Below is a copy/pasted post from a while back from me: (new reply bottom-posted) >> >The board of this game is a regular n-gon drawn (as well as >possible) on paper, n being whatever the 2 players decide. Each player uses a differently colored pencil and a straightedge to >take turns drawing 1 straight line per turn >between any 2 NONadjacent vertexes of the n-gon. >color), but vertexes can be used more than once. (I suggest that there >be a limit of 2 or 3 lines per vertex, as to make things more >interesting.) >After a fixed number, m, of moves per player, scoring is as follows: Player gets a point for each of her/his lines where the total number >of lines (of both players) crossing that particular line is an EVEN >integer. >(A shared vertex is not considered a line-interesection.) The winner is the player with the most points, as you would guess. If anyone decides to actually play this game, what values do you >suggest for m and n?? >(Possible m's are, of course, bounded above by a function of n and the >maximum number of lines which may share one vertex.) >> I have tried playing, solitarily, a few rounds of this game (left >> brain vs right brain...). >> >> A few suggestions: >> >> I think it might be best that, if n(k) = number of segments (drawn by >> either player) crossing the k_th segment (drawn by the player in >> question), then that player would get a point if n(k) is ODD, instead >> of even. >> >> I also think, for a more interesting game, that the number of sides, >> m, of the polygon should be higher, at least 12. (m is limited by the >> practical limitations of drawing all of this by hand, however.) >> >> I also suggest that there be a maximum of 2 segments (total, >> irrespective of which players drew the segments) that may be connected >> to any particular vertex. >> >> And I think that, in some ways, it would be best to have the number of >> moves per player (before ending play) equal to floor((m-1)/2) >> (ie. floor((m-1)/2)*2 total moves in a game), given a limit of 2 >> segments per vertex. >> This leaves one or two vertex-pairs unconnected at the game's end, >> which might have an effect on strategy (since the players cannot be >> absolutely certain that there will be, at the game's end, 2 segments >> connecting to each vertex that has not yet had 2 segments connected to >> it at some point during play). >> >> Leroy Quet >Please forgive my confusing lack of continuation with the uses of the >variable m (in 'm-gon' in the bottom) and the variable n (in 'n-gon' >in the first quoted post). >And then I use 'n(k)' to further confuse you all... Sorry... > I think a interesting variation might be to, instead of each player trying to get as many of their segments as possible having an even (or odd) number of crossings, just have one player win if there are an even number of *total* crossings, and the other player win if their are an odd number of crossings. (A 'crossing' is where 2 lines come together. For our purposes, if n {n = 3 or more} lines come together at a single point, this counts as n(n-1)/2 crossings.) One advantage of this new variation is that we can play this game with just one color of pencil/pen... (which seems to actually be an issue when playing the game in-practice...) In this variation there are no scores, just a winner and loser, which may be a disadvantage or an advantage, depending on your own opinion. thanks, Leroy Quet ==== >Prove or disprove: if two matrices commutes (AB=BA) then they can be >expressed with polynomials in terms of the same matrix P ( A=f(P), >B=g(P) ). What would P, f, and g be if A,B = [0 1 0 0] [0 0 1 0] [ ] [ ] [0 0 0 0] [0 0 0 1] [ ], [ ] [0 0 0 1] [0 0 0 0] [ ] [ ] [0 0 0 0] [0 0 0 0] By considering the potential Jordan canonical forms for P, and noting that A^2 = B^2 = 0, I think one can show there is no possible f and g no matter what P is. For example, if P were diagonalizable, f(P) and g(P) would be too (but A and B are not). dave ==== It's well known that if a real valued function f is monotonic on a compact interval I then f is integrable over I and the set of discontinuities of f on I is countable. I'm not sure, but it seems to me that if we combine this fact with the Fundamental Theorem of Integral Calculus (in the form that deals with the derivative of an integral)and consider Lebesgue Integrability Criterion, then we come to the conclusion that the set of elements of I at which f is not differentiable has measure zero.Is this conclusion true? Amanda ==== >It's well known that if a real valued function f is monotonic on a >compact interval I then f is integrable over I and the set of >discontinuities of f on I is countable. I'm not sure, but it seems to >me that if we combine this fact with the Fundamental Theorem of >Integral Calculus (in the form that deals with the derivative of an >integral)and consider Lebesgue Integrability Criterion, then we come >to the conclusion that the set of elements of I at which f is not >differentiable has measure zero.Is this conclusion true? The _conclusion_ is indeed true - a monotonic function is differentiable except on a set of measure zero. But I don't follow how you think you've proved this at all. (if you can give a proof that's that simple you can publish it. But you're going to need to explain the argument a little more carefully first...) >Amanda ************************ David C. Ullrich ==== > i looked into my drawer where i keep all my written notes, and i saw > and it was about large numbers. and i copied down a sentence definition from > amount of finite numbers; the number ÁˇbeforeÁ± infinity (Á[CapitalThorn]). now i find > this very interesting. can anyone confirm it to me? there is also another > number called the End Number want to know about it? > Your special symbols came across garbled in my newsreader. Can you rephrase using ASCII please? In any event there is no last finite number. Given any natural number N, the number N+1 is also a natural number (that's one of the Peano axioms) and N < N+1, so N isn't the largest number. Since N was arbitrary, we can conclude that there is no largest natural number. What you may have heard in school is that there are infinite numbers, two flavors of infinite numbers in fact, cardinals and ordinals. Aleph-0 (note use of ASCII rather than an attempt to render the Hebrew letter aleph) is the smallest infinite cardinal. The ordinal that describes the entire ordered set of natural numbers is lower-case omega, usually denoted 'w' in ASCII. ==== If you have the assistance of some people to take a 2 minute quiz and get some sample data this should be an easy exercise suitable for a small statistics question on determining causality. Just treat it like a question although the data is real. The hypothesis is that when people reply to me on usenet, I can more often than anyone else spot their alias *by what they write*. These are 3 small posts that all occured on the same day. See if you can tell which author was responsible for each post. If you can then it agrees with my claim since they are all replies to me on the same day, so evidenctly there is no data mining. i.e. if you can guess the author of the 1st post from 100 names then that shows better than 100 to 1 againts chance that the correlation is not random, 3 running on the same say I'm hoping someone can determine a confidence value for this data that the associations are *THERE*, not random or typical from writing style issues. Have a quick go yourself, spoiler at the end. ----1----------------------------------------------------------------------- ---- Randi will test you when you properly apply to be tested. Sign up here: http://www.randi.org/research/challenge.html ----2----------------------------------------------------------------------- ----- It really all depends on the situation. ----3----------------------------------------------------------------------- ----- If ever I actually found myself in that situation, I'd hold it upright, with the intent of attacking my assailant's knife hand. ----------------- A cliff86 B Rust C Shanx D NormDePloom E Rich Shewmaker F CNote G Jeff H See You In Hell My Friend. I Someone J Greg Neill spoiler ANSWERS E Rich Shewmaker 1 rich showmaker James Randi http://tinyurl.com/nd52 H See You In Hell My Friend. 2 It all depends http://tinyurl.com/nd53 B Rust 3 attacks metal http://tinyurl.com/nd56 Since I've lost most of you with claims of non classical nature, I will show the 1st Herc > Randi will test you when you properly apply to be tested. Sign up here: http://www.randi.org/research/challenge.html --Rich It should be evident there is something peculiar about this post! Noone will even state that, the whole of 100,000 people on usenet are all blind skeptics or wont speak up amongst the crowd of them. It *is* evident, several people have admitted its coincidental and some have selected his name Rich Shewmaker from a list of random names as the author of the post. James Randi offers $1,000,000 for proof of paranormal, you have to beat 1,000,000 to one odds on call. I have a degree in Computer Science with a strong mathematical background and I know these 3 posts are a million to one longshot being on the same day. Can anyone put a confidence interval on the *causality*? Can anyone go the other way and prove there is no correlation? I know for a fact given a small group of people to take the 'guess the name test' it would show over 10,000 to 1. And I know from general principles that not any person here can collate 3 posts that are all replies to them or anyone else on the same day that have this property to that degree. 10% of the million dollar prize goes to the 1st person who gets some recognition of On 02 02 2002 I posted PROOF OF PARANORMAL to a dozen newsgroups in the title, one year before. Herc -- www.StealthHostiing.com You rule Truman. http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah. Webmasters help the TRUEman by joining www.theBanner.net Current:1 Goal:1000 ---------------------------------------------------------------------------- ------ ==== > HI ALL! > > > > Can someone give an example of two sequences (X_n) and (Y_n) > such that: > > 00 00 00 > ---- ---- ---- > | | | | | | > / | X_n + Y_n | =/= / | X_n | + / | Y_n | > ---- ---- ---- > 1 1 1 > > > > David > I'll give a big clue: It does no good to consider the case that X and Y are both positive (or both negative) sequences. In that situation, mere convergence forces equality. The rest is pretty much trivial, and if you come up with any example that takes the clue into account, where both sides converge, I would almost be willing to wager you'd have an example of the inequality of the two sides. ==== > HI ALL! > > > > Can someone give an example of two sequences (X_n) and (Y_n) > such that: > > 00 00 00 > ---- ---- ---- > | | | | | | > / | X_n + Y_n | =/= / | X_n | + / | Y_n | > ---- ---- ---- > 1 1 1 > > When abs(x_n + y_n) <> abs(x_n) + abs(y_n) ? ==== I am reposting this from March: http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fdf .0303281748.762fa20d%40posting.google.com&rnum=7&prev= I am reposting because, given that my Mathematica is busted, I was wondering if someone could calculate numerically 1 of the 2 constants I describe below. (If the sums converge, which they do if the world is not insane...) Are either of the constants recognized by the Plouffe Inverter? I am also wondering again if the sequence mod 2 is in the EIS as what I had suspected. (see below) thanks, Leroy Quet Let A(1) = 1, a sequence with only a single element. Let A(m+1) = >{A(m), a(m)+1, A(m)}, a concatenation, where a(m) is the m_th term of the sequence A(m). ( A(m) has 2^m -1 elements. So a(m) is not the last element of A(m) >for m >= 2, it should be emphasized.) The sequence begins: 1,2,1,3,1,2,1,2,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,2,1,2,1,3,1,2,1,... >Noteworthy: The average of the terms approaches: 1 + sum{k=1 to oo} a(k)/2^(k+1) Again, a(k) is the k_th term of the original sequence, the sequence >whose terms we are calculating the average of the terms of... A closed-form for this sequence is, I think: If e(m) is a nonnegative integer such that: >2^e(m) is the highest power of 2 dividing m; then: a(m) is the number of e()'s where: e(e(e(...e(m)...))) = 0. -- >It might be more natural, for some purposes, to define A(1) = 0. >Then we will have the same sequence minus 1. (In this case, a(m) would be the number of e()'s needed to achieve an >odd integer, instead of 0.) And, in this case, the average of the terms will by 1 less, but so >will the a()'s. So the average, with a'(k) = a(k)-1, a(k) = original >a(k): 1/2 + sum{k=1 to oo} a'(k)/2^(k+1) -- I THINK that the a(k)-sequence, taken mod 2, might be likely to be the >absolute values of the first differences of the Thue-Morris sequence >according to the OEIS. (paraphrased title??) > PS: the sums are: 1 + sum{k=1 to oo} a(k)/2^(k+1) and/or 1/2 + sum{k=1 to oo} a'(k)/2^(k+1), which are the differing by exactly 1. ==== > 21002977+4. **** > 21002981+2. 21002983+4. almost prime quadruple???****. world record prime number+keno lucky nos. FROM 20990970 TO 21021000 (PRESS ESCAPE) > PROC SIFT / SIFT OUT FACTORS OF SIEVE > PROC PRT / PRINTOUT *twin*PRIME* SEQUENCE 20991049+4. 20991097+4. 20991239+2. 20991317+2. 20991319+4. 20991359+2. > ... 20995279+4. 20995409+2. 20995631+2. 20995717+4. > 20995781+2. 20995783+4. prime triple. > 20995939+4. 20996009+2.*** 40th known Mersenne prime exponent. 20996047+4. 20996299+4. 20996473+4. 20996891+2. > 20997101+2. 20997103+4. prime triple. 20997139+4. 20997311+2. 20997517+4. 20997521+2. prime triple 20997539+2. 20997913+4. 20997971+2. 20998139+2. 20998259+2. 20998529+2. > 20998727+2. 20998841+2. 20999261+2. 20999269+4. 20999357+2. 20999477+2. > 20999593+4. 20999659+4. 20999789+2. 20999939+2. 21000037+4. 21000139+4. > 21000191+2. 21000229+4. 21000599+2. 21000673+4. 21000737+2. 21000821+2. > 21000823+4. 21000851+2. > 21000907+4. 21000911+2.**** prime triple rearrange, 2001 Sept 11th. World Trade Centre.** 06.12.03.=3*2^n. > 21002587+4. 21002621+2. 21002743+4. 21002897+2. 21002953+4. > 21002977+4. **** > 21002981+2. 21002983+4. almost prime quadruple???****. 101 103 107 109. prime quadruple 30k+ 11,13,17,19.*** 21004807+4. 21004811+2. prime triple... > 21005291+2. 21005293+4. prime triple. > ------- > ... > 21020053+4. 21020057+2. prime triple PR%(X%=293)= 0. XTOT% PRIMES SO FAR= 293 fTIME = 14.86 SECS. > FROM 21021000 TO 21051030 (PRESS ESCAPE) > PROC SIFT / SIFT OUT FACTORS OF SIEVE Escape E N D PROG SIEV7/twin close *sp. I had one of these once, but the chocolate melted on it. My wife had one too, but the wheels fell off hers! -- Mark The opinions expressed here are not those of my employer, my wife, my church, or myself... But they are the opinions of Elvis as revealed to me through the medium of my pet hamster, Lee Harvey Oswald ==== Also available at http://math.ucr.edu/home/baez/week199.html This Week's Finds in Mathematical Physics - Week 199 John Baez I've had a really busy quarter, teaching 3 courses that all require serious thought on my part, so it's been a long while since I've been able to write an issue of This Week's Finds. But, back in September I went to a conference on homotopy theory and its applications at the University of Western Ontario, run by Dan Christensen and Rick Jardine. There were some really cool talks at this conference - my favorite was one by Jack Morava about elliptic cohomology, and I'm really sorry I missed his lectures on Galois theory. But, instead of trying to describe the talks, I think it would be better if I said a bit about spectra, which are an important tool in homotopy theory. The word spectrum has a lot of different meanings in mathematics and physics. In experimental physics it refers to the frequencies of light, sound or any other sort of wave emitted by an object. For example, if you send the light emitted by hydrogen through a spectrometer, you'll see a bunch of sharp lines at specific frequencies - the discrete spectrum - along with a diffuse glow at all frequencies - the continuous spectrum. The German high school teacher Balmer noticed that the sharp lines correspond to light with frequencies proportional to 1/n^2 - 1/m^2 where n,m = 1,2,3,... These days, in theoretical physics the spectrum of something is the set of frequencies at which it can vibrate - or in quantum theory, the set of energies it can have, since an energy is just a frequency times Planck's constant. For example, Bohr took Balmer's formula and realized that a hydrogen atom must have a discrete set of allowed energy levels -1/n^2 When the atom hops from one energy level to another, it emits or absorbs light with energy equal to the difference of two such numbers! This accounts for the discrete spectrum of light emitted by hydrogen. The atom can also have any *positive* energy, and this accounts for the continuous spectrum. In quantum mechanics, observables like energy are described as observable A is the set of values it's allowed to have, and mathematically this is the set of numbers x such that the operator A - x has no inverse. For example, if A is a Hamiltonian, the operator that describes the energy of a quantum system, its spectrum is just the set of allowed energies! The simplest case is when x is an eigenvalue of A: the eigenvalues of an operator form its discrete spectrum. But, there can also be numbers in the spectrum that aren't eigenvalues, and these form the continuous spectrum. In mathematical physics, people talk about the spectrum not just of one observable but of a whole bunch of commuting observables, since commuting observables can be measured simultaneously without the Heisenberg uncertainty principle kicking in to limit the precision. The nice way to think of the spectrum of a bunch of operators uses the concept of space that's closed under addition, multiplication and scalar multiplication, closed under taking adjoints and also closed in the norm topology, it's called a C*-algebra. The spectrum of a C*-algebra A is the set of all homomorphisms x: A -> C, where C is the complex numbers. Though it's not immediately obvious, this sort of spectrum reduces to the previous one when A is the C*-algebra of operators generated by a single self-adjoint operator. So, it's a nice way to define the spectrum of a whole bunch of observables. This generalization is not very useful when the C*-algebra is noncommutative, since then it may not have many homomorphisms to the complex numbers. But if it's commutative, we know *everything* about it once we know its spectrum! This amazing fact is called the Gelfand-Naimark theorem. Here's the idea. There's an easy way to make the spectrum of a commutative C*-algebra A into a topological space: we say x_i -> x precisely when x_i(a) -> x(a) for all elements a of A. With this topology any element a of A gives a continuous complex function on the spectrum, defined by this clever formula: a(x) = x(a). The physicist Chris Isham says he couldn't sleep all night when he first saw this formula, it's so darn clever! And, it turns out that *any* continuous function on the spectrum comes from an element of A via this formula! So, if you hand me the spectrum Spec(A) of a commutative C*-algebra A, I can recover A (up to isomorphism) by forming the C*-algebra of all continuous functions on Spec(A). As you can see, the concept of spectrum is getting more abstract - but it still has close ties to the original idea. What once was a bunch of lines on a spectrometer has now become a topological space associated to a commuting collection of observables. The idea is that each point in this space is a way of assigning values to all these observables... just like each line in the spectrometer represents a particular frequency of light! But the abstraction process doesn't stop here. In algebraic geometry, people want to think of *any* commutative ring as consisting of functions on some sort of space. For example, the commutative ring of real polynomials in two variables mod the relation x^2 + y^2 = 1 is just another way of thinking about polynomial functions on the circle. How do we get the circle back from this commutative ring? Simple: just form the space of all homomorphisms from it to the real numbers! It would be nice to have a recipe to take any commutative ring A and extract a space from it: its spectrum. As we've seen, one option is to take the spectrum to consist of all homomorphisms to the complex numbers: x: A -> C Another would be to use the real numbers: x: A -> R. Both the real and complex numbers are fields: commutative rings where we can divide by anything nonzero. But there are a lot of other fields, like Z/p where p is any prime number. So, instead of picking one field, a more evenhanded approach is to use *all possible* fields, and say a homomorphism to any one of these should give a point of the spectrum. Actually, since there are zillions of fields out there, a more manageable option is to look not at the homomorphism itself but its kernel: the set of elements a in A with x(a) = 0. The kernel of a homomorphism from A to any other ring is an ideal: a set closed under addition and also multiplication by all elements of A. Even better, the kernel of a homomorphism from A to a *field* is a prime ideal, meaning it's not not all of A, and whenever the product of two elements of A lies in the ideal, at least one of them must be in the ideal. Conversely, given a prime ideal in A, there's always a field k and a homomorphism x: A -> k whose kernel is that prime ideal. So, it's reasonable to define the spectrum of A, Spec(A) to be the set of all prime ideals in A. This turns out to exactly match the previous definition of spectrum when A is a C*-algebra. But why the word prime? Well, in the commutative ring of integers, Z, most prime ideals come from prime numbers. If we take all the multiples of any prime number, we get a prime ideal, which is the kernel of the obvious homomorphism x: Z -> Z/p There's just one other prime ideal in Z, namely all the multiples of 0. In other words, the set consisting of just 0 alone! This is the kernel of the homomorphism from Z into the rationals. For some fascinating reason I'd rather not explain now, this prime ideal is often called the prime at infinity. It's different from all the rest, but the wise know it's usually good to keep it in. So, the spectrum of the integers is just the set of ordinary primes together with the prime at infinity: Spec(Z) = {2, 3, 5, 7, 11, ... infinity} We seem to have gotten pretty far from physics by now, but in fact many people believe that taking this spectrum seriously from a *physical* viewpoint will be crucial to proving the Riemann hypothesis - a famous open conjecture related to the distribution of prime numbers. I don't have time to do justice to this, but the basic idea goes as follows. Suppose we have a quantum system whose Hamiltonian has this spectrum: {ln 2, ln 3, ln 5, ln 7, ln 11, ....} primon. Now let's second quantize this system. The idea of second quantization is that we form a new system consisting of an arbitrary finite collection of noninteracting indistinguishable copies of the original system. For this sort, treated as identical bosons. If second quantize our primon, we'll get a system with energy levels that are arbitrary sums of entries from the above list. If we write them in increasing order, they look like this: {0, ln 2, ln 3, ln 2 + ln 2, ln 5, ln 2 + ln 3, ln 7, ln 2 + ln 2 + ln 2, ....} or in other words, just {ln 1, ln 2, ln 3, ln 4, ln 5, ln 6, ln 7, ln 8, ....} since every whole number can be built from primons in a unique way! Bernard Julia calls this new system the free Riemann gas, since it's made of noninteracting primons - and in a minute we'll see it's related to the Riemann hypothesis. To see this, let's do some statistical mechanics with the free Riemann gas! As usual, at any temperature T the probability that this system will be in a state of energy E is proportional to exp(-beta E) where beta = 1/kT and k is Boltzmann's constant. But to get these numbers to add up to one as probabilities should, we have to normalize them, dividing by their sum, which goes by the name of the partition function. The partition function for the free Riemann gas is: sum_n exp(-beta ln n) = sum_n n^{-beta} the so-called Riemann zeta function. It's well-defined for beta > 1 - that is, low temperatures - but it blows up when beta = 1. This means that the free Riemann gas has a Hagedorn temperature: a temperature that it can't go above, because doing so would take an infinite amount of energy. Nonetheless we can analytically continue the Riemann zeta function around beta = 1, and the Riemann hypothesis says that it can only vanish if beta is a negative even integer or a number with real part equal to 1/2. And, precisely because the free Riemann gas is made of primons, this hypothesis has a lot to do with prime numbers! For example, it's equivalent to the assertion that the number of primes less than x differs from Li(x) = integral_2^x dt/ln t by less than some constant times sqrt(x) ln(x). All this is lots of fun. I urge the physicist reader to compute the free energy and specific heat of the free Riemann gas, of also to investigate the system where we treat the primons as fermions. But, the big question is whether we can use physics-inspired reasoning to prove the Riemann hypothesis! In 1995, a step in this direction was taken by Bost and Connes. I'm not ready to really explain it, so I'll just tantalize you by dangling their abstract in front of you: In this paper, we construct a natural C*-dynamical system whose partition function is the Riemann zeta function. Our construction is general and associates to an inclusion of rings (under a suitable finiteness assumption) an inclusion of discrete groups (the associated ax + b groups) and the corresponding Hecke algebras of bi-invariant functions. The latter algebra is endowed with a canonical one-parameter group of automorphisms measuring the lack of normality of the subgroup. The inclusion of rings Z provides the desired C*-dynamical system, which admits the zeta function as partition function and the Galois group Gal(Q^{cycl}/ Q) of the cyclotomic extension Q^{cycl} of Q as symmetry group. Moreover, it exhibits a phase transition with spontaneous symmetry breaking at inverse temperature beta = 1. Here's the reference: 1) J.-B. Bost and Alain Connes, Hecke Algebras, Type III factors and phase transitions with spontaneous symmetry breaking in number theory, Selecta Math. (New Series), 1 (1995) 411-457. The idea of the free Riemann gas was introduced most clearly by Julia, though there were many precursors: 2) Bernard L. Julia, Statistical theory of numbers, in Number Theory and Physics, eds. J. M. Luck, P. Moussa, and M. Waldschmidt, Springer Proceedings in Physics, Vol. 47, Springer-Verlag, Berlin, 1990, pp. 276-293. Summarized by Matthew Watkins in http://www.maths.ex.ac.uk/~mwatkins/zeta/Julia.htm Matthew Watkins has a lot of other fascinating material about prime numbers and physics on his website: 3) Matthew Watkins, http://www.maths.ex.ac.uk/~mwatkins/ so this is the best place to start if you're a beginner wanting to learn more about this stuff. There are also a bunch of new popular books on the Riemann hypothesis, so if you're looking for good Christmas gifts, you might try one of these: 4) Marcus du Sautoy, The Music of the Primes: Searching to Solve the Greatest 5) Karl Sabbagh, The Riemann Hypothesis: the Greatest Unsolved Problem 6) John Derbyshire, Prime Obsession: Bernhard Riemann and the Greatest I haven't read any of them, but from reviews it sounds like the third one focuses on Riemann while the first two talk more about modern developments. If you want something quite a bit more substantial but still not requiring a PhD, try this: 7) Jeffrey Stopple, A Primer of Analytic Number Theory: from Pythagoras This is the only introduction to analytic number theory that's so simple that I feel I have a good chance of reading it all the way through. There's also a lot of interesting work relating the Riemann zeta function to quantum chaos. Alas, I don't know how this is related to the free Riemann gas idea! But here's a nice easy introduction: 8) Barry Cipra, A prime case of chaos, in What's Happening in the Mathematical Sciences, vol. 4, American Mathematical Society. Also available at http://www.maths.ex.ac.uk/~mwatkins/zeta/cipra.htm Finally, if you get stuck on the fermionic version of the free Riemann gas, read Julia's paper or this one: 9) Donald Spector, Supersymmetry and the Moebius inversion function, Communications in Mathematical Physics 127 (1990) 239-252. Anyway, all this post up to now has been just a big joke - although everything I said is true. The joke is that all this stuff about different meanings of spectrum has nothing to do with the sort of spectra they were talking about at that conference on homotopy theory! Topologists like to study a completely different sort of spectrum... so now let me talk about those. In topology, a spectrum is defined to be a sequence of pointed topological spaces, each of which is homeomorphic to the space of all based loops in the next. So, each space in a spectrum is an infinite loop space: a space of loops in a space of loops in a space of loops in.... In week149 I described how this sort of spectrum gives a generalized cohomology theory, and I mentioned a bunch of examples. I gave some more examples in week150 and week197. But I never described the cool way to construct spectra that Graeme Segal came up with - so let me do that now. There's a cute way to get a space from a category that goes like this. First create a simplicial set from your category, with one 0-simplex for each object: . x one 1-simplex for each morphism: f .------>------. x y one 2-simplex for each composable pair of morphisms: y . / / / f/ g / / / / fg .