mm-237 === Subject: Re: Binomial Coefficient> I understand that I can use Pascal's Triangle to complete a binomial> expansion, but it's very long-winded. What is the quickest way of working> out the coefficient of x^5 in the expansion of (x^2 - (3/x))^7? Please bear in mind that I have a scientific calculator which I am > prepared> to use (Casio fx-991MS) and I have a test soon in which speed is > definitely> an issue (i.e. my working counts for nothing).Expressing the desired term two ways:a*x^5 = Binom(7,n)*(x^2)^n * (-3/x)^(7-n) = [7!/(n!*(7-n)!] * (-3)^(7-n)*x^(2*n -(7 - n))So, equating the exponents of x gives 5 = 3*n - 7, and n = 4And equating coefficients gives a === Please check or help....recurrence relations. Could you please check (and help) the following when trying to find a> solution (using recurrence relations):> A)> A_n = A_n-1 + 3 where A_0 = 1> ANS:> A_n = 3 + A_n-1> = 3+(3 + A_n-2) = (3+3) + A_n-2 = (3*3)+A_n-2> = (3*3)+(3 + A_n-3) = (3*3) + A_n-3> .> = (n*3)+A_(n-n) = (n*3)+A_0 = (n*3)+1> = 3n + 1I think Im corect on this prev one but the next one Im not sure what to do> because the initial form seems strange to me.>You're right about the first one. > B) A_n = -A_(n-1) where A_0 = 5> Do I do something like ...> -A_(-A_(n-2))....Im really not sure. = -5, A_2 = -A_1 = 5, ....Do you see a pattern?Moral: When in === and Non Rigid transformations of a functionThis question may seem a tad on the side of opinion, however those arewelcome.I have the question: Describe the nonrigid transformation of f(x)=.5(X+4) Now my opinion on the matter is that by the wording describe thenon-rigid transformation they simply are asking for the scalar hencethe graph is vertically scaled by half. While others have the opinionthat the question implies rigid transformations under the umbrella ofnon-rigid transformations and hence the answer including a shift tothe left 4 units.Does anyone know the correct interpretation of === 1-player game (much easier to play on computer,but not absolutely necessary):Start with a q-by-q grid, for q = a positive integer.(I would guess that q would be best in the range of 10 - 50,but this is only a guess. Perhaps q can be picked randomly somehow before play.)Also, the player has a collection of pieces which can bestacked on top of each other in individual squares of the grid.Before play, a random-number generator picks 3 positiveintegem, n, and r.(m and n are <= q, but I would suggest each be at least q/2.)(I suggest that r be in the vicinity of r^2/2, but this is onlya gut-instinct.)At each move, the random-number generator picks 4 positiveintegers,c(m), c(n), d(m), d(n), each under the following conditions:GCD(c(m),m) = 1,GCD(c(n),n) = 1,d(m) divides m,d(n) divides n,c(m)*d(n) <= r,c(n)*d(m) <= r,and neither the pair {c(m),d(n)} nor {c(n),d(m)} has been yetplayed by the player. (Although either pair may have beenpreviously picked by the computer{or by card-drawor by dice-roll or..}, just not played yet.) After the numbers are genera, the player can pick the {c(m),d(n)} pair or the {c(n),d(m)} pair.Then the player can choose to add or to subtract a groupof pieces to/from the pieces on the grid.The group is all pieces within a rectangle either ofc(m)-by-d(n) grid-squares or of c(n)-by-d(m) grid-squares(the rectangle being orien so the either c or d is thewidth, and {conversely} either d or c is the length).And the rectangle can be placed anywhere possible withinthe grid in which it fits. And all grid-squares must contain a nonnegative number ofpieces in them after a move.After a fixed number of moves, or after all allowable pairshave been played, the game is complete.The goal is to have the fewest number of pieces on the gridas possible at the game's end.--2-player version:2 grids of same dimensions.Same m and n and r for both players.Players can pick their opponents' c(m), c(n), d(m), d(n),but under the same restrictions as above.Goal is to simply have fewest number of pieces afterfixed number of moves.--Note: I am guessing that if m and n are squarefree,and a subtraction move is always available when necessaryto the strategy, then there is at least one strategy wherewhether to add or subtract is dependent only on d(m) or d(n)(whichever is played).This strategy should ALWAYS, if I am right, lead to ZEROpieces on the board at the end of play, if the number of playsin the game is such that every {d(m),c(n)} and {d(n),c(m)}pair is played exactly once.Puzzle: what would the stategy be?(I have not thought it out enough to know if such a stategyworks in all or any situations or if it is practical.)Leroy === 1-player game (much easier to play on computer,> but not absolutely necessary):> Start with a q-by-q grid, for q = a positive integer.> (I would guess that q would be best in the range of 10 - 50,> but this is only a guess. > Perhaps q can be picked randomly somehow before play.)Also, the player has a collection of pieces which can be> stacked on top of each other in individual squares of the grid.Before play, a random-number generator picks 3 positive> integem, n, and r.> (m and n are <= q, but I would suggest each be at least q/2.)> (I suggest that r be in the vicinity of r^2/2, but this is only> a gut-instinct.)Whoops, should be:(I suggest that r be in the vicinity of q^2/2, but this is only a gut-instinct.) ^...unless r is in the vicinity of 0 or 2, of course...;)LeroyAt each move, the random-number generator picks 4 positive> integers,> c(m), c(n), d(m), d(n), each under the following conditions:GCD(c(m),m) = 1,> GCD(c(n),n) = 1,> d(m) divides m,> d(n) divides n,> c(m)*d(n) <= r,> c(n)*d(m) <= r,> and neither the pair {c(m),d(n)} nor {c(n),d(m)} has been yet> played by the player. (Although either pair may have been> previously picked by the computer{or by card-draw> or by dice-roll or..}, just not played yet.) > After the numbers are genera, the player can pick > the {c(m),d(n)} pair or the {c(n),d(m)} pair.Then the player can choose to add or to subtract a group> of pieces to/from the pieces on the grid.> The group is all pieces within a rectangle either of> c(m)-by-d(n) grid-squares or of c(n)-by-d(m) grid-squares> (the rectangle being orien so the either c or d is the> width, and {conversely} either d or c is the length).> And the rectangle can be placed anywhere possible within> the grid in which it fits. And all grid-squares must contain a nonnegative number of> pieces in them after a move.> After a fixed number of moves, or after all allowable pairs> have been played, the game is complete.The goal is to have the fewest number of pieces on the grid> as possible at the game's end.> 2-player version:2 grids of same dimensions.Same m and n and r for both players.Players can pick their opponents' c(m), c(n), d(m), d(n),> but under the same restrictions as above.Goal is to simply have fewest number of pieces after> fixed number of moves.> --> Note: I am guessing that if m and n are squarefree,> and a subtraction move is always available when necessary> to the strategy, then there is at least one strategy where> whether to add or subtract is dependent only on d(m) or d(n)> (whichever is played).> This strategy should ALWAYS, if I am right, lead to ZERO> pieces on the board at the end of play, if the number of plays> in the game is such that every {d(m),c(n)} and {d(n),c(m)}> pair is played exactly once.Puzzle: what would the stategy be?> (I have not thought it out enough to know if such a stategy> works in all or any === dependency logic In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3> = [sqrt(1+x)+sqrt(1-x)-2] + 3Notice that has just subtrac and added 3.He's apparently lost the ability to reason, if he ever had it, on thissubject, as this is the third time I've seen a post of his pushingthat doing nothing as something.After all, if you add and subtract or subtract and add the samenumber, you get 0. He's just playing with 0. fall into the hands of JSH and his Core Errorian (not an attack on Koreans)> agents, but Arturo has saved us.Mathematicians are social creatures and they like to follow eachother.I consider it lemming like behavior.Some of you may have noticed my arguments and postings adjusting.My thinking is that if one mathematician, like , canlead many others down with him, that's not my problem.I will solve the problem, and with the truth, there will beconsequences.So follow lemmings. === Adjunct Assistant Professor at the University of Montana.> In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3Notice that has just subtrac and added 3.What is your point? This is exactly what you do as well.Your Lemma 1 in your Advanced Polynomial Factorization was thatany function f(x) can be written as f(x) = g(x) + f(0), whereg(0)=f(x)-f(0), remember?So, if f(x) = sqrt(1+x)+sqrt(1-x)+1,then the wqu to figure out what g(x) and what f(0) are is to plug inx=0,f(0) = sqrt(1)+sqrt(1) + 1 = 3and to find g(x), we take f(x)-f(0):g(x) = f(x)-f(0) = sqrt(1+x)+sqrt(1-x)+1-3 = sqrt(1+x) + sqrt(1-x) - 2.Are you claiming that it is NOT true that for f(x) = sqrt(1+x) +sqrt(1-x) + 1, ACCORDING TO YOUR DEFINITIONS, the constant term is 3and the nonconstant term is sqrt(1+x)+sqrt(1-x)-2?Are you claiming that it is NOT true that we get the constant term byevaluating at x=0?Are you cliaming that it is NOT true that we get the nonconstant termby taking f(x)-f(0)?or did you simply fail to notice that what YOU do is to writef(x) = (f(x)-f(0)) + f(0)and then claim that it is a big deal?>He's apparently lost the ability to reason, if he ever had it, on this>subject, as this is the third time I've seen a post of his pushing>that doing nothing as something.Was anything I did incorrect according to your definition? No.The ENTIRE content of your supposedly Earth Shattering(tm) discoveryis that you can add 0 by writing f(x) = (f(x)-f(0)) + f(0),so if you think that's trivial, then you are really commenting on thequality of YOUR work, not on the quality of mine. I'm following YOURdefinitions and YOUR process, as described BY YOU.Can you address the math? Or are you incapable of doing so except byrepeating your mistakes over and over?What IS the constant term of f(x) = sqrt(1+x) + sqrt(1-x) + 1?If we write f(x) = g(x)+c, with g(0)=0, what IS g(x)?Are the answers NOT 3 and sqrt(1+x)+sqrt(1-x)-2, respectively?If not, why not?>After all, if you add and subtract or subtract and add the same>number, you get 0. He's just playing with 0.It's amusing as hell to see you belittle your own argument whensomeone else uses it. Probably because (a) you don't really know whatyour argument is; and (b) you don't really care what everyone elsesays, all you care about is being right and calling everyone else anincompetent liar.Too bad that you're wrong and that nobody else is lying. It only makesyou look such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de === In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3Notice that has just subtrac and added 3.What is your point? This is exactly what you do as well.Your Lemma 1 in your Advanced Polynomial Factorization was that> any function f(x) can be written as f(x) = g(x) + f(0), where> g(0)=f(x)-f(0), remember?Yeah, and I noticed that posters like had a field dayconfusing the hell out of people with the more general argument, soI've simplified the argument in discussions.And, not surprisingly, the simplification leaves posters like ArturoMagidin less less room to obfuscate (judge for yourself if you don'tbelieve me), so wouldn't you know, he's trying to go backwards.He's trying to get me to go back to a MORE complica argument indiscussions, as that's where he could confuse people.But here's the *current* simplified argument.Notice how I'll be strongly emphasizing constant terms all the waydown.P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors areall 7, which can't be right, as the constant term of P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49gives me 22, as the new constant term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the constant terms keep === be strongly emphasizing constant terms all the way>down.P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078which has a constant term that is 1078.P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors are ^^^^^^^^^Does it? If it appears that way to you, then maybe you shouldwrite a_1(x), a_2(x), a_3(x) to make sure that you don't forgetthat the a's are functions in x. >all 7, which can't be right, as the constant term of P(x) is 1078, so>setting x=0, revealsThe a's do not have constant terms: because the a's are not polynomials,the concept constant term is meaningless in this context. You still havenot said what you mean by constant term. Most people are assuming thatyou just mean the value at 0. If that is true, why don't you say thatinstead. If that is false, what is the constant term of sqrt(x+1)?