Date  What happened  

4/14/2003  The review session on Tuesday evening, April 15, will be in
Hill 425 at 6:10 PM.
I began by doing problem #2 in section 5.4 Suppose
f(x)=1/x^{2}. Show that f is uniformly continuous on
A=[1,infinity) and is not uniformly continuous on B=(0,infinity).
We have now the following examples:
These examples might seem to support some sort of conjecture that if
the range is "big" (unbounded) then a continuous function is not
uniformly continuous, while if the range is "small" (bounded) then a
continuous function is uniformly continuous. Neither of those
statements is correct.
Theorem (Continuous functions on closed, bounded intervals are
uniformly continuous) Suppose f:[a,b]>R is
continuous. Then f is uniformly continuous.
If we knew more about calculus, then we could assert the following: if f is differentiable on the interval [a,b], and if f'(w)<=K for all w in [a,b], then the Mean Value Theorem would imply f(x)f(y)<=Kxy, because (f(x)f(y))/(xy)=f'(w) for some w in the interval between x and y. This is actually the way people frequently get estimates relating to uniform continuity. It is much more constructive than using the BolzanoWeierstrass Theorem as part of a proof by contradiction!
The picture displayed essentially attempts to show that a box which works at the origin for sqrt(x) will be satisfactory elsewhere. A "butterfly" won't "fly" too well at the origin, basically because the tangent line becomes perpendicular there. Uniform continuity will let us integrate continuous functions. Here is what we will do (this is essentially theorem 5.4.10 of the text). A step function will be defined by the following requirements: a "partition" of the interval [a,b] of this type:a=x_{0}<x_{1}<...<x_{n1}<x_{n}=b, and selection of n numbers y_{1},...,y_{n}. Then we will define S(x) to be y_{j} if x is in the interval [x_{j1},x_{j}). (If you are paying very close attention, you should see that the last interval needs special treatment, when x=b.) The idea is to approximate a continuous function f very closely by a step function. The steps of the step function and the boxes used in uniform continuity should seem to be related. The integral of a step function is easy: it will just be a sum of areas of rectangles. So this will approximate the integral of the continuous function. And if we can show that this approximation works well with area, then we will have defined the integral of a continuous function. Theorem 5.4.10 essentially describes one approach to this approximation. I will return to this later, and discuss it in more detail when we do integration.  
4/10/2003  I did some problems in the textbook and made some other
comments.
Section 5.3, problem #1 If I=[a,b] and f:I>R is
continuous, and if f(x)>0 for all x in [a,b], then there is
alpha>0 so that f(x)>=alpha for all x in [a,b]. Comment Consider [a,b]=[0,1] and f(x)=x if x>0 and f(0)=1. This f fails to be continuous at only one point (x=0) and its values are positive for all x. But the inf of f's values is 0 so there is no positive lower bound.
Section 5.3, problem #3 If I=[a,b] and f:I>R is
continuous, and if for every x in [a,b] there is a y in [a,b] so that
if f(y)<=(1/2)f(x), then f has a root in [a,b]: there must be r
in [a,b] with f(r)=0.
Section 5.3, problem #4 Show that every polynomial of odd
degree with real coefficients has at least one real root.
Section 5.3, problem #6 f:[0,1]>R and f(0)=f(1). Then
there is c in the interval [0,1/2] so that f(c)=f(c+1/2).
Now I restated the definition of continuity: In geometric terms, this means we have a rectangular box with center at (c,f(c)) and with horizontal sides +/ epsilon from y=f(c) and with vertical sides +/delta from x=c. The graph of y=f(x) is only "permitted" to "escape" from the box through the vertical sides. So when epsilon is specified, we can construct some box satisfying these geometric constraints.
Here's another definition: Therefore in this case, the same sized box will work for all points on the graph. We then considered the function f(x)=x^{2} and tried to slide boxes around on the graph. We came up with some conjectures which we then verified. Example 1 (uniform continuity satisfied) f(x)=x^{2} is uniformly continuous in A=[0,1]. Geometrically, this is because we can slide the "box that works" from its center at (1,1) down the graph to (0,0) and it will always "work". Let's see: f(x)f(c)=x^{2}c^{2}=x+c·x+c<=(x+c)·xc. If both x and c are in [0,1], then we know f(x)f(c)<=2xc. Therefore, if epsilon>0 is given, we could take delta=(1/2)epsilon, and this will satisfy the definition of uniform continuity for this f and A=[0,1].
Example 2 (uniform continuity not satisfied) f(x)=x^{2}
is not uniformly continuous in all of R. Geometrically, this is
because the function "wiggles" or "tilts" too much as x gets
large. If we take any box and move it so that its center is on the
graph, if the box is moved far enogh to the right or to the left,
eventually the graph will start poking out the top or the bottom of
the box.
Next time we will use a proof by contradiction to verify:  
4/9/2003  Propositions I and II and III are already known, and are
stated here in order to help people understand the lecture. Proposition I If (x_{n}) is a convergent sequence and if, for all n in N, x_{n} is in [a,b], then lim(x_{n}) is in [a,b]. Comments: This is an easy consequence of such sequential limit facts as: if L=lim(x_{n}) with x_{n}>=0 for all n, then L>=0. Just compare the x_{n}'s with a and with b (look at bx_{n} and x_{n}b). Proposition II Suppose (x_{n}) is a sequence with x_{n} in [a,b] for all n in N. Then there's a subsequence (x_{nk}) which converges. Comments: This is the BolzanoWeierstrass Theorem. It is generally impossible (or at least difficult!) to discover the subsequence. Proposition III If (x_{n}) converges and f is continuous, then the sequence (f(x_{n})) converges, and its limit is f(lim(x_{n})). Comments: This is the sequential statement which is equivalent to continuity. Theorem A (continuous functions on closed bounded intervals are bounded) If f is continuous on [a,b], then there is M>0 so that f(x)>M for all x in [a,b]. Comment: The phrase "f is continuous on [a,b]" means f is continuous at each x in [a,b]. Example A1 A function not continuous at one point of a closed bounded interval and the function is not bounded: take [a,b]=[0,1] and f(x)=1/x for x>0 and f(0)=0. f is not bounded, since f(1/n)=n for n in N and we have the Archimedean property. Example A2 A function continuous on an open interval which is not bounded: take (a,b)=(0,1), and f(x)=1/x. Proof of Theorem A: Suppose f is not bounded. Then for n in N, there exists x_{n} in [a,b] with f(x_{n})>=n. By Proposition II, the sequence (x_{n}) has a subsequence (x_{nk}) which converges. Also, f(x_{nk})>=n_{k}>=k. By Proposition I, the limit of the subsequence is in [a,b]. We have assumed that f is continuous at all points of [a,b], so it must be continuous at the limit of the subsequence, and therefore by Proposition III, (f(x_{nk})) converges. But a convergent sequence of real numbers is bounded, and this sequence is not bounded since its k^{th} element is at least k. This is a contradiction. Theorem B (Extreme Value Theorem) If f is continuous on [a,b], then there are elements v and w of [a,b] so that f(v)<=f(x)<=f(w) for all x in [a,b]. Comment: So the function "achieves" its maximum and minimum values. And the range of the function must be a subset of the interval [f(v),f(w)]. Example B1 A function which is continuous at all but one point of a closed bounded interval and the function is bounded and the function does not "achieve" its sup and its inf: we decided that if [a,b]=[1,1] and f(x)=x+1 for x in [1,0), f(0)=0, and f(x)=x1 for x in (0,1], then the range of f is the open interval (1,1). f is discontinuous only at 0. The sup of f's values is 1 and the inf of f's values is 1, and f never attains either 1 or 1: a fairly weird example. Proof of Theorem B: Since f is continuous on [a,b], Theorem A applies and f must be bounded. Therefore the set S={f(x): x in [a,b]} (the set of values of f) is nonempty (f(a) is in S) and is bounded. By the Completeness Axiom, S has both a sup and an inf. We will work with the sup, which will produce f(w). Parallel work with the inf will get f(v). Let q=sup S. Then given n in N, there is an element y_{n} of S so that q(1/n)<y_{n}<=q (this is the alternative characterization of sup). But since y_{n} is in S, there is x_{n} in [a,b] with f(x_{n})=y_{n}. By Proposition II, the sequence (x_{n}) has a subsequence (x_{nk}) which converges. Also, q<=f(x_{nk})>q(1/(n_{k}))>=q(1/k). By Proposition I, the limit of the subsequence is in [a,b] and is a point, w. We have assumed that f is continuous at all points of [a,b], so it must be continuous at the limit of the subsequence, and therefore by Proposition III, (f(x_{nk})) converges to f(w). But f's values on this subsequence are squeezed to q, and therefore f(w)=q, the sup of the values of f([a,b]). Corollary Values of continuous f on [a,b] always lie inside some closed bounded interval: f([a,b]) is a subset of [min value of f on [a,b],max value of f on [a,b]]. Theorem C (Intermediate Value Theorem) If f is continuous on [a,b], and if f(a)<0 and if f(b)>0, there is a element r of [a,b] with f(r)=0. Example C1 A function not continuous at only one point, with only two values, which doesn't have any roots: here [a,b]=[0,1] and f(x)=1 for x=0 and f(x)=1 for x>0. This is a silly example, but it works. Example C2 How many roots can such a function have? We saw that such a function can actually have any finite number of roots, and it even can have an infinite number of roots.
