\documentclass[12pt]{math23a}
\problemset{6}
\year1999\month11\day12
\author{Li-Chung Chen}

\DeclareMathOperator{\Img}{Img}
\DeclareMathOperator{\Mat}{Mat}
\newcommand{\R}{{\mathbb R}}
\def\theenumii{\Alph{enumii}}

\begin{document}

\item[1.5.4]
\begin{enumerate}
\item % (a)
Let $S_n$ be the $n$th partial sum of the series
$\sum^{\infty}_{k=0}\frac{1}{k!}A^k$.  Then
\begin{align*}
\abs{S_n} & = \abs{\sum^{n-1}_{k=0}\frac{1}{k!}A^k} & \\
& \leq \sum^{n-1}_{k=0}\abs{\frac{1}{k!}A^k} & \text{(triangle inequality)} \\
& \leq \abs{I} + \sum^{n-1}_{k=1}\frac{1}{k!}{\abs{A}}^k &
(\abs{A^k}\leq{\abs{A}}^k\text{ when }k\geq 1).
\end{align*}
But $\sum^{n-1}_{k=1}\frac{1}{k!}{\abs{A}}^k$ is just the $(n-1)$th partial
sum of the series
$\sum^{\infty}_{k=1}\frac{1}{k!}{\abs{A}}^k$, which converges to
$e^{\abs{A}}-1$.  Hence by the comparison test, the original series
$\sum^{\infty}_{k=0}\frac{1}{k!}A^k$ also converges, and
$\abs{e^A} = \abs{\sum^{\infty}_{k=0}\frac{1}{k!}A^k}$ is bounded above
by $\abs{I}+e^{\abs{A}}-1=\sqrt{n}+e^{\abs{A}}-1$.

\item % (b)
\begin{description}
\item[(1)]
$e^A = \sum^{\infty}_{k=0}\frac{1}{k!}
\begin{bmatrix}
a^k & 0 \\
0 & b^k
\end{bmatrix}
= \begin{bmatrix}
\sum^{\infty}_{i=0}\frac{1}{i!}a^i & 0 \\
0 & \sum^{\infty}_{j=0}\frac{1}{j!}b^j
\end{bmatrix}
= \begin{bmatrix}
e^a & 0 \\
0 & e^b
\end{bmatrix}$.
\item[(2)]
$e^A = I +
\begin{bmatrix}
0 & a \\
0 & 0
\end{bmatrix}
+
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
+
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
+ \dots
= 
\begin{bmatrix}
1 & a \\
0 & 1 
\end{bmatrix}$.
\item[(3)]
$e^A =
\begin{bmatrix}
1 - \frac{1}{2!}a^2 + \frac{1}{4!}a^4 - \dots &
\frac{1}{1!}a - \frac{1}{3!}a^3 + \frac{1}{5!}a^5 - \dots \\
-\frac{1}{1!}a + \frac{1}{3!}a^3 - \frac{1}{5!}a^5 + \dots & 
1 - \frac{1}{2!}a^2 + \frac{1}{4!}a^4 - \dots
\end{bmatrix}
=
\begin{bmatrix}
\cos{a} & \sin{a} \\
-\sin{a} & \cos{a}
\end{bmatrix}$.
\end{description}