------->---------. x z and so on ad infinitum. This is called the nerve of the category. Then, think of this simplicial set as a topological space - i.e., take its geometric realization. The result is called the classifying space of the category. By the way, I described this construction in a lot more detail in week117. I also explained how you can get *every* space, up to homotopy equivalence, as the classifying space of some category! But what I didn't say is this: If you start with a monoidal category, the group completion of its classifying space will be a loop space. You can get any loop space this way. If you start with a braided monoidal category, the group completion of its classifying space will be a double loop space. You can get any double loop space this way. If you start with a symmetric monoidal category, the group completion of its classifying space will be an infinite loop space. You can get any infinite loop space this way. Huh? There are lots of terms here that I haven't defined yet.... For starters, a loop space is the space of based loops in some pointed topological space. A double loop space is the space of based loops in the space of based loops in some pointed topological space, and so on. Secondly, all the above statements are only true up to homotopy equivalence. Third, I'm talking about various sorts of category here. A monoidal category is roughly a category with a tensor product. This gives its classifying space a product, making it into a topological monoid; turning this into a group by throwing in inverses is called group completion. A braided monoidal category is roughly a monoidal category with an isomorphism B_{x,y}: x tensor y -> y tensor x for any pair of objects; we require this isomorphism satisfy some rules motivated by thinking it as a braiding, like this: x y / / / / / / / / / / / / y x A symmetric monoidal category is roughly a monoidal category for which B_{x,y} is the inverse of B_{y,x}. Some more details on these category-theoretic notions can be found in week121. Symmetric monoidal categories abound in mathematics, so we can easily use them to get lots of nice infinite loop spaces - and thence spectra and generalized cohomology theories! For example, if we take the category of finite sets, with disjoint union as the tensor product, and the obvious braiding, its classifying space will be Omega^infinity S^infinity = lim_{k -> infinity} Omega^k S^k the limit of taking the kth loop space of the k-sphere! The corresponding spectrum is called the sphere spectrum and the corresponding generalized cohomology theory is called stable homotopy theory. If we take the category of finite-dimensional complex vector spaces, with direct sum as the tensor product, and the obvious braiding, its classifying space will be BU(infinity) = lim_{k -> infinity} BU(k) where BU(k) is the classifying space of the group of k x k unitary matrices! The corresponding spectrum is called the spectrum for connective complex K-theory and the corresponding generalized cohomology theory is called connective complex K-theory. (Here connective refers to the fact that unlike some other K-theory you may be familiar with, the cohomology groups K^i with i negative have been set to zero.) More generally, we can take the category of modules of a ring R, again with direct sum as the tensor product and the obvious braiding. This gives something called algebraic K-theory. More precisely, the homotopy groups of the resulting infinite loop space are called the algebraic K-theory groups K_i(R). Yet another example comes from taking the category of finite CW complexes, with disjoint union as the tensor product and the obvious braiding. This gives a generalized cohomology theory called A-theory, due to Waldhausen. I would like to say more about this stuff sometime. There's a lot more to say! For example, there are some cool relations between the algebraic K-theory groups of the integers, K_i(Z), and he Riemann zeta function at odd integers, zeta(2n+1). (Hmm, so maybe the different sorts of spectra *are* related!) There's also a lot of nice stuff about how algebraic K-theory is related to topology. You can learn about that here: 10) Jonathan Rosenberg, K-theory and geometric topology, available at http://www.math.umd.edu/users/jmr/geomtop.pdf But, I'll stop here for now. For more on how different sorts of category can be used to get ahold of n-fold loop spaces, see: 11) C. Balteanu, Z. Fiedorowicz, R. Schwaenzl, and R. Vogt, Iterated monoidal categories, available at math.AT/9808082. Quote of the week: Riemann's insight followed his discovery of a mathematical looking-glass through which he could gaze at the primes. Alice's world was turned upside down when she stepped through her looking-glass. In contrast, in the strange mathematical world beyond Riemann's glass, the chaos of the primes seemed to be transformed into an ordered pattern as strong as any mathematician could hope for. He conjectured that this order would be maintained however far one stared into the harmony on the far side of the mirror would explain why outwardly the primes look so chaotic. The metamorphosis provided by Riemann's mirror, where chaos turns to order, is one which most mathematicians find almost miraculous. The challenge that Riemann left the mathematical world was to prove that the order he thought he could discern was really there. - Marcus du Sautoy ----------------------------------------------------------------------- mathematics and physics, as well as some of my research papers, can be obtained at http://math.ucr.edu/home/baez/ For a table of contents of all the issues of This Week's Finds, try http://math.ucr.edu/home/baez/twf.html A simple jumping-off point to the old issues is available at http://math.ucr.edu/home/baez/twfshort.html If you just want the latest issue, go to http://math.ucr.edu/home/baez/this.week.html ==== > norm topology, it's called a C*-algebra. The spectrum of a=20 > C*-algebra A is the set of all homomorphisms=20 > x: A -> C,=20 > where C is the complex numbers. Though it's not immediately obvious, ... > the idea. There's an easy way to make the spectrum of a commutative > C*-algebra A into a topological space: we say x_i -> x precisely when > x_i(a) -> x(a) > for all elements a of A. With this topology any element a of A gives=20 > a continuous complex function on the spectrum, defined by this clever=20 > formula: I don't understand what you mean by x_i -> x x_i(a) -> x(a) What has the index i to do with this? x_i and x are the above mentioned homomorphisms, right? Could you explain a little more details? Rene. --=20 Ren=E9 Meyer Student of Physics & Mathematics Zhejiang University, Hangzhou, China ==== > >> norm topology, it's called a C*-algebra. The spectrum of a=20 >> C*-algebra A is the set of all homomorphisms=20 >> x: A -> C,=20 >> where C is the complex numbers. Though it's not immediately obvious, > ... >> the idea. There's an easy way to make the spectrum of a commutative >> C*-algebra A into a topological space: we say x_i -> x precisely when >> x_i(a) -> x(a) >> for all elements a of A. With this topology any element a of A gives=20 >> a continuous complex function on the spectrum, defined by this clever=20 >> formula: > > I don't understand what you mean by > > x_i -> x > x_i(a) -> x(a) > > What has the index i to do with this? x_i and x are the above > mentioned homomorphisms, right? Could you explain a little more > details? > > Rene. > > --=20 > Ren=E9 Meyer > Student of Physics & Mathematics > Zhejiang University, Hangzhou, China x_i is a sequence of morphisms indexed by the natural numbers, the topology on the space of morphisms is then given by saying the sequence x_i converges to x iff for every a in the C* algebra x_i(a) (an element of the complex numbers) converges to x(a) I want to use the phrase 'topology of pointwise convergence'. can someone tell me if I've misremembered, please; I could search for my functional analysis notes... ==== > the idea. There's an easy way to make the spectrum of a commutative > C*-algebra A into a topological space: we say x_i -> x precisely when > x_i(a) -> x(a) > for all elements a of A. With this topology any element a of A gives=20 > a continuous complex function on the spectrum, defined by this clever=20 > formula: >> >> I don't understand what you mean by >> >> x_i -> x >> x_i(a) -> x(a) >> >> What has the index i to do with this? x_i and x are the above >> mentioned homomorphisms, right? Could you explain a little more >> details? >x_i is a sequence of morphisms indexed by the natural numbers, the >topology on the space of morphisms is then given by saying the sequence >x_i converges to x iff for every a in the C* algebra x_i(a) (an element >of the complex numbers) converges to x(a) Be careful! The index set does not have to be the natural numbers. This is defining convergence of *nets* of homomorphisms, not just sequences of such. If the spectrum is not metrizable, sequences are not enough to determine the topology. There is another way of defining this topology: Given a morphism x, and an element a of the algebra and e>0, require the set of morphisms {y: |x(a)-y(a)|I want to use the phrase 'topology of pointwise convergence'. can someone >tell me if I've misremembered, please; I could search for my functional >analysis notes... Yes, this is right, although it's also called the weak* topology. --Dan Grubb ==== > >> the idea. There's an easy way to make the spectrum of a commutative >> C*-algebra A into a topological space: we say x_i -> x precisely when >> x_i(a) -> x(a) >> for all elements a of A. With this topology any element a of A gives=20 >> a continuous complex function on the spectrum, defined by this clever=20 >> formula: > > I don't understand what you mean by > > x_i -> x > x_i(a) -> x(a) > > What has the index i to do with this? x_i and x are the above > mentioned homomorphisms, right? Could you explain a little more > details? > >>x_i is a sequence of morphisms indexed by the natural numbers, the >>topology on the space of morphisms is then given by saying the sequence >>x_i converges to x iff for every a in the C* algebra x_i(a) (an element >>of the complex numbers) converges to x(a) > > Be careful! The index set does not have to be the natural numbers. > This is defining convergence of *nets* of homomorphisms, not just > sequences of such. If the spectrum is not metrizable, sequences are > not enough to determine the topology. > rather than Moore-Smith stuff. > There is another way of defining this topology: > > Given a morphism x, and an element a of the algebra and e>0, require > the set of morphisms {y: |x(a)-y(a)| set for the topology. > >>I want to use the phrase 'topology of pointwise convergence'. can someone >>tell me if I've misremembered, please; I could search for my functional >>analysis notes... > > Yes, this is right, although it's also called the weak* topology. > > --Dan Grubb Which, I seem to remember, is not the same as the *weak topology. ==== >I want to use the phrase 'topology of pointwise convergence'. can someone >tell me if I've misremembered, please; I could search for my functional >analysis notes... >> >> Yes, this is right, although it's also called the weak* topology. >> >Which, I seem to remember, is not the same as the *weak topology. Actually, you probably mean the weak topology (without the *), which is essentially the weak* topology from the double dual. To make matters worse, the topology on the spectrum is sometimes called the weak topology. --Dan Grubb ==== Idea in progress.... Perhaps we should post here definitions of infinite integer sequences, then rank our sequences by their randomness and by their simplicity of description somehow. The goal of the contest would be to post the sequence with the highest (amount of randomness) divided by, say, (the length of its decription in some language). By randomness, perhaps I mean the minimum order of a polynomial needed to duplicate the sequence for a fixed number of terms (say, 100). There might be a much better definition of randomness. Feel free to post any definition you feel might be better. As for the sequences' simplicity of description, we could subjectively evaluate each sequence's description-length as written in English or in Mathematica/Maple code. Again, since the above method is HIGHLY subjective, if you have a better description of simplicity of description for the sequences to be evaluated by, please reply. And post your own sequences here. I guess I can objectively define own as, the sequence is not in the EIS unless if you are the author anyway. (EIS: http://www.research.att.com/~njas/sequences/index.html#L) And we will need some computer expert out there to evaluate the sequences for randomness, perhaps. (Despite my apparent tone, this is SUPPOSE to be fun!) My entry, as an example: (A057176 in the EIS) a(0) = 1; for n >= 1, a(n) = sum{j=0 to a(n-1)(mod n)} a(j) (1,1,2,4,1,2,4,9,2,4,9,30,15,4,9,30,97,84,84,26,15,127, 308,30,15,127,898,24,913,97,24,913,308,69,2,4,9,30,2996, 4217,308,560,97,69,1040,11,69,868,9,30,2996,7327,14566,..) thanks, Leroy Quet ==== > Idea in progress.... Perhaps we should post here definitions of infinite integer sequences, > then rank our sequences by their randomness and by their simplicity > of description somehow. The goal of the contest would be to post the sequence with the highest > (amount of randomness) divided by, say, (the length of its decription > in some language). By randomness, perhaps I mean the minimum order of a polynomial needed > to duplicate the sequence for a fixed number of terms (say, 100). There might be a much better definition of randomness. Feel free to > post any definition you feel might be better. See Knuth, SemiNumerical Algorithms -- 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 ==== > Idea in progress.... > > The goal of the contest would be to post the sequence with the highest > (amount of randomness) divided by, say, (the length of its decription > in some language). > This sounds like a fun idea to me. As suggested, Kolmogorov complexity is essentially what the description length should be. However, it is a tad difficult to pin this length down. A specific language needs to be decided upon. I propose the C programming language, using minimal libraries (perhaps only allowing math.h). We could interpret the length of the program to be the number of nodes in its syntax tree. I think this is a little more natural (perhaps counterintuitively) to use than Maple/Mathematica code because those languages are too mathematically powerful. I think in Maple it would be far too easy to describe an extraordinarily esoteric sequence very tersely. The question of measuring randomness is certainly more delicate - essentially because, it is inevitable that this, too, is some sort of derivation of description length. Polynomials are no good. A good intuitive idea might be how quickly a human could recognize the pattern. But how could that be captured mathematically? For increasing sequences which go to infinity, we could look at either the Kolmogorov complexity of the asymptotics, or more interestingly, how quickly the sequence converges to this asymptotic. It seems to me that more random sequences would be harder to approximate accurately. It might be possible to extend this idea to other sequences, but I'm not quite sure how. We could turn any non-negative sequence into an increasing one by looking at the sum of the first n terms. Another thought is to use the time or space complexity of the above-mentioned description as a measure of randomness. Finally, I came up with four dashingly clever sequences, but alas they were _all_ already in EIS. I'll try to submit something original later. -Tyler ==== |Perhaps we should post here definitions of infinite integer sequences, |then rank our sequences by their randomness and by their simplicity |of description somehow. | |The goal of the contest would be to post the sequence with the highest |(amount of randomness) divided by, say, (the length of its decription |in some language). | |By randomness, perhaps I mean the minimum order of a polynomial needed |to duplicate the sequence for a fixed number of terms (say, 100). | |There might be a much better definition of randomness. Feel free to |post any definition you feel might be better. if you don't find a better definition of randomness then my sequence is f(n) = 2^n. -- ==== > Idea in progress.... > > Perhaps we should post here definitions of infinite integer sequences, > then rank our sequences by their randomness and by their simplicity > of description somehow. > > The goal of the contest would be to post the sequence with the highest > (amount of randomness) divided by, say, (the length of its decription > in some language). > > By randomness, perhaps I mean the minimum order of a polynomial needed > to duplicate the sequence for a fixed number of terms (say, 100). > > There might be a much better definition of randomness. Feel free to > post any definition you feel might be better. > > As for the sequences' simplicity of description, we could subjectively > evaluate each sequence's description-length as written in English or > in Mathematica/Maple code. > > Again, since the above method is HIGHLY subjective, if you have a > better description of simplicity of description for the sequences to > be evaluated by, please reply. > > > And post your own sequences here. > I guess I can objectively define own as, the sequence is not in the > EIS unless if you are the author anyway. > (EIS: http://www.research.att.com/~njas/sequences/index.html#L) > > > And we will need some computer expert out there to evaluate the > sequences for randomness, perhaps. > > > (Despite my apparent tone, this is SUPPOSE to be fun!) > > > My entry, as an example: > > (A057176 in the EIS) > > a(0) = 1; for n >= 1, a(n) = sum{j=0 to a(n-1)(mod n)} a(j) > > (1,1,2,4,1,2,4,9,2,4,9,30,15,4,9,30,97,84,84,26,15,127, > 308,30,15,127,898,24,913,97,24,913,308,69,2,4,9,30,2996, > 4217,308,560,97,69,1040,11,69,868,9,30,2996,7327,14566,..) > > thanks, > Leroy Quet How about this little 5th degree poly --- 17,26,4,5,13,46,14,63,7,88,20,59,79,94,54,13,85,92,204,19,89,80,190,35, 71,132,290,299,71,12,172,65,295,36,146,203,23,122,32,235..etc.. Bounded 1-300 to keep it simple! Don't ask me how to solve it! Dan ==== Idea in progress.... > >Perhaps we should post here definitions of infinite integer sequences, >then rank our sequences by their randomness and by their simplicity >of description somehow. > Suppose s is a positive integer. Set n(0)=1 and for i>=1 let n(i)=s-1-((s*i-1) mod n(i-1)). See a post of mine Is this sequence eventually periodic? from a year or so ago for more info on how it was derived (not by me) from a weird way of representing the reals. rich ==== > Again, since the above method is HIGHLY subjective, if you have a > better description of simplicity of description for the sequences to > be evaluated by, please reply. Kolmogorov complexity, IIRC. Kolmogorov introduced the idea of describing the complexity (or entropy, or information content) of a sequence as the minimum length of an algorithm needed to generate it. Of course, this definition is relative to a chosen way of describing algorithms, you can derive many interesting facts about Kolmogorov complexity that are true for *any* way of describing algorithms. ==== How are the expected results shaping up for this year? I think I got 10 questions (I missed A3 and B6) (though I forgot the 0 case for A2 and may have made some small mistakes) and I was wondering where that could place in the results. I was also shocked to see B5 on there. It was a problem in my first olympiad book: Mathematical Olympiad Challenges. Coincidently it was also the one question that I could not solve in that section back then, so I was not at much of an advantage for the area part. Ralph Furmaniak ==== > How are the expected results shaping up for this year? I think I got > 10 questions (I missed A3 and B6) (though I forgot the 0 case for A2 > and may have made some small mistakes) and I was wondering where that > could place in the results. Complete, correct, solutions of 10 problems would be a score of 100 and make you the top contestant in the world in a typical year. > > I was also shocked to see B5 on there. It was a problem in my first > olympiad book: Mathematical Olympiad Challenges. Coincidently it > was also the one question that I could not solve in that section back > then, so I was not at much of an advantage for the area part. > > > Ralph Furmaniak -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > >In time I will collect >the questions and good responses I see into a file which will be at > http://www.math.niu.edu/~rusin/problems-math/ > > In that directory there are now two new files, > putnam.03probs -- the questions (as I posted them previously, sans typos) > putnam.03 -- collected answers > > I am happy to report that there now seem to be answers for each of the > questions, some very slick and elegant, others hinting at broader contexts. > (I'm always on the lookout for alternative proofs, so if your solution > is distinctly different from any in the files, let me know and I can > add your answers!) > > dave I proved A1 as follows. The idea is to show that for each integer k, with 1 < k < n, there is exactly one sequence of length k which sums to n. We will use induction on n. So, suppose that for some N > 0 each of these, we construct one of length N+1, by adding 1 to the smallest element of the N-sum sequence. For our length N+1 sequence, use all 1's. QED. ==== > > > OK. But I was given another problem similar to this where I was asked > to consider the vectors [1, -2, k], [5, -2k, 25] and [k, -10, 25] and > find all values of k for which these vectors are linearly independent. > I solved it using row reduction and got that k not equal to 5 or -10. > If I take the determinant of the matrix and say that it is not equal to > zero, I get k^2 - 75k +250 not equal to zero whic is clearly not the > same answer. I do not understand where I am going wrong. Can anyone > > Zack > > I get a determinant of 2*k^3 - 150*k +500 = 2*(x-5)^2*(x+10), which > is non-zero under the same conditions as your row reduction result. > a lot. Change all 'x' to 'k' and it will become clear. This newsgroups is not intended to help each and every one with their homework. You have asked a lot of questions here, but they do not go beyond that level. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== Does anyone know if there is an efficient one-to-one mapping function from a subset of integers (say Z_n) to primes? ==== > Does anyone know if there is an efficient one-to-one mapping function > from a subset of integers (say Z_n) to primes? You posted this question the other day. Have you seen the answers you got? -- ==== > Does anyone know if there is an efficient one-to-one mapping function > from a subset of integers (say Z_n) to primes? The problem Is there an efficient 1-1 mapping from the integers to the times might be a difficult research problem. But if you restrict the problem to some finite subset, it is clearly trivial. ==== in which I need to map an integer identity (ID) to a prime number. I want h( ): Z_n -> Z-m (m > n) s.t. for every integers in Z_n, I can easily find a prime in Z_m that is distinct from the image of another integer. i.e. h(ID_i)=h(ID_j) => ID_i = ID_j. Please note that I don't want a lookup table for the mapping --- that's what I meant efficient. Aldar > Does anyone know if there is an efficient one-to-one mapping function > from a subset of integers (say Z_n) to primes? The problem Is there an efficient 1-1 mapping from the integers to > the times might be a difficult research problem. But if you restrict > the problem to some finite subset, it is clearly trivial. ==== > in which I need to map an integer identity (ID) to a > prime number. > > I want h( ): Z_n -> Z-m (m > n) s.t. for every integers in > Z_n, I can easily find a prime in Z_m that is distinct from > the image of another integer. > > i.e. h(ID_i)=h(ID_j) => ID_i = ID_j. > > Please note that I don't want a lookup table for the mapping --- > that's what I meant efficient. What's not efficient about table lookup? Do you want to do this for one particular n? or do you want something that works for all n? If the latter, you're looking for a simple (efficient) formula f(a) that returns a different prime for each positive integer a. You're unlikely to find anything to your liking. Certainly there is no polynomial that works. If n < 40 then you can use Euler's polynomial, n^2 + n + 41 or something like that. -- ==== What do you think about the Putnam this year, as compared to that last year and in 2001? I did much better this year than I did last year as a freshman. I think A1 and A2 are easy. A4 is good. A3 is interesting. I didn't have a chance to think about A5 and A6. B1 is good for a B1 :) B3 and B5 are not too hard. B6 looks like a very elegant problem. In general, I think the problem set this year is a little bit harder than last year. What do you think? Look forward to hearing your opinions. T Phan (newcomer) ==== Anyone knows if there is any good algorithm to find the least number of equal-size discs to cover an arbitrary polygon? The polygon can be concave or convex. How to find the center points of these discs? Pipi, ==== > There are some programs now, like one in Boston that heard about called The > Math Circle, where kids are learning advanced math. The two important > elements of the program are (1) no rigid structure and high pressure like > regular schools and (2) teachers that know about and care about the > material. And it's not just that the kids can understand abstract stuff, > they can also absorb knowledge much faster than they would at regular > school. An important fact here is that in addition to having good > teachers, which we all agree is important, you also need good > administrators who will let these good teachers teach. This all sounds good. But I've wondered whether these programs would succeed with unselected children -- all of the cases I've heard sounded like they were children who were to some degree interested in school (or at least their parents were). But I may be wrong. > They sent out a questionaire asking everywho who had taught calculus > about various details of their experience but they were very clear > that they were NOT interested in anyone's thoughts about the > committee asked about were pretty much irrelevant and that the real > problem WAS the curriculum. The committee was still not interested. Did you ever poke around and find out why they didn't want criticism of the curriculum? ==== >> There are some programs now, like one in Boston that heard about called The >> Math Circle, where kids are learning advanced math. The two important >> elements of the program are (1) no rigid structure and high pressure like >> regular schools and (2) teachers that know about and care about the >> material. And it's not just that the kids can understand abstract stuff, >> they can also absorb knowledge much faster than they would at regular >> school. An important fact here is that in addition to having good >> teachers, which we all agree is important, you also need good >> administrators who will let these good teachers teach. >This all sounds good. But I've wondered whether these programs would >succeed with unselected children -- all of the cases I've heard >sounded like they were children who were to some degree interested in >school (or at least their parents were). But I may be wrong. Nothing will succeed with unselected children. The idea that all children are capable of learning the same material, even at different rates, is false. >> They sent out a questionaire asking everywho who had taught calculus >> about various details of their experience but they were very clear >> that they were NOT interested in anyone's thoughts about the >> committee asked about were pretty much irrelevant and that the real >> problem WAS the curriculum. The committee was still not interested. >Did you ever poke around and find out why they didn't want criticism >of the curriculum? One major problem is that those who teach physics, and even more so engineering, want their unprepared students to be able to use mathematics in their courses yesterday. -- 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 ==== Here is an unoriginal game (inspired by Dots and Boxes{name?}, Go, and by Scrabble somewhat). But this post is not just a game-idea, but a question about strategy. The game can be played with any number(m) of players, m>=2, where each player has a black pencil/pen and a colored pencil/pen of a color different than the player's opponents' colors. The game is played on an n-by-n grid, where n^2 is divisible by m. The players take turns each writing (in black) 1, 2, 3, ...(n^2/m), each integer being written into any still-unoccupied grid-square. (with his/her colored pencil) all grid-squares within any rectangle in the grid, IF, and only if, each grid-square in the rectangle has an integer written in it, the SUM of every integer in the rectangle is any PRIME number, and all the squares in the rectangle are not yet colored in. After each player has moved (ie written an integer and perhaps colored in a rectangle) (n^2/m) times, the winner is the player with the most grid-squares of that player's pencil-color. Variations: Players can only fill in rectangles which contain the square they just Players can write their integers into the grid in any order they wish. (I like this variation, but it requires remembering which integers have already been played.) Players draw at-random their integers from a deck of cards. (Either from 4 decks of 1 through (n^2/m), or from one master deck of 1 through n^2.) The advantage of this is that it guards against the second player, in a 2-person game, simply matching the first players' moves. - In this game, with or without any of the variations (or with any variations of your own), what would be a good strategy??? By the way, ideally, the game board will look aesthetically interesting when the game is done, especially with the right m and n. (Any suggestions for m and n from this standpoint?) Leroy Quet ==== > Here is an example of an emerging digital conceptual art form. Kleitman once reviewed a paper for Mathematical Reviews, starting This paper fills a much-needed void in the literature. I believe a similar statement could be made about Golden Section as Psychovisual Archetype, if that is its actual name. ==== > >I wanted to look at this problem more generally. As in, if we select k >integers randomly from a set of n integers, what value of k will give us >>= 50% chance that 2 of those integers will be the same? Or, >What value of k gives p <= 0.5 in the general equation >p = n!/((n - k)! * n^k) > >Is there a way to solve this with respect to n? ie, Is there a formula >that I can plug n into and get a value of k that gives me p <= 0.5? > > If you're interested in a rule of thumb rather than an exact answer, then > k = 1.177 * sqrt(n) is a pretty good approximation. > > -- Erick this might give for large values of n? I would like to apply this to very large numbers indeed, certainly ones over 20 digits. Since my calculator stops calculating the original equation shortly after 400! I can't check to see how close the approximation is for larger values. -David C. ==== >What value of k gives p <= 0.5 in the general equation >p = n!/((n - k)! * n^k) Is there a way to solve this with respect to n? ie, Is there a formula >that I can plug n into and get a value of k that gives me p <= 0.5? >> >> If you're interested in a rule of thumb rather than an exact answer, then >> k = 1.177 * sqrt(n) is a pretty good approximation. this might give for large values of n? I would like to apply this to >very large numbers indeed, certainly ones over 20 digits. Since my >calculator stops calculating the original equation shortly after 400! I >can't check to see how close the approximation is for larger values. I see...Well, in relative terms it is an extremely accurate approximation for large n. The value of k / sqrt(n) converges to exactly sqrt(2 log 2) ~= 1.17741002, so using this constant for n that large will give you a k that is probably within 0.001% of the true value. I'm less confident about this, but I think it's pretty close to the mark in absolute terms too. I think for very large n the estimate for k is off by about 0.04, but empirical evidence for smaller n (up to 1000000) seems to suggest that the discrepancy might be closer to 0.30. In any case it appears to be quite minor. -- Erick ==== > (Lurkers who wonder who I am to > be dispensing this sort of advice should look > up some recent history: I'm the supreme > ruler of sci.math (explaining how I attained > this status would get us in trouble with the > censors again...)) > > I imagine you're feeling pretty smug right now, > thinking you caught me. Here's a hint: official > spokesman for the maths community and > supreme ruler of sci.math are not the same thing... You should remember that I voted against you on both occasions. Yeah, that is the reason of your negative reactions to all my ramblngs on this newsgroup. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== ... > I get help from mathematicians, much of which I don't post about, > though I have posted about some, but consistently when it comes time > to deliver, they run away. > > C++ program that I modified and posted a lot, who then took off. Yes, why did he take off? > Like I just had yet another mathematician go point by point through my > key argument showing the problem with algebraic integers. He said > he'd get back to me, and maybe he will, but I also wouldn't be > surprised if he didn't. Don't hold your breath. It is your understanding that is lacking. > Oh yeah, a paper of mine is *still* under peer review last I heard at > a math journal. Again, don't hold your breath. > Usually I mention that I'm *not* a mathematician, and am in need of > help or guidance. on the subjects that you are about to attack. You are just unwilling to do so. And are insulting the people that offer help. How long did it take before you used the term algebraic integer? How many people suggesting that way had you insulted along the road? > However, when I talk about corresponding with Barry Mazur, he actually > commented on a paper of mine, and offered up advice as well as > questioned a part of it. I answered, answering his question, and he > did not reply further. Perhaps you did not answer his question, but merely gave examples (as you do in this newsgroup)? > Or way back with Kenneth Ribet, which you should remember. He offered > one of his graduate students to consider one of my, unfortunately > flawed attempts at proving Fermat's Last Theorem, as he was intrigured > by that stupid bet that I ended up losing. Until now I have not seen an unflawed attempt. > Or Andrew Granville's several replies in different areas, like one > where he claimed that one of his graduate students had talked recently > about a prime counting function that he'd found, and his now famous on > sci.math statement that 20% of his graduate students came up with > them. > > Later he entered the picture *again* by referring my paper on > algebraic integers up to the chief editor of the New York Journal of > Mathematics. I suspect the reference was negative. > It has been a useful process. Apparently not. You flatly refuse to read about the subject you write about. Also you flatly refuse to write it in a format that is comprehensible to a professional reader. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== [snip] > > Like I just had yet another mathematician go point by point through my > > key argument showing the problem with algebraic integers. He said > > he'd get back to me, and maybe he will, but I also wouldn't be > > surprised if he didn't. Don't hold your breath. Funny, my advice would have been for James to hold his breath. -- 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 ==== > Why are you carrying on with this James Harris thing, I think the poor guy > obviously has problems, maybe you are making him worse by carrying it on, > but then perhaps you have your own complex reasons for bothering to reply to > him. Do you think it is in the best interests of public perception of the > maths community to be partially responsible for this exchange continuing. > Or, do you see it as entertainment at his expense? In the past Harris has ordered Dr. Ullrich to stop posting in his > threads. He's also contacted Dr. Ullrich's employer (more than once) > in an attempt to get him fired and/or force him to stop posting. > Dr. Ullrich has made it clear on several occasions that he believes > Harris's tactics of censorship and intimidation should not be allowed > to succeed, and that demonstrating that point is a major part of the I see, that sounds reasonable. ==== > > Why are you carrying on with this James Harris thing, I think the poor > guy > obviously has problems, maybe you are making him worse by carrying it > on, > but then perhaps you have your own complex reasons for bothering to > reply to > him. Do you think it is in the best interests of public perception of > the > maths community to be partially responsible for this exchange > continuing. > Or, do you see it as entertainment at his expense? > > In the past Harris has ordered Dr. Ullrich to stop posting in his > threads. He's also contacted Dr. Ullrich's employer (more than once) > in an attempt to get him fired and/or force him to stop posting. Ullrich likes to aggravate by repeatedly posting negatives in threads I create, usually deleting out any math that I presented. I've told him to stay out of my threads, but obviously, I can't force him. I did not contact Oklahoma State University to get him fired, but did contact them several times to register complaints about public statements by Ullrich that I thought reflected badly on him and that institution. > Dr. Ullrich has made it clear on several occasions that he believes > Harris's tactics of censorship and intimidation should not be allowed > to succeed, and that demonstrating that point is a major part of the Which belies the fact that Ullrich has been obsessively replying to my posts for YEARS, including a long period well before my complaints. Those curious about why I complained should go to http://www.google.com/advanced_group_search?hl=en and put David Ullrich in the author field and racial slur in the Exact Phrase field. There you can see messages that David Ullrich himself has posted on the subject. > > I see, that sounds reasonable. What Wayne Brown does is lie. Ullrich's behavior is very odd, its obsessive quality is disturbing, and his public comments have at times been reprehensible. Wayne Brown can't get away with his claims if you check the actual record, and I hope you read more than just one link in that search I mentioned above to get the full picture. James Harris ==== I was wondering about the following equation: a^b == c (mod n) {for 0 <= a <= n-1} For a given value of _b_, is there a heuristic that can give the max/min/average of how many unique values of _c_ there are in terms of n? (I'm most interested in the average, but max and min might be nice too) I know that if _n_ is prime and _b_ is 2 then the max&min&avg is about 1/2. [that is, ~(1/2)*n] However, I also want to consider composites. And when b=2, it looks like the avg ~ 1/4. [ie, ~(1/4)*n] However, I didn't look at very many values of n so this may go up or down. If anyone can think of a formula that would give exact (best, but might not be possible) or approximate (less than best but still good, and help you can provide. -David C. ==== Underwood Dudley > What I find interesting about the cranks in crank.net is that they all > seem to be able to see the crankiness of the other, but not in > themselves. I would be curious to find out what James' reaction is to > the other FLT proofs (and disproofs) found on that site. But I > suspect James will find their proofs/disproofs just as flawed as they > would find James'. > Yes. Once, in a malicious moment that I now regret, I put two > angle trisectors in contact with each other. Both concluded that the > other one was mistaken. No doubt this is a manifestation of Hammick's Exclusion Principle, an inexorable law of nature which says that no two megalomaniacs can coexist in the same system at the same time. Years ago, without trying to, Martin Gardner became an authority on four color theorem had come to outnumber proofs of FLT by three to one. But to be a crank in _math_, of all things! _Any_ other field would be an easier sell. But check this out: http://www.bearnol.pwp.blueyonder.co.uk/Math/ LH ==== that I could tell that you turned to the ultimate chapter without really reading the rest, to paraphrase. any comment, now? > Yes. Once, in a malicious moment that I now regret, I put two > angle trisectors in contact with each other. Both concluded that the > other one was mistaken. --Give the Gift of Dick Cheeny -- out of office, finally! ==== >> RICHARD KOEBERLEIN >> TOM ADEMA I get 2 choices since I asked for only 4 options :) note this is my 1st attempt at someone elses reply so don't get > excited. Herc I never got a response. Time's up, test is a failure. -- Enkidu ----- Of all things, good sense is the most fairly distributed: everyone thinks he is so well supplied with it that even those who are the hardest to satisfy in every other respect never desire more of it than they already have. Ren.8e Descartes, Discours de la M.8ethode. 1637. ==== >> RICHARD KOEBERLEIN >> TOM ADEMA > I never got a response. Time's up, test is a failure. admit it you're impressed I guessed him Herc ==== > RICHARD KOEBERLEIN >> TOM ADEMA > I never got a response. Time's up, test is a failure. admit it you're impressed I guessed him Herc Well, you gave me two names rather than one, which seriously ups your chances on a lucky hit. However, you are wrong. The correct name is in the ------------------------------------------------------------------------ I was referred to you by Nancy B*** (BTSA) for classroom observation. If at all possible I would like to spend the day at H********e to observe yourself and a few other teachers on Wednesday 12/10/03. I promise not to be a burden, and will assist any teacher that needs some help. I am looking for delivery procedures. Jim Freeman CVMS 7th Grade Math ------------------------------------------------------------------------ I've wiped the personal contact information. Otherwise, this is the message. JAMES FREEMAN is the name you were looking for. Absolute failure. -- Enkidu ----- Of all things, good sense is the most fairly distributed: everyone thinks he is so well supplied with it that even those who are the hardest to satisfy in every other respect never desire more of it than they already have. Ren.8e Descartes, Discours de la M.8ethode. 1637. ==== > >> RICHARD KOEBERLEIN >> TOM ADEMA > >> I never got a response. Time's up, test is a failure. > > admit it you're impressed I guessed him > > Herc > > Well, you gave me two names rather than one, which seriously ups your > chances on a lucky hit. However, you are wrong. The correct name is in the > ------------------------------------------------------------------------ > > I was referred to you by Nancy B*** (BTSA) for classroom observation. If > at all possible I would like to spend the day at H********e to observe > yourself and a few other teachers on Wednesday 12/10/03. I promise not > to be a burden, and will assist any teacher that needs some help. > > I am looking for delivery procedures. > > > > > Jim Freeman > CVMS > 7th Grade Math > ------------------------------------------------------------------------ > I've wiped the personal contact information. Otherwise, this is the > message. JAMES FREEMAN is the name you were looking for. > > Absolute failure. > -- Hell. We knew that herc was an absolute failure you have once more proved the fact... :-))) It looks as if reading and comprehension were never his strong points :-) ==== Herc was just testing you. Obviously, he *would* have guessed James Freeman if he'd been serious. You failed the test, and will thus not be invited to Paradise on your mortal demise. Just look at the evidence: > RICHARD KOEBERLEIN >> TOM ADEMA >> >> I never got a response. Time's up, test is a failure. admit it you're impressed I guessed him Herc >> >> Well, you gave me two names rather than one, which seriously ups your >> chances on a lucky hit. However, you are wrong. The correct name is in >> ------------------------------------------------------------------------ >> >> I was referred to you by Nancy B*** (BTSA) for classroom observation. If >> at all possible I would like to spend the day at H********e to observe >> yourself and a few other teachers on Wednesday 12/10/03. I promise not >> to be a burden, and will assist any teacher that needs some help. We already can guess that the message is from a man, but this bit evokes the idea of someone trying to find when the recipient is free, so it must be Freeman. >> I am looking for delivery procedures. In this line, all by itself, you can find James. The line starts with 'I', which is a lot like 'J' in many ways. It then has 'am', and end with 'es'. How much more obvious can you get?! This line identifies the author as James. Herc's paranormal power caused Enkidu to post this. >> >> >> >> Jim Freeman >> CVMS >> 7th Grade Math >> ------------------------------------------------------------------------ >> I've wiped the personal contact information. Otherwise, this is the >> message. JAMES FREEMAN is the name you were looking for. >> >> Absolute failure. >> -- > > Hell. We knew that herc was an absolute failure you have once more > proved the fact... :-))) > It looks as if reading and comprehension were never his strong points > :-) You're just jealous that you're not The Truman, subject of The Truman Show host of other interesting characters. You should be glad you aren't, because then you'd also be tortured constantly for years by major governments and corporations, as Herc has been. For the benefit of those in groups other than ROM, I should use an emoticon or two, but this statement should suffice instead. (first posting using Knode on RedHat 9, I hope it's configured correctly) ==== Nice one, seeing when you are free maybe, still a lot harder on the receiving end I admit Herc -- www.StealthHostiing.com You rule Truman. http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah. Webmasters help the TRUEman by joining www.theBanner.net Current:1 Goal:1000 ---------------------------------------------------------------------------- ------ > >> RICHARD KOEBERLEIN >> TOM ADEMA > >> I never got a response. Time's up, test is a failure. > > admit it you're impressed I guessed him > > Herc Well, you gave me two names rather than one, which seriously ups your > chances on a lucky hit. However, you are wrong. The correct name is in the > ------------------------------------------------------------------------ I was referred to you by Nancy B*** (BTSA) for classroom observation. If > at all possible I would like to spend the day at H********e to observe > yourself and a few other teachers on Wednesday 12/10/03. I promise not > to be a burden, and will assist any teacher that needs some help. I am looking for delivery procedures. > Jim Freeman > CVMS > 7th Grade Math > ------------------------------------------------------------------------ > I've wiped the personal contact information. Otherwise, this is the > message. JAMES FREEMAN is the name you were looking for. Absolute failure. > -- > Enkidu > ----- > Of all things, good sense is the most fairly distributed: everyone > thinks he is so well supplied with it that even those who are the > hardest to satisfy in every other respect never desire more of it than > they already have. Ren.8e Descartes, Discours de la M.8ethode. 1637. ==== >Nice one, seeing when you are free maybe, still a lot harder on the receiving end I admit Herc Well, what happened to your paranormal powers *THEN*, kook? -- Find out about Australia's most dangerous Doomsday Cult: http://users.bigpond.net.au/wanglese/pebble.htm You can't fool me, it's turtles all the way down. ==== > posted in alt.atheism: >WTF? > >Geez, I can't even follow th thing.... > >Starting off with a sig? If you actually try to follow |-|erc too closely you risk losing your > sanity. Ah, but that permits success. ==== ---------------------------------------------------------------------------- ------ > Any reasonable intelligent person can do better than chance matching > names > > wait a minute, its a false positive, *must* be a fraud > > Herc > > Wow, guessing names. That's an amazing power you have there. You deserve all the ridicule that you get. > Just the tip of the iceberg, usenet is a text medium so I have a text demonstration. 100,000 witnesses know I'm paranormal in Townsville Queensland Australia. You don't think the only paranormal man on the planet is just a spelling bee now? Herc ==== --------------------------------------------------------------------------- ------- > Any reasonable intelligent person can do better than chance matching >> names > wait a minute, its a false positive, *must* be a fraud Herc > Wow, guessing names. That's an amazing power you have there. >> You deserve all the ridicule that you get. Just the tip of the iceberg, usenet is a text medium so I have a text demonstration. 100,000 witnesses know I'm paranormal in Townsville Queensland Australia. You >don't think the only paranormal man on the planet is just a spelling bee now? > No, you mean 100,000 people in Queensland know you are *ab* normal. -- Find out about Australia's most dangerous Doomsday Cult: http://users.bigpond.net.au/wanglese/pebble.htm You can't fool me, it's turtles all the way down. ==== The following nice book on pi contains a small error, or so I believe.It is about the Apollonius circle given in a figure of the book which relates to the locus of a point such that the ratio of distances between two fixed points d1/d2=constant but it is given there as 1/d1 + 1/d2 = constant. 10010. 1971, Golem Press. pp116, Chap 11. There is a later Dover edition. Volunteers after checking above are kindly requested to approach the publishers for correction in next edition and notify the newsgroup here before hand to avoid duplication. My attempt in this regard was initially unsuccessful. thanks. ==== ---------------------------------------------------------------------------- ------ > ----------------------------------------------------------------------------- ----- > > CSICOP are another biased crowd of closed minded publicity stunt artists > And evidence is not forthcoming because the 'paranormal' is a scam... These 5 posts consitute my claimed evidence. I can't do anything about it if its just ignored. Can you identify the nature of the 1st 3? Can this feat be duplicated by anyone else? Very simple questions.. very quiet newsgroup. B Rust 3 attacks metal http://tinyurl.com/nd56 E Rich Shewmaker 1 rich showmaker James Randi http://tinyurl.com/nd52 H See You In Hell My Friend. 2 It all depends http://tinyurl.com/nd53 predictions that paranormal will be demonstrated on 02 02. www.tinyurl.com/gutr http://tinyurl.com/nw1r ----------------------------- Then I have 100,000 witnesses that I'm paranormal, here they admit that I am also monitored by the govt. because of it. http://tinyurl.com/iky5 http://tinyurl.com/iky8 http://tinyurl.com/iky9 http://tinyurl.com/iky4 http://tinyurl.com/rv5f http://tinyurl.com/v1yf Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! We're sick of you You rule Truman. >Is the truman is living in Townsville? I've been hearing stuff, yeah. and watch here live in Google as I predict back in 2000 Jim Carrey's next romance costar Laurie Holden with http://tinyurl.com/fuf8 she looks exactly like Laurie Holden of the x-files with bright green eyes. What would you consider evidence? When it goes on the news channel? Herc ==== : -------------------------------------------------------------------------- -------- : : -------------------------------------------------------------------------- -------- : > : > CSICOP are another biased crowd of closed minded publicity stunt artists paranormal events. : > : > And evidence is not forthcoming because the 'paranormal' is a scam... : : These 5 posts consitute my claimed evidence. I can't do anything about : it if its just ignored. Can you identify the nature of the 1st 3? Can this : feat be duplicated by anyone else? Very simple questions.. very quiet newsgroup. : : : B Rust 3 attacks metal http://tinyurl.com/nd56 : E Rich Shewmaker 1 rich showmaker James Randi http://tinyurl.com/nd52 : H See You In Hell My Friend. 2 It all depends http://tinyurl.com/nd53 : : : predictions that paranormal will be demonstrated on 02 02. : : www.tinyurl.com/gutr : http://tinyurl.com/nw1r : : : ----------------------------- : : Then I have 100,000 witnesses that I'm paranormal, here they admit : that I am also monitored by the govt. because of it. : : http://tinyurl.com/iky5 http://tinyurl.com/iky8 http://tinyurl.com/iky9 : http://tinyurl.com/iky4 http://tinyurl.com/rv5f http://tinyurl.com/v1yf : : Hey Trueman...love the show. YOU ARE the Truman I heard him. : Very spooky! We're sick of you You rule Truman. : >Is the truman is living in Townsville? I've been hearing stuff, yeah. : : and watch here live in Google as I predict back in 2000 Jim Carrey's next romance costar : Laurie Holden with http://tinyurl.com/fuf8 she looks exactly like Laurie Holden of the x-files : with bright green eyes. : : : : : What would you consider evidence? When it goes on the news channel? : : Herc i am convinced....you are not normal -- Steve The Earth is degenerating these days. Bribery and corruption abound. Children no longer mind their parents, every man wants to write a book, and it is evident that the end of the world is fast approaching. - Assyrian Stone Tablet, c.2800BC ==== How hard is it for anyone here to find someone to sit a 3 questions multi choice test? Guess your best answer, collate some polls and subjectivity is no longer in the realms of the unexaminable! I have hit a real stubling block for over 1 year on this, 1000s of posts asking for 2 minutes of assistance instead of rebuke. Herc this is proof of paranormal beyond 1,000,000 to 1, you actually have to use your mind to see it ----1----------------------------------------------------------------------- ---- Randi will test you when you properly apply to be tested. Sign up here: http://www.randi.org/research/challenge.html ----2----------------------------------------------------------------------- ----- It really all depends on the situation. ----3----------------------------------------------------------------------- ----- If ever I actually found myself in that situation, I'd hold it upright, with the intent of attacking my assailant's knife hand. ----------------- A cliff86 B Rust C Shanx D NormDePloom E Rich Shewmaker F CNote G Jeff H See You In Hell My Friend. I Someone J Greg Neill ==== |How can the following proposition be proven? | |Let P be a n-dimensional polytope that is not a simplex, i.e. has at least |(n+2) vertices. Then the representation as convex combination of vertices |is not unique for any inner point of P. | |I need this for my seminar talk tomorrow, was thinking about it all day |without getting anywhere. clearly the representation as _affine_ combination of vertexes is non-unique for any point of the polytope. given a representation of an interior point x as convex combination of vertexes, a small perturbation of that representation in the direction of some other representation of x as affine combination of vertexes will be another representation of x as convex combination of vertexes. (hope this actually makes sense; i didn't quite think it all the way through.) -- ==== > > > What is det(A)^2 or A^3. Is it the same as working with real numbers? > Zack > > If A is a square matrix and p(t) is a polynomial then the simple consequence > of det(AB)=det(A)det(B) is that det(p(A))=p(det(A)). Hmm, didn't think that one out very well, did you? Sure, det(AA) = det(A)det(A), so the answer to the OP's question is clear. But your statement about polynomials goes way beyond this, and is false. Not only do you get the rule for scalar multiplication wrong, but how do you propose to deduce a rule for the determinant of a matrix *sum* from the product rule? Remember, polynomials in general have more than one term. If you don't believe me, just try A = [(2,0),(0,2)] and p(t) = 2t. ==== > message > What with the war on terror, and Canada's puny military and corrupt > police force, the only way to insure peace and safety for that > country, is for it to be annexed by America. > > It is the only way foward for Canada aka Canuckistan. --------------------------------- Well, another solution is the liberation of U.$.A by Canadians and the > return of democracy to the Americans peoples. > I am sure Canadians will be welcome with (french) kisses and flowers when > they will topple the Little Dictator and his gang of corporate crooks. > Patriot act will be scraped; free press will be restaured, Health Care > will be available to all citizens, Yeah, like it is in Canada. If you wait. And wait. And wait. And... oops, died already. But at least the bigtime politicians and bureaucrats get their healthcare! (By paying cash in Yankee hospitals...) -- zimriel sbc dot at global net . http://pages.sbcglobal.net/zimriel/blog/zimblog.html because everyone else is doing it ==== > > What with the war on terror, and Canada's puny military and > corrupt police force, the only way to insure peace and safety for > that country, is for it to be annexed by America. > > It is the only way foward for Canada aka Canuckistan. > > --------------------------------- > > Well, another solution is the liberation of U.$.A by Canadians > and the return of democracy to the Americans peoples. > I am sure Canadians will be welcome with (french) kisses and > flowers when they will topple the Little Dictator and his gang of > corporate crooks. Patriot act will be scraped; free press will be > restaured, Health Care will be available to all citizens, Yeah, like it is in Canada. If you wait. And wait. And wait. And... > oops, died already. But at least the bigtime politicians and bureaucrats get their > healthcare! (By paying cash in Yankee hospitals...) Too bad the life expectancy and infant mortality rates do not back up your claim... but then maybe the exurbanite number of deaths by guns (in the US) is skewing the results on the American side? By the way, not loosing my life savings is well worth a few months wait for elective surgery! Some might dispute this but the majority of life threatening health concerns are treated quickly and professionally. When there are claims that care was not provided in a timely matter it is big news in Canada, and this does not happen very often. ==== Yeah, like it is in Canada. If you wait. And wait. And wait. And... oops, > died already. You morons harping on this bullshit know so little of what you speak. Yes, it happens, but certainly not to the extent that you like to believe, and certainly not to the extent that poor people get turned away at the emergency room doors of hospitals in the good ol' USA. I'll tell you, I was born with a CHD, and have had nothing but the best care my whole life. I now need a heart transplant, and would NOT want to have to be in the States for it. Idiot. -- x being the smartest window-licker on the shortbus doesn't make you a genius By the way, I'm petty and snide. Dumbass Dentistry - it's all fun and games until vegetation starts growing in someone's pulmonary valved conduit. ==== I am not sure about this, but I was just wondering if anyone knows anything about this: if a function is bounded above, is the derivative bounded above? I am thinking this can't be true in general, but I am not sure; I haven't tried to many cases yet. Are there any conditions where this is true? Does anyone know of anything related to this? TIA, Lurch ==== > > I am not sure about this, but I was just wondering if anyone knows anything > about this: if a function is bounded above, is the derivative bounded > above? I am thinking this can't be true in general, but I am not sure; I > haven't tried to many cases yet. Are there any conditions where this is > true? Does anyone know of anything related to this? > > TIA, > > Lurch > > f(x) = sin(x^2) is clearly bounded above and below. but f'(x) = 2*x*cos(x^2) is clearly not. ==== > I am not sure about this, but I was just wondering if anyone knows anything > about this: if a function is bounded above, is the derivative bounded > above? I am thinking this can't be true in general, but I am not sure; I > haven't tried to many cases yet. Are there any conditions where this is > true? Does anyone know of anything related to this? > No, sqr x and sin 1/x for example. I'll guess the condition you want is f' exists on compact domain. ==== >> I am not sure about this, but I was just wondering if anyone knows anything >> about this: if a function is bounded above, is the derivative bounded >> above? I am thinking this can't be true in general, but I am not sure; I >> haven't tried to many cases yet. Are there any conditions where this is >> true? Does anyone know of anything related to this? >No, sqr x and sin 1/x for example. >I'll guess the condition you want is f' exists on compact domain. Not necessarily, as the OP only stated bounded above. For example f(x)=-ln(x) is both bounded above and has a derivative that is bounded above. ==== >> I am not sure about this, but I was just wondering if anyone knows anything >> about this: if a function is bounded above, is the derivative bounded >> above? I am thinking this can't be true in general, but I am not sure; I >> haven't tried to many cases yet. Are there any conditions where this is >> true? Does anyone know of anything related to this? >No, sqr x and sin 1/x for example. >I'll guess the condition you want is f' exists on compact domain. Not necessarily, as the OP only stated bounded above. For example > f(x)=-ln(x) is both bounded above and has a derivative that is bounded > above. Yes, that was the function which originally got me thinking of this, then I all your help! Lurch ==== >> I am not sure about this, but I was just wondering if anyone knows anything >> about this: if a function is bounded above, is the derivative bounded >> above? I am thinking this can't be true in general, but I am not sure; I >> haven't tried to many cases yet. Are there any conditions where this is >> true? Does anyone know of anything related to this? >No, sqr x and sin 1/x for example. >I'll guess the condition you want is f' exists on compact domain. No, that's not enough. For example let f(x) = x^2 sin(1/x^10) for x <> 0, f(0) = 0. Then f' exists on all of [-1,1] but it's not bounded on [-1,1]. ************************ David C. Ullrich ==== sin(1/x) I am not sure about this, but I was just wondering if anyone knows anything > about this: if a function is bounded above, is the derivative bounded > above? I am thinking this can't be true in general, but I am not sure; I > haven't tried to many cases yet. Are there any conditions where this is > true? Does anyone know of anything related to this? TIA, Lurch ==== I am not sure about this, but I was just wondering if anyone knows anything > about this: if a function is bounded above, is the derivative bounded > above? I am thinking this can't be true in general, but I am not sure; I > haven't tried to many cases yet. Are there any conditions where this is > true? Does anyone know of anything related to this? TIA, Lurch of functions, but I can think of several functions where it does work: sin(x), Cos(x), etc... It is really just a fleeting thought, it may mean nothing, but I thought I would run it by some of you. It would help me in a problem I am solving. TIA, Lurch ==== > > I am not sure about this, but I was just wondering if anyone knows anything > about this: if a function is bounded above, is the derivative bounded > above? I am thinking this can't be true in general, but I am not sure; I > haven't tried to many cases yet. Are there any conditions where this is > true? Does anyone know of anything related to this? What can you say about the derivative of x^1/3, near x=0? Geometrically, does this curve have a tangent line at x=0? If so, what's its slope? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hB8LtB803947; ==== >use of limits. Here is his argument. Please critique and respond. Division by 0 is currently undefined, but consider defining it as infinity. >The reciprocal of infinity is defined as 0, so why would the reciprocal of 0 >not be infinity. If this is so, it creates an interesting problem. The >reciprocal of 0 could be represented by 1 over 0, which, as stated, could be >defined as infinity, but what about numbers other than one. When any real >number (except 0) is divided by a number closer and and closer to 0, the result >is always a number closer and closer to infinity. Therefore, when 0 is >reached, the result should be infinity, so any real number (except 0) divided >by 0 should be infinity. This is where the problem occurs. If both sides of >the equation are multiplied by 0, the result is infinity times 0, which is also >currently undefined (not 0), but if 0 and infinity are reciprocals of one >another, then infinity times 0 should yield one. Therefore, any real number >(except 0) divided by 0 should be defined as infinity, and infinity times 0 >should be defined as one. > ZERO CAN NOT I REPEAT CAN NOT BE A DENOMINATOR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ==== >>
Help me!
Did it occur to you that the person who asked for help _over five years 
ago_
is unlikely to still need assistance? Please do not resuscitate dormant
threads at Math Forum.
>ZERO CAN NOT I REPEAT CAN NOT BE A 
DENOMINATOR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Did it further occur to you that simply saying no is not likely to be
very helpful? Adding many factorial signs is unlikely to improve the 
situation.
dave
====
>>
>Help me!
 Did it occur to you that the person who asked for help _over five years
> ago_ is unlikely to still need assistance?
Yikes. When I responded to Amanda, I didn't notice the date of the post
to which she had responded.
> Please do not resuscitate dormant threads at Math Forum.
I suppose that's a reasonable request, but I'm not really sure why. OTOH,
it seems to me that there could be times when it would be desirable to post
to an old thread. Suppose, for example, that you posted something which,
much later, was found to be erroneous. Of course, you could post a
correction as a new thread, but it then would be unlikely that someone
reading the old thread would ever be aware that it had been corrected.
Surely, inserting the correction in the old thread would be better.
>ZERO CAN NOT I REPEAT CAN NOT BE A
>DENOMINATOR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
>!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 Did it further occur to you that simply saying no is not likely to be
> very helpful? Adding many factorial signs is unlikely to improve the
> situation.
Right. Many factorials signs are not needed, one is enough because...
TAH-DAH
nobody has trouble dividing by 0!
David
====
>Help me!  I have a student that insists that one can divide by zero
>without the use of limits.  Here is his argument.  Please critique and
>respond.
>  >Division by 0 is currently undefined, but consider defining it as
>infinity. The reciprocal of infinity is defined as 0, so why would the
>reciprocal of 0 not be infinity.  If this is so, it creates an
>interesting problem.  The reciprocal of 0 could be represented by 1 over
>0, which, as stated, could be defined as infinity, but what about
>numbers other than one.  When any real number (except 0) is divided by a
>number closer and and closer to 0, the result is always a number closer
>and closer to infinity.  Therefore, when 0 is reached, the result should
>be infinity, so any real number (except 0) divided by 0 should be
>infinity.
All fine so far.
>This is where the problem occurs.
Yes, perhaps it's fair to call it a problem. But I'd prefer to say 
merely
that this is where one must be careful.
>If both sides of the
>equation are multiplied by 0, the result is infinity times 0, which is
>also currently undefined (not 0), but if 0 and infinity are reciprocals
>of one another, then infinity times 0 should yield one.
And that conclusion is wrong. Suppose, for example, that the system you're
using is the one-point extension of the reals, R*. In that extension,
x/0 = oo, _unsigned_ infinity, for all nonzero x, and x/oo = 0 for all
finite x. But these results are not based _directly_ on the notion of
multiplicative inversion. So, although I think it was perfectly fair for
you to say 0 and infinity are reciprocals of one another, it would be
incorrect to conclude that they are then multiplicative inverses of each
other. In fact, 0 and oo cannot (at least in any system I'd call
reasonable) have multiplicative inverses. In such systems as the one-point
extension of the reals, we find that reciprocal of x = multiplicative
inverse of x iff x is finite and nonzero.
That's why I'd said that one must be careful above. When we extend a
number system in the hope of gaining something, we invariably loose
something too, it seems. Of course, in one sense, we didn't loose anything
in this instance, for we may still say in R*, just as we did in R, that
reciprocal of x = multiplicative inverse of x for all nonzero real 
x.
>Therefore, any
>real number (except 0) divided by 0 should be defined as infinity,
Yes. Unsigned infinity.
>and infinity times 0 should be defined as one.
No. In extensions such as R*, oo * 0 is often just undefined. (There are
other possibilities which are good, but taking that product to be 1 is
definitely not one of them!)
BTW, you might find

