Allearia Posted March 10, 2013 Share Posted March 10, 2013 My ds11 has suddenly become passionate about math. He has always liked it, but for the past year he has been working so hard on Greek and Latin he hasn't done as much. We started AOPS Algebra a few months ago, and at the same time got some Life of Fred books. He loves LOF and got Geometry, Trig, and Advanced Algebra and begged me for weeks to buy Calculus which the library didn't have. I finally bought it and he has been doing it every spare minute. I have no background in calculus at all and am wondering, is he really understanding the calculus? He got through ch 4 of AOPS algebra, then skipped to ch 12 and 13, then he really just read through the other LOF books. I am making him go back and do the other algebra chapters he skipped. It seems a bit silly to me to make him stop working on calculus so he can do his algebra. So maybe I should just let him do whichever of the math books he wants. But I don't want him to miss stuff he should know from algebra, geometry, algebra 2. Will he hit a wall in the calculus and go back? Or maybe he can just get what he can out of it and then when he gets to calculus again he will get more out of it. I also think that the idea of being in calculus already appeals to him a lot. I am fine with letting him just enjoy all this math, and I don't know exactly what question I am asking. How much should I insist he go back to chapter 4 of algebra? Or do I just turn him loose and let him go as long as he wants with whichever of these topics he wants to do? Since I have no idea what he may have missed in between that he would need for calculus i am a bit lost. We are going to be starting in a math circle this week so hopefully that will feed some of this. Any thoughts? Quote Link to comment Share on other sites More sharing options...

regentrude Posted March 10, 2013 Share Posted March 10, 2013 In order to understand the basic ideas behind calculus, he needs to know what a function is. Calculus is based on two fundamental ideas: the derivative, i.e. finding the slope of a tangent on a function, and the integral, i.e. finding the area under a curve. A young student can understand these two ideas if he has a concept what a function is and that it can be represented by a curve on an x-y graph. In order to actually perform calculations with these concepts, algebra, geometry and trigonometry are needed. So, as soon as he understands the concept of a function, he can understand the basic idea of calculus - but he will not be able to do anything with it until he has completed the other math courses. I am making him go back and do the other algebra chapters he skipped. It seems a bit silly to me to make him stop working on calculus so he can do his algebra. No, this is not silly at all. He must master algebra first if he wants to be able to do any calculus, beyond a rough understanding of the basic idea. I teach calculus based physics at the university. The students who struggle with math do so because their algebra and prealgebra skills are insufficient, not because they don't know calculus. Quote Link to comment Share on other sites More sharing options...

8filltheheart Posted March 10, 2013 Share Posted March 10, 2013 Ummmm, I think he is reading the story in LOF and maybe "thinks" he understands how to do the math. You can't do calculus with a strong foundation in alg 1 and alg 2 as well as trig. Going back to chpt 4 of alg would be non-negotiable in our house. ;) Quote Link to comment Share on other sites More sharing options...

dbmamaz Posted March 10, 2013 Share Posted March 10, 2013 i suspect he will notice that he cant do parts of it, and he will realize he needs to go back and review concepts. i tend to let kids plow forwards when they want to . .. i dont want to discourage passion. plus, my kids are very argumentative. i would probably warn that there are some things you might not understand because of what you skipped, and that when you find that, you might want to go back - but then let them decide if its time to go back or not Quote Link to comment Share on other sites More sharing options...

regentrude Posted March 10, 2013 Share Posted March 10, 2013 i suspect he will notice that he cant do parts of it, and he will realize he needs to go back and review concepts. i tend to let kids plow forwards when they want to . .. i dont want to discourage passion. plus, my kids are very argumentative. i would probably warn that there are some things you might not understand because of what you skipped, and that when you find that, you might want to go back - but then let them decide if its time to go back or not I disagree. Many of my students (science and engineering majors!) have never realized that their algebra skills are insufficient - until they struggle in college. Letting the student decide when he wants to go back and learn the foundations of math is a very bad idea, IMO, because the student has no idea what he is not understanding. It is the job of the teacher (or parent in this case) to make sure there are no holes; the learner will not be able to detect them. ETA: I would not discourage the student from pursuing his studies of calculus, if he is interested. I would, however, insist that this happens in addition to the algebra instruction, not instead. Quote Link to comment Share on other sites More sharing options...

8filltheheart Posted March 10, 2013 Share Posted March 10, 2013 ETA: I would not discourage the student from pursuing his studies of calculus, if he is interested. I would, however, insist that this happens in addition to the algebra instruction, not instead. I agree, but I would offer other resources like http://www.amazon.com/Calculus-Young-People-Worksheets-Donald-Cohen/dp/0962167452 There is another source but I cannot remember the name......something like calculus for children or hands on calculus?? I am drawing a complete blank. ETA: Found it: http://www.amazon.com/Calculus-Without-Tears-Learning-Students/dp/0976413809/ref=rec_dp_1 (this is a multi-book series.) Quote Link to comment Share on other sites More sharing options...

Alte Veste Academy Posted March 10, 2013 Share Posted March 10, 2013 I think perhaps the worst potential consequence of going ahead before he is ready is that he will inadvertently extinguish his passion. I would find ways to work sideways before I let my kid put the cart before the horse. There are oodles of wonderful, beautiful, exciting things that you can load onto the cart, but I do think the cart should always be behind the horse. :D Quote Link to comment Share on other sites More sharing options...

mumto2 Posted March 10, 2013 Share Posted March 10, 2013 I recently bought the rest of abeka's math program for just that reason. If they are missing the foundation their knowledge is never going to be complete. Ds is not happy with me about this and it is boring but needed. My ds has always been great at math. Done exceedingly well in some competions. Big math ideas he loves. He has read most of the LOF books which is very different from doing them. He has done halfway through Algebra 2. He has also picked up much from dd who does her work and is advanced but complete in knowledge. He likes to jump around. AoPS, LOF, NEM, and other topics from the library. What this makes him is a kid with big math ideas missing basic skills. Sometimes just missing terminolgy but usually the one lttle thing that makes his work 100% right. I also have to fight with him over writing his work down. At best I have a few chicken scratches. He is also careless with his daily work even at home tests. After a couple of frustrating months where I tried to figure out what was missing in algebra work(NEM tests were his downfall--daily exercises were great) I gave up and ordered the abeka. Right now just working through the quizzes and tests mainly. Going to the text for more when one is missed. I have found a couple of missed concepts. I know there will be more. Is abeka the best for this? Honestly probably not but for me since I am most comfortable with it and great price used the job will get done. Quote Link to comment Share on other sites More sharing options...

Mukmuk Posted March 11, 2013 Share Posted March 11, 2013 We had a month of this at home last year. Over December, ds found a copy of the Cartoon Guide to Calculus at the bookstore, insisted I buy it, and read it non stop. He was very excited about f(x) - he understood the idea behind it and could do some of the easy ones- but as the book progressed, he came to the conclusion that he couldn't quite grasp it. I thought of it as more as a supplement than anything, like his Murderous Math books. It was very nice to see him so excited about the idea, but really, in my son's case, he doesn't have the basics, and the book is too light to be a text. We didn't stop with AOPS. I have no doubt that he'd be taking the Cartoon Guide out again in future when the does get to calculus, He still does this with his Murderous MAth books - when he comes across something he's seen in the MM books but didn't have a full grasp back then, he returns to read for the upteenth time. ETA: Oh wow, there are other resources too. Tks! Quote Link to comment Share on other sites More sharing options...

brownie Posted March 11, 2013 Share Posted March 11, 2013 I would not want to extinguish his passion. I would make a deal that if he got the math work you assigned done 4 days a week, he could use the 5th day's math time to do whatever he wants. Don't tell him he's not ready for calculus, just that you want him to do algebra. Also, I would of course allow him to do whatever math he wants to do with his free time! Quote Link to comment Share on other sites More sharing options...

Kathy in Richmond Posted March 11, 2013 Share Posted March 11, 2013 It's terrific that your son is intrigued by advanced math topics, and I'd strive to keep that spark alive. If he were my kid, I'd have him continue working through the AoPS text thoroughly, but I'd set aside an hour or two a week (or whatever time you have) for math explorations. The Calculus by and for Young People book would be very nice at your son's age. It's not a complete calculus course, but rather a mathematically appropriate introduction to the important concepts of calculus that a motivated & smart young student can understand. It works best if there's an interested adult to work through the ideas & worksheets with the child. Also, take a look at Martin Gardner's books. Aha, Insight! and Aha! Gotcha!are especially good for his age & interest level. Quote Link to comment Share on other sites More sharing options...

