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diff --git a/lacey/hw3.tex b/lacey/hw3.tex new file mode 100644 index 0000000..93a275a --- /dev/null +++ b/lacey/hw3.tex @@ -0,0 +1,339 @@ +\font\bigbf=cmbx12 at 24pt +\newfam\bbold +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv +\def\bb#1{{\fam\bbold #1}} +\input color + +\newcount\pno +\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}} +\def\endtf{\egroup\medskip} +\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}} +\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}} + +\newcount\qno +\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}} + +\def\proof{\medskip\noindent{\it Proof.}\quad} +\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt +depth 0pt{\parfillskip0pt\medskip}} + +\let\bu\bullet + +{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad HOMEWORK \#3}} + +\question{Answer True or False. No justification is needed.} + +\tf +A sequence of reals that is bounded above converges. +\endtf + +\false + +\tf +A monotone sequence of reals converges. +\endtf + +\false + +\tf +A sequence of reals that is bounded above and is monotone converges. +\endtf + +\false + +\tf +If $\{x_n\}$ is a sequence, and $p_1=2, p_2=3, p_3=5,\ldots$ are the +primes listed in increasing order, then $\{x_{p_k}\}$ is a subsequence. +\endtf + +\true + +\tf +If $\{x_n\}$ is a sequence which converges, and +$p_1=2,p_2=3,p_3=5,\ldots$ are the primes listed in increasing order, +then $\{x_{p_k}\}$ converges. +\endtf + +\true + +\tf +If $\{x_n\}$ is a sequnce, then $\{x_{n+(-1)^n}\}$ is a subsequence. +\endtf + +\false + +\tf +The rationalso are closed under addition, multiplication, and division. +And a sequence of rationals that is convergent converges to a rational. +\endtf + +\false + +\tf +The supremum of the empty set is $\sup\emptyset = -\infty.$ +\endtf + +\true + +\tf +If non-empty set $S$ has supremum $\pi,$ then there are infinitely many +elements $s\in S$ with $\pi-1/100<s\leq\pi.$ +\endtf + +\false + +\question{Show directly, from the definition, and using elementary +methods that the three sequences below are Cauchy.} + +\item{(1)} $\{1/n : n\in\bb N\}$ + +\proof +Let $\epsilon > 0.$ +Choose $N \in\bb N$ such that $N > 1/\epsilon > 0.$ + +Let $m,n > N.$ +$$0 < 1/m, 1/n < 1/N < \epsilon.$$ + +Subtracting, +$$-\epsilon < 1/n - 1/m < \epsilon \Longrightarrow |1/n-1/m| < +\epsilon,$$ +giving us that the sequence is Cauchy. +\endproof + +\item{(2)} $\{{3n-1\over 2n+5} : n\in\bb N\}$ + +\proof +$$a_n := {3n-1\over 2n+5} = {3n+7.5-8.5\over 2n+5} = {3\over 2} - +{8.5\over 2n+5}.$$ + +Let $\epsilon > 0.$ +We choose $N = {4.25\over\epsilon}-2.5,$ and for all $n,m>N,$ +$$|a_n-a_m| = \left|\left({3\over 2} - {8.5\over 2n+5}\right) - +\left({3\over 2} - {8.5\over 2m+5}\right)\right| = \left|{8.5\over 2n+5} +- {8.5\over 2m+5}\right| < {8.5\over 2N+5} = \epsilon,$$ +with the inequality found by maximizing the absolute value variable (we +might also show this inequality by starting from $n,m > N$ and +rearranging). +%TODO: may be worth improving + +We have shown the sequence is Cauchy. +\endproof + +\item{(3)} Let $\{s_n\}$ be any sequence of digits with +$s_n\in\{0,1,2\}$ for all integers $n.$ Show that the sequence $\{x_n\}$ +is Cauchy, where $x_n = \sum_{m=1}^n {s_m\over 3^m}.$ + +\proof +Let $\epsilon > 0.$ +Pick $N$ such that $3^{-N} < \epsilon.$ +We will show that $|x_n-x_m| < \epsilon$ for all $n,l > N.$ +$$x_n = \sum_{m=1}^n {s_m\over 3^m} = \sum_{m=1}^N {s_m\over 3^m} + +\sum_{m=N}^n {s_m\over 3^m} \leq \sum_{m=1}^N {s_m\over 3^m} + +\sum_{m=1}^N {2\over 3^m} = \sum_{m=1}^N {s_m\over 3^m} + 3^{-N} - +3^{-n} \leq \sum_{m=1}^N {s_m\over 3^m} + 3^{-N},$$ +and, eliminating the second term, we also get $x_n \geq \sum_{m=1}^N.