mm-1086 === 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 >fathers instructions with care, the sons found that they had shared their >inheritance equally. How many sons were there, and how large was the estate? Its 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 thats 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 http://mygate.mailgate.org/mynews/sci/sci.math/ 1726d436d37f2c3aef4fe0fdbf1499 91.48257%40mygate.mailgate.org > #6 gets 6 which is all thats 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 siblings 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 doesnt 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 (dedekinds) or Cauchy Sequence element cannot be but a localized elements in the real-line, and the minimal scructure of any localized element cant 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 dont know >>what it is doesnt make it indefinite. >Without the axiom of choice, how does one define cardinality? Well first, I didnt mean to be saying anything about what happens without AC. AC _is_ part of the default these days - people _dont_ 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 Davids 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 isnt 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. ...??? Couldnt find anything in a quick look at the Borweins 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 shouldnt 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. > Didnt 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. >> Didnt 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 cant 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 hasnt. 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, >>Im 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 Einsteins 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 Ill try later. >It works in whatever units you want. No, you need the more general form with another constant if you arent 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 dont 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 = kmc 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 BTUs, I dont 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 dont 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 Ômatching 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 dont 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? > doesnt E =m*10^17 where E is in joules and m is in kg makes more > mathematical sense? No, youd need to replace the symbol with the whole constant, which is not a dimensionless number. It would be E = m x (89875517873681764 m/s) 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 ms. 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 = mc doesnt just work with the International System of Units. It doesnt 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 youd need to remember would be: 89875517873681764 m/s 898755178736817640000 cm/s 967412023047703971.7968... ft/s 139307331318869371938.7... in/s 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 = mc 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 = kmc Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: help with series needed Im 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)} Its 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. Its quite common to teach material not in the text, especially in graduate school. >3 Dont skimp on books - its 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 dont 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 === Subject: Re: How best to study math? >1. How should I take notes from the text? Im 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 ßexible. >2. How should my notes from lectures and notes from texts be >integrated? Again, whatever works for you. >2b. What if the instructor doesnt teach in a manner that allows me >to take a cohesive and coherent set of notes? Then dont. If hes teaching material that isnt 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 === 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 === 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, were stuck. The only answers we get are greater than 1 and thats 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 --------------------------------------------------- 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, were stuck. The only answers we get are > greater than 1 and thats 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 doesnt 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 Yohns 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, were stuck. The only answers we get are > greater than 1 and thats 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? http://mygate.mailgate.org/mynews/sci/sci.math/ 593ac82d8b2dc5ee5a1192de3d7581 86.48257%40mygate.mailgate.org [query] By the way, you have a typo in the last line of your siggie, its senSe in English. xanthian. -- === Subject: Re: What is probability of this? http://mygate.mailgate.org/mynews/sci/sci.math/ 0c176687247b074afcef714fd22260 6f.48257%40mygate.mailgate.org > 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, were stuck. The only answers we get are > greater than 1 and thats 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 Im being dull, but whats 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? Isnt that what conditional probability means? How on earth did you get an answer over 1, to satisfy my morbid curiosity, even if Im 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.Qs, 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 http://mygate.mailgate.org/mynews/sci/sci.math/ 7fff0d702958629ef99762a9eb658d 6b.48257%40mygate.mailgate.org [racist drivel] Groups that are over-represented in any field of endeavor, from mugging to astrophysics, probably work harder to participate in those fields. Its 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 dont, 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; Im not sure how any post on Usenet can be said to be thrust anything in ones 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 Blatts 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 Favards 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 doesnt >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 cant decide the question. (Im not an expert Problem is Unsolvable in American Mathematical Monthly 80 (1973) 233-269, and I dont 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 cant 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 cant 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 cant 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 doesnt 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 ? youve got it. I found an excellent exposition on this by a French mathematician (on his site), but Ill 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 dEnron! === 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 dont 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). Thats 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 dont even remember where I heard about them first, but its 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 zeros 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) arent in GF(p^n). > If the roots cannot > be found in GF(2^m), look for roots in GF(2^h), where h > m ? Im 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... 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_ reßexive. 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: Thats what identity _means_: A > thing is identical to itself and to nothing > > For a contrasting standpoint, see > ************************************************************** > David Ullrich asks: > > Whats an example of something thats 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 Ôblunder by saying that equality is reßexive 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 Im stupid, can anyone tell me whats the point of this discussion? Because youre stupid, why would anyone want to bother? === Subject: Re: David Ullrich on Identity > Because Im stupid, can anyone tell me whats the point of this discussion? Dont 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 ÔDavid 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 Cantors 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