Mukmuk Posted March 11, 2013 Share Posted March 11, 2013 We kind of enrich math like what Kathy suggests. We do AOPS, followed by a second segment that has its own spin. Right now, that second segment is Patty Paper Geometry and an AOPS class about math that is familiar to him, but requires different thinking. He loves the energy of the class interactions :). These are further enrichment that your son may like (not necessarily calculus though). Sorry to veer from your question, OP, but I recall where ds got his interest, which may be helpful to others. Ds got his first peek from Zacarro - Primary Grade Challenge Math uses the concept of function machines constantly, and there is a little chapter on calculus in Challenge Math. This is tantalising fruit for young children :). (I've never done calculus too, but I'm working on my own mindset and will (y.e.s.) work alongside him when the time comes. ) Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 12, 2013 Share Posted March 12, 2013 I agree with others here that my experience as a college math teacher has taught me that algebra mastery is the essential to success in working with calculus. As to the ideas, there is some confusion to me as to just what the word "calculus" actually refers. There are two fundamental problems, finding tangent lines to curves in the plane, and finding areas under (bounded by) such curves. These two problems are often called the subjects of differential and integral calculus respectively. But the first problem was studied and largely solved by Descartes and Fermat, before the time of the men usually credited with inventing the calculus, namely Newton and Leibniz. The second problem was largely solved in some sense even earlier by Archimedes. Both problems involve the method of "limits", or approximations, and some people think of calculus as concerned with the method of limits, even though this method was used thousands of years ago by the ancient Greeks. Moreover there are two aspects to each problem, first that of defining a solution precisely, i.e. of defining a tangent or an area precisely as a limit, and second that of computing such a limit in particular cases. Even the famous limit definition of a tangent line usually attributed to Newton, can be found clearly stated for tangents to circles in Prop. III.16 of Euclid. Archimedes also seems to me to have clearly understood the limiting definition of area as well as of arc length. Euclid also discusses volumes as limits. The essential new insight, it seems to me, was the realization that the two problems are linked. That if we start from one curve C1, and create a new curve C2 (or graph), where the height of C2 at a point is the cumulative area up to that point under the curve C1, then the height of C1 at a given point will be the slope of C2 at that same point. Thus if we start from curve C1 and create a new curve C2 whose height at each point is the cumulative area under C1, and then create a third curve C3 whose height at each point is the slope of C2, then C3 will have the same height at every point as C1. I.e. performing an area calculation followed by a slope calculation brings us back where we started. Thus the problems of computing area and slope are in a sense inverse to each other. This last insight it seems to me is the essence of the calculus. I.e. since it turns out to be rather easy to calculate slopes for curves whose heights we know, by reversing this calculation one can often compute areas for curves whose heights we know as well. Thus some people may feel that learning the theory of slopes alone, or of areas alone, constitutes an aspect of calculus, while others may feel that the calculus is the link between the two theories. If a curve represents the graph of a moving particle the slope represents velocity, and the curvature is acceleration. Then a related calculus problem is to compute the total distance traveled by the particle given the acceleration of that particle. Incidentally this problem was solved by Galileo, in the case of constant acceleration, using only geometry! In the last 15 pages of my epsilon camp Euclidean geometry notes from 2011: http://www.math.uga....camp2011/10.pdf I discuss the ancient Greek method of computing volumes and areas using limits, and relate it to the modern approach using calculus. I do not claim this is an explanation of "calculus", but it certainly makes a beginning on some of the main problems a calculus student learns to solve. The basic idea behind the "calculus" approach to areas is that two figures which meet every vertical line in segments of the same length, should also have the same area. Similarly, two solids which meet every horizontal plane in regions of the same area should have the same volume. This insight was known to Archimedes, and he used it to bootstrap from knowing the volume of one solid to deducing that of another. Archimedes thus knew that the volume of a solid is determined by knowing its slice areas, but he does not say just how to determine it in general. The modern advance made by Newton and Leibniz hundreds of years later goes from knowing a formula for the areas of horizontal slices of a figure to giving a formula for the volume of the figure. Perhaps my notes may be useful to some kids. I would be happy to answer questions about them. There is at least one misprint on line 18 of page 50, where the formula Ï€R^2 - Ï€x^2 for the slice area of a hemisphere at height x, is mistakenly said to be that of a sphere. Advice: read Euclid's Elements (esp.books 1-4), Euler's Elements of algebra, selected accessible passages of Archimedes, and Galileo: Two new sciences, esp. the second. Remark: What comes out of the modern formula based approach to calculus is that if a curve has height x^n above the x axis at point x, then its slope at that point is nx^(n-1) [reduce the exponent by one and multiply by the old exponent]. Consequently, the area under that curve (between the curve and the x axis) between the point 0 and the point x on the x axis, is (1/[n+1])x^(n+1), [raise the exponent by one and divide by the new exponent]. Thus the slope of the curve y = x^2 at the point x=1, is 2, and the area under it, from x=0 to x=1, is 1/3. Now does knowing these formulas mean one knows calculus?, or does that require understanding some of the principles Archimedes knew? or both? and their derivations? Well you have to start somewhere, so take your pick. Well, the person who wrote the wikipedia article seems to think that calculus is the abstraction of these ideas to the two processes differentiation and integration, applied to abstract "functions". He/she also seems to think calculus is about "change" and "infinitesimals". To me those are just ways of thinking about calculus, and not ways that I myself normally use. But they have their value. As to abstract formalisms derived from actual meaty examples, I tend not to think that way either, as it strips the ideas of their meaning for me. At least the wiki article writer does acknowledge that the person he credits with the first systematic treatment of calculus methods, namely Cavalieri, was anticipated in those ideas by many hundreds of years by Archimedes. A friend of mine wrote a wonderful calculus book that attempts to explain what calculus really means, along the lines of the wikipedia article, explaining the two inverse processes of differentiating and integrating, and also using infinitesimals, and I think it succeeds very well. The book is Calculus: the elements, by Michael Comenetz. Here is even a link to a free copy of the first chapter where that explanation occurs: http://www.worldscie.../4920_chap1.pdf and of course there is always the classic: Calculus made easy, by Sylvanus P. Thompson, which I was introduced to by another helpful student in the elite honors calculus class at Harvard, math 11 Fall 1960, who sympathized that the in class presentation was rather over my head. http://www.abebooks....easy&prevpage=3 On first pass, I felt the need to respectfully disagree with the author of Calculus without tears, aimed at 4th grade and up. In some sense there seems to be no calculus at all in that book, since the book restricts itself to the constant velocity case in discussing motion. That is the case that can be handled with algebra, hence is the sort of motion that is studied in a "physics without calculus" course. Calculus is needed precisely when the velocity is not constant. On further reflection, I think I know what the author means, since he does discuss the two processes of slope and area, or density and volume, but not in a case actually needing limits. In fact in trying to give an easy introduction to the fundamental ideas of calculus, I was led to imitate his approach in the post just below this one. In fact, searching it further, it seems that it is only volume 1, and there is a volume 2, in which motion due to gravity, the first case of non constant velocity is studied. So maybe this book is just going verrr..ry slowly, and perhaps there are yet other volumes doing more calculus. ( The author attributes to Newton the solution of the problem of motion due to force of gravity, whereas that problem was solved much earlier by Galileo in his Two New Sciences. Perhaps the author is thinking of some aspect of Newton's treatment, such as its use of more modern algebraic notation? as being more properly what we know as calculus today.) One can view many pages online of Calculus by and for Young People by Donald Cohen, and this book seems to examine the ideas behind the calculus very well. Especially chapters 13 and 14 introduce the study of areas under curves as they are introduced in a basic college calculus course, but more painstakingly and with careful use of examples. He very nicely allows the student to gather experience from several graduated examples which lead him/her to guess the insights to come, rather than just imposing them as formulas to memorize. In chapter 14 he seems to introduce what I would consider actual calculus by carefully exhibiting and allowing the student to discover, the connection between the area problem and the slope problem. This book seems to offer an excellent introduction to both ideas and computations of calculus. Again it seems to me that if "calculus" refers to the modern theory of Newton, it logically should mean the explanation of the link between the problems of area and slope. The theory of infinite sequences of approximations, and which gives a good understanding of each of the two problems on its own, was known to Archimedes, at least in the case of area. Moreover Euclid had essentially Newton's limiting definition of a tangent line, at least for the special case of a circle, in Book 3, Prop. 16 of the Elements. Hence it may be a little confusing to a reader who is told he is learning calculus as invented by Newton, when the book spends the first 13 chapters discussing mainly infinite series, and even citing Archimedes for the source of some of the discussion. In fact, once the link between the two