mm-1349 === Subject: RE: HELP ME by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i16FHXB07791; Hello Liz baby, Whats up with your question girl? I don't think you are asking the right question because I can't imagine anything with two eyes that lives in one. Anyway, If you find the answer, please let me know. have a nice weekend. Adache KCmth === Subject: Re: Abstracting out the method, non-polynomial factorization > Some corrections... > Despite all the controvery over my method, basically what I've done is > a balancing act--simple against complex--and here's an abstraction of > the technique: > > Consider f_1(x) f_2(x) = g F(x), and gab = gc, where > > (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c) > > where f_1(0) = f_2(0) = F(0) = G(0) = 0. > > > Now I use the fact that if you're in the ring of algebraic integers > then dividing g from both sides *must* give > > ab = c, > Actually that factorization is available but so are an infinity of > other unit factor factorizations like > (-a)(-b) = c > so I went for a strong condition and got it wrong, when the proper > condition that works well is simply that the factorization is > available in the ring being considered. > whereas, if you're in some other ring, like the field of algebraic > numbers, then you have an *infinity* of factorizations on the left, > like > > (sqrt(g) a)(b/sqrt(g)) = c. > Correcting following up from before, the issue isn't the number of > factorizations but availability of a given factorization. > Here the proper point is that the factorization shown is NOT available > in the ring of algebraic integers. Let's be direct about this. Your methods, if correct, would imply that Q(x) = (25*x^2 + 30*x + 2) cannot be factored in the form (5 a_1 + c_1)*(5 a_2 + c_2), when x > 0 and a_1, c_1, a_2, and c_2 are all algebraic integers. You say it's impossible. However, Rick Decker shows that when x = 1, Q(x) = (5 sqrt(-2) + sqrt(7))*(-5 sqrt(-2) + sqrt(7)). All the numbers in sight are algebraic integers. What you keep saying is impossible is clearly, unambiguously possible. Not just for x = 1, but for any x > 0. It is strange that you keep worrying about Rick's example, but you never quote what it actually says. You just keep saying it is impossible. Anyone who can do the arithmetic can see that you are wrong. Now you are taking your wrong conclusion and trying to generalize it into a wrong *method*. Even worse, you are trying to glorify it into some kind of astounding (but wrong!) GREAT DISCOVERY. Generalizing bad math still leaves you with just ... bad math. You know what counterexamples do. They disprove methods. That's what has happened here. Your attempt to deal with it, running at top speed in the wrong direction, changes nothing. You are still stuck with a hard-core fact: what you keep saying is impossible is sitting there, unrefutable, like a chunk of concrete. Nora B. > James Harris === Subject: Re: Abstracting out the method, non-polynomial factorization >> Some corrections... >> Despite all the controvery over my method, basically what I've done is >> a balancing act--simple against complex--and here's an abstraction of >> the technique: >> >> Consider f_1(x) f_2(x) = g F(x), and gab = gc, where >> >> (f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c) >> >> where f_1(0) = f_2(0) = F(0) = G(0) = 0. >> >> >> >> Now I use the fact that if you're in the ring of algebraic integers >> then dividing g from both sides *must* give >> >> ab = c, >> Actually that factorization is available but so are an infinity of >> other unit factor factorizations like >> (-a)(-b) = c >> so I went for a strong condition and got it wrong, when the proper >> condition that works well is simply that the factorization is >> available in the ring being considered. >> whereas, if you're in some other ring, like the field of algebraic >> numbers, then you have an *infinity* of factorizations on the left, >> like >> >> (sqrt(g) a)(b/sqrt(g)) = c. >> Correcting following up from before, the issue isn't the number of >> factorizations but availability of a given factorization. >> Here the proper point is that the factorization shown is NOT available >> in the ring of algebraic integers. > Let's be direct about this. > Your methods, if correct, would imply that > Q(x) = (25*x^2 + 30*x + 2) >cannot be factored in the form > (5 a_1 + c_1)*(5 