Teaching blog http://caicedoteaching.wordpress.com Andrés Caicedo Sun, 08 Nov 2009 23:11:26 +0000 http://wordpress.com/ en hourly 1 http://www.gravatar.com/blavatar/897844b926b51566c79036956ebd8973?s=96&d=http://s.wordpress.com/i/buttonw-com.png Teaching blog http://caicedoteaching.wordpress.com 502 – Cantor-Bendixson derivatives http://caicedoteaching.wordpress.com/2009/11/08/502-cantor-bendixson-derivatives/ http://caicedoteaching.wordpress.com/2009/11/08/502-cantor-bendixson-derivatives/#comments Sun, 08 Nov 2009 23:08:03 +0000 andrescaicedo http://caicedoteaching.wordpress.com/?p=2412 ]]>

Given a topological space X and a set B\subseteq X, let B' be the set of accumulation points of B, i.e., those points p of X such that any open neighborhood of p meets B in an infinite set.

Suppose that B is closed. Then B'\subseteq B. Define B^\alpha for B closed compact by recursion: B^0=B, B^{\alpha+1}=(B^\alpha)', and B^\lambda=\bigcap_{\alpha<\lambda}B^\alpha for \lambda limit. Note that this is a decreasing sequence, so that if we set B^\infty=\bigcap_{\alpha\in{\sf ORD}}B^\alpha, there must be an \alpha such that B^\infty=B^\beta for all \beta\ge\alpha. 

[The sets B^\alpha are the Cantor-Bendixson derivatives of B. In general, a derivative operation is a way of associating to sets B some kind of ``boundary.'']

For concreteness, suppose that B\subseteq[0,1] is compact. Then B^\infty is either empty or perfect (i.e., every point of B^\infty is an accumulation point of B^\infty). It is easy to see that every perfect subset X of {\mathbb R} has the same size as {\mathbb R}. For example, define (U_s\mid s\in 2^{<{\mathbb N}}) by recursion on |s| as follows:

  1.  U_\emptyset is an arbitrary open (in the relative topology) nonempty subset of X of diameter at most 1=2^-0.
  2. Given U_s, let x\ne y be distinct points of U_s (these exist since inductively U_s is open nonempty in X and X has no isolated points). Let U_{s0} and U_{s1} be disjoint open neighborhoods of x and y, respectively, whose closures are contained in U_s, and have diameter at most 2^{-(|s|+1)}. 

Then, for each x\in{}^\omega2, the set \bigcap_{n<\omega}U_{x\upharpoonright n}=\bigcap_{n<\omega}\bar U_{x\upharpoonright n} is a singleton, say \{p_x\}. Moreover, the map f:{}^\omega2\to X given by f(x)=p_x is injective (and continuous).

[By the way, the above is an example of a Cantor scheme.]

It follows that if B\subseteq[0,1] is countable, then B^\infty is necessarily empty. Let \alpha be least such that B^{\alpha+1}=\emptyset, and call \alpha the Cantor-bendixon rank of B. (Note that the first \alpha such that B^\alpha=\emptyset cannot be a limit ordinal.) Note that \alpha is necessarily countable.

It is a nice exercise to show that for all \alpha<\omega_1 there is a countable compact subset of {}[0,1] of rank precisely \alpha.

In a sense, set theory began with the study of these derivatives. Cantor used them to prove (by induction on the rank) that any countable compact subset of {\mathbb T}={\mathbb R}/2\pi{\mathbb Z} is a set of uniqueness for trigonometric series. See for example the introduction by Philip Jourdain to the English version of Cantor’s Contributions to the founding of the theory of Transfinite numbers, or Alekos Kechris’s nice article Set theory and uniqueness for trigonometric series.

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In this note I sketch the proof of the Löwenheim-Skølem (or Löwenheim-Skølem-Tarski) theorem for first order theories. This basic result of model theory is really a consequence of a set theoretic combinatorial lemma, as the proof will demonstrate.

Let {{\mathcal L}} be a first order language, understood as a set of constant, function, and relation symbols. Let

\displaystyle \kappa_{\mathcal L}=|{\mathcal L}|+\aleph_0,

so {\kappa_{\mathcal L}} is {|{\mathcal L}|,} unless {{\mathcal L}} is finite, in which case we take {\kappa_{\mathcal L}=\omega.} Talking about {\kappa_{\mathcal L}} rather than {|{\mathcal L}|} simplifies the presentation slightly.

The Löwenheim-Skølem theorem is concerned with the possible infinite sizes of models of first order theories. Of course, a theory {T} could only have finite models; the result does not say anything about {T} if that is the case.

