mm-4109 === Subject: Re: Fibonacci Number Problem >We have been posed a question which looks incredibly familiar to a problem I >have found on a site about Fibonacci and his history. The problem is from >Liber Abbaci, and reads as follows: >PROBLEM 4 (An Inheritance). A man whose end was approaching summoned his >sons and said: Divide my money as I shall prescribe. To his eldest son, he >said, You are to have 1 bezant and a seventh of what is left. To his >second son he said, Take 2 bezants and a seventh of what remains. To the >third son, You are to take 3 bezants and a seventh of what is left. Thus >he gave each son 1 bezant more than the previous son and a seventh of what >remained, and to the last son all that was left. After following their >father's instructions with care, the sons found that they had shared their >inheritance equally. How many sons were there, and how large was the estate? It's quite easy with a bit of algebra. Let y be the estate, and x what each son gets. So for the first son: x = 1 + (y-1)/7 = (y+6)/7 For the second: x = 2 + (y-(y+6)/7-2)/7 = (6y+78)/49 For these to be equal: (y+6)/7 = (6y+78)/49 y/49 = 36/49 so y = 36. Then x = (36+6)/7 = 6, and there must be 36/6 sons. Check that it works for all the sons: #1 gets 1 + 35/7 = 6, leaving 30 #2 gets 2 + 28/7 = 6, leaving 24 #3 gets 3 + 21/7 = 6, leaving 18 ... #6 gets 6 which is all that's left. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Fibonacci Number Problem > #6 gets 6 which is all that's left. I found it much easier doing it in my head to start from the other end; since the next to the last sibling grabbed 1/7th of what was left, the last sibling got 6/7ths. So the next to the last sibling's take before that grab, for the shares to be equal, had to be 5/7ths, making 6/7ths in all. A simple guess that we are talking in whole units lets the answer be intuited from there. xanthian. -- === Subject: Re: functions... It is not as simple as I thought ... but now I understand everything. wald === Subject: Re: A structural point of view on the Continuum and the Discreteness concepts First, thank you for your reply. >By Real Analysis the continuum is infinitely many elements with no >gaps between them. > No. That statement doesn't even appear to have any meaning. Any R member has a 1 to 1 correspondence whith some point in the real-line. If R is complete then there cannot be gaps between these points and we get infinitely many elements with no gaps between them. Any point, cut (dedekind's) or Cauchy Sequence element cannot be but a localized elements in the real-line, and the minimal scructure of any localized element can't be anything but = {.} . A continuous line, which is a non-localized element can never be constructed from localized elements, therefore its minimal structure = {__} . The non-localized element is the correspondence itself, and we can find these non-localized element between any two R members. Therefore the correspondence has the power of the continuum and R or any set that built from localized elements, can never reach the power of the continuum. === Subject: defining conditions of a solution was: engineer needs help. THIS HAS BEEN REVISED! I have 2 equations which yield 2 values. These values must obey certain conditions and I want to minimally iterate the variables from their initial values until the conditions are met. I think however that it is a little beyond me and would sure appreciate some help with this. eq1:= ((-w^2+y^2+z*(w^2-x^2))^0.5)/(z*(z-1))= v must be real and following on from eq1 --> eq2:= (w^2+v-x^2)/(2*v*w)=c must lie between -1 and 1 and be real Note: all variables w, x, y, z are real (they are in fact the length of phasors in a phasor diagram) and c is the cosine of an angle and hence must lie within -(-1)-->(+1) sorry for the errors in my original posting === Subject: Re: Indefinite sets. >DCU: >>No, any set has a definite cardinality. The fact that we don't know >>what it is doesn't make it indefinite. >Without the axiom of choice, how does one define cardinality? Well first, I didn't mean to be saying anything about what happens without AC. AC _is_ part of the default these days - people _don't_ say AC implies the following, they simply use AC in proofs without comment. Second, what Dave Seaman already said: in various ways, for example one can define the cardinal of a set to be the class of all sets which are bijectively equivalent to it. >I am familiar only with definitions in ZFC: A cardinal number is an >ordinal number which is not bijective with any smaller ordinal. The >cardinality of a set is the (unique) cardinal number bijective with the set. >Without choice, it is consistent that all ordinals are countable. Nope. Consider S, the set of all well-orderings of N. Show that if x and y are in S then either x is order-isomorphic to y, x is order -isomorphic to a proper initial segment of y or vice-versa. Use that to construct a certain equivalence relation ~ on S and an order on the set of equivalence classes S/~. Then it follows that S/~ is well-ordered and uncountable, no choice required. (Now show by induction in S/~ that the order on S/~ is equivalent to some ordinal.) >So >the above approach would not even define the cardinality of R. I know >there are other approaches. In these, is David's statement still true? ************************ David C. Ullrich === Subject: Re: Curve Fitting? I think what I need to learn is how to populate the vandermonde matrix. and the proper form of the polynomial.... i.e. t = f + ax + by + cz + dxy + exyz .... Then I should be able to solve it as a simple set of equations... If anyone knows the form of the polynomial or has a loop that will populate the vandermonde matrix I would really appreiate it. The matlab polyfit uses a QR decompostion that relies on a vandermond matrix. > I have a tensor whos values need to be expressed as: > I = fn(x,y,z). The function is third order on each axis. > Given a set of samples from a tensor I need to derive the co-efficents of > the polynomial that discribes the function fn. > Least squares worked well for me in one dimension but how do I go about > doing it in three dimesions. or is it four? > B. === Subject: Re: Interesting Inequality >> Show that >> a^b + b^a <= a^a + b^b >> for all positive real numbers a and b. > Show the nonpositivity of the first derivative of x |-> a^x + x^a - > x^x for x >= a. > I hope this isn't homework. So, for what f(x,y) is f(a,b) + f(b,a) <= f(a,a) + f(b,b)? True for f(x,y) = x*y. True for x+y. ...??? Couldn't find anything in a quick look at the Borwein's Pi and the AGM. Martin Cohen === Subject: Re: Volume > Hi~ > I was hoping someone could help with a math problem that I am can not seem > to solve. I have a volume in spherical coordinates defined as > r = F(theta,phi) where r is the radial distance, theta is the angle from the > z axis (0..Pi) and phi is the angle in the xy plane (0..2*Pi). Now I think > that the volume of this should be: vol = integral(integral(integral(s^2 * > sin(theta) ds*dtheta*dphi, s = 0..F(theta,phi) ), theta = 0..Pi ), phi = > 0..2*Pi ) But when I try and do this, in maple, for some given F (such as > F=cos(theta)*cos(phi) ) I get the vol = 0. Can someone please give me the > correct formula for the volume? My second problem is a little more > complicated. Suppose that I have now drawn a line circumscribing my volume. > The line being defined as r = F( theta(phi),phi) where theta(phi) is some > combination of cosines of phi say for example theta(phi) = cos(phi) > +cos(2*phi). Now since the same line is drawn for negative phi I can > connect the points ( F(theta(phi),phi) , theta(phi) , phi) and ( > F(theta(-phi),-phi) , theta(-phi) , -phi) with a line. This defines a > surface which is dividing the volume (r = F(theta,phi)). I now wish to seek > the volume of the part above this surface. How do I do this?? > Any help would be greatly appreciated. Your formula is correct if F > 0 everywhere, in which case the surface defined by r = F(theta, phi) bounds a well-defined solid. In your example, F=cos(theta)*cos(phi), F is not positive everywhere, the surface it defines is not embedded, and you shouldn't expect your volume formula to work. Your second problem seems complicated. You may need to calculate the answer numerically. John Mitchell === Subject: Re: God Bless Osama bin Laden >>Democracy: The triumph of popularity over principle. > Didn't Churchill say: > Many forms of Government have been tried, and will be tried in this world > of sin and woe. No one pretends that democracy is perfect or all-wise. Indeed, > it has been said that democracy is the worst form of Government except all > those others that have been tried from time to time. Yes, Churchill said it, but so what? Just another privileged