>P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the constant terms keep matching.(Assuming you mean constant term to mean value at 0) This is false,there is no reason why a_1(x) should be divisible by 7 for all algebraicintegers (or all integers). This has been poin out to you with several examples. You keep iterating it, but fail to give any argumentation atthis point.-- Peter van Rossum, | Universal law of linearity: for allDept. of Mathematics, New Mexico | f : R -> R and for all x, y in R:State University, Las Cruces, NM, USA. | f(x + y) = f(x) + === example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3Notice that has just subtrac and added 3.Oh, good grief. A new low. You want to revolutionize algebraic numbertheory, but you're flunking eighth-grade algebra.>He's apparently lost the ability to reason, if he ever had it, on this>subject, as this is the third time I've seen a post of his pushing>that doing nothing as something.After all, if you add and subtract or subtract and add the same>number, you get 0. He's just playing with 0.Three comments.1. Do you think that it is incorrect or represents an abandonment ofreason to say that x = x + 3 - 3?2. Do you think that there's no reason ever to add a constant andsubtract off the same constant? Before answering that, can youremember the technique called completing the square? How would youwrite x^2 - 2x in the form (x-a)^2 + b?Hint: It will involve adding b and subtracting b. Does this representa loss of reason?3. More to the point, this gets to the heart of your trumpeting aboutconstant terms of arbitrary functions. You say every function f(x)can be written in the form f(x) = g(x) + c, where g(0) = 0, so c isthe constant term of f.Fine. Please write f(x) = sqrt(x^2 + 9) in that form. What === > In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3Notice that has just subtrac and added 3.Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.Readers should note that what I said is indeed correct as ArturoMagidin *did* just subtract and add 3.And, of course, -3 + 3 = 0.Now here's the argument that and Randy Poe are tryingto confuse YOU the reader about as they work to convince you offalsehoods.Notice how I'll be strongly emphasizing constant terms all the waydown.P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors areall 7, which can't be right, as the constant term of P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49gives me 22, as the new constant term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the === logic >> In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>Notice that has just subtrac and added 3.Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.And, of course, -3 + 3 = 0.Now here's the argument that and Randy Poe are trying> to confuse YOU the reader about as they work to convince you of> falsehoods.Notice how I'll be strongly emphasizing constant terms all the way> down.P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078which has a constant term that is 1078.> I would like to see your method applied to x^3-x+6.> Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors are> all 7, which can't be right, as the constant term of P(x) is 1078, so> setting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2> for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49> gives me 22, as the new constant term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the constant terms keep matching.> === example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>Notice that has just subtrac and added 3.Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.And, of course, -3 + 3 = 0.Certainly. Did I claim that wasn't true? You snippedboth my comments and YOUR comment that caused me tojump in.Here's yours:>> He's apparently lost the ability to reason, if he ever had it, on this>> subject, as this is the third time I've seen a post of his pushing>> that doing nothing as something.Actually, I wondered if you'd wake up in the morning andregret saying this, if you'd remember that much ofelementary algebra consists of doing nothing in thesense of adding zero or multiplying by 1. I suspectyou did realize what an idiotic comment that was, andthat's why rather than defend it, you snipped it and chose to pretend that both you and I said somethingdifferent than what was actually said.Now here's the argument that and Randy Poe are trying> to confuse YOU the reader about as they work to convince you of> falsehoods.Arturo can speak for himself.If you're going to pretend to be answering MY questions,wouldn't it be more honest to include those questions,rather than snip them and pretend I asked somethingdifferent?What is your answer to the three questions I actually asked:1. Do you think that it is incorrect or represents an abandonment ofreason to say that x = x + 3 - 3?2. Do you think that there's no reason ever to add a constant andsubtract off the same constant? Before answering that, can youremember the technique called completing the square? How would youwrite x^2 - 2x in the form (x-a)^2 + b? Hint: It will involve addingb and subtracting b. Does this represent a loss of reason?3. More to the point, this gets to the heart of your trumpeting aboutconstant terms of arbitrary functions. You say every function f(x)can be written in the form f(x) = g(x) + c, where g(0) = 0, so c isthe constant term of f. Fine. Please write f(x) = sqrt(x^2 + 9) inthat form. What === logic>> In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>Notice that has just subtrac and added 3.> Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.And, of course, -3 + 3 = 0.Certainly. Did I claim that wasn't true? You snipped> both my comments and YOUR comment that caused me to> jump in.The truth is that I no that had just subtrac andadded 3.Then *you* Randy Poe said, immediately thereafter:Oh, good grief. A new low. You want to revolutionize algebraic numbertheory, but you're flunking eighth-grade algebra.So I replied after that paragraph, which was followed by a space.I no that, in fact, all had done was just subtractand add 3, which is what I said.For other readehere's what Randy Poe and are tryingto convince others is wrong.Notice how I'll be strongly emphasizing constant terms all the waydown.P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors areall 7, which can't be right, as the constant term of P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49gives me 22, as the new constant term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the constant terms keep === your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>Notice that has just subtrac and added 3.> Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.> Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.> And, of course, -3 + 3 = 0.Certainly. Did I claim that wasn't true? You snipped> both my comments and YOUR comment that caused me to> jump in.> The truth is that I no that had just subtrac and> added 3.Then *you* Randy Poe said, immediately thereafter:Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.Yes, you're right. That's the position I put this remark.It was a stylistic choice, meant to be an introduction tothe meat of my comments. It was a predecessor to yourridiculous comment which came NEXT. It's an ordering Iuse sometimes, that (I hope) is not usually so misinterpre.A remark that says Hey reader, check this out. Beingintroductory in nature, I feel there's no problem inputting it before the stuff it introduces. That feels morenatural to me.However, you chose to focus on the introductory sentenceand pretend that nothing actually follows it.Nevertheless, I did feel it was reasonable to interject*some* comment right at that point. Let's look at yourstatement again:>Notice that has just subtrac and added 3.You haven't said anything wrong, but I have to wonder whyyou felt the need to say anything at all. You are askingthe reader to notice something which is not noteworthy.Why do you think it is?> So I replied after that paragraph, which was followed by a space.I no that, in fact, all had done was just subtract> and add 3, which is what I said.For other readehere's what Randy Poe and are trying> to convince others is what I'msaying, which are two different things. Interesting that youkeep trying to convince people I'm pursuing an argumentwhich has not appeared in any of my posts. Why is that?And why is it that rather than answering my trivialquestion, you keep clipping it and substituting yourcut-and-paste text?I'm just asking you to apply your technique of rewriting f(x) = g(x) + cwhere g(0) = 0 and c = f(0).to the function f(x) = sqrt(x^2 - 9). That's simpler thanC. Bond's similar question (which you also haven't answered)but raises the same issue.I would say g(x) = sqrt(x^2-9) - 3, wouldn't you?I would say c = 3, wouldn't you? === apply your technique of rewriting> f(x) = g(x) + c> where g(0) = 0 and c = f(0).to the function f(x) = sqrt(x^2 - 9). That's simpler than> C. Bond's similar question (which you also haven't answered)> but raises the same issue.I would say g(x) = sqrt(x^2-9) - 3, wouldn't you?I certainly would not; you eithr meant sqrt(x^2+9), or you meant 3i above..., sans === example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>Notice that has just subtrac and added 3.> Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.> Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.> And, of course, -3 + 3 = 0.> Certainly. Did I claim that wasn't true? You snipped> both my comments and YOUR comment that caused me to> jump in.> The truth is that I no that had just subtrac and> added 3.Then *you* Randy Poe said, immediately thereafter:Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra. Yes, you're right. That's the position I put this remark.> It was a stylistic choice, meant to be an introduction to> the meat of my comments. It was a predecessor to your> ridiculous comment which came NEXT. It's an ordering I> use sometimes, that (I hope) is not usually so misinterpre.> A remark that says Hey reader, check this out. Being> introductory in nature, I feel there's no problem in> putting it before the stuff it introduces. That feels more> natural to me.It hardly matters as the essential point remains that all ArturoMagidin did was subtract and add 3. And, of course, -3 + 3 = 0, so hedid NOT DO ANYTHING worth mentioning.So it hardly matters what Randy Poe's style is, as the discussion isnonsensical from the start, as Randy Poe apparently wishes to act asif it's material to just subtract and add 3, to the extent that heinsul me at the start by saying,Oh, good grief. A new low. You want to revolutionize algebraic numbertheory, but you're flunking eighth-grade algebra.Which I think tells you that Randy Poe is not a person to be takenseriously when it comes to mathematics, as he was replying to myhaving said:Notice that has just subtrac and added 3.Which is the truth. But he's STILL arguing. And I'm replying topoint out just how silly Randy Poe is. It's not a surprise to me, butI want to make the point === your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 =3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>Notice that has just subtrac and added 3.> Oh, good grief. A new low. You want to revolutionize algebraicnumber> theory, but you're flunking eighth-grade algebra.> Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.> And, of course, -3 + 3 = 0.> Certainly. Did I claim that wasn't true? You snipped> both my comments and YOUR comment that caused me to> jump in.> The truth is that I no that had just subtrac and> added 3.> Then *you* Randy Poe said, immediately thereafter:> Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra. Yes, you're right. That's the position I put this remark.> It was a stylistic choice, meant to be an introduction to> the meat of my comments. It was a predecessor to your> ridiculous comment which came NEXT. It's an ordering I> use sometimes, that (I hope) is not usually so misinterpre.