\item % (c)
\begin{description}
\item[(1)]
The condition $AB=BA$ is called ``$A$ and $B$ commute.''
We claim that if $A$ and $B$ commute, then indeed $e^{A+B} = e^Ae^B$.  Let us see
how much $e^Ae^B$ differs from the $(r+1)$th partial sum of
$\sum^{\infty}_{p=0} \frac{1}{p!}{(A+B)}^p$.
We have
\begin{align*}
e^Ae^B - \sum^r_{p=0} \frac{1}{p!}{(A+B)}^p &
= \sum^{\infty}_{i=0} \frac{1}{i!}A^ie^B -
\sum^r_{i=0} A^i\left(\sum^{r-i}_{j=0} \frac{1}{(i+j)!}\binom{i+j}{i}B^j\right) \\
& = \sum^{\infty}_{i=r+1} \frac{1}{i!}A^ie^B
+ \sum^r_{i=0} A^i\left(\frac{1}{i!}e^B - 
\sum^{r-i}_{j=0}\frac{1}{i!j!}B^j\right) \\
& = \sum^{\infty}_{i=r+1} \frac{1}{i!}A^ie^B
+ \sum^r_{i=0} \frac{1}{i!}A^i\left(e^B -
\sum^{r-i}_{j=0}\frac{1}{j!}B^j\right) \\
& = \sum^{\infty}_{i=r+1} \frac{1}{i!}A^ie^B
+ \sum^r_{i=0} \frac{1}{i!}A^i
\left(\sum^{\infty}_{j=r-i+1}\frac{1}{j!}B^j\right)
\end{align*}
where the top equality uses the fact that $A$ and $B$ commute (why?).
Assume for now $r$ is even.  Let $r=2m$.  Then
\begin{align*}
& \abs{e^Ae^B - \sum^r_{p=0} \frac{1}{p!}{(A+B)}^p} \\ 
= & \abs{\sum^{\infty}_{i=2m+1} \frac{1}{i!}A^ie^B
+ \sum^{2m}_{i=0} \frac{1}{i!}A^i
\left(\sum^{\infty}_{j=2m-i+1}\frac{1}{j!}B^j\right)} \\
\leq & \abs{\sum^{\infty}_{i=2m+1} \frac{1}{i!}A^ie^B}
+ \abs{\sum^{2m}_{i=0} \frac{1}{i!}A^i
\left(\sum^{\infty}_{j=2m-i+1}\frac{1}{j!}B^j\right)} \\
\leq & \sum^{\infty}_{i=2m+1} \abs{\frac{1}{i!}A^ie^B}
+ \sum^{2m}_{i=0} \abs{\frac{1}{i!}A^i
\left(\sum^{\infty}_{j=2m-i+1}\frac{1}{j!}B^j\right)} \\
\leq & \sum^{\infty}_{i=2m+1} \abs{\frac{1}{i!}A^i}\abs{e^B}
+ \sum^{2m-1}_{i=m} \abs{\frac{1}{i!}A^i
\left(\sum^{\infty}_{j=2m-i+1}\frac{1}{j!}B^j\right)}
+ \sum^{m-1}_{i=0} \abs{\frac{1}{i!}A^i
\left(\sum^{\infty}_{j=2m-i+1}\frac{1}{j!}B^j\right)} \\
\leq & \sum^{\infty}_{i=2m+1} \abs{\frac{1}{i!}A^i}
(e^{\abs{B}}+\sqrt{n}-1)
+ \sum^{2m-1}_{i=m} \abs{\frac{1}{i!}A^i}
\left(\sum^{\infty}_{j=2m-i+1}\abs{\frac{1}{j!}B^j}\right)
+ \sum^{m-1}_{i=0} \abs{\frac{1}{i!}A^i}
\left(\sum^{\infty}_{j=2m-i+1}\abs{\frac{1}{j!}B^j}\right) \\
\leq & \sum^{\infty}_{i=2m+1} \frac{1}{i!}\abs{A^i}
(e^{\abs{B}}+\sqrt{n}-1)
+ \sum^{2m-1}_{i=m} \frac{1}{i!}\abs{A}^i
\left(\sum^{\infty}_{j=2m-i+1}\frac{1}{j!}\abs{B}^j\right)
+ \sum^{m-1}_{i=0} \frac{1}{i!}\abs{A}^i
\left(\sum^{\infty}_{j=m}\frac{1}{j!}\abs{B}^j\right) \\
= & \sum^{\infty}_{i=m} \frac{1}{i!}\abs{A}^i
(e^{\abs{B}}+\sqrt{n}-1)
+ \left(\sum^{m-1}_{i=0} \frac{1}{i!}\abs{A}^i\right)
\left(\sum^{\infty}_{j=m}\frac{1}{j!}\abs{B}^j\right) \\
\leq & \sum^{\infty}_{i=m} \frac{1}{i!}\abs{A}^i
(e^{\abs{B}}+\sqrt{n}-1)
+ \left(\sum^{\infty}_{i=0} \frac{1}{i!}\abs{A}^i\right)
\left(\sum^{\infty}_{j=m}\frac{1}{j!}\abs{B}^j\right) \\
\leq & \sum^{\infty}_{i=m} \frac{1}{i!}\abs{A}^i
(e^{\abs{B}}+\sqrt{n}-1)
+ (e^{\abs{A}}+\sqrt{n}-1)
\left(\sum^{\infty}_{j=m}\frac{1}{j!}\abs{B}^j\right) 
\end{align*}
Note that in the fifth inequality above, we used the fact that
when $0\leq i\leq m-1$, $2m-i+1>m$.

