mm-1023 === Subject: Re: there is no such thing as infinity > Dave Seaman wibbled: >> Last time I used unit 3, it was a card READER on an ICL1906A. >> Lots of programs use unit 3. Some programs work with dozens of files and need >> a unit number for each one. If unit 3 is not mentioned in an OPEN statement, >> then it will probably be connected with a file named Ôfort.3 or something >> similar. > When I learnt FORTRAN, we were advised that anything up to 9 was likely > to be reserved by the compiler for various h/w devices. Not any more. There is a standard input (unit=*) and a standard output (also unit=*). Card readers and card punches are gone, and printing and plotting are done by sending files to the appropriate device in a postprocessing step. Even if unit 3 had a predefined association in some implementation (which it doesnt in any modern Fortran that I am aware of), you can still use an OPEN statement to connect unit 3 to a file. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: discrete mathematics fat sets >> Given a set S {1,2,3,...,n} >> A subset P is said to be a fat set if every element in it is >= the >> cardinality of the subset P. >> The problem is to find the number of such fat sets. >Infinitely Countable > No. He _started_ with Given a set S {1,2,3,...,n}; > this set has only finitely many subsets. nulset {j}, j in S {j,k}, distinct j,k in S, min j,k >= 2 ... S - {j,k}, distinct j,k in S, min S-{j,k} >= n-2 Sj, j in S, min Sj >= n-1 S === Subject: Re: 1/0 now allowed [...] V. -- email: lastname at cs utk edu homepage: cs utk edu tilde lastname === Subject: Re: Bound of a sum >>Ive found the following inequality at a book: if a=1/sqrt(2), then >>a^((n + 1)(n + 2)) + a^((n + 2)(n + 3)) + a^((n + 3)(n + 4)) + ... < >>< a^(n(n + 3)). >>This is stated as a matter of fact, without any hint of a proof; it >>is not even suggested that the reader tries to prove it as an >>exercise. So, my guess is that it should be quite obvious, but >>the fact is that I have been unable to do it. Any idea? > Dividing both sides by a^(n(n + 3)), you want to prove > a^2 + a^(2n + 6) + a^(4n + 12) + a^(6n + 20) + ... > If n >= -1, then this is bounded by a^2 + a^4 + a^6 + ... < a^2/(1 - a^2) = 1. Like I said, it was quite obvious. Unfortunately, not for me. :-) Jose Carlos Santos === Subject: Re: JSH: Research question answered > > I think it interesting as an advanced question to figure out how to > construct one of the imperfect factorizations from a given > tautological space, though it seems that would require a tautological > space only valid in the complex plane. Hmmm...therefore, it seems logical that an imperfect factorization > cannot be constructed from a tautological space. The formalism necessary to prove that quick deduction must be rather > impressive. > > This *does* sound like an interesting and advanced question, if only I > knew what it means. > > So, following Jamess lead, I delved into advanced mathematical > space. I got three hits: two regarding some performance art hoohah > and one on Postethnic Narrative Criticism. Surprisingly, neither of > these topics have any mention of imperfect factorization, so it does > appear that Jamess proposed research is groundbreaking. > > I eagerly await development of the formalism that relates Jamess > advanced polynomial factorization to postethnic performance art. > Should be fascinating stuff. > The phrase tautological space is one I came up with from my other > math research. > It refers to regions other than > x=x, which is x = 0(mod x), > as thats the basic tautological space from which mathematicians > traditionally operate, and its used for most research, Correction, that should be 1=1, which is 1 = 0(mod 1), and yes, I know its trivial but its the base space or explicit space that mathematicians usually use. James Harris === Subject: Re: Plotting a 7 vertice graph in which every vertex has degree 4 David, I came up with three sub-graphs, which I have uploaded the images of with Mathematica. The first of the four images is just the complete graph for K_7, and doesnt relate. Do I have the right idea? http://home.earthlink.net/~diana53/mathematica/1-1-17.html Diana > I am trying to determine the number of isomorphism classes of simple > 7-vertex graphs in which every vertex has degree 4. > It is much easier if you consider their complements instead: 7-vertex > 2-regular graphs. A 2-regular graph is just a disjoint union of cycles, > and there are only two ways of doing this with seven vertices: a single > 7-cycle, or the disjoint union of a 3-cycle and a 4-cycle. > -- > David Eppstein http://www.ics.uci.edu/~eppstein/ > Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: JSH: Dont talk to me > I dont necessarily mind mocking responses. > Exactly. It isnt mockery or abuse that gets to you, it is simple and > clear proof that you are wrong. Thats the unforgiveable offence, isnt it? > Gib Hell no!!! Thats a relief!!! Im not a mathematician. Im some guy who decided hed go looking for something that might have been missed in the great rush of math society to build upon itself. And I found it. Like dont try the bullshit of saying I dont admit when Im wrong, when time after time over a period of years I have. And besides theres my prime counting function which any person out there with the balls to go do a Google search on can see is unique in that it uses a partial difference equation, and it doesnt take long to find out that no one else in recorded history managed to find such a gem. I think the problem is that todays mathematicians are pencil pushers--and not in a good way--who do NOT have guts. Freaking cowards are running as fast as they can. So Im in the process of chasing them down. James Harris === Subject: Math/Physics Valentines It is time to repost this again: MATH/PHYSICS VALENTINES Remember those cheesy valentines you used to get when you were in elementary school? Well I give something similar to my Math Physics students on Valentines day. I tried it last year with good success and I was hoping that I could get some more ideas for cards. So it was recommended that I tap some of the greatest minds around, which is why Im here. Im not sure if anyone can help me, but if you have any more ideas or Mathew The following is a list I have come up with already: * Youre one of the fundamental forces in my life. Be my Valentine. * Youre the net force that makes my heart accelerate. Be my Valentine. * Theres an attraction between us, I think its gravity. Be my Valentine. * Youre so great they should name a constant after you. Be my Valentine. * Were like opposite charges. Be my Valentine. * The lines of force point me towards you. Be my Valentine. * Like resonance tubes, were in harmony together. Be my Valentine. * We add up to a good team. Be my Valentine. * Youre a positive exponent in my life. Be my Valentine. * If they plotted you and I on a scatter-plot, they would find a positive correlation. Be my Valentine. * Youre a positive exponent in my life. Be my Valentine. * Youre the only variable for me. Be my Valentine. * Like functions, youre the only value for me. Be my Valentine. * Looking for an affractionate girl. Be my Valentine. * Met you at the decimall. Be my Valentine. * Talking about you, I told my best friend I would never lever! .Be my Valentine * I dont care if your breasts pendulum, be my Valentine < not for young students ;-) > * Do you also feel the attraction? Be my Valentine * Do you see the gravity of this situation? You have to be my Valentine * When I first saw you I felt the Big Bang ! Be my Valentine * Ion the other hand, would love for you to be my Valentine * The only predicate: be my Valentine * At absolute zero you would still move me. Be my Valentine * Be my Valentine, even if its only Faraday * My theorem is : youd be great as my Valentine * Im attracted, dont repel me. Be my Valentine * Wed make a nice tuple on Valentine, be mine * I want our relationship to be Ex-Static, be my Valentine * Dont be square, be my Valentine * Love hertz, be my Valentine * Wave if you accept to be my Valentine * The frequency of our dating would amplify if you are my Valentine * Oh, you could try the nerdy approach... My love for you is incalculable. * Or you could try Tweety Bird talk... Im equate-y for you! * Or you could get a bit racy... Algebras in the world cant measure up to your curvilinear structure. Beta Valentine of my life. * Ive finally worked up the courage to ask ... be a joule and Be My Valentine. * Would you be inclined to Be My Valentine? * It would matter to me if youd agree to Be My Valentine. * End the chaos in my life. Be My Valentine. * I feel an impulse to ask you to Be My Valentine. * Ive finally overcome the inertia of shyness to ask you to Be My Valentine. * I love your body with naked singularity. Be My Valentine. * I hope that my boldness in asking you to Be My Valentine wont cause friction in our relationship! * I think of you with more and more frequency. Be My Valentine. * I cant resist asking you to Be My Valentine. * You generate excitement in my life. Be My Valentine. * I hope that from the smile on my face, you can extrapolate that I want you to Be My Valentine. * If youll agree to Be My Valentine from across the room, please signify by giving me a standing wave. * If I ask you to Be My Valentine on February 15th, will you overlook that relative deviation? * I think our relationship has potential. Be My Valentine. * My heart is sad. You can rectify that by agreeing to Be My Valentine. * My affection for you will never decay. Be My Valentine. * Are you going to Be My Valentine, or watt? * Every second I think of you riding that Schwinn, and it hertz. Be My Valentine. * Be my valen[cy]-tine. * Be my valentine and I square Ill be yours. * Youre the root of my affection. I really mean it! * In case you didnt hear, Ill theta gain... Be mine! * Give me a sine... Will you be my Valentine? * My heart and my foot-pounds when you are around. * We could be dynamic together! * Ill give you a moment to decide if youll be mine. * I need to ask yaw... will you be mine? * If youre looking for synthesis your opportunity, Be my Valentine. * I cant wait to explore your fuzzy boundaries. Be my Valentine. * It was a magnetic moment when we met. Be my valentine. * Youre the Great Attractor. Be my valentine. Roses reßect a light frequency at one end of the visible electromagnetic spectrum, Violets reßect a light frequency at the other end of the visible electromagnetic spectrum, Sugar is C12H22O11, And you release the endorphins in my brain. * I would like to make you an eigenfunction of my Hamiltonian. * May I Lorentz boost into your centre of mass? * I expect your wave function to be degenerate under this operator. I am a positron spiralling in your electric field. Let my electron tunnel trough your barrier acquering negative energy defying space-time quantization. There is a force (F = r + 1/(r^4), r=distance) between us that gets larger with distance.,,, sigma(me)/me = sigma(you)/you -- Be my Valentine! The sum of my divisors equals the sum of yours -- Be my Valentine! You = K_f(Me) -- Be my Valentine! (K_f is the Love transform) We differ by multiplication of a unit -- Be my Valentine! My module is faithful -- Be my Valentine! We are connected by a natural homomorphism -- Be my Valentine! You are a purely inseparable extension of me -- Be my Valentine! My love for you is an invariant under the transformations of others -- Be my Valentine! You are a primitive element of my life -- Be my Valentine! You are the sum of your divisors -- Be my Valentine! Every polynomial is separable over you -- Be my Valentine! All our loops are contractible to a point -- Be my Valentine! My atoms are attracted to your electrons. -- This post is free post; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later version. Joachim Verhagen (jcdverha@xs4all.nl) WWW http://www.xs4all.nl/~jcdverha/ (Science Jokes) === Subject: Re: there is no such thing as infinity Dave Seaman wibbled: > Dave Seaman wibbled: >> Last time I used unit 3, it was a card READER on an ICL1906A. >> >> Lots of programs use unit 3. Some programs work with dozens of files and need >> a unit number for each one. If unit 3 is not mentioned in an OPEN statement, >> then it will probably be connected with a file named Ôfort.3 or something >> similar. > When I learnt FORTRAN, we were advised that anything up to 9 was likely > to be reserved by the compiler for various h/w devices. > Not any more. There is a standard input (unit=*) and a standard > output (also unit=*). Card readers and card punches are gone, and > printing and plotting are done by sending files to the appropriate device > in a postprocessing step. * was default input or output 1 and 2 were input and output, which might, depending on whether it was the ICL, CDC, or Minnesota compiler, have been the same as 5 and 6, which were always the tty in and out. iirc 3 and 4 were card reader/punch, and 7 and 8 were tape reader/punch. Might have been the other way around. We didnt have a direct connection to the plotters afair. > Even if unit 3 had a predefined association in some implementation (which > it doesnt in any modern Fortran that I am aware of), you can still use > an OPEN statement to connect unit 3 to a file. I knew that really, I was just being silly. I mean, the whole Ôno such thing as infinity thing was a joke anyway. I hope. -- Wanted: 24 === Subject: Re: about vector bundle proof tern ha scritto nel messaggio > i ask you kindley to read this paper: > I am persuaded it is correct. > Tern > Tp M denotes the tangent space to M at p ; > xi denotes a vector bundle over M ; > I will denote the differential of f at p with d(f)p . More > precisely, d(f)p : Tp M ---> Tf(p) xi > Now, suppose that f : M ---> xi belongs to Z^0_p (xi) = the set of section > of xi such that f(p) = 0. Then it follows that f vanishes at p and hence > (using Taylor series in a local coordinate system for M at p and a local > trivialization of xi near p) we can write f near p as a finite sum > f = g_1 f_1 + ... + g_n f_n where the f_i are sections of xi and the g_i > are smooth complex-valued functions(defined on M) THAT VANISH AT p. > Then it follows that d(f)p = d( g_1 )p tensor f_1 (p) +g_1(p) tensor d > (f_1)p + ... + d( g_n )p tensor f_n (p) + g_n(p) tensor d(f_n)p , > where tensor denotes the tensorial product , > g_i :M ---> C , > d( g_i )p is the differential of g_i at p , > d(g_i)p : Tp M ---> Tg_i (p) C ; C denotes the complex field, Tg_i (p) C > is canonically isomorphic to C, so d(g_i)p is an element of T*M_p ; > f_i : M ---> xi , > d( f_i )p is the differential of f_i at p > d(f_i)p : Tp M ---> Tf_i (p) xi > Observation: g_i(p) tensor d(f_i)p is trivially g_i(p) * d(f_i)p > where * denotes the product of the scalar g_i(p) by the vector d(f_i)p > Since g_i vanish at p , we have d(f)p = d( g_1 )p tensor f_1 (p) + ... + > d( g_n )p tensor f_n (p) > Conclusions: > d(f)p : Tp M ---> T0 xi where T0 xi denotes the tangent space to the > vector bundle xi at 0 . > d(f)p : Tp M --->Span( f_1 (p) , ..., f_n (p) ) ; > Span( f_1 (p) , ..., f_n (p) ) is contained in T0 xi_p === Subject: Re: James I always have the last word; so, with utmost finality, >Thats all from me, the model of a Newsgroup Personality. >> I think the last line is missing a two-syllable word near the end. Apart >> from that the whole thing scanned beautifully. > Yes, it would sound better, for example, with have the final word. > But wait! Compare it to how the original ends: > But still in matters vegetable, animal, and mineral, > He is the very model of a modern Major-General. > Reading this, it is clear that there is a caesura between vegetable > and animal, [You correctly guessed that I meant *second*-last line] But actually Ive always heard the original second-last line performed with 4 very clear syllables to the word ve-ge-ta-ble, so that the line has the same constant rhythm as all the other lines. This is called a patter song by the way, because it just patters along at a constant rhythm with no let-up. Gilbert and Sullivan included one in most of their shows. === Subject: Re: Plotting a 7 vertice graph in which every vertex has degree 4 <7GtVb.16201$F23.14567@newsread2.news.pas.earthlink.net David, > I came up with three sub-graphs, which I have uploaded the images of with > Mathematica. > The first of the four images is just the complete graph for K_7, and doesnt > relate. > Do I have the right idea? > http://home.earthlink.net/~diana53/mathematica/1-1-17.html Im not sure why you have the complete graph on 7 vertices, since its not 4-regular. The rest of the graphs look like different drawings of the same graph (the complement of the 7 cycle), as Professor Eppstein suggested. His other suggestion is the complement of a 4 cycle and 3 cycle; i.e. you will have a graph whose vertices can be partitioned into two sets, one set with 3 independent vertices and another set with 4 vertices and two disjoin edges, and then form all edges between these two sets. J === Subject: Re: there is no such thing as infinity @reader2.panix.com: Hush. Were enjoying this. === Subject: Re: JSH: Dont talk to me > I think the problem is that todays mathematicians are pencil > pushers--and not in a good way--who do NOT have guts. > Freaking cowards are running as fast as they can. > So Im in the process of chasing them down. Be awfully careful not to catch any of them or you just might get a little of those humongous loads of crap squeezed out of you. Then you might be too tiny to reach the keyboard and amuse us any more. === Subject: NaturalNumbers are the P-adics and why Kissing density jumps in KPP Re: applying RiemannHypothesis modification to Kepler Packing Problem > The below is an old post talking about the Kepler Packing Problem as === > Subject: Re: PROOF OF THE KEPLER PACKING > PROBLEM > <1993Aug19.021638.256@rp.CSIRO.AU Radiophysics/Australia > Telescope > National Facility > (snipped to save space) > example in 9 > dimensions, where the densest known > sphere packing is > that produced by > the Lamda_9 lattice (packing density > 0.14577, kissing > number 272), > whilst the greatest known kissing number > is that > achieved by the > non-lattice packing P_{9a} (packing > density 0.12885, > max kissing number > 306). For more details, an excellent book Okay, well, if the NaturalNumbers are the P-adics, and if RH implies that there are no straightlines at infinity because the P-adics compose the 1/2 Realline. Then what an application of RH would do to the Kepler Packing Problem is to first ask the question of does a p-adic dimensional space make much sense. Is there a ...99999 dimensional space in 10-adics? Is there a ....11111 dimensional space in 2-adics? Then further, a RH application of p-adics to the KPP of kissing points versus densest-nonkissing plan of attack to prove would then ask the Very Important Question: Question: does the above quoting suggest that the divergence of 9th dimension becomes even more divergent when in the 10th dimension. Then the 11th dimension, how much of a divergence if any from the previous dimensions. You see, if NaturalNumbers are really the P-adics, then in KPP there should be a linear increase in divergence as we go higher in dimensions with the density of packing. For example: the writer above noted that in 9th dimension the kissing diverges from regular KPP, then the kissing should also diverge in 10th dimension, and also in 11th dimension and so forth. But, if the KPP does not diverge in 10th dimension from that of kissing in 10th dimension Suggests or Implies that the P-adics are involved. If Straightlines exist out to infinity and if NaturalNumbers are the FiniteIntegers then the KPP should not be a pockmarked gapping of kissing points divergence as we increase in dimensions. On the other hand, if NaturalNumbers are the P-adics and that all straightlines curve as they approach infinity (i.e. straightlines do not exist), then the divergence of the KPP from that of kissing points versus densest pack would not be a smooth linear relationship as we increase in dimensions, and instead have gaps where in say dimension 22 the kissing points is the densest pack and where dimension 23 the kissing points are not the densest. Demonstration: If we take oranges to pack and we had a square box (Euclidean Space) and a similar volumed sphere and asked to pack those oranges in which container could we get the densest packing? The cube or the sphere? So that in the KPP, applying the RH would suggest that the divergence of kissing is because of the fundamental reason that NaturalNumbers are really the P-adics. Because if space is Euclidean and that straightlines remain straight out to infinity and that NaturalNumbers are FiniteIntegers then as you increase in dimensions from say 9 to 10 to 11 to 12 etc etc, that the divergence from packing should also be a Smooth and linear progression. But it is not. It is gap ridden and swinging back and forth between kissing as the densest and kissing not the densest. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: JSH: Research question answered > Its not clear to me whether James is thinking deconstructionist or > constructivist. Its not clear to me whether James is thinking. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Research question answered > > The phrase tautological space is one I came up with from my other > math research. > > It refers to regions other than > > x=x, which is x = 0(mod x), > > as thats the basic tautological space from which mathematicians > traditionally operate, and its used for most research, > Correction, that should be 1=1, which is 1 = 0(mod 1), and yes, I know > its trivial but its the base space or explicit space that > mathematicians usually use. I can imagine that JSH might use it, as he is capable of any sort of idiocy, but unless there is more to it than meets the eye, no one else will ever bother with it. In fact, this seems like a sterling example of a case where there is less to it than meets the eye. === Subject: need help!!! > Mark im sorry for being vague-im exploring the early Greek system of > using acrophonics and then the alphanumerical usage-in general im > looking at exploring how the greeks acquired their systems-how they > were modified and how they were superseded > neil > . > Perhaps you have a specific thing in mind by the ÔGreek system, > not just ancient Greek mathematics in general, Neil? > > Sounds intriguing. Can you say more? > > Mark > > Hi i require information wrt the above-im writting an essay on how the >> greek system came into being-and what impact it had on mathematics-can >> anyone direct me to good sites etc > hi every body can plot this function please say me with www,hupo19@yahoo.com it is y=arcsin^-1(3/cosx),thank you === Subject: Re: Derivative of a sum = sum of derivative? >Does a derivative of a sum always equal a sum of a derivative? I know for >integrals its not true.... For finite sums, yes. For infinite sums, consider that the pointwise convergent sum of infinitely many differentiable functions need not even be continuous at a point of convergence. === Subject: Re: JSH: Research question answered >> >> I think it interesting as an advanced question to figure out how to >> construct one of the imperfect factorizations from a given >> tautological space, though it seems that would require a tautological >> space only valid in the complex plane. >> Hmmm...therefore, it seems logical that an imperfect factorization >> cannot be constructed from a tautological space. >> [...] >> The phrase tautological space is one I came up with from my other >> math research. >> It refers to regions other than >> x=x, which is x = 0(mod x), >> as thats the basic tautological space from which mathematicians >> traditionally operate, and its used for most research, >Correction, that should be 1=1, which is 1 = 0(mod 1), and yes, I know >its trivial but its the base space or explicit space that >mathematicians usually use. Correction, calling 1 = 0(mod 1) a space, tautological or otherwise, is just meaningless nonsense. >James Harris ************************ David C. Ullrich === Subject: Re: straightlines curve at infinity; Riemann H. connects with Poincare Conjecture Re: when NaturalNumbers = p-adics what alters in the Riemann Hypothesis If we accept as true that the NaturalNumbers are the P-adics, then the 1/2 Realline in the RiemannHypothesis must be a curved line and that no straightlines ever exist but curve as the further we go out. In 1993 or 1994 I claimed this number of p-adics in the 10-adics of ....999999 as the largest number that exists. I claimed it was infinity itself. I am proud of that claim for it has not diminished in stature in these intervening years. And today I can put further use to that number .....99999 for in the Poincare Conjecture of a point compactification at infinity where you want to take the infinite Euclidean Plane and sort of take its 4 edge-points and like a sheet of wrapping paper want to join those four edge points and make a sphere. Well, I am proud to say that Euclidean geometry at infinity is a fiction a illusion and purely imaginery just as ghosts and witches are imaginary. That Euclidean Geometry is curved lines at infinity because the NaturalNumbers are the P-adics and that no-one needs to point compact the Euclidean Plane because it is already forming into a sphere and that this number .....999999 is the point that is the 4-pointedge of the infinite Euclidean plane. I suppose if you take just the 10-adics then it is a infinite circle. But if you take collectively all the p-adics of 2-adics, 3-adics etc etc they form an infinite sphere. I am guessing that the Collective P-adics is similar (I do not know if they are equal) to the geometry formed by the positive-Reals which is Riemannian Geometry. I do not know the relationship between the geometry formed by the Collective P-adics and the geometry formed by the positive Reals as Riemannian geometry. Both have positive curvature. But the positive Reals seem to have numbers such as pi and e which the Collective P-adics do not have and vice versa. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Differential operetor and variable changing I have an important question for my studies. I have a function, for example x=Rsin{theta}cos{phi}, so the differential dx became dx=sin{theta}cos{phi}dR+Rcos{theta}cos{phi}dtheta+(- Rsin{theta}sin{phi}dphi). In other words the differential of an n variables function is the sum of differentials depending each only by a variable. How became the operator frac{partial}{partial x} knowing that x is function of other variables like in the precedent example? -- Ah, contact me to fedelemail@yahoo.it === Subject: Re: JSH: Tautological spaces >> Ive spent a lot of time working the tautological space >> x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2) >> which youll notice has 4 elements. > No doubt they are earth, air, fire and water. Given the breadth and versatility of James imagination and intellect, theyre probably more like earth, dirt, dust and soil. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock Your inanity is truly unparalleled. I humble myself before your greatness. > Are you crazy? > I bet, you do not have enough power of imagination to understand magnitude > of my craziness. > Physics, this field is of crazy people. If you are sane, better to do 9 to 5 > job and never look at this NG. > -Abhi. === Subject: Re: help me please > I know the notions of pull back bundle, tensor product of two bundles over > I dont know how to construct the notion of tensor product of two > different bundles over two distinct manifolds, say M_1 and M_2 starting (i > suppose from the previous one notions). How about this? Consider M = M_1 x M_2. Let B_1 and B_2 be the bundles on M_1 and M_2. If pi_1: M -> M_1 is the projection consider the pullback pi_1^*(B_1) from M_1 to M. Similarly consider pi_2^*(B_2) and form the tensor product pi_1^*(B_1) (x) pi_2^*(B_2). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: JSH: Dont talk to me > Freaking cowards are running as fast as they can. > So Im in the process of chasing them down. > James Harris Hey Jim. What would you do if you caught one? === Subject: A little help is needed I am not quite sure how to solve this problem. Can someone give me a little help, It would greatly be appreciated. The question is as follows: Determine whether f: Q x Q -> Q given by f(a/b,c/d)=(a+c)/(b+d) is a well-defined function. Again thank you very much for your help. === Subject: Re: Point-set Topology Hints Wanted . Adjunct Assistant Professor at the University of Montana. >> >>I am trying to show that Cl(A) is a subset of A. >> >>A is the set of all limit point of A . >> limit pts A = derived A = A = { x | x in cl Ax } >> So what you actually want is that every punctured neighborhood >> intersect A. >> When space is T1, A is closed. >> Now for all A >> A subset cl A >> so >> A subset cl A = A >> Is A subset A even when space is not T1 ? >> Let X = {x,y}, with the indiscrete topology T = {empty, X}. Let A >> ={x}. Then x is not in A, but y is. So A = {y}. And by a symmetric >> argument, A = {x}. So A is disjoint from A. >Yes. Perhaps for an encore you have counter example for T0 space? As a matter of fact, yes, I figured one out yesterday while trying to learn to cross-country ski... Let X be the naturals union a single point, called infinity. The topology on X consists of the empty set, and all subsets U of X that satisfy the following two conditions: (1) infinity is in U; and (2) There exists k in N such that for all n>k, n is in U. This is a topology: the empty and total sets are there. An arbitrary union of sets satisfying (1) and (2) satisfies (1) and (2); and the intersection of two sets satisfying (1) and (2) also satisfies (1) and (2). It is also a T0 space: given any two natural numbers r and s, the sets U_r = X-{r}; U_s = X-{s} are open sets containing, respectively, r and not s; and s and not r. And given a natural number r and infinity, the set U_r = X-{r} is an open set containing infinity but nor r. There is no open set containing r and not infinity, so the topology is T0 but no T1. Now let A = {infinity}. Then A is all of N. And A is all of X, so A is not contained in A. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: appeal to check argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Juxt17918; Fair point, Toni. Well here it is then - I posted it last night already. Ta, Mark G. >>Does anyone have a couple of minutes to point out the errors if I >>post a two-screen-length argument about properties odd perfect >>numbers need to have? >What makes you think if people havent got time to do that, then they >have time to respond to you telling this is the case? >Oh, and if you post something to sci.math, someone will point out the >errors. Dont worry about that. >-- >Im not interested in mathematics that might have anything >to do with reality. -- Russell Easterly, in sci.math === Subject: Re: Silly question for someone with a big calculator. Theorem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Juwl17910; My Guess is as Good as Yours Theorem 2.1: Choose any alpha, beta in Z (whole numbers) such that 2|alpha| > |beta| > 1 in Z (whole numbers). Define x_m = coefficient of x in (alpha x + alpha y + beta e)^m y_m = coefficient of y in (alpha x + alpha y + beta e)^m e_m = coefficient of y in (alpha x + alpha y + beta e)^m Then x_m = y_m for all m in naturals and the sequence (2x_m/y_m) converges to some number z in reals such that 1,4142 < z < 1,4200. Note: Surely, it should work as well when alpha, beta in reals (instead of Z) fullfill the above condition. But since I am only working on the ring at the moment, that is how I thought I`d formulate it. //The rules of the game are: x*x = y*y = e*e = e and // x*y = -y*x C. Dement === Subject: Correction, reply to lemma by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv2L18055; like a nice name for a woman) could check her result by noting that the sum of the coefficients of x and y divided by the coefficient of e is always a constant. This is incorrect. However, perhaps more interestingly, it appears to approach a constant: For ^2, below, 12*2/17 = 1,411764... For ^3, 408*2/577 = 1,414211... For ^4, 470832*2/665857 = 1,414213... >For example (2 x + 2 y + (-3) e) ^ 2 = > = 4x*x + 4 x*y - 6x +4 y*x + 4y*y - 6y - 6x - 6y + 9 = > = 17e - 12x - 12y > Proceeding, I get > (2 x + 2 y + (-3) e) ^ 4 = -408x - 408y + 577e > > (2 x + 2 y + (-3) e) ^ 8 = -470832x -470832y +665857e === Subject: Re: JSH: Pattern argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv1Y17985; No it doesnt have a BS in physics but I dont think it will be hard for it to get one(after all, JSH got one). As for being a veteran, *packs the rock and mails it to a military base in Iraq*. Be patient for umm two days and youll get yourself a veteran rock. (Yes I am a drop of water in the Pacifist Ocean) === Subject: Re: 1/0 now allowed by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv1d17995; I define you to be a nutcase. === Subject: Re: JSH: Tautological spaces by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv1w18002; My hunch tells me that to construct M you will need elements from tautological space theory. *giggle* === Subject: Re: JSH: Dont talk to me by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv2s18029; >> Back in the 70s when it came out that Nixon had a secret shit list of >> journalists he disliked, it became a badge of honor to be on the list. >> In fact, many journalists NOT on the list were embarassed to have been >> omitted. >I dont find your analogy very convincing. Posts from the cult of JSH >detractors are often sad documents that should prove embarrassing to their >authors when and if their heads are ever removed from their posteriors. And >I dont think many sci.math posters are embarrassed at being ommited from >JSHs list, do you? It is true that there have been many nasty retorts to JSHs posts but oddly enough he has never seemed to be disturbed by those (except perhaps for reply as nastily). The three people that JSH lists here are people he specifically dislikes, apparently, because they are respond honestly and intelligently- attacking his statements rather than him. That appears to be what JSH cant stand. === Subject: Re: Silly question for someone with a big calculator. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv2518042; > For any given m in naturals which is a power a 2, we would like >> to find the coefficient (i.e., a whole number by definition) >> of the unit e in (2 x + 2 y + (-3) e) ^ m raised to the power >> of m, assuming that the associative/distibutive laws hold. The >> task is to find for what m this coefficient is not a prime number. >> The rules of the game are: x*x = y*y = e*e = e and >> x*y = -y*x >> For example (2 x + 2 y + (-3) e) ^ 2 = >> = 4x*x + 4 x*y - 6x +4 y*x + 4y*y - 6y - 6x - 6y + 9 = >> = 17e - 12x - 12y >> Proceeding, I get >> (2 x + 2 y + (-3) e) ^ 4 = -408x - 408y + 577e >> >> (2 x + 2 y + (-3) e) ^ 8 = -470832x -470832y +665857e >> which I quickly checked to be prime from >> http://www.num bertheory.org/php/prime_generator.html: 886731088897 = 257 * 1409 * 2448769. if it is correct or not: notice the sum of the coefficents of x and y divided by the coefficient of e is always a constant: Namely 1,414211... (! - I just saw this, myself. Of course, there will probably be some trivial reason for this.). C.Dement === Subject: Re: A little help is needed > I am not quite sure how to solve this problem. Can someone give me a > little help, It would greatly be appreciated. The question is as follows: > Determine whether f: Q x Q -> Q given by > f(a/b,c/d)=(a+c)/(b+d) > is a well-defined function. > Again thank you very much for your help. It would be well-defined if you described some unique way of expressing an arbitrary rational number in the form a/b. For example, you might say that and a and b are integers, b is positive and that a and b have no common factor greater than one. -- Clive Tooth http://www.clivetooth.dk === Subject: strange identification spaces We are talking about identification spaces in my topology class, and it was shown, with a rectangle with its bottom left corner on the origin, four units in length and 2 units in height, how to create a cylinder, Moebius strip, torus, and Klein bottle. For example, identifying (0, y) with (4, y) was a cylinder and (0, y) with (4, 1 - y) was a Moebius strip, while a Klein bottle was made by identifying (0, y) with (4, y) and then (x, 1) with (1-x, 0). So the question was brought up about what would the object be if we did both of the following things: identify (0, y) with (4, 1-y) identify (x, 1) with (1-x, 4) Our professor had no idea how to visualize this or if it was recognized topological entity. So does anyone know what this is, what it looks like, if it has a name, etc? Matt Lafer University of Michigan === Subject: Re: Number Theory Problem [quote:0f5f20a1f4] > while c=N-1 is not so representable. I can only prove this trivial case. Suppose c=N-1 is representable in the form that c=ax+by with x,y>=0 while N-1=(a-1)(b-1)-1=ab-a-b. Then ab-a-b=ax+by and hence the equality ab=(x+1)a+(y+1)b ...(1) holds. Since gcd(a,b)=1 and therefore x+1 is divisible by b while y+1 is divisible by a, which implies that (x+1)a+(y+1)b>=2ab, in contradiction to (1).[/quote:0f5f20a1f4] Now I can complete the whole proof. For any integer c>=N, it can be written as c=ab-a-b+m while m is an integer greater than 1. Since gcd(a,b)=1, there exists some integers s,t such that as-bt=1. Multipy m on both sides and we have ams-bmt=m ...(2) According to euclidean division, there exists some integers q,r such that mt=aq+r, 0<=r=0. Thus ax*=by*+m>0, which implies x*>0 and it follows that x=x*-1>=0. Thus ax+by=a(x*-1)+b(a-1-y*) =ax*-by*+ab-a-b =ab-a-b+m=c ( by (4) ), indicating that x,y are the coefficients required. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: laplace transforms and multiple intergration Does anyone know of any sites that you can enter the laplace transform and get the answer, and again the same for multiple intergration. neil === Subject: Calculus help This seems like it should be easy, but it is not doing it for me> the lim as x approaches 0 for the function (sin x) / x If x= -0.1, the answer is supposed to be .9983. The answer book says Ensure you use radian mode. Why?? Why does this problem require radian mode vice degree mode??? Stumped === Subject: Re: discrete mathematics fat sets > Given a set S {1,2,3,...,n} > A subset P is said to be a fat set if every element in it is >= the > cardinality of the subset P. > The problem is to find the number of such fat sets. > Infinitely Countable > nulset > {n}, n >= 1 > {n,m}, min n,m >= 2 > ... > {n1,.. nj }, min (n1,.. nj) >= j > ... Back to the question... GIven n, how many fat subsets of S = {1,2,3,...,n} exist? === Subject: Re: there is no such thing as infinity > That is a problem, but there also is no definition of M. M is the largest number. The computer will stop when it hits M. I dont need to define it as the computer will know it when it finds it. Thats why I am using a computer for the calculations as they are much more powerful than human beings. Dr. Ben Zona === Subject: Re: strange identification spaces Correction: The rectangle is *one* unit in length. === Subject: Re: strange identification spaces argh! I meant one unit in *height*. sorry for the multiple posts. So to reiterate: the rectangle is one unit in height, and four units in length. === Subject: Re: JSH: Dont talk to me > I dont necessarily mind mocking responses. > Exactly. It isnt mockery or abuse that gets to you, it is simple and > clear proof that you are wrong. Thats the unforgiveable offence, isnt it? > Gib > Hell no!!! Thats a relief!!! Im not a mathematician. Im some guy > who decided hed go looking for something that might have been missed > in the great rush of math society to build upon itself. > And I found it. > Like dont try the bullshit of saying I dont admit when Im wrong, > when time after time over a period of years I have. Yeah, after youve been smacked with the 2x4-of-truth enough times to leave a dent. > And besides theres my prime counting function which any person out > there with the balls to go do a Google search on can see is unique in > that it uses a partial difference equation, and it doesnt take long > to find out that no one else in recorded history managed to find such > a gem. > I think the problem is that todays mathematicians are pencil > pushers--and not in a good way--who do NOT have guts. > Freaking cowards are running as fast as they can. > So Im in the process of chasing them down. You look more like a clown stumbling around in a McDonalds play room. > James Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: a infinite integration problem > HI, > Can someone tell me the results of the following integration? > int_{-infty+j*epsilon}^{infty+j*epsilon}frac{1}{v(v-j*a)(v+j* b)}d v > where v is complex variable,a,b,epsilon are real varialbe and >0, > j=sqrt{-1}. > Tao According to Maple... > with(student): > assume(epsilon>0); > assume(b>0);additionally(b>epsilon); > assume(a>0);additionally(a>epsilon); > assume(t,real); > int(1/((t+I*epsilon)*(t+I*epsilon-I*a)*(t+I*epsilon+I*b)), > t=-infinity .. infinity); / I I 2 I Pi |- ------------- - --------------| a (I a - I b) b (-I a + I b)/ with assumptions on a and b > simplify(%); 2 I Pi ------ a b with assumptions on a and b But if a if it is correct or not: notice the sum of the coefficents > of x and y divided by the coefficient of e is always a constant: > Namely 1,414211... (! - I just saw this, myself. Of course, there > will probably be some trivial reason for this.). Define always a constant: 4/3 = 1.333333... 24/17 = 1.411764... 816/577 = 1.414211... 941664/665857 = 1.414213... If we let k represent that ratio for one of the exponents, the ratio for the next in this list is (4k)/(k^2+2); it can be shown that the ratios will converge to sqrt(2). 