mm-3399 === Subject: Re: Bessel functions of Large Orders and Large Arguments : >: It should be easy to modify a standard library to do arbitrary precision : >: bessel function computation (say by using mpfr etc.) : > : >Both precision and speed are important. Using MPFR or any other : >method to achieve arbitrary precision will slow down the evaluatoin : >time by a few orders of magnitude. : > : >I'm curious, can anyone here evaluate J1000000(999995) with a : >precision of 13 decimals or more, and how fast? : > : >Dan Baruth : >x87[]iging.com : : Good. Then how to download teh software?.. : : A.L. Sorry, _JaX is not meant to be downloaded. Do you need it for particular ranges of the input parameters? Please respond by e-mail (replace [] with @). DB -- === Subject: Re: Bessel functions of Large Orders and Large Arguments On Sun, 13 Aug 2006 14:23:06 -0700, D. Baruth > >: >: Good. Then how to download teh software?.. >: >: A.L. Sorry, _JaX is not meant to be downloaded. Do you need it for >particular ranges of the input parameters? Please respond by >e-mail (replace [] with @). DB Then what was the purpose of your initial posting?... A.L. === Subject: eigenvalues of product of orthogonal and diagonal matrix Hi All, I am learning Linear Algebra by self-study. I am trying to relate eigenvalues of product of two matrices to the eigen values of individual matrices. Are there any results regarding this ? Inparticular Suppose A is an orthogonal matrix and say V is a diagonal matrix. How are the eigen values of AV related to eigen values of A and V. ? Gov === Subject: Re: eigenvalues of product of orthogonal and diagonal matrix >Hi All, >I am learning Linear Algebra by self-study. I am trying to relate >eigenvalues of product of two matrices to the eigen values of >individual matrices. Are there any results regarding this ? 1) det(AB) = det(A) det(B), and the determinant is the product of the eigenvalues (counting multiplicity). 2) ||AB|| <= ||A|| ||B|| (for a matrix norm induced by a norm on vectors), and the norm is an upper bound on the absolute values of the eigenvectors. Not much other than that. >Inparticular Suppose A is an orthogonal matrix and say V is a >diagonal >matrix. How are the eigen values of AV related to eigen >values of A and >V. ? Not very well. However, the singular values of V and AV are the same. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: eigenvalues of product of orthogonal and diagonal matrix Gov >Hi All, >I am learning Linear Algebra by self-study. I am trying to relate >eigenvalues of product of two matrices to the eigen values of >individual matrices. Are there any results regarding this ? 1) det(AB) = det(A) det(B), and the determinant is the product > of the eigenvalues (counting multiplicity). 2) ||AB|| <= ||A|| ||B|| (for a matrix norm induced by a norm > on vectors), and the norm is an upper bound on the absolute > values of the eigenvectors. Not much other than that. >Inparticular Suppose A is an orthogonal matrix and say V is a >diagonal >matrix. How are the eigen values of AV related to eigen >values of A and >V. ? Not very well. However, the singular values of V and AV are > the same. Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Formula for alfa I promised this to someone still today; You probably wont even calculate the formula as it will be idiotic ,but that is Your problem. My duty is to inform about knew knowledge as I have tried to do in all posts, in vain:). This formula is valid for any organism ( alfa depends on development stage); but let us take The rotating Universe: Angular moment L universe = (1/h) ^ ((4th root ( 1/alfa)) - 2) With todays h, alpha values L universe = 6,027620093*10^22 =((N avogadro surface of universe ) ^2)/ (N avogadro volume universe ( normal N A)) And (1/alpha) = (1/alpha matth) * ((ln ( 1/h) /ln ( 1/2*pi*h)) ^2) (1/alpha math) = ln (( L/2*pi*h)/ln ( 1/h) ) ^4 = (2^4)* ( 1+G/10^-12) ^4 )= = (ln((N avogadro surface of universe ) ^2)/ (N avogadro volume universe)) -1))^4 D universe = 4th root ( 1/alpha) = 3,421437946 = f ( t) this formula demonstrates the fractal nature of 3rd/4th mass space dimension. Its value today is 2G= 1,42. 