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5/5/2003 |
THE LAST DAY!!!Let's analyze and prove one version of the Fundamental Theorem of Calculus (FTC). This is discussed in section 7.3 of the text.
This version of FTC has to do with how the integral behaves as a
function of its upper parameter. Perhaps an example will make the
difficulties clearer. Let's look at a function defined in [0,3] by the
"piecewise" formula f(0)=2, f(x)=1 for 0<x<=1, f(x)=0 for
1<x<=3. We first observed that f was indeed Riemann integrable
on [0,3]. We could use partitions such as {0,B,H,C,3} where B is
slightly bigger than 0, H is slightly less than 1, and C is slightly
larger than 1. The difference between the resulting upper and lower
sums is "very small". And consideration of such sums allows one to
actually compute the Riemann integral. Of course f is also Riemann
integrable on subintervals of [0,3], so we could define F by
F(x)=int
In what follows, we will need to "recall" some facts about the Riemann integral. - If f is Riemann integrable in an interval, then it is Riemann
integrable in any subinterval.
Proof: Discussed in the writeup for the lecture of 4/28/2003. - If f is Riemann integrable in an interval, then it is bounded, say
by M (so |f(x)|<=M for all x in the interval). Also the absolute
value of the integral of f on the interval is less than or equal to M
multiplied by the length of the interval.
Proof: Discussed in the writeup for the lecture of 4/24/2003.
Theorem (continuity of the integrated function) Suppose f is
Riemann integrable on [a,b]. Then f is Riemann integrable on [a,c] for
all c in [a,b], and if F(x)=int_{a}^{x}f for x in
[a,b], F is continuous on [a,b]. Indeed, if f is bounded by M, F
satisfies a Lipschitz condition with Lipschitz constant M: for all x
and y in [a,b], |F(x)-f(y)|<=M|x-y|.Proof: If x<y, then F(y)=int _{a}^{y}f=
int_{a}^{x}f+int_{x}^{y}f by
additivity on intervals (4/28 lecture). Since
F(x)=int_{a}^{x}f we see that
F(y)-F(x)=int_{x}^{y}f. If we know that
-M<=f(x)<=M, then -M(y-x)<=F(y)-F(x)<=M(y-x) so that
|F(x)-f(y)|<=M|x-y|.
Comment: the function g(x) which is 0 if x<0 and is sqrt(x) if
x>=0 does not satisfy a Lipschitz condition in any interval which
includes the "right side" of 0 because sqrt(x) doesn't satisfy the
Lipschitz property in such an interval (see the lecture on 4/14,
please). So this g
Although this result is almost always used where f is continuous in
the whole interval, so F is differentiable in the whole interval with
F'=f, we actually don't "need" continuity of f. Here's a fairly simple
example of discontinuities in f not "noticed" by F. Suppose f(x) is 1
if x=1/n for n in Of course if we make f non-zero on a "thicker" set then we may run into trouble. We have already seen examples where f is not Riemann integrable as a result (f(x)=1 if x is rational, or even f(x)=x if x is rational).
If we had also verified the Mean Value Theorem, then we would know
that two differentiable functions defined on the same interval with
the same derivative would differ by a constant. That would be enough
when combined with our version of FTC to prove that if G'=f on [a,b]
and f is Riemann integrable, then
int Please look at the review material for the final. | ||||||||||||||||||||||||||||||||

5/1/2003 |
The final is scheduled for
## Tuesday, May 13, 12:00-3:00 PM in SEC 205Review session on Saturday,
May 10, at 1 PM in Hill 525office hours in Hill 542 on Monday, May 12, from 1
PM to 5 PM. Almost surely I'll be in my office most days next
week, and I will also respond to e-mail. I will try to produce some
review material to hand out on Monday. We will cover a version of the
Fundamental Theorem of Calculus in the last class. Professor Cohen has created even more notes on Riemann sums, a total of 21 pages now. Please take a look. Material related to what I discuss today is covered in the textbook in section 5.6 (see 5.6.1 through 5.6.4) and in section 7.2 (see 7.2.7). In probability one builds models of "chance". The cdf (cumulative distribution function) of a random variable X, which is defined by f(x)=probability{X<=x} contains most of the useful probability information. Quantities such as the mean (expectation) and the variance can be computed from it (usually involving various integrals). The function I'd like to study today has the essential properties of a cdf which are listed below. I won't discuss a probability "model" that this function might come from. - f's values are in [0,1].