$ +Taking these together, +$|x_m-x_n| \leq 3^{-N} = \epsilon.$ + +\endproof + +\question{} + +\item{(1)} Show that the sequence below converges + +$$\sqrt 2, \sqrt{2+\sqrt 2}, \sqrt{2 + \sqrt{2 + \sqrt 2}}, \ldots$$ + +\proof +Let $a_1 = \sqrt 2$ and for $n\geq 1,$ that $a_{n+1} = \sqrt{2+a_n}.$ + +We will show by induction that it is bounded by 2 and, later, that it is +monotonically increasing. + +Clearly, $\sqrt 2 < 2,$ so $a_1 < 2.$ +For $k\geq 1,$ we assume $a_k < 2,$ then $2+a_k < 4,$ so $\sqrt{2+a_k} < +2,$ or in other words, $a_{k+1} < 2.$ +We have shown the sequence is bounded above. + +$0 < a_k < 2,$ so $a_k-2 < 0$ and $a_k+1 > 0.$ +Therefore, +$$(a_k-2)(a_k+1) = a_k^2 - a_k - 2 < 0,$$ +Rearranging, and noting that both sides are positive, so the square root +function can be used and is monotonically increasing so as not to change +the direction of the inequality, +$$a_k < \sqrt{a_k+2} = a_{k+1},$$ showing that our sequence is +strictly monotonically increasing, so the set is bounded below by the +first element. + +By the Monotone Convergence Theorem, the sequence converges. +\endproof + +\item{(2)} Does the sequence below converge? If so, to what? + +$$\sqrt 2, \sqrt{2\sqrt 2}, \sqrt{2\sqrt{2\sqrt 2}}, \ldots$$ + +\proof +Yes, this sequence converges to 2 (heuristically, $\sqrt{2x} = x$ has +roots 0 and 2, and all elements are positive) + +We will show it is monotone, bounded by 2, and that the limit of the set +cannot be less than 2. +Let $x_1 = \sqrt 2$ and $x_n = \sqrt{2x_{n-1}}$ otherwise. + +$x_1 = \sqrt 2 < 2,$ and assuming $x_k < 2,$ then $2x_k < 4,$ and +$\sqrt{2x_k} = x_{k+1} < 2.$ +This shows the sequence is bounded by 2. + +All elements are positive. +Specifically, $x_1 = \sqrt 2 > 0,$ and $\sqrt{2x_n}>0$ where $x_n>0,$ giving +the inductive step. +We have also already shown $2 > x_n,$ so $2x_n > x_n^2,$ so finally +$\sqrt{2x_n} = x_{n+1} > x_n,$ giving monotonicity. + +To show the limit cannot be less than 2, assume for the sake of +contradiction that the limit is $1.5 < 2-\mu < 2$ (or, phrased +otherwise, $0 < \mu < .5$) + +By the definition of convergence, we may find $x_n > 2-1.1\mu.$ +We will show $\sqrt{2x_n} = x_{n+1} > 2-\mu.$ +Note that $\mu < .9,$ so $.9-\mu > 0,$ and $\mu > 0.$ +Therefore, +$$(.9-\mu)\mu = .9\mu-\mu^2 > 0 \Rightarrow 2-1.1\mu > 2-2\mu+\mu^2/2 +\Rightarrow 2(2-1.1\mu) > (2-\mu)^2 \Rightarrow \sqrt{2(2-1.1\mu)} = +x_{n+1} > 2-\mu.$$ +we have thus shown that the sequence may exceed the limit, instructing +us that the limit is 2. + +\endproof + +\question{} + +\item{(1)} In Section 1.4, we used the Axiom of Completeness (AoC) to +prove the Archimedean Property of $\bb R$ (Theorem 1.4.2). Show that the +Monotone Convergence Theorem can also be used to prove the Archimedean +Property without making any use of AoC. + +\proof + +The sequence $a_n = n$ is monotone. + +We will show the Archimedean Property by contradiction. +Let there be $x\in\bb R$ such that there is no $n\in\bb N$ satisfying +$n>x.$ +Therefore, $a_n$ is bounded and converges to some number $a < n$ by the +Monotone Convergence Theorem. +Let $0 < \epsilon < 1/2.$ + +From the definition of convergence, there exists $N$ such that for all +$m\geq N,$ that $|a-a_N| < \epsilon.$ +Rewriting, $a_N-\epsilon < a < a_N+\epsilon,$ and from our inequality +about $\epsilon,$ +$$a < a_N+1/2 \to a < a_{N+1}-1/2,$$ +which contradicts our convergence statement. + +\endproof + +\item{(2)} Use the Monotone Convergence Theorem to supply a proof for +the Nested Interval Property (Theorem 1.4.1) that doesn't make use of +AoC. These two results suggest that we could have used the Monotone +Convergence Theorem in place of AoC as our starting axiom for building a +proper theory of the real numbers. + +\proof +Let $I_i = [a_i,b_i]$ be a sequence of nonempty closed intervals such +that $$I_1\supseteq I_2 \supseteq I_3 \supseteq \cdots.