20170915, 18:58  #23  
Feb 2017
Nowhere
2^{2}·5·257 Posts 
Quote:
To be fair, it was xilman's question first. I have merely been pressing the OP for an answer. Instead, the response has been increasingly smarmy evasions. I therefore conclude this thread is for trolling, and will herewith stop feeding the troll. 

20170915, 21:25  #24  
Aug 2006
3×1,993 Posts 
Quote:


20170916, 01:20  #25  
Apr 2012
Gracie on alert.
3^{4}×5 Posts 
Quote:
I logged in again to delete the above..but no, this fits. Because you need spoon feeding, here is a source that may help you but probably won't: https://www.quora.com/Historicallyh...sforminvented Last fiddled with by ewmayer on 20170916 at 03:17 Reason: Flamage control 

20170916, 03:21  #26 
∂^{2}ω=0
Sep 2002
República de California
2^{2}·3·7·139 Posts 
Please cool it, you two  the S/N ratio of the thread has dropped to near zero, further posts which do not appreciably raise it will be dealt with harshly.

20170916, 04:38  #27 
Apr 2012
Gracie on alert.
625_{8} Posts 
Sorry.

20170918, 22:31  #28 
Apr 2012
Gracie on alert.
3^{4}·5 Posts 
Due to the sour note in some prior posts within this thread I'm including this avant propos.
If anyone has an issue with how or why I ask questions the way I do and has a need to trash talk, please PM me. As always, constructive criticism is welcomed. This is a twopart post. 1. I had hoped that someone would notice that a scalar multiplier and constant simplified the initial polynomial. factor((2*x^3+(19/260*a60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c)*27000+12803); >(30*x+900*b*y+11+900*a)*(6300*a+6073+6300*b*y+9210*x+27000*d*y1800*x^2+27000*c) Second, that treating the cubic as a constant will allow generalization of the polynomial to a general conic equation. Without reservation, this polynomial will resolve any number N mod 27000 = 12803 as a product of sums. Modular sieves and methods to solve Thue equations can be used to solve for the variables but these are not deterministic. Please note that this is a starting point. 2. The following approximation is a curious result that I worked out in the early 90's. I believe the equivalence is dimensionally correct and numerically suggestive. The relation can be expanded according to the Euler/de Moivre formula. If possible, is there a simple physical experiment that can be performed to meassure this? (Maxwell noted that resistance has the same units of velocity, "The Scientific Papers of James Clerk Maxwell, Volume 2, p.126") The following link, http://vixra.org/pdf/1410.0105v1.pdf, "A brief essay on numerology of the mass ratio of proton to electron." is instructive concerning this genre of inquiry. F=96 485.3329 s A / mol h=6.62607004 × 1034 m2 kg / s G=6.67408 × 1011 m3 kg1 s2 c=299,792,458 m/s A=273.15 Celcius (ad hoc) A~=274.212.. Celcius (ad hoc) e=2.71828.. Pi=3.1415.. e^(Pi)=(h*c^2)/(2*G)*F*(A~/273.15) I'm presuming that this is a system at complete rest. Again, please note that this is a starting point. "The Remarkable Sine Functions" by A. I. Markushevich is a reference text that I have used for 1. and 2. "The Beat of a Different Drum: The Life and Science of Richard Feynman" by Jagdish Mehra is a reference for 2. Last fiddled with by jwaltos on 20170918 at 22:33 
20170919, 14:31  #29  
Aug 2006
3×1,993 Posts 
Quote:
Quote:
Quote:
That sort of numerology looks like a job for ries (in addition to Plouffe's inverter, which is already mentioned). CODATA has an updated value for mp/me but nothing is really affected, it's just a tweak to the trailing digits. 

20170919, 15:56  #30  
Apr 2012
Gracie on alert.
625_{8} Posts 
Quote:
Tzanakisde Weger method: I was unaware. Thanks. Regarding the paper, it's vixra  great for nonpeer reviewed works and sometimes helpful. I included it because what I have is pure speculation but it is curious. I'm sending a PM. Last fiddled with by jwaltos on 20170919 at 16:06 

20170920, 03:39  #31  
Apr 2012
Gracie on alert.
3^{4}×5 Posts 
Quote:
The Algorithmic Resolution of Diophantine Equations: A Computational Cookbook By Nigel P. Smart which contained the above method. The above method did not apply or I did not know how to apply it. I found the following paper which helped. I would not have found it had you not made your comment. On elliptic Diophantine equations that defy Thue's method: the case of the Ochoa curve Benjamin M. M. de Weger and Roel J. Stroeker As usual, attempting to clearly explain one's ideas to others usually helps one's own thought process. In this case, in addition to searching for viable algorithmic processes I will also be looking for identities. 