to be interesting.
> ZERO CAN NOT I REPEAT CAN NOT BE A
> DENOMINATOR![snip of many exclamation points]!
Well, at least you're right if we restrict ourselves, say, to the real or
complex number systems. (But there's no need to shout.)
David
====
> No. In extensions such as R*, oo * 0 is often just undefined. (There are
> other possibilities which are good, but taking that product to be 1 is
> definitely not one of them!)
Yes.  For one thing, you lose the associative property of
multiplication:
(2 * 0) * oo = 1
2 * (0 * oo) = 2
-- 
Daniel W. Johnson
panoptes@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
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====
> On sci.math, I think the consensus was that *mathematicians* and math
>> students were the only real group members, and that they had the right
>> to command postings from others, so that if you pissed them off, you
>> should leave if they commanded you to leave the group.
I did not know that.
>
  Well, now you do.  Clear out your desk and turn in your
keys.  Consider yourself commanded.  We are pissed.  If you
ever post here again, your right to use the letters QED
will be revoked.  Now get out before we have to sic David Ullrich 
on you.
  A.M.
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====
>My fourth graders are always asking why?  Why do we need to learn that?  
Why is this important?
Thye want to know if there is a profession that does not require math?
Actually, there is no ultimately important field of study.
What they need, in my opinion and my grandfather's, is the passion,
the desire to be extremely good at one thing they are interested.
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====
 I am actually working on the same problem, but I am using the boundary 
value problem function in MATLAB...i am having a hard time getting this to 
work though. if you have any ideas, please let me know, i can really use  
your help. thanks. manish 
>
I've recently been assigned a computer project to solve the
>Taylor-Maccoll Diff. EQ.  The equation has been normallized to
G*(1-Vr^-Vt^2)*(2*Vr+Vt*cot(s)+y)-Vt*(Vr*Vt+Vt*y)=0
where
G=(1.405-1)/2,
Vt=dVr/ds and dVt/ds=d^2Vr/ds^2=y.
The initial conditions are set by V^2=Vr^2 + Vt^2.
>V is found by the Mach number which is guessed and s is a guess
>initially.
The idea to solve for this equation is that there is an initial
>Vr and Vt value.  For each initial s, there corresponds a cone
>angle c.  Vr is the radial velocity and Vt is the polar velocity.
>Therefore, Vt is 0 when s=c.
This means that each step of the Runge-Kutta routine, s=s-h, where
>h is the step size.  We stop when Vt=0.
My problem is that I have never solved an equation by marching
>backwards.  Is there anything special I need to do in my routine?
>I am using a positive step size, and my k values in Runge-Kutta
>are somewhat like
k2=h*f(s+h/2,Vr+0.5*k1,Vt+0.5*l1) and l2 is similar with g(s,Vr,Vt)=y
> from above. f(s,Vr,Vt) is simply equal to Vt.
Do I need to negativize the terms in the k and l terms also?
I have finished the program but my values are incorrect somewhat.
Please help me if anyone has had experience marching backwards with
>numerical methods.

