@@ -65,6 +65,7 @@ clean: # ... done. rm -f calc_lulu.pdf rm -f calc.pdf + rm -f temp.pdf post: cp calc.pdf /home/bcrowell/Lightandmatter/calc Makefile @@ -1,5 +1,5 @@ %%chapter%% 01 -\chapter{Rates of Change} +\chapter{Rates of Change}\label{ch:rates-of-change} \section{Change in discrete steps} ch01/ch01.tex @@ -815,7 +815,7 @@ computation, as in the bogus result generated on line 9 of the example. \section{Continuity} Intuitively, a continuous function is one whose graph -has no sudden jumps in it; the graph is all a single connected piece. Formally, a function real function $f(x)$ is defined to +has no sudden jumps in it; the graph is all a single connected piece. Formally, a function $f(x)$ is defined to be continuous if for any real $x$ and any infinitesimal $\der x$, $f(x+\der x)-f(x)$ is infinitesimal.\label{def-continuity}\index{continuous function} \begin{eg}\label{eg:discontinuous} @@ -866,6 +866,46 @@ the properties of the underlying number system. For the reader with a interest i mathematics, I've discussed this in more detail on page \pageref{detour:intermediate-value} and given an abbreviated proof.\label{intermediate-value-ref-to-detour} +\begin{eg} +\egquestion Prove that every odd-order polynomial $P$ with real coefficients has at least one real root $x$, i.e., a +point at which $P(x)=0$. + +\eganswer +To see that the restriction to odd orders is necessary, consider the polynomial $x^2+1$, which has no real roots +because $x^2>0$ for any real number $x$. + +To fix our minds on a concrete example for the odd case, consider the polynomial $P(x)=x^3-x+17$. +For large values of $x$, the linear and constant terms will be negligible compared to the $x^3$ term, +and since $x^3$ is positive for large values of $x$ and negative for large negative ones, it follows +that $P$ is sometimes positive and sometimes negative. + +Making this argument more general and rigorous, +suppose we had a polynomial of odd order $n$ that always had the same sign for real $x$. Then by the transfer principle +the same would hold for any hyperreal value of $x$. Now if $x$ is infinite then the lower-order terms +are infinitesimal compared to the $x^n$ term, and the sign of the result is determined entirely by the +$x^n$ term, but $x^n$ and $(-x)^n$ have opposite signs, and therefore $P(x)$ and $P(-x)$ have opposite signs. +This is a contradiction, so we have disproved the assumption that $P$ always had the same sign for real $x$. +Since $P$ is sometimes negative and sometimes positive, we conclude by the intermediate value theorem that +it is zero somewhere. +\end{eg} + +In chapter \ref{ch:rates-of-change}, we saw that locating maxima and minima of functions may in general +be fairly difficult, because there are so many different ways in which a function can attain an extremum: +e.g., at an endpoint, at a place where its derivative is zero, or at a nondifferentiable point. The following +theorem allows us to make a very general statement about all these possible cases, assuming only continuity. + +The \emph{extreme value theorem}\label{extreme-value-theorem} +states that if $f$ is a continuous real-valued function on the real-number +interval defined by $a \le x \le b$, then $f$ has maximum and minimum values on that interval, which are +attained at specific points in the interval.\index{extreme value theorem} + +Let's first see why the assumptions are necessary. If we weren't combined to a finite interval, then +$y=x$ would be a counterexample, because it's continuous and doesn't have any maximum or minimum value. +If we didn't assume continuity, then we could have a function defined as $y=x$ for $x < 1$, and $y=0$ for +$x \ge 1$; this function never gets bigger than 1, but it never attains a value of 1 for any specific value of $x$. + +The extreme value theorem is proved, and generalized somewhat, on page \pageref{detour:extreme-value}. + \section{Limits}\label{sec:limits} Historically, the calculus of infinitesimals as created by Newton and Leibniz was reinterpreted ch02/ch02.tex @@ -314,6 +314,102 @@ also true if we take $y=x^2$ as the definition of a function on the hyperreals. in this way, the intermediate value theorem makes a statement that the transfer principle applies to, and it is therefore true for the hyperreal version of the function as well. +\detour{extreme-value}{Proof of the extreme value theorem}\index{extreme value theorem!proof} + +The extreme value theorem was stated on page \pageref{extreme-value-theorem}. Before we can prove +it, we need to establish some preliminaries, which turn out to be interesting for their own sake. + +Definition: Let $C$ be a subset of the real numbers whose definition can be expressed +in the type of language to which the transfer principle applies. Then $C$ is \emph{compact}\index{compact set} +if for every hyperreal number $x$ satisfying the definition of $C$, the standard part of $x$ exists +and is a member of $C$. + +To