Proof of Theorem C: The text uses the bisection method which leads
naturally to an algorithm used for rootfinding. Please look at the
text and see if you like that proof. I'll try another
technique here.
Of course the function takes on all "intermediate" values, not just
0. And we have the important: All the results of this class session are almost part of our subconscious: probably Piaget showed that young babies "learned" them at a very early age. Is the intellectual structure of the course worth the difficult trip we have gone through? That is to be judged by each individual, naturally. I like it. There will be an exam on Thursday, April 17. Here is information about the exam, and a collection of review problems.  
4/7/2003 
Many fine students showed up for class today in the midst of an
unseasonal snow storm. I proved that composition of continuous
functions was continuous.
Theorem (Composition and continuity) Suppose f:I>R and
g:J>R, and we know that f(I) is a subset of J. Also suppose
that f is continuous at c and that g is continuous at f(c).
Then the composition of g with f is continuous at c. Comments We can't generally drop any of the hypotheses of the theorem and expect the conclusion to remain valid. For example, if f(x)=x, any discontinuity of g will be "transmitted" to the composition. Similarly, if g(x)=x, the composition will have f's discontinuities. I then announced that an EXAM would be given on Thursday, April 17. My current intention is to cover the major results of section 5.3 on Wednesday, and then the definition and simple examples about uniform continuity on Thursday (section 5.4). The exam would include coverage of that material. I hope to give out a review sheet on Wednesday, and may try to schedule a review session next week. We mostly did textbook homework problems for the remainder of today's class.
Section 5.2, #3 Give an example of functions f and g both
discontinuous at c in R such that a) the sum f+g is continuous at c b)
the product f·g is continuous at c.
Section 5.2, #7 Give an example of a function f from [0,1] to
R which is not continuous at every point of [0,1] but
fis continuous at every point of [0,1].
Section 5.2, #8 Suppose f and g are continuous from R to
R. If f(r)=g(r) for all r in Q (that is, for all
rational numbers r) then show that f(x)=g(x) for all x in R.
#8 and a half (Not from the text!) Can we find an example of
two functions f and g defined and continuous on all of R so
that the set S of numbers s where f(s)=g(s) is infinite but there are
x's in R where f(x) is not equal to g(x)?
#8 and threequarters Can we find an example like the previous
one where the set of agreement of the two continuous functions is just
(0,infinity)?
#8 and seveneights (?) Suppose that f and g are continuous on
all of R and that for x>0 we know
f(x)g(x)<33sqrt(x). What can we say about f(0) and g(0)?
Section 5.2, #12 A function f from R to R is said
to be additive if A f(x+y)=f(x)+f(y) for all x and y in
R. Prove that if f is continuous at one x_{0} in
R then f is continuous at every point of R. In fact, an additive function f, satisfying f(x+y)=f(x)+f(y), actually has some further properties. For example, f(2x)=f(x+x)=f(x)+f(x)=2f(x). It isn't hard to verify by math induction that f(nx)=nf(x) for all x in R and n in N. Also, if in the equation f(2x)=2f(x) we take x=(1/2)y we get f(2·(1/2)y)=2f((1/2)y), so that f(y)=2f((1/2)y) and (1/2)f(y)=f((1/2)y). Again we can verify that if m is in N, (1/m)f(y)=f((1/m)y). Combining what we have observed, you may see that we have proved the following result: if f is additive and if r is rational, then f(rx)=rf(x) for all x in R. That is, f is linear with respect to the rational numbers, Q. If c=f(1), the f(r)=f(r·1)=rf(1)=cr. So, when restricted to the rational numbers, an additive function must be multiplication by a constant. Combine the results of problems 8 and 12: an additive function which is continuous at one point must be multiplication by a constant. This is the content of problem #13. Weird example It is amazing and almost unbelievable that there are additive functions which are not of this kind. To see this, I need to quote some results from linear algebra. Q is a field. Multiplication of real numbers by elements of Q establishes that R is a vector space over Q. This is very very weird to any sane human being, but it is correct. Of course, students need to know what a vector space is to be sure it is correct, but it is correct. Actually, the dimension of the vector space is infinite (!). We can define linear transformations from the vector space R to itself just by looking at what happens to a basis. Here is part of a basis: 1 and sqrt(2). They cannot be linearly dependent over Q because that would imply that sqrt(2) is rational. So here I will define a "linear transformation": its value on the basis element sqrt(2) is sqrt(2) (that is, multiplication by 1) and its value on the basis element 1 is 0 (that is, multiplication by 0). Also its value on all other basis elements is 0. This "linear transformation" is certainly additive. Can it be continuous? If it is continuous at any point, then it is continuous at all points. And if it is continuous at all points, then it multiplies all numbers by the same number (problem #13) which this additive function does not. So this function is not continuous at any point (!). Since f(0)=0, given epsilon>0, there is no delta>0 so that x0<delta implies f(x)f(0)=f(x) is less than epsilon. Therefore there must be x's in the interval (delta,delta) with f(x)>epsilon. If one considers closely what this means for the graph of such an f, it turns out that there are "dots" (points on the graph) in a dense subset of the whole plane: dots inside every box of the plane! This is very, very weird to me. Linear algebra is hard. Today's quote was, of course, from W. Shakespeare (15641616). Here is sonnet #18:
Just part of the Rutgers effort to educate the whole person.  