- The limit of f(x) as x-->infinity is 1.
- The limit of f(x) as x-->-infinity is 0.
- If x<y, then f(x)<=f(y).
- If a is in
**R**, then the limit of f(x) as x-->a^{-}exists. In the language of 311, this is the limit of f(x) as x-->a when the domain of f is (-infinity,a). - If a is in
**R**, then the limit of f(x) as x-->a^{+}exists and equals f(a). In the language of 311, this is the limit of f(x) as x-->a when the domain of f is (a,infinity).
is Math 311, I should prove something. So I
will show that if property 3 holds (so if x<y, then f(x)<=f(y)),
then property 5 is true (If a is in R, then the limit of f(x)
as x-->a^{-} exists.).Theorem Suppose for all x, y in R, if x<y, then
f(x)<=f(y). Then the limit of f(x) as x-->a^{-} exists and
is equal to sup{f(x):x<a}.Proof: Let's call the set {f(x) : x< a}, W, and call its sup, S. Why should S exist? We will use the completeness axiom. W is not empty (since, for example, f(a-33) is in W). W is bounded above, and one upper bound is f(a) (this uses the increasing assumption, of course). Therefore S exists using the completeness axiom on the set W, which is non-empty and bounded above. I claim that the stated limit exists. This is a one-sided limit, so the following implication must be verified: if epsilon>0 is given, then there is a delta>0 so that if a-delta<x<a, then S-epsilon<f(x)<=S. Since S is sup{f(x) : x< a}, given epsilon>0, there will be w in W so that S-epsilon<w<=S. But w is a value of f, so that there is v<a with f(v)=w. Take delta to be a-v. Then if a-delta<x<a, we know that a-(a-v)<x<a, so v<x<a. Since f is increasing, we may "apply" f to this inequality and get f(v)<=f(x)<=f(a). This means that w<=f(x)<=f(a). But w>S-epsilon, and since x<a, f(x)<=S because S is the sup of W. Now we have S-epsilon<f(x)<=S, which is what we wanted and the proof is done. Of course a similar statement is true about limits "on the right" with sup replaced by inf.
f(x)=the sum of 1/2
This is a weird definition. Since the range of B, the bijection, is
only the rationals in (0,1), if x is less than or equal to 0, there
are Things will get even more interesting when we look at #4. If x<y and x and y are in the unit interval, then the interval (x,y) has infinitely many rational numbers between 0 and 1 in it. Therefore f(x) must be strictly less than f(y). Thus, on [0,1], f is a strictly increasing function: f(x)<f(y) if x<y. This is more than #4 requires.
We saw that #4 implies that the left and right hand limits exist. So
all we need to do is investigate where f is continuous. Well, since
lim
Let's imagine an example, where, say B(17)=3/7. If x<3/7, the sum
for f(x) would not have the 1/2
In fact, f has a jump of 1/2
In [0,1], this f is continuous at every irrational number and is
continuous at 0 and at 1, and is
Is f Riemann integrable on [0,1]? Of course this is the same as asking
if, given epsilon>0, we can find a partition
What gen:=rand(1..99999);This asks Maple to create a "random" integer between 0 and 99,999. A:=[seq((gen()/100000),j=1..30)];n:=30;This asks Maple to create a sequence of "random" rationals (the integers divided by 100,000) in the open unit interval. It assigns this sequence to the name A. The next statement creates the variable n with value 30. bin := proc(x) local y, inc, j; global A, n; y := 0; inc := .5; for j to n do if A[j] < x then y := y + inc fi; inc := .5*inc od; RETURN(y) end;The procedure bin uses the global variables A and n to get
values of the function f(x) depending on the specifications of A and
n.
