mm-1011 === Subject: Decker example: Another clue It turns out that there's a big, giant clue within the Decker example (source information at bottom) itself, which should spark your curiousity. Remember he gave (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Decker clearly wanted an example where factors in common with 7 of the roots of that quadratic could be seen, as when x=1, *both* have sqrt(7) as a factor within the ring of algebraic integers. But what Decker didn't elaborate on is the fact that whenever x = 1 mod 7, both have sqrt(7) as a factor as well!!! I emphasize that to that degree to impress upon you how much that should jump out at you. After all, there are an *infinity* of integers covered by 1 mod 7, so sqrt(7) as a factor is selected out for an infinite number of values. James Harris Source Information ------------------ === Subject: Re: Mathematical consistency, courage Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). === Subject: Re: Decker example: Another clue > It turns out that there's a big, giant clue within the Decker example > (source information at bottom) itself, which should spark your > curiousity. > Remember he gave > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > Decker clearly wanted an example where factors in common with 7 of the > roots of that quadratic could be seen, as when x=1, *both* have > sqrt(7) as a factor within the ring of algebraic integers. > But what Decker didn't elaborate on is the fact that whenever x = 1 > mod 7, both have sqrt(7) as a factor as well!!! > I emphasize that to that degree to impress upon you how much that > should jump out at you. > After all, there are an *infinity* of integers covered by 1 mod 7, so > sqrt(7) as a factor is selected out for an infinite number of values. Decker chose x = 1 because the computations were easy there and he knew that even you would be able to understand them, and obviously - given your obsessive posting on this topic - you did. And as you correctly note, the quadratic formula implies that the computations are also relatively easy when x = 8, 15, etc.. What this means of course is that instead of one easily verified counterexample to what you claim, there are infinitely many. Is that what you think should jump out at us ? Then there is the fact that there is a counterexample to what you claim for EVERY positive integer x. Here is how it goes for any x such that x^2 + x is relatively prime to 7: Given x > 0, let f_1(x) = GCD(a_1(x), 7) and f_2(x) = GCD(a_2(x), 7), where a_1(x) and a_2(x) are the roots of a^2 - (x - 1)*a + 7*(x^2 + x). Both f_1(x) and f_2(x) are algebraic integers. Note that by definition of a GCD, a_1(x)/f_1(x), a_2(x)/f_2(x), 7/f_1(x), and 7/f_2(x) are all algebraic integers. Also note that since a_1(x)*a_2(x) = 7*(x^2 + x), and I am assuming that 7 and x^2 + x are relatively prime, it must be true that f_1(x)*f_2(x) = 7. Therefore, P(x)/7 = (25*x^2 + 30*x + 2) can be factored as (5 a_1(x)/f_1(x) + 7/f_1(x))*(5 a_2(x)/f_2(x) + 7/f_2(x)). All the terms in sight are algebraic integers as noted above. x = 1 is a special case of this in which f_1(1) = f_2(1) = sqrt(7). Even you can probably see that: since a_1(1) = sqrt(-14), it is pretty clear that GCD(a_1(1), 7) = GCD(sqrt(-14), 7) = sqrt(7) = f_1(1). Yes, f_1(x) and f_2(x) are *dependent on x*. In general the expressions for f_1(x) and f_2(x) are much more complicated than just sqrt(7). When x = 2, Keith Ramsay showed that f_1(x) and f_2(x) are roots of an 11-th degree polynomial! As noted, the argument above works whenever 7 and x^2 + x are coprime. You say it's impossible. Rick Decker blasted away that assertion with his original factorization, P(1)/7 = (5 sqrt(-2) + sqrt(7))*(-5 sqrt(-2) + sqrt(7)), which you have refused to quote in all your many threads on this topic. You have said x = 1 is a special case. Now, even in your present post, you have backed down. You recognize there are INFINITELY MANY special cases: any x = 1 mod 7 will do. Infinitely many exceptions to what you have claimed makes it seem a lot less special, eh? But it's even worse than that. Not just x = 1, 8, 15, ... etc., but also x = 2, 3, 4, 5, 9, ... The argument I just gave shows that MOST integers are special cases. x = 0 and x = -1 are the rare exceptions. THOSE are the special cases where your version of the factorization applies: only two out of infinity. You thought that what happened at x = 0, the singular point, was true in general. That is almost the exact opposite of the truth. You