alcholic. === Subject: Re: God Bless Osama bin Laden >Democracy: The triumph of popularity over principle. >> Didn't Churchill say: >> Many forms of Government have been tried, and will be tried in this world >> of sin and woe. No one pretends that democracy is perfect or all-wise. Indeed, >> it has been said that democracy is the worst form of Government except all >> those others that have been tried from time to time. >Yes, Churchill said it, but so what? Just another privileged alcholic. He also said, I have taken more out of alcohol than alcohol has taken out of me. Josh === Subject: Re: Improved prime counting function > I can't test anything higher with my current driver. However, > I was able to modify it from using atoi() to strtoull() without > difficulty: > pi(10^10): 0.11s > pi(10^11): 0.51s > pi(10^12): 2.26s > pi(10^13): 10.24s > pi(10^14): 46.88s > pi(10^15): 3m35.97s (215.97s) > which tells me that your Mac is about twice as fast as my x86 > after adjusting for clock cycles. That is actually quite interesting. My latest version runs slightly faster on a 733MHz Macintosh than on a 1700MHz Celeron PC using basically the same compiler. A major part of the total execution time is spent dividing 64 bit numbers by primes not much larger than N^(1/3). On the Macintosh, this without any divisions using some precalculated values. On the Macintosh, this makes the whole code about twice as fast. On the PC, it makes it slower. The difference is: Pentium has a 64 bit / 32 bit divide instruction, PowerPC hasn't. But on a Pentium, multiplication takes 14 cycles, and only four on a PowerPC. Makes it hard to compare. But things also depend very much on the compiler. I have gcc on the Macintosh as well, and that produces code that is about three times slower. No idea why. === Subject: Re: number of state > Suppose a state denoted as x=(i,j_1,j_2,...,j_M), the legal state must > satisfy > 0 <= i + j_1 + j_2 + ... + j_M <= C > 0 <= j_m <= int( C/m) ; m=1,2, ..., M > where M is positive integer, C is a positive integer. int ( C/m) > represents > the maximum integer less than or equal to C/m. > I hope to find the expression for the number of state. Plz give some tips. > If i >= 0, the upper bound on j_m is redundant. Introduce a slack variable > s >= 0 so that > i + j_1 + j_2 + ... + j_M + s = C > The number of nonnegative integer solutions to this equation is > binom(C + (M + 2 - 1), C) = binom(C + M + 1, C). > (Think of placing C dots among C + (M + 2 - 1) dots and dividers.) > If i is allowed to be negative, the answer is more complicated. You could > condition on the value of i and then sum the numbers of solutions of > j_1 + j_2 + ... + j_M + s = C - i. > But now the upper bound on j_m may not be redundant. You might try a > generating function approach. > Rob Pratt Oops! I was thinking the upper bound on j_m was C, not int(C/m). In that case, things are more complicated, and the generating function approach is probably the way to go. Assuming -M int(C/M) <= i <= C and 0 <= j_m <= C, you can obtain the count by summing the coefficients of x^t for t = 0 to C in the following generating function: (sum [r = -M int(C/M) to C] x^r) * (sum [k = 0 to int(C/M) x^k])^M (If i >= 0, just take r = 0 to C in the first sum.) For example, if M = 2 and C = 4, I get (1/x^4 + 1/x^3 + 1/x^2 + 1/x + 1 + x + x^2 + x^3 + x^4) * (1 + x + x^2)^2 = 1/x^4 + 3/x^3 + 6/x^2 + 8/x + 9 + 9 x + 9 x^2 + 9 x^3 + 9 x^4 + 8 x^5 + 6 x^6 + 3 x^7 + x^8 Summing the coefficients of x^t for t = 0 to C yields 9 + 9 + 9 + 9 + 9 = 45. This figure can be checked by explicitly enumerating the solutions. Curiously, these coefficients seem to be equal for fixed C, suggesting that a simpler method exists... Rob Pratt === Subject: Re: E=mc^2 >>OK, >>I'm starting to get it. >>I found a worked example. (same thread title this group a few years back) >>H-->He there is a shortfall in the atomic weight of 5*10^-27 kg >>the speed of light is 3*10^8 m/s >>Substituting in Einstein's equation >>E= 5*10^-27 kg *3*10^8m/s *3*10^8m/s >>=4.5*10^-10 kg(m/s)^2 >>=4.5*10^-12Kgm/s >>=4.5*10^-12 joules. >>Amazing! >>Does it work in imperial too? >>Or is it specific to those units! >>I think it must work. Same units either side. >>Maybe I'll try later. >It works in whatever units you want. No, you need the more general form with another constant if you aren't using a coherent system of units, if you want to use particular units for all three quantities.. >If c is meters per second and m is kilograms, then E is >kg-m^2/c^2=joules. >If c is centimeters per second and m is grams, then E is g-cm^2/c^2=ergs. >If c is feet per second and m is pounds mass, then E is foot-pounds. Wrong. It is in foot-poundals. >If c is football fields per fortnight and m is six-packs, then E is >sixpacks-fooballfields^2/fortnights^2, although this unit of energy is not >in common use. Yes, it will work with any units for mass and for speed, if you don't specify ahead of time what energy units you want. Or if you are similarly willing to use strange units for either time or distance or mass. My comments above about the more general form E = kamacO apply if you are going to specify units for all three quantities (this k is equal to 1 for a coherent system of units). >Likewise if you want E in BTU's, I don't think there are convenient units >of length and mass that correspond to it. If you want E in jellydonuts, >one jellydonut has a food value of around 250 kilocalories, or a >megajoule, which makes conversion from mks units easy. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Re: E=mc^2 I don't understand the point of the squared. >Seems to me:- >Since c is a constant, and the speed of light is not measured in the >same units as mass or energy, c is just a 'matching' constant. and >c^2 is just another matching constant So the equation >(E =mc) ,without the squared, is equally as valid No. c is measured in meters/second, and m is measured in kilograms. Therefore, > mc^2 is in units of kilograms*meters^2/seconds^2, which happens to be > equal to joules, the units associated with energy. Without the squares, you don't get that. Doug > So... > E= m c^2 where E is in joules, m in kg and c in ms^-1 > E= m x 300 000000^2 > E = (90,000,000,000,000,000)m where E is in joules and m is in kg > is that right? > If so, why not forget the c^2 term and simply use the static number? > doesn't E =m*10^17 where E is in joules and m is in kg makes more > mathematical sense? No, you'd need to replace the symbol with the whole constant, which is not a dimensionless number. It would be E = m x (89875517873681764 mO/sO) where the E and the first m would normally be in italics and the second m and the s upright, so there would be a visible difference in the m's. A harder to remember number, and harder to remember units, and most people using it will likely already know the speed of light in the unit system they want to use, or know how to figure it out or at the very least where to look it up. But more importantly, then you have to learn a particular number which depends on the system of units you are using. The formula E = macO doesn't just work with the International System of Units. It doesn't work with any arbitrarily chosen set of units either, but it will work with any system of units that is coherent as that word is used in the jargon of metrology. For example, it will work for any of these: m in grams and c in cm/s gives E in ergs m in pounds and c in ft/s gives E in foot-poundals m in slinches and c in in/s gives E in foot-pounds force m in slugs and c in ft/s gives E in foot-pounds force m in metric tons and c in m/s gives E in sthene-meters The would differ in each of these systems, so depending on which system you used, the constant you'd need to remember would be: 89875517873681764 mO/sO 898755178736817640000 cmO/sO 967412023047703971.7968... ftO/sO 139307331318869371938.7... inO/sO I think it was NASA that expects us to use some really strange units for energy, on the web page where they give this formula E = mcO and then tell us that the speed of light is 186,000 mi/s. Do they expect us to have energy in pound miles squared per second squared? Or in slinch (NASA engineers were the ones who gave this name to the unit of mass in the gravitational ips system of units) miles squared per second squared? But what can we expect from the people who turned the Mars Climate Orbiter into a Crash Lander by screwing up the units of measure? To use other systems such as those with both pounds and pounds force, or with both kilograms and kilograms force, you need to use a more general form of the equation which adds another constant dependent on the units used: E = kamacO Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: help with series needed I'm trying to find out