>

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====
>It's well known that if a real valued function f is monotonic on a
>compact interval I then f is integrable over I and the set of
>discontinuities of f on I is countable.  I'm not sure, but it seems
>to me that if we combine this fact with the Fundamental Theorem of
>Integral Calculus (in the form that deals with the derivative of an
>integral) and consider Lebesgue Integrability Criterion, then we
>come to the conclusion that the set of elements of I at which f is
>not differentiable has measure zero.  Is this conclusion true?
>Amanda
Perhaps this is a counter example?
Define:
f(0)=0
f(x+1)=(f(x)+1)/2
f(1/x)=1-f(x)
This defines f(x) for rational x>=0.  The irrationals can be filled
in using limits to define a continuous monotonic function which has
derivative equal to zero at all rationals as well as any other point
at which the function is differentiable.
====
>It's well known that if a real valued function f is monotonic on a
>compact interval I then f is integrable over I and the set of
>discontinuities of f on I is countable.  I'm not sure, but it seems
>to me that if we combine this fact with the Fundamental Theorem of
>Integral Calculus (in the form that deals with the derivative of an
>integral) and consider Lebesgue Integrability Criterion, then we
>come to the conclusion that the set of elements of I at which f is
>not differentiable has measure zero.  Is this conclusion true?
>Amanda
> 
> 
> Perhaps this is a counter example?
> 
> Define:
> 
> f(0)=0
> f(x+1)=(f(x)+1)/2
> f(1/x)=1-f(x)
> 
> This defines f(x) for rational x>=0.  The irrationals can be filled
> in using limits to define a continuous monotonic function which has
> derivative equal to zero at all rationals as well as any other point
> at which the function is differentiable.
I think this function is defined for every integer n and for every
number of the form 1/n, n<>0, n integer. It's increasing for x>=0, but
I don't see why it's defined for every rational >=0.
Anyway, this is not a counter example, you didn't show a monotonic
function such that the set of points where it's not differentiable is
not null.
Amanda
====
>>It's well known that if a real valued function f is monotonic on a
>>compact interval I then f is integrable over I and the set of
>>discontinuities of f on I is countable.  I'm not sure, but it seems
>>to me that if we combine this fact with the Fundamental Theorem of
>>Integral Calculus (in the form that deals with the derivative of an
>>integral) and consider Lebesgue Integrability Criterion, then we
>>come to the conclusion that the set of elements of I at which f is
>>not differentiable has measure zero.  Is this conclusion true?
>>Amanda

>Perhaps this is a counter example?
There are no counterexamples - any monotonic function is
in fact differentiable except on a set of measure zero.
(See any book on real analysis, meaning a book that
starts with measure theory...)
>Define:
f(0)=0
>f(x+1)=(f(x)+1)/2
>f(1/x)=1-f(x)
This defines f(x) for rational x>=0.  The irrationals can be filled
>in using limits to define a continuous monotonic function which has
>derivative equal to zero at all rationals as well as any other point
>at which the function is differentiable.
I don't follow your definition of f. But there are in fact 
non-constant strictly increasing functions that have
derivative equal to 0 almost everywhere; how does that
give a counterexample?
************************
David C. Ullrich
====
nice.
> 
> I want to implement an extended Euclidean algorithm in MATLAB that can
> compute the following: GCD(A(x),B(x)) = u(x)A(x) + v(x)B(x), where
> A(x), B(x), u(x) and v(x) are from GF(2^m).
> 
> I have a basic understanding how this works for integers, but don't
> want to re-invent the wheel for the above case.  I did a search on
> IEEEXPLORE, and the closest hit that I got was from Mandelbaum,
> although this was a hardware implementation.  Does anyone have perhaps
> pointers to literature where this is clearly explained?
> 
> (I once got a glimpse of such a technique in a textbook, but I spent
> the whole morning trying to find the book in the library without
> success.  It did something with a matrix)
> 
> I will greatly appreciate any suggestions and help
> Jaco
====
One method to compute the inverse of a polynomial in GF(2^m)[x] is to
use the extended Euclidean algorithm.  Are there perhaphs other
methods to compute the inverse?  Can one simplify the Euclidean
algorithm, as one already knows that the GCD should be 1?
Another challenge that I have is the following:
Suppose I work with the ideal , where g(x) divides x^n + 1 (both
are polynomials in GF(2^m)[x]).  If I have a factor ring a(x) + H,
where H = , all the polynomials in the factor ring a(x) + H will
have the same inverse a^{-1}(x) (In fact, the factor ring a^{-1}(x) +
H should be all inverses of the factor ring a(x) + H, but using the
extended Euclidean algorithm always yield a(x), where degree(a(x)) <
degree(g(x)) ).  (Actually, we work with a ring
GF(2^m)[x]/[x^n+1]/[g(x)], i.e. all polynomials in GF(2^m)[x] of
degree less than g(x), and these elements form a ring)  How can I find
a polynomial b(x) in a(x)+ H, if I have the degree of b(x), and the
inverse of a(x) + H is known (a^{-1}(x))?  The extended Euclidean
algorithm will give only a(x), where degree(a(x)) < degree(g(x)).
Any help and/or suggestions will be greatly appreciated
Jaco
====
> 
> One method to compute the inverse of a polynomial in GF(2^m)[x] is to
You mean GF(2^m)[x]/ methods to compute the inverse? 
Yes, but are they as good? Let g have degree d. Write the
putative inverse of f(x) in GF(q)/ as
h(x) = a_0 + a_1 x + ... + a_{d-1}x^{d-1}.
Reducing fh modulo g one gets a system of d linear equations
for the a_j. Is solving these preferable to using the EA?
> Another challenge that I have is the following:
> Suppose I work with the ideal , where g(x) divides x^n + 1 (both
> are polynomials in GF(2^m)[x]).  If I have a factor ring a(x) + H,
a what? That looks like a coset to me!
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
====
> 
> 
> One method to compute the inverse of a polynomial in GF(2^m)[x] is to
> use the extended Euclidean algorithm.  Are there perhaphs other
> methods to compute the inverse?  Can one simplify the Euclidean
> algorithm, as one already knows that the GCD should be 1?
I assume that you mean to compute inverses in a quotient ring 
GF(2^m)[x]/,
as very few polynomials have inverses in the ring GF(2^m)[x].
I don't think you can do an awful lot. IIRC something called an almost
inverse algorithm gives some speed enhancement, in particular, when
m=1, and the degree of g(x) is relatively large.
> Another challenge that I have is the following:
> Suppose I work with the ideal , where g(x) divides x^n + 1 (both
> are polynomials in GF(2^m)[x]).  If I have a factor ring a(x) + H,
> where H = , 
FYI, a(x)+H is called a coset. It is an element of the quotient ring 
GF(2^m)[x]/H.
>                 all the polynomials in the factor ring a(x) + H will
> have the same inverse a^{-1}(x) (In fact, the factor ring a^{-1}(x) +
> H should be all inverses of the factor ring a(x) + H, but using the
> extended Euclidean algorithm always yield a(x), where degree(a(x)) <
> degree(g(x)) ). 
Your use of terminology is slightly confusing. Strictly speaking it is
the coset a(x)+H that has the inverse, not any particular polynomial
within this coset. More often than not we equate a coset with any of its
representatives (i.e. any of its elements), but I can't help getting the
impression that you are not completely familiar with the distinction.
 (Actually, we work with a ring
> GF(2^m)[x]/[x^n+1]/[g(x)], i.e. all polynomials in GF(2^m)[x] of
> degree less than g(x), and these elements form a ring)  How can I find
> a polynomial b(x) in a(x)+ H, if I have the degree of b(x), and the
> inverse of a(x) + H is known (a^{-1}(x))?  The extended Euclidean
> algorithm will give only a(x), where degree(a(x)) < degree(g(x)).
Well, the inverse of the coset a(x)+H is another coset f(x)+H, where
f(x)a(x)-1 will be in H. (so my f(x) is your a(x)^{-1}). Surely you
can produce a lot of elements of the coset f(x)+H, once you know
a single polynomial, e.g. the lowest degree guy, in this coset:
When H=, then f(x)+H consists of polynomials of the form
f(x)+g(x)p(x), where p(x) is an arbitrary polynomial.
In other words, this coset has plenty of polynomials of any chosen
degree at least deg g(x), but only one of degree less than deg g(x).
Jyrki Lahtonen
====
Arturo Magidin  scribbled the following:
>>Jos.8e Carlos Santos  scribbled the following:
>>  
>By that standard you could insist that anybody who says there's 
>no highest prime number can't be an agnostic. All of math is based 
>on persuasive definition, not evidence.
>> 
>>  
>> Not being a mathematician, I don't know the best way to respond to
>> this; so I was hoping that someone with more knowledge of the subject
>> could address it.
> It is *false* that the statement there is no highest prime is true
> by definition. If some number p was the highest prime, then p! + 1 
would
> not be prime, since it would be greater. But, not being prime, it can
> be written has a product of several prime numbers. If p' is one such
> prime number, then p! + 1 is a multiple of p'. But then so is p! and
> therefore 1 is a multiple of p', which is not possible, since every
> prime number is greater than 1.
>>I don't understand this proof. Are you saying that if p is prime, then
>>p! + 1 must also be prime?
> Not at all. He is saying that there is no prime p such that every
> number strictly larger than p is composite. 
I thought that must not have been what he was saying. (I.e. that if p
was prime, so was p! + 1.)
> Since the existence of a highest prime would imply the existence of
> a prime p such that every number larger than p is not a prime (hence
> composite), it follows that there is no highest prime.
> The argument is that if p were a prime with the property that for all
> n, if n>p then n is composite, then we would have that p!+1 is
> composite (by being larger than p), and so must be divisible by some
> prime q. Since q>p   implies q composite, we must have q<=p, and
> therefore, q|(p!); since q also divides p!+1, you deduce that q
> divides 1, which is impossible. The contradiction arose form assuming
> that there existed a prime p such that every number strictly larger
> than p is a composite.
OK... q divides p!+1 because we just said so (it's how we chose q), and
q divided p!, because it's a prime number bigger than 1 and smaller
than p, so it's bound to appear as a factor of p! (every such number,
prime or composite, does). But how do you get the conclusion that q
should divide 1? This is the only thing I'm still not understanding.
-- 
/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland --------
-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
I am lying.
   - Anon
====
> Arturo Magidin  scribbled the following:
>>Jos.8e Carlos Santos  scribbled the following:
>> > 
>By that standard you could insist that anybody who says there's 
>no highest prime number can't be an agnostic. All of math is based 
>on persuasive definition, not evidence.
>> 
>> > 
>> Not being a mathematician, I don't know the best way to respond to
>> this; so I was hoping that someone with more knowledge of the 
subject
>> could address it.
>  
> It is *false* that the statement there is no highest prime is 
true
> by definition. If some number p was the highest prime, then p! + 1 
would
> not be prime, since it would be greater. But, not being prime, it can
> be written has a product of several prime numbers. If p' is one such
> prime number, then p! + 1 is a multiple of p'. But then so is p! and
> therefore 1 is a multiple of p', which is not possible, since every
> prime number is greater than 1.
>>I don't understand this proof. Are you saying that if p is prime, then
>>p! + 1 must also be prime?
>  
> Not at all. He is saying that there is no prime p such that every
> number strictly larger than p is composite. 
> 
> I thought that must not have been what he was saying. (I.e. that if p
> was prime, so was p! + 1.)
> 
> Since the existence of a highest prime would imply the existence of
> a prime p such that every number larger than p is not a prime (hence
> composite), it follows that there is no highest prime.
>  
> The argument is that if p were a prime with the property that for all
> n, if n>p then n is composite, then we would have that p!+1 is
> composite (by being larger than p), and so must be divisible by some
> prime q. Since q>p   implies q composite, we must have q<=p, and
> therefore, q|(p!); since q also divides p!+1, you deduce that q
> divides 1, which is impossible. The contradiction arose form assuming
> that there existed a prime p such that every number strictly larger
> than p is a composite.
> 
> OK... q divides p!+1 because we just said so (it's how we chose q), and
> q divided p!, because it's a prime number bigger than 1 and smaller
> than p, so it's bound to appear as a factor of p! (every such number,
> prime or composite, does). But how do you get the conclusion that q
> should divide 1? This is the only thing I'm still not understanding.
For starters, it's not true that p! + 1 is prime if p is prime, since
this can be blatantly shown to be false for p = 5. (p!+1) = 121 = 11 *
11
The original claim was for the product of all PRIMEs between 1 and p
[which I'll call f(p)]. And remember in the following that p is the
largest prime, so there can be no prime greater that p.
So the statement is that f(p)+1 must be a prime. The counterexample is
that if f(p)+1 is composite, it must have a prime factor q.
Now, this prime factor must have been included in the f(p) function,
so we have the following being true:
q divides f(p) from the definition of f(p)
q divides f(p)+1 from our assertion that q exists
if q divides f(p) and f(p)+1 then it must divide (f(p)+1) - f(p).
Which means that q must divide 1. There's our contradiction. So f(p)+1
cannot have any prime factors.
====
> I thought that must not have been what he was saying. (I.e. that if p
> was prime, so was p! + 1.)
I just cannot understand how can you possibly have thought that I was
saying that.
> OK... q divides p!+1 because we just said so (it's how we chose q), and
> q divided p!, because it's a prime number bigger than 1 and smaller
> than p, so it's bound to appear as a factor of p! (every such number,
> prime or composite, does). But how do you get the conclusion that q
> should divide 1? This is the only thing I'm still not understanding.
Since q divides p! + 1, you can write p! + 1 as q.m for some integer m and
since q divides p! you can write p! as q.n for some integer n. Therefore
   1 = (p! + 1) - p! = q.m - q.n = q.(m - n).
QED
Jose Carlos Santos
====
Jose Carlos Santos  scribbled the following:
>> I thought that must not have been what he was saying. (I.e. that if p
>> was prime, so was p! + 1.)
> I just cannot understand how can you possibly have thought that I was
> saying that.
I misread your proof. You said that if p is the largest prime, then all
numbers >p are composite, which would mean that all the prime factors
of p! + 1 are <=p. I misread that as claiming that if p is prime, then
so is p! + 1.
>> OK... q divides p!+1 because we just said so (it's how we chose q), and
>> q divided p!, because it's a prime number bigger than 1 and smaller
>> than p, so it's bound to appear as a factor of p! (every such number,
>> prime or composite, does). But how do you get the conclusion that q
>> should divide 1? This is the only thing I'm still not understanding.
> Since q divides p! + 1, you can write p! + 1 as q.m for some integer m 
and
> since q divides p! you can write p! as q.n for some integer n. Therefore
>    1 = (p! + 1) - p! = q.m - q.n = q.(m - n).
-- 
/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland --------
-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
'It can be easily shown that' means 'I saw a proof of this once (which I 
didn't
understand) which I can no longer remember'.
   - A maths teacher
====
> OK... q divides p!+1 because we just said so (it's how we chose q), and
> q divided p!, because it's a prime number bigger than 1 and smaller
> than p, so it's bound to appear as a factor of p! (every such number,
> prime or composite, does). But how do you get the conclusion that q
> should divide 1? This is the only thing I'm still not understanding.
q divides p! and q divides p! + 1. Therefore it divides the difference 
(p! + 1) - p! = 1.
====
Christian Bau  scribbled the following:
>> OK... q divides p!+1 because we just said so (it's how we chose q), and
>> q divided p!, because it's a prime number bigger than 1 and smaller
>> than p, so it's bound to appear as a factor of p! (every such number,
>> prime or composite, does). But how do you get the conclusion that q
>> should divide 1? This is the only thing I'm still not understanding.
> q divides p! and q divides p! + 1. Therefore it divides the difference 
> (p! + 1) - p! = 1.
-- 
/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland --------
-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
To doo bee doo bee doo.
   - Frank Sinatra
====
> 
>There is no largest prime, not by definition (directly) but rather
>by a clear chain of reasoning from definitions. It's not a hard proof
>to understand: assume there are a finite number of primes; multiply
>them all together and add one; no prime divides this new number, 
>so it's a new prime: contradiction
>>None of the primes *you already know about* divide this number. Either
>>it is a new prime, or it is composite and divisible by a new prime.
> 
> 
> Jim's proof is in fact correct (but not as clear as it could be).
> Your criticism applies only to an alternative presentation of
> Euclid's proof, namely: given a finite set S of known primes 
> there exists a prime not in S, i.e. a new prime. 
snip
> By the way, Euclid's original proof may be found online at
> http://aleph0.clarku.edu/~djoyce/java/elements/bookIX/propIX20.html
> 
> -Bill Dubuque
Mark Atherton
====
...
 > No.  The proof was complete.  It was not you know about that was
 > mentioned.  When you assume finitely many primes, and multiply *all*
 > those primes together you did not leave any one out.  So the product
 > with 1 added to is is not divisible by *any* existing prime.
 > 
 > Agreed, but why must it be true that the resulting number is itself 
prime?
By definition of the term prime. (*)
 > For example, assuming that the only primes are the first seven leads to:
 > 
 >    2*3*5*7*11*13*17+1 = 510510+1 = 510511 = 17*26869
 > 
 > Here 17 is the prime not in the original list, and contradicts the 
 > assumption that the original list of primes is complete.
But the conclusion that 510511 is a prime is already a contradiction;
it is not on the original list.  Why factor?
(*) Let's assume the following definitions and conclusions in the positive
integers:
1.  A number is prime when it is only divisible by 1 or the number itself.
2.  When a number is not prime, it is either 1 or divisible by a prime.
3.  Let the list of all primes be [2,3,5,7,11,13,17].
4.  Multiply the primes and add 1 to get n (= 510511).
5.  This number can not be divided by any of the primes given in [3], so
    by [2] it is a prime.
6.  Contradiction...
So there is a false assumption, and it must be [3].
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> (*) Let's assume the following definitions and conclusions in the 
positive
> integers:
 1.  A number is prime when it is only divisible by 1 or the number 
itself.
> 2.  When a number is not prime, it is either 1 or divisible by a prime.
> 3.  Let the list of all primes be [2,3,5,7,11,13,17].
> 4.  Multiply the primes and add 1 to get n (= 510511).
> 5.  This number can not be divided by any of the primes given in [3], so
>     by [2] it is a prime.
> 6.  Contradiction...
> So there is a false assumption, and it must be [3].
Yes, but why did you include the clause When a number is not prime
in [2]? If you omit that, you get a simpler proof:
1.  A number is prime when it is only divisible by 1 or the number itself.
2.  Every number is either 1 or divisible by a prime.
3.  Let the list of all primes be [2,3,5,7,11,13,17].
4.  Multiply the primes and add 1 to get n (= 510511).
5.  This number can not be divided by any of the primes given in [3],
    and it is not 1.
6.  Contradiction...
The question, whether 510511 is prime or composite, does not arise.
====
 > 
 > (*) Let's assume the following definitions and conclusions in the 
positive
 > integers:
 >
 > 1.  A number is prime when it is only divisible by 1 or the number 
itself.
 > 2.  When a number is not prime, it is either 1 or divisible by a 
prime.
 > 3.  Let the list of all primes be [2,3,5,7,11,13,17].
 > 4.  Multiply the primes and add 1 to get n (= 510511).
 > 5.  This number can not be divided by any of the primes given in [3], 
so
 >     by [2] it is a prime.
 > 6.  Contradiction...
 > So there is a false assumption, and it must be [3].
 > 
 > Yes, but why did you include the clause When a number is not prime
 > in [2]? If you omit that, you get a simpler proof:
But my goal was not the simplest proof.  Only a proof where
510511 is a prime is an immediately valid conclusion.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> ...
>  > No.  The proof was complete.  It was not you know about that was
>  > mentioned.  When you assume finitely many primes, and multiply *all*
>  > those primes together you did not leave any one out.  So the product
>  > with 1 added to is is not divisible by *any* existing prime.
>  > 
>  > Agreed, but why must it be true that the resulting number is itself 
prime?
> 
> By definition of the term prime. (*)
> 
>  > For example, assuming that the only primes are the first seven leads 
to:
>  > 
>  >    2*3*5*7*11*13*17+1 = 510510+1 = 510511 = 17*26869
>  > 
>  > Here 17 is the prime not in the original list, and contradicts the 
>  > assumption that the original list of primes is complete.
> 
> But the conclusion that 510511 is a prime is already a contradiction;
> it is not on the original list.  Why factor?
> 
> (*) Let's assume the following definitions and conclusions in the 
positive
> integers:
> 
> 1.  A number is prime when it is only divisible by 1 or the number 
itself.
> 2.  When a number is not prime, it is either 1 or divisible by a prime.
> 3.  Let the list of all primes be [2,3,5,7,11,13,17].
> 4.  Multiply the primes and add 1 to get n (= 510511).
> 5.  This number can not be divided by any of the primes given in [3], so
>     by [2] it is a prime.
> 6.  Contradiction...
> So there is a false assumption, and it must be [3].
Okay, I see what you mean.
What's more, I think I see why proofs in most texts include the 
possibility that the number obtained is composite. Silverman (since 
that's what I've got in front of me) only starts with a *finite* list of 
primes, not the assumption that the list is *complete*. He then goes on 
to show that any finite list can have another prime added to it. 
Similarly http://mathworld.wolfram.com/EuclidsTheorems.html assumes a 
finite list of consecutive primes, again not assumed to be complete.
Mark Atherton
====
>What's more, I think I see why proofs in most texts include the 
>possibility that the number obtained is composite. Silverman (since 
>that's what I've got in front of me) only starts with a *finite* list of 
>primes, not the assumption that the list is *complete*.
Wasn't this exact same topic covered a few weeks earlier, with exactly
the same arguments? One side thinks the proof must be written as X and
the other side that it must be written as Y and both sides agree the
other side should go back to freshman algebra class.
====
> 
>What's more, I think I see why proofs in most texts include the 
>possibility that the number obtained is composite. Silverman (since 
>that's what I've got in front of me) only starts with a *finite* list of 

