(0))][Df(0)]\vec{h}$$
In order for $(g \circ f)(\vec{x})=\vec{x}$, we would have $[D(g \circ f(0)]=I$. Thus:
$$[Dg(f(0))][Df(0)]=I$$
But by definition this would mean that $[Df(0)]$ is invertible as it is square and has a left inverse. This is a contradiction, thus no such $g$ exists and the statement is true.
\item {\bf T/F:} A set $F$ is closed implies that all points in $F$ are accumulation points.
------------------ \\
{\bf Solution: FALSE}
It is tempting to say this, since the definition we are going with for a closed set is one which contains all its accumulation points, but there is a big distinction here - that \textbf{does not} mean that all points are accumulation points. Take $F=\{0\}$. This is clearly a closed set, but $F$ contains no other points than $0$, so it is not an accumulation point of the set.
\item {\bf T/F:} For $f$ and $g$ differentiable functions from $\R^2 \to \R$ $$\frac{\partial f}{\partial x}\frac{\partial g}{\partial y} = \frac{\partial f}{\partial y}\frac{\partial g}{\partial x}$$
------------------ \\
{\bf Solution: FALSE}
This is so terribly false, I just threw it on here to make sure no one thinks about partial derivatives as fractions. As a counter example take $f(x,y)=x$ and $g(x,y)=y$. Then the left product is 1 and the right is 0.
\item {\bf T/F:} Every convergent subsequence of a sequence $\{a_i\}$ defined on a compact set $C$ converges to the same limit $a \in C $.
------------------ \\
{\bf Solution: FALSE}
Consider the sequence $x_i=\frac{1}{2}\cos{i\pi}$ for $i \in \N$. This infinite sequence lives in the compact interval $[-1,1]$, and it thus has a convergent subsequence. There are, however, \emph{two} possible sequences - all the even terms for which $x_{2k}=+\frac{1}{2}$ and all the odd terms for which $x_{2k+1}=-\frac{1}{2}$. They, of course, do not converge to the same limit.
\item Suppose $g: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is a function such that the directional derivatives at $x_0$ along $v_1$, $v_2$, and $v_3$ are $e_1$, $e_2$, and $e_3$ respectively. What can we say about $g$? Prove your claim.
------------------ \\
{\bf Solution:}
In $\mathbb{R}^2$, three vectors can't be independent. Thus $\{\vec{ v}_1, \vec{v}_2, \vec{v}_3\}$ must be dependent. \\
Without loss of generality let $\vec{v}_3 = a_1\vec{v}_1 + a_2\vec{v}_2$. Then,
$$[Dg(x_0)]\vec{v}_3 = a_1[Dg(x_0)]\vec{v}_1 + a_2[Dg(x_0)]\vec{v}_2$$
But, $\vec{e}_3 \neq a_1\vec{e}_1+a_2\vec{e}_2$, because they are independent. \\
Thus the derivative can not be a linear function, and $g$ is not differentiable at $x_0$!
\item
\begin{enumerate}
\item Define the zero-locus in $\R^2$ of the polynomial $f(x,y)=y-x^2$ to be the set of points $G_f=\{x_0,y_0 \in \R | f(x_0,y_0)=0\}$. Is the set $G_f$ open or closed as a subset of $\R^2$?
------------------ \\
{\bf Solution:}
Similarly to a straight line, the curve is closed as its complement is open. Alternatively, it contains all of its accumulation points, as any accumulation point necessarily satisfies the above locus equation.
\item It is one thing to claim that there is an open set, and another thing to actually construct one. Given a point $\bold{a}=\begin{pmatrix} 3 \\ 1 \end{pmatrix}$, we wish to construct the largest ball $B_r(\bold{a})$ around this point such all points in this ball are in the complement of $G_f$ - ie $B_r(\bold{a}) \subset \R^2 - C$. Find an equation that will determine the maximum $r$ of this ball in \emph{two} ways:
\begin{enumerate}
\item Finding an extreme value of the distance function between the point $\bold{a}$ and the parabola
------------------ \\
{\bf Solution:}
A general point on the parabola will be of the form $(a,a^2)$. So the distance we are looking for is
$$|\begin{bmatrix} 3-a \\1-a^2 \end{bmatrix}|$$
To make out lives easier, we know that the distance function will achieve a maximum or a minimum at the same value of $a$ as its square (why?), so we can choose to optimize for the square:
$$d(a)^2=(3-a)^2+(1-a^2)^2=a^4-a^2-6a+10$$
$$\Rightarrow \frac{d (d(a))}{da}=4a^3-2a-6$$
So we are looking for solutions to the equation $2a^3-a-3=0$
\item Use the concept that the shortest line between a point and a line will be a perpendicular.