problems is understood, it can be exploited to essentially remove the need for calculating with infinite series in the solution of the area problem. I.e. as long as one is using actual limits of infinite series to compute areas under curves one is not using the "fundamental theorem of calculus" at all. Conversely once one learns that fundamental theorem, one no longer needs any infinite series to calculate most elementary areas. The result is that most college students of calculus come away at best with a memorized technique for computing areas, depending on the fundamental theorem, but no understanding of how areas are analyzed precisely using infinite series. This allows yet another confusion in use of language. Does knowing "calculus" mean knowing the easy way to solve area problems while bypassing infinite series? or can it mean understanding infinite series without knowing the easy way provided by the fundamental theorem? Ideally one learns both. Unfortunately many students who think they know "calculus", do not know the limit definition either of the derivative or of the integral, but only know some examples such as: "the derivative of x^3 is 3x^2, so the area under the graph of y=3x^2, from x=0 to x=a, is a^3." This is indeed knowledge, but not understanding. I know I am going on far too long, and I apologize, but there is also a distinction between infinite approximations carried out by geometry, versus those studied by means of more modern symbolism, using algebra as it was not available to the Greeks. Galileo e.g. devoted the entire second book of his famous work Two New Sciences, to solving the problem of motion with non constant velocity, i.e. motion of falling bodies, purely by geometry. He thus arrived at the first instance of solving a problem of "integral calculus", by means of inverting the theory of rates of change, i.e. differential calculus. This was well before Newton. I would suggest these dialogues of Galileo as suitable for some young people. The language also is delightful, as when he justifies the hypothesis that acceleration of falling bodies must be as simple as possible, since that is what we observe in nature, e.g. in the flight of birds or the swimming of fishes. But as always, one should read "just as inclination leads" one, for understanding, without worrying too much about the terminology, which unfortunately I have obsessed over here. I have enjoyed thinking about all these aspects of calculus in response to this excellent question, and apologize if I have not been able to manage a useful answer. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 17, 2013 Share Posted March 17, 2013 â€œCalculusâ€ what does it mean? (This an attempt to actually explain something simply, instead of just critique others' works. Hopefully this may help one choose which of the suggested works linked above to look at.) There are traditionally considered to be two branches of the calculus, comprising two processes associated with solving the two related problems: given curve in the x,y plane, 1) to find the slope of the curve at one point, and 2) to find the area of that portion of the plane region bounded above by the curve, below by the â€œx axisâ€ , and lying between the vertical lines meeting the x axis at two given points. The first problem, deducing the slope of the curve above one point on the x axis, from a knowledge of the heights of the curve at all nearby points of the x axis, is called the problem of differential calculus. The problem of finding the area (between the curve and the x axis and) lying between two points on the x axis, again from knowing the heights of the curve at all intermediate points, is called the problem of integral calculus. The power of the method is based on a connection between these two problems called the fundamental theorem of calculus. This says that, given a first curve C1, and a point a on the x axis, if a second curve C2 is constructed such that the height of C2 above each point x, equals the area under C1 and lying between a and x, then the slope of C2 at the point above x, will equal the height of C1 above x. â€œThe slope of the area is the height.â€ The easiest examples to see the inverse relationship between these two problems is in the case of slope of curves which do not actually curve, i.e. straight lines, and area under curves which are not only straight but also horizontal, i.e. areas of rectangles. These cases of the two problems are so easy that it is stretching the meaning of the word â€œcalculusâ€ to describe their solution process. I.e., we usually use that term to describe the solution of these problems for curves that really do change direction, but it may be useful to see the link between them in the simplest possible case. This is what is done in the book Calculus Without Tears. Of course it may be that the tears have been avoided by also avoiding the actual calculus. But regardless, one can see something in this case. Namely the slope of a straight line is computed by choosing two points and considering the segment between them as the hypotenuse of a right triangle with sides parallel to the x and y axes. Then the slope equals height/base, for that right triangle, or â€œriseâ€/â€runâ€. In the area problem for a horizontal straight line, the area between two points a and b on the x axis, and bounded also by the x axis and the straight line segment above (and parallel to) it, is base x height. Thus in these two special cases, slope is computed by dividing height by base, and area is computed by multiplying height by base. Since multiplying and dividing are in some sense inverse operations, the two computations, slope and area are in some sense inverse. But we chose two different examples to do the calculations: we chose a horizontal line for the area calculation and a possibly tilted line for the slope calculation. How are these examples related? The relation is seen by doing both calculations for a varying choice of points on our lines. Start with a horizontal line at height 2 units above the x axis, and beginning from x = 0. then for each point x on the x axis, the area under this line, and lying between 0 and x, will equal base x height, or 3 times (x-0) = 3x. If we now â€œgraphâ€ this solution of the area problem, we get a line starting from (0,0) and at each point x, having height 3x. That is a tilted line. Now we solve the slope problem for this tilted line, (the graph of the area problem for the horizontal line). Choosing any two points on the tilted line, say (0,0) and (a,3a), we get a right triangle with base equal to the segment of the x axis from 0 to a, and height the vertical segment from the point (a,0) on the x axis, to the point (a,3a) on the graph. The length of this segment is 3a, so the slope = height/base = 3a/a = 3. Note that the slope of this area graph equals the height of the original horizontal line. thus computing the slope of the area graph for our horizontal line, gives back the height of the horizontal line. the miracle of calculus is that this inverse relationship remains true for curves that are not straight lines. I.e. given any smooth curve C1, if C2 is the graph of the area under C1 at various points, then the slope of C2 above a point x, equals the height of the curve C1 above that same point. To see this we have to make more precise what is meant by area bounded by, and slope of, actual curves. I.e. generalizing the formula slope = height/base to actual curves is called differential calculus, generalizing the formula area = base x height to actual curves is called integral calculus, and verifying the inverse relationship between them is called the fundamental theorem of calculus. So we have taken here the approach used in Calculus Without Tears, which might be more accurately called â€œa prelude to calculusâ€. Next I mean to present the next case, the first non trivial one, the problem solved by Galileo of the motion of an object under the influence of gravity. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 19, 2013 Share Posted March 19, 2013 Area under a linear graph versus slope of a parabola. Let us begin from the equation y = 2x, for all values of x â‰¥ 0, a straight line in the x,y plane starting from (0,0) and sloping upwards towards the right forever. Then we want to construct the area curve for this curve, i.e. the graph of the "area function" for y=2x. This will bne a new curve in the x,y plane whose height above the point x=a on the x axis is equal to the area between the previous line y=2x and the x axis, and lying between the points x=0 and x=a on the x axis. This is the area of the right triangle with vertices (0,0), (a,0), and (a,2a), as you can see by drawing the figure in the x,y plane. Now we know the formula for area of a triangle from Euclid as (1/2)(base).(height). The base of our triangle is the segment on the x axis from (0,0) to (a,0), hence has length = a. The right triangle has as vertical side the segment from (a,0) to (a,2a), hence has length 2a. Thus the area we seek equals (1/2)(a)(2a) = a^2. So the area curve, or rather the graph of the "area function", has height x^2 above each point xâ‰¥0 on the x axis. Thus its equation is y = x^2. We did not need calculus for this. Next we claim that the slope of that "area curve" above each point x, equals the height of the original curve at that same point. To check this we must show the slope of the curve y = x^2 above the point x=a, equals 2a. Equivalently the slope of the curve with equation y=x^2, must be shown to equal 2a at the point (a,a^2) of that curve. To do this we must figure out how to find the slope of a curve. This is our first genuine (differential) calculus problem. It will help if you draw pictures to illustrate this example. In pre-calculus we only learned how to find the slope of lines. The secret to finding the slope of a curve was given for circles by Euclid in Prop. 3.16 of The Elements, and generalized by Newton to all curves. The idea is that the slope of a curve at a point is close to the slope of a line joining that point to another nearby point. Taking many nearby points, getting nearer and nearer, we have a sequence of approximations to the slope of the curve, so we just have to figure out what number all those approximations are getting closer to. The number ultimately being approximated is called the "limit" of the approximations. E.g. suppose we have a sequence of approximations like .3, .33, .333, .3333, and so on forever, with each approximation adding one more 3, and we ask what number they are all approximations to. The natural guess is the infinite decimal .33333....