mm-1004 === Subject: Math notation - set theory posting-account=IdLdYQ0AAADr9le9LRiemALFGX2mA5dM Could someone please decipher the meaning of the equation at: http://psommerfeld.freehomepage.com/union_ex.GIF I am trying to understand a whitepaper which contains this, but this has got me caught up. I *think* it means: F = the union of all e which belong to epsilon_M, where e is not in M. Is this correct? In particular, Ive never seen the \ backslash symbol used before so Im assuming it means is not in. -- Pete === Subject: Re: Math notation - set theory > Could someone please decipher the meaning of the equation at: > http://psommerfeld.freehomepage.com/union_ex.GIF REMOTE LINKING FORBIDDEN This host does not support remote linking of images or files for FREE accounts. PLEASE UPGRADE YOUR ACCOUNT > I am trying to understand a whitepaper which contains this, but this > has got me caught up. I *think* it means: > F = the union of all e which belong to epsilon_M, where e is not in > M. > Is this correct? In particular, Ive never seen the backslash symbol > used before so Im assuming it means is not in. For set A and B, AB is the set of things in A but not in B. === Subject: Re: Math notation - set theory days. My association with the Department is that of an alumnus. >Could someone please decipher the meaning of the equation at: >http://psommerfeld.freehomepage.com/union_ex.GIF >I am trying to understand a whitepaper which contains this, but this >has got me caught up. I *think* it means: >F = the union of all e which belong to epsilon_M, where e is not in >Is this correct? In particular, Ive never seen the backslash symbol >used before so Im assuming it means is not in. The backslash symbol is usually taken to be the set difference: A B = { a in A : a is not in B}. Your interpretation is almost right. What you are missing is the realization that the e are actually sets, not elements. So, epsilon_M is a set, whose elements are sets. You are taking the union of all (eM) for e in epsilon_M. That is, the union will consist of all x which satisfy both the following: (i) x is not in M; and (ii) x is in some e, where e is an element of epsilon_M. (It is possible that the expression was meant to be: (Union e) M that is, do the union first, then subtract M; which yields the same final set, but may be easier to think about). -- Its not denial. Im just very selective \ about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Math notation - set theory posting-account=IdLdYQ0AAADr9le9LRiemALFGX2mA5dM === Subject: Sums of Two Cubes of Integers posting-account=0ykiNAwAAACsolrstEGiZ52jIZPhRYmp I have some questions about runs of integers expressible as a sum of two cubes of integers. It is obvious, by looking at things modulo 9, that there is no instance of 6 consecutive integers which are all expressible as sums of cubes of integers. There are runs of 5 integers: the trivial (-2,-1,0,1,2) and the slightly less trivial (+/-1)*(5^3+(-1)^3,5^3+0^3,5^3+1^3,7^3+(-6)^3,4^3+4^3). My question is: are there any others? If so, what are they? If not, why not (likely the proof will involve elliptic curves, but I am not too comfortable with them and not good with getting them to appear in different situations; Id like an explicit answer to that question)? Also, while it is trivial to get infinitely many instances of \ 3 consecutive integers expressible as a sum of 2 cubes of integers, what happens with runs of 4 such integers, and why (if this is known)? ---- David To send me email, move the r from the beginning to the end of the part before the @ and insert alum. at the beginning of the part after the @. === Subject: Re: Sums of Two Cubes of Integers posting-account=pwtqtw0AAACw59dt7zOnp9M9tGBpr6Lp -- David has asked interesting questions, and Christian Bau got the solution, but for a minor (:-) typo or something, of one of them. Let me try to present it cleanly. Each case of (*) a^3 + b^3 + c^3 = 2 gives us three qudruples of consecutive four integers, each representable as a sum of two cubes (occasionally some of these quadruples coincide, when a=b or similar): a^3-2 = (-b)^3+(-c)^3, a^3-1, a^3, a^3+1 and the same for b and c--two more similar quadruples (one can also take their negatives to get three more). Identity: (6*x^3+1)^3 - (6*x^3-1)^3 = (6*x^2)^3 + 2 provides us with infinitely many solutions of (*), hence with an infinitude of the required quadruples. -- Wlod === Subject: Re: Sums of Two Cubes of Integers posting-account=pwtqtw0AAACw59dt7zOnp9M9tGBpr6Lp -- David has asked interesting questions, and Christian Bau got the solution, but for a minor (:-) typo or something, of one of them. Let me try to present it cleanly. Each case of (*) a^3 + b^3 + c^3 = 2 gives us a qudruple of consecutive four integers, each representable as a sum of two cubes (occasionally some of these quadruples coincide, when a=b or similar): a^3-2, a^3-1, a^3, a^3+1 where a^3-2 = (-b)^3 + (-c)^3 (and the same for b and c--two more similar quadruples; one can also take their negatives to get three more). Identity: (6*x^3+1)^3 - (6*x^3-1)^3 = (6*x^2)^3 + 2 provides us with infinitely many solutions of (*), hence with an infinitude of the required quadruples. -- Wlod === Subject: Re: Sums of Two Cubes of Integers > I have some questions about runs of integers expressible as a sum of > two cubes of integers. > It is obvious, by looking at things modulo 9, that there is no instance > of 6 consecutive integers which are all expressible as sums of cubes of > integers. There are runs of 5 integers: > the trivial (-2,-1,0,1,2) and the slightly less trivial > (+/-1)*(5^3+(-1)^3,5^3+0^3,5^3+1^3,7^3+(-6)^3,4^3+4^3). > My question is: are there any others? If so, what are they? If not, why > not (likely the proof will involve elliptic curves, but I am not too > comfortable with them and not good with getting them to appear in > different situations; Id like an explicit answer to that question)? > Also, while it is trivial to get infinitely many instances of 3 > consecutive integers expressible as a sum of 2 cubes of integers, what > happens with runs of 4 such integers, and why (if this is known)? Take any integer n >= 0. Then (6n^3 - 1)^3 + (6n^2)^3 (6n^3 + 1)^3 + (-1)^3 (6n^3 + 1)^3 + 0^3 (6n^3 + 1)^3 + 1^3 are four consecutive integers. === Subject: Re: Sums of Two Cubes of Integers posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g I must be missing something. For n = 0 these give: -1^3 -1^3 0^3 1^3 === Subject: Re: Sums of Two Cubes of Integers > I must be missing something. For n = 0 these give: Indeed you are. The courtesy of including adequate context. > -1^3 > -1^3 > 0^3 > 1^3 Without any context included, I wont explain but just complain, all error was not excluded. === Subject: Re: Sums of Two Cubes of Integers posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g Oops, my mistake: -1, 0, +1, +2 -- chip === Subject: Math notation - set theory posting-account=IdLdYQ0AAADr9le9LRiemALFGX2mA5dM As per my previous message, if more background about the equation is needed, it can be found at: http://ballade.cs.ucla.edu/~cong/papers/mpl2.pdf The first paragraph of section 3. -- Pete === Subject: Re: Separable,not Banach,without Schauder basis by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBKMlYX30264; >> Hi all, >> Knowing that it is necessary for a normed space with a >> Schauder basis to be separable,its not \ difficult to give >> an example of a space without Schauder basis. >> l^oo is such an example as l^oo is not separable. >> For a long time it wasnt known if there existed \ separable Banach >> space without Schauder basis.Yes,there is such a space. >> That example is beyond me,so Im asking if someone knows \ if >> an example of separable space which is *not* Banach and >> doesnt have Schauder basis would be easier,and if \ possible >How about L^p[0,1], where 0functionals, and thus there are no Schauder bases. This is interesting and now Im even more curious. L^p[0,1],0 G. A. Edgar >Bourbaki did not think first category and >>second category were good terms, so he >>invented some new ones: meagre, residual. >>Do you think they are better terms? > Actually, Denjoy introduced the term residual > around 1912 or 1913. I believe he was trying to > distinguish between a set being second category > (i.e. the set is big) and the set having a first > category complement (i.e. the set is so big that > whats left over is small). So the contemporary term for the second concept is comeager. It has the advantage of providing a parallel term conull for a set whose complement is null. Null means measure zero in whatever measure is currently being considered, or more generally, that the set is an element of whatever ideal is currently under consideration. > The term generic is also used for residual, This use of generic matches up nicely with forcing terminology, but rather than saying a big set is generic, you more usually say that something happens generically often. === Subject: Re: Set theory question from beginner >> Its very unusual notation in any case to use BC to mean >> B intersected with C (or anything else, for that matter). >> If B and C are subsets of a common group, or other structure >> in which multiplication is defined, BC usually means the >> set of all products bc where b is taken from B and c from C. >> But if B and C are general sets, BC just doesnt mean anything >> at all, in most peoples notation. >> Are you following some strange or antiquated textbook? > In probability, one uses juxtapostion as an indication for intersection > all the time. Interesting. I didnt know that (though I probably had known it at one time). For general set theory, though, its clearly nonstandard notation. === Subject: Re: Differentiation of a matrix expression posting-account=W-ZPFwwAAAAmQHmZAkwVdHjTgwfSxtcM > I have an expression > J= GAATGT > where the T denotes transpose ie it is G times A times AT times GT > I wish to differentiate the expression wrt G (in fact all these > matrices are function of discrete time k ie G(k) and A(k),k=0,1,2...) > I need dJ/dG(k) and I am unsure how to find it.I think the answer is > -2AATGT > but then again it could be > -2GAAT > The transpose is confusing me. The A matrix I just take to be a > constant. > Would be grateful for any help! Call J = X Y, X = G A, Y = X = A \ G. (Ô means transpose) The partial of X Y with respect to a square matrix Z is (A. Graham, Matrix Calculus, Table 7) p (X Y) /p Z = (p X/p Z) (I * Y) + (I * X) (p Y/p Z) where p is the P.D. operator and * the Kronecker product. (This is usually denoted by an x inside a circle, when that symbol can be typed). Replace X and Y by G A and A G and use the matrix calculus chain rule, \ with care. === Subject: Re: Differentiation of a matrix expression posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs > I have an expression > J= GAATGT > where the T denotes transpose ie it is G times A times AT times GT OK. Matlab notation (Ô) works well for that: J = GAAG = (GA)(GA) A little more readable. > I wish to differentiate the expression wrt G What does that mean? Lets say all of these matrices are n x n. That means that G has n^2 components, and J has a derivative (which is an n x n matrix) with respect to each one of them. Alternately, each component of J has a derivative d(J_ij)/dG, which is a matrix whose pq-th component is d(J_ij)/d(G_pq). So what do you mean by the derivative with respect to G? The most reasonable thing to define would seem to be a matrix of matrices, i.e. a matrix of dimension n^2 x n^2. > (in fact all these > matrices are function of discrete time k ie G(k) and A(k),k=0,1,2...) > I need dJ/dG(k) and I am unsure how to find it.I think the answer is > -2AATGT No. Wrong size. Think about the above. Perhaps work it out component wise, youll probably see a pattern. - Randy === Subject: zeros of tructated taylor series for exp(x) posting-account=sTf06g0AAAAGLCRy98JDnEhgRX96AGXS Suppose we have a truncated taylor series for exp(x), truncated at the term, just beyond the N-th power. Lets call these truncated series trunc(N, x). Some examples: trunc(0, x) = 1 trunc(1, x) = 1 + x trunc(2, x) = 1 + x + x^2/2 ... trunc(N, x) = 1 + x + x^2/2! + .... x^N/N! Here N! means the factorial of the integer N. Here, trunc(N, x) is an N-th degree polynomial over the complex numbers with variable x. The zeros of trunc(N, x) are on a very regularly shaped curve. The shape of the curve hardly depends on N. As an example Ill show the zeros for trunc(400, x) as a gif image, computed by means of my polynomial solving routine (which uses adaptive multiple precision arithmetic). http://www.woelen.nl/zeros-trunc400-exp.gif The image shows the distribution of the zeros of trunc(400, x) in the complex plane. Each red dot is a single zero. As the image shows, the zeros are nicely distributed along a curve. My question is, is there a closed analytic expression for the curve, as function of N, or is there a closed analytic expression for the limiting curve, for N going towards infinity? When a plot is made for other values of N, then the curve looks very similar. For increasing N, the curve tends to blow up, but its basic shape hardly changes. If some of you has any idea about an expression for the curve, then I would be pleased to read about that. Wilco === Subject: Re: The consise Cantors proof ? Ive posted this 100 times, STOP using T10 as an argument already. > 20 times youve brought it up and its nothing to do with \ the post. > The diag of T10 has a CHARACTERISTIC EXPANSION that > distinguishes it from the members of the list, regardless if all digit > sequences are present. That argument simple does no apply to any > of the examples Ive been in the last 6 months. Correct, it does not. TX_10 contains all reals, no bones about it, > in any practical sense of contains. Not that mathematicians are all that practical. :-) > I believe my argument was (6 months ago) it contains all possible >> sequences of digits, hence changing digits is not a valid step. >> So does TX_10. > Thats what Im talking about here! 6 months \ ago I was using T10 as > an example right? > Yes. However, TX_10 doesnt contain all real numbers; it doesnt > even contain 1/3. > Oh, but 1/3 = {.3, .33, .333, .3333, ...}! Its clearly just an > omission of some sort. > Or some such. there is a function map(1/3) = {0.3, 0.33,0.333} and an inverse. 4GL use it all the time to process infinite streams in segments. REPEAT: That argument does not apply to any > of the examples Ive been using in the last 6 months Each row in T10 terminates! Each row in infinite coin tossers does not terminate. Most rows in UTM(x,y) dont terminate. SEE THE DIFFERENCE? Your T10 argument does not apply. If you have another list in mind, fine; produce it. R(x) = UTM(x,y) mod 10 >> I take it you forfeit from producing a new real then? >> Do you understand the difference between UTM and T10 here? >> Sounds benign but for some reason every time I suggest a list >> of reals you refute it with T10. >> Infinite real numbers computable by a UTM have \ infinite expansions. >> Not every UTM creates an infinite real. Consider this UTM, for instance: >> state 1: >> on input Ô Ô: write \ Ô0 shift right new state 2 >> on input Ô0: write \ Ô1 shift right new state 2 >> on input Ô1: write \ Ô0 shift right new state 2 >> state 2: >> on input Ô Ô: write \ Ô1 shift left new state 1 >> on input Ô1: write \ Ô0 shift left new state 1 >> on input Ô0: write \ Ô1 shift left new state 1 >> This UTM has two problems: >> [1] it will never halt. >> [2] it will never write a consistent number. > Some computer programs crash, but nobodys recalled all \ PCs have they? > It doesnt matter. I can (and do, below) diagonalize the problem. Right, your complaint here doesnt matter. >> As for a non-computable real, I have one (its not all that original). >> Consider the real number X represented by [d0] . [d1] [d2] [d3] ... >> where dx is Ô1 if UTM(x,x) halts, and \ Ô0 if UTM(x,x) does not halt. >> (Im assuming UTM(x,y) means a universal turing machine whose state >> is encoded somehow in number x (the machines stored in some sort >> of order), and whose input tape is y. The particulars arent that >> important.) > Right, UTM is just a particular TM or program that has 2 inputs. > 1 - the TM it emulates > 2 - the parameter it feeds to that TM. > In this case the parameter is the digit position you want calculated. > If the TM halts, you take mod 10 and thats the digit. > The only particulars are encoding 2 parameters on the one tape. >> X cannot be computed by a UTM, as the halting problem is undecidable. > Its not well defined then is it? > Isnt it? > Let h be a TM that takes two inputs: an encoding of the machine, and > an encoding of the tape fed into the machine. h will write 1 in the > first cell if the run terminates, a 0 if the run will never terminate, > after a finite time. > Let j be another TM that is a modification of h; j will write 1 in the > Hows that? because it reduces to let x = !x, print x. > Im thinking of a number between 1 and 10 but \ Im not going to tell you, > Ill call it Omega1to10. Is that noncomputable? > Say you line up all the stars from Sol, Alpha Centurai, ... > Then if that star explodes, that digit place has a 1, else a 0. > Since Sol and Alpha Centurai are big stars and should explode, > the number most likely starts out as 0.11_ _ _ _ _ _.. > Stellar cosmology doesnt seem to be your strong suit either. :-P I passed PH224. Why is my real OmegaStars different to your OmegaPrograms? Almost identical definition, you use indexed programs halting, I use closest stars exploding. Neither of us know which will terminate for sure. Is OmegaStars a valid real? _________ > You DONT KNOW what the values are, when the star \ Ôhalts you can > fill in a 1, if the star fizzles to a vapour you \ can put a 0. > That doesnt define a real number. The futures \ not ours to see. >> Theres one real. >> A second real might be the comparatively simple one >> .0123456789101112131415161718192021... >> which is occasionally referred to as Champernownes Constant. >> I would be mildly surprised if any TM can generate this constant, >> despite its very simple specification, if the alphabet is limited >> to [0-9] and blank. (If the alphabet is extended the machine >> might use a marker digit to indicate which digit it is currently >> processing as it generates the next part of the constant. >> Interspersing blanks in the number -- .0 1 2 3 4 5 6 7 8 9 10 11 ... -- >> is considered cheating. :-) ) > Since you can specify the digit required as input to the tape it > should be trivial. Probably found on some UTM(c, digit) where c < 10^20 > The example you were supposed to give was the diagonal of UTM(x,y)mod10 > I did that one already, using the halting problem. Did you have > some other in mind? Right, missed the (x,x) as part of Omega. Strangely enough the diag I had in mind was just r(x) = UTM(x,x)+1mod10 and the disproof that it exsits used similar functions to those created in the halting proof. Let TM-v(d) = TM-u(d,d) - 1 > Its not well defined then is it? > Isnt it? > Let h be a TM that takes two inputs: an encoding of the machine, and > an encoding of the tape fed into the machine. h will write 1 in the > first cell if the run terminates, a 0 if the run will never terminate, > after a finite time. > Let j be another TM that is a modification of h; j will write 1 in the > Hows that? You assume a TM exists, TM-h(a, b) = 0 IFF TM-a(b) does not halt This is clearly a bad definition since any TM that determines if another TM halts can invert the outcome and be applied to itself. Let TM-z(x)=1 IFF TM-h(x,x)=0 IF TM-h(z,z) = 0 then TM-z(z) = 1 That means, if Z(z) does not halt, then Z(z) outputs 1. Contradiction, obviously follows from such a POOR DEFINTION for a function. Even if you allow F(x)=!F(x) as a good defintion of a function, but is LATER found to have a contradiction, you then go ahead and use a contradictory function with UNKOWN values for a definition of a real. Let Real_F = 0. F(1) F(2) F(3).... where F(x) = !F(x) This is EXACTLY the same step you used to make Omega. GARBAGE IN GARBAGE OUT Herc === Subject: Re: JSH: Those Ullrich defenders > Well that goes to public standards of proper conduct, as part of my > point is that Ullrich is a professor at a state university, and as a > tax paying American, I say, I dont want professors \ getting my tax > dollars, even indirectly by way of federal funds, if theyre going to > go out and be publicly racist. Except that *you* are the one who keeps being publicly racist, not Dr. Ullrich. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Those Ullrich defenders http://mygate.mailgate.org/mynews/sci/sci.math/ 2a065996343c8352d9df960dfca434 11.48257%40mygate.mailgate.org > Except that *you* are the one who keeps being publicly racist, not > Dr. Ullrich. Yes, poor James happens, it seems, to be of African descent, and is of the astonishing opinion that his ethnicity makes his contemptable behavior less contemptable, and his bogus mathematics less bogus. Sadly, thats not how morality works, thats \ not how math works either. Both are quite color blind. xanthian. -- === Subject: Re: mathematicians utter contempt for common sense | It is also true that mathematicians (and logicians) prefer to use the word |valid rather than true. I dont think so. Paul Cohen: It is now known that the truth or falsity of the continuum hypothesis and other related conjectures cannot be determined by set theory as we know it today. Knapp: Assuming Fermats Last Theorem is false.... Logicians prefer to use the word valid when they *mean* valid, as in logical validity, which is usually in a different context than where they would use true. But the concept of logical validity isnt usually mentioned in ordinary mathematics. I would consider it bad usage to say, Assuming FLT is invalid.... Keith Ramsay You stupid common sense! === Subject: Re: mathematicians utter contempt for common sense > | It is also true that mathematicians (and logicians) prefer to use the word > |valid rather than true. [correction] It is also valid that mathematicians (and logicians) prefer to use the word valid rather than true. Herc === Subject: Re: mathematicians utter contempt for common sense |No statements practically used in mathamatics have their formula considered |IN A SYSTEM. After reading too many people who claim all do, its almost refreshing to read someone claim the opposite! In any case, Goedels theorem makes no claim about the informal notion of proof. Keith Ramsay === Subject: Re: mathematicians utter contempt for common sense > In any case, Goedels theorem makes no claim about the > informal notion of proof. > Keith Ramsay You didnt comment on your for a system point that I \ refuted. For every statement this has no proof in system S, there is an equivalent statement this has no proof in any system whatsoever. And your PROOF of the truth of this macro godel statement cannot be applied. ______________________________________________________ &&&&&&&&&& L I S T E N E V E R Y O N E &&&&&&&&&& ------------------------------------------------------------- ----------- THERE IS NO SUCH THING AS TRUE IN SUCH A SYSTEM Its the wool over your eyes hiding the PROOF of G that has no PROOF. PROOF OF G { f1 f2 f3 f4 G } Theres no G IS TRUE without some proof!! You know why? WHO IS GOING TO SAY ITS TRUE? NOTHING IS TRUE UNLESS ITS OBJECTIVELY TRUE For a mass of plebians denying PLATO you sure like invisible things to EXIST without any substance. Herc === Subject: Re: mathematicians utter contempt for common sense > |No statements practically used in mathamatics have their formula considered > |IN A SYSTEM. > After reading too many people who claim all do, its \ almost > refreshing to read someone claim the opposite! rubbish. in what system does the successor of 0 = 2? in what system is it necessary to *clarify* the formula succ(0) = 1. You are basically saying no fact will stand outright, but its only applied as a philosophy, 99.999% of the time facts are stated and used. > In any case, Goedels theorem makes no claim about the > informal notion of proof. Rubbish. You all have 2 versions of the claims in your heads, when the overpowering_irrefutable_ proof_ of_ all_ proofs claims are questioned you run into your mouse hole saying.... Ôfor these \ particulars.. Herc === Subject: Re: mathematicians utter contempt for common sense > The only one showing a contempt for commonsense is HERC. thats just classic, cant defeat the \ argument, attack the spelling I doubt you even followed 20% of the argument, since its a popular arguing technique of dummies. Herc === Subject: Re: Dominoes on a chessboard. Edge method. posting-account=pwtqtw0AAACw59dt7zOnp9M9tGBpr6Lp As an application of my method I will provide an algorithm which easily decides about the parity of the numbers of vertical dominoes and of the horizontal dominoes covering a given region. Ill show in the process that these parities do not depend on the configuration itself but only on its union, i.e. two dominoe configurations which have the same union have the same vertical and horizontal parities. More generally, instead of the unions of configurations I will consider the algebraic sums (defined below) of arbitrary families of dominoes. Remember that Tor(2 2) := {0 1}x{0,1} First lets define the mod 2 reduction Odd : ZxZ --> Tor(2 2) Odd : Tor(h v) --> Tor(2 2) of the infinite standard lattice ZxZ and of the toroidal boards Tor(h v) with both dimensions h v even: Odd(a s) := (Odd(a) Odd(s)) We extend this reduction onto finite sets of points, in particular onto dominoes, by applying the reduction to each point of the set, and then leaving only those points of Tor(2 2) which occured the odd number of times. (Technically: identify finite sets with their characteristic functions, while interpreting 0 1 as in GF(2) = {0 1}. Thus we interpret a finite set as a finite linear combination of points, with coefficients from GF(2). Thus reduction Odd on sets is just a linear extension of the reduction from the linear bas of single points onto the whole linear space of finite sets). Now, instead of unions of (paiwise disjoint) configurations of dominoes we may consider algebraic sums of arbitrary families of finite point-sets. A point belongs to the sum of a family iff it belongs to the odd number of the members of the family. In the case of pairwise disjoint families, like dominoe configurations, the union and the algebraic sum coincide. The Odd reduction of finite point-sets is a linear homomorphism: the reduction of an algerbnaic sum is the respective algebraic sum of the reductions. Moreover, THEOREM 6 The sum of boundary labels of a finite family of finite point-sets is the boundary label of the algebraic sum of the family. This leads for instance to the following generalization of the familiar theorem: THEOREM 7 The standard 8x8 chessboard minus two diagonally opposite corners, is not a sum of any finite family of dominoes. This time the dominoes of the family are allowed to overlap and to be outside of the chessboard, anywhere in the infinite square lattice. And a similar theorem holds for the toroidal board Tor(8 8). Furthermore, we may extend the Odd reduction onto the finite families of finite point-sets by reducing each set of the family and then by removing the sets which occured an even number of times while the other reduced sets are taken just once). Thus the Odd reduction of a finite family of finite point-sets is a family of subsets of Tor(2 2). The Odd reduction of a horizontal (resp. vertical) dominoe is a horizontal (resp. vertical) dominoe in Tor(2 2). Thus in general: THEOREM 8 The parity of the number of the horizontal (resp. vertical) dominoes of a set is the same as the same number of the Odd reduced family. THEOREM 9 If two finite sets of dominoes have the same algebraic sum then they have the same parities of the number of vertical and of the horizontal dominoes. PROOF It is enough to check it for the families of dominoes of Tor(2 2). It is thus an easy, finite task. Or, conceptually simpler, you may refer to the boundary label. END OF both PROOFs. Three algorithms for parities of a finite family of dominoes (I) Add the boundary labels of each point of each point-set of the family. Thats the worst (the least efficient) \ algorithm. (II) Add the boundary labels of each point of the algebraic sum of the family. Thats better but the best is: (III) Add the boundary labels of each point of the algebraic sum of the Odd reduced family (this time you have at the most 4 points to check--you may use a look-up table :-). REMARK When you know the algebraic sum of your finite family, as in many problems where the algebraic sum (or the union of a family of pairwise disjoint finite sets) is known a priori--is assumed, then of course simply reduce that sum, since the details of the family itself are irrelevant(!). Furthermore, it may be to our advantage contain the algebraic sum S in a board Tor(h v) with even dimensions h v. Then the boundary label of S is the same as of Tor(h v)S, while the later set can be easier to deal with, e.g. it may be small or more regular. In the case of dominoes the horizontal and vertical parities for S and for Tor(h v) are the same (for even h v). Wlod PS. I am quickly jotting the seeds of the idea, but it goes much further in several directions. E.g. one would use different groups or fields for labeling and in general different labelings for different polyominoes. One can use funny shapes for boards, and one can glue their edges in different manners, etc. On the other hand, this method does not tell you how to construct the required configuratuions; it can only make certain hints (e.g. about parities). === Subject: Re: Dominoes on a chessboard. Edge method. posting-account=pwtqtw0AAACw59dt7zOnp9M9tGBpr6Lp Quite a bit of what I have presented so far (except for the edge labeling and boundary labeling) is just a shallow reßection of the (serious :-) work done jointly years ago (in the 70ties) by J. Slawny and myself, in Statistical Mechanics. This means that this is a promising direction. Wlod === Subject: Re: looking for a certain kind of instructional toy > Suppose youre looking at a couple of small \ finite graphs and asking > yourself whether theyre isomorphic. \ Wouldnt it be cool to be able to > look for the answer graphically? Somewhere there must be a computer > program that allows you to draw a small graph with dots and lines, and > then move a given dot around while keeping it connected to its > neighbors. Does anybody know where to find such a thing? Id fire up MolipDraw for that. \ http://www.molips.com -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: How to visualize Riemann surfaces >http://library.wolfram.com/examples/riemannsurface/Links/ TalkGD99_lnk_1.htm l. >But they just take real{f(z)} or imag{f(z)} and plot it as a surface in >CxR. >Is that what you call as _picking_ a projection Dr. Rusin? Exactly so; the actual surface is a certain set of points (x+iy, u+iv) in C^2; on the web site they are showing you the sets of points (x,y,u) or (x,y,v) or (x,y, sqrt(u^2+v^2)) or whatever. Pretty much anything you do to make three numbers out of four will work just as well, I guess, as long as there are only rare occasions where two different points of the surface in C^2 get sent to the same point in R^3. (Maybe you can visualize the problem better by dropping down one dimension; what would you do to present to the residents of Flatland what the curve X = { ( t, t^2, t^3 ) } (which lives in R^3) looks like? Or maybe you should ask yourself how well you like the traditional self-intersecting picture of the Klein bottle. Thats a projection