know what's really striking here? The fact that you verified the arithmetic in Decker's example, and even extended it a little! You cannot possibly have done that without understanding what the example is really saying. That is, your smokescreen of refusing to state and denying Decker's result and trying to find a way around it is coming undone. It is clear that you know that your claims are disproved. Yet you continue acting as though you had a leg to stand on. Amazing! Most people's brains would be torn apart trying to continue believing such contradictory facts. You must be very accustomed to lying to yourself on a regular basis. Nora B. > James Harris > Source Information > ------------------ === > Subject: Re: Mathematical consistency, courage > Decker put forward the quadratic > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). === Subject: Re: JSH: Research question answered JSH doesn't mind insults, but he cannot stand a clear and simple proof (i.e. one that he can understand) that he is wrong. That brings out his rage. Gib === 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 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. :-) I think it's more like I want X to be true so badly that I am wearing blinders whenever anybody explains that it's not, so X must be true, and therefore I am a misunderstood genius. Or possibly, X is true. Rick Decker (a professor at Hamilton College) has given a specific numerical counterexample that I refuse to mention, and here is why the example (by Rick Decker of Hamilton College) cannot even exist. Hence X must be true and therefore I am a misunderstood genius. Or possibly, If A and B are algebraic integers such that A + B is divisible by C, then A/C and B/C must also be algebraic integers, since otherwise A/C + B/C would be a sum of two non-algebraic-integers, and therefore I am a misunderstood genius. Or possibly, If f(x) = g(x) + c, where g(0) = 0 and c is coprime to p, then f(x) must be coprime to p for any integer value of x, and therefore I am a misunderstood genius. Or possibly, Knowing who you are is enormously important to evaluating your arguments. If you don't have academic credentials it is safe for me to ignore you, since I don't have any academic credentials either and I know that we should listen only to people with academic credentials, especially if they are at Hamilton College. If you do have academic credentials, I can try to belittle you on the basis that you are a tool of the Establishment with a vested interest in protecting the status quo, as well as being a useless drain on taxpayers like me. If I don't know who you are then I don't know which of these arguments to use, so I can ignore anything you say no matter how mathematically correct it may be. Therefore, etc. Nora B. === 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. > 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. > 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. > elucidate the Factorization Tool yesterday. > It's neat. I'll start with the Decker example in explaining my > results. > 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: === > Subject: Re: Mathematical consistency, courage > Decker put forward the quadratic > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > Usually I focused on the fact that the middle coefficient goes to 0 > with an integer x, which is similar to my focus with the previous > examples from the others. > However, I now realize that the issue is the change in modular residue > with respect to 7 of *any* coefficient that matters. As a note to myself, and the future, it's clear now that the underlying functions are NOT algebraic integers for another reason, as at the base you have something like (5f_1(x) + 1)(5f_2(x) + 2) which means that sqrt(7) does have to be a unit in a ring where f_1(x) is always valid as at x=1, f_1(1) = sqrt(2)/sqrt(7). > So the issue here is that -(x-1) mod 7 is not constant. That seems to be a characteristic of cases where such functions are used, the residue of the coefficients of their defining polynomials vary with respect to the constant factor being considered. That's interesting. > Further research which I won't go into in detail indicates that in > using the Factorization Tool for such an example you have to calculate > constant terms for *each* of the coefficient's possible residues, so > you need 7. I think it interesting as an advanced question to figure out how to construct one of the imperfect factorizations from a given