Theorem 1 If {T} is a first order theory in a language {{\mathcal L},} and there is at least one infinite model of {T,} then there are models of {T} of size {\lambda,} for all {\lambda\ge\kappa_{\mathcal L}.}

We will prove a more precise statement. Before stating it, note that it is possible to have a theory {T} in some uncountable language {{\mathcal L}} such that {T} has models of certain infinite sizes, but not all. Theorem 1 does not say anything about infinite models of {T} of size {<\kappa_{\mathcal L}.} What cardinals in this range are the possible sizes of models of {T} is actually a rather difficult problem, and we will not address it.

To state the more precise version that we will prove, we need to review the notion of elementary substructure. Thoughout the note we use the notational convention that if {{\mathcal M}} is a {{\mathcal L}}-structure, its underlying universe is {M.}

Recall that if {{\mathcal M}} and {{\mathcal N}} are {{\mathcal L}}-structures, we say that

\displaystyle {\mathcal M}\subseteq{\mathcal N}

({{\mathcal M}} is a substructure of {{\mathcal N}}) iff:

  1. {M\subseteq N}.
  2. {R^{\mathcal M}=R^{\mathcal N}\cap M^j} for any {j<\omega} and any {j}-ary relation symbol {R} in {{\mathcal L}.}
  3. {c^{\mathcal M}=c^{\mathcal N}} for all constant symbols {c} in {{\mathcal L}.}
  4. {f^{\mathcal M}=f^{\mathcal N}\upharpoonright M^j:M^j\rightarrow M} for any {j<\omega} and any {j}-ary function symbol {f} in {{\mathcal L}.}

For example, a subgroup of a group {(G,*,e)} is simply a subset {H\subseteq G} such that {e\in H} and {H} is closed under {*.} A subgroup may have significantly different properties than the group it comes from. Consider, for example, {{\mathbb Z}\subseteq{\mathbb Q}.}

A substructure {{\mathcal M}} of {{\mathcal N}} is elementary, in symbols

\displaystyle {\mathcal M}\preceq{\mathcal N},

iff for any {{\mathcal L}}-formula {\varphi(x_1,\dots,x_n)} and any {a_1,\dots,a_n\in M,} we have that

\displaystyle {\mathcal M}\models\varphi(a_1,\dots,a_n)

iff

\displaystyle {\mathcal N}\models\varphi(a_1,\dots,a_n).

This is a much more strict notion than simply being a substructure. It is easy to check that {{\mathbb Z}} is not elementary in {{\mathbb Q},} for example.

Note that if {{\mathcal M}\preceq{\mathcal N}} then, for any {{\mathcal L}}-theory {T,} we have that {{\mathcal M}\models T} iff {{\mathcal N}\models T.} (Simply consider sentences {\varphi} in the definition above.)

Here is a natural example of how to obtain elementary extensions of a given model {{\mathcal M}:} Expand the language {{\mathcal L}} into a larger language {{\mathcal L}'} by adding to {{\mathcal L}} a set of {|M|} many new constant symbols {c_m} for {m\in M.} We can expand {{\mathcal M}} to an {{\mathcal L}'}-structure simply by setting {c_m^{\mathcal M}=m.} Let {{\mathcal M}'} denote this expanded structure. Consider the theory {T={\rm Th}({\mathcal M}'),} where the theory is of course considered in the language {{\mathcal L}'.} Let {{\mathcal N}} be any model of {T.} Then there is a natural way of identifying {{\mathcal N}} with an elementary extension of {{\mathcal M}.} More precisely, the map {j:{\mathcal M}\rightarrow{\mathcal N}} given by {j(m)=c_m^{\mathcal N}} is elementary, meaning that

\displaystyle {\mathcal M}'\models\varphi(a_1,\dots,a_n)

iff

\displaystyle {\mathcal N}\models\varphi(j(a_1),\dots,j(a_n))

for any {{\mathcal L}'}-formula {\varphi(x_1,\dots,x_n)} and any {a_1,\dots,a_n\in M.}

It easily follows that the pointwise image {j[M]} is the universe of an elementary substructure of {{\mathcal N}} that is isomorphic to {{\mathcal M}',} so we might as well assume that {j} is the identity. By considering the restriction of {{\mathcal N}'} to language {{\mathcal L}} (i.e., by “forgetting” about the new constant symbols), we obtain this way an elementary extension of {{\mathcal M}.}

In fact, it is easy to check that if {{\mathcal M}\preceq{\mathcal N},} then there is a natural expansion of {{\mathcal N}} to an {{\mathcal L}'}-structure that models {{\rm Th}({\mathcal M}'),} so all elementary extensions of {{\mathcal M}} arise in this fashion. From the observations above, Theorem 1 then follows from the following more precise version, where we write {|{\mathcal M}|} for {|M|,} and say that {{\mathcal M}} is infinite iff {M} is.

Theorem 2 Let {{\mathcal M}} be an infinite {{\mathcal L}}-structure.