> A remark that says Hey reader, check this out. Being> introductory in nature, I feel there's no problem in> putting it before the stuff it introduces. That feels more> natural to me. It hardly matters as the essential point remains that all Arturo> Magidin did was subtract and add 3. And, of course, -3 + 3 = 0, so he> did NOT DO ANYTHING worth mentioning. So it hardly matters what Randy Poe's style is, as the discussion is> nonsensical from the start, as Randy Poe apparently wishes to act as> if it's material to just subtract and add 3, to the extent that he> insul me at the start by saying, Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra. Which I think tells you that Randy Poe is not a person to be taken> seriously when it comes to mathematics, as he was replying to my> having said: Notice that has just subtrac and added 3. Which is the truth. But he's STILL arguing. And I'm replying to> point out just how silly Randy Poe is. It's not a surprise to me, but> I want to make the point clearly.> James,We subtract and add the same number all the time in mathematics.Say I want to write the quadratic y=x^2+6x+8 into the form y=a(x-h)^2+k.Looking at the first two terms, I see I can make a perfect square if I add9. What about the 8? Since I add 9 to x^2+6x, I must subtract it from the 8.y=x^2+6x+9+8-9=(x+3)^2-1This is used all the time especially when doing trig substitutions === >message>> In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 => 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>Notice that has just subtrac and added 3.> Oh, good grief. A new low. You want to revolutionize algebraic> number> theory, but you're flunking eighth-grade algebra.> Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.> And, of course, -3 + 3 = 0.> Certainly. Did I claim that wasn't true? You snipped> both my comments and YOUR comment that caused me to> jump in.>statement.> The truth is that I no that had just subtracand> added 3.> Then *you* Randy Poe said, immediately thereafter:> Oh, good grief. A new low. You want to revolutionize algebraicnumber> theory, but you're flunking eighth-grade algebra.> Yes, you're right. That's the position I put this remark.> It was a stylistic choice, meant to be an introduction to> the meat of my comments. It was a predecessor to your> ridiculous comment which came NEXT. It's an ordering I> use sometimes, that (I hope) is not usually so misinterpre.> A remark that says Hey reader, check this out. Being> introductory in nature, I feel there's no problem in> putting it before the stuff it introduces. That feels more> natural to me. It hardly matters as the essential point remains that all Arturo> Magidin did was subtract and add 3. And, of course, -3 + 3 = 0, so he> did NOT DO ANYTHING worth mentioning. So it hardly matters what Randy Poe's style is, as the discussion is> nonsensical from the start, as Randy Poe apparently wishes to act as> if it's material to just subtract and add 3, to the extent that he> insul me at the start by saying, Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra. Which I think tells you that Randy Poe is not a person to be taken> seriously when it comes to mathematics, as he was replying to my> having said: Notice that has just subtrac and added 3. Which is the truth. But he's STILL arguing. And I'm replying to> point out just how silly Randy Poe is. It's not a surprise to me, but> I want to make the point clearly.> James, We subtract and add the same number all the time in mathematics. Say I want to write the quadratic y=x^2+6x+8 into the form y=a(x-h)^2+k. Looking at the first two terms, I see I can make a perfect square if I add> 9. What about the 8? Since I add 9 to x^2+6x, I must subtract it from the8. y=x^2+6x+9+8-9=(x+3)^2-1 This is used all the time especially when doing trig substitutions with> integrals. I'd like to state for the record, so James doesn't accuse me, that by doingthe above transformation, I do not change the equation. It is just in a moreuseful form for determining characteristics and integrating when === example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>Notice that has just subtrac and added 3.> Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.> Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.> And, of course, -3 + 3 = 0.Certainly. Did I claim that wasn't true? You snipped> both my comments and YOUR comment that caused me to> jump in.> The truth is that I no that had just subtrac and> added 3.Then *you* Randy Poe said, immediately thereafter:Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.So I replied after that paragraph, which was followed by a space.I no that, in fact, all had done was just subtract> and add 3, which is what I said.For other readehere's what Randy Poe and are trying> to convince others is wrong.Notice how I'll be strongly emphasizing constant terms all the way> down.P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors are> all 7, which can't be right, as the constant term of P(x) is 1078, so> setting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2> for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49> gives me 22, as the new constant term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the constant terms keep matching.> Is it ? Let a1 = 1, a2 = 1562, and b3 = 25. Then (5*a1 + 7)*(5*a2 + 7)*(5*b3 + 22) = 12 * 7817 * 147 = 13789188 Now it so happens that P(1) = 14706125 - 900375 - 17640 + 1078 = 13789188. Note that a1 = 1 is not divisible by 7, and a2 = 1562 is also not === Assistant Professor at the University of Montana. In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3> = [sqrt(1+x)+sqrt(1-x)-2] + 3>>Notice that has just subtrac and added 3. Oh, good grief. A new low. You want to revolutionize algebraic number>> theory, but you're flunking eighth-grade algebra. Readers should note that what I said is indeed correct as Arturo>> Magidin *did* just subtract and add 3. And, of course, -3 + 3 = 0. Certainly. Did I claim that wasn't true? You snipped>> both my comments and YOUR comment that caused me to>> jump in.>The truth is that I no that had just subtrac and>added 3.What I did is what you SAID we SHOULD do:figure out the constant term c by evaluating at 0;then figure out the nonconstant term g(x) by taking f(x)-f(0);then write f(x) = g(x) + c.Is that NOT what you said to do?Here is what YOU INSERT ---Factorization Lemma:A factor g of a polynomial P(x) within a given commutative ring, thatis not a field, can always be written as g = r + cwhere r=0, or is some other factor of x, and c is a factor of theconstant term P(0).Proof:Given g is a factor of the polynomial P(x).Set x=0, then if r=0, c is a factor of the constant term P(0).If r does not equal 0, then r + c must be a factor of the constantterm P(0) and can be replaced with c1, so then I can write g = c1 = r+c, with a corresponding r1=0.Lemma proof complete. -- End insert --So, what did you do to get r and c?You set x = 0, figured out g(0), and you set c=g(0).Then to get r, you set it to g-c. Is that NOT what I did?I did it because that's what you said one SHOULD do to figure outconstant and non-constant terms.OF COURSE it amounts to doing nothing. OF COURSE it amounts to addingzero. All you are doing is saying that a function f(x) can be writtenasf(x) = r(x) + f(0), where r(x) = f(x)-f(0). That is,f(x) = f(x) - f(0) + f(0).THAT'S WHAT YOU'VE BEEN DOING FOR MONTHS.And we've been telling ->you<- that you cannot get out of this simplething what you claim to get.So glad you finally agree that it really amounts to nothing.That means that you AGREE that your approach is silly, that anyoneusing it can be accused of having lost all ability to reason.Good of you to finally come around.So, now that you agree that all your entire approach is built on atriviality which amounts to nothing much, what else is === your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3> = [sqrt(1+x)+sqrt(1-x)-2] + 3>>Notice that has just subtrac and added 3. Oh, good grief. A new low. You want to revolutionize algebraic number>> theory, but you're flunking eighth-grade algebra. Readers should note that what I said is indeed correct as Arturo>> Magidin *did* just subtract and add 3. And, of course, -3 + 3 = 0. Certainly. Did I claim that wasn't true? You snipped>> both my comments and YOUR comment that caused me to>> jump in.>The truth is that I no that had just subtrac and>added 3.What I did is what you SAID we SHOULD do:figure out the constant term c by evaluating at 0;then figure out the nonconstant term g(x) by taking f(x)-f(0);then write f(x) = g(x) + c.Is that NOT what you said to do?I'm not interes in silly attempts at head games . After all, readers can just look at the math to see what role constantterms play.Notice how I'll be strongly emphasizing constant terms all the waydown.P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors areall 7, which can't be right, as the constant term of P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49gives me 22, as the new constant term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the constant terms keep === <3c65f87.0310291406.20db9c1d@ f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3>> = [sqrt(1+x)+sqrt(1-x)-2] + 3>>Notice that has just subtrac and added 3.Oh, good grief. A new low. You want to revolutionize algebraic number> theory, but you're flunking eighth-grade algebra.Readers should note that what I said is indeed correct as Arturo> Magidin *did* just subtract and add 3.And, of course, -3 + 3 = 0.Certainly. Did I claim that wasn't true? You snipped> both my comments and YOUR comment that caused me to> jump in.>>The truth is that I no that had just subtrac and>>added 3. What I did is what you SAID we SHOULD do: figure out the constant term c by evaluating at 0; then figure out the nonconstant term g(x) by taking f(x)-f(0); then write f(x) = g(x) + c. Is that NOT what you said to do?I'm not interes in silly attempts at head games . > After all, readers can just look at the math to see what role constant> terms play.Notice how I'll be strongly emphasizing constant terms all the way> down.P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)That may be where the problem is. The constant term is 1078;any factorization of P(x) of the formP(x) = (c_1 x + d_1)(c_2 x + d_2)(c_3 x + d_3)has to have c_1 c_2 c_3 = 14706125 and d_1 d_2 d_3 = 1078.7^3 = 343.Note that I make no assumptions here about whether the c andd values are integealgebraic integeor what. Theproduct has to be as sta, by the simple but obviousobservation that(c_1 x + d_1)(c_2 x + d_2)(c_3 x + d_3) = c_3*c_2*c_1*x^3+ ((d_3*c_2 + c_3*d_2)*c_1 + c_3*c_2*d_1)*x^2+ (d_3*d_2*c_1 + (d_3*c_2 + c_3*d_2)*d_1)*x+ d_3*d_2*d_1(courtesy of GP/Pari).where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors are> all 7, which can't be right, as the constant term of P(x) is 1078, so> setting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2> for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49> gives me 22, as the new constant term.P(x)/49 = 300125*x^3 - 18375*x^2 - 360*x + 22Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the constant terms keep matching.I don't see that. I see a number of other possibilities, contingenton the solution of the above cubic (did you really mean toexpress the 'a' roots as a function of 'x'?):P(x)/49 = (5 a_1/7 + 7)(5 a_2/7 + 1/7)(5 b_3 + 22)P(x)/49 = (5 a_1/7 + 22)(5 a_2/7 + 1)(5 b_3 + 1)P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 22^(1/2))(5 b_3 + 22^(1/2))As it is, I'd have to dig. It is worth noting that the rootsof P(x)/49 = 0 does not define any integers:- 0.034613309107198208140335507980.034560090548514510139311848530 .06127770835460206534796243496(I can't do much better without a lot of work, as cubics with3 reals involve transcendentals.)P(x)/49 also is irreducible over Q (except for the constantfactor 300125, perhaps); the roots thereof thereforecannot be algebraic integewhich require polynomialswith integer coefficients and a coefficient of the highestterm (in this case, x^3) of +1 or -1.