about the following power series' behaviour on the boundary of its disc of convergence, |z|=1. sum_{n=1}^infty ((-1)^n)/n z^{n(n+1)} It's easy to see that it converges for z = 1, i, -1, -i, etc, but a general proof eludes me. Any help appreciated. nojb. === Subject: Re: How best to study math? at 06:05 PM, Mark Atherton said: >2 Never take notes in a lecture. Whatever the lecturer says is >written down more coherently in a good book. It's quite common to teach material not in the text, especially in graduate school. >3 Don't skimp on books - it's a false economy. Buy the books you >need rather than trying to get the previous edition from the >library when you can. Then read them! :-) And don't loan them out. I learned this the hard way :-( -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: How best to study math? >1. How should I take notes from the text? I'm with Charlie and Justin on this; do what works for you. Note that an approach that works in one class may not work in another, so try to be flexible. >2. How should my notes from lectures and notes from texts be >integrated? Again, whatever works for you. >2b. What if the instructor doesn't teach in a manner that allows me >to take a cohesive and coherent set of notes? Then don't. If he's teaching material that isn't in the text, talk to him about your problem. >3. What should notes contain? Whatever is useful for you. It might be summaries of material, cross references, hints at proof techniques, or whatever else you consider appropriate. You might also want to include worked out proofs in your notes. >Do notes even matter much since I can >always go back to the text? They matter if the instructor is giving material not in the text, or if you have better retention reading your notes than reading the text. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: infinity dimensional vector space at 03:58 PM, tern said: >Can you prove that the vector space of real-valued function over a >differentiable manifold M is infinity-dimensional Not without additional assumptions. With additional assumptions it is trivial. Is that a homework assignment? What was the original wording of the question? I suspect that the question that you asked is not the question you were supposed to answer. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: What is probability of this? Suppose you have an urn containing 6 white and 9 black balls. What is the probability of picking up (without any replacement) following sequence: W W B B? We tried with conditional probability and concluded that P(A1) = 6 / 15 and P(A2|A1) = P(A1A2) / P(A1) = [C(6,2)/C(15,2)] / (6/15) After that, we're stuck. The only answers we get are greater than 1 and that's no good for a probability. Any good hint? (A1 = first ball white, A2 = second ball white, A3 = third ball black, A4 = fourth ball black) -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: What is probability of this? >Suppose you have an urn containing 6 white and 9 black >balls. What is the probability of picking up (without any >replacement) following sequence: W W B B? The probability of getting the first white ball is 6 out of 15. (Assuming that the first ball was white), the probability of getting the second white ball is 5 out of 14. (Assuming that the first two balls are white), the probability of getting the black ball third is 9 out of 13 (since there are still nine black balls, and only four white balls, left). (Assuming that the first three balls were WWB), the probablity of getting the black ball fourth is 8 out of 12. Combining the four using the multiplication rule, we see that 6 * 5 * 9 * 8 2160 P (W,W,B,B) = --------------- = ----- = 6.6% 15*14*13*12 32760 Doug === Subject: Re: What is probability of this? > Suppose you have an urn containing 6 white and 9 black > balls. What is the probability of picking up (without any > replacement) following sequence: W W B B? > We tried with conditional probability and concluded that > P(A1) = 6 / 15 > and > P(A2|A1) = P(A1A2) / P(A1) = [C(6,2)/C(15,2)] / (6/15) > After that, we're stuck. The only answers we get are > greater than 1 and that's no good for a probability. > Any good hint? > (A1 = first ball white, A2 = second ball white, > A3 = third ball black, A4 = fourth ball black) Conditional probability doesn't look particularly helpful here. How many different sequences of four balls can be chosen from the original 15? How many of those sequences fit the pattern W W B B? (How many choices for the first ball? How many for the second? ...) -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: What is probability of this? > Suppose you have an urn containing 6 white and 9 black > balls. What is the probability of picking up (without any > replacement) following sequence: W W B B? > We tried with conditional probability and concluded that > P(A1) = 6 / 15 > and > P(A2|A1) = P(A1A2) / P(A1) = [C(6,2)/C(15,2)] / (6/15) > After that, we're stuck. The only answers we get are > greater than 1 and that's no good for a probability. > Any good hint? > (A1 = first ball white, A2 = second ball white, > A3 = third ball black, A4 = fourth ball black) It seems to me the probability of picking the first W should be 6/15. And the probability of picking the second white ball, given that the first ball was white should be 5/14. And the probability of picking the third ball as black, given that the first two balls are white is 9/13. And the probability of the following sequence is: W W B is (6/15)(5/14)(9/13) which is lower than one. Have I applied any formulas about conditional probability? Do you see how this extends to your sequence of four balls? -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: What is probability of this? [query] By the way, you have a typo in the last line of your siggie, it's senSe in English. xanthian. -- === Subject: Re: What is probability of this? > Suppose you have an urn containing 6 white and 9 black > balls. What is the probability of picking up (without any > replacement) following sequence: W W B B? > We tried with conditional probability and concluded that > P(A1) = 6 / 15 > and > P(A2|A1) = P(A1A2) / P(A1) = [C(6,2)/C(15,2)] / (6/15) > After that, we're stuck. The only answers we get are > greater than 1 and that's no good for a probability. > Any good hint? > (A1 = first ball white, A2 = second ball white, > A3 = third ball black, A4 = fourth ball black) I suppose I'm being dull, but what's wrong with: (6/15)*(5/14)*(9/13)*(8/12) or using the probability at each selection, given that your being successful to that point has altered the number of each color of balls in the jar appropriately, and taking the product of those unlikely events as the overall likelihood? Isn't that what conditional probability means? How on earth did you get an answer over 1, to satisfy my morbid curiosity, even if I'm in error with the above? xanthian. -- === Subject: SheerPower 4GL Discussion Forum now available!!! TOUCH TECHNOLOGIES, INC. RELEASES SHEERPOWER 4GL -- BEYOND BASIC FORUM SAN DIEGO -- SheerPower 4GL -- Beyond BASIC is delighted to offer a discussion board. The board is open to any of the over 10,000 SheerPower 4GL users. The discussion board includes areas of interest,such as -- Updates, Polls, F.A.Q's, General Topics, Announcements and MORE. Please ask feel free to visit and ask your questions, make suggestions or post code. We will be monitoring this discussion board daily !! http://www.sp4gl.com/forum.htmlx What is SheerPower 4GL --- BEYOND BASIC? SheerPower 4GL -- Beyond BASIC is a full development language, but shares many advantages with popular scripting languages -- such as ultra-fast development speeds and ease of learning the language. SheerPower can be used to write programs of any size, from simple web-enabled programs to vast database applications. Join thousands of others and download SheerPower 4GL today at http://www.sp4gl.com -- the download is 100% free. === Subject: Re: Fundamental Reason for High Achievements of Jews [racist drivel] Groups that are over-represented in any field of endeavor, from mugging to astrophysics, probably work harder to participate in those fields. It's called merit, something racists fail to find lacking in themselves, but all the rest of the world notices easily by its omission in their works. xanthian. -- === Subject: Re: Death Rattle Of Neoclassical Theory Of Value <8370fa18ff28632870b371a462223444.48257@mygate.mailgate.org>, Kent > Please don't, as you have not yet taken the point. A mathematician > who wants to read about economic theory has no need for you to thrust > it in his or her face; I'm not sure how any post on Usenet can be said to be thrust anything in one's face. I quite understand that some may choose to skip certain posts, threads, etc. > that