------------------ \\
{\bf Solution:}
We can use the derivative as a linear approximation to the function locally. Thus at any point $(a,a^2)$, the parabola looks like a linear function described by a vector in the direction $\begin{bmatrix} 1 \\ 2a \end{bmatrix}$. So we take the ``displacement" vector between our point and this arbitrary point on the parabola to be $\begin{bmatrix} 3-a \\ 1-a^2 \end{bmatrix}$.
The shortest distance is going to be when these two are perpendicular, i.e. when the dot product is 0. In other words:
$$\begin{bmatrix} 1 \\ 2a \end{bmatrix} \cdot \begin{bmatrix} 3-a \\ 1-a^2 \end{bmatrix}=0$$
$$\Rightarrow 3-a+2a-2a^3=0$$
Or simply $2a^3-a-3=0$ - the same equation as defined above!
\end{enumerate}
\end{enumerate}
\item Let
$$f\begin{pmatrix} x \\y
\end{pmatrix} = \frac{x ^3y^2+x^4y}{x^4+y^4}.$$
$f$ is defined to be 0 at $\begin{pmatrix} 0 \\0
\end{pmatrix}$.
\begin{enumerate}
\item Using the definition of continuity, prove that $f$ is continuous at $\begin{pmatrix} 0
\\0\end{pmatrix}$.
\item Show that both partial derivatives are zero at
$\begin{pmatrix} 0
\\0\end{pmatrix}$ but that the function is not differentiable there.
Note: You must work from the definition of ``differentiable.''
\end{enumerate}
------------------ \\
{\bf Solution:}
$$D_1f(0,0) = \lim_{h \to 0} \frac{1}{h} (f(0+h,0) -f(0,0)) = \lim_{h \to 0} \frac{0}{h}=0$$
$$D_2f(0,0) = \lim_{h \to 0} \frac{1}{h} (f(0,0+h) -f(0,0)) = \lim_{h \to 0} \frac{0}{h}=0$$
But for the directional derivative $\vec{h} = (h,h)$:
$$[Df(0,0)]\vec{h} = \lim_{h \to 0} \frac{1}{h} (\frac{h^5}{h^4})=1$$
As we know, $0 +0 \neq 1$, so it is not linear and the derivative does not exist.
\item The topological definition of a continuous function is as follows:
$$f:D \subset \R \to C \subset \R \text{ is continuous } \Leftrightarrow \forall \text{ open sets } A \subset C, f^{-1}(A) \text{ is open in } D$$
Using the standard metric topology of open sets with which you are familiar, show that this definition is equivalent to the $\epsilon$-$\delta$ definition with which you are familiar (\emph{Hint: you may want to use the formulation given in a true/false question above}). Why does this not contradict the counter-examples you found on HW 7 - i.e. the open sets that are not mapped to open sets?
------------------ \\
{\bf Solution:}
First we show that the standard definition implies the topological definition. So we assume that for a given continuous function $f$, for all $x$
$$\forall \epsilon >0 , \exists \delta>0 \text{ such that } |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$$
This is equivalent to saying that
$$\forall \epsilon>0, \exists \delta>0, f(B_{\delta}(x)) \subset B_{\epsilon}(f(x))$$
Given any $A$ open set in $C$, we want to show that $f^{-1}(A) \subset D$ is open. Or more precisely, given any $y \in A$, as $A$ is open, we know that $B_{\gamma}(y) \subset A$. So we want to show that $f^{-1}(B_{\gamma}(y))$ is open in $D$. \\
It suffices for us to show that given any $x \in f^{-1}(B_{\gamma}(y))$ there exists a $r$ such that $B_r(x) \subset f^{-1}(B_{\gamma}(y))$. \\
We simply find an $\epsilon >0$ such that $B_{\epsilon}(f(x)) \subset B_{\gamma}(y)$. This implies that there exists $\delta>0$ such that
$$f(B_{\delta}(x)) \subset B_{\epsilon}(f(x)) \subset B_{\gamma}(y)$$
$$B_{\delta}(x) \subset f^{-1}(B_{\epsilon}(f(x)) \subset f^{-1}(B_{\gamma}(y))$$
$$\Rightarrow \text{ let } r=\delta, \Rightarrow B_{r}(x) \subset f^{-1}(B_{\gamma}(y)$$
which completes the proof.
