mm-1082 === Subject: Polynomials i want to show, that if r is a root in a splitting field of f=x^3+x^2-2x-1 over Q, then r'=r^2-2 is another root. I split f into factors f=(x-r)(x^2+(r+1)x+(r^2+r-2)), but how can I see, that r^2-2 is a root of the quadratic factor? Johannes === Subject: Re: Polynomials > i want to show, that if r is a root in a splitting field of f=x^3+x^2-2x-1 > over Q, then r'=r^2-2 is another root. > I split f into factors f=(x-r)(x^2+(r+1)x+(r^2+r-2)), but how can I see, > that r^2-2 is a root of the quadratic factor? factorization of f(x) by just dividing it by (x - r). If that's the case, then you should have _no_ trouble finishing this -- just use polynomial long division to compute (x^2 + (r_1)x + (f^2+r-2))/(x - (r^2-2)) ... it'll work out nicely _because_ f(r) = 0 -- kinda neat, actually :-) (As a bonus, you'll also determine the third root of f in terms of r: it's r^2+r-1 ... ) > Johannes === Subject: Re: Polynomials > i want to show, that if r is a root in a splitting field of f=x^3+x^2-2x-1 > over Q, then r'=r^2-2 is another root. r being a root of f(x) = x^3 + x^2 - 2x - 1 implies that r^3 = 1 + 2r - r^2. Then r^4 = r + 2r^2 - r^3 = r + 2r^2 - (1 + 2r - r^2) = -1 - r + 3r^2. Similarly you can express r^5 and r^6 in terms of 1, r, and r^2. Now multiply out f(r^2 - 2) = (r^2 - 2)^3 + ... and reduce all the high powers of r as we've just seen is possible, and see what happens. Alternatively, let r'' = (r')^2 - 2; this must also be a root. r'' = (r^2 - 2)^2 - 2 = r^4 - 4 r^2 + 2 = (-1 - r + 3r^2) - 4 r^2 + 2 = 1 - r - r^2. Now show that r, r', and r'' are zeros of f. The easy part is r + r' + r'' = -1. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: 2-manifold metric spaces with many symmetries > [...] >> Here, I hadn't scanned up to see the use of n as the dimension >> of the ambient space, and used it incorrectly as the dimension of >> the submanifold; I've replaced n by k in this paragraph, and note >> that it refers to the dimension of the submanifold: > The result you want requires the manifold to have a transitive > action of isometries by some Lie group G; in addition, this > action must contain the rotation group SO(k) in the isotropy > subgroup of G at x for each element x. That means that your > k-manifold has a symmetry group of dimension at least > k + k(k-1)/2 = k(k+1)/2. I note that this is the dimension > of the group SO(k+1), the group of isometries of the k-sphere. > I'll note that among the various differential structures on > S^k (for k >= 7, these are non-diffeomorphic smooth structures > on the same topological space S^k), the only one that has a > group of isometries of this dimension is the standard, round > sphere. > I suspect that this condition poses a severe constraint on > the manifold, since it's as symmetric as the k-sphere. > Am I right that you have made a generalization to > k-manifolds, emmbedded in some euclidean space, with k>=2? Yes, I suppose that's correct. I had thought you were using the term surface to refer to a closed manifold, rather than the strict interpretation as a 2-dimensional object. The question does make sense in higher dimensions, so it's a reasonable generalization. > I'm trying to formulate the problem for a k-manifold M: > Condition (i): For any x, y in M, there is an isometry of M sending > x to y. [ Transitivity ] > Condition (ii): For any x in M, the subgroup of isometries that leave > x fixed is such that ___________ {something}. > I don't know what to put in place of ___________. This depends on your intent. If all you want is for any triangle to be mappable to any other congruent triangle via an isometry, then the following is sufficient: the action is transitive on the space of orthonormal 2-frames in TM_x An n-frame F in the vector space V is an ordered n-tuple of linearly independent vectors in V. If V has an inner product, an orthonormal n-frame is an n-frame for which the members are mutually orthogonal, and each has norm 1. If you have a k-manifold and want all congruent d-simplices (that is, the d-dimensional analogue of triangles [d=2]) to be mappable to one another via isometries, then you need this: the action is transitive on the space of orthonormal d-frames in TM_x For d=k, that is you want all congruent k-simplices to be mappable to one another via isometries, this is (obviously) the action is transitive on the space of orthonormal k-frames in TM_x But the space of orthonormal k-frames is the full orthogonal group O(k). Since the isotropy subgroup associated to a fixed point of an isometry must be orthogonal (i.e., a subgroup of O(k)), you get this: the group is O(k) You may restrict to orientation-preserving maps, in which the group would be SO(k). I'll note that having the condition met for d-simplices forces the condition to hold for all dimensions <= d. So, here's what this all amounts to: 1. Your group of isometries acts transitively on M. That means that M is what's called a *homogeneous space*, that is, a manifold with a transitive action by a Lie group. Given any homogeneous space M, one can construct a metric for which the action is via isometries (well, I know how to do this for actions by discrete groups, by averaging the metric over orbits; I think this still works for actions by compact Lie groups in general, by integrating the metric wrt Haar measure). 2. The isotropy subgroup at x acts transitively on the space of orthonormal d-frames in TM_x. The clasification of subgroups of O(k) that act transitively on the space of orthonormal d-frames in R^k (alias the Stiefel manifold V(d,k-d)) may be well understood; it isn't well-understood by me, so this poses a bit of a problem. However, if d=k, then you get the isotropy subgroups all being equal to the orthogonal group O(k), or if you only want orientation preserving transformations, then the special orthogonal group SO(k). In that case, here's how to construct an example: Take a compact Lie group L of dimension k(k+1)/2, that contains SO(k) as a subgroup. Then form the quotient L/SO(k), and that will be your manifold M. The action of L on M is via multiplication: L x L ---> L except you need to divide out by the subgroup SO(k): L x (L/SO(k)) ---> L x (L/SO(k)) This is not so hard, since the cosets of SO(k) in L are of the form x = a SO(k), for a being any element of the group L. Multiplication by any b in L is then well-defined: b (a SO(k)) = ba SO(k) and you find that if b fixes a particular element x of M, then b x = b (a SO(k)) = (ba) SO(k) so we find that (ba) SO(k) = a SO(k) or rather a' b a in SO(k), (where a' is a^(-1)) or b in a SO(k) a' That is, the isotropy subgroup at x = a SO(k) is conjugate (and thus isomorphic) to the subgroup SO(k). For the orientation-preserving case, and requiring oriented congruent k-simplices to be mapped isometrically to one another, this construction will produce all examples. === Subject: Re: inverse modulos by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2JbUg32369; >If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is >true, then x has a unique multiplicative inverse modulo m, a, so that ax >= 1 (mod m). >Knowing only the multiplicative inverse, a modulo m, and m, is it >possible to find x? >Is it true that a*x - 1 = k*m, for some k, possibly negative? Or >multiple k's? How to find the multiple k's so that x could be found? >Is that possible at all, so that the candidate k's can be found knowing >only the inverse a modulo m and m? >If there is no single way to find it, is there a reasonable trial and >error process to go through to find x given a and m? Using the ideas from extended GCD or continued fraction. Take the continued fraction expansion of the fraction a/m, and truncate it at various stages to get good rational approximations to a/m with small numerator and denominator. One of those will be k/x. === Subject: Re: inverse modulos > Using the ideas from extended GCD or continued fraction. > Take the continued fraction expansion of the fraction a/m, > and truncate it at various stages to get good rational > approximations to a/m with small numerator and denominator. > One of those will be k/x. Is there somewhere on the web that explains how I can carry out these use it for this... it seems to be only for trig functions). When should I truncate? And then how would I separate the k and the x in k/x? === Subject: sci.physics , sci.math is nonsense shut up when i m talking to you. shut up!! === Subject: Re: sci.physics , sci.math is nonsense > shut up when i m talking to you. shut up!! Speak louder we can't hear you. === Subject: Re: sci.physics , sci.math is nonsense shut up when i m taking to you. shut up, shut up. shut up when i m taking to you!!! === Subject: Re: sci.physics , sci.math is nonsense is that what your mama said to you over and over? > shut up when i m taking to you. shut up, shut up. shut up when i m taking to > you!!! === Subject: Re: sci.physics , sci.math is nonsense hello punk, certified punk > is that what your mama said to you over and over? >> shut up when i m taking to you. shut up, shut up. shut up when i m taking > to >> you!!! === Subject: Re: sci.physics , sci.math is nonsense reply-type=response > hello punk, certified punk are you boring on purpose? === Subject: Re: sci.physics , sci.math is nonsense <319evkF38gkbvU1@individual.net> <319v8eF2qj9e4U1@individual.net> <31bnb4F3agmorU1@individual.net> > hello punk, certified punk > are you boring on purpose? You know, for some reason, Lord of Chaos seems... 'Frazir-like'. -Mark Martin === Subject: Re: sci.physics , sci.math is nonsense your mom again >> hello punk, certified punk > are you boring on purpose? === Subject: Re: Personal conjectures > Hello sci.math readers, > . I'm sure that every mathematician has one or more > personal conjectures that he/she would give his/her soul to prove. > I'd like to see the conjectures that you have thought up, and that > you think about whenever you have a spare moment. > Asger. > I love Conjectures about prime numbers. This one I framed many years > ago: > For each integer N >= 1 there is an X>=0 that produces a prime > number with the formula: > 6(N - X^2) + 1 > Also for each integer N>=1 there is an Y>=0 that produces a prime with > the formula: > 6(N - Y^2) - 1 > That is, before X or Y attain sqr(N) it will result a prime number . > Ludovicus Naturally the value of X augments with N, but what about the relation : r = X / [log(N)]^2 for the first prime found? In this case my record in the formula 6(N - X^2) + 1 is 0.585 for the values N = 71160 , X = 73. And r = 0.528 in the formula 6(N - Y^2) - 1 N = 16892, Y = 50. Ludovicus === Subject: i clean up sci.physics, sci.math, rec.mensa.org gentleman bets taken!!! === Subject: Re: i clean up sci.physics, sci.math, rec.mensa.org gentleman bets taken!!! plonk === Subject: Re: i clean up sci.physics, sci.math, rec.mensa.org gentleman bets taken!!! pink > plonk === Subject: talk science and mathematics, you faggots. i m supreme. i alone command you. === Subject: Re: talk science and mathematics, you faggots. reply-type=response > i m supreme. i alone command you. still boring... === Subject: Re: talk science and mathematics, you faggots. >> i m supreme. i alone command you. > still boring... Lunacy is never boring. BTW CHAOS stands for Could (do to) Have Another Olanzapine Soon. === Subject: Re: talk science and mathematics, you faggots. In sci.math, Lord of Chaos(Suresh Devanathan) <319fl1F36rekiU1@individual.net>: > i m supreme. i alone command you. [1] A sultan commands his driver to get from point A to point B 60 miles distant in 1 hour. The driver, for various reasons, drives the first 30 minutes at 30 miles per hour. How fast would he have to drive the rest of the distance in order to make it, and how far would he have to travel? [2] A proposed US single-stage rocket craft has a thrust of 36 megaNewtons for 120 seconds, and an exhaust velocity of 4 km/s. Assuming that the fuel tanks comprise 3% of the rocket's mass (sans payload), the radius of the Earth is 6378 km, and the surface g is 9.805 m/s/s, can the rocket make it to a circular orbit 200 km above the surface, and if so, how much payload could it carry? Neglect rotation and air friction in the computation. (Hint: Tsiolkovsky.) [3] Same as [2], only this time the question is whether it can escape Earth's gravitational pull altogether, and how much payload it can carry. [4] A construction company has been contracted to build a dam out of cement, and the specification requires a certain thickness of cement at the bottom of the dam. This thickness can hold back 100 megaNewtons of pressure. However, the thickness also has to contend with its own weight, which is 9.805 m/s/s = N/kg. The mass of the concrete is 2400 kg/m^3. Assuming that one can fashion a triangular dam wedge, what is the maximum depth of water one can hold back with this concrete, and how thick is the dam at its base? Assume a water density of 1000 kg/m^3. [5] A flat iron bar of indefinite length and square cross section of 1 cm is propped up on at least two round rods, in such a fashion that it can travel freely. If the round rods are each 1 mm in diameter, how far apart can they be spaced without the iron bar touching the ground? Assume a rigidity modulus of 82 gigaPascal and a density of 7874 kg/m^3; neglect distortion in the support rods. Have fun, especially since I don't know the answers either! (Except to #1. No, it's not infinite speed.) -- #191, ewill3@earthlink.net -- no, I don't do this for a living It's still legal to go .sigless. === Subject: Re: talk science and mathematics, you faggots. We ARE discussing science and mathematics! > i m supreme. i alone command you. === Subject: Re: talk science and mathematics, you faggots. Yes you are... yes you are > We ARE discussing science and mathematics! >> i m supreme. i alone command you. === Subject: Re: talk science and mathematics, you faggots. > i m supreme. i alone command you. Hey stooopid, 1) Turbulence scales as the cube of the distance. 2) Graph your pathetic little brainfart on hyperbolic axes. 3) Your trailer park called - its garbage is missing. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: talk science and mathematics, you faggots. shut up! shut up when i m talking to you! >> i m supreme. i alone command you. > Hey stooopid, > 1) Turbulence scales as the cube of the distance. > 2) Graph your pathetic little brainfart on hyperbolic axes. > 3) Your trailer park called - its garbage is missing. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: talk science and mathematics, you faggots. > shut up! shut up when i m talking to you! > i m supreme. i alone command you. >> Hey stooopid, >> 1) Turbulence scales as the cube of the distance. >> 2) Graph your pathetic little brainfart on hyperbolic axes. >> 3) Your trailer park called - its garbage is missing. >> -- >> Uncle Al >> http://www.mazepath.com/uncleal/ >> (Toxic URL! Unsafe for children and most mammals) >> http://www.mazepath.com/uncleal/qz.pdf Lord of boring. -- Sl.87inte, Fletch === Subject: Re: talk science and mathematics, you faggots. lord of you === Subject: Re: talk science and mathematics, you faggots. you are not talking to him, you are typing at him. === Subject: Re: talk science and mathematics, you faggots. > you are not talking to him, you are typing at him. Not me, I bozobinned him. Pepe le Pew -- PT Barnum was right ! === Subject: Re: talk science and mathematics, you faggots. > i m supreme. i alone command you. Chaos = 1/0 Lord of Chaos = (1/0)^2 -- Will Twentyman email: wtwentyman at copper dot net === Subject: Universal sentences of Z and ZxZ How to prove that the universal sentences of Z are the one of ZxZ, both considered as additive groups ? A proof using ultraproduct will be much appreciated. And to avoid discimination, others also! === Subject: Re: Staff Announcement Actech advertised: > * IBM Processor for amazing speed and performance Which IBM processor? === Subject: how to measure entropy of music? I have vaguely heard about this method... and I am very interested in it... could anybody give me some pointers? After learning how to measure entropy of music... I can begin to measure entropy of texts, etc.. that's going to be fun! === Subject: Re: how to measure entropy of music? > I have vaguely heard about this method... and I am very interested in it... > could anybody give me some pointers? > After learning how to measure entropy of music... I can begin to measure > entropy of texts, etc.. that's going to be fun! Well, theoretically speaking, you should just treat the notes of the music as symbols for the computation of the entropy. Then, there is no difference between finding the entropy of text and finding the entropy of music. But if you sample the signal of music, encode it, and compute the entropy, I doubt if it makes sense to do that given the fact that you already have the notes of the music. If you just want to know what is about entropy of data, you might just start with any textbook for Information theory or some websites for some ideas, e.g., http://www.ScienceOxygen.com/signal.html http://www.ScienceOxygen.com/electrical.html Have fun, === Subject: Re: how to measure entropy of music? > I have vaguely heard about this method... and I am very interested in it... > could anybody give me some pointers? > After learning how to measure entropy of music... I can begin to measure > entropy of texts, etc.. that's going to be fun! Compress it using a suitable lossless compressor. Provided that the compressor knows how to detect (some amount of) musical redundancy, this will give you the Kolmogorov complexity with respect to a particular non-universal computer, e.g. algorithmic entropy which is more fundamental than Shannon entropy. This has already been exploited using the information distance of Vitanyi et al. Precise clustering of music files has been achieved with ordinary compressors like bzip2. Find the paper Algorithmic Clustering of Music by Rudi Cilibrasi and Paul Vitanyi. Needless to say, you have to compress uncompressed wav files... -- Eray Ozkural === Subject: Re: how to measure entropy of music? (snip, and previously snipped discussion of music entropy) > Compress it using a suitable lossless compressor. Provided that the > compressor knows how to detect (some amount of) musical redundancy, > this will give you the Kolmogorov complexity with respect to a > particular non-universal computer, e.g. algorithmic entropy which is > more fundamental than Shannon entropy. > This has already been exploited using the information distance of > Vitanyi et al. Precise clustering of music files has been achieved > with ordinary compressors like bzip2. Find the paper Algorithmic > Clustering of Music by Rudi Cilibrasi and Paul Vitanyi. Needless to > say, you have to compress uncompressed wav files... I would think a pure sine wave should be low complexity, so maybe you should compress the Fourier transform. Maybe that won't quite do it, but I can imagine wav files of low complexity audio signals not compressing very well. Has anyone ever done an FFT of a whole CD? -- glen === Subject: Re: how to measure entropy of music? > I would think a pure sine wave should be low complexity, so maybe > you should compress the Fourier transform. I think a vocoder does something similar: linear predictive coding tries to fit an AR model to the input, effectively searching for the formants (resonant peaks) of the vocal tract transfer function. If you initialize an IIR filter with the LPC coefficients and the states (delays) with the signal, you can let it ring like an oscillator - the output will converge to a sum of pure damped sine waves at the estimated frequencies of the formants. However, instead of storing the sine waves, the vocoder stores the LPC coefficients and the residue (prediction error). To take this back to entropy, the energy of the prediction error could well be regarded as the entropy of a vocal signal - it gives a good measure on how unexpected the signal is, given its past history. Recent work has shown that this method is also suited for general audio signals, not just speech. The order of the AR model just increases drastically (for interpolation, orders of 1000 - 3000 are used). The same applies here for the prediction error and the entropy. > Has anyone ever done an FFT of a whole CD? If you take a music CD, I would expect a linear trend of the form 1/f^a. This is a common model for music signals, and it agrees well with findings of long-range correlated time series. For speech, I would guess a rectangular pulse response. The formants vary across a certain frequency range, but are limited from above and below. > -- glen Andor === Subject: Re: how to measure entropy of music? > I have vaguely heard about this method... and I am very interested in it... > could anybody give me some pointers? > After learning how to measure entropy of music... I can begin to measure > entropy of texts, etc.. that's going to be fun! Learn all you want and more at Brian Whitman's site... Whitman is also the producer of Eigenradio (their motto: Statistically their iTunes and WMP links... -- Remove _me_ for e-mail address === Subject: Re: how to measure entropy of music? is this some joke ?! >I have vaguely heard about this method... and I am very interested in it... >could anybody give me some pointers? > After learning how to measure entropy of music... I can begin to measure > entropy of texts, etc.. that's going to be fun! === Subject: Re: how to measure entropy of music? >> After learning how to measure entropy of music... I can begin to measure >> entropy of texts, etc.. that's going to be fun! >is this some joke ?! The model of entropy that can be readily recognized: In music--how often can an informed listener infer the next note in a phrase; ie how many bits are needed to specifiy the next note. In sound--how many bits are needed to specify the next value of the signal. For written language, the analog is how many bits are needed to confirm an informed guess as to the next letter in a text. Perhaps no more than three. For some TV shows these days, its only two. For music, the question gets really challenging as one considers the entropy of scores--the parts played by accompanying instruments and the choices of these instruments requires a lot of encoding. In this case, the number of bits needed to feed a high level orchestral synthesizer might establish a lower bound. Using a predictor such as a Hidden Markov Model using the Viterbi algorithm or one of its descendants might go a long way in evaluating the predictability/aka entropy of sounds. (This is already the case in speech recognition.) A really interesting question would be to rank famous composers by the average entropy of their music--eg Bach, Mozart, the Beatles at the top, Andrew Lloyd Webber, Elton John, Cole Porter next--well, you get the idea. The basics are at: http://www.music-cog.ohio-state.edu/Music829D/Notes/Infotheory.html key words: hmm, entropy, and music I have learned the question is of significance today for serveral reasons: 1) Recognizing when TV commercials are playing. 2) Separating out background music behind speech in speech recognition. 3) Locating specific music contained in a large database of sound Try: http://crl.research.compaq.com/publications/techreports/techreports.html search the page for Logan or music. Also: http://www.idiap.ch/publications/ajmera-rr-01-26.bib.abs.html Speech/Music Discrimination using Entropy and Dynamism Features in a HMM Classification Framework Speech/Music Discrimination Using Discrete Hidden Markov Models http://crl.research.compaq.com/publications/techreports/reports/2000-1.pdf MUSIC SUMMARY USING KEY PHRASES M. Brand, .8bStructure Discovery in Conditional Probability Models via an Entropic Prior and Parameter Extinction,.8a Neural Computation, July 1999 John Bailey http://home.rochester.rr.com/jbxroads/mailto.html === Subject: Re: how to measure entropy of music? > For written language, the analog is how many bits are needed to > confirm an informed guess as to the next letter in a text. Perhaps no > more than three. For some TV shows these days, its only two. > For music, the question gets really challenging as one considers the > entropy of scores--the parts played by accompanying instruments and > the choices of these instruments requires a lot of encoding. In this > case, the number of bits needed to feed a high level orchestral > synthesizer might establish a lower bound. Very much a lower bound, I suspect, because even a fantastic orchestral synthesizer would be to an orchestra as a typesetter would be to the handwritten word: you don't only have the notes, you have the playing of each of them, and the interplay of the playing with the acoustics of the hall, etc. Still, it's a neat thought experiment. And you could indeed do an interesting analysis of musical scores, rather than muscial recordings. -- Andrew === Subject: Re: how to measure entropy of music? > The model of entropy that can be readily recognized: > In music--how often can an informed listener infer the next note in a > phrase; ie how many bits are needed to specifiy the next note. > In sound--how many bits are needed to specify the next value of the > signal. found music to be interesting to listen to when he had a low success rate at predicting what would happen next. I think he was thinking of higher level structures than single notes, but couldn't the concept of entropy apply to these higher structures as well? === Subject: Re: how to measure entropy of music? >As a matter of interest, >do you consider high entropy good or bad? >found music to be interesting to listen to when he had a low success >rate at predicting what would happen next. I think he was thinking of >higher level structures than single notes, but couldn't the concept of >entropy apply to these higher structures as well? >Presumably a completely random series of notes would have very high entropy, >while absolute silence has very low entropy. >I wouldn't have thought either was very enjoyable. Both of your comments are quite stimulating. Of course! The idea that good music is neither too rote nor too random is not new but the idea that entropy as a concept allows a deeper investigation of what people like is exciting. Conjecture: classes of composers cluster around various levels of entropy. On further thought--what is predictable changes with audience familiarity with a style. Music that is on the leading edge of predictability is the most interesting. Its probably all about what gives our music neurons a good workout. John Bailey http://home.rochester.rr.com/jbxroads/mailto.html === Subject: Re: how to measure entropy of music? > Try: > http://crl.research.compaq.com/publications/techreports/techreports.html > search the page for Logan or music. Some of Beth's papers are available from her now-HPLabs web-site: http://www.hpl.hp.com/research/crl/publications/papers.html Ciao, Peter K. === Subject: Re: how to measure entropy of music? > A really interesting question would be to rank famous composers by the > average entropy of their music--eg Bach, Mozart, the Beatles at the > top, Andrew Lloyd Webber, Elton John, Cole Porter next--well, you get > the idea. As a matter of interest, do you consider high entropy good or bad? Presumably a completely random series of notes would have very high entropy, while absolute silence has very low entropy. I wouldn't have thought either was very enjoyable. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: how to measure entropy of music? > The basics are at: > http://www.music-cog.ohio-state.edu/Music829D/Notes/Infotheory.html > key words: hmm, entropy, and music I have learned the question is of > significance today for serveral reasons: > 1) Recognizing when TV commercials are playing. > 2) Separating out background music behind speech in speech > recognition. I wonder if they could figure a why to identify background music and separate it from the sound reaching my ears :-) Steve === Subject: Re: how to measure entropy of music? >> The basics are at: >> http://www.music-cog.ohio-state.edu/Music829D/Notes/Infotheory.html >> key words: hmm, entropy, and music I have learned the question is of >> significance today for serveral reasons: >> 1) Recognizing when TV commercials are playing. >> 2) Separating out background music behind speech in speech >> recognition. >I wonder if they could figure a why to identify background music and >separate it from the sound reaching my ears :-) Good point! If the predictor works then you can make a pretty good Or better yet, a transmogriphier; if it's punk metal coming out of the speaker and you want classical, you could put some filter and adaptation circuits in that take the punk metal energy and convert it based on the predictor rules for classical. Punk metal out of the speaker, classical into your ear. Or you could buy an MP3 player and some headphones, either way. ;) Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions. http://www.ericjacobsen.org === Subject: Re: how to measure entropy of music? >> The basics are at: >> http://www.music-cog.ohio-state.edu/Music829D/Notes/Infotheory.html >> >> key words: hmm, entropy, and music I have learned the question is of >> significance today for serveral reasons: >> 1) Recognizing when TV commercials are playing. >> 2) Separating out background music behind speech in speech >> recognition. >I wonder if they could figure a why to identify background music and >separate it from the sound reaching my ears :-) > Good point! If the predictor works then you can make a pretty good Not sure how you can come to that conclusion. If you have one microphone and a perfect algorithm you could determine what is music and what is sum of music + speech and without a second sensor this would not be possible - unless you assume the statistics of the music do not change rapidly so that the music during the speech was approx the same as during non-speech - an assumption which may be true for some music maybe? On the other hand if we used multiple microphones then maybe we would have something worthwhile. Tom === Subject: Re: Prestigious mathematics journals > In biology we have publications such as Nature which are generally > considered to be very prestigious, which magazine in mathematics holds > the same distinction? > Aside from subjective opinions, there is the impact factor computed > by ISI Web of Knowledge (www.isiknowledge.com). > Who wants to guess the top mathematics journal on their list? What is the scale of the impact factor? and what sort of impact does it measure? I guess what I am trying to say is that are the calculating the impact in the scientific community or general public (this wouldn't make much sense) === Subject: Re: Prestigious mathematics journals >> In biology we have publications such as Nature which are generally >> considered to be very prestigious, which magazine in mathematics holds >> the same distinction? >> Aside from subjective opinions, there is the impact factor computed >> by ISI Web of Knowledge (www.isiknowledge.com). >> Who wants to guess the top mathematics journal on their list? > What is the scale of the impact factor? and what sort of impact does > it measure? I guess what I am trying to say is that are the > calculating the impact in the scientific community or general public > (this wouldn't make much sense) Impact factors are essentially citation rates. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Unstoppable Force vs Immovable Object > Yes, I know. In that case, it's also physically impossible that there > should be an immovable object. I was ignoring the laws of physics and > going for logic alone. > Logic, as such, has very little to say about physical forces. As I said > you are just playing a word game. > Bob Kolker Why? What's wrong with the argument? === Subject: Re: Unstoppable Force vs Immovable Object > Why? What's wrong with the argument? Force has a physical meaning, not a logical meaning. A force is some kind of action that changes the momentum of a body. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object Why? What's wrong with the argument? > Force has a physical meaning, not a logical meaning. A force is some > kind of action that changes the momentum of a body. > Bob Kolker I don't think you've identified a problem with my reasoning. === Subject: Re: Unstoppable Force vs Immovable Object > I don't think you've identified a problem with my reasoning. Your reasoning addresses a physical concept in a non-physical way. As I said, you are playing word games. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object I don't think you've identified a problem with my reasoning. > Your reasoning addresses a physical concept in a non-physical way. As > I said, you are playing word games. > Bob Kolker What's that supposed to mean, I address a physical concept in a non-physical way? I conclude that it's logically impossible for an unstoppable force to meet an immovable object. Do you disagree? Do you think it's logically possible? === Subject: Re: Unstoppable Force vs Immovable Object > What's that supposed to mean, I address a physical concept in a > non-physical way? > I conclude that it's logically impossible for an unstoppable force to > meet an immovable object. Do you disagree? Do you think it's logically > possible? If you consider what the word force means you would conclude there is no unstopable (i.e. infinite) force in the physical universe. Likewise, an immovable object would have infinite momentum (physically impossible). In either case you are talking about things which not only don't exist, but can't exist. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object What's that supposed to mean, I address a physical concept in a > non-physical way? I conclude that it's logically impossible for an unstoppable force to > meet an immovable object. Do you disagree? Do you think it's logically > possible? > If you consider what the word force means you would conclude there is no > unstopable (i.e. infinite) force in the physical universe. Likewise, an > immovable object would have infinite momentum (physically impossible). > In either case you are talking about things which not only don't exist, > but can't exist. > Bob Kolker So you conclude that it's logically impossible for an unstoppable force or an immovable object to exist. I'm not sure that's correct, but in any event it's consistent with my conclusion. So what's to argue about? === Subject: Products of Sines? Does the following product have a nice closed-form solution? sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) -- Daryl McCullough Ithaca, NY === Subject: Re: Products of Sines? Daryl McCullough escribi.97: > Does the following product have a nice closed-form solution? > sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) Let P(z) = (z^n - 1)/(z - 1) = 1 + z + z^2 + ... + z^(n-1) (#1) But also P(z) = Prod(z - z_k, k, 1, n-1) (#2) being z_k = cos(k*2pi/n) + i*sen(k*2pi/n), k = 1, 2, ..., n -1 the n-1 n-roots of unity distinct of 1. |P(1)| = |Prod(z - z_k, k, 1, n-1)| = Prod(|z - z_k|, k, 1, n-1) = Prod(d_k, k, 1, n-1) where d_k is the distance from (1, 0) to (cos(k*2pi/n), sen(k*2pi/n)). But d_k = sqrt((1 - cos(k*2pi/n))^2 + sen^2(k*2pi/n)) = sqrt(2 - 2cos(k*2pi/n)) = 2sqrt((1 - 1cos(k*2pi/n))/2) = 2*sen(kpi/n) Then |P(1)| = n = Prod(2sen(k*pi/n), k, 1, n-1) = 2^(n-1)*Prod(sen(k*pi/n), k, 1, Prod(sen(k*pi/n), k, 1, n-1) = n/2^(n-1) Curiously, it is the same that the probability of n points choosen at ramdom in a circle lie in the same semicircle. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Products of Sines? > Does the following product have a nice closed-form solution? > sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) This is very well-known. Set w = exp(pi i/n). The product is then (2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} - w^{-n+1}) = (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 - w^{-2n+2}). Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Products of Sines? Robin Chapman says... >> Does the following product have a nice closed-form solution? >> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) >This is very well-known. Set w = exp(pi i/n). The product >is then >(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} - w^{-n+1}) >= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 - w^{-2n+2}). >Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) >= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}. Okay, I'm puzzled by the step (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X - 1) -- Daryl McCullough Ithaca, NY === Subject: Re: Products of Sines? > Robin Chapman says... > Does the following product have a nice closed-form solution? sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) >>This is very well-known. Set w = exp(pi i/n). The product >>is then >>(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} - w^{-n+1}) >>= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 - w^{-2n+2}). >>Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) >>= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}. > Okay, I'm puzzled by the step > (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X - 1) What is the factorization of X^n - 1 into linear factors? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Products of Sines? > Does the following product have a nice closed-form solution? > sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) Yup. It is n/2^{n-1}. Pawel Gladki === Subject: Re: Products of Sines? >> Does the following product have a nice closed-form solution? >> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) >Yup. It is n/2^{n-1}. How do you derive that? -- Daryl McCullough === Subject: Re: Products of Sines? >Does the following product have a nice closed-form solution? > sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n) >>Yup. It is n/2^{n-1}. > How do you derive that? See Robin Chapman's answer. Pawel Gladki === Subject: Re: Vector invariants >Yes, I think I do need rotational invariance too. Oh, then the problem is trivial! There are natural one-to-one correspondences between these categories of thing: * lists of three vectors, up to permutations, translations, & rotations * sets of three points, up to translations, & rotations * congruence classes of triangles * sets of three positive lengths (subject to triangle inequality) * solutions to cubic polynomials (of some restricted type) In other words, all the invariants of the type you're looking for will be naturally built from the three symmetric functions of the three distances between the ends of the vectors. You can use the squares of the distances just as well; they can be expressed as polynomials in the coordinates. In other words, try using just these three invariants (and polynomials in them): -2*a1^2+2*a1*b1-2*b1^2-2*a2^2+2*a2*b2-2*b2^2+2*a1*c1-2*c1^2 +2*a2*c2-2*c2^2+2*c1*b1+2*c2*b2 (-a2*b1+a1*b2+a2*c1-b2*c1-a1*c2+b1*c2)^2 (a1^2-2*a1*b1+b1^2+a2^2-2*a2*b2+b2^2)*(a1^2-2*a1*c1+c1^2+a2^2-2*a2*c2+c2^2) *(c1^2-2*c1*b1+b1^2+c2^2-2*c2*b2+b2^2) > a sentence. dave === Subject: Re: So you want to count an infinite power set ? > P_1(SET) = { > {1000000000000.. AND SET}, > {0100000000000.. AND SET}, > {1100000000000.. AND SET}...} That is my only introduced notation, but I manually calculate the step straight after. P_1(SET) = { {1 AND <0,3>, 0 AND <1,1>, 0 AND <2,4> ...}, {0 AND <0,3>, 1 AND <1,1>, 0 AND <2,4> ...}, {1 AND <0,3>, 1 AND <1,1>, 0 AND <2,4> ...}, ...} 1 = true, 0 = false. TRUE AND X = X Look up 'bit masking The original SET is { <0,3>, <1,1>, <2,4>, ...} i.e. pi 314159.. put into <0 > <1 > <2 > <3 > ... Its the quickest way I could think of to make infinite distinct members that werent trivial N. I use { } for set and < > for sequence, fairly standard for programmers atleast. Herc === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Yes. Dwayne >My bad. I called it by the wrond name but it is the right function. > So > C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., > Prentice > Hall, Copyright 1999, p. 46-50, 210-211 > really defines the Laplace Transform as an integral from -infinity > to infinity, instead of the more usual 0 to infinity? >Dwayne > ************************ > David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >The limits of the bilateral LT are from -oo to +oo. >The limits of the unilateral LT are from 0 to +oo. >Both are usual. >I thought you _BOASTED_ of being a mathematician >of 20 years' standing? I don't think I BOASTED of that, but yes, I mentioned it. I have enough experience to know that something I've never seen before is not necessarily wrong. >> C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., >> Prentice >> Hall, Copyright 1999, p. 46-50, 210-211 >> really defines the Laplace Transform as an integral from -infinity >> to infinity, instead of the more usual 0 to infinity? ************************ David C. Ullrich === Subject: Probability Of Ultimate Extinction. Let's say that you have an amoeba which has a probability q of dying without divinding, and probability p of spliting into two. What is the probability of ultimate extinction? I know that the recursion formula for extinction after n cycles is P(n)=q+p*P(n-1)^2 where P(0)=q. of 3 or 4) is that the probability of ultimate extinction involves using the quadratic formula in the following way. 1±(1-4*p*q)^0.5 --------------- 2q where the minimum of the two values achieved by the quadratic equation is the desired probability. What I'd like to know is, why bother? I've discovered that in the above problem, the probability of ultimate extinction is p/q = 1 if p>=q. Why bother with the quadratic formula? -- ------------------------------- Patrick D. Rockwell === Subject: Re: Probability Of Ultimate Extinction. >Let's say that you have an amoeba which has a probability q of dying without >divinding, >and probability p of spliting into two. What is the probability of ultimate >extinction? >I know that the recursion formula for extinction after n cycles is >P(n)=q+p*P(n-1)^2 where P(0)=q. >of 3 or 4) is that >the probability of ultimate extinction involves using the quadratic formula >in the following way. >1±(1-4*p*q)^0.5 >--------------- >where the minimum of the two values achieved by the quadratic equation is >the desired probability. The general statement is that if p_j is the probability that the amoeba has j children, the probability of ultimate extinction is the least nonnegative solution of s = G(s), where G(s) = sum_{j=0}^infinity p_j s^j. In your case G(s) = q + p s^2 where q + p = 1. Since G(s) - s = p s^2 - s + q = (s - 1)(p s - q), the answer is min(1, q/p). >What I'd like to know is, why bother? I've discovered that in the above >problem, the probability >of ultimate extinction is p/q = 1 if p>=q. Huh?? Since when is p/q = 1 if p >= q? I think you mean q/p if p >= q, 1 if p <= q. >Why bother with the quadratic >formula? No need to do so in this case since the polynomial factors so easily (s=1 is always a solution). Who does bother with it? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Probability Of Ultimate Extinction. >>Let's say that you have an amoeba which has a probability q of dying >>without >>divinding, >>and probability p of spliting into two. What is the probability of >>ultimate >>extinction? >>I know that the recursion formula for extinction after n cycles is >>P(n)=q+p*P(n-1)^2 where P(0)=q. >>of 3 or 4) is that >>the probability of ultimate extinction involves using the quadratic >>formula >>in the following way. >>1±(1-4*p*q)^0.5 >>--------------- >>2q >>where the minimum of the two values achieved by the quadratic equation is >>the desired probability. > The general statement is that if p_j is the probability that the amoeba > has j children, the probability of ultimate extinction is the least > nonnegative solution of s = G(s), where G(s) = sum_{j=0}^infinity p_j s^j. > In your case G(s) = q + p s^2 where q + p = 1. Since > G(s) - s = p s^2 - s + q = (s - 1)(p s - q), the answer is > min(1, q/p). >>What I'd like to know is, why bother? I've discovered that in the above >>problem, the probability >>of ultimate extinction is p/q = 1 if p>=q. > Huh?? Since when is p/q = 1 if p >= q? I think you mean > q/p if p >= q, > 1 if p <= q. You're right. I should have said q/p. Sorry. :-) >>Why bother with the quadratic >>formula? > No need to do so in this case since the polynomial factors so easily > (s=1 is always a solution). Who does bother with it? I did a Google search and found this .PDF file I found a few other .PDF files which show the quadratic equation used in cases where you can't just divide p/q, but in the case of the amoebe which either divides into two, or just dies, it seems easier to just calculate p/q. -- ------------------------------- Patrick D. Rockwell === Subject: Re: Probability Of Ultimate Extinction. > Let's say that you have an amoeba which has a probability q of dying without > divinding, > and probability p of spliting into two. What is the probability of ultimate > extinction? As a side note, some say that cells have an inherent average number of divisions before senisence, or the inability to subdivide. Some research seems to indicate that it may be possible to optimize that average, reducing the standard deviation. === Subject: Re: Probability Of Ultimate Extinction. >Let's say that you have an amoeba which has a probability q of dying without >divinding, >and probability p of spliting into two. What is the probability of ultimate >extinction? >I know that the recursion formula for extinction after n cycles is >P(n)=q+p*P(n-1)^2 where P(0)=q. You have not said what a cycle is, but assusming p+q=1 then if you are looking for the probability of ultimate extinction, u, then it is the probability of going extinct without dividing plus the probability that it divides and both halves go extinct: u = q +p*u^2 >of 3 or 4) is that >the probability of ultimate extinction involves using the quadratic formula >in the following way. >1±(1-4*p*q)^0.5 >--------------- You should be dividing by 2p not 2q. >where the minimum of the two values achieved by the quadratic equation is >the desired probability. >What I'd like to know is, why bother? I've discovered that in the above >problem, the probability >of ultimate extinction is p/q = 1 if p>=q. Why bother with the quadratic >formula? Perhaps you have swapped p and q somewhere If p+q=1 then the quadratic gives the lower of (1-p)/p and 1, depending on whether p>=1/2 or p<=1/2 which makes sense and in the latter case confirms what is obvious. === Subject: Speaking of Attacking the Conclusions JSH constantly whines that everyone is attacking his conclusions and that this is a fallacy because his suppositions and logic are intact. Aside from the fact that this isn't true (people have been attacking his logic and pointing out gaping holes) it's funny to point out that he does the same thing. All he'll do is attack the conclusions posters have drawn based on his inacuracies. He won't actually directly attack or even acknowledge the points they make. In essence he's attacking their conclusions. === Subject: Re: Help soving this problem? still think the problem is way too hard for liberal arts admitance test :) Fletch > So each point must solve the line equation (since they both lie on it) > I.e. > For point (M,N), we have: > M=N/2 - 2/5 > And for (M + P, N+4), we have: > M+P = (N+4)/2 -2/5 > We substitute the value for M from our first equation into our second > equation: > (N/2 - 2/5) + P = (N+4)/2 -2/5 > implies > N/2 - 2/5 + P = N/2 + 2 - 2/5 > implies > P = 2 > -Darren > Trying to get back into school after many years away. I have to take > an admitance exam which I am currently studying for. Here is a problem > on a sample exam that is stumping me . I belive it is a y intercept > problem Any ideas on how to solve? I am thinking you solve for y but > that didn't seem to work for me. > M and N are the X and Y coordinates, respectively, of a point in a > coordinate plane. If the points (M,N) and (M + P, N +4) both lie on > the line defined by the equation X = y/2 - 2/5, what is the value of > P? === Subject: Re: Help soving this problem? I don't think I follow you... Fletch >Trying to get back into school after many years away. I have to take >an admitance exam which I am currently studying for. Here is a problem >on a sample exam that is stumping me . I belive it is a y intercept >problem Any ideas on how to solve? I am thinking you solve for y but >that didn't seem to work for me. > >M and N are the X and Y coordinates, respectively, of a point in a >coordinate plane. If the points (M,N) and (M + P, N +4) both lie on >the line defined by the equation X = y/2 - 2/5, what is the value of >P? > M = N/2 - 2/5 > M+P = (N+4)/2 - 2/5 > Subtract the first from the second. === Subject: Re: Help soving this problem? >Trying to get back into school after many years away. I have to take >an admitance exam which I am currently studying for. Here is a problem >on a sample exam that is stumping me . I belive it is a y intercept >problem Any ideas on how to solve? I am thinking you solve for y but >that didn't seem to work for me. >M and N are the X and Y coordinates, respectively, of a point in a >coordinate plane. If the points (M,N) and (M + P, N +4) both lie on >the line defined by the equation X = y/2 - 2/5, what is the value of >P? > 1. Solve your equation for y to get the form y = mx + b and read off > the slope m. > 2. Compute the slope between your two points and set it equal to your > value of m. > 3. Solve for P > --Lynn === Subject: Question about numbers in arithmetic progression Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. What is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? Rich === Subject: Re: Question about numbers in arithmetic progression >Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. What >is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? Regardless of gcd, the trivial solution applies. The minimum value is 1/max(|x_1|,|x_n|). --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Question about numbers in arithmetic progression >>Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. >What >>is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? >Regardless of gcd, the trivial solution applies. >The minimum value is 1/max(|x_1|,|x_n|). I intended that the a_i be integers. My apologies for not making this clear. Rich === Subject: Re: Question about numbers in arithmetic progression > Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. What > is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? Well, that would depend on x_1, n, and the common difference d, so I suppose you want some sort of bound (I don't think you'll get an exact expression) in terms of those parameters, yes? If x_1 & x_2 have any common factor then all the x_i have it, so it's enough to assume x_1 & x_2 are relatively prime. Let's further assume 0 < x_1 < x_2. Then certainly there's a solution with |a_1| < x_2 and |a_2| < x_1 and a_i = 0 for i = 3, .... - that gives 2 x_1 + d - 2 as a bound, independent of n. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Question about numbers in arithmetic progression >> Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. >What >> is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? >Well, that would depend on x_1, n, and the common difference d, >so I suppose you want some sort of bound (I don't think you'll get >an exact expression) in terms of those parameters, yes? Add the condition 0> Suppose x_1,..,x_n are in arithmetic progression and gcd(x_1,..,x_n)=1. > What >> is the minimum value of |a_1|+...+|a_n| where a_1*x_1+...+a_n*x_n=1? >Well, that would depend on x_1, n, and the common difference d, >so I suppose you want some sort of bound (I don't think you'll get >an exact expression) in terms of those parameters, yes? > Add the condition 0 to above. Here is what I get for m: > m(2001,2,5)=501 [251*2001-250*2009=1] > m(20001,2,5)=5001 [2501*20001-2500*20009=1] > m(200001,2,5)=50001 [25001*200001-25000*200009=1] > m(1001,3,5)=169 > m(10001,3,5)=1669 > m(100001,3,5)=16669 > m(1001,10,7)=35 > m(10001,10,7)=335 > m(100001,10,7)=3335 > So it seems as though there *may* be an exact expression for these parameters. > Others, like m(1001,31,7), ... are not at all obvious. > I have a method that seems to compute m(x_1,...,x_n) and am looking for inputs > with known minimums to test against, hence the question. If you apply the Jacobi method to a_1*x_1+...+a_n*x_n=1 you get an equation of the form x_1(a_1 + a_2 + .. a_n) + d(a_2 + 2a_3 + + (n-1)a_m) =1 x_1 A +dB = 1 If P/Q is the penultimate convergent of x_1/d then x_1 A +dB = 1 has a solution if you choose appropriate signs for P and Q. a_2 + 2a_3 + + (n-1)a_n = P => |a_n| = INT(p/(n-1)) and some |a_j| (2<= j |a_1| = |Q| + |1| +|INT(P/(n-1)) This gives minima which agree with your results. What values did you get for m(1001,31,7) etc. ? What is the basis of your method ? === Subject: Re: Question about numbers in arithmetic progression I. M. Davidson wote: >This gives minima which agree with your results. But doesn't prove these are the actual minima, right? >What values did you get for m(1001,31,7) etc. ? I get: m(1001,31,7)=87 m(10001,31,7)=1417 m(100001,31,7)=6462 m(1000001,31,7)=107548 m(10000001,31,7)=322585 and, curiously, m(10^12+1,31,7)=64516129040 m(10^22+1,31,7)=322580645161290322585. >What is the basis of your method ? Casting in a small pond full of big fish and hoping for the best? Seriously, I don't have a answer for your question. I will try to write a good description and post it here, though. Rich === Subject: Re: Question about numbers in arithmetic progression > I. M. Davidson wote: >This gives minima which agree with your results. > But doesn't prove these are the actual minima, right? Precisely. I was thinking that the CF method gives the smallest (in your sense) solution to AX + dB =1 and that the smallest solution to a_2 + d*a_2 + + (n-1)da_n must involve the a_j wth the largest coefficient and some other a_k =1. Bit I'm not sure if this necessarily follows. >What values did you get for m(1001,31,7) etc. ? > I get: > m(1001,31,7)=87 I get 83 as the minimum here a_1 = 45, a_5 =-1,a_6 = -37 i.e 45*1001 +1125*(-1) +1187(-37) =1 So your method does not find the minima in all cases, even though the actual minimum may be less than 83. > m(10001,31,7)=1417 Here I get 1954. A considerable difference. What are the actual values you get for the a's ? Whatever they are they must satisfy the Jacobi reduction. > m(100001,31,7)=6462 I also have a different value here. What were the a's in this case ? === Subject: Re: Question about numbers in arithmetic progression >> I. M. Davidson wote: >> >>This gives minima which agree with your results. >> But doesn't prove these are the actual minima, right? >Precisely. I was thinking that the CF method >gives the smallest (in your sense) solution >to AX + dB =1 and that the smallest solution >to a_2 + d*a_2 + + (n-1)da_n must involve the >a_j wth the largest coefficient and some >other a_k =1. Bit I'm not sure if this >necessarily follows. >>What values did you get for m(1001,31,7) etc. ? >> I get: >> m(1001,31,7)=87 >I get 83 as the minimum here >a_1 = 45, a_5 =-1,a_6 = -37 >i.e 45*1001 +1125*(-1) +1187(-37) =1 >So your method does not find the minima >in all cases, even though the actual >minimum may be less than 83. Yes. This is not a real suprise as the problem is quite difficult. I'll have to work on this one a bit... >> m(10001,31,7)=1417 >Here I get 1954. A considerable >difference. >What are the actual values you >get for the a's ? 1417: -715*10001+689*10187+5*10125+8*10156=1 1433: -723*10001+689*10187+13*10125+8*10032=1 1433: -715*10001+689*10187-8*10094+21*10125=1 1433: -715*10001+689*10187+13*10125+8*10094=1 1443: -728*10001+18*100063+689*100187+8*10094=1 >Whatever they are they must satisfy >the Jacobi reduction. >> m(100001,31,7)=6462 >I also have a different value here. >What were the a's in this case ? 