2 other space dimensions are on spheric surface of Universe; Fractal dimension 0,42 characterizes thickness of this surface layer measured in length units of surface topographical defects ( black holes) in relation to maximum possible length; Fractal dimension grows as universe gets older. This formula also links continuos and discrete dimensions via fractal dimenesion; Continuous mass time dimension only happens at death moment when Universe will disperse in non rotating masses to be further digested and freed from structure by worms of universe and serve as point mass food. D can also be considered an integral measure of Universe complexity; If laws of nature are the same in all universes, probably it is enough to look at this 1 value and it time evolution which could be calculated from system dynamics equations. G, N avogadro also evolves with universe Time scale of Universe if derivative of mass space; such that: (1/(G*c))* ( h/N) * (S/M^2)* (1/T) = const= (1/2*pi) G- gravitation constant c-speed of light S- surface of universe M- mass of universe T - age of universe, longest time scale in universe. N= N avogadro surf ^2/ N avogadro Volume I may have made some mistakes writing here of formulas is damn difficult. For other stuff, see my other posts. Do not expect ready theories. Use what You like. === Subject: Re: Formula for alfa Can't say whether you are raving or whether you have something. The appropriate way is to submit a paper and have 3 independent mathematicians review it then provide a reference. Anyhow it needs more work.. (99% of the ideas I come up with turn out to be rubbish on further inspection and that is a pretty good ratio..) > I promised this to someone still today; You probably wont even > calculate the formula as it will be idiotic ,but that is Your > problem. My duty is to inform about knew knowledge as I have tried to > do in all posts, in vain:). > This formula is valid for any organism ( alfa depends on development > stage); but let us take The rotating Universe: Angular moment L universe = (1/h) ^ ((4th root ( 1/alfa)) - 2) > With todays h, alpha values L universe = 6,027620093*10^22 =((N > avogadro surface of universe ) ^2)/ (N avogadro volume universe ( > normal N A)) > And > (1/alpha) = (1/alpha matth) * ((ln ( 1/h) /ln ( 1/2*pi*h)) ^2) > (1/alpha math) = ln (( L/2*pi*h)/ln ( 1/h) ) ^4 = (2^4)* ( 1+G/10^-12) > ^4 )= > = (ln((N avogadro surface of universe ) ^2)/ (N avogadro volume > universe)) -1))^4 > D universe = 4th root ( 1/alpha) = 3,421437946 = f ( t) > this formula demonstrates the fractal nature of 3rd/4th mass space > dimension. > Its value today is 2G= 1,42. 2 other space dimensions are on spheric > surface of Universe; Fractal dimension 0,42 characterizes thickness of > this surface layer measured in length units of surface topographical > defects ( black holes) in relation to maximum possible length; Fractal > dimension grows as universe gets older. > This formula also links continuos and discrete dimensions via fractal > dimenesion; Continuous mass time dimension only happens at death moment when Universe will disperse in non rotating masses to be further > digested and freed from structure by worms of universe and serve as > point mass food. > D can also be considered an integral measure of Universe complexity; If laws of nature are the same in all universes, probably it is enough to > look at this 1 value and it time evolution which could be calculated > from system dynamics equations. > G, N avogadro also evolves with universe > Time scale of Universe if derivative of mass space; such that: > (1/(G*c))* ( h/N) * (S/M^2)* (1/T) = const= (1/2*pi) > G- gravitation constant > c-speed of light > S- surface of universe > M- mass of universe > T - age of universe, longest time scale in universe. > N= N avogadro surf ^2/ N avogadro Volume > I may have made some mistakes writing here of formulas is damn > difficult. > For other stuff, see my other posts. Do not expect ready theories. Use > what You like. > -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/ === Subject: Re: Formula for alfa No one wants to answer because they assume such ideas must be rubbish from beginning. That is why I have become slighly unpolite in beginning. Excuse me. If You check the formula, You should get within 1% of N avogadro/10. I have some thinking behind it, it is not pure numerology, though it helps. Basic idea is conservation of local moment so that h should be conserved if it is minimum moment quanta, it can not dissappear; Which leads to possibility to sum them to get L; Which leads to a possibility to have fractal dimension which evolves ( of course, net mass of such universe grows and time scale(age) is also evolving, a derivative of other dimensions. Max potential number of dimensions in such Universe is 5 ( including mass); however , full 5 are reached probably only in the moment of collapse. If You do not get it right I can send Word file where there are all nice formulas, easy to read. Where do I get 3 mathematicians? Can't say whether you are raving or whether you have