 

  1. For any cardinal {\lambda\ge\kappa_{\mathcal L}+|{\mathcal M}|,} there is {{\mathcal N}\succeq{\mathcal M}} with {|{\mathcal N}|=\lambda.}
  2. For any cardinal {\lambda\in[\kappa_{\mathcal L},|{\mathcal M}|]} and any {X\subseteq M} with {|X|\le\lambda,} there is {{\mathcal N}\preceq{\mathcal M}} such that {|{\mathcal N}|=|\lambda|,} and {X\subseteq N.}

We note first that item 1 follows from item 2 and compactness: Consider the language {{\mathcal L}'} obtained by adding to {{\mathcal L}} a set of {\lambda} many new constant symbols {c_\alpha,} {\alpha<\lambda,} and {|M|} many new constant symbols {c_m,} {m\in M.} In this language, consider the theory {T} obtained by adding to {{\rm Th}({\mathcal M}')} the axioms {c_\alpha\ne c_\beta} for all {\alpha<\beta<\lambda,} where {{\mathcal M}'} is as above. This theory is consistent, by compactness (using that {M} is infinite), so it has models. But any model {{\mathcal A}} of {T} can be naturally identified with an elementary extension of {{\mathcal M},} and has size at least {\lambda.} By item 2, there is an elementary substructure of {{\mathcal A}} that contains {M} and has size precisely {\lambda.} Since {{\mathcal A}\models{\rm Th}({\mathcal M}'),} it follows that the reduct of {{\mathcal A}} to language {{\mathcal L}} is indeed an elementary extension of {{\mathcal M}} of size {\lambda.}

To prove item 2, we expand the language {{\mathcal L}} in a different manner.

For this, we need the notion of Skølem functions. Syntactically, a Skølem function for a formula {\varphi(x,y_1,\dots,y_n)} is simply a new function symbol {f_\varphi} for an {n}-ary function. Semantically: Given a structure {{\mathcal M}} in a language containing {\varphi} and {f_\varphi,} we call the interpretaion {f_\varphi^{\mathcal M}} a Skølem function iff, whenever {a_1,\dots,a_n\in M,} and

\displaystyle {\mathcal M}\models\exists x\,\varphi(x,a_1,\dots,a_n),

then in fact

\displaystyle {\mathcal M}\models\varphi(f_\varphi^{\mathcal M}(a_1,\dots,a_n),a_1,\dots,a_n).

We say that a language {{\mathcal L}} is closed under Skølem functions if a function symbol {f_\varphi} is in the language for each {\varphi} in the language. We say that a structure {{\mathcal M}} in this language has Skølem functions if it satisfies all axioms of the form

\displaystyle \forall x_1\dots\forall x\bigl(\exists x\,\varphi(x,x_1,\dots,x_n)\quad\rightarrow\quad\varphi(f_\varphi(x_1,\dots,x_n),x_1,\dots,x_n)).

Note that using the axiom of choice it is straightforward to define interpretations of all the function symbols {f_\varphi} so that these axioms are satisfied: Simply fix a well-ordering of {M,} and take {f_\varphi(\vec a)} as the least {b} (in the sense of the well-ordering) that witnesses {{\mathcal M}\models\exists x\,\varphi(x,\vec a),} if such is the case, or as the least element of {M} otherwise.

It is straightforward to expand a language to one closed under Skølem functions: Given {{\mathcal L},} let {{\mathcal L}_0={\mathcal L},} and set {{\mathcal L}_{n+1}={\mathcal L}_n\cup\{f_\varphi\mid\varphi} is an {{\mathcal L}_n}-formula{\}} for all {n<\omega.} Finally, set {{\mathcal L}_\omega=\bigcup_n{\mathcal L}_n,} and note that {{\mathcal L}_\omega} is as wanted.

Moreover, {\kappa_{{\mathcal L}_\omega}=\kappa_{\mathcal L},} as is easily verified by induction on {n.} This means that to prove item 2, we may as well assume that {{\mathcal L}} is closed under Skølem functions and {{\mathcal M}} has Skølem functions to begin with.

The point of all this is that if {{\mathcal M}} has Skølem functions and {{\mathcal N}\subseteq{\mathcal M}}, then automatically {{\mathcal N}\preceq{\mathcal M}.} This follows from the following test for elementarity:

Lemma 3 (Tarski-Vaught criterion) Suppose {{\mathcal N}\subseteq{\mathcal M}} and that whenever {a_1,\dots,a_n\in N} and

\displaystyle {\mathcal M}\models\exists x\,\varphi(x,\vec a),

then there exists {b\in N} such that

\displaystyle {\mathcal M}\models\varphi(b,\vec a).

Then {{\mathcal N}\preceq{\mathcal M}.}

Proof: By induction in the complexity of {\varphi.} \Box

The usefulness of the test derives from the fact that it only mentions satisfaction in {{\mathcal M}} rather than in both {{\mathcal M}} and {{\mathcal N}.}

To see how Lemma 3 allows us to conclude the claim made above, suppose that {{\mathcal M}} has Skølem functions, and that {{\mathcal N}} is a substructure of {{\mathcal M}.} Then, automatically, the hypothesis of Lemma 3 is satisfied, since {{\mathcal N}} must be closed under the restrictions of all the functions {f_\varphi^{\mathcal M},} being a substructure, meaning that {f_\varphi^{\mathcal M}(\vec a)\in N} whenever {\vec a\in N.} It then follows that {{\mathcal N}\preceq{\mathcal M},} as claimed.

