mm-235 === Subject: Re: Sylow Groupin message remember if G is a finite cyclic group of order n, then G has> exactly one subgroup of order d for each divisor d of n, and G has no> other subgroups. Can I generalize that for any finite group G of order n,> G has exactly one subgroup of order d for each divisor d of n, and G> has no other subgroupsOthers have told you why this is wrong. You might find itinteresting though that a converse of your first paragraph aboveholds: If G is a finite group of order n with exactly onesubgroup of order d for each divisor d of n, then G is cyclic.-- Jim === is not divided by rin message : You should always include the hypothesis in the text, not just the> header or subject. So, Let G be a group of order p^m*r. It seems to me that you mean> where p does not divide r, rather than where p is not divided by> r, since presumably p is a prime number?Almost certainly.>Let X be the set of all subsets of G of size p^m. Then>the textbook says |X| = C_{p^m}^{p^mr} but I don't know (p^m*r choose p^m,> perhaps)? Because my first impulse was to say that that did not make> sense, since C is the centralizer.Almost certainly the choice coefficient. The OP just staranother thread on Sylow's Theorems, and this coefficient is thefirst step in a common permutation-group-orien proof of them.Namely: |X| is coprime to p, so the action of G on X has anorbit of length coprime to p, so if r > 1 the stabilizer of anelement of that orbit is a proper subgroup G' of G such that|G:G'| is coprime to p, so by === random oracle model; replace the (fixed)| hash function by a random function, do your computations assuming| you're dealing with a random function, and then hope that your| computations carry over to the real, fixed hash function.| | This is only a heuristic -- but it seems to be a darn effective one.I think I see why that method works in the present case:Considering functions that map finite set X to finite set Y,the average of |g(X)|/|Y| (over all functions g: X -> Y) is 1 - (1 - 1/|Y|)^|X|. Now if X and Y are sufficiently large, the corresponding *variance* ~= 0, implying that for nearly all functions g, |g(X)|/|Y| is approximately equal to the average ~= 1 - exp(-|X|/|Y|). More precisely, if f is uniformly distribu on the set of all functions that map X to Y, and |X|->oo and |Y|->oo with fixed ratio r = |X|/|Y|, then |f(X)|/|Y| ->(in distribution) 1 - exp(-r)E |f(X)|/|Y| = 1 - exp(-r) (1 - (r/2)/|X|) + O(1/|X|^2) -> 1 - exp(-r)V |f(X)|/|Y| = r exp(-r) (1 - (r+1)exp(-r)) / |X| + O(1/|X|^2) -> 0.A proof is sketched at === Subject: Re: hash/SHA-1 questionsOriginator: daw@mozart.cs.berkeley.edu (David Wagner)>Considering functions that map finite set X to finite set Y,>the average of |g(X)|/|Y| (over all functions g: X -> Y) is >1 - (1 - 1/|Y|)^|X|. Now if X and Y are sufficiently large, >the corresponding *variance* ~= 0, implying that for nearly >all functions g, |g(X)|/|Y| is approximately equal to the >average ~= 1 - exp(-|X|/|Y|). Yeah, that's the sort of justification I had in mind -- but I was toolazy to work out the details. Nicely put.That said, there is one major caveat here. In practice, we don't pickour hash functions uniformly at random; not even close. For instance, weprobably select only from the subset of functions that can be implemenin at most a million gates. However, almost all functions require morethan a million gates; thus, we're selecting from a negligible subset ofthe whole space. If there were some correlation between having smallcircuits and having an unusually large (or small) value of |g(X)|/|Y|,then all bets would be off. The random oracle heuristic implicitlyassumes that any such correlations can be neglec, but I don't knowof any way to test this === Variables> Hi Mr. Koopman,And my point also remains unchanged: that you should start by considering> what happens to the mean and variance when you increase n and decrease p> in a way that leaves n*p constant, which is what your values approximate.Are you making a reference to the relation between the Poisson and the> Binomial distribution?If that's the case, there are two interesting sites about this fact:http://mathworld.wolfram.com/PoissonDistribution.htmlhttp === Re: Equivalent Binomial Random Variables> Unless I misunderstood your point, the mean and variance will remain> unchanged. That is, X1, X2 and X3 will have approximately the same> mean and variance, but is enough to let me state that they are> equivalent? If not, what are the conditions in order to state so?X1, X2, and X3 will have approximately the same mean, variance,> skewness, kurtosis, etc; that is, their PMFs will be approximately> the same (except, as you have no) in the tails. How close they> must be before you declare them equivalent is a subjective matter:> equivalence of this sort is in the eye of the beholder.Are you making a reference to the relation between> the Poisson and the Binomial distribution?Yes.I want to thank you for your comments and === rZxVEQZJwevoDFqxC7Sa7XF4dkjn7mylwdPwzrEsir0sDIyZjI0Hg0there recently were discussions in this group regarding automorphisms of finite simple groups. I left University ten years ago and am just curious if meanwhile anything happened with Schreier's Conjecture (the outer automorphism group of a finite simple group is soluble). Has anyone meanwhile succeeded (or even attemp) to find a proof of this fact that does not use the classification?By the way - how complex can those outomorphism groups get? I vaguely recall that the === Automorphisms of finite simple groups there recently were discussions in this group regarding automorphisms of >finite simple groups. I left University ten years ago and am just >curious if meanwhile anything happened with Schreier's Conjecture (the >outer automorphism group of a finite simple group is soluble). Has >anyone meanwhile succeeded (or even attemp) to find a proof of this >fact that does not use the classification?Schreier's Conjecture is true, but I am almost certain that the only proofuses the classification.>By the way - how complex can those outomorphism groups get? I vaguely >recall that the nilpotent length is at most three...Yes, that's right. An example with nilpotent length 3 is O^+(8,3), === Goodness of fit for a chi-squared distributionI have data that may be chi-squared distribu. I want to test thishypothesis with the Anderson Darling (AD) statistic. My currentapproach is to estimate the confidence limits through Monte-Carlosimulations, which is a lengthy procedure.Is it possible to use the relationship between the normal distributionand chi-squared to use the AD test for a normal distribution? Is thereanother approach I might try?The Kolmogorov-Smirnov test is not suitable because I estimate thechi-squared order from the data. The chi-squared-test needs a goodlyamount of data and is === generating pseudo-random numbers using galois fieldsI have an application in which I want to generate a pseudo-random id, x(i) tomatch an item's numeric id i i.e. a series X s.t i->x(i).Ideally, I would like the series X to be determinate, and furthermore, Iwant the series X to be bounded by some value p s.t. x(i)

4,31,13,22,211,121,112,1111> 5,41,14,32,23,311,131,113,221,212,122,2111,1211,1121,1112,11111 Looks like a simple 2**(n-1). Is it?Yes. In order to form all the 'unsor partitions' of n elements, you canpreced with '1' that ones of (n-1) elements, with 2 that ones of n-2elements, ... and with (n-1) the partition of one element. Adittionally, youmust take 'n'.UP(n) = UP(n-1) + UP(n) + ... + UP(1) + 1, if n > 1UP(1) = 1-- Best Ignacio Larrosa Ca.96estroA Coru.96a === Unsor partitions1> 2,11> 3,21,12,111> 4,31,13,22,211,121,112,1111> 5,41,14,32,23,311,131,113,221,212,122,2111,1211,1121,1112,11111 Looks like a simple 2**(n-1). Is it?Yes. Think about it this way (for the inductive step):You get the ordered partitions of k+1 from those of kby either letting the last 1 form a group by itselfor adding it to the last group of an ordered partitionof === solve this? Any ideas? > In other words, k = (n+1)/2 + y n for some integer y.>> Or 2 k - 1 = (2y+1) n>> So n can be any factor (> 1) of 2k-1, and 2y+1 = (2k-1)/n.>> Of course n=2k-1 always works. That is not very helpful. Y is an external variable. You had to createy>in order to solve the problem. I already did this except I used 'A'. I don't see what's unhelpful about it. You don't actually need the y.> You wan to know how to find the n's that work, and I told you:> n can be any factor (> 1) of 2k-1.I already knew that. Didn't care about that solution. I care about theother solutions. If that is the best Canada can produce, God help us all. Just kidding>(laugh). Who do you want, Wayne Gretzky?Do you know him? Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of === Re: Two coin flip/ clarification for C Bond>> The two questions are obviously equivalent. (That doesn't say>> that Two coins were flipped and at least one is a head>> and Two coins were flipped and at least one is a tail are >> equivalent, and it _certainly_ doesn't say that >> Two coins were flipped and at least one is a head>> and Two coins were flipped and at least one is a tail>> are _the same statement_, which is the idiocy you>> said I'd said.)> Okay, you say that the two questions are equivalent.>Then, you say that the two statements are not equivalent.> Whole. Part. Whole. Part. Whole Part.> Change heads for tails in part of the question: different.> Change heads for tails in the whole question: same.> Whole. Part. Whole. Part. Whole. Part.> Get it yet?> Let's get away from any questions involving the words heads and> tails, which seem to put you into an obsessive state where you can> not parse the English language.> Consider the following question:> Let x = 3 in x^2y + 7y^2 -17x = 3. What is y?> Now consider the following question:> Let y = 3 in y^2x + 7x^2 - 17x = 3. What is x?> See what happened there? I exchanged x for y throughout the entire> question. This relabelling does not change the question. Clearly the> problems are equivalent.> Now consider the following:> Let x = 3.> Let y = 3.> Do you understand that these are not necessarily equivalent because I> haven't sta what the rest of the question is after this? We could say that x = y.No, we can't say that x = y, since I haven't told you what> follows.>You said:Let x=3Let y=3We can say: x=y > If you'd read anything prior to this last bit, you would> have understood the context (I hope). But if you ignore > everything, as you apparently did, up to Now consider> the following, then my formatting is ambiguous.Let me make it more clear.Are the following two algebra problems equivalent?1. Let x = 3. Solve the following equation for y (equation> A).2. Let y = 3. Solve the following equation for x (equation> B).I haven't told you what equations A and B are. Are these> equivalent questions?We could say they are different in that x has two lines that c; y> has two lines that don't.That does not appear to be English. I can't make any connection> between your discussion of two lines and anything I was> saying. Mathematically speaking, they are equal.What are?>x and y, when they are both equal to 3. > Are they equivalent? Depends on the definition of equivalent.They're equivalent if equivalent means bear no resemblance> to each other in any way. Is that your definition?This> argument obvuscates the real argument.No, this argument IS the argument. At least it's the one> I walked in on: When does exchanging two labels change> the question, and when does it not?>If you had read anything prior to this last bit, you would have atleast known what we are arguing about. To refute my argument, you mustat least know what it(my argument) is. > Answer: Exchange two labels throughout a question, same.> Exchange two labels in only part of a question, different.Whole. Part. Whole. Part. Whole. Part.[More incomprehensible nonsense snipped]> You can't do me that way Randy. You snipped some of my best stuff asincomprehensible nonsense.Understand it, or stay out of the thread.I made a working model which works.You can't make a working model for the 1/3 answer which works. Notwithin the confines of our problem statement.Consider this:a)A coin was flipped. It landed heads.b)A coin was flipped. It landed tails.How are those two statements rela?1)A coin was flipped until it landed heads. How is that statementdifferent from a and b?c)Two coins were flipped. At least one is a head.d)Two coins were flipped. At least one is a tail.Do c and d have the same relationship as a and b? If not, why not?2)Two coins were flipped until at least one is a head. Does theuntil serve the same function here as it did in 1?Can the until be assumed into a and b?Can the until be assumed into c and d?You snipped my working model out of your post, but it's still in mine.Look it over. If you can understand it, you might learn something.If you can't, we can take it bit by bit, and maybe I can help youunderstand it. If you sincerely, definitely, cannot. I willunderstand.If you can say,I understand your argument, and it's wrong because?We can argue about the because.We should finally get it down to one statement. Then all we would haveto do is decide if that statement is true, or false.Two coins were flipped and at least one is a head. What are thechances for two heads? That's our argument.Two coins were flipped and at least one is a tail. What are thechances for two tails? Or, that's our argument. Take your pick. To === flip/ clarification for C Bond>> No, we can't say that x = y, since I haven't told you what>> follows.> You said:>Let x=3>Let y=3These are two separate questions. As I said:>> If you'd read anything prior to this last bit, you would>> have understood the context (I hope). But if you ignore >> everything, as you apparently did, up to Now consider>> the following, then my formatting is ambiguous.misunderstood what I was saying.Here's the part you completely missed that followed before:Consider the following two algebra questions:1. Let x = 3 in x^2y + 7y^2 -17x = 3. What is y?2. Let y = 3 in y^2x + 7x^2 - 17x = 3. What is x?1 is equivalent to 2.And consider the following two algebra questions:3. Let x = 3 in... (remainder unspecified)4. Let y = 3 in... (remainder unspecified)3 is not equivalent to 4, because we don't know what's in remainderunspecified. But I already explained what I meant. And again youcan be interpre in a different way. Even after I told you theintention of the question I was asking.>We can say: x=yIf you'd read the post, you'd understand I was comparing separatealgebra questions, not giving two facts for a single algebra question.But then what is the point of conversing with someone who doesn't readwhat you write? End of === C Bond> No, we can't say that x = y, since I haven't told you what>> follows.> You said:>Let x=3>Let y=3These are two separate questions. As I said:>> If you'd read anything prior to this last bit, you would>> have understood the context (I hope). But if you ignore >> everything, as you apparently did, up to Now consider>> the following, then my formatting is ambiguous.misunderstood what I was saying.Here's the part you completely missed that followed before:Consider the following two algebra questions:> 1. Let x = 3 in x^2y + 7y^2 -17x = 3. What is y?> 2. Let y = 3 in y^2x + 7x^2 - 17x = 3. What is x?1 is equivalent to 2.> And consider the following two algebra questions:3. Let x = 3 in... (remainder unspecified)> 4. Let y = 3 in... (remainder unspecified)3 is not equivalent to 4, because we don't know what's in remainder> unspecified. But I already explained what I meant. And again you> can be interpre in a different way. Even after I told you the> intention of the question I was asking.We can say: x=yIf you'd read the post, you'd understand I was comparing separate> algebra questions, not giving two facts for a single algebra question.But then what is the point of conversing with someone who doesn't read> what you write? End of conversation.Randy,I'll assume that you agree with the parts you ignored.A coin was flipped, it landed heads.A coin was flipped, it landed tails.Those two questions are rela. The proper name for the relationshipis moot. The relationship exists.Two coins were flipped, they landed HH.Two coins were flipped, they landed TT. Those two statements have the same relationshiop. Call it what youwish, but they're rela.Congratulations, you went farther than most. You have enoughcredentials, you should be able to understand my argument. Keep yournaive position, and just pass it off that === clarification for C Bond> But then what is the point of conversing with someone who doesn't read> what you write? End of conversation.> Randy,them, either deliberately or because you ignored whatpreceded it AND what I said about it in a followup.You continue to quote my entire texts, obviously unread asyou make no response to anything there, up to the last coupleof lines, which you again read out of context.As I === f(x) + 1/f(x) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9L4nrs16285;>What is the family of real -> real analytic functions f(x), such that: f(x+1) = f(x) + 1/f(x) for all real x? >Leroy Quet(retry: I pos this yesterday, but it seems to have been lost.)It's probably easier to find a function g(x,y), such that:g(x+1,y) = 1/( g(x,y) + 1/g(x,y) )and g(0,y) = y.I suspect that g(x,y) is analytic near the origin.f(x) can be 1/g(x,y) for some choice of y, and g(x,y) not zero.The function looks like it has some connection to the Mandelbrot set.I once experimen a little with itera functions by trying to find:e(x) such that e(e(x)) = exp(x) - 1e(x) = (x/b1)*(a1 + (x/b2)*(a2 + ... )) an bn n------------------------------- 1 1 1 1 4 2 1 12 3 0 8 4 1 10 5 -7 24 6 1 7 7 159 16 8 -281 3 9 -1231 480 10 2359233 44 11 -13303471 12 12and h(x) such that h(h(x)) = log(x+1).h(x) = (x/d1)*(c1 + (x/d2)*(c2 + ... )) cn dn n------------------------------- 1 1 1 -1 4 2 5 12 3 -5 2 4 109 40 5 -497 8 6 2032 7 7 -57845 48 8 312757 9 9 -6096275 32 10 165677473 44 11 -6200116573 60 12The same strategy of matching coefficients in the power seriescould be used to find g(1/2,y) and g(1/4,y), etc.In this way, derivatives could be found using limits, producingthe coefficients for the two === Question on algebras by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9LCG0S11162; >> What is an example of an algebra that does not have the following>> property: forall x, y in A exists z in A: x * z = y.> Every algebra A which contains more than one element is an example.>Just let x = 0 and y be any non-zero element. For example, let A be the>algebra of all real numbers.>Yes, I see I must reformulate the above question to avoid two trivial cases:1. forall x, x not 0, y in A exists z in A: x * z = y2. I am looking for examples of algebras other than those arising from matrix multiplication / direct sum of rings, etc. For example, I am aware that if K_2 is the ring of all matrices of order two over the field of real numbers and U is the set of elements of K_2 of the form [a b | 0 0], then U is a (right) ideal and thus the above condition is never === on algebras>> What is an example of an algebra that does not have the following > property: forall x, y in A exists z in A: x * z = y.>>Every algebra A which contains more than one element is an example.>>Just let x = 0 and y be any non-zero element. For example, let A be the>>algebra of all real numbers.Yes, I see I must reformulate the above question to avoid two >trivial cases: 1. forall x, x not 0, y in A exists z in A: x * z = y 2. I am looking for examples of algebras other than those> arising from matrix multiplication / direct sum of rings, > etc. For example, I am aware that if K_2 is the ring of > all matrices of order two over the field of real numbers and> U is the set of elements of K_2 of the form > [a b | 0 0], then U is a (right) ideal and thus the> above condition is never satisfied if x in U and y not in > U. Yes, you do need to reformulate the question. So tellus, what _is_ the reformulation you mean to ask about?************************ === (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9LCGKW11223; >> What is an example of an algebra that does not have the following>> property: forall x, y in A exists z in A: x * z = y.> Every algebra A which contains more than one element is an example.>Just let x = 0 and y be any non-zero element. For example, let A be the>algebra of all real numbers.>Yes, I must include the following two points to avoidtrivial cases . 1. forall x, x not 0, y in A exists z in A: x * z = y.2. I am primarily interes in examples of such algebras not arising from matrix multiplication or from the direct some of algebras. For example, if K_2 is the ring of all matrices of order two over the field of real numbers and U is the set of all elements of the form [a b | 0 0], then U is a (right) ideal and thus x *z = y cannot be fulfilled if x in U and y not in U. (I thought of this example while reading last night and would have excluded the case if I'd thought of it before little to do with my example algebra. ). It is imprtant to notice also that although any algebra with a nonzero ideal cannot have property 1 (examples??- other than those described), it is -at least theoretically- possible for an algebra to possess only a single ideal (the zero ideal) without property 1 being valid. Since my example algebra appears to only have the zero ideal, the existence of an algebra with only the zero ideal and without property 1 should === coloring graph edges by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9LCGSV11241;>Where should I look to find out about coloring edges of complete and>bipartite-complete graphs with the condition that adjacent edges can't have the>same color?>Coloring the edges of a complete bipartite graph K(n,n) is the samething as finding an n by n latin square.Coloring the edges of a complete graph K(n) is not much harder.The n by n latin square needs to be the same as its transpose andhave the === Western people stole our numbers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9LCGHY11206;Hey dude,lemme remind you that you stole the numbers from the indians and have been falsely claiming the credit for it.Right from zero to the hindu arabic numbers all can be historically traced to india and have exis for long in india before the arrival of Islam in Middle east.By your rationale you arabics should not use numbers at all since one hasnt heard of any ingenious number system originating in === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9LCGwx11305; >> 1. First the problematic definition:> Algebraic integers are defined to be roots of monic polynomials with>> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where>> monic refers to the leading coefficient.> My assertion is that the over hundred year old definition excludes>> numbers that have to be included to keep from having contradiction>> i.e. mathematical inconsistency. Stupid, evil definition... It's time to destroy it!>No definition - no problem! That's the easiest way to fix everything. Let's never use the term algebraic integers again, let's just call them>roots of monic polynomials with integer coefficient! All right, no core error from now on! While your at it, why not just define the word cool (in the sense of correct) to be anything JH sais and uncool as anything anyone else says. Gran, there are two problems with such a definition:1.Since the primary objective is fame, now anyone giving admiration in the form of a complement is uncool. 2. Since Herry Monster is not JH... and the new definition of cool has come from Herry Monster... cool = uncool (?). This posesa problem which may only be correc if JH takes this messageas advice for making his own definition.Herry === (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9LHTm201660; >Let A be an any algebra over the field of real numbers.>Choose x,y in A.>>Call a finite sequence (x,x_1,x_2, ...,x_n) a (finite) direc path >>from x to y if x*x_1*...*x_n=y and x_j in A forall 1 <= j <=n.>>Let P_{xy} be the set of all direc paths from x to y. >>Let P'_{xy} subset P_{xy} be the set of all direc paths from>>x to y without y being a member of any of the finite sequences>>in P_{xy}.>Question: Does there exist a set A which is a (possibly non-divisor)>>algebra and has two elements x, y such that P'_{xy}, or>>even P_{xy} = empty?. >I have found an example algebra A' for which P_{xy} = not empty >>forall x, y in A' No you haven't. If x = 0 and y <> 0 then P_{xy} is empty.I addressed this mishap in a previous message. To avoid trivial cases the question must berephrased:What is an example of an algebra that does not have the propertyforall x, x not 0, y in A exists z in A: x*z = y and which only has the trivial (zero) ideal? C. Dement >(to say the least) seemingly needs proving >>-but, of course, why prove it if its already part of some more>>general theorem?>Warning: Please go easy on me if I've overlooked something obvious >>as I am a physics student and not a mathematician. I'm sure path,>>for example, probably === be an any algebra over the field of real numbers.>Choose x,y in A. >Call a finite sequence (x,x_1,x_2, ...,x_n) a (finite) direc path >from x to y if x*x_1*...*x_n=y and x_j in A forall 1 <= j <=n. >Let P_{xy} be the set of all direc paths from x to y. >Let P'_{xy} subset P_{xy} be the set of all direc paths from >x to y without y being a member of any of the finite sequences >in P_{xy}.>Question: Does there exist a set A which is a (possibly non-divisor) >algebra and has two elements x, y such that P'_{xy}, or >even P_{xy} = empty?. >I have found an example algebra A' for which P_{xy} = not empty >forall x, y in A' >No you haven't. If x = 0 and y <> 0 then P_{xy} is empty. I addressed this mishap in a previous message. >To avoid trivial cases the question must be>rephrased: What is an example of an algebra that does not have the >property forall x, x not 0, y in A exists z in A: x*z = y and which only has the trivial (zero) ideal? In that previous message you didn't say exactly what thereformula question was. The answer to the currentquestion is again, there is no such algebra. Or at leastthere is no such algebra if you replace the wordideal with right ideal, as in your previous post:Suppose that the condition fails: there is an x in A,x <> 0, and there is y in A such that y is not equalto xz for any z. Then the set of all xz is a non-trivialright ideal. >C. Dement >>(to say the least) seemingly needs proving >-but, of course, why prove it if its already part of some more >general theorem?>Warning: Please go easy on me if I've overlooked something obvious >as I am a physics student and not a mathematician. I'm sure path, >for example, probably has another name. >C. Dement>************************> ************************ === (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9LHTqe01677;Use the fact: Distance equals Rate times Time.And === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9LHUix01971; how to derive analytic solution of ax^3+bx^2+cx+d=0 and change 0f 1% of which coefficient is going to change the solution === analytic solution of ax^3+bx^2+cx+d=0 and change 0f 1% of > which coefficient is going to change the solution most> One way to derive the analytic solution is: first, translate the roots to x' = x - b/(3 a) so that the new equation has no (x')^2 term, i.e., a' x'^3 + c' x' + d' = 0 with b' = 0second, assume that the (new) roots will be of formx_1 = u + vx_2 = u*w + v/wx_3 = u/w + v*wwhere w is either of the (complex) roots of w^2 + w + 1 = 0.Then equate the coefficients of like powers of x' in a' x'^3 + c' x' + d' = a' (x'-x_1`) (x' - x_2) (x' - x_3)It is now possible to solve for u^3 and v^3. Caution: your choice of a particular, possibly complex, cube root of u^3 determines which cube root of v^3 you must choose, and === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9M4BCm11511;dear sir, i am a post graduate student in mechanical engg.i ma interes in doing dessertation work in optimization with kangaroo algorithm. so please provide me the concept and rela details of the algorithm.i will be very much thankful for your help. thanking === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9M6UqE19820;>