tautological space, though it seems that would require a tautological space only valid in the complex plane. Hmmm...therefore, it seems logical that an imperfect factorization cannot be constructed from a tautological space. The formalism necessary to prove that quick deduction must be rather impressive. James Harris === Subject: Another Question - Integrals I have another question, this time regarding integrals. I dont have a Calculus book and the book that I am using now assumes that you know calc and doesnt explain anything.. So I am trying to search the web and look up the different integral tables and examples. I am trying to obtain an integral of: 0.06cos(10^4)t I know that the integral of cosx is sinx I couldnt find any example for this form - Can anyone provide me with the formula/method to obtain the integral of this function === Subject: Re: Another Question - Integrals > I have another question, this time regarding integrals. > I dont have a Calculus book and the book that I am using now assumes that > you know calc and doesnt explain anything.. > So I am trying to search the web and look up the different integral tables > and examples. > I am trying to obtain an integral of: 0.06cos(10^4)t > I know that the integral of cosx is sinx > I couldnt find any example for this form - Can anyone provide me with the > formula/method to obtain the integral of this function I don't think that the integral of cosine is relevant because the cosine of 10^4 is a constant. Looks to me what you're really asking for is the integral of (0.06cos*(10^4)t). In that case, the integral is .03cos(10^4)t^2 by the Power Rule. David Moran === Subject: Re: Another Question - Integrals > I have another question, this time regarding integrals. > I dont have a Calculus book and the book that I am using now assumes that > you know calc and doesnt explain anything.. > So I am trying to search the web and look up the different integral tables > and examples. > I am trying to obtain an integral of: 0.06cos(10^4)t > I know that the integral of cosx is sinx > I couldnt find any example for this form - Can anyone provide me with the > formula/method to obtain the integral of this function > I don't think that the integral of cosine is relevant because the cosine of > 10^4 is a constant. Looks to me what you're really asking for is the > integral of (0.06cos*(10^4)t). In that case, the integral is .03cos(10^4)t^2 > by the Power Rule. David, I think you misplaced a *. I suspect Steve meant 0.06*cos(10^4*t) whose integral is 6*10^(-6)*sin(10^4*t). To Steve: for goodness sake get a Calculus book (any old one will do) and read up on integration by substitution. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Another Question - Integrals I have another question, this time regarding integrals. > I dont have a Calculus book and the book that I am using now assumes that > you know calc and doesnt explain anything.. > So I am trying to search the web and look up the different integral tables > and examples. I am trying to obtain an integral of: 0.06cos(10^4)t I know that the integral of cosx is sinx > I couldnt find any example for this form - Can anyone provide me with the > formula/method to obtain the integral of this function I don't think that the integral of cosine is relevant because the cosine of > 10^4 is a constant. Looks to me what you're really asking for is the > integral of (0.06cos*(10^4)t). In that case, the integral is .03cos(10^4)t^2 > by the Power Rule. > David, I think you misplaced a *. > I suspect Steve meant 0.06*cos(10^4*t) whose integral is > 6*10^(-6)*sin(10^4*t). > To Steve: for goodness sake get a Calculus book (any old one will do) > and read up on integration by substitution. > -- > Paul Sperry > Columbia, SC (USA) David Moran === Subject: Re: 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.. Get on to your favorite used-books site, like http://www.abebooks.com, and get a copy of /Calculus for the Practical Man/. Sooner or later you're going to have to (re)learn calculus if your courses assume you know it. -- 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: help with one-dimensional dynamic iteration I'm having trouble proving something that should be rather simple, and would appreciate help (feel free to reply or email me at nageeb@stanford.edu). Let g(x)=d*x if x>1/d or g(x)=1+y-d*y*x where y >=0. Now, g has a single fixed point, which is a sink (attracting) if d*y<1 or a source (repelling) if d*y>1. Now we can consider iterating the