(infinitely many 3's continuing forever), which we may know equals 1/3 if we have ever done the problem of dividing out 1 by 3 as a decimal. Notice that none of the approximations is itself equal to 1/3, but they are getting closer and closer to 1/3, so 1/3 is the limit that is being approximated. Here is another infinite sequence of approximations 3 = 2+1, 2+(1/2), 2+(1/3), 2+(1/4),....., so on forever, each approximation being 2 plus the reciprocal of the next larger integer. Can you guess the limiting value, not of any one of these approximations, but the value they are all approximations to, taken as a whole? It should be a number which all of these are ultimately getting closer and closer to. I claim that limit is equal to 2. Notice the first number is within 1 of the number 2, the second number is within 1/2 of 2, the third number is within 1/3 of 2, and so on. Thus although none of these numbers ever quite reaches the limit 2, still we can see that eventually they will get as close to it as desired. Thus 2 is the number which is being approximated. E.g. the number being approximated is not 2.1, since although the first ten approximations do get closer to 2.1, beginning with the approximations 2+(1/11), 2+(1/12), 2+(1/13),... they are all getting further away from 2.1, but all are getting ultimately closer and closer to 2. Now try to find the slope of the curve y=x^2 at the point P = (a,a^2), (the "area curve" associated above to y = 2x). If we choose a second point near it, such as Q = (a+1, (a+1)^2), then from pre-calculus, the line segment joining the two points P and Q has slope (draw a picture, and use the two point form for the slope of a line) equal to ((a+1)^2 - a^2)/ ((a+1)-a)) = (2a+1)/1 = 2a+1. This is the first approximation to our slope. If we take the second point Q closer, e.g. Q = (a+(1/2), (a+(1/2))^2), the slope of the segment joining P and Q is now ((a+(1/2))^2 - a^2)/ ((a+(1/2))-a)) = a+(1/4)/(1/2) = 2a + 1/2. If we take Q to be (a+(1/3), (a+(1/3))^2), the slope of the segment joining P and Q is (work it out) now 2a + 1/3. If we take the second point Q to be (a+(1/n), (a+(1/n))^2), the slope of the segment joining P and Q is ((2a)(1/n) + (1/n)^2)/(1/n) = 2a + (1/n). As we continue with Q getting closer to the point P in this way, the slopes 2a+(1/2), 2a+(1/3), 2a+(1/4),....... are approximations to what number? If you agree it is 2a, then we have determined the slope of the curve y=x^2 at the point P = (a,a^2) to equal 2a. Thus the slope at the point (x,x^2) is 2x. If you do not agree, this is the time to think about it more, ask questions, and discuss with your teacher. This shows the slope of the curve y = x^2, which is the "area curve" for y= 2x, above the point x, is again exactly 2x. Thus the height of the curve y=2x at every point, equals the slope of its "area curve". This example is the subject of Flannery's vol. 2 of Calculus without Tears, and of the second Book of Galileo's Two New Sciences, where it is done purely by geometry with no algebraic formulas at all. Indeed Galileo represents a number by two line segments, whose ratio of lengths represents the desired number! In general, one of the fundamental theorems of calculus (the easier one) says that for a "continuous (unbroken) curve", the height above every point x, always equals the slope of its area curve above that same point. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 19, 2013 Share Posted March 19, 2013 Slope of y = x^n. Donald Cohen works this out, in Calculus by and for Young People, after over two hundred pages, for several curves with equations y = x^n. The derivation can be written down quickly, maybe too quickly, as follows. (If nothing else maybe this quick computation will motivate the student to prepare for calculus by reviewing algebra.) Let us consider the general curve y = x^n, for some positive integer n. We will approximate the slope of this curve at a point (a,a^n), by the slope of the segment joining the two points (a,a^n) and (x,x^n), where x is a point close to a on the x axis. That line segment has slope (x^n-a^n)/(x-a) by pre-calculus. By algebra this fraction equals x^(n-1)+ x^(n-2)a+x^(n-3)a^2+....+xa^(n-2) + a^(n-1). When x gets closer and closer to a, this is getting closer and closer to a^(n-1)+a^(n-2)a+a^(n-3)a^2+.....+a.a^(n-2)+a^(n-1) = n.a^(n-1). Thus the slope of the curve y=x^n at the point (a,a^n) equals n.a^(n-1). To complete this case of the fundamental theorem of calculus, we would need to show that the area between the curve nx^(n-1) and the x axis, and between x=0 and x=a, equals a^n. I.e. we would want to show that the curve y = x^n is the "area curve" for the curve y=n.x^(n-1). Equivalently, the curve y = x^n has area curve y = (1/(n+1))x^(n+1). That is a problem for another post (which Archimedes essentially did for n=2, i.e. he found the area under the curve y=x^2, and above the x axis, and between the points x=0 and x=a, as (1/3) the area of the circumscribed rectangle with vertices (0,0), (a,0), (a,a^2), (0,a^2)). Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 19, 2013 Share Posted March 19, 2013 Area under the parabola, y=x^2. We know the area of a rectangle equals base times height, so we approximate the area under a curve by a sequence of rectangles whose bases we take smaller and smaller, to get better and better approximations. Then we try to guess what number is being approximated as we did with slopes of curves approximated by slopes of secant lines. If we look at the parabola y= x ^2 , between x=0 and x=1, and put one rectangle around it, fitting as closely as possible, we get area = 1 for the rectangle, since the vertices are (0,0), (1,0), (1,1^2), and (0,1). That is our first (too big) approximation. Next we take two rectangles each enclosing part of our curve, the first with vertices (0,0), (1/2, 0), (1/2, (1/2)^2), and (0, (1/2)^2); and the second with vertices (1/2,0), (1,0), (1,1^2), and (1/2, 1^2). Both these two rectangles have base length = 1/2, and one has height (1/2)^2 while other ahs height 1^2. So together they have area (1/2)(1/2)^2 + (1/2)(1^2) = 1/8 + 1/2 = 5/8, our second (too big) approximation. We need to be more systematic to see the pattern of these approximations so we can guess where they are headed. If we chop up the base interval [0,1] into n pieces of equal length 1/n, we get n rectangles all of base 1/n, but they all have different heights. The first has height (1/n)^2, the second has height (2/n)^2, the third has height (3/n)^2, and so on until the last, the nth, has height (n/n)^2. Draw a picture. So the rectangles together have area: (1/n)(1/n)^2 + (1/n)(2/n)^2+.....+(1/n)(n/n)^2 = (1/n^3)(1^2+2^2+...+n^2). Did you follow the algebra when we factored out (1/n)^3 from each term in the sum? If so, then what we need is a formula for that sum of squares 1^2+2^2+...+n^2. Now this is not something everyone has seen at all, but there is a mechanism for finding such formulas using the binomial theorem. It is derived fully in a footnote at the bottom of page 27 of a fine calculus book by Richard Courant. When I saw this in college I knew I wasn't in Tennessee any more to paraphrase Dorothy in Oz. I will simply state the formula: (1^2+2^2+...+n^2) = (1/6)(n)(n+1)(2n+1). But it is more informative to notice this equals (n^3/3) + q(n), where q is a quadratic polynomial. I.e. a formula in n of degree two. Thus our area approximation above looks like (1/n^3)(1^2+2^2+...+n^2) = (1/n^3)( (n^3/3) + q(n)) = (1/3) + q(n)/(n^3). Now as we take more and more rectangles with smaller and smaller bases, to get a better and better approximation, I claim the fraction q(n)/(n^3) is getting closer and closer to zero, so that the area approximation is getting closer and closer to 1/3. To see that a little better, recall q(n) is quadratic in n, so looks like an^2 + bn + c. Thus dividing by n^3 gives a/n + b/n^2 + c/n^3. Now as n gets bigger, the tops of these fractions all stay the same while their bottoms get bigger and bigger. That makes all the fractions get smaller and smaller, so their sum gets closer and closer to zero. If we do the area computation from x=0 to x=a, the division points of the interval [0,a] are a/n, 2a/n, 3a/n, and so on, and so we get as area approximation: (a/n)(a/n)^2 + (a/n)(2a/n)^2+.....+(a/n)(na/n)^2 = (a^3/n^3)(1^2+2^2+...+n^2) = (a^3/n^3)( (n^3/3) + q(n)) = a^3/3 + (a^3q(n))/n^3. As before, this approximates a^3/3 better and better as n gets larger and larger, so the area under the curve which is being approximated must be a^3/3, Thus the area formula for the curve y=x^2 is x^3/3. If we knew similarly the fact that (1^k +2^k + .....+n^k) = n^(k+1)/(k+1) + f(n) where f is a polynomial of degree k or less, the same argument would show the area formula for y=x^k, between 0 and x, is x^(k+1)/(k+1). E.g. the ratio of the area under the curve y=x^n, and between 0 and a, to that of the rectangle with vertices (0,0), (a,0),(a,a^k),(0,a^k), is exactly 1/(k+1). This fact, the magic formula for the sum of the kth powers of the first n positive integers, is usually not proved in a college course. So instead of doing the area computation directly this way, which needs that formula, one usually just uses the fundamental theorem of calculus, that the area formula is the opposite of the slope formula. Since finding slope formulas is easier, this makes finding areas easier as well. Of course in most typical college courses the fundamental theorem is also not fully proved, but is taken largely for granted. You have just seen the first part of a typical college calculus course, assuming mostly just algebra, or roughly the last 60 pages of Donald Cohen's calculus book. Whew! Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 19, 2013 Share Posted March 19, 2013 Here's a little tidbit about how to compute logs using calculus. Using the formula a^(u+v) = a^u.a^v, it is easy to show that the slope formula for the exponential function a^x equals c.a^x where c is some constant. Since the slope changes continuously as you change the base a, there is a base called e for which the constant c=1. Thus the slope formula for e^x equals e^x, i.e. the slope of y=e^x is the same as the height. Now log is the inverse function of the exponential, i.e. the graph is the same but with y and x interchanged. Thus the slope of the log is the reciprocal of the slope of the exponential, but the letters change too; so since the slope of y = e^x is y, then the slope of y = log(x) is not 1/y but 1/x. Then by the fundamental theorem of calculus, the area formula for that slope formula is the log formula back again. I.e. (natural) log(x) = the area between the graph of y = 1/x, and the x axis, and between the points 1 and x of the x axis. This lets you compute logs as areas, or at least approximate them. The number e is the point on the x axis such that the area between 1 and e, and under y=1/x, equals 1. This lets you approximate e as roughly 2.71828.... As an incentive to never give up, my calculus teacher gave an assignment to compute e to at least enough places to see it begins 2,718281818.....(but it does not repeat after this). I did not have the stamina to do this in college and whiffed on that assignment. As a professor I was embarrassed to admit this to my students, so I completed the assignment and sent it in to my ex - prof (the famous John Torrence Tate Jr.) 40 years late. He was duly pleased and amused. Quote Link to comment Share on other sites More sharing options...