6462: 3234*100001-1*100094-3222*100187-5*100125=1 6472: 3223*100001+16*100063-3222*100187-11*100156 Rich === Subject: Re: Question about numbers in arithmetic progression >> I. M. Davidson wote: >> m(1001,31,7)=87 >I get 83 as the minimum here >a_1 = 45, a_5 =-1,a_6 = -37 >i.e 45*1001 +1125*(-1) +1187(-37) =1 >So your method does not find the minima >in all cases, even though the actual >minimum may be less than 83. > Yes. This is not a real suprise as the problem is quite difficult. I'll have > to work on this one a bit... realise that I had forgotten about the ambiguity arising when a CF ends in a 1 - as is the case with 1001/31 and 100001/31. This gave me values much greater than they should have been. >> m(10001,31,7)=1417 >Here I get 1954. A considerable >difference. >What are the actual values you >get for the a's ? > 1417: -715*10001+689*10187+5*10125+8*10156=1 > 1433: -723*10001+689*10187+13*10125+8*10032=1 > 1433: -715*10001+689*10187-8*10094+21*10125=1 > 1433: -715*10001+689*10187+13*10125+8*10094=1 > 1443: -728*10001+18*100063+689*100187+8*10094=1 >Whatever they are they must satisfy >the Jacobi reduction. In this case I now get 1411 (-712)*10001 + 699*10187 = 1 >> m(100001,31,7)=6462 >I also have a different value here. >What were the a's in this case ? > 6462: 3234*100001-1*100094-3222*100187-5*100125=1 > 6472: 3223*100001+16*100063-3222*100187-11*100156 Here too, I get the much lower value than before of 6458 3232*100001 +(-1)*100156) + (-3225)*10187 =1 As the Jacobi reduction is identical to the original equation, i.e. whenever the original equation is equal to 1 the Jacobi reduction is also equal to 1 and vice versa. As the CF expansion gives the minimum solution to AX +dY =1,x',y' the minimum a's must be given by the minimum solution of x' = a_1 +a_2 +.. + a_n y' = a_2 + 2*a_3 +.. + (n-1)*a_n choosing the appropriate signs for x',y' Given the minimum a's they must satisfy the Jacobi reduction and x',y'must be minima. So your initial intuition was correct - there is a formula (of sorts) for the minimum solutions. === Subject: Re: Question about numbers in arithmetic progression >So your initial intuition was correct - >there is a formula (of sorts) for the minimum > solutions. Rich === Subject: Re: Question about numbers in arithmetic progression > I. M. Davidson wote: >This gives minima which agree with your results. > But doesn't prove these are the actual minima, right? Precisely. I was thinking that the CF method gives the smallest (in your sense) solution to AX + dB =1 and that the smallest solution to a_2 + d*a_2 + + (n-1)da_n must involve the a_j wth the largest coefficient and some other a_k =1. Bit I'm not sure if this necessarily follows. >What values did you get for m(1001,31,7) etc. ? > I get: > m(1001,31,7)=87 I get 83 as the minimum here a_1 = 45, a_5 =-1,a_6 = -37 i.e 45*1001 +1125*(-1) +1187(-37) =1 So your method does not find the minima in all cases, even though the actual minimum may be less than 83. > m(10001,31,7)=1417 Here I get 1954. A considerable difference. What are the actual values you get for the a's ? Whatever they are they must satisfy the Jacobi reduction. > m(100001,31,7)=6462 I also have a different value here. What were the a's in this case ? === Subject: A first countable separable space that is not second countable I need to see an example of a topological space that is first countable and separable but not second countable. Are there any? === Subject: Re: A first countable separable space that is not second countable > I need to see an example of a topological space that is first countable > and separable but not second countable. Doesn't R topologized with basis the sets [a,b) satsify this? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: A first countable separable space that is not second countable Never mind. I found one. > I need to see an example of a topological space that is first countable and > separable but not second countable. > Are there any? === Subject: spherical triangle How is it possible to compute this: a spherical triangle is given by its three sides (three angles between the three vectors). Somewhere in the triangle lies the origin (North (angle between vector to point and North pole) to the North pole. How can one computes the distance (angle) to the third point? The coordindates of the points are not given. S. Nurbe === Subject: Re: spherical triangle This problem is easy to solve in spherical trig. Draw triangle ABC containing the North Pole N inside it. Use the Law of Cosines for Arcs to solve for angle ABC: cos(AC) = cos(AB)cos(BC)+sin(AB)sin(BC)cos(ABC) where three letters indicate an angle and two letters indicate an arc. All arcs in the above are known so the equation can be solved for the angle ABC. Now draw triangle ABN where the arcs AN and BN are also known. Solve for angle ABN by the same method as above. Subtract angle ABN from angle ABC getting angle NBC. Now draw triangle NBC and apply the Law of Cosines once more: cos(CN) = cos(BC)cos(BN)+sin(BC)sin(BN)cos(NBC) where this time BC, BN, and NBC are known and CN is to be calculated. --OL === Subject: Re: spherical triangle <22415-41B3BF7B-38@storefull-3254.bay.webtv.net> Now I have just one more problem. I have to put the computation of CN in one equation (which is not that problem) in order to put this equation in equation system which I let compute by matlab. The problem is that it computes to infinity as long as there are arccos terms in it (I don't no why). But I can't find a way to express the equation of CN without arccos! Anybody know? === Subject: Re: spherical triangle ETAsAhQxPOrKX1RFHCAvF1XT7U84eqlb8gIUNuLRyTsa85hodw7irKB4c7Gnt5w= Suppose you have cos(ABC) and cos(ABN). If NBC isthe difference between ABC and ABN, then an elemenary trigonometric formula gives: cos(NBC) = cos(ABC)cos(ABN) + sin(ABC)sin(ABN) where the sines are positive and figured out in the usual way from the cosines. That avoid using inverse trig functions to get the angles. But you still must take the inverse cosine of cos(CN) to get CN itself; no way around that. Condensing things into one quation will ive a monstrously complicated result. Make Matlab figure out the cosines of the angles first and then go after the arc CN. --OL === Subject: Re: spherical triangle > a spherical triangle is given by its three sides (three angles between > the three vectors). Somewhere in the triangle lies the origin (North > (angle between vector to point and North pole) to the North pole. How > can one computes the distance (angle) to the third point? > The coordindates of the points are not given. Hint: Stereographic conformal projection from South pole rays on to a plane tangent to North pole. === Subject: Re: spherical triangle Thought that it would be solvable by the conformally equivalent triangle obtained by projecting the geodesic sides on tangent plane at one pole from another pole. It is not so, and sorry for this indication.. The straight forward spherical trigonometry calculation outlined by Oscar Lanzi is easier.. often the old problem (sin(a)/sin(A) equalling to no known property of the traiangle, that it is only the Jacobian sin(a)~sin(A) etc.) seems to clog up an easy solution (to me), had posted about it earlier also. === Subject: Re: spherical triangle > a spherical triangle is given by its three sides (three angles between > the three vectors). Somewhere in the triangle lies the origin (North > (angle between vector to point and North pole) to the North pole. How > can one computes the distance (angle) to the third point? > The coordindates of the points are not given. > Hint: Stereographic conformal projection from South pole rays on to a > plane tangent to North pole. Do you have maybe some more information about this (because I can't find s.th. proper)? Or maybe another way? === Subject: Re: Real Analysis >> If f is a continuous function and e >0, can you find a linear combination >> of >> 1, x, x^2, x^3, ... which approximates f on [-1,1]. >> Wouldn't the Taylor series expansion for f about x=0 fit the bill? >No. Not every continuous function has a Taylor series. > Even if the function has a Taylor series, that series doesn't > necessarily converge anywhere. And you'd want the series to converge > uniformly to f on the whole interval [-1,1]: that's not true even > for functions that are real-analytic. >What is needed >is a proof of Weierstrass's approximation theorem. > Indeed. But notice that the question was can you find, not > prove that there is... Oh! I gave a proof of WAT that shows how to construct an appropriate poly. === Subject: JSH: Look at it backwards I usually start with one polynomial and then talk about dividing 49 off from it, but here I'll start *after* 49 has been divided off: S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 and consider the factorization S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) and the assertion that the d's must have factors w_1(x), w_2(x), and w_3(x) that multiply to give w_1(x) w_2(x) w_3(x) = 7. But there's no factors of 7 anywhere. So why should the d's have functions that are factors of 7? Ok, maybe that seems unfair, as there could be LOTS of different functions you can plug in for the c's and d's, but why should ANY of them have functions of x that are factors of 7? The equation has no memory. You have a memory, so if I start with the polynomial multiplied by 49, then you can say to yourself that there's some dependency on 7. But it's a mirage. Mathematically a constant multiple is not a big deal. It's just a constant multiple that you can divide off, leaving a result that--you guessed it--has no memory of the multiple! One of the weirder things about the discussions here, which I assume escapes most of you is some fascinating belief that the equation has a memory. You see something like P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 and your brain apparently HOLLERS at you that 49 is still there, but no, it's gone. When I show something like P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7) you brain insists that 7 is still there. I mean just LOOK! You can see 7 dividing two of the a's and for God's sake!!! There's a 7 in that last factor, you see it, don't you? In 5a_3(x) + 7. Come on, there's a 7 RIGHT THERE! Of course 7 is still there!!! So some poster comes at you claiming that 49 divides off from P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) in a way that varies dependent on the value of x, and your brain tells you, OK!!! You have a memory. To you 7 is still there, even though the factor is divided off. Follwing the sci.math'ers insistent raving you get P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) + 7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x)) where the w's are factors of 7, so that you're left with functions of 7. But they would STILL be there with the factorization of S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 inexplicably still there despite the polynomial not having any factors of 7. Your eyes fool you. Your memory betrays you. The math doesn't bother with such nonsense. A multiple of a polynomial can be divided off and it's gone. It doesn't leave a trace. In a way what's happening now is an instructive lesson in the limitations of the human brain. Your brains SEE something, and insistently tell you that 7 is STILL THERE, and so the arguments go on for years. After all, you can SEE the 7's. Come on, who's fooling who, right? Dammit. You can see the 7's in there, can't you? James Harris === Subject: Re: JSH: Look at it backwards > I usually start with one polynomial and then talk about dividing 49 > off from it, but here I'll start *after* 49 has been divided off: > S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 > and consider the factorization > S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) > and the assertion that the d's must have factors > w_1(x), w_2(x), and w_3(x) > that multiply to give w_1(x) w_2(x) w_3(x) = 7. Let w_1(x)=w_2(x)=w_3(x) = 7^(1/3). Then the w_i are factors of the d_i in the ring of algebraic numbers. (Are they factors in any other ring? Who knows? Who cares? No one except you has made such a claim.) > But there's no factors of 7 anywhere. So why should the d's have > functions that are factors of 7? the d's have functions that are factors of 7 is gibberish. -William Hughes === Subject: Re: JSH: Look at it backwards >I usually start with one polynomial and then talk about dividing 49 >off from it, but here I'll start *after* 49 has been divided off: >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >and consider the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >and the assertion that the d's must have factors >w_1(x), w_2(x), and w_3(x) >that multiply to give w_1(x) w_2(x) w_3(x) = 7. >But there's no factors of 7 anywhere. So why should the d's have >functions that are factors of 7? You are getting maximally confused. In your original approach, you are factoring P(x) as a polynomial in 5, not as a polynomial in x. The constant term with respect to 5 and the constant term with respect to x are different. It was a stupid mistake to think you were simplifying by treating 5 as if it were the polynomial variable in the first place. And that is what is causing your confusion here. >Ok, maybe that seems unfair, as there could be LOTS of different >functions you can plug in for the c's and d's, but why should ANY of >them have functions of x that are factors of 7? >The equation has no memory. The memory problem here is, you have forgotten what you started with. To see this, you need to go back to the way you were doing things originally: for example, P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) To put this in terms of what you are doing now: replace x in this expression by 5, and replace m by x, and replace f by 7. Also replace u by 1. This gives P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), and you can clearly see that, as a polynomial in 5, the constant term is 49 * 7 = 7^3, not 49 * 22. Go back to the expression above for P(m). The constant term with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to 49*22 when u = 1, f = 7, and x = 5. But the constant term with respect to x is u^3 f^3. You have forgotten this and gotten mixed up, that's all. >You have a memory, so if I start with the polynomial multiplied by 49, >then you can say to yourself that there's some dependency on 7. But >it's a mirage. >Mathematically a constant multiple is not a big deal. It's just a >constant multiple that you can divide off, leaving a result that--you >guessed it--has no memory of the multiple! >One of the weirder things about the discussions here, which I assume >escapes most of you is some fascinating belief that the equation has a >memory. >You see something like >P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 >and your brain apparently HOLLERS at you that 49 is still there, but >no, it's gone. No - my brain does no such hollering. >When I show something like >P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7) Key thing to note here: as I said, the polynomial variable with respect to which you are factoring is 5, not x. See? >you brain insists that 7 is still there. >I mean just LOOK! You can see 7 dividing two of the a's and for God's >sake!!! >There's a 7 in that last factor, you see it, don't you? >In 5a_3(x) + 7. Come on, there's a 7 RIGHT THERE! Of course 7 is >still there!!! >So some poster comes at you claiming that 49 divides off from >P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) >in a way that varies dependent on the value of x, and your brain tells >you, OK!!! >You have a memory. To you 7 is still there, even though the factor is >divided off. As a polynomial in 5, it's still there. It is not a coincidence that 22 = 3 * 5 + 7. >Follwing the sci.math'ers insistent raving you get >P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) + >7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x)) >where the w's are factors of 7, so that you're left with functions of >But they would STILL be there with the factorization of >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >inexplicably still there despite the polynomial not having any factors >of 7. >Your eyes fool you. Your memory betrays you. No, it's YOUR memory that is the problem. You have forgotten how you were factoring: not with respect to x, but with respect to 5. >The math doesn't bother with such nonsense. A multiple of a >polynomial can be divided off and it's gone. >It doesn't leave a trace. >In a way what's happening now is an instructive lesson in the >limitations of the human brain. Your brains SEE something, and >insistently tell you that 7 is STILL THERE, and so the arguments go on >for years. >After all, you can SEE the 7's. Come on, who's fooling who, right? You are hopelessly tangled in your own dimwitted oversimplification. You have forgotten what you were doing. >Dammit. >You can see the 7's in there, can't you? Sure. On *both* sides. Just look at: P(x)/49 = (x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7, and note that the constant term is ... 