## 502 – Completeness

September 25, 2009

1. Preliminaries

Just as with propositional logic, one can define a proof system for first order logic. We first need two definitions.

Definition 1 If ${\phi}$ is a formula, a generalization of ${\phi}$ is any formula of the form

$\displaystyle \forall x_1\forall x_2\dots\forall x_n\,\phi.$

(We are not requiring in the definition above that the ${x_i}$ are different or that they are present in ${\phi.}$)

Definition 2 Let ${t}$ be a term, let ${x}$ be a variable, and let ${\phi}$ be a formula. We say that ${t}$ is substitutable for ${x}$ in ${\phi}$ iff no free occurrence of ${x}$ in ${\phi}$ is within the scope of a quantifier ${\exists z}$ or ${\forall z}$ with ${z}$ a variable occurring in ${t.}$

## 175 – Midterm 1

September 25, 2009

Here is the midterm.

1. Hooke’s law says that the force required to stretch a spring a distance of $x$ units from its natural length is given by $F(x)=kx,$ where $k$ is a constant that only depends on the spring.

We are told that for the spring of problem 1 we have $F(2)=20,$ where units of distance are measured in meters, and forces in Newtons. This means that $k2=20,$ or $k=10.$ The work required to stretch the spring 10 m beyond its natural length is then $\displaystyle \int_0^{10}10x\,dx=10x^2/2|^{10}_0=500 \,N\cdot m.$

2. We are given a curve with equation $x=\sqrt{2y-1}$ and asked to rotate about the $y$-axis the segment where $1/2\le y\le 2.$ To find the area of the resulting surface, we can consider splitting the surface into little shells. A typical `thin’ shell, at height $y$ has surface  area $2\pi r\,ds$ where $r$ is the radius of the shell, in this case just the value of $x$ corresponding to $y,$ i.e., $\sqrt{2y-1}.$ Also, $ds$ is here the arc length element, given by $\sqrt{(x')^2+1}\,dy.$

We have: $\displaystyle x'=\frac12(2y-1)^{-1/2}2=\frac1{\sqrt{2y-1}},$ so $\displaystyle (x')^2+1=\frac1{2y-1}+1=\frac{2y}{2y-1}.$

Hence the requested area is given by $\displaystyle\int_{1/2}^2 2\pi\sqrt{2y-1}\sqrt{\frac{2y}{2y-1}}\,dy=\int_{1/2}^2 2\pi\sqrt{2y}\,dy.$

This expression reduces to $\displaystyle 2\pi\sqrt2\,\frac23\,y^{3/2}|^2_{1/2}=2\pi\sqrt2\frac23\left(2\sqrt2-\frac1{2\sqrt2}\right)=\frac{14\pi}3.$

3. We are given the region bounded by the curve $y=\sqrt x$ and the lines $y=2$ and $x=0.$ We rotate it about the $x$-axis, and are asked to find the volume of the resulting solid. A first attempt is to use the washer method. A typical thin washer consists of a slice of the solid for a fixed value of $x$ and with thickness $dx.$ Its volume is $\pi (r_1^2-r_2^2)\,dx.$ Here, $r_1$ is the exterior radius, which in this case is always 2, and $r_2$ is the interior radius, given by $y=\sqrt x.$ This means the washer contributes $\pi(4-x)\,dx$ of the full volume, and adding all of them together we obtain $\int_0^4 \pi(4-x)\,dx.$ The upper limit of 4 comes from observing that the curves $y=2$ and $y=\sqrt x$ cross precisely at the point $(4,2).$

Unfortunately, this integral is not one of the given expressions.

So we attempt a different approach, using now the shell method. Now a typical thin shell is obtained by fixing a height $y$ and rotating about the $x$-axis the slice of the figure of thickness $dy$ and height $y.$ Its volume is $2\pi r\,l\,dy,$ where $r$ is the radius of the shell, in this case $y;$ and $l$ is the length of the shell, in this case $x.$ We thus proceed to express $x$ in terms of $y:$ Since $y=\sqrt x,$ then $x=y^2.$ Hence the thin shell contributes $2\pi y\,y^2\,dy=2\pi y^3\,dy$ of the full volume, and the volume is obtained by adding all these contributions, obtaining $\int_0^2 2\pi y^3\,dy.$ This is expression (c).