too.) If youve got graphing software like the wolfram guys do, \ you can graph any other projection. For example, the points on the surface you mentioned, w = sqrt(z) (really meaning z = w^2) can be parameterized with two real parameters a and b as the set of points in C^2 of the form (x+iy,u+iv) = ( (a^2 - b^2)+i(2ab), a +i b). You can plot any projection you want now. For example you might try to study how the angles change in z and w by plotting (x^2+y^2, arctan(y/x), arctan(v/u)), each of those three coordinates given by expressions in a and b, which you then allow to vary. Thats just another projection to R^3. >But that is not what I want. What I want is a version where I can also >include points at infinity. Some kind of stereographic projection is >possible? I dont know, this is not clear. Please help me out. Oh, sure. Simply collapse lines (or rays) through a point P into single points; this is another way to lower dimension (although you would project from R^4 to S^3 this way, not to R^3. Youd have to compose this with a stereographic projection to R^3.) If youve already parameterized your surface you can get pretty pictures if you like, just by working out the formulas for what the projection is from R^4 to R^3. Since youre drawing surfaces in S^3 here, they have compact closures, which is to say you can see the points at infinity. I should probably observe at some point that not every Riemann surface is part of a compact Riemann surface; take for example an infinitely long chain of tori. And exactly what points at infinity means is not completely clear to me if the context is not specified. Usually I see that phrase when were talking about algebraic curves in C^2, and then I know what you mean when you compactify, but someone asked recently over in sci.math.research about the surface exp(w) = exp(z) - 1 and there this compactification \ doesnt make sense. Heck, for that matter, Im not sure I know what you mean by Riemann surface. Usually I think of nonsingular complex curves but in practice people are also interested in singularities too. Not sure just which one were talking about here. Do you know? I \ mean, if I just drew any old thing and colored a few points blue and called those the points at infinity, would you be able to explain why my representation is obviously wrong? (If not, what difference would it make whether you drew things right or not?) Anyway, I think Im coming back to the stance that \ its not clear just what you want to see (without a clear choice of projection). At the simplest possible level, if you way Please just tell me what the compact Riemann surface looks like, one answer is easy: Riemann surfaces are 1-dimensional complex manifolds, hence they are oriented 2-dimensional real manifolds, and hence they are homeomorphic to a sphere with g handles, for some integer g. So you get your picture just by drawing a sphere/doughnut/pretzel/... with g holes. All you need to do is to figure out which g applies to your example. How you do that depends on how the RS is given to you. For example, if you are presented with a homogeneous polynomial of degree d in three variables and asked to compute the genus g of the zero locus, then thatll be (d-1)(d-2)/2 if the curve is non-singular (and there are corrections you can make for singular points). If instead your surface is given as the quotient of the upper half-plane by some discrete group of Moebius transformations, you need something altogether different. But a problem with this just-tell-me-the-value-of-g-and-Ill-draw-the-picture approach is that you have a picture of the RS, but no real way of knowing how the picture is related to the actual surface. I am asserting here the _existence_ of a homeomorphism, but I am not _providing_ the homeomorphism. Also this is just a homeomorphism; it does not preserve any other information such as the differentiable structure or curvature or whatever which you might have put on the RS. If you want the picture to reßect this level of detail, you have to specify what detail you _do_ want to see, and you have to give a hint about how to show it to you. (OK, theres not much to show about differentiable structure, but thats just because \ youve got such a low dimension. By dimension 7, certainly, there are compact real manifolds which are homeomorphic to each other but not diffeomorphic to each other.) >I cannot take a course on geometry of manifolds to >understand the basic principles behind Riemann surfaces. Nor can I >rigorously go through all the definitions and theorems the books offer >before making any sense at all. >One important aspect of doing research is to simplify theories as much >as possible is not it? Strip any part of mathematics of its technical detail and all you have is pretty pictures that dont mean very much, sorry. dave PS -- no one calls me Dr. Rusin except my mom, who gets a kick out of it. === Subject: Conjugate spaces and notation? Hi Could someone help out with an answer for this situation? Let Vr (2n-D) be a real vector space with complex structure J. Then if Vr is complexified to Vc (2n-D) this can be written as a direct sum Vc=V^{1,0} O+ V^{0,1} where V^{1,0}={v in Vc|Jv=iv}={v in Vr|v-iJv} and V^{0,1}={v|Jv=-iv}={v in Vr|v+iJv}. Are the above statements correct? Usually the letter J is still used when it is Ôcomplexified, to say Jc. But \ should one strictly be writing Jv=iv or Jcv=iv? Are these latter two spaces Ôconjugate spaces? \ Is conjugate space=antispace? Could you give me standard/conventional/convenient bases for Vr,V^{1,0},V^{0,1} and Vc and their duals? This is what Id really like? Ron Jones === Subject: Re: Wikipedia on Decision Problems > Important undecidable decision problems include the halting problem > and Goodsteins theorem; > It seems to me that this is mixed up: > Goodsteins Theorem is a true arithmetical statement that is unprovable > in Peano Arithmetic. (Why would it be called a decision problem?) > The Halting problem is a decision problem where no algorithm > exists to correctly decide all cases (an unsolvable problem, > just as the word problem for groups is unsolvable.) > http://en.wikipedia.org/wiki/Decision_problem In any consistent formalization of mathematics that is sufficiently strong to axiomatize the natural numbers -- that is, sufficiently strong to define the operations that collectively define \ the natural numbers -- one can construct a statement that can be neither proved nor disproved within that system. G.9adels theorems are theorems in first-order \ logic, and must ultimately be understood in that context. What G.9adel showed is that in most cases, such as in number theory or real analysis, you can never discover the complete list of axioms. Each time you add a statement as an axiom, there will always be another statement out of reach. In first-order logic, theorems are recursively enumerable: you can write a computer program that will eventually generate any valid proof. The theorem only applies to systems that allow you to define the natural numbers as a set. It is not sufficient that the system contain the natural numbers. You must also be able to express the concept x is a natural number using your axioms and first-order logic. There are plenty of systems that contain the natural numbers and are complete. For example, both the real numbers and complex numbers have complete axiomatizations. Ive been hoping a sufficiently motivated expert \ would fix it one of these days. Chris Menzel === Subject: Re: Eulers identity >> I thought that: >> exp(j*theta) = cos(theta) + j*sin(theta) >> was due to de Moivre. > I used to think the same way, but now I belive that the de Moivre is > the identity > (cos(theta) + j*sin(theta))^n = cos(n theta) + j*sin(n theta) Yes, Stephen: you are correct, and I am wrong. Apologies to you, and to the group. However, I think my _other_ point - what is being sought from the original idea - is still in need of a /little/ thought ... John johnDOTmorrisonATtescoDOTnet -- Sex is like air. Its not important unless you \ arent getting any. - Paul Oldham === Subject: Re: Eulers identity at 01:23 PM, George.Ivey@gallaudet.edu (G.E.Ivey) said: > exp(j*ln(theta))= exp(j)exp(ln(theta)= theta* exp(j) ITYM exp(j+ln(theta))= exp(j)exp(ln(theta)= theta* exp(j) > Of course, from exp(j*theta)= cos(theta)+ j*sin(theta) we have >exp(h)= cos(1)+ j sin(1) > Of course, from exp(j*theta)= cos(theta)+ j*sin(theta) we have >exp(h)= cos(1)+ j sin(1) so ITYM exp(j)= cos(1)+ j sin(1) > exp(j*ln(theta))= theta(cos(1)+ j sin(1)). ITYM exp(j+ln(theta))= theta(cos(1)+ j sin(1)) -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Disappointed Differential <41BFB75A.8E80A14F@tiki-lounge.com> <41C004B8.B8258CAC@tiki-lounge.com> posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8 > So Im hoping you have that brief survey that you could send to me a > copy. > There is NO brief survey. There is NO management summary. There never > will be any. I can not and will not send you that copy. The hand of God > is in the tiny details. And not so much in generalities. > Han de Bruijn Theres not? I seem to recall someone several years ago posting a note to sci.math explaining that the general quintic and indeed higher order polynomials were solvable. Yet, then I hear there is some Abels Impossibility Conjecture that a polynomial of order n>=5, which has n many complex roots or zeros, is not solvable. Abels Impossibility Theorem, or Ruffini-Abel, \ refers to an algebraic solution with finitely many additions, subtractions, multiplies, divides, or roots. http://mathworld.wolfram.com/QuinticEquation.html The general quintic can be solved in terms of Jacobi theta functions, as was first done by Hermite in 1858. Kronecker subsequently obtained the same solution more simply, and Brioschi also derived the equation. So, I think I have heard of the way to solve for the roots of the general polynomial of arbitrary rank with complex \ coefficients. Where thats quite handy in determining solutions to systems of linear ordinary differential equations, its not the management summary of differentials that would help introduce the motivated and interested layman to the state-of-the-art of differential methods. Im reading Coddingtons introduction to \ ordinary differential equations, its good in starting from the beginning with the generalization to complex numbers, although its rather archaic, being a Dover reprint of a book from the 1950s. If there might \ not be a brief summary, where might I turn to a single source for known differential methods? It seems the nonlinear methods are very fragmented. Partially thats because many are relatively new. Consider Fubini and the development of group theory a hundred years ago, and the prevalence of algebraic groups in modern analysis. It reminds me of graph theory in the categorization and classification. For a given graph, from a list of about a hundred descriptive technical words, phrases, and values, all of the graphs properties that have to do with proving various notions about the graphs are encapsulated. For example, a planar graph fits into any proof that requires only a planar graph. The same would be so for various constructs f the differential functions and equations. Then, in for example a computer program, a method could determine the properties of a given equation, and present to the user all the possible things that could be derived from that meaningfully. Then towards some eventual goal, the combinatoric algorithm application becomes a graph problem, optimization, operations management. I think partially I think that way because I think numerical methods are not the way to go about solving things. When I see dx I think it means iota. Then, with something like d^2 x / d y^2, that leads to more consideration of iota in the nonstandard model where iota is the least positive real, like an atom is the Then, I hope to use the tools of EF and similar notions to determine some aspect of why, say, gravity affects photons. Ross F. === Subject: Re: Disappointed Differential > Where thats quite handy in determining solutions to systems of linear > ordinary differential equations, its not the management summary of > differentials that would help introduce the motivated and interested > layman to the state-of-the-art of differential methods. Its very simple, Ross. As a directive, you can not accept that there _doesnt exist_ a shortcut to that state-of-the-art of differential methods. This is not uncommon in science and technology, though, and youd better get used to it. *Real* science is abundant with details; theres no way around it. Most of us hate those management summaries, for the simple reason that they rape the truth. Show some respect for those who are at the working ßoor. Dont try to \ understand them in a cheap and easy way. True craftsmanship isnt cheap and never will. > Im reading Coddingtons introduction to \ ordinary differential > equations, its good in starting from the beginning with \ the > generalization to complex numbers, although its rather archaic, being > a Dover reprint of a book from the 1950s. If there might not be a > brief summary, where might I turn to a single source for known > differential methods? The year 1950 is recent enough. The old books are often better than the new ones. I have a book by Max Planck dating from 1912 in my collection and still read it with much pleasure. Its \ difficult to out-date *true* science anyway. > constructs f the differential functions and equations. Then, in for > example a computer program, a method could determine the properties of > a given equation, and present to the user all the possible things that > could be derived from that meaningfully. Then towards some eventual > goal, the combinatoric algorithm application becomes a graph problem, > optimization, operations management. Perhaps you should acquire Maple or Mathematica. Of course, *a lot* of knowledge about differentials is embedded in these packages. If *that* is what you are searching for ... > I think partially I think that way because I think numerical methods > are not the way to go about solving things. Numerical methods _cannot be separated_ from differential and integral calculus. But why should I bother you with the details, if you are not going to absorb them anyway? It seems that you are a friendly person. That makes it quite difficult to ignore you. :-) Han de Bruijn === Subject: Re: Disappointed Differential > I seem to recall someone several years ago posting a note to sci.math > explaining that the general quintic and indeed higher order polynomials > were solvable. Yet, then I hear there is some Abels Impossibility > Conjecture that a polynomial of order n>=5, which has n many complex > roots or zeros, is not solvable. Thats only because Abel doesnt read Usenet. \ As soon as he hears of your results, hell jump right out of his grave and admit \ his error. === Subject: Re: Poisson Process Probability >I have a probability question: >let Tk denote the waiting time until the Kth event, for a Poisson >process with intersity lamda. Find the distrubution of Tk/Tn for >1<=K<=n. (note: the conditional distribution of Tk/Tn, give Tn=t, is >indep of lamda and t.) >This is what I am thinking of quotient dist. of P >P( Tk/Tn <= s) = Integral (oo to 0) P (Tk<= st | Tn=t)dt >This just pop up to my mind, and I am not sure that logic is right or >wrong. Then I looking to the note, it says Tk/Tn is conditional dist. >Can anyone help me to clear my mind. I am stuck here. Confused, can >anyone tell me whats wrong with my approach? Not quite. What is the conditional joint distribution of T1, T2,..., T_(n-1) given T_n? Have you seen this in your reading or lecture? If not, can you derive it? The latter might be a little hard - see Ross, for example. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) > But he also assume implicit the existence of a _real_ diagonal number > which could not be in the list. This second assumption is not valid. > The diagonal number is a rational number, never a real number. Your claim is completely false. The way the diagonal sequence of digits was constructed, theres no reason whatsoever that the \ digits would repeat exactly over and over endlessly, which is the requirement that the real number represented be rational. And even in the rare chance it turned out to in fact be rational, itd still be real, and the proof would still be valid, having produced a real number not on the original list claimed to contain EVERY real number whatsoever. Whether that purported list of EVERY real number in fact was missing a rational or an irrational makes no difference, its still missing some real number, hence didnt contain EVERY real number as claimed. > More about this in de.sci.mathematik :Hat Cantor doch geirrt? Ive already seen quite enough of your false statements here in English, dont need to read more of the same in German too. === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) > By a real number r in R, I mean here a function (algorithm) > r:N->{0,1} which we can understand as the binary decimal expansion of > the real number r; i.e., with r(i) being the ith digit in rs > binary expansion. Thats a bit sloppy. You seem to be defining a \ real number r and the real number r to be two different things, the former being a binary decimal expansion of the latter. Note that the two sets \ dont match up, because there can be more than one a real number r which correspond to a single the real number r. For example, the real number r can be 5, and corresponding to that a real number r can be either of these: 101.00000000... 100.11111111... (... in that notation means forever same binary digits as just preceding) Accordingly it isnt correct to say the binary decimal expansion of .... > We _define_ the following term: the statement two real numbers r and > s are equal is equivalent to saying for all naturals m, r(m) = > s(m). It follows that, if there exists a natural m with r(m) not > equal to s(m) then r does not equal s. Thats most definitely faulty, because the two \ binary decimal expansions of 5 are not equal by that definition. Cantors theorem is about real numbers, not binary decimal expansions of real numbers. Binary/decimal/whatever expansions of real numbers are used as a means to an end, but thats not what the theorem \ is supposed to prove. Your overall logic is correct, but your details are wrong, so youve set up a strawman for the trolls to attack as if your false proof represented Cantors correct proof. === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) > Any proof of Cantors theorem either requires \ infinitary reasoning or > abstraction of actual infinity. (Whatever it is!) Thats not a problem! Cantors \ reals-not-countable theorem merely says the following cant all be true: (1) Theres a set of natural numbers satisfying \ Peanos postulates. (2) Theres a set of integers, constructable from the \ natural numbers in the usual way, and with operations + and * defined on that set in the usual way we have an ordered ring. (3) Theres a set of rational numbers, constructable from \ the integers in the usual way, and with operations + and * extended from the integers in the usual way we have an ordered field. (4) Theres a set of real numbers, constructable from the rationals in any of various usual ways (Dedekind cuts, Cauchy sequences, for example), such that with the operations + and * extended from the rationals in the usual way we have an ordered field. (5) Theres a bijection (1-1 correspondence) between the natural numbers and the reals. The proof is to assume all those are true and reach a contradiction. If you dont like anything infinite, and \ dont accept any set that isnt finite, then (1) is obviously false to \ you, so already the theorem is proven for you without needing any infinitary reasoning. On the other hand if infinite stuff is acceptable to you, then the regular proof applies. So the theorem is true regardless of whether you accept infinitary reasoning or not. (Now if infinitary \ reasoning means something *other* than reasoning about non-finite sets, please say what it is supposed to mean, because thats what it seems to me \ to mean as used above, OK?) === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) >>Any proof of Cantors theorem either requires \ infinitary reasoning or >>abstraction of actual infinity. (Whatever it is!) > Thats not a problem! Cantors \ reals-not-countable theorem merely says > the following cant all be true: > (1) Theres a set of natural numbers satisfying \ Peanos postulates. > (2) Theres a set of integers, constructable from the natural numbers > in the usual way, and with operations + and * defined on that set in > the usual way we have an ordered ring. > (3) Theres a set of rational numbers, constructable from the integers > in the usual way, and with operations + and * extended from the > integers in the usual way we have an ordered field. > (4) Theres a set of real numbers, constructable from the rationals in > any of various usual ways (Dedekind cuts, Cauchy sequences, for > example), such that with the operations + and * extended from the > rationals in the usual way we have an ordered field. > (5) Theres a bijection (1-1 correspondence) between the natural > numbers and the reals. > The proof is to assume all those are true and reach a contradiction. > If you dont like anything infinite, and \ dont accept any set that > isnt finite, then (1) is obviously false to \ you, so already the > theorem is proven for you without needing any infinitary reasoning. Also, if I recall correctly (2)-(4) are implied by (1). It occurred to me some time ago that most of the objections to Cantors proof are really objections to the axiom of infinity. There are several equivalent versions of the axiom, i.e. there exists an infinite set, the natural numbers are a set, and (1). You cant prove any of them, thats why it is called an axiom. One potential way of making sense of the discussion is that by potential infinity at least some people secretly mean the class of natural numbers and by actual infinity they mean the *set* of natural numbers which they reject (even if they dont have clear idea of what the difference is). It is perfectly consistent to assume the converse: all sets are finite, or the natural numbers are a proper class. There is still no largest number, but taking the power set is only a legal operation on a set, so the reals are not even defined. Nor are infinite sums, \ limits, etc. In this version of set theory it is impossible to *state* Cantors theorem, to say nothing of proving it. Most people (at least most mathematicians) consider giving up the reals altogether too high a price. I dont know if there is a weaker axiom for which something like the reals (but without many of the properties of the ordinary reals) exists. If so, Cantors theorem would not even be expressible in that axiom system, because bijections are something that apply to *sets*. Of course, that some people may reject the axioms on which Cantors proof depends in no way makes the proof any less valid. Ralph Hartley === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) I previously commented directly on the paper full of mostly nonsense. Now Ill comment on the second authors \ response to questions about it: > If you know (in contrast to Hilbert, Kleene, Cohen, Boubaki and so on) other > basis to differentiate infinite sets by their cardinality (unlike Cantors > theorem), please let us know and the meta-mathematical community will be > very grateful to you for ever. I strongly believe you are mis-characterising their knowledge of math. I believe most/all of them would disagree with your statement that the diagonal method is the only basis to differentiate infinite sets by their cardinality, and each/most of them would agree with me that testing for 1-1 correspondence between sets/subsets is the key basis for such differentiation by cardinality. The meta-mathematical community already knows what I know, and which you dont \ seem to know, so they dont need my help, but you surely do need my help, \ or somebodys help anyway. > Yes, I never say some more deep then [sic] (1) Cantors Theorem on the > uncountability of continuum is unprovable, but (2) its traditional CDM-proof > is invalid. Of what you say there, (1) is nonsense, (2) is incorrect. > My proof is a strict isomorfic deductive model (in Tarski sense) of one > well known invention. The invention is called the paradox of Grand Hotel. Its not really a paradox. Its just a \ property of infinite sets which is not a property of finite sets, so when a beginning math student first starts studying infinite sets after having \ previously studied only finite sets, its a bit surprizing. But \ once the surprize is over, it makes perfect sense. For example, a countably-infinite set is like a listing (enumeration) that just goes on and on and on, never is finished. The thing to realize is that if you put one item in front of it, it is retarded by one step compared to before, but still it goes on and on and on, never is finished, just like before, so in all senses the set with one additional element works just like the original set without that original element. This is unlike a finite set, where the original enumeration reaches the end at some point, but if an extra element were inserted before it, then it hasnt yet reached the end at that same point, it must go one more step before reaching the end. Once you get used to the idea, maybe some of this math will make sense to you. You quote somebody I never heard of: > It is awful to think what kind of pressure the Bourbakists put on > (evidently nonsilly) students to reduce them to formal machines! This kind > of formalized education is completely useless for any practical problem and > even dangerous, leading to Chernobyl-type events. ... That person quoted sounds as stupid as you do. I dont suppose you considered the fact he may be stupid and talking nonsense before you quoted him?? If you dont believe his remark is stupid, please provide some evidence to support it. How does the study of formal mathematics increase the likelihood of Chernobyl-type events? === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html Sometimes the English is not quite standard, but: > e-mail: alexzen@com2com.ru . > Voprosy Filosofii (Philosophy Problems), 2000, No. 2, 163-168. I would guess English is not the authors native language. Accordingly where I criticize the English usage below, please dont take it as chastisement, merely as feedback to help you clean up your paper to be better English and thus more easily understandable. ... the problem of the foundations of that sets theory ... Its usually called set theory (without the s). Modern (axiomatic) theory of sets is the only mathematical discipline, which knows how to differentiate infinite sets one from another according to their powers, i.e., basing on the number of elements making up that set. powers is a poor choice of word. Why not use the word cardinality? power in set theory usually refers to the power set or something related to it, so the word is misleading here. basing should read based. The one and only basis for such differentiation of infinities is the famous George \ Cantors theorem about the uncountablity of the set of all real numbers. No no no. The one and only basis for differentiation of infinities on the basis of size is 1-to-1 mappings between two sets or between a set and a subset of another. If there is such a mapping between two sets, they are of the same cardinality. If there is such a mapping between one set and a subset of the other, but not vice versa (second set and subset of first set), then the former set is smaller than the latter. Cantors theorem about uncountablity of real numbers is merely one tiny application of that methodology, not the basis of the methodology. binary digit G.Cantor, using the diagonal (*), builds a new infinite binary sequence: ... according to his famous diagonal rule: for any i: [if x[ii ]= 0, then y[1i ]= 1] and [if x[ii ]= 1, then y[1i ]= 0]. (***) Ignoring the problem with the first subscript being 1 again, using binary digits has a well-known problem that one two different representations of the same number may appear, one in the original list and one as the newly-constructed anti-diagonal number. To avoid that ßaw, you need to use a base larger than one, either explicitly, or by grouping the binary digits somehow. the inapplicability of Cantors diagonal method to finite enumerations is so obvious that nobody ever examined that issue. Its not a problem! Just as an infinite \ enumeration (mapping from natural numbers to reals) produces an anti-diagonal not among the enumeration hence the enumeration couldnt have been all the reals, a finite enumeration produces an anti-diagonal not among the enumeration etc. But we already knew the reals werent a \ finite set, so the second (finite) case can be eliminated from the very start. Thus, Cantors diagonal method proved to be applicable, without any changes, to both infinite and finite \ enumerations. Correct. We come to the following, very significant for the mathematics philosophy conclusion: The \ only method, which hitherto allowed meta-mathematicians to differentiate sets according to the number of their elements, i.e. by their power-cardinality, does not differentiate (distinguish) finite sets from infinite sets just by their power! Nope. The way to distinguish the size of two sets, as I said before, two sets can be put into 1-1 correspondence with each other, and if not then by checking which can be put into 1-1 correspondence with subset of other. For example a set can be compared with the set of natural numbers to find whether its smaller (hence \ finite), same size (hence countably infinite), or larger (hence uncountably \ infinite). Cantors diagonal argument is merely a tool to prove that its impossible for certain types of 1-1 correspondences to exist. Its obvious that it is only the actuality of those enumerations to which the method is applied, ... At this point the text starts making no sense whatsoever. REMARK 2. Using the second popular version of Cantors proof by assuming the existence of a 1-1-correspondence between infinite set, say U, and the set, P(U), of all its subsets (see for instance Hausdorff F., Theory of Sets. M.-L.: ONTI, 1937, p. 33-34) equally does not change anything, ... Thats a different theorem, but indeed the same kind of \ logic applies: Assumption that U and P(U) are in 1-1 correspondence produces a contradiction, thereby proving there is no such correspondence. And all your nonsense regarding enumeration of reals (starting where I said at this point .. no sense whatsoever), converted to apply to enumeration of power set, likewise remains nonsense. As to your claim that infinite sets are impossible: Well then the natural numbers are impossible to you, and therefore you cant even contemplate that kind of simple arithmetic. So questions whether an infinite set, which you dont believe can exist \ in the first place, are countable or not, are moot to you, and you have no business posting anything about them. By the way, if you dont believe there are an infinite number of natural numbers, then please tell us which natural number you (Alexander Zenkin, or Voprosy Filosofii) believe is the largest of them all. === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) > As to your claim that infinite sets are impossible: Well then the > natural numbers are impossible to you ... > By the way, if you dont believe there are an > infinite number of natural numbers, then please tell us which natural > number you believe is the largest of them all. One can consistently believe both that infinite sets are impossible, and that there is no largest natural number. Simply deny that there is a set containing all the natural numbers. In ordinary set theory (ZFC), there is no largest set, and there is no set that contains all sets (either of those would lead to a contradiction). Numbers are unlike sets in that you cant *prove* that there is no set of all natural numbers, you must assume it (and most of mathematics assumes the converse, that there *are* infinite sets). Many of the ordinary properties of the natural numbers do *not* depend on the existence of infinite sets. Look up reverse mathematics. Ralph Hartley === Subject: Re: What is meant by a meager subset?? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBL2R6u17134; Mike Oliver http://mathforum.org/discuss/sci.math/m/664050/664408 > So the contemporary term for the second concept > is comeager. It has the advantage of providing > a parallel term conull for a set whose complement > is null. Null means measure zero in whatever > measure is currently being considered, or more generally, > that the set is an element of whatever ideal is currently > under consideration. In the past few years Ive grown fond of using the prefix co as in co-meagerly, co-measure zero, co-countable, etc. A lot of the kinds of results Im interested in can be described nicely using quantifiers modulo some notion of smallness: There exist co-meagerly many bounded derivatives from [0,1] into R (sup norm) that are discontinuous at co-measure zero many points, (Weils Zbl 377.26005), and every bounded derivative from [0,1] into R is continuous at co-meagerly many points (Baire, 1899). For co-meagerly many compact subsets E of [0,1] (Hausdorff metric) and for co-meagerly many continuous measures mu on [0,1] (weak * topology via Riesz, I think), mu(E) = 0. (Dubins/Freedmans MR 30 # 4887, 3.11 and 3.12 on p. 1216) If C is a closed nowhere dense subset of R, then at co-meagerly many points of C (meager relative to C) both the left and right lower Lebesgue densities are zero and at co-measure zero many points of C (usual Lebesgue measure in R) both the left and the right lower Lebesgue densities are one. (Denjoys MR 8,260i, pp. 195-196 for first half; Lebesgue density theorem for second half) For co-meagerly many continuous f:[0,1] --> R (sup norm), we have d- = d+ = -oo and D- = D+ = oo (lower and upper Dini derivates of f) at co-meagerly many and co-measure zero many points, and we have d- = D- = oo at c-densely (in R) many points. (Jarniks Zbl 7.40102 and Sakss Zbl 5.39105) Incidentally, Alexander Kechris makes use of cardinality, category, and measure versions of there exists and for all in his book Classical Descriptive Set Theory. I think what he does there leads to a neat way of stating these kinds of results. Dave L. Renfro === Subject: Re: how to solve the problem? <23829-41C2F55F-448@storefull-3254.bay.webtv.net> posting-account=y_8zEgwAAADgSV7JOUCWfc4H-PfoU2q_ > If you sum the nondiagonal entries of (say) the 1st row > and multiply the result by 2, you generate the linear > equation > (n-1)x_1 + x_2 + ... + x_n = what you just calculated ; > doing the same for all of the other rows, you end up with a > system of n linear equations in n unknowns with the coefficient > matrix A having (n-1)s down the diagonal and \ 1s elsewhere. > If n = 2, the matrix A is singular, but for n > 2 it is _not_ > (so the problem has a unique solution for n > 2). We can solve the equations as follows. Let (n-1)x_1 + x_2 + ... + x_n = c_1, x_1 + (n-1)x_2 + ... + x_n = c_2, x_1 + x_n + ... + (n-1)x_n = c_n. Adding all these equations, we get 2(n-1)(x_1 + x_2 + ... + x_n) = (c_1 + ... + c_n), so that x_1 + x_2 + ... + x_n = (1/(2(n-1)))(c_1 + ... + c_n). Subtracting this equation from each of the above equations, we get (n-2)x_i = c_i - (1/(2(n-1)))(c_1 + ... + c_n), so that x_i = (1/(n-2))c_i - (1/(2(n-1)(n-2)))(c_1 + ... + c_n). Takao === Subject: Subset and subspace Hello I have a problem, canÍt seam to understand why; The subset S of R^2 consisting of all vectors x = (x1, 1)^T. S is not a subspace of R^2 Ok the notes go to prove that the subset is not closed under vector addition or scalar multiplication. But not withstanding the prove. imaging all vectors x=(x1,1)^T which would be a horizontal line in R^2 space, surly it is a subspace of R^2 plane, or what am I missing? === Subject: Re: Subset and subspace >Hello >I have a problem, can\.89t seam to understand why; >The subset S of R^2 consisting of all vectors x = (x1, 1)^T. S is not a >subspace of R^2 >Ok the notes go to prove that the subset is not closed under vector >addition or scalar multiplication. But not withstanding the prove. >imaging all vectors x=(x1,1)^T which would be a horizontal line in R^2 >space, surly it is a subspace of R^2 plane, or what am I missing? What youre missing is that words mean what the \ definitions say they mean, not what they surely seem to mean to you. To be a subspace a set has to be closed under addition. Here we have (1,1) in S and (2,1) in S but (1,1) + (2,1) = (3,2) is not in S. So S is not closed under addition, so its not a \ subspace. ************************ David C. Ullrich === Subject: Re: Subset and subspace Vector spaces are too restricted to capture this kind of subsets in a satisfactory way. The key concept is affine geometry. This is the geometry belonging to the affine transformation group, i.e. the group of projective transformations that transform the line at infinity into itself as a whole. Parallelism of lines in R2 (more generally: of affine subspaces of Rn) is preserved. Keywords: Erlanger Program (Felix Klein, 1872); classical 2D and 3D analytic geometry; transformation groups. IHTH: Johan E. Mebius > Hello > I have a problem, canÍt seam to understand why; > The subset S of R^2 consisting of all vectors x = (x1, 1)^T. S is not > a subspace of R^2 > Ok the notes go to prove that the subset is not closed under vector > addition or scalar multiplication. But not withstanding the prove. > imaging all vectors x=(x1,1)^T which would be a horizontal line in R^2 > space, surly it is a subspace of R^2 plane, or what am I missing? === Subject: Re: Subset and subspace > I have a problem, canÍt seam to understand why; > The subset S of R^2 consisting of all vectors x = (x1, 1)^T. S is not > a subspace of R^2 > Ok the notes go to prove that the subset is not closed under vector > addition or scalar multiplication. But not withstanding the prove. > imaging all vectors x=(x1,1)^T which would be a horizontal line in R^2 > space, surly it is a subspace of R^2 plane, or what am I missing? A line which does not pass through the origin is a *translate* of a subspace. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Subset and subspace > I have a problem, can\.89t seam to understand why; > The subset S of R^2 consisting of all vectors x = (x1, 1)^T. S is not a > subspace of R^2 > Ok the notes go to prove that the subset is not closed under vector > addition or scalar multiplication. But not withstanding the proof, imagine > all vectors x=(x1,1)^T which would be a horizontal line in R^2 space, > surly it is a subspace of R^2 plane, or what am I missing? All subspaces of R^2 have to contain the zero vector. Geometrically, the line has to pass through the origin. Otherwise the space would not be closed under vector addition. For subspaces of R^2 under the standard operations: 0) The only zero-dimensional subspace is the zero vector itself. 1) The only one-dimensional subspaces are vectors whose coordinates lie along any given line through the origin. 2) The only two-dimensional subspace is R^2 itself. === Subject: Re: JSH: Funding, real world, not fantasy ... [Tim Peters] >> It requires a determined imagination to find racism in Davids remark, >> and by all accounts JSH got nowhere trying to convince Davids >> employer or the state attorney general that the remark was actionable >> in any way. Score one for sanity. [David C. Ullrich] > Indeed. I didnt think wed have an argument about \ that one . >> But what does it matter now, given that it failed? Ive seen the >> claim that its important to keep rehashing this because it reveals >> something important about JSHs character [but \ dont buy it, because >> he reveals himself anew with every other post] ... > _My_ view is not that its important to keep rehashing this. But I > do, reasonably or not, feel its important to make sure he doesnt > get the idea that his behavior has succeeded in getting me to shut > up. I respect that. At the same time, its sad -- trying to dissuade JSH from embracing a fantastic idea seems an undertaking of uncertain outcome even if pursued for decades. FWIW, if you grow weary of it, I think youve earned the right to declare victory on this point. After all, if you withdraw and he does get that idea, he wont be able to resist trumpeting his imagined triumph. > Important because that sanity we celebrate above is not universal > in a world where people get in trouble for using the word niggardly. Yes, that is important. Still, while the world is far from wholly sane in this respect, I think getting in trouble for using the phrase some sort of racial slur can still only happen in JSHs mind. \ Ive been places where niggardly could get you killed -- luckily for you, they didnt have internet access . === Subject: Re: JSH: Funding, real world, not fantasy === >Subject: Re: JSH: Funding, real world, not fantasy >Message-id: <87fz21vzaf.fsf@phiwumbda.org> Hint: Nobody has ever disputed that making that complaint was >> within your rights as a citizen. >Well, lately Mensanator has disputed that. I think. I am the voice of one crying in the wilderness as I see through a glass, darkly, the pearls that are cast before swine. >-- >Jesse F. Hughes >But nothings being Dr. Ullrich is a particular case of somethings >being such that nothing is it: (Ex)~(Ey)(y = x) > -- John Correy on the failings of first order logic -- Mensanator Ace of Clubs === Subject: Re: JSH: Funding, real world, not fantasy |Making willful bogus complaints can be pursued as either libel or |slander (Im not too fine into the English \ language to be able to |differentiate the two). Libel is written; slander is spoken. I have no expertise in it, but I believe that under U.S. law its only a relatively narrow set of cases in which the slander or libel is actionable. For instance, I believe the plaintiff has to prove actual harm was caused. Keith Ramsay === Subject: Re: JSH: Funding, real world, not fantasy Discussion, linux) > |Making willful bogus complaints can be pursued as either libel or > |slander (Im not too fine into the English \ language to be able to > |differentiate the two). > Libel is written; slander is spoken. Really? Are you *sure* thats the difference? I dont recall the distinction myself, but Id \ be pretty surprised if thats it. -- Jesse F. Hughes [Iota]s the smallest infinitesimal, Russell, \ there are smaller infinitesimals. -- Ross Finlayson === Subject: Re: JSH: Funding, real world, not fantasy |> |Making willful bogus complaints can be pursued as either libel or |> |slander (Im not too fine into the English \ language to be able to |> |differentiate the two). |> Libel is written; slander is spoken. | |Really? Are you *sure* thats the difference? | |I dont recall the distinction myself, but \ Id be pretty surprised if |thats it. Well, did you try looking it up? Thats the distinction I remember, and I checked a couple of dictionaries before posting. Checking more now, they seem to agree for the most part. Some define libel as published defamation, where defamation is a term covering both libel and slander. Defamation through printed *images* is counted as libel. Defamation through gestures is counted as slander. :-) I saw a web page that defined them similarly, except that they exchanged the meanings of defamation and libel, which Im skeptical of. http://injury-law.freeadvice.com/libel_and_slander/ agrees with how I defined them, although it claims that radio is counted as libel despite being in some sense oral. http://www.infoplease.com/ce6/society/A0829656.html says that the terms were defined as I said in common law, but that the line is no longer so simple. The issue is that there is less of a burden of proof to show that a libel actually was harmful than there is for a slander. It appears now partly to be a matter of the permanence of the message and the extent of the audience. It claims too that whether defamations on radio and TV are libel or slander has been argued in the courts. So perhaps you remembered some of the complexities that are often glossed over? It still seems to be a question of whether the legal system in its wisdom has decided to treat a slander in the manner that common law had treated written and spoken slanders respectively. The fact that we now can record speech or broadcast it live to audiences of a size that formerly could only be reached by distributing printed text has presumably helped to muddle what once was a relatively straightforward distinction. Keith Ramsay === Subject: Re: JSH: Funding, real world, not fantasy posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK === >>Subject: Re: JSH: Funding, real world, not fantasy >>Message-id: <87d5x5q3ky.fsf@phiwumbda.org>I like my newsgroup politics nice and tidy. So that you never have to take a stand? So that as long as Harris > never intrudes into _your_ life, hes just a harmless \ clown? >>Damn near harmless, perhaps? >>Or, in deference to Douglas Adams, mostly harmless? >>Harriss behavior in the OkState affair was reprehensible, but (1) not >>particularly effective in this case and (2) not repeated in any >>obvious way since. >I was under the impression that at least one other person claims that >Harris contacted their employer. > Thats correct. >And Harris still claims that he was >within his rights in doing what he did. > Of course he was within his rights! >If you try to murder someone and >fail to pull it off, you still go to jail, dont you? > What planet are you from? Murder is illegal. Making > bogus complaints is not. Making bogus complaints in the presence of a restraining order would be illegal, would it not? How then, can such orders be enforced if those actions are within his rights? The simple answer is that they were never within his rights to start with. Your rights end where anothers rights begin. You are never within your rights when it comes to harassing someone. If you choose not to make a Federal case out of it, thats your business. But not everyone is a tenured professor who can laugh off such harassment. Frankly, that attitude coming from Jesse Hughes doesnt surprise me, but somehow I thought better of you. >> I dont think the JSH threat requires any stand >>from Jim Burns or you. Or me. >Yeah, what do I care, I dont even put my name on these posts. Let Harris >fuck with someone elses life. No skin off my nose. >>-- >>Jesse F. Hughes >>All Chinese are Confucianists when successful, and Taoists when they >>are failures. >> -- Lin Yutang, /My Country and My People/ > ************************ > David C. Ullrich === Subject: Re: JSH: Funding, real world, not fantasy Discussion, linux) >> What planet are you from? Murder is illegal. Making >> bogus complaints is not. > Making bogus complaints in the presence of a restraining order > would be illegal, would it not? How then, can such orders be > enforced if those actions are within his rights? Lamest attempt at sophistry in months. On a group populated by JSH, Finlayson, Lester Zick and others, thats something. Incarcerated convicts arent allowed to vote. Therefore, no one has the right to vote. They arent allowed to move freely. Therefore moving freely is not within your rights. They arent allowed to \ bear arms, etc. Some people have received court orders not to use the internet. Therefore, you are not acting within your rights by posting this. So cut it out. When we say that James was within his rights, we are not committed to the claim that those rights are inalienable. That a right may in some cases be restricted is utterly irrelevant to whether James behaved legally when he complained to OkState. -- Jesse F. Hughes Mathematicians dont fit in with a consistent \ view, unless you accept that to a strangely large extent they are acting under the inßuence of something very powerful, dark, and negative. -- James S. Harris === Subject: Re: JSH: Funding, real world, not fantasy <874qigcgvl.fsf@phiwumbda.org> posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK >> What planet are you from? Murder is illegal. Making >> bogus complaints is not. > Making bogus complaints in the presence of a restraining order > would be illegal, would it not? How then, can such orders be > enforced if those actions are within his rights? > Lamest attempt at sophistry in months. On a group populated by JSH, > Finlayson, Lester Zick and others, thats something. > Incarcerated convicts arent allowed to vote. Therefore, \ no one has the > right to vote. They arent allowed to move freely. Therefore moving > freely is not within your rights. They arent allowed to bear arms, > etc. > Some people have received court orders not to use the internet. > Therefore, you are not acting within your rights by posting this. So > cut it out. There is a distinction between losing your rights (when incarcerated) and never having the right in the first place. When you are slapped with a restraining order to stop harassing someone, you \ dont have any rights taken away. You cannot take away what never existed. The restraining order merely enforces the right of your victim to not be harassed. You never had a right to harass someone before the restraining order was issued, and you have no right afterwards. The examples you give above are not applicable. Luckily for you, I dont know what sophistry means. > When we say that James was within his rights, we are not committed to > the claim that those rights are inalienable. You guys are so good at spotting his math bullshit, how come you fall for this bullshit about justifying harassment as its my right as a citizen? > That a right may in some > cases be restricted is utterly irrelevant to whether James behaved > legally when he complained to OkState. The OkState people didnt fall for his bullshit and neither did the Attorney General. Why do you? > -- > Jesse F. Hughes > Mathematicians dont fit in with a consistent \ view, unless you accept > that to a strangely large extent they are acting under the inßuence > of something very powerful, dark, and negative. -- James S. Harris === Subject: Re: JSH: Funding, real world, not fantasy <874qigcgvl.fsf@phiwumbda.org> Discussion, linux) > Have you ever seen a restraining order? Ill show you mine if you show me yours. -- Jesse F. Hughes Why do the dirty villains always have to tie your hands *behind* ya? Thats what makes them villains. --Adventures by Morse (old radio show) === Subject: Re: JSH: Funding, real world, not fantasy <874qigcgvl.fsf@phiwumbda.org> Discussion, linux) > What planet are you from? Murder is illegal. Making > bogus complaints is not. >> Making bogus complaints in the presence of a restraining order >> would be illegal, would it not? How then, can such orders be >> enforced if those actions are within his rights? >> Lamest attempt at sophistry in months. On a group populated by JSH, >> Finlayson, Lester Zick and others, thats something. >> Incarcerated convicts arent allowed to vote. Therefore, no one has > the >> right to vote. They arent allowed to move freely. Therefore moving >> freely is not within your rights. They arent allowed to bear arms, >> etc. >> Some people have received court orders not to use the internet. >> Therefore, you are not acting within your rights by posting this. So >> cut it out. > There is a distinction between losing your rights (when incarcerated) > and never having the right in the first place. When you are slapped > with a restraining order to stop harassing someone, you dont have > any rights taken away. You lose the right to move about as freely as those of us without restraining orders hampering our movement. > You cannot take away what never existed. The restraining order > merely enforces the right of your victim to not be harassed. You > never had a right to harass someone before the restraining order was > issued, and you have no right afterwards. But *clearly* you had the right to travel in certain areas that you no longer can travel. You certainly have lost certain rights when a restraining order binds you. > The examples you give above are not applicable. Luckily for you, > I dont know what sophistry means. It means Im unimpressed by your (attempts at) reasoning. >> When we say that James was within his rights, we are not committed to >> the claim that those rights are inalienable. > You guys are so good at spotting his math bullshit, how come you fall > for this bullshit about justifying harassment as its my right as > a citizen? Because he didnt break any law of which I know. Name the law. >> That a right may in some >> cases be restricted is utterly irrelevant to whether James behaved >> legally when he complained to OkState. > The OkState people didnt fall for his bullshit and \ neither did the > Attorney General. Why do you? Did the Attorney General file charges against JSH? Threaten him with charges? If not, why not? Mere jurisdiction? Or maybe because Jamess behavior was not illegal. I dont fall for his bullshit (well, Im kinda \ enamored with it, but not fall in the way you mean). Theres a difference between saying JSH shouldnt behave like that and saying he \ hasnt the right to behave like that. -- Jesse F. Hughes Would you please stop talking and start talking? -- Vincent Price as the Saint === Subject: Re: JSH: Funding, real world, not fantasy > David, you are the one who evidently brought race into the conversation > ... in a sly, backhanded way. You should apologize and move on. JSH is > right on this one. What is taboo about race? Who are you to use the word Ôshould to another human? === Subject: Re: JSH: Funding, real world, not fantasy > David, you are the one who evidently brought race into the conversation > ... in a sly, backhanded way. You should apologize and move on. JSH is > right on this one. Not even close. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Funding, real world, not fantasy > I acted well within my rights as a citizen. Yes, its well within your rights as a citizen to \ demonstrate yourself to be an ignorant, offensive idiot in public. You seem to enjoy exercising that right quite often. > It was part of my training as an Army officer. And what a thoroughgoing disgrace to the uniform youve proven yourself to be... -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Funding, real world, not fantasy >I think you need to check the program. Youre mixing up the >characters. /// Yes... I think so....... === Subject: Re: JSH: Funding, real world, not fantasy >> Hint: Nobody has ever disputed that making that complaint was >> within your rights as a citizen. > Well, lately Mensanator has disputed that. I think. When? === Subject: Re: JSH: Funding, real world, not fantasy <87fz21vzaf.fsf@phiwumbda.org> Discussion, linux) > Hint: Nobody has ever disputed that making that complaint was > within your rights as a citizen. >> Well, lately Mensanator has disputed that. I think. > When? In the thread containing Message-ID In his followup to my post in this thread, he didnt \ complain about my characterization[1]. Id guess he agrees with it. Footnotes: [1] Actually, he was pretty funny in that followup. -- Yup, as far as Im concerned, if you live out your lives smiling the entire time full of pride in your *believed* accomplishments, when you never had any, well thats ok with me. --James Harris, a man of remarkable accomplishments. === Subject: Re: Question about Presidents Social Security plan >..... > In 1975, former Secretary of Commerce and former Director of the > Budget Maurice Stans hit the nail on the head when he stated: Social > Security payments rest upon the general credit of the Government of > the United States, upon its taxing power, and not upon any > accumulations in a trust fund. In 1976, then-Secretary of the > contributors to the system have not been building a fund at all. The > taxes they are paying into Social Security are merely being > handed over as benefits to other people. In other words, Social > Security is a huge Ponzi scheme. Exactly right. Social security is a tax collected and used by the government. The real problem is not the solevency of Social Security : it is the government budget deficit. However, being unable or unwilling to confront the long term budget deficit problem, W Bush and his republican cohorts find it more expedient to engineer and try to tackle the real problem of Social security. Fix the budget deficit (ie. balance the books of the government) and the social security crisis will disappear overnight! === Subject: Re: Question about Presidents Social Security plan <8ngbs05g00gg1bkqo94fjun6kq0rg8kumr@4ax.com> posting-account=HFsrmgwAAAAPyu963mcWzzWXBCE1q6Oi According to the SS trustees your claim is in error Long-Range Results Under the intermediate assumptions the combined OASI and DI Trust Funds are projected to become exhausted in 2042. For the 75-year projection period, the actuarial deficit is 1.89 percent of taxable payroll, 0.03 percentage point smaller than in last years report. The \ open group unfunded obligation for OASDI over the 75-year period is $3.7 trillion in present value, $0.2 trillion more than the obligation estimated a year ago. The OASDI annual cost rate is projected to increase from 11.07 percent percent in 2078, or to a level that is 5.91 percent of taxable payroll more than the projected income rate for 2078. Expressed in relation to the projected gross domestic product (GDP), OASDI cost is estimated to rise from the current level of 4.3 percent of GDP, to 6.3 percent in 2030, and to 6.6 percent in 2078. Between about 2010 and 2030, OASDI cost will increase rapidly due to the retirement of the large baby-boom generation. After 2030, increases in life expectancy and relatively low fertility rates will continue to increase Social Security system costs, but more slowly. Annual cost will exceed tax income starting in 2018 at which time the annual gap will be covered with cash from redeeming special obligations of the Treasury, until these assets are exhausted in 2042. Separately, the DI fund is projected to be exhausted in 2029 and the OASI fund in 2044. Solvency The combined OASDI Trust Funds are projected to become insolvent (i.e., unable to pay scheduled benefits in full on a timely basis) when assets are exhausted in 2042 under the long-range intermediate assumptions. For the trust funds to remain solvent throughout the 75-year projection period, the combined payroll tax rate could be increased during the period in a manner equivalent to an immediate and permanent increase of 1.89 percentage points, benefits could be reduced during the period in a manner equivalent to an immediate and permanent reduction of 12.6 percent, general revenue transfers equivalent to $3.7 trillion (in present value) could be made during the period, or some combination of approaches could be adopted. Significantly larger changes would be required to maintain solvency beyond 75 years. http://www.ssa.gov/OACT/TR/TR04/II_highlights.html#wp76455 T.Carr === Subject: Re: More on the State-of-the-Art in Physics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBL686j02013; >1)/.../ calculated that 95% of our universe is dark matter. >>You probably meen dark energy? As far as I know 20% is dark matter. >So 115% of our universe is dark? >Lee Rudolph Physicists agree on the figure 75$ dark energy and 25% but energy and matter are convertible to each other. So I just refer to both as dark matter. E. E. Escultura === Subject: Re: More on the State-of-the-Art in Physics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBL686J01997; >>1)/.../ calculated that 95% of our universe is dark matter. >You probably meen dark energy? As far as I know 20% is dark matter. >>2) /.../The discovery of the basic constituent of dark matter, the superstrings. >Where can you find this? To my knowledge this question is still highly open to debate and has been set to one of the 25 agendas to be set during the coming 25 years by KITP. >joccis Yes, physicists agree on the figure of 70% dark energy and 25% dar matter. But this is ambiguous because energy is motion of matter and it is difficult to separate the two. Moreover, matter is convertible to energy. So, I just lump the up together as matter. E. E. Escultura === Subject: Re: More on the State-of-the-Art in Physics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBL686n02001; >>MORE ON THE STATE-OF-THE-ART IN PHYSICS > I guess you didnt even notice that I did a complete overhaul of physics. >Whats wrong you dont want to give me credit \ for anything? > You want to rename it something else like the others did to steal my >theories on the Helix Spiral Spinning Field Theory showing internal and >external structure of the atom and and the universe. > Oh, you want ot call it the QM or String Vortex Model? My work is about 10 >years ahead of millions of the best minds in physics put together working >constantly together and they still dont even know what \ mass is at the >sub-atomic level. Whats wrong with you people you need a job that bad? So you >keep me out of it. > Even the Japanese technology in electronic engineering is just barely >catching up with me at the sub-atomic level in physics. Oh just think by the >just barely understand where I am at. >Smarts Alt. Physics News Group >ht tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv =1http://smart1234. s-enterprize.com / Im interested in your theory. Could you give me sample \ ideas like: What is the basic constituent of matter and prove that this is indestructible so that it would be consistent with energy conservation or even the first law of thermodynamics. Prove also that every piece of matter is reducible to or consists of that basic constituent of matter. E. E. Escultura University of the Philippines === Subject: Re: More on the State-of-the-Art in Physics >MORE ON THE STATE-OF-THE-ART IN PHYSICS >> I guess you didnt even notice that I did a complete overhaul of physics. >>Whats wrong you dont want to give me \ credit for anything? >> You want to rename it something else like the others did to steal my >>theories on the Helix Spiral Spinning Field Theory showing internal and >>external structure of the atom and and the universe. >> Oh, you want ot call it the QM or String Vortex Model? My work is about 10 >>years ahead of millions of the best minds in physics put together working >>constantly together and they still dont even know what mass is at the >>sub-atomic level. Whats wrong with you people you need a job that bad? So >you >>keep me out of it. >> Even the Japanese technology in electronic engineering is just barely >>catching up with me at the sub-atomic level in physics. Oh just think by the >>just barely understand where I am at. >>Smarts Alt. Physics News Group >>http://pub39.bravenet.com/forum/show.php?usernum=3320272813 &cpv=1 >>S. Enterprize (Science Journal) >>http://smart1234.s-enterprize.com/ >Im interested in your theory. Could you give me sample ideas like: >What is the basic constituent of matter and prove that this is indestructible >so that it would be consistent with energy conservation or even the first law >of thermodynamics. Prove also that every piece of matter is reducible to or >consists of that basic constituent of matter. >E. E. Escultura >University of the Philippines Just check out the Website. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: More on the State-of-the-Art in Physics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBL687u02019; >>1)/.../ calculated that 95% of our universe is dark matter. >You probably meen dark energy? As far as I know 20% is dark matter. >> So 115% of our universe is dark? >> Lee Rudolph >Its only 115% dark if youre a \ mathematician. >If youre a physicist its 100% dark. >If youre an engineer its 99.99% dark +/- 5% \ dark. >-- >Replace Roman numerals with digits to reply by email This makes sense. === Subject: Re: The State-of-the-Art in Mathematics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBL686L02009; >>The problem here is not the choice of values of sqrt(-1) but the concept i >>itself. It is not well-defined because the mapping sqrt is well-defined only >>on perfect square such as 1, 4, etc. >>E. E Escultura >Ok can it be represented like this? >n=1 to oo >with cartesian coordinates x,y,z >Use only integer powers of n >otherwise its an indeterminate >[(-1)^n/2] x >[(-1)^n/2] y >[(-1)^n/2] z >For x coordinate, >[(-1)^n/2] x >at n=1 --> indeterminate >at n=2 --> -1 >at n=3 --> indeterminate >at n=4 --> 1 >at n=5 ---> indeterminate >at n=6 ---> -1 >at n=7 ---> indeterminate >at n=8 ---> 1 >etc... > In cases where you have sqrt(-2), etc.., >for n=1 --> oo >Integer Powers of n Only >sqrt (-2) = sqrt(2) [(-1)^n/2)] > = (1.41421...) [(-1)^1/2)] > = (1.4142...) ( indeterminate) >And, >(sqrt (-2))^2 = (1.41421...)^2 [(-1)^2/2)] > = (2)(-1) > = -2 >And, >(sqrt(-2))^3 = (1.41421...)^3 [(-1)^3/2)] > = (2.8284...) (indeterminate) >And, >(sqrt(-2))^4 = 3.9998... >And so on... > Does this map sqrt correctly with negative sqrts? >Smarts Alt. Physics News Group >ht tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv =1http://smart1234. s-enterprize.com / The problem remains that sqrt is not well-defined on any number that is not a perfect square. === Subject: Re: The state-of-the-art in mathematics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBL686q02005; >>Hi Folks, >>I would like to share with you the latest findings on foundations, >>number theory and the real number system. They are all published in my >>papers but some of them have been resolved in MathForge.net and others >>are discussed in my websites.. >> (1) The subject matter of mathematics cannot be the concepts of >>thought which are subjective; they must be symbols (I also call them >>concepts) well-defined by a set of axioms that specify the existence >>and properties of and operations on and relations among the concepts; >>the axioms must well-define the given mathematical space. To insure >>validity of proof the rules of inference or logic must be specific to >>the given mathematical space and well-defined by the axioms. In >>particular, universal rules of inference such as formal logic must be >>rejected. Moreover, ambiguous sets such as infinite sets must be >>avoided because they are sources of contradictions. Therefore, only >>finite spaces, which can be unbounded, are free from contradiction >>provided the axioms are consistent. It follows that distinct >>mathematical spaces are independent. Therefore, any propositions >>involving concepts from distinct spaces or mapping between them is >>nonsense. In this regard, Godel\.89s incompleteness theorems are >>nonsense. >>axiomatization of the integers. I remedied this problem by embedding >>them in the new real number system that I have constructed. The new >>real number system is finite but unbounded, free from contradictions >>and paradoxes, has natural ordering, enriched by two new real numbers, >>namely, dark and unbounded numbers, and adequate for scientific and >>practical purposes. >> (3) The real numbers are decimals and it is well-defined if every >>digit is known or computable. Therefore, a nonperiodic real number or >>irrational is ill-defined or nonsense unless there is some algorithm >>for computing any of its digits. Other nonsense in the real number >>system includes classical curves and surfaces. Consequently, Wiles >>ëproof\.89 of FLT is wrong and his conclusion is also false because I >>have constructed countable counterexamples to FLT in several of my >>papers and in my websites. Curves and surfaces were fixed by L. C. >>Young in a series of papers from 1931 to 1969 where developed the >>theories of generalized curves and surfaces. Two of the axioms of the >>real number system see Royden\.89s Real Analysis, p. 31), namely, the >>completeness and dichotomy axioms, are false. Counterexamples to them >>were constructed by Banach-Tarski and Brouwer. Brouwer\.89 counterexample >>also implies that the irrationals are ill-defined and the real numbers >>have no natural ordering. In fact, these findings imply that the real >>number system is ill-defined, nonsense. The remedy is to reconstruct >>the real number system on finite set, namely, the set of basic >>integers 0, 1, á, 9, without these axioms using only three axioms. I >>have done this in several papers and part of it are posted in >>MathForge.net and discussed in my websites. >> (4) To illustrate how a wrong concept may wreck havoc on mathematics >>and physics, consider this Ullrich ßip-ßap: i = sqrt(-1) = >>sqrt(1/-1) = 1/i = -i or i = -i; dividing both sides of the last >>equation by i, I obtain 1 = -1 or 1 = 0 and the real number system >>goes down the drain. If I add i on both sides instead, I obtain 2i = 0 >>or i = 0 and the complex number system vanishes in thin air. The >>remedy is to take i as an operator on plane vectors, that is, rotation >>of a vector by pi/2 radians counterclockwise. (This resolution is >>shown in my paper, Exact solutions of Fermat\.89s equations (Definitive >>resolution of FLT), Nonlinear, Studies, Vol. V, pp. 227 \.9a 254) In >>physics this concept i has brought in such nonsense as negative or >>imaginary metric and energy. >>(5) My websites: http:// www.users.bigpond .com/pidro/home.htmhttp:// home.iprimus.com.au/pidro/< /a>E. E. Escultura >i = sqrt(-1) = sqrt(1/-1) = 1/i = -i or i = -i >... every first-semester-student of mathematics knows >that the square root of a number is 2-valued. >If the rest of your work is of the same >>depth<< ... Yes, but what made you choose one over the other? Which axiom of the real numbers support your claim that your choice is the right one? The problem here is that the mapping sqrt is well-defined only on perfect square; therefore i is ill-defined and does not exist. E. E. Escultura === Subject: Re: The state-of-the-art in mathematics posting-account=W9drAQ0AAAAwnUCtDicSi-THtOHRPCY8 If Im not mistaken, the square root is \ well-defined for all real values, not just perfect squares, since the square root function is just a representation of x^(1/2), which in turn can be represented in terms of e^x, which in turn is just an infinite sum. apologize if it isnt yours, Im not sure \ where the original thread is), is that you state that i is in the set of real numbers, and use it as a counterexample to show that the set of real numbers is faulty--but i isnt a real number at all! It is a complex number and is not an element of the set of reals, so your counterexample is irrelevant. A. === Subject: numerical solution to odes vs systems of odes posting-account=arnIEA0AAAAiqq71lhC1Y7JkPVXrK0Dm can you guys please help me with this? consider an ode. y(t) = y(t) with solution, y(t)= c * e^ t where c is the constant lets say you want to get a numerical solution of that using y(t_i +1 ) = y(t_i+1) etc etc (above i smy extent ot numerical solution of ode...) now.. 1. what does that mean? how do i do that on a calculator? now.. for more difficult problem... consider a system of odes y(t)= y(t) + x(t) x(t)= x(t) - y(t) 2. what will the numerical solution for above look like? and how will I solve that using a calculator? I really need soem help with this.. I dont understand it. ( \ i shouldve studied math not bio) === Subject: Re: numerical solution to odes vs systems of odes > can you guys please help me with this? > consider an ode. > y(t) = y(t) > with solution, y(t)= c * e^ t where c is the constant > lets say you want to get a numerical solution of that using > y(t_i +1 ) = y(t_i+1) Huh? Your notation is somewhat ambiguous. Do you mean t_(i+1) or (t_i) + 1? Anyway, a simple approach is to approximate the derivate: y(t) = (y(t+h) - y(t)) / h, where h is a small step size. Substituting that into the original equation and solving for y(t+h) gives: y(t+h) = y(t) + h * y(t) = (1+h) * y(t). The equation expresses the value of the solution at a later time (t+h), given the value now (i.e. at time t). > now.. for more difficult problem... > consider a system of odes > y(t)= y(t) + x(t) > x(t)= x(t) - y(t) > 2. what will the numerical solution for above look like? and how will > I solve that using a calculator? Here you do the same thing. You now have two equations: y(t+h) = y(t) + h * (y(t) + x(t)) x(t+h) = x(t) + h * (x(t) - y(t)) Given all the values at time t, evaluate the right hand side using a small value for h. This now gives the values at a slightly later time t+h. Now just repeat this process over and over. More sophisticated methods exist. You might try consulting -Michael. === Subject: Re: what is maximal rank of this matrix? >Let n > 2. >Let v(n) and w(n) be any two n-tuplets of real numbers. >Let v(n) and w(n) denote the transposes of \ these n-tuplets. >Let a, b, and c be any 3 real numbers. >Construct the nxn matrix >M(n) = a * v(n) v(n) + b * w(n) w(n) + c * \ { w(n) v(n) + v(n) >w(n) } >What is the maximum rank of the nxn matrix M(n)? >[I know the answer is 2, but what is a clever way to PROVE this?] If w is a constant multiple of v then M(n) is just a multiple of a projection operator on the subspace of v, thus of rank 1. If v and w are independent then, without loss of generality, you can assume that they are ortonormal (since, if theyre not, you can orthonormalize the base v,w without changing the form of M(n), only changing the values of the constants. So, assuming that theyre orthonormal, you can create the projection operator P(v,w) = vv + \ ww and show that PM = MP = 1 The continuation is trivial. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Homotopy classes of figures made of 5 lines in \ R^2 posting-account=SljyZA0AAAAif_9GYE3PwF4tgqwl4tkb > Consider figures consisting of 5 lines in the plane. Two such figures > are topologically equivalent if there there is a homotopy between them. > How many equivalence classes are there for figures consisting of five lines? > How many equivalence classes are there for figures consisting of N lines? > Is there a general theory for solving this problem? > Niels > PS: This is not homework, but somebody asked me for the number of > different solutions to the following problem, and I was embarrased not > to be able to answer. http://rec-puzzles.org/new/sol.pl/geometry/construction/5. lines.with.4.point s This is an unsolved problem. Integer sequence number A006245 has known counts of rhombic tilings, which is a related problem. http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A006245 62 is listed for tiling a decagon. Of these, there are 2 positions of the star arrangement and 10 rotational positions of each of 4 other arrangements. The last arrangement differs from its mirror image and adds 20 to the count. I dont know enough counts to search for a sequence of \ counts of equivalence classes, and A006245 doesnt provide a link \ to such a sequence either. Any arrangement of lines that intersect pair-wise can be converted into a rhombic tiling, but not every rhombic tiling can be converted into a line arrangement. About a year ago, Ian Andlewin posted a problem about finding a particular arrangement of 8 lines, and about a month after that, Alan Curry posted an impossibility proof. But, there is a rhombic tiling for that impossible line arrangement. Sequences A048872 and A048873 say they are about line arrangements. I think one of these counts projectively equivalent arrangements. http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A048872 http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A048873 === Subject: Publishing a paper FAQ Only one little question... Is it possible to publish a paper on some journal even if you e-mail address doesnt end with .edu? Because Im not an academic (I am only graduated in math), but I think I have a good idea for a paper submission. If is it possible what are the standard procedure (I mean anonymous reviewers? How many time does the process last?) Piero Giacomelli === Subject: Re: Publishing a paper FAQ > Is it possible to publish a paper on some journal even if you e-mail > address doesnt end with .edu? What about reading first the replies you got the other times youve posted the *same* question here: Jose Carlos Santos === Subject: Re: Publishing a paper FAQ posting-account=W2DCTA0AAAAlbhDMl3GrysSnPy1IK_7f Anyone can submit a paper, regardless of whether they are affiliated with a university or not. The submission process varies slightly from journal to journal; you will usually be able to find instructions on the journals website, or in a hard copy of the journal. / ftp / hardcopy depending on the journals preferences); \ the editors will then forward it to a referee of their choosing. When they receive the referees report, they will let you know. This will usually take several months (and if your paper is to be accepted, and is thus given a careful reading, may well take much longer). The full cycle between submission and the time the paper actually appears in print varies as well, but often takes a couple of years. However, the best thing may be to first speak with a professional mathematician (such as one of the lecturers at the university you studied at), about your paper. They should be able to tell you whether you have anything new, and whether and where you should publish it. Hope this helps, Lasse --- === Subject: b(xy) = b(x)b(y) --> b=a^H, H>=0 ?? posting-account=zKr99gsAAACCG7HWd7bdCw3Mpdn53Yvi Hi all! I have a simple question. Given a function b:R->R such that 1. b is nondecreasing 2. b satisties: b(xy) = b(x)b(y) Why this does implies that b(a)=a^H for some unique constant H>=0 ? I found this step while studying selfsimilar stochastic processes. Diego Andres === Subject: Re: b(xy) = b(x)b(y) --> b=a^H, H>=0 ?? posting-account=zKr99gsAAACCG7HWd7bdCw3Mpdn53Yvi > Given a function b:R->R such that > 1. b is nondecreasing > 2. b satisties: b(xy) = b(x)b(y) SORRY I FORGOT SOMETHING:!!!! 2. b satisties: b(xy) = b(x)b(y) for x,y>0 > Why this does implies that b(a)=a^H for some unique constant H>=0 ? === Subject: Re: b(xy) = b(x)b(y) --> b=a^H, H>=0 ?? posting-account=sAS5-AwAAABlKnmtMjBbYHvhxI6W0cAg Sorry, I was slightly inaccurate. There is one case where the function isnt odd, and thats where b(x)=1 for all x. \ But as I say, I think really you just wanted the function to be defined for positive numbers. === Subject: Re: b(xy) = b(x)b(y) --> b=a^H, H>=0 ?? posting-account=sAS5-AwAAABlKnmtMjBbYHvhxI6W0cAg > Hi all! > I have a simple question. > Given a function b:R->R such that > 1. b is nondecreasing > 2. b satisties: b(xy) = b(x)b(y) > Why this does implies that b(a)=a^H for some unique constant H>=0 ? > I found this step while studying selfsimilar stochastic processes. > Diego Andres If b(1)=0, then b(x)=0 for all x. Otherwise b(1)=1, so b(2)>=1. Easily prove that b(2^r)=[b(2)]^r for all rational r, then prove this for all real numbers r by the nondecreasing assumption. So for positive x, b(x)=[b(2)]^(log_2 x)=x^(log_2 b(2)). You probably didnt actually want the function to be \ defined for non-positive x - but you can easily prove b(-1)=-1 and so the function is odd. === Subject: Re: b(xy) = b(x)b(y) --> b=a^H, H>=0 ?? * Diego Andres Alvarez > Hi all! > I have a simple question. > Given a function b:R->R such that > 1. b is nondecreasing > 2. b satisties: b(xy) = b(x)b(y) > Why this does implies that b(a)=a^H for some unique constant H>=0 ? b(0) = b(0*x) = b(0)*b(x) So, either b(0)=0 or b(o)=c, for some constant c. Then b(x)=b(0)/b(0) = c/c = 1 for all x. Neither solution implies that b(a)=a^H, unless H=0. Did I miss something? -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 === Subject: Re: b(xy) = b(x)b(y) --> b=a^H, H>=0 ?? posting-account=sAS5-AwAAABlKnmtMjBbYHvhxI6W0cAg b(0)=0 doesnt give you a single solution. It gives you lots of different solutions. === Subject: Re: b(xy) = b(x)b(y) --> b=a^H, H>=0 ?? * Jon Haugsand > * Diego Andres Alvarez > Hi all! I have a simple question. Given a function b:R->R such that 1. b is nondecreasing > 2. b satisties: b(xy) = b(x)b(y) Why this does implies that b(a)=a^H for some unique constant H>=0 ? > b(0) = b(0*x) = b(0)*b(x) > So, either b(0)=0 or b(o)=c, for some constant c. Then > b(x)=b(0)/b(0) = c/c = 1 for all x. > Neither solution implies that b(a)=a^H, unless H=0. > Did I miss something? Yes, that b(0)=0^H=0 is perfectly reasonable. -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 === Subject: Re: Cantors diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> <41ad5139$15$fuzhry+tra$mr2ice@news.patriot.net> <41b2427f$15$fuzhry+tra$mr2ice@news.patriot.net> <41c22df7$7$fuzhry+tra$mr2ice@news.patriot.net> <41c6141c$37$fuzhry+tra$mr2ice@news.patriot.net> posting-account=htRwYA0AAACUC1yg4djqvdjZ_SB9JXGq >There are no infinite natural numbers in all mathematics. > Well, there are in NSA, but that is not the issue. A set of finite > numbers need not be finite any more than a box of red spoons need be > red or a spoon. Thats not the issue too. We can definie \ sequences and IN can be given the form of a sequence too. Card(IN) = aleph_0 is said to be larger than any natural number. But the number of natural numbers is counted by natural numbers. Define: A_n = {1,2,3,...,n}. Card(A_n) = n. Card(A_n) < aleph_0 for any n of IN. Card(A_omega) = aleph_0, but omega is not a narural number. It does not belong to IN. Hence Card(IN) < aleph_0. Merry Christmas to all participants. I will be absent for a while, spending my holidays in the Harz-mountains where Cantor liked to spend his holidays too. === Subject: Re: Cantors diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> <41ad5139$15$fuzhry+tra$mr2ice@news.patriot.net> <41b2427f$15$fuzhry+tra$mr2ice@news.patriot.net> <41c22df7$7$fuzhry+tra$mr2ice@news.patriot.net> posting-account=htRwYA0AAACUC1yg4djqvdjZ_SB9JXGq > Prof. Dr. Mueckenheim, I disagree with you. First, the universe is > infinite. It has not yet been deided whether the universe is finite or not. It is expanding. But its actual shape doesnt matter. The universe accessible to us, is finite and will remain so forever. Secondly, if 10^100 is the largest number you can imagine, > then it would be infinity. Otherwise, 10^100+1 would be a number, and > 10^100 not the largest. 10^100 is not the largest number, because 10^10^10^10 is by far larger. 10^100 is an upper bound for the means capable of storing bits. It is impossible to store more than 10^100 bits. > The counting numbers are each a finite integer, taken together theyre > the infinite integer. All the natural numbers are finite. Each n gives the cardinal number of its sequence {1,2,3,...,n}. The is no infinite number n, hence there is no infinite number of numbers. [...] > Consider Plancks constant, h, and Plancks \ constant, h bar = h / 2 pi. > Something measured in a rational number of Planck distances (or Planck > lengths) is some irrational number of modified, or corrected, Planck > distances, because pi is irrational. Thats about that irrational > quantities exist. Irrational numbers dont exist, because none of them can be represented. I require that a number (which is said to exist) can be identified unambigously. As you are not able to name the \ first 10^100 digits of pi (and nobody will ever be able to do so) this number does not exist and it will never exist. Its the same with every irrational number. Merry Christmas to all participants. I will be absent for a while, spending my holidays in the Harz-mountains where Cantor liked to spend his holidays too. === Subject: Re: Cantors diagonal proof wrong? >> Prof. Dr. Mueckenheim, I disagree with you. First, the universe is >> infinite. > It has not yet been deided whether the universe is finite or not. It is > expanding. But its actual shape doesnt matter. The \ universe > accessible to us, is finite and will remain so forever. Forever? So the universe is infinite in temporal extent? Q.E.D. John Briggs === Subject: Re: Cantors diagonal proof wrong? > Irrational numbers dont exist, because none of them can \ be > represented. Then squares dont have diagonals? === Subject: Re: Cantors diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> <41ad5139$15$fuzhry+tra$mr2ice@news.patriot.net> <41b2427f$15$fuzhry+tra$mr2ice@news.patriot.net> <41c22df7$7$fuzhry+tra$mr2ice@news.patriot.net> <41C25DB8.EF7C5F2F@tiki-lounge.com> posting-account=htRwYA0AAACUC1yg4djqvdjZ_SB9JXGq >> But the good thing is that we only need the space for storing the laws >> that _all_ numbers obey. Like a Shakespearean play: the words of it >> are spoken by players, but the essence of the play is in the book, and >> it remains there even when a play is not being performed. >> And the literary critics can talk about the play without actually >> seeing a single performance of it. > Whats in a name? that which we call a number by another name would > count as well. > Could you give me the sum of Integer(pi*10^10^100) and > Integer(sqrt(2)*10^10^100), please? > Integer(pi*10^10^100) + Integer(sqrt(2)*10^10^100) > Or are your laws insufficient to describe what primarily counts with > numbers, namely counting? > Where is the problem? I havent yet been able to figure out whether it \ is an even number. Seems to be an odd problem. Could you help me? Merry Christmas to all participants. I will be absent for a while, spending my holidays in the Harz mountains where Cantor liked to spend his holidays too. === Subject: Re: .99999... still=/= 1 >>I won this debate years ago. >So what is the reason to start it again? To make me feel better by about this much, 99.999...% . Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 Lets look at another approach to this. You say, 1 = .999... well lets see, function1 f(n)_1 = 1 function2 f(n)_2 = 9/10^n Does, f(n)_1 = f(n)_2 ? If we integrate (from 0 to oo) it should show that each function is equal, if 1 = .999... . Integrate the function f(n)_2 = 9/10^n from 0 to oo and compare this with the Integration of the function f(n)_1 = 1 from 0 to oo, both with respect to n. Integral ( 0 to oo) 1 dn = n ( 0 to oo) = oo - 0 = oo Integral ( 0 to oo ) 9/10^n dn 9/10^n ( 0 to oo) = 9/10^oo - 9/10^0 = 0 - 9 = -9 ( assume 9/10^oo -->0) If 1 = .999.., then the integration value from 0 to oo for both functions should be the same and they are not. -9 =/= oo .999... =/= 1 f(n)_1 =/= f(n)_2 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 Lets look at another approach to this... If .999... = 1 then, 2.999... X 10^8m/sec This should occur. 2.999... X 10^8m/sec ---> 3.0 X 10^8m/sec reach the speed of light and cause .999... = 1, which would make, 2.999... X 10^8m/sec --> 3 X 10^8 m/sec ------------------------------------------------------- Another approach to this in energy aspects of .999... and 1 is : Write a computer program. 10 n = 1 20 print n run program 1 Very little energy needed to run this program. Now try this. 5 a = 0 10 For n = 1 to oo 20 a = a + 9/10^n 30 If a = 1 then 50 40 next n 50 print a run It will take an infinite amount of energy to run this program. It will never reach 1. Just like accelerating matter near the speed of light, you need an infinite amount of energy to reach the speed of light, according to Einstein. So with respect to matter and energy, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > reach the speed of light and cause .999... = 1, which would make, This has nothing to do with dividing by zero. Besides there is only a finite amount of energy in the entire physical cosmos. So the upperbound Bob Kolker === Subject: Re: .99999... still=/= 1 >if there is a difference between 0.999... and 1, then what is that >difference? >is the difference 0.000...(infinity)...1 ? because thats not possible >x = 0.999... >10x = 9.999... >- x = 0.999... >--------------------- >9x = 9 Nope, 10 x = 9 x = .9 >9x/9 = 9/9 >x = 1, x=0.999..., 1=0.999.... 9/9 = 1 not .999... If 9/9 = .999... then 8/8 = .999... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 In sci.math, S. Enterprize Company if there is a difference between 0.999... and 1, then what is that >>difference? >>is the difference 0.000...(infinity)...1 ? because thats not possible >>x = 0.999... >>10x = 9.999... >>- x = 0.999... >>--------------------- >>9x = 9 > Nope, > 10 x = 9 > x = .9 If youre going to refute it at least do it properly. 10x = 9.999... x = .999... 10x - x = 8.999... >>9x/9 = 9/9 >>x = 1, x=0.999..., 1=0.999.... > 9/9 = 1 not .999... > If 9/9 = .999... > then 8/8 = .999... And it does, in base 11. :-) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: .99999... still=/= 1 >> Interesting. So even in the hyperreals, S. Enterprize is simply wrong. :-) >I told you. >Bob Kolker .999... < 1 x is said to be infinitesimal iff |x| < 1/n for all integers \ n. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > x is said to be infinitesimal iff |x| < 1/n for all integers n. By the transfer principle the limit of the partial sums still is 1. Bob Kolker === Subject: Re: .99999... still=/= 1 >> This is not to say S. Enterprizes arguments are any \ good; >> theyre extremely sloppy, in fact. >His arguments are not even wrong. They are non-existent. Stringing words >together doth not an argument make. >Bob Kolker Thou shouldth have understandeth by nowth. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > And in regard to 1/3 = .333... . Well at least it can be > expressed as a fraction. So it may be classified as a real > number. But .999... can not be represented as a fraction >>Help me understand what you mean by .333... and .999...: >>Which of the following statements are true? >>a) 3 * .111... = .333... >>b) 2 * .333... = .666... >>c) 3 * .333... = .999... >> Big friggin deal. >> i^2 = -1 >> So does that mean i is a real number too. >Well, the big friggin deal is that with your definition >of numbers, it doesnt seem possible to do arithmetic. >For example, does 3 * .111... equal .333? Yes? No? Maybe? >It doesnt seem like youre able to say. This \ isnt very >practical. >>d) 10 * .999... = 9.999... >>e) .111... + .222... = .333... >>f) .333... + .666... = .999... >>g) 9 + .999... = 9.999... >>h) .111... = 1/9 >>i) .333... = 1/3 >And again: you cant say which are true and which are \ false. >The fact that you dont know if you can add and multiply these >numbers suggests that you shouldnt be lecturing about numbers. >>And what is .999... + .999... ? >> Its an indeterminate, unless you use Partial Sums Convergence >to give you >> an approximation of what it is. >Hmmm. You call .999... a number, though not a real number. You call >.999... + .999... an indeterminate, which I gather is not a number. >> Express .999... as a fraction. DUH???? What you waiting on? >> 9/9 = 1 not .999... >> 9/10 = .9 >I cannot express .999... as a fraction that does not equal 1. >You know, people here would a agree that you can define an ordered set >of strings used to denote real numbers (with certain conventions like, >say, always using 0.3 instead of .3 and 5. instead of 5). For >this ordered set we have, for example, >0. < 0.9 < 0.99 < 0.999... < 1. < 1.999... < 2. >Note that 0.999... does not equal 1 here. >Yep, 0.999... =/= 1. And, whats more, I can convince at least 80% >of sci.math that this is true, because Im able to say what I mean. >> DUH????? Whats wrong, you cant admit you \ are mistaken? >People use symbols to represent mathematical ideas. For example, + >is used to mean plus. They use strings of digits to represent real >numbers. There is a convention which says which strings mean which >numbers. Under this convention, 0.999... and 1. and represent >the same number. >The strings themselves are not the numbers. They are ways of >representing the numbers. Four and 4 represent the same number. >0.999... and 1. represent the same number. We could have a >convention that 0.999... does not represent a number: this wouldnt >really be a problem because we could just use 1. to represent the >number one. >But the convention that people have agreed upon is that .999... >does represent a (real) number. Its not rational or irrational. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 posting-account=uMDgiw0AAAANoAxJOs_DnrtdjhRMBFah >But the convention that people have agreed upon is that .999... >does represent a (real) number. > Its not rational or irrational. When I write .999... I mean the number 1, which is a rational number. I could explain to you why I mean the number 1 when I write .999..., but it involves defining the real numbers and a convention about interpreting strings. You do not interpret the string .999... to correspond to any real number. Okay. Thats fine. When someone says .999... = 1 this means that .999... and 1 are both strings that, under some convention, have interpretations as numbers, and that the numbers they correspond to are equal. The standard convention is that .999... means the number 1. You seem to disagree, but in what sense to you disagree? * Do you think that .999... means 1 is *not* the standard convention? * Do you think that the convention itself is wrong? And, if so, what does it mean for a convention to be wrong? === Subject: Re: .99999... still=/= 1 > When someone says .999... = 1 this means that .999... and > 1 are both strings that, under some convention, have > interpretations as numbers, and that the numbers they > correspond to are equal. 1 is the limit of a sequence of partial sums of a series idiomatically written as .999... . Strictly speaking .999... is not a well formed mathematical expression, but it can be associated with one. Showing that lim (N >= 1) Sum ( 1 <= k <= N) [9/10^k] = 1 takes a little work (not much though). Bob Kolker === Subject: Re: .99999... still=/= 1 > Its not rational or irrational. Contradiction. It violates the law of the excluded middle. Bob Kolker === Subject: Re: .99999... still=/= 1 >> Using the surreal symbol { | } >> (.5_ -) | -- >> then sqrt (-1) = (.5_-)(1) = (1_-) >> ^ >> | >> This shows half negative sign. >> then, >> (.5_- + .5_-)(1)^2 = -1 >> or >> (1_-)^2 = -1 >> or >> (1_+) - (+) = (1_-) >> or, >> ( 1 ) - (1_+) = (1_+) >> or, >> (1_+) + (1_-) = 0 >> This would be sign and number value annihilation. >> etc... >This looks interesting. >Why not create a web page (a central repository) for this and other >math researches? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 >In sci.math, S. Enterprize Company > | >> last digit seen is zero right before infinity. It never reaches 1. >There is no right before infinity, numbskull. >> Hey, there are probably about 1 million supporters using MathCAD >> Professional. This is an industry standard for math for scientists and >> engineers. A non-standard approach using MathCAD clearly shows that, >> n-->oo -1 >> lim 9/10^n ---> 90/10^n >> The digit right before oo for the hyperreal number or >> series .999... is 0. Therefore, there is a space existing between, >> .999... and 1 so, >> as the definition of a hyper-real number implies, >> .999... < 1 >An interesting notion, that. So D[.999..., w-1] = 0, eh? [*] >What is D[.999..., w-2]? How about D[.999..., w/2]? Using MathCAD I get, n-->oo - 2 lim 9/10^n ---> 900/10^oo How can this be interpreted? In a non-standard analysis, in my opinion, it could be mean we are moving further away from 1, but we still maintain the surreal number or hyperreal number. For n digits less than infinity .999..n | 1 you still get, .999... < 1 For the case 1/2 a digit less than infinity, MathCAD calls this a bidirectional limit. n-->oo/2 lim 9/10^n --> 9/((10^oo)^1/2) But I dont think this applies to a surreal or a hyperreal number. .999... doesnt really exist anymore. So I dont think \ you can say this. >Its easily proven that, if D[.999..., n] = 9, then D[.999...., n+1] = 9 Ok, because, If the mth digit less than oo = oo n--> oo - m lim 9/10^oo-oo ---> 9 What can this be interpreted as in regard to a surreal number? It looks like to me, in this case we have totally removed the original hyperreal or surreal number. I think there is a limit to how far you can go minus digits before you change, .999... totally from it original form. In this case, you no longer have, .999..., you have something else. You have just 9. This isnt surreal or hyperreal. >as well (the simplest method arguably is to evaluate D[x*10, n]), >for any finite n. Not sure if w-1 is finite or \ not -- or even meaningful. >As for MathCAD: thats a program, an approximation of reality. Not that >real numbers are all that real, anyway -- theyre mathematical/symbolic >abstractions, there because Dedekind, Cauchy, and Cantor and others needed >more numbers for set theory. In light of what Ive written before >regarding 1/3, one might have to verify the results carefully. >> A hyper-real number causes a space to exist between it and a real number. >> 1 is the real number >> .999... is the hyper-real number that forms the space between the two. >> So, >> .999... =/= 1 >> .999... < 1 >Your logic is extremely sloppy, though your conclusion is interesting. Why would it be sloppy to you? Do you mean as far as accuracy is concerned? >Im just not sure which realm it exists in, although the standard >real realm does not contain it (the standard realm doesnt >contain any numbers between 0 and all 1/n, n > 0, n in J: the hyperreal >realm, however, does). >[.sigsnip] >[*] I dont have an omega, so Im using \ Ôw here to indicate the > first transfinite ordinal. Is there a w_0, \ analogous to > the cardinal aleph_0? This gets a bit messy. > D[r,n] = the digit associated with the nth decimal place > after the decimal point (e.g., D[.98765, 4] = 6). >-- >#191, ewill3@earthlink.net >Its still legal to go .sigless. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > Its necessary to observe: As the cardinality of reals R > cardinality of > integers N, so the cardinality of N_inf > N. As you count the limit > N --->oo instead of N-inf, then you certainly omit something - namely > hyperreal part. >> All very confused. Assuming you mean N_inf = *N, the set of >> hyperintegers, then its true that *N as an external set has the >> cardinality of the reals, but the internal set *N has a *cardinality >> equal to *aleph_0. > This is of-course OK! >> This certainly doesnt mean that the hypernaturals >> are the same as transfinite cardinals. For one thing, there is a >> smallest transfinite cardinal (aleph_0), > Ok! >> but there is no such thing as a >> smallest infinite hypernatural. If n is an \ infinitely large integer, >> then so is n-1. > Omega was defined in this discussion earlier as follows: > The number that is greater than any infinite integer. That is the smallest > transinfinite number. Omega is the successor of ...999, i.e. n+1. The > precessor (n-1) of ...999 is ...998. Its possible to define the hyperordinals and to \ talk about *omega in NSA, but this *omega has no relation to anything you said in that paragraph. For one thing, *omega is not a member of *N. (In fact, *omega is identical to the set *N itself). For another, there is no connection between the hyperordinals and strings (even hyperstrings!) of decimal digits. I dont accept your definition. Are you under the impression that ...999 is a hyperinteger? It isnt, unless you explain which equivalence class of sequences of integers you are talking about. I can think of ways to make such a correspondence, but you havent said what you mean. >> And none of this has anything to do with limits. You seem to think that >> infinity has only one meaning in all of mathematics. Wrong. > Your opinion. :-) Infinity has many meanings. It can refer to a \ compactification of the real line or of the complex plane, or to transfinite ordinals or cardinals, or to hyperreals, or to surreals. All of those are different. And this is not intended to be a complete list. > N (finite integers) does not cover hyperreal area, i.e. the numbers > smaller > than reals, because N hardly covers real area as discussed under the > thread > Are reals well-ordered. > As we count the real limit in NSA, then the hyperreals exist, but they > are > not counted in limit as N -->oo, but not as N_inf -->oo. Hyperreals are > omitted and therefore 0.999...<1 in hyperreals, though the real part of > the > limit equals to 1. >> I take it you mean the standard part of the limit is 1. > Well, we can talk about standard part and non-standard part, if you > prefer > this. >> Im trying to guess what you might mean by the real part, since you >> have not defined your meaning. > the real part was the decimal part - as you certainly knew. No, I didnt know what you meant, and I still am not sure. What is the real part of sqrt(2), for example. It sounds like you trying to say the real part is sqrt(2) - 1, or approximately 0.41421. That was not one of my first two guesses. I would call that the fractional part. >> We are not talking about complex numbers >> here. > Exactly! That was my first guess, but my second guess was standard part. That is evidently not what you meant, either. In a similar fashion, you keep assuming that I must know what you mean by ...999. However, I assure you that I dont. > 2)Yes, the limit is exactly 1only if you count the limit including the > standard part as N --->oo. Then your reference set is N. >> Are you talking about standard analysis here, or nonstandard? > Standard analysis as You correctly observed. Then you are not discussing the value of 0.999... in NSA at all, as I previously thought. Looks like yet another example of miscommunication. Lets summarize: (1) We have 0.999... = 1 in standard analysis, because the sum is over N. (2) We have 0.999... = 1 in NSA, because the sum is over *N. (3) It is not possible to sum over *N in standard analysis, because *N is not a set in standard analysis. (4) It is not possible to sum over N in NSA, because N is not a set in NSA. Do you agree? >> The set N >> (consisting of the finite naturals) is not a set in the internal set >> theory of NSA. > Coorect, but it should be as *N is the extension of the set N. No, it should not be. There is an important principle involved, known as the transfer principle. This says that every theorem of standard analysis is also a theorem of NSA, provided we make the appropriate substitutions (such as substituting *N for each occurence of N). The transfer principle is extremely important. Without it, NSA loses most of its value as a tool of analysis. It turns out that if all sets in standard analysis are allowed to be internal sets in NSA, then we lose the transfer principle. Thats why things are defined the way they are. >> Its counterpart is *N, which includes the infinitely >> large naturals. If you are forming a sum in NSA, then you can sum over >> *N, but not over N. > We can sum over subset too. Yes, but we cant sum over things that fail to be sets at all. Thats the point. >> There is no case 3 above. I have explained that the sum over N is not >> a sum in NSA, since N is not an internal set in NSA. Thats why we sum >> over *N instead. > Because You dont see N as subset of *N. :-) No, that is not the reason. The reason is that Abraham Robinson, the founder of NSA, did not see N as an internal set. And he had very good reasons. >> Nonsense. If { d_k } is a decimal digit string, then the number >> represented by the string is sum_{k in *N} d_k * 10^(-k). No point of >> reference is needed. > You cut (=snip) out the climax of my explanation. What does ^(-k) refer? > Please answer to that simple question! Why minus - what does it refer? The symbol ^ means that what follows is an exponent. Thus, 10^(-k) means ten to the power of (-k), which is the same as 1/10^k. We can write 0.999... as sum_{k=1}^oo 9*10^(-k) = sum_{k=1}^oo 9/10^k. >> [ snip nonsense about point of reference ] > Your opinion, because You could not consider the successor of ...999. What > is the successor of ...999 as all the placeholders are infinitely occupied > with the maximal digit 9? You havent said what ...999 is. How am I supposed to answer questions about it if you dont define it? -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: .99999... still=/= 1 (snip) > but there is no such thing as a > smallest infinite hypernatural. If n is an \ infinitely large integer, > then so is n-1. >> Omega was defined in this discussion earlier as follows: >> The number that is greater than any infinite integer. That is the >> smallest >> transinfinite number. Omega is the successor of ...999, i.e. n+1. The >> precessor (n-1) of ...999 is ...998. > Its possible to define the hyperordinals and \ to talk about *omega in > NSA, but this *omega has no relation to anything you said in that > paragraph. For one thing, *omega is not a member of *N. (In fact, > *omega is identical to the set *N itself). Correct, because that omega above (maybe some other name is better - lets use your symbol *w ?) is in the next infinite area. The reasons are: Every number has a successor, i.e. you can always add 1. All the placeholders in the infinite long area *N were already occupied. > For another, there is no > connection between the hyperordinals and strings (even hyperstrings!) of > decimal digits. I dont accept your \ definition. I could not follow you above, sorry. :-( > Are you under the impression that ...999 is a hyperinteger? It isnt, > unless you explain which equivalence class of sequences of integers you > are talking about. I can think of ways to make such a correspondence, > but you havent said what you mean. OK, lets try again. First of all, I think we should start a new thread from the beginning so that everything is constructed and we use the same concepts and definitions. 1) ...999 is not N as it is not in a classic way finite integer. Lets call it infinite integer (N_inf) over one infinity. 2) I define it as I have done earlier sum (k 0 --> oo) 9*10^k. It does not have classic limit as the k refers now the standard point of reference, but it has a limit as you hopefully noticed as we have another point of reference. 3) It is defined and it exists, now - how do You like to name it? Hyperinteger or something else? (snip) > Im trying to guess what you might mean by the real part, since you > have not defined your meaning. >> the real part was the decimal part - as you certainly knew. > No, I didnt know what you meant, and I still am not sure. What is the > real part of sqrt(2), for example. It sounds like you trying to say > the real part is sqrt(2) - 1, or approximately 0.41421. That was not > one of my first two guesses. I would call that the fractional part. Ok, that suits for me. (snip) > In a similar fashion, you keep assuming that I must know what you mean by > ...999. However, I assure you that I dont. Uh! I have tried to explain so simple as possible. I had once a fealing that You understood very well. There must be some miscommunication. >> 2)Yes, the limit is exactly 1only if you count the limit including the >> standard part as N --->oo. Then your reference set is N. > Are you talking about standard analysis here, or nonstandard? >> Standard analysis as You correctly observed. > Then you are not discussing the value of 0.999... in NSA at all, as I > previously thought. Looks like yet another example of miscommunication. > Lets summarize: > (1) We have 0.999... = 1 in standard analysis, because the > sum is over N. Yes, the limit is over N. Im voluntary to point You that \ the sum and the limit are not the same thing. > (2) We have 0.999... = 1 in NSA, because the sum is over > *N. Yes, the limit is over *N. > (3) It is not possible to sum over *N in standard analysis, > because *N is not a set in standard analysis. Yes. I consider *N is the extension of N. > (4) It is not possible to sum over N in NSA, because N is > not a set in NSA. > Do you agree? I agree 1,2 and 3 but in the case 4 I disagree. Maybe we have to discuss about that topic more accurate. I consider finite integers are a subset of infinite integers over the first one \ infinity. > The set N > (consisting of the finite naturals) is not a set in the internal set > theory of NSA. >> Coorect, but it should be as *N is the extension of the set N. > No, it should not be. There is an important principle involved, known as > the transfer principle. This says that every theorem of standard > analysis is also a theorem of NSA, provided we make the appropriate > substitutions (such as substituting *N for each occurence of N). > The transfer principle is extremely important. Without it, NSA loses > most of its value as a tool of analysis. > It turns out that if all sets in standard analysis are allowed to be > internal sets in NSA, then we lose the transfer principle. Thats why > things are defined the way they are. I cannot see where the transfer principle fails, except in the case of sum and limit. But in this case there are natural reasons as I have tried to explain. And the reason is not NSA. (snip) >> Your opinion, because You could not consider the successor of ...999. >> What >> is the successor of ...999 as all the placeholders are infinitely >> occupied >> with the maximal digit 9? > You havent said what ...999 is. How am I supposed to answer questions > about it if you dont define it? I assume You can now answer after the definition - above. I propose anyway to start a new thread. Tapio > -- > Dave Seaman > Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. > === Subject: Re: .99999... still=/= 1 > (snip) >> but there is no such thing as a >> smallest infinite hypernatural. If n is an \ infinitely large integer, >> then so is n-1. > Omega was defined in this discussion earlier as follows: > The number that is greater than any infinite integer. That is the > smallest > transinfinite number. Omega is the successor of ...999, i.e. n+1. The > precessor (n-1) of ...999 is ...998. >> Its possible to define the hyperordinals and \ to talk about *omega in >> NSA, but this *omega has no relation to anything you said in that >> paragraph. For one thing, *omega is not a member of *N. (In fact, >> *omega is identical to the set *N itself). > Correct, because that omega above (maybe some other name is better - lets > use your symbol *w ?) is in the next infinite area. The reasons are: > Every number has a successor, i.e. you can always add 1. All the > placeholders in the infinite long area *N were already occupied. I dont know what you mean by an infinite area \ or by a placeholder. >> For another, there is no >> connection between the hyperordinals and strings (even hyperstrings!) of >> decimal digits. I dont accept your \ definition. > I could not follow you above, sorry. :-( >> Are you under the impression that ...999 is a hyperinteger? It isnt, >> unless you explain which equivalence class of sequences of integers you >> are talking about. I can think of ways to make such a correspondence, >> but you havent said what you mean. > OK, lets try again. First of all, I think we should start a new thread from > the beginning so that everything is constructed and we use the same concepts > and definitions. > 1) ...999 is not N as it is not in a classic way finite integer. Lets call > it infinite integer (N_inf) over one infinity. As far as I am concerned, ...999 does not mean anything at all. It has no connection with NSA or anything else as far as I can see. You keep throwing that string around as if you think it has a meaning, but you have never said what that meaning is. I am not aware of any useful way of representing the members of *N as decimal digit strings. Any scheme I can think of suffers from one of two defects: either there are members of *N that dont \ fit into the naming scheme at all, or the coding scheme is so cryptic that you cant even compare the sizes of two numbers just by looking at the digits. Perhaps you can think of some scheme that I have overlooked, but you will have to define your terms very carefully before I will be convinced. > 2) I define it as I have done earlier sum (k 0 --> oo) 9*10^k. That sum diverges, even in NSA. It is not a number. > It does not > have classic limit as the k refers now the standard point of reference, but > it has a limit as you hopefully noticed as we have another point of > reference. There you go with your point of reference again. Sorry, but you cant use one undefined term to define another. > 3) It is defined and it exists, now - how do You like to name it? > Hyperinteger or something else? > (snip) Its a string of digits. Nothing more. It is not a number at all, at least in any sense that you have yet defined. And now, I would like to ask a counter-question. I will describe a certain member of *N, and I would like you to tell me what decimal digit string you think corresponds to it. First, some background. You may recall from other discussions that the (standard) real numbers can be defined as equivalence classes of Cauchy sequences of rationals. That means I can identify a real number by presenting you with a Cauchy sequence, and it is understood that any other sequence that happens to fall into the same equivalence class is an equally good representation of that number. For example, the sequences < 9/10, 99/100, 999/1000, ... > and < 1, 1, 1, 1, ... > happen to be two different representations of the same real number. Now, lets consider a similar construction that lies at the heart of nonstandard analysis. To describe a member of *N, for example, I can present you with a sequence of natural numbers (members of N). Unlike the real-number construction, this one doesnt require the sequences to be Cauchy. Technically, I also need to describe to you the equivalence relation that will be used, but thats a bit more complicated. It involves something called a free ultrafilter on N. You can \ find an explanation of the concept at . For our purposes, its enough to know that if I give you a sequence in N, there is a unique member of *N that is represented by that sequence. Ok so far? Here is the sequence I have in mind. Let A(x,y) be the Ackermann Function, as described at . Now, let a_k = A(k,k) for each k. This sequence starts out: a_0 = A(0,0) = 1 a_1 = A(1,1) = 3 a_2 = A(2,2) = 7 a_3 = A(3,3) = 2^6 - 3 = 61 a_4 = A(4,4) = 2^2^2^2^2^2^2 - 3 = (too big to write out here) and after that the sequence starts to grow rather quickly. :-) Let a be the member of *N that is associated with this sequence. My question is: (1) what decimal digit string do you think represents a? (2) what decimal digit string do you think represents A(a,a)? My point is that your decimal digit strings are woefully inadequate in this context. They cannot even begin to describe the numbers in *N in any useful way. >> Im trying to guess what you might mean by the real part, since you >> have not defined your meaning. > the real part was the decimal part - as you certainly knew. >> No, I didnt know what you meant, and I still am not \ sure. What is the >> real part of sqrt(2), for example. It sounds like you trying to say >> the real part is sqrt(2) - 1, or approximately 0.41421. That was not >> one of my first two guesses. I would call that the fractional part. > Ok, that suits for me. > (snip) >> In a similar fashion, you keep assuming that I must know what you mean by >> ...999. However, I assure you that I dont. > Uh! I have tried to explain so simple as possible. I had once a fealing that > You understood very well. > There must be some miscommunication. Answer my questions (1) and (2) above, and then well see whether there is any purpose at all in discussing decimal digit strings in connection with *N. >> (4) It is not possible to sum over N in NSA, because N is >> not a set in NSA. >> Do you agree? > I agree 1,2 and 3 but in the case 4 I disagree. Maybe we have to discuss > about that topic more accurate. I consider finite integers are a subset of > infinite integers over the first one \ infinity. Each of the finite naturals is a member of *N, but it does not follow from this that N is a subset of *N. Thats part of why \ its called nonstandard analysis. Not everything is a set in this model. >> The set N >> (consisting of the finite naturals) is not a set in the internal set >> theory of NSA. > Coorect, but it should be as *N is the extension of the set N. >> No, it should not be. There is an important principle involved, known as >> the transfer principle. This says that every theorem of standard >> analysis is also a theorem of NSA, provided we make the appropriate >> substitutions (such as substituting *N for each occurence of N). >> The transfer principle is extremely important. Without it, NSA loses >> most of its value as a tool of analysis. >> It turns out that if all sets in standard analysis are allowed to be >> internal sets in NSA, then we lose the transfer principle. Thats why >> things are defined the way they are. > I cannot see where the transfer principle fails, except in the case of sum > and limit. But in this case there are natural reasons as I have tried to > explain. And the reason is not NSA. Lets consider the case of *R, for example. We know *R is \ the extension (the enlargement) of R. We also know that R satisfies the least upper bound property. By the transfer principle, *R must satisfy the least upper bound property. However, we also know that the collection R of finite reals is nonempty and is bounded above in *R by any infinitely large number. But R does not have a least upper bound in *R, because if x is infinitely large, then so is x-1. The conclusion is that R cannot be a set in NSA. The least upper bound principle is preserved because it applies only to nonempty *subsets* of *R that are bounded above, and R is not a *subset* at all. > (snip) > Your opinion, because You could not consider the successor of ...999. > What > is the successor of ...999 as all the placeholders are infinitely > occupied > with the maximal digit 9? >> You havent said what ...999 is. How am I supposed to answer questions >> about it if you dont define it? > I assume You can now answer after the definition - above. I propose anyway > to start a new thread. You have not defined ...999. If you think you know what it is, then why dont you try answering your question and explaining your answer? -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: .99999... still=/= 1 >In sci.math, S. Enterprize Company > So 1/9 = 0.1111... >> and 8/9 = 0.8888... >> then what is 1/9 + 8/9? >9/9. expressions like .111... .888.... and .999.... are not proper >mathematical expressions. If you mean to write down series, then do so. >Bob Kolker >> I guess you dont know what a hyper-real number is. Here we go again and >> again. >A hyperreal number is a number less than any 1/n (n in J, n > 0) but >greater than 0. It is not entirely clear to me how many hyperreal >numbers there are. >[.sigsnip] >-- >#191, ewill3@earthlink.net >Its still legal to go .sigless. Thats a good question. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 >I dont know about you guys, but I actually like the character S.E. >quite a bit, if only for the humour and entertainment his posts >provides. >And all this .999... arguing back and forth is cracking me up. I >dont know whats funnier (or more pathetic), \ S.E. making himself >look like a fool, or you guys actually taking him seriously and >making yourselves look like retards. >My one and only advice to S.E. is better to keep your mouth shut and >let others suspect you are a fool, than to open it and forever remove >all doubts. Too late now, I guess. >You know, most newsgroups need an idiot of some kind, and our math >genius is serving that role quite nicely. He reminds me of Beavis >(you know, the imbecile in the Beavis and Butthead cartoons -- not >that Butthead was any less imbecilic, but in a different way) after he >had ingested an overly large dose of caffein and sugar. >A drugged-up Beavis would pull his T-shirt over his head, raise his >arms surrender-style, and spew complete nonsense at strangers. All >offers of help, logical arguments, insults, even physical threats were >totally ignored because, well, actually I dont know why, but I >suspect mostly because he had gone nuts, or, less likely, they werent >what he was looking for (he usually made it abundantly clearly that he >was looking for TP for his bunghole, but everyone nearby would fail to >pick up that glaringly obvious point). >S.E.: I am the one and only mighty bunghooooole...you must bow down >before the great bunghole...and .9999... =/= 1...bungholio.... >Sci.Math: Chill out, heres the proof. Take a look, blah blah blah... >S.E.: Are you threatening me? I am cornholio! ...I need TP for my >bunghole...Bunghole!! And rational doesnt converge partial >divergent sum limit test Mathcad Rababababa 99999999 genius >hyperfilterunrealnumber HOOOOLLLIOOOOOO... >Sci.Math: Now you are being completely non-sensical...do you even know >what a rational number is? Heres the \ definition of limits, >for any epsilon > 0 blah blah blah... >S.E.: My bunghole it goes Raaabbabababa yeeeedadadadada Bungholioooo! >(singing) Would you like to see my bunghoooole? You have TP? >TP for my bunghole? I am a gringo, and I have no bunghole... >and analysis asymptote (huh huh huh...I said ass) non standard >pi irrational dimensionless infinite 99999999 hyperreal ... >And it goes on and on. The only solution to the Beavis problem was to >wait long enough for the caffein and sugar rush to run its course and >he would calm down. I suppose if you all ignore S.E. for a few days >hell move on to troll in other groups \ (didnt he say he was a physics >genius too?), or maybe hell have found TP for his bunghole and his >quest would come to a successful end. >-- L. You even make me laugh. HAHAHAHHHHHAHAHHAHHHH LOL Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: what is maximum eigenvalue of this matrix? >Let v(n) is some n-tuplet whose every element is finite, >and let v(n) be the transpose of v(n). >Define the nxn matrix >M(n) = v(n) v(n) . >What is the necessary and sufficient conditions on v(n) such that >the maximum eigenvalue of M(n) grows UNBOUNDEDLY in proportion to n as >n --> infinity? Well, m(n) has only one non-zero eigenvalue. Calculate it. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: what is maximum eigenvalue of this matrix? >Let v(n) is some n-tuplet whose every element is finite, >and let v(n) be the transpose of v(n). >Define the nxn matrix >M(n) = v(n) v(n) . >What is the necessary and sufficient conditions on v(n) such that >the maximum eigenvalue of M(n) grows UNBOUNDEDLY in proportion to n as >n --> infinity? The characteristic values of M(n) are all 0s except one which is the sum of squares of the elements of v(n). -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: what is maximum eigenvalue of this matrix? posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g The maximum eigenvalue of your matrix is the Euclidean norm squared of v(n). All the other eigenvalues are zero; the matrix is rank one if v(n) is nonzero. You can see this by computing: M(n) v(n) = ||v(n)||^2 v(n) === Subject: FORGET ABOUT TRUE IN A SYSTEM Lend me your ears, just for the next 30 seconds, imagine all propositions, all statements are either TRUE of FALSE. in what system is Successor(0) = 2? NO LOCAL VARIABLES EVERYTHING IS GLOBALLY TRUE OR FALSE, or neither. Now look at this statement, there is no SYSTEM, there is only the one universe, what is it? This statement has no proof doesnt matter if youre in system Alpha \ Centaurie, its the same statement. WHO SAID TRUE? Well you better have a darn good reason to say so, if so then you must have PROVEN it true, so it cant be. Herc -- YOU CANT PROVE ME If you prove its true then it has a proof, which makes it false If you dont prove it, then its true 10,000 people in sci.math ALL believe this is irrefutable that mathematics will always be incomplete. === Subject: Re: FORGET ABOUT TRUE IN A SYSTEM >all statements >are either TRUE of FALSE. But that is false. The statement This statement is false. is a counterexample. Abandon the law of the excluded middle, which is true only for first-order statements, and there are no paradoxes. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: FORGET ABOUT TRUE IN A SYSTEM John Savard says... >Abandon the law of the excluded middle, which is true only for >first-order statements, and there are no paradoxes. Thats not true. The Liar paradox does not depend on \ excluded middle. -- Daryl McCullough Ithaca, NY === Subject: Re: FORGET ABOUT TRUE IN A SYSTEM > Lend me your ears, > just for the next 30 seconds, imagine all propositions, all statements > are either TRUE of FALSE. in what system is Successor(0) = 2? > NO LOCAL VARIABLES > EVERYTHING IS GLOBALLY TRUE OR FALSE, or neither. You need a system to define what Successor(0) means. For example, in the even whole numbers, successor(0)=2 would be TRUE under a natural interpretation. -- Will Twentyman email: wtwentyman at copper dot net === Subject: theoretical physics and Goldbach To what extent is the statement Goldbachs conjecture is unprovable a scientific observation about the cosmos? It is clearly verifiable, and falsifiable because mathematicians might one day prove Goldbachs conjecture, so the remark satisfies at least some of the criteria for a full-blown theory of physics. The statement might be as fundamental a property of reality as the unobservable behavior of sub-quarks, and with equally important consequences to the nature of the cosmos. BSP === Subject: Re: PROOF that 0.99999... = 1 >> Not quite rigorous enough, but pretty close. One first has >> to make sure 3 * (3*10^-1 + 3*10^-2 + ... + 3*10^(-n) + ...) >> makes sense. >Do you accept 1/3 = (3*10^-1 + 3*10^-2 + ... + 3*10^(-n) + ...) ? It >pretty much follows that the series converges.... >Anyway, if you REALLY want proof, the series is Cauchy therefore its >OK. QED. Hah! >> Not that I expect S. Enterprize to understand much of this, as >> he seems to think ... = oo and have some very weird notions >> regarding series -- and no understanding at all regarding limits. >True... why did we even bother... >--D .444... = .444 + (oo repeating 4s) .999... = .999 + (oo repeating 9s) oo = ... = infinite digits to the right of the decimal Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >In sci.math, Dan Weiner >: >> [megasnip] >> 1/3 = .33333... = 3x10^-1+3x10^-2+3x10^-3+3x10^-4+3x10^-5+... >> thus: >> 3*1/3 = >> 3*(3x10^-1+3x10^-2+3x10^-3+3x10^-4+3x10^-5+...) >> which is: >> 9x10^-1+9x10^-2+9x10^-3+9x10^-4+9x10^-5+... = .99999... >> therefore .99999... = 3*1/3 = 3/3 = 1 >> [] >Not quite rigorous enough, but pretty close. One first has >to make sure 3 * (3*10^-1 + 3*10^-2 + ... + 3*10^(-n) + ...) >makes sense. >To illustrate, I shall prove the nonsense result below: >Let x = 1 + 2 + 4 + 8 + ... + 2^n + ... >Then 2 * x = 2 * (1 + 2 + 4 + 8 + ... + 2^n + ...) > = 2 + 4 + 8 + ... + 2^(n+1) + ... > = x - 1 >Since 2 * x = x - 1, x = -1. QED. >The main ßaw of course is the divergent series x. This >ones obvious but there are some subtle ones such as >the alternate-sign harmonic series that arent. >Fortunately, Cauchy proved that, under certain conditions >(which are met by your series, but not by mine :-) ) the >distribution of the product does make sense. >One can go the long way, if one wishes; define >S3_n = .333...3 (n 3s) = (3 * 10^-1 + ... + 3 * 10^-n). >Multiplying S3_n by 3 is perfectly valid, and leads to >S9_n = .999...9 = (9 * 10^-1 + ... + 9 * 10^-n). >As n tends to the limit, one can easily prove that S3_n -> S3 = 1/3, >and therefore S9_n -> S9 = 3 * S3 = 3/3 = 1. >Not that I expect S. Enterprize to understand much of this, as >he seems to think ... = oo and have some very weird notions >regarding series -- and no understanding at all regarding limits. >#191, ewill3@earthlink.net >Its still legal to go .sigless. (1.151253165325315326525...) * 1/1.151253165325315326525... =/= .999... 