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## Any good calculus places to start?

i tried to make my notes at most one page per lecture, so they comprise a less than 45 page differential calculus book. They are quite hard and slow to read, but i hope that someone with patience can actually learn more from those 2 pages I posted than from whole chapters of some books. But that person must accept that they should be read very slowly. Also, i am here to answer questions on them. Probably they should be taken as thought provoking starts for discussion with friends.
 You did supremum on your second day? That's insane.
 Recognitions: Homework Help Science Advisor well i may be insane, but my position is that we are brainwashed into thinking certain simple ideas are hard by the way they are presented. ask a class why .9999.... is supposed to be equal to 1. the simplest way to say this i know of is to say that 1 is the smallest number larger than all the finite decimals of form .999......9. calling it the "sup" just makes it sound hard. if we expect people to use words like "limit" we need them to begin with easier ideas. what you call "sup" is just the limit of an increasing sequence. i.e. general limits are much harder than "sup" 's. I claim that you can teach anything you actually understand, but maybe not so well if you just parrot formal definitions. of course this one page of notes is a very short version of a one hour lecture full of examples. I have also taught euler characteristics to 2nd graders, quite successfully, not of course by defining them in terms of homology groups, but by handing out colored cardboard polyhedrons. I also suffer from the belief that some of my students are very bright and will benefit from being told what is actually going on behind the jargon. Usually some of them prove me right, and they deserve all i can give them. i do not penalize the others for not getting it. I try to teach ambitiously and grade generously.
 I never said it was hard. I just thought it may be a little dense for the second day of calculus. :) I like teachers who reveal the inner workings to those ready to learn them. Not many do though, for fear of confusing the crap out of those not ready While self-studying it, I've noticed a lot of calculus stuff that I learned in Alg II. I'm taking it now, *so far*, the only difference between Alg II(HS algebra) and Calculus is that leaving off the remainder of long division of polynomials has a new name, "removing discontinuity," and it all has to have "lim" in from of it. From my self-studying, I know there's plenty to be covered that wasn't in Alg II, but I'm speaking in terms of so far.
 Recognitions: Homework Help Science Advisor you seem like a sharp guy tyler, and i apologize for throwing out these 2 minute synopses of a one hour lecture, but here is day three, using least upper bounds again to prove the area formula for a circle. of course this is just reinforcement of an hour lecture with lots of pictures and entertaining singing and dancing (just kidding). sometimes kids (i.e. it happened at least once) come up after this lecture and say this is the first time they have ever understood why this formula is true. are you a high school student? you seem very advanced for that. I think I'll try tossing you some of my more creative stuff on limits, if it will copy here, but i doubt it will. Day 3, 2200, jan.18, 2000. Area of circle as a least upper bound Recall the definition of the “least upper bound” of a set S of real numbers, is “the smallest real number not smaller than any number in the set S”. Then 1 is the least upper bound of the infinite set of real numbers of form {.9, .99, .999, .9999, ...........}, since 1 is larger than all these numnbers, and yet these numbers eventually become larger than any number less than 1. Indeed this is what it means to say the infinite decimal .9999999..... represents (i.e. equals) the number 1, since the real number represented by an infinite decimal is the least upper bound of the set of all finite decimals obtained by truncating the infinite decimal at all finite stages. E.g. the real number represented by the infinite decimal .333333....., is by definition the least upper bound of the sequence of finite decimals {.3, .33, .333, .3333, .......}, and equals 1/3. Recall that area is measured in “square units” and that the area of a plane region is intuitively the number of unit squares that will fit inside the region. We make sense of the area of regions such as rectangles, parallelograms, triangles, and polygons, by cutting up a certain number of unit squares and reassembling them to fit inside the region. Then consider a region such as a circle, where no amount of cutting and reassembling can ever make a finite number of unit squares fit perfectly inside the region. One way to make sense of the concept of area here is to approximate the area of the circle by the areas of an infinite number of simpler regions for which we understand area better, such as inscribed polygons. We define the area of a circle as the least upper bound of the areas of all (convex) polygons which can be inscribed in the circle. We will use the Pythagorean “distance formula” to measure the length of a line segment and we define the length of a circle to be the least upper bound of the lengths (i.e. perimeters) of all polygons which can be inscribed in the circle. We define the number π to be the ratio π = C/2r of the circumference to twice the radius of any circle, (accepting that this ratio is the same for all circles). We argue that the area of a circle of radius r is πr^2 as follows: the area of the circle is the lub of the areas of all inscribed polygons. Looking only at regular polygons, i.e. ones with all sides the same length, seems acceptable. Then we see by dividing the polygon up into triangles that the area of the polygon is half the product of the perimeter of the polygon times the common height of each triangle, (where that common height is the distance from the center of the polygon to the center of any one edge of the polygon). We agree the lub of this product of increasing factors, is the product of the lub’s of the factors. Thus the area of the circle is 1/2 the product of the lub of the heights of the triangles times the lub of the perimeters of the polygons. But the lub of the heights of the triangles is the radius of the circle, and the lub of the perimeters of the polygons is the circumference of the circle. Thus the lub of the areas of the inscribed polygons is 1/2 the product of the radius of the circle times its circumference, i.e. A = (1/2)(2πr)(r) = πr^2.