DETAILS AND SEND IT TO MY EMAIL. TELL ME ABOUT THE
SOLUTION OF ORDINARY DIFFERENTIAL EQUATION USING>LAPLACE
TRANSFORMS IN PHYSICS IN DETAILS AND SEND IT TO MY EMAIL.>
===
problem which is probably the cutest one I've seen inQM, pages
149-150 of Quantum Mechanics by Landau and Lifz, youhave an
hydrogen atom and you hit the proton or the electron with
anspecific momentum p, what's the probability of keeping them
bound?Momentum is just the space derivative of the wave
function so youmultiply it by exp(ipx) and do the scalar
product of both, since theoriginal is real you don't need to
worry about complex conjugation,just the product of both
integra over all space. Here's the damnproblem, the original
wave function depends on the distance to thecenter so you have
exp(-r/a) exp(-r/a+ipx)=exp(-2r/a+ipx), nowr=sqrt(x^2+y^2+z^2)
so it's finally exp(-2(x^2+y^2+z^2)/a+ipx)integra over all
space (probability being the square of it's modulelike usual),
Mathematica 3.0 doesn't want to integrate that, I thoughtof
doing it 1 dimentional and I got 1/(1+p^2 a^2/4) when in the
bookit says 1/(1+p^2 a^2/4)^4, is that 4th power because of
integratingover all space? Is there any software that can do
integration over allspace? Just how do I integrate simple
functions over all space or flatspacetime. What software or
what methods do I use? Some link to somePDF text? Any help is
===
spaceOriginator: jbell@presby.edu (Jon Bell)> [...]>
exp(-2(x^2+y^2+z^2)/a+ipx) integra over all space This factors
into a product of three integrals, one apiece in x, y, and
z.The y and z integrals have integrands of the form exp
(-alpha * x^2) which are well known as Gaussian integrals and
should be found in mostany table of definite integrals, or
with a Google search. In particularthis one, integra over all
x, is sqrt(pi/alpha).The x integrand is of the form exp
(-alpha*x^2 + beta*x) which you can solve by defining a new
variable u such that you can complete the square in the
exponent and get exp (-gamma*u^2 + constant). Move the
constant part out of the integral and you're back to a
Gaussian integral again.-- Jon Bell 
Presbyterian CollegeDept. of Physics and Computer Science
===
all space[...]> exp(-2(x^2+y^2+z^2)/a+ipx) integra over all
spaceThis factors into a product of three integrals, one
apiece in x, y, and z.That's right, but unfortunately that's
not the integral the poor guy hasto compute... he dropped a
square root; the integral he has to calculateisthe following:
exp(-2 sqrt(x^2+y^2+z^2)/a + ipx)And that doesn't
factorize...As I already said: he should try using spherical
===
over all space> center so you have exp(-r/a)
exp(-r/a+ipx)=exp(-2r/a+ipx), now> r=sqrt(x^2+y^2+z^2) so it's
finally exp(-2(x^2+y^2+z^2)/a+ipx)That's *not* the same as
exp(-r/a + ipx).> integra over all space.Using spherical
polars I got 16 pi a^3 p/(4 + a^2 p^2)^2(probably wrong).--
===
problem which is probably the cutest one I've seen in> QM,
pages 149-150 of Quantum Mechanics by Landau and Lifz, you>
have an hydrogen atom and you hit the proton or the electron
with an> specific momentum p, what's the probability of
keeping them bound?> Momentum is just the space derivative of
the wave functionStrange wording. Better: The momentum
*operator*, acting on the wavefunction, gives the gradient of
the wave function.> so you multiply it by exp(ipx) and do the
scalar product of both> since the original is real you don't
need to worry about complex> conjugation, just the product of
both integra over all space.Sorry, but how did you jump to
this conclusion?you want p to be real here?), moving in x
direction (or did you meanvectors x and p here?). So the
scalar product gives you the probabilityhas this to do with
the probability of keeping the atom bound after> Here's the
damn> problem, the original wave function depends on the
distance to the> center so you have exp(-r/a)
exp(-r/a+ipx)=exp(-2r/a+ipx), now> r=sqrt(x^2+y^2+z^2) so it's
finally exp(-2(x^2+y^2+z^2)/a+ipx)> integra over all
spaceSomehow the square root got lost here...> (probability
being the square of it's module> like usual), Mathematica 3.0
doesn't want to integrate that,Have you tried Mathematica's
functions for Fourier transforming? (if youdidn't notice, the
integral you are trying to do here is essentially aFourier
transform).> I thought of doing it 1 dimentionalHow,
specifically, did you do it?> and I got 1/(1+p^2 a^2/4) when
in the book> it says 1/(1+p^2 a^2/4)^4, is that 4th power
because of integrating> over all space? Probably. Try for
yourself. See below.> Is there any software that can do
integration over all> space?Mathematica can do this quite well
in general.OTOH, a little work with paper and pencil should
solve this, too. If pand x are vectotry spherical coordinates;
if they aren't, argue thatthe result should be independent of
the direction of p (I leave thereason for this to yourself to
find out...), write exp(ipx) as exp(ivec{p} vec{x}) and
proceed as before.> Just how do I integrate simple functions
over all space or flat> spacetime. With lots of tricks,
usually.> What software or what methods do I use?See above. I
hope you know how to use spherical coordinates inintegrals?>
Some link to some PDF text? Any help is welcome.> Roman Arce
===
Integration over all space> so you multiply it by exp(ipx) and
do the scalar product of both> since the original is real you
don't need to worry about complex> conjugation, just the
product of both integra over all space.Sorry, but how did you
jump to this conclusion?you want p to be real here?), moving
in x direction (or did you mean> vectors x and p here?). So
the scalar product gives you the probability> has this to do
with the probability of keeping the atom bound
aftermultiplying by exp(ipx) changes the phase of each
amplitudeproportionally to the position x, that's giving it
momentum so thefunction with momentum isphi exp(ipx), that's
not scalar product, I didn't integrate overspace, the scalar
product of that and phi gives the amplitude ofkeeping them
bound phi exp(ipx) times phi integra over space=phi^2exp(ipx)
integra over space,in dirac notation  where a is
the wave function of theelectron before giving it momentum and
b is after, the amplitud forthe atom to remain bound is the
component of the state vector b in thea axis in Hilbert space,
just the scalar product of  Here's the
damn> problem, the original wave function depends on the
distance to the> center so you have exp(-r/a)
exp(-r/a+ipx)=exp(-2r/a+ipx), now> r=sqrt(x^2+y^2+z^2) so it's
finally exp(-2(x^2+y^2+z^2)/a+ipx)> integra over all
spaceSomehow the square root got lost here...that was just a
typo when writing to post it in the usenet> (probability being
the square of it's module> like usual), Mathematica 3.0 doesn't
want to integrate that,Have you tried Mathematica's functions
for Fourier transforming? (if you> didn't notice, the integral
you are trying to do here is essentially a> Fourier
transform).the problem is the sqrt(x^2+y^2+z^2) in the
exponential > I thought of doing it 1 dimentionalHow,
specifically, did you do it?I replaced sqrt(x^2+y^2+z^2) by
sqrt(x^2)=x in the exponential, exp(x)is easy > and I got
1/(1+p^2 a^2/4) when in the bookanother typo, it should say
the square of that cos I'm talking aboutprobabilities and that
would be just the amplitude> it says 1/(1+p^2 a^2/4)^4, is that
4th power because of integrating> over all space? Probably. Try
for yourself. See below.Is there any software that can do
integration over all> space?Mathematica can do this quite well
in general.it doesn't like exp(sqrt(x^2+y^2+z^2)) integra over
x > OTOH, a little work with paper and pencil should solve
this, too. If p> and x are vectotry spherical coordinates; if
they aren't, argue that> the result should be independent of
the direction of p (I leave the> reason for this to yourself
to find out...), write exp(ipx) as exp(i> vec{p} vec{x}) and
proceed as before.r=sqrt(x^2+y^2+z^2) is better in spherical
coordinates, but I wouldneed to convert exp(ipx) too and x is
better in cartesian ones> Just how do I integrate simple
functions over all space or flat> spacetime. With lots of
tricks, usually.What software or what methods do I use?See
above. I hope you know how to use spherical coordinates in>
integrals?Some link to some PDF text? Any help is welcome.>
+54-11-4956-1576I hope what I said above helped a
===
multiply it by exp(ipx) and do the scalar product of both>
since the original is real you don't need to worry about
complex> conjugation, just the product of both integra over
all space. Sorry, but how did you jump to this conclusion? you
want p to be real here?), moving in x direction (or did you
mean> vectors x and p here?). So the scalar product gives you
the probability> has this to do with the probability of
keeping the atom bound aftermultiplying by exp(ipx) changes
the phase of each amplitude> proportionally to the position
x,Right.> that's giving it momentumHuh???> so the function
with momentum is phi exp(ipx),Sorry, that makes no sense to
me. What is a function with momentum? Door something like
that?> that's not scalar product, I didn't integrate over
space,Err, you said above do the scalar product of both;
that's what I wasreferring to.> the scalar product of that and
phi gives the amplitude of> keeping them bound phi exp(ipx)
times phi integra over space=phi^2> exp(ipx) integra over
space,Oh, so you aren't taking the scalar product of phi and
exp(ipx) (as youoriginally said!!! See above!), but you are
calculating the *expectationvalue* of the operator exp(ipx)?
But I still don't understand what thishas to do with giving
momentum to it - and what this phrase issupposed to mean.> in
dirac notation  where a is the wave function of
the> electron before giving it momentum What on earth does
this giving momentum to it mean? Do you think thatexp(ipx)
times phi describes an electron to which momentum has
beenadded? If yes, why do you think so? I can imagine a
momentumtranslation operator, analogously to the translation
operator in space,but this operator wouldn't be exp(ipx), but
more something like exp(ihat{p} x), where hat{p} is the
momentum *operator*.> and b is after, the amplitud for> the
atom to remain bound is the component of the state vector b in
the> a axis in Hilbert space, just the scalar product of  |exp(ipx)|a>*If* exp(ipx) describes a wave function for
an electron with momentum padded, then you are right. But why
should this be so? See above.> Here's the damn> problem, the
original wave function depends on the distance to the> center
so you have exp(-r/a) exp(-r/a+ipx)=exp(-2r/a+ipx), now>
r=sqrt(x^2+y^2+z^2) so it's finally
exp(-2(x^2+y^2+z^2)/a+ipx)> integra over all space Somehow the
square root got lost here...that was just a typo when writing
to post it in the usenet> (probability being the square of
it's module> like usual), Mathematica 3.0 doesn't want to
integrate that, Have you tried Mathematica's functions for
Fourier transforming? (if you> didn't notice, the integral you
are trying to do here is essentially a> Fourier transform).the
problem is the sqrt(x^2+y^2+z^2) in the exponentialErr, yes, I
noticed that. However you didn't answer my question: Haveyou
tried Mathematica's functions for Fourier transforming?> I
thought of doing it 1 dimentional How, specifically, did you
do it?I replaced sqrt(x^2+y^2+z^2) by sqrt(x^2)=x in the
exponential, exp(x)> is easyDid you suppose that x > 0 or
what? What were your limits ofintegration?You *do* know that
in general sqrt(x^2) = |x| and not x, don't you?> and I got
1/(1+p^2 a^2/4) when in the bookanother typo, it should say
the square of that cos I'm talking about> probabilities and
that would be just the amplitudeWell, you were talking about
the value of the integral here, hence aboutthe amplitude.> it
says 1/(1+p^2 a^2/4)^4, is that 4th power because of
integrating> over all space? Probably. Try for yourself. See
below. Is there any software that can do integration over all>
space? Mathematica can do this quite well in general.it doesn't
like exp(sqrt(x^2+y^2+z^2)) integra over xWhat does it say to
this?> OTOH, a little work with paper and pencil should solve
this, too. If p> and x are vectotry spherical coordinates; if
they aren't, argue that> the result should be independent of
the direction of p (I leave the> reason for this to yourself
to find out...), write exp(ipx) as exp(i> vec{p} vec{x}) and
proceed as before.r=sqrt(x^2+y^2+z^2) is better in spherical
coordinates, but I would> need to convert exp(ipx) too and x
is better in cartesian onesdirection of p (do you know why?),
and therefore you can replace px withvec{p} vec{x}. That, on
the other hand, can be replaced by |vec{p}| rcos(theta), and
the remaining integral is then easily done in
===
Integration over all space> so you multiply it by exp(ipx) and
do the scalar product of both> since the original is real you
don't need to worry about complex> conjugation, just the
product of both integra over all space. Sorry, but how did you
jump to this conclusion? you want p to be real here?), moving
in x direction (or did you mean> vectors x and p here?). So
the scalar product gives you the probability> has this to do
with the probability of keeping the atom bound after>
multiplying by exp(ipx) changes the phase of each amplitude>
proportionally to the position x,> Right.> that's giving it
momentum> Huh???Below> so the function with momentum is phi
exp(ipx),> Sorry, that makes no sense to me. What is a
function with momentum? Do> or something like that?Yes> that's
not scalar product, I didn't integrate over space,> Err, you
said above do the scalar product of both; that's what I was>
referring to.first the original wave function times exp(ipx),
then the scalarproduct of all that with the original function
again.> the scalar product of that and phi gives the amplitude
of> keeping them bound phi exp(ipx) times phi integra over
space=phi^2> exp(ipx) integra over space,Oh, so you aren't
taking the scalar product of phi and exp(ipx) (as you>
originally said!!! See above!), but you are calculating the
*expectationI said the scalar product of (phi times exp(ipx))
and phi.> value* of the operator exp(ipx)? But I still don't
understand what this> has to do with giving momentum to it -
and what this phrase is> supposed to mean.> in dirac notation
 where a is the wave function of the> electron
before giving it momentum > What on earth does this giving
momentum to it mean? Do you think that> exp(ipx) times phi
describes an electron to which momentum has been> added? If
yes, why do you think so? I can imagine a momentum>
translation operator, analogously to the translation operator
in space,> but this operator wouldn't be exp(ipx), but more
something like exp(i> hat{p} x), where hat{p} is the momentum
*operator*.??? in coordinate space the momentum operator is -i
hbar nabla, howcan you take the gradient of a variable,
operators act on functions,multipliying by exp(ipx) is an
operator.> and b is after, the amplitud for> the atom to
remain bound is the component of the state vector b in the> a
axis in Hilbert space, just the scalar product of 
|exp(ipx)|a>*If* exp(ipx) describes a wave function for an
electron with momentum p> added, then you are right. But why
should this be so? See above.Let's take the wave function for
the electron in the ground state inan hydrogen atom with
Bohr's radius and hbar being equal to 1, also 1dimentional,
the momentum operator is-i d/dx (hbar=1)the wave function is
exp(-x)applying that operator you obtaini exp(-x)the wave
function with added momentum is exp(-x+ipx)applying that
operator you obtain(i+p) exp(-x+ipx)in the first case the
momentum operator scales the wave function by i,in the second
case by i+p, that means that any measument of momentumwill
give a value of p extra, as a general rule multiplying ANY
wavefunction by exp(ipx) gives it momentum p> Here's the damn>
problem, the original wave function depends on the distance to
the> center so you have exp(-r/a)
exp(-r/a+ipx)=exp(-2r/a+ipx), now> r=sqrt(x^2+y^2+z^2) so it's
finally exp(-2(x^2+y^2+z^2)/a+ipx)> integra over all space
Somehow the square root got lost here...that was just a typo
when writing to post it in the usenet> (probability being the
square of it's module> like usual), Mathematica 3.0 doesn't
want to integrate that, Have you tried Mathematica's functions
for Fourier transforming? (if you> didn't notice, the integral
you are trying to do here is essentially a> Fourier
transform).the problem is the sqrt(x^2+y^2+z^2) in the
exponentialErr, yes, I noticed that. However you didn't answer
my question: Have> you tried Mathematica's functions for
Fourier transforming?I'll try it.> I thought of doing it 1
dimentional> How, specifically, did you do it?> I replaced
sqrt(x^2+y^2+z^2) by sqrt(x^2)=x in the exponential, exp(x)>
is easy> Did you suppose that x > 0 or what? What were your
limits of> integration?> You *do* know that in general
sqrt(x^2) = |x| and not x, don't you?Yes, I used absolute
value, the limits were -inf to inf> and I got 1/(1+p^2 a^2/4)
when in the book> another typo, it should say the square of
that cos I'm talking about> probabilities and that would be
just the amplitude> Well, you were talking about the value of
the integral here, hence about> the amplitude.> it says
1/(1+p^2 a^2/4)^4, is that 4th power because of integrating>
over all space?> Probably. Try for yourself. See below.> Is
there any software that can do integration over all> space?>
Mathematica can do this quite well in general.> it doesn't
like exp(sqrt(x^2+y^2+z^2)) integra over x> What does it say
to this?cannot solve something.> OTOH, a little work with
paper and pencil should solve this, too. If p> and x are
vectotry spherical coordinates; if they aren't, argue that>
the result should be independent of the direction of p (I
leave the> reason for this to yourself to find out...), write
exp(ipx) as exp(i> vec{p} vec{x}) and proceed as
before.r=sqrt(x^2+y^2+z^2) is better in spherical coordinates,
but I would> need to convert exp(ipx) too and x is better in
cartesian onesdirection of p (do you know why?), and therefore
you can replace px with> vec{p} vec{x}. That, on the other
hand, can be replaced by |vec{p}| r> cos(theta), and the
remaining integral is then easily done in spherical>
coordinates.But I would need to convert one specific axis (cos
the extra momentumis only in one direction) to polar
coordinates, and the integralbecomes hard to solve
===
everybody,I am trying to solve the following homework
(Esercise 6.1.C. inEngelking's book):Let $X$ be a connec
space, $Y$ be a connec subspace of $X$ andsupposethat
$XsetminusY = A cup B$ where $A$ and $B$ are disjoint open
setsin $Xsetminus A$.Show that the sets $Y cup A$ (and
analogously $Y cup B$) isconnec.--------------I tryed to do it
by counterassuming that $Ycup A$ is not connec.So there are
$U,V$ disjoint (non trivial) open sets in $Ycup A$
suchthat$Ycup A = U cup V$.Since $Y$ is connec, $Y$ must be
completely contained in either $U$or $V$.Suppose $Y subseteq
U$. Then $Vsubseteq A$ and $V$ is open $A$.<<< at this point I
didn't find nothing to finish the proof >>In particular i don't
understand how to use the hypotesis that $X$ isconnec (I need
some open set in $X$ but it seems I have not it!!)Could
===
> I am trying to solve the following homework (Esercise 6.1.C.
in> Engelking's book):Let $X$ be a connec space, $Y$ be a
connec subspace of $X$ and> suppose> that $XsetminusY = A cup
B$ where $A$ and $B$ are disjoint open sets> in $Xsetminus
A$.> Show that the sets $Y cup A$ (and analogously $Y cup B$)
is> connec.--------------I tryed to do it by counterassuming
that $Ycup A$ is not connec.> So there are $U,V$ disjoint (non
trivial) open sets in $Ycup A$ such> that> $Ycup A = U cup V$.>
Since $Y$ is connec, $Y$ must be completely contained in either
$U$> or $V$.> Suppose $Y subseteq U$. Then $Vsubseteq A$ and
$V$ is open $A$.> <<< at this point I didn't find nothing to
finish the proof > In particular i don't understand how to use
the hypotesis that $X$ is> connec (I need some open set in $X$
but it seems I have not it!!)Could someone help me in solving
this exercise? I will use X = A + B + C, Y = C
------------------- | | | | A | | | | | |--------| C | | | | |
B | | | | | --------------------You are given open sets H and K
that contain A and B and only meet in C,Contemplate a few
seconds to realize that even their closures only meetin C.You
are assuming open U containing C and open V containing BU that
onlymeet in A.Again even their closures only meet in A.Open
sets minus closed sets are open. Open sets plus open sets are
open.Perhaps you can do a little manipulation using H, K, U, V
and theirclosures to come up with disjoint open sets that cover
the whole space.As an aside, there is no need to argue by
contradiction. Let U and V be disjoint open sets covering B +
C, wlog U containing apoint of C.As you no, U must contain all
of C. Hence V is a subset of B. So far you have not used the
fact that V is not empty and there is noneed to. Just aim
towards showing V cap B is empty.
