mm-3709 === Subject: Re: Question about a book? [...] > I have seen a website with a page How To > Learn Mathematics that very strongly recommends reading Polya How to Solve > It. [...] Although many recommend it, I am not a fan of Polya. I think that reading about how to do math is like reading about how to lay bricks: interesting perhaps but not very useful when it comes to actually doing it. A competent instructor in whom you have confidence is indispensable. It is almost impossible for a beginner to know whether his answer to Show that ... is correct. -- === Subject: Re: What should I do? A resorting to name calling, again. How mature. And didn't you just complain about people who try to tell others when or when not to post? HMMM, very interesting, inf intellectually inconsistent. >> Well, you could quit wating my time by positng to this group for >> starters. >> And then, with all the time you save, you could learn some math and maybe >> PROVE something REAL. > You waste your own time. > It's your choice to read or post. > Usenet is about making ADULT decisions about what you read and respond > to, and not childishly whining when someone says something you're not > interested in. > Just ignore them. > Now with that said, get out of my threads. LOL. > See? I can be childish too, you immature twerp. > James Harris === Subject: Re: What should I do? I just couldn't help myself on this one Mr. Sperry. The inconsistency in his argument created when he resorts to the kind of ad hominem attacks of which he so blatantly accuses others and the paranoid schizophrenia exhibited by the mathematician conspiracy around the world is just too sad (or is it humorous) to leave alone in this case. It makes his argument point to the fact that he shouldn't post here. But looking back, you are right. JSH posts are being added to my rules wizard. So that I am not tempted again. Dustin >> Well, you could quit wating my time by positng to this group for >> starters. >> And then, with all the time you save, you could learn some math and >> maybe >> PROVE something REAL. >> You waste your own time. >> It's your choice to read or post. >> Usenet is about making ADULT decisions about what you read and respond >> to, and not childishly whining when someone says something you're not >> interested in. >> Just ignore them. >> Now with that said, get out of my threads. LOL. >> See? I can be childish too, you immature twerp. > standelds, > Welcome to the wild, wacky and rude world of James Harris. If you are > misguided enough to continue to correspond with him (I gave it up long > ago), immature twerp will be the least of what he will call you. > Since I find Harris' English awful to read and his attempts at > mathematics even worse, I haven't been following his current screeds in > any detail. > He sets 7(A' + 1)(B' + 1) = (A + 7)(B + 1), uses the distributive > propery - with which he is infatuated - to get > (7A' + 7)(B' + 1) = (A + 7)(B + 1) then leaps to the conclusion that > 7A' + 7 = A + 7 and B' + 1 = B + 1. When challenged he gives examples > of the distributive property and ignores all examples showing that his > conclusion is false. > Unless you just enjoy doing it, it is futile to argue with him. > -- > Paul Sperry > Columbia, SC (USA) === Subject: URGENT! SPSS Analysis for Research Project Needed Am a business consultant doing a short course in sociology and have completed a research assignment recently. I want to append some SPSS analysis to the report but don't have access to SPSS. if someone is interested in doing some math analysis for this particular project with fully annotated references made to your work, please contact me. All SPSS work will be fully referenced as to the author of the analysis. There are two sets of questions, One is broken into about 4 Tasks of between 5-10 questions and the other is a straight set of questions based upon another set of data to the tune of approximately 10 questions. I'm a BDM doing part-time study in Brisbane, Australia. If anyone is interested I can send you the questions and datasheet via email. my address is detectivened@yahoo.com === Subject: Re: graphing 'y=cosech x' & 'y = coth x' > Hello all, > I'm new to the topic of hyperbolics and am having some difficulty with the > cosech and coth curves. > y = cosech x = 1 / (sinh x) = 2 / (e^x - e^-x) > if [1 / (sinh x)] equals [cosech x] then shouldn't the inverse of sinhx > curve be the curve of cosech x? Well I know it's not but I'm wondering why. As another poster has pointed out, you hav misunderstood the definition of an inverse function - it does not mean 1/f(x), even though it is maybe confusingly written f^-1(x). if we have a function y = f(x), then this takes values that are in the set 'x' (ie. the domain of the function) and 'maps' them to the set 'y' (the co-domain or range). Now what if we wanted to go the other way, ie. start with a value in 'y' and find its corresponding value in 'x' ? This is where the inverse function comes in as it allows us to do that as x = f^-1(y). > (i.e. why is the graph of cosech x not equal to arsinh x?) Which must mean > that (sinh x)^-1 is not equal to (sinh^-1 x) - right? This question come to > mind when I'm trying to sketch the curves of cosech x and cothx... > Any tips on how to plot the curves of cosech x and cothx... Any guidance > will be much appreciated. > Also, I would be thankful for any defintions on the difference of odd and > even graphs. === Subject: proving theta relationship I am reviewing discrete math and found this problem on the web from a Princeton discrete math class. The problem is: Show that ln n = theta(log n). The given answer is: Proof: ln n = log n / log e. To me, this proof isn't sufficient because I thought in order to prove a big theta relationship, you must prove that ln n = O(log n) and log n = O(ln n), in other words, you must prove both big-oh relationships. The proof given seems to me just