something. > The appropriate way is to submit a paper and have 3 independent > mathematicians > review it then provide a reference. > Anyhow it needs more work.. > (99% of the ideas I come up with turn out to be rubbish on further > inspection > and that is a pretty good ratio..) I promised this to someone still today; You probably wont even > calculate the formula as it will be idiotic ,but that is Your > problem. My duty is to inform about knew knowledge as I have tried to > do in all posts, in vain:). > This formula is valid for any organism ( alfa depends on development > stage); but let us take The rotating Universe: Angular moment L universe = (1/h) ^ ((4th root ( 1/alfa)) - 2) > With todays h, alpha values L universe = 6,027620093*10^22 =((N > avogadro surface of universe ) ^2)/ (N avogadro volume universe ( > normal N A)) > And > (1/alpha) = (1/alpha matth) * ((ln ( 1/h) /ln ( 1/2*pi*h)) ^2) > (1/alpha math) = ln (( L/2*pi*h)/ln ( 1/h) ) ^4 = (2^4)* ( 1+G/10^-12) > ^4 )= > = (ln((N avogadro surface of universe ) ^2)/ (N avogadro volume > universe)) -1))^4 > D universe = 4th root ( 1/alpha) = 3,421437946 = f ( t) > this formula demonstrates the fractal nature of 3rd/4th mass space > dimension. > Its value today is 2G= 1,42. 2 other space dimensions are on spheric > surface of Universe; Fractal dimension 0,42 characterizes thickness of > this surface layer measured in length units of surface topographical > defects ( black holes) in relation to maximum possible length; Fractal > dimension grows as universe gets older. > This formula also links continuos and discrete dimensions via fractal > dimenesion; Continuous mass time dimension only happens at death moment when Universe will disperse in non rotating masses to be further > digested and freed from structure by worms of universe and serve as > point mass food. > D can also be considered an integral measure of Universe complexity; If laws of nature are the same in all universes, probably it is enough to > look at this 1 value and it time evolution which could be calculated > from system dynamics equations. > G, N avogadro also evolves with universe > Time scale of Universe if derivative of mass space; such that: > (1/(G*c))* ( h/N) * (S/M^2)* (1/T) = const= (1/2*pi) > G- gravitation constant > c-speed of light > S- surface of universe > M- mass of universe > T - age of universe, longest time scale in universe. > N= N avogadro surf ^2/ N avogadro Volume > I may have made some mistakes writing here of formulas is damn > difficult. > For other stuff, see my other posts. Do not expect ready theories. Use > what You like. > -- > Using Opera's revolutionary e-mail client: http://www.opera.com/mail/ === Subject: Re: Formula for alfa You are in the right place! Three best mathematicians frequently visit this site. , Archimede Plutonium, and J. Sarfatti. Truly cream of the crop. > Where do I get 3 mathematicians? === Subject: Re: Formula for alfa > You probably wont even > calculate the formula as it will be idiotic ,but that is Your > problem. Oh, no, that is *YOUR* problem. === Subject: Hi all! I need help hi! I am tring the apply gaussian smoothing using x,y coordinates. E.g: I have curve that is created using simple (x,y) coordinates and I would Like to perform gaussian smoothing. Maybe using MATLAB? can anyone Help me? Shawn === Subject: Re: Hi all! I need help > hi! > I am tring the apply gaussian smoothing using x,y coordinates. E.g: I > have curve that is created using simple (x,y) coordinates and I would > Like to perform gaussian smoothing. Maybe using MATLAB? can anyone Help > me? Shawn Perhaps you can try: comp.soft-sys.matlab There are also other choices like Mathematica or Matlab and there are probably some libraries (C, C++, et. al.) that can be used. Not sure that helps - A === Subject: Re: Hi all! I need help > hi! > I am tring the apply gaussian smoothing using x,y coordinates. E.g: I > have curve that is created using simple (x,y) coordinates and I would > Like to perform gaussian smoothing. Maybe using MATLAB? can anyone Help > me? Shawn Perhaps you can try: comp.soft-sys.matlab There are also other choices like Mathematica or Matlab and there are > probably some libraries (C, C++, et. al.) that can be used. To OP: In Matlab, this would be done with a filtering command. Check the help for FILTER. Outside of Matlab, you want to search for smoothing or blurring. - Randy === Subject: Re: Weather Forecast > > The local weatherman says No Rain, and his record is 2/3 accuracy of >prediction. But the Federal Meteorological Service predicts rain, and >their record is 3/4. With no other data available, what is the chance of rain? >SPOILER SPACE >40 Is Litton's Problematical Recreations still in print? 