area := proc() local y, j, inc; global A, n; y := 0; inc := .5; for j to n do y := y + (1. - A[j])*inc; inc := .5*inc od; RETURN(y) end;This Maple procedure gets the area of the approximation to the function. I then plotted the approximation with the command plot(bin,0..1,thickness=3,color=black);One value of A is this: 8193 79919 3341 63119 38091 1436 18757 8339 69373 [-----, ------, ----, ------, -----, ----, ------, -----, ------, 50000 100000 6250 100000 50000 3125 100000 20000 100000 4197 12881 29181 23023 41 89573 79983 287 4481 -----, -----, ------, -----, ----, ------, ------, ---, -----, 10000 50000 100000 50000 1000 100000 100000 800 12500 12149 2917 51331 13887 327 23963 9167 1109 ------, ------, ------, ------, -----, -----, -----, ------, 100000 100000 100000 100000 20000 25000 12500 100000 29977 94989 13559 9011 ------, ------, -----, -----] 100000 100000 50000 20000The associated graph is shown and it has approximate area .57585733. One surprising aspect of the graph to me was the enormous flatness of most of it: but of course the graph is not "flat" anywhere (any interval has infinitely many rationals, so the graph must increase). The amount of increase is mostly very very very small. Maple also displays the vertical jumps with vertical line segments.
I don't know or understand very much about the possible values of
int
Problem Is there a bijection B which has
int_{0}^{1}f=1/2? I don't know. In fact I don't know
any specific value (or non-value!) of
int_{0}^{1}f. Certainly fairly easy reasoning (moving
around the big blocks) shows that the values of
int_{0}^{1}f are dense in (0,1), but I really
don't know the answer to the question just asked. I suspect it is "yes".
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4/30/2003 |
Tomorrow I will give out student evaluation forms. Also tomorrow I
will request information on when I can usefully be available before
the final exam. The final is scheduled for
## Tuesday, May 13, 12:00-3:00 PM in SEC 205
Historically probability has been the inspiration of many of the more intricate results about integration. I now take a small detour to present a complicated example of a function. First I tried to give some background on probability.
Probability originated
in the 1600's in an effort to predict gambling odds. Here's the
basic idea, as it is now understood. One plays a game "many" times and
observes the outcomes. A quotient called the "relative frequency" is
computed: this is the (number of outcomes of a desired type) divided
by (total number of times the game has been played). Of course
relative frequency is a number between 0 and 1. Now the idea or hope
or desire is that as the(total number of times the game has been
played) gets large (approaches infinity?) this relative frequency
should somehow "stabilize" or approach a limit. This limit is called
the
^{n} as n goes from 1 to infinity is 1), so its probability
must be 0! So here is a conceivable event which happens hardly ever,
according to this model.
The most famous results of probability deal with repeated experiements and the tendency of random variables to have nice "asymptotic" properties. One such result is the Central Limit Theorem, which essentially states that the normal curve rules every repeated experiment. Here are two applets simulating the CLT, one with dice and one with a sort of pachinko-like "game". Such results are usually understood and investigated using the cumulative distribution function, cdf, of a random variable X. So cdf's are extremely important in probability.
If X is a random variable, the Some effort is needed to be acquainted with this definition. Let's look at our three random variable examples, and graph their cdf's.
- f's values are in [0,1].
- The limit of f(x) as x-->infinity is 1.
- The limit of f(x) as x-->-infinity is 0.
- If x<y, then f(x)<=f(y).
- If a is in
**R**, then the limit of f(x) as x-->a^{-}exists. In the language of 311, this is the limit of f(x) as x-->a when the domain of f is (-infinity,a). - If a is in
**R**, then the limit of f(x) as x-->a^{+}exists and equals f(a). In the language of 311, this is the limit of f(x) as x-->a when the domain of f is (a,infinity).
Sometimes people say that properties 5 and 6 mean that the cdf is a
In the case of continuous random variables, another function is sometimes studied, the density function. This turns out to be the derivative of the cdf, and its utility for discrete random variables is not immediately clear. (What should the derivative of a mostly horizontal function be?) So I will just look at cdf's here, today and tomorrow. | ||||||||||||||||||||||||||||||||

4/28/2003 | Again we are going through the technicalities on integral and order,
integral and linearity, and additivity of the integral over intervals.
This takes effort and discipline, but Math 311 is the course
whose total object is constucting calculus (also called "analysis") with all
the interconnections showing. So let's move on and finish up these
technicalities.