function, (g^n)(x) = d*(g^(n-1))(x) if (g^(n-1))(x)>1/d or (g^n)(x) = 1 + y - d*y* (g^(n-1))(x). If d*y > 1 but (d^2) * y <1, then it is straightforward to show that (g^2) has one repelling fixed point and two attracting fixed points. I'd like to generalize this to n. i.e., find how many attracting and repelling fixed points there are for arbitrary n. Any suggestions? -- Nageeb === Subject: Re: JSH: Abstracting the tool jstevh@msn.com (James Harris) sobered up enough to type > snip JSH garbage Nice subject line. Since you are a tool, I thought you were abstracting yourself. Be careful, you could go blind if you do. Have a nice day, Jay === Subject: Re: JSH: Abstracting the tool In sci.math.num-analysis, James Harris You might have noticed that I've abstracted out the mathematical tool >> that I've been using, as well, it's fun to do so, and probably very >> useful. >> The full abstraction now is at my blog: >> http://mathforprofit.blogspot.com/ >> It's a lot easier to correct there, versus making posts and having to >> go back when I make mistakes as I just did, with follow-ups. >> Now discussion can be over the *tool* versus over where it's used. > And using it is straightforward. I'll use the Decker example to > explain. > At the front end you have > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > The Factorization Tool will tell you that factors of 7 cannot in > general be separated from its roots--a_1(x) and a_2(x)--within the > ring of algebraic integers. > Because the Tool requires functions that equal 0, when x=0, you need > that linear substitution b_2(x) = a_2(x) - 1, where a_2(0) = 1. > The indices are arbitrary, but you know that *one* of the a's must > equal 1 when x=0 because then you have > a^2 + a = 0. Try a different x. For x = 3, for example, one gets: a^2 - 2*a + 35 = 0. with a = 1 +/- sqrt(-34) for the roots. Which one of these roots is divisible by 7? Or, if you prefer, you can compute any of the following: 5 * a_1(3) + 7= 5 * b_1(3) + 2 = 12 + 5 * sqrt(-34) 5 * a_2(3) + 7= 5 * b_2(3) + 2 = 12 - 5 * sqrt(-34) Which one of those is divisible by 7? Here's a hint. If we take c = (12 + 5*sqrt(-34)) / 7, we can deduce what equation it solves by simply completing the square in the obvious fashion: c - 12/7 = 5/7 * sqrt(-34) (c - 12/7)^2 = 25/49 * -34 = -850/49 c^2 - 24/7*c + 144/49 = -850/49 c^2 - 24/7*c + 142/7 = 0. or 7*c^2 - 24*c + 142 = 0. This polynomial is not reducible. Therefore, c is not an algebraic integer; therefore *neither* of the two quantities is divisible by 7. In short, James, you got lucky in picking x = 0 (or x = -1, which also works). That's all. > So now you have > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > which is the Tool with n=2, and the other function and variable values > are obvious enough. > You can actually prove that the factorization is pushed outside of the > ring of algebraic integers with x=1, which is an easy enough value to > check. > Notice that the factorization a_1(x) a_2(x) = 7(x^2 + x), is not a > perfect factorization in the ring of algebraic integers, as at x=1, > a_1(x) has a_2(x) as a factor, and a_2(x) has a_1(x) as a factor, but > they aren't factors of each other for all algebraic integer x. > Not surprisingly, in the field of algebraic numbers, as the Tool still > has validity as it doesn't care, you *can* divide 7 off, and in that > ring, the factorization is perfect, and the field is a complete ring. > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Factorial question I am working on proving that 1*1! + 2*2! + ... n*n! = (n + 1)! - 1 by induction. I started with my base step as being true, since P(1) =1 which is true. For my inductive steps, I am trying to prove that P(k + 1) would equal (n + 2)! -1. So: 1*1! + 2*2! +...n*n! + (n + 1)*(n + 1)!=(n + 1)! - 1 + (n + 1)*(n + 1)! Getting past this point is what is getting me messed up. I know that I need to arrive at (n+2)! - 1. However, in trying to work out the algebra here, I get (n + 1)*(n + 2)! - 1. I did find an answer to this problem online, which went like this: =(n + 1)! - 1 + (n + 1)*(n + 1)! =(n + 1)*(n + 2)! - 1 =(n + 2)! - 1 My question is, how do you get from =(n + 1)*(n + 2)! - 1 to =(n + 2)! - 1? === Subject: Re: Factorial question >I am working on proving that 1*1! + 2*2! + ... n*n! = (n + 1)! - 1 by >induction. [For the induction step, need to prove: ] >1*1! + 2*2! +...n*n! + (n + 1)*(n + 1)!