The last expression above tends to $0$ as $m$ tends to infinity.  But let us
be more precise.  Suppose we are given $\epsilon > 0$.  Let
$\epsilon_1 = \frac{\epsilon}{2(e^{\abs{B}}+\sqrt{n}-1)} > 0$ and
$\epsilon_2 = \frac{\epsilon}{2(e^{\abs{A}}+\sqrt{n}-1)} > 0$.
Then there exist $M_1$ and $M_2$ such that
$$
\sum^{\infty}_{i=m} \frac{1}{i!}\abs{A}^i < \epsilon_1
\text{ when }m\geq M_1,\text{ and}
$$
$$
\sum^{\infty}_{j=m} \frac{1}{j!}\abs{B}^j < \epsilon_2
\text{ when }m\geq M_2.
$$
Let $M=\max\{M_1,M_2\}$.  Then whenever $m\geq M$,
$\abs{e^Ae^B - \sum^{2m}_{p=0} \frac{1}{p!}{(A+B)}^p}
< \epsilon_1(e^{\abs{B}}+\sqrt{n}-1)
+ \epsilon_2(e^{\abs{A}}+\sqrt{n}-1)
= \epsilon$.
This only takes care of the case that $r$ is even.  However, it shows
that the sequence of partial sums
$$\left\{\sum^2_{p=0} \frac{1}{p!}{(A+B)}^p,
\sum^4_{p=0} \frac{1}{p!}{(A+B)}^p,
\sum^6_{p=0} \frac{1}{p!}{(A+B)}^p, \dots \right\}$$
converges to $e^Ae^B$.  The above sequence is a subsequence of
$$\left\{\sum^1_{p=0} \frac{1}{p!}{(A+B)}^p,
\sum^2_{p=0} \frac{1}{p!}{(A+B)}^p,
\sum^3_{p=0} \frac{1}{p!}{(A+B)}^p, \dots \right\},$$
which converges to $e^{A+B}$.  Hence by problem 1.5.17,
the sequence of even partial sums converges to $e^{A+B}$, implying
that $e^Ae^B = e^{A+B}$ (if a sequence converges, it converges to
an unique value).

\item[(2)]
Because $A$ commutes with itself, by (c)(1) we see that
$e^{2A} = e^{A+A} = e^Ae^A = {(e^A)}^2$.
\end{description}
\end{enumerate}
\begin{rmk}
The exponential of a matrix is useful because it turns out to solve
\emph{matrix} differential equations.  Consider an $n\times n$ matrix
$F(t)$, each of whose entries is a real-valued function of $t$.  We
can take the derivative $F'(t)$ by differentiating each entry.  Then
there is a theorem that given constant $n \times n$ matrices $A$ and
$B$, the only $F(t)$ that satisfies
$$F'(t) = AF(t) \text{ and } F(0) = B$$
is $F(t) = e^{tA}B$.  For instance, if
$A = \begin{bmatrix}0 & a \\-a & 0\end{bmatrix}$ and
$B = \begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix}$, then
$e^{tA}B = e^{tA} =
\begin{bmatrix}
\cos(ta) & \sin(ta) \\
-\sin(ta) & \cos(ta)
\end{bmatrix}$.
\end{rmk}