20170920, 14:43  #32 
Feb 2017
Nowhere
5140_{10} Posts 
OK, I'll take your word for it. Your posted apology for your flamage is accepted. I withdraw my accusation, and apologize for it.
Of course, I had noticed that F = 2*x^3+(19/260*a60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c was a quadratic in y, but had broken off pursuit of the matter because its discriminant was not a square, so that F had no polynomial factorization into factors which were linear in y, and because, in answer to my first question, you had indicated you were looking for representations by F of the number to be factored. Be that as it may, the question of finding linear transforms of a quadratic which have linear factors is an intriguing one. And, as it turns out, there is a curious feature of F when viewed as a quadratic in y, which allows one to obtain such a transform. Its discriminant D is a quartic in x, with lead coefficient a perfect square. Thus there is a Laurent series for sqrt(D) beginning with the square root of the lead term. If R is the polynomial truncation of this series ("polynomial square root" of D), D  R^2 is a polynomial in x of degree at most 4/2  1 = 1. But in this case (and this is the curious thing), D  R^2 is a constant, D  R^2 = 89621/225*b^2 + 25606/15*d*b = 12803/225*b*(7*b+30*d). Alas, setting this constant equal to 0 (b = 0 or 7*b + 30*d = 0) makes the coefficient of y^2 equal to 0, so does not lead to an algebraic factorization of F. Trying again with G = p*F + Q, we again find the same curious thing with the discriminant (call it D2) and its "polynomial square root" R2. In this case, D2  R2^2 = (89621/225*p^2  840*q*p)*b^2 + (25606/15*p^2  3600*q*p)*d*b, or D2  R2^2 = b*p*(7*b + 30*d)*(12803*p  27000*q)/225 If we set D2  R2^2 = 0, D2 = R2^2 is a perfect square. As with F, setting p = 0 or 7*p + 30*d = 0 makes the coefficient of y^2 equal to 0, so does not lead to an algebraic factorization. However, setting (*) 12803*p  27000*q = 0 does lead to an algebraic factorization of G = p*F + q. One can take p = 27000, q = 12803. In this case, one could obtain a factorization of N via a representations by F of (N  q)/p (rather than a representation of N by F). So this is an actual derivation of a linear transform p*F + q of F with an algebraic factorization into polynomials that are linear in y. It is unique in the sense that (*) is the only linear relation between p and q which leads to a factorization by making the discriminant of p*F + q (viewed as a quadratic in y) the square of a polynomial, without making the coefficient of y^2 equal to 0. It hinges on an unusual feature of the discriminant D2 of p*F + q, viewed as a function of y: D2 is a quartic in x whose lead term is a perfect square. Thus, there is a "polynomial square root" R2. And D2  R2^2 is a constant, rather than a linear function in x. Setting this constant equal to 0 makes the discriminant equal to the square of its "polynomial square root," which in one case leads to an algebraic factorization. I'm not sure what the fact that D2  R2^2 is a constant (rather than linear in x) means, but I am not smart enough to know that it doesn't mean anything. 
20170920, 22:52  #33 
Apr 2012
Gracie on alert.
3^{4}·5 Posts 
Thanks for attempting to solve this and showing your methods.
One thing which you may not have tried is to find sequential solutions for F, that is, ...3,2,1,0,1,2,3... or ...3i,2i,1i,0,1i,2i,3i...(complex), and others. You will find trivial factorizations of prime numbers which you can attempt to sequence as a solution subset..etc.... I found that by working with a pencil and paper only, I was able to establish patterns, boundary and initial conditions so that I could encapsulate the little mathematical universe I was working within, Mandelbrot sets et al. Once done I could then use things like cellular automata and other computational tools to model the behaviour of what I was looking at. After enough poking around I could then start to formalize things. This would be like looking at a stream then attempting to describe what you're looking at with NavierStokes. It's messy stuff to the untrained eye but chaotic dynamics exists as a science for a reason ;) Just added this as an aside: http://www.ams.org/notices/200902/rtx090200226p.pdf Last fiddled with by jwaltos on 20170920 at 23:07 Reason: Free Willy. 
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