> -- #191, ewill3@earthlink.netIt's still legal to go === Assistant Professor at the University of Montana. In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so>> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3> = [sqrt(1+x)+sqrt(1-x)-2] + 3>>Notice that has just subtrac and added 3. Oh, good grief. A new low. You want to revolutionize algebraic number>> theory, but you're flunking eighth-grade algebra. Readers should note that what I said is indeed correct as Arturo>> Magidin *did* just subtract and add 3. And, of course, -3 + 3 = 0.Certainly. Did I claim that wasn't true? You snipped>both my comments and YOUR comment that caused me to>jump in.Here's yours:> He's apparently lost the ability to reason, if he ever had it, on this> subject, as this is the third time I've seen a post of his pushing> that doing nothing as something.Actually, I wondered if you'd wake up in the morning and>regret saying this, if you'd remember that much of>elementary algebra consists of doing nothing in the>sense of adding zero or multiplying by 1. I suspect>you did realize what an idiotic comment that was, and>that's why rather than defend it, you snipped it and >chose to pretend that both you and I said something>different than what was actually said.I wondered if he would realize that it is what HE does. It's the basisof his Advanced Polynomial Factorization, after all. His Lemma 1.>> Now here's the argument that and Randy Poe are trying>> to confuse YOU the reader about as they work to convince you of>> falsehoods.Arturo can speak for himself.Actually, I'm on my way out of town to give some talks, and then Ihave a lot of work to do when I get back. So I won't for at least afew days, if not such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de === > In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so > f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3 > = [sqrt(1+x)+sqrt(1-x)-2] + 3 > Notice that has just subtrac and added 3.Well, as it was my example, I feel a comment is needed. In the form ofquestions again (you apparenty refuse to answer my original question).I asked you what is the constant term in sqrt(1+x) + sqrt(1-x) + 1.But do you agree that the constant term in that expression is 3?If not, why not. If so, do you agree that the part varying withx is sqrt(1-x)+sqrt(1+x)-2? > He's apparently lost the ability to reason, if he ever had it, on this > subject, as this is the third time I've seen a post of his pushing > that doing nothing as something.He just tries to ascertain what you mean by the words constant term.As it is not a mathematical term (except in the context of polynomials)that is a good question. > I will solve the problem, and with the truth, there will be > === dependency logic> In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so> f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3> = [sqrt(1+x)+sqrt(1-x)-2] + 3> Notice that has just subtrac and added 3.Well, as it was my example, I feel a comment is needed. In the form of> questions again (you apparenty refuse to answer my original question).> I asked you what is the constant term in sqrt(1+x) + sqrt(1-x) + 1.> But do you agree that the constant term in that expression is 3?> If not, why not. If so, do you agree that the part varying with> x is sqrt(1-x)+sqrt(1+x)-2?I don't worry about the part varying with x as I concentrate on theconstant term, which in the example given is either 3, or -1 dependingon how you take the square root. You *do* remember that the squareroot operator gives *two* answeright Dik Winter? > He's apparently lost the ability to reason, if he ever had it, on this> subject, as this is the third time I've seen a post of his pushing> that doing nothing as something.He just tries to ascertain what you mean by the words constant term.> As it is not a mathematical term (except in the context of polynomials)> that is a good question.That's rather rank bulls.For instance 2 is a constant. It is not a constant only in thecontext of polynomials as it's a number.A constant is a constant.> I will solve the problem, and with the truth, there will be> consequences.What consequences?People will realize that you tried to b.s. them, and convince them ofstupid things like that numbers such as 2 or 3 are constant only inthe context of polynomials for one thing.Some people might be a little bit angry at you playing them for foolsDik Winter.Here's the actual argument for those wondering why constant terms areso important.Notice how I'll be strongly emphasizing constant terms all the waydown.P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors areall 7, which can't be right, as the constant term of P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49gives me 22, as the new constant term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the === logic... > Well, as it was my example, I feel a comment is needed. In the form of > questions again (you apparenty refuse to answer my original question). > I asked you what is the constant term in sqrt(1+x) + sqrt(1-x) + 1. > But do you agree that the constant term in that expression is 3? > If not, why not. If so, do you agree that the part varying with > x is sqrt(1-x)+sqrt(1+x)-2? > I don't worry about the part varying with x as I concentrate on the > constant term, which in the example given is either 3, or -1 depending > on how you take the square root. You *do* remember that the square > root operator gives *two* answeright Dik Winter?No, I do not remember that, as it is false. In mathematics a functioncan not give you two possible answethat is the definition of function. > He's apparently lost the ability to reason, if he ever had it, on this > subject, as this is the third time I've seen a post of his pushing > that doing nothing as something. > > He just tries to ascertain what you mean by the words constant term. > As it is not a mathematical term (except in the context of polynomials) > that is a good question. > That's rather rank bulls. > For instance 2 is a constant. It is not a constant only in the > context of polynomials as it's a number. > A constant is a constant.Yup, a constant is a constant. But what is a constant term outside thecontext of a polynomial? Note the additional word term. > I will solve the problem, and with the truth, there will be > consequences. > What consequences? > People will realize that you tried to b.s. them, and convince them of > stupid things like that numbers such as 2 or 3 are constant only in > the context of polynomials for one thing.I have sta no such thing. I think you have a reading problem.When I write sqrt(2 + x) + sqrt(3 - x) + 1, there are three constants.but what is the constant term? Mathematics does not define that. > Some people might be a little bit angry at you playing them for fools > Dik Winter.Pray improve your reading. > Here's the actual argument for those wondering why constant terms are > so important. > Notice how I'll be strongly emphasizing constant terms all the way > down. > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has a constant term that is 1078. > Well P(x) can also be written out as > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > so I can factor to get > P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Notice it *appears* that the constant terms for the three factors are > all 7, which can't be right, as the constant term of P(x) is 1078, so > setting x=0, reveals > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's with x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2 > for 0, so that leaves a_3 with a value of 3 when x=0. > So let a_3 = b_3 + 3, where I keep indices matched. Then I have > P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7) > P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) > and now my constant terms work out correctly. > But P(x) has 49 as a factor as every term in > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 > gives me 22, as the new constant term. > Well that means that > P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) > is the only way that the constant terms keep matching.But *why* should that also be a factorisation in the === Abelian groups and K is a subgroup of G, then G is a torsiongroup iff both K and G/K are torsion groups.I can easily prove the => direction.(Assume G is a torsion group, then T(G) = G and hence G is finite. As well,all k in K are finite, since K is contained in G, and so by Lagrange theorem|G/K| = |G|/|K| means |G/K| is finite.However, I'm having trouble proving the other direction.(i.e.: Assuming K and G/K are torsion groups.)My best idea so far was to say:We know that |G/K| = |G|/|K| iff |G| is finite. Hence, if |G/K| = n and |K|= m thenn = |G|/m ==> nm = |G|. ==> |G| is finite.I'm pretty sure I can't do that.Could someone please give me a hint as how to proceed.(A hint! It's an === 8bit, Greg> If K and G are Abelian groups and K is a subgroup of G, then G is a torsion> group iff both K and G/K are torsion groups.> You want to think about the order of an element of the group not theorder of the group - two different things. [As has been poin outelsewhere, torsion does *not* imply finite.]-- Paul SperryColumbia, SC === groups and K is a subgroup of G, then G is a torsion> group iff both K and G/K are torsion groups.I can easily prove the => direction.> (Assume G is a torsion group, then T(G) = G and hence G is finite. As well,> all k in K are finite, since K is contained in G, and so by Lagrange theorem> |G/K| = |G|/|K| means |G/K| is finite.> Finite abelian groups are torsion groups.Torsion groups are not necessarily finite. Notice that Q/Z is torsion(given a rational number r = p/q, there is an integer N for which Nris an integer [q itself works, by the way: q*r = p]).You'd be better off using the actual definition of torsion rather thanthe presumed [and non-existent] finiteness of such groups.In other words, your argument based on finiteness is flawed.However, the ==> direction is absolutely trivial, if you rely only on 1. The definition of torsion (g is a torsion element of the abelian group G iff there is an integer n for which ng = 0). The abelian group G is a torsion group iff all its elements are torsion elements. 2. For any subgroup K of the abelian group G, the quotient homomorphism G --> G/K is well-defined.> However, I'm having trouble proving the other direction.> (i.e.: Assuming K and G/K are torsion groups.)> My best idea so far was to say:> We know that |G/K| = |G|/|K| iff |G| is finite. Hence, if |G/K| = n and |K|> = m then> n = |G|/m ==> nm = |G|. ==> |G| is finite.I'm pretty sure I can't do that.Could someone please give me a hint as how to proceed.> (A hint! It's an assignment!)> Given the SES (Short Exact Sequence: the arrows are homomorphisms,the image of each arrow is the kernel of the following arrow): 0 ---> K ---> G ---> G/K --->0let g be an element of G. You want to show that g is a torsion element.So, 1. Map g to [g] = g+K in G/K. a. G/K is torsion, so what can you say about [g]? b. Given exactness (Im = Ker) at G, what does that imply about g? are rela to the subgroup K. You also know (by hypothesis) that K is a torsion group. The obvious === step is the correct step at this point.> GREGDaleSubject: G are Abelian groups and K is a subgroup of G, then >> G is a torsion group iff both K and G/K are torsion groups.>The definition of torsion (g is a torsion element of the >abelian group G iff there is an integer n for which ng = 0). >The abelian group G is a torsion group iff all its elements >are torsion elements.Q/Z torsion group. If r+Z = p/q+Z in Q/Z: qr+Z = p+Z = ZDon't these definitions extend to non-Abelian groups?Thus Finite groups are torsion groups and theorem extends toK normal subgroup G ==> G torsion iff K and G/K torsionI see no reason why the proof below for Abelian groupswouldn't extend immediately as is, to the extended theorem.Proof ==> directionIf g+K in G/K: some n in N with ng = 0; n(g+K) = ng+K = KIf k in K: k in G some n in N with nk = 0 subgroups of torsion groups are torsionProof <== directionIf g in G: some n in N with K = n(g+K) = ng+K some k in K with === Re: Torsion Groups>> If K and G are Abelian groups and K is a subgroup of G, then>> G is a torsion group iff both K and G/K are torsion groups. >The definition of torsion (g is a torsion element of the>abelian group G iff there is an integer n for which ng = 0).>The abelian group G is a torsion group iff all its elements>are torsion elements.> Q/Z torsion group. If r+Z = p/q+Z in Q/Z: qr+Z = p+Z = ZDon't these definitions extend to non-Abelian groups?> Thus Finite groups are torsion groups and theorem extends to> K normal subgroup G == G torsion iff K and G/K torsionI see no reason why the proof below for Abelian groups> wouldn't extend immediately as is, to the extended theorem.