4/3/2003 
We discussed a few more "facts" about limits of functions.
I tried to find a limited converse to Fact 3. That is, if a function
is positive and it has a limit, must the limit also be positive?
Mr. Benson suggested the following example:
Fact 4: nonnegativity is inherited by the limit
Suppose c is a cluster point of a set A, and f is defined on
A. Suppose also that the limit of f as x>c for x in A exists and
is equal to L. If f(x)>=0, then L>=0.
Fact 5: squeeze theorem Suppose c is a cluster point of a set
A, and f and g and h are functions defined on A and, for x in A,
f(x)<=g(x)<=h(x). If the limit of f(x) as x>c for x in A
exists, and the limit of h(x) as x>c for x in A exists, and if these
two limits are equal, then the limit of g(x) as x>c for x in A exists,
and equals the common value of the other two limits. Theorem (Cluster points and sequences) c is a cluster point of a set A if and only if there is a sequence (x_{n}) with the following properties:
On the other hand, suppose c is a cluster point of A. Then take delta=1/n. We know there must be an element of A (which I will call x_{n}!) satisfying 0<x_{n}c<delta. The first inequality implies that x_{n} is not c (so property 2 above is true). And certainly each x_{n} is in A. And we also know (unrolling the second inequality) that c(1/n)<x_{n}<c+(1/n). But the sequence version of the squeeze theorem then implies that (x_{n}) converges to c. And we are done.
Theorem (Limits and sequential limits) Suppose that c is a
cluster point of a set A, and f is defined on A. Then the limit of
f(x) as x>c is L if and only if for every sequence (x_{n})
with each x_{n} in A and no x_{n} equal to c and
which satisfies lim(x_{n})=c, then (f(x_{n}))
converges and its limit is L.
A return to the proof of fact 5: we verify that the limit of g as x>c
exists by checking on sequences, which is what the previous theorem
allows us to do. So if x_{n}) is a sequence
with each x_{n} in A and no x_{n} equal to c, we know
by the hypotheses of fact 5 that
f(x_{n})<=g(x_{n})<=h(x_{n}) for all
n. We also know (using one implication of the theorem above) that
both sequences (f(x_{n})) and (h(x_{n})) converge with
a common limit. But then the sequence version of the squeeze theorem
applies, and we know that (g(x_{n})) converges with the same
limit. But then the other implication of the theorem above
implies that the limit of the function g exists and is as desired. Informally, we should remember from calculus that a function will be continuous at a point if it is defined at the point, and if the limit of the function at that point exists and equals the value of the function there. This will be quite complicated if I continue using the setting "c is a cluster point of A", so I won't. From now on (at least for a while) f will be defined on an interval I. My examples of I will be R (all of the reals) or [0,1] (a closed interval) or (0,1) (an open interval). Here is a formal definition of continuity for our course. Definition (Continuity) Suppose f:I>R. If c is an element of I, then f is continuous at c if, given any epsilon>0, there is delta>0 so that if xc<delta with x in I, then f(x)f(c)<epsilon.
One can try to understand this in terms of input and output
tolerances, as I previously explained. One difference between this
definition and the definition of limit is that we just have
"xc<delta" and not "0<xc<delta". This is because we are
assuming that the limit of f at c is actually f(c): it is the "L" in
the previous definition of function limit.
Example 1, a rational function (I did not do 1/x which I was
assured had been verified in the text!). Let us consider
f(x)=1/(x^{2}+1). I claim that f is continuous for every
number c. Since this function is defined by an
algebraic formula, my first step is to try to get some algebraic
connection between f(x)f(c) and xc. This is easy, but a bit
tedious:
Example 2, sqrt(x) Here f(x)=sqrt(x) and the domain, I, is
[0,infinity). I claim that f is continuous for every number c in its
domain. Here again I want to find a connection between f(x)f(c) and
xc. Now f(x)f(c)=sqrt(x)sqrt(c)=xc/sqrt(x)+sqrt(c). Now
some observations about this equation: first,
sqrt(x)+sqrt(c) is the same as sqrt(x)+sqrt(c). Second, the equation
is undefined if both x and c are 0. So I will first analyze what
happens when c>0.
Then all of our work with sequences easily verifies the following
result:
Proposition (nonzero behavior of continuity)
If f is continuous at c and f(c) is not 0, then there is
delta>0 so that for all x in I (the domain of f) satisfying
xc<delta, f(x) is not zero.
Theorem (more algebra and continuity) If f and g are continuous
at c, and if f(c) is not 0, then 1/f and g/f are both defined in some
intervals containing c, and are continuous at c.
The most important result about continuity is composition. The nicest
thing about the definition of continuity that we settled on is that it
makes the following result very easy to prove.  
4/2/2003  I wrote again the provisional definitions #1 and #2 from the
last lecture. I copied examples 1 through 4 from the last lecture. I verified that example 1 (f(x)=x^{2}) is continuous at c exactly as the textbook does. That is, I wrote x^{2}c^{2}=(xc)(x+c) and stated that I wanted to "control" x^{2}c^{2} with some restriction on xc. I first said that maybe we should ask xc<1. Ms. Guthrie verified that this implied x<c+1 (you can see this by "unrolling" the inequality as we have already done many times). Then xc<=x+c<2c+1. Therefore if we knew that xc<epsilon/(2c+1), we see x^{2}c^{2}=xc·x+c<epsilon/(2c+1)·(2c+1)=epsilon. So take delta to be the minimum of 1 and epsilon/(2c+1). We need the 1 in order to get some bound on the x in xc. Feeling silly, I did a similar thing for x^{5}, which is more than any sane person might do: x^{5}c^{5}=(xc)(x^{4}+x^{3}c+x^{2}c^{2}+xc^{3}+c^{4}). So if x<c+1, the multiplier of xc can be bounded by an expression not involving x (replace each x by c+1). This can in turn be used to produce a delta given an epsilon. And so it goes ....
I then turned my attention to example 4, and tried to verify that the
f given there was not continuous at any irrational c. After drawing
a picture, Mr. Benson suggested that I try c/(327) as my epsilon. If c>0
I needed to show that given delta>0, the implication: if
xc<delta then
f(x)f(c)<c/(327) was false some x. If c>0,
I took x to be
a rational number in the open interval (c,c+delta) (possible since the
rationals are dense). Then f(x)=x, and f(c)=0
so that f(x)f(x)=x>c>c/(327) certainly. If c<0, we could
take x rational in the interval (cdelta,c). Example 5 f(x)=1 if x=0 or if x=1, and f(x)=0 otherwise. This f is not continuous at 0 and 1 and is continuous at all other x's.
Example 6 The Question of the day!
Find an example of f which is continuous at 0 and at 1, and is not
continuous at other numbers. Then I tried to analyze the reasons the text is so "particular" about domains in its discussion in chapter 4.
We probably want to have the following results true:
An element c of a subset A of R is called an isolated
point of A if there is delta>0 so that the only point of A in
(cdelta,c+delta) is c.