=(n + 1)! - 1 + (n + 1)*(n + 1)! >I know that I need to arrive at (n+2)! - 1 Try rearranging the RHS and factoring: = (n+1)! - 1 + (n+1)*(n+1)! = (n+1)! + (n+1)*(n+1)! - 1 = (n+1)! * [1 + (n+1)] - 1 = (n+1)! * (n+2) - 1 (See note 1.) = (n+1)! - 1 >However, in trying to work out the algebra here, I >get (n + 1)*(n + 2)! - 1. I don't see how you got that. Are you sure you don't mean (n+1)! * (n+2) - 1? That would be the next-to-last step in my derivation above. Note 1: (n+1)! * (n+2) = (n+2)! because: (n+1)! * (n+2) [1 * 2 * 3 * ... * n * (n+1)] * (n+2) (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: Factorial question wrong. One question though. Looking at step 3 of your derivation, how did you come up with [1 + (n + 1)] - 1? I'm specifically curious about the 1+ part of the derivation. I guess I don't see that. > Try rearranging the RHS and factoring: > = (n+1)! - 1 + (n+1)*(n+1)! > = (n+1)! + (n+1)*(n+1)! - 1 > = (n+1)! * [1 + (n+1)] - 1 > = (n+1)! * (n+2) - 1 (See note 1.) > = (n+1)! - 1 Also, would you or anyone else know of a good web site that has a tutorial on factorials and how to derive them? === Subject: Re: Factorial question >> Try rearranging the RHS and factoring: >> = (n+1)! - 1 + (n+1)*(n+1)! >> = (n+1)! + (n+1)*(n+1)! - 1 >> = (n+1)! * [1 + (n+1)] - 1 >> = (n+1)! * (n+2) - 1 (See note 1.) >> = (n+1)! - 1 >Looking at step 3 of your derivation, how did >you come up with [1 + (n + 1)] - 1? I'm specifically curious about the 1+ >part of the derivation. Okay, let's expand it a bit (n+1)! + (n+1)*(n+1)! - 1 Now write A for (n+1)! and you have A + (n+1)*A - 1 Insert a redundant 1 * for clarity 1*A + (n+1)*A - 1 Factor A from first two terms A*[1+(n+1)] - 1 Undo the substitution (n+1)! * [1 + (n+1)] - 1 When you quote someone, it is good manners -- and helps everyone follow discussions better -- if you (1) attribute the quote properly (2) trim out as much irrelevant stuff as possible and (3) put your comments after the quoted material you are commenting on. See for details. See how much more sense your comment makes when _not_ upside down? -- 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: Factorial question >> Try rearranging the RHS and factoring: >> = (n+1)! - 1 + (n+1)*(n+1)! >> = (n+1)! + (n+1)*(n+1)! - 1 >> = (n+1)! * [1 + (n+1)] - 1 >> = (n+1)! * (n+2) - 1 (See note 1.) >> = (n+1)! - 1 >Looking at step 3 of your derivation, how did >you come up with [1 + (n + 1)] - 1? I'm specifically curious about the 1+ >part of the derivation. > Okay, let's expand it a bit > (n+1)! + (n+1)*(n+1)! - 1 Now write A for (n+1)! and you have > A + (n+1)*A - 1 Insert a redundant 1 * for clarity > 1*A + (n+1)*A - 1 Factor A from first two terms > A*[1+(n+1)] - 1 Undo the substitution > (n+1)! * [1 + (n+1)] - 1 > When you quote someone, it is good manners -- and helps everyone > follow discussions better -- if you (1) attribute the quote properly > (2) trim out as much irrelevant stuff as possible and (3) put your > comments after the quoted material you are commenting on. See > for details. See how much > more sense your comment makes when _not_ upside down? > -- > 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 Ok, I understand. The answer is just plain obvious, and I was too dense to Also, thank you for the web link about proper quoting. I will read it and try to do better next time. === Subject: Re: Factorial question > I am working on proving that 1*1! + 2*2! + ... n*n! = (n + 1)! - 1 by > induction. > I started with my base step as being true, since P(1) =1 which is true. For > my inductive steps, I am trying to prove that P(k + 1) would equal (n + > 2)! -1. > So: > 1*1! + 2*2! +...n*n! + (n + 1)*(n + 1)!=(n + 1)! - 1 + (n + 1)*(n + 1)! > Getting past this point is what is getting me messed up. I know that I need > to arrive at (n+2)! - 1. However, in trying to work out the algebra here, I > get (n + 1)*(n + 2)! - 1. > I did find an answer to this problem online, which went like this: > =(n + 1)! - 1 + (n + 1)*(n + 1)! = (n + 1)!*(1 + n + 1) - 1 = (n + 1)!