$$ +$a_i$ is a weakly monotone increasing sequence because if $a_{i+1}<a_i,$ +then $a_{i+1}\in I_{i+1},$ but $a_{i+1}\not\in I_i,$ violating the +interval condition. +It is clearly bounded by $b_1$ because if $a_i > b_1,$ then +$I_i\not\subseteq I_1,$ and $I_1$ is nonempty. + +The Monotone Convergence Theorem tells us that $a_i$ converges to some +$x\geq a_i$ because $a_i$ is strictly increasing. + +For the sake of contradiction, let there be $b_k < x.$ +Let $\epsilon = x-b_k.$ +Because $a_i$ converges to $x,$ there must be for some $i>k,$ an $a_i > +x-\epsilon = b_k,$ violating the definitions of our intervals. +We have thus obtained $x\in [a_i, b_i],$ for all $i\in\bb N,$ and +therefore in the intersection of these intervals. +\endproof + +\question{Exercise 2.4.7 (Limit Superior). Let $(a_n)$ be a bounded +sequence.} + +\item{a)} Prove that the sequence defined by $y_n = \sup\{a_k : k\geq +n\}$ converges. + +\proof +$y_n \geq y_{n+1},$ because $\{a_k : k\geq n\}\supseteq\{a_k : k \geq +n+1\},$ so any upper bound for the former set is an upper bound for the +latter. +It is monotone and bounded by whatever bound $(a_n)$ is bounded by, so +it must converge by the Monotone Convergence Theorem. +\endproof + +\item{b)} The limit superior of $(a_n)$ or $\limsup a_n,$ is defined +by $\limsup a_n = \lim y_n,$ where $y_n$ is the sequence from part +$(a)$ of this exercise. Provide a reasonable definition for $\liminf +a_n$ and briefly explain why it always exists for any bounded sequence. + +\proof +We can define $$\liminf a_n := -\limsup (-a_n).$$ +If $|a_n| < b$ for some $b,$ then $|-a_n| = |a_n| < b$ too. +Clearly, this exists because $(-a_n)$ is also a bounded sequence where +$\limsup$ exists. + +Also, this is equivalent to the more intuitive definition +$$\liminf a_n := \lim(\inf\{a_k : k\geq n\})$$ because the lower bound of that set maps +onto the upper bound of its negation. +This would also exist because the infimum would be monotonically +increasing and bounded in the same way. +\endproof + +\item{c)} Prove that $\liminf a_n \leq \limsup a_n$ for every bounded +sequence, and give an example of a sequence for which the inequality is +strict. + +\proof +For all sets of bounded numbers $S_k = \{a_k : k\geq n\},$ +we know $\inf S_k \leq \sup S_k$ (the lower bound must be below the +upper bound). +By the Nested Interval Property, and because the infimum and supremum +should create a set of nested intervals (they are increasing and +decreasing respectively), we should have some $x\in [\liminf a_n, +\limsup a_n],$ meaning $\liminf a_n \leq \limsup a_n.$ +The inequality is strict for the sequence $(0)_{n\in\bb N}.$ +\endproof + +\item{d)} Show that $\liminf a_n = \limsup a_n$ if and only if $\lim +a_n$ exists. In this case, all three share the same value. + +\proof +$(\Rightarrow)$ +Let $x := \liminf a_n = \limsup a_n.$ +Let $\epsilon > 0.$ + +Using the definition of convergence, for some $N,$ +$$|\inf\{a_k : k \geq N\} - x| < \epsilon \Rightarrow x > \inf\{a_k : +k\geq N\}-\epsilon \Rightarrow x > a_j-\epsilon,$$ +for all $j\geq N.$ +Similarly, we can obtain $a_j+\epsilon > x.$ +This means for all $j>N,$ the sequence converges as $|a_j-x| < +\epsilon,$ so $\lim a_n = x$ exists. + +$(\Leftarrow)$ +Let $\lim a_n = x$ exist. +Let $\epsilon > 0.$ +We must have $N$ such that for all $n > N,$ $|a_n - x| < \epsilon.$ +That means $$\limsup a_n \leq \sup\{a_n : n > N\} < x+\epsilon,$$ and a +similar lower limit for $\liminf.$ +$|\limsup a_n - \liminf a_n| < 2\epsilon,$ and by the "give me some +room" principle, $\limsup a_n = \liminf a_n = x.$ +\endproof + +\bye |