Examples:
Another motivating example if the limit of ((f(x)f(c))/(xc)) as x>c. The "natural domain" of this quotient, wellknown for being part of the definition of the derivative, will always exclude c. So we must be a bit more careful about limits.
Definition Suppose c is a cluster point of A, and f is defined
on A. Then we say that the limit as x>c with x in A of f is L
if the following is correct: Notice that there are several differences between what we wrote here and Provisional Definition #2. First we have the additional layer of "A" and "cluster point of A". In this course, A will almost always be an interval (closed or open) or a finite union of intervals. Second, and more interesting, is the additional restriction on x: 0<xc. This is to take care of such situations as the definition of derivative. We can now merrily (?) prove a collection of facts.
Fact 1: uniqueness of limits Suppose c is a cluster point of a
set A, and f is defined on A. If L_{1} and L_{2} are
both limits of f at c, then L_{1}=L_{2}.
Fact 2: limit existence implies local boundedness Suppose c is a cluster point of a set A, and f is defined on A.
Suppose also that the limit of f as x>c for x in A exists.
Then f is bounded on A near c: there is delta>0 and M>0 so that
if x is in A intersect (cdelta,c+delta), then f(x)<M.
Fact 3: functions locally inherit the signs of their limits
Suppose c is a cluster point of a set A, and f is defined on
A. Suppose also that the limit of f as x>c for x in A exists. If L
is the limit and L is positive, then f is positive for x in A near c:
there is delta>0 so that if x is in A intersect (cdelta,c+delta)
with x not equal to c, then f(x)>0. Essentially I am "converting" sequence/limit facts to function/limit facts. We could continue like this for a while. I will do a few more tomorrow.  
3/31/2003 
The lecture primarily dealt with two provisional definitions of
continuity. They were "provisional" or temporary principally because I
wanted to delay thinking about the intricacies of domain questions
involving functions. Provisional definition #1 A function f:R>R is said to be continuous at c if for all sequences (x_{n}) satisfying lim(x_{n})=c, then (f(x_{n})) converges, and lim(f(x_{n}))=f(c). Provisional definition #2 A function f:R>R is said to be continuous at c if, given any epsilon>0, there is a delta>0 so that if xc<delta, then f(x)f(c)<epsilon.
#2 is frequently briefly presented in calculus 1. There are various
interpretations of it. The function via its graph Consider the graph of y=f(x). If we draw horizontal lines y=f(c)+epsilon and y=f(c)epsilon, then locally near (c,f(c)) the graph is trapped between the two horizontal lines. When we look for delta, we are asking if there is an interval centered around c, an interval of the form (cdelta,c+delta) so that the portion of the graph over the interval will lie within the horizontal strip specified. Perhaps one of these "explanations" helps to comprehend PD#2.
We then proved an essential chunk of the course: PD#1 and PD#2 are
logically equivalent. We had to prove two implications:
If PD#1 is true, then PD#2 is true.
If PD#2 is true, then PD#1 is true. I find this pair of proofs logically adequate but somehow lacking in definiteness. I can only write or discuss this feeling approximately and awkwardly, but somehow, we contradict these complexly quantified statements in order to construct a "house of cards" which then, by design, collapses at the end. To me this seems, emotionally, somewhat unsatisfactory. Example 1 f(x)=x^{2}. We saw using PD#1 that this f was continuous at every c in R. We should also check this with PD#2. Example 2 f(x)=1 if x=0 and f(x)=0 otherwise. We saw using PD#1 that f was not continuous at 0. If c is not 0, then f is continuous at c. Here one can take delta=c (the delta will not depend on epsilon). Then f(x)=0 for xc<delta, since x can't be 0 (for if x=0, then xc=c<c which is false. So f(x)f(c)=00=0<epsilon for any positive epsilon. Example 3 f(x)=1 if x is rational and f(x)=0 if x is rational. This function is continuous nowhere. That is, if c is in R, we can find an "output tolerance" epsilon which can be satisfied by no input tolerance. Ms. Guthrie suggested that we use epsilon=1. The proof naturally divides into two cases.
Example 4 f(x)=x if x is rational and f(x)=0 if x is
irrational. This function is continuous only at 0. Indeed, we verify
PD#2 at 0: given epsilon>0, take delta=epsilon. Then f(0)=0, and we
must show: if x0<epsilon, then f(x)f(0)<epsilon. There are
two cases: if x is rational, then f(x)=x, so x0<epsilon
certainly implies f(x)0=x0<epsilon; if x is irrational, then
f(x)=0, so x0<epsilon certainly implies
f(x)0=00<epsilon. Now we wanted to show that f is not
continuous at c which was not 0. If we want to show that PD#2 is not
true, we need to find one epsilon for which no delta will
serve. The suggestion to take epsilon=c was made. I think this is
inspired partially by example 3 and even example 2. I first looked at
two specific c's to check the logic. Textbook problems due Thursday: 3.7: 8, 9 and 4.1: 2, 12.
 
3/27/2003 
Absolutely convergent series can be rearranged without changing their
"sums". Theorem Suppose that sum_{j=1}^{infinity}a_{j} converges absolutely. If sum_{j=1}^{infinity}a_{j} converges with sum L, then any rearrangement of the series, sum_{j=1}^{infinity}a_{f(j)} will converge and its sum will be L. Proof: (A proof of this result appears in section 9.1.) Since sum_{j=1}^{infinity}a_{j} converges with sum L, if we define x_{n}=sum_{j=1}^{n}a_{j} we know: Given epsilon>0, there is K(epsilon) in N so that for n>=K(epsilon), x_{n}L<epsilon. Now let J(epsilon) be equal to the maximum of the numbers f(1),f(2),...,f(K(epsilon)). If we define the partial sums for the "rearranged" series to be y_{n}=sum_{j=1}^{n}a_{f(j)} then when n is at least J(epsilon), every y_{n} has all the "pieces" of x_{K(epsilon)}. So we could "separate" the y_{n} sum into sum_{f(j) is one of [1,...,K(epsilon)]}a_{f(j)}+sum_{the other terms}a_{f(j)}. Now consider y_{n}L. We can use the triangle inequality to estimate this: y_{n}L<=sum_{f(j) is one of [1,...,K(epsilon)]}a_{f(j)}L+sum_{the other terms}a_{f(j)}. The first term is less than epsilon because of the specification of K(epsilon). As for the second term, a different strategy is needed. sum_{the other terms}a_{f(j)}<=(Triangle!)sum_{the other terms}a_{f(j)}. Note that each of these a_{f(j)} is "far out": at least more than K(epsilon).