Initially I want to show that h is Riemann integrable. I will use the
necessary and sufficient condition with epsilon. What do I mean? I
must show that given epsilon>0, there is a partition
We need to show that
int
The last three propositions can be abbreviated by writing that:
_{0}^{5}g=13, and that |g(x)|<=3 for all x in
[0,2]. What over and underestimates can you make about
int_{2}^{5}g and why?
Several (dubiously) interesting (?) linguistic comments were made during today's class. First was MEGO. The web site www.acronymfinder.com reports that this means "My Eyes Glaze Over (during a boring speech or briefing)". I asked for the source of the quotation "sup of the evening, beautiful sup" -- this was a misspelling of the word "soup" and the phrase comes from chapter 10 of Lewis Carroll's "Alice in Wonderland" where it is the first line of a song that the Mock Turtle sings. John Tenniel's historic illustration is shown. The Mock Turtle also discusses its education, and remarks that it studied `the different branches of Arithmetic-- Ambition, Distraction, Uglification, and Derision.' Lewis Carroll was actually an academic mathematician at Oxford University named Charles Lutwidge Dodgson. Biographical information is abundant. | ||||||||||||||||||||||||||||||||

4/24/2003 |
The Question of the DaySuppose f: R-->R is defined by f(x)=5 when x=3 and
f(x)=-9 when x=6, while f(x)=0 for all other x's. Is f Riemann
integrable on [2,7], and, if it is, what is the Riemann integral of f
on that interval?Answers: Yes, and 0.
I began by observing that the special arguments last time actually
proved more than I stated.
We reconsidered an example discussed on 4/14/2003 (see the material on boxes
and butterflies): the function sqrt(x) on the interval [0,1]. Since
f'(x)=x
We will use a simple partition again. So
An example or twoI will assume the standard properties of sine for these examples (we could instead get functions whose graphs would be polygons with similar properties, but would it be worth the trouble to define them?).
Here is a picture of the function f(x)=sin(1/x) for x not 0, and
f(0)=0. This function is continuous in [0,1]. If (x
Now consider the function sin(1/x) on (0,1]. A first observation is
that there's no way to define this function at 0 so it will be
continuous there. That's because it is possible to find sequences
(x
Given epsilon>0, we need a partition This web page has links to notes of Professor Cohen which we are (approximately) following. Please look down the page and find "Riemann Integral, Section 1".
Now apply these observations to the collection of upper sums of g and
the upper sums of f. Note that US(g, | ||||||||||||||||||||||||||||||||

4/23/2003 |
We used the lemma proved last time to verify the following Theorem (upper sums dominate lower sums) If S and
T are any partitions of [a,b], then
US(f,S)>=LS(f,T).Proof: Here's the proof, which is very witty. Last time we proved that more points in the partition may make the upper sum decrease, but can't make it increase. A similar result (reversing directions, though!) is true for lower sums. Therefore, if P is the union
of the partitions S and T, we have the following
sequence of inequalities:LS(f, T)<=LS(f,P)<=US(f,P)<=US(f,S).The central inequality (between LS and US for P) is true
because sups are bigger than infs, always.
The theorem just stated presents us with a situation which should be
familiar from earlier work in the course (a month or more ago). The
sets A and B have the following properties: if a is in A, then a is a
lower bound of B, and if b is in B, then b is an upper bound of A. It
is natural to look at the sup of A and the inf of B. Here we will use
special phrases: If f is example 3 of the last lecture (0 on the irrationals, and 1 on the rationals, and the interval is [0,1]) then A={0} and B={1}, not very big sets, and not very complicated!
How can we tell if the Riemann integral exists, and how can we get
interesting examples? We will begin with this theorem: Now we need to verify the "epsilon condition" implies Riemann integrability. Remember what the sets A and B are. Since every element of B is an upper bound of A, and every element of B is a lower bound for A, we already know that sup A<=inf B. Why is this true? Well, if sup A>inf B, then take epsilon= sup A-inf B>0. We can find (sup chracterization) a in A so that sup A>=a>sup A-epsilon. a is then greater than inf B=sup A-epsilon. But we could then (inf characterization) find b in B with a>b>=inf B, which contradicts the known fact that a<=b for all choices of a in A and b in B. How can we prove Riemann integrability? This condition is exactly the same as proving sup A=inf B. Since we already know sup A<=inf B, let us see what happens when sup A<inf B. Then take epsilon=inf B-sup A>0. The assumption in the statement of the theorem says we can find a in A and b in B with b-a<"this" epsilon. So b-a<inf B-sup A. But certainly a<=sup A and inf B<=b, yielding (since -sup A<=-a) inf B-sup A<=b-a. This is a contradiction! Whew. (The logic in all this is a bit intricate, but is very similar to lots of proofs we did a month or two ago.)