Moreover, note that if {N\subseteq M} and {N} is (nonempty and) closed under all the Skølem functions from {{\mathcal M},} then {N} is the universe of a (necessarily elementary) substructure of {{\mathcal M}.} To see this, note that from the definition of substructure given above, all that we require is that {c^{\mathcal M}\in N} for all constant symbols {c} in {{\mathcal L},} and that {f^{\mathcal M}(\vec a)\in N} whenever {\vec a\in N,} for all function symbols {f} in {{\mathcal M}.}

But consider {\varphi(x,y)\equiv y=y\land x=c.} Then {f_varphi^{\mathcal M}(y_0)=c^{\mathcal M}} for any {y_0\in M,} so if {N} is nonempty and as above, then {N} contains the interpretaions of all the constant symbols from the language. (The only reason we added {y} to the formula {\varphi} rather than just taking {\varphi(x)\equiv x=c} is to avoid having {f_\varphi} to be {0}-ary, since our official definition of function symbols requires this not to be the case. Similarly, consider {\varphi(x,y_1,\dots,y_n)\equiv x=f(y_1,\dots,y_n).} Then {f_\varphi^{\mathcal M}=f^{\mathcal M}.}

We have finally arrived at the combinatorial core of Theorem 2. Some notation is useful (we follow Kunen Set theory. An introduction to independence proofs in the remainder of this note).

For {n<\omega} say that {f} is an {n}-ary function on a set {A} iff {n>0} and, as usual, {f:A^n\rightarrow A,} or {n=0} and {f\in A.} (This is a natural extension of the notion for {n>0:} Note that {A^0=1} and any {f:1\rightarrow A} is simply picking an element of {A;} here we are just identifying {f} and this element.) Say that {f} is a finitary function on {A} iff {f} is {n}-ary on {A} for some {n<\omega.}

Say that {B\subseteq A} is closed under a finitary function {f} iff {f\cdot B\subseteq B,} where {f\cdot B=\{f\}} if {f} is {0}-ary, and {f\cdot B=f[B^n]} if {f} is {n}-ary, {n>0.} If {S} is a set of finitary functions on {A,} we say that {B} is closed under {S} iff {f\cdot B\subseteq B} for all {f\in S.} The closure of {B} under {S} is the {\subseteq}-smallest {C\subseteq A} such that {B\subseteq C} and {C} is closed under {S.}

From the preliminaries above, it is clear that the following completes the proof of Theorem 2, since we can take as {B} the set {X} in item 2 (or rather, a superset of {X} of size exactly {\lambda}), as {\kappa} the cardinal {\lambda,} as {A} the set {M} and as {S} the set of (interpretations of) Skølem functions on {{\mathcal M}.}

Lemma 4 Suppose that {\kappa} is infinite, that {B\subseteq A,} {|B|\le\kappa,} and that {S} is a set of at most {\kappa} many finitary functions on {A.} Then the closure of {B} under {S} exists and has size at most {\kappa.}

Proof: Given {C\subseteq A,} write {S\cdot C} for {\bigcup\{f\cdot C\mid f\in S\}.} Set {B_0=B} and {B_{n+1}=B_n\cup S\cdot B_n} for all {n<\omega.} Finally, set {B_\omega=\bigcup_n B_n.} Then, by induction on {n,} each {B_n} is contained in any subset of {A} closed under {S} that contains {B.} Since {B_\omega} itself is closed under {S,} it follows that {B_\omega} is the closure of {B.} By induction, {|B_n|\le\kappa} for all {n,} and so the same holds for {B.} \Box

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Here is quiz 5.

Problem 1 asks for the partial fractions decomposition of {\displaystyle\frac1{x^4-x^2}.} To find this, first we factor the denominator:

{x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1).} This means that there must be constants {A,B,C,D} such that

\displaystyle \frac1{x^4-x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{x+1}.

To find these constants, we add the fractions on the right hand side, and obtain

\displaystyle \frac1{x^4-x^2}=\frac{Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)}{x^4-x^2}.

This means that

\displaystyle 1=Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1).

Setting {x=0} gives us

\displaystyle 1=-B\mbox{ or }B=-1.

Setting {x=1} gives us

\displaystyle 1=2C\mbox{ or }C=1/2.

Setting {x=-1} gives us

\displaystyle 1=-2D\mbox{ or }D=-1/2.

Setting {x=2} and using the values of {B,C,D} just found, gives us

\displaystyle 1=6A-3+6-2\mbox{ or }1=6A+1\mbox{ or }A=0.