Please note that we have not used part of the hypothesis yet, that the
series converges absolutely. So let's use it now. We can label the
partial sums of this series
z_{n}=sum_{j=1}^{n}a_{j}. Since
(z_{n}) converges, it must be a Cauchy sequence (here is
that fact again!). Therefore, given epsilon>0, there is
W(epsilon) in N so that for n>m>W(epsilon),
z_{n}z_{m}<epsilon. But each z_{stuff}
is a sum of a bigger chunk of a series with nonnegative terms, so
that we now know
sum_{j=m+1}^{M}a_{j}<epsilon. In fact,
every finite "chunk" of this series summed up (if you have indices at
least W(epsilon)) will be less than epsilon. In fact, any
finite sum of such terms can be "embedded" into a (maybe larger) sum
of successive terms, so that any such finite sum is less than epsilon,
provided that the indices are all at least W(epsilon). This gives us a
way to control the other terms, the ones in purple above. This theorem helps those new to series from making mistakes. That's because many if not most of the series these people encounter are power series, sum_{j=1}^{infinity}a_{j}x^{j}. Power series have intervals of convergence. Outside these intervals, the series diverges. Inside these intervals, the series converges absolutely, so any sort of rearrangement leaves convergence and the specific sum unchanged. However, Fourier series and series derived from wavelets typically do not have such behavior, and so more care is needed to deal with them. Both Fourier series and especially series derived from wavelets are used in "real life". In particular, I believe that a few years ago the FBI fingerprint files were converted to storage based on wavelets. I further challenged the class to something very specific. If one believes in the Cauchy criterion, then it should be possible, perhaps even easily possible, to create a list of infinitely many positive numbers so that the sum of any finite subset of them is less than .0001 (1/(10,000)). After some effort, we did create such a "list": the n^{th} number (for n in N) would be, perhaps, 1/[(10,001)2^{n}]. Since we know that the sum of 1/2^{n} as n goes from 1 to infinity is 1, our example is complete. Onward to CALCULUS! What do we know about continuity, and what can it be used for? Most nice functions (functions defined by simple formulas) are continuous, or, at least, are continuous most places. sine, cosine, polynomials, rational functions, logs, exponentials. You can add, subtract, etc. and compose such functions to get further continuous functions. So graphs of such functions are unbroken: important in curve sketching and root finding. And areas "underneath" such curves are defined and maybe computable or approximable (the definite integral "exists"). If a function models some sort of physical "process", then we can think of x as input and f(x) as output. f's continuity is roughly reflected in the fact that small changes in input (x) should imply small changes in output (f(x)).
Here is a candidate for a Math 311 definition of continuity. I warn
you that it is different from what's at the beginning of chapter 4. So
please let us agree that it is only a preliminary definition of
continuity. Also, in order to avoid annoying technicalities at the
beginning, I will assume that the function f is defined
everywhere and not worry for now about domain questions. We
will need to modify this definition a bit later.
Example 1 Suppose f(x)=x^{2}. Then f is continuous at
every c in R. We can prove this: if (x_{n}) is a
sequence which converges to c, then (theorem about limits and
arithmetic) ((x_{n})^{2}) must also converge, and its
limit will be c^{2}. So we have shown that (f(x_{n})
converges, and lim(f(x_{n})=f(c).
My next "job" will be to contrast this preliminary definition of
continuity with the one given in the text. I will also need to deal
with the complexities which occur when domain problems are included.
 
3/26/2003  We discussed infinite series. This material is somewhat
contained in sections 3.7 and 9.1 of the text. This is basic material
about infinite series, but we will also cover material which is not in
the text. Definition The infinite series sum_{j=1}^{infinity}a_{j} converges if the sequence of partial sums x_{n}=sum_{j=1}^{n}a_{j} converges. If L is the limit of (x_{n}), L is called the sum of the infinite series. You will see that it is very important to think only about this definition, and not to get "distracted" by the idea of "adding up infinitely many numbers". This is rather different.
Theorem If sum_{j=1}^{infinity}a_{j}
converges, then
the sequence (a_{n}) itself must converge to 0. The converse is false. That is, we can get series sum_{j=1}^{infinity}a_{j} which do not converge, but where lim(a_{j})=0. The bestknown example is probably the harmonic series: sum_{j=1}^{infinity}1/j, which diverges, but whose individual terms >0. But signs (+/) can matter. The alternating harmonic series, sum_{j=1}^{infinity}(1)^{j+1}/j, converges. This is a simple consequence of the alternating series test. So there's another definition.
Definition The series
sum_{j=1}^{infinity}a_{j} converges
absolutely if sum_{j=1}^{infinity}a_{j}
converges.
Theorem If sum_{j=1}^{infinity}a_{j}
converges absolutely, then it must converge. Many examples in calc 2 use this result. For instance, since we know that sum_{j=1}^{infinity}1/2^{j} converges, we therefore know that sum_{j=1}^{infinity}(sin(j^3+5j))/2^{j} converges, since sin(j^3+5j)<=1. Of course, convergence does not imply absolute convergence. The alternating harmonic series, mentioned above, shows this. And indeed we have the following additional definition. Definition The infinite series sum_{j=1}^{infinity}a_{j} is conditionally convergent if it converges but sum_{j=1}^{infinity}a_{j} does not. Example: The alternating harmonic series converges conditionally. Let us "discuss" such conditionally convergent series. For the remainder of this lecture, sum_{j=1}^{infinity}a_{j} will be a conditionally convergent series. So
Now we know that N is "divided" into two infinite sets, and the conditionally convergent series has infinitely many positive and infinitely many negative terms. In fact, let's look at say the positive terms. If the sum of the positive terms converges, then we could subtract (cancel out!) the positive terms from the convergent series sum_{j=1}^{infinity}a_{j}. Since the difference of two convergent series must also converge (theorems on arithmetic and limits) we then see that the series of negative terms must converge. But then since the series of a_{j}'s is the difference of the positive terms and the negative terms (not profound: it is always true that x=x for x>=0 and x=x for x<0), it would follow that the sum of the absolute values would converge! But this is a contradiction. So if you can follow this chain of hypothetical reasoning we have seen that the sum of the positive terms must be divergent (and that means that the partial sums must be unbounded above). Similar reasoning establishes that the sum of the negative terms must be unbounded below. Now here is a wonderful result, which is totally unintuitive to me. Ooops: just as I did in class, I forgot to state a definition first.
Definition Suppose
sum_{j=1}^{infinity}a_{j} is a series. Then a
rearrangement of this series is gotten by specifying a
bijection f:N>N to obtain an infinite series
sum_{j=1}^{infinity}a_{f(j)}.
Riemann's Rearrangement Theorem Suppose
sum_{j=1}^{infinity}a_{j} converges
conditionally. If c is a real number, then there exists a
rearrangement so that
sum_{j=1}^{infinity}a_{f(j)} converges, and
the limit of the rearranged series is c.
I could not "believe" this result when I first saw it. I needed to think about it quite a bit. Notice that if the rearrangement bijection, f, only changes finitely many of the integers (that is, if there is N so that f(n)=n for n>=N), then the {condi}vergence of the rearranged series and its sum do not change, because the partial sums of the original series and the rearranged series are identical after the first N such. Again, "only tails matter". The preceding theorem shows that we shouldn't think of infinite series as sums. They are just a different sort of creature. As I mentioned in class, if we called sum_{j=1}^{infinity}a_{j} something totally new and ludicrous, like GLUMP(a_{j}), we then would have less "intuition" or preconceptions (prejudices!) to get rid of. If someone told us that GLUMP(a_{j}) and GLUMP(a_{f(j)}) were not necessarily the same, well, then, I guess they would not necessarily be the same: not a tragedy. I'll do just a bit more on series tomorrow before starting chapter 4.  