> Sure. However, I was really only interes in addressing theOP's question. I was mainly drawn in by the OP's interpretationof torsion as being synonymous with finite.> Proof ==> direction ... proof dele ...> Proof <== direction ... proof dele ...----Here, I was trying to honor the OP's request for a hint only,as this was a homework problem. It wasn't really necessary,or even desirable, to solve the === Groups Adjunct Assistant Professor at the University of Montana.>> If K and G are Abelian groups and K is a subgroup of G, then>> G is a torsion group iff both K and G/K are torsion groups.>The definition of torsion (g is a torsion element of the>abelian group G iff there is an integer n for which ng = 0).>The abelian group G is a torsion group iff all its elements>are torsion elements.>Q/Z torsion group. If r+Z = p/q+Z in Q/Z: qr+Z = p+Z = ZDon't these definitions extend to non-Abelian groups?>Thus Finite groups are torsion groups and theorem extends to>K normal subgroup G == G torsion iff K and G/K torsionI see no reason why the proof below for Abelian groups>wouldn't extend immediately as is, to the extended theorem.Yes. Say G is a group, K a normal subgroup, and both K and G/K aretorsion (where being torsion means that for all x in G there existsn>0 such that x^n=e).Let x in G. Then since G/K is torsion, there exists N>0 such that x^Nin K. Since K is torsion, there exists M>0 such that (x^N)^M=e; sox^{NM}=e, proving that G is torsion.The converse is trivial because torsion groups are closed underquotients and === known that the zeta function only converges for re(s) > 1. Myunderstanding is that this is due to the pole at s = 1.The alternating zeta function (eta function) is stable because a term that addsthe zero at s=1 also adds complex zeros along the line re(s) = 1.Now if we look at 1/zeta(s), the pole becomes a zero. Does this affectthe region of convergence of 1/zeta(s)?If we take 1/eta(s), the complex zeroes that were complex poles that causethe function to diverge. Is this correct?If we use Eulers product form of zeta(s) instead of the infinite sumform, is rge region of convergence changed? What happens to the ROC ifwe look at 1/(zeta-product-form(s)) === Convergence of 1/zeta(s)> It is known that the zeta function only converges for re(s) > 1. My> understanding is that this is due to the pole at s = 1.The alternating zeta function (eta this pole. However, the same term that adds> the zero at s=1 also adds complex zeros along the line re(s) = 1.Now if we look at 1/zeta(s), the pole becomes a zero. Does this affect> the region of convergence of 1/zeta(s)?> If we take 1/eta(s), the complex zeroes that were introduced by the> term used to function to diverge. Is this correct?If we use Eulers product form of zeta(s) instead of the infinite sum> form, is rge region of convergence changed? What happens to the ROC if> we AdamsHere is an old post (from Feb 02) of mine which is rela to 1/eta:>Let a[1] = 1;For m >= 2, a[m] = sum{k|m, k>1} (-1)^k a[m/k],where the sum is over the divisors,k, k > 1, of m.Then:If m is odd, a[m] = mu(m), where mu() is the Mobius function.If m is even, a[m] = mu(n) 2^(k-1), >where m = n *2^k, n = odd integer, k = integer.The sequence {a[m]} begins:>1, 1, -1, 2, -1, -1, -1, 4, 0, -1, -1, -2, -1, -1, 1, 8, -1, 0,>-1, -2, 1, -1, -1, -4, 0, -1, 0, -2, -1, 1, -1, 16, 1, -1, 1, 0, >-1, -1, 1, -4, -1, 1, -1, -2, 0, -1, -1, -8, 0, 0, 1, -2, -1, 0,>1, -4, 1, -1, -1, 2, -1, -1, 0, 32, 1, 1, -1, -2, 1, 1, -1, 0,...The closed form for a[m] I found with the identity:>sum{m=1 to oo} a[m]/m^r =>1 /((1-2^(1-r)) zeta(r)).If x = -1 in my Reciprocal Of Polylogarithm post:>http://www.mathforum.com/epigone/sci.math/trimpfleldgroo >Then we have -a[-1,m] = a[m].I do not know if this post can at all help answer any of yourquestions, but it might be interesting to some sci.math readersinteres in 1/eta === 1/zeta(s)> It is known that the zeta function only converges for re(s) > 1.As Gerry poin out this is a category mistake: zeta is a function,not a series, and so cannot converge.> Now if we look at 1/zeta(s), the pole becomes a zero. Does this affect> the region of convergence of 1/zeta(s)?Again, I presume you mean the Dirichlet seriessum_{n=1}^infinity mu(n)/n^s when you say 1/zeta(s).This converges for Re(s) > 1 as is obvious. It alsoconverges (to the right value 0) for s = 1. This is aconsequence of the prime number theorem. I think(not 100% sure) that it converges to the right value on theline Re(s) = 1.But to the right of the line Re(s) = 1 we don't know. If itconverged for some s_0 with Re(s_0) < 1, it would convergeto 1/zeta(s) for all s with Re(s) > Re(s_0) nd this wouldimply that zeta was zero-free in the strip Re(s_0) < Re(s) < 1.No zero-free region of this type has ever been proved === It is known that the zeta function only converges for re(s) > 1. No, it isn't. What is known is that the series sum n^(-s) converges only for re(s) > 1. The rest of your post doesn't make much sense until you learn to distinguish between a function and a formula that represents the function in some region. In particular, your phrase, the region of convergence === solution to a alternating binomial sumHey Everyone,I've been trying to figure out the closed form solution to this sum,but have failed so far. Can anyone help? sum (-1)^{i+1} * C(j, i) * C( a, a/2 )^i * C( n-i*a, n/2 - i*a/2) i=1..jwhere a, n, j === 10:45 PM, john_correy@yahoo.com (John) said:--Nevertheless not all collections satisfy Standard Extensionality.>>It is a commonplace that *clubs* don't, for different clubs may be>>identical in membership. A club is not a collection, or at least not a collection of members. A>> club has other attributes besides its membership. To put this into the context of set theory, if x and y are not>> identical then neither are (x,y) and (y,x). Does that violate>> extensionality?I'm not clear on what you're asking.He's asking whether the fact that (1,2) <> (2,1) violates> extensionality - those ordered pairs have the same> elements but they're not equal.Hint: the answer is of course not.Hint: It was a rhetorical question, meant to illustrate why the> fact that two different clubs can have the same members> is not a violation of extensionality either.The ordered pairs (x,y) and (y,x)--that is, {{x},{x,y}}and {{y},{y,x}}--haven't the same elements. Is thissomething you didn't AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)ClassificationC4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Exhibit of proof of Ex~(x=x) from C1-C4 and === MKC Set Theories <3f98510e$10$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Cuser: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 05:59 PM, john_correy@yahoo.com (John) said:>The ordered pairs (x,y) and (y,x)--that is, {{x},{x,y}}>and {{y},{y,x}}--haven't the same asking. Two clubs with the same members don't havethe same elements either. Was that something you didn't know?-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolici bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === at 05:59 PM, john_correy@yahoo.com (John) said:The ordered pairs (x,y) and (y,x)--that is, {{x},{x,y}}>and {{y},{y,x}}--haven't the same elements. Is this>something you members don't have> the same elements either. Was that something you didn't know?Standard extensionality says that classes are determinedby their membersAxAy[Az(z in x <-> z in y) -> x=y] Standard ExtensionalityThis means that if x and y have the same membethen x=y.Because the axioms of MK include Standard Extensionality, in MKclasses *and* sets are determined by their members. As aresult, in MK the class of self-identicals and the class of setsare the same class. However, the axioms of MKC include Correy Extensionality instead ofStandard Extensionality.Correy Extensionality AxAy[Az(z in x <-> z in y) -> {(set x & set y) <-> x = y}] According to Correy Extensionality, classes with the same membersare distinct unless these are sets. So, because the class ofself-identicals and the class of sets are not sets, these aredistinct--even though they have the same members.How do we know that the class of sets and the class of self-identicals are not sets? Because this follows from CorreyExtensionality and MK Classification.MK Classification EyAx[x in y <-> set x & A] (with y not free in A) I hope this === <3f98510e$10$fuzhry+tra$mr2ice@news.patriot.net> <3fa02ca0$15$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Cuser: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS>Standard extensionality says that classes are determined by their>membersThis is just one of many cases where a word is overloaded. The peoplewho belong to a club are not members in the sense in which the term isused in set theory. Similarly, x and y are not members of (x,y) in thesense of set theory.>I hope this helps.No, you resta your original description, but did not state anymotivation. In particular, you did not address the fact thatextensionality does not prevent you from using set theory to describeclubs.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolici bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === at 05:59 PM, john_correy@yahoo.com (John) said:>The ordered pairs (x,y) and (y,x)--that is, {{x},{x,y}}>and {{y},{y,x}}--haven't the same elements. Is this>something you same members don't have>>the same elements either. Was that something you didn't know?> Standard extensionality says that classes are determined> by their membersAxAy[Az(z in x <-> z in y) -> x=y] Standard ExtensionalityThis means that if x and y have the same membethen x=y.Because the axioms of MK include Standard Extensionality, in MK> classes *and* sets are determined by their members. As a> result, in MK the class of self-identicals and the class of sets> are the same class. However, the axioms of MKC include Correy Extensionality instead of> Standard Extensionality.Correy Extensionality> AxAy[Az(z in x <-> z in y) -> {(set x & set y) <-> x = y}] > Three quick questions.(1) I asume that your in corresponds to the element of or epsilon notation used in MK, i.e. that when reading a type set version of atreatment of MK, I can substitute in for epsilon.(2) When I think of set theories (NGB for example), I usually think of first order theories without equality. Equality is then defined, essentially via your Standard Extensionality formula. In this case, the axiom of extensionality is slightly different, (X=Y) -> (X in Z <-> Y in Z)With this axiom in hand, one can then prove that the theory is in fact a theory with equality. I assume that this is not what you are doing. Would it be possible for you to translate your axiom into something along these lines? I just find it easier to grok when equality isdefined and then it is proven that equality satisfies the usual axioms.(3) What do you mean by set x? Is this just the standard way of distingushing between classes and sets, i.e. set X for (EY)(X in Y)or do you have some other definition in mind.> According to Correy Extensionality, classes with the same members> are distinct unless these are sets. So, because the class of> self-identicals and the class of sets are not sets, these are> distinct--even though they have the same members.How do we know that the class of sets and the class of self-> identicals are not sets? Because this follows from Correy> Extensionality and MK Classification.MK Classification> EyAx[x in y <-> set x & A] === T87eNBZOge9is6p33tBNOseFygUkgNUMye5KcLfYXA+VWIiKDkHpg0 Three quick questions.> Three quick answers. ;-) (1) I asume that your in corresponds to the element of or epsilon > notation used in MK, i.e. that when reading a type set version of a> treatment of MK, I can substitute in for epsilon.> Sure. (2) When I think of set theories (NGB for example), I usually think of > first order theories without equality. Equality is then defined, > essentially via your Standard Extensionality formula. In this case, the > axiom of extensionality is slightly different, (x = y) -> (x in z <-> y in z)With this axiom in hand, one can then prove that the theory is in fact a > theory with equality.>Right. This is well known. [Even in this NG.] I assume that this is not what you are doing.>Right.Would it be possible for you to translate your axiom into something > along these lines? I just find it easier to grok when equality is> defined and then it is proven that equality satisfies the usual axioms.