Putting this together, the partial fractions decomposition is

\displaystyle \frac1{x^4-x^2}=-\frac1{x^2}+\frac{1/2}{x-1}-\frac{1/2}{x+1}.

Problem 2 asks to determine whether {\displaystyle\int_1^2\frac{dx}{x^4-x^2}} converges. Note that the expression we are integrating is defined in {(1,2],} but not at {x=1,} so this is an improper integral of Type II and to evaluate it we use the definition:

\displaystyle \int_1^2\frac{dx}{x^4-x^2}=\lim_{a\rightarrow1^+}\int_a^2\frac{dx}{x^4-x^2}.

To evaluate this last expression, we use the result from Problem 1:

\displaystyle \begin{array}{rl} \displaystyle \lim_{a\rightarrow1^+}\int_a^2\frac{dx}{x^4-x^2}&=\displaystyle\lim_{a\rightarrow1^+}\int_a^2-\frac1{x^2}+\frac{1/2}{x-1}-\frac{1/2}{x+1}\,dx\\ &=\displaystyle\lim_{a\rightarrow1^+}\left[\left.\frac1x+\frac12\ln(x-1)-\frac12\ln(x+1)\right]\right|_a^2\\ &=\displaystyle\lim_{a\rightarrow1^+}\left[\frac12+\frac12\ln(1)-\frac12\ln(3)\right]\\ &\hspace{1cm}\displaystyle-\left[\frac1a+\frac12\ln(a-1)-\frac12\ln(a+1)\right]. \end{array}

This expression diverges because {\displaystyle \lim_{a\rightarrow1^+}\frac1a-\frac12\ln(a+1)=1-\frac12\ln2,} but

\displaystyle \lim_{a\rightarrow1^+}\frac12\ln(a-1)=-\infty.

Problem 3 asks to determine whether {\displaystyle\int_2^\infty\frac{dx}{x^4-x^2}} converges. Note that the expression we are integrating is defined in {[2,\infty)} but, of course, the interval of integration is infinite, so this is an improper integral of Type I and to evaluate it we use the definition:

\displaystyle \int_2^\infty\frac{dx}{x^4-x^2}=\lim_{N\rightarrow\infty}\int_2^N\frac{dx}{x^4-x^2}.

To evaluate this last expression, we proceed as in Problem 2:

\displaystyle \begin{array}{rl} \displaystyle \lim_{N\rightarrow\infty}\int_2^N\frac{dx}{x^4-x^2}&=\displaystyle\lim_{N\rightarrow\infty}\left[\left.\frac1x+\frac12\ln(x-1)-\frac12\ln(x+1)\right]\right|_2^N\\ &=\displaystyle\lim_{N\rightarrow\infty}\left[\frac1N+\frac12\ln(N-1)-\frac12\ln(N+1)\right]\\ &\hspace{1cm}\displaystyle-\left[\frac12+\frac12\ln(1)-\frac12\ln(3)\right]. \end{array}

To evaluate the expression within the first set of parentheses, we use that

\displaystyle \frac12\ln(N-1)-\frac12\ln(N+1) \displaystyle =\ln\sqrt{N-1}+\ln\left(\frac1{\sqrt{N+1}}\right) \displaystyle=\ln\left(\sqrt{\frac{N-1}{N+1}}\right).

The limit of this expression as {N\rightarrow\infty} is {\ln\sqrt1=0,} because {\displaystyle\lim_{N\rightarrow\infty}\frac{N-1}{N+1}=\lim_{N\rightarrow\infty}\frac{1}{1}=1,} using l’Hôpital’s rule. This means that

\displaystyle \int_2^\infty\frac{dx}{x^4-x^2}=-\frac12+\frac12\ln3.

(In particular, the integral converges.)

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Leming Qu, Wed. November 11, 2:40-3:30 pm, MG 120.

Wavelet Image Restoration and Regularization Parameters Selection

For the restoration of an image based on its noisy distorted observations, we propose wavelet domain restoration by a scale-dependent L^1 penalized regularization method (WaveRSL1). The data-adaptive choice of the regularization parameters is based on the Akaike Information Criterion (AIC) and the degrees of freedom (df) are estimated by the number of nonzero elements in the solution. Experiments on some commonly used testing images illustrate that the proposed method possesses good empirical properties.

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Here is the second midterm.

Question 1: Find the volume of the solid obtained by rotating about the {y}-axis the region between the {x}-axis and the curve {y=\sec^2x} for {0\le x\le\pi/4}.

Here is a graph of {y=\sec^2x} in the specified region. To compute the volume, we use the shell method, and see that it is given by

\displaystyle V=\int_0^{\pi/4} 2\pi x\cdot\sec^2x\,dx.

We evaluate this integral using integration by parts.

