mm-1789 === Subject: invariant differential forms in a lie group I'm looking for proofs (of references) using the least possible lie theory of the following fact: if w is a left and right invariant differential form in a lie group G then w is closed. nojb. PS: is the following correct? Let G be a connected lie group, g in G and let L_g be the G -> G diffeomorphism given by left translation by g, then L_g s = s in homology for every singular chain. Proof: H(g,t) = f(t).g is a homotopy between id : G -> G and L_g where f is a path between e and g in G. === Subject: cone cone intersection hi, how can I compute the two intersection lines, when two cones intersect each other? Erich === Subject: Re: cone cone intersection >how can I compute the two intersection lines, when two cones intersect >each other? Write the equations for the two cones, and solve for two components in terms of the other. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Hilbert 16th Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3IkHx27203; Hello I Search new Interpretations For the Limit Cycle's Phenomena and unusuall formulation for the hilbert 16th problem: (any comments and new suggestions is appreciated), I begin with Following interpretations 1)The Number of attractores of a vector field X is equall to fredholm index of functional operator X(g)=X.g please review : http://www.arxiv.org/abs/math.DS/0408037 2)let g be a riamanian metrics compatible to a vecter field X:that is the trajectories of X are geodesics,so the limit cycles of X are closed geodesics,we can try to find various metrics g which curvature-function has an appropriate behaviour (please See Limit Cycles And complex Geometry,Romanovski St Ptrs Jour Of Math 1997) 3)Quantum Mechanics, however I have no background In Math. Aspects Of Quantum Mechanics,But I Have found The Following two Papers(with Intersting titles) in The Web,last year: i)IS QUANTUM MECHANICS A LIMIT CYCLE THEORY,By A.M.Cetto and de la Penna ii)Limit Cycles In Quantum Theories,Glazek Wilson Physical review letter 2002 Question :What is a Quantization Of Hilbert 16th Problem: Let we have A Vec. Field X on A surface M,it generate a flow and this flow naturally defines a flowb on C^inf(M),now let's Quantize M,that is to embedd C^inf(M) in Some B(H),now the flow in C^inf(M) would be converted to a flow in a subspace Of B(H)....(what process you suggest for continuation....) === Subject: Re: Compute plane with angles and vectors >I have two known vectors with the same origin and a unknown plain >which goes also through this origin. The two angles between the two >vectors and the normal of the plane are known, but not the normal >itself. How can I compute the normal? >Is there another way than this: >a1*n1 + a2*n2 + a3*n3 = cos(alpha1) >b1*n1 + b2*n2 + b3*n3 = cos(alpha2) >n1^2 + n2^2 + n3^2 = 1 >Because the result is really huge (over pages)... >There can be two results, right? >Can I fix this with a third vector (and a third angle), such that I >have only one result? >How would I do this? Yes, with a third vector C it would be easy. You could compute both N cross (A cross B) and N cross (A cross C) from your information, each of which would be perpendicular to N, giving two vectors parallel to the plane. Cross them again to get a normal. --Lynn === Subject: Re: Compute plane with angles and vectors >I have two known vectors with the same origin and a unknown plain >which goes also through this origin. The two angles between the two >vectors and the normal of the plane are known, but not the normal >itself. How can I compute the normal? >Is there another way than this: >a1*n1 + a2*n2 + a3*n3 = cos(alpha1) >b1*n1 + b2*n2 + b3*n3 = cos(alpha2) >n1^2 + n2^2 + n3^2 = 1 >Because the result is really huge (over pages)... >There can be two results, right? >Can I fix this with a third vector (and a third angle), such that I >have only one result? >How would I do this? > Yes, with a third vector C it would be easy. You could compute both > N cross (A cross B) and N cross (A cross C) from your information, > each of which would be perpendicular to N, giving two vectors parallel > to the plane. Cross them again to get a normal. > --Lynn I'm sorry, but how would it look like? equationsystem? Because N is unknown. === Subject: Re: need proof!!! >i resently asked the following question >I'm looking for a continuous function f:R->R with discontinuity on >irrational domain and continuous on Q. >the good people who responded, told me that the answer is NO. >now, does anyone knows a formal proof for this... >but if you dont know just tell me why >Q is a subset of R, and so is the irrational numbers R - Q. If you >>have discontinuity on R - Q, then you have discontinuity on a >>subset of R, and hence on R as a whole. Therefore, no such >>function exists. > > ok, so why is the oppside function exist??? > if the discontinuity is on Q which is also a subset of R??? > instead of posting an answer to what you meant to write. Your problem is > function which was *discountinuous* at some points (namely, at the > irrational numbers). Obviously, no such function exists, by the reasons > stated by mike3. What you meant to write was: > I'm looking for a function f:R->R which is discontinuous on every > irrational point and continuous on every point of Q. > Jose Carlos Santos Though the reverse is quite possible, continuous at irrational points and discontinuous at rational ones. === Subject: bootstrap and Fisher's fiducial analysis Trying to understand the logic behind the bootstrap, it seems to me more and more that this is similar to Fisher's fiducial inference. Any opinion? Kjetil Halvorsen === Subject: Re: Takers and leavers: Synthesis > The way I see this, the Western society is torn between two lines of > thought. One is the unconditional growth of the civilization, whatever > the consequences for the planet and for the future generations. The > other is a reaction against it: a return-to-the-roots hippie > environmentalism that seeks to What old-fashioned hippie types (and their neoLuddite descendants) fail to take into account is the sheer self-serving strategic profligacy of Nature based on the policy of unconditional growth. Plants do not produce just enough seeds to ensure zero population growth for themselves, they produce enough for any given species to fill their local biome (in some cases, blanket the planet) in one generation. Similarly, animals do not produce just enough progeny to ensure ZPG for themselves. Fortunately, the Law Of Fang And Claw (for plants, chemical warfare and rapacious competition for water and sunlight) keeps any given species from taking over completely. So, who's actually acting in accordance with Nature, hippies or Capitalists? And before you bewail the sheer mass of human flesh that will occupy the planet after we've killed everything else off and have to resort to cannibalism because there's nothing else to eat, consider what happened to the poor, unsung anaerobes that formerly exclusively ruled the Earth. Now, us aerobes have to eat each other while trying to finish off the anaerobes. Nature is change in strobing neon caps, not any kind of imaginary idyllic steady state. Environmentalism, at its root, is narrow-minded arrogance. Mark L. Fergerson === Subject: Re: How to measure randomness of a deck of cards? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3KH3I03194; >I read several papers, there are few approaches such as variation >distances, birthday bound and markov chain. I need a >method/algorithm to determine the randomness of my deck of cards >(permutation) >I would be appreciated if anybody can give me some ideas or good solutions. Estimate how much information is lost due to each shuffle. Ideally, it is as much as 1 bit. This would occur if you couldn't tell which half of the deck contains a given card (assuming you knew which half before the shuffle). Assuming each shuffle is independent, each destroys one bit of information. In practice about 8 or 9 shuffles provides pretty good randomness. Google for more including experiments phil === Subject: nearest common ancestor on Stern-Brocot tree? For example, the nearest common ancestor for 5/9 and 3/4 is 2/3. Indeed 5/9 = LRLLL 3/4 = LRR 2/3 = LR The straightforward way to calculate NCA is by converting the rational number into LR representation applying (extended) Euclidian algirithm. Is there a formula for NCA? === Subject: Re: i alone > i fight the evil of psychiatry, psychology. surrender your vermin. What, so you can feed them your medication? === Subject: Re: JSH: Look at it backwards > I usually start with one polynomial and then talk about dividing 49 > off from it, but here I'll start *after* 49 has been divided off: > S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 > and consider the factorization > S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) > and the assertion that the d's must have factors > w_1(x), w_2(x), and w_3(x) > that multiply to give w_1(x) w_2(x) w_3(x) = 7. Let w_1(x)=w_2(x)=w_3(x) = 7^(1/3). Then the w_i are factors of the d_i in the ring of algebraic numbers. (Are they factors in any other ring? Who knows? Who cares? No one except you has made such a claim.) > But there's no factors of 7 anywhere. So why should the d's have > functions that are factors of 7? the d's have functions that are factors of 7 is gibberish. -William Hughes === Subject: Re: JSH: Look at it backwards >I usually start with one polynomial and then talk about dividing 49 >off from it, but here I'll start *after* 49 has been divided off: >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >and consider the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >and the assertion that the d's must have factors >w_1(x), w_2(x), and w_3(x) >that multiply to give w_1(x) w_2(x) w_3(x) = 7. >But there's no factors of 7 anywhere. So why should the d's have >functions that are factors of 7? You are getting maximally confused. In your original approach, you are factoring P(x) as a polynomial in 5, not as a polynomial in x. The constant term with respect to 5 and the constant term with respect to x are different. It was a stupid mistake to think you were simplifying by treating 5 as if it were the polynomial variable in the first place. And that is what is causing your confusion here. >Ok, maybe that seems unfair, as there could be LOTS of different >functions you can plug in for the c's and d's, but why should ANY of >them have functions of x that are factors of 7? >The equation has no memory. The memory problem here is, you have forgotten what you started with. To see this, you need to go back to the way you were doing things originally: for example, P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) To put this in terms of what you are doing now: replace x in this expression by 5, and replace m by x, and replace f by 7. Also replace u by 1. This gives P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), and you can clearly see that, as a polynomial in 5, the constant term is 49 * 7 = 7^3, not 49 * 22. Go back to the expression above for P(m). The constant term with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to 49*22 when u = 1, f = 7, and x = 5. But the constant term with respect to x is u^3 f^3. You have forgotten this and gotten mixed up, that's all. >You have a memory, so if I start with the polynomial multiplied by 49, >then you can say to yourself that there's some dependency on 7. But >it's a mirage. >Mathematically a constant multiple is not a big deal. It's just a >constant multiple that you can divide off, leaving a result that--you >guessed it--has no memory of the multiple! >One of the weirder things about the discussions here, which I assume >escapes most of you is some fascinating belief that the equation has a >memory. >You see something like >P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 >and your brain apparently HOLLERS at you that 49 is still there, but >no, it's gone. No - my brain does no such hollering. >When I show something like >P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7) Key thing to note here: as I said, the polynomial variable with respect to which you are factoring is 5, not x. See? >you brain insists that 7 is still there. >I mean just LOOK! You can see 7 dividing two of the a's and for God's >sake!!! >There's a 7 in that last factor, you see it, don't you? >In 5a_3(x) + 7. Come on, there's a 7 RIGHT THERE! Of course 7 is >still there!!! >So some poster comes at you claiming that 49 divides off from >P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) >in a way that varies dependent on the value of x, and your brain tells >you, OK!!! >You have a memory. To you 7 is still there, even though the factor is >divided off. As a polynomial in 5, it's still there. It is not a coincidence that 22 = 3 * 5 + 7. >Follwing the sci.math'ers insistent raving you get >P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) + >7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x)) >where the w's are factors of 7, so that you're left with functions of >But they would STILL be there with the factorization of >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >inexplicably still there despite the polynomial not having any factors >of 7. >Your eyes fool you. Your memory betrays you. No, it's YOUR memory that is the problem. You have forgotten how you were factoring: not with respect to x, but with respect to 5. >The math doesn't bother with such nonsense. A multiple of a >polynomial can be divided off and it's gone. >It doesn't leave a trace. >In a way what's happening now is an instructive lesson in the >limitations of the human brain. Your brains SEE something, and >insistently tell you that 7 is STILL THERE, and so the arguments go on >for years. >After all, you can SEE the 7's. Come on, who's fooling who, right? You are hopelessly tangled in your own dimwitted oversimplification. You have forgotten what you were doing. >Dammit. >You can see the 7's in there, can't you? Sure. On *both* sides. Just look at: P(x)/49 = (x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7, and note that the constant term is ... 7, just like it should be, if you regard 5 as the polynomial variable! Nora B. >James Harris === Subject: Re: JSH: Look at it backwards >I usually start with one polynomial and then talk about dividing 49 >off from it, but here I'll start *after* 49 has been divided off: >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >and consider the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >and the assertion that the d's must have factors >w_1(x), w_2(x), and w_3(x) >that multiply to give w_1(x) w_2(x) w_3(x) = 7. >But there's no factors of 7 anywhere. So why should the d's have >functions that are factors of 7? > You are getting maximally confused. In your original > approach, you are factoring P(x) as a polynomial in 5, > not as a polynomial in x. The constant term with > respect to 5 and the constant term with respect to > x are different. It was a stupid mistake to think you > were simplifying by treating 5 as if it were the > polynomial variable in the first place. And that is what > is causing your confusion here. There is only one variable Nora Baron, so what other variable would you have listed with S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) considering that S(x) = 300125x^3 - 18375 x^2 - 360 x + 22? So who's confused? >Ok, maybe that seems unfair, as there could be LOTS of different >functions you can plug in for the c's and d's, but why should ANY of >them have functions of x that are factors of 7? >The equation has no memory. > The memory problem here is, you have forgotten what you > started with. To see this, you need to go back to the way you > were doing things originally: for example, Now note, the polynomial S(x) doesn't have 7 as a factor, as this time I'm starting with the result of dividing off the multiple. So the poster Nora Baron (actually a guy as revealed in another post when he ended with a male name) is trying to show how 7 gets back into the expression. That is, he's trying to prove that the factorization DOES have a memory of the multiple 49. > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > To put this in terms of what you are doing now: replace x in this > expression by 5, and replace m by x, and replace f by 7. Also > replace u by 1. This gives > P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), > and you can clearly see that, as a polynomial in 5, the constant > term is 49 * 7 = 7^3, not 49 * 22. No. It can be factored with respect to 5 and 7, but 5 is not a variable. The polynomial variable is x. Now, I've explained lots of times to the Nora Baron poster, and what I want you all to consider is that the poster isn't really that dense, but instead knows you better than you know yourselves. Essentially the basic strategy is just to disagree. Time after time, and even in answer to surveys that I've done, readers who normally lurk will admit that they primarily rely on the fact that people argue with me, assuming that if I were right, then others wouldn't disagree! So, for sci.math'ers like Nora Baron, the strategy is clear--just disagree. They often don't even TRY to actually make mathematical senze. > Go back to the expression above for P(m). The constant term > with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to > 49*22 when u = 1, f = 7, and x = 5. But the constant term with > respect to x is u^3 f^3. You have forgotten this and gotten mixed > up, that's all. Now the expressions actually is S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 where the factorization S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) is being considered. Now how do you get 7 into that? Well, that's not the issue for the poster! All he has to do is disagree. For most readers that's what works. Well, that enough here. James Harris === Subject: Re: JSH: Look at it backwards > So the poster Nora Baron (actually a guy as revealed in another post > when he ended with a male name) is trying to show how 7 gets back into > the expression. What difference does it make what Nora's gender is? What does it have to do with math? Has anybody but you made note of the question of Nora's gender? No, because it is not relevant. Your obsession with her (or his) gender just screams mental illness. It is (probably) just a pseudonym, nothing more. So what? My pseudonym is o[CapitalYAcute]in, so do you claim that I am pretending to be a Norse god? Well, nobody cares!!! Idiot!!! === Subject: Re: The Size of Graham's Number > Trying to explain the sheer hugeness of Graham's Number in a popular > way to someone, I kind of came up short. I have read on the net > somewhere that even if all matter in the universe were converted to > pen and paper, it wouldn't be enough to write the number down. But > that's a pretty difficult concept to grasp as well. Is there an easier > way to a) estimate the number of digits in Graham's Number and b) > express this as something slighty less unfathomable (i.e. it > wouldrequier a hard drive 40 times the size of our galaxy to store it > or something) The number of digits it would take to write down Graham's number is just as unfathomable as the number itself. === Subject: Re: The Size of Graham's Number > Trying to explain the sheer hugeness of Graham's Number in a popular > way to someone, I kind of came up short. I have read on the net > somewhere that even if all matter in the universe were converted to > pen and paper, it wouldn't be enough to write the number down. But > that's a pretty difficult concept to grasp as well. Is there an easier > way to a) estimate the number of digits in Graham's Number and b) > express this as something slighty less unfathomable (i.e. it > wouldrequier a hard drive 40 times the size of our galaxy to store it > or something) Trouble is, those descriptions that you came up with also apply to relatively puny numbers such as googolplex (if, by writing the number down, you mean writing a one followed by a googol of zeroes). On the other hand, you can write googolplex = 10^googol = 10^10^100 = 10^10^10^2, but Graham's number is so mind-bogglingly huge that nothing you said above even comes close to expressing it. In fact, you could start writing 10^10^10^10^... until you run out of space in the universe, and you still wouldn't reach the required size. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: The Size of Graham's Number === >Subject: Re: The Size of Graham's Number === >>Subject: The Size of Graham's Number >>Message-id: Trying to explain the sheer hugeness of Graham's Number in a popular >>way to someone, I kind of came up short. I have read on the net >>somewhere that even if all matter in the universe were converted to >>pen and paper, it wouldn't be enough to write the number down. But >>that's a pretty difficult concept to grasp as well. Is there an easier >>way to a) estimate the number of digits in Graham's Number and b) >>express this as something slighty less unfathomable (i.e. it >>wouldrequier a hard drive 40 times the size of our galaxy to store it >>or something) >> Not likely. Even a tiny number like 10**600 is unfathonable. >> See How big is a 2000-bit number? >> http://members.aol.com/mensanator666/fun/2000_bit.htm >what is it? >a googol is 10^(10^100)) That's a googolplex. A googol is just 10^100. >think of a number with 100 zeroes, that's how many zeroes it has!! >a googol of monkeys tapping keys at a googol typewriters WILL come up >with the entire works of Shakespear. >Herc -- Mensanator Ace of Clubs === Subject: Re: Escultura affair: publication scandal >1271503.iARF35e16701@proapp.mathforum.org... Norm > Unfortunately, mathematics is not a popularity contest; it is a struggle for precision. One of the requirements of mathematics is that every concept must well-defined, that is, its existence, properties and relations with other concepts must be specified by the axioms. The axioms are the basis of proof. Proofs are nonsense unless they follow from the axioms. That is why a true mathematician looks at the foundations of a field before doing anything there. This is where Wiles failed miserably. Critique-rectification naturally involves new language. > E. E. Escultura > University of the Philppines No, it's not a popularity contest -- it's a game. Well, at least there was a game called IIRC Wiff-N-Proof that taught the creation of and deduction of correct logical propositions from axioms. If you're really a mathematician then you already know that there are several different logical systems based on differing foundational axiom sets and their semantics. Assuming your system is consistent, you've simply created another one that's neither more nor less correct than the others. What you see as errors in their system results from applying your axiom scheme to someone else's system -- which isn't valid. Norm === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3GGKw12516; >>Zentralblatt f.9fr Mathematik< 9 publications >>of E. Escultura are listed. For 5 of them the >>Zentralblatt just mentions >not reviewed<, whatever >>that means. >>This is an indexing service like so many others. So it simply notes some contributions unless someone wants to try uncharted course. >I don't agree. The Zentralblatt and the Math Reviews are the >two major indexing services on this planet. Indeed together >they cover all countries. >>>wonder why persons like E. Escultura or J. Harris >>provoke such strong reactions. >>The most likely reason is fear of new ideas. >>E. E. Escultura >>University of the Philippines >This is one possibility. Why is it the most >likely one? Another is feeling of inadequacy. Which one is most likely, I don't know. === Subject: Circle-line intersection Hi there! Don't tell me to look at other posts in this forum. This is a special problem aimed to they who know math. The question how the get 't', look below. I thought I solved the problem but the outcome was a failure. I got a correct equation from mathworld but that is not what I want though. A line(L) intersects with a circle(C), tell me if something is wrong. All positions orients from circle position which is (0,0). t is what I want. L = A + (B-A)*t where D = B-A Lx = cos(v)*R Ly = sin(v)*R A combination of equations: cos(v)*R = Ax + (Dx)*t sin(v)*R = Ay + (Dy)*t I do this because the points of the line intersects on the circle-line which distance is R from origo. 