7/7 = 1 7/7 =/= .999... (999...)/(999...) =/= .999... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >[megasnip] >1/3 = .33333... = 3x10^-1+3x10^-2+3x10^-3+3x10^-4+3x10^-5+... >thus: >3*1/3 = >3*(3x10^-1+3x10^-2+3x10^-3+3x10^-4+3x10^-5+...) >which is: >9x10^-1+9x10^-2+9x10^-3+9x10^-4+9x10^-5+... = .99999... >therefore .99999... = 3*1/3 = 3/3 = 1 4/4 = 1 also 4/4 =/= .999... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >In sci.math, S. Enterprize Company > Its still useless. >> let x = 2 >> 10( 2) = 9 + 2 >> 20 =/= 11 Congratulations! Youve proven that x=2 is not a root of >the above equation. Care to try another number, though? :-) [.sigsnip] -- >#191, ewill3@earthlink.net >Its still legal to go .sigless. >> Ok, lets look at it in this point of view. >> 10x = 9 + x >> x can equal 1 or .999... . >> There are two roots to the equation. >> x1 = 1 >> x2 = .999... >> but, >> x1 =/= x2 >If x1 != x2, is 10*x1 - 9 - x1 == 10*x2 - 9 - x2? >> x1 =/= x2 >> Two different roots. >Oh, so an equation of polynomial degree n (n=1 in this case) >can have more than n roots? >[rest snipped] >-- >#191, ewill3@earthlink.net >Its still legal to go .sigless. Theres an exception to the rule I guess. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. >Would you point out where he makes this assumption? I repeat the entire >proof, expanded a bit, with line numbers added for your convenience: >[1] 9=9 >[2] 9=9.9999999...-.9999999... >[3] 9(1)=10(.9999999...)-.9999999 right here >[4] Let x = .9999999... and substitute in [3] >[5] 9(1) = 10x - x right here >[6] 9(1) = 9(x) and right here >[7] Therefore x=1 >In which line is the assumption .99999... = 1 used? >--Mark Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving correct? . > And in fact these are the only continuous solutions on R. Ive been considering the case where f is not assumed continuous. I havent yet been able to construct an example that \ isnt f(x) = x or f(x) = -x/2, but havent been able to prove \ theyre the only solutions. I know f(0) = 0, and with h(x) = x + 2 f(x), f(h(x)) = h(x) for all x. There are many other restrictions on f(x) and the related function g(x) = f(x) - x such as periodicity and so forth. Also, if f(x) is continuous anywhere, then it is continuous everywhere. - Tim === Subject: Re: More on groups of order 16 > I am still having some trouble with 1 of the 14 groups of order > 16. It is the one with 2 generators G = xyxy = 1 ; xy^3 = yx^3 > Thinking about this some more, I came up with a more direct > approach. Since |G| = 16, || = || = 4 and ^ = 1, > we have that G = , i.e., every element of G can be > expressed as a power of x times a power of y, in that order. So > lets see if we can find a way to multiply any \ two such > elements, (x^k y^l)(x^m y^n), using the given presentation. What > we need is a way to Ôpull x^m through y^l, \ or alternatively to > Ôpush y^l through x^m. One obvious way to do \ this is to see if > we can determine the conjugation action of any power of y on any > power of x, and of course we can (if not, the group \ wouldnt > have order 16). Since xyxy = 1 then yxy^-1 = x^3 y^2, and since > xy^3 = yx^3 then y x^3 y^-1 = xy^2, so y^2 x y^-2 = > y (x^3 y^2) y^-1 = xy^2 y^2 = x, so y^2 commutes with x and thus > y^3 acts the same way on x that y does. Also, y x^2 y^-1 = > (yxy^-1)^2 = (x^3 y^2)^2 = x^2 (since y^2 commutes with x), so y > commutes with x^2, so y and y^3 act the same way on x^3 that > they do on x. Now that we know how to multiply any two elements in G = , > its straightforward to write down all the conjugacy classes and > cyclic subgroups, determine the complete subgroup structure, and > see that G is in fact (Z_2 x Z_2) x| Z_4. Jim, I am going thru this now, but what do you mean by ^ = > 1? > / or intersection [...] > {y^i x^m y^(-i) | i,m in Z_4} what you mean by ^ ? > And how can you say this is 1? > not the intersection, but the action of Y = on X = and . I thought context would be sufficient to disambiguate the two usages of the caret character. Sorry for the confusion. (I guess Ill just go back to / , which is what I used until recently.) > Putting this aside, the rest is true. > If yxy^(-1) = x = y^2 x^3, then y^2 x y^2 = = y^4 x^9 = \ x, > and y^2 is in the center Z. So is x^2, so Z = . Correct. > Also, one can write any element of G in the form x^i y^j, with i,j in > Z_4. > I hesitate to post this, as I have yet to work it out clearly, > but apart from the ^ = 1 statement, which I dont see, > I think the rest of what is said is true. > I too would like to say more about what this statement means: > G is in fact (Z_2 x Z_2) x| Z_4. > Usually this means that (Z_2 x Z_2) is a normal subgroup, and > that Z_4 is acting by conjugation, is this not right? Yes, thats right. > The only normal subgroup of order 4 I have found is the center Z, > in which case the x| would reduce to the direct product, and we > would be talking about an Abelian group, which we arent. > I question if this is true; G = (Z_2 x Z_2) x| Z_4 > If it is, I dont understand in what sense it is true, \ i.e., > what the normal subgroup of G is, and what the action of Z_4 on > the normal subgroups giving the structure is. > I have been fiddling with this for a while and this is still about > all I can say about this group. You need to determine the subgroup structure of G. Hints: First write down every element, its conjugates, and the cyclic subgroup it generates. Next look for subgroups of order 8. One obvious place to start is =~ Z_4 x Z_2, which has only one subgroup of order 4 thats normal in G, so G has exactly three subgroups of order 8. What are the other two? Now, every subgroup of order 4 must be contained in a subgroup of order 8, so list them all of the former. Finally, which of the subgroups of order 4 are isomorphic to Z_2 x Z_2 and normal, but non-central, in G, and thus are candidates for the subnormal group in the semidirect product? And which subgroups of order 4 are isomorphic to Z_4 and intersect the subnormal group(s) trivially, and thus are candidates for the other factor in the semidirect product? -- Jim Heckman === Subject: Re: More on groups of order 16 <10sdfcrtmei0125@corp.supernews.com> posting-account=jcZk7AwAAADXpPEyHtVyWC264SxtppRB I also want to say that neither or are normal in G, Its surprising that we can still write g = x^i y^j for the elements of G. For if A,B < G, AB is a group if A or B is normal. But I guess that G = AB does not mean that A or B must be normal. Van === Subject: Re: More on groups of order 16 <10sdfcrtmei0125@corp.supernews.com> posting-account=jcZk7AwAAADXpPEyHtVyWC264SxtppRB > What does x| mean? > Its an ASCII rendition of the semidirect product sign (like a > multiplication sign, but with a vertical line joining the right-hand > extremities of the cross). > Information on semidirect products here: > http://en.wikipedia.org/wiki/Semidirect_product > http://planetmath.org/encyclopedia/ SemidirectProductOfGroups.html I have deleted my stupid posts. Of course ^ = / = intersect = 1 My only excuse is that I recently corresponded with someone who instisted on writing yxy^(-1) as x^y. Of course, w.r.t. sets or subgroups it has no meaning. I have always disliked ^ for intersection for that reason, even though I have used / both for exterior product and for intersection, here it is clearly intersection as exterior product would be commutator for which most use [x,y]. But enought of my excuses. Clearly the fog is starting to cover my mind if I make posts like that. I am ashamed. Van === Subject: Re: one integration question > whats the general form for any nu? > QQ >whats the closed form of this integration >int_{-infty}^{infty} |x-u|^nu e^{-|x|^nu} dx >I cant get it >>According to Maple, for nu=1 you get: >>2*u+2*exp(-u) >>for nu=2 you get: >>(u^2+1/2)*sqrt(Pi) >>and for nu=3 you get: >>+3/4*u^4*exp(-1/2*u^3)*WhittakerM(1/6,2/3,u^3)/(u^3)^(1/6) >>+u*exp(-1/2*u^3)*WhittakerM(7/6,2/3,u^3)/(u^3)^(1/6) >>-9/10*u^4*exp(-1/2*u^3)*WhittakerM(1/3, 5/6, u^3)/(u^3)^(1/3) >>-3/2*u*exp(-1/2*u^3)*WhittakerM(4/3,5/6,u^3)/(u^3)^(1/3) >>-1/4*u*exp(-1/2*u^3)*WhittakerM(1/6,2/3,u^3)/(u^3)^(1/6) >>-- >>G. A. Edgar > http://www.math.ohio-state.edu/~edgar/ === Subject: Re: 1 -1/2 + 1/3..... Let g_n=1 + 1/2 + 1/3 + ... + 1/n - ln(n) This converges to Euler constant C = 0.552... (the value does not matter, we need that does converge; try to prove... Hint: monotone and bounded) Then your sequence l_n = 1 - 1/2 + 1/3 - ... + (-1)^n/n: l_{2n} = g_{2n} + ln(2n) - 2 (1/2 + 1/4 + ... + 1/(2n) )= g_{2n} - g_n + ln(2n) - ln(n) -> ln(2) l_{2n+1}= l_{2n}-1/(2n+1) -> ln(2) Since l_{2n} and l_{2n+1} converge to the same number l_n converges also and to the same number (try to prove this in details) Vitaliy > Hello > Id like some clues to prove that the alternating series Sum (n=1, > oo) ((-1)^n)/n)) converges to Ln(2). > Since 1/n -> 0 as n -> oo and 1/n is strictly decreasing, the series > does converge. We know that, for every real x satisfying |x| <1, > Taylors theorem implies that Ln can be expanded about 0, so that > Ln(x) =x - x^2/2 + x^3/3.... Since [0,a] is compact, for every 0 < a < > 1 this power series converges uniformly on [0,a] to Ln, but, even > though the series converges for x =1, this doesnt mean \ the > convergence is uniform on [0,1). If this were true, then 1 -1/2 + 1/3 > ... would converge to lim (x ->1) Ln(1+x) = Ln(2), but this argument > failed. > Ana === Subject: Re: 1 -1/2 + 1/3..... > Id like some clues to prove that the alternating series Sum (n=1, > oo) ((-1)^n)/n)) converges to Ln(2). > Since 1/n -> 0 as n -> oo and 1/n is strictly decreasing, the series > does converge. We know that, for every real x satisfying |x| <1, > Taylors theorem implies that Ln can be expanded about 0, so that > Ln(x) =x - x^2/2 + x^3/3.... Since [0,a] is compact, for every 0 < a < > 1 this power series converges uniformly on [0,a] to Ln, but, even > though the series converges for x =1, this doesnt mean \ the > convergence is uniform on [0,1). If this were true, then 1 -1/2 + 1/3 > ... would converge to lim (x ->1) Ln(1+x) = Ln(2), but this argument > failed. > If you define ln(1+x) as the integral of 1/(1+x) and develop 1/(1+x) > around 1, then differentiate... what then differentiate... means exactly). === Subject: Re: 1 -1/2 + 1/3..... posting-account=pwtqtw0AAACw59dt7zOnp9M9tGBpr6Lp Ana: > prove that the alternating series > Sum (n=1,oo) ((-1)^n)/n)) converges > to Ln(2). Divide [1;2] into n equal intervals (of length 1/n). Ln(2) is the integral of 1/x from 1 to 2, and it is approximated by the n-th partial sum: Sum( (1/(1+k/n))*(1/n) : k=1...n) = Sum(1/(n+k) : k=1...n) = Sum((-1)^(k+1)/k : k=1...(2*n)) --> Ln(2) for n --> oo. Wlod (Wlodzimierz Holsztynski) === Subject: Re: 1 -1/2 + 1/3..... >> | f(1) - c | <= | f(1) - f(x) | + | f(x) - s_n(x) | >Given e > 0 do you see how to make the right side small? You dont follow this?? Me neither. I must have been asleep when I --Lynn === Subject: Re: Spherical trigonometry question. >How can I divide a sphere into N equal and identical parts? > I start with the premise that the surface can be divided by an >arbitrary number N, and is enclosed by P points on the surface of the >sphere. I want to know the formula for the points, the angle of >separation and distribution. > One method I am considering is to assume that each point is >positively charged, and that if left to ßoat on the surface, the >points will reach an equilibrium being equidistant from each other. > It wont work, unless N = 1, 2, 3, 4, 6, 8, 12 or 20. \ There arent enough > regular polyhedra. You will reach an equilibrium (at least a local > minimum of the potential energy, but probably not a global one if N is > large), but the parts wont be identical. It wont work for 8 either. Phil -- The gun is good. The penis is evil... Go forth and kill. === Subject: Re: Is zero even or odd? >I know 0 is neither negative or positive but what about odd/even? I think > its even. And what about 2? Since its the only even prime, \ its definitely odd. Tim === Subject: Re: Is zero even or odd? >>Lim{x->0}[(-x^)2] = 0 >Either you mangled your parentheses, or this is a insidious way of >sneaking in a smilie. >> Nothing is mangled if you view plain text. >Perhaps you could explain what (-x^)2 means then? Thats easy, x^ is the so called Erd.9as constant. Later it was found that x^=0. Thomas === Subject: Re: Is zero even or odd? posting-account=YEgZ2gwAAABJXWwDrJ38p9qyq9A1Zi2G > I know 0 is neither negative or positive but what about odd/even? I think > its even. > Odd numbers start at 1 and go every other number 1,3,5,7;1,-1,-3,-5,-7 > Even starts at 2 and go every other number 2,4,6,8;2,0,-2,-4,-6,-8 > The usual definition of an even integer (non-integers are neither even > nor odd) is that N is an even integer iff n = 2*m for some integer m. > Is 0 = 2*m for some integer m? For convenience 0 must be even.As 1 is not a prime number. An odd number leaves a remaider when divided by 2. Zero doesnt. === Subject: Quantum Albebra by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLCoki02603; >QUANTUM ALGEBRA >TODAY, QUANTUM physicists don\.89t know the structure of the electron. Computation fails. It took qualitative mathematics to know it. Recall that the electron is a left primum that induces a vortex ßux of superstrings to form a magnetic field of left polarity. A right primum has right polarity. The poles of a primum are extremities of its cylindrical eye, a region of low pressure. Therefore, they suck, de-agitate and accumulate superstrings inside it. Their agitation and conversion to kinetic energy \.9a prima or photons \.9a provide the awesome power of nuclear explosion. > I apply qualitative mathematics to primal interaction. By ßux compatibility, two prima of opposite toroidal ßux spins attract and two prima of the same ßux spin repel at their equators. However, prima of the same torodal ßux spin, although repulsive at their equators, are joined by a primum of opposite toroidal ßux spin called connector. Quantum algebra determines the charge of coupled or composite prima (algebraic sum of component charges). > Being neutral, the neutrino cannot be simple primum; it is equatorially coupled pair of prima of opposite but numerically equal charges, say, +q and -q. Therefore, the sum is 0 and there is no coherent ßux around it. > The proton is a pair of right quarks joined equatorially by a left primum as connector. Their centers are collinear, by energy conservation, since the two right prima are repulsive. Therefore, the charge of the proton is +2/3 -1/3 +2/3 = +1. It has counterclockwise net ßux around it. The neutron is a pair of right quarks joined by an electron and a left quark between them. Being of the same charge, these connectors are far as apart as possible. The coherent ßuxes at the center between the electron and left quark and the two right quarks sucks a neutrino (being neutral), by ßux-low-pressure complementarity. Therefore, the neutron\.89s charge is: 2/3 -1/3 +2/3 -1 +0 = 0. > The photon is simple primum that has broken away from its loop when scooped up by a basic cosmic wave (propagated by vibration of atomic nuclei). Dark viscosity ßattens the photon\.89s cycles; it becomes sinusoidal. As long as its forward toroidal ßux speed equals the speed of its carrier basic cosmic wave, the photon is stable. Otherwise, it breaks and its toroidal ßux remains non-agitated superstring (dark). Thus, the photon has no rest mass. > These prima have known masses: 1.674754 x 1/1024 grams = 1.0087 atomic mass units (amu); Proton: 1.672648 x 1/1024 grams = 1.0073 amu; electron: 0.0009109535 x 1/1024 grams = 0.0005486 amu. Note that the proton is 1836 times as massive as the electron. > When the electron gets suitably close to its anti-matter they collide and convert and project each other into two photons in opposite directions (I shall discuss anti-matter interaction later). Thus, photon from electron has the mass of the electron because it carries its toroidal ßux (true of photons from other prima). > Physicists previously assumed the neutrino had 0 mass. Detecting it is quite difficult. They place a pool of water about two miles underground. When neutrino strikes the pool, bubble forms and that requires energy. Therefore, the neutrino has mass but that would be difficult to find without a theory. However, the few natural laws I have discovered suffice to calculate it. > The same laws allowed me to determine the neutron\.89s component prima and how they are joined together. It consists of a proton, an electron and a neutrino. Since the masses of the neutron, proton and electron are known I calculate the mass of the neutrino: 0.0013515 x 1/1024 grams = 0.0008514 amu obtained by subtracting from the neutron\.89s mass the sum of the masses of the electron and proton. Comparing it with the mass of the electron, it turns out to be heavier, that is, 1.55 times the electron\.89s mass. One can see here the power of a theory. Earlier, we saw how the search for the basic constituent shifted away from the atom since it is not there. The shift had the benefit of insights from the laws of nature. Huge resources were spent in that futile search. I eventually discovered it in dark matter at the cost of a few minutes of reßection under the guidance of the laws of nature. The prima were discovered at no material cost; only the right scientific orientation, a train! e! >d mind and analysis of data provided by the Hubble >E. E. Escultura >University of the Philippines Sorry for the ommision. The subject is Quantum Algebra. === Subject: Re: Set theory question from beginner by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLCokX02595; >All, >Just started learning set theory, so this is a question that arises >from profound ignorance. My apologies in advance. >Given the (disjoint) sets A, B, C in omega: >___________ >| ___ | >| | | | >| | A | | >| |___| | >| ___ | >| | | | >| | B | | >| |___| | >| ___ | >| | | | >| | C | | >| |___| | >|___________| >Isnt the expression ABC ambiguous as \ written, since >(AB)C= null >whereas >A(BC)= C >??? >-Euge Since BC=B,we have A(BC)=null === Subject: Re: State-of-the-Art in Physics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLCoke02590; > THE STATE-OF-THE-ART IN PHYSICS There is similarity and distinction between mathematics and physics. > They are similar in the sense that they both study the structure, > properties and behavior of their respective subject matter. The > difference lies in their subject matter: that of mathematics is the > representation of thought, making it a language, and that of physics > is nature, making it a science. Both suffer from similar defects: > mathematical spaces and their concepts are ill-defined and nature and > physical concepts are also ill-defined. > Thats one of the reasons The Smart Model Theory, was developed. >>Physics >is full of ill-defined things like, electron orbitals, and electron change >>of >orbital to show various lines on the line spectra, and uncertainty etc... . > The Smart Model has accurately predicted the structures now observed from >real atomic images, like for example oxygen. The Smart Model predicted >>oxygen >was a ring. QM said it was a sphere. Now they observe oxygen rings on, PdO. >>It >looks like this from real atomic images showing oxygen attached to Pd like >>this >on the outer layers of Pd, > ooooooooooooooooooooo > oooooo oooooooooooooo > oooooooooooo > ooooooo ooooooooo > oooooooooooooo > ooooooooooooooooo >oooooooooooooooooo >Smarts Alt. Physics News Group >ht tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv =1http://smart1234. s-enterprize.com /basic constituent of matter? >>E. E. Escultura >Fields, in an intelligent logic way to keep the proton perpetually moving for >an upper mean life of about 10^30 years. >with Helix Spiral Spinning Fields, > CW spin > ( ) sub-proton > ( ) > ((((( ))))) > (((((((((( )))))))))) Neutral Zone > ((((( ))))) made of 3 quarks > ( ) > ( ) sub-anti-proton > CCW spin > This is the internal view. The external view is another Helix Spiral >frequency. When additional energy is absorbed, a photon is released within the >Neutral zone. The Neutral Zone prevents the matter and anti-matter from >colliding. The quarks oscillate producing a constant line spectra. This >constant resonance also keeps the mass constant, which is just a high >sub-atomic pressure zone in the Neutral Zone. > You may ask what keeps the proton spinning. The EM fields produced within >high pressure neutral zone sets up a sub-atomic resistance. The quarks sets up >a sub-atomic capacitance, and the main internal Helix Spiral Field sets up a >sub-atomic inductance. So what you have is basically a RCL sub-atomic electric >oscillator circuit. The basic sub-atomic voltage is the potential difference >between the matter and anti-matter side which is for a proton at rest about 938 >MeV. > I made calculations comparing a mitochondria proto-motive potential with >this proton set up and I end up with the same basic current found in a >mitochondria, which is the power source for a cell and the RNA and DNA. > The oscillator field produces the same line spectra for hydrogen. The line >spectra is really the convergence of three oscillator interactions at the >Neutral Zone core. They produce certain lines when a cfft ( complex fast >fourier tranform) is taken for these oscillators. Using these oscillators fft, >the total line spectra can be produced for hydrogen. > And the same goes for helium and all the other elements. Except there is a >Periodic Table of Elements. > The Smart Model is really more accurate than the QM Model or the Standard >Model and is in agreement at all levels of analysis, including chemistry, math, >physics, classicle physics, stellar, galactic, and universal levels. > The Smart Model also has a unified universal \ field theory that unifies all >fields in the universe, not just the atomic level that Einstein was working on. >Smarts Alt. Physics News Group >ht tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv =1http://smart1234. s-enterprize.com / Please give us the central ideas of your theory. For example, (1) What is the basic constituent of matter and (2) prove that it is indestructibel . (3) Well-define the electron, proton and neutron. (4) What are the fundamental physical principles of your theory? E. E. Escultura === Subject: Re: MORE INFORMATION FROM THE NEW PHYSICS by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLCokZ02616; >> MORE INFORMATION FROM THE NEW PHYSICS >You know this is sci.math, not sci.physics? >-- >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 >Francis Wheen, _How Mumbo-Jumbo Conquered the World_ Physics is science; it belongs to sci-math === Subject: Re: The state-of-the-art in mathematics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLColH02624; E. E. Escultura >>i = sqrt(-1) = sqrt(1/-1) = 1/i = -i or i = -i >>... every first-semester-student of mathematics knows >>that the square root of a number is 2-valued. >>If the rest of your work is of the same >>depth<< ... >Yes, but what made you choose one over the other? Which axiom of the >real numbers support your claim that your choice is the right one? >The problem here is that the mapping sqrt is well-defined only on perfect square; therefore i is ill-defined and does not exist. >E. E. Escultura I did not choose anything! You choose ... namely at the 3rd equality sign. The square roots of -1 within C (the complex numbers) are the numbers i and -i. So already the equation i=sqrt(-1) is misleading, although most mathematicians I know use equations like that and treat them with the necessary caution. Thats the simple point where you go wrong - as has been pointed out by other posters already. H === Subject: Re: The state-of-the-art in mathematics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLColp02640; >If Im not mistaken, the square root is \ well-defined for all real >values, not just perfect squares, since the square root function is >just a representation of x^(1/2), which in turn can be represented in >terms of e^x, which in turn is just an infinite sum. >apologize if it isnt yours, Im not sure \ where the original thread >is), is that you state that i is in the set of real numbers, and use it >as a counterexample to show that the set of real numbers is faulty--but >i isnt a real number at all! It is a complex number and is not an >element of the set of reals, so your counterexample is irrelevant. Regarding i, we recall that this is supposed to be the solution of the equation x^2 + 1 = 0 in the real numbers which does not exist. Therefore,it is not well-defined. E. E. Escultura === Subject: Re: JSH: Somewhat puzzled > Now over a year ago I was talking about a paper at a math journal and > even at one point mentioned that the journal was the Southwest Journal > of Pure and Applied Mathematics. I was pleased to see little activity > over that on this newsgroup. > I should have known that didnt mean much. > Months went by and the journal told me they liked the paper and were > going to publish. > Someone posting about it on the newsgroup and some posters got together > to email the journal. > So some posters here were able to push Ioannis Argyros with a paper > that I guess gave him some wiggle room so that he thought he could get > away with yanking it. > I fixed several problems: > problem within the discipline of algebraic number theory. But you never stated what that problem *is*. > I covered the issue of the contradiction with what you get if you > believe that the ring of algebraic integers doesnt have problems, > covering the area where posters have claimed counterexamples. Except you didnt address the counterexamples at all. What you may have proven is that the algebraic integers dont have properties you expected them to have. Thats not a problem, just an interesting \ quirk. > And most importantly, I sent the paper to a top-ranked PRINT journal, > so that I dont have to worry about some ßaky editor telling me one > thing one day, and then doing something else another. > The paper is at the Annals in Princeton, having been accepted for > formal peer review. > However, I still see many of you behaving here as if you have no > concerns, and that puzzles me somewhat. > At best you may have a few months to keep playing these games, but it > could be all over tomorrow. > What makes you tick? How do you keep on waiting as if you can get > away? Ill apply Occams Razor: No one is worried. \ No one thinks theyve acted dishonestly. No one thinks the math theyve been teaching is wrong. Your detractors are acting in good faith with no ulterior motives. > Its MATHEMATICS and sure, you can lie about it if you wish, but that > doesnt change the mathematics. The editors at Princeton are top > mathematicians, and their reviewers are the best in the world. Occams Razor: no one here believes they are lying. > None of the various lies and strategies youve used on the newsgroup > over the years will work with them, and it doesnt matter if some of > you decide to email them like you did with that other journal. > So how do you go on? How will some of you reply to this post as if > youre safe? How will you keep moving for the next few weeks or months > until you are known for what you are, the story becomes huge all over > the world, and you have reporters at your schools asking you, why? > James Harris Occams Razor: we dont believe any of this is \ a matter of safety. We dont expect anything remarkable to happen. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Somewhat puzzled posting-account=ixf50QwAAABFW6BQ4fQR2F0f49XaC1bE | Let me qualify this, however. By the banana getting published | I mean that an actual, physical banana is shipped with every copy | of the issue. So let me get clear on this-- if sending someone a banana conveys a defamatory message about a third party, would publishing the banana in the Annals constitute libel or slander? Keith Ramsay === Subject: Re: JSH: Somewhat puzzled posting-account=uMDgiw0AAAANoAxJOs_DnrtdjhRMBFah > | Let me qualify this, however. By the banana getting published > | I mean that an actual, physical banana is shipped with every copy > | of the issue. > So let me get clear on this-- if sending someone a banana conveys > a defamatory message about a third party, would publishing the > banana in the Annals constitute libel or slander? Even contemplating it could be illegal. If were not careful to keep the press away from \ Jamess (clever pseudonym, James), many of the laws protecting the mathematical elite could be overturned. Think about it people. === Subject: Re: JSH: Somewhat puzzled posting-account=W2DCTA0AAAAlbhDMl3GrysSnPy1IK_7f > So there is some sense in which your work is more akin to a work of > mathematics than a banana is. > Furthermore, I believe that the probability of your work being > published in the Annals is *strictly greater* (yes > not >=) > than that of the banana getting published! > Let me qualify this, however. By the banana getting published > I mean that an actual, physical banana is shipped with every copy > of the issue. It would not suffice for the issue to have a Xerox > of the banana in it, for example. This latter scenario strikes me > as being roughly as likely as your work getting published there. > Both the Xerox of the banana and your work would be nice material > for an April Fools edition, for example: \ its hard to pick one as > better than the other. But an actual, physical banana: just think > of the distribution hassles that would entail! Why would anyone go > to the trouble? we are sorry to let you know that we will not be able to publish your banana in the Annals of Mathematics. We believe it may be more suited to a specialized journal. === Subject: Re: JSH: Somewhat puzzled posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW Princeton is about as clubby of a place as you could possibly submit some thing to, amongst all of the ivy league hangers-on. it doesnt exist to serve the public, of this or any other nation, other than its cruddy british/confederate meign -- its the place that gave us the schooling for The Birth of a Nation! you cant really be serious; its got to be some sort of meta-web-research into how long a guy calling himself Jimi can pull our legs. and counting! > doesnt change the mathematics. The editors at Princeton are top > mathematicians, and their reviewers are the best in the world. --4my of it? http://www.wlym.com === Subject: Re: JSH: Somewhat puzzled > Ah, right. I do recall the problem, but I never got the > definition of proper unit straight. David, are you *that* dense? It is not a proper unit, it is porperly a unit! Get with the program! === Subject: Re: JSH: Somewhat puzzled >> Ah, right. I do recall the problem, but I never got the >> definition of proper unit straight. >David, are you *that* dense? It is not a proper unit, it is porperly a >unit! Get with the program! Sorry. (Or maybe sorrily, or possibly srollily...) ************************ David C. Ullrich === Subject: Re: JSH: Somewhat puzzled > Ah, right. I do recall the problem, but I never got the > definition of proper unit straight. >>David, are you *that* dense? It is not a proper unit, it is porperly a >>unit! Get with the program! > Sorry. (Or maybe sorrily, or possibly srollily...) Its OK... I found it tricky at first too. You would think that becasue Ôunit is a noun, then \ Ôproper should be used, since it is an adjective. But it is properly a unit, and properly is an adverb mofdifyuing the verb Ôis as in is a unit. Work with it a while, and \ you will be OK. Perhaps James has found a fundamental problem with grammer as well as in algebra? === Subject: Re: JSH: Somewhat puzzled > Ah, right. I do recall the problem, but I never got the > definition of proper unit straight. >>David, are you *that* dense? It is not a proper unit, it is porperly a >>unit! Get with the program! > Sorry. (Or maybe sorrily, or possibly srollily...) > Its OK... I found it tricky at first too. You would think that becasue > Ôunit is a noun, then \ Ôproper should be used, since it is an adjective. > But it is properly a unit, and properly is an adverb mofdifyuing the verb > Ôis as in is a unit. Work with it a while, \ and you will be OK. > Perhaps James has found a fundamental problem with grammer as well as in > algebra? Or spelling, even? === Subject: Re: JSH: Somewhat puzzled posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY > Lets say theres a very simple argument, \ which quite logically shows > that some people made some mistakes in their mathematical reasoning. > The argument is basic, stepped out in extreme detail, and clearly > shows > that they made a mistake. > OK, fair enough. So lets suppose that Im \ extremely busy (which I > am), not a mathematician (which Im not), \ dont have time to look > through the archives of sci.math for your argument (which I dont), and > wouldnt be able to understand your highly technical argument anyway > (which I probably wouldnt). > Now, I have no stake in this argument that you seem to have with the > regulars at sci.math, Im just an innocent bystander, so would you > mind summarizing your key result for me? Ok. When considering polynomials, people usually solve for them, like if you have x^2 + 3x + 3 = 0, you want to know what x is. Or, they may factor them, like x^2 + 2x + 1 = (x+1)(x+1), and my work involves factoring polynomials, but not using polynomial factors. So at the heart of my work is a different way of looking at polynomials, which like much of my work is provably unique, while also being simple. That technique reveals a subtle problem in how others have looked at certain numbers for over a hundred years. For that reason, more than anything else, people argue with me. My work is revolutionary not only in that it changes the way people look at certain numbers, but it changes how you look at mathematicians over the last hundred years. Its like its a kick in the ego. How hard is it to verify my research? For a trained mathematician, its not hard. How much evidence supports my work? All of \ it, as I have example after example where the math shows Im right, versus nothing for people arguing with me. Think about the evidence that should make sense to you, even if youre not a mathematician, like the paper of mine that was pulled after some sci.mathers got together as a group to email the journal claiming it was in error: Supposedly a math paper that goes through a journals publication process is immune to people pressure, right? Yet despite having my paper for over nine months and assuring me that they would behave fairly, the editors yanked my paper IMMEDIATELY after sci.mathers started emailing them, and didnt even give me a chance to defend. Thats FEAR on a huge scale. The results I have are that \ huge. Think about what itd take for a journal editor to behave that way. How much social pressure must have been felt. Check the link. Consider what it takes for a paper to get yanked in such a way, for a journal to behave in that way, and try to find another SINGLE CASE in mathematical history where such a thing has occurred. Yet here on sci.math people will talk like its nothing, as if it happens every day because theyre in la-la land. Their \ denial is so great that they just give explanations that dont make \ sense. Mainly, no matter how detailed I go into proof, or how much I elaborate, posters will just disagree with me, even when theyre disagreeing with the most basic concepts in algebra itself. For me the exercise of arguing out my research is more of a way to make sure that it is absolutely correct and that I didnt miss anything. So now that Im satisfied Ive \ gone to a major journal. James Harris === Subject: Re: JSH: Somewhat puzzled posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn > Lets say theres a very simple argument, \ which quite logically > shows > that some people made some mistakes in their mathematical > reasoning. > The argument is basic, stepped out in extreme detail, and clearly > shows > that they made a mistake. OK, fair enough. So lets suppose that Im \ extremely busy (which I > am), not a mathematician (which Im not), \ dont have time to look > through the archives of sci.math for your argument (which I dont), > and > wouldnt be able to understand your highly technical argument anyway > (which I probably wouldnt). > Now, I have no stake in this argument that you seem to have with the > regulars at sci.math, Im just an innocent bystander, so would you > mind summarizing your key result for me? > Ok. When considering polynomials, people usually solve for them, like > if you have x^2 + 3x + 3 = 0, you want to know what x is. Or, they may > factor them, like x^2 + 2x + 1 = (x+1)(x+1), and my work involves > factoring polynomials, but not using polynomial factors. > So at the heart of my work is a different way of looking at > polynomials, which like much of my work is provably unique, while also > being simple. > That technique reveals a subtle problem in how others have looked at > certain numbers for over a hundred years. Mr. Loggins asked a straightforward question: summarize your key result. Your response is, I factor polynomials using nonpolynomial factors, and that reveals a Ôsubtle \ problem. You call that an answer ??? Here, Ill make it easy for you. Consider your favorite polynomial, P(x) = 65 x^3 - 12 x + 1. Assume P(x) is factored in the form P(x) = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1), where a_1, a_2, and a_3 are algebraic integers. Using an argument based on an incorrect application of your own definition of constant term, you claim that two of these coefficients MUST be divisible by sqrt(5). The third coefficient must therefore be coprime to 5 in the algebraic integers. This result contradicts elementary applications of (1) algebraic number theory, and (2) Galois theory. You now accept the fact that all three of a_1, a_2, and a_3 share nonunit factors with 5. You interpret this contradiction to mean that there is some kind of deficiency in the ring of algebraic integers. That is, you think a_1/sqrt(5) should be an algebraic integer, but it is not. You think this deficiency can be addressed by going to a larger ring - a ring which contains the algebraic integers, but one in which the only integers which are units are 1 and -1. Is that a fair summary? > For that reason, more than anything else, people argue with me. The reason people argue with you is because your math is wrong. > My work is revolutionary not only in that it changes the way people > look at certain numbers, but it changes how you look at mathematicians > over the last hundred years. Lets examine the claim that one of your big original ideas is non- polynomial factorization of polynomials. Consider f(x) = a x^2 + b x + c. Quadratic polynomial in x. You can factor it using the quadratic formula: f(x) = (x - (-b + sqrt(b^2 - 4 ac))/2a)(x - (-b - sqrt(b^2 - 4 ac))/2a). Wow, look at that. Those factors are not polynomials in the coefficients! In particular, sqrt(b^2 - 4 ac)/2a is not a polynomial in a, b, or c. Amazing! Imagine that, factoring a polynomial with nonpolynomial factors! Lets see, this may not be a real new idea, eh? How long has the quadratic formula been around? Now, you may say, Yes, but what I did was different! It wasnt. You were factoring your function P(m) as if it were a polynomial in x. The coefficients of the factorization were nonpolynomial functions of the coefficients of P(m). \ These coefficients were functions of m. What you were doing was exactly the same as what I just did using the quadratic formula, just a little more complicated. You called it P(m) rather than f(x). Otherwise, completely analogous. The idea that the coefficients of a factorization are complicated (nonpolynomial) functions of the coefficients of the original polynomial is NOT NEW. Moreover, what you have done with that idea is NOT REVOLUTIONARY, NOT HUGE, NOT EARTH-SHATTERING, and in fact, NOT EVEN INTERESTING. > Its like its a kick in the ego. > How hard is it to verify my research? Since it is wrong, it cannot be verified. However, it is easy to refute it with counterexamples and counterproofs. This has been done many times. > For a trained mathematician, > its not hard. Name ONE trained mathematician who has EVER stated that your results were correct. Just ONE! > How much evidence supports my work? None. Its wrong. > All of it, as I > have example after example where the math shows Im right, versus > nothing for people arguing with me. This is quite simply and out-and-out lie. > Think about the evidence that should make sense to you, even if youre > not a mathematician, like the paper of mine that was pulled after some > sci.mathers got together as a group to email the journal claiming it > was in error: > Supposedly a math paper that goes through a journals publication > process is immune to people pressure, right? Where do you get that dumb idea? If an incorrect result gets published and someone finds the error, the NORMAL thing to do is to write the editor. The editor may then take several actions: (1) publish the letter and write to the author to see if he wants to publish a rebuttal; (2) suggest to the author that he/she retract the incorrect results in the paper; (3) publish an acknowledgement that the results in the paper are incorrect, with or without the authors consent. Until recently, actually physically withdrawing the paper was not an option. That has changed with the advent of electronic journals. > Yet despite having my > paper for over nine months and assuring me that they would behave > fairly, the editors yanked my paper IMMEDIATELY after sci.mathers > started emailing them, and didnt even give me a chance to defend. > Thats FEAR on a huge scale. The results I have are that huge. It wasnt fear at all. Because of appalling sloppiness in \ the way that journal was managed, the paper got in the to be published file rather than in the reject file. Just to \ check: did you ever see a referees report *before* the paper was \ published? You are aware, of course, that in addition to the major central error in the paper, there were other low-level errors which were pointed out to you *over a year* after the paper was written. Remember that? Do you actually think any reviewer ever actually READ that paper? If they did, why did they overlook those simple errors? Not to mention the major ones. You yourself have said that the editor *did* send you a referees report. But you also said it was an excerpt of a letter from Dale Hall to the editor, written AFTER the paper was published! What that should tell you is the following: (1) there was never any review prior to the publication; (2) if the editor really told you that was a referees report, he was a liar as well as a coward. The whole thing was just an incredible editorial screw-up. The editor should have openly and honestly explained what happened and apologized to you. > Think about what itd take for a journal editor to behave that way. > How much social pressure must have been felt. The only role social pressure played was the following. The editor took the cowardly way out - to avoid embarrassing himself, he simply yanked the paper with no explanation, no apology, nothing. And he got away with it. Its his journal and he is still editor. Most of his readers have no idea what happened. He has gotten away with it at your expense. > Check the link. Consider what it takes for a paper to get yanked in > such a way, for a journal to behave in that way, and try to find > another SINGLE CASE in mathematical history where such a thing has > occurred. > Yet here on sci.math people will talk like its nothing, \ as if it > happens every day because theyre in la-la land. Their denial is so > great that they just give explanations that dont make sense. They whole thing was a fiasco. There is \ significant danger that electronic journals are going to degrade the whole business of peer review. There are already totally unedited, unreviewed electronic journals. This one supposedly goes through the motions, but the content and quality level is quite bad. The papers published in it are barely edited if at all. Its a \ deplorable trend. > Mainly, no matter how detailed I go into proof, or how much I > elaborate, posters will just disagree with me, even when theyre > disagreeing with the most basic concepts in algebra itself. This is a continuation of the lie started above. You never really answer rigorous proofs that your work is wrong. You skip over the hard parts, call your critics liars, start a new thread, and play to some imaginary grandstand of people who dont know much math. Just like youre doing now. > For me the exercise of arguing out my research is more of a way to make > sure that it is absolutely correct and that I didnt miss anything. > So now that Im satisfied Ive \ gone to a major journal. And will you accept their decision? Nora B. > James Harris === Subject: Re: JSH: Somewhat puzzled Discussion, linux) > The only role social pressure played was the following. The > editor took the cowardly way out - to avoid embarrassing himself, > he simply yanked the paper with no explanation, no apology, nothing. > And he got away with it. Its his journal and he is still editor. > Most of his readers have no idea what happened. He has gotten away > with it at your expense. It was recently reported that the journal has folded due to budgetary constraints[1]. Got away with it my butt. I wouldnt be surprised if the editor has been disappeared. Boy, are you naive or what? Footnotes: [1] See . -- Come on people!!! The US just blew up a lot of people in Iraq, dont you realize that a person with my exposure might just end up dead, by mysterious circumstances? --James Harris, on the dangers of proving Fermats last theorem === Subject: Re: JSH: Somewhat puzzled <877jncch8i.fsf@phiwumbda.org> posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn Its part of the Conspiracy. First he publishes the Harris paper, then he yanks it (discrediting Harris), then he uses the methods to break RSA codes and blackmail the Russian government. He is now basking in the sun at his villa on the Black Sea. Meanwhile Harris is reduced to selling used hammers, dog food, and dioxin pellets in a shabby little kiosk outside the CNN tower in Atlanta. N.B. === Subject: Re: JSH: Somewhat puzzled Discussion, linux) >> For a trained mathematician, >> its not hard. > Name ONE trained mathematician who has EVER stated that your > results were correct. Just ONE! Fear, baby, fear. His results are just that HUGE. Duh. -- Jesse F. Hughes When you try to kiss a girl, its hard not to get spit on \ the girl. -- Quincy P. Hughes, age 3 (almost 4) === Subject: Re: JSH: Somewhat puzzled > Now, I have no stake in this argument that you seem to have with the > regulars at sci.math, Im just an innocent bystander, so would you > mind summarizing your key result for me? > Ok. You say, OK. But apparently you misunderstood his question. He wasnt asking you to affirm how \ Ômagnificent you are, or how Ôrevolutionary your work is. He was asking you to summarize it -- this implies he is interested in explanations and *content*, not subjective evaluations and megalomaniac diatribes. > When considering polynomials, people usually solve for them, like > if you have x^2 + 3x + 3 = 0, you want to know what x is. Or, they may > factor them, like x^2 + 2x + 1 = (x+1)(x+1), and my work involves > factoring polynomials, but not using polynomial factors. > So at the heart of my work is a different way of looking at > polynomials, which like much of my work is provably unique, while also > being simple. If its so simple, why not explain it? Nowhere in this post do you do so. > That technique reveals a subtle problem in how others have looked at > certain numbers for over a hundred years. What technique? > For that reason, more than anything else, people argue with me. Nope. They argue with you because you are not only provably wrong, hence incompetent, but are also dumber than a sack of hammers. (Get it? Hammer....) > My work is revolutionary not only in that it changes the way people > look at certain numbers, but it changes how you look at mathematicians > over the last hundred years. So far it does not seem to have changed anything -- especially the way I look at mathematicians. But what numbers are you referring to? You have repeatedly claimed that you have discovered numbers which *should be* algebraic numbers, but which are not. In spite of dozens of requests, even pleadings, you have never produced a single such number. Tsk, tsk... For something so simple, it should have been done long ago. > Its like its a kick in the ego. If this technobabble of yours is Ôa kick in the \ ego, go for it. But any respectable troll gets as many kicks. > How hard is it to verify my research? For a trained mathematician, > its not hard. How much evidence supports my work? All of it, as I > have example after example where the math shows Im right, versus > nothing for people arguing with me. Right about what? That the ring of algebraic integers is ßawed. Then what, exactly, is the ßaw? If its not hard to verify your so-called research, why not answer the posters question? Zzzzzz.... -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Somewhat puzzled > How hard is it to verify my research? I think it has been done. -- email: lastname at cs utk edu homepage: www cs utk edu tilde lastname === Subject: Re: JSH: Somewhat puzzled > there are numbers that are properly units of the algebraic > integers without being algebraic integers. That doesnt make any sense to me. Is there a difference between units and properly units? If so, whats the difference? If theres no difference, then: The word unit is defined in reference to a particular ring. A unit of that ring is an element of that ring x such that x*y=1 has a solution where 1 is the multiplicative identity of that ring and y is any element of the ring. For example, in the ring of ordinary integers, -1 is a unit because (-1)*(-1)=1, but 1/2 is not a unit because its not even in the ring. Something cant be a unit of a ring unless its in the ring in the first place! So if you are talking about the ring of algebraic integers, its impossible for anything to be a \ unit of the algebraic integers if its not an algebraic integer in the first place. properly units is different from units, or if youre talking about units of some *other* ring, not the ring of algebraic integers, then you need to explain. > take (1/2) which is properly a unit in the integers ... No, 1/2 isnt even in the integers in the first \ place, so it cant be mumble in the integers whatever mumble is supposed to mean. Please tell us what the phrase <> is supposed to mean, and then maybe well be able to \ figure out a better way of expressing the concept that doesnt appear blatantly self-contradictory. === Subject: Re: JSH: Somewhat puzzled I wonder how come there was no reactions yet to the extremely pertinent ) Reminds me, by the way, of remarks by Borges (in The Library of Babel : http://jubal.westnet.com/hyperdiscordia/library_of_babel.html ) about the possibility that a book be also a ladder (http://jubal.westnet.com/hyperdiscordia/library_of_ babel.html#footnote3) ... === Subject: Re: Ring problem, integral elements by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLDZcP06863; >The obtained polynomial does not work for the case x=y=-1 and a=1 (in >the ring of complex numbers) that is, in this case >f(1-1)=f(0)=/=0 Yes, of course. I should have noticed that in the universal E = Z[x, y, a]/(x^5 = -a = y^5), is not an integral domain (so that my ensuing calculations were a bit fast and loose). It would be clearer to say we have an embedding of polynomial rings, Z[a] --> Z[x]: a |--> -x^5 (adjoin a fifth root of -a) and then to consider the extension of Z[x]: E = Z[x, y]/(x^5 - y^5) where x^5 - y^5 = (x - y)g(x, y). Now D = Z[x, y]/(g(x, y)) is an integral domain, and if we find a monic polynomial h(u) with coefficients in Z[a] which is zero in D for u = x - y, i.e., if h(x-y) = g(x, y)k(x, y) in Z[x, y], then (x-y)h(x-y) = (x^5 - y^5)k(x, y) in Z[x, y], so that uh(u) is a monic polynomial over Z[a] which is zero for u = x - y in E. Then uh(u) is the desired monic polynomial for your problem. Now the solution I outlined before (which is carried out in Q(x)[w] where w is a fifth root of unity) should in principle work to find this h(u), although I havent found \ time to recheck my calculations for arithmetic mistakes. Just now Im a bit busy with holiday preparations :-) Todd Trimble >also for x= 1/4+1/4*5^(1/2)-1/4*I*2^(1/2)*(5-5^(1/2))^(1/2) >and y=1/4+1/4*5^(1/2)+1/4*I*2^(1/2)*(5-5^(1/2))^(1/2) >(I=sqrt(-1)) >and where a=1 (in fact x and y are two distinct fifth root of unity) >we have >f(x-y)=-2069375/2+910525/2*5^(1/2)=/=0 >These show that, maybe somthing is wrong in your calculations, but >perhaps I can find out it ...........??? >Anyway thank you very much for your information. >Alireza >>Excuse-me in my question R is not assumed to be commutative. and >the >>correct (complete) question is the following (I think!!): >>Suppose that R is a ring with identity such that the centre Z(R) of >>is a field. Assume that x and y are two commuting >>elements of R (i.e. xy=yx) such that >>x^5+a=0 >>y^5+a=0 >>for some a in Z(R). In fact x and y are integral over >>Z(R). >>Question: Construct a non-zero polynomial f with coeficceint in >Z(R) >>such that f(x-y)=0. >> We may restrict attention to the commutative ring Z(R)[x, y], >> and solve the problem universally by replacing Z(R) by Z[a], >> a an indeterminate (Z the integers), and Z(R)[x, y] by the >extension >> E = Z[x, y, a]/(x^5 = -a = y^5) over Z[a]. >> If we find a monic polynomial for x - y with \ coefficients in >> Z[a], then we are done (since we can just apply some obvious >> rings homomorphisms Z[a] --> Z(R), E --> R to the equation >> f(x-y) = 0 in E). >> Obviously y/x is a fifth root of unity (denoted w) over the >> field Q(a), so >> (x - y)^5 = a(w - 1)^5 >> and the problem reduces to finding a monic polynomial (over Z) >> for (w - 1)^5. This is routine if a bit tedious; one works >> over the complex numbers and calculates the coefficients of >> (u - r1)(u - r2)(u - r3)(u - r4) >> to the fact that the ri are pure imaginary and come in conjugate >> pairs; my back-of-envelope calculations give the last polynomial >> as >> u^4 + 3936u^2 + 3125 >> Substituting u = (x - y)^5/a and clearing denominators, I get >> (x - y)^20 + 3936a^2(x - y)^10 + 3125a^4 = 0. >> Todd Trimble === Subject: Re: Need help on my educational website > Hi. Im building a nonprofit educational \ website with the wikimedia > software to serve everyone in the world who would like to see example > problems and solutions for all kinds of mathematics. Please, for the > good of everyone taking a math class next semester, contribute one or > two problems and solutions to the site. Follow the format Ive laid > www.exampleproblems.com > -Todd Smith > UCF, Mathematics It looks like a good idea. Could you add a section for arithmatic as well? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Need help on my educational website posting-account=W9drAQ0AAAAwnUCtDicSi-THtOHRPCY8 Im trying to put in a couple of calculus problems on your site, but I cant figure out how to make integrals in your \ format. Could you post, either here on or you site a brief list of how to get these and other basic functions to look right? A === Subject: Re: Need help on my educational website posting-account=LSltBg0AAACI7ITc0n2uQOmc7-4a5kwQ > Im trying to put in a couple of calculus problems on your site, but I > cant figure out how to make integrals in your \ format. Could you post, > either here on or you site a brief list of how to get these and other > basic functions to look right? It works the same as TeX. I usually go to a page on wikipedia like this: http://en.wikipedia.org/wiki/Partial_derivative or this http://en.wikipedia.org/wiki/Integral to see how they did it. heres some TeX symbols: http://www.astro.uiuc.edu/~bima/proposal/TeXsymbols.html And heres how you did it: http://www.exampleproblems.com/wiki/index.php?title=Calc1.1 === Subject: Re: algebra, geometry, trigonometry books > Hi everyone, > I am looking for a book that has many algebra, geometry and trigonometry > problems. The books, that I have found, have very easy problems. I am > looking for a book that has many challenging problems as well. Please, let > me know if you know any good math book that has many practice problems. Consider some older books e.g. Hobson: Trigonometry Hall and Knight: Higher Algebra Chrystal: Algebra (2 vols I think) You may have to go to abebooks to find these used http://www.abebooks.com The above are not exact titles, do author/keyword search. Jim Buddenhagen === Subject: Re: algebra, geometry, trigonometry books >> Hi everyone, >> I am looking for a book that has many algebra, geometry and trigonometry >> problems. The books, that I have found, have very easy problems. I am >> looking for a book that has many challenging problems as well. Please, >> let >> me know if you know any good math book that has many practice problems. > Consider some older books e.g. > Hobson: Trigonometry > Hall and Knight: Higher Algebra > Chrystal: Algebra (2 vols I think) Then theres Basic Algebra I & II by Jacobson :=) DM === Subject: Re: algebra, geometry, trigonometry books > Hi everyone, > I am looking for a book that has many algebra, geometry and trigonometry > problems. The books, that I have found, have very easy problems. I am > looking for a book that has many challenging problems as well. Please, let > me know if you know any good math book that has many practice problems. Are you looking for yourself? (Im guessing you are a fully grown adult.) For my children, I got the books from the Gelfand Correspondence Series, published by Birkhauser (or Dover). http://gcpm.rutgers.edu/books.html Those books are excellent. I first learned about them in a book review by Professor Richard Askey. I also got the books The Art of Problem Solving by Richard Rusczyk and Sandor Lehoczky. http://www.artofproblemsolving.com/Books/AoPS_B_About.php The problems in the Art of Problem Solving books are mostly competition problems for young people. The books also include exposition about all the main topics of secondary school mathematics. The problems are MUCH more challenging than those I have found in any other English-language textbook for young people. I think the problems will be challenging to most adults. Hope this helps! Please tell us more about why you are looking. -- Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 Learn in Freedom (TM) http://learninfreedom.org/ remove .de to email === Subject: Trigonometry and the Moebius function posting-account=oTYsyw0AAACT9cOtqRsOt-ohbN8pStcb Hi all, In the webpage http://en.wikipedia.org/wiki/Uses_of_trigonometry#Number_ theory there is an interesting statement under the Number Theory heading. The claim sum{x is not a divisor of 42, from x=1 to 41} cos(2*pi*x/42) the sum is equal to the Moebius function evaluated for 42. I have read as much as I can on the Moebius function and the Moebius inversion formula, but I cannot see how this statement is derived. Could anyone help me? The Brush Turkey === Subject: Combinatorial design? posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs I need to construct N x N blocks such that each number from 1 to N occurs exactly once in each row and each column. I have this vague memory that such constructions fall under the keyword mentioned in the subject line, but I dont have a combinatorics book at hand. Are there well-known algorithms for constructing such arrangements? How many are there? One thought I had was to make the trivial one: 1 2 3 ... N 2 3 ...N 1 3 ..N 1 2 . . . N 1 2 ...N-1 and them permute the rows and columns. Will that generate all possible legitimate arrangements? - Randy === Subject: Re: Combinatorial design? |I need to construct N x N blocks such that each number |from 1 to N occurs exactly once in each row and each |column. I have this vague memory that such constructions |fall under the keyword mentioned in the subject line, |but I dont have a combinatorics book at hand. These are known as latin squares. You should be able to find quite a bit about them knowing what \ theyre called. |Are there well-known algorithms for constructing such |arrangements? I assume so. |How many are there? The number is Sloans sequence A002860. One of the order 11 there are. Im assuming that theres \ no known nice formula for how many there are. Let X=Y=Z={1,...,N}. A latin square can be thought of as a function f:XxY->Z, but a function can be thought of as a special kind of relation, r on X, Y, and Z, which brings out a symmetry in the definition: a relation r on three sets of order N is a latin square if each pair in XxY, XxZ, and YxZ occurs once, i.e., its a function from XxY to Z, from XxZ to Y, and YxZ to X. The sets X, Y, and Z are called the constraints. |One thought I had |was to make the trivial one: | |1 2 3 ... N |2 3 ...N 1 |3 ..N 1 2 |. |. |. |N 1 2 ...N-1 | |and them permute the rows and columns. Will that generate |all possible legitimate arrangements? No. Two are considered isomorphic if one can obtain one from the other by permuting rows, columns, and symbols. Thinking of it as a relation, two are isomorphic if one can be obtained from the other by applying permutations to the constraints. (The six permutations of the constraints with each other are also sometimes relevant....) Heres a latin square not isomorphic to the trivial one: 1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1. The permutations needed to obtain one row from another in your trivial one are the elements of a cyclic group of order n. For this one, they form a group isomorphic to a Klein 4-group; each has order 2. In general, the multiplication table of a group is a Latin square. Not all latin squares are isomorphic to group tables, though. For example, 12345 21453 35124 43512 54231. Keith Ramsay === Subject: Re: Combinatorial design? >I need to construct N x N blocks such that each number >from 1 to N occurs exactly once in each row and each >column. I have this vague memory that such constructions >fall under the keyword mentioned in the subject line, >but I dont have a combinatorics book at hand. >Are there well-known algorithms for constructing such >arrangements? How many are there? One thought I had >was to make the trivial one: >1 2 3 ... N >2 3 ...N 1 >3 ..N 1 2 >N 1 2 ...N-1 >and them permute the rows and columns. Will that generate >all possible legitimate arrangements? No. Permuting the rows and columns, and even permuting the labels inside, keeps one in the same isotopy class. As soon as N=4, there is more than one isotopy class. What you have written is the multiplication table of a cyclic group. For groups, isotopy is isomorphism, but for quasigroups, it is not. The keyword is correct. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Combinatorial design? posting-account=oTYsyw0AAACT9cOtqRsOt-ohbN8pStcb No, it wont generate all possible arrangements. For \ example, consider 2 1 3 ... N, where I just interchanged the first two elements of the \ first row. In general, there will be N! combinations, so not all arrangements will fit into one NxN matrix. The Brush Turkey === Subject: Re: Combinatorial design? posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs > No, it wont generate all possible arrangements. For example, consider > 2 1 3 ... N, > where I just interchanged the first two elements of the \ first row. In > general, there will be N! combinations, so not all arrangements will > fit into one NxN matrix. > The Brush Turkey I think perhaps I didnt explain my problem very well. every number from 1 to N occurs once in each row and each column. If I permute the rows or the columns, the block will still have this property. There are as many as (N!)^2 such blocks reachable by permutation of rows and columns. If you just interchange the first two elements of the first row, then the number 1 now occurs twice in column 2, and the number 2 occurs twice in column 1. This is not a valid block. Id like to know how many blocks there are with the property, and if there is a good way to generate them, especially if there are any not reachable by permutation of my original block. - Randy === Subject: Re: Combinatorial design? posting-account=oTYsyw0AAACT9cOtqRsOt-ohbN8pStcb > every number from 1 to N occurs once in each row and each > column. Sorry. I missed the and each column part in your original post. Forget everything I said. > If I permute the rows or the columns, the block will still > have this property. There are as many as (N!)^2 such blocks > reachable by permutation of rows and columns. > If you just interchange the first two elements of the > first row, then the number 1 now occurs twice in > column 2, and the number 2 occurs twice in column 1. > This is not a valid block. Indeed. > Id like to know how many blocks there are with the > property, and if there is a good way to generate them, > especially if there are any not reachable by permutation > of my original block. I dont know a good algorithm offhand. The Brush Turkey === Subject: Re: Spherical trigonometry question. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLG0vr20590; >There is a trivial way to divide a sphere into N equal and identical >parts, but Im afraid it will not be satisfying to you. >Draw N semicircular arcs from one pole to the other of the sphere and >cut along these lines. >Of course these pieces are not triangles, as each will have two edges >and two vertices. Chip, I know of a good paper which discusses, among other things, generating uniformly distributed points on the sphere. Is your e-mail adress real? If so, I will forward a copy of the paper to you. - MO === Subject: Re: WAR IMMINENT betweeen Mathematicians and Physicists! >Identically? Not unless youre talking Quaternions, which you hadnt >previously[1] mentioned. >[1] You brought them up after the text I quoted. Well, yes. I figured that the people reading my post would be familiar with quaternions, and would know which j and k I was talking about. This reminds me... I had thought to replace the E in numbers like 1.35E-2 (which stands for 1.35 * (10 ^ (-2))) with # for easier parsing of a computer language, and where # would be a standard dyadic operator; a # b would mean a * (10 ^ b) even for a and b both real. Continuing that trend, I thought to use ! as a unary operator, not meaning not (~ would do that: bitwise when applied to INTEGER variables, logical when applied to LOGICAL variables) but instead meaning i times. So complex numbers would also be standard expressions; 3+!5 standing for 3 + 5i. Unary * would be the complex conjugate, although prefix instead of postfix, not the sign function used in APL for unary multiply. For j and k, one would use an extended character set, including Spanish punctuation marks. (Or, of course, compound symbols: unlike those of C, my hypothetical language strictly obeyed the prefix property.) John Savard John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Trigonometry and the Moebius function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLGtvJ26983; >Hi all, >In the webpage >http ://en.wikipedia.org/wiki/Uses_of_trigonometry#Number_theory there >is an interesting statement under the Number Theory heading. The claim >sum{x is not a divisor of 42, from x=1 to 41} cos(2*pi*x/42) >the sum is equal to the Moebius function evaluated for 42. I have read >as much as I can on the Moebius function and the Moebius inversion >formula, but I cannot see how this statement is derived. Could anyone >help me? >The Brush Turkey Let g(d) = sum of primitive d-th roots of unity (which are of the form exp(2k*pi*i/d) where k is relatively prime to d, 1 <= k <= d). Then f(n) = Sum_{d|n} g(d) = Sum of all n-th roots of unity = 1 if n = 1 = 0 otherwise. Apply Moebius inversion to get g(d) = mu(d). Taking real parts in the case d = 42, mu(42) = Sum_{k} cos(2k*pi*i/42) summing over k between 1 and 42, k relatively prime to 42. every such k is a nondivisor of 42. But it is the same as which have no common factor with 42, which is the same as saying k is relatively prime to 42. Todd Trimble === Subject: Re: Trigonometry and the Moebius function posting-account=oTYsyw0AAACT9cOtqRsOt-ohbN8pStcb [snip derivation] > summing over k between 1 and 42, k relatively prime to 42. > every such k is a nondivisor of 42. Im not terribly experienced in number theory, but would \ that be the other way around? That is, not every nondivisor of 42 is relatively prime to 42? > But it is the same as > which have no common factor with 42, which is the same as > saying k is relatively prime to 42. divisor so often it has become automatic, even when I am thinking of relative primality. But your telling me this explicitly will hopefully jolt my head into using better terminology. The Brush Turkey === Subject: Re: Trigonometry and the Moebius function > summing over k between 1 and 42, k relatively prime to 42. > every such k is a nondivisor of 42. > Im not terribly experienced in number theory, but would that be the > other way around? That is, not every nondivisor of 42 is relatively > prime to 42? 40 is not a divisor of 42, but 40 is not relatively prime to 42. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: matrix inverse of the sum of two inverse matrices >-The most general problem I need to solve is to find the inverse of the >sum of the inverses of these 2 matrices. How do you find the inverse of 1/a+1/b where a and b are real numbers? >-In a slightly more restricted problem, one of the matrices is diagonal >in Ôreal space, and the other >is diagonal in frequency space (i.e., depends only on |i-j|, where i,j >are indices over real space. Thats an interesting restriction, but even in the 3x3 case it does not seem to make the problem much less complicated than inverting a general 3x3 matrix. (In higher dimensions, of course, there has to be _some_ simplification -- your sums lie in a 2N-dimensional subspace of an N^2-dimensional space.) >-In an even more restricted case (this is the one where I really hope >to get some help), all the elements >of one of the two matrices are multiplied by a very small number, i.e., >Inv(C) = a_i*delta_ij + epsilon*b_ij >I can assume b_ij=f(|i-j|), so this matrix is diagonal in frequency >space, epsilon is very small and I would like to know C. There must be >some kind of power expansion one can do... Well, sure. You want to invert M = A + B with A diagonal; it suffices to be able to invert M = A^{-1} M = I \ + A^{-1} B, and that one can do with a power-series expansion: (I+X)^{-1} = I - X + X^2 - ... In your case X = A^{-1} B will be small because B is, so this series will converge (quickly, since epsilon is very small). Of course this X does not have the special structures of either A or B. (As noted above, its not a _completely_ structureless matrix, but I dont think you can get much mileage out of the structure that it does have.) Im guessing these remarks are not very helpful to you but I am not optimistic that there _is_ a good answer here. dave === Subject: Re: Intersection Puzzle/Grid-Solitaire Game posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE > You start with an 8-by-8 grid, which is 9-by-9 lines, lightly drawn on > paper. > Start on any corner of the grid. > You draw darker lines along the light lines from vertex to vertex > (vertex = intersection of light lines), alternating drawing horizontal > and vertical lines, not drawing any darker lines where darker lines > already exist. > You can make your line any length in one direction (up, down, left, or > right) as long as the line goes from the end of the last line to a > vertex of the grid. > No darker lines should coincide, except where the darker lines > intersect (and continue past each other). Your path of connected darker > lines should not even meet itself at any of its corners (meeting at a > single vertex). > You get a point every time an intersection is Ôperpendicular to the > last intersection made (AS the darker lines are being drawn). > (And you get a point for your first intersection.) > By Ôperpendicular intersection, I mean that \ if your previous > intersection was a horizontal newer dark line crossing a vertical older > dark line, then your current intersection is a vertical newer dark line > crossing a horizontal older dark line, and vice versa. > Your path of connected darker lines starts and stops at the same > corner. > I got 13 points. Can you get 13 points or better? > I also got 14 points by drawing a completely different path. Can you > get 14 points or better? > Example of perpendicular crossings: > (View with fixed-width font.) > ----+ +---+ > ! ! ! > +---+---+---+ > ! ! ! > +---+ ! > Example of non-perpendicular crossings: > ----+ > +---+---+ > ! ! ! > +---+ ! > ---+---+ > ! ! > +---+ > Leroy Quet On Google at least, the above ascii art is messed up, both on sci.math and rec.puzzles (but messed up in different ways, depending on the group). Here is the ascii art again with periods replacing the spaces. HOPEFULLY the art comes out right this time on most newsreaders and Usenet portals. View with fixed-width font, of course. Example of perpendicular crossings: .----+...+---+ .....!...!...! .+---+---+---+ .!...!...!.... .+---+...!.... Example of non-perpendicular crossings: .----+........ .....!........ .+---+---+.... .!...!...!.... .+---+...!.... .........!.... ......---+---+ .........!...! .........+---+ Leroy Quet === Subject: Re: Intersection Puzzle/Grid-Solitaire Game posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE I apologize for reposting the post below, but MathForum at least seems to not be having posts appear if they were made to multiple groups from Google. So I will repost this only to sci.math in hope it does not get censored. ---- > You start with an 8-by-8 grid, which is 9-by-9 lines, lightly drawn > on > paper. > Start on any corner of the grid. > You draw darker lines along the light lines from vertex to vertex > (vertex = intersection of light lines), alternating drawing > horizontal > and vertical lines, not drawing any darker lines where darker lines > already exist. > You can make your line any length in one direction (up, down, left, > or > right) as long as the line goes from the end of the last line to a > vertex of the grid. > No darker lines should coincide, except where the darker lines > intersect (and continue past each other). Your path of connected > darker > lines should not even meet itself at any of its corners (meeting at a > single vertex). > You get a point every time an intersection is Ôperpendicular to the > last intersection made (AS the darker lines are being drawn). > (And you get a point for your first intersection.) > By Ôperpendicular intersection, I mean that \ if your previous > intersection was a horizontal newer dark line crossing a vertical > older > dark line, then your current intersection is a vertical newer dark > line > crossing a horizontal older dark line, and vice versa. > Your path of connected darker lines starts and stops at the same > corner. > I got 13 points. Can you get 13 points or better? > I also got 14 points by drawing a completely different path. Can you > get 14 points or better? > Example of perpendicular crossings: > (View with fixed-width font.) > ----+ +---+ > ! ! ! > +---+---+---+ > ! ! ! > +---+ ! > Example of non-perpendicular crossings: > ----+ > ! > +---+---+ > ! ! ! > +---+ ! > ! > ---+---+ > ! ! > +---+ > Leroy Quet > On Google at least, the above ascii art is messed up, both on sci.math > and rec.puzzles > (but messed up in different ways, depending on the group). > Here is the ascii art again with periods replacing the spaces. > HOPEFULLY the art comes out right this time on most newsreaders and > Usenet portals. > View with fixed-width font, of course. > Example of perpendicular crossings: > .----+...+---+ > .....!...!...! > .+---+---+---+ > .!...!...!.... > .+---+...!.... > Example of non-perpendicular crossings: > .----+........ > .....!........ > .+---+---+.... > .!...!...!.... > .+---+...!.... > .........!.... > ......---+---+ > .........!...! > .........+---+ > Leroy Quet === Subject: 2 Permutations Defining Themselves posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE I have recently submitted the following sequences to the EIS: >%S A000001 1,2,1,4,1,3,1,8,1,5,1,9,1,6,1,16,1,7,1,14,1,10,1,21,1,11,1 >%N A000001 a(2n-1) = 1; >a(2n) = a(n)th lowest positive integer not among the earlier terms of the >sequence. >%C A000001 The sequence {a(2k)} forms a permutation of the integers >= 2. >%e A000001 a(12) = the a(6)th (the 3rd) lowest positive integer not among >the first 11 terms of the sequence. Not among the \ first 11 terms are 6, 7, >9, 10,... >The 3rd of these is 9, which is a(12). >%O A000001 1 >%K A000001 ,easy,more,nonn, >%S A000001 1,2,4,3,7,9,6,5,13,15,19,17,11,14,10,8 >%N A000001 a(1)= 1; a(n) = a(2^ceiling(log_2(n))+1-n)th lowest positive >integer not yet in the sequence. >%C A000001 Sequence is a permutation of the positive integers. >2^ceiling(log_2(n)) +1 -n is sequence A080079 with a change of offset. >%e A000001 Since 2^ceiling(log_2(n)) +1 -n = 3 at n = 6, a(6) = the a(3)th >(the 4th) lowest positive integer not among the first 5 terms of the >sequence. >The positive integers not among the first 5 terms are 5,6,8,9,10,... >The 4th of these is 9, which is a(6). >%O A000001 1 >%K A000001 ,easy,more,nonn, Is there a direct (nonrecursive) way to calculate the terms of either sequence? Leroy Quet === Subject: Complex Analysis Problem The problem is, Let W = {x : Re z > 0} and let F = {f : f analytic in W, |f(z)|<1 for z in W, |f(1)| = 1/2}. Give the value of sup{ |f(1)| : f in F}. I use Cauchy integral forumla. Let C_r be the circle centered at 1 and radius r. Let f in F. Then, by Cauchy integral forumla f(z) = 1/(2 Pi i) int_{C_r} f(z)/(z-1)^2 dz Now, f(z) is bounded by 1. On W, C_r cannot have radius more than 1 and so, 1/(z-1)^2 is bounded by 1. So, I arrive at the conclusion, |f(1)| <= 1/(2 Pi) But the problem is, I never used |f(1)| = 1/2 for f in F. Is it like real functions where the derivative of the function does not depend on the value at the point? Or |f(z) = 1/2| limit on what f I can choose from f? === Subject: Re: Complex Analysis Problem > The problem is, > Let W = {x : Re z > 0} and let > F = {f : f analytic in W, |f(z)|<1 for z in W, |f(1)| = 1/2}. Sorry this should read z not x. Let W = {z : Re z > 0} and ... === Subject: Criss-Cross Grid Game (repost) posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE I apologize for reposting (yet again!) the post below, but MathForum at least seems to not be having posts appear if they were made to multiple groups from Google. So I will repost this only to sci.math in hope it does not get censored. ---- Here is yet another of my games played on an n-by-n grid drawn on paper. 2 players,each with a different colored pencil. Players alternate placing an X or an O of their color in the grids squares as follows: *Player 1 places the first O in the upper-left square. *Each player places their symbol immediately next to the symbol last drawn by the other player in the directions of vertically, horizontally, or diagonally. *If the square a player draws his symbol in is empty, he/she draws an O. O. *A player may not draw in (move to) a square already occupied by an X (and an O). Play continues until moving is impossible or when one player first gets a predetermined number of points. Scoring: *If a player places an X in the same square as an O, the player who did *not* draw the O in that square gets a point. *If a player places an O in an empty square diagonally adjacent to the last square drawn in by the other player, and the other 2 squares in the 4-by-4 square -- formed by the empty square being drawn in and the last drawn-in square -- are both filled, then the player Ôcompleting the square (the player whose move it is) gets a point. Example: +---+---+ ! O ! ! +---+---+ !OX ! O ! +---+---+ Last move put an X in lower left square (getting a point for the player who did *not* place the O there earlier). Player now wants to place an O in upper right square, which would get him/her a point. What is a good strategy for this game? Leroy Quet === Subject: radicals radicals radicals Let a,b be in Q ( set of rational numbers), b is not equal to zero. Let x be in R ( set of real numbers) x ^ (a/b) = x ^ ( 2 * a/(2b)) = (x ^ 2) ^ (a/2b) = | x ^ (a/b) | for example, let x = -27; a = 1; b = 3; (-27) ^ (1/3) = (-27) ^ ( 2 * 1/6) = ((-27) ^ 2) ^ (1/6) = 729 ^ (1/6) = 3; But (-27) ^ (1/3) = -3; there should be some mistake i did, suppose a,b,c,d be in Q such that b,d are not equal to zero; let x be in R; x ^ ( (a/b) * (c/d) ) is not always equal to ( x ^ ( a/b) ) ^ (c/d) is that true? in mathworld.com , http://mathworld.wolfram.com/SquareRoot.html , they talk about principal square root, but anyway thats only in the case of square root. === Subject: Re: radicals radicals radicals > x ^ ( (a/b) * (c/d) ) is not always equal to ( x ^ ( a/b) ) ^ (c/d) > is that true? Yes, it is not always equal. For x>0 it is OK. === Subject: Re: radicals radicals radicals http://mygate.mailgate.org/mynews/sci/sci.math/ 09d93424533352aa39ec405819afbc 35.48257%40mygate.mailgate.org > Let a,b be in Q ( set of rational numbers), b is > not equal to zero. > Let x be in R ( set of real numbers) > x ^ (a/b) = x ^ ( 2 * a/(2b)) = (x ^ 2) ^ (a/2b) = | x ^ (a/b) | Lets rewrite that for ease of annotation as: x ^ (a/b) = x ^ ( 2 * a/(2b)) = (x ^ 2) ^ (a/2b) = | x ^ (a/b) | > for example, > let x = -27; a = 1; b = 3; > (-27) ^ (1/3) = (-27) ^ ( 2 * 1/6) = ((-27) ^ 2) ^ (1/6) = 729 ^ (1/6) = 3; > But (-27) ^ (1/3) = -3; > there should be some mistake i did, > suppose a,b,c,d be in Q such that b,d are not equal to zero; > let x be in R; > x ^ ( (a/b) * (c/d) ) is not always equal to ( x ^ ( a/b) ) ^ (c/d) > is that true? > in mathworld.com , > http://mathworld.wolfram.com/SquareRoot.html , > they talk about principal square root, but anyway > thats only in the case of square root. Well, no, thats in the case of _any_ even powered radical, and your mistake is that when you shifted from an odd powered radical to an even powered radical, you introduced an extraneous root, and then siezed on it as if it were the _only_ root. x ^ (a/b) = [Radicals arent this nicely behaved.] x ^ ( 2 * a/(2b)) = [By now, with your example x, a, and b, you have added an extraneous root +3.] (x ^ 2) ^ (a/2b) = [This is sheer magic on your part. You have assumed that because the left hand parenthesized expression is positive, then the whole expression is positive, and arbitrarily on that incorrect assumption, wrapped absolute value delimiters around the whole expression. You cant do that. The appropriate sixth root of (-27*-27) is still -3, not +3, so the value of the expression is still negative, not positive.] | x ^ (a/b) | Whenever working with radicals, you have to keep track of any extraneous roots you introduce by your manipulations, separately, and hand-eliminate them from the final result. In this case, instead, you eliminated the legitimate root, leaving only the extraneous one. When you see some startling result, it isnt because you have overturned the foundations of arithmetic, it is because you have made a mistake and not seen it yet. HTH xanthian. -- === Subject: Direction-Based Grid-Game (& Questions)(repost) posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE I apologize for reposting (yet again!) the post below, but MathForum at least seems to not be having posts appear if they were made to multiple groups from Google. So I will repost this only to sci.math in hope it does not get censored. ---- a question about a related integer sequence. ---- Here is yet another game of mine played on a grid drawn on paper. It is a somewhat unusual in that the game is a cooperative game, the 2 players get the same number of points. Start with an n-by-n grid drawn on paper, where n is odd (and I suggest n be at least 7, maybe much higher). Players alternatingly take turns filling in, on each move, an *empty* square immediately next to the last square filled in by the other player. Player 1 can only fill in an empty square immediately either to the left of, to the right of, above, or below the last square filled. Player 2 can only fill in an empty square diagonally adjacent to the last square filled. Fill in the center square. Player 1 then fills in a square above,below, left of, or right of the center square. Player 2 then fills in a square diagonally touching the previously filled square. Etc. Play continues until no more squares can be filled in. The score (for both players) is the number of squares filled in before play must stop. (And there should be very little communication between players.) If you insist on playing a competitive game, you can do so with 3 or more players. Just have each pair of players play a game, for a total (if m= number of players) of m(m-1)/2 games, and add every players particular scores to get a total score for each player. You can also play this game solitaire, where you alternate Ômoving non-diagonally (up, down,left,or right) \ and then diagonally; but I believe that perhaps playing with 2 people might be more fun, if only because you try to guess each others strategies. ---- Questions: Is it possible, for any odd n >= 3, to visit once EVERY square of an n-by-n grid (starting at the center square)? What is the maximum possible score, in theory, for a n-by-n grid? (How does the sequence start where the nth term is the maximum possible score for an n-by-n grid?) Leroy Quet === Subject: integrating (sin x)^(1/3) Can someone remind me how this indefinite integral is solved? === Subject: Group Theory Question posting-account= UsMTzhMAAAAITFFPlj9rUB2S3BtXg0OQGdaE3UVKRbgY8da85xMcNA I found this question hanging around this newsgroup in 1995, but no one seemed to post a solution. Let G be a group with no subgroups of index 2. Prove that every subgroup of index 3 is a normal subgroup. TIA, === Subject: Re: Group Theory Question days. My association with the Department is that of an alumnus. >I found this question hanging around this newsgroup in 1995, but no one >seemed to post a solution. >Let G be a group with no subgroups of index 2. Prove that every >subgroup of index 3 is a normal subgroup. Let H be a subgroup of index 3, and let G act on the left cosets by having g send aH to gaH. This gives a map from G to S_3. Since S_3 has a subgroup of index 2, the map cannot be surjective. The map can also not be trivial, because if g is any element not in H, then g does not act trivially on the coset H. And the image cannot be a subgroup of order 2, because then the kernel would be of index 2. Thus, the only possibility is for G to map onto a subgroup of S_3 of order 3. We claim that the kernel of this map is H. To see this, note that H acts trivially on the coset H, so it must either fix the other two cosets, or swap them. But swapping them would correspond to a permutation of order 2, and a subgroup of S_3 of order 3 has no such element. So H must act trivially on all three cosets. Therefore, H is contained in the kernel. Since the kernel is of index 3, same as H, that means that H equal the kernel, and in particular it must be normal. -- Its not denial. Im just very selective \ about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: The structure of the Universe posting-account=4v2pmg0AAABs-rnZhu2ZIbgA6Ot7XddN The structure of the Universe Contained in human chest The orderly and mutinous The motion and the rest That no one can explain Creates the ßying galaxies And all that they contain. The magnetism and gravity The forces weak and strong Eternity and brevity Combine into a song That rings through all the galaxies Through all of space and time And with inspired melodies Nurtures the human mind. The structure of the Universe Contained in human heart In each stop, a continuance In each end, a new start And as blood passes through the lungs With every passing breath And birth, from every death. The universal passion Contained in human brain - Recombinance and clashing Of Galilean frames Eternity in brevity - They pass each other by Reshaping with their gravity Each others space and time, Consigning to oblivion Each others sense of self And bringing the dominion Of wherefrom they all delve. The structure of the universe Contained in human eye The protons and the neutrons dressed In cloud that whizzes by That wont give you the answer How fast they ßy or where, But like a ballet dancer Command and fix the stare - That soar dispelling fear Where nobody can find And are the see-through mirror Into the human mind. The structure of the Universe Contained in human soul The ever-present superstrings Connecting each to all Containing like a hologram The whole in each part, Incorporating all of them In universal heart - The ultimate affinity Of each and every soul Approaches infinity And merges with the whole That manifests through timespaces In fractal twists and swirls The curvatures and chaoses Contained in peoples souls. The electromagnetic field Within the human voice Contains the power people wield In mechanism of choice That shape, create and structure The world in which they live, Make wholesome or else rupture Engender or deceive Contain the inspiration For newer better worlds, Or, lacking integration, Remain just empty words. Ill find the truth in sunshine And tease it out of night Ill feel it with my passion And know it with my mind Combining and refining, Destroying what is wrong Explaining and defining And crafting a new song - A song that rings through timespace With message that is new, That solves each passing crisis And makes the dreams come true, Through neutrons and through protons And electronic fields Through forces, weak and strong ones, And through the superstrings, Through universal whole And through each precious part Will manifest mans soul Contained within Gods heart. Ilya Shambat. === Subject: The structure of the Universe posting-account=4v2pmg0AAABs-rnZhu2ZIbgA6Ot7XddN The structure of the Universe Contained in human chest The orderly and mutinous The motion and the rest That no one can explain Creates the ßying galaxies And all that they contain. The magnetism and gravity The forces weak and strong Eternity and brevity Combine into a song That rings through all the galaxies Through all of space and time And with inspired melodies Nurtures the human mind. The structure of the Universe Contained in human heart In each stop, a continuance In each end, a new start And as blood passes through the lungs With every passing breath And birth, from every death. The universal passion Contained in human brain - Recombinance and clashing Of Galilean frames Eternity in brevity - They pass each other by Reshaping with their gravity Each others space and time, Consigning to oblivion Each others sense of self And bringing the dominion Of wherefrom they all delve. The structure of the universe Contained in human eye The protons and the neutrons dressed In cloud that whizzes by That wont give you the answer How fast they ßy or where, But like a ballet dancer Command and fix the stare - That soar dispelling fear Where nobody can find And are the see-through mirror Into the human mind. The structure of the Universe Contained in human soul The ever-present superstrings Connecting each to all Containing like a hologram The whole in each part, Incorporating all of them In universal heart - The ultimate affinity Of each and every soul Approaches infinity And merges with the whole That manifests through timespaces In fractal twists and swirls The curvatures and chaoses Contained in peoples souls. The electromagnetic field Within the human voice Contains the power people wield In mechanism of choice That shape, create and structure The world in which they live, Make wholesome or else rupture Engender or deceive Contain the inspiration For newer better worlds, Or, lacking integration, Remain just empty words. Ill find the truth in sunshine And tease it out of night Ill feel it with my passion And know it with my mind Combining and refining, Destroying what is wrong Explaining and defining And crafting a new song - A song that rings through timespace With message that is new, That solves each passing crisis And makes the dreams come true, Through neutrons and through protons And electronic fields Through forces, weak and strong ones, And through the superstrings, Through universal whole And through each precious part Will manifest mans soul Contained within Gods heart. Ilya Shambat. === Subject: ? S-L problem in more vars Hi: Found in several books about the Sturm-Liouville problems, but they only talk the S-L problems in one variable. How about cases with more variables? by Cheng Cosine Dec/21/2k4 Ut === Subject: Re: Trigonometry and the Moebius function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBLJOp409379; >[snip derivation] >> summing over k between 1 and 42, k relatively prime to 42. >> every such k is a nondivisor of 42. >Im not terribly experienced in number theory, but would that be the >other way around? That is, not every nondivisor of 42 is relatively >prime to 42? Arghh. Yes. My bad. But it is the same as >> which have no common factor with 42, which is the same as >> saying k is relatively prime to 42. >divisor so often it has become automatic, even when I am thinking of >relative primality. But your telling me this explicitly will hopefully >jolt my head into using better terminology. >The Brush Turkey === Subject: Re: integrating (sin x)^(1/3) > Can someone remind me how this indefinite integral is \ solved? If you want an answer that can be written as a finite combination of elementary functions, you wont be able to do this. You can make the substitution z = (sin x)^(2/3) to transform this integral into int(3z/(2*sqrt(1 - z^3)) dz), but this is an elliptic integral. ________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817