to prove the first one. Shouldn't another line be given that proves the second big oh relationship? On the other hand, I tend to believe that a professor of a Princeton Discrete Math class knows more than I do... === Subject: Re: proving theta relationship > I am reviewing discrete math and found this problem on the web from a > Princeton discrete math class. The problem is: Show that ln n = > theta(log n). The given answer is: > Proof: ln n = log n / log e. > To me, this proof isn't sufficient because I thought in order to prove > a big theta relationship, you must prove that ln n = O(log n) and log n >= O(ln n), in other words, you must prove both big-oh relationships. > The proof given seems to me just to prove the first one. Shouldn't > another line be given that proves the second big oh relationship? > On the other hand, I tend to believe that a professor of a Princeton > Discrete Math class knows more than I do... proving that they differ by a constant ratio seems sufficient to me -- Bye. Jasen === Subject: Re: proving theta relationship On 11 Mar 2006 23:51:23 -0800, Foster Bell Princeton discrete math class. The problem is: Show that ln n = >theta(log n). The given answer is: >Proof: ln n = log n / log e. >To me, this proof isn't sufficient because I thought in order to prove >a big theta relationship, you must prove that ln n = O(log n) and log n >= O(ln n), in other words, you must prove both big-oh relationships. >The proof given seems to me just to prove the first one. Shouldn't >another line be given that proves the second big oh relationship? _Formally_ yes, another line is required. Here the second line is utterly immediately incredibly awesomely obvious: (i) ln n = log n / log e (ii) log n = ln n * log e. If someone is making a claim that depends on (i) and (ii) he's not going to state (ii) explicitly. Are you _actually_ saying that once (i) was stated, and you realized you needed something like (ii), you really didn't realize that (ii) followed from (i)? >On the other hand, I tend to believe that a professor of a Princeton >Discrete Math class knows more than I do... === Subject: Re: proving theta relationship <1c2812l2g683t0se055p36se103etc8ten@4ax.com> Are you _actually_ saying > that once (i) was stated, and you realized you needed something > like (ii), you really didn't realize that (ii) followed from (i)? No, what I'm saying is that the proof didn't seem complete. The algebra is obvious; the formal method(s) are not. In learning the formal methods, I need to learn what needs to be stated explicitl and what does not. In this case, it is apparently not necessary to state both lines individually, if a simple algebra step can is all that is needed to get from one to another. I have never seen this shorthand rule stated anywhere, but am willing to accept it. It is not at all obvious what needs to be formally proven in general and what does not though. I got points off for putting x > log x for x greater than or equal to 1. This wasn't proof enough for the problem. Understanding how much detail needs to be provided in each step is a crucial part of doing proofs. === Subject: Re: Idea of Usenet, controlling posters >Usenet is generally considered to be a free speech area, but it's nice >to have some general rules, like on a math newsgroup, you'd hope people >mostly talk about mathematics Curious that you should say this, given what a large amount of space in _your_ posts is dedicated to things other than mathematics. >while sometimes it's necessary, like >here, to talk about how Usenet operates as well. >On a math newsgroup that has undergrad in the name, you'd also think >that mathematics relevant to undergraduates would be discussed as well. >But hey, beyond that--those are the rules--there really isn't any basis >for going after any particular person to order them to not post, or in >trying to control their postings. And it's simply _disgusting_ that _you_ would say _this_, given the way _you_ continually tell people to stop posting in your threads. Like standelds just now. >It's just not part of the idea of Usenet, it's not in the spirit of it, >and it doesn't actually work. >[...] === Subject: Re: Indefinite integrals - I'm stuck you should try treasuretrooper it pays great! I got paid 35 dollars in about 3 hours of work! Its free no cost! no credit card needed! [url]http://www.treasuretrooper.com/18552[/url] === Subject: Re: Indefinite integrals - I'm stuck >y = integral [ f(x) dx] >Think of the differentials dy and dx as telling you what >variable you're integrating with respect to. On the left >side, the integral of dy is y since there is an >understood 1 in front ot the dy and the derivative of >y wrt y is 1. On the right, since we have only a >general expression we can just leave it that way for >now. >On the left side, the integral of dy is y >since there is an understood 1 in front ot the dy Is 1 in front of dy because sum of all infinitesimally small terms f(x) dx gives 1 ( 100 percents of a whole )? >and the derivative of y wrt y is 1 This I don't understand. === Subject: Re: Indefinite integrals - I'm stuck >y = integral [ f(x) dx] >>Think of the differentials dy and dx as telling you what >>variable you're integrating with respect to. On the left >>side, the integral of dy is y since there is an >>understood 1 in front ot the dy and the derivative of >>y wrt y is 1. On the right, since we have only a >>general expression we can just leave it that way for >>now. >>On the left side, the integral of dy is y >>since there is an understood 1 in front ot the dy > Is 1 in front of dy because sum of all infinitesimally > small terms f(x) dx gives 1 ( 100 percents of a whole )? When finding an antiderivative, you are not summing anything. With respect to other posters, frankly unless you're talking about DEFINITE integration, which you are not, speaking of the limit of a sum at this point is premature in this context. 