'cos(v)*R' is the horizontal distance and sin(v)*R is the vertical distance. Hope you understand my thinking here. Only if could draw with a pen :). And no we do 'power of two' at both sides of equation. This I do because of a forumla which I've to merge with this equation, see below. cos^2(v)*R^2 = (Ax + Dx*t)^2 which also are ((Ax + Dx*t)^2) cos^2(v) = -------------------- R^2 sin^2(v)*R^2 = (Ay + Dy*t)^2 which also are ((Ay + Dy*t)^2) sin^2(v) = -------------------- R^2 Ok now to the hard part. This is a math rule: sin^2(v) + cos^2(v) = 1 and this gives cos^2(v) = 1 - sin^2(v) Now we combine the first combination with the formula above: ((Ax + Dx*t)^2) cos^2(v) = -------------------- R^2 will be ((Ax + Dx*t)^2) 1 - sin^2(v) = -------------------- R^2 And know we just take away the sin-function: ((Ay + Dy*t)^2) ((Ax + Dx*t)^2) 1 - -------------------- = -------------------- R^2 R^2 Now I can get 't'. The equation will be, after some flipping and flopping of the variables, look like this: t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 If you don't see what this is I tell you this: ax^2 + bx + c = 0 Now if you know what math is you know what it is. Now the question: Why DOESN'T THIS WORK!!! :| === Subject: Re: Circle-line intersection days. My association with the Department is that of an alumnus. >A line(L) intersects with a circle(C), tell me if something is wrong. >All positions orients from circle position which is (0,0). You might want to say what it is you are trying to do... >t is what I want. >L = A + (B-A)*t >where D = B-A There was no D so far.... You mean: L is a line going through the points A and B, and we obtain it (by vector addition) as A + Dt, where D = (B-A) (A, B being vectors with initial point at the origin and terminal point at the point A and B, presumably, and t being a real number parameter...) Is that right? >Lx = cos(v)*R >Ly = sin(v)*R What are Lx and Ly supposed to mean? What is R? What is v? I assume Lx means the x coordinate of the point on the line, and Ly means the y coordinate (as below you have Ax, Dx, Ay, Dy, presumably being x coordinate of A, x coordinate of D, etc). Meaning... you are trying to figure out the intersection? (x,y) are the coordinates of the intersection? And then v is simply the angle, R the radius of the circle? Sounds reasonable to me... >A combination of equations: cos(v)*R = Ax + (Dx)*t > sin(v)*R = Ay + (Dy)*t >I do this because the points of the line intersects on the circle-line >which distance is R from origin. 'cos(v)*R' is the horizontal distance >and sin(v)*R is the vertical distance. Hope you understand my thinking >here. Only if could draw with a pen :). Let's see. We are working on the plane. You are describing the circle as being given by all points of the form (R*cos(v),R*sin(v)), as v ranges over, say, 0 to 2pi. You are describing the line as going through the points A=(a_1,a_2) and B=(b_1,b_2), so you let D = B-A = (b_1-a_1,b_2-a_2) = (d_1,d_2), and the line becomes (x,y) = (a_1,a_2) + t(d_1,d_2) where t ranges over all real numbers. So any intersection between the circle and the line would have to satisfy a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) for some value of t and some value of v. >And no we do 'power of two' at both sides of equation. You are squaring both equations... Okay... HOWEVER: remember that because you squared the original equations, you may have introduced new solutions that were not part of your original intersections. For example, if you find a point where R*(cos(v)) = - (a_1+td_1) R*(sin(v)) = (a_2 + td_2) then these values of v and t will ALSO satisfy R^2*cos^2(v) = (a_1+td_1)^2 R^2*sin^2(v) = (a_2+td_2)^2 even though they do not satisfy the original equation. In fact, the solutions to the following four systems: ORIGINAL: a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) Plus a_1 + td_1 = -R*cos(v) a_2 + td_2 = R*sin(v) a_1 + td_1 = R*cos(v) a_2 + td_2 = -R*sin(v) and a_1 + td_1 = -R*cos(v) a_2 + td_2 = -R*sin(v) will also satisfy (a_1 + td_1)^2 = R^2*cos^2(v) (a_2 + td_2)^2 = R^2*sin^2(v) So ->not every solution you find after squaring needs to be a solution of your original equation<-. Every solution to your original equation will also be a solution to the new one, but the new one may have more solutions. (It's like, if you start with x= 3, there is only one solution; but if you square it, you get x^2 = 9, and now both x=3 and x=-3 are solutions). >This I do >because >of a forumla which I've to merge with this equation, see below. > cos^2(v)*R^2 = (Ax + Dx*t)^2 > which also are > ((Ax + Dx*t)^2) > cos^2(v) = -------------------- > R^2 > > sin^2(v)*R^2 = (Ay + Dy*t)^2 > which also are > ((Ay + Dy*t)^2) > sin^2(v) = -------------------- > R^2 Fine. Since a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) then cos(v) = (a_1 + t*d_1)/R sin(v) = (a_2 + t*d_2)/R >Ok now to the hard part. >This is a math rule: > sin^2(v) + cos^2(v) = 1 Identity. Yes. >and this gives > cos^2(v) = 1 - sin^2(v) Yes. >Now we combine the first combination with the formula above: > ((Ax + Dx*t)^2) > cos^2(v) = -------------------- > R^2 > > will be > > ((Ax + Dx*t)^2) > 1 - sin^2(v) = -------------------- > R^2 >And know we just take away the sin-function: > ((Ay + Dy*t)^2) ((Ax + Dx*t)^2) > 1 - -------------------- = -------------------- > R^2 R^2 Easier to just add the two equations you had: a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) we have R^2 = R^2(cos^2(v) + sin^2(v) = (a_1+td_1)^2 + (a_2 + td_2)^2. >Now I can get 't'. The equation will be, after some flipping and >flopping of the variables, look like this: >t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 >If you don't see what this is I tell you this: ax^2 + bx + c = 0 >Now if you know what math is you know what it is. >Now the question: Why DOESN'T THIS WORK!!! :| Looks right. What do you mean by This does not work? Maybe you mean you sometimes get wrong answers? That's because of the observation above. You have to make sure that whatever answers you get actually satisfy the original equations. Plug them in. They should work fine. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Shiing-shen Chern (Chen Xingshen) passes away at 93 Just grabbed off Xinhua web site: Noted mathematician passes away in Tianjin TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a world-renowned overseas Chinese mathematician, 93, died of illness at his home at Nankai University in north China's Tianjin Municipality at around 7:15 p.m. Friday, the university announced. Chern, a US citizen, is best known for his achievements in the study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang Province, east China. He graduated from Nankai University in 1930 and received further education at Qinghua University and the University of Hamburg in Germany. He taught at several Chinese and US universities -- including Princeton University, the University of Chicago, and the University of California, Berkeley -- and is the only Chinese to win the Wolf Prize -- the most distinguished award in the international mathematics field. The International Astronomical Union officially named asteroid No. 1998CS2 after the noted mathematician in November for his outstanding contributions to human society. === Subject: Re: Shiing-shen Chern (Chen Xingshen) passes away at 93 > Just grabbed off Xinhua web site: > Noted mathematician passes away in Tianjin > TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a > world-renowned overseas Chinese mathematician, 93, died of illness at > his home at Nankai University in north China's Tianjin Municipality at > around 7:15 p.m. Friday, the university announced. > Chern, a US citizen, is best known for his achievements in the > study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang > Province, east China. He graduated from Nankai University in 1930 and > received further education at Qinghua University and the University of > Hamburg in Germany. > He taught at several Chinese and US universities -- including > Princeton University, the University of Chicago, and the University of > California, Berkeley -- and is the only Chinese to win the Wolf Prize -- > the most distinguished award in the international mathematics field. > The International Astronomical Union officially named asteroid No. > 1998CS2 after the noted mathematician in November for his outstanding > contributions to human society. I had the privilege of taking an upper division class (upper division = 3rd or 4th year undergrad math major) from Professor Chern in 1976 at U.C Berkeley. I remember him as a warm, down to earth man. Approachable to us students. He cared about his students and about people. I'm sad to hear of his passing. === Subject: Re: Shiing-shen Chern (Chen Xingshen) passes away at 93 Moment of Silence. W. Dale Hall ??? > Just grabbed off Xinhua web site: > Noted mathematician passes away in Tianjin > TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a > world-renowned overseas Chinese mathematician, 93, died of illness at > his home at Nankai University in north China's Tianjin Municipality at > around 7:15 p.m. Friday, the university announced. > Chern, a US citizen, is best known for his achievements in the > study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang > Province, east China. He graduated from Nankai University in 1930 and > received further education at Qinghua University and the University of > Hamburg in Germany. > He taught at several Chinese and US universities -- including > Princeton University, the University of Chicago, and the University of > California, Berkeley -- and is the only Chinese to win the Wolf Prize -- > the most distinguished award in the international mathematics field. > The International Astronomical Union officially named asteroid No. > 1998CS2 after the noted mathematician in November for his outstanding > contributions to human society. === Subject: Re: sin(x^x) Hi. Is there a formula that describes the density of > the graph of sin(x^x) for real x>0 as x increases > without bound? > [...] Topologically, it is nowhere dense. In terms of two-dimensional area (say, > Lebesgue measure), it has measure 0. Fractal-wise, it is of Hausdorff dimension 1. (And it is a real-analytic submanifold of the plane, > perhaps with some care taken about the missing left endpoint). Any other criteria? If it means how close the subsequent roots (or the increasing and > decreasing segments) are, sit down and do the arithmetic, using the > Lambert W-function. For your convenience, the explicit formula for y^y=x (x sufficiently > large) is y = ln(x) / lambertW(ln(x)) > By density I meant distance between the peaks of the waves > as x -> inf. Obviously, the distance -> 0 (and so the density -> inf) as > x -> inf but what is the equation that describes that density? I would guess it's something like this: sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak > whenever x^x = npi/2, with n a positive integer. Caution: you have listed not only local maxima, but also roots and > local minima. Sort them out: > Ohh... because of n = 2, then you get sin(pi) = 0, then n = 3 -> sin(3pi/2) > = -1, > then n = 4 -> sin(2pi) = 0, n = 5 -> sin(5pi/2) = 1, etc. > u = (4*k+1)*pi/2 (k integer) gives local maxima of sin(u). > You can complete the list. > So it should be > D(n) = 1/(ln((4n+2)pi/2)W(ln((4n+2)pi/2)) - > ln((4n+1)pi/2)/W(ln((4n+1)pi/2))). Replace (4n+2) by (4n+5), because you need to use 4(n+1)+1, rather than (4n+1)+1. > [nothing added by me below] > Then the nth peak is at x = ln(npi/2)/W(ln(npi/2)) So the 'density' function I'm looking for would then > be D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))). Would that work? > === Subject: Re: Graphing polynomial equations >xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 >Obviously you would loop through x and y and calculate >z, and we'd use a numerical approximation technique to >do that. I'm trying to write a program that would plot >the graph. The problem boils down to applying a >numerical algorithm to a quintic like this: >z^5 + a4 z^4 + a3 z^3 + a2 z^2 + a1 z + a0 = 0. >But the problem I'm having is this: Most numerical >approximation algorithms require an initial guess. So, what >would be a good algorithm to find a good initial >guess? You probably realize that a degree-5 polynomial can have up to 5 zeroes, so you would want 5 (or maybe more) initial guesses. The high and low guesses should be fairly easy, because there's no such thing as too high or too low for the tangential method (sorry I forgot who discovered that or I'd give him credit!). Even if you're using some other numerical method you can still use the tangential results to seed it. You can get the other possibilities by looking at the zeroes of the derivative, which are potentially relative mimima and maxima. Any zeroes other than the first and last will be between a relative maximum and a relative minimum. You may have to apply this process recursively until you get down to degree 2 where you can use the quadratic equation. Once you're started, you can use results from neighboring (x,y) as guesses. They won't always work out, but when they do it will save time. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Fermat Fermat's Last Theorem Ben Ito 12-1-04 I will show that Fermat's n=4 is invalid. Fermat's n=4 proof is based on a deception that alters the variables to imply that the n=2 and n=4 equations are identical equations.However, the n=2 and n=4 equations are completely different equations; therefore, the integer solution equations of n=2 cannot be used to prove n=4 as Fermat has done. Fermat uses the elliptical curve equation to prove that the area of a right triangle (XY)/2 = d^2 (equ 1) does not form an integer value of d. Fermat uses the integer solution equation of n=2 in the area equation yet the area equation (equ l) is completely different form the n=2 length equation X^2 + Y^2 = Z^2. (equ 2) Fermat is using the same method used in his n=4 proof where he uses the n=2 solutions in the n=4 equation; however, both equations are completely different. Consequently, Fermat's area proof, that uses an elliptic curve equation, is invalid. Wiles proof of Fermat's Last Theorem is based on the elliptic curve equation , Y^2 = X(X - a^n)(X + b^n) = X^3 + X^2(b^n - a^n) - X(ab)^n (equ 3) however, this equation does not include the c^n variable of Fermat's equation a^n + b^n = c^n, (equ 4) and (ab)^n is also not part of Fermat's equation (equ 4). Therefore, Wiles' proof of Fermat's last theorem is invalid. I will prove Fermat's last theorem using mathematical quantitative analysis. The n=1 and n=2 solutions, and near solutions of n=3 are described; I will then show a pattern using mathematical quantitative analysis to assess the distribution of the integer solutions, of n=2, only the n=1 and n=2 form integer solutions using the solution patterns formed by the distribution of the integer solutions of n=1 and n=2 and the near solutions of n=3. l. Introduction Fermat studied law at a univerity in Orl.8eans. He received a degree in civil law and in 1631 was a councillor at the parliament in Toulouse, France. In his spare time Fermat was allegedly a mathematician. Fermat attempt at proving that the n=4 equations do not form integer solutions is based on a deception where he alters the variables that are being used, and also does not include all integer combinations in his proof of n=4. Therefore, Fermat's n=4 proof violates logic and is incomplete; consequently, Fermat's n=4 proof is invalid. Fermat's last theorem states that X^n + Y^n = Z^n, (equ 5) when n>2 does not form integer solutions of X, Y and Z. However, elliptic curve derivation, it is apparent that he did not have a proof for n>2. In all of his attempts at forming a proof, n=4 and area proof, Fermat always used the integer solution equations of n=2; therefore, Fermat did not understanding the basic underlining principle of the problem that there are an infinite number of possible integer solution combinations that form an infinite number of equations; therefore, it is impossible to prove n=4 or n>2. Fermat's contradiction proof is base on a set of equations that form solutions to the n=4 equation yet does not form integer solutions; however, the equation used are derived from the integer solutions of n=2. Fermat acheives this contradicts by altering the variables. I will show the reader that foundation principles of Fermat's n=4 and Wiles proofs are invalid; consequently, if the foundation of any proof is found to be invalid than the proof of this invalid foundational statement supported by the remaining of the proof is also invalid. I will show that Fermat's n=4 and Wiles' proofs are invalid, using this principle, then prove Fermat's last theorem using mathematical quantitative analysis based on experimental physics. 2. Fermat's n=4 Proof Fermat's n=4 proof is described. Fermat implies that by proving that, X^4 + Y^4 = Z^2 (equ 6). does not form integer solutions also proves that X^4 + Y^4 = Z^4 (equ 7) does not form integer solutions (Shanks, p. 144). However, using n=4 in equation 5 forms equation 7 yet Fermat uses equation 6 to prove n=4; however. equations 6 and 7 are completely different equation; therefore, Fermat does not prove n=4. Fermat uses the integer solution equations of n=2, a = 2uv, b = u^2 - v^2, and c = u^2 + v^2 (equ 8a,b,c), and, X^2 = 2uv, Y^2 = u^2 - v^2, and Z = u^2 + v^2 (equ 9a,b,c), (Shanks, p.141). Equation 5a,b,c are then used to prove n=4; however, Fermat squares the right sides of equations 8a,b,c without squaring the left side. Using equations 8a,b and 9a,b, a = X^2, b = X^2 and c = Z. (equ 10a,b,c) Inserting equations 10a,b,c in equation 6, a^2 + b^2 = c^2 (equ 11) Fermat is implying that the equation of n=2 is equal to the equation of n=4. The formation of equations 9a,b,c is a key step in Fermat's contradiction method (Osserman, p. 18). Fermat is implying that when n=4, a^2 = X^4, (equ 12) Fermat is implying that when the value of n increases the value of X can be altered. The alteration in the variables in the key to Fermat's proof. However, using a = 9^(1/n), b = 16^(1/n) and c = 25^(1/n) (equ 13a,b,c) in a^n + b^n = c^n----------------> 9 + 16 = 25 (14a,b) Equation 14 is a solution since the variables (a,b,c) of equation 13a,b,c are being alter when n increases. However, when n>2 equations 13a,b,c never form integer values of a,b and c. The solution of equation 14 are essential in Fermat's proof; however, Fermat is altering the variables to form the solution of equation 6. Fermat n=4 proof is not a proof but a manipulation of the variables. In addition, Fermat is using equations (equ 9a,b,c), that right side of the equation is equivalant to the n=2 integer solution equation, are derived from equation 5a,b,c to prove n=4; however, X^2 + Y^2 = Z^2 (equ 11) completely different from, X^4 + Y^4 = Z^2 (equ 12). Using Z = 6, equation 9 forms a circle of radius 6, X^2 + Y^2 = 36.(equ 13) Using Z = 6 in equation 10, X^4 + Y^4 = 36,(equ 14) equation 12 does not form a circle; therefore, equation 12 is completely different from equation 11. The integer solutions of n=2 cannot be used to prove n=4 since equations 11 and 12 are completely different equations. Therefore, equations 6a,b,c cannot be used to prove n=4. Fermat proof is implying that the equations of n=2 and n=4 are equal using equations 7a,b. Therefore, Fermat's n=4 proof is invalid. Also, Fermat's n=4 proof does not testing all integer combinations of X and Y. Fermat's proof is only proving equations 5a,b,c, that are derived from n=2, do not form integer solutions. However, there are an infinite number of integer combinations that are not tested in Fermat's n=4 proof; example, the integers X = 22, Y = 23 and C=24 are not included in Fermat's n=4 contradiction proof; Fermat does not prove that (22,23,24) does not form integer solution. Fermat is implying that by manipulating the integer solutions of n=2 he has proven that n=4 does not form integer solutions; however, n=2 solutions do not include all possible integer combinations of X and Y. Example, using n=4 (X^4 + Y^4)^(1/4)=C (equ 15) Let Y = X + 1, (equ 16) inserting equation 13 in equation 12, [(X + 1)^4 + X^4]^(1/4) = C --------> (2X^4 + 4X^3 + 6X^2+ 4X + 1)^(1/4) = C (equ 17) Using X = 1,2,3,4,5,6,7,8,9,.................,(equ 18) There are an infinite number of equations that describe integer combinations that are not included in Fermat's n=4 proof. Example, using f(X) = X + 2, f(X) = X^2 + 1, f(X) = X^2 + X, f(X) = X^2 + X + 3 etc.............(equ 19) in [(f(A))^4 + A^4]^(1/4) = C (equ 20) Fermat's n=4 proof uses (equ 9a,b,c) that represents the integer solutions of n=2;therefore, not all possible integer combinations are represented in Fermat's n=4 proof. The proof for n=4 must include all possible integer combinations of X and Y. However, there are an infinite number of integer combinations that are not included in Fermat's n=4 proof. Fermat is justifying the non-existence of integer solutions of n=4 using a contradiction method where a single group of equations (equ 5a,b,c) are used; however, Fermat's n=4 proof does not include all possible integer combinations of X and Y; therefore, Fermat's n=4 proof in incomplete and therefore, invalid. Fermat realized that the n>2 was not possible to prove using his n=4 or any other algebraic method. There are two possibilities in solving Fermat's proof; one is by algebraically proving n>2 does not form solutions and the second method would be to show that only a right triangle forms integer solution; therefore, Fermat released that the n>2 prove was algebraically impossible; therefore, he went for the second possibility by using the right triangle area and the elliptic curve. 3. Wiles' Proof Wiles uses elliptic curve equation to prove n>2 since probably Fermat used an elliptic curve. Fermat's use of the elliptic curve equation is described. Fermat states that a right triangle has integer sides, X and Y; (XY)/2 = d^2 (equ 12) the area described with (XY)/2 does not form an integer value of d. Fermat then uses the integer solution equations of n=2, X = (m^2 - n^2), Y = 2mn, Z = (m^2 + n^2),(equ 13) in equation 12, the second equation takes the form: 2d^2 = XY = (m^2 - n^2) x 2mn, (equ 14) or d^2 = (nm^3 - mn^3).(equ 15) We now let X = m/n, Y = (d/n^2).(equ 16) Then X^3 - X = (m^3)/(n^3) - m/n = (nm^3 - mn^3)/n^4 = (d^2)/n^4 = Y^2 (equ 17) which forms the elliptic curve equation, Y^2 = X^3 - X, (equ 17) However, equation 12 is not equal to equation 6.Equation 12 represent the area of a right triangle; whereas, equation 6 represents the length of the sides of the right triangles The n=2 equation is not the area equation; therefore, Fermat violates logic by using the integer solution equation (equ ) in the area equation to prove that the area equation (equ ) does not from integer solutions. Fermat is using the same method used in his n=4 proof where he uses the n=2 solutions in a completely different equation (equ 12); therefore, Fermat's area integer proof is invalid. Mathematician don't stray to far from home and this is also the case with Wiles. Wiles implied that since Fermat uses elliptic curve equation that the elliptic curve would be used to prove n>2. Wiles proof of Fermat's Last Theorem is based on the elliptic curve equation (Poorten, p. 196-7), Y^2 = X(X - a^n)(X + b^n) (equ 18). However, Fermat's equation is a^n + b^n = c^n (equ 19). it is not mathematical possible to derive equation from equation 18 since equation 18 becomes, Y^2 = X^3 + X^2(b^n - a^n) - X(ab)^n (equ 20) therefore, Wiles' proof is invalid since equation 20 does not include the c^n variable of Fermat's equation (equ 19). In addition, equation is the sum of a^n and b^n; whereas, equation 20 forms b^n - a^n that are not sums. Also. the (ab)^2 of equation 20 is not part of equation 19;therefore, Wiles proof ( Wiles, p. 448) that uses equation 18 is invalid. 4. Proof of FLT. I will prove Fermat's last equation using the pattern formed by the solutions formed by n=1 and n=2, and the near solutions of n=3 and n=4. Using n=1 in equation 1, X + Y = Z (equ 21) All integers of equation 21 form solutions. An integer solution set occurs at (1,2,3). The integer solutions of n=2 are described with the following equations, X = 2uv, Y = u^2 - v^2 and Z = u^2 + v^2 (equ 22) Not all integers form integer solutions for n=2. An integer solution set occurs at (3,4,5). The n=1 and n=2 solutions show a pattern where integer solutions are formed when a solution set is forms close to the origin. In addition, when n=1, the strength of the solution set formation is maximum since all integers form solutions; when n=2 not all integers form solutions; therefore, the strength of the solution formation is weakening. When n=3, 6^3 + 8^3 = 728 where 9^3 = 729 (equ 23) 9^3 + 10^3 = 1729 where 12^3 = 1728 (equ 24) an integer solution set does not form close to the origin for n=3;however, equations 23 and 24 are off by one. The solution set is weakening as n increases; at n=4 7^4 + 8^4 = 6497 where 9^4 = 6561(equ 25) consequently, since n=3 does not from solutions near the origin, the integer sequence ends at n = 3. Therefore, only n=1 and n=2 form integer solutions. 5. Conclusion I have shown that Fermat's derivation of n=4 is base on deception since Fermat uses non-integer solutions of A', B' and C to prove that Fermat's n=4 equation does not from integer solution. In addition, Fermat is using the integer solution of n=2 to prove n=4 which violates logic since n=2 and n=4 form completely different equations. Fermat's n=4 proof does not testing all integers. Fermat's proof is only proving equations 5a,b,c do not form integer solutions. It is impossible to prove n=4, using Fermat's method, since it would require an infinite number of equations; therefore, Fermat's proof of n=4 is invalid. Wiles proof of Fermat's Last Theorem is based on the elliptic curve equation , Y^2 = X(X - a^n)(X + b^n) = X^3 + X^2(b^n - a^n) - X(ab)^n (equ 26) However, equation 26 does not include the c^n variable of Fermat's equation a^n + b^n = c^n (equ 27). Therefore, Wiles' proof of Fermat's last theorem using equation 26 is invalid. I will prove FLT by showing a pattern forms from the n=1 and n=2 solution sets. All of the integer of n=1 form solutions; when n=2 the number of solutions in a finite range is less then that of n=2; therefore, the strength of the solutions is decreasing as n increases. When n=3 the solution sequence ends and solutions are only formed when n = 1 and n = 2 . 