**************************************************************
***********************One very useful characterization of
connec is no non-trivial clopensets.When you finish what you
will have actually done is to start with anon-emptyclopen set
U in B + C and show that it is all of B + C(by showing V = (B
===
apologies for butting into a group that I don't normally post
to,but I was involved in a thread that was cpos here earlier
thismonth, and here is the result!This is an invitation to
participate in a paranormal challenge.A person posting under
the usenet name of |-|erc has been claiming forthe last few
years that, he has the power to influence responses tohis
usenet posts such that he can divine the identity of the
author ofa post given a selection of authors to choose from,
with a successrate better than 1 in 10,000. He further claims
that, because of hisinfluence, that others are able to perform
this feat, but that he cando it better than them.|-|erc has
persistently pos challenges to sceptics and others provethis
power but has been variously dismissed as not being
validlytestable or downright trivial.In response to his latest
challenge I have crea a members-onlymodera group in Yahoo, and
|-|erc has agreed in an usenet post thatthis forum, given
enough messages and author base to choose from,could be used
to test his claim.I invite you to view the aim and protocol of
the challenge
at:We
also need Group Members to build a data set to test |-|erc
with, soyou are further invi to join the Yahoo Group either by
followinglink at the website, or emailing me direct.--Eric
===
Eigenfunctions of the Fourier transform>I heard an interesting
talk today from a guy who is searching (so far>unsuccessfully)
for a new class of eigenfunctions of the Fourier>Transform, of
a certain type. I was aware that gaussians transform
to>gaussians, so that exp(-ax^2) transforms to itself for
suitable a.>What I didn't realize was that functions of the
form (Hermite>polynomial)*(exp(-ax^2/2)) are also
eigenfunctions. Apparently that is a consequence of a certain
differential operator>(d^2/dx^2 - x^2) commuting with the
Fourier Transform operator. My questions for the newsgroup: Is
anyone aware of other classes of eigenfunctions of the
Fourier>Transform?You should take this post with a _huge_
grain of salt. I _think_that the closed span of the
eigenfunctions you mention (for agiven eigenvalue) is the
entire eigenspace. I could very wellbe totally wrong - the
following is very fuzzy:Say T = (d^2/dx^2 - x^2) and F is the
Fourier transform. ThenF and T are normal operators (T is
unbounded densely defined)and FT = TF, so the spectral theorem
says that F and T aresimultaneously diagonalizable: There
exists a unitary equivalenceof L^2(R) with L^2(mu) for a
certain measure mu on a spaceX, such that T and F correspond
to multiplication operatorson L^2(mu): F becomes
multiplication by m, where m takesonly the four values 1, -1,
i, -i, and T becomes multiplicationby an unbounded function
(defined on the set of functionssuch that the product is in
L^2(mu).) I _think_ that T haspure point spectrum, and I
_think_ that it follows that theeigenfunctions of T span L^2;
I also think that the eigenfunctions of T are exactly the
functions you mention.Perhaps.>Are there other known operators
that commute with the Fourier>Transform? Especially is there a
characterization of any such>operator?
===
Fourier transform>>I heard an interesting talk today from a guy
who is searching (so far>>unsuccessfully) for a new class of
eigenfunctions of the Fourier>>Transform, of a certain type. I
was aware that gaussians transform to>>gaussians, so that
exp(-ax^2) transforms to itself for suitable a.>>What I didn't
realize was that functions of the form
(Hermite>>polynomial)*(exp(-ax^2/2)) are also
eigenfunctions.>Apparently that is a consequence of a certain
differential operator>>(d^2/dx^2 - x^2) commuting with the
Fourier Transform operator.>My questions for the newsgroup:>Is
anyone aware of other classes of eigenfunctions of the
Fourier>>Transform?>You should take this post with a _huge_
grain of salt. I _think_>that the closed span of the
eigenfunctions you mention (for a>given eigenvalue) is the
entire eigenspace. I could very well>be totally wrong - the
following is very fuzzy:>Say T = (d^2/dx^2 - x^2) and F is the
Fourier transform. Then>F and T are normal operators (T is
unbounded densely defined)>and FT = TF, so the spectral
theorem says that F and T are>simultaneously diagonalizable:
There exists a unitary equivalence>of L^2(R) with L^2(mu) for
a certain measure mu on a space>X, such that T and F
correspond to multiplication operators>on L^2(mu): F becomes
multiplication by m, where m takes>only the four values 1, -1,
i, -i, and T becomes multiplication>by an unbounded function
(defined on the set of functions>such that the product is in
L^2(mu).) I _think_ that T has>pure point spectrum, and I
_think_ that it follows that the>eigenfunctions of T span L^2;
I also think that the >eigenfunctions of T are exactly the
functions you mention.T does have a point spectrum. The
spectrum of T is {-n : n odd, positive}, i.e. the spectrum of
T is {-1, -3, -5, ...}. This is a version of a reasonably
well-known fact (at least amongst people working in quantum
mechanics), i.e. the spectrum of thequantum mechanical simple
harmonic oscillator Hamiltonian.And the eigenfunctions of T
are indeed those mentioned by Randy.This fact would also be
demonstra in most, if not all, introductory quantum mechanic
texts. The simplest approach tothe identification of the point
spectrum of T is to work with the ladder operators d/dx + x and
d/dx - x.Also, exp(iT/8) = exp(-i/8) F (or exp(-i/8) F^{-1}),
depending on the convention used for the Fourier transform.
This means that,depending on convention, F is the closure of
exp(i(T+1)/8) orthe closure of exp(-i(T+1)/8).David
McAnally>Perhaps.>>Are there other known operators that
commute with the Fourier>>Transform? Especially is there a
characterization of any such>>operator?>
===
the Fourier transform>You should take this post with a _huge_
grain of salt. I _think_>that the closed span of the
eigenfunctions you mention (for a>given eigenvalue) is the
entire eigenspace. I could very well>be totally wrong - the
following is very fuzzy:>Say T = (d^2/dx^2 - x^2) and F is the
Fourier transform. Then>F and T are normal operators (T is
unbounded densely defined)>and FT = TF, so the spectral
theorem says that F and T are>simultaneously diagonalizable:
There exists a unitary equivalence>of L^2(R) with L^2(mu) for
a certain measure mu on a space>X, such that T and F
correspond to multiplication operators>on L^2(mu): F becomes
multiplication by m, where m takes>only the four values 1, -1,
i, -i, and T becomes multiplication>by an unbounded function
(defined on the set of functions>such that the product is in
L^2(mu).) I _think_ that T has>pure point spectrum, and I
_think_ that it follows that the>eigenfunctions of T span L^2;
I also think that the >eigenfunctions of T are exactly the
functions you mention.Yes. See e.g. Reed and Simon, Methods of
Modern Mathematical Physicsvol II (Fourier Analysis,
Self-Adjointness), ch. IX problem 6 or 7,or Glimm and Jaffe,
Quantum Physics: A Functional Integral Point of View,sec.
1.5.Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israelUniversity of British
===
Eigenfunctions of the Fourier transform>>I heard an
interesting talk today from a guy who is searching (so
far>>unsuccessfully) for a new class of eigenfunctions of the
Fourier>>Transform, of a certain type. I was aware that
gaussians transform to>>gaussians, so that exp(-ax^2)
transforms to itself for suitable a.>>What I didn't realize
was that functions of the form
(Hermite>>polynomial)*(exp(-ax^2/2)) are also eigenfunctions.
>Apparently that is a consequence of a certain differential
operator>>(d^2/dx^2 - x^2) commuting with the Fourier
Transform operator. Known to physicists as the Hamiltonian of
the harmonic oscillator. >Is anyone aware of other classes of
eigenfunctions of the Fourier>>Transform? The Fourier
transform F on L^2(R) (with the normalization that makes >it
unitary) satisfies F^4 = I, so its eigenvalues are fourth
roots of >unity. For any f in L^2, g1(x) = f(x) + F(f)(x) +
f(-x) + F(f)(-x),>g2(x) = f(x) - i F(f)(x) - f(-x) + i
F(f)(-x),>g3(x) = f(x) - F(f)(x) + f(-x) - F(f)(-x),>g4(x) =
f(x) + i F(f)(x) - f(-x) - i F(f)(-x)>are eigenfunctions for
1, i, -1 and -i respectively.One might also point out that
since f = (g1 + g2 + g3 + g4)/4it follows that every
eigenfunction is as above - since thoseeigenspaces span L^2
there's no room for any more eigenvectorsto sneak in.One might
also point out that this doesn't quite answer thequestion of
whether the eigenvectors that Randy mentionsare the only ones.
I _think_ that the answer is the closedspan of the eigenvectors
he mentions are the only ones,really not sure.>>Are there other
known operators that commute with the Fourier>>Transform?
Especially is there a characterization of any such>>operator?
All operators that leave the four eigenspaces invariant.
Robert Israel israel@math.ubc.ca>Department of Mathematics
http://www.math.ubc.ca/~israel >University of British Columbia
>Vancouver, BC, Canada V6T 1Z2************************
===
prove the following:Show that if f:R^k -> R is lebesgue
integrable and with compactsupport then(*) Mf (x) <= c/|x|^k
for large |x| (with c positive constant).where Mf is the
maximal function of hardy-littlewood of f : Mf(x) =sup_Q 1/|Q|
int_Q |f| where the sup is taken over cubes Q centeredin x and
contracting towards it.I've been told that it should be easy
once one proves:If E subset R^k is bounded then(**) c|E|/|x|^k
<= Mchi_E <= C|E|/|x|^k for large |x| (with c,Cpositive
constants)I've already proven this... my plan to prove (*)
was:Given e > 0, there exists a > 0 such that |{|f|>a}| Show that if f:R^k -> R is
lebesgue integrable and with compact> support then(*) Mf (x)
<= c/|x|^k for large |x| (with c positive constant).where Mf
is the maximal function of hardy-littlewood of f : Mf(x) =>
sup_Q 1/|Q| int_Q |f| where the sup is taken over cubes Q
centered> in x and contracting towards it.Here's a simpler
problem. Let f be the unit point mass at the origin and let M*
be the maximal function defined in terms of balls centered at
x. Then it's easy to see M*f(x) = 1/|B(x,|x|)| = c/|x|^kfor
all nonzero x. Back to your problem: the cubical maximal
function is bounded by a constant times M*, and f basically
looks like a point mass from the point of view of a large x.
===
to prove the following: Show that if f:R^k -> R is lebesgue
integrable and with compact>support then (*) Mf (x) <= c/|x|^k
for large |x| (with c positive constant). where Mf is the
maximal function of hardy-littlewood of f : Mf(x) =>sup_Q
1/|Q| int_Q |f| where the sup is taken over cubes Q
centered>in x and contracting towards it.The sup is over _all_
cubes centered at x, period. Saying thecubes contract to x
makes sense if you're talking about a _limit_instead of a
sup...>I've been told that it should be easy once one proves:
If E subset R^k is bounded then (**) c|E|/|x|^k <= Mchi_E <=
C|E|/|x|^k for large |x| (with c,C>positive constants)Well
forget this. It's easy to prove the inequality directlyfrom
the definition, just using the fact that int_Q |f| isno larger
than the L^1 norm of f (and the fact that iff is suppor in {x :
|x| < A} then int_Q |f| = 0 if |x| > 2Aand the side length of Q
is < cA.)>I've already proven this... my plan to prove (*) was:
Given e > 0, there exists a > 0 such that |{|f|>a}|{|f|<=a}, E is bounded and |f|<= achi_E for all x except on
the set>of measure < e. If I could conclude that Mf = M|f| <=
M(achi_E) then>I would be finished because of (**). But I
cannot prove this>inequality... Any ideas?
>nojb.************************
> have
one, could you test this out and tell how many seconds>> and
how many keystrokes it takes to calculate this hypothetical>>
thing:>> (9+7)(8-3)(5+4)>> ------------------ + (6-4)(3-1)>>
(9+2)(9-4)(7+3)>> ?>> Of course you're not allowed to use your
knowledge of simple>> arithmetic ;-)>> With RPN (and without
practicing first) it this takes 39 keystrokes>> and approx. 13
seconds. And without using memory other than the>> implicit
stack.>> How does this go with an algebraic with parentheses?
Surely I don't have to conclude that it can't be done?> I
wouldn't be surprised though ;-) Can anyone with an algebraic
parentheses calculator> please try this? I'm really interes in
the number of> keystrokes (exact) and seconds (approx.) it
had a TI-59, which implemen> multiply-before-add and had nine
levels of parentheses. On that> calculator the evaluation
would have looked like:
(9+7)*(8-3)*(5+4)/((9+2)*(9-4)*(7+3))+(6-4)*(3-1)= I count 51
keystrokes (50, if you omit the ) just before the =, since>
all unclosed parentheses were automatically closed). Also, I
didn't like> having to wonder in the middle of a calculation
just how many levels of> parentheses I had used, or whether I
had used enough or perhaps too many.So it's rather simple then
===
the masses: Free the math!> A science journal, instead of
charging readers a subscription rate, Mathematics journals
published by universities or societies (as opposedto those
published by commercial publishers) traditionally have
pagecharges ... the sponsor of the research (such as
government orindustry) pays the journal when that research is
published.But individual researchers without sponsors to pay
for the research arenot expec to pay for
===
science journal, instead of charging readers a subscription
rate,Mathematics journals published by universities or
societies (as opposed> to those published by commercial
publishers) traditionally have page> charges Do they? I don't
===
math!Leroy Quet> Stephen J. Herschkorn >>A science journal,
instead of charging readers a subscription rate,I am all for
this, except that the authors seem, to me, to be
charged>>a>>huge amount. That is called vanity publishing. Few
regard it as a Good Thing. > economics/finance has told me of
first-tier journals that charge a fee> when one submits a
paper to be refereed. The journal keeps the fee,> regardless
of whether the paper is accep or not. I don't know if the>
journal charges for the resubmission of a revision. Sounds
like a good scam.> Yeah. Just submit your paper to our
journal, and we will ,surely,> certainly consider it for
publication,..certainly, we REALLY will. You> ONLY have to pay
a *small* consideration-fee.... > On those few occasions, when
I have had a math-problem of mine> published in either the
Mathematics Magazine or the American> Mathematical Monthly, my
friends sometimes ask, So, are they paying> you?...Hold the
phone, guys. The above online journal _is_ peer-reviewed, or
so itsays on the CNN international edition that I looked at.
But the journal isnot about math, and US$1500 is not a big
deal if the paper is about someexpensive work in biomedical
===
to Tell About It!> Thoughtful comments will be welcomed,
especially suggestions forPerhaps an unsigned infinity would
be a better addition to the reals?After all, if an unsigned
zero is valid why not infinity?I'm not sure whether such a
concept could be _mathematically_ valid,but it seems to fit in
with some of the things you were saying. Onething you mentioned
was the dichotomy of which of -oo and +oo would bethe defined
value of 1/0. With an unsigned infinity this is no longera
problem...Here X is any element of these extended reals and U
is the unsignedinfinity: X / 0 = U, 0 / X = 0, and X / X = 1,
with the exception that 0 / 0 is still undefined X / U = 0, U
/ X = U, and X / X = 1, with the exception that U / U is
undefined X - U = U, U - X = U, and X - X = 0, with the
exception that U - U is undefined X * U = U with the exception
that 0 * U is undefined X + U = U without exceptionOf course
this creates as many problems as it solves: It might
resolvethe fact that now (1/x), abbrevia by both
lim[x->0+](1/x) andlim[x->0-](1/x), share the same result, U,
but now seems we need newterminology to describe lim[x->0+](ln
x) where the result is quitedefinitely signed.So, since limits
are used when signed infinities and infinitesimalsare
required, perhaps we should define a new limit terminology for
theunsigned infinity:As[x->0+](ln x)->U-i.e. As x approaches 0
from the positive side, ln x approachesinfinity from the
negative side.As[x->0](1/x)->Ui.e. As x approaches 0, 1/x
approaches infinity.where the above is 'shorthand' for the
more specific:As[x->0-](1/x)->U- and As[x->0+](1/x)->U+So we
===
Re: How I Divided by Zero, and Lived to Tell About It!
Thoughtful comments will be welcomed, especially suggestions
infinity would be a better addition to the reals?It depends on
what you want to do. Note that I did briefly mention suchan
extension, and one of the links I gave contained a link
to,which covers almost all
of the stuff you've mentioned below.> After all, if an
unsigned zero is valid why not infinity?Of course.> I'm not
sure whether such a concept could be _mathematically_
valid,Certainly it is.> but it seems to fit in with some of
the things you were saying. One> thing you mentioned was the
dichotomy of which of -oo and +oo would be> the defined value
of 1/0. With an unsigned infinity this is no longer> a
problem...Right.> Here X is any element of these extended
reals and U is the unsigned> infinity: X / 0 = U,> 0 / X = 0,>
and> X / X = 1, with the exception that> 0 / 0 is still
undefined X / U = 0,> U / X = U,> and> X / X = 1, with the
exception that> U / U is undefined X - U = U,> U - X = U,>
and> X - X = 0, with the exception that> U - U is undefined X
* U = U with the exception that> 0 * U is undefinedThat's all
standard.> X + U = U without exceptionActually, I have sugges
that before in this newsgroup, but it iscertainly not
standard. Rather, one almost always sees U + U as
beingundefined. (BTW, I didn't mention my preference for U + U
= U in theanything controversial.) And of course, if U + U = U,
then I would suggest,correspondingly, that U - U be U as well,
rather than undefined.> Of course this creates as many
problems as it solves: It might resolve> the fact that now
(1/x), abbrevia by both lim[x->0+](1/x) and> lim[x->0-](1/x),
share the same result, U, but now seems we need new>
terminology to describe lim[x->0+](ln x) where the result is
quite> definitely signed.In the one-point extension of the
reals, that limit is just U. No newterminology is required.
But it could be useful nonetheless, so that thereis no loss of
information, so to speak.Here's an example having nothing to do
with 0 or U:lim[x -> -2^+] x^2 = 4^-which is saying that, as x
approaches -2 from the right on the number line,x^2 approaches
4 from the left.> So, since limits are used when signed
infinities and infinitesimals> are required, perhaps we should
define a new limit terminology for the> unsigned infinity:
As[x->0+](ln x)->U-> i.e. As x approaches 0 from the positive
side, ln x approaches> infinity from the negative side.Oops.
Your notation is backward. Here's how the sides of
unsignedinfinity, U, compare with the signed infinities, -oo
and +oo, of thetwo-point extension of the reals: U^+ = -oo U^-
= +ooIt helps, I think, to remember that the diagram of the
one-point extensionSo we would say that lim[x -> 0^+] ln(x) =
U^+ David Cantrell> As[x->0](1/x)->U> i.e. As x approaches 0,
1/x approaches infinity. where the above is 'shorthand' for
the more specific: As[x->0-](1/x)->U- and As[x->0+](1/x)->U+
So we can also say: As[x->0]((cos x)/x)->U etc. etc.