>primes, not the assumption that the list is *complete*.
> 
> Wasn't this exact same topic covered a few weeks earlier, with exactly
> the same arguments? One side thinks the proof must be written as X and
> the other side that it must be written as Y and both sides agree the
> other side should go back to freshman algebra class.
The original Euclid proof was, I believe, of the form that any 
(finite) list of primes is incomplete, and did not involve proof by 
contradiction.
Both the incompleteness proof and the contradiction proof are 
valid, but as a matter of style, many prefer to avoid proofs by 
contradiction when it is as easy to do as in this case.
A similar situation (and a similar argument over which proof is 
best) arises in Cantor's diagonal proof that there is no 
surjection from the integers to the reals. Cantor's original version 
shows only that any list (image of the naturals) of reals is 
incomplete, but many people recast the proof to state it as a proof 
by contradiction.
Now there probably do exist proofs by contradiction for which no 
equally simple direct proof is known, so proof by contradiction 
remains a viable technique, but, as a matter of style, shouldn't it 
be avoded when it is so easy to do so?
====
>The original Euclid proof was, I believe, of the form that any 
>(finite) list of primes is incomplete, and did not involve proof by 
>contradiction.
Both the incompleteness proof and the contradiction proof are 
>valid, but as a matter of style, many prefer to avoid proofs by 
>contradiction when it is as easy to do as in this case.
It seems many proofs can be stated as pairs of of incompleteness-
contradiction proofs - in effect it's the exactly same construction.
For another example Google for the recent discussion over the proof
that any infinite set of reals that sums to a finite value must have
at most countably many non-zero elements.
>A similar situation (and a similar argument over which proof is 
>best) arises in Cantor's diagonal proof that there is no 
>surjection from the integers to the reals. Cantor's original version 
>shows only that any list (image of the naturals) of reals is 
>incomplete, but many people recast the proof to state it as a proof 
>by contradiction.
And some people recast it as evidence of a massive conspiracy by
mathematicians, but the less spoken about that the better.
>Now there probably do exist proofs by contradiction for which no 
>equally simple direct proof is known
I tried looking for a simple proof that involved proof by
contradiction and about the first thing I could find involved proving
that 0 < 1:
Since we can prove 0 =/= 1 based on the axioms of reals and 1 < 0
leads to a contradiction because of (again by the axioms of reals):
1 < 0 = > 1 + (-1) < 0 + (-1) => 0 < -1 => 0 < (-1)(-1) => 0 < 1 
we conclude that the only option is 0 < 1. Maybe this doesn't count as
a true proof by contradiction but rather as exhausting all other
options? Still, it relies on a contradiction to prove a certain case
impossible.
====
> 
> 
>>What's more, I think I see why proofs in most texts include the 
>>possibility that the number obtained is composite. Silverman (since 
>>that's what I've got in front of me) only starts with a *finite* list of 
>>primes, not the assumption that the list is *complete*.
> 
> 
> Wasn't this exact same topic covered a few weeks earlier, with exactly
> the same arguments?
Don't recall it. Can't find it on Google.
> One side thinks the proof must be written as X and
> the other side that it must be written as Y and both sides agree the
> other side should go back to freshman algebra class.
I'm not so stupid as to suggest that Dik T. Winter go back to basic 
algebra (shouldn't that be number theory?) - I'll leave that to JSH. To 
quote myself:
> I appreciate that I am probably missing some subtle point here so please 
enlighten me! 
> 
> Mark Atherton
Mark Atherton
====
>> 
>What's more, I think I see why proofs in most texts include the 
>possibility that the number obtained is composite. Silverman (since 
>that's what I've got in front of me) only starts with a *finite* list of 

>primes, not the assumption that the list is *complete*.
>> 
>> Wasn't this exact same topic covered a few weeks earlier, with exactly
>> the same arguments?
Don't recall it. Can't find it on Google.


I've heard of long-running Usenet threads but this particular argument
must have been going on since Euclid himself so it must be some kind
of record.
>> I appreciate that I am probably missing some subtle point here so please 
enlighten me! 
The problem often is that even the simplest proof has one person state
it elegantly but so that the reader cannot understand it or
misunderstands it, then somebody corrects the elegant proof to be more
illustrative at which point a third person corrects the second person
but makes a mistake so a fourth person then corrects the third person
and so on...
The definition of rigorous mathematics is, when given a solution to a
problem one understands the solution but no longer understands the
problem.
====

>I would gather
>(broadly) that deF's approach takes the intuitive meaning of
>probability and then derives axioms upon which the theory may be
>developed, trying to avoid the artificial math that goes with the
>traditional approach, such as the limit as n goes to infinity, etc. 
>This strikes me as sensible insofar as probability theory seeks
>application to real-world events.  Even though the theory based on
>Kolmogorov's Axioms is fascinating in itself and certainly worth
>studying, which approach do you think would be more fruitful in terms
>of applications?
>
Just study probability as it usually is done (i.e., with countable 
additivity).  It is still a coherent system, even if not strictly 
implied by more basic axioms.  You can think about relaxation of the 
probability axioms later, if you wish.  As Herman has pointed out, this 
is not necessarily an easy task.
Personally, I (a Bayesian) am pretty much in agreement with De Finetti's 
interpretation of probability, but I am not bothered by countable 
additivity, since it does make the analysis much easier.
-- 
Stephen J. Herschkorn                        herschko@rutcor.rutgers.edu
====
Today I landed a pretty idea. Whether it flys or not is unknown as
yet. It came to me whilst thinking of those cylindrical farm bins used
to store grains. Suppose one were to build one not out of steel but
instead from concrete block or brick. Expensive I know but consider
it. And suppose the builder was given a 3/8 joint for mortar.
Pretty Idea: then given a standard size brick would dictate the
circumference and diameter of this building. So that given the brick
size would determine the size of the building where no sawing of brick
is allowed. Granted of course the building is not really a circle but
we can consider the midpoint inside each brick once the building is
completed sribes a circle.
What is pretty about this idea is the fact that a rectangle determines
a circle. So that the circle is quantized from a rectangle.
I have not thought of the reverse where given a circle determines or
quantizes
a rectangle or a square. It maybe even possible that a circle just
does not quantize a square or rectangle and if that is the case gives
deep mystery and deep implications as to why this Cosmos is built
where rectangles can quantize circles but not the reverse. And if that
is the case would go back to the idea that the cosmos is one gigantic
plutonium atom which is cosmically spherical and cylindrical and that
life sees Euclidean geometry not as a reflection of reality but as a
result of the mind trying to make sense of the world.
Euclidean Geometry in this sense does not exist in Nature but only as
a figment of the imagination of a mind. Euclidean Geometry would be
like those many Optical Illusions we see such as the artist Escher.
I doubt that Euclidean Geometry is a illusion for zero is a true
existing number and that Euclidean Geometry is the zero curvature
geometry.
But, if rectangles and squares can generate circles but the reverse is
not true would have deep implications upon Euclidean Geometry.
Archimedes Plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
====
> Today I landed a pretty idea. Whether it flys or not is unknown as
> yet. It came to me whilst thinking of those cylindrical farm bins used
> to store grains. Suppose one were to build one not out of steel but
> instead from concrete block or brick. Expensive I know but consider
> it. And suppose the builder was given a 3/8 joint for mortar.
> 
> Pretty Idea: then given a standard size brick would dictate the
> circumference and diameter of this building. So that given the brick
> size would determine the size of the building where no sawing of brick
> is allowed. Granted of course the building is not really a circle but
> we can consider the midpoint inside each brick once the building is
> completed sribes a circle.
> 
> What is pretty about this idea is the fact that a rectangle determines
> a circle. So that the circle is quantized from a rectangle.
> 
> I have not thought of the reverse where given a circle determines or
> quantizes
> a rectangle or a square. It maybe even possible that a circle just
> does not quantize a square or rectangle and if that is the case gives
> deep mystery and deep implications as to why this Cosmos is built
> where rectangles can quantize circles but not the reverse. And if that
> is the case would go back to the idea that the cosmos is one gigantic
> plutonium atom which is cosmically spherical and cylindrical and that
> life sees Euclidean geometry not as a reflection of reality but as a
> result of the mind trying to make sense of the world.
> 
> Euclidean Geometry in this sense does not exist in Nature but only as
> a figment of the imagination of a mind. Euclidean Geometry would be
> like those many Optical Illusions we see such as the artist Escher.
> 
> I doubt that Euclidean Geometry is a illusion for zero is a true
> existing number and that Euclidean Geometry is the zero curvature
> geometry.
> 
> But, if rectangles and squares can generate circles but the reverse is
> not true would have deep implications upon Euclidean Geometry.
> 
> Archimedes Plutonium
> whole entire Universe is just one big atom where dots
> of the electron-dot-cloud are galaxies
.....
A circle is a complete circuit and can conduct vibrations in sets of
standing waves where lamba(n) = 2(pi)v. A circle can therfore conduct
a system of waves with four peaks and troughs perfectly spaced which
would establish precise four corners of a square.
Your inquiry is on the right track for identifying basic rules of
quantization which determines size and mass of the electron, proton
and neutron. If you integrate the concept of tensile strength. A lomg
steel rod could be bent around to form a circle. A very short piece
could never be bent into a circle before it breaks. The tensile
strength would in a way, determine similar units as to your bricks.
If someone threw a ball bearing at the quantum observer and hit him in
the head when he wasn't looking, I think he would give up the idea
that his observation is determining his reality. The path of the ball
bearing can be described by Euclidean geometry. The vectors of the
force applied are real and are in a straight line. It is not the
mental picture of the person throwing the ball bearing either. It is
the adenine tri-phosphate in his muscles and coordination of the
transmiters and inhibitors in a complex nervous system that does it.
In analysing the paths of orbital bodies, only Euclidean geometry
works. All vectors are in a straight line. The vector at the apex of
the ellipse, (the highest velocity), if plotted on a graph, according
to the pull of gravity at any given moment which is also in a staight
line, determines the entire orbit. This velocity and the distance from
the scource of gravitation that this velocity occurs, determines the
eccentricity. All orbits are identical according to; a cubed = p
squared. This law is a direct product of the inverse square law for
gravitational attraction and the second law of radial vectors is
derived from the third law.
The vectors are always in a straight line. No momentum to rotate in a
curve exists. The angular momentum is a combined product of the
momentum(velocity x mass) and the continueing effect of the
gravitation.
Gausian coordinates are valuable however since the force of gravity is
relative to the distance. Equal force is described by concentric
circles.
Both systems of coordinates apply.
http://home.earthlink.net/~kdthrge
====
Obviously if we go the other way of starting with a circle or spheroid
in 3D that we can construct a rectangle or square. For it is simply a
Kepler Packing such as oranges in a grocery aisle.
But there is an asymmetry of quantization. For it is more difficult to
construct a circle out of a rectangle or square than to construct a
rectangle or square given circles. There is the asymmetry that the
rectangle determines a unique circle. But the reverse is that the
circle determines an infinity of rectangles or squares. So the
rectangle quantization makes a unique circle but the reverse of
starting with a circle can determine an infinity of squares or
rectangles. Example: given a circle then 4 of them make a square and 6
of them make a rectangle and 16 of them make a different square.
I am guessing that there exists a Mathematical Theorem already stated
and proved of what I am saying in the above. If not, then I have
discovered another new Math theorem, perhaps for geometry or
projective-geometry.
And as stated in my other post last night on this subject regarding
Euclidean Geometry as a optical illusion. Apparently it is not that.
But the Asymmetry of the above must tie into or link with the
asymmetry of Euclidean geometry versus Riemannian and Lobachevskian.
Euclidean Geometry is equivalent to the number zero on the Real Number
line which then says that Riemannian geometry has an infinity of
curvatures for every positive number on the Real Number line is a
specific Riemannian geometry, likewise for Lobachevskian with negative
Reals. So the infinity of circles to craft rectangles or squares, but
the reversal of where a rectangle or square can only craft a unique
circle is reflected in the fact that Euclidean Geometry is a unique
Real of zero whereas Riem and Loba are infinite.
Archimedes Plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
====
> A finite group G is defined p-solvable if it has a chief series with
>> every factor a p-group or a p'-group. A trivial example of a
>> p-solvable group is any G simple with p a prime not dividing
>> |G|. Another rather trivial example is C_pxG, with G as before and C_p
>> the cyclic group of order p. I'm looking for an example of a
>> p-solvable group with O_p(G)=1, where O_p(G) is the largest normal
>> p-group. 
>> 
>> What about G = S_3? It is 2-solvable and O_2(G) = 1.
> If you want it non-solvable, than your wreath-product of A_5 with C_7 
is
> probably of minimal order.
>>the condition that p divides |G| (since Even G=C_3 is 2-solvable and
>>O_2(G)=1).
>>Rafael
A still smaller example is P Gamma L (2,32) (an extension of the simple 
group
>PSL(2,32) by a field automorphism of order 5), which has order 33.32.31.5 
=
>163680. It is 5-solvable with O_5(G) = 1.
Derek Holt.
And a smaller one still is Sz(8).3, an extension of the Suzuki group
Sz(8) by a field automorphism of order 3. This has order 87360 and is
3-solvable with O_3(G) = 1. Any further improvements?
Derek Holt.
====
----------------------------------------------------------------------------
------
> JS: Astronauts float in zero g all the way round and round the timelike
> geodesic passing to different LIFs like Duchamp's Nude Descending a
> Staircase........
 [Hammond]
> Jack... IMHO there are poets who don't know it and
> there are physicists who understand God BUT don't know it.
> Duchamp's descending image concatenations, I
> find you in the latter catagory.
>   Your analogy is however a bit sloppy.  Duchamp's
> nude descending the staircase is more a case of
> adiabatic transitions between frames of differing
> gravitational potential (differing curvature), with
> the descent of the stairway indicating the gradient
> of the curvature.
>   Anyway, it's nice to know there is someone else
> out there besides me and Chris Isham who think
> there is a scientific explanation of God.
> =================================
> HANMMOND'S PROOF OF GOD WEBSITE
> http:geocities.com/scientific_proof_of_god
> =================================
I have a proof of God website at www.adamskingdom.com
Will be greatly expanded soon :
Did you know that I broadcast to half dozen newsgroups on 02 02 2002
with the subject line PROOF OF GOD.
www.tinyurl.com/gutr     G.U.T. Religion by 'coincidence' of tinyurl 
database!
One year to the day before the shuttle disaster!
Does your theory of distributed awareness involve the seperate (non local) 
causality effect
of quantum entanglement?
Herc
====
[snip]
>             Basic GR works well in all four of its classical
> predictions: light-bending, gravitational redshift, Shapiro delay, and
> perihelion advance. 
             perihelion advance
Please, Tom point out planets for which one this type of corrections is 
applied. 
> The problem areas are in 2-significant-mass
> pericenter advance, frame dragging, and higher order effects. Also, some
> alternative interpretations predict phenomena that GR does not, such as
> gravitational shielding, a MOND-like effect to replace dark matter,
> heat generation in massive bodies, etc. The next ten years will probably
> see the definitive testing of most of these. -|Tom|-
> 
> 
> Tom Van Flandern - Washington, DC - see our web site on replacement
> astronomy research at http://metaresearch.org
====
> Suppose f(x) is in Z[x], is irreducible of degree n and has all real
> roots.
> 
> 
> Is there anything in the literature about the degree over Q of the
> splitting field of an irreducible polynomial of degree n that has all 
real
> roots? In particular, I was hoping that there might be some result out
> there someplace that gives a better upper bound than n!.
no: the generic irreducible polynomial with n real zeroes
has Galois group S_n.
 
> Is there an algorithm that can determine the degree over Q of the
> splitting field of a polynomial f(x) in Z[x] from its coefficient, 
without
> actually computing the splitting field itself?
Yes, see Van der Waerden's _Modern Algebra_ for a totally impractical
algorithm.
> I have a family of polynomials that derive from certain Thue equations.
> For low degrees n their roots are all real (this could probably be easily
> proved for the polynomial of arbitrary degree n) and the degree of the
> splitting field (for the polynomial of degree n) is n*(n-1). This is a
> computational result output by pari/gp and magma, but there is a low 
limit
> on the degree of the polynomial that these computer algebra systems can
> work with.
What are these polynomials?
For instance the splitting field of X^n - a (OK it has nonreal roots)
has degree n(n-1) (usually). To get an upper bound for your
polynomials, one must use their structure in some nontrivial way ---
and more than knowing the number of real roots is required!
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
====
> A quote from:
>> The New Fowler's Modern English Usage, Third Edition, Edited by R. W.
>> Burchfield, The acknowledged authority on English usage
>> [all of that from the front of the dust jacket...]
>> 
>> Under the topic billion:
>> It is best now to work on the assumption that the word means 'a
>> thousand millions' in all English-speaking areas...
But that is a false assumption. My home is an English-speaking area in 
which 
>the word does not mean that.
If the book suggests that it is best to work on false assumptions, then 
>perhaps it isn't that good a book.
Which is more useful :
The big billion (hardly any practical uses at present except
maybe in astroniomy)
The usual billion (1000 million) - many uses especially in
the fields of computing etc.
The big billion is, and always has been, a nonsense, so
the Americans changed it to something more sensible.
Good for them.
====
) Which is more useful :
)
) The big billion (hardly any practical uses at present except
) maybe in astroniomy)
) The usual billion (1000 million) - many uses especially in
) the fields of computing etc.
)
) The big billion is, and always has been, a nonsense, so
) the Americans changed it to something more sensible.
) Good for them.
There was a perfectly good word for a thousand million, which is milliard.
SaSW, Willem
-- 
            made in the above text. For all I know I might be
            drugged or something..
            No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
====

 
> There was a perfectly good word for a thousand million, which is 
milliard.
        ^^^
Still is.
-- 
Richard Heathfield : binary@eton.powernet.co.uk
Usenet is a strange place. - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
====
Iain:
> The big billion is, and always has been, a nonsense, so
> the Americans changed it to something more sensible.
> Good for them.
Actually, that was the French.  (Silly people, they later changed back.)
American English copied it from them, but I don't think there was a
deliberate decision to use one method or the other; it just happened
that they got the right one.  Credit for *choosing* the better system
goes to the British, when they dumped the big billion in the 20th century.
 