a_2 + c_2), >when x > 0 and a_1, c_1, a_2, and c_2 are all >algebraic integers. > You say it's impossible. > However, Rick Decker shows that when x = 1, > > Q(x) = (5 sqrt(-2) + sqrt(7))*(-5 sqrt(-2) + sqrt(7)). > All the numbers in sight are algebraic integers. > What you keep saying is impossible is clearly, unambiguously >possible. Not just for x = 1, but for any x > 0. Where for any x > 0 means ___ ? > It is strange that you keep worrying about Rick's example, >but you never quote what it actually says. You just keep >saying it is impossible. > Anyone who can do the arithmetic can see that you are >wrong. Now you are taking your wrong conclusion and trying >to generalize it into a wrong *method*. Even worse, you >are trying to glorify it into some kind of astounding (but >wrong!) GREAT DISCOVERY. > Generalizing bad math still leaves you with just ... bad math. > You know what counterexamples do. They disprove methods. >That's what has happened here. Your attempt to deal with it, >running at top speed in the wrong direction, changes nothing. >You are still stuck with a hard-core fact: what you keep saying >is impossible is sitting there, unrefutable, like a chunk of concrete. > Nora B. >> James Harris ************************ David C. Ullrich === Subject: Re: Abstracting out the method, non-polynomial factorization > Some corrections... > > [deletia] >>whereas, if you're in some other ring, like the field of algebraic >>numbers, then you have an *infinity* of factorizations on the left, >>like >> >>(sqrt(g) a)(b/sqrt(g)) = c. > > > Correcting following up from before, the issue isn't the number of > factorizations but availability of a given factorization. > > Here the proper point is that the factorization shown is NOT available > in the ring of algebraic integers. > what if g=4 and b is even? Special cases aren't of interest here as it's the generalization that's important. Like, (x+2)/3 is not in general in the ring of integers, but at x=1 it is, so it's like your special case with g=4. If it will help you understand, instead use (sqrt(3) a)(b/sqrt(3)) = c which is a factorization also available in the field of algebraic numbers but NOT in the ring of algebraic integers. James Harris === Subject: Re: Abstracting out the method, non-polynomial factorization > Special cases aren't of interest here as it's the generalization > that's important. Special cases are often used to refute a generalization -- or didn't you know that? > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Research question answered > One of the reasons I've been so fascinated by the Decker example is > that it revealed to me that I wasn't exactly certain about > methodologies that I'd discovered, as it seems I didn't understand > exactly how everything worked. Could it possibly be because it _doesn't_ work? Could it be that several people have proven conclusively that you're wrong, and you can _see_ that their objections are valid, but you can't accept your own failure, so you force yourself to go on believing in spite of the evidence to the contrary? > Recently Rick Decker, a professor at Hamilton College, apparently > trying to refute my research came up with a quadratic example, which I > like because it's a quadratic, and easier to manipulate than the > cubics I've used before. > If you wish to see his original post here are some headers which also > show that he posts from Hamilton College: James, do you sit around your hovel thinking of ways to look like a moron? Because every time you re-post this weird obsession with Hamilton College you make yourself appear _even more_ moronic. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Research question answered >> One of the reasons I've been so fascinated by the Decker example is >> that it revealed to me that I wasn't exactly certain about >> methodologies that I'd discovered, as it seems I didn't understand >> exactly how everything worked. >Could it possibly be because it _doesn't_ work? Could it be that several >people have proven conclusively that you're wrong, and you can _see_ >that their objections are valid, but you can't accept your own failure, >so you force yourself to go on believing in spite of the evidence to >the contrary? Well now let's be fair. Just because there are counterexamples to what he's proved doesn't actually imply that his proofs are