*(n + 2) - 1 > =(n + 1)*(n + 2)! - 1 There's either a typo or you miscopied. > =(n + 2)! - 1 > My question is, how do you get from > =(n + 1)*(n + 2)! - 1 to =(n + 2)! - 1? -- Paul Sperry Columbia, SC (USA) === Subject: Found the lost formula of doom I canĒt post it here, because it would cause your death. The one who can solve this formula will rule the world. === Subject: Re: JSH: Pattern argument James, Is there any reason why you keep including sci.cognitive in your threads? -- mail1dotstofanetdotdk === Subject: Re: JSH: Pattern argument >[...] >The more astute of you should see a pattern. Uh, we can see a pattern doesn't count as a proof of anything unless you can _prove_ the pattern exists. Phrasing that as The more astute of you should see a pattern in the hope that nobody will be willing to admit that he's so stupid he can't see the pattern is just pathetic. ************************ David C. Ullrich === Subject: Re: JSH: Pattern argument >[...] >The more astute of you should see a pattern. > Uh, we can see a pattern doesn't count as a proof of > anything unless you can _prove_ the pattern exists. > Phrasing that as The more astute of you should see a pattern > in the hope that nobody will be willing to admit that he's so > stupid he can't see the pattern is just pathetic. Oh, there is a pattern! The same pattern he established about 8 years ago of displaying his ignorance and arrogance to the world. === Subject: Re: JSH: Pattern argument > So I've put out the Decker quadratic (source information at bottom), > but replies from various people indicate there's still room for > confusion, so I'll give a pattern argument. > The original quadratic is > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > You can modify it easily enough to get > b^2 - (x - 1)b + 17(x^2 + x), where you then have > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 25x) + 17(5)(x-1) + 17^2 Just a suggestion, to reduce confusion... before you post all this again, why not change the things that have a, b or c as roots from the expressions that they are at present (which do not have roots) into equations (that DO have roots) it would be a start, anyway... === Subject: Re: JSH: Pattern argument In sci.math, James Harris but replies from various people indicate there's still room for > confusion, so I'll give a pattern argument. > The original quadratic is > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > You can modify it easily enough to get > b^2 - (x - 1)b + 17(x^2 + x), where you then have Here's a thought for you James. Let's generalize your problem a bit. Pick an integer N such that gcd(N,5) = 1. This integer can be 7, 17, or 177, or anything you like. Then take (5 * a_1(x) + N) * (5 * a_2(x) + N) = N * (25 * x^2 + 30*x + y) where the a's are roots of a^2 - (x - 1) * a + N * (x^2 + x). and y has yet to be determined. Obviously I'll need to find y. Time for a side jaunt. We know the a_i() are roots, so we compute and equate: (a - a_1(x)) * (a - a_2(x)) = a^2 - (x - 1) * a + N * (x^2 + x) = a^2 - (a_1(x) + a_2(x)) * a + a_1(x) * a_2(x). Therefore, Therefore, a_1(x) + a_2(x) = x - 1 a_1(x) * a_2(x) = N * (x^2 + x) and (5 * a_1(x) + N) * (5 * a_2(x) + N) = 25 * a_1(x) * a_2(x) + N * 5 * (a_1(x) + a_2(x)) + N^2 = N * 25 * (x^2 + x) + N * 5 * (x - 1) + N^2 = N * (25 * x^2 + 25 * x + 5 * x - 5 + N) so y = N - 5. I can generalize *again*. What's so special about 5? Let's replace 5 by M, and see what we get. (M * a_1(x) + N) * (M * a_2(x) + N) = M^2 * a_1(x) * a_2(x) + N * M * (a_1(x) + a_2(x)) + N^2 = N * M^2 * (x^2 + x) + N * M * (x - 1) + N^2 = N * (M^2 * x^2 + (M^2 + M) * x + N - M) So we have, for arbitrary M and N, (M * a_1(x) + N) * (M * a_2(x) + N) = N * (M^2 * x^2 + (M^2 + M) * x + N - M) where the a's are roots of a^2 - (x - 1) * a + N * (x^2 + x). To check this problem we pick M = 5, N = 7, and get 7 * (25 * x^2 + 30 * x + 2) for the right side, as we should. And now, of course we divide by N. The right side is easy, of course. The left side is slightly arbitrary. Of course we can set b_2(x) = a_2(x) + 1 and then get (M * a_1(x) + N) * (M * b_2(x) + N-M) / N = (M^2 * x^2 + (M^2 + M) * x + N - M) Where does the '/N' go? Good question. The traditional place is of course the first factor, yielding (M/N * a_1(x) + 1) * (M * b_2(x) + N-M) = (M^2 * x^2 + (M^2 + M) * x + N - M) Now let x = 0, and then a^2 - (x - 1) * a + N * (x^2 + x) = a^2 + a with roots 0, -1. a_1(0) = 0. a_2(0) = -1; b_2(0) = 0. We substitute, and get for the grand finale the identity (M/N * 0 + 1) * (M * 0 + N-M) = (M^2 * 0^2 + (M^2 + M) * 0 + N - M) or N-M = N-M. Useful? Good question. Beware of drawing any conclusions because of that special case 0. > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 25x) + 17(5)(x-1) + 17^2 > which is > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12) > or you can go down to > c^2 - (x - 1)c + 2(x^2 + x), which gives you > (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 25x) - 2(5)(x-1) + 2^2 > which is > (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x - 3). > So I have > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12) and > (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x - 3). > The more astute of you should see a pattern. Clearly the constant > factor on the right, which shows up as a coefficient on the left must > equal 2 mod 5 for this particular setup. > But why? > It turns out that I'm exploiting a symmetry with *integer* > coefficients in the factorization. > Imagine that in each case there's a base equation > (5y_1(x) + 1)(5y_2(x) + 2) > where you can linearly tranform the second factor (5y_2(x) + 2) rather > easily leaving you free to multiply the first factor by various > integers as long as they equal 2 mod 5. > Now if you beliee that *any* of the explanations given by various > posters out there can relate that to reducibility over Q or Galois > Theory, then you're not good with a basic pattern. > The problem is that y_1(x) and y_2(x) can't both be algebraic integer > functions. > Now if you *still* are having a problem understanding, ask yourself, > how can anyone force a constant factor like 7, or 2, or 17 to only > work with *one* factorization? > Like consider again > b^2 - (x - 1)b + 17(x^2 + x), where you have > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12). > Why does that 17 lead to *one* factorization on the left? How is it > possible for it to be limited? > The answer is that there's only one way to distribute it and have > *integer* coefficients in the factorization on the left. > Mathematicians here need to be at least a little curious, or you'll > never figure it out!!! > Think logically, and force people disagreeing with me to answer basic > questions, like how do they explain the pattern: > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12) and > (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x - 3)? > James Harris > Decker Quadratic Source Information > --------------------- > 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: === > Subject: Re: Mathematical consistency, courage > Decker put forward the quadratic > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: JSH: Pattern argument [snip handwaving pseudo-math/crap attempting to establish a proof by observation of some alleged pattern] Hey, James, I'm *great* with patterns, too. Here's one: 3 is an odd number, 3 is a prime 5 is an odd number, 5 is a prime (see the pattern?...) 7 is an odd number, 7 is a prime... Therefore all odd numbers are primes! By the way, don't bother posting any 'special cases' as counter-examples -- they don't matter because it's the generalization that counts! Harris' Proof Method: 1) Ready! 2) Place cross-hairs squarely over foot, 3) Fire! -- 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: Pattern argument > So I've put out the Decker quadratic (source information at bottom), > but replies from various people indicate there's still room for > confusion, so I'll give a pattern argument. In JSH's mind, there has always been room for confusion. It seems as if his mind is capable of containing infinite amounts of confusion. James S. Harris, having again failed to refute the proofs of his errors posted in other threads, does his usual thing of ignoring those disproofs and starting another thread with the same old tired and disproved crap restated as if new and true. The pattern is clear even before reading, that JSH relies again on the same false assumptions about algebra that any reasonably bright highschooler would by now have given up. Obsesive-compulsive masterbation of his ego! === Subject: Re: JSH: Pattern argument >>So I've put out the Decker quadratic (source information at bottom), >>but replies from various people indicate there's still room for >>confusion, so I'll give a pattern argument. > In JSH's mind, there has always been room for confusion. It seems as if > his mind is capable of containing infinite amounts of confusion. > James S. Harris, having again failed to refute the proofs of his errors > posted in other threads, does his usual thing of ignoring those > disproofs and starting another thread with the same old tired and > disproved crap restated as if new and true. I think his strategy is this: repost false argument until everyone gets bored with responding in detail. At that point his argument goes unrefuted, and he can then tell himself that it has been accepted. It isn't the way mathematicians work, but as JSH has frequently stated, he is not a mathematician. Gib === Subject: JSH: Howard Aiken quote, my situation Oh hey, I came across this great quote by a guy called Howard