We take: \displaystyle u=2\pi x,\mbox{ so }du=2\pi\,dx, and dv=\sec^2x\,dx, so v=\tan x; so \displaystyle V=\left.2\pi x\tan x\right|_0^{\pi/4}-\int_0^{\pi/4}2\pi\tan x\,dx

\displaystyle=\left.2\pi x\tan x+2\pi\ln(\cos x)\right|_0^{\pi/4}

\displaystyle=\left(2\pi\cdot\frac{\pi}4\cdot1+2\pi\ln\left(\frac1{\sqrt2}\right)\right)-(0+2\pi\ln(1))

\displaystyle=\frac{\pi^2}2-\pi\ln2.

Question 2: Solve the initial value problem {\displaystyle (t^2+3t+3)\frac{dx}{dt}=1}, {x(3)=0}.

We use the technique of separation of variables, and rewrite the given differential equation as:

\displaystyle dx=\frac{dt}{t^2+3t+3},

so

\displaystyle x=\int \frac{dt}{t^2+3t+3}.

To solve this integral, we begin by completing the square in the denominator of the given fraction:

\displaystyle t^2+3t+3=\left(t+\frac32\right)^2+\left(3-\frac94\right)=\left(t+\frac32\right)^2+\frac34.

This indicates that to solve the integral we want to use the trigonometric substitution

\displaystyle t+\frac32=\frac{\sqrt3}2\tan\theta,

or

\displaystyle \theta=\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right).

This gives: \displaystyle \left(t+\frac32\right)^2+\frac34=\frac34\sec^2\theta,

\displaystyle dt=\frac{\sqrt3}2\sec^2\theta\,d\theta,\mbox{ and}

\displaystyle \int \frac{dt}{t^2+3t+3}=\int\frac{(\sqrt3/2)\,d\theta}{3/4}

\displaystyle=\frac2{\sqrt3}\theta+C

\displaystyle=\frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right)+C. 

We conclude by finding the value of {C}, for which we use that {x(3)=0.} This gives us

\displaystyle 0=x(3)= \frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}3+\sqrt3\right)+C,

or

\displaystyle C=-\frac2{\sqrt3}\tan^{-1}(3\sqrt3).

Putting this together, we have

\displaystyle x(t)= \frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right)-\frac2{\sqrt3}\tan^{-1}(3\sqrt3).

Question 3: Find {\displaystyle \int\frac{x^3+5}{x^4-x^2}\,dx}.

To solve this integral, we use the technique of partial fractions decomposition, and begin by factoring

\displaystyle x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1).

This means we need to find constants {A,B,C,} and {D} such that

\displaystyle \frac{x^3+5}{x^4-x^2}=\frac Ax+\frac B{x^2}+\frac C{x-1}+\frac D{x+1}.

As seen in class, there are several ways of doing this. For example, we can begin by adding the fractions on the right hand side, to obtain

\displaystyle \frac{Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)}{x^4-x^2}.

This means that

\displaystyle x^3+5=Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)

for all values of {x}. When {x=0,} this gives

\displaystyle 5=-B,\mbox{ or }B=-5.

When {x=1}, we have

\displaystyle 6=2C,\mbox{ or }C=3.

When {x=-1}, we have

\displaystyle 4=-2D,\mbox{ or }D=-2.

Finally, when {x=2} we have

\displaystyle 13=6A-5\cdot3+3\cdot12-2\cdot4,

or {A=0.}

Putting this together, we have

\displaystyle \frac{x^3+5}{x^4-x^2}=-\frac 5{x^2}+\frac 3{x-1}-\frac 2{x+1},

and

\displaystyle \int\frac{x^3+5}{x^4-x^2}\,dx=\frac5x+3\ln|x-1|-2\ln|x+1|+C.

[Actually, although this was not addressed in lecture, we can use different values of {C} in each of the intervals {(-\infty,-1),} {(-1,0),} {(0,1),} and {(1,\infty)}.]

Question 4: We have that {\displaystyle\int_0^1\frac4{1+t^2}\,dt=\pi}. Suppose we do not know the value of {\pi}. We can use the trapezoidal rule to approximate it. Find a value of {n} such that the theoretical error for the trapezoidal rule using {n} subdivisions is strictly smaller than {1/10}. Find the approximation given by the trapezoidal rule using this {n}.

The error {E_T} using the trapezoidal rule with {n} subintervals is bounded above by

\displaystyle \frac{M_2(b-a)^3}{12n^2},

where {a=0,} {b=1,} and {M_2=\max_{0\le t\le 1}|f''(t)|} where {\displaystyle f(t)=\frac4{1+t^2}.}

To compute a bound on {M_2,} we first find {f''(t):} \displaystyle f'(t)=\frac{-8t}{(1+t^2)^2},

\displaystyle f''(t)=\frac{32t^2-8(1+t^2)}{(1+t^2)^3}

\displaystyle =\frac{24t^2-8}{(1+t^2)^3}.