understand the content of this definition, we need to look at the two ways in which a set could +fail to satisfy it. + +First, suppose $U$ is defined by $x \ge 0$. Then there are positive infinite +hyperreal numbers that satisfy the definition, and their standard part is not defined, so $U$ is not +compact. The reason $U$ is not compact is that it is unbounded. + +Second, let $V$ be defined by $0 \le x < 1$. Then if $dx$ is a positive infinitesimal, +$1-dx$ satisfies the definition of $V$, but its standard part is 1, which is not in $V$, so $V$ +is not compact. The set $V$ has boundary points at 0 and 1, and the reason it is not compact +is that it doesn't contain its right-hand boundary point. A boundary point\index{boundary point} +is a real number which is infinitesimally close to some points inside the set, and also to some +other points that are on the outside. + +We therefore arrive at the following alternative characterization of the notion of a compact +set, whose proof is straightforward. + +Theorem: A set is compact if and only if it is bounded and contains all of its boundary points. + +Intuitively, the reason compact sets are interesting is that if you're standing inside a compact +set and start taking steps in a certain direction, without ever turning around, you're guaranteed to +approach some point in the set as a limit. (You might step over some gaps that aren't included in the set.) +If the set was unbounded, you could just walk forever at a constant speed. +If the set didn't contain its boundary point, then you could asymptotically approach the boundary, but +the goal you were approaching wouldn't be a member of the set. In other words, every increasing +sequence of numbers within a compact set approaches a limit within that set, and similarly for every +decreasing sequence. + +The following theorem turns out to be the most difficult part of the discussion. + +Theorem: A compact set contains its maximum and minimum.\\ +Proof: Let $C$ be a compact set. We know it's bounded, so let $M$ be the set of all real numbers +that are greater than any member of $C$. By the completeness property of the real +numbers, there is some real number $x$ between $C$ and $M$. Let $^{*}C$ be the set of hyperreal numbers that satisfies the same +definition that $C$ does. + +Every real $x'$ greater than $x$ fails +to satisfy the condition that defines $C$, and by the transfer principle the same must be true if $x'$ is any hyperreal, +so if $dx$ is a positive infinitesimal, $x+dx$ must be outside of $^{*}C$. + +But now consider $x-dx$. +The following statement holds for the reals: there is no number $x'<x$ that is greater than every +member of $C$. By the transfer principle, we find that there is some hyperreal number $q$ in $^{*}C$ +that is greater than $x-dx$. But the standard part of $q$ must equal $x$, for otherwise $\zu{st} q$ would +be a member of $C$ that was greater than $x$. +Therefore $x$ is a boundary point of $C$, and since +$C$ is compact, $x$ is a member of $C$. We conclude $C$ contains its maximum. A similar argument shows that +$C$ contains its minimum, so the theorem is proved. + +There were two subtle things about this proof. The first was that we ended up constructing the +set of hyperreals $^{*}C$, which was the hyperreal ``big brother'' of the real set $C$. This +is exactly the sort of thing that the transfer principle does \emph{not} guarantee we can do. +However, if you look back through the proof, you can see that $^{*}C$ is used only as a notational +convenience. Rather than talking about whether a certain number was a member of $^{*}C$, we could +have referred, more cumbersomely, to whether or not it satisfied the condition that had originally +been used to define $C$. The price we paid for this was a slight loss of generality. There are so +many different sets of real numbers that they can't possibly all have explicit definitions that can +be written down on a piece of paper. However, there is very little reason to be interested in +studying the properties of a set that we were never able to define in the first place. The +other subtlety was that we had to construct the auxiliary +point $x-dx$, but there was not much we could actually say about $x-dx$ itself. In particular, it +might or might not have been a member of $C$. For example, if $C$ is defined by the condition $x=0$, then +$^{*}C$ likewise contains only the single element 0, and $x-dx$ is not a member of $^{*}C$. +But if $C$ is defined by $0 \le x \le 1$, then $x-dx$ is a member of $^{*}C$. + +The original goal was to prove the extreme value theorem, which is a statement about continuous functions, but so far we haven't said anything about functions. + +Lemma: Let $f$ be a real function defined on a set of points $C$. Let $D$ be the image of $C$, i.e., +the set of all values $f(x)$ that occur for some $x$ in $C$. Then if $f$ is continous and $C$ is +compact, $D$ is compact as well. In other words, continuous functions take compact sets to compact sets.