FairProspects Posted March 20, 2013 Share Posted March 20, 2013 Would the recommendations be essentially the same for a kid who instinctively "gets" algebra but is missing some arithmetic algorithms (operations with fractions, decimals, etc.)? Basically, devote one day a week to algebra or algebraic concepts and the rest of the time plug away at 5th & 6th grade math concepts? Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 20, 2013 Share Posted March 20, 2013 I would recommend practicing until those operations are instinctively correct. I.e. just "intuitively getting it" does not help make calculations if the operations are flaky. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 20, 2013 Share Posted March 20, 2013 Here is the power of calculus. Once you know that area and slope problems are inverse to each other, you can do harder area problems just because you know how to do easier slope problems. The hardest post I made above is probably the area for curves of form y = x^n, by taking limits, and requiring formulas for sum of powers of integers. On the other hand the slope formulas for curves like y=x^n were relatively "easy" since they needed only the binomial formula for expanding, or dividing (x^n-a^n) by (x-a). But the FTC says knowing slope formulas also gives you area formulas. I.e. if i want the area formula for the curve y = x^n between x=a and x=b, I just guess a formula whose slope is x^n, namely I guess (1/(n+1))x^(n+1), which is easy from our slope calculations above. Then since the area between 0 and a is (1/(n+1))a^(n+1), by subtracting, the area under y=x^n between x=a and x=b is simply given by the difference of my guessed formula for those values, i.e. it is (1/(n+1))(b^(n+1) - a^(n+1)). Thus modern calculus makes it easier for us to compute that area than it was for Archimedes. That is the power given to us by modern calculus, i.e. by knowing the link between the slope and area problems. Now that I say this, I realize that even the fine book of Donald Cohen does not contain this result, the fundamental theorem of calculus. I.e. he treats the two problems of area and slope separately and shows that in many cases they come out inverse to each other, but he never shows this must be true in general. Without that, one cannot use the easier slope problem to compute the harder area problem. So the main point of the modern calculus is not there. Hence everything he has there is due to the ancients, except perhaps the observation that the processes appear to be inverse. So in a picky strict sense one could also say this book too stops short of presenting the real power of calculus. I.e. the main (and the most difficult) point is that a function is determined by knowing its slope function everywhere and its height at one point, so two functions with the same slope function differ at most by a constant. Then the easier result that the slope of the area function is the height function, tells us that any function we can guess which also has the height function as its slope function, must indeed equal the area function except possibly for a constant. This lets us recognize the area function without computing it by limits, just by guessing a function with the right slope. I.e. the fact that the slope of x^3 is 3x^2, guarantees that x^3 is the area function for y=3x^2 up to a constant. So the area between x= a and x=b and under y=3x^2, must equal b^3 - a^3, (since the unknown constant gets canceled out when we subtract.) Maybe I'll make an attempt to explain these two results, since I did not find in the two calculus books for children linked here (but maybe Flannery is heading toward this gradually). Maybe some parents will decode them for their mathy kids. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 21, 2013 Share Posted March 21, 2013 (first) Fundamental theorem of Calculus, part i) ( slope and area functions): The "easy direction" of the fundamental theorem of calculus: i.e. "the slope of the area is the height ". Here the "slope of the area" is the change in the area compared to the change in the base. Hence in the simplest case of a rectangle, this is the fact cited by Flannery, in Calculus Without Tears, that for rectangles, (area/base) = height, or equivalently that area = (base).(height)! In the general case we approximate areas by rectangles and get the same result. Flannery's point is that this simple fact is clear when the original graph is a horizontal line. I.e. then the regions whose areas are measured are all rectangles, the area graph is a sloped straight line, and we know all about areas of rectangles and slopes of straight lines, so we can easily compare them. In the case of graphs that actually curve, we approximate curved areas by areas of rectangles, and slopes of curves by slopes of straight lines, and thus the equation between height and slope of area, even for curves, is true because it is approximated by these simpler true equations for rectangles and lines. Recall that the slope of a line equals "rise"/"run", i.e. (vertical change)/(horizontal change), measured between any two distinct points on the line. For a curve in the (x,y) plane, since the x coordinates measure horizontal distances and y coordinates measure vertical distances, slope = "change in y"/"change in x". Hence the "two - point form" of the slope of the line segment joining points P = (a, b ) and Q = (x,y), is slope = (y-b)/(x-a). Starting from a curve y = f(x), where f(x) is some formula in x, also called a "function of x", defined say for x â‰¥ 0, we construct the graph of the area function corresponding to it. This is a curve y = A(x), whose height above each point x on the x axis, is the area A(x) lying between the original curve y = f(x) and the x axis, and also between the vertical lines meeting the x - axis at the points 0 and x. (We assume the curve y = f(x) is always above the x axis, i.e. that f(x) is positive for all x â‰¥ 0.) Thus a point lies in the region whose area we consider, if its x - coordinate lies between 0 and x, and its y coordinate lies between 0 and f(x). So three sides of the region are straight lines and only the top is (allowed to be) a curve. For example, starting from the curve y = 3x^2, for 0 â‰¤ x, we saw the area function A(x) = x^3. In that case we also saw the slope function of this area function is again f(x) = 3x^2. We want to show this happens in general, i.e. that taking the slope of the area function gives back the height function of the original curve. As is always the case some hypothesis is needed, since there are exceptions to virtually everything, and here the hypothesis is that f is a "continuous function". So we explain that concept next. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 21, 2013 Share Posted March 21, 2013 first FTC part ii): (continuity): The function f is continuous at the point a if f(x) can be made as near f(a) as desired by taking x near enough to a. E.g. if we have f(x) = x^2, and a = 1, then f(1) = 1, and suppose we want to guarantee that x^2 is within 1/10 of 1^1. It suffices to take x within 1/100 of 1. I.e. since (.99)^2 = .9801, and (1.01)^2 = 1.0201, and since .99 < x < 1.01 guarantees (.99)^2 < x^2 < (1.01)^2, we are ok. I.e. we only wanted to be sure that x was close enough to 1 to force .9 < x^2 < 1.1. Generalizing this argument one can show x^2 is a continuous function at every point. Roughly, f is continuous at a, if the values of f at points nearby a, are all close to the value at a; slightly more precisely, as x gets closer and closer to a, the values f(x) get closer and closer to f(a). We write this as follows: "whenever x--->a, then also f(x)--->f(a)". Intuitively a function is continuous if its graph has no breaks or gaps in it. Theorem: If f is continuous, then f is the slope function of its area function. (In fact to make life simpler,) we will also assume that f is increasing. Then we intend to show that for any point x=a, the slope of the curve y = A(x) at the point (a,A(a)) lying above x=a, equals f(x). This is obvious from a picture, so please draw one, but I will attempt to render it into words in the next post. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 21, 2013 Share Posted March 21, 2013 first FTC part iii): (computing the slope of the area function): Slope of a curve at a point is approximated by the slopes of lines joining nearby points, so the slope of the area function A(x) at the point (a,A(a)) is approximated by the slope of a line joining that point to another point (x,A(x)), where x is a point near a. The slope determined by the two points (a,A(a)), (x,A(x)), is the fraction (A(x) - A(a))/(x-a). Now A(x) - A(a) is the area under that part of the graph of f lying between the points a and x on the x-axis. The base of this region is the segment between a and x on the x -axis. Since f is increasing the height of this region rises from its lowest value f(a) at the left end of this segment, to its highest value f(x) at the right end of this segment. Hence the area of the region lies between the area of the rectangle with base (x-a) and height f(a), and the area of the rectangle with base (x-a) and height f(x). By the formula base times height for area of a rectangle, this inequality is true: f(a)(x-a) â‰¤ A(x)-A(a) â‰¤ f(x)(x-a). If you are looking a picture (and I recommend that) this just says that since the region under the curve lies between two rectangles, one entirely below the curve and one slightly above it, then the area of the region also lies between the areas of those two rectangles. Dividing by (x-a) we can estimate the slope fraction above as follows: f(a) â‰¤ (A(x)-A(a))/(x-a) â‰¤ f(x). Now recall that the fraction in the middle gets closer and closer to the slope of the graph of A as x gets closer and closer to a. But by the hypothesis of continuity of f, when x gets closer and closer to a, the value of f(x) gets closer and closer to f(a). If f(x) is getting closer and closer to f(a), then the slope fraction caught in between is forced to also get closer and closer to f(a). Thus the slope of the area function A at the point a, which is the number being approximated by those slope fractions, equals f(a), the height of the original graph whose area was being considered. Tataa!! Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 21, 2013 Share Posted March 21, 2013 If other calculus teachers or learners want to chime in about how this material might be made more palatable or understandable by early (or later) learners, I would benefit from the comments. Odd as it may seem from this dense mass, my goal, and every teacher's goal, is to disseminate what they love more widely. In my experience I have found that when a student is willing to give feedback in the form of questions, it is eventually possible to remedy the failings of the first attempt at explanation. This is something to remind ones young students to do in classes of every sort. By the way, are home schooled kids more open to asking questions than the silent ones in schools? Is it just peer pressure that stifles questions, or does it depend more on how we respond to them, or is it a facet of the learner's personality, or something else? Quote Link to comment Share on other sites More sharing options...