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here is a detailed actual lecture from the beginning of the theory of area.
Attached Files
 2210 2003 01 areas, vols.pdf (91.6 KB, 12 views)

Thanks. I'm going to keep a local copy. :)

Yeah, I'm a junior in high school. I took Pre-Calulus last semester(with a 100, I might add :P), and am taking Calculus at the local community college this semester.

 Recall the definition of the “least upper bound” of a set S of real numbers, is “the smallest real number not smaller than any number in the set S”. Then 1 is the least upper bound of the infinite set of real numbers of form {.9, .99, .999, .9999, ...........}, since 1 is larger than all these numnbers, and yet these numbers eventually become larger than any number less than 1.
I'm going to have to call circular argument here. You are assuming 1 <= .999... to prove 1 = .999... :)

Instead, take a simple property of the reals: There are infinitely many members between any two members. What's a number between .999... and 1? There is none. Therefore, by the contrapositive of the aforementioned property, .999... = 1.

EDIT: I just noticed you never explicitly said that it was a proof .999... = 1. But, I get the feeling it was implied, no?
 What happened to the simple old 1=(1/3)*3=(.333...)*3=.999...?
 How do you prove 1/3 = .333... ? That is, without using something not "simple" and "old". :P
 That's a trivial property of fractions. Some can be represented as infinite decimals. Are you going to heckle me to prove what the equal sign means too?
 Recognitions: Homework Help Science Advisor I don't understand your objection to my proof Tyler. I used the fact that 1 is greater than each of the finite decimals .9999. That follows from the definition of the ordering on the finite decimal numbers. Then it follows from the definition of the infinite decimal .99999...... as the smallest such number, that 1 is at least as great as .9999.......

The least upper bound, as you know, is the least number greater than all the numbers in the set. In an argument about whether .999... = 1, you must assume "1 <= .999..." to say "sup{.9, .99, .999, ...} = .999... = 1." If you assume, as the opposition would likely be arguing, that ".999... < 1", then "sup{.9, .99, .999, ...} = .999...," because .999... is larger than all the numbers in the set, but as assumed, is smaller than 1. In other words, the result relies on the assumption(ie uses circular logic).

Your assumption that .999... = 1 would be correct, so your result is correct. It's just not a valid proof of .999... = 1. I realize you never said it was a proof that .999... = 1, but if it was, it uses circular logic. If it was never intended as a proof that .999... = 1, then that was a fallacious assumption on my part. :)

These notes are awesome. Do you have them on a website somewhere?

 That's a trivial property of fractions. Some can be represented as infinite decimals. Are you going to heckle me to prove what the equal sign means too?
It wasn't my intention to heckle. I'm sorry if that is how my post seemed.
 Recognitions: Homework Help Science Advisor The harder part is to show that no number smaller than 1 is as large as all those finite decimals. Note that those finite decimals of form .9999. differ from 1 by a finite decimal of form .0001. So if there were a number lying strictly between 1 and all those finite decimals, it would differ from 1 by a positive number which is less than every number of form .000000......0001. I quit there in an elementary class saying that cannot happen. But for you, here is a sketch of the slightly tedious argument: It amounts to showing there is no positive number smaller than all those numbers of form .000000......0001. Now any non zero finite decimal has a first non zero digit in some position, and if we put a 0 in that position and follow it by a 1, we have a smaller number of the form .000000......0001. Thus no finite decimal can be smaller than all those. As for an infinite decimal, it is at least as large as all its finite truncations by definition, and we can find a number of form .000000......0001. that is strictly smaller than one of those truncations. Thus also every infinite decimal is larger than some number of form .000000......0001. That does it. I once taught a class on real numbers as finite and infinite decimals to a high school class, and wrote 50 pages of notes just on these ideas. One of my ex - students from that class is now a professor of mathematics at Brown University.
 Recognitions: Homework Help Science Advisor I still don't understand your objection. To prove that 1 is the lub of all the numbers of form .99...9, all I need to do is show two things: 1) 1 is at least as great as all those finite decimals, 2) no smaller number than 1 is also as great as all of them, i.e. any number smaller than 1 is also smaller than one of those finite decimals, or equivalently no positive number is smaller than all the differences 1 - .999.....9. = .000...001. But part 1) is true by definition of the lexicographic order on finite decimals. Part 2) is proved in post 30. If you like, I am showing that 1 is less than or equal to .99999........, not directly, but by showing it is less than or equal to EVERY upper bound of all the finite numbers .9999....9. In particular, since .999....... is such an upper bound, 1 is less than or equal to it too.
 Recognitions: Homework Help Science Advisor I have lots of notes just sitting on my computer, and some on my website, but not any of the calculus notes. The web site has mostly abstract algebra notes and a few advanced algebraic geometry notes.
 Recognitions: Homework Help Science Advisor Ah I see your objection, you are assuming as an honest person would do, that I will give a direct proof. But I am giving instead a very roundabout one. To show 1 is the lub, I show that 1 is itself an upper bound, and then that no number smaller than 1 is an upper bound, hence any other upper bound is at least as large as 1, so 1 is the least upper bound. Tedious, but I claim correct.

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