\\ +Proof: Let $y=f(x)$ be any hyperreal output corresponding to a hyperreal input $x$ in $^{*}C$. +We need to prove that the standard part of $y$ exists, and is a member of $D$. Since $C$ is compact, +the standard part of $x$ exists and is a member of $C$. But then by continuity $y$ differs only +infinitesimally from $f(\zu{st} x)$, which is real, so $\zu{st} y=f(\zu{st} x)$ is defined and is a member +of $D$. + +We are now ready to prove the extreme value theorem, in a version slightly more general than the one +originally given on page \pageref{extreme-value-theorem}. + +The extreme value theorem: Any continuous function on a compact set achieves a maximum and minimum value, +and does so at specific points in the set. + +Proof: Let $f$ be continuous, and let $C$ be the compact set on which we seek its maximum and minimum. +Then the image $D$ as defined in the lemma above is compact. Therefore $D$ contains its maximum and +minimum values. + \detour{mean-value-proof}{Proof of the mean value theorem}\index{mean value theorem!proof} Suppose that the mean value theorem is violated. Let $L$ be the set of all $x$ in the interval from $a$ to $b$ @@ -324,9 +420,17 @@ the definition of the definite integral that when one function is less than anot than the other's. Since $y$ takes on values less than and greater than $\bar{y}$, it follows from the intermediate value theorem that $y$ takes on the value $\bar{y}$ somewhere (intuitively, at a boundary between $L$ and $M$). +\pagebreak + \detour{fn-thm-alg-proof}{Proof of the fundamental theorem of algebra}\index{fundamental theorem of algebra!proof} -Theorem: In the complex number system, an nth-order polynomial has exactly $n$ +We start with the following lemma, which is intuitively obvious, because polynomials don't have asymptotes. +Its proof is given after the proof of the main theorem. + +Lemma: For any polynomial $P(z)$ in the complex plane, its magnitude $|P(z)|$ achieves its minimum +value at some specific point $z_\zu{o}$. + +The fundamental theorem of algebra: In the complex number system, a nonzero nth-order polynomial has exactly $n$ roots, i.e., it can be factored into the form $P(z)=(z-a_1)(z-a_2)\ldots(z-a_n)$, where the $a_i$ are complex numbers. @@ -338,9 +442,8 @@ we can show that every polynomial of degree greater than zero has at least one root. Suppose, on the contrary, that there is an nth order polynomial $P(z)$, with $n>0$, that has -no roots at all. Then $|P(z)|$ must have some minimum value, which is achieved -at $z=z_\zu{o}$. (Polynomials don't have asymptotes, so the minimum really does -have to occur for some specific, finite $z_\zu{o}$.) To make things more +no roots at all. Then by the lemma $|P|$ achieves its minimum value +at some point $z_\zu{o}$. To make things more simple and concrete, we can construct another polynomial $Q(z)=P(z+z_\zu{o})/P(z_\zu{o})$, so that $|Q|$ has a minimum value of 1, achieved at $Q(0)=1$. This means that $Q$'s constant term is 1. What about its other terms? Let $Q(z)=1+c_1z+\ldots+c_nz^n$. @@ -354,3 +457,17 @@ less than one for appropriately chosen values of $z$. Continuing this process, w find that $Q(z)$ has no terms at all beyond the constant term, i.e., $Q(z)=1$. This contradicts the assumption that $n$ was greater than zero, so we've proved by contradiction that there is no $P$ with the properties claimed. + +Uninteresting proof of the lemma: Let $M(r)$ be the minimum value of $|P(z)|$ on the disk defined by $|z| \le r$. +We first prove that $M(r)$ can't asymptotically approach a minimum as $r$ approaches infinity. +Suppose to the contrary: for every $r$, there is some $r'>r$ with $M(r')<M(r)$. +Then by the transfer principle, the same would have to be true for +hyperreal values of $r$. But it's clear that if $r$ is infinite, the lower-order terms of $P$ +will be infinitesimally small compared to the highest-order term, and therefore $M(r)$ is +infinite for infinite values of $r$, which is a contradiction, since by construction $M$ is +decreasing, and finite for finite $r$. We can therefore conclude +by the extreme value theorem that $M$ achieves its minimum for some specific value of $r$. +The least such $r$ describes a circle $|z|=r$ in the complex plane, and the minimum of $|P|$ +on this circle must be the same as its global minimum. Applying the extreme value function +to $|P(z)|$ as a function of arg $z$ on the interval $0 \le \zu{arg} z \le 2\pi$, we establish +the desired result. ch99/detours.tex f17af1959837e2faa523f8f9ba8cb04e05ebcc01 Ben Crowell githubcrowell09@lightandmatter.com http://github.com/bcrowell/calculus/commit/2af053c0131e12df44ddd806d9aebe05bbe471f0 2af053c0131e12df44ddd806d9aebe05bbe471f0 2009-03-12T23:17:37-07:00 2009-03-12T23:17:37-07:00 extreme value theorem 93d762c195e286d176b0684861a2121f5954e1fc Ben Crowell githubcrowell09@lightandmatter.com