===
function.q is the number of the elements in the field, and
would usually be greaterthan 1.It sounds as if the series
doesn't converge for large q.I had misinterpre a problem,
someone help me simplyfy the power series expression: The
summation as t goes from 0 to infinity of q^t - 1/t * the
summation for divisors d of t of MoebiusMu(t/d)*q^d ? Diana I
believe that you want t to go from 1, not 0, right? For at t =
0,> all right of the minus is undefined. Now,> sum{t= 1 to
infinity} 1/t^r sum{d|t} mu(t/d) q^d = polylog(q,r) /zeta(r),
where polylog(q,r) = sum{t=1 to infinity} q^t/t^r (for those r
and q where the sum converges), and where I have otherwise
shortened your notation. (And zeta(r) = polylog(1,r), of
course {for r > 1}.) So,> sum{t= 1 to infinity} (q^t - 1/t
sum{d|t} mu(t/d} q^d) = q/(1-q) - polylog(q,1)/zeta(1). But
1/zeta(r) -> 0 as r -> 1 from above.> And, for -1 <= q < 1,>
polylog(q,1) = -log(1-q) (but this ends up not mattering). So,
*IF* -1 <= q < 1, your sum (if t is from 1) is simply> q/(1-q).
I have avoided being rigorous, however, and I may have
===
approximationHi all,Perhaps it's a bit offtopic here, but I
hope someone can help me out.I'm writing a small computer
program (Linux kernel module that is), whichneeds a pow
function (x to the y; in computer-language that is
pow(x,y)).Both x and y are floats (real numbers that is).Does
anybody know a usable routine to calculate this? Should be
fast too,because it has to run on a slow processor.Currently,
I already found a sin replacement.My current function is:float
D-science lab, Hogeschool AntwerpenReal time Linux for
===
Re: pow(x,y) approximation> Hi all,Perhaps it's a bit offtopic
here, but I hope someone can help me out.I'm writing a small
computer program (Linux kernel module that is), which> needs a
pow function (x to the y; in computer-language that is
pow(x,y)).> Both x and y are floats (real numbers that
is).Does anybody know a usable routine to calculate this?
Should be fast too,> because it has to run on a slow
processor.> Currently, I already found a sin replacement.My
current function is:float pow(float x, float y) {> // Still
empty :(> }> Well x^y can be calcula as Exp( y * ln( x ) ).
Assuming that youhave access to exp and ln functions. If not
you can at least get theln() as an assembly instrcution using
the intel processor and expthrough a taylor series. Their
exist a lot of speed-up tricks insteadof a plain taylor series
of Exp. A good source I found was
===
it's a bit offtopic here, but I hope someone can help me
out.I'm writing a small computer program (Linux kernel module
that is), which> needs a pow function (x to the y; in
computer-language that is pow(x,y)).> Both x and y are floats
(real numbers that is).Does anybody know a usable routine to
calculate this? Should be fast too,> because it has to run on
a slow processor.> Currently, I already found a sin
replacement.My current function is:float pow(float x, float y)
Amazon):Elementary Functions: Algorithms and Implementationby
Jean-Michel Muller, Jean-Michael MullerList Price:
$69.95Software Manual for the Elementary Functionsby William
===
program (Linux kernel module that is), which> needs a pow
function (x to the y; in computer-language that is pow(x,y)).>
Both x and y are floats (real numbers that is).What's wrong
with #include ? Why are you redefining all
===
computer program (Linux kernel module that is), which>> needs
a pow function (x to the y; in computer-language that is
pow(x,y)).>> Both x and y are floats (real numbers that
is).What's wrong with #include ? Why are you
redefining all these> functions? only gives me:*
unresolved symbols* non-optimzed math-routines (at least not
for PPC)Most routines are written in asm here, to get the
fastest result.-- Groeten, Toni Van RemortelWetenschappelijk
medewerker, D-science lab, Hogeschool AntwerpenReal time Linux
===
Re: pow(x,y) approximation> Hi all,Perhaps it's a bit offtopic
here, but I hope someone can help me out.I'm writing a small
computer program (Linux kernel module that is), which> needs a
pow function (x to the y; in computer-language that is
pow(x,y)).> Both x and y are floats (real numbers that
is).Does anybody know a usable routine to calculate this?
Should be fast too,> because it has to run on a slow
processor.> Currently, I already found a sin replacement.My
current function is:float pow(float x, float y) {> // Still
empty :(> }> If you do exp() and log(), then pow(x,y) =
===
Perhaps it's a bit offtopic here, but I hope someone can help
me out. I'm writing a small computer program (Linux kernel
module that is),> which needs a pow function (x to the y; in
computer-language that is> pow(x,y)). Both x and y are floats
(real numbers that is). Does anybody know a usable routine to
calculate this? Should be fast> too, because it has to run on
a slow processor.> Currently, I already found a sin
replacement. My current function is: float pow(float x, float
y) {> // Still empty :(> } > If you do exp() and log(), then
pow(x,y) = exp(y*log(x))A couple of years ago, G. A. Edgar
sugges the same thing, and Ibasically agreed with him. (See
copy below.) I then got (and, IIRC, GAEdid too) a private
email from someone whom I suspect is better informedthan I
about such mattestating that, in a floating-point system, it
isa bad idea to try to compute pow(x,y) by using
exp(y*log(x)). The troublein doing so would perhaps be best
addressed by a numerical analyst. David
Cantrell---------------------------------------------------->
,
Artraze 1) evaluation of exponents. I need a way to evaluate
the expression> x^y for any real values x and y.If by real you
really mean floating-point, then the answer is perhapsnot
simple. Have you checked for references on the web? You might
look atsomething like
. In its Annex
F,IEC 559 floating-point arithmetic is discussed. You would be
interesparticularly in F.9.4.3 The pow function, page 566. (For
what it's worth,I think some of the mathematics there is highly
questionable! See below.)> For positive x, use x^y =
exp(y*ln(x))Yes.> For negative x, you get into complex values
and multiple values, but> use the same formula...If we are
talking about floating-point arithmetic, then we don't want
tobe dealing with complex or multiple values. Personally, I
think that thesafest and simplest thing to do is to sayFor
positive x, x^y = exp(y*ln(x)) and essentially just leave it
atthat by saying Otherwise, x^y = NaN. Of course, the
===
Re: pow(x,y) approximation> I'm writing a small computer
program (Linux kernel module that is),> which needs a pow
function (x to the y; in computer-language that is> pow(x,y)).
Both x and y are floats (real numbers that is).The issue here
is that the general floating point operationsare not available
in kernel mode. While I'm not an expert,my memory is that the
general advice is not to do floating pointoperations in kernel
mode. One big problem, for example, is thatthe floating point
registers might contain data from a user process,so you would
have to explicitly save these.This may not be the correct
newsgroup. Are you absolutely sureyou want to do this kernel
mode? Can you give a hint what youare trying to
===
small computer program (Linux kernel module that is),>> which
needs a pow function (x to the y; in computer-language that
is>> pow(x,y)). Both x and y are floats (real numbers that
is).The issue here is that the general floating point
operations> are not available in kernel mode.Not if you go the
'normal' way, but you can use float in kernel space whenusing
extra libraries linked to the module.It might render slow, but
when I use fast math routines, I should be ableto get good
results.> This may not be the correct newsgroup. Are you
absolutely sure> you want to do this kernel mode?It's my only
solution so far.> Can you give a hint what you are trying to
do?Calculating the SuperFormule (geniaal.be) for projection
with a laser.This projection-control will happen in real time
(with RTAI), and can be'controlled' by a non real time GUI
interface (just pass variables throughthe /proc fs to the
module).The module should calculate about 360 points per
figure, and project 25figures per second at least. This gives
me a calculation ratio of approx.10kHz, which is very high for
a 200MHz PPC without FPU.More info on our website (see sig).--
Groeten, Toni Van RemortelWetenschappelijk medewerker,
D-science lab, Hogeschool AntwerpenReal time Linux for
===
embedded systems: http://linemb.d-sciencelab.comSubject: Re:
program (Linux kernel module that is),>> which needs a pow
function (x to the y; in computer-language that is>>
pow(x,y)). Both x and y are floats (real numbers that is).> If
you do exp() and log(), then pow(x,y) = exp(y*log(x))>A couple
of years ago, G. A. Edgar sugges the same thing, and
I>basically agreed with him. (See copy below.) I then got
(and, IIRC, GAE>did too) a private email from someone whom I
suspect is better informed>than I about such mattestating
that, in a floating-point system, it is>a bad idea to try to
compute pow(x,y) by using exp(y*log(x)). The trouble>in doing
so would perhaps be best addressed by a numerical
analyst.Software Manual for the Elementary Functions by
Cody.Fine little cookbook for doing things like this,
carefulattention to preserving all the accuracy feasible,
withexplanations for why he was doing what he was doing,
===
of Jews ... snip incredible nonsense from a dedica anti-semite
... > Careful, Kaz, these days the ones who always cry
anti-Semite> are now perceived to be the real bigots and
racists at heart. Reflect on that and ask yourself why this
came about.> Hint: 50+ years of incessant and loud Jew
propaganda about> Jewish moral and intellectual superiority,
......perhaps?> All good things do come to an end,
Kaz,.......ahahahaha.... One has to wonder why some people do
this.> Are they brainwashed,> or do they support the people
who instigate> conflict and war for power and riches?????> Tom
Potter > I don't know, nor care, what post preceded this one,>
but since ever mankind exis some people of/in any> tribe have
always and will always instigate conflict and> war for power
and riches...... or what other reasons are> there for killing
each other?......and if the conditions were> right, Tom, you'd
be in the forefront doing the same, given> the personality you
exhibit on the Usenet......ahahahaha... > Gee, I don't get
this. How do I get into this power elite? I'm usually>
considered quite bright and I am Jewish. But I have never been
in the> company of any Jew who was in this power elite. My late
brother-in-law> was rich, Jewish, but not one of the power
elite. How do I get there.> Why have the other Jews kept this
a secret from me and all my relatively> poor ancestors? Even
the atomic bomb had those who blabbed about it. Why> haven't I
heard about this grand scheme? I want in. I could contribute a>
lot. I have lots of Christian relatives who would provide cover
for me.> I know a few fairly powerful politicians and could
influence them if it> were worth my while. Why has all this
been hidden from me for so many> years? > Very astute and
SEEMINGLY factual, Kaz, BUT.....> See, Kaz, apparently you
really don't get it. It's very simple.> It is not so-called
riches and power that gets Jews into deep > periodically and
epically. A lot of goys have far more of each, yet> it
normally does not get them into trouble. Most have learned
how> to handle it: Discretely, smoothly and QUIETLY. But not
the Jews. With the Jews it is the opposite. It is their ing,>
incessant **loudness about themselves**, advocating that they
are> richer, better, smarter and more lovable BECAUSE THE ARE
JEWISH,> from private events to general media stuff, (like
i.e. reporting on the> front page or in headlines that it
rained in Tel Aviv, but shoving aschool> bus accident report
that killed 20 US-goy kids into the background......)> that is
what makes the goyim, which outnumber Jew almost a million>
fold, so irate and belligerent towards them. --- It is a
simple as that.> We won't even have to touch that for 50 years
US taxpayers got> forced to pay billions, annually, to the
Jewish State... with NOTHING to> show for in return! Will it
change? No. Why not?> Because that would mean for Jews to
become Un-Jewish....ahahahaha....> Of course Potter will come
along now and give recital of his> Not all Jews are... song
.........aahahahaha........ahahahaha......... Could it be that
Tom Potter is a troll who makes this all up out> of whole
cloth? Naw, not Tom Potter.> Everybody knows he is honest to a
fault.> FK ahahahaha............fault, Kaz, what
fault?.........honesty and moralityis> Potterian by
DEFAULT...........ahahahahaha........but then it becomes> more
evident with each post of yours that you, Kaz, are the Jewish
mirror> image of PotterHow so?Are you asserting that fkasneris
against the instigation of conflict and warfor power and
riches, by non-Jews like Bush, Blair and Rumfeld,and that
fkasner thinks that the root causes of how conflictcomes about
should be discussed, researched, and somethingdone to eliminate
the root causes?In other words, are you asserting that
fkasnerthinks that the root cause of conflict and war
likeinstigation for power and riches,should be trea, rather
than symptoms likeboys throwing rocks at tanks demolishing
theirhomes and neighborhoods, and girls becoming human
bombers,because family members were murdered by soldiers?--Tom
===
[fkasner] snipped incredible nonsense > from a dedica
anti-semite ...[Tom Potter] > Careful, Kaz, these days the
ones who always cry anti-Semite> are now perceived to be the
real bigots and racists at heart. Reflect on that and ask
yourself why this came about.> Hint: 50+ years of incessant
and loud Jew propaganda about> Jewish moral and intellectual
superiority, ......perhaps?> All good things do come to an
end, Kaz,.......ahahahaha.... One has to wonder why some
people do this.> Are they brainwashed,> or do they support the
people who instigate> conflict and war for power and
riches?????> Tom Potter > I don't know, nor care, what post
preceded this one,> but since ever mankind exis some people
of/in any> tribe have always and will always instigate
conflict and> war for power and riches...... or what other
reasons are> there for killing each other?......and if the
conditions were> right, Tom, you'd be in the forefront doing
the same, given> the personality you exhibit on the
Usenet......ahahahaha... > Gee, I don't get this. How do I get
into this power elite? I'm usually> considered quite bright and
I am Jewish. But I have never been in the> company of any Jew
who was in this power elite. My late brother-in-law> was rich,
Jewish, but not one of the power elite. How do I get there.>
Why have the other Jews kept this a secret from me and all my
relatively> poor ancestors? Even the atomic bomb had those who
blabbed about it. Why> haven't I heard about this grand scheme?
I want in. I could contribute a> lot. I have lots of Christian
relatives who would provide cover for me.> I know a few fairly
powerful politicians and could influence them if it> were worth
my while. Why has all this been hidden from me for so many>
years? > Very astute and SEEMINGLY factual, Kaz, BUT.....>
See, Kaz, apparently you really don't get it. It's very
simple.> It is not so-called riches and power that gets Jews
into deep > periodically and epically. A lot of goys have far
more of each, yet> it normally does not get them into trouble.
Most have learned how> to handle it: Discretely, smoothly and
QUIETLY. But not the Jews. With the Jews it is the opposite.
It is their ing,> incessant **loudness about themselves**,
advocating that they are> richer, better, smarter and more
lovable BECAUSE THE ARE JEWISH,> from private events to
general media stuff, (like i.e. reporting on the> front page
or in headlines that it rained in Tel Aviv, but shoving a>
school > bus accident report that killed 20 US-goy kids into
the > background......)> that is what makes the goyim, which
outnumber Jew almost a million> fold, so irate and belligerent
towards them. --- It is a simple as that.> We won't even have
to touch that for 50 years US taxpayers got> forced to pay
billions, annually, to the Jewish State... with NOTHING to>
show for in return! Will it change? No. Why not?> Because that
would mean for Jews to become Un-Jewish....ahahahaha....> Of
course Potter will come along now and give recital of his> Not
all Jews are... song
.........aahahahaha........ahahahaha......... Could it be that
Tom Potter is a troll who makes this all up out> of whole
cloth? Naw, not Tom Potter.> Everybody knows he is honest to a
fault.> FK ahahahaha........fault, Kaz, what
fault?.........honesty and morality> is Potterian by
DEFAULT.....ahahahahaha........but then it becomes> more
evident with each post of yours that you, Kaz, are the Jewish
> mirror image of PotterHow so?> Tom Potter
AHAHHAHAH..........ahahahahaha........Tom, Tom.......until you
answer this yourself, to yourself, about yourself and for
yourselfyou are as blind to the solution of the problem as are
your instigators of conflict and war for power and
riches.......ahahahha.....ahahahaAre you asserting that
fkasner> is against the instigation of conflict and war> for
power and riches, by non-Jews like Bush, Blair and Rumfeld,>
and that fkasner thinks that the root causes of how conflict>
comes about should be discussed, researched, and something>
done to eliminate the root causes?Suit yourself Tom, but let
me repeat for your benefit, Tom:> Of course Potter will come
along now and give recital of his> Not all Jews are... song
.........aahahahaha........ahahahaha.........which he just did
with a variation thereof. In other words, are you asserting
that fkasner> thinks that the root cause of conflict and war
like> instigation for power and riches,> should be trea,
rather than symptoms like> boys throwing rocks at tanks
demolishing their> homes and neighborhoods, and girls becoming
human bombers,> because family members were murdered by
soldiers?> Tom Potter Suit yourself, Tom, whether in your,
mine or other words.You are reading different things off/in my
post to Kaz then I do.You come ac to me here just as arguing
for argument's sake.Let me repeat for your benefit, Tom:> Of
course Potter will come along now and give recital of his> Not
all Jews are... song
.........aahahahaha........ahahahaha.........which he just did
===
Reason for High Achievements of Jews> [fkasner] snipped
incredible nonsense >>from a dedica anti-semite ...[Tom
Potter] >>Careful, Kaz, these days the ones who always cry
anti-Semite >are now perceived to be the real bigots and
racists at heart.>Reflect on that and ask yourself why this
came about. >Hint: 50+ years of incessant and loud Jew
propaganda about >Jewish moral and intellectual superiority,
......perhaps? >All good things do come to an end,
Kaz,.......ahahahaha....>>One has to wonder why some people do
this. >Are they brainwashed, >or do they support the people who
instigate >conflict and war for power and riches????? >Tom
Potter >I don't know, nor care, what post preceded this one,
>but since ever mankind exis some people of/in any >tribe have
always and will always instigate conflict and >war for power
and riches...... or what other reasons are >there for killing
each other?......and if the conditions were >right, Tom, you'd
be in the forefront doing the same, given >the personality you
exhibit on the Usenet......ahahahaha...>>Gee, I don't get
this. How do I get into this power elite? I'm usually
>>considered quite bright and I am Jewish. But I have never
been in the >>company of any Jew who was in this power elite.
My late brother-in-law >>was rich, Jewish, but not one of the
power elite. How do I get there. >>Why have the other Jews
kept this a secret from me and all my relatively >>poor
ancestors? Even the atomic bomb had those who blabbed about
it. Why >>haven't I heard about this grand scheme? I want in.
I could contribute a >>lot. I have lots of Christian relatives
who would provide cover for me. >>I know a few fairly powerful
politicians and could influence them if it >>were worth my
while. Why has all this been hidden from me for so many
>>years? >>Very astute and SEEMINGLY factual, Kaz, BUT.....
>See, Kaz, apparently you really don't get it. It's very
simple. >It is not so-called riches and power that gets Jews
into deep >periodically and epically. A lot of goys have far
more of each, yet >it normally does not get them into trouble.
Most have learned how >to handle it: Discretely, smoothly and
QUIETLY.>But not the Jews. With the Jews it is the opposite.
It is their ing, >incessant **loudness about themselves**,
advocating that they are >richer, better, smarter and more
lovable BECAUSE THE ARE JEWISH, >from private events to
general media stuff, (like i.e. reporting on the >front page
or in headlines that it rained in Tel Aviv, but shoving a
>school > bus accident report that killed 20 US-goy kids into
the >background......) >that is what makes the goyim, which
outnumber Jew almost a million >fold, so irate and belligerent
towards them. --- It is a simple as that. >We won't even have
to touch that for 50 years US taxpayers got >forced to pay
billions, annually, to the Jewish State... with NOTHING to
>show for in return!>Will it change? No. Why not? >Because
that would mean for Jews to become Un-Jewish....ahahahaha....