An easy way of bounding this expression is to note that {24t^2-8} is increasing for {0\le t\le 1,} with value {-8} at {t=0,} and value {16} for {t=1,} so the maximum of its absolute value is {16.} Similarly, {\displaystyle\frac1{(1+t^2)^3}} is decreasing, so it maximum occurs at {t=0}, where it is 1.

Hence {M_2} is bounded above by {\displaystyle \frac{16}1=16,} and

\displaystyle E_T\le \frac{16}{12n^2}=\frac4{3n^2}.

If we want {E_t<1/10,} it suffices that

\displaystyle \frac4{3n^2}<\frac1{10},

or

\displaystyle n^2>40/3=13.33\dots,

so {n\ge4.}

The approximation {T_4} is given by

\displaystyle \frac {1/4}2\left(4+\frac8{1+\frac1{16}}+\frac8{1+\frac4{16}}+\frac8{1+\frac9{16}}+2\right)\approx3.131

which is indeed an accurate approximation to {\pi} within {1/10.}

[Here is a graph of {f''(t)} for {0\le t\le 1.} A more careful computation of {M_2} would have revealed that {M_2=8,} which gives {n\ge3.} (Note that {\max_{0\le t\le 1} f''(t)=2,} but {M_2} looks at the maximum of {|f''(t)|,} not just {f''(t)}.)

The approximation {T_3} gives the value {3.123\dots}. A less careful computation, replacing {24t^2-8} with {24t^2+8} (so we do not have to worry about both positive and negative numbers) gives {n\ge6}, with {T_6\approx3.137.} Actually, {T_2=3.1} works as well but the theoretical error does not predict this.]

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Jodi Mead, Wed. November 4, 2:40-3:30 pm, MG 120.

Non-smooth Solutions to Least Squares Problems

In an attempt to overcome the ill-posedness or ill-conditioning of inverse problems, regularization methods are implemented by introducing assumptions on the solution.  Common regularization methods include total variation, L-curve, Generalized Cross Validation (GCV), and the discrepancy principle. It is generally accepted that all of these approaches except total variation unnecessarily smooth solutions, mainly because the regularization operator is in L^2. Alternatively, statistical approaches to ill-posed problems typically involve specifying a priori information about the parameters in the form of Bayesian inference. These approaches can be more accurate than typical regularization methods because the regularization term is weighted with a matrix rather than a constant. The drawback is that the matrix weight requires information that is typically not available or is expensive to calculate.

The \chi^2 method developed by the author and colleagues can be viewed as a regularization method that uses statistical information to find matrices to weight the regularization term.  We will demonstrate that unique and simple L^2 solutions found by this method do not unnecessarily smooth solutions when the regularization term is accurately weighted with a diagonal matrix.

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Two homework problems. The first one is easier, so you can consider the second one to be extra credit. A proof of these results can be found in different places, for example, the paper Division by three, by Conway and Doyle. (Please don’t look at the paper while working on the homework, of course.) Unfortunately, the paper could use a serious trimming and editing, so I cannot really recommend it, but the proof is carefully written there.

  1. Without using the axiom of choice, show that if A and B are sets, and |A\times 2|=|B\times 2|, then |A|=|B|.
  2. Same as 1., but now with 3 instead of 2. 
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Jens Harlander, Wed. October 28, 2:40-3:30 pm, MG 120.

Introduction to Computational Complexity

Complexity theory provides ways of measuring the difficulty of computational mathematics problems. Some problems are indeed impossibly difficult (your Math 108 and 143 students are right after all!). For example, there does not exist an algorithm that decides whether a polynomial (in an arbitrary number of variables) with integer coefficients has integer roots. However for many difficult problems, simple strategies work well in practice as long as one is willing to ignore a hopefully sparse set of inputs. I will discuss basic features of the theory, give you more examples of impossibly hard problems and tell you about the relevance of all of this to Internet security.

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Here is the list of speakers for the rest of the term. 

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Here is quiz 4.

Problem 1 is Exercise 7.1.36 from the book. Here is a graph showing the curve {y=x\sin x} for {0\le x\le\pi.}

We rotate about the line {x=\pi} the region bounded by the {x}-axis and this curve.

To find its volume, a first natural attempt would be to use the washer method. We would then attempt to compute the volume as

\displaystyle \int_0^{Max}\pi((\pi-x_1)^2-(\pi-x_2)^2)\,dy,

where {Max} is the maximum of {y} for {0\le x\le \pi} and, for any given value of {y} with {0\le y\le Max,} {x_1} and {x_2} are the values of {x} with {0\le x_1<x_2\le\pi} such that {x\sin x=y.}

Unfortunately, this does not seem to be the best approach, as there does not seem to be a reasonable way of solving for {x} the equation {x\sin x=y.}

(In fact, this equation cannot be solved in terms of elementary functions.

Similarly, there is no way of finding exactly what the value of {Max} is in terms of elementary functions.)