\item[1.5.11]
Suppose $\bar{X}$ is not closed.  Let $Y = \R^n - \bar{X}$.  Then
$Y$ is not open, so there exists $y\in Y$ such that for any
$\epsilon > 0$, $B_{\epsilon}(y)$ is not contained in $Y$.  By letting
$\epsilon = \frac{1}{n}$, we see that for
any positive integer $n$, there exists $\bar{x}_n \in B_{\frac{1}{n}}(y)$
not contained in $Y$.  But this means $\bar{x}_n \in \bar{X}$, so there
exists a sequence $x_{n1},x_{n2},\dots$ of points in $X$ that converges
to $x_n$.  Due to convergence, there exists $x_n \in
\{x_{n1},x_{n2},\dots\}$ such that $\abs{x_n-\bar{x}_n} < \frac{1}{n}$.
Now $\abs{\bar{x}_n-y} < \frac{1}{n}$ because $\bar{x}_n\in
B_{\frac{1}{n}}(y)$.  Hence by the triangle inequality,
$$\abs{x_n-y}\leq\abs{x_n-\bar{x}_n}+\abs{\bar{x}_n-y} < \frac{2}{n}.$$
We claim that the sequence $x_1,x_2,\dots$ converges to $y$.  For a
given $\epsilon > 0$, we can find $N$ such that $\frac{2}{N}<\epsilon$.
Then for any $n\geq N$,
$\abs{x_n-y} < \frac{2}{n} \leq \frac{2}{N} < \epsilon$.  By definition,
the sequence $x_1,x_2,\dots$ converges to $y$.  Because the $x_i's$ are all
in $X$, $y$ is in $\bar{X}$, a contradiction.  Therefore, $\bar{X}$
is closed.

\item[1.5.12]
\begin{enumerate}
\item % (a)
First suppose the sequence $a_1,a_2,\dots$ converges to $a$.  Then for
any $\epsilon' > 0$, there exists $N$ such that for $n > N$,
$\abs{a_n-a} < \epsilon'$.  Now, for a given $\epsilon > 0$, we apply
the above to $\epsilon' = \phi(\epsilon)$ (which is positive)
to obtain an $N$ such that
for $n > N$, $\abs{a_n-a} < \phi(\epsilon)$.  This is exactly what we
want to prove.

Next, suppose that for any $\epsilon' > 0$, there exists $N$ such that
for $n > N$, we have $\abs{a_n-a}\leq\phi(\epsilon')$.  Fix an
$\epsilon > 0$ (e.g. the $\epsilon$ our opponent has given us).  Because
$\lim_{\epsilon'\rightarrow 0}\phi(\epsilon') = 0$, there exists
$\epsilon' > 0$ such that $\phi(\epsilon') < \epsilon$.  For this
$\epsilon'$ (which depends on $\epsilon$), by assumption we can find
an $N$ (which now also depends on $\epsilon$) such that for $n > N$,
we have $\abs{a_n-a}\leq\phi(\epsilon')$.  Since
$\phi(\epsilon')<\epsilon$, we know that when $n > N$,
$\abs{a_n-a}<\epsilon$.  This is exactly the definition of convergence,
so the sequence $a_1,a_2,\dots$ converges to $a$.
\item % (b)
Let $\phi$ be as in part (a).  Then a function $f$ converges at $x_0$
iff for $\epsilon > 0$, there exists $\delta > 0$ such that
$\abs{x-x_0} < \delta$ implies $\abs{f(x)-f(x_0)} < \phi(\epsilon)$.
\end{enumerate}

\item[1.5.14]
\begin{description}
\item[(a)]
Let $f\from \R^2 \to \R$ and $g\from \R^2 \to \R$ be given by
$f\binom{x}{y} = x$ and
$g\binom{x}{y} = y$.  Our problem
is to determine whether $\frac{f^2}{f+g}$ has a limit at
$\binom{1}{2}$.  By corollary 1.5.28,
$f$ and $g$ are continuous.  Theorem 1.5.27(a) and (d) tell us that
$f^2$ and $f+g$ are continuous.  Since
$(f+g)\binom{1}{2} = 1+2 = 3 \neq 0$,
theorem 1.5.27(c) tells us that $\frac{f^2}{f+g}$ is continuous at
$\binom{1}{2}$.  Hence the given
limit exists and is equal to $(\frac{f^2}{f+g})
\binom{1}{2} = \frac{1}{3}$.