mathematician is quite capable of finding > the sci.econ* desmene on his or her own. Mathematicians, strange as > it may seem to your desire to wallpaper the net with your words, come > to sci.math to read about math, not economics. sci.math seems to me to be a healthy community. And it seems to me that many threads are not about serious math. In fact, some of the regular posters here seem to me seem to participate in certain threads for amusement. By the way, do you think John Blatt's paper How Economists Misuse Mathematics might have something to do with math? How about the demonstration of logical problems with certain mathematical models used in applied fields. Might that have something to do with math? -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Orthogonal Polynomials Weight Functions > Given a three term recursion relation in the canonical form > The fact that such a measure exists is known as Favard's theorem and all the moments of the moment functional. Inverting the moments, === Subject: Re: Decidability of diophantine equations >Corresponding to any given axiomatization of number theory, one can >explicitly >construct a Diophantine equation which has no solutions, but such that >this >fact cannot be proved within the given axiomatization. This is not quite right. >OK. So say P(x,y,z,...,w)=0 is a diophantine equation that doesn't >have a >solution, and say that this cannot be proven in the axiomatic system >A. If we >_prove_ this fact, namely prove that >(*) in system A, P=0 cannot be proven to have no solutions >then this statement obviously implies that P=0 indeed has no >solutions, since >verifying a solution is straightforward (and should be possible to >carry out in >any aximoatic system that purports to describe the natural numbers, >such as A). >Therefore the statement (*) itself also cannot be proven in system A. >So when >one builds a diophantine equation such as this, and proves that it >does the >job, one needs to specify two axiomatic systems: the system A, which >one shows >is too weak to prove that P has no solutions, and the stronger system >B in >which one is proving this very fact. What Matiyasevic (building on the earlier contributions of Davis, Robinson and others) actually showed is that any recursively enumerable set is Diophantine. Roughly speaking, a set S of n-tuples of positive integers is recursively enumerable if there is an algorithm that will eventually write down each member of S, and does not write down any n-tuple that is not a member of S. A set S of n-tuples of positive integers is Diophantine if there is a polynomial P(x_1,...,x_n,y_1,...,y_m) with integer coefficients such that (x_1,...,x_n) is in S iff there are positive integers y_1,...,y_m such that P(x_1,...,x_n,y_1,...,y_m)=0. It is easy to show that a Diophantine set is recursively enumerable: you just have to search through all possible (x_1,...,x_n,y_1,...,y_m), and when you find one where P = 0 you write down (x_1,...,x_n). The argument continues by showing that there is a particular, rather complicated, polynomial P(x,y,z_1,...,z_n) such that there is no algorithm such that, given x and y, the algorithm will decide correctly in finite time whether there exist z_1,...,z_n such that P(x,y,z_1,...,z_n) = 0. In fact, I _think_ that given an algorithm T you can actually construct an x and y for which T can't decide the question. (I'm not an expert Problem is Unsolvable in American Mathematical Monthly 80 (1973) 233-269, and I don't think it really makes this point clear) Now consider a formal system A. We can imagine an algorithm that, given x and y, searches through all possible z_1,...,z_n, computes P(x,y,z_1,...,z_n), and if the result is 0, halts with the answer YES. At the same time, it searches through all finite strings in the alphabet of A, checking to see if the string is a proof in A of the statement that there is no z_1,...,z_n for which P(x,y,z_1,...,z_n) = 0. If it finds one, it halts with the answer NO. According to the theorem, this algorithm can't always halt in finite time with the correct answer. Since an answer of YES will always be correct, there are two logical possibilities: 1) an answer of NO which is incorrect. That is, there is a proof in A that there is no such z_1,...,z_n, but in fact there is such a z_1,...,z_n. And (assuming the system A is strong enough to do the calculation P(x,y,z_1,...,z_n) = 0 for this z_1,...,z_n and conclude that such a z_1,...,z_n exists), this implies that A is inconsistent. or 2) the algorithm will not halt. That is, there is no solution z_1,...,z_n, but there is no proof of that fact in A. The consistency question is important. If we know that A is consistent, we can say that A can't prove the existence of a solution to P(x,y,z_1,...,z_n). Of course if A is inconsistent, it can prove that statement just as it can prove every well-formed statement in its language. It is possible that the proof we used can also be done within A, with one exception: A can't prove its own consistency (or else, by Goedel, it is inconsistent). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Is ...9999.9999... = 0 ? > A real contradiction? > Well, the 10-adic numbers has zero-divisors, > is that bad enough? In case someone would like to see explicit examples ... The 10-adic integer solutions of x*x = x are 0, 1, a, b, where a = ...57423423230896109004106619977392256259918212890625 b = ...42576576769103890995893380022607743740081787109376. Both a and b (b = 1 - a) are zero-divisors, with a*b = 0. --r.e.s. === Subject: Re: Is ...9999.9999... = 0 ? > Is ...9999.9999... = 0 ? > It doesn't converge in the real number system. Did you have in mind > some other number system? If so, you should specify it in the > question! > ...9999. = -1 in the 10-adic numbers > .999... = 1 in the real numbers. > But there is no known way to add these together... > A real contradiction? Well, the 10-adic numbers has zero-divisors, > is that bad enough? I am not specifying a metric or topology although It should be easy to rig one up for which both sides would converge, although the standard operations might not be continuous. The real question is: can you used what would appear to be normal algebra/arithmetic and obtain a contradiction if you take limits in both directions. I realize that in standard valuation theory there is no obvious choice for the algebra, it might be just some weird topology on the rationals. === Subject: Re: Is ...9999.9999... = 0 ? you've got it. I found an excellent exposition on this by a French mathematician (on his site), but I'll recall it, later. actually (and this is not his result), *any* repeating sequence that crosses the POINT, will sum to zero. can you prove it? > ...9999. = -1 in the 10-adic numbers > .999... = 1 in the real numbers. > But there is no known way to add these together... > A real contradiction? Well, the 10-adic numbers has zero-divisors, > is that bad enough? --les ducs d'Enron! === Subject: Re: Is ...9999.9999... = 0 ? > Is ...9999.9999... = 0 ? > The expression...9999.9999... does not represent any real number, > in any usual sense, since a (presumably decimal) representation of a > real number must have a most significant digit, which > ...9999.9999... does not have. > This thread gets recycled about three time a year. > Have fun, guys!! > Bob Pease 0.999...=1 stuff which was not interesting. there might have been some ...9999.0 =-1 stuff which is standard p-adic/q-adic stuff not really interesting. I did find one posting that mentioned ...9999.9999... = 0 which seemed to consider it standard stuff which I don't think it is, since the metrics are incompatable. But is there some ready contradiction === Subject: Re: NTT transform for vectors of arbitrary length in GF(2^m)? Hi Jaco, > I am looking for a NTT transform that can transform vectors in > GF(2^m), but the vectors should not be constrained to length 2^m -1. > The vectors that I want to investigate are seldom of length (2^m) - 1, > as is required with the NTT transforms I have found thus far. To do an N-point NTT with elements in some field GF(p^n), you need a primitive Nth root of 1 in that field, in the same way that e^-(2ipi/N) is a primitive root of 1 in the complex number field. This implies that N has to be a factor of (p^n)-1. Because the algorithm is a lot easier when N is a power of two, you may want to find a field that supports these lengths as well as meeting your other criteria. I have successfully used GF(2^32-2^20+1). That's a prime number, and as a field characteristic it supports power-of-two NTT lengths up to 2^20. It also fits snugly into a 32-bit word, which was my other primary criteria at the time. > Does anyone know where I can find any literature on these transforms > (also known as finite field Fourier