7, just like it should be, if you regard 5 as the polynomial variable! Nora B. >James Harris === Subject: Re: JSH: Look at it backwards >I usually start with one polynomial and then talk about dividing 49 >off from it, but here I'll start *after* 49 has been divided off: >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >and consider the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >and the assertion that the d's must have factors >w_1(x), w_2(x), and w_3(x) >that multiply to give w_1(x) w_2(x) w_3(x) = 7. >But there's no factors of 7 anywhere. So why should the d's have >functions that are factors of 7? > You are getting maximally confused. In your original > approach, you are factoring P(x) as a polynomial in 5, > not as a polynomial in x. The constant term with > respect to 5 and the constant term with respect to > x are different. It was a stupid mistake to think you > were simplifying by treating 5 as if it were the > polynomial variable in the first place. And that is what > is causing your confusion here. There is only one variable Nora Baron, so what other variable would you have listed with S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) considering that S(x) = 300125x^3 - 18375 x^2 - 360 x + 22? So who's confused? >Ok, maybe that seems unfair, as there could be LOTS of different >functions you can plug in for the c's and d's, but why should ANY of >them have functions of x that are factors of 7? >The equation has no memory. > The memory problem here is, you have forgotten what you > started with. To see this, you need to go back to the way you > were doing things originally: for example, Now note, the polynomial S(x) doesn't have 7 as a factor, as this time I'm starting with the result of dividing off the multiple. So the poster Nora Baron (actually a guy as revealed in another post when he ended with a male name) is trying to show how 7 gets back into the expression. That is, he's trying to prove that the factorization DOES have a memory of the multiple 49. > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > To put this in terms of what you are doing now: replace x in this > expression by 5, and replace m by x, and replace f by 7. Also > replace u by 1. This gives > P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), > and you can clearly see that, as a polynomial in 5, the constant > term is 49 * 7 = 7^3, not 49 * 22. No. It can be factored with respect to 5 and 7, but 5 is not a variable. The polynomial variable is x. Now, I've explained lots of times to the Nora Baron poster, and what I want you all to consider is that the poster isn't really that dense, but instead knows you better than you know yourselves. Essentially the basic strategy is just to disagree. Time after time, and even in answer to surveys that I've done, readers who normally lurk will admit that they primarily rely on the fact that people argue with me, assuming that if I were right, then others wouldn't disagree! So, for sci.math'ers like Nora Baron, the strategy is clear--just disagree. They often don't even TRY to actually make mathematical senze. > Go back to the expression above for P(m). The constant term > with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to > 49*22 when u = 1, f = 7, and x = 5. But the constant term with > respect to x is u^3 f^3. You have forgotten this and gotten mixed > up, that's all. Now the expressions actually is S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 where the factorization S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) is being considered. Now how do you get 7 into that? Well, that's not the issue for the poster! All he has to do is disagree. For most readers that's what works. Well, that enough here. James Harris === Subject: Re: JSH: Look at it backwards >snip >There is only one variable Nora Baron . . . Here I agree with Mr. Harris. I once posted that there are evidently a number of real people living in the United States that really are named Nora Baron, but those individuals are constant while our Nora Baron may indeed be the only one that is a variable -- although that property has not yet been established for sure. === Subject: Re: JSH: Look at it backwards >snip >There is only one variable Nora Baron . . . > Here I agree with Mr. Harris. I once posted that there are evidently > a number of real people living in the United States that really are > named Nora Baron, but those individuals are constant while our Nora > Baron may indeed be the only one that is a variable -- although that > property has not yet been established for sure. LOL. I've been having a good time today. Lots of fun stuff, and some neat accomplishments. And sci.math posters are actually getting a LOT more entertaining with some of their replies. It's a hoot. The real work though is in working through the details of *explaining* my result. Which is an interesting problem, which is also why I need so many posts as I just throw a bunch of ideas at the problem until I find something that sticks. It's basically how I found my math results, brainstorming, but with expository style. I think I'm almost done. I've got that mathematicians ignoring my work are passive-aggressive, and using a passive-aggressive style. I have that the work itself resolves down to acceptance of some VERY BASIC mathematics known for thousands of years. And I have the explanation for why some posters obsessively reply to my posts as they've learned that simply disagreeing is their most potent tactic. The information can be played out in many different combinations, and as each one plays out, I use various tools to look for impact. Eventually I'll get the right combination. The key is what I'm looking for, and I'm hot on the hunt now. It's just so, terribly exciting! James Harris === Subject: Re: JSH: Look at it backwards > The key is what I'm looking for, and I'm hot on the hunt now. It's > just so, terribly exciting! Be sure and keep your zipper up. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Look at it backwards >>I usually start with one polynomial and then talk about dividing 49 >off from it, but here I'll start *after* 49 has been divided off: >> >>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >> >>and consider the factorization >> >>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >> >>and the assertion that the d's must have factors >> >>w_1(x), w_2(x), and w_3(x) >> >>that multiply to give w_1(x) w_2(x) w_3(x) = 7. >> >>But there's no factors of 7 anywhere. So why should the d's have >>functions that are factors of 7? >> >> You are getting maximally confused. In your original >> approach, you are factoring P(x) as a polynomial in 5, >> not as a polynomial in x. The constant term with >> respect to 5 and the constant term with respect to >> x are different. It was a stupid mistake to think you >> were simplifying by treating 5 as if it were the >> polynomial variable in the first place. And that is what >> is causing your confusion here. >There is only one variable Nora Baron, I didn't say there were other variables in your present polynomial. I was describing its origin, when there were 4 variables, which explains why you are now confused about the whole thing. > so what other variable would >you have listed with >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >considering that >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22? >So who's confused? No question about that. You are. >>Ok, maybe that seems unfair, as there could be LOTS of different >>functions you can plug in for the c's and d's, but why should ANY of >>them have functions of x that are factors of 7? >> >>The equation has no memory. >> >> The memory problem here is, you have forgotten what you >> started with. To see this, you need to go back to the way you >> were doing things originally: for example, >Now note, the polynomial S(x) doesn't have 7 as a factor, I DIDN'T SAY IT DID, ASSHOLE > as this time >I'm starting with the result of dividing off the multiple. >So the poster Nora Baron (actually a guy as revealed in another post >when he ended with a male name) is trying to show how 7 gets back into >the expression. It gets back in in the constant term, because you have been factoring this function as if it were a polynomial in 5. >That is, he's trying to prove that the factorization DOES have a >memory of the multiple 49. Not at all. The 49 is obviously gone. But the constant term at the end, when P(x)/49 is viewed as a polynomial in 5 [your idea, not mine!], is still there, and it is 7, not 22. >> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) >> To put this in terms of what you are doing now: replace x in this >> expression by 5, and replace m by x, and replace f by 7. Also >> replace u by 1. This gives >> P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), >> and you can clearly see that, as a polynomial in 5, the constant >> term is 49 * 7 = 7^3, not 49 * 22. >No. It can be factored with respect to 5 and 7, but 5 is not a >variable. You have treated it exactly as such. When you factor P(x) in the form P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7), you are NOT factoring it as a polynomial in x. If you were, the terms a_1(x), etc., would themselves be polynomials in x. As you yourself have pointed out MANY TIMES, this is not a polynomial factorization when considered as a function of x. (In fact you have claimed that such 'nonpolynomial factorization' was one of your great conceptual breakthroughs. Now you want to deny it ???). However it CLEARLY has the FORM of a factorization in 5, where 5 is treated as a polynomial variable. >The polynomial variable is x. It most certainly is not [yes, I know you are going to howl about this, but I stand by what I say]. When you factor a b x^2 + (a + b) x + 1 as (a x + 1)(b x + 1), THEN you are factoring in terms of the polynomial variable x. The coefficients of x in the factors are functions of a and b, NOT functions of x, as in your present factorization of P(x). Have you really no recollection of how you got started on this track? Originally, x was m, and 5 was x. At that time you were considering factors like a_1(m) x + u f, etc. This was a factorization in terms of the polynomial variable x, with coefficients which were algebraic integer functions of m. Just plug in x = 5, u = 1, f = 7, and m = x, and you have your present factorization - as a polynomial in 5, not in x !! It's easy, actually, to see how you have gotten confused, especially if (unlike real mathematicians) you have a poor memory. >Now, I've explained lots of times to the Nora Baron poster, and what >I want you all to consider is that the poster isn't really that dense, >but instead knows you better than you know yourselves. >Essentially the basic strategy is just to disagree. I only disagree when you are wrong. >Time after time, and even in answer to surveys that I've done, readers >who normally lurk will admit that they primarily rely on the fact that >people argue with me, assuming that if I were right, then others >wouldn't disagree! I don't recall people saying that. What lurkers do seem to think, when they speak up, is that YOUR math is vague, hand-waving and illogical. That's not my fault. Plus they seem to think that your habit of making nasty personal attacks damages your credibility. That's not my fault either. >So, for sci.math'ers like Nora Baron, the strategy is clear--just >disagree. Again, I only disagree when you are wrong. Of course, that happens to be virtually all the time. >They often don't even TRY to actually make mathematical senze. This is gratuitous and false in my case, as you well know. Also in the cases of Dik Winter, Rick Decker, Arturo Magidin, 'Rupert', Will Twentyman, Dale Hall, and others. >> Go back to the expression above for P(m). The constant term >> with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to >> 49*22 when u = 1, f = 7, and x = 5. But the constant term with >> respect to x is u^3 f^3. You have forgotten this and gotten mixed >> up, that's all. >Now the expressions actually is >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >where the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >is being considered. >Now how do you get 7 into that? Well, that's not the issue for the >poster! It was explained in what I posted. You need to consider how you arrive at S(x), and for that you need to consider the form of the original function P(x): as a polynomial in 5, it has constant term 7^3. I thought you were simply confused on this point, having forgotten how you got here. I am now inclined to think you are trying to deny the obvious correct explanation, though I am not sure why. Plus throughout your entire reply you seem to be playing to the grandstand. You whine, exaggerate, and attempt to discredit. The actual math you have hardly addressed at all. You seem to think that you can win mathematical arguments by an appeal to some great silent majority of people out there who you can cajole into thinking that we critics are evil, cheating liars who are jealous of your discoveries. (If that were true, one of us evil cheating liars would have published it long ago, of course giving no credit to James S.Harris. Funny, that hasn't happened. Wonder why? Why isn't anyone stealing your ideas?) You seem to have the impression that math is a popularity contest. It's not. The only way to win is through rigorous proof. You don't have one. That's the main reason you are not winning. >All he has to do is disagree. >For most readers that's what works. Right. Go ahead and ask your faithful admiring grandstand if that's what works. >Well, that enough here. Yuh - Duh - that enough ! Nora B. >James Harris === Subject: Re: JSH: Look at it backwards > So the poster Nora Baron (actually a guy as revealed in another post > when he ended with a male name) is trying to show how 7 gets back into > the expression. What difference does it make what Nora's gender is? What does it have to do with math? Has anybody but you made note of the question of Nora's gender? No, because it is not relevant. Your obsession with her (or his) gender just screams mental illness. It is (probably) just a pseudonym, nothing more. So what? My pseudonym is o[CapitalYAcute]in, so do you claim that I am pretending to be a Norse god? Well, nobody cares!!! Idiot!!! === Subject: Re: JSH: gametes So... James Harris got me wondering... which type of gametes do you actually produce, Nora Baron? === Subject: Re: JSH: gametes > So... James Harris got me wondering... which type of gametes do you actually > produce, Nora Baron? See how that works? You start by mocking JSH, and after a while you start to think like him. He's addictive. He's actually making some very good points about the social aspect of proof. Philosophers of math make the same arguments ... that a proof is whatever convinces a majority of working mathematicians; that what's considered a proof in one era is regarded as utterly lacking in rigor in another. And now he's got you caring about the gender of a palindromic poster. === Subject: Re: JSH: gametes > See how that works? You start by mocking JSH, and after a while you > start to think like him. He's addictive. He's actually making some very > good points about the social aspect of proof. Philosophers of math make > the same arguments ... that a proof is whatever convinces a majority of > working mathematicians; that what's considered a proof in one era is > regarded as utterly lacking in rigor in another. > And now he's got you caring about the gender of a palindromic poster. Yes. You are 100 correct in all that... but now I really do want to know... Damn it!!! === Subject: Re: JSH: gametes Nora B. Baron comes from a long line of palindromic forearmed folks; if you can make that stick for two generations, that's farther than I've gotten! > And now he's got you caring about the gender of a palindromic poster. --Advice, 0.05; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: JSH: Look at it backwards > I usually start with one polynomial and then talk about dividing 49 > off from it, but here I'll start *after* 49 has been divided off: > S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 > and consider the factorization > S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) Ok, at this point there is no indication as to what these c_i(x) and d_i(x) might be, aside from the obvious necessary properties to generate the coefficients of S(x) when the factors are multiplied together. Are they meant to be continuous functions? Continuous almost everywhere? What are their domain/range? > and the assertion that the d's must have factors > w_1(x), w_2(x), and w_3(x) > that multiply to give w_1(x) w_2(x) w_3(x) = 7. If they have such factors, then the range must not be in the algebraic integers. Note: I mention that since you usually say something about everything is in the algebraic integers. > But there's no factors of 7 anywhere. So why should the d's have > functions that are factors of 7? They can quite easily in the appropriate ring. For example, 22 has 7 as a factor in the rational numbers, algebraic numbers, reals, etc. What makes you say there's no factors of 7 anywhere? > Ok, maybe that seems unfair, as there could be LOTS of different > functions you can plug in for the c's and d's, but why should ANY of > them have functions of x that are factors of 7? *IN WHAT RING?* Do you not see that you are making some sort of assumption that you have not shared with us? > The equation has no memory. Huh? Why would you even speak of such a thing? > You have a memory, so if I start with the polynomial multiplied by 49, > then you can say to yourself that there's some dependency on 7. But > it's a mirage. Actually, you are the one who keeps talking about 7 with such a polynomial. > Mathematically a constant multiple is not a big deal. It's just a > constant multiple that you can divide off, leaving a result that--you > guessed it--has no memory of the multiple! Clue: you are think of things as being processes. Try thinking of them as being equivalent statements. You can talk about two different equations being equivalent, but they are different equations. The notion of a process, or memory of prior steps in the process, is an artifact of the work a person does to work towards a goal. The result in something like the above is simply a set of equivalent statements and some justification for them being equivalent. > One of the weirder things about the discussions here, which I assume > escapes most of you is some fascinating belief that the equation has a > memory. Then why are you the only one who talks about it? > You see something like > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > and your brain apparently HOLLERS at you that 49 is still there, but > no, it's gone. I see it quite clearly. On the left side of the equation. Where it was in the previous equation (if at all) is irrelevent. > When I show something like > P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7) > you brain insists that 7 is still there. > I mean just LOOK! You can see 7 dividing two of the a's and for God's > sake!!! I see three 7s, actually. > There's a 7 in that last factor, you see it, don't you? > In 5a_3(x) + 7. Come on, there's a 7 RIGHT THERE! Of course 7 is > still there!!! Precisely. > So some poster comes at you claiming that 49 divides off from > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) > in a way that varies dependent on the value of x, and your brain tells > you, OK!!! Wrong. Simple examples make it clear that, in general, this is how things work when viewed from the perspective of working on *factors* of a polynomial with a goal of keeping the terms of each factor in a particular ring. If you focus on terms of the factors or don't care about staying within a ring at any particular level, then it doesn't matter how you write it. > You have a memory. To you 7 is still there, even though the factor is > divided off. I see only /7, /7, and +7. > Follwing the sci.math'ers insistent raving you get > P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) + > 7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x)) > where the w's are factors of 7, so that you're left with functions of > 7. Functions of 7? Huh? > But they would STILL be there with the factorization of > S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 > inexplicably still there despite the polynomial not having any factors > of 7. So, it would appear that an appropriate definition for your functions at the top of your post, to tie it in with what's down here would be: given the a_i(x)s and w_i(x)s, then: c_i(x) = 5a_i(x)/w_i(x) and d_i(x)=7/w_i(x). However, that would mean that d_i(x)*w_i(x)=7, which does not suggest anything about d_i(x) having a factor of 7 as a factor. So, assuming that the d_i(x) was arrived at as above, and that the w_i(x) are the same in both cases, where is the notion of w_i(x) being a factor of d_i(x) coming from? > Your eyes fool you. Your memory betrays you. > The math doesn't bother with such nonsense. A multiple of a > polynomial can be divided off and it's gone. > It doesn't leave a trace. > In a way what's happening now is an instructive lesson in the > limitations of the human brain. Your brains SEE something, and > insistently tell you that 7 is STILL THERE, and so the arguments go on > for years. Apparently you did not understand any of the arguments, or you wouldn't be saying any of this. > After all, you can SEE the 7's. Come on, who's fooling who, right? > Dammit. > You can see the 7's in there, can't you? I see two division by 7's and a +7. Now, which 7s are supposed to still be there? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Look at it backwards > The equation has no memory. This is new! What the are you on about now? > You have a memory, so if I start with the polynomial multiplied by 49, > then you can say to yourself that there's some dependency on 7. But > it's a mirage. Mirage? Ummmmm..... Holy moly... I have no words for this... Ummmm.... OK, go on.... > Mathematically a constant multiple is not a big deal. Can you prove that assertion, and can you first define mathematically what a big deal is. Just joking, I know you are talking figuratively... but really James, do you have to add in all this goofy non-math talk in every post? > It's just a > constant multiple that you can divide off, leaving a result that--you > guessed it--has no memory of the multiple! What does divide off mean? I think I know what you mean, but can you not try to use standard terminology? > Your eyes fool you. Your memory betrays you. And Wiles was tricked in this way in his FLT proof, right? === Subject: Area between these curves.. A friend recently posed this question to me, and I'd like to be able to explain the answer: integrate[ (sec x)^2 - tan x ] from x=0 to x=1 Simple enough. the antiderivative of (sec x)^2 = tan x and the antiderivative of tan x = -ln(cos x) So, after calculating F(1) - F(0), the exact answer is: tan (1) + ln [cos(1)] Ok, but now say we were to integrate the same function, this time from x=0 to x=Pi. There is an asymptote in the given domain shared by both functions, at Pi / 2. 2 to Pi, the area is also infinite. Plugging x=Pi into -ln(cos x) results in an imaginary. Would it be correct to say, that the area between the curves from 0 to Pi is infinite? Or is it really undefined because of the imaginary and/or the asymptote (non-continuous, therefore non-integrable)? Or, should the answer be infinity, except at x = Pi / 2? --Samantha === Subject: Re: Area between these curves.. I wasn't aware that an absolute value was taken in the -ln(|cos x|) part, but this makes sense to me now upon looking closely at it. And I will be looking into complex integration. --Samantha === Subject: Re: Area between these curves.. >A friend recently posed this question to me, and I'd like to be able to >explain the answer: >integrate[ (sec x)^2 - tan x ] from x=0 to x=1 >Simple enough. >the antiderivative of (sec x)^2 = tan x >and >the antiderivative of tan x = -ln(cos x) If you wish to find the antiderivative of 1/u in domains where u may be negative, you get ln(|u|) + c. So you would have -ln(|cos(x)|). >So, after calculating F(1) - F(0), the exact answer is: >tan (1) + ln [cos(1)] >Ok, but now say we were to integrate the same function, this time from x=0 >to x=Pi. >There is an asymptote in the given domain shared by both functions, at Pi / >2 to Pi, the area is also infinite. The integral is called an improper integral because of the asymptote, and you are correct, it diverges to infinity. >Plugging x=Pi into -ln(cos x) results in an imaginary. No, it doesn't, see the note above. >Would it be correct to say, that the area between the curves from 0 to Pi is >infinite? That is common terminology, or you could say the integral is divergent. > Or is it really undefined because of the imaginary No. > and/or the asymptote (non-continuous, therefore non-integrable)? It is not Riemann integrable, nor does the improper Riemann integral converge. >Or, should the answer be infinity, except at x = Pi / 2? No. >--Samantha --Lynn === Subject: Re: Area between these curves.. | Ok, but now say we were to integrate the same function, this time from x=0 | to x=Pi. | | There is an asymptote in the given domain shared by both functions, at Pi / | 2. | | 2 to Pi, the area is also infinite. Intepreting it physically (area) then it's infinite (or undefined!) | | Plugging x=Pi into -ln(cos x) results in an imaginary. Not too sure about this one. Try read up something in the field of complex integration. eg, treat x as z. This is university level maths ok!! :) === Subject: JSH: Limitations in thinking If you follow logic, then there's no room to argue with me. But if you rely on people relying on their intuitions and what they think you see, then you can argue with me for years, as some sci.math'ers have done. For me the realization of what was happening came just a while back when I talked with Professor McKenzie at Vanderbilt University. I traced out the argument explaining the paper Advanced Polynomial Factorization. his blackboard, and asked pointed questions at key areas. When he was satisfied, we'd move on. It took about 40 minutes to go through it all. I didn't hear *any* of the objections that many of you have relied on for years from Professor McKenzie. I don't get those objections anywhere but on Usenet. Think about it. I thought about it for a while, and I concluded that certain posters were *deliberately* putting up objections that relied on most of you readers not knowing any better. They were and are deliberately fooling you, and must be doing it consciously as some of them apparently are actually rather well-versed in mathematics. That makes it easier for them to fool you. It was over a year ago when I realized that and it was extremely depressing. So these people understood what I was talking about, understood the mathematics, understood why it was important, but for them, more importantly, they understood that they could lie about it successfully to people like you, who would either get confused or just believe them. I mean, look at how easily they do it? Not just with my work on algebraic number theory, but also with the prime distribution. That's with freaking PRIME NUMBERS, and still they confuse many of you, and convince many of you with statements that are just crazy. Even there, I contacted leading mathematicians, and corresponded by email with Lagarias and Odlyzko, mathematicians noted in the area, and didn't get the objections that sci.math'ers give, until Odlyzko replied to me finally, claiming that there were many results that while true were not of mathematical interest. Lagarias never said that or anything like it. He'd just suggest things like sending it to certain areas, like posting my prime counting on arxiv. Well, I told him I couldn't get on arxiv. He didn't reply. The way this works is that mathematicians I contact by email or in person basically just walk away. I explain things to them, they may offer a question or two, but that's it. Actually