3/24/2003 
Sequences which satisfy the Cauchy criterion and convergent sequences
are the same sequences! I tried to give "real" examples illustrating
the usefulness of this "coincidence". I began by restating the definition of convergence, and the definition of the Cauchy criterion. Of course, the Cauchy criterion does not include knowledge of the limit. But we can state the following: since we know that given epsilon>0, there is M(epsilon) in N so that when n and m are at least M(epsilon), then x_{n}x_{m}<epsilon, we do know that all x_{n}'s with n>=M(epsilon) satisfy x_{n}x_{M(epsilon)}<epsilon. But that means that those x_{n}'s lie inside the interval (x_{M(epsilon)}epsilon,x_{M(epsilon)}+epsilon). The infinite tail of the sequence is in that open interval. Then the work we have done on limits and inequalities will tell us that the limit must be in the closed interval [x_{M(epsilon)}epsilon,x_{M(epsilon)}+epsilon].
With that behind us, I introduced my example: an infinite series. So I
want to analyze the sum as n goes from 1 to infinity of
+/(1/n^{n}) where the sign is given in the following weird
way: one + sign, then two  signs, then three + signs, then four 
signs, etc. So the series begins:
Much of what we're going to do here could also be done in a second semester calc course, but I want to do it from the viewpoint of Math 311. The basic idea is to compare this series with a more familiar series. And the most familiar series are probably geometric series. Consider the series 1+1/2+1/4+1/8+...+1/2^{n1}+... where a_{n}=1/2^{n1}. Does this series converge? Well, we need to reach back to second semester calculus. We say that the series converges if the sequence of partial sums converges. Here the sequence of partial sums is defined by x_{n}=sum_{j=1}^{n}(1/2^{j1}). The nicest thing about geometric series is that there are simple formulas for their partial sums. Here we can multiply x_{n} by 1/2 and subtract from the defining sum for x_{n}. Lots of things cancel out, and we have x_{n}(1/2)x_{n}=11/2^{n}, so that x_{n}=21/2^{n1}. Now let's go back to 311. The sequence (x_{n}) is obtained by adding positive numbers. So it is a monotone increasing sequence. For such sequences, there is a dichotomy (the online dictionary says that this means "a division into two, esp. a sharply defined one.") Either a monotone increasing sequence is bounded and converges, or it is unbounded and diverges. But we have a nice formula for x_{n}, so we know that the terms are all bounded above by 2. And, indeed, since we know that 1/2^{n}>0 we even know that 2 is the limit of this sequence.
The Cauchy criterion can be applied to this sequence. That is, given
epsilon>0, there is M(epsilon) in N so that for n and m at least M(epsilon),
x_{n}x_{M(epsilon)}<epsilon. Suppose m>n. We will
actually be able to "compute" M(epsilon) in this case. Here we know that
x_{n}=sum_{j=1}^{n}(1/2^{j1}) and
x_{m}=sum_{j=1}^{m}(1/2^{j1}) so
that
x_{m}x_{n}=sum_{j=n+1}^{m}(1/2^{j1}).
We can multiply by 1/2 and subtract, just as we did above. The result
gets us a nice formula:
x_{m}x_{n}=(1/2^{n})(1(1/2^{mn})
(remember that m>n). Therefore we can simplify (get rid of one
variable on the righthand side, and give up strict equality): Now let's go back to the original weird sequence. Let me call b_{n} the n^{th} term, which is 1/n^{n} with a weird sign. And let me call y_{n}, the n^{th} partial sum of the b_{n}'s. What can we say about the sequence (y_{n})? It is certainly not monotone, because of the + and  signs distributed among the b_{n}'s. But can we compare the two sequences? Since b_{n}=1/n^{n}, and we know that 1/n^{n}<=1/2^{n1} (we verified this for n=1 and n=2, and I bet it is true for all n>=2), we know that y_{m}y_{n}=sum_{j=n+1}^{m}{weird sign}1/j^{j}, and the triangle inequality breaks this up to <=sum_{j=n+1}^{m}1/j^{j}<= sum_{j=n+1}^{m}1/2^{j1}=x_{m}x_{n}. But now if n and m are at least M(epsilon) for the (x_{n}) sequence, we know x_{m}x_{n}<epsilon. This implies that y_{m}y_{n}<epsilon also for those n and m's. Therefore, (y_{n}) is also a Cauchy sequence. And further therefore (!), we now know that (y_{n}) must converge. Even more, we know that y_{11}, which we could compute, would be within .001 of the sum of the whole series. So this strategy allows us to conclude that a certain infinite series converges, and to get a good approximation of the sum of the series. Now let me write down some of the theory we have actually in effect proved:
Theorem Suppose (x_{n}) and (y_{n}) are
sequences. If we know that there is some positive constant C so
y_{m}y_{n}<=Cx_{m}x_{n} and
if we know that (x_{n}) converges, then (y_{n})
converges.
Theorem If an infinite series converges absolutely, then
it converges.
I used some Maple instructions to compute the sum of the weird series
I started out with. Here are the Maple instructions, and the result. I wanted to do another example, but I sort of ran out of time. I wanted to show an "iterative" way to get a root of a polynomial. This is discussed in section 3.5, pages 84 and 85.  
3/13/2003  We began the lecture by contemplating a page of a final exam
I gave in 1996 to a second semester calculus class (o.k.: now I'll
admit it. It was a small class, an honors second semester calc course,
and the students were quite good). The worksheet had four true/false
statements. We analyzed those statements. For a): the sequence (a_{k}) defined by a_{k}=(1)^{k} has the property that (a_{k}) converges, but (a_{k}) does not. So the assertion in a) is false. For b): this is a more subtle question. If L is not zero, and if (a_{k}) and (b_{k}) both converge to L, then ("eventually", which means, there exists K in N so that a_{k} is not 0 for k>=K) the sequence (a_{k}/b_{k}) converges to 1. If L=0, then (using the examples 1/k and 1/k^{2}) we can get sequences (a_{k}/b_{k}) which maybe diverge or which converge to 0. We can even get a sequence (use 1/k and 37/k) which converges to 37. So the statement as given is false. For c): the statement is true. Here I tried to give a proof. Since (a_{k}) converges (say to L) we know that given epsilon>0, there is K(epsilon) in N so that for k>=K(epsilon), a_{k}L<epsilon. Now we need to create M(epsilon) so that for n>=M(epsilon), (a_{k+1}a_{k})0<epsilon. Here we can take M(epsilon)=K(epsilon/2) (in class we used L(epsilon/100) which also works but this answer is more traditional). Then if k>=M(epsilon), we know that a_{k+1}L<epsilon/2 and a_{k}L<epsilon/2. Now use the triangle inequality: (a_{k+1}a_{k})0<=(a_{k+1}L)+(La_{k})<=a_{k+1}L+La_{k}<epsilon/2+epsilon/2=epsilon. So we are done: we have used the K(epsilon) "machine" to "build" an M(epsilon) "machine". For d): this is probably the most subtle question. We have in fact addressed it before. If we consider the sequence a_{k}=sum_{j=1}^{k}1/j then the logic behind the Integral Test of calculus allows us to underestimate a_{k} by ln(k) or ln(k+1) (see the diary entry for 3/5/2003). So in fact this sequence is unbounded. Mr. Hedberg suggested a sequence from a homework problem: a_{k}=sqrt(k). The problem showed that (a_{k+1}a_{k}) converged to 0 when a_{k}=sqrt(k). This example is more in the spirit of Math 311: the integral test is way more advanced than what we can use now. I altered the example. I asked: suppose that you know that both (a_{k+1}a_{k}) and (a_{k+2}a_{k}) converge to 0. Can you conclude that (a_{k}) converges? And, indeed, the same example (a_{k}=sqrt(k)) shows that the answer is "No." Even more, we can add any finite number of such requirements and still the answer will be "No". So getting any criterion for convergence which doesn't seem to depend either on already knowing a limit or on some special structure (such as monotonicity) seems difficult, which makes the Cauchy criterion yet more amazing.