4. The length of the curve $y=\sqrt{1-x^2}$ for $-1\le x\le 1$ is given by the integral $\displaystyle \int_{-1}^1\sqrt{1+(y')^2}\,dx.$ In this case, $\displaystyle y'=\frac12\,\frac{-2x}{\sqrt{1-x^2}}=-\frac x{\sqrt{1-x^2}},$ so $\displaystyle (y')^2=\frac{x^2}{1-x^2},$ and $\displaystyle 1+(y')^2=\frac1{1-x^2},$ so the integral reduces to $\displaystyle\int_{-1}^1\frac 1{\sqrt{1-x^2}}\,dx,$ which is expression (b).

[There is a small problem with this integral, though, because the denominator vanishes at both endpoints. Later, in Chapter 7, we will learn how to handle integrals of this kind. Note that if the question had been to actually find the length, there is an easier method: The curve is half a circumference of radius 1, so the length is $\pi.$]

5. (a) The differential equation $\displaystyle \frac{dy}{dx}=\frac{e^{3x+4y}}{e^{x^2-7y}}$ can be rewritten as $\displaystyle\frac{dy}{dx}=e^{3x-x^2}e^{11 y},$ which is clearly separable. (T)

(b) The area swept by $r=2f(\theta)$ for $\alpha\le\theta\le\beta,$ is four times the area swept by $r=f(\theta).$ Intuitively, we are making lengths twice as long, and areas carry two dimensions of length. Think of the area of a square of side 2, for example. Formally, we can check that this is the case: The area swept by the first curve is given by $\displaystyle\int_\alpha^\beta\frac12 (2f(\theta))^2\,d\theta=4\int_\alpha^\beta\frac12(f(\theta))^2\,d\theta,$ while the area swept by the second curve is given by $\displaystyle\int_\alpha^\beta\frac12 (f(\theta))^2\,d\theta.$ (T)

(c) The half life of a given element is 3 days. This means that 50% of the given amount of the element has decayed after 3 days. After 3 more days, another 25% would have decayed. After 3 more days, another 12.5%. After 3 more days, another 6.25%. Since $50+25+12.5+6.25>90,$ the given statement is clearly true.

A longer way of checking the same is by recalling that the total amount of the element that remains after time $t$ if originally the amount is $y_0$ is given by $y=y_0e^{-kt},$ where $k$ is a given constant. Measure $t$ in days. We are told that $y(3)=y_0/2,$ so $e^{3k}=2.$ Then $e^{27 k}=(e^{3k})^9=2^9=512,$ and $y(27)=y_0/512 (T)

## 502 – Compactness

September 25, 2009

First, two exercises to work some with the notion of ultrapower: Check that $|\prod_n M_n/{\mathcal U}|=|{\mathbb R}|$ whenever ${\mathcal U}$ is a nonprincipal ultrafilter on the natural numbers, and

1. $M_n={\mathbb N}$ for all $n,$ or
2. $\lim_{n\to\infty}|M_n|=\infty.$

Our argument for compactness required the existence of nonprincipal ultrafilters. One might wonder whether this is a necessity or just an artifact of the proof. It is actually necessary. To see this, I will in fact show the following result as a corollary of compactness:

Theorem.  If ${\mathcal F}$ is a nonprincipal filter on a set $I,$ then there is a nonprincipal ultrafilter on $I$ that extends ${\mathcal F}.$

(Of course, this is a consequence of Zorn’s lemma. The point is that all we need is the compactness theorem.)

Proof. Consider the language ${\mathcal L}=\{\hat X\mid X\subseteq I\}\cup\{c,\hat\in\}.$ Here, each $\hat X$ is a constant symbol, $c$ is another constant symbol, and $\hat\in$ is a symbol for a binary relation (which we will interpret below as membership).

In this language, consider the theory $\Sigma=\{c\hat\in\hat X\mid X\in {\mathcal F}\}\cup{\rm Th}(I\cup{\mathcal P}(I),\in,X\mid X\subseteq I).$ A model $M$ of this theory $\Sigma$ would look a lot like $I\cup{\mathcal P}(I),$ except that the natural interpretation of ${\mathcal F}$ in $M,$ namely, $\{\hat X^M\mid X\in{\mathcal F}\}$ is no longer nonprincipal in $M$, because $c^M$ is a common element of all these sets.

Note that there are indeed models $M$ of $\Sigma,$ thanks to the compactness theorem.

If $M\models\Sigma,$ let ${\mathcal U}=\{X\subseteq I\mid M\models c\hat\in \hat X\},$ and note that ${\mathcal U}$ is a nonprincipal ultrafilter over $I$ that contains ${\mathcal F}.$ $\Box$