transforms, if I am right)? I don't even remember where I heard about them first, but it's just a DFT in a different field, and all the DFT results that you would expect to apply, like the convolution theorem, do apply and are pretty easy to prove to yourself. > The nice property of the transform that I have used is that the > transformed vector have zero's in the coefficients corresponding to > the roots of the original vector. Is this true for all the NTTs? Like a DFT, an NTT is equivalent to polynomial evaluation at multiple abscissas simultaneously. The abscissas are the consecutive powers of that primitive root. If you use a maximum length NTT, then the set of abscissas inlcudes all non-zero elements of the underlying field, so it will identify all the roots of the input polynomial that are elements of the field. > Also, can a NTT be used to factor a polynomial? It only finds roots in the field, so an NTT in GF(2^m) would only find simple factors of the form (x-a) where a is in the field. In general, most roots of polynomials over GF(p^n) aren't in GF(p^n). > If the roots cannot > be found in GF(2^m), look for roots in GF(2^h), where h > m ? I'm not sure what you mean here. === Subject: Re: Why poles so named? This was asked before. Then (as now) there were wild guesses. But no information... -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Why poles so named? > Zeros so named makes sense but what has infinity got to do with a pole? > If you graph the functions with the z-coordinate being abs(f), you get > telephone poles. Another possibility: At such a point, f(z) -> the north pole on the Riemann sphere. === Subject: Re: David Ullrich on Identity > David Ullrich says: And yes, identity is in _fact_ reflexive. To > refute that statement you need to give an > example of something which is not identical > to itself. The idea that there is something > which is _not_ identical to itself is simply > ludicrous: That's what identity _means_: A > thing is identical to itself and to nothing For a contrasting standpoint, see > ************************************************************** > David Ullrich asks: What's an example of something that's not identical ************************************************************** > David Ullrich dares: Exhibit of proof of Ex~(x=x) from > C1-C4 and someone will point out the error. >C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1 >>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2 >>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) >>Classification >>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] > Weak Would someone be kind enough help David out with a proof? > ************************************************************* > David Ullrich remonstrates: I 'blunder' by saying that equality is reflexive by definition? > Huh. Do you have any idea what the word definition means? Homework for David Ullrich: 1) What philosopher said: ...definitions are available only for transforming > truths, not for founding them. 2) In your own words, explain why (or why not) you think > this is true. --John > Because I'm stupid, can anyone tell me what's the point of this discussion? Because you're stupid, why would anyone want to bother? === Subject: Re: David Ullrich on Identity > Because I'm stupid, can anyone tell me what's the point of this discussion? Don't worry about responses to my reply here. My main detractor is quite expert with slander. In point set topology, the singleton {x} is often shortened to x. Consequently, there is an ambiguity of notation in mathematics associated with reference to objects. This presents no problem for mathematics because of how invariance is characterized in algebraic topology. However, set theory--and its metaphysics of paradox--is a particular place where logic and mathematics share interests. Unfortunately, the people who develop an understanding of set-theoretic identity within the topological framework find their reasonable intuitions continually subject to appeal-to-ridicule in logic communities. If you have any interest in pursuing this matter sufficiently to understand what I mean by *reasonable intuition* here is an excerpt from a citeseer abstract that I found with a quick Google search on 'David Lewis' mereology: Just as mereotopology can be seen as an extension of mereology through the addition of some topological primitive such as connection or interior part, so also set theory can itself be seen as an extension of mereology through the addition of the primitive set theoretic notion of singleton. David Lewis (1991) has shown how, with the help of this one single notion, all the standard axioms of set theory can be derived within a mereological framework. The theory of sets and the 3 theory of mereological sums (or fusions ) of singletons are, it turns out, formally indistinguishable. David Lewis is not a mathematician. He is recently deceased and was a tenured professor in philosophy at Princeton University. The 1991 publication being referenced here is entitled Parts of Classes. The advocates of this mereological perspective cite Husserl as originator. Husserl was a student of Weierstrass along with Cantor. In his third Logical Investigation, he discusses a theory of parts and wholes and--being aware of Cantor's work--actually distinguishes between parts and pieces. In any case, this is part of the reason why the revision is coming from philosophy departments even though it seems to be a mathematics/logic issue. So, what is going on here is fairly straightforward. John falls into that group of people who understand the identity predicate within that topological framework. He is trying to defend his ideas in the place where everyone else thinks the expertise lies. lol Have a good day. Hope that helps a little. :-) mitch === Subject: Re: Fibonacci Try this site: http://www.amazon.com/exec/obidos/tg/detail/-/0767908155/102-1535541-3413723 ?v=glance Aubrey === Subject: Re: Fibonacci Kavon P.S. (I'm on my way as soon as I click send. > Try this site: http://www.amazon.com/exec/obidos/tg/detail/-/0767908155/102-1535541-3413723 ?v=glance > Aubrey === Subject: Re: Seeking SVD code It is C++ and Free http://math.nist.gov/tnt/download.html Roland > Can someone point me towards some portable source code which will > efficiently perform singular value decompositions of large matrices? I am > looking for something freely available, preferably in the public domain or > under a Free or Open Source licence such as the GPL. Ideally the program > will be in C, though any languages compilable under Unix/Linux and MS > Windows without proprietary tools such as Matlab are OK. > Tristan > -- > _ > _V.-o Tristan Miller [en,(fr,de,ia)] >< Space is limited > / |`-' -=-=-=-=-=-=-=-=-=-=-=-=-=-=-= <> In a haiku, so it's hard > (7_ http://www.nothingisreal.com/ >< To finish what you === Subject: Re: Seeking SVD code > It is C++ and Free > http://math.nist.gov/tnt/download.html Does it work on any common compilers? It uses a lot of tricky C++ templates, and compiler support for those features tends to be spotty. === Subject: Re: transferring electrical energy through metal wall Well it seems to me that the best answer is actually a compromise (or a combination) of some of the other answers you have received so far. 1) Several people suggested high pressure feedthroughs (I agree) 2) You said that the parts receiving the energy inside the pressure vessle must move (I take it they will rotate continuously?) and therefore wires are not paractical (and for that matter I don't think brushes are practical either since you will be operating the device in a high pressure environment.) so you feel that some sort of inductive coupling of the energy to the moving parts is required. (Once again I agree) So the answer is to put both your primary and your secondary on the inside of the device and to power the primary via the high pressure feed through and allow the secondary to move relative to the primary to accomidate your necessary movement. It sounds an awful lot like you are going to test a deepwater submersible motor to me? Unless you are going to use a high pressure gas? This plan allows you to accomidate both the high pressure and still have high electromagnetic conversion efficiency because you don't have to resort to turning a bulkhead wall into a crappy transformer core. Use a better and much more standard transformer/motor design instead. Although this solution still avoids the basic problem of how to pass a large amount of energy through a solid ferrous wall with resonable efficiency and not drill holes? That in it'self is an interesting question and I suggest that the discussion continue on that problem. Perhaps you could do it with with X-rays? I don't think so but I could be wrong? === Subject: GMRES/CGS/QMR for Singular Matrices Hi I have been looking for techniques based on GMRES or Transpose-Free methods like CGS/QMR that can be utilized for singular matrices. I understand all these techniques require nonsingular matrices. Any one knows if these techniques have been extended to singular and/or very ill-conditioned systems? Alien+ === Subject: Re: GMRES/CGS/QMR for Singular Matrices >Hi >I have been looking for techniques based on GMRES or Transpose-Free methods >like CGS/QMR that can be utilized for singular matrices. I understand all >these techniques require nonsingular matrices. Any one knows if these >techniques have been extended to singular and/or very ill-conditioned >systems? no. but LSQR should do. title LSQR for overdetermined or underdetermined sparse systems of linear equations, , sparse least squares problems, and damped sparse least squares problems by C.C. Paige and M.A. Saunders ref ACM TOMS 8 (1982) 195-209 this is http://www.netlib.org/toms/583 also Kaczmarc's method owrks for singular systems (this is cyclic projection on the planes ai'*X-B_i=0 where ai' are the rows of the matrix and B_i the i-th component of the right hand side. but this is very slow. hth peter === Subject: Re: GMRES/CGS/QMR for Singular Matrices >Hi >I have been looking for techniques based on GMRES or Transpose-Free methods >like CGS/QMR that can be utilized for singular matrices. I understand all >these techniques require nonsingular matrices. Any one knows if these >techniques have been extended to singular and/or very ill-conditioned >systems? > no. but LSQR should do. > title LSQR > for overdetermined or underdetermined sparse systems of linear equations, > , sparse least squares problems, and damped sparse least squares problems > by C.C. Paige and M.A. Saunders > ref ACM TOMS 8 (1982) 195-209 > this is http://www.netlib.org/toms/583 > also Kaczmarc's method owrks for singular systems (this is cyclic projection > on the planes ai'*X-B_i=0 where ai' are the rows of the matrix and B_i the i-th > component of the right hand side. but this is very slow. > hth > peter here you can get LSQR in three different languages: http://www.stanford.edu/group/SOL/software/lsqr.html Hans Mittelmann === Subject: Re: steepest slope estimate >No. I asked you to fit a spline, not to interpolate one. >See the newknot algorithm in de Boor's book. >You begin with a spline fit through 4 points (two end points and >two pionts in between), and then you add the two points >with the biggest positive and negative error, and repeat this until >your fit is satisfactory. In each step you need to solve a >linear least squares problem for the spline coefficients, >and if set up correctly (add basis function (x-x_k)_+^3 at the >new interpolation points x_k), you can update the QR factorization >used to solve it, thus getting a cheap solution algorithm. >Everything is spelled out in detail in deBoor's book on splines. >Arnold Neumaier yes, o.k., but he must not reenvent the wheel: the code is in http://www.netlib.org/dierckx one can also use the smoothing spline from http://www.netlib.org/toms/642 which does a very good job. hth peter === Subject: Re: steepest slope estimate === Subject: Re: steepest slope estimate === Subject: Re: Kahan summation and matrix multiply >A recent post pointed me towards Kahan summation as a way of reducing >roundoff errors. I was just wondering - is this type of method used >in, say, the BLAS matrix multiply routines, which would seem to be >just the type of place it's needed most. it goes as follows: s_0=0; w_0=0; for i=1 to m s_i = s_{i-1}+a_i; w_i=w_{i-1}+((s_{i-1}-s_i)+a_i); s=s_n+w_n; %s=sum_{i=1 to m} a_i; this has a considerable overhead and hence is not used in blas. hth peter === Subject: Re: Kahan summation and matrix multiply > this has a considerable overhead and hence is not used in blas. IMO, this point of view is way too common among numerical analysts. A 4x penalty might be acceptable to a lot of people. I have numerical analysis texts from the 1960s and 1970s that sometimes give a method and then say it is impractical because of a 4x speed penalty. Computers gain 4x in speed every 5 years or so. So if a method was 4x too slow 5 years ago, then it should be accept today, for all those problems that were being run 5 years ago. === Subject: Re: Kahan summation and matrix multiply > A recent post pointed me towards Kahan summation as a way of reducing > roundoff errors. I was just wondering - is this type of method used > in, say, the BLAS matrix multiply routines, which would seem to be > just the type of place it's needed most. They sometimes use extra precision, such as using an 80-bit accumulator on Intel CPUs. That is much cheaper. === Subject: Help: Needed approximations of exp(x) and/or pow(x,y), fast and quite accurate Hello all, I'm somewhat versed in math but no expert by far. I've been browsing the internet, news-groups and all, but still I have not found a good answer to my question. I'm looking for a Java (J2ME) implementation (or something from which i can create this implementation) for the approximations (only using additions, subtractions, multiplications, divisions and mayb a square-root function and logarithm) for one or both of these two functions: exp(x) e^x (or pow(2,y), ... raising the power of 2 on computers can be faster) and/or pow(x,y) x^y where both x and(!) y can be real numbers (not necessarily integers). The approximations need to be relatively fast (J2ME, MIDP1.0... limited device capabilities, no floating-point support) but quite accurate. I've already have good implementations for the log(x) (using the Maclaurin series for 1/(ln[x+1])) and squareroot(x) (using Newton's Iteration). I need the approximations where y is between 0 and 1 and where y is a large number: (e^x = e^(largenum+fraction) = e^largenum * e^fraction) PS: The x and y are real numbers represented in J2ME by a class representing some form of fixed-point values. === Subject: Inverse of De L'Hopital for solving equations Suppose we have a real function f(x) and we must find zero-crossing points. So this is equal to say that we have an equation in the form: f(x) = 0 and we want to find its solutions. If lim (x->t) f(x) / g(x) = 0 / 0 De L'Hopital says: lim (x->t) f(x) / g(x) = lim (x->t) f '(x) / g'(x) = lim (x->t) f ''(x) / g''(x) and so on until f(x) or g(x) become a constant value. But if f(t)=0 (as hypotesis) then t must be a zero-crossing point, and so a solution of the equation f(x)=0. Now I must define a particular function g(x) that tends to 0 as x tends to t. I also need to eliminate the term t from inside the limit (and so from g(x) ) in order to calculate its value. So a function that initially tends to 0 as x->t can be: g(x)= e^x - e^t Applying De L'Hopital: lim (x->t) f(x) / (e^x - e^t) = lim (x->t) f '(x) / (e^x) = lim (x->t) f ''(x) / (e^x) ....... In the first limit I can't calculate the value of t because I have it also inside the limit. So I consider the 2nd and the 3rd limit: lim (x->t) f '(x) / (e^x) = lim (x->t) f ''(x) / (e^x) Applying the properties of limits: lim (x->t) (f '(x) - f ''(x) ) / (e^x) = 0 I define a new function as the argument of the limit: h(x) = (f '(x) - f ''(x) ) / (e^x) At this point suppose we have a limit in the form: lim (x->t) h(x) = 0 then if t exists in h(x), t = h^(-1) (0) (h^(-1) is the inverse function of h) where h^(-1) (0) is a solution of the equation f(x)=0. Anyone can help me find the mistakes? I tried it with Derive 5, but it won't work properly. === Subject: Re: Factorization lemma, example I saw the picture, and the guy *looks* intelligent, or he must try to fit into the charicature of four-eyes. find a doctor of philosophy, and get a prescription for a nice Club Med, Mister Monster Math. or, at the least, make a hypothesis and try to break it, or prove it. I read a great definition in the paper, the other day, but I don't recall the example, humorous, just that it linked was necessary, but not sufficient. that's all that you have to do, grapple with those two words o'Leibniz. > y_1 = (r^2 + 2r)(-1 + sqrt(2))^{1/3} y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1 + sqrt(2))^{1/3} y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1 + sqrt(2))^{1/3} and r^2/2 *should* be an algebraic integer, but provably is not. > You keep saying *should*. Deal with the fact that it's not and draw > your conclusions from that. Either that or define more clearly *why* > you think it should be something it isn't. However, notice that you run into a problem with r^2/2 not having a > factor of 2 that is 2. > Only through an artificial construction. > However, notice that it is true that y_2 and y_3 must be coprime to 2. > Or not, as others have shown. --les ducs d'Enron! http://members.tripod.com/~american_almanac === Subject: Re: Lines Crossing In A Nonagon: puzzle I don't grok the difficulty, as there are only three (well-known) star possibilites, other than the simple enneagon: skip every-other vertex (nine total crossings); skip over two (three trigona, each line crosses four others); skip over three (each line crosses six others, but combinatorics isn't my specialty; I just drew them .