1 is in front of dx for the exact same reason there is an understood 1 in front of any expression, for ex. x. Its the same as 1 times x. So what function (of y) has 1 as its derivative? y, right? >>and the derivative of y wrt y is 1 > This I don't understand. Then you really are stuck. -- Darrell === Subject: Re: Indefinite integrals - I'm stuck <10291330.1142126333268.JavaMail.jakarta@nitrogen.mathforum.org> y = integral [ f(x) dx] >>Think of the differentials dy and dx as telling you what >>variable you're integrating with respect to. On the left >>side, the integral of dy is y since there is an >>understood 1 in front ot the dy and the derivative of >>y wrt y is 1. On the right, since we have only a >>general expression we can just leave it that way for >>now. >>On the left side, the integral of dy is y >>since there is an understood 1 in front ot the dy > Is 1 in front of dy because sum of all infinitesimally > small terms f(x) dx gives 1 ( 100 percents of a whole )? > When finding an antiderivative, you are not summing anything. With respect > to other posters, frankly unless you're talking about DEFINITE integration, > which you are not, speaking of the limit of a sum at this point is premature > in this context. Look at the original post. The OP wanted to know what the dx in the indefinite integral notation is doing there, and what relation it has to differentials. That is the question I answered. The integral notation literally means that we are taking a sum of infinitely many infinitesimal terms - whether it is a indefinite integral OR an indefinite integral. It so happens that the indefinite integral IS the antiderivative. It is true that you often may not think of the indefinite integral as being a sum - you think of it as just an antiderivative. But if you want to understand why the notation is the way it is then you must understand what it literally means. === Subject: Re: Indefinite integrals - I'm stuck > Look at the original post. I always do before I even reply. > The OP wanted to know what the dx in the > indefinite integral notation is doing there, and what relation it has > to differentials. That is the question I answered. The integral > notation literally means that we are taking a sum of infinitely many > infinitesimal terms - whether it is a indefinite integral OR an > indefinite integral. You're obfuscating the issue. In all probability the OP is being exposed to differentiation, indefinite integration, and definite integration, in THAT order, despite Archimedes. This is 2006. > It so happens that the indefinite integral IS the > antiderivative. It is true that you often may not think of the > indefinite integral as being a sum - you think of it as just an > antiderivative. But if you want to understand why the notation is the > way it is then you must understand what it literally means. An indefinite integral is a family of functions with the property that they are the derivative of a certain function. Save sums for definite integration. THAT is where the connection between dx and its role with a sum will be discovered. This is standard calc series surely, not any advanced real analysis stuff, so remember you need to wave the hand a little when speaking of dx, ESPECIALLY in the notation of the indefinite integral. The fact is many elementary calc texts simply DEFINE the notation is similar terms as I did in my previous post, ie in terms of a general solution to a general differential equation: dy/dx = f(x) leads directly to y=INT f(x)dx BY DEFINITION. -- Darrell === Subject: Re: Indefinite integrals - I'm stuck <10291330.1142126333268.JavaMail.jakarta@nitrogen.mathforum.org> I always do before I even reply. > The OP wanted to know what the dx in the > indefinite integral notation is doing there, and what relation it has > to differentials. That is the question I answered. The integral > notation literally means that we are taking a sum of infinitely many > infinitesimal terms - whether it is a indefinite integral OR an > indefinite integral. > You're obfuscating the issue. In all probability the OP is being exposed to > differentiation, indefinite integration, and definite integration, in THAT > order, despite Archimedes. This is 2006. > It so happens that the indefinite integral IS the > antiderivative. It is true that you often may not think of the > indefinite integral as being a sum - you think of it as just an > antiderivative. But if you want to understand why the notation is the > way it is then you must understand what it literally means. > An indefinite integral is a family of functions with the property that they > are the derivative of a certain function. Save sums for definite > integration. THAT is where the connection between dx and its role with a > sum will be discovered. This is standard calc series surely, not any > advanced real analysis stuff, so remember you need to wave the hand a little > when speaking of dx, ESPECIALLY in the notation of the indefinite integral. > The fact is many elementary calc texts simply DEFINE the notation is similar > terms as I did in my previous post, ie in terms of a general solution to a > general differential equation: > dy/dx = f(x) leads directly to y=INT f(x)dx BY DEFINITION. If asked to integrate f(x) then I usually view the integral purely as the antiderivative: I am trying to find a function that, when differentiated, gives f(x). However, I find it helpful to *also* understand what the integration notation literally means. It helps to link a lot of things together which can otherwise seem like they're coming out of the blue. For example, given f(x)dx = f(y)dy, how do we get to #f(x)dx = #f(y)dy? Just a bit of magic to do with antiderivatives? No, it's because we are summing the terms on each side (in the limit). Why is the definite integral between a and b calculated by evaluating the indefinite integral at b and subtracting the indefinite integral at a? Another