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 >Assuming the two sources are independent: P(rain) = P(weatherman is wrong and FMS is right) = 1/3 * 3/4 = 1/4 P(no rain) = P(weatherman is right and FMS is wrong) = 2/3 * 1/4 = 1/6 Actual probability of rain = P(rain) / (P(rain) + P(no rain)) = (1/4) / (1/4 + 1/6) = 3/5. >-- Don >>doesn't (P(rain) + P(no rain)) = 1? > > > Not when P(rain) is redefined as P(weatherman is wrong and FMS is > right), and P(no rain) is similarly changed. Perhaps I should have > used something like P'(rain). > > -- Don In that case, doesn't P(weatherman is wrong and FMS is right) + P(weatherman is right and FMS is wrong) = 1? :P (I can't see how both of them could be wrong..) === Subject: Re: Weather Forecast <36eqd25rva6bb7fc8hm9i55e73fnkofeqd@4ax.com> <890vd2h9c94agulk6s5tmqt3dga6153i4q@4ax.com> <44dfa8bc$1@news.maxnet.co.nz >>The local weatherman says No Rain, and his record is 2/3 accuracy of >prediction. But the Federal Meteorological Service predicts rain, and >their record is 3/4. With no other data available, what is the chance of rain? >SPOILER SPACE >40 Is Litton's Problematical Recreations still in print? 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 >Assuming the two sources are independent: P(rain) = P(weatherman is wrong and FMS is right) = 1/3 * 3/4 = 1/4 P(no rain) = P(weatherman is right and FMS is wrong) = 2/3 * 1/4 = 1/6 Actual probability of rain = P(rain) / (P(rain) + P(no rain)) = (1/4) / (1/4 + 1/6) = 3/5. >-- Don >>doesn't (P(rain) + P(no rain)) = 1? > Not when P(rain) is redefined as P(weatherman is wrong and FMS is > right), and P(no rain) is similarly changed. Perhaps I should have > used something like P'(rain). -- Don In that case, doesn't P(weatherman is wrong and FMS is right) + > P(weatherman is right and FMS is wrong) = 1? :P (I can't see how both of > them could be wrong..) does nick atty's post have something to do with this? do unrealizable cases have some bearing on the probability of the outcome? i'd guess no. the strange thing about this problem is that weatherguy can only be right when fms is wrong and visa versa. so the 2 random variables are entangled. just like a qubit. === Subject: Quick Question Hey all On the real line, if I have a function continuous on [0,1] interval, the derivitive at 0 exists f '(0) exists f(0) =0 how could I prove that f ' exists for all x in some ball around the point 0? mr === Subject: Re: Quick Question > Hey all On the real line, if I have a function continuous on [0,1] interval, > the derivitive at 0 exists f '(0) exists > f(0) =0 how could I prove that f ' exists for all x in some ball around the > point 0? > > mr lol wot the hell is this about? === Subject: Re: Quick Question >On the real line, if I have a function continuous on [0,1] interval, >the derivitive at 0 exists f '(0) exists >f(0) =0 how could I prove that f ' exists for all x in some ball around the >point 0? > It's not true. Let f(x) = x^2 if x is irrational, 0 if x is rational. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Quick Question <44DFAF31.9030507@netscape.net >On the real line, if I have a function continuous on [0,1] interval, >the derivitive at 0 exists f '(0) exists >f(0) =0 how could I prove that f ' exists for all x in some ball around the >point 0? > It's not true. Let f(x) = x^2 if x is irrational, 0 if x is > rational. > Yeah, but your f is not continuous on [0,1]!! > -- > Stephen J. Herschkorn sjherschko@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Quick Question >On the real line, if I have a function continuous on [0,1] >interval, >>the derivitive at 0 exists f '(0) exists >>f(0) =0 >>how could I prove that f ' exists for all x in some ball >around the >>point 0? >> It's not true. Let f(x) = x^2 if x is irrational, 0 >if x is >> rational. Yeah, but your f is not continuous on [0,1]!! Try f(x) = x^2 arcsin(sin(1/x)) for x <> 0, 0 for x = 0. Note f is non-differentiable at x = 2/(pi n) for odd integers n. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Quick Question <44DFAF31.9030507@netscape.net> >On the real line, if I have a function continuous on [0,1] >interval, >>the derivitive at 0 exists f '(0) exists >>f(0) =0 >>how could I prove that f ' exists for all x in some ball >around the >>point 0? >> It's not true. Let f(x) = x^2 if x is irrational, 0 >if x is >> rational. Yeah, but your f is not continuous on [0,1]!! Try f(x) = x^2 arcsin(sin(1/x)) for x <> 0, 0 for x = 0. > Note f is non-differentiable at x = 2/(pi n) for odd integers > n. Note that in this case the derivative at 0 (f ' (0)) does not exist because of the 1/x. Also, if you take the derivative of f then try to take the limit as x->0, you can see that we'll oscilate between -1 and 1. mr > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: Quick Question <44DFAF31.9030507@netscape.net> >On the real line, if I have a function continuous on [0,1] >interval, >>the derivitive at 0 exists f '(0) exists >>f(0) =0 >>how could I prove that f ' exists for all x in some ball >around the >>point 0? >> It's not true. Let f(x) = x^2 if x is irrational, 0 >if x is >> rational. Yeah, but your f is not continuous on [0,1]!! Try f(x) = x^2 arcsin(sin(1/x)) for x <> 0, 0 for x = 0. > Note f is non-differentiable at x = 2/(pi n) for odd integers > n. Note that in this case the derivative at 0 (f ' (0)) does not exist because of the 1/x. Also, if you take the derivative of f then try to take the limit as x->0, you can see that we'll oscilate between -1 and 1. mr > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: Quick Question On the real line, if I have a function continuous on [0,1] >>interval, the derivitive at 0 exists f '(0) exists f(0) =0 >how could I prove that f ' exists for all x in some ball >>around the point 0? >> It's not true. Let f(x) = x^2 if x is irrational, 0 >>if x is rational. Yeah, but your f is not continuous on [0,1]!! >> Try f(x) = x^2 arcsin(sin(1/x)) for x <> 0, 0 for x = 0. >> Note f is non-differentiable at x = 2/(pi n) for odd integers >> n. >Note that in this case the derivative at 0 (f ' (0)) does not exist >because of the 1/x. No, f'(0) does exist. The techniques you learn in a calculus class don't work here, but f'(0) exists nonetheless. This is very easy to see from the _definition_ of the derivative. > Also, if you take the derivative of f then try to >take the limit as x->0, you can see that we'll oscilate between -1 and >1. mr >> Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada ************************ David C. Ullrich === Subject: Re: Quick Question > ... Try f(x) = x^2 arcsin(sin(1/x)) for x <> 0, 0 for x = 0. Note f is non-differentiable at x = 2/(pi n) for odd integers n. >>Note that in this case the derivative at 0 (f ' (0)) does not exist >>because of the 1/x. No, f'(0) does exist. The techniques you learn in a calculus class >don't work here, but f'(0) exists nonetheless. This is very easy to >see from the _definition_ of the derivative. And, as we all know (to our sorrow), the technique of using the definition of the derivative (to compute the derivative of a given function, say) is *not* one of the techniques you learn in a calculus class, except for very small values of you. Lee Rudolph === Subject: Re: Quick Question >... > Try f(x) = x^2 arcsin(sin(1/x)) for x <> 0, 0 for x = 0. > Note f is non-differentiable at x = 2/(pi n) for odd integers > n. Note that in this case the derivative at 0 (f ' (0)) does not exist because of the 1/x. >>No, f'(0) does exist. The techniques you learn in a calculus class >>don't work here, but f'(0) exists nonetheless. This is very easy to >>see from the _definition_ of the derivative. And, as we all know (to our sorrow), the technique of using the definition >of the derivative (to compute the derivative of a given function, say) >is *not* one of the techniques you learn in a calculus class, except >for very small values of you. It might be kinder to say except for very few values of 'you'. >Lee Rudolph ************************ David C. Ullrich === Subject: Re: Quick Question On the real line, if I have a function continuous on [0,1] >>interval, the derivitive at 0 exists f '(0) exists f(0) =0 >how could I prove that f ' exists for all x in some ball >>around the point 0? >> It's not true. Let f(x) = x^2 if x is irrational, 0 >>if x is rational. Yeah, but your f is not continuous on [0,1]!! >> Try f(x) = x^2 arcsin(sin(1/x)) for x <> 0, 0 for x = 0. >> Note f is non-differentiable at x = 2/(pi n) for odd integers >> n. >Note that in this case the derivative at 0 (f ' (0)) does not exist >because of the 1/x. Also, if you take the derivative of f then try to >take the limit as x->0, you can see that we'll oscilate between -1 and >1. The derivative at 0 does exist and is 0. There is no limit of the derivative as x -> 0: that is a different question. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Quick Question On the real line, if I have a function continuous on [0,1] >>interval, the derivitive at 0 exists f '(0) exists f(0) =0 >how could I prove that f ' exists for all x in some ball >>around the point 0? >> It's not true. Let f(x) = x^2 if x is irrational, 0 >>if x is rational. Yeah, but your f is not continuous on [0,1]!! >> Try f(x) = x^2 arcsin(sin(1/x)) for x <> 0, 0 for x = 0. >> Note f is non-differentiable at x = 2/(pi n) for odd integers >> n. > Note that in this case the derivative at 0 (f ' (0)) does not exist > because of the 1/x. Also, if you take the derivative of f then try to > take the limit as x->0, you can see that we'll oscilate between -1 and > 1. lim_{x->0} | x^2 arcsin(sin(1/x)) - 0|/|x| = 0, b/c | x^2 arcsin(sin(1/x)) - 0|/|x| <= Pi|x|/2, so lim_{x->0} x^2 arcsin(sin(1/x)) - 0/x = 0, proving that f ' (0)=0. mr >> Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada > === Subject: Re: Need help with computing volumes using the divergence (Gauss) theorem that, it will ultimately affect the numerical value of the surface >integral over the boundary of the given volume. No, it won't. Any continuously differentiable vector field > will do, and you'll get the same numerical value (assuming you do > the integration correctly). > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Yes, of course, you are right. I can't understand what I was doing wrong all along. I was so confused with it, that -from some point on- I was almost certain that it was a matter of bad selection of f. Obviously, I wasn't replacing the surface integral with the appropriate symbolic solution... I can't explain it otherwise... now it works fine for any f, as long as its div is 1. (In fact the result seems to be scaled by f 's divergence) === Subject: Re: finding which fraction is bigger without a calculator robert, yeah i think your method is akin to what i was trying to do in my first reply... > Hi please help me, i am trying to figuree out which one of the two > fractions is bigger between 11/45 and 12/50. When I was in 5th or 6th grade I invented the following method: > If the denominators are larger, subtract the numerators, and vice > versa, until you have reduced the fractions to something obvious, or > until one numerator is larger and the other denominator is larger. > Applying the method to your numbers: > 11/45 and 12/50 > 11/34 and 12/38 > 11/23 and 12/26 > 11/12 and 12/14 > 11/1=11 and 12/2=6 (obviously 11 larger than 6 so we're done) === Subject: Re: finding which fraction is bigger without a calculator Hi please help me, i am trying to figuree out which one of the two >fractions is bigger between 11/45 and 12/50. >In general, how would i do this without trying to work it out >percentage wise? In October 2005, someone using the same name (jennifer) and > the same e-mail address posted to sci.math: I am trying to show that given a step function g on [a,b], there is a > continuous function h such that g(x) = h(x) except on a set of measure > less that epsilon/3. (and so on). Did you forget to wear your bicycle helmet once too often, or what??? Lee Rudolph My first thought when I saw this posting was Duh!! This is just two > people using the same account (presumably with permission from the > account owner). What on earth is Lee's problem? However, when I turned to the Oct 2005 Jennifer posting [don't get > judgmental about my time-wasting -- the whole process took me less > than 10 minutes or so], I saw that there were indeed stylistic > similarities between both Jennifer postings, making my first thought > unlikely. The Oct 2005 posting goes on: How do i start this please? I just need hints on this please and i will finish it. > Very similar in prose style. This seems to be just an incredibly extreme example of someone > attempting to understand something (in Oct 2005 with step-functions, > continuity, etc.) without having established (anything like) the > foundations to make the (relatively) advanced study meaningful. While you're probably right, I'll play devil's advocate and ask: isn't it possible for someone to have an adequate understanding of measure theory and not be the least bit competent in terms of actual calculation techniques? I'm reminded of one of my calc professors who (literally) needed help adding and subtracting two digit numbers. She would stare at the board for about 8-10 seconds if no one spoke up and would finally arrive at the answer. She was an instructor, though, and not a professor, so I don't know how much advanced math, if any, she knew. Jason === Subject: contour integration non uniform grid Hello Guys, Here is my problem. I have a set of datapoint i got from measures. Basicly for each set of X and Y i have a value Z, it's constitute a non uniform grid of about 2000 Z values. I need to integrate the Z values along X and Y, but knowing it's a non uniform grid i was thinking to