incorrect. Could be he's found an inconsistency in core mathematics, or Tool mathematics or whatever it is... >> Recently Rick Decker, a professor at Hamilton College, apparently >> trying to refute my research came up with a quadratic example, which I >> like because it's a quadratic, and easier to manipulate than the >> cubics I've used before. >> If you wish to see his original post here are some headers which also >> show that he posts from Hamilton College: >James, do you sit around your hovel thinking of ways to look like a moron? >Because every time you re-post this weird obsession with Hamilton College >you make yourself appear _even more_ moronic. My guess is at's an attempt at intimidation, as though Rick is going to read this and say to himself Oh my, it's that obvious that I'm posting from Hamilton College? I better cut that out before I get in trouble! (Not that I understand what Rick would be worried about, that's just my guess at the point to this particular bit of moronitiude; James is known to have curious ideas regarding what people are going to get in trouble for...) ************************ David C. Ullrich === Subject: Re: JSH: Research question answered > Well now let's be fair. Just because there are counterexamples to what > he's proved doesn't actually imply that his proofs are > incorrect. Could be he's found an inconsistency in core > mathematics, or Tool mathematics or whatever it is... Actually, JSH's proofs work, once you understand his terminology. For example, when he says X, so Y that doesn't mean Y follows from X like it does in standard terminology. In JSH terminology, it means X is true, and now, for something completely different, here is Y, which has nothing to do with X, and probably has a counterexample in small integers, and I'm a genius. Once you realize this, and a couple other places where he has his own, special, terminology, his proofs make sense, in the sense that everything in them is a correct statement. :-) -- --Tim Smith === Subject: Re: JSH: Research question answered > One of the reasons I've been so fascinated by the Decker example is > that it revealed to me that I wasn't exactly certain about > methodologies that I'd discovered, as it seems I didn't understand > exactly how everything worked. An understatement of monumental proportions ... If the Decker example is so fascinating to you, why is it that, in the 15-20 separate threads you have started on this topic, you have never ONCE quoted Decker's conclusion ? Are you so terrified that other people might read it and realize that Decker was totally, obviously right, and your claims to the contrary were totally, obviously so much hot air? > I mention Decker the most but the first person I remember giving such > an example was Nora Baron, followed by Andrezj Kolowski, then Dik > Winter, and then Rick Decker. Those examples might bear quoting as well. > In response to them I came up with various answers that ultimately > didn't satisfy me, though I could satisfy myself that my own work was > correct, I didn't have a handle on how their examples fit into the > picture. This is true. You just sort of stopped talking about them. You never once handled them in any substantive way. > elucidate the Factorization Tool yesterday. No - all you have done is try to abstract and generalize your previous incorrect and inadequate explanations. The fact remains: you say that for x > 0, Q(x) = (25*x^2 + 30*x + 2) cannot be factored in the form in the form (5 c_1 + d_1)*(5 c_2 + d_2), where c_1, d_1, c_2, and d_2 are algebraic integers, and a_1 = 7 c_1 and a_2 = 7 c_2 are roots of a^2 - (x - 1)*a + 7*(x^2 + x). You keep saying that no such factorization is possible. However, Rick Decker showed that for x = 1, Q(1) = 57 = (5 sqrt(-2) + sqrt(7))*(-5 sqrt(-2) + sqrt(7)), where ALL the numbers on the right side are algebraic integers. You say it's not possible. You say it over an over again, and now you say you have a GENERAL method for showing it's not possible. But there it is, plain as day. It is simple arithmetic to verify it: it IS possible. Counterexamples trump claims. Specific counterexamples always trump general, abstract claims. You can lie to yourself all you want. That's fine. Why you keep trying to lie to everyone else here, given this obvious simple arithmetic refutation of your claim, is quite mystifying. Nora