Arcadia Posted March 21, 2013 Share Posted March 21, 2013 By the way, are home schooled kids more open to asking questions than the silent ones in schools? Is it just peer pressure that stifles questions, or does it depend more on how we respond to them, or is it a facet of the learner's personality, or something else? All of the above except kids can be silent in school because they are bored or have insufficient sleep :) My older does not ask questions in school because his questions are out of topic/scope for that grade level and the teachers either have no time or not enough knowledge. He ask plenty of questions at the tech museum/science centers since nothing is out of topic and there are usually subject experts around when we go. Also he can ask until the museum close for the day. My younger however ask questions even if they don't get answered; it is his personality. Peer pressure in a class environment can help or hinter asking questions. In a class where most people ask questions, it is conducive and less threatening to ask a question. In a class where being brainy is "uncool", kids will tend to be quiet. Some teachers are good at leading class discussions and that prompts more questions from students. Some teachers just talk and students driff off. So the style of teaching and tone of the teachers also affect the atmosphere of the class. One of my calculus teacher could discuss and debate any topic competently and patiently explain concepts. The other is not good in calculus and can't answer any questions. We reserved all calculus questions to the "better" teacher. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 21, 2013 Share Posted March 21, 2013 You remind me that when a professor sits in a class for students, the professor is requested not to ask questions because those advanced questions hinder the students from asking theirs. In a sense your advanced childrens' questions are like those of a professor in an elementary class, so it is perhaps beneficial to other children, although not to your advanced child, if he/she remains mostly silent. There are exceptions however, and an advanced child, even an astute professor, can learn to ask questions that benefit everyone. This is tricky though, and every question should be genuine. I have learned it seems patronizing to ask a question one knows the answer to, under the assumption someone else may need to know the answer. It is a thrill sometimes in college when an advanced student asks you the question they could never get an answer to in high school. If you nail it, the ("we are not in Kansas any more") look on the student's face is a sweet reward indeed. But you also have 35+ others to care about. I am also reminded of my experience sitting in colloquia at university with professors who were decidedly not requested to be silent. Their questions could be instructive, amusing, and intimidating. One of my friends paid me the (intended) compliment by recalling his impression of me there: "you became a mathematician while some others did not, because you were not afraid to ask stupid questions". In honesty, I add that I was then in my late thirties and had learned the hard way (over two decades) the consequences of not asking them. Quote Link to comment Share on other sites More sharing options...