use Contour Integartion but i don't know how to code that, so if someone could put me on the right track .... hope i'm clear, Michael === Subject: Recursive model sensitivity analysis I have a recursive model: a discrete multivariate Kalman filter. I currently have 30 variables in the model. I need to implement a model sensitivity analysis to: 1- Make sure that one of these variables has not become irrelevant in the state equation 2- Check for new variables that would increase the fit of the filter. I have a large pool of potential variables. These variables were not relevant in my filter in the past, but they could become in the future. To give you an idea of the magnitude, I have a pool of 500 potential variables and a maximum of 1 or 2 could become relevant in my filter each day. Theorically, with a Kalman filter I *could* put in 530 variables, and let the filter adjust. But it would probably give me false positives and include variables just to improve the fit. I'd like to identify the variables that have the most chance of being relevant on a daily basis. I think 30-40 variables would be manageable. If my model was not recursive, I would do a stepwise regression. I am aware of the dangers of using a stepwise regression, but in my case there would be no problem. I may be wrong, but I don't think stepwise regressions would work for a large number of variables and short time frame (1-2 days max). This would lead to a degrees of freedom problem. Anyway, I have never seen that. How would you approach the problem? === Subject: help with properties of logarithms hi im new to logaritms and i was confused on how to simplify an expression with the properties of logrrithms. i have a problem that contains both a log and a natural log ln(ex) + log3(1/3) how would i go about simplifing that, i break it down and then i get stuck === Subject: Re: help with properties of logarithms > hi im new to logaritms and i was confused on how to simplify an > expression with the properties of logrrithms. i have a problem that > contains both a log and a natural log ln(ex) + log3(1/3) how would i go about simplifing that, i break it down and then i get > stuck > You need to use the change of base formula for logarithms. Any logarithm of a base b, where b>0, can be expressed in terms of logarithms of any base a, where a>0. In your case, a and b could be e or 10, for instance. Hope this helps, Bill === Subject: Re: help with properties of logarithms > >hi im new to logaritms and i was confused on how to simplify an >>expression with the properties of logrrithms. i have a problem that >>contains both a log and a natural log >>ln(ex) + log3(1/3) >>how would i go about simplifing that, i break it down and then i get >>stuck >> You need to use the change of base formula for logarithms. Any logarithm >of a base b, where b>0, can be expressed in terms of logarithms of any base >a, where a>0. In your case, a and b could be e or 10, for instance. Hope >this helps, Bill > ????? 1/3 = 3^(-1), so log3(1/3) = -1. Assuming ex = e*x, the expression simplifies to ln x (= ln e + ln x - 1 = 1 + ln x - 1). -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: help with properties of logarithms <44DFEA22.7070708@netscape.net >>hi im new to logaritms and i was confused on how to simplify an >>expression with the properties of logrrithms. i have a problem that >>contains both a log and a natural log >>ln(ex) + log3(1/3) >>how would i go about simplifing that, i break it down and then i get >>stuck You need to use the change of base formula for logarithms. Any logarithm >of a base b, where b>0, can be expressed in terms of logarithms of any base >a, where a>0. In your case, a and b could be e or 10, for instance. Hope >this helps, Bill > ????? 1/3 = 3^(-1), so log3(1/3) = -1. Assuming ex = e*x, the expression > simplifies to ln x (= ln e + ln x - 1 = > 1 + ln x - 1). > -- > Stephen J. Herschkorn sjherschko@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhattan no the expression is ln(ex) not ln(e^x) === Subject: Re: help with properties of logarithms <44DFEA22.7070708@netscape.net >>hi im new to logaritms and i was confused on how to simplify an >>expression with the properties of logrrithms. i have a problem that >>contains both a log and a natural log >>ln(ex) + log3(1/3) >>how would i go about simplifing that, i break it down and then i get >>stuck You need to use the change of base formula for logarithms. Any logarithm >of a base b, where b>0, can be expressed in terms of logarithms of any base >a, where a>0. In your case, a and b could be e or 10, for instance. Hope >this helps, Bill > ????? 