B. > It's neat. I'll start with the Decker example in explaining my > results. [b.s. deleted] > James Harris === Subject: Re: JSH: Research question answered > One of the reasons I've been so fascinated by the Decker example is > that it revealed to me that I wasn't exactly certain about > methodologies that I'd discovered, as it seems I didn't understand > exactly how everything worked. > > An understatement of monumental proportions ... > If the Decker example is so fascinating to you, why is it > that, in the 15-20 separate threads you have started on this > topic, you have never ONCE quoted Decker's conclusion ? Are > you so terrified that other people might read it and realize > that Decker was totally, obviously right, and your claims to > the contrary were totally, obviously so much hot air? off. I will repeat that you should off. If you wish to continue replying to me, you will find you are giving me more opportunities to tell you to OFF!!! I know now that you're not a woman. What woman would so *publicly* stalk a guy like this and repeatedly be told to off and keep at it? Who are you really Nora Baron? James Harris === Subject: Re: JSH: Research question answered >> If the Decker example is so fascinating to you, why is it >> that, in the 15-20 separate threads you have started on this >> topic, you have never ONCE quoted Decker's conclusion ? Are >> you so terrified that other people might read it and realize >> that Decker was totally, obviously right, and your claims to >> the contrary were totally, obviously so much hot air? > off. > I will repeat that you should off. That doesn't answer the question. Try again. -- --Tim Smith === Subject: Re: JSH: Research question answered > Who are you really Nora Baron? I heard a really juicy rumour that she works at Hamilton College. Does that interest you? Doug === Subject: Re: JSH: Research question answered > Who are you really Nora Baron? > I heard a really juicy rumour that she works at Hamilton College. Does > that interest you? And me thinking it was OK State. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Research question answered >> Who are you really Nora Baron? > I heard a really juicy rumour that she works at Hamilton College. Does > that interest you? I heard she was at Egello College. -- --Tim Smith === Subject: Re: JSH: Research question answered > Who are you really Nora Baron? > I heard a really juicy rumour that she works at Hamilton College. Does > that interest you? > Doug *giggle* It should, since every post (well, almost) lately has had something to do with Hamilton College. David Moran === Subject: Re: JSH: Research question answered > One of the reasons I've been so fascinated by the Decker example is > that it revealed to me that I wasn't exactly certain about > methodologies that I'd discovered, as it seems I didn't understand > exactly how everything worked. > > An understatement of monumental proportions ... > If the Decker example is so fascinating to you, why is it > that, in the 15-20 separate threads you have started on this > topic, you have never ONCE quoted Decker's conclusion ? Are > you so terrified that other people might read it and realize > that Decker was totally, obviously right, and your claims to > the contrary were totally, obviously so much hot air? > off. > I will repeat that you should off. > If you wish to continue replying to me, you will find you are giving > me more opportunities to tell you to OFF!!! > I know now that you're not a woman. What woman would so *publicly* > stalk a guy like this and repeatedly be told to off and keep at > it? > Who are you really Nora Baron? > James Harris post, as you do. David Moran === Subject: Re: JSH: Research question answered >One of the reasons I've been so fascinated by the Decker example is >that it revealed to me that I wasn't exactly certain about >methodologies that I'd discovered, as it seems I didn't understand >exactly how everything worked. Guffaw! You know usually when the rest of us prove something we _do_ understand how the proof works. Truly fascinating: The fact that not one mathematician on the planet can understand your stuff proves that they're all part of a global conspiracy to suppress the Truth. Even though it turns out you didn't understand it either. ************************ David C. Ullrich === Subject: Re: i need help with 3 