Arcadia Posted March 21, 2013 Share Posted March 21, 2013 There are exceptions however, and an advanced child, even an astute professor, can learn to ask questions that benefit everyone. This is tricky though, and every question should be genuine. That is the thin line that my older is trying to figure out how to navigate. People can often sense that there is a question in his eyes but he is not going to vocalised the questions/doubts unless he feels comfortable with its appropriateness. It would probably come with maturity hopefully because I don't know how to teach that skill. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 21, 2013 Share Posted March 21, 2013 I think often that as long as the question is genuine, i.e. conveys honestly that the questioner does not know the answer and would like to, it can benefit. Others learn what kind of question interests the strong student, and may be helped to turn their own attention in a helpful direction. I know it fascinates me to get a glimpse of how a more insightful person thinks. Even just the feeling that the strongest among us does not know it all, and is trying to learn, is reassuring. But it is subtle, and I am not master of it myself. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 22, 2013 Share Posted March 22, 2013 The second part of the fundamental theorem of calculus:(i) Why is there something to prove? I.e. we know the slope of an area function gives back the original function, so why shouldn't the area of a slope function give back the original function? Isn't that the same thing said backwards? Well not quite. If we start with a function G, we do not know whether G can be an area function. If it were, we would be done. I.e. if G is the area function of some other (continuous) function f, then taking the slope of G would give f, by the first part of the FTC. Then taking the area function of f would indeed give back G, by our assumption. But if there were some G that is not itself an area function for anything, then it cannot of course be the area function for its own slope function. If we just start from a function G having a (continuous) slope function f, and form the area function A for f, then the slope function of A is f again by part 1 of FTC. Notice this does not say that G is equal to A, only that G and A have the same slope function. So what we get from part 1 of FTC is not quite that G is an area function, but only that G has the same slope as an area function. So what can we deduce about G just from knowing something about its slope function? In fact we claim that if G and A have the same slope function, then A and G differ at most by a constant. We cannot do better than that, as is shown already by looking at examples of constant functions. I.e. every constant function, no matter what the constant value, has the same slope function as every other, namely zero, because all the graphs are horizontal. So just knowing the slope of a graph at every point cannot tell us the height anywhere. But the fundamental theorem says that knowing the slopes everywhere, and knowing the height at just one point, does completely determine the graph, and the function. So we start from a given function (or formula) G(x), form the slope function f(x) for this G, and then form the area function A(x) associated to f(x). By the first part of the FTC, we know A(x) is a function whose slope function is f(x) again. Thus both G and A at least have the same slope function. We claim G and A must differ at most by a constant. Why? Note that their graphs are curves that always slope in the same direction, i.e. the curves are "parallel" at every point in the x,y plane. It is then sort of clear that they must always be a constant distance apart, so that indeed the values G(x) and A(x) always differ by a constant. I.e. the original function G equals the area function of its slope function, except possibly for a constant added on. But we want to try to give some argument for this. I.e. we learn in Euclidean geometry that lines with the same slope stay a constant distance apart, and we want to prove this also for curves. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 23, 2013 Share Posted March 23, 2013 second FTC ii): How shall we prove it? We want to show that two functions whose graphs have the same slope everywhere must differ at most by a constant. The first simplification comes from using the wonderful trick of subtraction. I.e. if G,H are any two functions, and we subtract them, then the slope of G-H equals the slope of G minus the slope of H. Thus if G and H have the same slope, then G-H has slope zero, and then we want to deduce that G-H is a constant, i.e. that G and H differ by a constant. So we are reduced to proving the simpler statement that a function whose graph has slope zero everywhere, must be a constant. This is so obvious intuitively, I hesitate to belabor it with a proof. However, in fact it is actually false as stated. I.e. consider a function f defined only for non zero numbers x, and with value f(x) = 1 whenever x>0, and f(x) = -1 whenever x < 0. For example take f(x) = |x|/x, for x â‰ 0. Then the graph of f is horizontal near every point, but f is not constant, since it has two different values! Notice to measure the slope of a graph at a given point, we only look at the part of the graph which is very near that point. I.e. the slope of f at x = a, is the limiting value of the fraction (f(x)-f(a))/(x-a), as x becomes extremely close to a. If we look at a point a>0 for instance, then all points x close enough to a, also have x > 0, and this fraction will always equal zero for x close enough to a. The problem here is that this f is not defined at x = 0 where the value changes from -1 to 1. So the theorem needs a hypothesis that the function is defined on a connected interval on the x axis, not two disjoint intervals. But even then, how do we know that an interval on the x axis does not have any gaps in it where our function could jump up? For example if we had a function defined only at rational numbers, then we could set it equal to 1 when x > sqrt(2), and equal to -1 when x < sqrt(2). Since sqrt(2) is not rational, there would not be a point of the graph over sqrt(2), and the graph would still be horizontal near each (rational) point, although not constant everywhere. So we must use the assumption that our function is defined on a connected interval of the real line, irrational numbers and all. The fact that this makes the interval complete and unbroken, with no gaps anywhere, is a deep idea, which was perhaps not fully understood by the Greeks and became clarified only in the 19th century. Now lets try to prove that a graph with slope zero everywhere over an interval of the x - axis, has constant height over that interval. We will prove an equivalent statement that is stated differently, the "contrapositive" statement. There are always two ways to state any assertion. If your mom says she will buy you a new iphone provided you get all A's next semester, another way to say the same thing is to say that the only way she will not buy you a new iphone is if you do not get all A's. So we will prove if a function (which has a slope) is not constant on an interval, then it cannot have zero slope everywhere; i.e. if our function is not constant everywhere on an interval, then its slope must be non zero somewhere on that interval. More precisely, if f is a function defined and having a well defined slope for all x in the interval [a,b], and if f(a) â‰ f(b ), then the graph of f has non zero slope at some point between a and b. E.g. imagine that f(b ) > f(a). Then the graph of f goes up from height f(a) over x = a, to the greater height f(b ) over x = b. We claim if this happens then the slope of f must be positive at some point x between a and b. To see this, connect the point P = (a,f(a)) to the point Q = (b,f(b )) by a straight line segment. Then this line has positive slope since Q is higher than P. The graph of f also joins these two points, and if you draw it and look at the graph, it should look as if the graph of f must be parallel somewhere to the line segment joining P and Q. At that point where the graph of f is parallel to the line segment PQ, the slope of the graph will also be positive, hence not zero. That will do what we want. So where is that point where the graph of f looks parallel to the line segment PQ? I claim it occurs at the place where the graph is as far as possible away from the line segment. To prove all this we can simplify our lives again by using subtraction. I.e. we claim that if G and H are two functions whose graphs pass through the same two points P = (a,G(a)) = (a,H(a)), and Q = (b, G(a)) = (b,H(b )), then the graphs of G and H must have the same slope at least at one point x between a and b. By subtracting H from G we are reduced to proving that if a function F = G-H has the same value at two different points a and b, e.g. F(a) = G(a)-H(a) = 0 = G(b )-H(b ) = F(b ), then the graph of F has slope zero at some point between a and b. To prove this we claim two things: i) any continuous function F on an interval [a,b] has a maximum (or a minimum) value at some point c between a and b; ii) if F has a maximum or minimum value on the interval [a,b] at a point c between a and b, then graph F has slope zero at x = c, (assuming the slope of F is defined). Part ii) is elementary, since if F has a maximum value say, at x=c, then the fraction (F(x)-FÂ©)/(x-c) is positive when x < c, and negative when x >c. Since it approximates the slope of F at c, that slope must be a number that can be approximated by both positive and negative numbers and the only such number is zero. So at a maximum occurring between a and b, the slope must be zero. We take part i) on faith, unless requested otherwise. Quote Link to comment Share on other sites More sharing options...