> You might adopt with the following definition (instead of the usualone): a = b :<-> (set a & set b) & Ax(x e a <-> x e b).Your axiom (from above) may still be used.(This is the whole secret behind Correy's approach: he is talkingabout something one might call set equality, instead of the usualclass equality. Set equality holds for sets only.)This leads to the following rather alien theorems: ~Ax(x = x)or Ex~(x = x).(...since there are classes in MK which are no sets.) (3) What do you mean by set x? Is this just the standard way of > distinguishing between classes === TheoriesYou might adopt with the following definition (instead of the usual> one): a = b :<-> (set a & set b) & Ax(x e a <-> x e b).Your axiom (from above) may still be used.(This is the whole secret behind Correy's approach: he is talking> about something one might call set equality, instead of the usual> class equality. Set equality holds for sets only.)> Hmm... So this is all that John is doing? Do you know off the top ofyour head if this definition is in fact a equality. By this I meanare the usual axioms for a theory with equality theorems here?> This leads to the following rather alien theorems: ~Ax(x = x)> or> Ex~(x = x).(...since there are classes in MK which are no sets.)Well, since according to Mendelson, equality in a first order theory with equality must satisfy(x)(x = x)and(x = y) -> (A(x,x) -> A(x,y))where A(x,x) is an wf and A(x,y) arises from A(x,x) by replacing some,but not necessarily all, free occurences of x by y, with the proviso that y is free for the occurrences of x which it replaces.this implies that it is not a first order theory with === S1RSs+ZdQeR+4fqCRxB9NIPwz+59df0zATfQH4uFLTjtXCu5fNbhYjYou might adopt with the following definition (instead of the usual> one): a = b :<-> (set a & set b) & (x)(x e a <-> x e b).Your axiom (from above) may still be used.(This is the whole secret behind Correy's approach: he is talking> about something one might call set equality, instead of the usual> class equality. Set equality holds for sets only.)> Hmm... So this is all that John is doing?>Not *all*, I guess; but it's certainly the essence of his (settheoretic) approach. Do you know off the top of your head if this definition is in fact a equality. > [...]> If we restrict our considerations to sets (only) then the answer is yes.(In general no, of course). This leads to the following rather alien theorems: ~(x)(x = x)> or> (Ex)~(x = x).(...since there are classes in MK which are no sets.)Well, since according to Mendelson, equality in a first order theory > with equality must satisfy (x)(x = x)> In Correy's theory '=' is some sort of weak identity. H e r e we onlyhave: (x)(set x -> x = x). [...] this implies that it is not a first order theory with equality in the > usual sense.> === with the following definition (instead of the > usual one): > a = b :<-> (set a & set b) & (x)(x e a <-> x e b). > Your axiom (from above) may still be used. > (This is the whole secret behind Correy's approach: he is > talking about something one might call set equality, instead of > the usual class equality. Set equality holds for sets only.) >>> Hmm... So this is all that John is doing? >> Not *all*, I guess; but it's certainly the essence of his (set > theoretic) approach. >By *all* I was (incorrectly) thinking the following: In this modifiedapproach, equality is redefined, but the axioms are unchanged. Thus anyinterpretation for MK would also be an interpretation for MKC. Now I'min business, I actually have some intuitions about interpretations ofMK. Of course, this is incorrect. Equality is an ingredient in severalaxioms (for instance paring and replacement as I know these axioms).John would probably want these to change too. It may be the case thatthese occurances of equality apply only to set variables and so changingthe definition would actually have no effect on the axioms, but this isfar from clear. This is always the problem when redefining a notation Iknow and love (or hate) -- === TheoriesX-DMCA-Notifications: http://www.giganews.com/info/dmca.html> at 10:45 PM, john_correy@yahoo.com (John) said:--Nevertheless not all collections satisfy Standard Extensionality.>It is a commonplace that *clubs* don't, for different clubs may be>identical in membership.A club is not a collection, or at least not a collection of members. A> club has other attributes besides its membership.To put this into the context of set theory, if x and y are not> identical then neither are (x,y) and (y,x). Does that violate> extensionality?>>I'm not clear on what you're asking. He's asking whether the fact that (1,2) <> (2,1) violates>> extensionality - those ordered pairs have the same>> elements but they're not equal. Hint: the answer is of course not. Hint: It was a rhetorical question, meant to illustrate why the>> fact that two different clubs can have the same members>> is not a violation of extensionality either.The ordered pairs (x,y) and (y,x)--that is, {{x},{x,y}}>and {{y},{y,x}}--haven't the same elements. Is this>something you didn't know?They don't have the same _elements_ - that's why Iput the word elements in quotes. The way you insist on missing points is remarkable.The _point_ was that (x,y) <> (y,x) does not violateextensionality, precisely _because_ (x,y) is nota set with elements x and y. The point to the pointbeing that in exactly the same way your commenton clubs does not say anything about extensionality(or about the need for a new and improved set theorywithout extensionality) because a club is not a setwith elements equal to its y)]>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)>Classification>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point out>the error.>--David === Theories at 10:45 PM, john_correy@yahoo.com (John) said:--Nevertheless not all collections satisfy Standard Extensionality.>It is a commonplace that *clubs* don't, for different clubs may be>identical in membership.A club is not a collection, or at least not a collection of members. A> club has other attributes besides its membership.To put this into the context of set theory, if x and y are not> identical then neither are (x,y) and (y,x). Does that violate> extensionality?>>I'm not clear on what you're asking. He's asking whether the fact that (1,2) <> (2,1) violates>> extensionality - those ordered pairs have the same>> elements but they're not equal. Hint: the answer is of course not. Hint: It was a rhetorical question, meant to illustrate why the>> fact that two different clubs can have the same members>> is not a violation of extensionality either.The ordered pairs (x,y) and (y,x)--that is, {{x},{x,y}}>and {{y},{y,x}}--haven't the same elements. Is this>something you didn't know?They don't have the same _elements_ - that's why I> put the word elements in quotes. The way you insist on missing points is remarkable.> The _point_ was that (x,y) <> (y,x) does not violate> extensionality, precisely _because_ (x,y) is not> a set with elements x and y. The point to the point> being that in exactly the same way your comment> on clubs does not say anything about extensionality> (or about the need for a new and improved set theory> without extensionality) because a club is not a set> with elements equal to its members. David and _______ Debauched dual And David, --What else!-- A Mathies are(collectively) the most arrogant, unimaginative and short-sighsame-sayers ever to have prowled the face of the earth. You inparticular embody a plethora of distasteful mathietraits. You distort, you lie, you taunt, you belittle, and youconceal. In addition to which you have no more capacity for...mathematics than Herbert Simon's bot.--John AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)>Classification>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point out>the error.>--David === Theories>>The ordered pairs (x,y) and (y,x)--that is, {{x},{x,y}}>>and {{y},{y,x}}--haven't the same elements. That is only if one accepts the Kuratowski definitionof an ordered pair. The Kuratowski definition is simplyon (of many) ways of modelling the primitive notionof ordered pair within set theory.> Is this>>something you didn't know?Correy shouldn't assume that since he is stupidand ignorant, that other people are too.> They don't have the same _elements_ - that's why I> put the word elements in quotes.The way you insist on missing points is remarkable.Deliberately obtuse.> The _point_ was that (x,y) <> (y,x) does not violate> extensionality, precisely _because_ (x,y) is not> a set with elements x and y. The point to the point> being that in exactly the same way your comment> on clubs does not say anything about extensionality> (or about the need for a new and improved set theory> without extensionality) because a club is not a set> with elements equal to its members.Obvious to all save Correy, no doubt.-- === irrational?>At some level of these iterations, will (e) become an irrational?No. As you've noticed, at each approximation of e, the approximationis rational. It's only when you take the limit (i.e. as the error term goesto zero) that the limit === become an irrational?At some level of these iterations, will (e) become an irrational? No. As you've noticed, at each approximation of e, the approximation> is rational. It's only when you take the limit (i.e. as the error termgoes> to zero) that the limit becomes irrational.It's going to zero every step of the way. It's when it gets to 0 that theexpansion === What are computable numbers?A real number r is computable if there is a computable functionf from the natural numbers to the rationals, such that|f(n)-r|<1/n for every n. Take any reasonable model of computationand way of representing natural === What is Advanced Calculus?I am a bit confused.Maybe it is because I self taught myself elementary calculus, and inthe process of doing this exhaus not 1 but 4 calculus textbooks,including Stewart's huge e and Swokowski's excellent book (myfavorite of the 4). So maybe I have acquired an understanding ofepsilon-delta and Riemann sums that most college students are justassumed to skim through and then memorize via rote only what is neededto pass the exams?Because I have looked at several advanced calculus texts and theyseem to be utterly the same as elementary texts, except all 3 coursescondensed into 1 and one single Fourier section added to theinfinite series chapter. Aside from that Fourier stuff, I findbasically no differences between advanced calculus and normalcalculus!It seems as though Multivariable Mathematics is just another namefor Advanced Calculus, and the same applies to it as well.So my calculus has hit a rut because any time I pick up an advancedcalc text I end up becoming terribly bored as it's basically nothingbut review (and one section on === tanbanso@iinet.net.auX-CompuServe-Cuser: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 06:06 PM, dreamvigile@hotmail.com (dreamvigile) said:>I am a bit confused.Welcome to the Tower of Babel. The description Advanced Calculus hasno standardized meaning. It's used for a grab bag of topics that theauthor doesn't believe are covered in the standard Calculus courses,which are themselves highly variable.Some topics that are often covered in Advanced Calculus courser are: Analysis Multi-variable Calculus[1] Vector Analysis in 3-space[1] Series Transcendental functions Introduction to differential equationsThe bot line is that it's whatever the author wants it to be.[1] There's no standard content for that, either.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolici bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Calculus?dreamvigile> I am a bit confused.> Maybe it is because I self taught myself elementary calculus, and in> the process of doing this exhaus not 1 but 4 calculus textbooks,> including Stewart's huge e and Swokowski's excellent book (my> favorite of the 4). So maybe I have acquired an understanding of> epsilon-delta and Riemann sums that most college students are just> assumed to skim through and then memorize via rote only what is needed> to pass the exams?> Because I have looked at several advanced calculus texts and they> seem to be utterly the same as elementary texts, except all 3 courses> condensed into 1 and one single Fourier section added to the> infinite series chapter. Aside from that Fourier stuff, I find> basically no differences between advanced calculus and normal> calculus!