bit of magic? No, it's because the indefinite integral carries with it the same notion of summation as the definite. It's the *same thing*: it's the sum up to x from some starting point that depends on the arbitrary constant of integration. And, last but not least, it explains why the dx in an indefinite integral is the same dx as in differentiation - for example, why it can be replaced by another equivalent expression (involving, say, du in substitution). It shows that dx it is not just an arbitrarily-chosen variable-marker to show the variable we are integrating with respect to. If you don't find it helpful to think of it this way then fair enough. === Subject: Re: Indefinite integrals - I'm stuck <10291330.1142126333268.JavaMail.jakarta@nitrogen.mathforum.org> notation literally means that we are taking a sum of infinitely many > infinitesimal terms - whether it is a indefinite integral OR an > indefinite integral. I did of course mean whether it is a indefinite integral OR a definite integral. === Subject: Re: Indefinite integrals - I'm stuck 1) >>So we could say that notations F = #f(x)*dx, dF = >>f(x)*dx and [integral] f(x)*dx are like words in >>chinese language( meaning we have to translate it in >>our minds to our native language, but the fact is >>they are telling us the same thing ) telling us that >>F is antiderivative of f(x)? >yes, although the >two equations say different things, the implication of >both is that f is the derivative of F, hence F is the >antiderivative of f. What two equations did you have in mind? Is the following correct: F = #f(x)*dx ->tells or just IMPLIES that sum of f(x)dx gives antiderivative? dF = f(x)*dx ->tells what change_in_function is when [h-->0] equals f(x)*dx, but also IMPLIES that F is anti-derivative? [integral] f(x)*dx says the exact same thing as F = #f(x)*dx? ******************************************** 2) >lim(delta-F -> 0) = lim(delta-x -> 0, f(x)*delta-x) >You are right that strictly this gives us 0 = 0, and the >same uninteresting result would be achieved for any F >and f, regardless of whether F' = f. In fact, >dF=f(x)*dx says something far stronger and more >interesting than 0 = 0. Loosely speaking, it says that >both sides tend to zero equally. Uh, can you be more specific about both sides tend to zero equally? Does it mean that if delta-x gets 10 times smaller, then both dF and f(x)*dx will get 10 times smaller? I didn't noticed that pattern with example you gave: >Take the real example f(x) = 3*x^2, F(x) = x^3. >Let x = 1 and delta-x = 0.01. Calculate delta-F and >f(x)*delta-x and compare them. Now try delta-x = 0.001 >and do the same again. See what happens? *********************************************** 3) >Integrating both sides of dF=f(x)*dx gives #dF = #f(x)*dx What exactly does it mean when you say that we integrate dF and get #dF? Just that we're summing infinitesimally small terms and thus INDIRECTLY that F is anti-derivative? *************************************************** >Loosely speaking, we are summing an infinite number of >On the left side, the integral of dy is y >since there is an understood 1 in front ot the dy , >to give a non-zero quantity on both sides. Each >infinitesimally small quantity dF is equal to the >corresponding infinitesimally small quantity f(x)*dx, >so the sums must also be the same. (This is not a >rigorous explanation, just a way of visualising what is >going on.) I get it. === Subject: Re: Indefinite integrals - I'm stuck <155714.1142126389806.JavaMail.jakarta@nitrogen.mathforum.org 1) >>So we could say that notations F = #f(x)*dx, dF = >>f(x)*dx and [integral] f(x)*dx are like words in >>chinese language( meaning we have to translate it in >>our minds to our native language, but the fact is >>they are telling us the same thing ) telling us that >>F is antiderivative of f(x)? >yes, although the >two equations say different things, the implication of >both is that f is the derivative of F, hence F is the >antiderivative of f. > What two equations did you have in mind? The two that you mentioned in the passage I was replying to. > Is the following correct: > F = #f(x)*dx ->tells or just IMPLIES that sum of f(x)dx gives antiderivative? > dF = f(x)*dx ->tells what change_in_function is when > [h-->0] equals f(x)*dx, but also IMPLIES that F is > anti-derivative? You're getting unnecessarily confused about some perceived difference between tells and implies, or something. Let's start again. You understand that dF/dx = f means that f is the derivative of F (with respect to x), and hence F is the antiderivative of f, right? You can see how dF/dx = f leads to dF = f*dx, right? Then take dF = f*dx and integrate both sides, to get F = #f*dx, OK? We have ended up with F = integral of f (with respect to x). And the integral of f *is* the antiderivative of f, so all is well. The integration turns the relationship between dx and dF into a relationship between the functions as a whole, if you want to think of it that way. The three equations are three different ways of indicating the relationship between f and F. The first, dF/dx = f, emphasises that the derivative of F w.r.t. x is f. The second, dF = f*dx, is written in a sort of intermediate naked differentials form, for want of a better way of putting it. The third, F = #f*dx, emphasises that F is the integral (or antiderivative) of f. > [integral] f(x)*dx says the exact same thing as F = #f(x)*dx? The latter is an equation but the former is merely an expression (that doesn't even mention F), so how can they possibly say the exact same thing? > ******************************************** > 2) >lim(delta-F -> 0) = lim(delta-x -> 0, f(x)*delta-x) >You are right that strictly this gives us 0 = 0, and the >same uninteresting result would be achieved for any F >and f, regardless of whether F' = f. In fact, >dF=f(x)*dx says something far stronger and more >interesting than 0 = 0. Loosely speaking, it says that >both sides tend to zero equally. > Uh, can you be more specific about both sides tend to zero equally? Does it mean that if delta-x gets 10 times > smaller, then both dF and f(x)*dx will get 10 times smaller? It means more than that. > I didn't noticed that pattern with example you gave: What numbers did you get then? >Take the real example f(x) = 3*x^2, F(x) = x^3. >Let x = 1 and delta-x = 0.01. Calculate delta-F and >f(x)*delta-x and compare them. Now try delta-x = 0.001 >and do the same again. See what happens? > *********************************************** > 3) >Integrating both sides of dF=f(x)*dx gives #dF = #f(x)*dx > What exactly does it mean when you say that we integrate > dF and get #dF? Just that we're summing infinitesimally > small terms and thus INDIRECTLY that F is anti-derivative? #dF means #1*dF, i.e. the the integral of 1 w.r.t F. It means that function of F which, when differentiated with respect to F, equals one. This function is obviously F. With regard to the infinite summation. Originally you wanted to know what the dx was doing in the integral notation and how it squared with the use of dx as a differential. The answer, as I have tried to explain, has to do with understanding an integral as an infinite sum. However, when presented with an integral to solve it is more usual to just think of it as an antiderivative. For example, you see #g(x)dx, and you just think right, the answer to this integral is that function of x which, when differentiated w.r.t. x, gives g(x). In fact, once you are familiar with this stuff, you can happily go directly, in one leap, from dF/dx = f to F = #f dx. You just think f is the derivative of F, so F is the integral of f, and you're done. === Subject: Re: JSH: Updated quick quide, quadratics up front > Unlike with my first quick math guide, which has been useful for years > now, I screwed up with my second and introduced a flawed argument, so I > need to update. > This quick guide supersedes the previous one. I use these guides to > make it easier on me in replying to posts as I can just copy and paste > out arguments already worked out. > Possibly I just tried to create a quick guide too soon, but whats done > is done, so I may as well just update and move on. > As I did years ago when I had key results that I wanted easy reference > to at any time in the future, I'm creating a quick quide, where this > time it is to the simple proof in the complex plane and the quadratic > with which examples can be found that eradicate all reasonable > objections. > In the complex plane, given > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > it must be true that 7 multiplied through the first factor A(x) + 7. Of course if you're in the context of the complex numbers, then your result is trivially true (however, in that context 7 also multiplies through the second factor as well). Perhaps you want to restrict divisibility to mean divisibility in the ring of algebraic integers. If you insist on avoiding the algebraic integers, try C(x) = (x + 1)^2 A(x) = 3x + 2sqrt(-10x^2 - 14x) B(x) = (3x - 2sqrt(-10x^2 - 14x)) / 7 Obviously, A(0) = B(0) = 0 and with a bit of work you can verify that 7C(x) = (A(x) + 7)(B(x) + 1) Take a close look at A. Is it divisible by 7? It's sort of like slight of hand, eh? Where'd the 7 go? Rick === Subject: Re: JSH: Updated quick quide, quadratics up front A completely non-trickery counterexample to what JSH probably *really* means by all this 7 multiplies through crap, followed by this: > Take a close look at A. Is it divisible by 7? It's sort of like > slight of hand, eh? Where'd the 7 go? obligatory nit to pick. Although pronounced slight of hand, it's really sleight of hand. > Rick Dale. === Subject: Re: JSH: Updated quick quide, quadratics up front > A completely non-trickery counterexample to what JSH probably *really* > means by all this 7 multiplies through crap, followed by this: >> Take a close look at A. Is it divisible by 7? It's sort of like >> slight of hand, eh? Where'd the 7 go? > obligatory nit to pick. > Although pronounced slight of hand, it's really sleight of hand. Quite correct, Dale. I can't even claim it was a typo. It was a thinko. I'm shagreened. Rick === Subject: Re: JSH: Updated quick quide, quadratics up front means by all this 7 multiplies through crap, followed by this: >> Take a close look at A. Is it divisible by 7? It's sort of like >> slight of hand, eh? Where'd the 7 go? > obligatory nit to pick. > Although pronounced slight of hand, it's really sleight of hand. > Quite correct, Dale. I can't even claim it was a typo. It was a thinko. > I'm shagreened. > Rick I read your post. I'm not interested. Major work is done dude. The typo is just one of those things. I hardly even acknowledge my own anymore. Years of experience, and many typos under the bridge, you might say. James Harris === Subject: Re: JSH: Updated quick quide, quadratics up front > > > A completely non-trickery counterexample to what JSH probably *really* >means by all this 7 multiplies through crap, followed by this: > >> Take a close look at A. Is it divisible by 7? It's sort of like >> slight of hand, eh? Where'd the 7 go? > obligatory nit to pick. > > Although pronounced slight of hand, it's really sleight of hand. > Quite correct, Dale. I can't even claim it was a typo. It was a thinko. > I'm shagreened. > Rick > I read your post. I'm not interested. Major work is done dude. > The typo is just one of those things. I hardly even acknowledge my own > anymore. That explains a lot. === Subject: Re: JSH: Updated quick quide, quadratics up front > A completely non-trickery counterexample to what JSH probably *really* >means by all this 7 multiplies through crap, followed by this: > >> Take a close look at A. Is it divisible by 7? It's sort of like >> slight of hand, eh? Where'd the 7 go? > obligatory nit to pick. > > Although pronounced slight of hand, it's really sleight of