6. References Robert Osserman. Fermat's Last Theorem (a supplement to the video). MSRI Berkeley. 1994 Marilyn vos savant. The World's Most Famous Math Problem. St Martin's Press. 1993 Daniel Shanks. Solved and Unsolved Problems in Number Theory. Chelsea Pub. 1985. A. J. Van Der Poorten. Notes on Fermat's Last Theorem. John Wiley. 1996 Andrew J. Wiles. Annals of Mathematics. Editor Andrew Wiles. Princeton University Press. May 1995. 7. Acknowledgment forum, Best Science forum, and About Physics forum, HSU, CSUS, CR, SCC, USC, Hiram Johnson HS Sacramento (Mrs Larson), UCD, Stanford, MIT, Harvard, ASU, Rutgers and UCLA mathematics Dept. Memory of Yasser Arafat 11-11-04 *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Number of prime factors of an odd perfect number Content-Length: 215 Originator: rusin@vesuvius What is the known upper bound on the number of prime factors / distinct prime factors of an odd perfect number? Does anyone have a conjecture on what the tighest lower bound would be? Edmond === Subject: Re: Number of prime factors of an odd perfect number > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? > Edmond You've already received some excellent replies, but it might be of interest to note that any constant upper bound here would imply there are only finitely many odd perfect numbers: Dickson showed that there are at most finitely many odd perfect numbers with a prescribed number of distinct prime factors. Dickson's result can be made effective. A striking result of Heath-Brown is that an odd perfect number with k distinct prime factors is bounded by 4^{4^k}. (The proof is entirely elementary.) This has been improved to 2^{4^k} by Nielsen in [1]. Hope this helps, Paul [1] Nielsen, Pace An Upper Bound for Odd Perfect Numbers http://www.emis.de/journals/INTEGERS/papers/d14/d14.pdf === Subject: Re: Number of prime factors of an odd perfect number >> What is the known upper bound on the number of prime factors / > distinct >> prime factors of an odd perfect number? Does anyone have a conjecture > on >> what the tighest lower bound would be? >> Edmond > You've already received some excellent replies, but it might be of > interest to note that any constant upper bound here would imply there > are only finitely many odd perfect numbers: Dickson showed that there > are at most finitely many odd perfect numbers with a prescribed number > of distinct prime factors. > Dickson's result can be made effective. A striking result of > Heath-Brown is that an odd perfect number with k distinct prime factors > is bounded by 4^{4^k}. (The proof is entirely elementary.) This has > been improved to 2^{4^k} by Nielsen in [1]. > Hope this helps, > Paul > [1] Nielsen, Pace > An Upper Bound for Odd Perfect Numbers > http://www.emis.de/journals/INTEGERS/papers/d14/d14.pdf fact, but I lately realised that if the non-existance of odd perfect numbers is proven, then by virtue of the recent proof that primality testing can be done in polynomial time, it would follow that perfect number testing can also be done in polynomial time. Does anyone think I'm mistaken or does anyone have any other corollaries that would follow from the proof of non-existance of odd perfect numbers. Edmond === Subject: Re: Number of prime factors of an odd perfect number > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? I don't know much about recent advances, but there are a few classical results: J. J. Sylvester, L. E. Dickson and H. J. Kanold proved that there is no odd perfect number with 4 prime divisors. Next, I. S. Gradshtein, U. K.9fhnel and G. C. Weber disproved existence of numbers with 4 divisors and C. Pomerance and N. Robbins showed the same for 6 divisors. P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 divisors and I don't know about any newer results. Pawel Gladki === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous > theorem for 8 divisors and I don't know about any newer results. Of course we are talking about _distinct_ prime factors. For the bound of prime factors counted with repetitions it is known that there are at least 47 factors - see Kevin Hare's preprint: http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf Pawel Gladki === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 >> divisors and I don't know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound of > prime factors counted with repetitions it is known that there are at least > 47 factors - see Kevin Hare's preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf interested in known UPPER BOUNDS and a CONJECTURE on what would be the ACTUAL LOWER BOUND. Edmond === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 >> divisors and I don't know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound of > prime factors counted with repetitions it is known that there are at least > 47 factors - see Kevin Hare's preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the > ACTUAL LOWER BOUND. Known upper bounds? You mean a proof that if there is an odd perfect number it can't have more than so many prime factors? I don't think I've ever seen a result along those lines - have you? As for conjectured lower bounds, I think those would be infinite, as I think the conjecture is that there aren't any odd perfect numbers. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for > 8 > divisors and I don't know about any newer results. >> Of course we are talking about _distinct_ prime factors. For the bound >> of >> prime factors counted with repetitions it is known that there are at >> least >> 47 factors - see Kevin Hare's preprint: >> http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf >> interested in known UPPER BOUNDS and a CONJECTURE on what would be the >> ACTUAL LOWER BOUND. > Known upper bounds? You mean a proof that if there is an odd perfect > number it can't have more than so many prime factors? Exactly > I don't think > I've ever seen a result along those lines - have you? No, I haven't, but I'd like to know at least a reasonably tight upper bound. > As for conjectured lower bounds, I think those would be infinite, > as I think the conjecture is that there aren't any odd perfect numbers. Let us assume that someone does prove this conjecture. What would be the important corollaries thereof ? Edmond === Subject: Re: Number of prime factors of an odd perfect number days. My association with the Department is that of an alumnus. >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for >> 8 >> divisors and I don't know about any newer results. Of course we are talking about _distinct_ prime factors. For the bound > of > prime factors counted with repetitions it is known that there are at > least > 47 factors - see Kevin Hare's preprint: http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the > ACTUAL LOWER BOUND. >> Known upper bounds? You mean a proof that if there is an odd perfect >> number it can't have more than so many prime factors? > Exactly >> I don't think >> I've ever seen a result along those lines - have you? > No, I haven't, but I'd like to know at least a reasonably tight upper >bound. Myerson's point is that, as far as he is aware (and as far as I am aware), there are ->no<- results on upper bounds on the number of prime factors for an odd perfect number. I'm sure you would like to know one. But if there are none to be found, you'll have to prove one if you want to know one. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem > for > 8 > divisors and I don't know about any newer results. >> Of course we are talking about _distinct_ prime factors. For the >> bound >> of >> prime factors counted with repetitions it is known that there are at >> least >> 47 factors - see Kevin Hare's preprint: >> http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf >> interested in known UPPER BOUNDS and a CONJECTURE on what would be the >> ACTUAL LOWER BOUND. > Known upper bounds? You mean a proof that if there is an odd perfect > number it can't have more than so many prime factors? >> Exactly > I don't think > I've ever seen a result along those lines - have you? >> No, I haven't, but I'd like to know at least a reasonably tight upper >>bound. > Myerson's point is that, as far as he is aware (and as far as I am > aware), there are ->no<- results on upper bounds on the number of > prime factors for an odd perfect number. > I'm sure you would like to know one. But if there are none to be > found, you'll have to prove one if you want to know one. I guess I have to set out to prove one myself. But I'm still curious as to how ( or maybe why ) so much work has been done on the lower bound, whereas we have no results regarding the upper bound. Also, is there a good example of a similar upper bound proof, i.e., an upper bound on some property of a hypothetical number ? Edmond === Subject: Re: Number of prime factors of an odd perfect number days. My association with the Department is that of an alumnus. >> Myerson's point is that, as far as he is aware (and as far as I am >> aware), there are ->no<- results on upper bounds on the number of >> prime factors for an odd perfect number. >> I'm sure you would like to know one. But if there are none to be >> found, you'll have to prove one if you want to know one. >I guess I have to set out to prove one myself. But I'm still curious as to >how ( or maybe why ) so much work has been done on the lower bound, whereas >we have no results regarding the upper bound. Because... how would you prove an upper bound on the number of distinct prime factors? The lower bounds are proven by using the fact that if m and n are relatively prime, then sigma(m*n) = sigma(m)*sigma(n), with sigma(k) = sum of all divisors of k, and showing that if q1,...,qs are distinct prime powers (for which it is easy to calculate sigma), then the product of the sigma(qi) simply does not add up to twice q1*...*qs, for small values of s. But how would you prove an upper bound? How would you even ->approach<- such a problem? The upper bound in the even case comes by showing that if you write the number as 2^n*q, with q odd, then the only way this works out is if q is a prime (and, more specifically, a prime of the form 2^{n}-1). It is not really a generalizable argument. > Also, is there a good example >of a similar upper bound proof, i.e., an upper bound on some property of a >hypothetical number ? Yes. Ramsey numbers have well known upper and lower bounds, but many of them are not exactly known. But here, you prove existence by proving an upper bound. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Number of prime factors of an odd perfect number > Myerson's point is that, as far as he is aware (and as far as I am > aware), there are ->no<- results on upper bounds on the number of > prime factors for an odd perfect number. > I'm sure you would like to know one. But if there are none to be > found, you'll have to prove one if you want to know one. >>I guess I have to set out to prove one myself. But I'm still curious as to >>how ( or maybe why ) so much work has been done on the lower bound, >>whereas >>we have no results regarding the upper bound. > Because... how would you prove an upper bound on the number of > distinct prime factors? The lower bounds are proven by using the fact > that if m and n are relatively prime, then sigma(m*n) = > sigma(m)*sigma(n), with sigma(k) = sum of all divisors of k, and > showing that if q1,...,qs are distinct prime powers (for which it is > easy to calculate sigma), then the product of the sigma(qi) simply > does not add up to twice q1*...*qs, for small values of s. > But how would you prove an upper bound? How would you even > ->approach<- such a problem? The upper bound in the even case comes > by showing that if you write the number as 2^n*q, with q odd, then the > only way this works out is if q is a prime (and, more specifically, a > prime of the form 2^{n}-1). It is not really a generalizable argument. participants. Edmond === Subject: Re: Smullyan's Quiz Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3LRdl09441; >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justified. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. perhaps one should think seriously what unexpected means. to do this one could try to formalise expectations. but this is readily done in elementary game theory, and from there one also gets the wiseness of formalising the situation as a game. with reasonable pay-off functions it is clear that there are no pure strategy equilibria but the professor uses a mixed strategy to determine the date of the exam. the students' expectations about the mixed strategy are, of course, correct in equilibrium but they do not know the realisation of the mixed strategy, i.e. the true date of the exam. in most cases then the exam will be a surprise exam in the sense that the studentd cannot predict the date with probability one. only if the realisation is friday they will be able to predict on thursday evening that the exam will be on friday with probability one. the point here is that this is not so much a problem of logic as a problem of formalising the situation correctly. and when one does! formalise it it turns out to be a game. === Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere AB is a fixed geodesic great circle arc, say a latitude line,on a >> sphere radius R.C moves so that area between arcs AB,BC and CA >> {(A+B+C-pi)*R^2} is constant. What is locus of C? >I think you mean longitude line. Latitude lines are not great circles, >except for the equator. Either way, I would guess that the answer is 2 circles centered on the same axis as AB. For example, if AB is a segment of the equator, C would be a pair of latitude lines, 1 north and 1 south. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Question about numbers in arithmetic progression >Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. What >is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? Regardless of gcd, the trivial solution applies. The minimum value is 1/max(|x_1|,|x_n|). --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Question about numbers in arithmetic progression >>Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. >What >>is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? >Regardless of gcd, the trivial solution applies. >The minimum value is 1/max(|x_1|,|x_n|). I intended that the a_i be integers. My apologies for not making this clear. Rich === Subject: Re: Question about numbers in arithmetic progression >> Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. > What >> is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? >Well, that would depend on x_1, n, and the common difference d, >so I suppose you want some sort of bound (I don't think you'll get >an exact expression) in terms of those parameters, yes? > Add the condition 0 to above. Here is what I get for m: > m(2001,2,5)=501 [251*2001-250*2009=1] > m(20001,2,5)=5001 [2501*20001-2500*20009=1] > m(200001,2,5)=50001 [25001*200001-25000*200009=1] > m(1001,3,5)=169 > m(10001,3,5)=1669 > m(100001,3,5)=16669 > m(1001,10,7)=35 > m(10001,10,7)=335 > m(100001,10,7)=3335 > So it seems as though there *may* be an exact expression for these parameters. > Others, like m(1001,31,7), ... are not at all obvious. > I have a method that seems to compute m(x_1,...,x_n) and am looking for inputs > with known minimums to test against, hence the question. If you apply the Jacobi method to a_1*x_1+...+a_n*x_n=1 you get an equation of the form x_1(a_1 + a_2 + .. a_n) + d(a_2 + 2a_3 + + (n-1)a_m) =1 x_1 A +dB = 1 If P/Q is the penultimate convergent of x_1/d then x_1 A +dB = 1 has a solution if you choose appropriate signs for P and Q. a_2 + 2a_3 + + (n-1)a_n = P => |a_n| = INT(p/(n-1)) and some |a_j| (2<= j |a_1| = |Q| + |1| +|INT(P/(n-1)) This gives minima which agree with your results. What values did you get for m(1001,31,7) etc. ? What is the basis of your method ? === Subject: Re: Question about numbers in arithmetic progression I. M. Davidson wote: >This gives minima which agree with your results. But doesn't prove these are the actual minima, right? >What values did you get for m(1001,31,7) etc. ? I get: m(1001,31,7)=87 m(10001,31,7)=1417 m(100001,31,7)=6462 m(1000001,31,7)=107548 m(10000001,31,7)=322585 and, curiously, m(10^12+1,31,7)=64516129040 m(10^22+1,31,7)=322580645161290322585. >What is the basis of your method ? Casting in a small pond full of big fish and hoping for the best? Seriously, I don't have a answer for your question. I will try to write a good description and post it here, though. Rich === Subject: Re: Question about numbers in arithmetic progression > I. M. Davidson wote: >This gives minima which agree with your results. > But doesn't prove these are the actual minima, right? Precisely. I was thinking that the CF method gives the smallest (in your sense) solution to AX + dB =1 and that the smallest solution to a_2 + d*a_2 + + (n-1)da_n must involve the a_j wth the largest coefficient and some other a_k =1. Bit I'm not sure if this necessarily follows. >What values did you get for m(1001,31,7) etc. ? > I get: > m(1001,31,7)=87 I get 83 as the minimum here a_1 = 45, a_5 =-1,a_6 = -37 i.e 45*1001 +1125*(-1) +1187(-37) =1 So your method does not find the minima in all cases, even though the actual minimum may be less than 83. > m(10001,31,7)=1417 Here I get 1954. A considerable difference. What are the actual values you get for the a's ? Whatever they are they must satisfy the Jacobi reduction. > m(100001,31,7)=6462 I also have a different value here. What were the a's in this case ? === Subject: Re: Question about numbers in arithmetic progression >> I. M. Davidson wote: >>This gives minima which agree with your results. >> But doesn't prove these are the actual minima, right? >Precisely. I was thinking that the CF method >gives the smallest (in your sense) solution >to AX + dB =1 and that the smallest solution >to a_2 + d*a_2 + + (n-1)da_n must involve the >a_j wth the largest coefficient and some >other a_k =1. Bit I'm not sure if this >necessarily follows. >>What values did you get for m(1001,31,7) etc. ? >> I get: >> m(1001,31,7)=87 >I get 83 as the minimum here >a_1 = 45, a_5 =-1,a_6 = -37 >i.e 45*1001 +1125*(-1) +1187(-37) =1 >So your method does not find the minima >in all cases, even though the actual >minimum may be less than 83. Yes. This is not a real suprise as the problem is quite difficult. I'll have to work on this one a bit... >> m(10001,31,7)=1417 >Here I get 1954. A considerable >difference. >What are the actual values you >get for the a's ? 1417: -715*10001+689*10187+5*10125+8*10156=1 1433: -723*10001+689*10187+13*10125+8*10032=1 1433: -715*10001+689*10187-8*10094+21*10125=1 1433: -715*10001+689*10187+13*10125+8*10094=1 1443: -728*10001+18*100063+689*100187+8*10094=1 >Whatever they are they must satisfy >the Jacobi reduction. >> m(100001,31,7)=6462 >I also have a different value here. >What were the a's in this case ? 6462: 3234*100001-1*100094-3222*100187-5*100125=1 6472: 3223*100001+16*100063-3222*100187-11*100156 Rich === Subject: Re: Question about numbers in arithmetic progression >> I. M. Davidson wote: >> m(1001,31,7)=87 >I get 83 as the minimum here >a_1 = 45, a_5 =-1,a_6 = -37 >i.e 45*1001 +1125*(-1) +1187(-37) =1 >So your method does not find the minima >in all cases, even though the actual >minimum may be less than 83. > Yes. This is not a real suprise as the problem is quite difficult. I'll have > to work on this one a bit... realise that I had forgotten about the ambiguity arising when a CF ends in a 1 - as is the case with 1001/31 and 100001/31. This gave me values much greater than they should have been. >> m(10001,31,7)=1417 >Here I get 1954. A considerable >difference. >What are the actual values you >get for the a's ? > 1417: -715*10001+689*10187+5*10125+8*10156=1 > 1433: -723*10001+689*10187+13*10125+8*10032=1 > 1433: -715*10001+689*10187-8*10094+21*10125=1 > 1433: -715*10001+689*10187+13*10125+8*10094=1 > 1443: -728*10001+18*100063+689*100187+8*10094=1 >Whatever they are they must satisfy >the Jacobi reduction. In this case I now get 1411 (-712)*10001 + 699*10187 = 1 >> m(100001,31,7)=6462 >I also have a different value here. >What were the a's in this case ? > 6462: 3234*100001-1*100094-3222*100187-5*100125=1 > 6472: 3223*100001+16*100063-3222*100187-11*100156 Here too, I get the much lower value than before of 6458 3232*100001 +(-1)*100156) + (-3225)*10187 =1 As the Jacobi reduction is identical to the original equation, i.e. whenever the original equation is equal to 1 the Jacobi reduction is also equal to 1 and vice versa. As the CF expansion gives the minimum solution to AX +dY =1,x',y' the minimum a's must be given by the minimum solution of x' = a_1 +a_2 +.. + a_n y' = a_2 + 2*a_3 +.. + (n-1)*a_n choosing the appropriate signs for x',y' Given the minimum a's they must satisfy the Jacobi reduction and x',y'must be minima. So your initial intuition was correct - there is a formula (of sorts) for the minimum solutions. === Subject: Re: Question about numbers in arithmetic progression >So your initial intuition was correct - >there is a formula (of sorts) for the minimum > solutions. Rich === Subject: Re: Question about numbers in arithmetic progression > I. M. Davidson wote: >This gives minima which agree with your results. > But doesn't prove these are the actual minima, right? Precisely. I was thinking that the CF method gives the smallest (in your sense) solution to AX + dB =1 and that the smallest solution to a_2 + d*a_2 + + (n-1)da_n must involve the a_j wth the largest coefficient and some other a_k =1. Bit I'm not sure if this necessarily follows. >What values did you get for m(1001,31,7) etc. ? > I get: > m(1001,31,7)=87 I get 83 as the minimum here a_1 = 45, a_5 =-1,a_6 = -37 i.e 45*1001 +1125*(-1) +1187(-37) =1 So your method does not find the minima in all cases, even though the actual minimum may be less than 83. > m(10001,31,7)=1417 Here I get 1954. A considerable difference. What are the actual values you get for the a's ? Whatever they are they must satisfy the Jacobi reduction. > m(100001,31,7)=6462 I also have a different value here. What were the a's in this case ? === Subject: How Many M&Ms? Hello All. I am writing in reference to a contest a friend of mine is having. The rules are as follows. I bought a brand new clear plastic jar. It is filled with the standard size M & M candies. The jar is round and 8 inches tall. The diameter of the jar inside is 4 inches across. I bought over 4 pounds of M & Ms to fill it. I counted every M & M before filling the jar to the very top.. You can probably guess what I am seeking: How many M & Ms are in the jar as described? If anyone can help me on this it would be wonderful as I am not a good mathmatician. Joey === Subject: Re: How Many M&Ms? > Hello All. I am writing in reference to a contest a friend of mine is > having. The rules are as follows. > I bought a brand new clear plastic jar. It is filled > with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of > the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to fill it. I counted > every M & M before filling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? > If anyone can help me on this it would be wonderful as I am not a good > mathmatician. > Joey The only precise method is to count them. Probably the next most accurate method is to esimate the number by weight. Count out as large a fraction of the total as you have patience for and weigh them to find an average weight per piece. Then weigh the total (less jar weight) and divide by the average weight per piece. Or maybe the manfacturer will give you the average weight per piece. === Subject: Re: How Many M&Ms? I don't get this post.... Joey counted the M&Ms but wants help knowing how many? oh she counted the 4 pounds of M&Ms then threw in a portion of them into the jar. If you're running the competition and people have to guess, they might expect you to count the exact number in the jar. Of course you could just take the median of the guesses and select them as winner, quoting the prize amount as a few different from what they guessed, but if the jar is the prize they might check, and be honest and hand it over to the actual best guesser, this could result in all sorts of bad publicity, you might find your next M&M competition you'll be eating them yourself. A few questions, the jar you bought was empty right? You and your friend are both running the competition? You counted the M&Ms of all the bags you bought? Can you count the remaining M&Ms after you put them in the jar and substract? Is this like that puzzle, if there are 5 apples and I take away 2 how many do I have? Are we eligible for the prize to calculate our guesses? 