> It seems as though Multivariable Mathematics is just another name> for Advanced Calculus, and the same applies to it as well. So my calculus has hit a rut because any time I pick up an advanced> calc text I end up becoming terribly bored as it's basically nothing> but review (and one section on Fourier).To begin at the end, Fourier analysis (or harmonic analysis) is a _very_powerful theme, and well worth a good look.Like the other respondeI recommend starting with books intendended forengineers and physicists -- less formal but more intelligible. I still haveSpiegel's Advanced Calculus from the 60's. I hope the Schaum peoplehaven't changed it too much. It's good for self-study in that it has a lotof fully worked-out exercises. Also in the exercises are many substantial,well-chosen applications: the evaluation of zeta(2) and zeta(4), a proof ofWallis's product and some of its relatives, things about the Gaussiandistribution, and a bunch more.The pedagogy of calculus is a hardy annual, indeed. IMO Advanced Calculusis an example of what I call textbook language. Another example isFundamental Theorem of Such-and-such. These names are just loosely definedjargon deriving from pedagogical or institutional === now called Analysis used to be called Advanced calculus. It ismerely an archaic term. Advanced calc. these days can take on bothmulti-variable calculus and Analysis.Lurch> dreamvigile> I am a bit confused.> Maybe it is because I self taught myself elementary calculus, and in> the process of doing this exhaus not 1 but 4 calculus textbooks,> including Stewart's huge e and Swokowski's excellent book (my> favorite of the 4). So maybe I have acquired an understanding of> epsilon-delta and Riemann sums that most college students are just> assumed to skim through and then memorize via rote only what is needed> to pass the exams?> Because I have looked at several advanced calculus texts and they> seem to be utterly the same as elementary texts, except all 3 courses> condensed into 1 and one single Fourier section added to the> infinite series chapter. Aside from that Fourier stuff, I find> basically no differences between advanced calculus and normal> calculus!> It seems as though Multivariable Mathematics is just another name> for Advanced Calculus, and the same applies to it as well. So my calculus has hit a rut because any time I pick up an advanced> calc text I end up becoming terribly bored as it's basically nothing> but review (and one section on Fourier).> To begin at the end, Fourier analysis (or harmonic analysis) is a _very_> powerful theme, and well worth a good look.> Like the other respondeI recommend starting with books intendended for> engineers and physicists -- less formal but more intelligible. I stillhave> Spiegel's Advanced Calculus from the 60's. I hope the Schaum people> haven't changed it too much. It's good for self-study in that it has a lot> of fully worked-out exercises. Also in the exercises are many substantial,> well-chosen applications: the evaluation of zeta(2) and zeta(4), a proofof> Wallis's product and some of its relatives, things about the Gaussian> distribution, and a bunch more.> The pedagogy of calculus is a hardy annual, indeed. IMO AdvancedCalculus> is an example of what I call textbook language. Another example is> Fundamental Theorem of Such-and-such. These names are just looselydefined> jargon deriving from pedagogical or institutional motives.> === you might enjoy working through a text - and there are many,look for Advanced Mathematics for Engineeit is usually a 3 or 4 semestercourse taught from the same book, similar to a decent 3 or 4 semester coursein Calculus. It will cover alot of applied stuff, Fourier Series,differential equations, etc.Advanced Calculus could be a more thorough calc, or it could be RealAnalysis depending on your school. Real Analysis is a further refinement ofepsilon delta, proofs, clarification on open-ness, closedness, convergence,sets, domains, intgrability. It is much more rigorous than the 1st 4semesters of calculus.I have a copy of Swokowski and I agree - this is a most excellent book withthe exception of a few errors here and there. But if you like Swokowski, getyourself a copy of Salas, Hill. You wont regret it.> I am a bit confused.> Maybe it is because I self taught myself elementary calculus, and in> the process of doing this exhaus not 1 but 4 calculus textbooks,> including Stewart's huge e and Swokowski's excellent book (my> favorite of the 4). So maybe I have acquired an understanding of> epsilon-delta and Riemann sums that most college students are just> assumed to skim through and then memorize via rote only what is needed> to pass the exams?Epsilon Delta is really very simple, but annoying to most people and so Ithink that you are doing yoursefl a huge favor by looking at it.> Because I have looked at several advanced calculus texts and they> seem to be utterly the same as elementary texts, except all 3 courses> condensed into 1 and one single Fourier section added to the> infinite series chapter. Aside from that Fourier stuff, I find> basically no differences between advanced calculus and normal> calculus!> It seems as though Multivariable Mathematics is just another name> for Advanced Calculus, and the same applies to it as well. So my calculus has hit a rut because any time I pick up an advanced> calc text I end up becoming terribly bored as it's basically nothing> but review === taught myself elementary calculus, and in> the process of doing this exhaus not 1 but 4 calculus textbooks,> including Stewart's huge e and Swokowski's excellent book (my> favorite of the 4). So maybe I have acquired an understanding of> epsilon-delta and Riemann sums that most college students are just> assumed to skim through and then memorize via rote only what is needed> to pass the exams?> Because I have looked at several advanced calculus texts and they> seem to be utterly the same as elementary texts, except all 3 courses> condensed into 1 and one single Fourier section added to the> infinite series chapter. Aside from that Fourier stuff, I find> basically no differences between advanced calculus and normal> calculus!> It seems as though Multivariable Mathematics is just another name> for Advanced Calculus, and the same applies to it as well. So my calculus has hit a rut because any time I pick up an advanced> calc text I end up becoming terribly bored as it's basically nothing> but review (and one section on Fourier).I think this is one of those terms that has been overloaded in usage so thisquestion is difficult to answer. Does this help? Not sure it is an easyquestion to answer as fractions are quite advanced for some.1. http://oregonstate.edu/~peterseb/mth311/docs/311note_ whatis.html2. http://www.math.rochester.edu/courses/164/home/ description.htmlHere is an course outline at a university:ADVANCED CALCULUSThis course is taught on Tuesday evenings throughout the autumn term withapproximately 10 evenings of lectures.Course outlineFunctions of more than one variable.Partial differentiation and its applications.Multiple Integrals.Higher order differential equations.Laplace transforms.Applications of Calculus in the Social Sciences.Recommended textbooks:Adams, R. A. Calculus of several variables, Addison-Wesley.Adams, R. A. Calculus: A complete course, Addison-Wesley.Harris, K. Discovering Calculus with Maple, Wiley.Goldsmith, C. & Nelson, D. Extensions of Calculus, Cambridge.Does that constitue advanced? Hard to answer as it is perspective Isuppose.Here is another course outline:http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/ notes.htmlNot sure this answers your question, but hope it === anti-correlation? determinantapparently p must be greater than -1/(N-1) because the determinant>of the highly symmetric matrix is >det = (1-p)^(N-1) (1 + (N-1)p). So >det > 0 if, and only if, p > 1/(N-1). Now, what is the simplest way to prove that my formula for det is correct? Take v to be the (column) vector of dimension n ...Since you apparently recognized this question right off the bat, mayone out of curiosity ask where === anti-correlation? determinantX-SessionID: 5bc2b5e9>>apparently p must be greater than -1/(N-1) because the determinant>>of the highly symmetric matrix is >>det = (1-p)^(N-1) (1 + (N-1)p). So >>det > 0 if, and only if, p > 1/(N-1). >>Now, what is the simplest way to prove that my formula for det is correct?>> Take v to be the (column) vector of dimension n ...Since you apparently recognized this question right off the bat, may>one out of curiosity ask where you recognized it from?I've had a lot to do with projection operatoover the year. My favorite geometry tool. So, when I see one I can recognize it. And representing a symmetric operator as a combination of projections usually pays off. It sure did this time.Mati Meron | When you argue with a fool,meron@cars.uchicago.edu | chances are he is doing just === the sameSubject: Re: JSH: More advanced topic with my latest people who like to lie> about mathematics, so I've just talked about resolving an issue that> I'd noticed brought up by Nora Baron, Dik Winter, and possibly> others.But I decided to talk about it now, so here's quick explanation, as I> make sure that I'm right.I have an expression that is P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3where the key term for this discussion is- 3(-1 + 49 x )(5)(7^2)because if x=1/49, it goes to 0.Ok.Some people noticed I guess that you can just take the 49 away, and> have7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3so now, x=1, will work to 0 out that term.Huh? This is a different polynomial entirely.But then you have7^2(2401 - 147 + 3) (5^3) + 7^3which factors easily enough.The answer is that what may seem like a simple maneuver gives a result> that is also given with a different expression by something like> x=1/49 with the first.But this *isn't* going to tell you anything about what happens in the first polynomial.That is, there exists an expression for which x=1/49 will give you7^2(2401 - 147 + 3) (5^3) + 7^3so mathematically, that possibility has precedence because for> consistency, you can have 7 a unit in the ring, and cover both> possibilities.Where is this coming from?That is, given7^2(2401 - 147 + 3) (5^3) + 7^3there's no way to tell mathematically whether or not it came from one> expression or the other, so the math defaults to the more general> condition, which is that 7 is a unit.> James, pick a polynomial and work with it. I don't care which one.-- === technique> I've been somewhat wary of giving fodder to people who like to lie> about mathematics, so I've just talked about resolving an issue that> I'd noticed brought up by Nora Baron, Dik Winter, and possibly> others.But I decided to talk about it now, so here's quick explanation, as I> make sure that I'm right.I have an expression that is P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3where the key term for this discussion is- 3(-1 + 49 x )(5)(7^2)because if x=1/49, it goes to 0.Some people noticed I guess that you can just take the 49 away, and> have7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3so now, x=1, will work to 0 out that term.But then you have7^2(2401 - 147 + 3) (5^3) + 7^3which factors easily enough.> By which you mean, I think, it factors into linear factors where 5 plays the role of a polynomial variable - that is, 7^2*2257*5^3 + 7^3 = 7^2(2257*5^3 + 7),can be factored as (a1*5 + 7)*(a2*5 + 7)*(a3*5 + 7),where you know that each of a1, a2, and a3 has a factorof 7^{2/3} because it is easy to solve for them explicitly.[Again, you are really thinking of 5 here not as a number, butas a polynomial variable.]> The answer is that what may seem like a simple maneuver gives a result> that is also given with a different expression by something like> x=1/49 with the first.That is, there exists an expression for which x=1/49 will give you7^2(2401 - 147 + 3) (5^3) + 7^3> Well, no, because here you substitu x = 1/49 in onlyone place. In your original expression,P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3notice that x occurs in SEVERAL places. What you are doingwhen you let x = 1/49 does not give you the same answer as youget when you simply 'take the 49 away' and set x = 1 throughout.You seem to be trying to do both things: setting x = 1/49 in oneplace, and then also setting it equal to 1 in other places in your expression: inconsistent.