hand. > Quite correct, Dale. I can't even claim it was a typo. It was a thinko. > I'm shagreened. > Rick > I read your post. I'm not interested. Major work is done dude. > The typo is just one of those things. I hardly even acknowledge my own > anymore. And when Gerald Ford got tired of the press constantly publishing photos of him banging his head on the doorway every time he exited his helicopter, his solution was to ban photographers from covering his helicopter landings. > Years of experience, and many typos under the bridge, you might say. > James Harris === Subject: Re: JSH: Updated quick quide, quadratics up front >Unlike with my first quick math guide, which has been useful for years >now, I screwed up with my second and introduced a flawed argument, No. Really? _You_, introduce a flawed argument? Let's see, that would be the argument such that anyone who denied it was disputing the distributive property, right? Now it turns out we didn't need to deny the distributive property, the argument was actually flawed? Who'da thunk it? (A: Anyone who's been paying even the slightest bit of attention. Everyone but _you_ realizes that _all_ your arguments are flawed. Which, by the way, is not an attempt to control your posting, as though that were possible. Far from it - your fans miss you during your periodic absences.) > so I >need to update. >This quick guide supersedes the previous one. I use these guides to >make it easier on me in replying to posts as I can just copy and paste >out arguments already worked out. >Possibly I just tried to create a quick guide too soon, but whats done >is done, so I may as well just update and move on. >As I did years ago when I had key results that I wanted easy reference >to at any time in the future, I'm creating a quick quide, where this >time it is to the simple proof in the complex plane and the quadratic >with which examples can be found that eradicate all reasonable >objections. >In the complex plane, given >7C(x) = (A(x) + 7)(B(x) + 1) >true for all x, where A(0) = B(0) = 0 >it must be true that 7 multiplied through the first factor A(x) + 7. Exactly what that statement means is not at all clear. Recently you've given statements that presumably are supposed to mean the same thing, but which unfortunately _were_ clear enough that it's incredibly easy to see they're false. Switching back to the unintelligible version is probably a good idea. >Proof: >1. The distributive property acts without regard to value, so the value >of the functions has no impact on how it operates, for instance >a(f(x) + b) = af(x) + ab >without the distributive property acting in different ways dependent on >the value of f(x). >2. Then setting the functions to 0 will NOT AFFECT the distributive >property. >So by 2. I can let x=0 which sets the functions to 0, giving >f C(0) = (0 + f)(0 + 1) >proving that f multiplied through the first factor and then by 1. I >have that it does so for all x. >That result valid over the complex plane allows me to cement the case >for the full argument proving I have been right with my research where >you can directly SEE the result with integer roots of >a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0. >There if you pick x and f integers and f not 1 or -1 and coprime to x, >if you find rational roots one root will have f as a factor while the >other will be coprime to f, and that rests on the distributive property >as shown in the previous proof. >It's a remarkable case where the distributive property itself is the >linchpin to an important result. >It is the kind of simple result that shatters all reasonable objections >and reveals the social nature of resistance to these results from the >mathematical community. >Note, the complex plane is an area known to most mathematicians, while >the ring of algebraic integers while known is more of an esoteric area >important to number theorists, by proving my key result on the complex >plane, I remove the excuse of other mathematicians that the results >were all out of their area. >Proving that there is a systemic problem across the entire discipline, >and not just in number theory. >James Harris === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? > hello > I understand the chain rule and I know > [integral] F'(g(x))*g'(x)*dx > represents antiderivative F(g(x)), but when we create > new variable u = g(x) my mind goes blank. > F(g(x)) = [integral] F'(g(x))*g'(x)*dx > u = g(x) => du = g'(x)*dx > so > [integral] f(u)*du = F(u) = F(g(x)) and F'(u) = f(u) > I simply can not see the use for variable 'u', except for > it to save us some writting and complicate the matters! A substitution is intended to make things easier. It makes it easier to recognize antiderivatives. Substitution, at first, is a difficult topic. However, once you master it, it makes a lot of integration tasks easier. I had a lot of trouble with it when I first learned it; and now I am a degreed mathematician! Dave === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? <8595721.1142126506841.JavaMail.jakarta@nitrogen.mathforum.org>, > 1) >hf'(x) = f(x + h) - f(x) + r(h) >where r(h) -> 0 as h -> 0. > What is r(h)? Doesn't the word derivative mean value > f(x + h) - f(x)/h as [h-->0]? Yes, f'(x) = the derivative of f at x = lim((f(x + h) - f(x))/h, h -> 0). That says, informally, that f'(x) and (f( x + h) - f(x))/h get close as h gets small. So their difference approaches 0 as h approaches 0. That difference is r(h)/h. The requirement is actually lim(|r(h)/h|, h -> 0) = 0. What may be throwing you off is that hf'(x) = f(x + h) - f(x) + r(h) is a _property_ of the derivative not a definition. > Judging by your formula the word derivatives also stands > for values when h isn't yet infinitesimally small? No, it is just another way of saying that f'(x) and (f(x + h) - f(x))/h get close in value as h gets close to 0. > ********************************************** > 2) >Note, we are