2601. Herc === Subject: Re: How Many M&Ms? Fortunately, there has been some research on M&M packing: The net is that the density of packing is around 71% (if packed in an irregular fashion). Further, it looks like the someone has actually measured an individual M&M: http://www.kleinbottle.com/Bernie_Tao.htm Which a result of 0.45239 cm. so, volume of jar is: PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 cm3 So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 Now the result is probably only good to around 2 significant figures, and there will probably be some rounding down due to the sides of the container, so I would say aound 2500 M&Ms ? Am I close? -Darren > Hello All. I am writing in reference to a contest a friend of mine is > having. The rules are as follows. > I bought a brand new clear plastic jar. It is filled > with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of > the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to fill it. I counted > every M & M before filling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? > If anyone can help me on this it would be wonderful as I am not a good > mathmatician. > Joey === Subject: Re: How Many M&Ms? In sci.math, darrenn : > Fortunately, there has been some research on M&M packing: > The net is that the density of packing is around 71% (if packed in an > irregular fashion). > Further, it looks like the someone has actually measured an individual M&M: > http://www.kleinbottle.com/Bernie_Tao.htm > Which a result of 0.45239 cm. Pedant point. An M&M is generally a flattened round thing and probably should be measured in all three dimensions, or perhaps one can pour a number into a liquid (although water will melt off the coatings, methinks) and measure the displacement. > so, volume of jar is: > PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 > cm3 > So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 > Now the result is probably only good to around 2 significant figures, and > there will probably be some rounding down due to the sides of the container, > so I would say aound 2500 M&Ms ? > Am I close? > -Darren >> Hello All. I am writing in reference to a contest a friend of mine is >> having. The rules are as follows. >> I bought a brand new clear plastic jar. It is filled >> with the standard size M & M candies. >> The jar is round and 8 inches tall. The diameter of >> the jar inside is 4 inches across. >> I bought over 4 pounds of M & Ms to fill it. I counted >> every M & M before filling the jar to the very top.. >> You can probably guess what I am seeking: >> How many M & Ms are in the jar as described? >> If anyone can help me on this it would be wonderful as I am not a good >> mathmatician. >> Joey -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: How Many M&Ms? > In sci.math, darrenn > Fortunately, there has been some research on M&M packing: > The net is that the density of packing is around 71% (if packed in an > irregular fashion). > Further, it looks like the someone has actually measured an individual M&M: > http://www.kleinbottle.com/Bernie_Tao.htm > Which a result of 0.45239 cm. > Pedant point. An M&M is generally a flattened round thing and > probably should be measured in all three dimensions, or perhaps > one can pour a number into a liquid (although water will melt > off the coatings, methinks) and measure the displacement. > so, volume of jar is: > PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 > cm3 > So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 > Now the result is probably only good to around 2 significant figures, and > there will probably be some rounding down due to the sides of the container, > so I would say aound 2500 M&Ms ? > Am I close? > -Darren >> Hello All. I am writing in reference to a contest a friend of mine is >> having. The rules are as follows. >> I bought a brand new clear plastic jar. It is filled >> with the standard size M & M candies. >> The jar is round and 8 inches tall. The diameter of >> the jar inside is 4 inches across. >> I bought over 4 pounds of M & Ms to fill it. I counted >> every M & M before filling the jar to the very top.. >> You can probably guess what I am seeking: >> How many M & Ms are in the jar as described? >> If anyone can help me on this it would be wonderful as I am not a good >> mathmatician. >> Joey what does it matter if the water removes the coating? the coating will still contribute to the displacement. === Subject: Re: How Many M&Ms? In sci.math, David Bandel [snip for brevity -- discussing # of M&M's in a certain sized jar, and using water for one measurement of volume] > what does it matter if the water removes the coating? the coating will > still contribute to the displacement. There's a few issues regarding solubility that may complicate the analysis. I'd frankly have to look. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: How Many M&Ms? >[snip for brevity -- discussing # of M&M's in a certain sized jar, > and using water for one measurement of volume] >> what does it matter if the water removes the coating? the coating will >> still contribute to the displacement. >There's a few issues regarding solubility that may complicate >the analysis. I'd frankly have to look. M&M's require solving a polynomial of degree greater than 5 ? === Subject: Re: How Many M&Ms? > Which a result of 0.45239 cm. > Pedant point. An M&M is generally a flattened round thing and > probably should be measured in all three dimensions, or perhaps > one can pour a number into a liquid (although water will melt > off the coatings, methinks) and measure the displacement. True. This should say 0.45239 cm^3. -Darren === Subject: Re: Novice: Pi in other base systems > Hello... > My apologies if this is a silly question. How does pi behave in > different base systems, for instance binary, base 8 or base 16? > Is it still a mysterious number with endless non-repeating > numbers? Is there any base system where there's some especially > interesting results? > (I guess the correct way to ask this is how the ration of > circumference to diameter behaves in different number systems, > but still you get the point.) > Just curious. > Thx, > D. It is possible to compute the n-th digit of pi in hexadecimal --- and thus also in binary --- notation without computing all the previous digits. (I don't have the reference here right now, but I'm sure you will not be eventually repeating since pi is not rational. Hope this helps, Lasse --- === Subject: Re: Novice: Pi in other base systems In sci.math, piter <1jRrd.45$wy5.6266@news.uswest.net>: > Hello... > My apologies if this is a silly question. How does pi behave in > different base systems, for instance binary, base 8 or base 16? > Is it still a mysterious number with endless non-repeating > numbers? Is there any base system where there's some especially > interesting results? > (I guess the correct way to ask this is how the ration of > circumference to diameter behaves in different number systems, > but still you get the point.) > Just curious. > Thx, > D. The digits differ, but the nature of pi never changes. :-) However, this may interest you; it can be used to implement progressive digits in base 16. Look at equations (29), (31), and (32) in http://mathworld.wolfram.com/PiFormulas.html -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: JSH: James, you be the judge... Here is how James Harris operates: 1. Start with some goofy polynomial that was a leftover from one of his many failed FLT proofs. No explanation of motivation or reasoning provided for choosing that particular polynomial or explanation of why it might be meaningful in his new context. 2. Avoid being specific about such important details as which ring he intends to work in. 3. Dream up a bunch of weird non-standard terminology like properly unit, dividing off, etc. Use terms incorrectly, such as referring to multiple when he should use the term factor. Also, make use of standard terminology like distributive property and constant term without actually understanding it or using it properly. Also, add lots of vague distractions such as the equation has no memory, you can see the 7's in there, can't you? to act as a smoke screen. 4. Apply a bunch of algebraic manipulations, some of which are just wrong. Make inappropriate generalizations from specific cases to general cases, etc. 5. Get a result that seems contradictory, and assume that the error was with the core of algebra rather than the much more likely explanation that he himself might have made an error. 6. Through out a bunch of paranoid ad hominem attacks on critics. Then hypocritically claim that all his critics resort to social crap and personal attacks rather than sticking to mathematics, logic, and reasoning. We are supposed to believe that his arbitrarily chosen goofy polynomial just happens to one that, when probed by the genius of James Harris, give results that shake mathematics to its core! James, you be the judge. Are you a pile of crap? === Subject: Re: Smullyan's Quiz Problem http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Smullyan.html Looking at his history the exam question isn't paradoxical at all, but expected. unexpected has the connotation of Random: If you substitute at random for unexpected , it becomes an probability problem with an expectation distribution , and not a paradox or logical question anymore. It becomes equally probable that he will give the exam on the class days that week ( including the day he announces it). Here's an alternative: The Swedish civil defense paradox on the chalk board. He, then, announces he with give an unexpected exam at some time that week. The week passes and he gives no exam. At the first day of the next week when no one has responded to the phrase which he has left on the chalkboard, he announces that they all failed the exam: no one has told him what the The Swedish civil defense paradox is about. Dirty trick? Or a clever way to make the students pay better attention? >>This is the Swedish civil defense paradox: >>page 34Mathematical Fallacies and Paradoxes, Bryan Bunch, Dover books,1982 >>It was analyed by Williard Van Orman Quine in 1953 >>The analysis was that the 4th logical possiblity is best : >>The test will occur this week but it will be unexpected >>Martin Gardner also has a version of this in an Scientific American >>Bunch concludes that reasoning isn't enough to solve the paradox. >> >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > can't get the quiz on Friday because if I haven't gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? >Now if the professor had stated to his class on Friday, I will give >you a surprise examiniation some day next week, then there would be >inconsistency problems for all days of the week, but that's not the >way the problem is stated. >Comments anyone? > > >Agreed, the professor can give the exam on Monday and >it will indeed be unexpected. However, that really doesn't >deal with the paradox. >This paradox has been presented under many names: >the Swedish civil defense paradox, the prediction paradox, >the unexpected exam, the unexpected egg >to name a few. It gets my vote as the best paradox. >I have not seen a resoltuion that I find satisfactory. >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justified. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. > -William Hughes -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: Smullyan's Quiz Problem windows-nt) >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > can't get the quiz on Friday because if I haven't gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? > > One of the hypothesis is that an examination will be given during the > week. The argument shows that the hypothesis lead to a contradiction: > that no examination will be given during the week. That is, > H->not(H). But > > (H -> not(H)) -> not(H) > > is a tautology. That means that the only thing we can deduce from the > professor's statements are that his conditions will not be met. > > Since we can only conclude that his conditions will not be met, none > of the original argument showing that no test can occur is valid. The > test may occur at any time. >>Arturo, >>point is that, ON MONDAY, the professor stated You will not know that there is >>an examination when the class begins. When the Monday morning class began, the >>students had no idea that there was going to be a surprise examination that week >>since the professor had yet to make the announcement. Thus giving the exam on >>Monday would not contract his statement - it would be a surprise examination >>and the students won't have known that there is an examination when the >>Monday morning class began. > What makes you think that the announcement was made after class that > begun? In fact, one might argue that the announcement occurred before > class had started, as evidenced by the fact that the student tried to > use induction all the way to Monday. > Yes, that's being a bit persnickety, but that is math, after all. Let me also add (to my adjacent post) that this is a question of the problem definition. The definition of the problem should be clear. If it's not, let's make define it where it lacks clarity. THEN let's work on the solution. I really would like this point to be addressed and resolved before moving forward with discussion of the problem, otherwise we know not what we're discussing. -- % Randy Yates % Ticket to the moon, flight leaves here today %% Fuquay-Varina, NC % from Satellite 2 %%% 919-577-9882 % 'Ticket To The Moon' %%%% % *Time*, Electric Light Orchestra http://home.earthlink.net/~yatescr === Subject: Re: Smullyan's Quiz Problem days. My association with the Department is that of an alumnus. >Let me also add (to my adjacent post) that this is a question of the >problem definition. The definition of the problem should be clear. If >it's not, let's make define it where it lacks clarity. THEN let's work >on the solution. >I really would like this point to be addressed and resolved before moving >forward with discussion of the problem, otherwise we know not what we're >discussing. I think your argument is as follows: let us call the times when the test may be administered the admissible range of the statement by the professor. Let x be the time when the test begins. The admissible range is divided into days; think of them as being integers, so that floor(x) represents the beginning of the day at which the test begins. The professor's qualifying statement says something about the knowledge of the students at floor(x) (the start of the day) where x is the day in which the test will occur. The professor is making a statement at time t. The statement says: (1) x>t, and x is in the admissible range. (2) Something will be true at floor(x). Let us denote the admissible range by [Y,Z]. I agree with you that we have two different situations. In the classical Paradox of the Unexpected Hanging/Exam/Egg, we have floor(Y)>t. Your interpretation of Smullyan's presentation has floor(Y)t, all statements made are about future events; since the statements are about knowledge, the information provided at instant t may affect the situation at instant floor(Y). That is in fact what the student's argument is. On the other hand, if floor(Y)> One of the hypothesis is that an examination will be given during the >> week. The argument shows that the hypothesis lead to a contradiction: >> that no examination will be given during the week. That is, >> H->not(H). But >> (H -> not(H)) -> not(H) >> is a tautology. That means that the only thing we can deduce from the >> professor's statements are that his conditions will not be met. >> Since we can only conclude that his conditions will not be met, none >> of the original argument showing that no test can occur is valid. The >> test may occur at any time. >I agree with your analysis of the purely logical version of >the paradox. The professor can be said to be asserting a contradiction. >Thus, either the professor must be considered fallible, or >we have a truth teller asserting a contradiction. In this case >we can prove anything, (including the facts that a test on Wednesday >is both expected and unexpected). >However, I think that the paradox is much deeper than this. Consider >four statements the professor could make: > i) there will be an unexpected test tomorrow > ii) there will be an unxepected test in the next three days > iii) there will be an unexpected test next week > iv) there will be an unexpected test this semester >All four have the same logical structure. However, i) seems absurd, >ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >A full resoution of the paradox must explain the difference. The paradox here is literally that they contradict our intuition. But the only explanation needed is that ->our intuition is wrong<-. Just because some of them seem absurd and others do not does not mean that the statements actually are or are not wrong. These statements have a whole bunch of unstated assumptions. For example, I disagree with you that (i) seems absurd prima facie. It only becomes absurd if you assume that you know what time and what context such a test would take place in. The reason (iv) do not seem so absurd is that the latitude for those unknowns seems so much wider that you cannot lull yourself into a false sense of knowledge about the statement. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Smullyan's Quiz Problem > [.snip.] >> One of the hypothesis is that an examination will be given during the >> week. The argument shows that the hypothesis lead to a contradiction: >> that no examination will be given during the week. That is, >> H->not(H). But >> >> (H -> not(H)) -> not(H) >> >> is a tautology. That means that the only thing we can deduce from the >> professor's statements are that his conditions will not be met. >> >> Since we can only conclude that his conditions will not be met, none >> of the original argument showing that no test can occur is valid. The >> test may occur at any time. >> >I agree with your analysis of the purely logical version of >the paradox. The professor can be said to be asserting a contradiction. >Thus, either the professor must be considered fallible, or >we have a truth teller asserting a contradiction. In this case >we can prove anything, (including the facts that a test on Wednesday >is both expected and unexpected). >However, I think that the paradox is much deeper than this. Consider >four statements the professor could make: > i) there will be an unexpected test tomorrow > ii) there will be an unxepected test in the next three days > iii) there will be an unexpected test next week > iv) there will be an unexpected test this semester >All four have the same logical structure. However, i) seems absurd, >ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >A full resoution of the paradox must explain the difference. > The paradox here is literally that they contradict our intuition. Agreed, the arguments are now informal, so paradox has exactly this meaning. > But the only explanation needed is that ->our intuition is wrong<-. True, but the bald statement our intuintion is wrong is not very helpful. A full resolution of the paradox requires an explanation as to why our intuition is wrong >Just > because some of them seem absurd and others do not does not mean > that the statements actually are or are not wrong. > These statements have a whole bunch of unstated assumptions. For > example, I disagree with you that (i) seems absurd prima facie. It > only becomes absurd if you assume that you know what time and what > context such a test would take place in. Yes, there are ways to make (i) sensible (e.g. An unxepected exam is one on a different colour of paper), however, these do not seem sensible to me. If you take (i) to mean (and this seems to me the most obvious meaning) You will have an exam tomorrow but you don't know this, (i) is Bertram's paradox. > The reason (iv) do not seem > so absurd is that the latitude for those unknowns seems so much wider > that you cannot lull yourself into a false sense of knowledge about > the statement. Indeed. However, both the professor and students agree that information has been communicated, and appear to agree on what has been communicated. So the question's are: Do the two parties actually agree on what has been communicated? and If the professor's statement is self contradictory, what should he have said? -William Hughes === Subject: Re: Smullyan's Quiz Problem days. My association with the Department is that of an alumnus. [.snip.] >>I agree with your analysis of the purely logical version of >>the paradox. The professor can be said to be asserting a contradiction. >>Thus, either the professor must be considered fallible, or >>we have a truth teller asserting a contradiction. In this case >>we can prove anything, (including the facts that a test on Wednesday >>is both expected and unexpected). >>However, I think that the paradox is much deeper than this. Consider >>four statements the professor could make: >> i) there will be an unexpected test tomorrow >> ii) there will be an unxepected test in the next three days >> iii) there will be an unexpected test next week >> iv) there will be an unexpected test this semester >>All four have the same logical structure. However, i) seems absurd, >>ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >>A full resoution of the paradox must explain the difference. >> The paradox here is literally that they contradict our intuition. >Agreed, the arguments are now informal, so paradox has >exactly this meaning. It should also be added that there is a fair amount of the Wilde-like paradox in this; you know, the The only thing worse than being talked about is not being talked about kind of statements, since we argue that if we expect the test then we don't expect the test (the old I got you by not getting you argument and variants thereof). >> But the only explanation needed is that ->our intuition is wrong<-. >True, but the bald statement our intuintion is wrong is not >very helpful. A full resolution of the paradox requires an >explanation as to why our intuition is wrong I think out intuition is wrong because we are making a lot of unstated assumptions about these statements. As I noted: there will be an unexpected test tomorrow seems absurd when we assume that 'unexpected' means something like 'I won't know exactly when', but that knowing the ->day<- of the exam narrows the window sufficiently for us to figure out exactly when. The other three statements don't lull us into that assumption because, prima facie, we have more than one option and no information about which option we should take. But that (i) if we know the day then we know when; and (ii) if we don't know which day in advance then we don't know exactly when; are common enough conclusions which are both unwarranted in this situation. >>Just >> because some of them seem absurd and others do not does not mean >> that the statements actually are or are not wrong. >> These statements have a whole bunch of unstated assumptions. For >> example, I disagree with you that (i) seems absurd prima facie. It >> only becomes absurd if you assume that you know what time and what >> context such a test would take place in. >Yes, there are ways to make (i) sensible (e.g. An unxepected exam >is one on a different colour of paper), however, these do not >seem sensible to me. I'm not even saying that. You can even make it sensible if unexpected refers to you won't know exactly when; I teach at three different hours; which hour will contain the exam? > If you take (i) to mean (and this seems >to me the most obvious meaning) You will have an exam tomorrow but >you don't know this, (i) is Bertram's paradox. And here we are making another assumption about what unexpected means. Does it mean, you do not expect an exam at all? Or does it mean, you won't know exactly when? It is in the latter sense that the Paradox of the Unexpected Hanging uses it, for example. >> The reason (iv) do not seem >> so absurd is that the latitude for those unknowns seems so much wider >> that you cannot lull yourself into a false sense of knowledge about >> the statement. >Indeed. However, both the professor and students agree >that information has been communicated, and appear to agree on >what has been communicated. Which, as we know, in the real world is a virtual guarantee that it is not true that what the professor meant to communicate is the communication that the students have received... (-: > So the question's are: Do the two >parties actually agree on what has been communicated? and >If the professor's statement is self contradictory, what should >he have said? Do you mean, how can he make a statement which communicates the same information but is not self-contradictory? You may be begging the question there, as in fact the point is that you cannot communicate that information without also communicating a contradiction. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Pi in space I was having an argument with a friend here and he claimed that pi doesn't really have all this significance it's attached to it, since it is transcendental only in the theoretical space of Euclidian geometry and all know that this geometry does not represent true space-time. Is pi transcendental in ANY space except in the theoretical space of Euclidian geometry? For example, is it transcendental in Riemannian geometry? Lobatchevskian geometry? or any other geometry? What about in the relativistic universe? Is the ratio of the circumference of a circle to its diameter inside the Einsteinian universe rational or irrational? Anyone knows? -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesn't really have all this significance it's attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? We don't know the true geometry of space. In particular, we don't know if it's continuous. Digital physicists think it is not: see www.digitalphysics.org The concept of Pi, however, should not be regarded transcendental or anything like that since it is represented perfectly with a small program. True, it is an idealization, but what is transcendental? In my opinion, nothing transcends physical reality, and if Pi is part of physical models, it is for a good reason: physics is mostly about ideal models that work in ideal conditions, etc. Note however, that, if the universe turns out to be discrete, then, obviously, the real number Pi does not describe any physical reality. The computation of Pi is still sensible, however, (e.g. computation of Pi up to the nth decimal place) and it would indeed describe physical reality in a discrete universe (if it's an actually infinite discrete universe, things get more complicated so just assume finite which seems to be the case for *our* universe;). Most mathematicians on this forum seem to be Platonists, or Platonists who lack the philosophical maturity to identify themselves as Platonists, so you could expect quite a lot of Platonist nonsense in response to your question. However, as one poster correctly pointed out, constructivists and formalists will most likely look at Pi as a useful mental construct of some sort, and no talk of transcending anything will be necessary. Pi is not in space: it's in your mind. [*] Indeed, constructivism is a more favorable position than Platonic Realism, in my opinion, and a point of view supported by great mathematicians, so it's an alternative you should think about. More recently, instrumentalism has been suggested as an alternative way to look at mathematics. In my opinion, that too is considerable. -- Eray Ozkural There is no perfect circle [*] A caveat, however, there are perfect discrete circles in the world. It just may be the case that there are no perfect continuous circles, except conceptually!!!!!! === Subject: Re: Pi in space |Most mathematicians on this forum seem to be Platonists, or Platonists |who lack the philosophical maturity to identify themselves as |Platonists, so you could expect quite a lot of Platonist nonsense in |response to your question. No, because such points of philosophy are thoroughly irrelevant to his question. |However, as one poster correctly pointed |out, constructivists and formalists will most likely look at Pi as a |useful mental construct of some sort, and no talk of transcending |anything will be necessary. Pi is not in space: it's in your mind. [*] I am disinclined to assume that you know that transcendental is a technical mathematical term meaning not a root of a nonzero polynomial with integer coefficients. |Indeed, constructivism is a more favorable position than Platonic |Realism, in my opinion, and a point of view supported by great |mathematicians, so it's an alternative you should think about. Kronecker, Brouwer, and Bishop were all outstanding mathematicians (whether we call them great sort of depends on how high a standard we set for greatness). Bishop supported constructivism; Brouwer supported some kind of constructivism; Kronecker appears to have supported something along those lines. I wouldn't say, however, that popularity among great mathematicians is a very good standard for judging such things. To the extent that it can be trusted at all, I see no obvious sign of the very best mathematicians differing very far in terms of the distribution of their opinions on Platonism (or realism), formalism, constructivism and so on from the run-of-the-mill. I also don't see sci.math as being much of a hotbed of Platonism or realism. Keith Ramsay === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. I think you need to look up the definition of transcendental number. You seem to be confusing it with some philosophical notion. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. > I think you need to look up the definition of transcendental number. > You seem to be confusing it with some philosophical notion. the concept of a transcendental number. AFAICT, the notion of a transcendental number is not relative to a give geometry, or the true geometry of our universe. Pi is a transcendental number, according to the definition of a transcendental number in mathematics. I do think his question ought to be philosophical. Why should he talk about relativity then? What difference is there between any continuous metric and a Riemann tensor regarding the fact that Pi is a transcendental number? Maybe, I think, his friend's intention was to give a better definition of what it means for a number to be transcendental than the ordinary usage. He might want to pick another term, though, or highlight the difference than the ordinary usage carefully. Otherwise, we become confused in argumentation. I thought so, and said that this philosophical sense is probably irrelevant to a computable real like Pi which captures a general geometric fact in a compact form! -- Eray Ozkural === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesn't really have all this significance it's attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. We know nothing whether space-time really exists and if it exists whether it has a given geometry of its geometry can be a matter of convention. You are asking metaphysical (ontological) questions that have no answers at this point. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? The geometries have relative consistency to Euclidean geometry. This, pi must be transcendental in all of them to maintain relative consistency. > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? You can draw a circle even on your pixelized TFT screen, consider it a perfect one and use pi to find its perimeter. I get the impression you are asking about the relation of abstract mathematical object to the actual universe. We do not know it. That's another metaphysical subject. Platonists would argue circles are really part of the ontology of space-time and Universe whereas Formalists and Constructivists will look at circles as useful geometrical models with no real existence apart from human mind. Mike === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi doesn't >really have all this significance it's attached to it, since it is >transcendental only in the theoretical space of Euclidian geometry and all >know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the circumference > of a circle to its diameter inside the Einsteinian universe rational or > irrational? Anyone knows? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ Defining pi to be the empirically measured ratio of a circle's circumference to its diameter, the only geometry in which this is ratio is a constant is Euclidean geometry. Pi doesn't exist as a constant that can be empirically measured in any other geometry, so its value in other geometries is a meaningless question. Of course, defining pi as the sum of a series, or the constant in a solution to a differential equation or proper integral, or any of the other billion ways of defining pi has absolutely nothing to do with the geometry of the Universe. So there is only one value of pi, and yes, it is transcendental. === Subject: Re: Pi in space |Defining pi to be the empirically measured ratio of a circle's circumference |to its diameter, the only geometry in which this is ratio is a constant is |Euclidean geometry. Pi doesn't exist as a constant that can be empirically |measured in any other geometry, so its value in other geometries is a |meaningless question. Obviously this is not a practical issue, but I thought as long as we were busy making points about what is true in principle, I would note that there often is a natural relationship between pi and other geometries. In some other geometries, pi is the limit of the ratio between the circumference and the diameter of a circle as the radius goes to 0, and could in principle (if you had such a space in your hands...) be measured that way. At a fixed radius, measuring the circumference and diameter would only be able to approximate pi up to a certain degree of precision, but measurements of lengths are always only up to a certain degree of precision anyway, so one is not any worse of in an essential way. On the hyperbolic plane, for example, you could find a length scale on which a triangle whose sides have lengths in a 3:4:5 ratio has an angle of between 90 and 91 degrees. One can then put an upper bound on the deviation of the ratio of the circumference to the diameter from pi, in terms of the ratio of the radius to the length of a side of this triangle. The circumference of a circle of radius r in the hyperbolic plane is 2*pi*a*sinh(r/a) where the length a depends on the curvature. The size of the 3:4:5 triangle with an angle of 91 degrees is likewise proportional to this length a. (I won't bother figuring out what the ratio is right here.) So to get n digits, we could make a circle whose radius is roughly 10^{-n/3} times the length of a side of this triangle, and measure it very precisely. Keith Ramsay === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi doesn't >really have all this significance it's attached to it, since it is >transcendental only in the theoretical space of Euclidian geometry and all >know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the circumference > of a circle to its diameter inside the Einsteinian universe rational or > irrational? Anyone knows? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ > Defining pi to be the empirically measured ratio of a circle's circumference > to its diameter, the only geometry in which this is ratio is a constant is > Euclidean geometry. It is not possible to empirically measure any irrational in any geometry, if by that you mean physical measurement. And if not, what do you mean? === Subject: Re: Pi in space >>I was having an argument with a friend here and he claimed that pi >>doesn't >>really have all this significance it's attached to it, since it is >>transcendental only in the theoretical space of Euclidian geometry and >>all >>know that this geometry does not represent true space-time. >> Is pi transcendental in ANY space except in the theoretical space of >> Euclidian geometry? >> For example, is it transcendental in Riemannian geometry? >> Lobatchevskian >> geometry? or any other geometry? >> What about in the relativistic universe? Is the ratio of the >> circumference >> of a circle to its diameter inside the Einsteinian universe rational or >> irrational? Anyone knows? >> -- >> I. N. G. --- http://users.forthnet.gr/ath/jgal/ >> Defining pi to be the empirically measured ratio of a circle's >> circumference >> to its diameter, the only geometry in which this is ratio is a constant >> is >> Euclidean geometry. > It is not possible to empirically measure any irrational in any > geometry, if by that you mean physical measurement. Sure it is. You can measure the diagonal of unit square just as easily as you can measure the length of one side. The fact that one is rational and the other irrational doesn't affect your ability to measure them. > And if not, what do > you mean? The reason that I talk about empirical measurement is assertion implicit in the OP that if the universe is not flat, then pi does not have the value everybody thinks it does. The only sense in which this is sort-of true is that if you measure the circumference of a circle and divide it by the diameter, then you will not get pi as we know it as the answer. I used the term empirically to suggest the act of phyiscally measuring (for example with a tape measure) circumferences and diameters of circles embedded in that space. === Subject: Re: Pi in space (1) Pi is a constant. It's value is 4 ATAN(1) regardless of whether space is Euclidean. Your question is like asking whether the number 1 remains rational in non-Euclidean space. A constant is a constant is a constant. (2) we all know that this geometry..... Who is we? We certainly does NOT include all because the recent super-nova evidence that revealed the existence of dark energy also shows that the universe is Euclidean. (3) Pi is transcendental in ALL geometries. You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: Pi in space ETAtAhUAuKZDfmVeTIVHWzPAf+1RbygxYcACFGivLjq2oYFNZgzALqH4m/0aKNiF Gee, the last time I looked, the transcendentality of a number was INDEPENDENT of geometry, being a property of the number itself. Whether pi, or other famously nonconstructible numbers like the cube root of 2 or the sine of 20 degrees, become constructible in non-Euclidean geometries is a different question. --OL === Subject: Re: Pi in space >(1) Pi is a constant. It's value is 4 ATAN(1) regardless >of whether space is Euclidean. But wasn't there a part of the muppets show called Pi's in space? Thomas >Your question is like asking whether the number 1 >remains rational in non-Euclidean space. >A constant is a constant is a constant. >(2) we all know that this geometry..... >Who is we? We certainly does NOT include >all because the recent super-nova evidence that >revealed the existence of dark energy also shows >that the universe is Euclidean. >(3) Pi is transcendental in ALL geometries. >You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: Pi in space > (1) Pi is a constant. It's value is 4 ATAN(1) regardless > of whether space is Euclidean. > Your question is like asking whether the number 1 > remains rational in non-Euclidean space. > A constant is a constant is a constant. Damn. I was somehow hoping that from context my question would be clear. Apparently it was not. Let me repeat it more accurately phrased: Define pi=Length of circumference/Length of diameter of a circle however the later terms are defined in the appropriate geometry we are talking about. In which geometries pi is transcendental? -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space >> (1) Pi is a constant. It's value is 4 ATAN(1) regardless >> of whether space is Euclidean. >> Your question is like asking whether the number 1 >> remains rational in non-Euclidean space. >> A constant is a constant is a constant. > Damn. I was somehow hoping that from context my question would be clear. > Apparently it was not. Let me repeat it more accurately phrased: > Define pi=Length of circumference/Length of diameter of a circle > however the later terms are defined in the appropriate geometry we are > talking about. > In which geometries pi is transcendental? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ As I explained below, the ratio of a circle's circumference to its radius is only a constant in Euclidean space. In non-Euclidean geometries, the ratio depends upon the size of the circle; there is no single value of pi. Consider, for example, the space as consisting of the surface of a sphere as big as the earth. If you measue the ratio of the circumference to the diameter of a circle that is only a few inches across, the ratio will be very close to 3.14159265... If you measure the ratio for the equator, you get pi=2 (the radius is the distance from the North Pole to the equator, which is one quarter the circumference). So in non-Euclidean geometries, the ratio of a circles circumference to its diameter can take on a wide range of values, depending upon the size of the circle. As the size of the circle approaches zero, the ratio approaches pi. So as the ratio can take on a continuous range of values, you can make it transcendental or not, simply by selecting the size of the circle you use to form the ratio. Which is, of course, different to saying pi varies - in all geometries, pi is the same; only the measured ratios change. === Subject: Re: Pi in space > As I explained below, the ratio of a circle's circumference to its radius is > only a constant in Euclidean space. In non-Euclidean geometries, the ratio > depends upon the size of the circle; there is no single value of pi. > Consider, for example, the space as consisting of the surface of a sphere as > big as the earth. If you measue the ratio of the circumference to the > diameter of a circle that is only a few inches across, the ratio will be > very close to 3.14159265... If you measure the ratio for the equator, you > get pi=2 (the radius is the distance from the North Pole to the equator, > which is one quarter the circumference). > So in non-Euclidean geometries, the ratio of a circles circumference to its > diameter can take on a wide range of values, depending upon the size of the > circle. As the size of the circle approaches zero, the ratio approaches pi. > So as the ratio can take on a continuous range of values, you can make it > transcendental or not, simply by selecting the size of the circle you use to > form the ratio. > Which is, of course, different to saying pi varies - in all geometries, pi > is the same; only the measured ratios change. changes. That would be ridiculous. I was talking about the ratio. Serves me right: If I cannot phrase a question correctly, then I end up having all the group's farts on my back :-( -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space <1102142838.264150@athnrd02 Define pi=Length of circumference/Length of diameter of a circle > however the later terms are defined in the appropriate geometry we are > talking about. > In which geometries pi is transcendental? Consider R^2 with the discrete metric and the circle at (0,0) with radius 1. Computing the circumference and dividing it by the diameter 2, I find pi, the ratio circumference/2, to be transcending. ;-) === Subject: Re: Pi in space > Your question is like asking whether the number 1 > remains rational in non-Euclidean space. Well, does it? DOES IT? How come you mathematicians never answer the IMPORTANT questions? === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi >doesn't really have all this significance it's attached to it, since it >is transcendental only in the theoretical space of Euclidian geometry >and all know that this geometry does not represent true space-time. Everything after since is incoherent. >Is pi transcendental in ANY space except in the theoretical space of >Euclidian geometry? >For example, is it transcendental in Riemannian geometry? Lobatchevskian >geometry? or any other geometry? >What about in the relativistic universe? Is the ratio of the >circumference of a circle to its diameter inside the Einsteinian >universe rational or irrational? Anyone knows? ...As are all those questions. Lee Rudolph === Subject: Re: JSH: Two sides > I've had months to wonder why mathematicians would work to ignore or > hide my results, Ignoring your results takes no effort, hiding them does not occur. -- 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: Two sides [snip delusion] What does properly a unit mean? Can you answer that simple question James Harris? === Subject: Re: JSH: Two sides > [snip delusion] > What does properly a unit mean? > Can you answer that simple question James Harris? JSH does not answer questions. === Subject: Re: Why exp(-st) in the Laplace Transform? Does anyone have an explaination why the kernel function exp(-st) was >>used in the definition of the Laplace transform? >>Is there a physical meaning to the use of this function? >>I know s is a complex frequency, how can we visualize what the Laplace >>transform is doing? > I believe the answer is no, there's no interpretation of what the > Laplace transform really means, analogous to the way one interprets, > say, the Fourier transform. The kernels are not orthogonal in any > obvious sense... > It's hard to know for certain that no is the correct answer, of > course. But I've never seen an answer of the sort I think you're > looking for, I've thought about it and never found one, and > I once posted more or less the same question to sci.math, and > had a few smart people agree the answer was no. Arthur Mattuck, an MIT professor, ''interprets'' the Laplace transform as a continuous analog of a power series expansion of a function. You can watch a video of him lecturing on this point at Click on Lecture 19. T. Monroe === Subject: Re: Why exp(-st) in the Laplace Transform? > heard something somewhat similar tho - that laplace transforms are the >> continuous generalization of power series. I can't recall exactly how >> the explanation went, it was in an ODE book by a guy named Simmons. > Perhaps you have seen something like this. > Let F(p) = integral exp(-px) f(x) dx (: Laplace Transform) > ==> F(0) = integral f(x) dx = area beneath f(x) > And F'(p) = integral (-x) exp(-px) f(x) dx > ==> F'(0) = - integral x f(x) dx = - first order moment > / center of gravity / midpoint > And F''(p) = integral x^2 exp(-px) f(x) dx > ==> F''(0) = integral x^2 f(x) dx = second order moment > / moment of inertia / variance > And then you have F(p) = F(0) + p.F'(0) + p^2/2.F''(0) + ... > Interpretation: subsequent terms of the series expansion of the Laplace > Transform involve an area, a midpoint, a moment of intertia .. In short: > the most important (physical/global) characteristics of a function come > first, when considered in the Laplace domain. > Han de Bruijn Arthur Mattuck, an MIT professor, ''interprets'' the Laplace transform as a continuous analog of a power series expansion of a function. You can watch a video of him lecturing on this point at Click on Lecture 19. T. Monroe === Subject: Re: Why exp(-st) in the Laplace Transform? >> heard something somewhat similar tho - that laplace transforms are the >> continuous generalization of power series. I can't recall exactly how >> the explanation went, it was in an ODE book by a guy named Simmons. > Perhaps you have seen something like this. > Let F(p) = integral exp(-px) f(x) dx (: Laplace Transform) > ==> F(0) = integral f(x) dx = area beneath f(x) > And F'(p) = integral (-x) exp(-px) f(x) dx > ==> F'(0) = - integral x f(x) dx = - first order moment > / center of gravity / midpoint > And F''(p) = integral x^2 exp(-px) f(x) dx > ==> F''(0) = integral x^2 f(x) dx = second order moment > / moment of inertia / variance > And then you have F(p) = F(0) + p.F'(0) + p^2/2.F''(0) + ... > Interpretation: subsequent terms of the series expansion of the Laplace > Transform involve an area, a midpoint, a moment of intertia .. In short: > the most important (physical/global) characteristics of a function come > first, when considered in the Laplace domain. > Han de Bruijn > Arthur Mattuck, an MIT professor, ''interprets'' the Laplace transform as a > continuous analog of a power series expansion of a function. You can watch > a video of him lecturing on this point at > Click on Lecture 19. > T. Monroe Excellent pointer. Dirk Vdm === Subject: Re: Gravitomagnetism > Ok, you don't need to respond to this comment, > but I have no major problem with non-integer > weight if you allow non-integer dimensionality. > > Non-integer? Don't understand this.. > > Well you know from SR as you move an object faster > and faster it's length contracts and it's time > slows down, and at c these dimensions vanish. > So in the intermediary, between v=0 and v=c, > can we say definitely that we have a fixed integer > dimensionality? > > To put that on a more rational mathematical foundation > one can integrate, (no integration constants) > > $ x dx = (1/2) x^2 (area) > > $$ x dx dx = (1/6) x^3 (volume) > > which shows how integration generates dimensions. > Well conformal weight is a concept that is independent of > dimensionality, that is, a covariant in a Weyl conformal space has > properties under both coordinate and scale changes. The idea of weight > embodies the latter. to check that out, but I think you are right. You know, moving from field to quantum physics where are those integers. > GR is quite clear, G_uv = T_uv is where I'll begin. > If I understand you correctly, you state > vanishing gravity (implying G_uv =>0) and > *inhomogeneous* electrodynamics > > (implying T(Maxwell)_uv =/=0 > > are compatible? > Yes! That is what is so surprising. Nowhere is a current posited - > there *is* no RHS here. In GR, as in Maxwell-Lorentz, one has a > field that is driven by a *posited* current (energy tensor, charge > current resp.), and then that field acts back on the current. Here, > there is *no* posited current at all, rather, the assumption of > strictly local metricity requires both the metric and calibration > field, and these *jointly* assume roles in a manner that appears as > symmetric Ricci driven by energy-momentum, and Maxwell driven by > charge-current! Of course they are really 6-d *vacuum* equations: > Rmn - (2R/W) Tmn + (1/2W) (DmDn + DnDm) W = 0 (Rmn = symmetric part > of CCT) I have a problem, I can read that in the usual 4d terms, but when you uprate to 6d are your mn over 4 or 6, ah...your pushin the envelope so please tell us your encryption. > 1/S d/dxm ( S R Fmn ) - 5/4 (Dn W) = 0 (S = sqrt det g) > The new physics would be found in the pure geometry terms 1/2W > {Dm,Dn} W and (Dn W), which are respectively, energy-momentum and > charge current. These are absent in flat space absent in flat space, that's what Einstein said when he doubled his prediction for the deflection of light. Would you be able to provide how your equations reduce to GR in the the weak field, that way giving us ameans to connect GRist's with the physicality of your 6D. > Yes, I'm trying to make the simplest possible metric > consistent with a nonorthogonal space, i.e. > > g_uv = g*g_uv + A_u B_v > ^ ^ > calibration EM antisymmetry => A_u B_v = - A_v B_u > (Weyl=>gauge) (Einstein) > > where det g = 1 - AB, (that's a bit crude, but close). > > Ken S. Tucker What paper? ken > -drl === Subject: Re: Probability Of Ultimate Extinction. >Let's say that you have an amoeba which has a probability q of dying without >divinding, >and probability p of spliting into two. What is the probability of ultimate >extinction? >I know that the recursion formula for extinction after n cycles is >P(n)=q+p*P(n-1)^2 where P(0)=q. You have not said what a cycle is, but assusming p+q=1 then if you are looking for the probability of ultimate extinction, u, then it is the probability of going extinct without dividing plus the probability that it divides and both halves go extinct: u = q +p*u^2 >of 3 or 4) is that >the probability of ultimate extinction involves using the quadratic formula >in the following way. >1±(1-4*p*q)^0.5 >--------------- You should be dividing by 2p not 2q. >where the minimum of the two values achieved by the quadratic equation is >the desired probability. >What I'd like to know is, why bother? I've discovered that in the above >problem, the probability >of ultimate extinction is p/q = 1 if p>=q. Why bother with the quadratic >formula? Perhaps you have swapped p and q somewhere If p+q=1 then the quadratic gives the lower of (1-p)/p and 1, depending on whether p>=1/2 or p<=1/2 which makes sense and in the latter case confirms what is obvious. === Subject: JSH: Good news, and bad news I'll give the good news first. Despite the venting I do at times in frustration, the reality is that I have an easygoing nature. So I thought I'd point out a few things along with that for any mathematicians who might wander by and read this post to consider. First of all, my research is not limiting but expanding. What I've found are new tools for deeper explorations into the properties of numbers. Given that mathematics is difficult for many people, it's unlikely that the status quo in mainstream math society would change too dramatically once the full story is out. Sure I'd no longer be a crank, but top mathematicians probably wouldn't move much. So let's say you are Barry Mazur, or Wiles or Taylor, and you find out that there is actually this problem with algebraic number theory that you learned and a LOT of what you thought you had proven, was not proven. So what? A lot of it you can go ahead and prove anyway in less space, without error, using the more advanced techniques. but very large work, you can move to a proof of a few pages. And if Wiles cares more about seeing the proof of Fermat's Last Theorem than the glory of being the person who found that proof, then it's gravy. Sure, it's embarrassing that the underlying theory was wrong, but the mistake entered the discipline over a hundred years ago, long before anyone now living was even born. If the desire is to see a proof of Fermat's Last Theore, then no problem! That's the good news. If the desire is the glory of having found a proof of Fermat's Last Theorem. Problem. So it's a measure of the men, as they are mostly men, how they react. The good news is that not much will change in many ways if they accept what is mathematically true. The bad news is that if they react only when forced, then clearly they were in it for the glory. And if glory hounds willing to fight against the truth, they get no pity. So, one way, not much changes. Professorships remain. There is just this amazing story of which they are a part. The other way, much can change, much can be lost, and a world can be severely disappointed, when the full story eventually comes out. The choice is yours. James Harris === Subject: Re: Good news, and bad news > The good news is that not much will change in many ways if they accept > what is mathematically true. Good, indeed. News, hardly. Have you ever read the fineprint? >The choice is yours. You have one claim (trivial) to a prime counting algorithm, and another (childish) denying basic algebra. At most one of them could stand. The choice is yours - is the bottle half full, half empty or fully empty? Barry === Subject: Re: Good news, and bad news Mr. Harris, Your level of delusion seems to grow without bound! The amount of crap spewage that you are propagating is beyond any level of decent mathematical discussion. You are a phony, a liar, a cheat and an ignoramus. Even if you managed to stumble into something in mathematics that actually made sense while in one of your drunken states, no one would give a damn! You just don't get it, do you? === Subject: Re: Good news, and bad news > What I've found are new tools for deeper explorations into the > properties of numbers. Yes, you found ing great new tools for deeper explorations... Right... And you did manage to proved FLT using slick 17th century style algebra, so why not move onto the Riemann Hypothesis with your new power tools? That is the next big one, right? I just know you can do it with your new deep math exploration tools. After all, it is a great challenge. It is kind off to analysis what FLT is to arithmetic, right. True, you have fought hard, and won your deservedg lory for great achievement with FLT. But why rest on your laurels? Move on you dumb little ........ Or was FLT just appealing to you only because the statement of the problem was so simple that you thought you had a clue about it. You figure you can follow the arithmetic because the problem statement looks like high school stuff, but you don't know head from your ass beyond high school, right. Well, that is probably why cranks like you always stick to FLT. === Subject: Re: Good news, and bad news > but very large work, you can move to a proof of a few pages. You often now make the claim that Wile's FLT proof is wrong. Can you point out the error he made? Did he make an error in determining those properly units thingy-ma-jiggies? If you make this claim and you cannot back it up, then why are we to believe you? === Subject: Re: Good news, and bad news > but very large work, you can move to a proof of a few pages. > You often now make the claim that Wile's FLT proof is wrong. Moreover, Wiles did not prove FLT directly, he proved a certain conjecture (the conjecture being that every rational elliptic curve is some kind of equivalent to a modular form), for a limited number of elliptic curves, namely for the semi-stable elliptic curves. By earlier work from Frey and Ribet this was sufficient to show that FLT was true. The full conjecture was proved true later by Conrad et al. Because JSH apparently attacks Wiles' proof, he should also consider the proof by Breuil, Conrad, Diamond and Taylor for the full conjecture... Or he should attack the work by Frey, of that by Ribet, or whatever. -- 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: Good news, and bad news > You often now make the claim that Wile's FLT proof is wrong. ... OOOOPS I mean Wiles', not Wile's... === Subject: Re: JSH: Good news, and bad news > I'll give the good news first. Despite the venting I do at times in > frustration, the reality is that I have an easygoing nature. So I > thought I'd point out a few things along with that for any > mathematicians who might wander by and read this post to consider. > First of all, my research is not limiting but expanding. That would be the bad news then. === Subject: Re: JSH: Good news, and bad news >I'll give the good news first. Despite the venting I do at times in >frustration, the reality is that I have an easygoing nature. Right. > So I >thought I'd point out a few things along with that for any >mathematicians who might wander by and read this post to consider. >First of all, my research is not limiting but expanding. >What I've found are new tools for deeper explorations into the >properties of numbers. >Given that mathematics is difficult for many people, it's unlikely >that the status quo in mainstream math society would change too >dramatically once the full story is out. Sure I'd no longer be a >crank, but top mathematicians probably wouldn't move much. >So let's say you are Barry Mazur, or Wiles or Taylor, and you find out >that there is actually this problem with algebraic number theory that >you learned and a LOT of what you thought you had proven, was not >proven. >So what? A lot of it you can go ahead and prove anyway in less space, >without error, using the more advanced techniques. >but very large work, you can move to a proof of a few pages. And if >Wiles cares more about seeing the proof of Fermat's Last Theorem than >the glory of being the person who found that proof, then it's gravy. Actually it's clear that you haven't been paying attention. Fermat's Last Theorem is _false_! E. E. Escultura has posted counterexamples right here in sci.math. I think the two of you need to get together and work this out. Because you're both light-years past anything that the rest of us can comprehend, but you can't both be right. Check out his posts. Contact the guy, and write back as soon as the two of you agree on which one of the two is wrong. >Sure, it's embarrassing that the underlying theory was wrong, but the >mistake entered the discipline over a hundred years ago, long before >anyone now living was even born. >If the desire is to see a proof of Fermat's Last Theore, then no >problem! >That's the good news. >If the desire is the glory of having found a proof of Fermat's Last >Theorem. >Problem. >So it's a measure of the men, as they are mostly men, how they react. >The good news is that not much will change in many ways if they accept >what is mathematically true. >The bad news is that if they react only when forced, then clearly they >were in it for the glory. And if glory hounds willing to fight >against the truth, they get no pity. >So, one way, not much changes. Professorships remain. There is just >this amazing story of which they are a part. >The other way, much can change, much can be lost, and a world can be >severely disappointed, when the full story eventually comes out. >The choice is yours. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Good news, and bad news >but very large work, you can move to a proof of a few pages. And if >Wiles cares more about seeing the proof of Fermat's Last Theorem than >the glory of being the person who found that proof, then it's gravy. > Actually it's clear that you haven't been paying attention. > Fermat's Last Theorem is _false_! > E. E. Escultura has posted counterexamples right here in sci.math. > I think the two of you need to get together and work this out. > Because you're both light-years past anything that the rest of > us can comprehend, but you can't both be right. > ^^^^^^^^^^^^^^^^^^^^^^^ Oh, that is so retrograde of you. Surely their mathematical skills are advanced enough that they can prove Fermat's Last Theorem AND its negation. Understand now? === Subject: Re: JSH: Good news, and bad news you're an easygoer in what respect ... til you were found in a cyberspace, and became marginally surlier. I missed the bad news, other than what the last-prior poster suggested. > Understand now? --Advice, 0.05; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: Good news, and bad news > What I've found are new tools for deeper explorations into the > properties of numbers. Name or describe one new tool for deeper exploration that you have found. All you have done is claim that there is a deep problem in algebraic integers, and as far as that goes, you have convinced a grand total of zero converts. You never explained what that problem is. You never showed any results of what that problem are. And you have certainly never found any new tools for mathematicians to work with! === Subject: Re: JSH: Good news, and bad news !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > I'll give the good news first. Despite the venting I do at times in > frustration, the reality is that I have an easygoing nature. Well, the _real_ good news is that the bad news turns out as hilariously absurd as the good news. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Simple proof > The poster Nora Baron is not even a girl. That poster is a guy, > playing a girl. A perfect match for yourself -- as I discovered when I found the anagram of your name which reveals your true identity: James S. Harris = Ms. J. Harrie Ass (note the gender) -- 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: Simple proof > The poster Nora Baron is not even a girl. That poster is a guy, > playing a girl. Why does that bother you so much, James? Is it too reminiscent of your dating experiences? -- 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: Simple proof > Quit with the drama. I'm not interested in histrionics from posters > like you. > I can create all the drama I won't by myself. > ^^^^^ > Dammit, James, you just ruined a perfect .sig. He's not adopted an explicitly stream-of-consciousness style before, but there's a first time for everything: perhaps this is to be read as I can create all the drama [if I want to]; [but] I won't by myself [so let's keep playing] -- Larry Lard Replies to group please === Subject: Re: JSH: Simple proof <87fz2nbuq4.fsf@phiwumbda.org> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Quit with the drama. I'm not interested in histrionics from posters >> like you. >> I can create all the drama I won't by myself. >> ^^^^^ >> Dammit, James, you just ruined a perfect .sig. > He's not adopted an explicitly stream-of-consciousness style before, > but there's a first time for everything: perhaps this is to be read > as > I can create all the drama [if I want to]; [but] I won't by myself > [so let's keep playing] Uh, what's this stream-of-conscious nonsense? His sentence makes perfect sense without any such tags. It is equivalent to I have the ability to also create by myself all the drama which I actually am not going to create. So that simply means that his potential for drama is not restricted at all, even though he might refrain from some drama deliberately. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: induction vs Cantor defines a function which given an infinite list of reals (that is, > an element of R^N) returns an element of R that is not on the list. > Cantor didn't prove that there were uncountably many such functions, > he only proved that there was one. That's all he needs to be able > to show that no list of reals contains every real. Do you think there are any counter-examples to Cantor's diagonal proof? If so, can you provide a valid modification that keeps the concept of providing a real that is proven not in the list and has no counter-example? === Subject: Re: induction vs Cantor > defines a function which given an infinite list of reals (that is, > an element of R^N) returns an element of R that is not on the list. > Cantor didn't prove that there were uncountably many such functions, > he only proved that there was one. That's all he needs to be able > to show that no list of reals contains every real. > Do you think there are any counter-examples to Cantor's diagonal > proof? No. > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? NA. === Subject: Re: induction vs Cantor > ... >... > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? > NA. There exists a countable model of your ZF set. === Subject: Re: induction vs Cantor >> ... >>... >> If so, can you provide a valid modification that keeps the concept >> of providing a real that is proven not in the list and has no >> counter-example? >> NA. >There exists a countable model of your ZF set. Sure, but inside such models it is still the case that |N| < |R|. Since |N| < |R| follows from the ZF axioms, it must hold in all models that satisfy those axioms, regardless of the size of the model (as viewed from the outside). -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: induction vs Cantor >> ... >>... >> If so, can you provide a valid modification that keeps the concept >> of providing a real that is proven not in the list and has no >> counter-example? >> NA. >There exists a countable model of your ZF set. > Sure, but inside such models it is still the case that |N| < |R|. Since |N| > < |R| follows from the ZF axioms, it must hold in all models that satisfy > those axioms, regardless of the size of the model (as viewed from the > outside). Hi Barb, That leads to a problem. The set N, the natural numbers, in any of those extensions, contains no elements not contained in each other copy of the set N, and it contains each of them. The copies are equal and identical. There is thus identity, a tautology, between any copies of N, and as well between any copies of P(N), Then, if there is a bijection between N and P(N), there is a bijection between N and P(N). theory bogosity is a reaction to Skolem, who otherwise says infinite sets are equivalent. Look outside. Do you use transfinite cardinals? Is it for anything besides transfinite cardinals? Some people use them as a definition as part of measure theory, but that's because they lack better tools, and the results of measure theory are largely upon continua. That is to say, the useful results of measure theory are derivable without the use of transfinite cardinals, and they should be. Of your ordinals, is there any that specifically represents the integer -1? Is not the order type of the powerset of the naturals the successor of the order type of the naturals? If it's infinite there's always one more. That's part of the basis of induction, that something can be proven to hold for each, thus that for each it is proven, as opposed to proved. When we get to the contradiction between each infinite set being equivalent inductively and the necessity of dual representation via Cantor's results, then accept that there is dual representation instead of that induction fails. That way, you're on the path to being inside first order logic, and nobody can prove you incomplete, inconsistent, and immaterial. If you want a consistent theory, start by removing the inconsistencies, not reusing them. Ross F. === Subject: Re: induction vs Cantor > ... >... > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? NA. >>There exists a countable model of your ZF set. >> Sure, but inside such models it is still the case that |N| < |R|. Since |N| >> < |R| follows from the ZF axioms, it must hold in all models that satisfy >> those axioms, regardless of the size of the model (as viewed from the >> outside). >Hi Barb, >That leads to a problem. >The set N, the natural numbers, in any of those extensions, contains no >elements not contained in each other copy of the set N, and it contains each of >them. The copies are equal and identical. >There is thus identity, a tautology, between any copies of N, I agree. N (via von Neumann ordinals) is categorical in ZF, so the N in different models would be isomorphic. >and as well between any copies of P(N), That looks like a serious misstep. AIUI (which admittedly isn't very deeply), the downward LST gives a model in which all the elements correspond to first-order-definable terms. So, the set of such elements of P(N) in the LST countable model is very much smaller than P(N) in the usual iterative-hierarchy models (which includes the uncountably-many elements which are not first-order definable). (It's rather like the difference between the set of computable infinite binary sequences and the set of all infinite binary sequences computable or not.) But |N| < |P(N)| even in the countable model, because the bijection between N and the model's P(N) is not itself first-order definable and therefore does not exist in the model! ISTM you have strong Constructivist leanings. That's fine, but it's important then not to mix constructive and non-constructive entities in the same argument. If you exclude all non-constructive entities then you can't argue that |N| = |P(N)| in the countable model, because the isomorphism would be non-constructive and therefore not talkable-about. > Then, if there is a bijection between N and P(N), >there is a bijection between N and P(N). >theory bogosity is a reaction to Skolem, who otherwise says infinite sets are >equivalent. I'm sure he doesn't actually say or imply that. >Look outside. OK. >Do you use transfinite cardinals? I don't see any outside, but then again I don't see ANY mathematical entities outside -- there are no natural numbers flitting through the trees. >Is it for anything besides transfinite cardinals? Sure. One needs at least P(N) for representing the real numbers when putting Analysis on a set-theory foundation. >Some people use them as a definition as part of measure theory, but that's >because they lack better tools, and the results of measure theory are largely >upon continua. That is to say, the useful results of measure theory are >derivable without the use of transfinite cardinals, and they should be. Giving up continua is asking a lot. Continuum Mechanics is a useful area with many practical applications. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: induction vs Cantor; nonstandard numbers in ZF >>... >> If so, can you provide a valid modification that keeps the concept >> of providing a real that is proven not in the list and has no >> counter-example? >> NA. There exists a countable model of your ZF set. > Sure, but inside such models it is still the case that |N| < |R|. Since |N| > < |R| follows from the ZF axioms, it must hold in all models that satisfy > those axioms, regardless of the size of the model (as viewed from the > outside). >>Hi Barb, >>That leads to a problem. >>The set N, the natural numbers, in any of those extensions, contains >>no elements not contained in each other copy of the set N, and it >>contains each of them. The copies are equal and identical. >>There is thus identity, a tautology, between any copies of N, > I agree. N (via von Neumann ordinals) is categorical in ZF, so the N > in different models would be isomorphic. That can't be right....*thinking*....nope. It's not right. You can just run familiar compactness arguments. In ZF we can define the finite von Neumann ordinals in a variety of ways, where define here means that we have a formula Nat(x) that is true of exactly the real finite von Neumann ordinals *in a standard model of ZF*. For instance, let Ord be your favorite definition of the von Neumann ordinals .9fberhaupt, say Ord(x) =df Trans(x) and Connected(x), and define x to be a natural number just in case it and all its elements are either 0 or successor ordinals: Nat(x) =df (y)(y=x v y in x -> (y=0 v (exists z)(y=zU{z}))). TeXnically speaking, we usually call the set {x | Nat(x)} omega, of course. But now add the constant k to the language of ZF and consider the following set of sentences K = {Nat(k), k != 0, k != 1, k != 2, ...