Willem:
> There was a perfectly good word for a thousand million, which is 
milliard.
Not perfectly good in English.  Because of the way that unaccented
syllables are slurred in English, it would sound too much like million.
And if you treat it as a French word and accent it at the end (unlike
the way we accent billiard when taking about the game), then there's
still a problem because the whole series of intermediate words, 
milliard
(10^9), billiard (10^15), trilliard (10^21), etc., would all sound
too similar.
-- 
Mark Brader, Toronto              These Millennia are like buses.
msb@vex.net                                           --Arwel Parry
====
> The big billion is, and always has been, a nonsense, so
> the Americans changed it to something more sensible.
> Good for them.
If a billion's between a million and a trillion, how come it isn't
named a 'pillion'?
====
> 
> The big billion is, and always has been, a nonsense, so
> the Americans changed it to something more sensible.
> Good for them.
> 
> If a billion's between a million and a trillion, how come it isn't
> named a 'pillion'?
Since it is closer, additively if not multiplicatively, to a 
million, shouldn't that be a nillion?
====
> If a billion's between a million and a trillion, how come it
>> isn't named a 'pillion'?
 Since it is closer, additively if not multiplicatively, to a
> million, shouldn't that be a nillion?
If we were being that rational about the names, we'd have
something like this:
aillion = 10^3
billion = 10^6
cillion = 10^9
dillion = 10^12
etc.
As far as the general question about whether American or
traditional British system is better, the latter is more
systematic, but the former is more convenient. As it is, the
current British practice has problems with the number 10^12 (and
higher), since I understand they've mainly only adopted the
American billion, but not the rest of the American system.
--
Dan Tilque
====
>I need a a regular space that is not a Tychonov space.
>I think Tychonov corkscrew is an exemple, but I don't know it.
  
>
The corkscrew is indeed such an example; a much siimpler one is
in the following paper:
A. Mysior, Proc. Amer. Math. Soc. {bf 81} (1981), no.~4, 652--653; MR 
82j:54026
KP
-- 
E-MAIL: K.P.Hart@EWI.TUDelft.NL       PAPER: Faculty EWI
PHONE:  +31-15-2784572                       TU Delft
FAX:    +31-15-2786178                       Postbus 5031
URL:    http://aw.twi.tudelft.nl/~hart       2600 GA  Delft
                                             the Netherlands
====
e^x is walking down the road one day when he meets several other
functions coming the other way. One of them, x^n, cries Run for your
life! A differential is coming! He's already eliminated some constants!
as he runs past. Hah! exclaimed e^x, I'm not afraid of any
differential; I'm e^x, and he can't affect me. So he walked on a little
further and, sure enough, spied a differential coming towards him. He
-- 
====
> e^x is walking down the road one day when he meets several other
> functions coming the other way. One of them, x^n, cries Run for your
> life! A differential is coming! He's already eliminated some constants!
> as he runs past. Hah! exclaimed e^x, I'm not afraid of any
> differential; I'm e^x, and he can't affect me. So he walked on a little
> further and, sure enough, spied a differential coming towards him. He
To blow some dust off that antique:
        Oh yea?  Let me introduce you to my husband dlog/dx.
>
====
> 
> 
>When is the inclusion of the wedge in the product a fibration ?
>>I'm not sure I understand your question. (I wonder whether you
>>meant a cofibration)
> 
> 
> Or, maybe, L.P means homotopic to a fibration (or a fibration
> in the homotopy category or whatever)?
> 
> Lee Rudolph
> 
> I suppose that could be the case, but given a map f : X --> Y between
> sufficiently nice spaces (homotopy type of a CW complex should be nice
> enough), the path-space construction:
> 
>       P_f = {(x,p) in X x map(I,X) | p(0) = x }
> 
> yields a space homotopy-equivalent to X under the inclusion
> 
>           i
>       X ----> P_f
>       x |--> (x, const(x))
> 
> where const(x) is the constant path at x, and the map
> 
>              j
>       P_f -------> Y
> 
>       (x,p) |--> p(1)
> 
> is a fibration. Further, the diagram
> 
>                 f
>       X --------------->_Y
>                        /|
>                       /
>                      /
>          i          /j
>                    /
>                   /
>                  /
>              _| /
>                 P_f
> 
> is commutative.
> 
> In other words, every map is a fibration, up to
> homotopy. Well, just as long as the spaces are
> sufficiently well-behaved to permit the above
> construction, and it's been long enough for me
> since I worked with that construction that I
> don't recall just what's needed.
> 
> 
This was actually how I should have posed my question. Sorry.
I should have said *homotopy* fibration. Just starting out,
bit confused at the moment. Thinking about it, this path-space
fibration construction, shoudn't it turn any map into a fibration (
that is any
map is a homotopy fibration) ?
L
====
        ... a buncha stuff ...
> 
> 
> 
> This was actually how I should have posed my question. Sorry.
> 
> I should have said *homotopy* fibration. Just starting out,
> bit confused at the moment. Thinking about it, this path-space
> fibration construction, shoudn't it turn any map into a fibration (
> that is any
> map is a homotopy fibration) ?
> 
> 
> L
Yes, every map is a homotopy fibration. Unfortunately, I neglected to
look at my copy of Spanier last night (it's at home, and I'm not), so
I still don't know what restrictinos there are on the spaces involved.
But just like any map (with some modest restrictions) is a cofibration
up to homotopy, via the mapping cylinder construction, so is any map
a fibration, again up to homotopy.
BTW, I was able to refresh my memory of the details of that construction
by Googling this set of buzzwords:
        path-space construction fibration
However, I wasn't able to discern those mysterious restrictions (and am
too far out of the game to produce them independently, sorry to say).
.
====
>I still don't know what restrictinos there are on the spaces involved.
>But just like any map (with some modest restrictions) is a cofibration
Hey, I like that: a modest restriction is a restrictino!
====
> 
> 
>>I still don't know what restrictinos there are on the spaces involved.
>>But just like any map (with some modest restrictions) is a cofibration
> 
> 
> Hey, I like that: a modest restriction is a restrictino!
Golly, I would have liked it better if I had thought of it.
.
====
Some of you may have noticed these free-wheeling mathematical
discussions I've been having with all these people, and wondered.
Maybe mostly you were wondering what they were doing on your
newsgroup.
Well I've been someone who has dabbled in mathematical research in
full knowledge that I'm NOT a mathematician, but partly because it's
fun, and partly because I have confidence that people can miss things.
Mathematicians don't believe they could miss anything simple enough
that a non-mathematician could find that's important.  And I'm not
just saying that as I've read it more than once from mathematicians.
Now as I've talked about my various ideas and research, math people
have talked a bit about the math, but mostly they've retaliated by
insulting me and saying nasty things about me on webpages.
Yes, it's odd but true.
One of their favorite labels is crank, where the idea is that you
have some person who is unreasonable, who has ideas that don't work,
who refuses to acknowledge the truth.
Only problem is that I enjoy tracing out the steps of my mathematical
arguments, and I can do things like give you a Java program that
works.
The truth is that I've dabbled in a lot of mathematical areas, poking
around looking for something dramatic, and pissed off some math
people.
Does that make me a crank?
It does to math people.  They're a sensitive bunch that seems to lack
a sense of humor.  They also take themselves WAY too seriously.
So I've been doing my thing, posting on Usenet, at times contacting
mathematicians about specifics, and over the years this weird hate
cult has grown and grown on the Internet and Usenet, where math people
spend a good bit of energy worrying about what I'm saying, but then
calling me a crank.
There's some guy who was going around quoting from me all the time in
his posts!
Now then, use your common sense.
Is it simpler that I'm really some crank, or that I've just upset a
lot of oversensitive people who think that negative labeling and
insults are the way to go?
Now the problem is bad lessons learned on Usenet.  I'm sure lots of
dabblers who had some ideas here or there were rapidly chased off of
the sci.math newsgroup when they were ganged up on and insulted.  So
posters learned that they could control people by insults!  They were
happy!!!  They could insult people and get rid of them!  But I came
along and decided that I didn't want to let people control my speech
by insults.
So because I refused to be insulted into silence you have a weird
religion that's developed in the math community which is based on
obsessively insulting me!
And I wanted others on other newsgroups to see that oddity.
It's rather freaky, eh?  But then again, who really thought math
people were normal?  And don't worry, this post is very unlikely to
change anything.
Yup, believe it or not, it will all just keep going.  I'll talk about
math, there will be some who will argue with the math, while most will
just lob insults.
Fascinating.
James Harris
====
James Harris, barely-literate yard-ape.
====
> 
> James Harris, barely-literate yard-ape.
Yard ape, curtain climber, carpet shark, ankle biter, nose miner,
sprog, linoleum lizard, loin monkey, fartling, crotch fruit,
shriekling, flesh loaf, womb dropping, little sticky person, crotch
trophy, howling shit-machine.
But give James Harris credit: he fought off the rusty coat hanger to
evenutally emerge as a twat lugie.  Hey stooopid loud troll James
Harris, put up or shut up,
http://www.rsasecurity.com/rsalabs/challenges/factoring/faq.html
http://www.rsasecurity.com/rsalabs/challenges/factoring/numbers.html
http://www.crank.net/harris.html
 It's not every braying jackass that gets a whole page at crank.net
Is a $10,000 prize no questions asked too small to justify your
submission of two little prime numbers?


-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
Quis custodiet ipsos custodes?  The Net!
====
> Some of you may have noticed these free-wheeling mathematical
> discussions I've been having with all these people, and wondered.
> 
> Maybe mostly you were wondering what they were doing on your
> newsgroup.
> 
> Well I've been someone who has dabbled in mathematical research in
> full knowledge that I'm NOT a mathematician, but partly because it's
> fun, and partly because I have confidence that people can miss things.
> 
> Mathematicians don't believe they could miss anything simple enough
> that a non-mathematician could find that's important.  And I'm not
> just saying that as I've read it more than once from mathematicians.
> 
> Now as I've talked about my various ideas and research, math people
> have talked a bit about the math, but mostly they've retaliated by
> insulting me and saying nasty things about me on webpages.
> 
> Yes, it's odd but true.
> 
> One of their favorite labels is crank, where the idea is that you
> have some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
> 
> Only problem is that I enjoy tracing out the steps of my mathematical
> arguments, and I can do things like give you a Java program that
> works.
> 
> The truth is that I've dabbled in a lot of mathematical areas, poking
> around looking for something dramatic, and pissed off some math
> people.
> 
> Does that make me a crank?
> 
> It does to math people.  They're a sensitive bunch that seems to lack
> a sense of humor.  They also take themselves WAY too seriously.
> 
> So I've been doing my thing, posting on Usenet, at times contacting
> mathematicians about specifics, and over the years this weird hate
> cult has grown and grown on the Internet and Usenet, where math people
> spend a good bit of energy worrying about what I'm saying, but then
> calling me a crank.
> 
> There's some guy who was going around quoting from me all the time in
> his posts!
> 
> Now then, use your common sense.
> 
> Is it simpler that I'm really some crank, or that I've just upset a
> lot of oversensitive people who think that negative labeling and
> insults are the way to go?
> 
> Now the problem is bad lessons learned on Usenet.  I'm sure lots of
> dabblers who had some ideas here or there were rapidly chased off of
> the sci.math newsgroup when they were ganged up on and insulted.  So
> posters learned that they could control people by insults!  They were
> happy!!!  They could insult people and get rid of them!  But I came
> along and decided that I didn't want to let people control my speech
> by insults.
> 
> So because I refused to be insulted into silence you have a weird
> religion that's developed in the math community which is based on
> obsessively insulting me!
> 
> And I wanted others on other newsgroups to see that oddity.
> 
> It's rather freaky, eh?  But then again, who really thought math
> people were normal?  And don't worry, this post is very unlikely to
> change anything.
> 
> Yup, believe it or not, it will all just keep going.  I'll talk about
> math, there will be some who will argue with the math, while most will
> just lob insults.
> 
> Fascinating.
> 
> 
> James Harris
fuffy
====
It's nice of James to make this little summary for us of his experiences
on USENET.  However, he seems to have overlooked some of his own best
posted almost exactly a year ago.
http://groups.google.com/groups?selm=3c65f87.0212222034.d5959fd%40posting.go
ogle.com
http://groups.google.com/groups?selm=3c65f87.0212251249.4b69d7c5%40posting.g
oogle.com
for yourself whether he succeeded this year in finding the change and
escape for which he claimed to be looking.
-- 
Wayne Brown  (HPCC #1104)  | When your tail's in a crack, you improvise
fwbrown@bellsouth.net      |  if you're good enough.  Otherwise you give
                           |  your pelt to the trapper.
e^(i*pi) = -1  -- Euler  |           -- John Myers Myers, 
Silverlock
====
> 
> Now then, use your common sense.
> 
> Is it simpler that I'm really some crank, or that I've just upset a
> lot of oversensitive people
All other things being equal, surely the former is more probable than
the latter...  The choice is between (i) making an assumption about
the character of a single individual and (ii) making a blanket
assumption about the characters of a lot of people, no?
====
>  > Now then, use your common sense.
>  > Is it simpler that I'm really some crank, or that I've just upset a
> lot of oversensitive people
 All other things being equal, surely the former is more probable than
> the latter...  The choice is between (i) making an assumption about
> the character of a single individual and (ii) making a blanket
> assumption about the characters of a lot of people, no?
It's simpler than that. You can *conclude* that Harris is a crank from the 
massive evidence he has
provided supporting that honor. No need to invoke the character of 
others...
--
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
====
> Some of you may have noticed these free-wheeling mathematical
> discussions I've been having with all these people, and wondered.
> 
> Maybe mostly you were wondering what they were doing on your
> newsgroup.
> 
> Well I've been someone who has dabbled in mathematical research in
> full knowledge that I'm NOT a mathematician, but partly because it's
> fun, and partly because I have confidence that people can miss things.
> 
> Mathematicians don't believe they could miss anything simple enough
> that a non-mathematician could find that's important.  And I'm not
> just saying that as I've read it more than once from mathematicians.
> 
> Now as I've talked about my various ideas and research, math people
> have talked a bit about the math, but mostly they've retaliated by
> insulting me and saying nasty things about me on webpages.
> 
> Yes, it's odd but true.
> 
> One of their favorite labels is crank, where the idea is that you
> have some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
> 
> Only problem is that I enjoy tracing out the steps of my mathematical
> arguments, and I can do things like give you a Java program that
> works.
> 
> The truth is that I've dabbled in a lot of mathematical areas, poking
> around looking for something dramatic, and pissed off some math
> people.
> 
> Does that make me a crank?
> 
> It does to math people.  They're a sensitive bunch that seems to lack
> a sense of humor.  They also take themselves WAY too seriously.
> 
> So I've been doing my thing, posting on Usenet, at times contacting
> mathematicians about specifics, and over the years this weird hate
> cult has grown and grown on the Internet and Usenet, where math people
> spend a good bit of energy worrying about what I'm saying, but then
> calling me a crank.
> 
> There's some guy who was going around quoting from me all the time in
> his posts!
> 
> Now then, use your common sense.
> 
> Is it simpler that I'm really some crank, or that I've just upset a
> lot of oversensitive people who think that negative labeling and
> insults are the way to go?
> 
> Now the problem is bad lessons learned on Usenet.  I'm sure lots of
> dabblers who had some ideas here or there were rapidly chased off of
> the sci.math newsgroup when they were ganged up on and insulted.  
If they were chased off, it's because they refused to admit their 
ignorance. Those who _learn_ from their mistakes (myself) are not chased 
off.
> So
> posters learned that they could control people by insults!  They were
> happy!!!  They could insult people and get rid of them!  
Those who think that stupidity is a virtue are not welcome in a science 
group.
> But I came
> along and decided that I didn't want to let people control my speech
> by insults.
And decided you didn't want to learn anything either. Why is that?
> 
> So because I refused to be insulted into silence you have a weird
> religion that's developed in the math community which is based on
> obsessively insulting me!
> 
> And I wanted others on other newsgroups to see that oddity.
The Difference Between Ignorance And Stupidity:
IGNORANT 
a person who is merely unaware of the TRUTH 
STUPID 
a person who has had the TRUTH revealed to him but does not embrace it 
> 
> It's rather freaky, eh?  But then again, who really thought math
> people were normal?  And don't worry, this post is very unlikely to
> change anything.
> 
> Yup, believe it or not, it will all just keep going.  I'll talk about
> math, there will be some who will argue with the math, while most will
> just lob insults.
As a dog returns to his spew, a fool returns to his folly.
> 
> Fascinating.
Yup.
> 
> 
> James Harris
====
>Is it simpler that I'm really some crank, 
yes.
-- 
Wolf Kirchmeir, Blind River ON Canada
Nature does not deal in rewards or punishments, but only in 
consequences.
(Robert Ingersoll)
====
> 
> Some of you may have noticed these free-wheeling mathematical
> discussions I've been having with all these people, and wondered.
[snip]
No wonder at all,
http://www.apa.org/journals/psp/psp7761121.html
http://insti.physics.sunysb.edu/~siegel/quack.html


http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav
http://w0rli.home.att.net/youare.swf
http://www.mazepath.com/uncleal/sunshine.jpg
http://www.you-moron.com/
Hey stooopid loud troll James Harris, put up or shut up,
http://www.rsasecurity.com/rsalabs/challenges/factoring/faq.html
http://www.rsasecurity.com/rsalabs/challenges/factoring/numbers.html
http://www.crank.net/harris.html
 It's not every braying jackass that gets a whole page at crank.net
Is a $10,000 prize no questions asked too small to justify your
submission of two little prime numbers?
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
Quis custodiet ipsos custodes?  The Net!
====
Educating James Harris has an expectation value even less than that of
administering medicine to the dead.  However, for everybody else who
stands to benefit,
http://www.maths.ex.ac.uk/~mwatkins/
Jeffrey Stopple, A Primer of Analytic Number Theory: from Pythagoras
Marcus du Sautoy, The Music of the Primes: Searching to Solve the
Karl Sabbagh, The Riemann Hypothesis: the Greatest Unsolved Problem
John Derbyshire, Prime Obsession: Bernhard Riemann and the Greatest 
http://www.maths.ex.ac.uk/~mwatkins/zeta/cipra.htm
http://www.maths.ex.ac.uk/~mwatkins/zeta/Julia.htm
(The foregoing list is obviously cribbed.  Uncle Al does not wish to
demean the original poster by associating his name with James Harris.)
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
Quis custodiet ipsos custodes?  The Net!
====
> 
[snip]
> Now then, use your common sense.
> 
> Is it simpler that I'm really some crank, or that I've just upset a
> lot of oversensitive people who think that negative labeling and
> insults are the way to go?
Judging from your posts (and the fact that you keep ignoring /
misunderstanding the counterarguments to your claims), the first
explanation is much simpler.
[snip rest]
Bye,
Bjoern
====
>Some of you may have noticed these free-wheeling mathematical
>discussions I've been having with all these people, and wondered.
I've been noticing this endless train of threads you've been starting 
about your crank status.
-- 
Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding. -- Friedrich August 
Kekul.8e
====
> One of their favorite labels is crank, where the idea is that you
> have some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
 Only problem is that I enjoy tracing out the steps of my mathematical
> arguments, and I can do things like give you a Java program that
> works.
Would you like to trace the steps of your proof that 1/3 is in Z[1/2]?
I don't think so.
Would you acknowledge the truth that 1/3 is not in Z[1/2]?
No way.
So what does it makes you?
====
> Some of you may have noticed these free-wheeling mathematical
> discussions I've been having with all these people, and wondered.
 Maybe mostly you were wondering what they were doing on your
> newsgroup.
 Well I've been someone who has dabbled in mathematical research in
> full knowledge that I'm NOT a mathematician, but partly because it's
> fun, and partly because I have confidence that people can miss things.
 Mathematicians don't believe they could miss anything simple enough
> that a non-mathematician could find that's important.  And I'm not
> just saying that as I've read it more than once from mathematicians.
 Now as I've talked about my various ideas and research, math people
> have talked a bit about the math, but mostly they've retaliated by
> insulting me and saying nasty things about me on webpages.
 Yes, it's odd but true.
 One of their favorite labels is crank, where the idea is that you
> have some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
 Only problem is that I enjoy tracing out the steps of my mathematical
> arguments, and I can do things like give you a Java program that
> works.
 The truth is that I've dabbled in a lot of mathematical areas, poking
> around looking for something dramatic, and pissed off some math
> people.
 Does that make me a crank?
 It does to math people.  They're a sensitive bunch that seems to lack
> a sense of humor.  They also take themselves WAY too seriously.
 So I've been doing my thing, posting on Usenet, at times contacting
> mathematicians about specifics, and over the years this weird hate
> cult has grown and grown on the Internet and Usenet, where math people
> spend a good bit of energy worrying about what I'm saying, but then
> calling me a crank.
 There's some guy who was going around quoting from me all the time in
> his posts!
 Now then, use your common sense.
 Is it simpler that I'm really some crank, or that I've just upset a
> lot of oversensitive people who think that negative labeling and
> insults are the way to go?
 Now the problem is bad lessons learned on Usenet.  I'm sure lots of
> dabblers who had some ideas here or there were rapidly chased off of
> the sci.math newsgroup when they were ganged up on and insulted.  So
> posters learned that they could control people by insults!  They were
> happy!!!  They could insult people and get rid of them!  But I came
> along and decided that I didn't want to let people control my speech
> by insults.
 So because I refused to be insulted into silence you have a weird
> religion that's developed in the math community which is based on
> obsessively insulting me!
 And I wanted others on other newsgroups to see that oddity.
 It's rather freaky, eh?  But then again, who really thought math
> people were normal?  And don't worry, this post is very unlikely to
> change anything.
 Yup, believe it or not, it will all just keep going.  I'll talk about
> math, there will be some who will argue with the math, while most will
> just lob insults.
 Fascinating.