Now we prove that the topological definition implies the standard definition. \\
For a given $x \in C$, and $\epsilon>0$, we know an open set in $C$ - namely $B_{\epsilon}(f(x))$. For $f$ continuous, we know this means that $f^{-1}(B_{\epsilon}(f(x)))$ is open in $D$.
Thus for $x \in f^{-1}(B_{\epsilon}(f(x)))$, $\exists r >0$ such that $B_r(x) \subset f^{-1}(B_{\epsilon}(f(x)))$. \\
Now we simply let $\delta=r$, and derive
$$B_{\delta}(x) \subset f^{-1}(B_{\epsilon}(f(x))) \rightarrow f(B_{\delta}(x)) \subset B_{\epsilon}(f(x))$$
which completes the proof.
\item {\bf T/F:} Let $X$ and $Y$ be sets with $A_1,A_2 \subset X$. For all functions $f: X \to Y$
$$f(A_1 \cap A_2) = f(A_1) \cap f(A_2)$$
as subsets of $Y$. And for $B_1,B_2 \subset Y$,
$$f^{-1}(B_1 \cap B_2)=f^{-1}(B_1) \cap f^{-1}(B_2)$$
as subsets of $X$.
--------------------\\
{\bf Solution: FALSE}
The first of these statements is false, while the second is true (one of the reasons why inverses images are considerably nicer to work with). The true statement is
$$f(A_1 \cap A_2) \subset f(A_1) \cap f(A_2)$$
When this fails to be equality, it will be because $f$ isn't injective. A counter-example for the inclusion in both directions is the following function $f:S \to S$ for $S:=\{1,2,3\}$ defined by $f(1)=2$, $f(2=2)$, $f(3)=3$. We can form 2 sets $A_1=\{1,3\}$ and $A_2=\{2,3\}$. Then $A_1 \cap A_2=\{3\}$ and $f(A_1 \cap A_2)=\{3\}$. But $f(A_1)=\{2,3\}$ and $f(A_2)=\{2,3\}$ so $f(A_1) \cap f(A_2)=\{2,3\}$. Clearly these are not equal.
\item {\bf T/F:} Let $h\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x^2y \\ y^2z \end{pmatrix}$. By the implicit function theorem, at the point $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ we have a continuously differentiable function $i$ locally expressing
$x$ and $z$ in terms of $y$ to give the locus of points whose image under $h$ is $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$. \\
--------------------\\
{\bf Solution: FALSE}
The point $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ is not in the zero-locus of $h$.
\item {\bf T/F:} If $a: \mathbb{N} \rightarrow \mathbb{R}$ is a sequence converging to 0, then for any $b: \mathbb{N} \rightarrow \mathbb{R}$, the sequence $(a_ib_i)$, for $i \in \mathbb{N}$ converges to 0\\
--------------------\\
{\bf Solution: FALSE}
This would be true if $b_i$ were bounded. \\
Out-with-a-bang counter example: $a_i = \frac{1}{i}, b_i = e^i$ \\
Subtle counter example: $a_i = \frac{1}{i}, b_i = i$
\item Give an example of a continuous function $f: \R \to \R$ such that the restriction of the domain of $f$ to $\Z \subset \R$ is a continuous map, or prove that one does not exist.
--------------------\\
{\bf Solution: EXAMPLE}
Perhaps continuous functions on $\Z$ make you uncomfortable (rightly so) but think about this, what you really need for continuity, is to be able to zoom in in the \emph{codomain}. This is you $\forall \epsilon>0$ part. However, if this doesn't actually force you to squish in on values in the domain, no problem. As an example, take the zero map.
\item Give an example of a function $g: \R \to \R$ that has a local inverse at \emph{every} point, but for which no global inverse exists, or prove that one does not exist.
--------------------\\
{\bf Solution: NO EXAMPLE}
If there is a local inverse at every point then $g'$ is never zero, meaning $g$ is monotonic. Thus there is a global inverse.