1 = 3^2-2^2-2^2 = 17^2-12^2-12^2 = 577^2-408^2-408^2 = ... -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: JSH: Tautological spaces > I did a post recently where I said the base tautological space that > mathematicians operate in is x = 0(mod x), and I realized later thats > wrong as its 1 = 0(mod 1). > Tautological space. That is very funny. I have noted this before. JSH is > a humorist. It basically means logical domain. Mathematicians reßexively work in a simple logical domain 1 = 1 which seems so intuitive, so obvious that most dont seem to realize that it IS a logical domain, where theres a dependency on a tautology. Unless of course youre talking about fallacious reasoning which leads to 1 = 2 as contradiction proves falsity. What happened to me is I ended up with a problem that pushed me into recognition of higher logical domains, which are *inclusive* of the base explicit domain. I call them tautological spaces. If you ever wonder where I get cubic examples of non-polynomial factorization, the answer is that theyre from the tautological space x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2, which workably, like if youre going to do some mathematics in that space, is x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2), which seems to be an idea that leaves most people gasping and retreating to the notion that tautologies are meaningless for proving anything. But tautological space link mathematics and physics in a way that you cant begin to grasp without some hint of their power and validity. Like the tautological space I use is a 4 dimensional one, as it has the 4 elements x, y, z, and v. Whether you realize it or not, I think it might actually limit our physical world in a way that I dont have the mental energy and focus--or maybe the ability--to deal with right now. James Harris === Subject: Re: Calculus help > This seems like it should be easy, but it is not doing it for me the lim as x approaches 0 for the function (sin x) / x > If x= -0.1, the answer is supposed to be .9983. > The answer book says Ensure you use radian mode. > Why?? Why does this problem require radian mode vice degree mode??? Lets see ... because degrees are not the same as radians? Consider for example sin(30 degrees)/(30 degrees), which numerically is (1/2)/30 = 1/60. Replacing degrees with radians we get sin(pi/6)/(pi/6) = (1/2)/(3.14../6), which is pretty close to 1. If you look at the geometry of the situation, youll see why radians are the natural candidate to insure the above limit is 1. === Subject: Re: :: towards a constructive education :: (news server friendly) > And stop grinning, until you put up a decent argument. Its just sad. What is it with imbeciles like you? decent argument ??? Look Daddy, I got on the debating team!!! ad nauseum :-) mitch === Subject: Re: there is no such thing as infinity In sci.math, N.LENN @ WKX.KM.EU : >>can use their computer time. The program in FORTRAN is simple: >>00001 n=1 >>00002 1 n=n+1 >>00003 print(3,4)n >>00004 if(n.eq.M) then print(3,4)M >>00005 else go to 1 >>00006 end if >>00007 end > What bastard version of FORTRAN is this? > Last time I used unit 3, it was a card READER on an ICL1906A. > Wheres your format definition for label 4? > Wheres the STOP ? > Why are the sequence numbers in columns 1 to 5 instead of 72 to 80? > Why is GO TO spelt as two words? > Why is ENDIF spelt as two words? > Why doesnt the compiler bork when it spots the use of an undefined > variable in the IF? > Jokes are supposed to be internally consistent, you know. Looks more like a variant of BASIC to me...and probably not a working one, as Ive not seen a BASIC with an ELSE statement on a separate line. The (3,4) could be a cursor reference, but the only BASIC I remember doing that is a TRS-80, which Im not that familiar with personally, and I think it used an Ô@ anyway. If it is FORTRAN, your objections for the most part are valid, and theres the issue that statements must begin in Column 7. BTW, older FORTRANS were happy with GOTO, GO TO, G O T O, if Im not mistaken. Im not sure theyd be happy with the IF statement as structured; if its straight FORTRAN it wouldnt know what THEN is, and if its WATFOR or equivalent the print() should be on a different line. The example is probably better written anyway. In this case Ill use C, since its what I generally use: void doit(int m) { int n = 1; do { n++; } while(n < m); printf(%dn, m); } Should give one an idea of how one might structure this, erm, problem. The encapsulation of this thing in a working main() Ill leave to the interested reader. For what all this is worth. :-) (Am I missing the joke? :-) ) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Number Theory Problem > > while c=N-1 is not so representable. > > I can only prove this trivial case. > Suppose c=N-1 is representable in the form that c=ax+by > with x,y>=0 while N-1=(a-1)(b-1)-1=ab-a-b. > Then ab-a-b=ax+by and hence the equality > ab=(x+1)a+(y+1)b ...(1) > holds. > Since gcd(a,b)=1 and therefore x+1 is divisible by b > while y+1 is divisible by a, which implies that > (x+1)a+(y+1)b>=2ab, in contradiction to (1). You can find integers m, n >= 0 so that am - bn = 1 (since gcd(a,b) = 1). Then, for any integer c, the possible representatiuons of c are all of the form c = a(mc - kb) - b(nc - ka) for some integer k. So the problem reduces to finding some choice for k that forces the inequalities mc - kb >= 0, nc - ka <= 0 to be satisfied. Thats equivalent to being able to choose k satisfying nc/a <= k <= mc/b . Note that (mc/b) - (nc/a) = ((am - bn)c)/(ab) = c/(ab) -- thus, the interval [nc/a, mc/b] contains at least one integer as soon as c >= ab (since, then, that interval has length >= 1). This proves the result for c >= ab. How about the remaining values for c ?? If N <= c < ab, consider c + a + b which is > ab (by the way in which N is defined). If c + a + b is not a multiple of either a or b, then c + a + b = ra + sb with r, s > 0 and its immediate that c = (r-1)a + (s-1)b which is a representation of the correct sort. If c + a + b = sa, then c = (s-1)a - b; since s > b, you can write s-1 = qb+r with q >= 1, 0 <= r < b and then verify that c = ra + (q-1)b, taking care of this case. The argument when c + a + b turns out to be a multiple of b is (pretty much) the same. [] === Subject: Re: Number Theory Problem >I found the following problem in an old number theory text by William >J. LeVeque. Neither I nor anybody I know can solve this problem. >Let N=(a-1)(b-1), where a,b are positive integers >and gcd(a,b)=1 >Show that every integer c>=N is representable in the form >c=ax+by with x,y>=0, while c=N-1 is not so representable. Since gcd(a,b)=1, we have some (u,v) so that au+bv=1. The set of all solutions to ax+by=c is { (cu+bk,cv-ak) : k in Z }. Thus, we have a non-negative solution (x,y), i.e. one in which both x and y are non- negative, precisely when there is an integer k so that k >= -cu/b (so that cu+bk >= 0) and k <= cv/a (so that cv-ak >= 0). Thus, ax+by=c has a non-negative iff there is an integer in [-cu/b,cv/a]. Since au+bv=1, [-cu/b,cv/a] has length c/(ab). Suppose there is no integer in this interval. This means that it must be contained in some interval (j,j+1). Since cx and cy are integers, we must have -cu/b-j >= 1/b [1a] and j+1-cv/a >= 1/a [1b] Adding [1a] and [1b] and multiplying by ab gives ab-cau-cbv >= a+b [2] Since au+bv=1, [2] becomes c <= ab-a-b [3] Therefore, if c >= ab-a-b+1 = (a-1)(b-1), then there is a non-negative solution (x,y) to ax+by=c. Suppose c = (a-1)(b-1)-1 = ab-a-b. Then, since au = 1-bv, -cu/b = -(ab-a-b)u/b = -u(a-1) - v + 1/b [4a] and since bv = 1-au cv/a = (ab-a-b)v/a = v(b-1) + u - 1/a [4b] Combining [4a] and [4b] and au+bv=1, we get [-cu/b,cv/a] = [-au+u-v + 1/b,-au+u-v + 1-1/a] [5] Thus, the interval in [5] contains no integers since it is contained in the interval (j,j+1) where j = -au+u-v. Therefore, if c = ab-a-b = (a-1)(b-1)-1, there are no non-negative solutions (x,y) to ax+by=c. Rob Johnson take out the trash before replying === Subject: Re: Combinatoric:4 groups of 4 Look up references on finite geometries or combinatorial designs, and affine planes in particular. > This is a planning problem for scheduling players in teams of 4. > Not sure if it is feasible. > 16 players, 5 days of play, 4 teams each day. > Each player teamed with each other player only once over the 5 days. . . . > Any suggestions or pointers to places of study are much appreciated. === Subject: Re: strange identification spaces >We are talking about identification spaces in my topology class, [...] >So the question was >brought up about what would the object be if we did both of the >following things: >identify (0, y) with (4, 1-y) >identify (x, 1) with (1-x, 4) If I understand the corrections in your other posts, you mean you want to identify the edges of a rectangle as follows: *----->>------* | | | | | | ^ v | | | | | | *-----<<------* This is often presented in other ways; for example, its equivalent to identifying antipodal points on the boundary of a disc, or to identifying antipodal points on a sphere. Theyre all the Projective Plane RP^2. It doesnt embed in R^3, which may be why you cant visualize it easily. On the other hand, >Our professor had no idea how to visualize this or if it was >recognized topological entity. Please tell me your professor is teaching topology as part of an exchange program with, I dont know, Control Theory or Mathematical Logic or something? This is sort of basic! dave === Subject: Recursive Expansion Is there a way to explicitly compute an arbitrary element of a list that is given by the expansion R_0 = null R_{n+1} = 0,R_n,R_n,R_n,1 where we are putting three copies of the previous list between 0 and 1. I want to know how to randomly access an element in that list without using recursion. It is also necessary that no requirements external storage space grow larger than n. === Subject: Re: JSH: Tautological spaces > > I did a post recently where I said the base tautological space that > mathematicians operate in is x = 0(mod x), and I realized later thats > wrong as its 1 = 0(mod 1). > > Tautological space. That is very funny. I have noted this before. JSH is > a humorist. > It basically means logical domain. > Mathematicians reßexively work in a simple logical domain > 1 = 1 > which seems so intuitive, so obvious that most dont seem to realize > that it IS a logical domain, where theres a dependency on a > tautology. Where mathematicians actually work has proven itself out of reach of JSHs intuition and intellect for lo these 8 years. > Unless of course youre talking about fallacious reasoning No. That is what JSH is always dealing with. Actual mathematicians avoid it as much as possible. > which leads to 1 = 2 A common consequence of JSHs chains of reasonning. > as contradiction proves falsity. > What happened to me is I ended up with a problem that pushed me into > recognition of higher logical domains, which are *inclusive* of the > base explicit domain. So far, nothing has been able to drag you, screaming and kicking, into any _logical_ domain at any level. One expects that JSHs perceptions here will again prove to be delusions. > I call them tautological spaces. We they should be called I am too plite to mention here. > If you ever wonder where I get cubic examples of non-polynomial > factorization, the answer is that theyre from the tautological space > x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2, I have often wondered why, but never where. > which workably, like if youre going to do some mathematics in that > space, is > x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2), > which seems to be an idea that leaves most people gasping and > retreating to the notion that tautologies are meaningless for proving > anything. Your tautological spaces have so far proved meaningless for proving anything, and those who have proved things, like those who have provided the many proofs of your many errors, have not relied of your tautological spaces to do so. > But tautological space link mathematics and physics in a way that you > cant begin to grasp without some hint of their power and validity. > Like the tautological space I use is a 4 dimensional one, as it has > the 4 elements x, y, z, and v. > Whether you realize it or not, I think it might actually limit our > physical world in a way that I dont have the mental energy and > focus--or maybe the ability--to deal with right now. What you think has so far always been proved nonsense. And it has been so long this way, over 8 years now I believe, that no one expects anything else from you, except some amusement from the NGs private clown. === Subject: Re: A little help is needed > I am not quite sure how to solve this problem. Can someone give me a > little help, It would greatly be appreciated. The question is as follows: > Determine whether f: Q x Q -> Q given by > f(a/b,c/d)=(a+c)/(b+d) > is a well-defined function. > Again thank you very much for your help. f(2/3,4/5) = 6/8 = 3/4 f(4/6,4/5) = 8/11 === Subject: Re: A little help is needed Adjunct Assistant Professor at the University of Montana. >I am not quite sure how to solve this problem. Can someone give me a >little help, It would greatly be appreciated. The question is as follows: >Determine whether f: Q x Q -> Q given by > f(a/b,c/d)=(a+c)/(b+d) >is a well-defined function. >Again thank you very much for your help. Whether it is well-defined can mean two different things, both of which are relevant here: (1) Whether given any input, the output belongs to the specified range (in this case, wether given any two rationals, the output is a rational); and/or (2) Whether given the same input, written in two different ways, will yield the same output. This because (a) there are many different ways to write a rational as a/b; and (b) the function is defined Here, you are being extremely imprecise in describing the elements of Q; for all I know, a, b, c, d may be rationals themselves! Though, presumably, you mean the to be integers, b and d nonzero. If thats all you ask, then (1) is equivalent to asking wether under those conditions, b+d will be nonzero. Alas, not if you are at liberty to choose a, b, c, d arbitrarily. For example, you might take a=1, b=-1, c =d =1, in which case the rules for your function would ask you to evaluate (1+1)/(-1+1). So in order for us to meet the requirement of (1), we must somehow restrict how the rationals are being written. One obvious way to avoid the pitfall above is to specify that b and d must be ->positive<- rather than just nonzero. In that case, the number (a+c)/(b+d) is a rational number for all integer values of a, b, c, and d, with b and d positive. So now we turn to (2). Your definition depends on how we represent the rationals r=a/b and s=c/d. If we choose some other representation of r and/or of s, say r=p/q, s=u/v, then we would need (a+c)/(b+d) = f( (a/b), (c/d) ) = f(r,s) = f( (p/q),(u/v) ) = (p+u)/(q+v). So we would need to check whether, if a, b, c, d, p, q, u, v are integers, b, d, q, v positive, such that: (i) aq = bp; and (ii) cv = du, whether this implies that (a+c)(q+v) = (p+u)(b+d) (so that (a+c)/(b+d) = (p+u)/(q+v)). Alas, no. Say a = b = 1, p = q = 2, c=u=1, d=v=2. Then f ( 1/1, 1/2) = 2/3 but f (2/2, 1/2) = 3/4 and 2/3 is not equal to 3/4. So your function is not well defined even if we ask that b and d both be nonzero. One easy way to make it well-defined then is to specify a ->unique<- way to write the rational inputs a/b and c/d, so that the value of f will not depend on how we write them: there will only be one valid way to write them, and thats the way we write them in order to evaluate the function. Of course, that means that to do an actual evaluation you first must check to see if your rationals satisfy these conditions. The easiest way to set this unique way is to say: f( (a/b), (c/d) ) = (a+c)/(b+d) where a, b, c, d are integers, b, d are positive integers, and gcd(a,b)=gcd(c,d)=1 (that is, written in lowest terms with positive denominators). Since every rational number x has a unique expression as a quotient of two coprime integers, with the divisor positive, this will guarantee that the value of f is uniquely determined. And since b and d are specified to be positive, the value is a rational number (the denominator is not zero). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Tautological spaces > I did a post recently where I said the base tautological space that > mathematicians operate in is x = 0(mod x), and I realized later thats > wrong as its 1 = 0(mod 1). > Thats the base tautological space where by tautological space I mean > a region of truth. > In mathematics its then a region of mathematical truth. The precision of your thought processes approximates the precision with which a demented goat could carve a small stellated dodecahedron out of Brie by using a broken plastic fork Scotch taped to one of its horns. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Dense Subset of Sobolev Space? > Hi Everybody, > Im interested in finding a dense set in the Sobolev >space of functions on the interval [-1,1], such that >f is in L2[-1,1], with the inner product > (f,g)=int( f(x)*g(x) + f(x)*g(x), x = -1..1). > I think -- but am not sure -- that the set of (finite) >linear combinations of x and exp(i*n*x) (where n runs over all >integers) is dense. > I think that the exp(i*n*x) are dense, but that has to do > with nonharmonic Fourier series and the fact that > 1 < pi; I think, but am not sure, that you simply left out > a pi somewhere. Im going to assume you meant > L2[-Pi, Pi] instead of L2[-1,1]; you could instead talk > about [-1,1] and put some Pis into the exponentials. > You also need to note that since youre clearly including > an inner product! You need to use the complex > conjugate of g and g. > With those modifications its clear that the span of > the exponentials is dense in the space; lets call > the space H. That seems false to me. Suppose P_n are trig polys converging to x in this space. Then P_n -> 1 in L2 = L2[-Pi, Pi]. But the trig poly P_n has no constant term, so the distance from P_n to 1 is at least 1 by orthogonality. (In fact no space of 2Pi-periodic functions in H can be dense in H; they all miss x.) Suppose we do this: For f in L2 let I(f)(x) = integral_[0,x] f(t) dt. If D is a dense subspace of L2, then {I(f) : f in D} + {constants} is dense in H. Proof: If f is in H, then f = I(f) + constant. So if f_n in D converge to f in L2, then I(fn) -> I(f) = f + c in L2, so that I(fn) - c -> f in L2 and (I(fn) - c) = fn -> f in L2 as desired. This shows that {trig polys} + span{x} is dense in H; same for ordinary polynomials, ie, span {1, x, x^2, ...}. === Subject: Re: discrete mathematics fat sets >Given a set S {1,2,3,...,n} >A subset P is said to be a fat set if every element in it is >= the >cardinality of the subset P. >The problem is to find the number of such fat sets. Also to come up >with a recurrsion logic. Hints: 1) S is a fat subset of {1,...,n-1} iff (S+1) union {n+1} is a fat subset of {1,...,n+1}, where S+1 = {s+1: s in S}. 2) Does Fibonacci ring a bell? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: there is no such thing as infinity >> >> I didnt know retarded people posted to newsgroups. In fact, I didnt >> know that retarded people could get phds. What mail-order diploma >> factory did you get your degree from again? > > I am not retarded and have an above average IQ, thank you. I got my > PhD from Univerisity of San Moritz, a non-acredited but well-respected > university in England. > ? > What part of England is San Moritz in? The University of San Moritz does not exist. It is a bogus name used by an Internet diploma mill. === Subject: Re: strange identification spaces >>Our professor had no idea how to visualize this or if it was >>recognized topological entity. >Please tell me your professor is teaching topology as part of an exchange >program with, I dont know, Control Theory or Mathematical Logic or >something? This is sort of basic! His teacher is (if I did a good job of cyberstalking) a dynamical systems type. My theory is that this is an instance of the Socratic Method (as they called it in Athens when the perfeseer trolled the stoonts). Lee Rudolph === Subject: Re: discrete mathematics fat sets >Given a set S {1,2,3,...,n} >A subset P is said to be a fat set if every element in it is >= the >cardinality of the subset P. >The problem is to find the number of such fat sets. Also to come up >with a recurrsion logic. Hint: How can you recursively generate the fat subsets of {1, 2, 3, ..., n+1} from those of {1, 2, 3, ..., n}? Letting c_n be the number of fat subsets of {1, 2, 3, ..., n}, I get the recursion c_(n+1) = 2 c_n - sum(k=1..ßoor((n+1)/2), C(n-k, k-1)), where C(i,j) is the binomail coefficient, i choose j. This checks for 0 <= n <= 5. Interestingly, empirically, it seems that c_n = F_(n+2), the (n+2)nd Fibonacci number (with F1 = F2 = 1). I havent verified this from the recursion, nor can I come up with an explanation why this is so. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: there is no such thing as infinity > In sci.math, N.LENN @ WKX.KM.EU >: >can use their computer time. The program in FORTRAN is simple: 00001 n=1 >00002 1 n=n+1 >00003 print(3,4)n >00004 if(n.eq.M) then print(3,4)M >00005 else go to 1 >00006 end if >00007 end >> What bastard version of FORTRAN is this? >> Last time I used unit 3, it was a card READER on an ICL1906A. >> Wheres your format definition for label 4? >> Wheres the STOP ? >> Why are the sequence numbers in columns 1 to 5 instead of 72 to 80? >> Why is GO TO spelt as two words? >> Why is ENDIF spelt as two words? >> Why doesnt the compiler bork when it spots the use of an undefined >> variable in the IF? >> Jokes are supposed to be internally consistent, you know. > Looks more like a variant of BASIC to me...and probably not > a working one, as Ive not seen a BASIC with an ELSE statement > on a separate line. The (3,4) could be a cursor reference, but > the only BASIC I remember doing that is a TRS-80, which Im not > that familiar with personally, and I think it used an Ô@ anyway. I think its intended to be Fortran, which almost works if you assume the numbers in columns 1-5 are an artifact of the text editor and are not actually seen by the compiler. > If it is FORTRAN, your objections for the most part are valid, > and theres the issue that statements must begin in Column 7. Thats true if you are using fixed format, but it could be free format, in which case it doesnt matter where the statements begin. > BTW, older FORTRANS were happy with GOTO, GO TO, G O T O, if > Im not mistaken. Im not sure theyd be happy with the IF statement > as structured; if its straight FORTRAN it wouldnt know what THEN > is, and if its WATFOR or equivalent the print() should be on a > different line. In fixed-format Fortran, blanks are not significant, and therefore the compiler sees no difference at all between GOTO, GO TO, and G O T O. In free-format source, the forms GOTO and GO TO are explicitly allowed, and similarly with ENDIF or END IF. However, the Fortran IF-THEN-ELSE would look like IF (n .eq. M) THEN print (3,4) M ELSE GO TO 1 END IF or perhaps the whole thing would be nested in a DO WHILE or a DO with EXIT that eliminates the need for a GOTO. Its legal to have a single-line IF statement, but such a statement would have no THEN and no ELSE. > (Am I missing the joke? :-) ) I think the joke is that the original program could be shortened to: print *, huge(1) end which is a standard-conforming way to print the largest value of type default integer that can be represented in a particular Fortran implementation. However, just because its the largest default integer doesnt mean its the largest integer, or even that its the largest integer that can be represented in that implementation. Fortran implementations are allowed to offer other integer types besides the default type, and there is a standard function call, SELECTED_INT_KIND(n), which tells whether there is an INTEGER kind that can represent integers in the range [-10^n,10^n]. The function returns a positive value if a suitable kind is available, and -1 if not. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Recursive Expansion > Is there a way to explicitly compute an arbitrary element of a list that is > given by the expansion > R_0 = null > R_{n+1} = 0,R_n,R_n,R_n,1 > where we are putting three copies of the previous list between 0 and 1. I > want to know how to randomly access an element in that list without using > recursion. It is also necessary that no requirements external storage space > grow larger than n. Generate the first few items in the list and look for a pattern. Determine how to generate the pattern, given n. You may need to keep track of a level counter as you go through the pattern, equivalent to a recursion level. You may also need to keep track of a modest amount of information in an array. Thad === Subject: Re: Series > What are the next ten characters in the following series? > 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 > I think you mean Ôsequence. > 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 6 > > Probably Ôsymbols vice Ôcharacters as well. > Either way, you original use of series is incorrect. > The 32nd Ôsymbol can either be Ô6 or Ô5 > It is 6 if you want to make it at least remotely interesting. But if you > meant that one can come up with a rule so that the 32nd symbol can be > either Ô5 or Ô6, then you might as well say that the 32nd symbol can be > any other real number as well. I was thinking of toggling bits. If there are only 5 bits to toggle, the 32nd number is 5. More then five bits, then the 32nd number is 6. === Subject: Re: there is no such thing as infinity > I didnt know retarded people posted to newsgroups. In fact, I didnt > know that retarded people could get phds. What mail-order diploma factory > did you get your degree from again? > I am not retarded and have an above average IQ, thank you. I got my > PhD from Univerisity of San Moritz, a non-acredited but well-respected Retarded and gullible. -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Tautological spaces Discussion, linux) >> I did a post recently where I said the base tautological space that >> mathematicians operate in is x = 0(mod x), and I realized later thats >> wrong as its 1 = 0(mod 1). >> Thats the base tautological space where by tautological space I mean >> a region of truth. >> In mathematics its then a region of mathematical truth. > The precision of your thought processes approximates the precision with > which a demented goat could carve a small stellated dodecahedron out of Brie > by using a broken plastic fork Scotch taped to one of its horns. Not fair! Hes since clarified this notion. Region of mathematical truth was only loose talk. A tautological space is *really* a logical domain. -- ...you are around so that I have something else to do when Im not figuring something important out. I was especially intrigued on this iteration by cursing, which I think Ill continue at some later date as its so amusing. --- James S. Harris === Subject: Re: Calculus help > This seems like it should be easy, but it is not doing it for me the lim as x approaches 0 for the function (sin x) / x > If x= -0.1, the answer is supposed to be .9983. > The answer book says Ensure you use radian mode. > Why?? Why does this problem require radian mode vice degree mode??? > Stumped Because x_radians is not the same as x_degrees in measuring an angle, and sin(x_radians) is not the same as sin(x_degrees) in numerical value, any more than x_feet is the same distance as x_meters. Usually, in evaluating trigonometric functions, values of the function are based on the underlying angles, and the choice between radian and degree measure is ony a matter of convenience. However, this difference cannot be ignored in evaluating expressions like sin(x)/x, in which x appears both inside and outside of the sine function. For use here, and generally in calculus, angular measure should always be in radians. === Subject: Re: there is no such thing as infinity > That is a problem, but there also is no definition of M. > M is the largest number. The computer will stop when it hits M. I > dont need to define it as the computer will know it when it finds it. > Thats why I am using a computer for the calculations as they are much > more powerful than human beings. > Dr. Ben Zona And does this zoned out doctor claim that all computers will stop at the same M? === Subject: Re: Recursive Expansion > Is there a way to explicitly compute an arbitrary element of a list that > is given by the expansion > R_0 = null > R_{n+1} = 0,R_n,R_n,R_n,1 > where we are putting three copies of the previous list between 0 and 1. I > want to know how to randomly access an element in that list without using > recursion. It is also necessary that no requirements external storage > space grow larger than n. This gets very big very quickly: R0 R1 0,,,,1 R2 0,0,,,,1,0,,,,1,0,,,,1,1 R3 0,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,, 1,0,,,,1,0,,,,1,1 ,1 R4 0,0,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,, 1,0,,,,1,0,,,,1 ,1,1,0,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,,1,0,,,,1,1,0,0, ,,,1,0,,,,1,0,,, ,1,1,1,0,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,,1,0,,,, 1,1,0,0,,,,1,0,,,,1,0, ,,,1,1,1,1 Are you sure about those commas? -- 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 === Subject: Re: there is no such thing as infinity >I didnt know retarded people posted to newsgroups. In fact, I didnt >know that retarded people could get phds. What mail-order diploma factory >did you get your degree from again? >>I am not retarded and have an above average IQ, thank you. I got my >>PhD from Univerisity of San Moritz, a non-acredited but well-respected > Retarded and gullible. ------- http://www.dailyfreepress.com/media/paper87/DFPArchive/ frontpage/1015982.htm l The diplomas, standard size at 101/2 by 151/2 inches, bestow all the merits of a degree from the University of San Moritz, Puerto Rico, a non-accredited school. The company offered a Free Press reporter a University of San Moritz transcript-- complete with credits transferred from BU-- for another $200. University Degree Program asked for no proof that the reporter, who initially called using an alias, took the BU classes or that the classes even exist. ------- No, he thinks everyone else is gullible. -- Suppose weve chosen the wrong god. Every time we go to church were just making him madder and madder --Homer Simpson === Subject: Re: 1/0 now allowed >ItÇs defined as 1.6367348238383838 >DonÇt ask why, but must be used or your calculations will be wrong. sight! -Staunch Giga Troll Defender === Subject: Sequence A001511 If anyone had looked up the sequence 1, 2, 1, 3, ... , 5 in the On-line Encyclopedia of Integer Sequences, they would have found it listed as sequence A001511. === Subject: Re: discrete mathematics fat sets * Stephen J. Herschkorn >Given a set S {1,2,3,...,n} >A subset P is said to be a fat set if every element in it is >= the >cardinality of the subset P. > Interestingly, empirically, it seems that c_n = > F_(n+2), the (n+2)nd Fibonacci number (with F1 = F2 = 1). I > havent verified this from the recursion, nor can I come up with an > explanation why this is so. Here is an attempt: For F_{n+1}, the set is S_{n+1}={1,2,...,n+1} we have that all the sets in S_n that are fat are still fat in S_{n+1}, there are F_n such sets. So we need to include all the sets where n+1 is a member. Look at set S_{n-1} and list all the fat sets: fat(n-1)={ø, [1], [2], ..., [n-1], [2,3], ...} We know that each such set U in fat(n-1) is such that any memaber x of U is such that x>=|U|. Make the familiy fatnew(n-1) that is equal to fat(n-1) with the element n+1 added. I.e: fatnew(n-1)={[n+1], [1,n+1], [2,n+1], ..., [n-1,n+1], [2,3,n+1], ...} We of course expect that some of the members of fatnew(n-1) are no longer fat. However, add 1 to all member not equal to n+1: fatnew(n-1)={[n+1], [2,n+1], [3,n+1], ..., [n,n+1], [3,4,n+1], ...} We claim that the collection fatnew(n-1) contains only fat sets. Why? Well, any set U in fatnew is such that x in U => x>=|U|-1. After every element is increased, we can conclude that x in U => x-1 >= |U|-1, or x>=|U|. (The new element n+1 does not destroy the property of any set since no set is greater that n.) The collection fatnew(n-1) has in fact only sets from S_{n+1} and fatnew(n-1) has no common elements with fat(n). We have now two collections of fat sets from S_{n+1}. Are there any more fat sets with the element n+1. No, but only handwaving can show it. :-) Therefore, F_{n+1} = |fat(n)| + |fatnew(n-1)| = F_n + F_{n-1} QED -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52 === Subject: Re: a infinite integration problem >> HI, >> Can someone tell me the results of the following integration? >> int_{-infty+j*epsilon}^{infty+j*epsilon}frac{1}{v(v-j*a)(v+j* b)}d v >> where v is complex variable,a,b,epsilon are real varialbe and >0, >> j=sqrt{-1}. >> Tao >According to Maple... >> with(student): >> assume(epsilon>0); >> assume(b>0);additionally(b>epsilon); Since the pole at -I*b is below the real axis and the path t + I*epsilon lies above the real axis, the relative sizes of b and epsilon do not affect the integral. That is, we dont need to assume b>epsilon. >> assume(a>0);additionally(a>epsilon); >> assume(t,real); >> int(1/((t+I*epsilon)*(t+I*epsilon-I*a)*(t+I*epsilon+I*b)), >> t=-infinity .. infinity); > / I I > 2 I Pi |- ------------- - --------------| > a (I a - I b) b (-I a + I b)/ > with assumptions on a and b >> simplify(%); > 2 I Pi > ------ > a b The residues at the poles of 1/(v(v-ja)(v+jb)) are 1 - ------ at ja a(a+b) 1 -- at 0 ab 1 - ------ at -jb b(a+b) With the assumption that a>epsilon, the path given for the integral can be viewed as a limiting case of a contour circling the pole at ja once counterclockwise, and so the integral should be 2 pi j - ------ a(a+b) The value given by Maple is for a contour circling 0 counterclockwise. I dont off hand see anything wrong with the formula you gave to Maple, so, assuming there was not a copy and paste error, it would seem that Maple is in error. > with assumptions on a and b >But if a take out the trash before replying === Subject: Re: Recursive Expansion >Is there a way to explicitly compute an arbitrary element of a list that is >given by the expansion >R_0 = null >R_{n+1} = 0,R_n,R_n,R_n,1 >where we are putting three copies of the previous list between 0 and 1. I >want to know how to randomly access an element in that list without using >recursion. It is also necessary that no requirements external storage space >grow larger than n. R_n has length 3^n-1. If we start numbering at 0, element number j of R_n is f(j,n) = { 0 if j = 0 { 1 if j = 3^n-2 { f((j-1) mod (3^(n-1)-1), n-1) otherwise This can be written using iteration instead of recursion. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Proof for Median Minimizing sum[] abs(x_i - M) I cant for the life of me find a proof of this on the net and I dont have any books to look it up in. Could someone please point me in the right direction? The median M of a collection of numbers minimizes the following: sum[i=1 to N] abs(x_i - M) === Subject: Fat subsets of ordinals >Given a set S {1,2,3,...,n} >A subset P is said to be a fat set if every element in it is >= the >cardinality of the subset P. >The problem is to find the number of such fat sets. Also to come up >with a recurrsion logic. Now lets consider fat subsets of ordinals. That is, taking the ordinal definition of cardinals and letting |.| denote cardinality, say a set A of ordinals is fat iff a >= |A| for all a in A. We work in ZFC. For an ordinal a, define S_a = {A subset of a: A is fat}; c_a = |S_a|, and d_a = |P(a) S_a|. (P(.) denotes power set.) Seems to me that, given an infinite cardinal k, k <= a < k2 (ordinal multiplication) implies c_a = sup {2^m: mYup, its quite true. Like I found a way to count prime numbers where >a partial difference equation is key. No one else in recorded human >history managed such a feat, but mathematicians refuse to give me the >time of day. The bastards!!! Hell yeah! ... ...what? I am little and innocent and emotionally charged speech thrills me =) >Because its a partial difference equation, it leads to a partial >difference equation, Because its a partial difference equation, it leads to a partial difference equation. teeheehee. I suppose the second equation should be function. Then again, it could be bzioung and the statement would be just as valid :) >which could be key in research to prove or >disprove the Riemann Hypothesis. Oh Lord I just had an inspiration! Using your partial difference equation i have managed to prove that the Riemann Hypothesis is probably either true or false! >But mathematicians today belong to a democracy, and they dont *like* >me, so they dont like my research. Try this: go anonymous or switch names, rename your functions etc etc and then post your research again. Chances are, while they wont know it is you and will not immediately recognize the research as yours(no comments please :P) , they still wont like it. >What boring people. I can be quite entertaining. === Subject: Re: James I always have the last word; so, with utmost finality, >Thats all from me, the model of a Newsgroup Personality. >> >> I think the last line is missing a two-syllable word near the end. Apart >> from that the whole thing scanned beautifully. > > Yes, it would sound better, for example, with have the final word. > But wait! Compare it to how the original ends: > > But still in matters vegetable, animal, and mineral, > He is the very model of a modern Major-General. > > Reading this, it is clear that there is a caesura between vegetable > and animal, > [You correctly guessed that I meant *second*-last line] > But actually Ive always heard the original second-last line performed > with 4 very clear syllables to the word ve-ge-ta-ble, so that the line > has the same constant rhythm as all the other lines. Ah! I hadnt thought of this (having never heard it performed). So we may indeed say that the parody is ßawed there. > This is called a patter song by the way, because it just patters along > at a constant rhythm with no let-up. Gilbert and Sullivan included one in > most of their shows. Always nice to learn a new bit of jargon. What a shame that the parody falls short of pure and pattering iambic octameter by a mere syllable. If this piece is indeed an anonymous bit of netlore, then Id be in favor of promulgating a version with final word instead of last word. -Jim Ferry === Subject: Re: Calculus help >Why?? Why does this problem require radian mode vice degree mode??? >Stumped If you will look in your calculus book at the argument where it shows the limit of sin(x)/x is 1, you will most likely find a geometric argument. That argument will usually involve the formula for the area of a sector of a circle A = (1/2)r^2*theta, probably with r = 1. Where did that formula come from? It comes from the proportion of the area of the sector to the area of the circle: AreaSector/AreaCircle = SectorCentralAngle/CircleCentralAngle AreaSector/Pi*r^2 = theta / 2*Pi (using radian measure here!!) Hence AreaSector = (1/2)r^2 theta. So this formula is incorrect if you dont use radian measure, and so would be the sin(x)/x limit. The same applies to any result using the s = r*theta formula for arc length. --Lynn === Subject: Re: Proof for Median Minimizing sum[] abs(x_i - M) >I cant for the life of me find a proof of this on the net and I dont >have any books to look it up in. Could someone please point me in the >right direction? >The median M of a collection of numbers minimizes the following: >sum[i=1 to N] abs(x_i - M) The sum is piecewise linear and convex in M, so it is minimized at M = x_i for some i. Show that if i < n-i, then the sum is greater when M = x_(i+1) than when M = x_i. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: JSH: Dont talk to me > So its ok for me to continue calling you a disloyal American who is a disgrace > to > the armed services of which you claim to be a veteran? Hah, thats our President. === Subject: Modular Arithmetic I was wondering if anyone knows a way to reduce 2^100 mod 5. I know the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what lot. === Subject: Proofs - Please help quickly! Im really having trouble with 4 trigonometric functions. Here they are: 1) Find an exact expression for: sin(pi/12) 2) Solve to 4 decimal places (0_I was wondering if anyone knows a way to reduce 2^100 mod 5. I know >the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what >lot. Look at the residues modulo 5: 2^1 = 2 (mod 5) 2^2 = 4 (mod 5) 2^3 = 3 (mod 5) 2^4 = 1 (mod 5) Therefore, 2^{4k} = (2^{4})^k = 1^k = 1 (mod 5). Since 100 = 4*25, you get 2^{100} = (2^{4})^{25} = 1^{25} = 1 (mod 5). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Modular Arithmetic Shawn Windle escribi.97: > I was wondering if anyone knows a way to reduce 2^100 mod 5. I know > the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what > lot. 2^100 = (2^2)^50 = (-1)^50 = 1 (mod 5) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Proofs - Please help quickly! >Im really having trouble with 4 trigonometric functions. Here they are: >1) Find an exact expression for: sin(pi/12) HINT: pi/12 = (1/2)(pi/6), and sin(pi/6) = 1/2. Use the half angle formulas. >2) Solve to 4 decimal places (0_sin2x + cosx = 0 HINT: Use the doulbe angle formulas: sin(2x) = 2sin(x)cos(x); so 0 = sin(2x)+cos x = 2sinx(x)cos(x) +cos(x) = cosx(2sin(x) + 1). This is zero if and only if either cos(x)=0, or 2sin(x)+1 = 0. >3) Solve sin^2 x - 2sinx - 1 = 0 and find the general solution. HINT: substitute y = sin(x). >4) Prove the following (this is a real toughy): tanx/secx = secx HINT: tan(x) = sin(x)/cos(x) sec(x) = 1/cos(x) (a/b)/(c/d) = (ad)/(bc) By the way: youll have a hard time proving it. I suspect you are meant to show that tan(x)/sec(x) = sin(x). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Proofs - Please help quickly! Istari escribi.97: > Im really having trouble with 4 trigonometric functions. Here they > are: > 1) Find an exact expression for: sin(pi/12) Use sin(a/2) = sqrt((1 - sin(a))/2 or sin(a - b) = sin(a)cos(b) + cos(a)sin(b) with suitable a (and b) > 2) Solve to 4 decimal places (0_ sin2x + cosx = 0 Use sin(2x) = 2sin(x)cos(x), then factorize > 3) Solve sin^2 x - 2sinx - 1 = 0 and find the general solution. Let sin(x) = t, and solve quadratic. > 4) Prove the following (this is a real toughy): tanx/secx = secx You mean tan(x)/sec(x) = sin(x). Express tan(x) and sec(x) in terms of sin(x) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: interval for continuously compounding interest? are learning about different ways to compound interest, interval and continuously. I had to look deeper, however, to say what is the interval for continuously compounding interest?. This is easily determined by setting the 2 well know equations equal to each other: (P(1+(R/x))^(x*T))=(P*e^(R*T)) However, I have not been able to solve this equation for X. Can anyone === Subject: Re: Recursive Expansion >Is there a way to explicitly compute an arbitrary element of a list that is >given by the expansion >R_0 = null >R_{n+1} = 0,R_n,R_n,R_n,1 >where we are putting three copies of the previous list between 0 and 1. I >want to know how to randomly access an element in that list without using >recursion. It is also necessary that no requirements external storage space >grow larger than n. First of all, it is easily checked by recursion that |R_n| = 3^n - 1. We seek R_n[k]. R_n[0] = 0 and R_n[3^n-2] = 1. If k is neither 0 nor 3^n-2, then R_n[k] = R_{n-1}[k-1 mod 3^{n-1}-1]. Continue until either k = 0 or k = 3^n-2. This will be true at least by the time n = 1. For example, R_3[10] = R_2[1] = R_1[0] = 0 R_10[32768] = R_9[13085] = R_8[6524] = R_7[2151] = R_6[694] = R_5[209] = R_4[48] = R_3[21] = R_2[4] = R_1[1] = 1 R_6[940] = R_5[213] = R_4[52] = R_3[25] = 1 I suspect that there might be a neater method involving the base 3 representation of the index. Rob Johnson take out the trash before replying === Subject: Re: Point-set Topology Hints Wanted . Well , great now you have jumped ahead of me in the book. >> >>I am trying to show that Cl(A) is a subset of A. >> >>A is the set of all limit point of A . >> limit pts A = derived A = A = { x | x in cl Ax } >> So what you actually want is that every punctured neighborhood >> intersect A. >> When space is T1, A is closed. >> Now for all A >> A subset cl A >> so >> A subset cl A = A >> Is A subset A even when space is not T1 ? >> Let X = {x,y}, with the indiscrete topology T = {empty, X}. Let A >> ={x}. Then x is not in A, but y is. So A = {y}. And by a symmetric >> argument, A = {x}. So A is disjoint from A. >Yes. Perhaps for an encore you have counter example for T0 space? > As a matter of fact, yes, I figured one out yesterday while trying to > learn to cross-country ski... > Let X be the naturals union a single point, called infinity. > The topology on X consists of the empty set, and all subsets U of X > that satisfy the following two conditions: > (1) infinity is in U; and > (2) There exists k in N such that for all n>k, n is in U. > This is a topology: the empty and total sets are there. An arbitrary > union of sets satisfying (1) and (2) satisfies (1) and (2); and the > intersection of two sets satisfying (1) and (2) also satisfies (1) and > (2). > It is also a T0 space: given any two natural numbers r and s, the sets > U_r = X-{r}; U_s = X-{s} are open sets containing, respectively, > r and not s; and s and not r. And given a natural number r and > infinity, the set U_r = X-{r} is an open set containing infinity but > nor r. There is no open set containing r and not infinity, so the > topology is T0 but no T1. > Now let A = {infinity}. Then A is all of N. And A is all of X, so > A is not contained in A. > -- > Its not denial. Im just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: Modular Arithmetic > I was wondering if anyone knows a way to reduce 2^100 mod 5. I know > the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what > lot. Hint: What is 2^4 (mod 5)? If you know that, then what is (2^4)^25 (mod 5)? === Subject: Re: JSH: Dont talk to me > Jim, my son, whats troubling you? > Let me know. > Im waiting. > God Eloi, Eloi, lama sabachthani? === Subject: Re: Modular Arithmetic >I was wondering if anyone knows a way to reduce 2^100 mod 5. I know >the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what >lot. As shown in other subthreads, there are quick ways using the fact that a^{p-1} = 1 mod p when p is a prime. Here is a method that is of some use when the modulus is not necessarily a prime and is still quite fast. 100 in base 2 is 1100100. Each time you add a bit to the exponent, you square the number if the added bit is a one, then you multiply by the base as well: 2^(0 base 2) = 1 mod 5 2^(1 base 2) = 1^2*2 = 2 mod 5 2^(11 base 2) = 2^2*2 = 3 mod 5 2^(110 base 2) = 3^2 = 4 mod 5 2^(1100 base 2) = 4^2 = 1 mod 5 2^(11001 base 2) = 1^2*2 = 2 mod 5 2^(110010 base 2) = 2^2 = 4 mod 5 2^(1100100 base 2) = 4^2 = 1 mod 5 So 2^100 = 1 mod 5. Rob Johnson take out the trash before replying === Subject: Re: James > >To be sung to the ÔPirates Of Penzance tune ÔModern Major General >(author >unknown): >> [snip most of it] I always have the last word; so, with utmost finality, >Thats all from me, the model of a Newsgroup Personality. >> >> I think the last line is missing a two-syllable word near the end. Apart >> from that the whole thing scanned beautifully. > Eh? The second last line seems to me to trip along better if last is > replaced by final, but the last line looks just fine; > Sorry, I meant the second last line. Replacing last with final is a > start, but you still need another syllable. Adding and after the > semi-colon completes the fix. (The rest of the lyric scans just as well as > Gilberts, who was an unparallelled master of scansion, no joke about it.) Well, the second last line scans OK for me with stresses on al, have, fin, word, with, most, nal. Perhaps you are stressing ut and shortening the vowel of fi to a schwa? -- Chris Henrich Those readers who are unacquainted with the mathematical technicalities will find that they can manage quite well by ignoring them. -- John Nash === Subject: Re: there is no such thing as infinity > Ive thought really hard about this one and came to the conclusion > that there is no scientific evidence of infinity existing. The highest > number that anyone has ever measured to according to Isaac Asimov in > his book Science and Human Thought is only about 5.0 x 10^48. No one > has ever gotten past that number. Doesnt this sound weird? > Whats to say that eventually there is a number where it is impossible > to count higher than? If someone were to find this number and prove > that it is in fact the highest number, then that person would > undoubtably be rich and famous. > I am currently running a computer program that will eventually find > this magic number (I hope and pray) that I call M for short. It > counts and counts and counts and my theory is that it will eventually > stop at M. I am looking for collaborators in this experiment so that I > can use their computer time. The program in FORTRAN is simple: > 00001 n=1 > 00002 1 n=n+1 > 00003 print(3,4)n > 00004 if(n.eq.M) then print(3,4)M > 00005 else go to 1 > 00006 end if > 00007 end > It has currently reached about 2.0 x 10^18. Just as Einstein proved > that there is no aether, I am convinced that I will prove that there > is no infinity and then write a book or two. > Dr. Ben Zona ....Dr of what??? === Subject: Re: there is no such thing as infinity > Ive thought really hard about this one and came to the conclusion > that there is no scientific evidence of infinity existing. The highest > number that anyone has ever measured to according to Isaac Asimov in > his book Science and Human Thought is only about 5.0 x 10^48. No one > has ever gotten past that number. Doesnt this sound weird? > Whats to say that eventually there is a number where it is impossible > to count higher than? If someone were to find this number and prove > that it is in fact the highest number, then that person would > undoubtably be rich and famous. > I am currently running a computer program that will eventually find > this magic number (I hope and pray) that I call M for short. It > counts and counts and counts and my theory is that it will eventually > stop at M. I am looking for collaborators in this experiment so that I > can use their computer time. The program in FORTRAN is simple: > 00001 n=1 > 00002 1 n=n+1 > 00003 print(3,4)n > 00004 if(n.eq.M) then print(3,4)M > 00005 else go to 1 > 00006 end if > 00007 end > It has currently reached about 2.0 x 10^18. Just as Einstein proved > that there is no aether, I am convinced that I will prove that there > is no infinity and then write a book or two. > Dr. Ben Zona Numbers dont exist outside of the human mind. They are abstractions which give us useful information about the outside world, but the number itself does not exist. (...Starblade Riven Darksquall...) === Subject: Re: JSH: Research question answered [snip incomprehensible pseudo-math about tautological space ] The only tautological space you have established is between your ears -- unfortunately, it is *not* tautological, but contradictory. Any given JSH identity lead to a contradiction. > James Harris JSH Motto: When making love to a beautiful woman, I enhance the pleasure by fantasizing I am masturbating. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: there is no such thing as infinity > Clue #1: Blatant Simpsons reference. How so? Which part of which episode? > Ben Zona, PhD > Clue #2: Yiddish Um... how is it Yiddish? (...Starblade Riven Darksquall...) === Subject: Re: there is no such thing as infinity > You should take a look at Cantors theory of sets. In it, Cantor does > treat infinities as numbers, cardinals and ordinals. Try a Google > search on it. Cantor was a rusha. He confused generations of mathematicians by telling them there are lots of infinities. It is impossible to get a BS in math without admitting that his theories are correct. The diagonal argument was a sham because it presupposes the existence of infinity without proving its existence. His theorem is proof that infinity does not in fact exist, because if it did, then there would be only one type, not aleph null, aleph one etc., because infinity by definition is as far as one can go. He even defiled the Hebrew alphabet by putting subscripts next to aleph when the gematria of aleph is not infinity or even M but one. His theory of infinities was in fact a precursor to the communist revolution. Marx believed in thesis and antithesis, both infinite premise of dialectical materialism is true. Had Cantor never been alive, there would be world peace and for sure the number M would have been determined. > One of the brilliant things Cantor did was to define an infinite set > as a set whose elements can be put into 1-1 correspondence with a > proper subset of itself (obviously something one cannot do with a > finite set). For instance, you can put the set of positive integers > into 1-1 correspondence with the set of positive even integers by the > correspondence > n -> 2n > It seems to go against common sense to say that these two sets have > the same cardinalities. Then again, we dont have any common > experience working with infinities. That is even more evidence for the ridiculousness of there being an infinity. Any math that completely contradicts the senses is wrong just as any scientific theory that contradicts observation is wrong. You just cannot have an infinite set. Dr. Ben Zona === Subject: Concentric-Circles Game (to help teach basic number theory) I made this simple game up today, in part so as to help teach kids basic number-theory concepts (such a congruences and divisibility), but it might be far more useful as useless entertainment...if even that... Start with m concentric circles drawn on paper, where n is from 5 to 12 (or higher perhaps). The rings between the circles are subdivided so that the inner circle is subdivided into 2 equal-area sections, the next ring out is subdivided into 3 equal sections, the next ring into 4, etc..until the outer ring is subdivided into (m+1) equal sections. (And the sections are aligned so that one boundry of each is along a single line.) Players start by each placing a game-piece (small enough to fit in the smallest section) into any unoccupied section on the board. A spinner or dice are used to pick a random integer n, where each n should be between 1 and somewhere near m. On each move, a player can place a new game-piece (leaving earlier pieces) n rings inward (if possible) from their last game-pieces position, or n rings outwards (if possible). (Outward and inward are along a chain of sections which touch. Do not pass center circle or outer circle.) OR the player can place a game-piece clockwise or counterclockwise (in the same ring) a *multiple* of n spaces from their last-placed piece. But in any case, a player cant change directions during his/her move. (And the players {counter}clockwise moves may indeed circle around a single ring as many times as needed.) A player can jump over occupied spaces when measuring the number of spaces from their last piece, but they must ONLY place a piece on an unoccupied section. And the winner is the last player able to place a piece on the board. Leroy Quet === Subject: Re: :: towards a constructive education :: (news server friendly) You ask me to cut many parts out, so I have. I did read them, and still find that I dont quite understand this game of question a topic followed by anger at me responding, somehow implying that I am bringing up the topics when they are obviously yours. But I wont reply to them anymore. I didnt find much about Heyting algebras in your response. You did mention [...] : Now lets get to what is wrong with your post. Youve already worn : down my rage with your stamina, (I never expected you to reply, I : presumed it would be : below you) and its Sunday, so my tone is softened down a bit. You : have had more negative responses than positive, and you really havent : started any constructive discussion anywhere. If you look to my replies to Sergio Roa Ovalle and John Wilkins (even somewhat Brian Scott) in the group 2 newgroups, you will find the start of a bibliography and detailed references to expand particular topics. : Some have openly : declared that they didnt even bother to read the whole lot of : nonsense. I actually printed it out and : read it through a few times. It really angered me a lot. Why? : Because of the hocus pocus with the W at the beginning. : Because of the endless list of assertions for which you neither : provide proofs, examples, or clues as to why they should be of : interdisciplinary interest. Im also being told that I am mentioning topics beyond the level to be expected by newsgroups. I am being told a lot of contradictory things about what people can comprehend and what they cant. I am detailing sources in the leafs of this thread that request them, and although they are necessarily only skeletal, they do form the foundations for the talk I am looking for. : Because of your explanation in terms of insecurity, (You choose not to : interpret insecurity as something negative. But that is your : subjective choice. I think most people would prefer not to be : insecure), which is arrogant and belittling even if your personal : opinion is another. Contrast this with: : Because you keep rubbing under ones nose, how naturally it all comes : to you, as opposed to everybody else. : Because I couldnt get rid of the feeling that you are biassed against : boolean logic for private reasons. : The last one, combined with your failure to provide compelling reasons : for considering your point of view, and your galathaea: prankster, : fablist, magician, liar give me the impression of someone showing of : her stylish chic new theory just to impress others, like a teeny with : a newly discovered worldview. There are moral implications in what : your attitude reßects, and I dont have the impression that I like : them. You have to see that this at least tells me you have, at least somewhat, a problem with non-Boolean logics. My signature is a tease, yes, but it has its explanation in my choice of the name galathaea for my usenet and elsewhere electronic persona, and Ive written about this on codeguru. There is something about the moral implications of all of these non-Boolean logics which you dont like. And thats all I really meant by that hunch of mine. It unsettles you in some way. You were angry that I might propose this, yet you use it in attack of my programme. So now Im back believing it with a little higher ordering in my belief hierarchy. : As for your intended topic, you mentioned the halting problem as an : example where one needs polyvalent logic. When I studied computer : science, there never : was a mention of the need for polivalent logic in this context. Maybe : I missed something. As I see it, for a given algorithm, and a given : input, you know its halting behaviour or you dont. That is pretty : boolean to me. So I would like you to elaborate more on this example. : Then I would like you to establish : a relationship based on heyting structures and another area of : knowledge from among the ngs you posted to, and to explain why this : relationship is important : to both areas. Actually, attempting a Boolean algebra (and deriving a contradiction) is key to the proof of the undecidability of the halting problem. A good introduction to the topic and proof is found in the cute Wikipedia, found at http://en.wikipedia.org/wiki/Halting_problem. I explain in the other leafs I mention above what the lambda calculus has to do with Heyting algebras including the Curry-Howard isomorphism. : My advice to you, if you want to be more successful in your quest to : gather a multidisciplinary group of people willing to discuss your : topic: : Try to establish a path of such examples, relationships and : explanations that goes through all of the disciplines you addressed. : Expose this path as : succinctly as you can, without compromising understandability. : Leave out all the hocus pocus stuff. : Leave out as much of your personal opinion as you can. : And consider dropping (this is again very personal, Im afraid) the : prankster, fablist, magician, liar it is no great recommendation of : yourself. Maybe it works wonders among intelectuals of the liberal : arts, but I think its a real turnoff for the more scientific oriented : mind. See, I thought building a path from vision research through conceptual structures and cognitive work, moving on to external models of reality would work to bring them all together, just as you suggest. Yet I found that many described this as too long, and because they were unaware that the vision research was legitimate, they blew it off as hocus-pocus. Some stopped reading, others stopped believing. I tried to build all of the connections in a summary fashion and have still been chastised for length. Others are telling me to include more references. relationships among those that already are familiar with the logical research, whatever field they concentrate in. But I probably did need to add some other path for those who may not be aware of the research and see my comments as hand-waving. My signature will not change. All of it is true. -- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: interval for continuously compounding interest? > I had to look deeper, however, to say what is the > interval for continuously compounding interest?. This is easily > determined by setting the 2 well know equations equal to each other: > (P(1+(R/x))^(x*T))=(P*e^(R*T)) > However, I have not been able to solve this equation for X. There is no interval for continuously compounding interest. Thats why we call it continuous. If there were such an interval, then x would be just an ordinary number and you would use the left hand side above; e^(R*T) would not enter the picture. === Subject: Re: interval for continuously compounding interest? > are learning about different ways to compound interest, interval and > continuously. I had to look deeper, however, to say what is the > interval for continuously compounding interest?. This is easily > determined by setting the 2 well know equations equal to each other: > (P(1+(R/x))^(x*T))=(P*e^(R*T)) > However, I have not been able to solve this equation for X. Can anyone Consider interest compounded every year, quarter, month, week, day, hour, minute, second, etc. Continuous compounding is the limit of this sequence as the interval shrinks to zero. To make your equation correct, you must take the limit of the left-hand side as x goes to infinity. As an example, let r=0.06, say, and see how money you have after one year with different frequencies of compounding. Gordon Everstine Gaithersburg, MD === Subject: Re: A little help is needed > I am not quite sure how to solve this problem. Can someone give me a > little help, It would greatly be appreciated. The question is as follows: > Determine whether f: Q x Q -> Q given by > f(a/b,c/d)=(a+c)/(b+d) > is a well-defined function. > Again thank you very much for your help. Just a hint. Well-defined means that f(q_1, q_2) is independent of the particular realizations of q_1 and q_2 as quotients of integers. For instance if q_1 = 1/2, it could also equal 2/4, 3/6, 4/8, etc. If you plug in all of these different values into your formula, you must get the same thing if you are to call f a well-defined function achava === Subject: [arccos((sqrt(5)-1)/2 )] / pi is irrational Hello How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect this has something to with the golden ratio, and maybe with Fibonacci numbers or continued fractions? But I couldt start. Amanda === Subject: Re: JSH: Dont talk to me >I dont find your analogy very convincing. Posts from the cult of JSH >detractors are often sad documents that should prove embarrassing to their >authors when and if their heads are ever removed from their posteriors. And >I dont think many sci.math posters are embarrassed at being ommited from >JSHs list, do you? > It is true that there have been many nasty retorts to JSHs posts but oddly > enough he has never seemed to be disturbed by those (except perhaps for > reply as nastily). The three people that JSH lists here are people he > specifically dislikes, apparently, because they are respond honestly and > intelligently- attacking his statements rather than him. That appears to > be what JSH cant stand. The idea that these posters always respond intelligently - attacking