> so mathematically, that possibility has precedence because for> consistency, you can have 7 a unit in the ring, and cover both> possibilities.That is, given7^2(2401 - 147 + 3) (5^3) + 7^3there's no way to tell mathematically whether or not it came from one> expression or the other, so the math defaults to the more general> condition, which is that 7 is a unit.> The math defaults to the more general condition ????This is your 'latest technique'? You just make up a phraseof some kind and claim it is a true principle ?? I have newsfor you. This is not how math is done. You start with axiomsand you construct proofs. The B.S. you just burped up here is not one of the axioms. It is totally off the reservation, and you know it. Not only that, but, as off-the-wall loony as your new principle is, it looks like you === More advanced topic with my latest technique> I've been somewhat wary of giving fodder to people who like to lie> about mathematics, so I've just talked about resolving an issue that> I'd noticed brought up by Nora Baron, Dik Winter, and possibly> others.But I decided to talk about it now, so here's quick explanation, as I> make sure that I'm right.I have an expression that is P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3where the key term for this discussion is- 3(-1 + 49 x )(5)(7^2)because if x=1/49, it goes to 0.Some people noticed I guess that you can just take the 49 away, and> have7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3so now, x=1, will work to 0 out that term.But then you have7^2(2401 - 147 + 3) (5^3) + 7^3which factors easily enough.> By which you mean, I think, it factors into linear factors where > 5 plays the role of a polynomial variable - that is,Hmmm...the latest insanity from math society. So you think that 5 is a variable?Or I guess as you'd === advanced topic with my latest technique> I've been somewhat wary of giving fodder to people who like to lie> about mathematics, so I've just talked about resolving an issue that> I'd noticed brought up by Nora Baron, Dik Winter, and possibly> others.> But I decided to talk about it now, so here's quick explanation, as I> make sure that I'm right.> I have an expression that is > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> where the key term for this discussion is> - 3(-1 + 49 x )(5)(7^2)> because if x=1/49, it goes to 0.> Some people noticed I guess that you can just take the 49 away, and> have> 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3> so now, x=1, will work to 0 out that term.> But then you have> 7^2(2401 - 147 + 3) (5^3) + 7^3> which factors easily enough.By which you mean, I think, it factors into linear factors where > 5 plays the role of a dele the rest of my post without havinga chance to read it. Here it is again for your convenience -since it includes substantive objections to which you mightwant to respond - 7^2*2257*5^3 + 7^3 = 7^2(2257*5^3 + 7),can be factored as (a1*5 + 7)*(a2*5 + 7)*(a3*5 + 7),where you know that each of a1, a2, and a3 has a factorof 7^{2/3} because it is easy to solve for them explicitly.> The answer is that what may seem like a simple maneuver gives a result> that is also given with a different expression by something like> x=1/49 with the first.That is, there exists an expression for which x=1/49 will give you7^2(2401 - 147 + 3) (5^3) + 7^3> Well, no, because here you substitu x = 1/49 in onlyone place. In your original expression,P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3notice that x occurs in SEVERAL places. What you are doingwhen you let x = 1/49 does not give you the same answer as youget when you simply 'take the 49 away' and set x = 1 throughout.You seem to be trying to do both things: setting x = 1/49 in oneplace, and then also setting it equal to 1 in other places in your expression: inconsistent.> so mathematically, that possibility has precedence because for> consistency, you can have 7 a unit in the ring, and cover both> possibilities.That is, given7^2(2401 - 147 + 3) (5^3) + 7^3there's no way to tell mathematically whether or not it came from one> expression or the other, so the math defaults to the more general> condition, which is that 7 is a unit.> The math defaults to the more general condition ????This is your 'latest technique'? You just make up a phraseof some kind and claim it is a true principle ?? I have newsfor you. This is not how math is done. You start with axiomsand you construct proofs. The B.S. you just burped up here is not one of the axioms. It is totally off the reservation, and you know it. Not only that, but, as off-the-wall loony as your new principle is, it looks like you === More advanced topic with my latest technique> I've been somewhat wary of giving fodder to people who like to lie> about mathematics, so I've just talked about resolving an issue that> I'd noticed brought up by Nora Baron, Dik Winter, and possibly> others.> But I decided to talk about it now, so here's quick explanation, as I> make sure that I'm right.> I have an expression that is > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> where the key term for this discussion is> - 3(-1 + 49 x )(5)(7^2)> because if x=1/49, it goes to 0.> Some people noticed I guess that you can just take the 49 away, and> have> 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3> so now, x=1, will work to 0 out that term.> But then you have> 7^2(2401 - 147 + 3) (5^3) + 7^3> which factors easily enough.> By which you mean, I think, it factors into linear factors where > 5 plays the role you dele the rest of my post without having> a chance to read it. Here it is again for your convenience -> since it includes substantive objections to which you might> want to respond -Readers should note that I questioned the sanity of claiming that 5can play the role of a variable, and the poster Andrzej Kolowskidele out what came after Hmmm....Here's the full sentence:Hmmm...the latest insanity from math society. So you === topic with my latest technique[apologies to the local googolplex warden,for butting-in again, when I'd just said a hail-fair-thee-wellto our Math Hero; please remove some of my googol priveleges,per the book.]you get awfully touchy, when someone ignores your peculiar usages, *or*takes them seriously (as in using a number as the variable). I mean,you get snippy about such a thing, but you don't botherto address the question. but, hey; it's *your* Fan Club. since you so deftly evade the tiny audience-participation (we,the proud & few), there has got to be another reason for it. or, you will answer a query, but onlyby starting a new topic, as with C.Bond's correction.above. on the other hand,I could never see why you use such complica formulae,when simple, two- or three-digit coefficients would do; why? > Readers should note that I questioned the sanity of claiming that 5> can play the role of a variable, and the poster Andrzej Kolowski> dele out what came after Hmmm....Here's the full sentence:Hmmm...the latest insanity from math society. So you think that 5 is> a === topic with my latest technique>> Gee, too bad you dele the rest of my post without having>> a chance to read it. Here it is again for your convenience ->> since it includes substantive objections to which you might>> want to respond -Readers should note that I questioned the sanity of claiming that 5>can play the role of a variable, and the poster Andrzej Kolowski>dele out what came after Hmmm....Yeah, readers can note that you snipped his explanation the firsttime, just after the words that is... where he explained what hemeant.Then readers can note that you snipped his explanation a second time,and again pretend that there is no explanation.Some readers might call this dishonest. === technique> I've been somewhat wary of giving fodder to people who like to lie> about mathematics, so I've just talked about resolving an issue that> I'd noticed brought up by Nora Baron, Dik Winter, and possibly> others.> But I decided to talk about it now, so here's quick explanation, as I> make sure that I'm right.> I have an expression that is > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> where the key term for this discussion is> - 3(-1 + 49 x )(5)(7^2)> because if x=1/49, it goes to 0.> Some people noticed I guess that you can just take the 49 away, and> have> 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3> so now, x=1, will work to 0 out that term.> But then you have> 7^2(2401 - 147 + 3) (5^3) + 7^3> which factors easily enough.> By which you mean, I think, it factors into linear factors where > 5 plays the role of a polynomial variable - that is,> having> a chance to read it. Here it is again for your convenience -> since it includes substantive objections to which you might> want to respond -Readers should note that I ... [delete] Readers should note that Mr. Harris dele out all the restof my reply to his post. It includes substantive objections to what he said, and to which, evidently, he does not care to respond. Here it is anyway:By which you mean, I think, it factors into linear factors - that is, 7^2*2257*5^3 + 7^3 = 7^2(2257*5^3 + 7),can be factored as (a1*5 + 7)*(a2*5 + 7)*(a3*5 + 7),where you know that each of a1, a2, and a3 has a factorof 7^{2/3} because it is easy to solve for them explicitly.> The answer is that what may seem like a simple maneuver gives a result> that is also given with a different expression by something like> x=1/49 with the first.That is, there exists an expression for which x=1/49 will give you7^2(2401 - 147 + 3) (5^3) + 7^3> Well, no, because here you substitu x = 1/49 in onlyone place. In your original expression,P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3notice that x occurs in SEVERAL places. What you are doingwhen you let x = 1/49 does not give you the same answer as youget when you simply 'take the 49 away' and set x = 1 throughout.You seem to be trying to do both things: setting x = 1/49 in oneplace, and then also setting it equal to 1 in other places in your expression: inconsistent.> so mathematically, that possibility has precedence because for> consistency, you can have 7 a unit in the ring, and cover both> possibilities.That is, given7^2(2401 - 147 + 3) (5^3) + 7^3there's no way to tell mathematically whether or not it came from one> expression or the other, so the math defaults to the more general> condition, which is that 7 is a unit.> The math defaults to the more general condition ????This is your 'latest technique'? You just make up a phraseof some kind and claim it is a true principle ?? I have newsfor you. This is not how math is done. You start with axiomsand you construct proofs. The B.S. you just burped up here is not one of the axioms. It is totally off the reservation, and you know it. Not only that, but, as off-the-wall loony as your new principle is, it looks like you have not even applied it === latest techniqueReaders should note that I questioned the sanity of claiming that 5> can play the role of a variable, and the poster Andrzej Kolowski> dele out what came after Hmmm....What is the role of a variable?Good question. Ask Andrzej Kolowski as he came up with the insanity.I want readers to see that in trying to cast doubt on my work, peoplelike Andrzej Kolowski necessarily have to attack mathematics. So payattention to the statements by Andrzej Kolowski.As, of course, 5 is a NUMBER, not a variable, so how === advanced topic with my latest technique> Readers should note that I questioned the sanity of claiming that 5> can play the role of a variable, and the poster Andrzej Kolowski> dele out what came after Hmmm....What is the role of a variable?Good question. Ask Andrzej Kolowski as he came up with the insanity.after play the role of a variable, that is... you'dunderstand what he means by the term play the role ofa variable.Here's the text you keep deleting:>> ...5 plays the role of a polynomial variable - that is,>> 7^2*2257*5^3 + 7^3 = 7^2(2257*5^3 + 7),>> can be factored as>> (a1*5 + 7)*(a2*5 + 7)*(a3*5 + 7),>> where you know that each of a1, a2, and a3 has a factor>> of 7^{2/3} because it is easy to solve for them explicitly.> As, of course, 5 is a NUMBER, not a variable, so how can it play the> role of a variable?In the English language, the phrase that is, usuallycomes between a phrase whose meaning you might wonder about,and the author's explanation of the meaning. Thus, if youwonder what he meant, keep reading past the key wordsthat is,... See? Isn't that