\item For $F_1,...,F_n...$ closed sets such that $F_i \subset \R^2$, let
$$F=\bigcap_i F_i$$
Give an example of a function $f: F \to \R$ which is differentiable, or prove that one does not exist.
--------------------\\
{\bf Solution: NO EXAMPLE}
No differentiable functions on closed sets, and the infinite intersection of closed sets is closed.
\item Give an example of a sequence of vectors $\vec{v}_1,\vec{v}_2,... \in \R^n$ which is convergent but the length of each vector is not uniformly bounded by some $M$, or prove that one does not exist.
--------------------\\
{\bf Solution: NO EXAMPLE}
Every such sequence will be uniformly bounded. To be convergent to $\vec{v}$ we need
$$\forall \epsilon>0, \exists N>0 \text{ s.t. } \forall n>N, |\vec{v}_n -\vec{v}|<\epsilon$$
which means that for any $\vec{v}_n$ with $n>N$:
$$|\vec{v}_n|=|\vec{v}_n-\vec{v}+\vec{v}|\leq |\vec{v}_n-\vec{v}|+|\vec{v}|<\epsilon +|\vec{v}|$$
And we only have finitely many vectors less than $\vec{v}_N$ so we simply take the max:
$$M_1=\max\{\vec{v}_i\}_{i\epsilon$.
\item Give an example of a surjective map $g:\Q \to \R$ which injects into its image, or prove that one does not exist.
--------------------\\
{\bf Solution: NO EXAMPLE}
Nope, that would mean $g$ was bijective, but $\Q$ and $\R$ don't have the same cardinality.
\item {\bf The Inverse and Implicit Function Theorems}
\begin{enumerate}
\item As I am sure you can imagine, the implicit function theorem has many important practical applications. Here is a somewhat whimsical one: \\
The school district's curriculum committee has accurately calculated that if they provide $x$ units of instruction in test-geared material and $y$ units of instruction in academic exploration they achieve test scores S and student happiness H given by the function
$$\begin{pmatrix} S \\ H \end{pmatrix} = f\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x^3 + xy^{1/2} \\ y^2 -4x^{1/2} \end{pmatrix}$$
Currently the curriculum is dictating educational time such that $x=4, y=1, S=68, H=-7$. Fearing trouble with reelection due to student unhappiness, the committee wants to moderately increase $H$ to -2 and only decrease $S$ to 63. Use the inverse function theorem to determine appropriate values of $x$ and $y$ that will approximate their desired output. How accurate is your answer? Why?
--------------------\\
{\bf Solution: }
The final answer is $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3.8 \\ 3.4 \end{pmatrix}$
\item Call $X$ the space of linear transformations$A: \R^2 \to \R^2$ with determinant $+1$. Show that locally you can parameterize the passive variable(s) of $X$ in terms of the active variable(s). \\
$X$ is the locus of the function $F:\R^4 \to \R$ given by
$$F:A \to \det(A)-1$$
The Jacobian is given by
$$[DF(A)]=\begin{bmatrix} d & -c & -b & a \end{bmatrix}$$
This is onto except for $a=b-c=d=0$, which is not even in $X$, so we are good!
\item Call $Y$ the space of linear transformations $B:\R^3 \to \R^3$ such that $B\circ B^T=I$ ($B$ is known as a \emph{orthogonal matrix}). How would you apply the implicit function theorem to locally parameterize $Y$?
\begin{enumerate}
\item Using the following statement:
\vspace{5pt}
\emph{All matrices $A\in Y$ satisfy the equation $AA^T-I=0$, thus I have a function $F:\R^9 \to \R^9$ given by $F:A \to AA^T-I$ for which $Y$ is the locus.}
\vspace{5pt}
Is the matrix $[DF]$ ever onto?
--------------------\\
{\bf Solution: }
As written, no, because $Y$ is not 9-dimensional and thus can not span $\R^9$.
\item How do you resolve this? Think about what you know about $AA^T$.
--------------------\\
{\bf Solution: }
Because $AA^T$ is symmetric we only need to consider the upper triangle. Our space $Y$ is the locus of the function $G:\R^9 \to \R^6$ given by this comparison. In this case, if you work out the Jacobian, you find that it is everywhere onto.
\end{enumerate}
\end{enumerate}
\end{enumerate}
\end{document}