\item[(g)]
This limit does \emph{not} exist because we can find a sequence
$\binom{x_1}{y_1},\binom{x_2}{y_2},\dots\rightarrow\binom{0}{0}$ such
that the sequence $({x_1}^2+{y_1}^2)(log\abs{x_1y_1}),
({x_2}^2+{y_2}^2)(log\abs{x_2y_2}),\dots$ does not converge.  Let
$x_n=\frac{1}{n}$ and $y_n=e^{1/{x_n}^3}=e^{-n^3}$.  Because
$x_n$ and $y_n$ both converge to $0$ as $n \rightarrow \infty$,
$\binom{x_n}{y_n}$ converges to $\binom{0}{0}$.  However,
\begin{align*}
({x_n}^2+{y_n}^2)(\log\abs{x_ny_n}) &
= \left(\frac{1}{n^2}+{y_n}^2\right)(-n^3+\log x_n) & \\
& \leq \left(\frac{1}{n^2}+{y_n}^2\right)(-n^3) & \text{since }
\left(\frac{1}{n^2}+{y_n}^2\right)\log x_n\leq 0\\
& < \frac{1}{n^2}(-n^3) & \text{since }-{y_n}^2n^3 < 0 \\
& = -n. &
\end{align*}
Hence $({x_n}^2+{y_n}^2)(\log\abs{x_ny_n})$ goes to $-\infty$ and
diverges.

Note that if we approach $\binom{0}{0}$ from any particular straight
line, our limit will be $0$.  The $\binom{x_n}{y_n}$ we defined above
hence must be approaching $\binom{0}{0}$ ``from a curve.''

\item[(h)]
We claim that the limit exists and is $0$.  We will show that for
any sequence $\binom{x_1}{y_1},\binom{x_2}{y_2},\dots$ converging to
$\binom{0}{0}$, the sequence
$\{({x_n}^2+{y_n}^2)\log({x_n}^2+{y_n}^2)\}$ converges to $0$.
Suppose we have such sequence $\{\binom{x_n}{y_n}\}$.  Because any
polynomial function is continuous (corollary 1.5.28), the sequence
$\{{x_n}^2+{y_n}^2\}$ converges to $0^2+0^2 = 0$.  So all we need
to show is that $\lim_{z\rightarrow 0^+} z\log z = 0$.  This can be
shown using L'Hopital's rule:
$$
\lim_{z\rightarrow 0^+} z\log z 
= \lim_{t\rightarrow +\infty} \frac{1}{t}\log{\frac{1}{t}}
= \lim_{t\rightarrow +\infty} -\frac{\log t}{t}
= \lim_{t\rightarrow +\infty} -\frac{1/t}{1}
= 0.
$$
\end{description}

\item[1.5.15]
\begin{enumerate}
\item % (a)
$\lim_{\binom{x}{y}\rightarrow\binom{0}{0}}f\binom{x}{y} = a$ means
that for any $\epsilon > 0$, there exists $\delta > 0$ such that
whenever $\abs{\binom{x}{y}-\binom{0}{0}} = \abs{\binom{x}{y}} < \delta$
\emph{and} $\binom{x}{y} \in D^*$, $\abs{f\binom{x}{y}-a} < \epsilon$.

\item % (b)
First we show that the limit for $f$ does not exist by approaching
$\binom{0}{0}$ from two different lines $y = 0$ and $y = x$.  If we
approach $\binom{0}{0}$ from the line $y = 0$ along the positive
$x$-axis, then the limit is
$\lim_{x\rightarrow 0^+} f\binom{x}{0}
= \lim_{x\rightarrow 0^+} \frac{\sin x}{x}
= \lim_{x\rightarrow 0^+} \frac{\cos x}{1}
= 1$, by using L'Hopital.  If we approach $\binom{0}{0}$ from the
line $y = x$ from the first quadrant, then the limit is
$\lim_{x\rightarrow 0^+} f\binom{x}{x}
= \lim_{x\rightarrow 0^+} \frac{\sin(2x)}{\sqrt{2}x}
= \lim_{x\rightarrow 0^+} \frac{2\cos(2x)}{\sqrt{2}}
= \sqrt{2}$, by using L'Hopital.  Because we get two different values,
the original limit must not exist.