>Of course Potter will come along now and give recital of his
>Not all Jews are... song
.........aahahahaha........ahahahaha.........>>Could it be
that Tom Potter is a troll who makes this all up out >>of
whole cloth? Naw, not Tom Potter. >>Everybody knows he is
honest to a fault. >>FK >>ahahahaha........fault, Kaz, what
fault?.........honesty and morality >is Potterian by
DEFAULT.....ahahahahaha........but then it becomes >more
evident with each post of yours that you, Kaz, are the Jewish
>mirror image of Potter>How so?>>Tom Potter >
AHAHHAHAH..........ahahahahaha........Tom, Tom.......> until
you answer this yourself, to yourself, about yourself and for
yourself> you are as blind to the solution of the problem as
are your instigators of > conflict and war for power and
riches.......ahahahha.....ahahaha>>Are you asserting that
fkasner>>is against the instigation of conflict and war>>for
power and riches, by non-Jews like Bush, Blair and
Rumfeld,>>and that fkasner thinks that the root causes of how
conflict>>comes about should be discussed, researched, and
something>>done to eliminate the root causes?> Suit yourself
Tom, but let me repeat for your benefit, Tom:>Of course
Potter will come along now and give recital of his >Not all
Jews are... song
.........aahahahaha........ahahahaha.........which he just did
with a variation thereof. >In other words, are you asserting
that fkasner>>thinks that the root cause of conflict and war
like>>instigation for power and riches,>>should be trea,
rather than symptoms like>>boys throwing rocks at tanks
demolishing their>>homes and neighborhoods, and girls becoming
human bombers,>>because family members were murdered by
soldiers?>>Tom Potter >Suit yourself, Tom, whether in your,
mine or other words.> You are reading different things off/in
my post to Kaz then I do.> You come ac to me here just as
arguing for argument's sake.> Let me repeat for your benefit,
Tom:>Of course Potter will come along now and give recital
of his >Not all Jews are... song
.........aahahahaha........ahahahaha.........which he just did
with another variation thereof. > Forget it. He's a dumb-ass
troll who thinks that he can continue to ring the bells of
those who as a point of ethics and morality see the grand
Jewish conspiracy as a mean spiri attempt to pick on a small
minority who as emancipation has left them free to succeed are
a wonderfully handy scapegoat. Why even in countries that have
almost no Jews they are now having their prime ministers
engage in blatant anti-semitism. What a deal for a troll. But
probably in point of fact Tom Potshard is an anti-semite. But
then again he probably hates almost anyone who he is convinced
won't punch out his lights for showing his hate. So he is what
===
proof about Sylow theoremsLet G be finite and suppose P is a
subset of G and it is a psubgroup,S in Syl_p(G). Then P is a
subset of S^x for some x in GProorf: Let H-{Sx|x in G} and let
P act on K by right multiplicationWe have |K|=|G:S| = |G|/|S|
and this is not divisible by p since S isa Sylow p-subgroup.It
follows that some orbits of the action of P on K must have size
notdivisible by P.(I am not very clear about why)Since P is a
p-group, however, all of these orbits have p-power size.I am
not sure of this tooCould anyone help me understand this
theorem?I am learning by myself, so sometimes I have
===
theorems> Let G be finite and suppose P is a subset of G and
it is a psubgroup,> S in Syl_p(G). Then P is a subset of S^x
for some x in G> Proorf: Let H-{Sx|x in G} and let P act on K
by right multiplicationYou mean K = {Sx : x in G}.> We have
|K|=|G:S| = |G|/|S| and this is not divisible by p since S is>
a Sylow p-subgroup.> It follows that some orbits of the action
of P on K must have size not> divisible by P.By p.> (I am not
very clear about why)The order of K is the sum of the lengths
of the orbits of P on K.If all of these lengths were divisible
by p then the orderof K would be divisible by p.> Since P is a
p-group, however, all of these orbits have p-power size.Don't
forget that 1 is a power of p> I am not sure of this tooThe
length of an orbit of P is |P| divided by the order of a
stabilizer.As |P| is a power of the prime p, then so is the
===
and n-dimensional topological manifolds respectively.Given
that I can use Brouwer's Theorem on the Invariance of Domain,
how canI show that if m does not equal n, then M and N cannot
be homeomorphic.The Brouwer theorem says let A,B be subsets of
R^n and homeomorphicsubspaces. Then if A is open in R^n, B is
also open in R^n.So I want to use the fact of an m-dimensional
manifold that for each p in M(the manifold), there is a
neighborhood U of p and a homeomorphism f : U -->V , where V
is an open subset of R^m.This sounds like an easy exercise,
but I don't think I can quite get it. SoI think the best way
would be to prove by contradiction. Suppose M and Nwere
homeomorphic. Now, every neighborhood of M is homeo to an open
subsetof R^m and every neighborhood of N is homeo to an open
subset of R^n. But,we can use the fact that R^m and R^n cannot
be homeomorphic when m does notequal n. It seems like this is
all I need, but I am really not convinced.Is this the right
===
and Brouwer> Let M and N be m and n-dimensional topological
manifolds respectively.> Given that I can use Brouwer's
Theorem on the Invariance of Domain, how> can I show that if m
does not equal n, then M and N cannot be> homeomorphic.It's
easier, as I poin out recently in another thread,to use the
fact that the local homology groups of an n-manifoldare all
nonzero in dimension n and vanish in all other dimensions.This
is easier in the sense that it relies on a bit less
homologytheory than Brouwer's invariance of domain.But if you
insist on Brouwer, suppose that M is both an m-manifoldand an
n-manifold. Take any point x in M; then there are
openneighbourhoods U and V of X homeomorphic to R^m and R^n
respectively.Consider U intersect V. This is homeomorphic to
an open subsetof R^m and to an open subset of R^n. This
===
factsThe algebra here is *simple* and mathematicians may be
able to go onfor a while attacking algebra itself, but you
destroy your future, aseventually, I'll make sure that the
truth is known, and then you willhave been shown in the act,
caught in the act of betraying your owndiscipline.The
polynomial I'm using now is not very complica:P(m) = 14706125
m^3 - 900375 m^2 + 17640 m + 1078Note: The constant term is
1078, and 1078 = 7(7)(22)The method I use may be extraordinary
but it's not impossible tounderstand, no matter what liars like
Jim Ferry say, with P(m) itinvolves shifting things around so
that I haveP(m)= 7^2(2401 m^3 - 147 m^2 + 3m) (5^3) - 3(-1 +
49 m )(5)(7^2) + 7^3which isP(m) = 49((2401 m^3 - 147 m^2 +
3m) 5^3 - 3(-1 + 49 m )5 + 7).Now I can factor to getP(m) = (5
a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).where the a's are the roots of
the cubica^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 +
3m).Looking at my factorization I find that the factors have
constantvalues of 7, 7, and 22 respectively.Now look at the
next phase:P(m)/49 = 300125 m^3 - 18375 m^2 + 360 m + 22which
can be specially grouped to getP(m)/49 = (2401 m^3 - 147 m^2 +
3m) 5^3 - 3(-1 + 49 m )5 + 7.Now the constant term is 22, which
you can *see*, and notice it'sCOPRIME to 7.Therefore, P(m)/(49)
= (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)follows from the
*constant* terms.Mathematicians can fight me, but I will
continue to refine myexposition, and then I'll destroy
===
a previous post; correc here.) > The polynomial I'm using now
is not very complica: > P(m) = 14706125 m^3 - 900375 m^2 +
17640 m + 1078Q(m) = 6125 m^3 + 6125 m^2 - 6370 m + 1078 >
Note: The constant term is 1078, and 1078 = 7(7)(22) > The
method I use may be extraordinary but it's not impossible to >
understand, no matter what liars like Jim Ferry say, with P(m)
it > involves shifting things around so that I have > P(m)=
7^2(2401 m^3 - 147 m^2 + 3m) (5^3) - 3(-1 + 49 m )(5)(7^2) +
7^3Q(m) = 7^2(m^3 + m^2 - m) (5^3) - 3(-1 + m)(5)(7^2) + 7^3 >
> which is > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 +
49 m )5 + 7).Q(m) = 49((m^3 + m^2 - m) (5^3) - 3(-1 + m)5 + 7).
> Now I can factor to get > P(m) = (5 a_1 + 7)(5 a_2 + 7)(5
a_3 + 7).Q(m) = (5 b1 + 7)(5 b2 + 7)(5 b3 + 7). > where the
a's are the roots of the cubic > a^3 + 3(-1 + 49m)a^2 -
49(2401 m^3 - 147 m^2 + 3m).b^3 + 3(-1 + m)b^2 - 49(m^3 + m^2
- m). > Q(0) = 22. > Looking at my factorization I find that
the factors have constant > values of 7, 7, and 22
respectively.(Eh?) > Now look at the next phase: > P(m)/49
= 300125 m^3 - 18375 m^2 + 360 m + 22Q(m)/49 = 125 m^3 + 125
m^2 - 130 m + 22 > which can be specially grouped to get >
P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 +
7.Q(m)/49 = (m^3 + m^2 - m) 5^3 - 3(-1 + m)5 + 7. > Now the
constant term is 22, which you can *see*, and notice it's >
COPRIME to 7. > Therefore, > P(m)/(49) = (5 a_1/7 + 1)(5
a_2/7 + 1)(5 a_3 + 7)Q(m)/49 = (5 b1/7 + 1)(5 b2/7 + 1)(5 b3 +
7) > follows from the *constant* terms.Eh, no. These are not
factors in the algebraic integers. (And also shouldnot be
factors in the algebraic integeunless 7 is a unit.) These
arefactors in the algebraic numbeand possibly in a whole lot
of otherrings, but in each such ring, 7 is a unit.Consider
with my example: b^3 + 3(-1 + m)b^2 - 49(m^3 + m^2 - m).m = 0:
b^3 - 3b, so b1=b2=0 and b3 = 3 are roots.m = 1: b^3 - 49, so
the three cube roots of 49 are the roots of the equation.m = 2
(here is the correction): b^3 + 3 b^2 - 490, one root is 7, the
other two have some complica algebraic integer in common with 7
( [-sqrt(5)+-3i]/sqrt(2) ). (The polynomial is reducible as (b
===
facts > (I made an error in a previous post; correc here.) >
The polynomial I'm using now is not very complica: > P(m) =
14706125 m^3 - 900375 m^2 + 17640 m + 1078 > Q(m) = 6125 m^3
+ 6125 m^2 - 6370 m + 1078 > Note: The constant term is
1078, and 1078 = 7(7)(22) > The method I use may be
extraordinary but it's not impossible to > understand, no
matter what liars like Jim Ferry say, with P(m) it > involves
shifting things around so that I have > P(m)= 7^2(2401 m^3 -
147 m^2 + 3m) (5^3) - 3(-1 + 49 m )(5)(7^2) + 7^3 > Q(m) =
7^2(m^3 + m^2 - m) (5^3) - 3(-1 + m)(5)(7^2) + 7^3 > which
is > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m
)5 + 7). > Q(m) = 49((m^3 + m^2 - m) (5^3) - 3(-1 + m)5 +
7). > Now I can factor to get > P(m) = (5 a_1 + 7)(5 a_2 +
7)(5 a_3 + 7). > Q(m) = (5 b1 + 7)(5 b2 + 7)(5 b3 + 7). >
where the a's are the roots of the cubic > a^3 + 3(-1 +
49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). > b^3 + 3(-1 + m)b^2
- 49(m^3 + m^2 - m). > Q(0) = 22.Carried away I think.
P(0) = Q(0) = 1078. > Looking at my factorization I find
that the factors have constant > values of 7, 7, and 22
respectively. > (Eh?)Now it makes more sense. > Now look
at the next phase: > P(m)/49 = 300125 m^3 - 18375 m^2 + 360
m + 22 > Q(m)/49 = 125 m^3 + 125 m^2 - 130 m + 22 > which
can be specially grouped to get > P(m)/49 = (2401 m^3 - 147
m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7. > Q(m)/49 = (m^3 + m^2 -
m) 5^3 - 3(-1 + m)5 + 7. > Now the constant term is 22, which
you can *see*, and notice it's > COPRIME to 7. > Therefore, >
> P(m)/(49) = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) > Q(m)/49
= (5 b1/7 + 1)(5 b2/7 + 1)(5 b3 + 7) > follows from the
*constant* terms. > Eh, no. These are not factors in the
algebraic integers. (And also should > not be factors in the
algebraic integeunless 7 is a unit.) These are > factors in
the algebraic numbeand possibly in a whole lot of other >
rings, but in each such ring, 7 is a unit. > Consider with
my example: > b^3 + 3(-1 + m)b^2 - 49(m^3 + m^2 - m). > m = 0:
> b^3 - 3b, so b1=b2=0 and b3 = 3 are roots. > m = 1: > b^3 -
49, so the three cube roots of 49 are the roots of the
equation. > m = 2 (here is the correction): > b^3 + 3 b^2 -
490, one root is 7, the other two have some complica >
algebraic integer in common with 7 ( [-sqrt(5)+-3i]/sqrt(2) ).
(The > polynomial is reducible as (b - 7)(b^2 + 10b + 70).) >
-- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland, +31205924131 > home: bovenover 215, 1025 jn
===
Re: JSH: Simple math factsThe algebra here is *simple* and
mathematicians may be able to go onfor a while attacking
algebra itself, but you destroy your future, aseventually,
I'll make sure that the truth is known, and then you willhave
been shown in the act, caught in the act of betraying your
owndiscipline.The polynomial I'm using now is not very
complica:P(m) = 14706125 m^3 - 900375 m^2 + 17640 m +
1078Note: The constant term is 1078, and 1078 = 7(7)(22)The
method I use may be extraordinary but it's not impossible
tounderstand, no matter what liars like Jim Ferry say, with
P(m) itinvolves shifting things around so that I haveP(m)=
7^2(2401 m^3 - 147 m^2 + 3m) (5^3) - 3(-1 + 49 m )(5)(7^2) +
7^3which isP(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49
m )5 + 7).Now I can factor to getP(m) = (5 a_1 + 7)(5 a_2 +
7)(5 a_3 + 7).where the a's are the roots of the cubica^3 +
3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m).I then
concentrate on constant terms. Looking at my factorization I
find that the factors have constantvalues of 7, 7, and 22
respectively.Now look at the next phase:P(m)/49 = 300125 m^3 -
18375 m^2 + 360 m + 22which can be specially grouped to
getP(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 +
7.Now the constant term is 22, which you can *see*, and notice
it'sCOPRIME to 7.Therefore, P(m)/(49) = (5 a_1/7 + 1)(5 a_2/7 +
1)(5 a_3 + 7)follows from the *constant* terms.Mathematicians
can fight me, but I will continue to refine myexposition, and
===
facts>> The algebra here is *simple* and mathematicians may be
able to go on>> for a while attacking algebra itself, but you
destroy your future, as>> eventually, I'll make sure that the
truth is known, and then you will>> have been shown in the
act, caught in the act of betraying your own>> discipline.The
polynomial I'm using now is not very complica:P(m) = 14706125
m^3 - 900375 m^2 + 17640 m + 1078Note: The constant term is
1078, and 1078 = 7(7)(22)The method I use may be extraordinary
but it's not impossible to>> understand, no matter what liars
like Jim Ferry say, with P(m) it>> involves shifting things
around so that I haveP(m)= 7^2(2401 m^3 - 147 m^2 + 3m) (5^3)
- 3(-1 + 49 m )(5)(7^2) + 7^3which isP(m) = 49((2401 m^3 - 147
m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7).Now I can factor to getP(m)
= (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).where the a's are the roots
of the cubic[*] a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 +
3m). I've been tossing that out for a while, but considering a
post by Dik>Winter in this thread, I began to reconsider it. It
turns out that the cubic is only valid within the ring of
algebraic>integeor my ring of objects, for specific values of
m, as>otherwise you're forced into a field.> What do you mean
by valid? There is a very simple andstraightforward proof that
a1, a2, and a3 are all roots ofthe cubic [*] above, and it is
easy to show also that theyare algebraic integers. You have
derived [*] yourself.Your derivation is *correct*. You are
*not* forced into afield. The coefficients of [*] are all
integers and it is monic. Its roots are algebraic integers. I
think what has happened here is this. You have finally by
someweird accident reproduced the computations that we have
beendescribing to you for many weeks. You have apparently
stumbled on the refutation of your claims that we have
beentrying to get you to understand. Dik must have said it in
exactlythe right way to get you to actually think about it
andabsorb it. Until now all of us, Dik included, were
categorizedas liars and incompetents for this. Finally you see
what we were talking about. And of course the result is instant
denial.>My assumption was that it didn't matter *how* you got
the cubic, so>I've been tossing it out there as if it didn't
matter that you need to>operate within the field algebraic
numbebut apparently, it does>matter.> The roots of the cubic
[*] are algebraic integers. >I will admit that the math here
is kind of freaky. This is astonishing! the math here is **
kind of freaky ** ?? What the hell does THAT mean? Might
another way todescribe that is, the math here is wrong, or the
math Ihave been using here is nonsense, or I don't
understandthe math here, but of course my 'proof' takes
precedence overall other mathematical arguments and is right
no matter whatit contradicts!> The situation is a> LOT like
quantum mechanics being introduced to the physics world, as>
how can it matter mathematically that you have to go to a
field to get> the cubic? What can you be talking about? The
coefficients of thecubic [*] are all ordinary integers. The
roots are algebraicintegers. There is nothing to makes you
have to go to a fieldto get the cubic. Oddly there IS however
a parallel here with quantum mechanics.Factoring polynomials
as done in elementary classes, by inspection,like (x^2 + 7*x +
12) = (x + 3)*(x + 4), is sort of like classicalphysics, and
divisibility properties of the factors do not
presentsurprises. Why? Because such polynomials are REDUCIBLE
over therationals. But IRREDUCIBLE polynomials behave in
really essentiallydifferent ways. Part of the difference is
due to the fact thatthere is no such thing as a prime in the
algebraic integers. Allalgebraic integers are infinitely
divisible. Ordinary integersare not. Irreducible polynomials
factor in really essentiallydifferent ways from reducible
ones; they are more like quantum mechanics, even in the sense
that things (i.e. factors) can be thoughtof as being in two
places at once! This parallelism is, to borrowone of your
favorite words, rather fascinating. Well, apparently it does,
as the simple principle that constant terms>are indeed
constant and independent of m, simply can't be
logically>broken.> NO ONE argues that constant terms are not
constant. Conversely,one would think that no one would argue,
as you have done, that NONCONSTANTterms are CONSTANT (at least
with respect to factorability). This looks like a very, very
===
Dik for somehow getting you to see it. Subject: Re: JSH: Simple
may be able to go on> for a while attacking algebra itself,
but you destroy your future, as> eventually, I'll make sure
that the truth is known, and then you will> have been shown in
the act, caught in the act of betraying your own> discipline.>
[same stuff dele]Have you considered trying to say something
===
Assistant Professor at the University of Montana. > The
algebra here is *simple* and mathematicians may be able to go
on>> for a while attacking algebra itself, but you destroy
your future, as>> eventually, I'll make sure that the truth is
known, and then you will>> have been shown in the act, caught
in the act of betraying your own>> discipline.>[same stuff
dele] Have you considered trying to say something *new*?Seems
unlikely, as he does not consider ->looking<- at something
new,either. Hell, he does not consider looking at something
->old<-, liketrying to figure out what the real 'nonconstant
terms' of hisfunctions are by doing the algebra, instead of
relying on his
expose 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
===
destination is the journey-- Peter van um, 
| Universal law of linearity: for allDept. of Mathematics, New
Mexico | f : R -> R and for all x, y in R:State University, Las
===
Simple math facts> The algebra here is *simple* and
mathematicians may be able to go on> for a while attacking
algebra itself, but you destroy your future, as> eventually,
I'll make sure that the truth is known, and then you will>
have been shown in the act, caught in the act of betraying
your own> discipline.The polynomial I'm using now is not very
complica:P(m) = 14706125 m^3 - 900375 m^2 + 17640 m +
1078Note: The constant term is 1078, and 1078 = 7(7)(22)The
method I use may be extraordinary but it's not impossible to>
understand, no matter what liars like Jim Ferry say, with P(m)
it> involves shifting things around so that I haveP(m)=
7^2(2401 m^3 - 147 m^2 + 3m) (5^3) - 3(-1 + 49 m )(5)(7^2) +
7^3which isP(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49
m )5 + 7).Now I can factor to getP(m) = (5 a_1 + 7)(5 a_2 +
7)(5 a_3 + 7).where the a's are the roots of the cubica^3 +
3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m).I've been tossing
that out for a while, but considering a post by DikWinter in
this thread, I began to reconsider it.It turns out that the
cubic is only valid within the ring of algebraicintegeor my
ring of objects, for specific values of m, asotherwise you're
forced into a field.My assumption was that it didn't matter
*how* you got the cubic, soI've been tossing it out there as
if it didn't matter that you need tooperate within the field
algebraic numbebut apparently, it doesmatter.I will admit that
the math here is kind of freaky. The situation is aLOT like
quantum mechanics being introduced to the physics world, ashow
can it matter mathematically that you have to go to a field to
getthe cubic?Well, apparently it does, as the simple principle
that constant termsare indeed constant and independent of m,
===
math facts... > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1
+ 49 m )5 + 7). > Now I can factor to get > P(m) = (5 a_1 +
7)(5 a_2 + 7)(5 a_3 + 7). > where the a's are the roots of
the cubic > a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 +
3m). > I've been tossing that out for a while, but
considering a post by Dik > Winter in this thread, I began to
reconsider it. > It turns out that the cubic is only valid
within the ring of algebraic > integeor my ring of objects,
for specific values of m, as > otherwise you're forced into a
field.When m is an integer (which I thought was the case), the
roots of the cubicin a are obviously algebraic integers (the
coefficients are integers).Your and mine cubic are both of
this form. Nevertheless with my cubic,when m = 0 the factors
of 49 distribute as you tell, with m = 1 allthree roots have a
factor of 7^{2/3} and when m = 2 one root has afactor 7, the
other two have both other algebraic integer, non-unitfactrs of
7. So what is the problem, where am I forced to go into afield?
> My assumption was that it didn't matter *how* you got the
cubic, so > I've been tossing it out there as if it didn't
matter that you need to > operate within the field algebraic
numbebut apparently, it does > matter.I *do* work in the
algebraic numbemore specific, I work in thealgebraic integers.