Aiken: http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Aiken.html Don't worry about people stealing your ideas. If your ideas are any good, you'll have to ram them down people's throats. Howard Aiken Source: http://www-gap.dcs.st-and.ac.uk/~history/Quotations/Aiken.html Yup, it's quite true. Like I found a way to count prime numbers where a partial difference equation is key. No one else in recorded human history managed such a feat, but mathematicians refuse to give me the time of day. The bastards!!! Because it's a partial difference equation, it leads to a partial difference equation, which could be key in research to prove or disprove the Riemann Hypothesis. But mathematicians today belong to a democracy, and they don't *like* me, so they don't like my research. What boring people. James Harris === Subject: Re: JSH: Howard Aiken quote, my situation >Oh hey, I came across this great quote by a guy called Howard Aiken: >http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Aiken.html >Don't worry about people stealing your ideas. If your ideas are any >good, you'll have to ram them down people's throats. >Howard Aiken >Source: http://www-gap.dcs.st-and.ac.uk/~history/Quotations/Aiken.html >Yup, it's quite true. Possibly, although I don't recall any mathematician needing to shove his ideas down people's throats. Regardless, even if it is true, it doesn't follow that if you have to shove your ideas down people's throats then they're any good. Remember, they also laughed at Bozo the clown. >Like I found a way to count prime numbers where >a partial difference equation is key. No one else in recorded human >history managed such a feat, Except for Legendre, and that famous 20% of the grad students working on such things you mean. > but mathematicians refuse to give me the >time of day. The bastards!!! Here in OK it's 1:06 PM right now. Happy? >Because it's a partial difference equation, it leads to a partial >difference equation, Presumably that last is a typo for partial differential equation? You've never shown that the solution to that pde has anything whatever to do with counting primes. >which could be key in research to prove or >disprove the Riemann Hypothesis. And you've never given any evidence that you even know what RH _states_. >But mathematicians today belong to a democracy, and they don't *like* >me, Does _anyone_ like you, other than I suppose your mother? Just curious. >so they don't like my research. No, that's not why they don't like your research. People are just wild about the research of some real slime balls. >What boring people. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Howard Aiken quote, my situation > Like I found a way to count prime numbers where > a partial difference equation is key. No one else in recorded human > history managed such a feat, but mathematicians refuse to give me the > time of day. The bastards!!! JSH is still suffering from his feat in mouth disease. Worse is that he makes us suffer from it, too. === Subject: Re: Howard Aiken quote, my situation > Oh hey, I came across this great quote by a guy called Howard Aiken: > http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Aiken.html > Don't worry about people stealing your ideas. If your ideas are any > good, you'll have to ram them down people's throats. > Howard Aiken > Source: http://www-gap.dcs.st-and.ac.uk/~history/Quotations/Aiken.html > Yup, it's quite true. Like I found a way to count prime numbers where > a partial difference equation is key. No one else in recorded human > history managed such a feat, but mathematicians refuse to give me the > time of day. The bastards!!! > Because it's a partial difference equation, it leads to a partial > difference equation, which could be key in research to prove or > disprove the Riemann Hypothesis. > But mathematicians today belong to a democracy, and they don't *like* > me, so they don't like my research. > What boring people. > James Harris Are you sure that a partial difference equation leads to a partial difference equation? What seminal work you are doing! Lurch === Subject: Re: Howard Aiken quote, my situation > Are you sure that a partial difference equation leads to a partial > difference equation? What seminal work you are doing! It's a typo. He thinks his partial difference equation leads to a partiel differential equation. Without Googling back to check for sure, I seem to recall that this was the step of his prime counter work that no one else could follow. -- --Tim Smith