Second we show that the limit for $g$ exists and is $0$.  It suffices to
show that
$\lim_{\binom{x}{y}\rightarrow\binom{0}{0}}\abs{x}\log(x^2+y^4)$ and
$\lim_{\binom{x}{y}\rightarrow\binom{0}{0}}\abs{y}\log(x^2+y^4)$ exist and
are both $0$.
Consider an arbitrary sequence $\{\binom{x_n}{y_n}\}$ that converges to
$\binom{0}{0}$.  Then ${x_n}^2 + {y_n}^4 \rightarrow 0$, so for sufficiently
large $n$, ${x_n}^2 + {y_n}^4 < 1$ and $\abs{x_n}\log({x_n}^2+{y_n}^4) \leq 0$.
On the other hand, $\abs{x_n}\log({x_n}^2+{y_n}^4) \geq \abs{x_n}\log({x_n}^2)
= 2\abs{x_n}\log(x_n)$.  So for sufficiently large $n$,
$$ 2\abs{x_n}\log(x_n) \leq \abs{x_n}\log({x_n}^2+{y_n}^4) \leq 0.$$
Now $x_n \rightarrow 0$ implies that $2\abs{x_n}\log{x_n} \rightarrow 0$
(see 1.5.14(h)), so by a squeezing lemma, we know that
$\lim_{\binom{x}{y}\rightarrow\binom{0}{0}}\abs{x}\log(x^2+y^4)$ must exist
and must be $0$.  (You should try to formulate an appropriate squeezing lemma
and prove it.)  A similar argument (reversing the roles of $x_n$ and $y_n$)
shows that 
$\lim_{\binom{x}{y}\rightarrow\binom{0}{0}}\abs{y}\log(x^2+y^4)$ exists and
is $0$.  Hence the original limit exists and is $0$.
\end{enumerate}

\item[1.5.17]
Suppose $a_{i(1)}, a_{i(2)}, \dots$ is an infinite subsequence of
$a_1, a_2, \dots$.  For a given $\epsilon > 0$, there exists $N$ such that
when $n > N$, $\abs{a_n - a} < \epsilon$.  Because $i(j) \rightarrow \infty$
as $j \rightarrow \infty$, there exists $N'$ such that $i(N') > N$.
(Alternatively, let $N' = N$.  Then $i(N') \geq N' > N$.)  For any
$j > N'$, $i(j) > i(N') > N$ and hence $\abs{a_{i(j)} - a} < \epsilon$.
Thus by definition, $a_{i(j)} \rightarrow a$.

\item[1.5.23]
Part (a) is a special case of part (c), so it suffices to show part (c).
Let $f$ be the given map.
Define maps $g,h\from \R^2 \to \R$ by
$g\begin{bmatrix}x \\ y\end{bmatrix} = ax+by$ and
$h\begin{bmatrix}x \\ y\end{bmatrix} = cx+dy$.  By corollary 1.5.28,
$g$ and $h$ are continuous.  The continuity of $f$ follows from the
following lemma, which follows directly from proposition 1.5.20 (why?):
\begin{lem}
Let $f_1,\dots,f_n$ be real-valued functions defined on a domain
$U \subset \R^n$, and define $f\from U\to\R^n$ by
$f(x) = \left(\begin{array}{c}f_1(x)\\ \vdots \\f_n(x)\end{array}\right)$.
Then $f$ is continuous at $a = 
\left(\begin{array}{c}a_1\\ \vdots \\)a_n\end{array}\right)$ if and only
if each $f_i$ is continuous at $a_i$.
\end{lem}

\item[1.5.24]
The problem statement is vague in several senses, so we will just say that
$\begin{bmatrix}3.14\\2.71\end{bmatrix}$ and
$\begin{bmatrix}3.1415\\2.7182\end{bmatrix}$ are the first two that
work, and that the $a_n$'s thereafter also work.  The latter is true
because $\abs{a_n-\begin{bmatrix}\pi\\ e\end{bmatrix}}$ decreases as
$n$ increases (why?).

\end{document}