Since this approach seems to lead us nowhere, we now try to compute the volume using the shell method. Now the volume is expressed as

\displaystyle \int_0^\pi 2\pi(\pi-x)x\sin x\,dx.

This expression looks approachable with the techniques we have studied. First, let’s rewrite the integral as

\displaystyle 2\pi \int_0^\pi\pi x\sin x\,dx-2\pi\int_0^\pi x^2\sin x\,dx.

We compute both expressions using integration by parts:

To find {\displaystyle \int x\sin x\,dx,} we use {u=x} and {dv=\sin x\,dx,} so {du=dx} and we can take {v=-\cos x.} Hence

\displaystyle \int_0^\pi\pi x\sin x=-\pi x\cos x\vert_0^\pi+\int_0^\pi \pi\cos x\,dx.

We recognize from the graph of {\cos x} that the second expression is zero, and we have:

\displaystyle \int_0^\pi\pi x\sin x=-\pi (-\pi-0)=\pi^2.

Similarly, for {\displaystyle \int x^2\sin x\,dx} we have {u=x^2} and {dv=\sin x\,dx,} so {du=2x\,dx} and {v=-\cos x,} and

\displaystyle \int_0^\pi x^2\sin x=-x^2\cos x\vert_0^\pi+\int_0^\pi 2x\cos x\,dx \displaystyle=\pi^2+2\int_0^\pi x\cos x\,dx.

The last expression {\displaystyle \int x\cos x\,dx} is once more computed using parts, now with {u=x} and {dv=\cos x\,dx,} so {du=dx} and {v=\sin x.} This gives

\displaystyle \int_0^\pi x\cos x\,dx=x\sin x\vert_0^\pi-\int_0^\pi\sin x\,dx=0+\cos x\vert_0^\pi=-2.

Hence

\displaystyle \int_0^\pi x^2\sin x\,dx=\pi^2-4.

Finally, the required volume is

\displaystyle 2\pi(\pi^2)-2\pi(\pi^2-4)=8\pi.

Problem 2 asked to evaluate {\displaystyle \int_0^3\sqrt{x^2+6x}\,dx.}

A first attempt may go by using integration by parts, with {u=\sqrt x} and {dv=\sqrt{x+6}\,dx.} Unfortunately, this approach would not lead to simpler expressions, as both integrals and derivatives of {\sqrt x} and {\sqrt{x+6}} carry radicals.

If the expression inside the square root were of the form {x^2+a^2} or {x^2-a^2} we could use a trigonometric substitution. However, {x^2+6x} is not of this form. On the other hand, in Chapter 9 we saw that it is sometimes useful to complete squares, so we may want to try that here. We have:

\displaystyle x^2+6x=x^2+6x+9-9=(x+3)^2-9.

This suggest trying the trigonometric substitution {x+3=3\sec\theta} for {0\le\theta<\pi/2.} We have {dx=3\sec\theta\tan\theta\,d\theta} and{\sqrt{x^2+6x}=3\tan\theta.} Also, when {x=0} we have {\sec\theta=1,} or {\theta=0,} and when {x=3} we have {\sec\theta=2,} or {\theta=\pi/3.}

In terms of {\theta,} the integral becomes

\displaystyle \int_0^{\pi/3}3\tan\theta\,3\sec\theta\tan\theta\,d\theta=9\int_0^{\pi/3}\sec\theta\tan^2\theta\,d\theta.

To evaluate expressions of this form, we use the identity {\tan^2\theta=\sec^2\theta-1,} and obtain

\displaystyle \int\sec\theta\tan^2\theta\,d\theta=\int\sec^3\theta\,d\theta-\int\sec\theta\,d\theta.

The second expression we recognize as {-\ln|\sec\theta+\tan\theta|+ C_1.} For the first, we use either the reduction formula found in lecture, or integration by parts:

\displaystyle \int\sec^3\theta\,d\theta=\int\sec\theta\sec^2\theta\,d\theta \displaystyle=\sec\theta\tan\theta-\int\tan\theta\sec\theta\tan\theta\,d\theta,

or {\displaystyle \int\sec^3\theta\,d\theta=}

\displaystyle \sec\theta\tan\theta-\int(\sec^2\theta-1)\sec\theta\,d\theta \displaystyle=\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|-\int\sec^3\theta\,d\theta,

from which we get

\displaystyle \int\sec^3\,d\theta=\frac{\sec\theta\tan\theta}2+\frac12\ln|\sec\theta+\tan\theta| + C_2.

Finally,

\displaystyle 9\int_0^{\pi/3}\sec\theta\tan^2\theta\,d\theta \displaystyle=9\left(\left(\frac{2\sqrt3}2-\frac12\ln(2+\sqrt3)\right)-\left(\frac02-\frac12\ln(1+0)\right)\right),

so the required integral equals {\displaystyle 9\sqrt3-\frac{9}2\ln(2+\sqrt3).}

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