> I will admit that the math here is kind of freaky. The
situation is a > LOT like quantum mechanics being introduced
to the physics world, as > how can it matter mathematically
that you have to go to a field to get > the cubic?Eh? > Well,
apparently it does, as the simple principle that constant
terms > are indeed constant and independent of m, simply can't
be logically > broken.Constant terms are constant. Your belief
that when you have a functionof m, with f(0) = 0, that (f(m) +
7) is divisible by 7 for all m, isalso constant. But the latter
===
Mathematicians can fight me, but I will continue to refine my>
exposition, and then I'll destroy them.Yawn...You tried
something like that with your prime counting function, and
===
math facts> The algebra here is *simple* and mathematicians
may be able to go on> for a while attacking algebra itself,
but you destroy your future, as> eventually, I'll make sure
that the truth is known, and then you will> have been shown in
the act, caught in the act of betraying your own>
discipline.The polynomial I'm using now is not very
complica:P(m) = 14706125 m^3 - 900375 m^2 + 17640 m +
1078Note: The constant term is 1078, and 1078 = 7(7)(22)The
method I use may be extraordinary but it's not impossible to>
understand, no matter what liars like Jim Ferry say, with P(m)
it> involves shifting things around so that I haveP(m)=
7^2(2401 m^3 - 147 m^2 + 3m) (5^3) - 3(-1 + 49 m )(5)(7^2) +
7^3which isP(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49
m )5 + 7).Now I can factor to getP(m) = (5 a_1 + 7)(5 a_2 +
7)(5 a_3 + 7).where the a's are the roots of the cubica^3 +
3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m).Looking at my
factorization I find that the factors have constant> values of
7, 7, and 22 respectively.Now look at the next phase:P(m)/49 =
300125 m^3 - 18375 m^2 + 360 m + 22which can be specially
grouped to getP(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 +
49 m )5 + 7.Now the constant term is 22, which you can *see*,
and notice it's> COPRIME to 7.Therefore, P(m)/(49) = (5 a_1/7
+ 1)(5 a_2/7 + 1)(5 a_3 + 7)follows from the *constant*
terms.Mathematicians can fight me, but I will continue to
refine my> exposition, and then I'll destroy them.> Mr.
Harris,Please keep refining. It is still not clear to me about
the relavance of the terms that include a_i's. I'm not a
mathematician, although I have maintained an interest in
mathematics since my Uncle Ralph showed me that 12345679 times
a single digit multiple of 9 yielded nine of the digit.
(Example 12345679 * 27 = 333333333). This is a fun way to
check out whether a calculator will handle more that eight
significant digits. The cheap calculators don't usually.my
impression that P(m) evalua at m = 0 is 1078, which factors
===
facts> The algebra here is *simple* and mathematicians may be
able to go on> for a while attacking algebra itself, but you
destroy your future, as> eventually, I'll make sure that the
truth is known, and then you will> have been shown in the act,
caught in the act of betraying your own> discipline.The
polynomial I'm using now is not very complica:P(m) = 14706125
m^3 - 900375 m^2 + 17640 m + 1078Note: The constant term is
1078, and 1078 = 7(7)(22)The method I use may be extraordinary
but it's not impossible to> understand, no matter what liars
like Jim Ferry say, with P(m) it> involves shifting things
around so that I haveP(m)= 7^2(2401 m^3 - 147 m^2 + 3m) (5^3)
- 3(-1 + 49 m )(5)(7^2) + 7^3which isP(m) = 49((2401 m^3 - 147
m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7).Now I can factor to getP(m)
= (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).where the a's are the roots
of the cubica^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 +
3m).> That's a mistake as it should be P(0) = 7(7)(22).>
Looking at my factorization I find that the factors have
constant> values of 7, 7, and 22 respectively. 7,7, and 22.Duane AllenYup, it was a mistake, and
===
facts>[...]>Mathematicians can fight me, but I will continue
to refine my>exposition, and then I'll destroy them.That's at
least three posts today that sound like the work ofa lunatic.
===
Re: JSH: Simple math facts> Mathematicians can fight me, but I
will continue to refine my> exposition, and then I'll destroy
===
fight me, but I will continue to refine my> exposition, and
then I'll destroy them.WWhat, still haven't figured it out yet
Jim Ferry? I react to*negative* emotions and
behaviors.Mathematicians are currently attacking mathematics
itself, and that isnot a good idea.It also will not be a
===
facts>Mathematicians can fight me, but I will continue to
refine my >exposition, and then I'll destroy them.>W> What,
still haven't figured it out yet Jim Ferry? I react to>
*negative* emotions and behaviors.Which is why sci.math takes
such a terrible toll on you. You're like amacho man who eats
plutonium because he's tough, because he can handle it.>
Mathematicians are currently attacking mathematics itself, and
that is> not a good idea.Mathematics is, by defintion, what
mathematicians create. You've creaa new structure which
rightly should be called by a different name, andwhich seeks
to replace mathematics. It seems that you expect it to
usurpmathematics in the same way that cro-magnons replaced the
neanderthals.> It also will not be a successful course of
action.Really? I doubt that strongly. The road you've chosen
leads nowhere butto your own despair. But you think it's the
only road, so you trudge on,fighting dramatic battles. Listen:
dry, sterile thunder.Some say that man wrests the secrets of
nature like a thief. You've stolena golden apple from the most
prized tree, and you expect to just waltz rightout? Or to fight
your way through?Boundless optimism.Where will you be ten years
===
Simple math facts>Mathematicians can fight me, but I will
continue to refine my >exposition, and then I'll destroy
them.>W> What, still haven't figured it out yet Jim Ferry? I
react to> *negative* emotions and behaviors.Which is why
sci.math takes such a terrible toll on you. You're like a>
macho man who eats plutonium because he's tough, because he
can handle it.What toll? As I type, my paper is under review
at a math journal. Some of my other work last time I checked
was under consideration at amajor American university, and I'm
here trying to figure out ifthere's something I've missed.I'm
troubled by the answer I gave in a recent post in this thread
asit still doesn't quite make sense to me.That post by Dik
Winter in this thread is using an idea that I firstsaw from
Nora Baron and it's bugging the hell out of me.While I'd like
to acquiesce if I'm wrong, the simple truth is thatCONSTANT
factors DO NOT CHANGE as if they're not constant, as that's
adirect contradiction.There MUST be a logical answer that
covers everything.And no stupid responses just claiming that
the logical answer is thatI'm wrong, as you have to handle
constant factors being INDEPENDENT. > Mathematicians are
currently attacking mathematics itself, and that is> not a
good idea.Mathematics is, by defintion, what mathematicians
create. You've crea> a new structure which rightly should be
called by a different name, and> which seeks to replace
mathematics. It seems that you expect it to usurp> mathematics
in the same way that cro-magnons replaced the
neanderthals.That's bull. I've emphasized that I'm using basic
ALGEBRA.I'm relying on mathematics that is SIMPLER than what
others keeptossing at me, so it's not like it's new math.Why
do you keep mouthing wacky stuff Jim Ferry instead of jumping
intothe fray?I need some serious mind power to figure out this
latest weirdness.If I'm wrong, I'll accept that, as LOGIC
dictates.But make no mistake, I won't let go until I figure it
out. > It also will not be a successful course of
action.Really? I doubt that strongly. The road you've chosen
leads nowhere but> to your own despair. But you think it's the
only road, so you trudge on,> fighting dramatic battles.
Listen: dry, sterile thunder.Hey, I have my fun, but that's no
reason for you to toss out logic. > Some say that man wrests
the secrets of nature like a thief. You've stolen> a golden
apple from the most prized tree, and you expect to just waltz
right> out? Or to fight your way through?Boundless
optimism.Where will you be ten years from now? Still on the
verge of victory?HELL NO!!! I'm in the thick of it trying to
figure out what's the truth, when Ithought it was all
handled.But that Nora Baron - Dik Winter example is bugging
the hell out ofme.There's a RATIONAL answer, so if one of you
===
facts Adjunct Assistant Professor at the University of
Montana.>The algebra here is *simple* and mathematicians may
be able to go on>for a while attacking algebra itself, but you
destroy your future, as>eventually, I'll make sure that the
truth is known, and then you will>have been shown in the act,
caught in the act of betraying your own>discipline. The
polynomial I'm using now is not very complica: P(m) = 14706125
m^3 - 900375 m^2 + 17640 m + 1078 Note: The constant term is
1078, and 1078 = 7(7)(22) The method I use may be
extraordinary but it's not impossible to>understand, no matter
what liars like Jim Ferry say, with P(m) it>involves shifting
things around so that I have P(m)= 7^2(2401 m^3 - 147 m^2 +
3m) (5^3) - 3(-1 + 49 m )(5)(7^2) + 7^3 which is P(m) =
49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7). Now I
can factor to get P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).
where the a's are the roots of the cubic a^3 + 3(-1 + 49m)a^2
- 49(2401 m^3 - 147 m^2 + 3m). Looking at my factorization I
find that the factors have constant>values of 7, 7, and 22
respectively. Now look at the next phase: P(m)/49 = 300125 m^3
- 18375 m^2 + 360 m + 22 which can be specially grouped to get
P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7.
Now the constant term is 22, which you can *see*, and notice
it's>COPRIME to 7. Therefore, P(m)/(49) = (5 a_1/7 + 1)(5
a_2/7 + 1)(5 a_3 + 7) follows from the *constant*
terms.Finally, I think I get a glimpse of your thought
process.Is this what you are thinking?[NOTE TO EVERYONE ELSE:
Please do not comment on the correctness orincorrectness of
what follows; I want to know, ahead of time, if thisis a
correct parsing of James's thought process.] 1. We have P(m) =
g_1(m)*g_2(m)*g_3(m). We can write g_i(m) = h_i(m) + g_i(0),
where h_i(m)=g_i(m)-g_i(0). Then P(m) =
(h_1(m)+g_1(0))*(h_2(m)+g_2(0))*(h_3(m)_3+g_3(0)) and
g_1(0)*g_2(0)*g_3(0) = 22*49. 2. h_1(0) = h_2(0)=h_3(0) = 0.
3. P(m)/49 =
(1/49)(h_1(m)+g_1(0))(h_2(m)+g_2(0))(h_3(m)+g_3(0)). We must
be able to factor out 49 from the terms; if w_1(m), w_2(m),
and w_3(m) are what we factor out from each, then we must have
w_1(m)*w_2(m)*w_3(m) = 49, and P(m)/49 = [(h_1(m)/w_1(m)) +
(g_1(0)/w_1(m))] * * [ (h_2(m)/w_2(m)) + (g_2(0)/w_2(m))] * [
(h_3(m)/w_3(m)) + (g_3(0)/w_3(m))] 4. At m=0, h_i(m) = 0
(since h_i(m) = g_i(m)-g_i(0)), so we just get g_i(0)/w_i(0)
So if we write (h_i(m)/w_i(m)) + (g_i(0)/w_i(m)) as H_i(m) +
constant term, then constant term = g_i(0)/w_i(0). 5. Then we
must have that g_1(0)/w_1(0) is the constant term, and
h_1(m)/w_1(m) is the non-constant term; which means that, we
have H_i(m) = h_i(m)/w_i(m) H_i(0) = g_1(0)/w_i(0)
(h_i(m)/w_i(m)) + (g_i(0)/w_i(m)) = H_i(m) + H_i(0) =
(h_i(m)/w_i(m)) + g_i(0)/w_i(0) So g_i(0)/w_i(m) =
g_i(0)/w_i(0), and therefore w_i(0)=w_i(m) for all m, proving
it is constant. Since w_1(0)=7, w_2(0)=7, and w_3(0)=1, that
means that h_1(m)+g_1(0) is always a multiple of w_1(m) = 7,
h_2(m)+g_2(0) is always a multiple of w_2(m)=7, and
h_3(m)+g_3(0) never contributes to the 49.Is this what is
behind your argument? Please state yes or no; if no,then
please say exactly what you disagree with and why. I'll
commentonly after your
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
===
(5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).[...]> Therefore, P(m)/(49)
= (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)follows from the
*constant* terms.It looks to me like it follows from simple
division.Ok, so P(m)/49 is that expression. Now what? What's
your point?V.-- email: lastname at cs utk eduhomepage: cs utk
===
(5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).[...]Therefore, P(m)/(49) =
(5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)follows from the
*constant* terms.It looks to me like it follows from simple
division.Ok, so P(m)/49 is that expression. Now what? What's
your point?V.(5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) = 300125
m^3 - 18375 m^2 + 360 m + 22where readers can see that the
constant term is 22, which is coprimeto 7.1 as their constant
term, while the third of the a's is 3 when m=0,which gives the
last factor as 22 as required.The math IS basic. Some may
believe that 5 a_1/7 forces 7 to be aunit in the ring, for
some value of a_1, but in fact, it's just notpossible to in
general express the factorization for all m, but 7 isnever a
===
JSH: Simple math factsP(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 +
7).[...]Therefore, P(m)/(49) = (5 a_1/7 + 1)(5 a_2/7 + 1)(5
a_3 + 7)follows from the *constant* terms.It looks to me like
it follows from simple division.Ok, so P(m)/49 is that
expression. Now what? What's your point?V. (5 a_1/7 + 1)(5
a_2/7 + 1)(5 a_3 + 7) = 300125 m^3 - 18375 m^2 + 360 m + 22
where readers can see that the constant term is 22, which is
coprime>to 7.And.....? Why is this important?>1 as their
constant term, while the third of the a's is 3 when m=0,>which
gives the last factor as 22 as required.I have no idea why you
think this is significant. I'm not talking aboutwhether or not
anyone agrees or disagrees, I'm just wondering why
yourstatement matters. So a_1(m) = 3, a_2(m) = a_3(m) = 0 when
m = 0. Sowhat? >The math IS basic. Some may believe that 5
a_1/7 forces 7 to be a>unit in the ring, for some value of
a_1, but in fact, it's just not>possible to in general express
the factorization for all m, but 7 is>never a unit, or the
constant terms would be different.Yah know, with most proofs I
can usually tell what the author is trying todo even if I don't
understand the proof itself. With what you've written Ican't
tell why you have written out the equations you have nor why
you thinkthey prove what you are trying to prove. Alan--
===
(5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).[...]Therefore, P(m)/(49) =
(5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)follows from the
*constant* terms.It looks to me like it follows from simple
division.Ok, so P(m)/49 is that expression. Now what? What's
your point?V. (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) = 300125
m^3 - 18375 m^2 + 360 m + 22 where readers can see that the
constant term is 22, which is coprime>to 7.And.....? Why is
this important?1 as their constant term, while the third of
the a's is 3 when m=0,>which gives the last factor as 22 as
required.I have no idea why you think this is significant. I'm
not talking about> whether or not anyone agrees or disagrees,
I'm just wondering why your> statement matters. So a_1(m) = 3,
a_2(m) = a_3(m) = 0 when m = 0. So> what?The constant terms are
independent of m, right? So any changes tothem must occur
without regard to the value of m.Here's a simpler example to
illustrate the principle.Q(m) = (7m + 7)(7m + 7)(5m + 22)and
notice that the constant term for Q(m) is 7(7)(22), and that
Q(m)has 49 as a factor. But dividing by 49 givesQ(m)/49 =
(m+1)(m+1)(5m + 22)and the constant term is now 22.It's such
an obvious thing that you probably NEVER thought through
themath principles of what just happened.However, the a's are
VERY complica functions and you can't justlook at them and
figure things out the way you can with that basicexample.So I
had to use the principle which you may have never
consideredbecause you can *physically* use your eyes with
polynomial factors. >The math IS basic. Some may believe that
5 a_1/7 forces 7 to be a>unit in the ring, for some value of
a_1, but in fact, it's just not>possible to in general express
the factorization for all m, but 7 is>never a unit, or the
constant terms would be different.Yah know, with most proofs I
can usually tell what the author is trying to> do even if I
don't understand the proof itself. With what you've written I>
can't tell why you have written out the equations you have nor
why you think> they prove what you are trying to prove.
AlanYou're probably trying to use your intuition and your
physical sight.That is you LOOK at the expression as if your
eyes can physically peelaway the details because that usually
works for you, but the situationrequires that you THINK about
what principles are being used.One problem I think is that
mathematicians are currently taught torely too much on their
intuition, which was a problem in physics aswell, until
===
quantum mechanics came along.Subject: Re: JSH: Simple math
forces 7 to be a >unit in the ring, for some value of a_1, but
in fact, it's just not >possible to in general express the
factorization for all m, but 7 is >never a unit, or the
constant terms would be different.>Yah know, with most proofs
I can usually tell what the author is trying to>>do even if I
don't understand the proof itself. With what you've written
I>>can't tell why you have written out the equations you have
nor why you think>>they prove what you are trying to prove.
>Alan> You're probably trying to use your intuition and your
physical sight.That is you LOOK at the expression as if your
eyes can physically peel> away the details because that
usually works for you, but the situation> requires that you
THINK about what principles are being used.One problem I think
is that mathematicians are currently taught to> rely too much
on their intuition, which was a problem in physics as> well,
until quantum mechanics came along.> You've obviously never
been in my classroom. I just spent this morning telling my
students that they *cannot* trust their intuition about
probabilities. They need to do the computations and, if done
correctly, trust the results regardless of their intuition.--
===
like 2300 years. Euclid proved there is an infinitude of
primes.> Naah, actually, he never mentions infinite at all.>
He just does: Given any [finite] set of primes, to construct a
prime> not in the set.> Does the usual proof actually construct
a prime?No, but it shows that such a prime must exist.> Since
nobody else sketched the argument, I will.Let p1, p2, p3, ...,
pn be the first n primes. Consider> the number
(p1*p2*...*pn)+1. This number has a remainder> of 1 when
divided by any prime up to pn. Therefore either> the number is
prime, or if it is composite then its prime> factors are all
larger than pn.Thus given the first n primes where n is
finite,> there exists another prime which is larger than all>
of them.If p1, p2, ... pn are any n primes (not the first n),>
then you've proved as Virgil says that there exists another>
prime not in this list, but it's not necessarily> a larger
one. For instance, (3*5*13)+1 has 2 and 7 as> prime factors.
Still, since that means no finite list of> primes is complete
it is sufficient to prove there are> infinitely many primes.
You showed that or why the set of primes is called infinite in
size,fine. However, you and others say that no prime (or
integer) isinfinite. Please consider this view to the
contrary:Each prime is unique. And, we can arrange them in
order of increasingmagnitude. Now, if we do so, and then count
these primes, in ascendingorder, the counting number n will
increase as we proceed. Of course,we can't continue this
forever to exhaust all the primes, but if wecould, as
necessary to count all the primes, wouldn't n becomeinfinite
in the same sense as the size of the set? And, thus, theprime
corresponding to n would also become infinite in the same
sense?I guess the answer is; by definition, we can't actually
count toinfinity; not even in our best wishes or wildest
dreams orimagination. But, then the meaning of infinity, as
simplyuncountable, must also prevail. i.e. to say that the set
of primes(or integers) is infinite, is simply to say the set is
uncountable.Yet, we do say the set is denumerable, i.e. its
elements can be placedin one-one correspondence with the
natural numbers. I thoughtdenumerable and countable were
synonyms. But, their usage hereseems contradictory or
inconsistant. Does this confusion resultbecause, as with
counting, we can't actually carry on a correspondenceprocedure
===
definition, we can't actually count to> infinity; not even in
our best wishes or wildest dreams or> imagination. But, then
the meaning of infinity, as simply> uncountable, must also
prevail. i.e. to say that the set of primes> (or integers) is
infinite, is simply to say the set is uncountable.In math, the
word uncountable has a technical meaning thatis different from
the way you are using the word here. You areright that it
would be impossible, in practice, to count allthe primes --
but still, if you had *infinite* time, you couldcount them
all. In such cases, we say the set is countableor denumerable.
(The actual definition is as you say below.)There are some sets
so big that they can't be coun even ininfinite time. Those sets
are called uncountable. The set ofprimes is not one of these
sets; neither is the set of integers.> Yet, we do say the set
is denumerable, i.e. its elements can be placed> in one-one
correspondence with the natural numbers. I thought>
denumerable and countable were synonyms. But, their usage
here> seems contradictory or inconsistant. Does this confusion
result> because, as with counting, we can't actually carry on a
correspondence> procedure to infinite extent, either?I think
the confusion is that you are trying to use the wordcountable
in its natural-language meaning, instead of itstechnical
meaning. Saying that a set is countable (in thetechnical
sense) does not necessarily mean that a real humanbeing can
===
prime really exist?> I think the confusion is that you are
trying to use the word> countable in its natural-language
meaning, instead of its> technical meaning. Saying that a set
is countable (in the> technical sense) does not necessarily
mean that a real human> being can ever count it completely.As
far as that goes, there are large finite integers that no
human could ever count to. The number of seconds in a human
life rarely exceeds 3,188,673,600 seconds (100 years). There
is no human who could ever count the number of dollars in the
USA's national debt, which is more that a thousand times as
large as that number of seconds. And scientificmeasurement can
===
prime really exist?> could ever count the number of dollars in
the USA's national debt, > which is more that a thousand times
as large as that number of > seconds. And
scientificmeasurement can generate much larger integers. The
national debt is a continually upda sum of other quantities.