-) >Draw a path consisting of connected straight >line-segments within the nonagon, so that each segment starts/ends on >the vertexes of the 9-gon, and where each vertex is visited exactly >once, and where the 1st and last segment are connected. >Now the path must be such that there are a total of exactly 13 crossings >inside the nonagon. > This rule generates the right sequence: > Starting at an arbitrary vertex, visit alternately the second and third vertex > from the last, when you find that you would be revisiting a vertex, skip to the > next empty one. --les ducs d'Enron! http://members.tripod.com/~american_almanac === Subject: Moving forward, generalizing with y^3 + 3y - 2 It occurred to me a while back that the claims against factors I'd used for certain factorizations would apply to *any* non unit factor, but I've hesitated in putting that forward, but now I might as well. Consider y^3 + 3y - 2 and its three roots. I've been arguing that only one of those roots is not coprime to 2, but now imagine that as others have argued, they each have some algebraic integer in common with 2, and let's call those factors f_1, f_2, and f_3. Now given a root r_1, you have that r_1/f_1 should be a unit. Generalizing, use r = fw, for a root, and substitute to get f^3 w^3 + 3fw - 2 = 0, so you can divide off f, and get f^2 w^3 + 3w - 2/f. That is non-monic and not in Q as f is not rational. However, no amount of algebraic manipulations can turn f^2 w^3 + 3w - 2/f into a polynomial with integer coefficients. For one thing, for any f, like f_1, you need f_2 or f_3 as well to get back to an integer. Basically there is NO way to find a polynomial with integer coefficients let alone a monic one with integer coefficients to prove that one of the roots of f^2 w^3 + 3w - 2/f is a unit. And in fact, as what should be the unit root is NOT expressible as the root of a monic polynomial with integer coefficients, it is not an algebraic integer. Several posters have been getting away with casting doubt on values for f that I've used that follow from advanced factorization techniques, but they've managed to hide the fact that if you divide off *any* non unit factor of 2 from the roots you run into the same thing. That is, their own argument traps them. The only conclusion that can be drawn is that what should be algebraic integer units are not, and now the challenge that has been directed my way, can go elsewhere as those who wish to dispute that conclusion can do a simple thing: Give the monic polynomial with integer coefficients. By their own claims at least you know the last coefficient should be 1 or -1. Given the polynomial y^3 + 3y - 2, one of these people should be able to give a monic polynomial with integer coefficients that define what should be units when you divide off factors in common with 2 from the roots, if they're telling the truth. James Harris === Subject: Re: Pure math, physics, and FLT James' bike has bigonal tires; if he understood trigona, he'd already have solved it -- or, understood where some of the elementary algebraic stuff that he proposes is hypothetically inadequate. mathematicians battled imaginary numbers for millenia, and they still havent' learned !?! > around on a triangle tired bicycle at worst. --les ducs d'Enron! http://members.tripod.com/~american_almanac === Subject: question about paths in the group of unitary operators on a Hilbert space Suppose a separable infinite dimensional Hilbert space H has closed subspaces U, V and W and two bounded hermitian operators q and r on H such that: (exp(i*q))(U) = V, (exp(i*r))(V) = W, does it follow that there exists a bounded hermitian operator s on H such that: (exp(i*s))(U) = W ? (The motivation is to try to define equivalence classes of closed subspaces of H based on continuous rotations of one into the other; I already have U ~= U, U ~= V => V ~= U and no proof of transitivity.) David Bernier === Subject: Re: extension of Minkowski's inequality Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > prove of the validity of Minkowski's inequality with parameter infty > < p < 0, I would appreciate very much if someone could tell me where > in the literature that can be found. Again, it would be great if the > result were formulated in the terminology of random variables. Both Hardy, Littlewood, and Polya's Inequalities and Beckenbach and Bellman's Inequalities have the theorems and the proofs. Just look in the table of contents. === Subject: Re: Which norm is compatible with Lie bracket? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > In particular, if f is constant [f,g] is the directional derivative > of g in the direction of f. Fix such a nonzero f, and let D_f be > the operator g -> [f,g]. Then D_f would have to be bounded. But > consider g(x) = exp(k f.x) f for constant k, which satisfies > D_f g = k (f.f) g. Thus ||D_f|| >= (f.f) |k|. Since k is arbitrary, > D_f must be unbounded. So there is no such norm. That is sad. I need some estimate. Let me refine the question: How can we take a subset of smooth maps and define a norm for f,g: R^n -> R^n so that the Lie bracket [f,g] = g'.f-f'.g satisfies || [f,g] || <= M ||f|| ||g|| where M is a constant? E.g. if we take n=1 and polynomials of order not more than 2, f(x) = f0 + f1 x + f2 x^2 g(x) = g0 + g1 x + g2 x^2 then [f,g] = f0 g1 - f1 g0 + 2(f0 g2 - f2 g0)x + (f1 g2 - f2 g1)x^2 (resembling a vector product in R^3) and with the Euclidean norm we have || [f,g] || <= 2 ||f|| ||g|| I would like to see some infinite dimensional subset, if possible. -- Pavel Pokorny Math Dept, Prague Institute of Chemical Technology http://www.vscht.cz/mat/Pavel.Pokorny === Subject: Re: Which norm is compatible with Lie bracket? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >> In particular, if f is constant [f,g] is the directional derivative >> of g in the direction of f. Fix such a nonzero f, and let D_f be >> the operator g -> [f,g]. Then D_f would have to be bounded. But >> consider g(x) = exp(k f.x) f for constant k, which satisfies >> D_f g = k (f.f) g. Thus ||D_f|| >= (f.f) |k|. Since k is arbitrary, >> D_f must be unbounded. So there is no such norm. >That is sad. I need some estimate. >Let me refine the question: >How can we take a subset of smooth maps >and define a norm for > f,g: R^n -> R^n >so that the Lie bracket > [f,g] = g'.f-f'.g >satisfies >|| [f,g] || <= M ||f|| ||g|| >where M is a constant? >E.g. if we take n=1 and polynomials of order not more than 2, >f(x) = f0 + f1 x + f2 x^2 >g(x) = g0 + g1 x + g2 x^2 >then >[f,g] = f0 g1 - f1 g0 + 2(f0 g2 - f2 g0)x + (f1 g2 - f2 g1)x^2 >(resembling a vector product in R^3) >and with the Euclidean norm we have >|| [f,g] || <= 2 ||f|| ||g|| >I would like to see some infinite dimensional subset, if possible. For simplicity, let's work in the case n=1. According to my example, there's no possibility of such a norm on any subspace that includes 1 and exp(kx) for arbitrarily large k. For polynomials, there's bad news too. Since [x, x^m] = (m-1) x^m, there is no such norm on a subspace that includes x and x^m for arbitrarily large m. Moreover, if your subset is a Lie algebra (i.e closed under addition, multiplication by scalars, and your Lie bracket operation) and contains x^2 and x^p for some integer p > 2, then it must contain x^m for all integers m >= p since [x^2, x^m] = (m-2) x^(m+1). However, if you avoid 1 and x the news is better. Consider the space of polynomials with no constant term and no x term, and use the norm || sum_{j=2}^d c_j x^j || = sum_{j=2}^d |c_j| exp(-j^2) Then this will work with M = 1/(2 e^2) since for j, k >= 2 || [x^j, x^k] || = || (k-j) x^(j+k-1) || = |k-j| exp(-(j+k-1)^2) = |k-j| exp(-2(j-1)(k-1)+1) ||x^j|| ||x^k|| <= M ||x^j|| ||x^k|| Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Is Hol(D) a nice subset of B(H)? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > ... > Since the original question wondered if Hol(D) is a > deformation > retract of Fred(H), that seems relevant, though maybe > it isn't. >As to whether this is a retract in Fred(H), I have > no idea, though it >seems possible. >Dan > Lee Rudolph Yes, my original question is to find a nice deformation retract of Fred(H). Let's,first, prove that each component of Fred(H) is not contractible. Do you have any idea? i heard that each component of Fred(H) is homotopy equivalent to B(U), the base space of Milnor construction of a universal principle bundle, corresponding to group U, where U is the union of U_n the space of unitary n by n complex matrix...let's check in detail... Ali Taghavi alitghv_@_yahoo.com === Subject: Re: Which TVS is compatible with Lie bracket? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >In fact the Example of Robert Israel shows that the following question 1 has negative answer,since the spectrum is unbounded,But what about the following second question 2(search for TVS structur compatible to the data?): in the following V is a linear space and T is a linear map from V to V Question 1:Is there a norm on V such that T is a bounded operator question 2:is there a topological vector space structure on V such that T is continious linear map? Ali Taghavi > According to my example, there's no possibility of > such a norm on > any subspace that includes 1 and exp(kx) for > arbitrarily large k. > Robert Israel > israel@math.ubc.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada === Subject: Finding all integral solutions of some Diophantine equations Originator: bergv@math.uiuc.edu (Maarten Bergvelt) A few years ago, I noticed a paper which finds =all= integral solutions of certain Diophantine equation. That Diophantine equation actually had a couple of non-trivial integral solutions. The proof used effective(and useless in almost all real situations) Baker's method + some jumping(in the author's own terms?) technique which made the upper bound for the search space so practically small that the author was able to find =all= integral solutions of certain Diophantine equation. Would anyone kindly guide me to any relevant paper(s) which explains the jumping technique? === Subject: Re: Finding all integral solutions of some Diophantine equations Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > A few years ago, I noticed a paper which finds =all= integral solutions > of certain Diophantine equation. > That Diophantine equation actually had a couple of non-trivial integral > solutions. > The proof used effective(and useless in almost all real situations) > Baker's method + some jumping(in the author's own terms?) technique > which made the upper bound for the search space so practically small > that the author was able to find =all= integral solutions of certain > Diophantine equation. > Would anyone kindly guide me to any relevant paper(s) which explains > the jumping technique? I don't think I've seen the term jumping technique, but there are a number of papers by deWeger, Tzanakis, Stroeker, Zagier... that combine Baker-style upper bounds with other methods (such as lattice reduction) to make the process practical. If one knows a basis for the group of rational points, then instead of estimates for linear forms in (ordinary) logs, one can use more practical bounds for linear forms in elliptic logs. That, too, is explained in these papers. Here are a few references, but it's by no means an exhaustive list: Stroeker, R. J.(NL-ROTT-EM); Tzanakis, N.(GR-CRET) Solving elliptic Diophantine equations by estimating linear forms in elliptic logarithms. Acta Arith. 67 (1994), no. 2, 177--196. Zagier, Don(D-MPI) Large integral points on elliptic curves. Math. Comp. 48 (1987), no. 177, 425--436. Tzanakis, N.(GR-CRET); de Weger, B. M. M.(NL-TWEN-A) On the practical solution of the Thue equation. J. Number Theory 31 (1989), no. 2, 99--132. The algorithm is based on a combination of Baker's method with the so-called basis reduction algorithm due to Lenstra, Lenstra and Lov.87sz and several skilful lemmas. As a remarkable application of the main result all the 22 integral points of the elliptic curve y^2=x^3-4x+1 are determined. So possibly it's the use of lattice reduction that is known as jumping. Joe Silverman === Subject: Re: Finding all integral solutions of some Diophantine equations Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I think that the first paper to use such a technique was: MR0248079 (40 #1333) Baker, A.; Davenport, H. The equations 3x^2-2=y^2 and 8x^2-7=z^2. Quart. J. Math. Oxford Ser. (2) 20 1969 129--137. In it they use diophantine approximations noting that they got lucky that certain convergents to a continued fraction of a number formed by logarithms of algebraic numbers were unusually large, so this allowed them to cut off a large chunk of the search space. A good reference for all of this is the book by Benne de Weger listed below. He goes into very great detail about these methods. MR1026936 (90m:11205) de Weger, B. M. M.(NL-TWEN) Algorithms for Diophantine equations. CWI Tract, 65. Stichting Mathematisch Centrum, Centrum voor Wiskunde en Informatica, Amsterdam, 1989. viii+212 pp. ISBN 90-6196-375-3 Victor Miller === Subject: Paper published by Geometry and Topology Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following paper has been published: Geometry and Topology, Volume 9 (2005) Paper no. 40, pages 1775--1834 URL: http://www.maths.warwick.ac.uk/gt/GTVol9/paper40.abs.html DOI: 10.2140/gt.2005.9.1775 Title: Squeezing in Floer theory and refined Hofer-Zehnder capacities of sets near symplectic submanifolds Author(s): Ely Kerman Abstract: We use Floer homology to study the Hofer-Zehnder capacity of neighborhoods near a closed symplectic submanifold M of a geometrically bounded and symplectically aspherical ambient manifold. We prove that, when the unit normal bundle of M is homologically trivial in degree dim(M) (for example, if codim(M) > dim(M)), a refined version of the Hofer-Zehnder capacity is finite for all open sets close enough to M. We compute this capacity for certain tubular neighborhoods of M by using a squeezing argument in which the algebraic framework of Floer theory, is used to detect nontrivial periodic orbits. We partially recover some existence results of Arnold nondegenerate magnetic field on a torus. We also relate our refined capacity to the study of Hamiltonian paths with minimal Hofer length. Secondary: 37J45 Keywords: Hofer-Zehnder capacity, symplectic submanifold, Floer homology Received: 22 March 2005 Revised: 11 September 2005 Accepted: 12 August 2005 Published: 25 September 2005 Proposed: Leonid Polterovich Seconded: Yasha Eliashberg, Eleny Ionel Author(s) address(es): Mathematics, University of Illinois at Urbana-Champaign Urbana, IL 61801, USA and Institute of Science, Walailak University Nakhon Si Thammarat, 80160, Thailand Email: ekerman@math.uiuc.edu URL: http://www.math.uiuc.edu/~ekerman/ === Subject: real points on an algebraic variety Originator: bergv@math.uiuc.edu (Maarten Bergvelt) A graduate student asked me the following question and it stumped me, although surely there is a relatively easy solution. Let V be an affine algebraic variety [i.e. a closed subspace of affine n-space] over the real numbers R, and let V_1,V_2,V_3,... denote countably infinitely many closed subvarieties of V. If V(R) (the real points of V) is the union of the V_i(R), 1<=i in the following V is a linear space and T is a > linear map from V to V > Question 1:Is there a norm on V such that T is a > bounded operator Counterexample: let V have the basis e_1, e_2, e_3, ... Define T by T(e_j) = j e_j Then for any norm N: N(T(e_j)) = j N(e_j) so T is not bounded. > question 2:is there a topological vector space > structure on V such > that T is continious linear map? Yes, let N be a norm on V and set s_k(v) = N(T^k(v)) Then the family s_k, k=0,1,2,... of seminorms defines a topology on V and one has s_k(T(v)) = s_{k+1}(v), hence T is continuous wrt this topology. Anton === Subject: Paper published by Geometry and Topology Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following paper has been published: Geometry and Topology, Volume 9 (2005) Paper no. 41, pages 1835--1880 URL: http://www.maths.warwick.ac.uk/gt/GTVol9/paper41.abs.html DOI: 10.2140/gt.2005.9.1835 Title: The Grushko decomposition of a finite graph of finite rank free groups: an algorithm Author(s): Guo-An Diao, Mark Feighn Abstract: A finitely generated group admits a decomposition, called its Grushko decomposition, into a free product of freely indecomposable groups. There is an algorithm to construct the Grushko decomposition of a finite graph of finite rank free groups. In particular, it is possible to decide if such a group is free. Secondary: 20E05 Keywords: Graph of groups, Grushko decomposition, algorithm, labeled graph Received: 5 February 2005 Revised: 11 September 2005 Accepted: 7 August 2005 Published: 25 September 2005 Proposed: Benson Farb Seconded: Martin Bridson, Joan Birman Author(s) address(es): School of Arts and Sciences, Holy Family University Philadelphia, PA 19114, USA and Department of Mathematics and Computer Science Rutgers University, Newark, NJ 07102, USA Email: gdiao@holyfamily.edu, feighn@andromeda.rutgers.edu