looking for a function whose derivative is >f(x) _not_ for a function whose differential is f(x)dx. > Is there a difference? I mean don't both functions have > same expresion? I know function whose differential is f(x)dx is basicly just > delta-F, but it is still F function ( it has same expression ) The problem is that dx is used in three different ways: It is one of the missing arguments in df(x,dx) = f'(x)dx. Here dx is just a number - usually thought of as being small. Part of the confusion comes from the habit of writing merely df = f'(x)dx. Now, what about df/dx? If I'm thinking differentials, I really should write df(x,dx)/dx which is, in fact, f'(x). Mostly, df/dx is interpreted as an instruction: Find the derivative of f - use x as the variable. Finally, dx is used in the integration symbols just to confuse things :-) - see below. > ************************************************* > 3) >Now consider integrals for a minute. As you know the >(Riemann) integral from a to b is the limit of sums and >the terms of each sum are of the form f(x)h where x is >in a subinterval of length h of [a, b]. As the >limit is taken, h -> 0. Again, it is tradition that >instead of h we write dx. Since we are adding up terms >of the form f(x)dx, we write (long skinny S)f(x)dx. >Note the only connection between the dx in the >differential and the dx in the integral is that they >both go to zero (sort of). > What do you mean by connection? Isn't dx the same variable just used in > different situation? I'll get to your question eventually. Definite integration preceded derivatives (and thus anti-derivatives) by quite a bit. It goes back at least to Archimedes. Leibniz and Newton more or less independently but concurrently developed the derivative as an instantaneous rate of change. Then it was noticed that there was a truly remarkable connection between the derivative and the (definate) integral: If we could find a function F whose derivative was f then we could calculate int(f(x),x=a..b) without tediously calculating the limit of the (Riemann) sums. Because of that connection and the notation for the definite integral which was intended, I suppose, to point out what was being summed, the same notation was used (without the limits) for the anti-derivative. Now, what about the integral notation? The dx in the indefinate integral and the dx in the definite integral are not the same because they are really not anything - they are just part of the notation. All that is needed is that there be no confusion about variables - you could just as well subscript the integral sign. In Maple, int(x*y^2,x) = (1/2)x^2*y^2; int(x*y^2,y) = (1/3)x*y^3. Mathematica works the same way. Suppose I'm looking for an anti-derivative of x^2. Am I looking for a function F such that F'(x) = x^2 or am I looking for a function F such that dF = dF(x,dx) = x^2dx? Take your choice. As another example suppose I ask you to solve xdx = dy (a _differential_ equation). I am asking for y = y(x) such that dy = dy(x,dx) = xdx. I might just as well ask you to solve x = y'. Two forms of the same question. -- === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? you should try treasuretrooper it pays great! I got paid 35 dollars in about 3 hours of work! Its free no cost! no credit card needed! [url]http://www.treasuretrooper.com/18552[/url] === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? <17339442.1142186262693.JavaMail.jakarta@nitrogen.mathforum.org you should try treasuretrooper it pays great! I got paid 35 dollars in about 3 hours of work! > Its free no cost! no credit card needed! > [url]http://www.treasuretrooper.com/18552[/url] All context is missing including reference, haven't time for back tracking the thread to reconstruct the discussion. Don't know if you're addressing this to me or not. Thus I consider your post not addressed to me, bye. -- To Google and MathForum users: Reply only if adequate context is included _within_ the reply. Otherwise all contexts are removed from my view, the flow of thought disrupted and chaos reigns. In particular for Google users: Instead of simply hitting the prominent Reply link, which doesn't include a copy of the post to which one is replying, click the Show Options link (toward the top of an item in the thread), which causes a shaded area of links to appear next to the top of the item, including Reply (first) that does introduce a copy of the previous text (offset by > signs in the usual fashion). ---- === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? >I understand the chain rule and I know >[integral] F'(g(x))*g'(x)*dx > > represents antiderivative F(g(x)), but when we create >new variable u = g(x) my mind goes blank. > I have a detailed example, of which I'm rather proud, at > http://www.acad.sunytccc.edu/instruct/sbrown/calc/usubst.htm Too many high bit ascii graphics too make much sense out of in addition to x6, u4, v2 possiblity for x^6, u^4, v^2. === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? > Fri, 10 Mar 2006 20:46:58 -0800 from William Elliot > : >I understand the chain rule and I know >[integral] F'(g(x))*g'(x)*dx > > represents antiderivative F(g(x)), but when we create > > new variable u = g(x) my mind goes blank. > I have a detailed example, of which I'm rather proud, at > http://www.acad.sunytccc.edu/instruct/sbrown/calc/usubst.htm > Too many high bit ascii graphics too make much sense out of in addition to > x6, u4, v2 possiblity for x^6, u^4, v^2. Looks fine to me. You might want to update your browser. -- === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? >I have a detailed example, of which I'm rather proud, at >http://www.acad.sunytccc.edu/instruct/sbrown/calc/usubst.htm > Too many high bit ascii graphics too make much sense out of in addition to > x6, u4, v2 possiblity for x^6, u^4, v^2. > Looks fine to me. You might want to update your browser. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? <120320060119163100%plsperry@sc.rr.com > Fri, 10 Mar 2006 20:46:58 -0800 from William Elliot > Too many high bit ascii graphics too make much sense out of in addition to > x6, u4, v2 possibility for x^6, u^4, v^2. > Looks fine to me. You might want to update your browser. Computer won't handle it and new computers come with toxic odor. I'm in the market, not for a stupid MicroStuff, but a Unix or Linex system on an old lap top prior to the excessive use of persist volatile toxic chemicals. Same problem with phones, this sick fad of drenching phones in addition to computers, with that anti-static chemical, even the wires and instructions they come with, makes them dangerous to use. Thus I hoar all old phones that I can get my hands on. Too bad, an excellent technology, made stupid by MicroStuff-It software companies and toxic by corporate don't give a damn about people attitide. === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? I feel like crying :( ************************ [integral] (x^3 + 1)^5*x^2*dx = 1/18 * u^6 + C where u = x^3 + 1 and du = 3x^2*dx => x^2*dx=1/3*du *********************************** First of all I know how to solve this problem without the use of variable u. But when we use u: u = x^3 + 1 du = 3x^2*dx => x^2*dx = 1/3du I'm really, really having troubles grasping why we can do the above thing. Why is x^2*dx equal to 1/3*du? I know you've said dx is there to tell us which variable we are integrating with respect to. But here we actually use it (together with du) in an equation????!!!! I'm really having hard time figuring out how to treat this dx and du when it comes to integrals. === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? <11967930.1142126711995.JavaMail.jakarta@nitrogen.mathforum.org I feel like crying :( > ************************ > [integral] (x^3 + 1)^5*x^2*dx = 1/18 * u^6 + C > where > u = x^3 + 1 and du = 3x^2*dx => x^2*dx=1/3*du > *********************************** > First of all I know how to solve this problem without the use of variable u. > But when we use u: > u = x^3 + 1 > du = 3x^2*dx => x^2*dx = 1/3du > I'm really, really having troubles grasping why we can do > the above thing. > Why is x^2*dx equal to 1/3*du? defined u that way because you hoped it would help to simplify the integral. If you'd defined u a different way (to help solve a different integral, say) then the relationship between dx and du would be different. === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? <11967930.1142126711995.JavaMail.jakarta@nitrogen.mathforum.org [integral] (x^3 + 1)^5*x^2*dx = 1/18 * u^6 + C > where > u = x^3 + 1 and du = 3x^2*dx => x^2*dx=1/3*du Rule: integral f(u(x)) du/dx dx = integral f(u) du u = x^3 + 1; du/dx = 3x^2; f(u) = u^5; f(u(x)) = (x^3 + 1)^5 Thus substituting those values in the rule integral (x^3 + 1)^5 x^2 dx = integral u^5 (1/3) du/dx dx = (1/3) integral u^5 du/dx dx = (1/3) integral u^5 du = (1/3) (1/6) u^6 + c = (1/18) u^6 + c > But when we use u: > u = x^3 + 1 > du = 3x^2*dx => x^2*dx = 1/3du 1/3 * du > I'm really, really having troubles grasping why we can do > the above thing. > Why is x^2*dx equal to 1/3*du? > I know you've said dx is there to tell us which variable > we are integrating with respect to. But here we actually > use it (together with du) in an equation????!!!! As you removed all context including even the references like what you're replying without backtracking the thread which I seldom do, even unlike tonight, I've the time to do so. Thus I'll skip your comment as without cogency. > I'm really having hard time figuring out how to treat this dx and du > when it comes to integrals. I've shown you a more exact method without using the mnemonics of the dx du stuff. === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? 1) >hf'(x) = f(x + h) - f(x) + r(h) >where r(h) -> 0 as h -> 0. What is r(h)? Doesn't the word derivative mean value f(x + h) - f(x)/h as [h-->0]? Judging by your formula the word derivatives also stands for values when h isn't yet infinitesimally small? ********************************************** 2) >Note, we are looking for a function whose derivative is >f(x) _not_ for a function whose differential is f(x)dx. Is there a difference? I mean don't both functions have same expresion? I know function whose differential is f(x)dx is basicly just delta-F, but it is still F function ( it has same expression ) ************************************************* 3) >Now consider integrals for a minute. As you know the >(Riemann) integral from a to b is the limit of sums and >the terms of each sum are of the form f(x)h where x is >in a subinterval of length h of [a, b]. As the >limit is taken, h -> 0. Again, it is tradition that >instead of h we write dx. Since we are adding up terms >of the form f(x)dx, we write (long skinny S)f(x)dx. >Note the only connection between the dx in the >differential and the dx in the integral is that they >both go to zero (sort of). What do you mean by connection? Isn't dx the same variable just used in different situation? === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? >The result will be a function of u. To get the answer in >terms of x, reverse the substitution u = x^3 + 1. How do I reverse it? :0 === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? The result will be a function of u. To get the answer in >terms of x, reverse the substitution u = x^3 + 1. > How do I reverse it? :0 In the original substitution you turned an expression involving x into one involving only u. By reversing this substitution I mean that, having found the answer to the integral in terms of u, turn it back into an answer involving only x. === Subject: Re: Substitution Rule for Indefinite Integrals-why we need new variable? The result will be a function of u. To get the answer in >terms of x, reverse the substitution u = x^3 + 1. > How do I reverse it? :0 Solve for x in terms of u, ie isolate x to one side of the equation.