I will call this CC for Cauchy criterion:
We will further contrast it with the following, which I will
temporarily label V (it is the definition of
convergence): The purpose of the remainder of this class is to verify that CC and V are equivalent. CC is used a great deal in practice and in theory, because it does not need a specific L to be designated, but guarantees convergence. I'll try to show in classes after vacation how CC is used. First, as always in Math 311, the proofs:
Theorem V implies CC. That wasn't hard. Now for the other way. We will sneak up (?) on the result.
Proposition If (x_{n}) has CC, then the sequence is
bounded.
Proposition If (x_{n}) has CC, then (x_{n}) has some
convergent subsequence.
Proposition If (x_{n}) has CC and if (x_{n}) has some
convergent subsequence, then (x_{n}) converges.
Take n>=K(epsilon) as defined above. Look at x_{n}L=x_{n}x_{nmax(W(epsilon/2),M(epsilon/2))}+x_{nmax(W(epsilon/2),M(epsilon/2))}L<x_{n}x_{nmax(W(epsilon/2),M(epsilon/2))}+x_{nmax(W(epsilon/2),M(epsilon/2))}L. Now look at each piece.
I tried diligently and perhaps (almost certainly!) unsuccessfully to "motivate" this elaborate strategy in class. It is certainly complicated. We have now proved a major theorem. Theorem A sequence converges if and only if the sequence satisfies the Cauchy condition. We will use this a lot.  
3/12/2003 
I began by advising students that I would "cover" sections 3.13.5 of
chapter 3 and requesting that students read these sections. I hope
that I will finish this material and that soon after vacation I will
begin chapter 4. From now on, many of the results (theorems, examples,
etc.) will have increasingly familiar statements. Most students should
recall versions of the results from calculus. But in Math 311 I will
be principally interested, almost solely interested, in the
proofs of these statements. For example, the "Ratio Test" in
the text: Suppose (x_{n}) is a sequence of positive numbers, and that lim(x_{n+1}/x_{n}) exists and is a number L. If $L<1, then lim(x_{n}) exists and is 0. I won't prove this but it is a part of what students should know (Theorem 3.2.11 of the text). One special case is interesting, however, since the text's verification of the result uses several of the concepts we've been looking at recently.
Example If 0<b<1, then the sequence defined by
x_{n}=b^{n} converges, and its limit is 0.
Here is one of the most important results of the course. Inductive step Here we are given the following elaborate ingredients: a subinterval I_{n} of [a,b] with the length of I_{n}=(ba)/2^{n}, and an infinite subset S_{n} of N so that if k is in S_{n}, then x_{k} is in I_{n}. (There will be lots of indices to keep track of in this proof!) Now I divide I_{n} into two equal halves, the left and right halves. I also create S_{n}^{L} and S_{n}^{}: these are two subsets of S_{n} defined by: k is in S_{n}^{L} if x_{k} is in the left half of I_{n} and k is in S_{n}^{R} if x_{k} is in the right half of I_{nn} is the union of these two subsets, at least one of them must be infinite. So S_{n+1} will be one of the subsets (either the "left" or the "right" one) which is infinite. And I_{n+1} will be the associated half interval. So I have done the inductive "step". What have we created? We have several "things":
Discussion I need to convince you of the claim that this is an important theorem. But I also mention that lots of people don't "like" it. They don't like it because it is nonconstructive, it is "ineffective": by that I mean that no mechanism is shown to create an explicit subsequence which converges. Somehow one "knows" that there are infinitely many points in one half of each interval, etc. etc. I don't think human beings are too comfortable contemplating infinity so directly. Humans tend to like things they can verify a step at a time, and jumping to "Hey, half the interval has infinitely many points" is quite a big jump. It is amusing, though, to look at what follows: we will soon be able to "construct" sequences that surely converge, and this will be, at least theoretically, a consequence of the BolzanoWeierstrass Theorem. The theorem applies to any sequence which is bounded: no other condition is needed. Here is the theorem again, stated simply: Please note that K. Weierstrass is a professional ancestor of the instructor of this course. The instructor's professional ancestors can be inspected. In the fashion of Math 311, yet another definition: Definition A sequence (x_{n}) is a Cauchy sequence if for all epsilon>0, there is K(epsilon) in N so that for n and m >=K(epsilon), x_{m}x_{n}<epsilon.
There are lots of quantifiers in this definition, which is somewhat
reminiscent of the definition of convergence. What does the definition
"say"? Well, let's try to particularize it. Suppose epsilon=46. And just
suppose that K(45)=400,000. Then, certainly, if n>=400,000,
x_{400,000}x_{n}<46. We can "unroll" this inequality. It
means all the elements of the
sequence after the first 399,999 must be in the interval
(x_{400,000}46,x_{400,000}+46). So in fact the
sequence must be bounded, because the "infinite tail" of the sequence
is caught in this interval and there are only finitely many (hey:
399,999 is just a finite number!) outside the interval. So we have
almost proved that  
3/10/2003 
Since I left my notes at home, I had to "wing it" a bit, and
spontaneous talk about the material. I went over 3.3, monotone
sequences, more abstractly. In particular, I proved the following
result:
Theorem Suppose (x_{n}) is a monotone increasing sequence. Then:
(x_{n}) converges if and only if the set {x_{n} : n in N} is bounded
above. If (x_{n}) converges, then its limit is the sup of the set {x_{n} : n
in N}. As I remarked last time, this class of sequences is interesting because of the large number of applications involving them. It is quite easy to create examples. Here, let me do it "spontaneously" (something like this arose in conversations with Mr. Oleynick after class): I know from "elementary" considerations that the following functions are increasing (at least for positive numbers):
I wrote in the background to the course (and said during various lectures) that the single most important definition was that of limit of a sequence. I can sort of try to understand sequences: they are a countable object, sort of a list, and I can try to use simpleminded (?) techniques like mathematical induction on sequences. Yet I am trying to investigate the real numbers. One of the horrors and the pleasures of the real numbers is that they are uncountable. How can I use the tool of sequences, basically what seems to be a countable tool, to investigate an huge, uncountable set? The method is to use subsequences. I will take for today a very formal view of sequences and subsequences. Partly this is an effort to keep students alert and away from the familiar standard notation (a subsequence is denoted by a subscript on a subscript). But also it is a way to keep us honest and rely only upon proving things. So here I go. First, let's repeat what a sequence is: A sequence is a function f from N to R. A function h:N > N is strictly increasing if for all a and b in N, a<b implies h(a)<h(b). A function g:N > R is a subsequence of a function f:N > R if there is some strictly increasing function h:N > N so that g=foh (the notation o stands for the "little circle" indicating composition).