riddels <4022AA35.3030809@microprizes.com> thusly: > Possible spoiler for the third... > Possible spoiler for the third... > Possible spoiler for the third... > Possible spoiler for the third... > >> On the wall I may be found or from heaven I'll come down, to teach to > >> warn or to astound, I'll help the lost if I am found......what am I? > >> its only 4 letters or less... > The the four points of a magnetic compass, arranged as the word NEWS > Carl G. You seem to have it all SEWN up. -- Paul Townsend I put it down there, and when I went back to it, there it was GONE! Interchange the alphabetic elements to reply === Subject: Re: middle term expansion > What is the middle term in the expansion of (x-1/x)^6? > > > Dennis > If you mean (x - (1/x))^6 it is -20. > If you mean ((x - 1)/x)^6 it is -20/(x^3) > [Parentheses are your friends.] the solution. I mean the first option(x-(1/x))^6. Dennis === Subject: Re: middle term expansion >the solution. I mean the first option(x-(1/x))^6. Have you mislaid your textbook? Look under binomial theorem. You should find something for the kth (or ith) term of the expansion of (a+b)^n. Here a=x, b=-1/x, n=6, and k (or i) is the number of the middle term, which is n/2. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com An expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: Re: middle term expansion Look up the binomial theorem. It seems to me looking at the questions you are asking you are in a Discrete Mathematics class? Or combinatorics? -Bill > Dennis > > What is the middle term in the expansion of (x-1/x)^6? > > > Dennis > > If you mean (x - (1/x))^6 it is -20. > If you mean ((x - 1)/x)^6 it is -20/(x^3) > > [Parentheses are your friends.] > the solution. I mean the first option(x-(1/x))^6. > Dennis -- Bill Stevenson Editor in Chief ACM Crossroads Magazine http://www.acm.org/crossroads http://www.billstevenson.org Applied Cognitive Science Laboratory Pennsylvania State University http://acs.ist.psu.edu === Subject: What is the value of k? If the remainder is 7 when x^3+kx^2-3x-15 is divided by x-2, then k= ? Would be gratefull if you drop a few words about the way the solution goes. Dennis === Subject: Re: What is the value of k? > If the remainder is 7 when x^3+kx^2-3x-15 is divided by x-2, then k= ? > Would be gratefull if you drop a few words about the way the solution goes. > Dennis Here's how I'd do it 2 1 k -3 -15 2 2k+4 4k+2 1 k+2 2k+1 4k-13 4k-13=7 4k=20 k=5 Check it 2 1 5 -3 -15 2 14 22 1 7 11 7 Checks! David Moran === Subject: Re: What is the value of k? d.siemens@myrealbox.com asks for help: >If the remainder is 7 when x^3+kx^2-3x-15 is divided by x-2, then k= ? >Would be gratefull if you drop a few words about the way the solution goes. Perform polynomial division (or synthetic division if you know the process) and find an expression for the remainder: Remainder express = 2(2k+1)-15 Set up the equation for the numeric value of the remainder of 7: 7 = 2(2k+1)-15 k=5 More of this is treated in PreCalculus or College Algebra. G C === Subject: Re: What is the value of k? > If the remainder is 7 when x^3+kx^2-3x-15 is divided by x-2, then k= ? > Would be gratefull if you drop a few words about the way the solution goes. If you work out the division normally, you get an answer in terms of k. I got a remainder of 4k-1 (I did the problem pretty quickly, so you don't want to assume that's correct :-)). If 4k-1=7, then 4k=8 and so k=2. === Subject: what expression represents the nth term of the sequence? If the first term of a geometric sequence is 3/2 and the second and third terms are -3/4 and 3/8 respectively, what is the expression for the nth term of the sequence? Dennis === Subject: Re: what expression represents the nth term of the sequence? > If the first term of a geometric sequence is 3/2 and the second and > third terms are -3/4 and 3/8 respectively, what is the expression for > the nth term of the sequence? > Dennis First, find some ratios Take the first term and divide by the second. This is the same as saying (3/2)(-4/3)=-2 Now do the same for the second and third. (-3/4)(8/3)=-2 Now we have a common ratio so the nth term is an=3/2(1/2)^n The one half is the reciprocal of the ratio because clearly the ratio must be less than 1. David Moran === Subject: Re: what expression represents the nth term of the sequence? >If the first term of a geometric sequence is 3/2 and the second and >third terms are -3/4 and 3/8 respectively, what is the expression for >the nth term of the sequence? In your textbook you will find that the general term is something like ar^n, where a is the first term and r is the ratio between successive terms. You should be able to find a and r quite easily in your example. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com An expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: Re: what expression represents the nth term of the sequence? Of course you are looking for a closed form expression, i.e., an equation, not a number. As a hint for you, the way to make the sign alternate between positive and negative is to have part of it be (-1)^k for example, (-1)^0 = +1 (-1)^1 = -1 (-1)^2 = +1 and so on. Given that the first value (3/2) is positive, you want to pick your k (or n or whatever) appropriately, seemingly nudging it by 1. So you've got (-1)^(n+1) [or n-1] as your first part. And given that the numerator is always some value of three, it's clear that the numerator is (-1)^(n+1) * 3 The denominator just goes 2/4/8 which are powers of two. It's my hope that you can capture that part, and once you put them together, you're sorted out. > If the first term of a geometric sequence is 3/2 and the second and > third terms are -3/4 and 3/8 respectively, what is the expression for > the nth term of the sequence? > Dennis -- Bill Stevenson Editor in Chief ACM Crossroads Magazine http://www.acm.org/crossroads http://www.billstevenson.org Applied Cognitive Science Laboratory Pennsylvania State University http://acs.ist.psu.edu === Subject: Re: what expression represents the nth term of the sequence? > If the first term of a geometric sequence is 3/2 and the second and > third terms are -3/4 and 3/8 respectively, what is the expression for > the nth term of the sequence? > Dennis Well, the sign seems to alternate and the denominator goes 2, 4, 8 while everything else stays the same. So what do you think? Bill === Subject: Re: Pre Calc book > Can anyone recommend a book which deals with precalculus material which > is analogous to Tom Apostol's Calculus? I would think any Pre-Calculus book would prep you for any Calculus book. I have a precalc book in my office by Demana, Waits, Clemens, and Foley that's pretty good. David Moran === Subject: Re: Abstracting the tool > Later, discussion can focus on specific examples, like the core > error that comes from interesting and intriguing properties of > algebraic integers: properties revealed by my tool. > James Harris > possessed of a very tiny tool. and he can't play the basketball either ... what a waste! === Subject: Re: What are all real values of x? If your second posting does not contain any mistakes, there are no real results, because it leads to x^2=-9, so x=+3i or -3i, where i represents the square root of minus one. To check this works, try for example x=+3i 2/(3 -3i) = 1/3 +(1/3)i = 1/3 - 1/(3i) However, if you are not starting on complex numbers at the moment and your question asks for real results, suspect a mistake in copying down or interpretation of your teacher's words. Good luck - Ian Hutcheson === Subject: Re: Integral involving exponential function The rule you need is INTEGRAL f'(x)/f(x) =ln [f(x) ] +c the simplest example being INTEGRAL 1/x =ln [x ] +c Now reinterpret your given function as e^x/(e^x+1) = [d/dx of (e^x+1)] /(e^x+1) and you get ln [e^x+1]+c Hope this makes it easy to see - Ian Hutcheson === Subject: Derivatives Help I am taking a class called Circuit Analysis for Engineering. It requires alot of Calculus concepts that I honestly forgot and the book is no help at all and doesnt even have any examples.. The problem that I am currently having problem with is something like this. i(t) = C * dv(t)/dt The given info is C = 0.2 F (farads) and v(t) = 4(1 - e^-10t) for t >0 They want you to determine the expression for i(t) The answer is 8 x 10^-6 e^-10t Now I have been trying to figure out how to do that and how they got that answer for a long time and I am just going crazy. I was thinking that 10^-6 is a conversion between Farads and MicroFarads.. I atleast want to find out what the derivative of v(t) is. Any help is appreciated -- Steve, I-Net+ Free PC Tech Support - http://www.webzila.com Sports Talk Forum - http://www.houstonsportstalk.com