}. For any finite subset K' of K, ZF U K' is obviously satisfiable (if ZF is). Hence, by compactness, so is ZF U K. But obviously omega (that is, the set containing exactly the elements satisfying Nat(x)) in any model of ZF U K can't be isomorphic to N. And any such model, of course, can be turned into a model of ZF by restricting it to the original language of ZF. The problem, of course, ZF being first-order and all, is that you can't rule out the possibility nonstandard finite ordinals that, e.g., are transitive in the model but not in the real world. I reckon, in fact that the sets playing the role of omega in models of ZF U K probably have properties similar if not identical to the usual nonstandard models of PA. Who's an expert around here? Herb Enderton, Mike Oliver, Aatu K, etc should be able to tell us more. Chris Menzel === Subject: Re: induction vs Cantor <8JqdnQxnYLcA6zrcRVn-3w@giganews.com> <41a7dbeb$0$20860$afc38c87@news.optusnet.com.au> <1fRrd.107121$T02.77638@twister.rdc-kc.rr.com> <41B40665.2DB637D2@tiki-lounge.com> <41B4DED8.947CA9BB@tiki-lounge.com> Hi Barb, Yeah, it is. The reals are great. Their simple assumption as a continuum offers a necessary tool for the pursuit of many and much mathematics. Most of the fundamental results about real numbers are in place without set theory, eg via Euclid. The (set of) real numbers is the indefinite contiguous sequence, the point set. You use zero, and iota, and integral multiples of iota, and those are all the non-negative real numbers. Mapping the integers to the reals in this way escapes the consequences of Cantor's first proof, which some apply to the rational numbers, and I to neither. Zorn's Lemma, the well-ordering principle, or the Axiom of Choice allow methods to guarantee enumerability of the sets. Well-order the reals and inductively select elements to inject into the integers, for the integers, the existence of integer n guarantees the existence of integer n+1. I agree that I'm constructivist, but I'm more quasi-intuitionist and not prejudiced. Also I'm a platonist because math is real. Math can be quite irrelevant. The natural numbers are the trees. Do you use ZF with classes? What's the class of all classes? If your answer is no, can there be more than one proper class? There are theories with a set of all sets, that set is its own powerset, there the identity and tautology is between the set and itself, its own powerset. That's for intuitionism to resolve, the singular excluded excluded middle and the double entendre. Double entendre: a double entendre. The proper class is the ur-element, it's at once all and nothing, and the void and null. Its dualistic nature is its singular nature, and vice versa. Ross F. === Subject: Re: induction vs Cantor <8JqdnQxnYLcA6zrcRVn-3w@giganews.com> <41a7dbeb$0$20860$afc38c87@news.optusnet.com.au> <1fRrd.107121$T02.77638@twister.rdc-kc.rr.com> <41B40665.2DB637D2@tiki-lounge.com> <41B4DED8.947CA9BB@tiki-lounge.com> RF> The reals are great. How the would YOU know THAT? You don't even know a basic definition of what they are. RF> Their simple assumption as a continuum There is nothing simple about it TO YOU, idiot. To the rest of us, there is a simple first-order axiomatization, but it is hardly a simple assumption as a continuum; it is an assumption that they are an ordered field that is complete in the sense of containing all of the things you can produce from it by the operation of taking limits of COUNTABLY infinite sequences of things already in it. RF> offers a necessary tool for the pursuit of many and much mathematics. RF> Most of the fundamental results about real numbers are in place RF> without set theory, Yes, there are axioms defining the reals without mentioning sets. RF> eg via Euclid. To prove that Euclid is an example of that, you would have to give an example FROM Euclid of some results about reals. RF> The (set of) real numbers is the indefinite contiguous sequence, In modern set theories, you CAN'T just gloss over that in parentheses: you have to PICK A SIDE: is the class of all reals proper, or is it a set???? RF> the point set. You use zero, and iota, and integral multiples of iota, RF> and those are all the non-negative real numbers. It is a consequence of the fact that first-order languages are countable that you can model any consistent first-order theory countably, so if there are denumerably many iota-terms, yes, there is a way you can make them serve as a model. BUt that does NOT mean that they really are the reals. RF> Mapping the integers to the reals in this way escapes the RF> consequences of Cantor's first proof, which some apply RF> to the rational numbers, and I to neither. NOthing escapes the proof. No matter how many iota's you use to represent the reals, they will STILL all have denumerably-long-bit-string representations as well, and all you have to do to reproduce the proof in the iota- context is define the function that returns the nth decimal place (or bit) of a real as output, give the right number of iotas as input. RF> Zorn's Lemma, the well-ordering principle, RF> or the Axiom of Choice allow RF> methods to guarantee enumerability of the sets. They DO NOT, dumbass. All these things are NON- constructive. You do get an enumeration or well-ordering from them but you NEVER get a METHOD! RF> Well-order the reals and inductively select elements to RF> inject into the integers, Obviously, this is not possible; you run out of natnums. RF> for the integers, the existence of integer n guarantees RF> the existence of integer n+1. And the existence of a term with n iota's guarantees the existence of a term with n+1 iota's. But nothing guarantees that you can inject the reals into either of those. RF> Do you use ZF with classes? NObody uses ZF with classes. ZF DOES NOT HAVE classes. In ZF, EVERYTHING IN THE ENTIRE UNIVERSE is a set. Proper classes DO NOT EXIST. RF> What's the class of all classes? A contradiction in terms, THAT'S what it is. RF> There are theories with a set of all sets, that set is its own RF> powerset, there the identity and tautology is between RF> the set and itself, its own powerset. You haven't actually studied any set theories with a universal set and you DON'T know WTF you are talking about. All these theories have some VERY counter-intuitive consequences at very simple levels. === Subject: Re: induction vs Cantor <8JqdnQxnYLcA6zrcRVn-3w@giganews.com> <41a7dbeb$0$20860$afc38c87@news.optusnet.com.au> <1fRrd.107121$T02.77638@twister.rdc-kc.rr.com> <41B40665.2DB637D2@tiki-lounge.com> <41B4DED8.947CA9BB@tiki-lounge.com> The reals are great. > How the would YOU know THAT? > You don't even know a basic definition of what they are. > RF> Their simple assumption as a continuum > There is nothing simple about it TO YOU, idiot. > To the rest of us, there is a simple first-order > axiomatization, but it is hardly a simple assumption > as a continuum; it is an assumption that they are > an ordered field that is complete in the sense of containing > all of the things you can produce from it by the operation of > taking limits of COUNTABLY infinite sequences of things > already in it. Erm, for every cauchy sequence... or for every dedikind cut... is second order. The first order theory of the real field is just the theory of real closed fields axiomatised via ordered field + square roots of positive elements + roots of odd degree polynomials. === Subject: Re: induction vs Cantor Poker Joker says... >Cantor's diagonalization procedure >> defines a function which given an infinite list of reals (that is, >> an element of R^N) returns an element of R that is not on the list. >> Cantor didn't prove that there were uncountably many such functions, >> he only proved that there was one. That's all he needs to be able >> to show that no list of reals contains every real. >Do you think there are any counter-examples to Cantor's diagonal >proof? What do you mean by a counter-example? >If so, can you provide a valid modification that keeps the concept >of providing a real that is proven not in the list and has no >counter-example? Sorry, I don't know what you mean by counterexample. -- Daryl McCullough Ithaca, NY === Subject: Re: induction vs Cantor > Poker Joker says... >> what I have called the Cantor function from R^N to R. >Describe this class of functions and show how Cantor used them in his >proof as you say. I'm looking for where he classified them as R^N to R >and that they were uncountable. > I think Virgil means the class of functions f which given > an infinite list of reals returns a new real that is not > on the list. Precisely! > Cantor came up with only one such function: his diagonalization > function. R^N just means the set of functions from N to R, or > equivalently, the set of infinite lists of reals, or a set of > reals indexed by the naturals. Cantor's diagonalization procedure > defines a function which given an infinite list of reals (that is, > an element of R^N) returns an element of R that is not on the list. > Cantor didn't prove that there were uncountably many such functions, > he only proved that there was one. That's all he needs to be able > to show that no list of reals contains every real. > It is easy enough to show that the cardinality of the set of such > functions is uncountable, although this fact is not used in Cantor's > proof. > -- > Daryl McCullough > Ithaca, NY === Subject: JSH: Simply fascinating The math here is so readily understandable that it's actually almost as interesting watching how people react to it, as anything else. For instance, I've given a polynomial P(x) repeatedly where I factor it into three factors. I point out that the factors must include factors of the constant term of the polynomial P(x). I note that the factors of the constant term are independent of x, as they are, in fact, constant. Mathematically it's easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the polynomial results from multiplying together the three factors which I've called g_1(x), g_2(x), and g_3(x), then you can get the factors of the constant term, by setting x=0. It's that simple. Notice (1) you have the factors of P(x), (2) the constant term of P(x) is determinable by setting x=0, (3) you also get the factors of the constant term. Mathematically, it's simple to the point of trivial. Now then, if you have *constants* which are factors of another constant then why would anyone try to argue that they are actually variables? You have x, as the variable. Besides x there are just these numbers. If you clear out x, then what's left are constants. Letting x=0, clears it out, leaving the constants visible. Some may say, yeah, sure, at x=0, but what about when x doesn't equal 0? Um, if the numbers are constant, and so are independent of x, then, duh, why should it matter what value x has? The logic is inescapable. In terms of difficulty, my proof is about as easy as it gets in algebraic number theory, in terms of the actual mathematics. But the concepts are where there is a problem, and the social hang-up is in accepting that there's this simple technique that shows a BIG problem, which can invalidate claims of proof for, well, over a hundred plus years. So the mathemtics is EASY for a trained mathematician to follow. The social implications are hard, if social stuff is important to you, and clearly from what I've seen it is to many of you. For instance, at this point I've removed all objections raised in detail. Like I can explain supposed counter examples to my work. I can give an actual example where you can see the factorization play out--just as the theory shows it must. And you probably know that my research is the work that can be said to have gone to a journal which at least claims it does formal peer review. They thanked me for the paper said the reviewers liked it, and then some sci.math'ers got together--actually literally conspiring online in posts on sci.math--sent them emails and the editors yanked my papers THE NEXT DAY. They had it for nine months. I'd corresponded with them for a while, even corrected them when they called me Dr. Harris as I don't have a Ph.d, and I told them I was an independent researcher with concerns about how my work would be handled. They kept saying no problem, ok, all that matters is what's correct. Then they yanked my paper after sci.math'ers emailed them: All the pertinent facts are in my favor. So what's the hold-up? My research shows that some mistakes were made over a hundred years ago, and a lot of people missed them, and gave proofs which were not, and are not proofs. Mathematics is unforgiving. It doesn't care about the social implications of the truth. So it doesn't matter mathematically that a LOT of people out there are terribly dependent on the false beliefs and incorrect results, but it DOES matter a lot to those people! I call their behavior passive-aggessive, as by dragging their feet, taking as long as possible before acknowledging my research, or worse, hoping to NEVER acknowledge it at all, they are passively hoping to escape mathematical truth, in what amounts to a very aggressive way. I liken their behavior to judges at a race, who watch a runner break a world record, and then lie about his time, refuse to admit he even finished the race, and some even call him names!!! They're turning the way it's supposed to work with a major discovery, upside down. And it's silly behavior as eventually the truth will come out, and you know what I'll do then? Probably go to the beach. I'll also hang out in some bars. Yup, I'll definitely hang out in some bars, preferably near a beach. Yup, you guessed it, I'll do my best to forget about them, as why bother worrying about silly people who do silly things. Life's too short. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Simply fascinating > I've given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). In general that's not true at all. For example, consider: (x+1)*(x+2)*(x+3) If you expand it out to get a polynomial in the usual form, you see that the constant term is 6. But 6 is not a factor of any of the three polymomial factors, in fact neither 2 nor 3 is a factor of any of them say what you really meant to say? Or maybe you were totally mistaken and need to just retract what you said. > My research shows that some mistakes were made over a hundred years > ago, and a lot of people missed them, and gave proofs which were > not, and are not proofs. It's true that some mistakes were made long ago, the worst in my opinion being the first developers of set theory who concocted a set of all sets, which was subsequently proven not to exist. But I've seen no evidence that *you* have ever done any research that showed any such past mistakes of others. Instead all I see is *you* making mistakes yourself and not admitting them. Do you know how to factor 7 in the ring of integers with sqrt(7) adjoined? That's easy, of course you do, right? But do you know how to factor 7 in the ring of integers with sqrt(2) adjoined? Now if you can figure out that simple arithmetic problem, here's something more interesting: Find all primes p (other than 2 and 7 which I've already shown you) such that 7 can be nontrivially factored in the ring of integers with sqrt(p) adjoined. Finally, find all finite sets of two or more primes p1,p2,...,pn such that 7 is composite in the ring you get when you adjoin all the sqrts of those primes but it's prime in the ring you get when you adjoin all but one of those sqrts. Same question if you allow adjoining sqrt(-1) in addition to adjoining sqrt of a prime. (Note: When I refer to a prime p or primes p1,p2..., I mean prime in the ring of integers. When I refer to 7 being prime or composite, I mean prime or composite in the ring of integers adjoined with the various sqrts.) === Subject: Re: JSH: Simply fascinating |It's true that some mistakes were made long ago, the worst in my |opinion being the first developers of set theory who concocted a set |of all sets, which was subsequently proven not to exist. As far as I know, this incident involved Frege and nobody else. If you have other developers of set theory in mind, who were they? Second, the set of all sets was only proven not to exist under some additional assumptions. The problem with Frege's system was that it implied there exists a Russell set, R = {x : x is not a member of x}. The existence of a Russell set is directly contradictory. There are systems of set theory that have a universal set and are consistent (although they are less used). Frege didn't directly postulate a universal set or a Russell set; he just stated axioms that he regarded as intuitive, that together implied that for any predicate P there exists a set {x : x satisfies P}. Keith Ramsay === Subject: Re: JSH: Simply fascinating > The problem with Frege's system was that it implied there exists a > Russell set, R = {x : x is not a member of x}. By what grammatical axiom(s) is the right side of that equation even a well-formed formula? I would guess two such grammatical axioms: If S and T are WFFs denoting sets, then S is a member of T is also a WFF denoting a truthvalue. If P is a WFF denoting a truthvalue, then not P is also a WFF denoting a truthvalue. x is not a member of x is shorthand for not (x is a member of x). If x is a WFF variable, and P is a boolean-valued WFF with x as the only free variable, then {x : P} is a WFF denoting a set. I personally accept all those grammatical axioms except the last as reasonable. The last I would replace with: If S is a WFF denoting a set, and x is a WFF variable, and P is a boolean-valued WFF with x as the only free variable, then {x in S : P} is a WFF denoting a set. The reason I don't like the original last grammatical axiom is that it begs the question what kind of Universe we're dealing with, whereas my alternative clearly says S is the universe we're dealing with. If we beg the question what Universe we're dealing with, then we aren't really making a definition, we aren't saying what is allowed to be in a set and what isn't allowed to be in a set, so we aren't saying whether some monstrosity in somebody's fantasy can be in our set or not. > There are systems of set theory that have a universal set and are > consistent (although they are less used). Do you know a WebPage that describes any of these axiomatic systems? Alternately, do you remember which commonsense axioms of set-theory grammar are *not* included, thereby avoiding writing a Russell set as a WFF in said system? === Subject: Re: JSH: Simply fascinating Discussion, linux) >> The problem with Frege's system was that it implied there exists a >> Russell set, R = {x : x is not a member of x}. > By what grammatical axiom(s) is the right side of that equation even a > well-formed formula? I would guess two such grammatical axioms: ^^^^^^^ term, not formula. > If S and T are WFFs denoting sets, > then S is a member of T is also a WFF denoting a truthvalue. > If P is a WFF denoting a truthvalue, then not P is also a WFF > denoting a truthvalue. > x is not a member of x is shorthand for not (x is a member of x). > If x is a WFF variable, > and P is a boolean-valued WFF with x as the only free variable, > then {x : P} is a WFF denoting a set. > I personally accept all those grammatical axioms except the last as > reasonable. The last I would replace with: > If S is a WFF denoting a set, and x is a WFF variable, > and P is a boolean-valued WFF with x as the only free variable, > then {x in S : P} is a WFF denoting a set. > The reason I don't like the original last grammatical axiom is that > it begs the question what kind of Universe we're dealing with, > whereas my alternative clearly says S is the universe we're dealing ^^^^^^^^^^^^^^ your alternative? You're very influential. > with. If we beg the question what Universe we're dealing with, then > we aren't really making a definition, we aren't saying what is > allowed to be in a set and what isn't allowed to be in a set, so we > aren't saying whether some monstrosity in somebody's fantasy can be > in our set or not. -- Who knows, maybe that may be the only way to settle this crap. It's not like it'd be that hard for me to go back and get a math degree. I can penetrate the math social group and then finish the takedown from inside. -- James S. Harris contemplates a new strategy. === Subject: Re: JSH: Simply fascinating > I've given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > In general that's not true at all. For example, consider: > (x+1)*(x+2)*(x+3) But he said, e.g. in your example, having factored your polynomial into (x+1), (x+2), and (x+3), 1, 2, and 3 must be factors of 6, the constant term of your expanded polynomial. > If you expand it out to get a polynomial in the usual form, you see > that the constant term is 6. But 6 is not a factor of any of the three > polymomial factors, in fact neither 2 nor 3 is a factor of any of them He didn't say, with respect to your example, that 6 must be a factor of 1, 2, and 3. Rather that 1, 2, and 3 must be factors of 6. KeithK > say what you really meant to say? Or maybe you were totally mistaken > and need to just retract what you said. > My research shows that some mistakes were made over a hundred years > ago, and a lot of people missed them, and gave proofs which were > not, and are not proofs. > It's true that some mistakes were made long ago, the worst in my > opinion being the first developers of set theory who concocted a set > of all sets, which was subsequently proven not to exist. But I've seen > no evidence that *you* have ever done any research that showed any such > past mistakes of others. Instead all I see is *you* making mistakes > yourself and not admitting them. > Do you know how to factor 7 in the ring of integers with sqrt(7) > adjoined? That's easy, of course you do, right? But do you know how to > factor 7 in the ring of integers with sqrt(2) adjoined? Now if you can > figure out that simple arithmetic problem, here's something more > interesting: Find all primes p (other than 2 and 7 which I've already > shown you) such that 7 can be nontrivially factored in the ring of > integers with sqrt(p) adjoined. Finally, find all finite sets of two or > more primes p1,p2,...,pn such that 7 is composite in the ring you get > when you adjoin all the sqrts of those primes but it's prime in the > ring you get when you adjoin all but one of those sqrts. Same question > if you allow adjoining sqrt(-1) in addition to adjoining sqrt of a > prime. (Note: When I refer to a prime p or primes p1,p2..., I mean > prime in the ring of integers. When I refer to 7 being prime or > composite, I mean prime or composite in the ring of integers adjoined > with the various sqrts.) === Subject: Re: JSH: Simply fascinating > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? If they are independent of 'x', why did you have to set 'x' to zero to uncover them? Evaluating a univariate polynomial at any numeric value of the variable produces a numeric result. Are all these results 'constants'? > The logic is inescapable. > Life's too short. I think yours is too long. -- 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: Simply fascinating > The math here is so readily understandable that it's actually almost > as interesting watching how people react to it, as anything else. > For instance, I've given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > I note that the factors of the constant term are independent of x, as > they are, in fact, constant. > Mathematically it's easy to show: > g_1(x) g_2(x) g_3(x) = P(x) > and > g_1(0) g_2(0) g_3(0) = P(0) > as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > I've called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. > It's that simple. Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. > Mathematically, it's simple to the point of trivial. Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? No one does. If as you say, you assume that your factors are constant, of course they are not variables. That is nothing but a dimwit tautology. And of course that is not what we say. We say that if you factor 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) by those variable functions, then you can obtain algebraic integer factors on both sides of the resulting equation. That is, g_1(x)/w_1(x), etc., are all algebraic integers. That is all you require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). No problem with that either. And the product of these three constant terms equals P(0)/49. It all works out. Factoring out 49 as a product of variable algebraic integer functions does not lead to ANY contradiction. It is only when you try to use a CONSTANT factorization that you arrive at what you think is a problem with algebraic integers. If you do the factorization in the right way there is no such problem. Suppose z is an integer variable, and you consider Q(z) = (x + 5)(x + 6). You notice that Q(z) is alway an even integer. Therefore you can always divide it by 2. You notice that when z = 0, Q(z) = Q(0) = 5 * 6. You notice that Q(0)/2 = 5 * (6/2). That is, when you divide by 2 to get an integer, you divide the constant term of the second factor, 6, by 2. If you tried to divide 5 by 2 you would not have an integer quotient. So by your logic, Q(z)/2 = (z + 5)*(z/2 + 6/2) = (z + 5)*(z/2 + 3) would be the only right, CONSTANT way to factor out 2. This is analogous to your division by 49 = 7*7*1: (5 a_1(x)/7 + 7/7)(5 a_2(x)/7 + 7/7)(5 b_3(x)/1 + 22/1) But, back to Q(z)/2 = (z + 5)(z/2 + 3). The problem is, z/2 is not always an integer. If you want integer quotients in both factors, you have to divide out 2 in a NONCONSTANT way. When z is even, you divide 2 out of (z + 6). When z is odd, you divide 2 out of (z + 5). Get it? Factoring 2 out in a constant fashion does not work. Factoring 2 out in a *variable* fashion does work. No problem arises with constant terms. > You have x, as the variable. Besides x there are just these numbers. > If you clear out x, then what's left are constants. Letting x=0, > clears it out, leaving the constants visible. > Some may say, yeah, sure, at x=0, but what about when x doesn't equal > 0? Some may say. > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? You MUST divide by a variable factorization of 49. If you assume a constant factorization you get into trouble, because a_1(x)/7 in general is not an algebraic integer. You know this, yet you persist in thinking that the factorization has to be constant. You conclude *not* that your own thinking is wrong (as you should) but that there is something fundamentally wrong with algebraic number theory. You try to justify using the constant 7*7*1 factorization by repeating your mantra that the constant terms must be constant. This is based on your incorrect conclusion that, if you divide (a_1(x) + 7) by w_1(x), then the constant term is 7/w_1(x). And right there is your central mistake. BY YOUR OWN DEFINTION, the constant term of (a_1(x) + 7)/w_1(x) is not 7/w_1(x). It is instead 7/w_1(0). You want to conclude that 7/w_1(x) must be equal to the constant, 7, because 1 * 1 * 22 = 22, the constant term of P(x)/49. If you do the *correct factorization*, you still get 22, but it is in the form (7/w_1(x)) * (7/w_2(x)) * (22/w_3(x)) = 22 because w_1(x)*w_2(x)*w_3(x) = 49. Now, you may howl, YES BUT 22/w_3(x) IS NOT AN ALGEBRAIC INTEGER!!! To which the reply is : yes, you're right. BUT THERE IS NO REASON IT SHOULD BE. All that is required is that (5 a_3(x) + 7)/w_3(x) is an algebraic integer, and this follows from the correct choice of w_1(x), w_2(x), and w_3(x). As it turns out, both 5 a_3(x)/w_3(x) AND 7/w_3(x) are algebraic integers. And as above: the constant term of (5 a_3(x) + 7)/w_3(x) is NOT 22/w_3(x), as you believe. It is merely 22/w_3(0) = 22. These two are not the same, unless you assume what you want to prove, i.e., that w_3(x) is a constant function. And you know it is not. Everything works out. There is no contradiction or problem involving the constant terms. It is YOU who has been trying to claim that the constant terms are not constant, e.g., when you say that 22/w_3(x) is the constant term of (5 a_3(x) + 7)/w_3(x). You are making such a rookie mistake. Why can't you see it? > The logic is inescapable. True enough. > In terms of difficulty, my proof is about as easy as it gets in > algebraic number theory, in terms of the actual mathematics. Easy, yes. Correct, no. > But the concepts are where there is a problem, and the social [tiresome pompous rant about social factors etc. deleted] Nora B. > Life's too short. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that it's actually almost > as interesting watching how people react to it, as anything else. > > For instance, I've given a polynomial P(x) repeatedly where I factor > it into three factors. > > I point out that the factors must include factors of the constant term > of the polynomial P(x). > > I note that the factors of the constant term are independent of x, as > they are, in fact, constant. > > Mathematically it's easy to show: > > g_1(x) g_2(x) g_3(x) = P(x) > > and > > g_1(0) g_2(0) g_3(0) = P(0) > > as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > I've called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. > > It's that simple. > > Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. > > Mathematically, it's simple to the point of trivial. > > Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? > > No one does. If as you say, you assume that your factors are > constant, of course they are not variables. That is nothing but > a dimwit tautology. > And of course that is not what we say. We say that if you factor > 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), > and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) > by those variable functions, then you can obtain algebraic integer > factors on both sides of the resulting equation. That is, > g_1(x)/w_1(x), etc., are all algebraic integers. That is all you > require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and > g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to > g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). > No problem with that either. And the product of these three > constant terms equals P(0)/49. Ok, I'm going to answer the Nora Baron poster yet again. And I want readers to understand that I've replied to this poster who you know lies anyway as it's a guy posting as a woman using a name that's a palindrome MANY TIMES explaining in detail. What happens is that when I shoot down these objections, the poster either just repeats later, or replies with nonsense. One telling time when I carefully refuted point-by-point the poster replied deleting out everything I'd said. Just deleted out everything, and you know what? I STILL so people replying about how supposedly I don't answer the objections from Nora Baron. It's partly a game for some people on sci.math, I'm sure, and partly a case where many readers are hoodwinked as *they* don't know it's a game. They seem incapable of realizing that there are people in this world willing to behave in such a way, on such a level. So why reply to this poster? Maybe I'm hopelessly naive, but I just have to believe that eventually people will just get tired of being made fools of by this poster and others in the group with him (or her). It hasn't happened yet from what I've seen, but I keep hoping. Ok, so what's wrong with the poster's assertion? Well, the w's the posters gives are factors of 49. But 49 comes into the picture because 49 is a multiple of the polynomial. But the w's STILL REMAIN after 49 has been divided off--after the multiple is divided off--as the poster actually claims. (Notice Nora Baron gives you clues, but somehow sci.math readers don't seem to get it.) Notice the poster's actually gives that when you divide off 49, you get g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x) which are TRACES, left of 49--after it has been divided off. So here we have mathematics. Here in an area where the truth can actually be determined, a poster who you can't be sure is a guy or a girl has confused many of you into questioning whether or not a multiple of a polynomial divides off as a variable or not. Did it ever occur to any of you that for some people that might be a great lark? Are you so trusting as to not consider that a poster who otherwise would probably be unknown can get off on confusing you not only on his or her gender, but on some of the most basic concepts in mathematics? Look now, even after the post where Nora Baron ended with a male name, people are STILL supporting the poster! It has occurred to me that this poster did so deliberately, ended that post with a male name, because the person knew you better than you still seem to know yourselves. In a way it's sad as you're giving up so much for nothing in return. James Harris === Subject: Re: JSH: Simply fascinating >> The math here is so readily understandable that it's actually almost >> as interesting watching how people react to it, as anything else. >> >> For instance, I've given a polynomial P(x) repeatedly where I factor >> it into three factors. >> >> I point out that the factors must include factors of the constant term >> of the polynomial P(x). >> >> I note that the factors of the constant term are independent of x, as >> they are, in fact, constant. >> >> Mathematically it's easy to show: >> >> g_1(x) g_2(x) g_3(x) = P(x) >> >> and >> >> g_1(0) g_2(0) g_3(0) = P(0) >> >> as P(0) gives the constant term of the polynomial, and since the >> polynomial results from multiplying together the three factors which >> I've called g_1(x), g_2(x), and g_3(x), then you can get the factors >> of the constant term, by setting x=0. >> >> It's that simple. >> >> Yes. Right so far. Trivial, but right. >> Notice (1) you have the factors of P(x), (2) the constant term of P(x) >> is determinable by setting x=0, (3) you also get the factors of the >> constant term. >> >> Mathematically, it's simple to the point of trivial. >> >> Correct. >> Now then, if you have *constants* which are factors of another >> constant then why would anyone try to argue that they are actually >> variables? >> >> No one does. If as you say, you assume that your factors are >> constant, of course they are not variables. That is nothing but >> a dimwit tautology. >> And of course that is not what we say. We say that if you factor >> 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), >> and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) >> by those variable functions, then you can obtain algebraic integer >> factors on both sides of the resulting equation. That is, >> g_1(x)/w_1(x), etc., are all algebraic integers. That is all you >> require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and >> g_3(x)/w_3(x) are BY YOUR OWN DEFINITION equal to >> g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_3(0). [a couple of misprints corrected here] >> No problem with that either. And the product of these three >> constant terms equals P(0)/49. >Ok, I'm going to answer the Nora Baron poster yet again. >And I want readers to understand that I've replied to this poster who >you know lies anyway as it's a guy posting as a woman using a name >that's a palindrome MANY TIMES explaining in detail. >What happens is that when I shoot down these objections, the poster >either just repeats later, or replies with nonsense. One telling time >when I carefully refuted point-by-point the poster replied deleting >out everything I'd said. Because it was the SOS, repeated for the nth time. I replied with a complete detailed and rigorous proof. You were enraged but never really provided any refutation. >Just deleted out everything, and you know what? I STILL so people >replying about how supposedly I don't answer the objections from Nora >Baron. You STILL so people replying ??? Really! >It's partly a game for some people on sci.math, I'm sure, and partly a >case where many readers are hoodwinked as *they* don't know it's a >game. They seem incapable of realizing that there are people in this >world willing to behave in such a way, on such a level. >So why reply to this poster? Maybe I'm hopelessly naive, but I just >have to believe that eventually people will just get tired of being >made fools of by this poster and others in the group with him (or >her). It hasn't happened yet from what I've seen, but I keep hoping. Whine, whine, whine. Why not go straight to the math rather and skip the propaganda ? >Ok, so what's wrong with the poster's assertion? >Well, the w's the posters gives are factors of 49. But 49 comes into >the picture because 49 is a multiple of the polynomial. But the w's >STILL REMAIN after 49 has been divided off--after the multiple is >divided off--as the poster actually claims. I claim no such thing. The product of the w's is 49. You divide P(x) by 49. You divide the product of the g's by the product of the w's. The 49 is gone from both sides. Note that here g_1(x) = (5 a_1(x) + 7), g_2(x) = (5 a_2(x) + 7), and g_3(x) = (5 a_3(x) + 7) = (5 b_3(x) + 22), where a_1, a_2, a_3 are all algebraic integer functions of x, and b_3(x) = a_3(x) - 3. Note that a_1(0) = a_2(0) = 0, and a_3(0) = 3. Also, note that w_1(0) = w_2(0) = 7, and w_3(0) = 1. When I divide the product of the g's by the product of the w's, the result is (5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x)), where c_i(x) = a_i(x)/w_i(x) and d_i(x) = 7/w_i(x). Here is what Mr. Harris is squawking about. The product of the d's is 7. (That happens to be true for any x.) Harris thinks that is a problem, because it then looks like the product of the constant terms of (5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x)) is 7, whereas the constant term of P(x)/49 is 22. That is, d_1(x)*d_2(x)*d_3(x) = 7, whereas P(0)/49 = 22. Puzzling! Is Harris right? I will now give my deceptive, lying, sleight-of-hand explanation. Watch very carefully, because as Mr. Harris says, I am going to try very hard to mislead you! So here is the explanation. The d's are NOT the constant terms of the factors that contain them. That is, d_1(x) is NOT the constant term of (5 c_1(x) + d_1(x)). Mr. Harris thinks it is. Certainly it is in the right position for being a constant term. Since Mr. Harris understands and mostly relies on only high-school level mathematics, he detects constant terms by *inspection*. The irony here is, he has provided a perfectly good, rigorous definition of constant term of a function, but he doesn't use it. It is the value that the function takes on when the argument is 0. Thus the constant term of (5 c_1(x) + d_1(x)) is NOT d_1(x) = 7/w_1(x). It is instead d_1(0) = 7/w_1(0) = 1. This is directly from Mr. Harris's own definition of constant term. Why is he so reluctant to apply that definition? Why does he instead rely on inspection, which gives the incorrect answer? What HE thinks is the constant term, namely d_1(x) = 7/w_1(x) is not even constant (unless he assumes what he wants to prove) ! Similary, the constant term of (5 c_2(x) + d_2(x)) is 1 also. However, the constant term of (5 c_3(x) + d_3(x)) is (5 c_3(0) + d_3(0)) = (5 a_3(0)/w_3(0) + 7/w_3(0)) = 5 * 3/1 + 7/1 = 15 + 7 = 22. Wow, look at that! The product of the constant terms is the constant term of the product! That is, d_1(0)*d_2(0)*(5 a_3(0) + d_3(0)) = 1*1*22 = 22 = P(0)/49. Amazing! Did I successfully trick you? Where's the trick? Everything works out just as it should: IF you correctly identify the constant terms. Where's the trick, Mr. Harris ? >(Notice Nora Baron gives you clues, but somehow sci.math readers >don't seem to get it.) I think not. I think that, with perhaps one exception, sci.math readers DO get it. In fact, I even think there is no exception. I think Mr. Harris actually gets it also, but he cannot stand to admit it. There is too much at stake. He clings to a delusion on the one-in-a trillion chance that he will be proven right. >Notice the poster's actually gives that when you divide off 49, you >get >g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x) >which are TRACES, left of 49--after it has been divided off. See above. There is no mystery. After division, the product of the constant terms is the constant term of the product, just as it should be. That is, IF you correctly identify what the constant terms actually are. If you don't, then you get the wrong answer. This is Mr. Harris's mistake. This is not rocket science. This is not Galois theory, or algebraic number theory, or field theory, or group theory. This is not abstract algebra at all. This is not even high-school level algebra. The mistake Mr. Harris is making is really a low-level one. A ROOKY mistake, as Mr. Harris used to say (about me and Arturo Magidin and others). He is simply not using his own definition. He is substituting in x when he should be substituting in 0. And incredibly, he seems unable to understand it. Math genius Harris, boy wonder Harris - from this you would think he needed help changing his own soiled underwear. [more propaganda deleted] Nora B. >James Harris === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that it's actually almost > as interesting watching how people react to it, as anything else. For instance, I've given a polynomial P(x) repeatedly where I factor > it into three factors. I point out that the factors must include factors of the constant term > of the polynomial P(x). I note that the factors of the constant term are independent of x, as > they are, in fact, constant. Mathematically it's easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > I've called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. It's that simple. Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. Mathematically, it's simple to the point of trivial. Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? No one does. If as you say, you assume that your factors are > constant, of course they are not variables. That is nothing but > a dimwit tautology. > And of course that is not what we say. We say that if you factor > 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), > and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) > by those variable functions, then you can obtain algebraic integer > factors on both sides of the resulting equation. That is, > g_1(x)/w_1(x), etc., are all algebraic integers. That is all you > require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and > g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to > g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). > No problem with that either. And the product of these three > constant terms equals P(0)/49. > Ok, I'm going to answer the Nora Baron poster yet again. > And I want readers to understand that I've replied to this poster who > you know lies anyway as it's a guy posting as a woman using a name > that's a palindrome MANY TIMES explaining in detail. Please provide a link to a message where nora baron says that she is a woman. <... Look now, even after the post where Nora Baron ended with a male > name, people are STILL supporting the poster! Please provide a link to a message where nora baron says that he is a man. === Subject: Re: JSH: Simply fascinating > ... 49 is a multiple of the polynomial. You keep saying this kind of stuoid stuff. But 49 is *not* a multiple of the polynomial. Read the definition of the word 'multiple'. > Look now, even after the post where Nora Baron ended with a male > name, people are STILL supporting the poster! That is because nobody really cares what Nora's geneder really is. They care about what he says... i.e. his math. Is that not what you claim is the only thing that counts? If Nora was a flaming tranvestite hemaphroditic cross-dresser, it would not make his math incorrect or your math correct. And it is not a lie to use a pseudonym, it is a personal choice. Most of us do not use our real names here. === Subject: Re: JSH: Simply fascinating What does Nora's gender have to do with math. You only point this out because you are immature, trying to create a distraction, and you know your work is wrong. Get a life. Dave === Subject: Re: JSH: Simply fascinating > What does Nora's gender have to do with math. You only point this out > because you are immature, trying to create a distraction, and you know > your work is wrong. Get a life. > Dave The poster lies repeatedly. I've argued with this poster for some time and noted the lying, and the gender lying is just another example of a trend. Some people don't like being lied to, and take it seriously. My position is that I can explain in detail, and with very simple concepts how my argument works, but that mathematicians might believe they have lots of reasons for avoiding the truth--all social ones. Unfortunately, they are aided by the poster who manages to maintain confusion about the issues, and maintain the lie that there is any vagueness or area of real mathematical doubt about my work, as many people trust. They trust that if I were right mathematicians wouldn't disagree with me, and there wouldn't be so much opposition to my work. So the poster Nora Baron can just lie for the sake of showing opposition, which helps to create the illusion of uncertainty about my work, helping to hide the truth from people who don't know better. The people who are well-trained mathematicians though, they are not fooled, and I know from my own experiences talking with well-trained mathematicians about my work. So what's happening now is sad in many ways, but part of it has to do with a basic contempt for the public, and a belief that they can be lead astray indefinitely by rather basic tactics. I say, the strategy is about to die, and the passive-aggessive strategy of hoping that indefinitely the public will be fooled will be shown to be one of professional suicide. James Harris === Subject: Re: JSH: Simply fascinating > What does Nora's gender have to do with math. You only point this out > because you are immature, trying to create a distraction, and you know > your work is wrong. Get a life. > Dave > The poster lies repeatedly. I've argued with this poster for some > time and noted the lying, and the gender lying is just another example > of a trend. Bullcrap. Nora Baron has repeatedly shown errors in the poster James Harris' math and THAT is why he hates her. (Oh he does! He said so not long ago). He appears to be particularly infuriated because she keeps pinning him back to the more involved polynomials from which his simpler examples are derived, and showing where his error lies (no pun intended). This poster James Harris historically responds to math rebuttals with diatribes like this when he is unable to respond successfully to the math with math. Perhaps what REALLY pisses him off is that a _female_ could so effectively demolish his erroneous math. > Some people don't like being lied to, and take it seriously. > My position is that I can explain in detail, and with very simple > concepts how my argument works, but that mathematicians might believe > they have lots of reasons for avoiding the truth--all social ones. Here it is: > Unfortunately, they are aided by the poster who manages to maintain > confusion about the issues, and maintain the lie that there is any > vagueness or area of real mathematical doubt about my work What this poster James Harris calls 'lies' are any math rebuttals that show that there is any vagueness or area of real mathematical doubt about his work. Since his work is correct they _must_ be lies. KeithK >, as many > people trust. The people who are well-trained mathematicians though, they are not > fooled, and I know from my own experiences talking with well-trained > mathematicians about my work. So what's your problem? If well-trained mathematicians have discussed your work with you, surely you don't need to keep banging your head against the lame brains at sci.math or any other newsgroup. Just have some of these well-trained mathematicians promote your work and publications, accolades, Field medals, parades, etc. will follow. -- 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: Simply fascinating > The math here is so readily understandable that it's actually almost > as interesting watching how people react to it, as anything else. > For instance, I've given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > I note that the factors of the constant term are independent of x, as > they are, in fact, constant. > Mathematically it's easy to show: > g_1(x) g_2(x) g_3(x) = P(x) > and > g_1(0) g_2(0) g_3(0) = P(0) > as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > I've called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. > It's that simple. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. > Mathematically, it's simple to the point of trivial. Yes if P(x) = g_1(x) g_2(x) g_3(x) then P(0) = g_1(0) g_2(0) g_3(0) why did you surround this trivial fact with all the verbiage. No one is disputing this. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? > You have x, as the variable. Besides x there are just these numbers. > If you clear out x, then what's left are constants. Letting x=0, > clears it out, leaving the constants visible. > Some may say, yeah, sure, at x=0, but what about when x doesn't equal > 0? > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? > The logic is inescapable. And by this logic the constant term of (a(0)=0, w(0)=1) h(x) = (a(x)/w(x) + 7/w(x)), a(0)=0, w(0)=1 is 7. The constant term is not 7/w(x). Indeed you can write h(x) so you can see the constant. Let t(x) = (a(x) -7w(x) +7)/w(x) Note t(0) = 0 and h(x) = (t(x) + 7) So the constant term of h(x) is 7. And yes 7 divides the constant term of P(x). No one is arguing that 7/w(x) divides the constant term of P(x). -William Hughes === Subject: Re: Simply fascinating Having fun following this thread, but I just want to make sure I understand it. So, James has the expression: P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) and he has shown that each a_*(x) evaluates to an algebraic integer for integer x. And he has shown that a_*(0) is an integer. Then he makes this leap by saying that a_1(0) = 7 implies a_1(x) = 7.b (b algebraic) for all integer x. Is that correct how I stated it, or am I missing something? Darren === Subject: Re: Simply fascinating > Then he makes this leap by saying that > a_1(0) = 7 implies a_1(x) = 7.b (b algebraic) for all integer x. > Is that correct how I stated it, or am I missing something? You got it all correctly. You see, 7 is a constant. It does not depend on x. So 7 cannot change just because x changes. So a_1(x) is still 7, since 7 has not changed..... get it? And by the way, the above equations have no memory of the 49 that got properly divided off as a multiple, or something like that. Get it? === Subject: Re: Simply fascinating Meant: > a_1(0) = 7 implies a_1(x) = 7.b (b algebraic Integer) for all integer x. === Subject: Re: Simply fascinating > But the concepts are where there is a problem, and the social hang-up > is in accepting that there's this simple technique that shows a BIG > problem, which can invalidate claims of proof for, well, over a > hundred plus years. That is not the hang-up. The hang-up is that you are 100% ed up. I would have no problem accepting something that would show a big problem, invalidating proofs for over a hundred years. I would just love it. I would think, wow, that is so cool! It would be a thrill. But you have not convinced me or anyone else that you have found anything interesting or correct. Just fluffy bogus claims. Lots of claims. Little valid math. Little valid logic. Lots of paranoia. Lots of grandiosity. Lots of non-standard terminology. Lots of smoke screen crap, like the gender of Nora. Look James... lots of people here would love to see the next crisis/revolution in math... another Russell shakeup, another G.9adel shakeup. So tell me... why has nobody been interested in any of your ideas? Could it be that your ideas are ? Or do you think that it is more likely that everyone else is nuts. Apply Occam's Razor here! Just a thought.