One is not counting the number of dollar bills in a very large
pile. But even if there are piles of paper containing as many
bills as the national debt contains, one builds a -machine- to
do the count. Building the machine takes much less time.Bob
===
exist?could ever count the number of dollars in the USA's
national debt, > which is more that a thousand times as large
as that number of > seconds. And scientificmeasurement can
generate much larger integers. The national debt is a
continually upda sum of other quantities. One > is not
counting the number of dollar bills in a very large pile. But
> even if there are piles of paper containing as many bills as
the > national debt contains, one builds a -machine- to do the
count. Building > the machine takes much less time.Bob Kolker>
But whatever counting machine you actually build, there are
natural numbers that it cannot count within its operating
===
exist?>You showed that or why the set of primes is called
infinite in size,>fine. However, you and others say that no
prime (or integer) is>infinite. Please consider this view to
the contrary: Each prime is unique. And, we can arrange them
in order of increasing>magnitude. Now, if we do so, and then
count these primes, in ascending>order, the counting number n
will increase as we proceed. Of course,>we can't continue this
forever to exhaust all the primes, but if we>could, as
necessary to count all the primes, wouldn't n become>infinite
in the same sense as the size of the set?No. There is no point
at which you add 1 to a finite number and get aninfinite
number. You do not exhaust the finite numbers. n does
notbecome infinite. It is always finite.>I guess the answer
is; by definition, we can't actually count to>infinity; not
even in our best wishes or wildest dreams or>imagination. But,
then the meaning of infinity, as simply>uncountable, must also
prevail. i.e. to say that the set of primes>(or integers) is
infinite, is simply to say the set is uncountable.>Yet, we do
say the set is denumerable, i.e. its elements can be placed>in
one-one correspondence with the natural numbers. I
thought>denumerable and countable were synonyms.Yes. Being
placed in one-to-one correspondence with the naturalnumbers
(which are all finite) means that you have mapped everyelement
in your set S being coun with some natural number, which
isfinite.Since every member of S has a finite-length label,
how does it followthat some member of S must be infinite?>
But, their usage here>seems contradictory or inconsistant.The
key word is seems, which is an error with your understanding
ofthe principles, not with the logic itself. Not to be harsh,
this isone of the most common fallacies I see when people
first try tounderstand how mathematics reasons about infinite
sets.> Does this confusion result>because, as with counting,
we can't actually carry on a correspondence>procedure to
infinite extent, either?*YOUR* confusion results because you
don't understand that the naturalnumbers are also all finite,
===
Re: Does an infinite prime really exist?>Each prime is unique.
And, we can arrange them in order of increasing>magnitude. Now,
if we do so, and then count these primes, in ascending>order,
the counting number n will increase as we proceed. Of
course,>we can't continue this forever to exhaust all the
primes, but if we>could, as necessary to count all the primes,
wouldn't n become>infinite in the same sense as the size of the
set?No. Every one of your n's will be finite, but there will be
aninfinite number of them. Just like the primes.Consider your
argument but with prime replaced by finite integer.Are you
going to conclude that some finite integer is infinite?--
Richard-- Spam filter: to mail me from a .com/.net site, put
===
an infinite prime really exist?> Consider your argument but
with prime replaced by finite integer.> Are you going to
conclude that some finite integer is infinite?It's quite
===
- July 24th 1982> Shannon was the 6th of...<<> What I can't figure
out is - the nerd who is posting this must figure it is
interesting to someone, so why does he keep changing his email
address? Surely if those not interes are kill-filing him and
those that are are filtering into a really interesting stuff
===
Jack - July 24th 1982>> Shannon was the 6th of...<<> What I
can't figure out is - the nerd who is posting this must figure
it > is interesting to someone, so why does he keep changing
his email > address? Surely if those not interes are
kill-filing him and those > that are are filtering into a
really interesting stuff folder, this > will defeat both.For a
while I was partly responsible, because I was getting his
I've lost trackof how many of Kabatoff's Hotmail and Google
after a while I realized thatit was making things harder for
those who wan to killfile him, soI stopped. It's been months
since I last repor him to Hotmail orGoogle. (Possibly others
are still reporting him, which might explainwhy he still
changes his addresses periodically.) Since then I've
beenworking on a more permanent solution: Getting him locked
up in a psychoward again so he can't bother anyone, on USENET
===
1982> What I can't figure out is - the nerd who is posting this
must figure it > is interesting to someone, so why does he keep
changing his email > address? Surely if those not interes are
kill-filing him and those > that are are filtering into a
really interesting stuff folder, this > will defeat both.For a
while I was partly responsible, because I was getting his new>
lost track> of how many of Kabatoff's Hotmail and Google
and after a while I realized that> it was making things harder
for those who wan to killfile him, so> I stopped. It's been
months since I last repor him to Hotmail or> Google. (Possibly
others are still reporting him, which might explain> why he
still changes his addresses periodically.) Since then I've
been> working on a more permanent solution: Getting him locked
up in a psycho> ward again so he can't bother anyone, on USENET
or elsewhere.It is always best to either kill or to lock away
and torture peoplewho say that the Egyptian obelisks on the
roofs of your filthychurches are representations of penises
and are in opposition to God'sSecond Commandment. Dr. Gene
Marcoux once bluntly told me that myposters were crazy, and if
I ever put up another poster in town, thenhe will personally
have me arres and returned to the psychiatricward where he
will treat me. Wayne Brown, if you seriously aretrying to get
me arres and returned for treatment, I suggest thatyou contain
Dr. Gene Marcoux and ask him to act on this. And if youcan do
this soon, then everybody can turn trees into idols this
winterwithout so much as a peep from me. -Daryl Shawn
===
proof>> What makes my situation especially frustrating is how
simple the>> argument is that proves there's this problem with
algebraic integers,>> yet still I have to keep explaining.The
basic principle is like withP(m) = 2(m^2 + 2m + 1) = (a_1 m +
2)(a_2 m + 1)where most of you can probably guess what factor
a_1 must have!!!Now the polynomial I use is more complica such
that I need to set>> m=0 to figure things out, but notice here
what happens:P(0) = 2, and dividing off 2 from P(m)
givesP(m)/2 = m^2 + 2m + 1and notice that P(0)/2 = 1, which
tell you that the independent term>> changed.Given (a_1 m +
2)(a_2 m + 1)it's clear that a_1 has a factor that is 2.>>
Tell me what happens when P(m) = 5*(m^2 + 7*m + 5) Readers
should note the challenge to the *reality* that setting
m=0>gives the independent, or constant term. But trivially
with the poster's example the constant term is 25. The
independent term is 25.> and you assume P(m) = (a_1*m +
5)*(a_2*m + 5). Andrzej There's nothing to assume as what I do
is focus on constant terms. Again, trivially, the constant term
is 25. The independent term is 25. I want readers to pay
*careful* attention as in fact several posters>have gotten
away with misrepresenting what I do when in fact I focus>on
constant or independent terms. The math is basic. These
postehowever, need to get you to>question basic algebra which
many of you learned while still children. And they've
apparently been VERY successful up to now. I focus on the
independent terms. That's how the argument works. To win the
war of confidence, posters like Andrzej Kolowski have to>get
you to question your own mathematical knowledge, to get you
to>lack confidence in your own mathematical understanding, so
that you>rely on them as the experts, and then they lie to
you.> I appreciate all the comments, but I don't think you
answered my question at all, though admitly it was a little
vague. What I wan to know was about the divisibility of a_1
anda_2 by 5 for various values of m. Is one or the other
ofthem always divisible by 5 regardless of m, or what?
Youranswer on this could greatly strengthen your credibility
asfar as I am concerned. Andrzej>It's not complica people, and
it's an old story in human history as>I'm challenging a
powerful group that'd rather not let the truth get>in its way.
===
same ****, different day. >> P(m) = 5*(m^2 + 7*m + 5) Readers
should note the challenge to the *reality* that setting
m=0>gives the independent, or constant term. But trivially
with the poster's example the constant term is 25. The
independent term is 25.> and you assume P(m) = (a_1*m +
5)*(a_2*m + 5). > What I wan to know was about the
divisibility of a_1 and> a_2 by 5 for various values of m. Is
one or the other of> them always divisible by 5 regardless of
m, or what? Your >It's not complica people, and it's an old
===
Simple principle in core error proof>> What makes my situation
especially frustrating is how simple the>> argument is that
proves there's this problem with algebraic integers,>> yet
still I have to keep explaining.The basic principle is like
withP(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)where most
of you can probably guess what factor a_1 must have!!!Now the
polynomial I use is more complica such that I need to set>>
m=0 to figure things out, but notice here what happens:P(0) =
2, and dividing off 2 from P(m) givesP(m)/2 = m^2 + 2m + 1and
notice that P(0)/2 = 1, which tell you that the independent
term>> changed.Given (a_1 m + 2)(a_2 m + 1)it's clear that a_1
has a factor that is 2.>> Tell me what happens when P(m) =
5*(m^2 + 7*m + 5) Readers should note the challenge to the
*reality* that setting m=0>gives the independent, or constant
term. But trivially with the poster's example the constant
term is 25. The independent term is 25.> and you assume P(m) =
(a_1*m + 5)*(a_2*m + 5). Andrzej There's nothing to assume as
what I do is focus on constant terms. Again, trivially, the
constant term is 25. The independent term is 25. I want
readers to pay *careful* attention as in fact several
posters>have gotten away with misrepresenting what I do when
in fact I focus>on constant or independent terms. The math is
basic. These postehowever, need to get you to>question basic
algebra which many of you learned while still children. And
they've apparently been VERY successful up to now. I focus on
the independent terms. That's how the argument works. To win
the war of confidence, posters like Andrzej Kolowski have
to>get you to question your own mathematical knowledge, to get
you to>lack confidence in your own mathematical understanding,
so that you>rely on them as the experts, and then they lie to
you.> I appreciate all the comments, but I don't think you >
answered my question at all, though admitly it was a > little
vague. What I wan to know was about the divisibility of a_1
and> a_2 by 5 for various values of m. Is one or the other of>
them always divisible by 5 regardless of m, or what? Your>
answer on this could greatly strengthen your credibility as>
far as I am concerned. AndrzejHere's what you have.>> P(m) =
5*(m^2 + 7*m + 5)and>> P(m) = (a_1*m + 5)*(a_2*m + 5).The
constant term is 25. The most you can divide through by with m
analgebraic integer m, is 5, which gives you a constant term of
5 foryour P(m)/5.So there's no relevance to my argument as you
still have 5 as a factorof the constant term.The simple and
basic algebra argument that proves my case, and showsthat
there IS an error in core because of the definition of
algebraicintegedepends on coprimeness of the constant term,
which is how Ifilled what had been called a gap.Remember that
Andrzej Kolowski? It was almost two years ago, when Iadmit to
having a gap, as several posters had claimed, and thenwhen I
filled it, you all just lied, so I know about credibility,
andyou have none.You've been lying for almost two years now,
over an issue that Ihandled back in November of 2001.People
like you've been torturing me, in an inhumane behavior
that'shard to explain, as I see you as evil incarnate, dedica
to doingevil until someone stops you.And I will stop
===
rkip2MZe8e6Zd8nYITinZLSOdMxad-76VOfYkqfIDikadhYqVNYTU-People
like you've been torturing me, in an inhumane behavior that's>
hard to explain, as I see you as evil incarnate, dedica to
doing> evil until someone stops you.> Go away, asshole, and
===
in core error proof> People like you've been torturing me, in
an inhumane behavior that's> hard to explain, as I see you as
evil incarnate, dedica to doing> evil until someone stops
you.Go away, asshole, and ask a psychiatrist for help.You are
foolish if you think that you're anonymous. None of you
aretruly anonymous as I'll be able to use the full resources
of theUni States of America to track any of you down, soon
enough.Let's see how brave you are then G. Frege when I have
your name andnumber, and agents are sent to knock on your
door.And in case you think it matters if you're not in the US,
remembermathematicians are fighting mathematics, and in doing
so, are fightingworld society. I'll have not just the
resources of the Uni States,but the resources of other
===
countries as well.Subject: Re: Simple principle in core error
behavior that's >hard to explain, as I see you as evil
incarnate, dedica to doing >evil until someone stops you.>>Go
away, asshole, and ask a psychiatrist for help.> You are
foolish if you think that you're anonymous. None of you are>
truly anonymous as I'll be able to use the full resources of
the> Uni States of America to track any of you down, soon
enough.I doubt that very many of us here believe that. All
box.Let's see how brave you are then G. Frege when I have your
name and> number, and agents are sent to knock on your
door.Why? Are you going to start stalking him? Are you going
to be stalking all of us?And in case you think it matters if
you're not in the US, remember> mathematicians are fighting
mathematics, and in doing so, are fighting> world society.
I'll have not just the resources of the Uni States,> but the
resources of other countries as well.You sound like a loon.
This does not lend credence to your mathematical arguments.
===
Re: Simple principle in core error proof>> People like you've
been torturing me, in an inhumane behavior that's>> hard to
explain, as I see you as evil incarnate, dedica to doing>>
evil until someone stops you.Go away, asshole, and ask a
psychiatrist for help. You are foolish if you think that
you're anonymous. None of you are>truly anonymous as I'll be
able to use the full resources of the>Uni States of America to
track any of you down, soon enough. Let's see how brave you are
then G. Frege when I have your name and>number, and agents are
sent to knock on your door.You really must have no idea how
utterly crazy this sounds, or youwouldn't talk this way in
public. So I'll give you a hint: The ideathat agents are going
to track him down and knock on his doorbecause he said Go away,
asshole, and ask a psychiatrist for helpis utterly crazy.There,
now you know.>And in case you think it matters if you're not in
the US, remember>mathematicians are fighting mathematics, and
in doing so, are fighting>world society. I'll have not just
the resources of the Uni States,>but the resources of other
===
Simple principle in core error proof>> People like you've
been torturing me, in an inhumane behavior that's>> hard to
explain, as I see you as evil incarnate, dedica to doing>>
evil until someone stops you.Go away, asshole, and ask a
psychiatrist for help. You are foolish if you think that
you're anonymous. None of you are>truly anonymous as I'll be
able to use the full resources of the>Uni States of America to
track any of you down, soon enough. Let's see how brave you are
then G. Frege when I have your name and>number, and agents are
sent to knock on your door.You really must have no idea how
utterly crazy this sounds, or you> wouldn't talk this way in
public. So I'll give you a hint: The idea> that agents are
going to track him down and knock on his door> because he said
Go away, asshole, and ask a psychiatrist for help> is utterly
crazy.But doesn't it have great potential as the premise fora
TV show? Think about it:1. The NSA hires a lot of
mathematicians.2. Writers who think CIA agents sounds too
passe' or notmysterious enough often use NSA agents
instead.They're mathematicians, they're NSA agents, they're
fightingfor a simple proof of FLT, for world society, and for
you!!!(OK, maybe the pitch needs a little work. Paging Jim
===
Gbmm6ZZJgeXWSU1F565C148Ub7heJtX4AU9-605wU-gYUqZduIipgp> People
like you've been torturing me, in an inhumane behavior that's>
hard to explain, as I see you as evil incarnate, dedica to
doing> evil until someone stops you.> Go away, asshole, and
ask a psychiatrist for help.You are foolish if you think that
you're anonymous. None of you are> truly anonymous as I'll be
able to use the full resources of the> Uni States of America
to track any of you down, soon enough.Let's see how brave you
are then G. Frege when I have your name and> number, and
agents are sent to knock on your door.And in case you think it
matters if you're not in the US, remember> mathematicians are
fighting mathematics, and in doing so, are fighting> world
society. I'll have not just the resources of the Uni States,>
but the resources of other countries as well.> PLEASE ask a
psychiatrist for help (and GO AWAY)! Hell, you REALLY need
===
People like you've been torturing me, in an inhumane behavior
that's> hard to explain, as I see you as evil incarnate,
dedica to doing> evil until someone stops you. Go away,
asshole, and ask a psychiatrist for help. You are foolish if
you think that you're anonymous. None of you are> truly
anonymous as I'll be able to use the full resources of the>
Uni States of America to track any of you down, soon enough.
Let's see how brave you are then G. Frege when I have your
name and> number, and agents are sent to knock on your
door.What did he do to have agents sent to his door?
Apparently, you forgot aboutthe threats you've made. And in
case you think it matters if you're not in the US, remember>
mathematicians are fighting mathematics, and in doing so, are
fighting> world society. I'll have not just the resources of
the Uni States,> but the resources of other countries as well.
===
Assistant Professor at the University of Montana. [.snip.]>The
simple and basic algebra argument that proves my case, and
shows>that there IS an error in core because of the definition
of algebraic>integedepends on coprimeness of the constant term,
which is how I>filled what had been called a gap. Remember that
Andrzej Kolowski? It was almost two years ago, when I>admit to
having a gap, as several posters had claimed, and then>when I
filled it, you all just lied, so I know about credibility,
and>you have none.You certainly must know about credibility.
Otherwise, how could youavoid it so systematically?>You've
been lying for almost two years now, over an issue that
I>handled back in November of 2001.The above has several major
gaps in accuracy and credibility.Back in November 2001, you
still did not know what algebraicintegers were; back in March
2002 you were still askingabout algebraic integers (saying you
did not know what they
of 2002 arguing about limitations you claimed thealgebraic
integers did NOT have (you claimed they could be roots
ofnon-monic, primitive, irreducible polynomials with
integercoefficients), until December 2002 when you finally
agreed to look atthe proof of a theorem we had sta and given
proofs of for months,and agreed you had been wrong. As far as
I know, you still claim thatthe theorem that states you can
factor any polynomial with integercoefficients into (not
necessarily monic) linear terms with algebraicinteger
coefficients is false (having said, at different times, thatit
is trivial, obvious, obviously false, and false; sometimes
bothinvoking it AND saying it was false at the same
time).specific instances of your claim were wrong. Finally,
quite recently,you finally agreed that the calculations DID
show that certainalgebraic integers could not be coprime.
2nd)Now, assume, for the sake of argument, that you are
correct in yourclaim that the definition of algebraic integer
somehow introduces acontradiction in mathematics. What you've
sta, explicitly, is thatthis means that one can present
->proofs<- both that certain algebraicintegers ARE divisible,
and that they are NOT divisible, by f. Thatmeans, you are
agreeing that the proofs presen to counter yourconclusions
were CORRECT PROOFS, a correct proof being one that startsfrom
the axioms and derives a conclusion based only on the axioms
andlogical consequences. What you are claiming is that the
AXIOMS arewrong.So, you are ->agreeing<- that the people, like
Adrzej and myself, whohave been saying you are wrong about your
claims since around themiddle of 2002 when you switched to
algebraic integers (NOT November2001, as you falsely claim
here) had correct proofs; it is only quiterecently (a couple
of months at most, since you agreed that the proofspresen were
correct; you've been talking about an error in corefor longer
than that, but what you were talking about THEN is NOT whatyou
are talking about NOW, despite using the same expression to
refereto it) that you've come to realize that the problem lies
in the->definition<-.But if that is true, then that means that
neither Adrzej, nor I, norany of the others were ->lying<-; we
were basing our argument onproofs you agree are CORRECT PROOFS!
How can someone, presenting acorrect proof, be lying? At worst,
you could accuse people of lyingonly from the time at which you
realized and publicised the error incore that you are currently
claiming, and that cannot be longer thanwere correct!As for
saying you were wrong BEFORE March of 2002, well, you
yourselfadmit your developement was no good, so we were
certainly not lyingthen ->either<-: you AGREE that you were
wrong.What's more: IF as you claim, there is an error in core,
then yourproof of November 2001 (which did NOT use algebraic
integer, butinstead relied on the factorization(v^3+1)X^3 -
3vX + 2 = (sqrt(v+1)X+b1)(sqrt(v+1)X+b2)((v^2-v+1)X+b3)), as
well as your proofs throughout most of 2002 (which relied on
theflawed concept of 'algebraic integer') are worthless: if an
area ofmathematics contains a contradiction, then it is
possible to proveEVERYTHING, and a proof of Fermat's Last
Theorem in an area thatcontains a contradiction is therefore
useless and worthless. So howcould there have been systematic
lies since November 2001 about yourproof, when there were no
lies, and your proof was useless?And before you were talking
about algebraic integeyou ->were<-talking about objects, but
the definition of objects from back thenyou have since
discarded and admit it was wrong! You changed theclaims of
2001 have been correct, if they were based EITHER on theflawed
definition of algebraic integer (which you did not know yet),OR
the flawed definition of object (which you had not correc
yet)?How can you possibly reconcile all your claims? They are
incompatible! You don't remember what the gap back then was.
You just rememberthere was a gap and that you filled it. You
don't even rememberHOW you filled it, only that you used
objects. But the objects ofthen are not the objects of now, so
you could not possibly have beencorrect back then, EVEN IF YOU
ARE CORRECT NOW. It's the same sort of self-deception you
engage in when you callpeople liars. You call them liars
simply because you have calledthem liars so often that you are
now convinced they are liars. Butevery time you called them
liait was because they stasomething that YOU later admit was
true; so they were notlying. But you have called them liars so
often and so vehemently, thatthis is the only thing you
remember: that they are liars; how do youknow they are liars?
Because you've said so in the past so often itmust be true. A
better example of the goebbelian principle ofrepeating
something so often until people become convinced it's true
Ihave not seen, except that rather than using repetition to
fool othersall you have accomplished is to fool
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
===
integers ... unless,we can find the cheat who fooled us with
such a travestyof mathematics; arggh! >
- 3vX + 2 = (sqrt(v+1)X+b1)(sqrt(v+1)X+b2)((v^2-v+1)X+b3)--les
===
proof [.snip.]The simple and basic algebra argument that
proves my case, and shows>that there IS an error in core
because of the definition of algebraic>integedepends on
coprimeness of the constant term, which is how I>filled what
had been called a gap. Remember that Andrzej Kolowski? It was
almost two years ago, when I>admit to having a gap, as several
posters had claimed, and then>when I filled it, you all just
lied, so I know about credibility, and>you have none.You
certainly must know about credibility. Otherwise, how could
you> avoid it so systematically?Part of my task was evaluating
mathematicians as a security risk.Let's say that certain
personality characteristics of mathematiciansas a group stood
out, and raised my concerns.Notably mathematicians have shown
signs of loyalties that they willfollow *against* national
interests. Since mathematicians are soimportant to the
national security of this and many nations, it wasimperative
that the full extent of the problem be ascertained.Certain
problems in this area will be fixed rather quickly, I'd
===
proof>[...] Part of my task was evaluating mathematicians as a
security risk.Uh, right. Were you assigned this task by the CIA
or by God or what?>Let's say that certain personality
characteristics of mathematicians>as a group stood out, and
raised my concerns. Notably mathematicians have shown signs of
loyalties that they will>follow *against* national interests.