I have written this in a very formal manner, to be darn near
incomprehensible: incomprehensible but honest. I gave a rather silly
example: f(n)=1/n, h(n)=n^{2}, so that the subsequence g(n)
was 1/n^{2}. In this example, we have a subsequence of a
convergent sequence which also converges, and the limits agree. This
is no accident:
Proposition If h:N>N is strictly increasing, then
h(n)>=n for all n in N. Back to the proof of the theorem: suppose the sequence f converges to x. This means: given epsilon>0. there is K(epsilon) in N so that if n>=K(epsilon), then f(n)x<epsilon. But h(K(epsilon))>=K(epsilon), so if m is in N and m>K(epsilon), then h(m)>=h(K(epsilon)>K(epsilon), so that f(h(m))x<epsilon, which is exactly the definition of "g=foh" converges to x. And we're done. Notation If g is a subsequence of f, then the traditional notation goes something like this: f(n) corresponds to x_{n}, and g(k) corresponds to x_{nk}. The text uses the more traditional notation, so that's what I will generally do in the work that follows. Example We saw in the last lecture that the sequence f(n)=(1)^{n} does not converge. The way we showed it in the last lecture seemed rather elaborate and difficult to motivate. Here's another method. If f(n)=(1)^{n} converges, then every subsequence must converge, and the limit of every subsequence must be the same. So we looked at the following strictly increasing functions from N to N: h_{1}(n)=2n (just "hits" the even integers) and h_{2}(n)=2n1 (gets the odd ones). Then g_{1}=foh_{1} and g_{2}=foh_{2} are two subsequences of f. Not very elaborate computation shows that g_{1} is the constant sequence 1 (1 raised to an even integer power is 1) and g_{2} is the constant sequence 1 (1 raised to an odd integer power is 1). So these two subsequences converge, and since they do not converge to the same number (1 is not equal to 1) the original sequence f cannot converge. Then I asked if it is possible to have a sequence with subsequences converging to three numbers. We decided that could happen (the sequence would alternate among 1 and 0 and 1). We could even have a sequence with subsequences converging to 17 different numbers. A more complicated question is the following:
Question Is there a sequence whose subsequences converge to
infinitely many distinct numbers?
Question Is there a sequence which, for each x in [0,1], has a
subsequence which converges to x?
 
3/5/2003 
We began the material of section 3.3 today. I decided to look at a
specific example and try to understand it well. I decided on the
following "process" in a fairly random way. Here is what I did: I started with 1, then I multiplied by 8, added 13 to the result, and took the square root of that. The result was 4.582575695. I took this, multiplied it by 8, added 13, and took the square root. The result was 7.0470281. I took this, multiplied it by 8, added 13, and took the square root. The result was 8.3292392. I took this, multiplied it by 8, added 13, and took the square root. The result was 8.9237836. I took this, multiplied it by 8, added 13, and took the square root. The result was 9.1864176. I took this, multiplied it by 8, added 13, and took the square root. The result was 9.3000721. I took this, multiplied it by 8, added 13, and took the square root. The result was 9.3488276. ETC. What is going on? People with experience in numerical analysis may recognize this. Let me try to explain (and, hopefully, interest!) other people.
First, we are looking at a sequence defined by the following recursive
rules:
What can happen to this sequence? Here is a simple sequence of
"lemmas" analyzing what goes on:
Theorem: The sequence (x_{n}) defined by So we look now for more information about the sequence (x_{n}). In fact, we wonder if (x_{n}) converges. Certainly an increasing sequence alone need not converge. A simple example is x_{n}=n, which, by the Archimedean Property, cannot converge. Mr. Goode observed that the sequence whose initial terms we computed seems to have smaller and smaller "jumps" between successive terms. Maybe this means it converges. I remarked that there is a sequence from calc 2 with interesting properties: x_{n}= the sum as j goes from 1 to n of 1/j. This is called the sequence of harmonic numbers. It is an increasing sequence since it is adding up more and more positive numbers as n increases. In this case, x_{n+1}x_{n} is 1/(n+1), so certainly the sequence is taking ever smaller "steps" up as n increases. However, we can use logic familiar from the Integral Test, and compare x_{n} with the lefthand Riemann sums of width 1 for the integral from 1 to n+1 of 1/x. x_{n} is larger than this integral, and the integral is ln(n+1), which certainly is unbounded as n grows. (I note that it is possible to get the result that the sequence of harmonic numbers is unbounded without using integrals [I think the logic is in the book] but this way is quicker.) .
What happens to our sequence (x_{n})? Mr. Hedberg suggested that the set
S={x_{n} : n in N} is bounded. He even suggested the bound of 45. We
checked this: So (x_{n}) is increasing and bounded. Does it converge, and, if it converges, what is its limit? Well, it can't converge to 46, say, since 45 is an upper bound of the x_{n}'s, and 46 is one more, so that the inequality x_{n}46<1 will never be satisfied. Recall that a bounded sequence need not converge (we considered ((1)^{n}) last time, an example of a bounded sequence which did not converge). But here we have additional "structure": the sequence increases. In fact, the sequence converges to the sup of the set S={x_{n} : n in N}. Why should this set even have a sup? We know that S is not empty (1 is in S!). And we also know that 45 is an upper bound of S. Therefore by the Completeness Axiom, S must have a least upper bound. Let me call that least upper bound, L.
Theorem: (x_{n}) converges to L.
But what is L? A whole bunch of moreorless unsatisfactory
answers can be given: Since x_{n+1}=sqrt(8x_{n+1}13), (x_{n+1})^{2}=8x_{n+1}13. We know that (x_{n}) converges and that the limit of (x_{n}) is L. We also can deduce that the sequence (x_{n+1}) converges, and its limit is also L (remember, "only tails matter."). The limit of the lefthand side is L^{2}. The limit of the righthand side is 8L+13. So we know L^{2}8L13=0. The quadratic formula is valid (I think it was an exercise earlier in the book) so that L must be (8+/sqrt(8^{2}+4(13)))/2=(8+/sqrt(116))/2=[approx](8+/10.77)/2. So there seem to be two choices for L. One choice is negative. But, golly, each of the x_{n}'s is positive, so from previous results on order, we know that the limit can't be negative. Therefore the limit must be [approx]9.385 which is certainly at least consistent with the previously computed numbers. What is going on here? Look at the picture. The line shown is just y=x. The parabola is y=sqrt(8x+13), or even y^{2}=8x+13. The point A is just an initial x_{1}. Go "up" to the parabola. The point B is (x_{1},x_{2}), because the "rule of formation" of the sequence is the same as y=sqrt(8x+13). Now "bounce" to the diagonal line, where C is (x_{2},x_{2}). Now up to D, which must be (x_{2},x_{3}), etc. The sequence is gotten by bouncing back and forth between the curves. The (x_{n}) of Math 311 is just the first coordinates of these points, marching to the right on the real line. The intersection of the curve and the straight line is the point (L,L), and the sequence "clearly" (well, maybe clearly geometrically) converges to L. It is interesting to note that this process is quite "robust"  perturbations of the initial "guess" x_{1} don't affect its convergence to L. If the initial guess is chosen "too large", so x_{1} is greater than L, the geometry shows that the sequence decreases, wiggling back and forth, down towards L. This is a very stable, very neat method of approximating a root. We have an "attractive fixed point". The main object of section 3.3 is to discuss:

Maintained by greenfie@math.rutgers.edu and last modified 3/12/2003.