kiwik Posted March 23, 2013 Share Posted March 23, 2013 It depends. Where I am maths have always been spiral so one year you do the basics of trig, geometry, algebra and statistics (and other stuff) the next year you do it all again at a higher level etc. In the final year of school maths is split into calculus and statistics and you take one or both (or none). If your son has done basic algebra he can probably manage basic calculus but he will have to spiral between books if he does it that way. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 23, 2013 Share Posted March 23, 2013 The explanation here of some fundamental theorems of calculus has attempted to use only algebra, plus the concept of graphing a function in the x,y plane. Of course if we want to apply those theorems to the case of trig functions we must know what those are, but that is easily presented in a few words. Actually to do trig functions carefully requires a little calculus itself. I.e. it is hard even to define the sin and cosine functions precisely without knowing some integral calculus first. But the point is that calculus is a technique which applies to functions. If you start from your favorite function and apply calculus to it, you generate new functions, namely slope and area functions from that original one. So that technique can be applied to any family of functions you are familiar with. if you only know algebra, you can apply it to those functions met with in algebra, namely polynomials. There are only two other types of functions normally encountered in elementary math, i.e. trig and exponential/log functions. These functions however cannot be precisely even defined without using calculus. I.e., carefully defined, a log function is inverse to a certain area function, the area function for the rational function 1/x. sine and cosine are inverse to the arc length function for a circle, and arc length is itself measured by an integral of a square root function, using the Pythagorean theorem, i.e. the integral of sqrt(1-x^2). So in fact, the logical order of the subjects, followed in some modern theoretical calculus books Iike that by Michael Spivak), is algebra/geometry (and graphing), then calculus, then exponentials and logs, and trig. of course in real life, not la-la land, it helps to know even an imprecise version of trig and exponential/logs before starting calculus. I myself however did not have this background (I had neither trig nor calculus in high school) when I started la-la land calculus in college, and I sort of enjoyed seeing things correctly the first time. Another point of view is possible and maybe preferable, but not usually seen in colleges today. that is Euler's approach which presents power series before calculus. This makes sense pedagogically because power series are like infinite polynomials, hence easy to motivate to students of algebra. Then trig and exponential functions can both be presented via certain power series, which makes them both theoretically precise and practically computable. E.g. then the function sin(x) is given by the simple formula x/1- x^3/(1.2.3) + x^5/(1.2.3.4.5) - x^7/(1.2.3.4.5.6.7) + - ..... the exponential function is even easier: e^x = 1+ x + x^2/(1.2) + x^3/(1.2.3) +x^4/(1.2.3.4)+..., and (natural) log is given for small x by ln(1+x) = x- x^2/2 + x^3/3 - x^4/4 + -..... Indeed motivating these formulas without calculus seems challenging. I have not consulted Euler lately on this but I remember he considers this precalculus. Come to think of it, my college calculus teacher just defined e^x and sin(x) this way, by power series, with no motivation at all. Actually he defined e^x this way but also for complex values of x, and then he defined sin(x) = (1/2i)(e^(ix) - e^(-ix)), the "odd part" of e^(ix), which is equivalent to the series above, as you can "check" by multiplying out and adding and canceling. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 24, 2013 Share Posted March 24, 2013 Actually one can do at least differential "calculus" entirely by algebra, without even limits. In fact this may help take away the mystery of the subject a bit. The idea is that a derivative is the slope of a tangent line to a graph, and that tangent line is the best straight line approximation to the graph. Equivalently we are looking for the best linear, or first degree, function or formula that approximates our given formula, or function, near a given point. Now that is really easy for polynomials, i.e. powers of x, provided the point we are interested in is x= 0. I.e. the best first degree approximation to say f(x) = 3 + 4x -6x^2 + 7x^3, near x= 0, is just the first degree part of this formula, i.e. just 3 + 4x! The derivative of this function f at x=0 is the coefficient of the first degree power of x, namely 4. if you take the derivative of f by the usual rules and evaluate at x=0 you will verify this is the same answer as one gets by limits or what have you. Let's check that in detail by taking a limit, as if we did not know the rules for derivatives. the derivative of the above f(x) at x=0 is by definition the limit of [3+4x-6x^2+7x^3 - 3]/{x) as x-->0. i.e. of [4x-6x^2+7x^3]/(x) as x-->0, and after canceling this is the limit of [4 -6x +7x^2] as x-->0, which is just 4, since the other terms approach closer and closer to zero as x does so. Thus it is entirely trivial to "take the derivative" of a polynomial function at x=0 without any effort at all, merely by plucking off the linear part of the formula. But what about other points x = a, with a â‰ 0? Well it's almost the same, but first we have to re expand the formula in powers of, you guessed it, x-a. E.g., to take the derivative of the same f at x=1, we substitute (x-1)+1 = x and re expand. Thus f(x) = 3 + 4x -6x^2 + 7x^3 = 3 + 4[(x-1)+1] -6[(x-1)+1]^2 + 7[(x-1)+1]^3 = 3 + 4(x-1) + 4 - 6(x-1)^2 -12(x-1)- 6 + 7(x-1)^3 +21(x-1)^2 +21(x-1) +7 = 8 + 13(x-1) +15(x-1)^2 +7(x-1)^3, (I hope, as I did it in my head). Now picking off the first degree part gives 8 + 13(x-1). So I claim the derivative of f(x) at x=1 is 13. We can check this derivative by using the original formula, since 8 should be f(1) = 3 + 4(1) -6(1)^2 +7(1)^3 = 14-6 = 8, yes! and the derivative should be 4 -12x +21x^2 evaluated at x= 1, which gives 4 -12 + 21 25-12 = 13, yes! So if f is any polynomial at all, then to get the derivative of f at x=a, just substitute (x-a)+a = x, into f(x), expand out in powers of (x-a) and pick off the first degree part. The constant term is the value f(a) and the coefficient of (x-a) is the derivative of f at x=a. Note you can see the general rule for derivatives of powers this way as well, since for f(x) = x^n, we get [(x-a)+a]^n = (x-a)^n + n(x-a)^(n-1).a +.....+n(x-a)a^(n-1) + a^n. So the linear, or first degree part, is just the last two terms, and the coefficient of (x-a) is na^(n-1), as expected from the usual "power rule" for derivatives. Thus differential calculus is completely algebraic for polynomials at least. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 24, 2013 Share Posted March 24, 2013 Oh by the way, you don't need to do all that heavy lifting of expanding out the whole thing in powers of (x-a), since at the end you are going to throw away all terms of degree higher than one. So you might as well throw them away sooner. I.e. let's re expand 3+4x-6x^2+7x^3 = 3 + 4[(x-a)+a] -6[(x-a)+a]^2 +7[(x-a+a]^3 only keeping first powers of (x-a). We get 3 +4(x-a).... well, we might as well throw away constants too, since those will add up to f(a),.... so we get f(a) + 4(x-a) -6(2a(x-a)) +7(3a^2(x-a)) = f(a) + (4-12a+21a^2)(x-a), and the derivative of f at a is (4-12a+21a^2). That checks. And to see geometrically why this is working, just graph the functions y = constant, y=bx, y=cx^2, y=dx^3,... and look at the graphs near x=0. Notice they all have slope zero except for the graph of y=bx. That's the linear or first degree term. So the slope of a formula like 3 + 4x -6x^2 +7x^3 at x=0, comes entirely from the linear term, i.e. from the 4x term. Perhaps any child that has learned algebra and graphing of polynomials may appreciate this phenomenon. So at least some of the ideas of calculus are very elementary, but not always taught that way. You might look in Flannery to see if he does some derivatives this way, in volume 2 or later. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 24, 2013 Share Posted March 24, 2013 Descartes' method: Another way to do derivatives purely by algebra is Descartes' way, by determining which line has a "double intersection" with our curve y =f(x) at the point (a,f(a)). So let y=f(x) be the graph of a function f(x), and we seek to determine the tangent line to that graph at the point (a,f(a)). That will be a line that passes through the point (a,f(a)) and also has the same slope as the graph has there. Now we claim the fact that line has the same slope as the graph, can be determined by counting the number of intersection points with the graph, i.e. by counting how many common points the line and the graph share. The secret is that the tangent line at P will have fewer total points in common with the graph than most other lines through P. I.e. if we look at a curve and a line through the point P on that curve, we will see that the line usually meets the curve again somewhere else. Indeed if f is a polynomial of degree n > 1, then the line will have at most n, and often exactly n points in common with the graph y = f(x). Following the idea of Euclid and then Newton, of approximating the tangent line through P by a secant line that meets the curve not only at P but also at another nearby point Q, we see that as that second nearby point Q moves closer and closer to P, the secant line comes closer and closer to the tangent line at P. Hence the tangent line at P is the result of taking a line meeting the graph at two points P and Q (and maybe others as well), and letting the point Q move closer to P until it actually equals the point P. Since two intersection points P ,Q have come together at the same point, the total number of intersections of the tangent line with the curve has gone down at least by one. I.e. if a general line through P had n points in common with the graph, the tangent line will have at most n-1 points in common. However as we move the line through P it could happen that it becomes tangent to the graph at some other point R away from P. if that happens, two intersection points will come together at R, so even though the total number of intersection points went down, it did not occur because of two of them collapsing at P. So we want to measure not the total number of intersection points, but the number of common points occurring at P. This is done by means of algebra, and the concept of the multiplicity of a root. You probably have noticed that although most quadratic equations have two distinct roots, some have none and others have two repeated roots, i.e. a "double root". In the same way we want to find a line passing through P that has a double intersection at P with the graph y = f(x). To find common points of the graph and the line, and assign multiplicities to them algebraically, we first write an equation for our line. Since we want our line to pass through the point (a,f(a)) on the graph, our line has equation L: m(x-a) + f(a), where m is the slope we want to determine. Note this equation does pass the point (a,f(a)), and in fact this is the point-slope form of the equation for the line. I.e. we know the point namely (a,f(a)), but we don't yet know the slope m. Now common points of this line with the graph y = f(x) are found by setting the two equations equal and solving. So we set f(x) = m(x-a)+f(a), or f(x)-f(a) -m(x-a) = 0, and try to solve for x, i.e. we try to find the roots. Well one root is x=a of course. Now by the super important "root -factor theorem", or just "factor theorem", the fact that x=a is a root is equivalent to the statement that (x-a) is a factor. In particular (x-a) must be a factor of f(x)-f(a), so in practice we can divide out by (x-a) getting (x-a)[(f(x)-f(a))/(x-a)] -m] = 0. Of course although we know that (x-a) does factor into f(x)-f(a) evenly, since we have not chosen a specific f(x) we cannot write down the precise formula for the factor [(f(x)-f(a))/(x-a)], which we only know is some polynomial. But now we know that for x=a to be a double root of the product (x-a).[(f(x)-f(a))/(x-a)] -m], then x=a must be a root of the second factor [(f(x)-f(a))/(x-a)] -m]. I.e. setting x=a must make [(f(x)-f(a))/(x-a)] -m] = 0. That means the fraction [(f(x)-f(a))/(x-a)], when simplified, must equal m when x=a. I.e. in order for our line to have a double intersection with the graph y = f(x) at the point (a,f(a)), the slope m of the line must equal the value of the quotient [(f(x)-f(a))/(x-a)], when x=a. Let's do the basic example f(x) = x^n. Then f(a) = a^n, so we have [x^n-a^n]/(x-a) =x^(n-1) + x^(n-2).a + ... + x.a^(n-1) + a^(n-1). When we set x=a, we get n terms, all equal to a^(n-1), so the slope m, i.e. the derivative of x^n at x=a, is na^(n-1). This is essentially the same as Newton's limiting method in this case, but without explicitly mentioning limits. So again we have done differential calculus purely by algebra, but only in the case of an algebraic function, i.e. a polynomial. As you can see, in a lifetime of teaching I have tried many many different approaches to trying to make calculus understandable! Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 28, 2013 Share Posted March 28, 2013 Here are links to my college course notes on basic ideas of differential calculus. The first set avoids limits and the second set explains and uses them. To Allearia, I would suggest the first set of notes below, as a test of whether your child knows enough algebra to understand a basic minimum of calculus. I.e. those notes presume only a grasp of basic manipulations with polynomials, including the concept of factoring, and multiplicity of roots. I wrote these notes specifically to explain differential calculus to students who knew only algebra. One doesn't even need the quadratic formula. More precisely, one needs to know what are often called the "remainder theorem" and the "factor theorem". Are these taught in AOPS algebra? http://www.math.uga....hout_limits.pdf http://www.math.uga....ngents_to_y.pdf Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 29, 2013 Share Posted March 29, 2013 Forgive me Allearia, but may I ask whether the first one of the two sets of notes I posted in the previous post seem accessible to your child? I am not quite up to speed on what is learned from AOPS algebra, the various chapters. I would appreciate help in learning how to mesh my lessons with what is currently being taught. I confess also that I have learned I must make a new post in order to learn whether anyone has viewed the previous post. I.e. this browser does not update views unless a new post is made. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted March 30, 2013 Share Posted March 30, 2013 I searched on LOF calculus to see what I could learn about it and found a brief sample on the author's website. It seemed to be a real calculus book with roughly the same topics usually found in typical (non honors) college calculus courses, as the author says, but written in a storybook style apparently aimed at very young children. The level of the stories seems aimed at much younger children than the math ideas. It differs from the presentation in my notes in that, instead of presenting simpler versions of the math concepts as I try to do, he tries to present the usual abstract concepts but in a friendly example oriented way. I.e. he thinks it worthwhile to actually teach first what functions and limits are, which were 19th century innovations, before teaching calculus ideas themselves, even though those ideas were thousands of years old before functions were introduced. I try to teach the ideas without the formality introduced more recently. Which way works best is something I cannot predict. He also charges a fair price, $40 for his book, but his price comparisons with other new edition prices costing hundreds of dollars, although perhaps correct, are a little misleading, since used copies of those other books, such as Thomas Calculus for example, are available (e.g. on abebooks.com) for under $4. including shipping. It looks as if he has made a real effort to explain the usual content of an average calculus course in a gentle way. My approach is different in that I try to present somewhat different takes on those ideas. His approach might be better for someone wanting to eventually mesh with a traditional course. Mine is intended to foster understanding without necessarily going over the same ground as usual. His is obviously much more detailed than mine since I have already covered above in a few pages, results such as the fundamental theorem of calculus, that he presents after over 400 pages. Mine of course is also free. If your child is enjoying LOF calculus, I think he will be very well served reading it and practicing the skills and ideas learned there. If he wants to read my notes I would be glad to offer answers to questions that arise. Good luck! Quote Link to comment Share on other sites More sharing options...

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