Uhh-huh. It's somehow crucial to national security that we
agreeyou're right about the algebraic integers being
inconsistent somehow.> Since mathematicians are so>important
to the national security of this and many nations, it
was>imperative that the full extent of the problem be
ascertained. Certain problems in this area will be fixed
rather quickly, I'd think.Here's something I've explained many
times before. Pay attentionthis time:You complain a lot about
people believing Magidin just becausehe has that degree, even
if they can't follow the math. Whoknows, it may be that some
people do believe he's right justfor that reason. Not
fair._But_ when you go on this way you _create_ the verysame
problem for yourself! Someone who knew nomath at all might
decide that he had no idea whetheryou are Magidin was right
about those details. But when that hypothetical person sees
you raving like thishe's going to decide you're a lunatic, and
that's _going_to have an effect on whether he believes you
mightbe right.Honest - it's really not good for your
===
principle in core error proof> Part of my task was evaluating
mathematicians as a security risk.Let's say that certain
personality characteristics of mathematicians> as a group
stood out, and raised my concerns.Notably mathematicians have
shown signs of loyalties that they will> follow *against*
national interests. Since mathematicians are so> important to
the national security of this and many nations, it was>
imperative that the full extent of the problem be
ascertained.Certain problems in this area will be fixed rather
quickly, I'd think.Looks like you are even more bonkers than
===
proof> What makes my situation especially frustrating is how
simple the> argument is that proves there's this problem with
algebraic integers,> yet still I have to keep explaining.The
basic principle is like withP(m) = 2(m^2 + 2m + 1) = (a_1 m +
2)(a_2 m + 1)where most of you can probably guess what factor
a_1 must have!!!Now the polynomial I use is more complica such
that I need to set> m=0 to figure things out, but notice here
what happens:P(0) = 2, and dividing off 2 from P(m)
givesP(m)/2 = m^2 + 2m + 1and notice that P(0)/2 = 1, which
tell you that the independent term> changed.Given (a_1 m +
2)(a_2 m + 1)it's clear that a_1 has a factor that is 2.> Tell
me what happens when P(m) = 5*(m^2 + 7*m + 5)Readers should
note the challenge to the *reality* that setting m=0gives the
independent, or constant term.But trivially with the poster's
example the constant term is 25.The independent term is 25. >
and you assume P(m) = (a_1*m + 5)*(a_2*m + 5). AndrzejThere's
nothing to assume as what I do is focus on constant
terms.Again, trivially, the constant term is 25.The
independent term is 25.I want readers to pay *careful*
attention as in fact several postershave gotten away with
misrepresenting what I do when in fact I focuson constant or
independent terms.The math is basic. These postehowever, need
to get you toquestion basic algebra which many of you learned
while still children.And they've apparently been VERY
successful up to now.I focus on the independent terms.That's
how the argument works.To win the war of confidence, posters
like Andrzej Kolowski have toget you to question your own
mathematical knowledge, to get you tolack confidence in your
own mathematical understanding, so that yourely on them as the
experts, and then they lie to you.It's not complica people, and
it's an old story in human history asI'm challenging a powerful
===
Re: Simple principle in core error proof > What makes my
situation especially frustrating is how simple the > argument
is that proves there's this problem with algebraic integers, >
yet still I have to keep explaining. > The basic principle is
like with > P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1) >
> where most of you can probably guess what factor a_1 must
have!!! > Now the polynomial I use is more complica such
that I need to set > m=0 to figure things out, but notice here
what happens: > P(0) = 2, and dividing off 2 from P(m) gives
> P(m)/2 = m^2 + 2m + 1 > and notice that P(0)/2 = 1,
which tell you that the independent term > changed. > Given
> (a_1 m + 2)(a_2 m + 1) > it's clear that a_1 has a
factor that is 2. > Tell me what happens when > P(m) =
5*(m^2 + 7*m + 5)>Readers should note the challenge to the
*reality* that setting m=0>>gives the independent, or constant
term.>But trivially with the poster's example the constant term
is 25.>The independent term is 25.>> and you assume > P(m)
= (a_1*m + 5)*(a_2*m + 5). > Andrzej>There's nothing to
assume as what I do is focus on constant terms.>Again,
trivially, the constant term is 25.>The independent term is
25.>I want readers to pay *careful* attention as in fact
several posters>>have gotten away with misrepresenting what I
do when in fact I focus>>on constant or independent terms.>The
math is basic. These postehowever, need to get you to>>question
basic algebra which many of you learned while still
children.>And they've apparently been VERY successful up to
now.>I focus on the independent terms.>That's how the argument
works.>To win the war of confidence, posters like Andrzej
Kolowski have to>>get you to question your own mathematical
knowledge, to get you to>>lack confidence in your own
mathematical understanding, so that you>>rely on them as the
experts, and then they lie to you.> I appreciate all the
comments, but I don't think you >> answered my question at
all, though admitly it was a >> little vague. What I wan to
know was about the divisibility of a_1 and>> a_2 by 5 for
various values of m. Is one or the other of>> them always
divisible by 5 regardless of m, or what? Your>> answer on this
could greatly strengthen your credibility as>> far as I am
concerned. Andrzej Here's what you have. >> P(m) = 5*(m^2 +
7*m + 5) and >> P(m) = (a_1*m + 5)*(a_2*m + 5). The constant
term is 25. The most you can divide through by with m
an>algebraic integer m, is 5, which gives you a constant term
of 5 for>your P(m)/5.> Which is somewhat parallel to what
happens in yourargument when you divide through by f^2: note
thatwhen you factor f^2 out of the first *two* terms of (a1*x
+ uf)*(a2*x + uf)*(a3*x + uf), you have f^2*((a1/f)*x +
u)*((a2/f)*x + u)*(a3*x + uf),and you still have uf in that
last term. I note that you didn't really answer my question,
whether one, both, or neither of a1 and a2 are divisibleby 5
when m <> 0. But I admit, I do see your point. You are talking
aboutthe constant term with respect to m, not the constantterm
with respect to x. My example, unlike your P(m),doesn't have
any x's in it. Let me modify my example a little bit to get
closerto your P(m): let Q(m) = f*(m^2*x^2 + 3*x + f). Then
Q(0) = f*3*x + f^2 = f*(3*x + f). This is the constant term,
independent of m. Note that Q(0)/f = 3*x + f. This is similar
to your polynomial P(m), where P(0)/f^2 = u^2*(3*x + u*f), or,
when u = 1, P(0)/f^2 = 3*x + f. Now assume that my polynomial
Q(m) is factored in the form Q(m) = (b1*x + f)*(b2*x + f). As
in your P(m), b1 and b2 are going to be functionsof m: b1(m)
and b2(m). The fact that Q(0) = f*(3*x + f) is consistent
withb1(0) = 0 and b2(0) = 3. Agree? So b1(0) is divisible by f
and b2(0) is relativelyprime to f (if f is a prime <> 3). This
too is similar to your polynomial P(m), where a1(0) = a2(0) =
0 and a3(0) = 3, the latter being coprime to f (if f is a
prime <> 3). Anyway, for my new polynomial Q, I would like to
see your answer to the question: for my polynomial, for m in
general, is b1(m) or b2(m) divisible by f? Both, only one, or
neither ? >So there's no relevance to my argument as you still
have 5 as a factor>of the constant term. The simple and basic
algebra argument that proves my case, and shows>that there IS
an error in core because of the definition of
algebraic>integedepends on coprimeness of the constant term,
which is how I>filled what had been called a gap. Remember
that Andrzej Kolowski? It was almost two years ago, when
I>admit to having a gap, as several posters had claimed, and
then>when I filled it, you all just lied, so I know about
credibility, and>you have none. You've been lying for almost
two years now, over an issue that I>handled back in November
of 2001. People like you've been torturing me, in an inhumane
behavior that's>hard to explain, as I see you as evil
incarnate, dedica to doing>evil until someone stops you. And I
===
What makes my situation especially frustrating is how simple
the > argument is that proves there's this problem with
algebraic integers, > yet still I have to keep explaining. >
The basic principle is like with > P(m) = 2(m^2 + 2m + 1) =
(a_1 m + 2)(a_2 m + 1) > where most of you can probably
guess what factor a_1 must have!!! > Now the polynomial I
use is more complica such that I need to set > m=0 to figure
things out, but notice here what happens: > P(0) = 2, and
dividing off 2 from P(m) gives > P(m)/2 = m^2 + 2m + 1 >
and notice that P(0)/2 = 1, which tell you that the
independent term > changed. > Given > (a_1 m + 2)(a_2 m +
1) > it's clear that a_1 has a factor that is 2. > Tell
me what happens when > P(m) = 5*(m^2 + 7*m + 5)>Readers
should note the challenge to the *reality* that setting
m=0>>gives the independent, or constant term.>But trivially
with the poster's example the constant term is 25.>The
independent term is 25.>> and you assume > P(m) = (a_1*m +
5)*(a_2*m + 5). > Andrzej>There's nothing to assume as what I
do is focus on constant terms.>Again, trivially, the constant
term is 25.>The independent term is 25.>I want readers to pay
*careful* attention as in fact several posters>>have gotten
away with misrepresenting what I do when in fact I focus>>on
constant or independent terms.>The math is basic. These
postehowever, need to get you to>>question basic algebra which
many of you learned while still children.>And they've
apparently been VERY successful up to now.>I focus on the
independent terms.>That's how the argument works.>To win the
war of confidence, posters like Andrzej Kolowski have to>>get
you to question your own mathematical knowledge, to get you
to>>lack confidence in your own mathematical understanding, so
that you>>rely on them as the experts, and then they lie to
you.> I appreciate all the comments, but I don't think you >>
answered my question at all, though admitly it was a >> little
vague. What I wan to know was about the divisibility of a_1
and>> a_2 by 5 for various values of m. Is one or the other
of>> them always divisible by 5 regardless of m, or what?
Your>> answer on this could greatly strengthen your
credibility as>> far as I am concerned. Andrzej Here's what
you have. >> P(m) = 5*(m^2 + 7*m + 5) and >> P(m) = (a_1*m +
5)*(a_2*m + 5). The constant term is 25. The most you can
divide through by with m an>algebraic integer m, is 5, which
gives you a constant term of 5 for>your P(m)/5.> Which is
somewhat parallel to what happens in your> argument when you
divide through by f^2: note that> when you factor f^2 out of
the first *two* terms of (a1*x + uf)*(a2*x + uf)*(a3*x + uf),
you have f^2*((a1/f)*x + u)*((a2/f)*x + u)*(a3*x + uf),and you
still have uf in that last term.I figured out that posters were
having a field day with all thevariables, and that it made it
*easy* for them to lie and confusepeople.But P(m) = 14706125
m^3 - 900375 m^2 + 17640 m + 1078takes that away.But notice
how they try to go back to more variables, like the twisand
despicable people they are.I'm telling you, these people are
EVIL. Now if they just admit that constant terms are constant,
then the restof the argument follows.That's because the
constant terms CANNOT change dependent on the valueof m.That
means that focusing exclusively on the constant terms and
howthey do change when 49 divides off ofP(m) = 14706125 m^3 -
900375 m^2 + 17640 m + 1078tells the tale.Now I'm informing
all of you that the people arguing against me areEVIL, yes
they are real, live EVIL people as mathematics is
thatimportant, so it's important enough for Evil itself to
===
error proof... > That means that focusing exclusively on the
constant terms and how > they do change when 49 divides off of
> P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078 > tells
the tale.Q(m) = 6125 m^3 + 6125 m^2 - 6370 m + 1078tells
===
proofso, just who *else* do you think is reading your
stuff,other than the Axis of Ones Who Argue with JSH? Arturo
gave a tutorial on indepeence of termsin another item, in
reply to you, buti have a feeling taht that's the last you'll
goto it. I'm not arguing *against* you, just with you! > Now
if they just admit that constant terms are constant, then the
rest> of the argument follows.That's because the constant
terms CANNOT change dependent on the value> of m.That means
that focusing exclusively on the constant terms and how> they
do change when 49 divides off ofP(m) = 14706125 m^3 - 900375
m^2 + 17640 m + 1078tells the tale.Now I'm informing all of
you that the people arguing against me are> EVIL, yes they are
real, live EVIL people as mathematics is that> important, so
it's important enough for Evil itself to send minions> like
===
Z7xwKEZT8eZ30myxIF4EwHKH+RmwQKzS6j7vIAVyp7ELj1X4P2e+cSNow I'm
informing all of you that the people arguing against me
areEVIL, yes they are real, live EVIL people as mathematics is
thatimportant, so it's important enough for Evil itself to send
===
principle in core error proof)Now I'm informing all of you
that the people arguing against me are> EVIL, yes they are
real, live EVIL people as mathematics is that> important, so
it's important enough for Evil itself to send minions> like
===
P(m) = 5*(m^2 + 7*m + 5) Readers should note the challenge to
the *reality* that setting m=0> gives the independent, or
constant term. But trivially with the poster's example the
constant term is 25. The independent term is 25. and you
assume P(m) = (a_1*m + 5)*(a_2*m + 5). Andrzej There's nothing
to assume as what I do is focus on constant terms. Again,
trivially, the constant term is 25. The independent term is
25.Your use of the qualifier 'independent' is incorrect. The
function, P(m), is a polynomial in 'm'. Presumably
*all*coefficients of the individual terms involving products
with powers of 'm' are 'independent' of 'm'. Each of
thecoefficients in the polynomial is a constant, including the
coefficient of the term corresponding to m^0 ( the
standardconstant term). If you *evaluate* the polynomial at
m=0, only the constant term remains, but not because it
is'independent' of 'm' -- is is neither more nor less
dependent on 'm' than any other coefficient -- rather that you
havesimply replaced 'm' with 0 to zero out all other
coefficients and evalua P(0).--There are two things you must
never attempt to prove: the unprovable -- and the
===
principle in core error proof>>Unfortunately, is a rather evil
person who lies about>>the math as if the core error should
stay hidden.>He's bad, he's evil, and I'm sick of his crap.>
Yet, when I offered to stop posting if you would only tell me
to do> so, you ->refused<-. Were you too scared to tell
someone to stop> posting, or are you just lying here about
being sick of [my] crap?> (No, I won't count as a lie the fact
that you have given no evidence> whatsoever of anything I have
pos being false, let along 'crap';> that's just your usual
lying to yourself).> Why, if as you claim I am bad, evil,
rather evil, someone who> lies about the math, and if you are
sick of [my] crap, did you> refuse my offer to 'step out of
your way', as it were, by no longer> posting to anything you
might say? No doubt the slimy little coward won't accept your
offer this time,either. He needs the attention to much.--
===
its not there, its nowhere else... =Plydia
 schrieb im Newsbeitrag> I have the
following problem and I would appreciate any help.> g(t,c_i,
f_i(t)), i=1..n where c_i is a parameter non-linear to g and>
f_i(t) known functions.> I want to find c_i so that
g(t)=g(t,c_i,f_i(t))=:g_i(t) for all i What I have tried so
far is to minimize> integral (sum_i,j (g_i(t)-g_j(t))^2) dt>
or equivalently minimize> integral ((sum_i
g_i(t)^2)-(1/n)*(sum_i g_i(t))^2) dt So Ive differentia the
above term with respect to c_j and set it> to 0:> int (g_j(t))
dt = 1/n * int (sum_i g_i(t)) dt.> But this is not very helpful
since I cant solve for the cs. Any suggetions? Or any
===
of mathematics or physics after age 50> I'm in my mid-fifties.
Will my age be (perceived as) an impediment to> either
undergraduate or graduate studies?Most definitely. As Hardy no
in A Mathematician's Apology, math is a > young man's game.
Yes, but keep in mind what Hardy meant by that statement. Age
isn'tmuch of a barrier to learning how to learn & build
mathematicalskills. It's somewhat like buying yourself a new
socket wrench set andlearning how to put it to good use. Hardy
meant the kind of cuttingedge mathematical exploration that he
===
algebra problemwhat do you mean by for? you could equate the
two expressions since nequals n1*n2 and it also equals kp^x+1
but what does that solve fo you, arelooking for
dinmensions?Jay> If n has the form: n = (p^x)k + 1, and n =
===
PS. Lest you feel I am being dishonest when you find
out later, let>me say that it will be obvious to many people
in Britain that>I am using a nom-de-keyboard.And do you live
near the ArchePenny?MarvinMarvin Sebourn>
osugeography@aol.comA literate person, I see. The Dower House,
Lakey Hill
actually.http://www.south-borsetshire-hunt.org.uk/
===
SiteI have the site back online after the move. I went ahead
been active and I hope for a successfulendevor.The new URL
-www.collatzconjecture.com-or-collatzconjecture.com-----------
-------------------I have been interes in this problem for
many years and have noticeda need, or at least a desire on my
part, for a community type websitededica to this problem. So,
I crea one.This site serves these features;1. A Links page for
you to post pages about this problem with theability to rate
the pages for their usefulness.2. A discussion board for a
focused discussion on this problem.3. A Books review section
to suggest published material to assist withthe topics rela to
this problem.4. An events announcement page, where events can
be put forward to atarge community of people interes in this
problem.5. A simple image gallery for images of interest about
this problem.6. A downloads section where users can get a
central listing of PDF,Mathematica Notebooks, Maple Files,
etc.. all rela to this problem.This site is brand new, but
should be fairly stable, though slow for awhile until I find a
good site to host it. I would be interes inanyone who would
like to be a moderator to keep the site clean andtarge and
assist with the sites overall design and content.Please check
it out. You can comment directly to me at the link on thesite
in the header.I think a site like this would serve as a needed
gathering place forany and all individuals interes in the
Collatz Conjecture.The url is a bit odd as I am borrowing one
I currently have and am notusing. If you have a suggestion for
the url, please let me know.The URL is:Collatz Conjecture
CommunityWeb Site(it was for my mother, who does not
===
Antipodal Points and S^1That is, what do you know about X
metric space such that if f:X-->Ris continuous then there are
Intermediate value theorem>> Let's say I have a continuous map
f : S^1 ---> R (where R is the real>> numbers). What is the
proof that there always exists an element x of S^1>> such
that>> f(x) = f(-x) regardless of the continuous map f?>
===
InfinityNo, it's not.Considering the unit interval, if some
subset of those points has ameasure of one half, then, half of
the points on the unit interval areelements of the subset.
Consider for example half the unit interval.A problem with
measure is that it says that many sets that are propersubsets
and infinitely proper subsets of the unit interval's
pointshave measure equal to one, in the above sense implying
that all theelements of the set are elements of the proper
subset, which would bea contradiction in terms, which is why
it is necessarily not definedthat way.Borel said that almost
all of the elements of reals are absolutelynormal, it's been
shown that uncountably many elements of the realsare abnormal,
not normal. I believe that this also applies to finiteintervals
in the reals. So we consider the unit interval, and
acharacteristic of it is that almost all of the elements are
absolutelynormal. To be absolutely normal the number has equal
zero and onedensity in its binary representation for it to be
normal to base two,four, eight, etcetera. There are further
conditions upon it but thatis a necessary condition. We evalua
the combinatoric expression ofhow many sequences have equal
numbers of zeros and ones and itevaluates to an asymptotic
expression of 1/sqrt(n) ~ 0. I wouldn'tcall that almost all
real numbebut then again as I've made clearmy thought was that
about half of the sequences would be of that form.So we have
two in a way opposing statements: 1. Almost all of the real
numbers are normal2. Almost none of the real numbers are
normalSimlarly:1. Almost all of the real numbers have equal
zero- and one-densities.2. Almost none of the real numbers
have equal zero- andone-densities.There are those and my
unfounded assumption:3. Half of the real numbers have equal
zero- and one-densities.A key element of that logical parallel
is that a normal number hasequal zero- and one-densities.Grams
and meters are SI units of mass and length. A gram is
basedupon the mass of a standard bar in France, or as well a
mole of H^1. A meter is based upon the distance traveled by an
unladen photoncesium atom, ie, the unit of length is a unit
derived from standardlyrecognized velocity and time.Then here
we are talking about how many elements of an infinite
binarysequence, in terms of an asymptotic value that in terms
of everincreasing finite lengths that are not infinite, the
variable n, howmany of those sequences exhibit certain
characteristics. Borel saysone thing about that, combinatorics
says another about that. Isuggest another.About irrationals, an
irrational is plainly a number that can not berepresen as the
ratio of two integers. That's its definition. About the
convergence, it reminds me of the discussion of the methodto
preserve the density of a canonical sequence: preserve the
densityin any finite sequence.There are a lot of rational
===
Infinity> There are a lot of rational numbers.But no rational
's?