> James Harris
What James doesn't understand is that he gets insulted because he insults.
He seems to think he can dish it out, but doesn't have to take it. He is
just way too arrogant to get anything done.
-- 
David Moran
Chief Meteorologist
Oklahoma Storm Team
====
> Now as I've talked about my various ideas and research, math people
> have talked a bit about the math, but mostly they've retaliated by
> insulting me and saying nasty things about me on webpages.
 Yes, it's odd but true.
 One of their favorite labels is crank, where the idea is that you
> have some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
James
    I have dipped into the discussion from time to time in the early days
and don't really have an opinion (my math specialities lie elsewhere),
however what hints crank to me is someone who insults others (you must 
admit
that you language reaches insulting levels), posts to (apparently)
non-relevant NGs like sci.cognitive & sci.physics and continually starts
very similar threads reposting the same assertions.
While I have stopped reading the threads in depth, I do note that you start
threads with alarming regularity.
Face it, you are not going to persuade some people that you are right. Life
is too short to spend persuading the unpersuadable. You posts started well
balanced but you appear to be tending towards a conspiracy and vendetta
tone. You are at risk of letting this seriously affect your well-being. 
Take
some time-off, smell the flowers and kick back, this is in danger of
stressing you out too much. Personally, I would drop it but at the very
least take a break - don't let this turn you bitter and twisted.
Andy
====
> One of their favorite labels is crank, where the idea is that you
> have some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
Let's see now...
...some person who is unreasonable, who has ideas that don't work,
who refuses to acknowledge the truth.
Sounds just like you, James!
-- 
Alec McKenzie
mckenzie@despammed.com
====
> 
> One of their favorite labels is crank, where the idea is that you
> have some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
> 
> Let's see now...
> 
> ...some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
> 
> Sounds just like you, James!
Your opinion is noted Alec McKenzie.
I do admit that if enough people think that I'm just some nut, then I
have to wonder.
However, I've looked through the literature, considered various
attacks on my work, and noticed that mathematicians rely first and
foremost on insults, and can't defend their own assertions against my
work without them.
Then rational people supposing that mathematicians are rational
themselves, consider the insults to be proof that I must be wrong, or
why would mathematicians be so insulting?
I *dare* any posters who disagree to give just a point-by-point attack
on my prime counting function, numbering out each point, and providing
all the information in their post rather than just posting links.
Obviously then I will go through each point, and refute it.  With
everything outlined in details readers can judge whether or not I'm
really avoiding facts or whether in fact they're being lied to by
mathematicians confident in their ability to dissemble successfully.
James Harris
My math discoveries, found for profit
http://mathforprofit.blogspot.com/
====
>> 
> One of their favorite labels is crank, where the idea is that you
> have some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
>> 
>> Let's see now...
>> 
>> ...some person who is unreasonable, who has ideas that don't work,
>> who refuses to acknowledge the truth.
>> 
>> Sounds just like you, James!
Your opinion is noted Alec McKenzie.
I do admit that if enough people think that I'm just some nut, then I
>have to wonder.
_If_ enough people thought that? What reason do you have to
think that _anyone_, with only one exception that I can think of
right now, doesn't think of you as just some kind of nut?
>However, I've looked through the literature, 
You have not looked at any of the _literature_ on number theory
except what you've found on the internet.
>considered various
>attacks on my work, and noticed that mathematicians rely first and
>foremost on insults, and can't defend their own assertions against my
>work without them.
Then rational people supposing that mathematicians are rational
>themselves, consider the insults to be proof that I must be wrong, or
>why would mathematicians be so insulting?
I *dare* any posters who disagree to give just a point-by-point attack
>on my prime counting function, numbering out each point, and providing
>all the information in their post rather than just posting links.
Obviously then I will go through each point, and refute it.  With
>everything outlined in details readers can judge whether or not I'm
>really avoiding facts or whether in fact they're being lied to by
>mathematicians confident in their ability to dissemble successfully.

>James Harris
My math discoveries, found for profit
>http://mathforprofit.blogspot.com/
************************
David C. Ullrich
====
[snip]
> I *dare* any posters who disagree to give just a point-by-point attack
> on my prime counting function, numbering out each point, and providing
> all the information in their post rather than just posting links.
You're a little late with issuing challenges to others. There is an 
unfinished challenge to you to provide proof that
your so-called 'partial differential equation' actually solves the prime 
counting function. Since *you* made the claim
(over and over and over and ...) it is your responsibility to provide the 
proof, or even some meager evidence, or even
some unstructured results from its application.
If you won't defend your own claims with supporting evidence, why should 
anyone bother with attacking something they
aren't interested in?
> Obviously then I will go through each point, and refute it.  With
> everything outlined in details readers can judge whether or not I'm
> really avoiding facts or whether in fact they're being lied to by
> mathematicians confident in their ability to dissemble successfully.
Obviously you have *not* proven your claim that your so-called 'partial 
differential equation' solves the prime
counting function. You haven't even posted any results whatsoever. That 
certainly qualifies as 'avoiding facts' and
suggests that we are 'being lied to by' you.
Put up or SHUT UP!
--
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
====
> ...some person who is unreasonable, who has ideas that don't work,
> who refuses to acknowledge the truth.
> 
> Sounds just like you, James!
> 
> Your opinion is noted Alec McKenzie.
> 
> I do admit that if enough people think that I'm just some nut, then I
> have to wonder.
> 
> However, I've looked through the literature, considered various
> attacks on my work, and noticed that mathematicians rely first and
> foremost on insults, and can't defend their own assertions against my
> work without them.
The trouble was, that those who posted mathematics to point out the 
flaws in your work had their mathematics ignored or insulted or 
both. 
They defended their own assertions well enough so that no one except 
JSH now accepts any of JSH's assertions.
====
Epilog
The Interesting Question has already been satisfactorily answered now.
But I thought that some people might enjoy a nice, pertinent passage
(which I only just now discovered) in the excellent _Concrete Mathematics_
by Graham, Knuth, and Patashnik.
On page 481 (2nd ed.), where m is the number of terms of Stirling's
asymptotic expansion of ln(n!) to be used, they say
... if n is fixed and m increases,
the error bound |B_(2m+2)|/((2m+2)(2m+1)n^(2m+1)) decreases to a certain
point and then begins to increase. Therefore the approximation reaches a
point beyond which a sort of uncertainty principle limits the amount by
which n! can be approximated.
David
 <3fd4b2c9$1@rutgers.edu>
====
That f is injection and A infinite isn't enuf to prove f not
surjection.  For example, f could be the identity map.
A is finite when for all injections f:A -> A, f is surjection.
The negation of that is
A is infinite when there's some injection f:A -> A that's not a
surjection. So you have to pick f with care at very the start.
 >then h(x) is not a surjection by the definition of h(x).
You might want to detail that.
 >But this goes against the assumption that B is finite
Thus by contradiction, A must be finite.
 >Is that any better?
Yes, one correctable mistake.
-- exercise
A finite when for injections f:A -> A, f bijection.
Show that definition is equivalent to
A finite when for surjections f:A -> A, f bijection.
That requires showing
If f:A -> A surjection and A finite, then f bijection.
and
If for all surjections f:A -> A, f bijection, then A finite.
--
>> Assume A is a subset of B and that B is finite.
>> Prove that A is finite.
> assume
> f:A -> A injection
> extend f to
> g:B -> B, x -> f(x) if x in A, x -> x if x in BA
> show g injection;  thus g surjection as B finite
> show restriction of g to A, g|A:A -> A surjection
> now as f = g|A, f surjection, QED
>
--
> Assume injection h:B -> B and bijection f:A -> B
> g = f^-1hf:A -> A
> injection, hence bijection as A finite.  Thus
> A = g(A) = f^-1hf(A) = f^-1h(B)
> B = f(A) = ff^-1h(B) = h(B)
> Hence h surjection.
>
----
====
> Curious readers who haven't yet looked should check
 http://www.crank.net/harris.html
 and see what evidence Erik Max Francis presents to justify his insulting
label.
 And remember, he used to be a regular sci.math poster.
 Now then, what does he put on his page to justify his insults?

> James Harris
Ummmm, read it dickhead.
It's self explanatory to anyone who isn't a complete fucking nut-job
K. Jones
====
> 
> James, did you read some of that web site? Do you agree that, you
> excepted, all those people on crank.net *are* cranks?
> 
> That's illogical Victor Eijkhout, as even if that were true, it
> wouldn't prove that I am, as your induction is false.
> 
> I'm sorry, what induction? I'm asking you a question.
Yes, that's correct, but to what purpose?
First of all, I have no interest in looking over Erik Max Francis's
other opinions on the website, as I don't think they are worth
considering, while I am, of course interested in insults directed at
me.
Second, even if I were so inclined to spend my time considering his
assessments, and concluded that in all other cases he was correct, it
would be false to conclude that proved that I deserved the insults he
delivers.
Later you *ask* whether or not others on his site think of themselves
as cranks, which shows your original line of thinking.
Your efforts were clumsy as well as transparent Victor Eijkhout.
However, if I made a mistake in assessing your intent, you can expand
on your questions and your point in asking them.
> 
> Do you suppose they themselves think they are cranks?
> 
> Now you're trying to imply that because a crank probably wouldn't
> accept the label that non-acceptance on my part would just be a sign
> that I'm a crank.
> 
> No, I'm asking you a question.
> 
Well maybe I got you wrong, eh?  So then, you were just asking
questions without any attempt at backhanded insults?
Then the answer is that Erik Max Francis has his opinions, and I'm not
that interested in reading through much of what he has to say, though
his insults directed at me at http://www.crank.net/harris.html did, of
course, get my interest.
Besides that, your questions are impractical, though they do imply
that *you* have spent the time to go over each and every one of the
pages where Erik Max Francis calls someone a crank on his website.
Did do that Victor Eijkhout?
> Victor leaving the final step as exercise to the reader
> 
> Now the full import of your post is clear.  You were speaking to the
> crowd trying to convince others that I am indeed a crank by using
> false induction and the implication that refusal to accept the label
> is just more proof.
> 
> No, the crowd already has their opinion of you. I'm just asking you a
> couple of simple questions.
> 
And I've given my reasons for doubting that you're telling the truth.
However, since you've firmly said that you were just asking questions,
you've also *firmly* said that you did not mean anything by them.
> What I find fascinating about your post though Victor Eijkhout is the
> rather clumsily false induction you used, since presumably someone
> trained as a mathematician would be more logical.
> 
> Oh, when I'm actually doing math I'm pretty logical. When I'm responding
> to you it's purely for amusement value since any correlation between you
> and mathematics is vanishingly small.
> 
> V.
Well at least it looks like you switched to being more upfront with
the attempts at insults!
I think it's better to just be honest from the start, instead of being
called out over time.
James Harris
====
>  > James, did you read some of that web site? Do you agree that, you
> excepted, all those people on crank.net *are* cranks?
>  > That's illogical Victor Eijkhout, as even if that were true, it
> wouldn't prove that I am, as your induction is false.
>  > I'm sorry, what induction? I'm asking you a question.
 Yes, that's correct, but to what purpose?
 First of all, I have no interest in looking over Erik Max Francis's
> other opinions on the website, as I don't think they are worth
> considering, while I am, of course interested in insults directed at
> me.
 Second, even if I were so inclined to spend my time considering his
> assessments, and concluded that in all other cases he was correct, it
> would be false to conclude that proved that I deserved the insults he
> delivers.
 Later you *ask* whether or not others on his site think of themselves
> as cranks, which shows your original line of thinking.
 Your efforts were clumsy as well as transparent Victor Eijkhout.
 However, if I made a mistake in assessing your intent, you can expand
> on your questions and your point in asking them.
  > Do you suppose they themselves think they are cranks?
>  > Now you're trying to imply that because a crank probably wouldn't
> accept the label that non-acceptance on my part would just be a sign
> that I'm a crank.
>  > No, I'm asking you a question.
> 
> Well maybe I got you wrong, eh?  So then, you were just asking
> questions without any attempt at backhanded insults?
 Then the answer is that Erik Max Francis has his opinions, and I'm not
> that interested in reading through much of what he has to say, though
> his insults directed at me at http://www.crank.net/harris.html did, of
> course, get my interest.
 Besides that, your questions are impractical, though they do imply
> that *you* have spent the time to go over each and every one of the
> pages where Erik Max Francis calls someone a crank on his website.
 Did do that Victor Eijkhout?
 > Victor leaving the final step as exercise to the reader
>  > Now the full import of your post is clear.  You were speaking to the
> crowd trying to convince others that I am indeed a crank by using
> false induction and the implication that refusal to accept the label
> is just more proof.
>  > No, the crowd already has their opinion of you. I'm just asking you a
> couple of simple questions.
> 
> And I've given my reasons for doubting that you're telling the truth.
 However, since you've firmly said that you were just asking questions,
> you've also *firmly* said that you did not mean anything by them.
 > What I find fascinating about your post though Victor Eijkhout is the
> rather clumsily false induction you used, since presumably someone
> trained as a mathematician would be more logical.
>  > Oh, when I'm actually doing math I'm pretty logical. When I'm 
responding
> to you it's purely for amusement value since any correlation between 
you
> and mathematics is vanishingly small.
>  > V.
 Well at least it looks like you switched to being more upfront with
> the attempts at insults!
 I think it's better to just be honest from the start, instead of being
> called out over time.
 James Harris
It might be even better to answer his question. You accused him of posting a 
false induction. He
responded, I'm sorry, what induction? You posted the above rambling, 
twisted monologue which wanders
around and never answers the question. Don't you get it? If you make a 
claim, you have an obligation to
support it. WHAT INDUCTION?
Put up or SHUT UP!
--
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
====
As can be seen by the number of posts in this thread,
and the references to his web site in thousands of other posts,
a computer programmer, who took some data processing classes
at a third rate California college, has become a highly regarded expert
in math, physics, and other science disciplines, and
many people, who pretend to be rational, intelligent, open-minded
scientists (Or at least, pretend to have a scientific mind.),
frequently use this programmer as a major reference.
I assert that this indicates that science,
is pretty much like show business and politics,
and that the ideas that get elevated to high status
are those that are promoted best.
I suggest that those folks,
who feel passionate about their ideas, and want to promote them,
should first set up a web site much like the highly regarded crank.net,
and after they become recognized as a  highly regarded expert,
to slowly incorporate their ideas into the web site,
and take sly shots at competing ideas.
As it would be helpful to new readers to know
who the sociopaths are in the newsgroups,
another web site that would be popular
would be one that puts the internet flamers in the spotlight,
by posting some of their posts, and their backgrounds.
--
Tom Potter     http://tompotter.us
====
>  > As can be seen by the number of posts in this thread,
> and the references to his web site in thousands of other posts,
> a computer programmer, who took some data processing classes
> at a third rate California college, has become a highly regarded expert
> in math, physics, and other science disciplines, and
> many people, who pretend to be rational, intelligent, open-minded
> scientists (Or at least, pretend to have a scientific mind.),
> frequently use this programmer as a major reference.
> 
> What third rate California college? Who rated it? What criteria?
>
Hey Wormley,
as you use this programmer's web site as your primary rederence,
it seems to me that you should know what college your resident expert
attended.
If you want to know how this college rates,
I suggest that you learn how to use Google.
I'll give you some hints.
Caltech and Stanford and first rate California colleges.
The college that Baez teaches at is a second rate college.
Your expert took some data processing classes at a third rate college.
> Most scientist are computer programmers... are you knocking us Potter?
Wormley,
why do you always try to identify yourself with some group?
Does identifying yourself with a group make you feel more secure,
or do you think [sic] that it lends strength to your position?
Do you have the courage to express any independent ideas you have
(Assuming that you have an independent idea.),
or the knowledge to address the point of a dichotomy,
rather than try to position an opponents point
against some group that you identify with?
In other words Wormley,
are you a man or a mouse?
--
Tom Potter     http://tompotter.us
===============
WHO instigates conflict and war for power and wealth?
WHO instigated the class wars of the 1900's?
WHO is instigating the religious wars of the 2000's?
WHO has a well organized propaganda machine?
WHO gang attacks all who expose their agenda and methods?
Visit my web site, and download the world's best physics tutorial!
===============
====
...
 > However, my request was for readers to go to the page and tell me what
 > Erik Max Francis presents to support his insulting charge.
 > 
 > It is a reasonable request.
 > 
 > I'm curious.
Why?  Read the links and see what is written.  Some are even links to
pages I have written in the course of times.  Note that the pages I
have written were based on proofs by you (yes, *now* I have the proof),
where your methodology leads to proofs of things that where obviously
false.  You never actually responded to them.  Actually you even never
succeeded in proving FLT for the power 4, the simplest of them all.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
it appeared to be too learned in numbertheory for me, but
it seemed like a reasonable, point-by-point treatment.
    what is there about it that *you* comprehend?
 
> http://www.crank.net/harris.html
> 
> and see what evidence Erik Max Francis presents to justify his insulting 
label.
 
> Most of us have.  It seems pretty cut and dried.
--Give the Gift of Dick Cheeny -- out of office, finally!
====
> it appeared to be too learned in numbertheory for me, but
> it seemed like a reasonable, point-by-point treatment.
>     what is there about it that *you* comprehend?
> 
I'll admit the FLT stuff is a bit arcane for me, and I
defer to mathematicians, some of whom have gone
to great and patient lengths to analyze Harris'
claims point-by-point - analyses which he ignores.
My judgement of Harris is based more on his
prime counting algorithm. In that case, anyone
with a basic knowledge of algebra can see that it's
functionally equivalent to the Legendre formula
http://mathworld.wolfram.com/LegendresFormula.html
except that he's replace the sum with a dS and
claimed that he's done something profound.
There's also the circumstantial evidence that
he thinks all of this stuff is appropriate
to physics and literature groups, not to
mention his delusions about the riches and
fame awaiting the discoverer of a new and
better prime counting algorithm.
-E
>  
> 
>http://www.crank.net/harris.html
>>and see what evidence Erik Max Francis presents to justify his insulting 
label.
> 
>  
> 
>>Most of us have.  It seems pretty cut and dried.
> 
> 
> --Give the Gift of Dick Cheeny -- out of office, finally!
====
> it appeared to be too learned in numbertheory for me, but
> it seemed like a reasonable, point-by-point treatment.
>     what is there about it that *you* comprehend?
> 
> 
> I'll admit the FLT stuff is a bit arcane for me, and I
> defer to mathematicians, some of whom have gone
> to great and patient lengths to analyze Harris'
> claims point-by-point - analyses which he ignores.
That is against the evidence.  Actually I've gone to great lengths to
explain my work, and consider objections.
And I'm one of my own biggest skeptics as I had *YEARS* of wrong
ideas, and attempts that failed.  Worse, for some of them it took
*MONTHS* before I figured out where I screwed up.
Unfortunately, mathematicians used my failures against me, and just
kept claiming that I was still failing even when I succeeded.  It's
bizarre behavior, but there's not a lot I can do about it.
They do have the power to get away with it...at least for a while.
> 
> My judgement of Harris is based more on his
> prime counting algorithm. In that case, anyone
> with a basic knowledge of algebra can see that it's
> functionally equivalent to the Legendre formula
> http://mathworld.wolfram.com/LegendresFormula.html
> except that he's replace the sum with a dS and
> claimed that he's done something profound.
Well functionally equivalent can mean a lot of things.  After all,
there is an exact count of prime numbers over a given interval, like
from 1 to 10 there are 4 prime numbers and they are 2, 3, 5, and 7.
So you can argue that ANY method that counts primes is functionally
equivalent and that's a fascinating dodge that reveals the reality of
mathematicians, math groupies and supposed love of pure math.
What's key in my discovery is that it does things that others do not,
and cannot do.
> 
> There's also the circumstantial evidence that
> he thinks all of this stuff is appropriate
> to physics and literature groups, not to
> mention his delusions about the riches and
> fame awaiting the discoverer of a new and
> better prime counting algorithm.
> 
> -E
I have a Bachelors of Science in Physics from Vanderbilt University.
Facing hostile and irrational reactions from *mathematicians* it makes
sense that I'd go to a group that I hope I understand better that I
know can understand the mathematics.
Physicists are quite cool.
However, I found that sci.physics didn't quite give me the reception I
desired, so I went to other groups to include with it, and have a MUCH
better reaction as math people seem to behave a little better when
they see the cross-posting.
I hope readers understand that I'm a guy who went looking for some
neat math results, found some, and then faced a mathematicians dead
set on preserving the belief that only mathematicians can find
anything important.
Mathematicians and math groupies clearly believe that insults,
innuendo, and various personal attacks are legitimate ways to deal
with information from some they don't like, and my point is that such
behavior is primarily a sign of weakness at best.
James Harris
My math discoveries, found for profit
http://mathforprofit.blogspot.com/