his statements rather than him - cannot be taken seriously. === Subject: Re: A little help is needed Let S and T be sets. A function f:S->T is given by associating to each element s in S a (unique) element f(s) in T. Thus to check that a function is well-defined we must check both (i) For every s in S, f(s) is in T. (ii) If s_1 and s_2 are in S with s_1=s_2 then f(s_1)=f(s_2). We note that equality in Q is determined by cross-multiplying; That is, a/b = c/d iff ad = bc. So for the example f:QxQ->Q given by f(a/b,c/d) = (a+c)/(b+d) we must show (i) and (ii). (i) Assume (a/b,c/d) is in QxQ. That is a,b,c,d in Z and neither b nor d is zero. Does (a+c)/(b+d) qualify for membership in Q? Well a+c and b+d are clearly integers, since a,b,c, and d are. Moroever, since neither b nor d is zero, b+d does not equal zero either. So (a+c)/(b+d) is in Q. (ii) Assume (a/b,c/d) and (e/f,g/h) are in QxQ and (a/b,c/d) = (e/f,g/h) That is a/b = e/f and c/d = g/h, or af=be and ch=dg. Now, f(a/b,c/d) = (a+c)/(b+d) and f(e/f,g/h) = (e+g)/(f+h). We must determine whether (a+c)/(b+d) = (e+g)/(f+h) Or (a+c)(f+h) = (e+g)(b+d) Or af + ah + cf + ch = be + de + bg + dg. This does not seem to hold in general since we only have af=be and ch=dg. In fact, we can exhibit a counterexample: (1/2,1/3) = (2/4,3/9) But f(1/2,1/3) = 2/5 whereas f(2/4,3/9) = 5/13. Since 26 > 25 these are not equal. So f holds (i) but not (ii), and so f is not well-defined. This is why one might decide to define g:QxQ->Q by g(a/b,c/d) = (ad+bc)/(bd). (Since b and d are nonzero bd is not zero). Which is well-defined. Hope this helps, Sam > I am not quite sure how to solve this problem. Can someone give me a > little help, It would greatly be appreciated. The question is as follows: > Determine whether f: Q x Q -> Q given by > f(a/b,c/d)=(a+c)/(b+d) > is a well-defined function. > Again thank you very much for your help. === Subject: Re: there is no such thing as infinity >> It seems to go against common sense to say that these two sets have >> the same cardinalities. Then again, we dont have any common >> experience working with infinities. >That is even more evidence for the ridiculousness of there being an >infinity. Any math that completely contradicts the senses is wrong >just as any scientific theory that contradicts observation is wrong. >You just cannot have an infinite set. Agreed that contradictory mathematics is fallacious. But ultimately you come down to saying that you just cant have those damn infinite sets. Where is your proof? Answer: You cant prove it. Its false. Your Fortran program is an exercise in futility because you want it to eventually REFUTE its own definition! You want it to stop counting at some point, even though the existence of any number n implies the existence of the number n+1. THAT is a contradiction. That is why your program will never stop counting. Yes, we DO live in a FINITE universe. Otherwise we could have no identity. But if what youre trying futilely to prove is true, then there is no such thing as calculus :-) Xevious === Subject: Mathematical Models and Methods in Applied Sciences - Vol 14 No 1 Mathematical Models and Methods in Applied Sciences View table-of-contents and abstracts at http://www.worldscinet.com/m3as.html Contents: An Image Segmentation Variational Model With Free Discontinuities And Contour Curvature Giovanni Bellettini And Riccardo March Homogenization Of Boundary Value Problems And Spectral Problems For Neutron Transport In Locally Periodic Media Mustapha Mokhtar-Kharroubi The Main Inequality Of 3d Vector Analysis Giles Auchmuty 3d-Shell Elements And Their Underlying Mathematical Model D. Chapelle, A. Ferent and K. J. Bathe Flux Spikes In Viscous Flows Bai-He Tan And Zi-Niu Wu For more information, go to http://www.worldscinet.com/m3as.html === Subject: Re: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i192Qsb15702; > Sorry, I forgot that e^x and arctan(x) are algebraic ... >:?? In what sense are e^x & arctan x algebraic? In usual mathematical >:parlance, these are called transcendental functions. >I suspect irony; however, there is a classification of transcendental >functions which puts e^x and arctan(x) in one compartment. Each satisfies >a differential equation (here even of first order), where the differential >expression equated to zero is a non-trivial algebraic function (can be >made a polynomial) in the variables and the derivative: > (d/dx) y - y = 0, satisfied by y=e^x and others, > (1+x^2) * (d/dx) y = 0 , satisfied by y=arctan(x). >I have seen a name algebraic transcendental function for these. It >sounds like an oxymoron, in the company of regular singular points. One >can make a hierarchy of algebraic transcendental functions by assigning >two integers to such an f: > (1) the minimal order p of an algebraic differential equation satisfied > by f, > (2) within the class of algebraic differential equations of order p, the > minimal degree of the polynomial in the variables which is satisfied > by f. >(Finer subdivisions are possible) . > Examples of higher order algebraic trancendental functions are Bessel >functions. > Exercise: Many know the second order linear differential equation >satisfied by sin(x): y+y=0. Show that sin(x) is actually an algebraic >transcendental function of order 1. >I think it was Charles Hermite who proved that the Gamma function, >although its reciprocal extends to an entire analytic function, is not >algebraic transcendental. I saw the proof in an old-fashioned Advanced >Calculus book. > hey do you guys have a copy of the proof of e is transcendental? ..id really appreciate if somebody could help me. ive been searching and found nothing. === Subject: Re: grateful for comments: argument > If an odd perfect number exists and has three prime > factors a,b,c then we can partition the total list of > factors, both prime and composite, into those dividing > by a [call this Ôa.sum], remaining factors dividing > by b [b.sum], remaining factors dividing by c [c.sum], > and 1. > For perfect number status, what we call Ôcomplements > of each sum must divide by that prime: > for example 1 + a.sum + b.sum = c.comp must divide by c. > In general x must divide x.comp for each of x = a,b,c. > By examining each of > a.comp = 1 + b.sum + c.sum, > b.comp = 1 + a.sum + c.sum, > c.comp = 1 + a.sum + b.sum > we find it is never the case that all three divide by > they primes they should, and hence there are no odd > perfect numbers. Further there are no even perfect > numbers dividing by more than two primes. What if n = 3^2 x 13^2 x 61? The divisors of n not divisible by 3 are (1 + 13 + 169) x (1 + 61) = 183 x 62, which is divisible by 3. The divisors of n not divisible by 13 are (1 + 3 + 9) x (1 + 61) = 13 x 62, which is divisible by 13. The divisors of n not divisible by 61 are (1 + 3 + 9) x (1 + 13 + 169) = 13 x 183 = 13 x 3 x 61, which is divisible by 61. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: James >> >>To be sung to the ÔPirates Of Penzance tune ÔModern Major General >>(author >>unknown): > [snip most of it] >>I always have the last word; so, with utmost finality, >>Thats all from me, the model of a Newsgroup Personality. > > I think the last line is missing a two-syllable word near the end. Apart > from that the whole thing scanned beautifully. >> Eh? The second last line seems to me to trip along better if last is >> replaced by final, but the last line looks just fine; >> Sorry, I meant the second last line. Replacing last with final is a >> start, but you still need another syllable. Adding and after the >> semi-colon completes the fix. (The rest of the lyric scans just as well as >> Gilberts, who was an unparallelled master of scansion, no joke about it.) > Well, the second last line scans OK for me with stresses on > al, have, fin, word, with, most, nal. > Perhaps you are stressing ut and shortening the vowel of fi to a > schwa? you know what, I found it so impossible to read utmost with the stress on most, that I thought there was another syllable missing. Youre right, changing last to final makes it the right number of syllables. The mis-stress on utmost is more distracting than anything else in this parody though, and I think itd be better if we kept last, but changed utmost to uttermost, which isnt really a word, but its rhythm is perfect... === Subject: Re: Combinatoric:4 groups of 4 > This is a planning problem for scheduling players in teams of 4. > Not sure if it is feasible. > > 16 players, 5 days of play, 4 teams each day. > Each player teamed with each other player only once over the 5 days. > Look up references on finite geometries or combinatorial designs, and > affine planes in particular. Or look for references on bridge rotations (or whist rotations), as this sort of thing has been studied in depth by tournament organizers over the years. Maybe even post the question to a bridge group. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Limit Of Sum Over Some Rationals First, a simple looking limit, where I have revealed the (somewhat unexpected) closed-form at the posts bottom. --- t 1 r limit ----- / ---- , m -> oo m^2 --- e^r r which is, in linear-mode: limit{m-> oo} (1/m^2) sum{r} r^t /e^r And the r-sum is over all distinct positive rationals r which have denominators, in their simplest forms, which are each <= pi*m. -- Okay, the lesson for today: Start with the finite double-sum identity: sum{k=1 to n} sum{j=1 to m, GCD(k,j)=1} f(k,j) = sum{i=1 to min(m,n)} mu(i) (sum{1<=k<=m/i} sum{1<=j<=n/i} f(ki,ji)) - Defining f(k,j) as a(j/m) b(k/m) c(k/j), and taking limits, after dividing both sides by m^2, we get: For positive q and s, --- 1 limit ----- / a(den(r)/m) b(num(r)/m) c(r) m -> oo m^2 --- r= distinct rationals, 1<=den(r)<=q*m 1<=num(r)<=s*m 6 /q /s = ---- | | pi^2 | | a(x) b(y) c(y/x) dy dx /0 /0 (limit{m->oo} (1/m^2) sum{r= distinct rationals,1<=den(r)<=q*m,1<=num(r)<=s*m} a(den(r)/m) b(num(r)/m) c(r) = (6/pi^2)* integral{0 to q}integral{0 to s} a(x) b(y) c(y/x) dy dx) where a(), b(),c() are such that the integral exists, is defined, and the identity above is correct...) ;) (And num(r) is numerator of reduced r, den(r) is denomiator of reduced r.) -- So, we might have, if I am right so far, limit{m-> oo} (1/m^(t+u+2)) sum{r} (den(r))^u (num(r))^t /e^r = t! q^(2+t+u) (6/pi^2)/(2+t+u), where the r-sum is over all positive distinct rationals r, where den(r) <= m*q, q = fixed positive real. -- So, in answer to the original question: limit{m-> oo} (1/m^2) sum{r} r^t /e^r, where the r-sum is over all distinct positive rationals r which have denominators, in their simplest forms, which are each <= pi*m, is 3 * t!. (?) (I find this more interesting than otherwise because of the fact that 3 is the integer closest to both pi and e, which are each part of the original limit.) Leroy Quet === Subject: Order/Chaos Question I would like to know of a number distribution generator that can be adjusted to like a rheostat to produce numbers between order or chaos. The problem is similar to mutating a perfect sine wave into less and less order, eventually turning it into random noise. I would like a randomness index built into it that goes from 0 to 1, where zero is chaos and 1 is order. All fractions in between can also be input. I would like it to generate integers within a specified range from 0 to r, where r is the ceiling integer. I am only interested in the simplest case of numbers along a one dimensional line segment. The distribution of integers is along the line segment. There should be another integer n that designates how many points are plotted on the line segment, with the option of the points being either in combination (no point generated more than once) or overlapping (the same point can be generated more than once). ++++ Is it possible for noise from a known source, to be tuned into resonance with noise from a reference signal? === Subject: Re: Series > What are the next ten characters in the following series? > > 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 > Lookup ID Number A001511 in the On-Line Encyclopedia of Integer Sequences It should be noted that this sequence is such that a(2^n -k) = a(k), for all k where 1<= k <= 2^n -1. In other words, each sequence of the first 2^n -1 terms, for n = any positive integer, is palindromic. Leroy Quet === Subject: Re: JSH: Tautological spaces > In a tautological space like x+y+z = 0(mod x+y+z), everything has > x+y+z as a factor, but also you have 3 distinct elements x, y and z, > which give form to the space without regard to their values. > Ive spent a lot of time working the tautological space > x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2) > which youll notice has 4 elements. > James Harris If f(x) is a homeomorphism from T onto S and for every point p in T, f(U(p)) = U(f(p)), and the monad is invariant under standard topological transformations, with the caveat that the definition also comprizes a type of dynamic situation sematics, where concepts, such as proper set, ordinal and cardinal are relativised to context, taking care of paradox at all levels via symmetry, and the top, would naturally not exist, of course, since there is nothing outside the universe. So it becomes an infinite chain or composition of ever more inclusive situated sets expressing an interesting informational - topological dynamic. Belief = B Knowing = K K(B) > B(K) [T|F = T ] ---> K ---> B Symmetry forms the basis of logic, thus symmetry forms the basis of truth. To know is to believe but to believe, is not necessarily, to know. Outside of Total Existence there is nothing. This is an irrefutable fact. Or we could say that there is no outside of Total Existence. Therefore the largest possible set does not exist, where does not exist is equivalent to nothing. Nothing contains everything. Before the beginning there is nothing. After the end, there is nothing. Therefore Alpha = Omega. If space is *quantized* yet also continuous, then it too, has the indivisible units, then a measurement of space means that Fermats last theorem holds, for it. According to the Pythagorean theorem: x^2 + y^2 = z^2 All possible integer solutions are then rerpresented as: [a^2 - b^2]^2 + [2ab]^2 = [a^2 + b^2]^2 a^4 -2(ab)^2 + b^4 + 4(ab)^2 = a^4 + 2(ab)^2 + b^4 = [a^2 + b^2]^2 all odd numbers can be represented as: [a^2 - b^2] or Z^p - Y^p if Y is an even natural n and Z is odd, same for a and b . Fermats last theorem, for integers a,b,Z,Y,p: [a^2 - b^2]^p + Y^p = Z^p [a^2 - b^2]^p = Z^p - Y^p a^2 - b^2 = [Z^p - Y^p]^[1/p] When Z^p - Y^p is a prime number, it cannot have an integer root. a^2 - b^2 is not an integer, for [Z^p - Y^p]^[1/p] , for a,b,Z,Y,p, unless p http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: JSH: Tautological spaces > In a tautological space like x+y+z = 0(mod x+y+z), everything has > x+y+z as a factor, but also you have 3 distinct elements x, y and z, > which give form to the space without regard to their values. > Ive spent a lot of time working the tautological space > x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2) > which youll notice has 4 elements. > James Harris If f(x) is a homeomorphism from T onto S and for every point p in T, f(U(p)) = U(f(p)), and the monad is invariant under standard topological transformations, with the caveat that the definition also comprizes a type of dynamic situation sematics, where concepts, such as proper set, ordinal and cardinal are relativised to context, taking care of paradox at all levels via symmetry, and the top, would naturally not exist, of course, since there is nothing outside the universe. So it becomes an infinite chain or composition of ever more inclusive situated sets expressing an interesting informational - topological dynamic. Belief = B Knowing = K K(B) > B(K) [T|F = T ] ---> K ---> B Symmetry forms the basis of logic, thus symmetry forms the basis of truth. To know is to believe but to believe, is not necessarily, to know. Outside of Total Existence there is nothing. This is an irrefutable fact. Or we could say that there is no outside of Total Existence. Therefore the largest possible set does not exist, where does not exist is equivalent to nothing. Nothing contains everything. Before the beginning there is nothing. After the end, there is nothing. Therefore Alpha = Omega. If space is *quantized* yet also continuous, then it too, has the indivisible units, then a measurement of space means that Fermats last theorem holds, for it. According to the Pythagorean theorem: x^2 + y^2 = z^2 All possible integer solutions are then rerpresented as: [a^2 - b^2]^2 + [2ab]^2 = [a^2 + b^2]^2 a^4 -2(ab)^2 + b^4 + 4(ab)^2 = a^4 + 2(ab)^2 + b^4 = [a^2 + b^2]^2 all odd numbers can be represented as: [a^2 - b^2] or Z^p - Y^p if Y is an even natural n and Z is odd, same for a and b . Fermats last theorem, for integers a,b,Z,Y,p: [a^2 - b^2]^p + Y^p = Z^p [a^2 - b^2]^p = Z^p - Y^p a^2 - b^2 = [Z^p - Y^p]^[1/p] When Z^p - Y^p is a prime number, it cannot have an integer root. a^2 - b^2 is not an integer, for [Z^p - Y^p]^[1/p] , for a,b,Z,Y,p, unless p http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Number Theory Problem > while c=N-1 is not so representable. > I can only prove this trivial case. > Suppose c=N-1 is representable in the form that c=ax+by > with x,y>=0 while N-1=(a-1)(b-1)-1=ab-a-b. > Then ab-a-b=ax+by and hence the equality > ab=(x+1)a+(y+1)b ...(1) > holds. > Since gcd(a,b)=1 and therefore x+1 is divisible by b > while y+1 is divisible by a, which implies that > (x+1)a+(y+1)b>=2ab, in contradiction to (1). > You can find integers m, n >= 0 so that am - bn = 1 (since gcd(a,b) = 1). Then, for any integer c, the possible representatiuons of c are all of the form c = a(mc - kb) - b(nc - ka) for some integer k. So the problem reduces to finding some choice for k that forces the inequalities mc - kb >= 0, nc - ka <= 0 to be satisfied. Thats equivalent to being able to choose k satisfying nc/a <= k <= mc/b . Note that (mc/b) - (nc/a) = ((am - bn)c)/(ab) = c/(ab) -- thus, the interval [nc/a, mc/b] contains at least one integer as soon as c >= ab (since, then, that interval has length >= 1). This proves the result for c >= ab. How about the remaining values for c ?? If N <= c < ab, consider c + a + b which is > ab (by the way in which N is defined). If c + a + b is not a multiple of either a or b, then c + a + b = ra + sb with r, s > 0 and its immediate that c = (r-1)a + (s-1)b which is a representation of the correct sort. If c + a + b = sa, then c = (s-1)a - b; since s > b, you can write s-1 = qb+r with q >= 1, 0 <= r < b and then verify that c = ra + (q-1)b, taking care of this case. The argument when c + a + b turns out to be a multiple of b is (pretty much) the same. [][/quote:3bb60765ba] My second post has *previouly* exhibited a proof > For any integer c>=N, it can be written as > c=ab-a-b+m while m is an > integer greater than 1. > Since gcd(a,b)=1, there exists some integers s,t such that as-bt=1. > Multipy m on both sides and we have > ams-bmt=m ...(2) > According to euclidean division, there exists some integers q,r such that > mt=aq+r, 0<=r Let x*=ms-bq,y*=r=mt-aq and we have > ax*-by*=m ...(4) > by (2). > Finally Let x=x*-1,y=a-1-y*. > Since y*=r<=a-1 by (3), y=a-1-y*>=0. > Thus ax*=by*+m>0, which implies x*>0 and it follows that > x=x*-1>=0. > Thus ax+by=a(x*-1)+b(a-1-y*) > =ax*-by*+ab-a-b > =ab-a-b+m=c ( by (4) ), > indicating that x,y are the coefficients required. for the whole problem. > Let N=(a-1)(b-1), where a,b are positive integers > and gcd(a,b)=1 > Show that every integer c>=N is representable in the form > c=ax+by with x,y>=0, while c=N-1 is not so representable. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: greek numerals by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1931MG18556; Really sounds very interesting - what is acrophonics? The Greek alphabet itself is usually regarded as an adaptation of the earlier Phoenician alphabet, but you probably knew that already. My website http://www.otherlanguages.org has some hobby stuff about languages and language resources if youre curious. Please fill me in on acrophonics. I cant find it in the dictionary. >Mark im sorry for being vague-im exploring the early Greek system of >using acrophonics and then the alphanumerical usage-in general im >looking at exploring how the greeks acquired their systems-how they >were modified and how they were superseded >neil >> . >> Perhaps you have a specific thing in mind by the ÔGreek system, >> not just ancient Greek mathematics in general, Neil? >> Sounds intriguing. Can you say more? >> Mark > Hi i require information wrt the above-im writting an essay on how the > greek system came into being-and what impact it had on mathematics-can > anyone direct me to good sites etc >> === Subject: Re: Plotting a 7 vertice graph in which every vertex has degree 4 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1931NF18564; . Or to be really crass, 1. draw a regular seven-sided polygon in pencil, then - 2. join two edges from one corner to the two corners furthest away with pencil where they are not already marked in pen - 3. go over the 2 or 1 new pencil lines from [1.] in pen if they look right - 4. repeat [2.] with the corner to the immediate right [unless you are left-handed, in which case to the immediate left] unless it is already joined to two corners by pen lines - 5. go over the outside seven edges in pen. Best wishes, Mark . >> David, >> I came up with three sub-graphs, which I have uploaded the images of with >> Mathematica. >> The first of the four images is just the complete graph for K_7, and doesnt >> relate. >> Do I have the right idea? >> http://ho me.earthlink.net/~diana53/mathematica/1-1-17.htmlnot 4-regular. The rest of the graphs look like different drawings of the >same graph (the complement of the 7 cycle), as Professor Eppstein >suggested. His other suggestion is the complement of a 4 cycle and 3 >cycle; i.e. you will have a graph whose vertices can be partitioned into >two sets, one set with 3 independent vertices and another set with 4 >vertices and two disjoin edges, and then form all edges between these two >sets. === Subject: Re: there is no such thing as infinity > Ive thought really hard about this one and came to the conclusion > that there is no scientific evidence of infinity existing. The highest > number that anyone has ever measured to according to Isaac Asimov in > his book Science and Human Thought is only about 5.0 x 10^48. No one > has ever gotten past that number. Doesnt this sound weird? > > Whats to say that eventually there is a number where it is impossible > to count higher than? If someone were to find this number and prove > that it is in fact the highest number, then that person would > undoubtably be rich and famous. > > I am currently running a computer program that will eventually find > this magic number (I hope and pray) that I call M for short. It > counts and counts and counts and my theory is that it will eventually > stop at M. I am looking for collaborators in this experiment so that I > can use their computer time. The program in FORTRAN is simple: > > 00001 n=1 > 00002 1 n=n+1 > 00003 print(3,4)n > 00004 if(n.eq.M) then print(3,4)M > 00005 else go to 1 > 00006 end if > 00007 end > > It has currently reached about 2.0 x 10^18. Just as Einstein proved > that there is no aether, I am convinced that I will prove that there > is no infinity and then write a book or two. > > Dr. Ben Zona > my week. (I did have to check my Yiddish lexicon, though.) Hebrew, I would have thought. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Concentric-Circles Game (to help teach basic number theory) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1931NP18568; . Sounds fun. Have you tried it out on some children? How did it go? Mark >I made this simple game up today, in part so as to help teach kids >basic number-theory concepts (such a congruences and divisibility), >but it might be far more useful as useless entertainment...if even >that... >Start with m concentric circles drawn on paper, >where n is from 5 to 12 (or higher perhaps). >The rings between the circles are subdivided so that the inner circle >is subdivided into 2 equal-area sections, the next ring out is >subdivided into 3 equal sections, the next ring into 4, etc..until the >outer ring is subdivided into (m+1) equal sections. >(And the sections are aligned so that one boundry of each is along a >single line.) >Players start by each placing a game-piece (small enough to fit in the > smallest section) into any unoccupied section on the board. >A spinner or dice are used to pick a random integer n, where each n >should be between 1 and somewhere near m. >On each move, a player can place a new game-piece (leaving earlier >pieces) n rings inward (if possible) from their last game-pieces >position, or n rings outwards (if possible). >(Outward and inward are along a chain of sections which touch. Do >not pass center circle or outer circle.) >OR the player can place a game-piece clockwise or counterclockwise (in >the same ring) >a *multiple* of n spaces from their last-placed piece. >But in any case, a player cant change directions during his/her move. >(And the players {counter}clockwise moves may indeed circle around a >single ring as many times as needed.) >A player can jump over occupied spaces when measuring the number of >spaces from their last piece, but they must ONLY place a piece on an >unoccupied section. >And the winner is the last player able to place a piece on the board. >Leroy Quet === Subject: Re: Modular Arithmetic > 100 > How to reduce 2 mod 5 4 25 HINT (1 == 2 ) (mod 5) -Bill Dubuque === Subject: Re: 1 + 1/3 + 1/5 + ... + 1/(2*N - 1) > Im interested in the value of the sum > 1 + 1/3 + 1/5 + ... + 1/(2*N - 1) > I can bound this sum above and below using integrals, but I wonder > if there are better ways to approximate its value for large N. > jill Related information: http://www.research.att.com/projects/OEIS?Anum=A025550 http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A004041 As well as the other replies representation involving logarithm, I could add that your sum is also: sum{k=1 to 2n-1} H(k) (-1)^(k+1), where H(k) = 1 +1/2 +...+ 1/k, as is noted in some other replies in this thread. And your sum has the infinite-sum representation: sum{k=1 to oo} (1/(k2) -1/(k+2n) +1/(2k-2n)) = n *sum{k=1 to oo} (n +k*3/2)/(k(k+n*2)(k+n)) (But I would not trust my last sum, since I am making lots of errors lately.) And with the infinite sum, wecan attribute a value to 1 + 1/3 + 1/5 +...+1/(2n-1) for non-integer n. Leroy Quet === Subject: Re: NaturalNumbers are the P-adics and why Kissing density jumps in KPP Re: applying RiemannHypothesis modification to Kepler Packing Problem > The below is an old post talking about the Kepler Packing Problem as > === > Subject: Re: PROOF OF THE KEPLER PACKING > PROBLEM > <1993Aug19.021638.256@rp.CSIRO.AU Radiophysics/Australia > Telescope > National Facility > example in 9 > dimensions, where the densest known > sphere packing is > that produced by > the Lamda_9 lattice (packing density > 0.14577, kissing > number 272), > whilst the greatest known kissing number > is that > achieved by the > non-lattice packing P_{9a} (packing > density 0.12885, > max kissing number > 306). For more details, an excellent book > Okay, well, if the NaturalNumbers are the P-adics, and if RH implies > that there are no straightlines at infinity because the P-adics > compose the 1/2 Realline. > Then what an application of RH would do to the Kepler Packing Problem > is to first ask the question of does a p-adic dimensional space make > much sense. Is there a ...99999 dimensional space in 10-adics? Is > there a ....11111 dimensional space in 2-adics? > Then further, a RH application of p-adics to the KPP of kissing points > versus densest-nonkissing plan of attack to prove would then ask the > Very Important Question: > Question: does the above quoting suggest that the divergence of 9th > dimension becomes even more divergent when in the 10th dimension. Then > the 11th dimension, how much of a divergence if any from the previous > dimensions. > You see, if NaturalNumbers are really the P-adics, then in KPP there > should be a linear increase in divergence as we go higher in > dimensions with the density of packing. > For example: the writer above noted that in 9th dimension the kissing > diverges from regular KPP, then the kissing should also diverge in > 10th dimension, and also in 11th dimension and so forth. But, if the > KPP does not diverge in 10th dimension from that of kissing in 10th > dimension Suggests or Implies that the P-adics are involved. > If Straightlines exist out to infinity and if NaturalNumbers are the > FiniteIntegers then the KPP should not be a pockmarked gapping of > kissing points divergence as we increase in dimensions. > On the other hand, if NaturalNumbers are the P-adics and that all > straightlines curve as they approach infinity (i.e. straightlines do > not exist), then the divergence of the KPP from that of kissing points > versus densest pack would not be a smooth linear relationship as we > increase in dimensions, and instead have gaps where in say dimension > 22 the kissing points is the densest pack and where dimension 23 the > kissing points are not the densest. > Demonstration: If we take oranges to pack and we had a square box > (Euclidean Space) and a similar volumed sphere and asked to pack those > oranges in which container could we get the densest packing? The cube > or the sphere? So that in the KPP, applying the RH would suggest that > the divergence of kissing is because of the fundamental reason that > NaturalNumbers are really the P-adics. > Because if space is Euclidean and that straightlines remain straight > out to infinity and that NaturalNumbers are FiniteIntegers then as you > increase in dimensions from say 9 to 10 to 11 to 12 etc etc, that the > divergence from packing should also be a Smooth and linear > progression. But it is not. It is gap ridden and swinging back and > forth between kissing as the densest and kissing not the densest. I know I shouldnt feed the trolls, but... The p-adics are a different completion of the rationals than the reals. The p-adics are not in the real numbers. They are a different class of objects. Ôcid Ôooh === Subject: Re: need help!!! > Mark im sorry for being vague-im exploring the early Greek system of > using acrophonics and then the alphanumerical usage-in general im > looking at exploring how the greeks acquired their systems-how they > were modified and how they were superseded > neil > > > . > Perhaps you have a specific thing in mind by the ÔGreek system, > not just ancient Greek mathematics in general, Neil? > > Sounds intriguing. Can you say more? > > Mark > > Hi i require information wrt the above-im writting an essay on how the >> greek system came into being-and what impact it had on mathematics-can >> anyone direct me to good sites etc hi every body can plot this function please say me with > www,hupo19@yahoo.com it is y=arcsin^-1(3/cosx),thank you dear Mark and Neil can u plot this function y=arcsin^-1(3/cosx) === Subject: Research Scientist in Knowledge Representation & Reasoning Researcher Level B: Knowledge Representation and Reasoning National Information and Communication Technologies Australia ------------------------------------------------------------- National ICT Australia Limited (NICTA) invites applications from highly motivated researchers for appointment as Researchers (Level B). The positions are to work within the Knowledge Representation and Reasoning Program (KRR) in traditional areas of knowledge representation for a fixed term period of two to three years. Research conducted in the KRR program includes, but is not limited to, the following: - Nonmonotonic reasoning - Belief revision and merging - Negotiation and games - Ontologies - Cognitive robotics and planning - Multi-agent systems - Logic programming Preference will be given to applicants with a strong background in logic. For more information about NICTA, go to http://www.nicta.com.au. The KRR program is described in http://nicta.com.au/programs/krr/kr.html. Essential criteria: A PhD in an appropriate area with a strong publication record. Evidence of impact in the relevant areas. Willingness to supervise graduate students, and to participate in graduate teaching. Desirable criteria: Strong communication skills. Interest in graduate teaching. Record of liaison with industry. Ability to work in a team. Experience in writing research proposals. Level B salary packages are in the range of $70,000 - $100,000 (inclusive of employer contribution to superannuation). Funds are available for research support and infrastructure, and travel support. Applications should be sent to jobs@nicta.com.au and should include a full CV, list of publications, names and contact details of three referees, and a short (at most 1 page) statement of how the applicants research interests, achievements and future goals align with NICTAs KRR program. Applicants should look at the general conditions (which apply to Level B researchers in all NICTA programs) in http://www.nicta.com.au to ensure that they are in agreement with, and can fulfill them. NICTA reserves the right to appoint by invitation or not to make any appointment. For further information, contact Professor Norman Foo (norman@cse.unsw.edu.au) or Dr Maurice Pagnucco (morri@cse.unsw.edu.au). === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational >How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect >this has something to with the golden ratio, and maybe with Fibonacci >numbers or continued fractions? But I couldt start. >Amanda expression? Does the expression contain a misprint? G C === Subject: Re: JSH: Pattern argument === >Subject: Re: JSH: Pattern argument >I went on a trip to the nearby beach to find rocks that have approximately >your IQ. It wasnt easy but I think I have found a sufficiently stupid one. > But does it have a BS in Physics and is it a veteran? > -- > Mensanator > Ace of Clubs Better yet, did James serve his country during the Korean War? I heard he helped the Marines escape from the Chosin Reservoir and served from 1950-1953. He also belongs to the Allied Seniors Recreational Vehicle Association - so please, afford him some respect. Please dont toss him a bone either; a man doesnt do that to a Korean War veteran. -ET === Subject: Re: 1/0 now allowed > ItÇs defined as 1.6367348238383838 > DonÇt ask why, but must be used or your calculations will be wrong. The phrase way too much free time just popped into my head. -- http://hertzlinger.blogspot.com === Subject: Re: A little help is needed In sci.math, Jeff little help, It would greatly be appreciated. The question is as follows: > Determine whether f: Q x Q -> Q given by > f(a/b,c/d)=(a+c)/(b+d) > is a well-defined function. > Again thank you very much for your help. There is an issue regarding how one defines negative fractions, but otherwise it looks fine. The main issue is incompleteness of the definition; to be full about it, one should probably do something like: f: Q x Q -> Q f(r1, r2) = (a+c)/(b+d) where r1 = a/b, r2 = c/d, a,b,c,d integer, b > 0, d > 0, gcd(a,b)=1, gcd(c,d)=1 [*] Obviously this would be very hard to characterize in terms of the standard arithmetic operations on Q, but a functions a function, and since every rational r can be uniquely written as the quotient of two integers a/b, b > 0, gcd(a,b) = 1, this definition works. Just try not to integrate it over the unit square... :-) [*] probably should also state that 0 = 0/1 is the unique representation of 0 as well. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Modular Arithmetic In sci.math, Shawn Windle the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what > lot. Well, the simplest method I can think of is to note that: 2 = 2 (mod 5) 2^2 = 4 (mod 5) 2^3 = 8 = 3 (mod 5) 2^4 = 16 = 1 (mod 5) Ah ha...if 2^4 = 1 (mod 5) then so is 2^8, 2^12, etc. which means 2^100 = 1 (mod 5) since 100 = 25 x 4. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Proofs - Please help quickly! In sci.math, Istari 1) Find an exact expression for: sin(pi/12) 180/pi deg/rad * pi/12 rad = 180/12 deg = 15deg cos^2 15deg - sin^2 15deg = cos 30deg = sqrt(3)/2 1 - 2*sin^2 15deg = sqrt(3)/2 2*sin^2 15deg = 1 - sqrt(3)/2 sin^2 15deg = 1 - sqrt(3)/2 sin 15deg = sqrt(1 - sqrt(3)/2) > 2) Solve to 4 decimal places (0_ sin2x + cosx = 0 sin2x = 2 sinx cosx 0 = 2 sinx cosx + cosx = cosx * (1 + 2 sinx) 0 = 1 + 2 sinx or 0 = cosx cosx = 0 at pi/2, 3pi/2 1 + 2 sinx = 0 => sinx = -1/2 sinx = -1/2 at 7 pi/6, 11 pi/6 > 3) Solve sin^2 x - 2sinx - 1 = 0 and find the general solution. Standard quadratic formula. sin x = (2 +/- sqrt(4 + 4))/2 = 1 + sqrt(2) or 1 - sqrt(2) 1 - sqrt(2) yields the approximate solution x = -0.42707858... 1 + sqrt(2) yields no real solution, but the complex solution x = Pi/2 + 1.528570919480 * i is available. (This according to GP/Pari.) It is not clear how the problem generalizes beyond the obvious addition of 2 * N * Pi for any integer N, or the specification of an arbitrary A, B, and C in the equation A * sin^2 x + B * sin x + C = 0, yielding x = arcsin( (-B +/- sqrt(B^2 - 4*A*C))/(2*A) ) + 2*N*Pi as one might expect. > 4) Prove the following (this is a real toughy): tanx/secx = secx tanx = sinx/cosx secx = 1/cosx tanx/secx = sinx/cosx * cosx = sinx The problem appears misspecified. If you wish to solve this equation, as opposed to proving it for general x, one can simply note that if cosx = secx, then cos^2x = cosx * secx = 1, cos x = 1 or -1, and therefore x=0 or pi. > Please help! Soon! Well, its late Sunday here; by the time you get this the homework is probably past due... :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Plotting a 7 vertice graph in which every vertex has degree 4 Guys, to your detailed help, have figured it out. Diana > David, > I came up with three sub-graphs, which I have uploaded the images of with > Mathematica. > The first of the four images is just the complete graph for K_7, and doesnt > relate. > Do I have the right idea? > http://home.earthlink.net/~diana53/mathematica/1-1-17.html > Im not sure why you have the complete graph on 7 vertices, since its > not 4-regular. The rest of the graphs look like different drawings of the > same graph (the complement of the 7 cycle), as Professor Eppstein > suggested. His other suggestion is the complement of a 4 cycle and 3 > cycle; i.e. you will have a graph whose vertices can be partitioned into > two sets, one set with 3 independent vertices and another set with 4 > vertices and two disjoin edges, and then form all edges between these two > sets. === Subject: Re: Plotting a 7 vertice graph in which every vertex has degree 4 Guys, to your detailed help, have figured it out. Diana > Or to be really crass, > 1. draw a regular seven-sided polygon in pencil, then - > 2. join two edges from one corner to the two corners furthest > away with pencil where they are not already marked in pen - > 3. go over the 2 or 1 new pencil lines from [1.] in pen if > they look right - > 4. repeat [2.] with the corner to the immediate right [unless > you are left-handed, in which case to the immediate left] unless it > is already joined to two corners by pen lines - > 5. go over the outside seven edges in pen. > Best wishes, Mark >> David, >> I came up with three sub-graphs, which I have uploaded the images of with >> Mathematica. >> The first of the four images is just the complete graph for K_7, and doesnt >> relate. >> Do I have the right idea? >> http://h om e.earthlink.net/~diana53/mathematica/1-1-17.htmlnot 4-regular. The rest of the graphs look like different drawings of the >same graph (the complement of the 7 cycle), as Professor Eppstein >suggested. His other suggestion is the complement of a 4 cycle and 3 >cycle; i.e. you will have a graph whose vertices can be partitioned into >two sets, one set with 3 independent vertices and another set with 4 >vertices and two disjoin edges, and then form all edges between these two >sets. >J === Subject: Re: need help!!! > hi every body can plot this function please say me with > www,hupo19@yahoo.com it is y=arcsin^-1(3/cosx),thank you > dear Mark and Neil can u plot this function y=arcsin^-1(3/cosx) The arcsine function is only real for arguments no larger that 1 in absolute value, but 3/cos(x) is never smaller than 3 in absolute value. So unless arcsin^-1 means something like inverse of arcsine, there are no real points on the graph. === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational >How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect >this has something to with the golden ratio, and maybe with Fibonacci >numbers or continued fractions? But I couldt start. >Amanda > expression? Does the expression contain a misprint? I was thinking the same thing, but then I looked at the subject title. === Subject: wronskian For some problems we need to evaluate the wronskian to show whether a couple of solutions are a fundamental set. Currently Im doing this by actually evaluating the determinant |f g | |f Ô g Ô| and it does fine. However, a theorem I have says that W(t) = ce^P(t), where P(t) = Integral( -p(t)*dt ), and p(t) comes from the homogeneous second order linear diferential equation: y +p(t)y +q(t)y = 0 This formula is much faster than doing the determinant, and so I would like to use it. But, I do not know how to get the constant c. I assume that c is the constant that pops out of the integral, but without some initial conditions I dont know how to get it. === Subject: Alternate representations of convex functions My question concerns convex, monotonically decreasing functions g(x) that can be represented as: g(x) = int f(x,y) dy. Specifically, is there a principled way of finding an alternate parameterization y2 such that, g(x) = int f_2(x,y2) dy2 = max_y2 f(x,y2). I have found specific cases where this can be done but I dont know any general principles for finding the requisite transformation or feasibility conditions. Any info would be greatly appreciated. D. Macricknu === Subject: Re: there is no such thing as infinity > Cantor was a rusha. He confused generations of mathematicians by > telling them there are lots of infinities. It is impossible to get a > BS in math without admitting that his theories are correct. > The diagonal argument was a sham because it presupposes the existence > of infinity without proving its existence. His theorem is proof that > infinity does not in fact exist, because if it did, then there would > be only one type, not aleph null, aleph one etc., because infinity by > definition is as far as one can go. He even defiled the Hebrew > alphabet by putting subscripts next to aleph when the gematria of > aleph is not infinity or even M but one. Okay, lets assume there is no infinity, which is your whole point, no? And now you have this magic number M. People have estimated that there are about 6x10^79 atoms in the entire universe (http://www.sunspot.noao.edu/sunspot/pr/answerbook/ universe.html#q70). I suppose this number 6x10^79 is your magic M. If you tried to write down all the numbers 1 through 6x10^79, you would run out of space to write them. (I wont even go into counting them, since even counting one number every picosecond means this would take ~2x10^61 years) Now, you might be wondering why I am requiring you to write all these umbers down (and why I think your computer program might be cheating) I can write 1 6x10^90, but that does not mean this number exists. It might just be too big of a number, much larger than your M. So, unless you have proof that all numbers up to your number M exists, proving M exists is all for naught. Okay, maybe that number is too big. Afterall, people guess the universe is only 13.7 billion years old (http://wiener.math.csi.cuny.edu/UsingR/Data/ age.universe.html), or roughly 4.3x10^42 yocto-seconds (10^-24). Maybe this should be M, since the age all of existence is less than this number. Personally, I prefer to think of 42 as infinity. Until you convince me otherwise, I will just continue believing this way. 43? Doesnt even exist. - Tim Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: JSH: Pattern argument <40270A45.47104F0B@osu.edu> Discussion, linux) > Better yet, did James serve his country during the Korean War? I > heard he helped the Marines escape from the Chosin Reservoir and > served from 1950-1953. He also belongs to the Allied Seniors > Recreational Vehicle Association - so please, afford him some > respect. Please dont toss him a bone either; a man doesnt do that > to a Korean War veteran. James says that he is around 35 years old. -- Jesse F. Hughes Leaving things always seems to fix me, Running seems to ease my worried mind. -- Bad Livers, Honey, Ive Found a Brand New Way === Subject: Re: wronskian For some problems we need to evaluate the wronskian to show whether a couple > of solutions are a fundamental set. > Currently Im doing this by actually evaluating the determinant > |f g | > |f Ô g Ô| > and it does fine. However, a theorem I have says that W(t) = ce^P(t), where > P(t) = Integral( -p(t)*dt ), and p(t) comes from the homogeneous second > order linear diferential equation: > y +p(t)y +q(t)y = 0 > This formula is much faster than doing the determinant, and so I would like > to use it. But, I do not know how to get the constant c. I assume that c > is the constant that pops out of the integral, but without some initial > conditions I dont know how to get it. c=W(t_0) where t_0 is the initial point. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: JSH: Dont talk to me Discussion, linux) >> So its ok for me to continue calling you a disloyal American who >> is a disgrace to the armed services of which you claim to be a >> veteran? > Hah, thats our President. James Harris is our president? President Garfield was the author of an original proof of the Pythagorean theorem[1]. I suppose that if James S Harris is the pseudonym of George W. Bush, the *real* prover of Fermats last theorem and other mathematical marvels, then this is a continuation of a great tradition (if Garfield by himself counts as a tradition, that is). Footnotes: [1] At least, thats what I understood from my copy of Proofs Without Words, which presents the proof. -- Jesse F. Hughes If anything is true in general about Usenet, its that people can go on and on about just about anything. -- James Harris speaks the truth. === Subject: Re: NaturalNumbers are the P-adics and why Kissing density jumps in KPP Re: applying RiemannHypothesis modification to Kepler Packing Problem > Oooh I see. While I couldnt understand the specifics of your post, I understand it belongs in the same research field with tautological spaces and the maximum natural number theory. Let me try again because my first post was garbled in some paragraphs. In the early 1990s I offered a proof of the Kepler Packing Problem based on Kissing points. That the densest packing would have the largest number of kissing points. And so I thought KPP was done and vanquished but when I posted that proof to the Internet circa 1993, objections were raised in that kissing-points diverged from fcc as the densest packing in higher dimensions such as 9th dimension. If geometry consisted only of 3rd dimension, I had proved KPP because kissing-points was equal to fcc in 3rd dimension. So, the only objection to my proof is that in higher dimension it seems as though kissing-points is no longer fcc. But I have a rejoinder to the higher dimensional objection. My rejoinder is that the NaturalNumbers are the P-adics. And the P-adics do not form a straightline Euclidean space. So the 9th dimensional objection is no longer a straightline space but is a curved space. And so where the kissing points in 9th dimensions yields a 0.12885 density yet the fcc yields a 0.14577 creates this divergence in 9th dimension. But if you allow that the NaturalNumbers are really P-adics and that this 9th dimension is a curved space and not a straightline Euclidean space then the kissingpoint density equals the fcc density in 9th dimension. I am claiming that if you apply P-adics to the dimensions such as 9th dimension then the kissingpoints become equal to the fcc packing just as they were equal in 3rd dimension. I claim the proof of KPP is based all upon Kissing Points and what needs altering is the fact that straightlines become curved the further you go out because the NaturalNumbers are the P-adics and so by the time one enters 9th dimensions it is not a 9th dimensional Euclidean cube to pack with balls but rather a sphere like object since the space is curved. In other words, the disparity between kissing points and fcc packing with higher dimensions is a means of gauging or measuring how much space has curved starting at 2 dimensions then 3rd dimension on up to 9th dimension and on higher. Kissing Points proves Kepler Packing and the reason there exists a divergence in higher dimensions is because they are curved and no longer Euclidean. KPP is a fine example of where we can measure the rate of curvature of Euclidean straight lines. We thus must see that starting at 9, a supposedly straightline has already begun to curve and is no longer straight. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Max to Min objective conversion I have an LP as follows... max v st v-0.5x-0.5y<=0 x=1 y=1 bounds 0<=v<=1 integers v end v,x and y can take on the value 0 or 1 only. v gets the value 1 if and only if both x and y are 1 respectively, in all other cases, v=0. How can we convert this problem into a minimiztion problem (min v) so that v again gets the value 1 if and only if x and y are 1 respectively ? For all other values of x and y, v must get the value 0. Can you please let me know the formulation to get the above solution ? Arun === Subject: Re: lattices, rings of algebraic integers |>Is it always the case that Z[alpha] is a lattice in C if and only if |>Z[beta] is a lattice? |>David Bernier | | No. Try the three cube roots of 2. This counterexample doesnt work because Z[u] where u is one of the complex roots isnt a lattice either. It contains the element 1+u+u^2 which lies inside the unit circle, as well as all of its powers which tend to 0. Keith Ramsay === Subject: Re: wronskian |For some problems we need to evaluate the wronskian to show whether a couple |of solutions are a fundamental set. | |Currently Im doing this by actually evaluating the determinant | ||f g | ||f Ô g Ô| | |and it does fine. However, a theorem I have says that W(t) = ce^P(t), where |P(t) = Integral( -p(t)*dt ), and p(t) comes from the homogeneous second |order linear diferential equation: | |y +p(t)y +q(t)y = 0 | |This formula is much faster than doing the determinant, and so I would like |to use it. But, I do not know how to get the constant c. I assume that c |is the constant that pops out of the integral, but without some initial |conditions I dont know how to get it. In general, determining the constant of integration amounts to computing W for one value of t, and using that as an initial value. Keith Ramsay === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational >How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect >this has something to with the golden ratio, and maybe with Fibonacci >numbers or continued fractions? But I couldt start. There are several possible methods. What do you know of either Chebyshev polynomials or cyclotomic polynomials? Or what can you say about the polynomial 1+2 t-2 t^2+2 t^3+t^4? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: classes of transcendental numbers ? >hey do you guys have a copy of the proof of e is transcendental? ..id >really appreciate if somebody could help me. ive been searching and >found nothing. You might look in Herstein, Topics in Algebra, section 5.2. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Infinite Dimensional Infinitesimal Any smooth connected 1 dimensional manifold is diffeomorphic either to the circle, or to some interval of real numbers. Take a line segment of length 1. It is one dimensional. A-----------B Find the midpoint of the line segment and rotate it into 2 dimensions A | | |------B Each leg is 1/2 Rotate into 3 dimensions, and each leg is 1/3 A | | |------C | | B rotate into N dimensions and each leg is 1/N Continue this process as a limit N---->oo By the above process, an infinite dimensional universe is a point. === Subject: Re: wronskian >For some problems we need to evaluate the wronskian to show whether a couple >of solutions are a fundamental set. >Currently Im doing this by actually evaluating the determinant >|f g | >|f Ô g Ô| I find it bafßing that the standard DE textbooks waste time doing this for solutions of second order DEs. f and g are linearly independent unless one is a constant multiple of the other. In any case where it isnt blatantly obvious whether that is the case, I suspect it also wont be obvious whether the Wronskian simplifies to 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: classes of transcendental numbers ? >>hey do you guys have a copy of the proof of e is transcendental? ..id >>really appreciate if somebody could help me. ive been searching and >>found nothing. >You might look in Herstein, Topics in Algebra, section 5.2. Or Stewart and Tall, Galois Theory have easily readable proofs of transcendality of both e and pi. (Thats the best feature of that book!) Derek Holt. === Subject: Re: classes of transcendental numbers ? >>hey do you guys have a copy of the proof of e is transcendental? ..id >>really appreciate if somebody could help me. ive been searching and >>found nothing. > You might look in Herstein, Topics in Algebra, section 5.2. Or type transcendence of e into Google. :-) -- --Tim Smith > === Subject: Re: Rationals are Uncountable > If q is rational and the original sequencing, x_n, contains every > rational in (-1,1), then any translation of (-1,1) by a rational, > x -> x + q, with |q| < 1, produces a new x_n which still > contains every rational in (-1+q,1+q), including 0. > 0 is provably not in x_n. I have simplified this proof. Let X be every rational in (-1,1) and x_n a sequence that contains every member of X. Construct sequences a_n and b_n as described by Cantor. Define a new sequence: x_n = x_n - a_i for some i Conside the sequnce a_n. Let a_i = x_m and a_j = x_n. If i And stop grinning, until you put up a decent argument. Its just sad. > What is it with imbeciles like you? decent argument ??? That must be an idiom with some funny connotations or the phrase is of a subtlety that is lost upon me, since Im not a native english speaker and also an imbecile. Come on Mitchie dont be a prick and let me in on the fun! === Subject: Re: there is no such thing as infinity > Cantor was a rusha. He confused generations of mathematicians by > telling them there are lots of infinities. It is impossible to get a > BS in math without admitting that his theories are correct. > > The diagonal argument was a sham because it presupposes the existence > of infinity without proving its existence. His theorem is proof that > infinity does not in fact exist, because if it did, then there would > be only one type, not aleph null, aleph one etc., because infinity by > definition is as far as one can go. He even defiled the Hebrew > alphabet by putting subscripts next to aleph when the gematria of > aleph is not infinity or even M but one. > > Okay, lets assume there is no infinity, which is your whole point, no? > And now you have this magic number M. > People have estimated that there are about 6x10^79 atoms in the entire > universe > (http://www.sunspot.noao.edu/sunspot/pr/answerbook/ universe.html#q70). > I suppose this number 6x10^79 is your magic M. If you tried to write > down all the numbers 1 through 6x10^79, you would run out of space to > write them. (I wont even go into counting them, since even counting > one number every picosecond means this would take ~2x10^61 years) > Now, you might be wondering why I am requiring you to write all these > umbers down (and why I think your computer program might be cheating) > I can write 1 6x10^90, but that does not mean this number exists. It > might just be too big of a number, much larger than your M. So, unless > you have proof that all numbers up to your number M exists, proving M > exists is all for naught. > Okay, maybe that number is too big. Afterall, people guess the universe > is only 13.7 billion years old > (http://wiener.math.csi.cuny.edu/UsingR/Data/ age.universe.html), or > roughly 4.3x10^42 yocto-seconds (10^-24). Maybe this should be M, since > the age all of existence is less than this number. > Personally, I prefer to think of 42 as infinity. Until you convince me > otherwise, I will just continue believing this way. 43? Doesnt even > exist. Mathematicians has always that though. And since Planks Constant has less to do with math than idiots with names like Feynmann, thats obviously why what you is say you is true, but its only true in North Carolina and Calthech. So its also only true on Tuesdays. > - Tim > Timothy M. Brauch > Graduate Student > Department of Mathematics > Wake Forest University === Subject: Re: there is no such thing as infinity > >Ive thought really hard about this one and came to the conclusion >that there is no scientific evidence of infinity existing. The highest >number that anyone has ever measured to according to Isaac Asimov in >his book Science and Human Thought is only about 5.0 x 10^48. No one >has ever gotten past that number. Doesnt this sound weird? > > You are trying to relate infinity to a quantity. Infinity is not a > quantity. Nor is it the absence of quantity. If it could be so > related, then it would be a number that is not a number, and hence > would have no identity. But infinity IS NOT A NUMBER. > You should take a look at Cantors theory of sets. In it, Cantor does > treat infinities as numbers, cardinals and ordinals. Try a Google > search on it. Most people already have. But Cantor never claimed that infinity was a number. Only recursive mathema-rejects have ever done that. > One of the brilliant things Cantor did was to define an infinite set > as a set whose elements can be put into 1-1 correspondence with a > proper subset of itself (obviously something one cannot do with a > finite set). For instance, you can put the set of positive integers > into 1-1 correspondence with the set of positive even integers by the > correspondence > n -> 2n But Cantor didnt do that. Ancient Greeks did that. > It seems to go against common sense to say that these two sets have > the same cardinalities. Then again, we dont have any common > experience working with infinities. > Patrick === Subject: Re: Math Too Advanced For Mainstream Economists I think this thread would amusing to those who find JSH threads amusing, but with a twist. I was delighted to find in a dictionary the word MUMPSIMUS, which means stubborn persistence in an error that has been exposed. -- Joan Robinson > ...the original poster was unable to find a single piece of > empirical support for the labor market model presented... The above, of course, is untrue, and it is hard to see how poor Mark Witte cannot know it is untrue, if he had any clue about what he is talking about. > A > demand curve is a relationship between price and quantity demanded. > If something else changes, such as prices of related goods in demand > (here the discount rate would be an example), this shifts the curve. Poor Mark Witte is confused. A discount rate is not a price. > To be charitable to the original poster, his confusion probably stems > from his notion that a change by the firm in question in the number of > workers it hires would change the discount rate for the economy. The above is a strawperson. No such claim is to be found at the above URL. Notice no claim is made about the discount rate for the economy being changed by a change in how many workers firm hires. In fact, an economy-wide market for financial capital is not described at my URL. Poor Mark Witte just doesnt understand accounting. If the best return a firm can obtain is 10%, then thats what the firm gives up in investing elsewhere. If costs are different for the best investment (e.g., because the level of wages is different), then the firm gives up some other rate of return than 10%. To help poor Mark Witte out, heres some explanations of a concept important for understanding Step 2 of the algorithm given at my URL: My URL is more a tutorial to results well-established in the literature. No claim is made to novelty there. For that matter, no claim is made there that the relationship shown is a demand curve for labor. But that claim is made in the literature. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Rationals are Uncountable > But what happens when you start with a sequence which does contain > all rationals in (-1,1) ? Please present one. Then demonstrate how to translate that sequence to contain every rational in (-1-r,1-r) where r is irrational and |r| < 1. Or do it the other way. Give a sequence that contains every rational in (-1-r,1-r) and show how to translate it to include every rational in (-1,1). Russell - 2 many 2 count === Subject: How far can one go with self-study nowadays? Just thinking about all the neat things available online, compared to when I was in school 20ish years ago, Im wondering how far one can go nowadays in mathematics with self-study. 20 years ago, it would have been hard. If you lived near a university that had a graduate program in math, you would have been able to at least figure out what books to read, but if you didnt live near a university, even figuring out what books to read might have been reasonably hard. And if you had needed help, finding someone to ask would have been hard. It would have been doable, I think, but it would have been very hard. Compare that to nowadays. You can find textbooks easily online, and find customer reviews of them, to get an idea of which ones are good. If you get stuck on something, there is a fair chance that a Google search will find a solution, and if not, there is usenet--a post to sci.math will probably get an answer. I think one should be able to reasonably get to at least a fair way into graduate level. When you get to the point that you need to regularly read journals, that could be a problem--those things are expensive, and without a university library you might be limited. Is there a way around this, or is that the first big barrier? -- --Tim Smith === Subject: Looking for primes of a particular form. I am trying to find primes numbers of the following forms: p = 6*(10^N)-1, q = 6*(10^N)+1 All I can find so far are: N=1, p= 5, q= 7 N=2, p= 59, q= 61 N=3, p=599, q=601 sieves possible values of N between 4 and 10000. It uses modular arithmetic as a substitute for trial division by each prime. That program left me with a manageable list of candidate values of N. (By the way, at my level of skill, this is the most advanced modular arithmetic problem Ive ever successfully programmed a solution for.) But how do I further test to see which of these candidates, (e.g., values of N such that p or q have no small prime factors) are really prime? I thought I heard once of a fast primality test that requires one to know the prime factors of n-1. Well in my case I have the luxury of knowing that the only prime factors of q-1 are 2,3, and 5. But I do not know how implement the test in my computer code. Is there any theoretical proof, that I may have missed, that tells me whether or not the primes I am looking for even exist? Also, a separate question is whether any twin primes exist in these forms for N>3? (E.g. both p and q are prime for the same value of N.) Is there be a proof of the (non)existence of such twin primes? Summary of my questions: 1. I want to know how to program the fast primality test that relies on knowing the factors of n-1. 2. I want to know if any theoretic considerations can be brought forth regarding number of primes that even exist for the two forms I gave. Same for twin primes. P.S. What primality tests are known that are optimized for numbers of the form a*(b^n) +/- 1?