Even better is the following result: With this result (whose proof I postponed until next time) we will actually be able to effectively recognize some Riemann integrable functions and maybe compute some integrals.
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4/21/2003 |
The problem of areaWhat is area? This is a serious geometric question and difficult to answer. Usually we would like "area" to be something satisfying the following rules: - The area of a piece of
**R**^{2}should be a non-negative number. - The area of bigger pieces should be a bigger number.
- The area of congruent pieces should be the same.
- If two pieces don't overlap or overlap only at the boundary, then the area of the union of the two pieces should be the sum of the areas of the pieces. More generally, if you split up pieces of the plane into subpieces, the areas of the subpieces should add up to the area of the original piece.
- (Normalization) The area of the unit square should be 1.
lots of
explanation. One book that I've read recently which discusses the
classical Euclidean approach to this problem and others is
Hartshorne's Geometry: Euclid and Beyond which I would
recommend for those who want to study the axiomatics of classical
geometry, now thousands of years old. The book isn't easy, but it has
a great deal of content. The idea of a "piece" of the plane is
certainly imprecise. "Bigger" in the case of numbers is <=. In the
case of pieces of the plane, it probably should mean "is a subset
of", so larger sets have larger areas. The word "congruent" probably
means, as was said in class, the same shape and size: this means that
if we translate or rotate or flip sets, the results should have areas
equal to the sets we started with. In the case of sets being "split
up" they should not have overlapping interiors: only the boundaries
are allowed to overlap. Whew!
Just an initial statement of these properties is awesome. What is more
distressing is the following statement, whose verification needs more
time than this course has to run: there is
We will follow the lead of Cauchy and Riemann in this. Bressoud's book
(referenced in the general background to the course) discusses some of
Cauchy's ideas and shows that some of what Cauchy wrote was just
We will start with a function f defined on [a,b]. We need to split up
the interval. The word "partition" is used both as a verb and as a
noun in this subject. As a verb, partition means to break up the
interval into subintervals. As a noun, currently more important,
"partition"
The
So the functions of examples 1 and 2 are Riemann integrable, with integrals of value 1/2 and 0 respectively. The function in example 3 is not Riemann integrable.
The following result plays an important part in any development of the
Riemann integral. On Wednesday I hope to give out some notes written by Professor Cohen which will outline this material. | ||||||||||||||||||||||||||||||||

4/16/2003 |
The instructor addressed the question, "Where do we go from
here?" and then answered even more inquiries in preparation for the
exam. I brought in the text (a standard calculus book) used for Math 151 and compared what's in Chapter 6 of the Math 311 textbook with that text. The approach and even many of the pictures, are the same. What happens? - The definition of derivative; a differentiable function is continuous.
- If a function has a local extremum and is differentiable there, then the derivative is 0 at that point.
- Rolle's Theorem (a special case of the MVT): a function which is differentiable in [a,b] with f(a)=f(b)=0 must have at least one c in (a,b) with f'(c)=0.
- Mean Value Theorem (tilted form of Rolle's Theorem): if f is differentiable in [a,b], there is at least one c in (a,b) with f'(c)=(f(b)-f(a))/(b-a).
- f' positive in an interval implies f is increasing there; f' negative in an interval implies f is decreasing there.
- More topics: l'Hopital's rule; Taylor's Theorem.
In the balance of time remaining for the class I tried to answer more questions in preparation for the exam. Maybe the only interesting comment was my remembering a "workshop" problem from calculus. The problem was something like this:
Aliens change and permute 10,000 values of the function
F(x)=x
The answer is that lim I worked on some other textbook and workshop problems, just as I had in the review session the night before. |

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