getting to Professor McKenzie for an in-depth discussion was a major event for me, as mathematicians usually just walk away, which probably happened because I'm an alumnus of Vanderbilt and professors are told to be nice to us! So, the difference between Usenet and the rest of the math world is that you talk to me, but you make up bizarre objections, which I know are really bogus and weird. Journals represent the area where mathematicians are supposed to listen, so I can send a paper to a journal, and if the editors play by the rules, it should get a fair hearing. The Annals at Princeton, will most likely I believe, follow the rules. By the rules that math journals follow they are to properly evaluate my paper for correctness and relevance. That's all I ask. So why keep posting on Usenet? I think it really is about boredom and also about a need just to talk about my research to someone, even hostile weird people who make up voodoo mathematics and lie a lot. I'm just short of people to talk to about my research. On Usenet, I can go on and on and on about it. James Harris === Subject: Re: JSH: Limitations in thinking > I'm just short of people to talk to about my research. > On Usenet, I can go on and on and on about it. If you really want to communicate, try the following: Write down clear, precise definitions for all the terms you use. Put it on the web somewhere, and reference it when you post. For example, you could write if a is an element of the (unitary) ring R, then we say a is a unit of (or in) R iff there exists some c in R so that ca=1. Then, when you make a post, be careful to always use the definitions you have set up and be precise. For example, never write is a unit, but is a unit of R. It really won't take up much time. > James Harris Tim Mellor === Subject: Re: JSH: Limitations in thinking > For me the realization of what was happening came just a while back > when I talked with Professor McKenzie at Vanderbilt University. > I traced out the argument explaining the paper Advanced Polynomial > Factorization. > his blackboard, and asked pointed questions at key areas. > When he was satisfied, we'd move on. It took about 40 minutes to go > through it all. Yes, yes... And what did Professor McKenzie say _at the end_? If I may be permitted a few words from the peanut gallery, Harris, old chap, I'll try to help you understand why you're not actually very convincing up here. Of course we love to hear you talk the way you do - past all the people who post objections, at some imaginary attentive audience hanging on your every word, yet dangerously close to the slippery slope of being persuaded by the arguments of your tormentors. Trouble is, that when one of your detractors says something that one of us (if I may presume to be one of 'us') doesn't understand, then we ask, and typically get a reply we can understand. What is a unit? Well, comes the answer, in some ring, dot dot dot. You, on the other hand, use inscrutable terminology we cannot find in books, and no-one else seems to be able to understand either. When you say something or other is not a unit in the ring of algebraic integers, but is properly a unit, what does this mean? May seem tough, since others can delegate the responsibility for answering elementary questions, but it looks like you are the only one who can answer this. So unless you do, we ain't believing anything you say. Brian Chandler http://imaginatorium.org > On Usenet, I can go on and on and on about it. > James Harris === Subject: Re: JSH: Limitations in thinking >[...] >Journals represent the area where mathematicians are supposed to >listen, so I can send a paper to a journal, and if the editors play by >the rules, it should get a fair hearing. >The Annals at Princeton, will most likely I believe, follow the rules. >By the rules that math journals follow they are to properly evaluate >my paper for correctness and relevance. No doubt. And they will _not_ be publishing the paper. Because they're going to follow the rules, one of which is that they're not supposed to publish obvious nonsense. What's not entirely clear is whether you'll get a letter explaining that it's obvious nonsense or a polite letter saying that it's not suitable, maybe some other journal. A letter saying it's obvious nonsense would really be the best thing for you, in terms of retaining some sort of contact with reality, but I doubt they'll send such a letter, because they won't think it would be polite. >That's all I ask. >So why keep posting on Usenet? I think it really is about boredom and >also about a need just to talk about my research to someone, even >hostile weird people who make up voodoo mathematics and lie a lot. >I'm just short of people to talk to about my research. >On Usenet, I can go on and on and on about it. Uh, we're aware that you can go on and on on usenet. If you'd actually _learn_ some math then you could make posts here that people would be interested in. >James Harris ************************ David C. Ullrich === Subject: Re: Limitations in thinking > If you follow logic, then there's no room to argue with me. Has anybody ever followed your logic? If not, do you think the problem just might be in your logic? > I didn't hear *any* of the objections that many of you have relied on > for years from Professor McKenzie. Perhaps he was being polite? Perhaps he was scared of you? Think back to that day... do you recall him telling you tat you were correct? If he thought you were correct, and he understood you logic, why did he not recognize the big consequences that you claim about destroying Galois and Wiles? Why has he not spread the gospel according to James? > didn't get the objections that sci.math'ers give, until Odlyzko > replied to me finally, claiming that there were many results that > while true were not of mathematical interest. Right... if you had what you claim you have, he would have seen lots of mathematical interest in your work. What can you infer from that James? > Well, I told him I couldn't get on arxiv. He didn't reply. Oh? He did not reply? Keeping his distance from you? What can you infer from that James? > The way this works is that mathematicians I contact by email or in > person basically just walk away. What can you infer from that James? > mathematicians usually just walk away, ... What can you infer from that James? > The Annals at Princeton, will most likely I believe, follow the rules. And they will reject your paper! > I'm just short of people to talk to about my research. What can you infer from that James? > On Usenet, I can go on and on and on about it. Yes, for years.... we know James... === Subject: JSH: Two sides I've had months to wonder why mathematicians would work to ignore or hide my results, and I think there are two sides to it. For one thing, my discoveries betray much of what they've been taught about mathematics and it's not just that I was able to find a problem in core, but also that there were results to find, using rather basic methods. Somehow, despite what has been often said, all those other mathematicians working over hundred of years didn't cover all the ground for use of simple arguments and simple ideas. Like my work with non-polynomial factors relies on something as simple as factoring a polynomial with respect to something other than the polynomial variable. Do that and an entire world opens up. Possibilities that didn't exist with other methods are suddenly easy. Factors of irrational numbers can be discerned, even if you can't tell *which* root might have a number like 7 as a factor, you can know that at least one of them must. And to add insult to injury (from a particular point of view) I went on to find a prime counting function that relies on a partial difference equation. Hints of it can be seen all over the area, and Meissel's Formula actually uses a variant of it, while part of it looks (as some sci.math'ers have noted) like the recurrence relation that follows from Legendre's Method, as in fact a relationship can easily be proven. So what was missed? It looks like no one figured on a multi-variable function, but instead kept focusing on the single variable function. After all IT MAKES SENSE that it be a single variable function, as you have one cout of prime numbers for a given value. The count of primes up to 10 is 4. The count of primes up to 100 is 25. The count of primes up to 1000 is 168. One variable seems to be all you need. Yet, here is a simple, compact prime counting function that captures the entire prime distribution in a small space, with, a three-dimensional p(x,y) function. So yet another simple idea, not far away, but so close. Yet losing what you had, your beliefs, your community's shared vision, is one thing, looking at a world that's not cognizant of it is another. The second even more brutal side of it for mathematicians may be in realizing that the rest of the world doesn't have a clue. After all, the discoverer was labeled a crank by sci.math'ers and the label seems to have stuck. Years have gone by, with so much overturned, so many beliefs shattered, and so much research shown to be invalid, but the public doesn't notice, or care. The classes keep coming. The papers are still being published. Awards and prizes still given out. Day follows day. Month follows month, and the world continues on. That may be the greatest insult of all that maybe none of it ever really mattered. That maybe the world never really cared. That mathematicians were just part of some play, some game, some empty soap opera without any real meaning. After all, some of the most dramatic events in world history are passing without a notice. The much vaunted world press hasn't a clue. The supposed intellectuals in all their corners prance, and preen, talk as much as ever, and don't realize. The world knows nothing, and that maybe the harshest lesson of all. Maybe then, after all, it never really mattered, and what matters now is to hold on to what you have. It's not really possible to completely understand how they feel, what they feel, unless you're in their shoes. To have so many beliefs shattered. To have so much that you thought true, shown to be false. To feel betrayed down to the depths of your core...that's something you have to live to understand. James Harris === Subject: Re: JSH: Two sides > I've had months to wonder why mathematicians would work to ignore or > hide my results, Ignoring your results takes no effort, hiding them does not occur. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Two sides [snip delusion] What does properly a unit mean? Can you answer that simple question James Harris? === Subject: Re: JSH: Two sides > [snip delusion] > What does properly a unit mean? > Can you answer that simple question James Harris? JSH does not answer questions. === Subject: Re: JSH: Two sides !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > I've had months to wonder why mathematicians would work to ignore or > hide my results, and I think there are two sides to it. Ignoring takes no work. > Like my work with non-polynomial factors relies on something as > simple as factoring a polynomial with respect to something other > than the polynomial variable. > Do that and an entire world opens up. Sure, if you have something more than polynomials in one variable, you have lots of opportunities for confusing constants with regard to one variable to constants with regard to another. It is something you have to be aware of as soon as you work with more than one variable. Very basic stuff. And where variables are not necessarily independent, things get more complicated (partial differential equations ooze all over with the intricacies). You don't get it. Not even the basics. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Two sides > I've had months to wonder why mathematicians would work to ignore or > hide my results, and I think there are two sides to it. Just one side to it.... you are wrong. === Subject: Re: Two sides says... > I've had months to wonder why mathematicians would work to ignore or > hide my results, and I think there are two sides to it. > Just one side to it.... you are wrong. and INCREDIBLY lonely. === Subject: Re: November 25 is Infinite Clause day!! <9kq772-t9i.ln1@sirius.athghost7038suus.net> <96c872-j7j.ln1@sirius.athghost7038suus.net> <87653og9oe.fsf@phiwumbda.org> huh! Go figger! Herc === Subject: Re: The origin of quaternions - Hamilton's 1844 paper > big snip > > In hindsight one may guess that Hamilton's motives for his quest were > (1) to generalise the algebra of complex numbers to 3D, and to maintain > in the process: (2) the concepts of quotient of directions and of > quotient of vectors, and (3) the law of moduli of complex numbers, and > of course (4) the often repeated questions at breakfast by his sons, > aged six and eight in the autumn of 1843: Well, Papa, can you multiply > triplets? (*) > Well, I don't claim to be on the order of Hamilton but I do have a > construction that extends complex properties to three dimensions and > beyond. Please see polysigned numbers in sci.math. > The solution is not orthogonal. It comes as a result the additive > identity in higher signs. > Complex numbers are equivalent to three-signed numbers. > In four-signed numbers you can interpret them graphically as rays > extending from the center of a tetrahedron to each corner. Labeling > them -,+,*,# and enforcing the additive identity: > - a + a * a # a = 0, where a is either an unsigned scalar or a > four-signed value. > Thence points in 3D space can be defined as sums in the four signs. > The algebraic product rules follows: Funny, When I view this table in the original it is all screwed up, but now it looks good again. Put some spaces on the end just in case. > | - + * # > ----------------- > - | + * # - > | > + | * # - + > | > * | # - + * > | > # | - + * # > This is simply a process of sign counting as the product of any two > components are taken, wrapping the count at the highest sign. This > obviously yields the rotational nature of the arithmetic. > And so a general 3D product like: > ( - a + b * c # d )( - e + f * g # h ) > yields: > + ae * af # ag - ah * be # bf - bg + bh # ce - cf + cg * ch - de + df > * dg # dh > This procedure in three-signed math yields the complex numbers. > This procedure in two-signed math yields the real numbers. > All of these can be looked at as number-line style algebra where each > sign has a ray emanating from an origin. Takin