1/3 = 3^(-1), so log3(1/3) = -1. Assuming ex = e*x, the expression > simplifies to ln x (= ln e + ln x - 1 = > 1 + ln x - 1). > -- > Stephen J. Herschkorn sjherschko@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhattan no the expression is ln(ex) not ln(e^x) pi10^x which is pi times 10 to the x is equal to e^x or e to the x how do i get both xs out with a natural log then a regular log, i tried doing it but i cant find an answer with algerbra, the answer is x = === Subject: Re: help with properties of logarithms expression with the properties of logrrithms. i have a problem that > contains both a log and a natural log ln(ex) + log3(1/3) how would i go about simplifing that, i break it down and then i get > stuck > You need to use the change of base formula for logarithms. Any logarithm > of a base b, where b>0, can be expressed in terms of logarithms of any base > a, where a>0. In your case, a and b could be e or 10, for instance. Hope > this helps, Bill but how do i simplify it like ln(e) + ln(x) + log3(1) - log3(3) where we have log base 3 of 1 and 3 === Subject: Re: help with properties of logarithms > hi im new to logaritms and i was confused on how > to simplify an expression with the properties of > logrrithms. i have a problem that contains both a > log and a natural log ln(ex) + log3(1/3) how would i go about simplifing that, i break it > down and then i get stuck > You need to use the change of base formula for > logarithms. Any logarithm of a base b, where b>0, > can be expressed in terms of logarithms of any base > a, where a>0. In your case, a and b could be e or > 10, for instance. Hope this helps, Bill > > but how do i simplify it like > > ln(e) + ln(x) + log3(1) - log3(3) where we have log > base 3 of 1 and 3 Yes, but keep going. Hint: log_b(b) = 1 and log_b(1) = 0 Kyle Czarnecki === Subject: Rsa type composites and pi. Yet another hair brained idea! Composite within pi that have two equal length prime factors --- floor(pi(10^17)) = 314159265358979323 = 317213509*990371647 The ratio between factors = 3.12209795264425513479.. Interesting here is that the ratio between the two factors ~ pi. It is more than likely just a coincidence! The next candidate that has possibly only two prime factors but not known if they are of equal length is --- floor(pi*(10^103)) = 31415926535897932384626433832795028841971693993751 05820974944592307816406286208998628034825342117067 9821 This is a novel idea but, if the prime factors are of equal length will their ratio be even closer to pi than the first two factors of floor(pi(10^17))? BTW that second composite is an excellent factoring challenge. Dan === Subject: Re: Finding large primes, quickly [Tim Peters] >> ... >> Hey, I /like/ his prime-counting formula :-) It's a nice little piece of >> work, although grossly obscured by the convoluted ways he presents it. >> If you let s(n, a) = the number of integers /not/ crossed off by the >> Sieve of Eratosthenes applied to the naturals <= n after sieving against >> the first `a` primes, his formula follows immediately from repeatedly >> expanding (and then obfuscating the result, alas) the easily-proved >> recurrence: >> s(n, a) = s(n, a-1) - (s(n/p_a, a-1) - (a-1)) >> where p_a is the a'th prime, and where p_a <= n is assumed. Oops! That assumes p_a^2 <= n, not the weaker p_a <= n stated. The a-1 correction assumes that the first a-1 primes are included in 1 .. n/p_a, and p_a^2 <= n is strong enough to ensure that n/p_a >= p_a > p_(a-1), but p_a <= n isn't strong enough. As noted later, whenever p_a^2 > n, s(n, a) = s(n, pi(sqrt(n))), and then the recurrence applies directly. >> The parenthetical term on the RHS is the number of naturals <= n whose >> smallest prime divisor is p_a, exluding p_a itself. Now pause to prove >> it -- James seems to believe it requires God-like mutant brain wiring to >> succeed :-) [jstevh@msn.com] > Nope. I have merely challenged people to re-derive the full formula, > from scratch. You tend to always make jumps without making anything like a > step-by-step lucid exposition of a derivation which ends with the FULL > equation, which does not need a prime list. That's because it's so simple most people who know about this stuff don't need to see a tedious derivation. For goodness sake, from: s(n, a) = s(n, a-1) - (s(n/p_a, a-1) - (a-1)) assuming p_a^2 <= n (which, as I said, can be trivially derived from Legendre's recurrence for phi -- or deduced directly from reasoning about how the Sieve of Eratosthenes works) it follows immediately (by applying the same recurrence to s(n, a-1), then to s(n, a-2), ..., and finally to s(n, 1)) that: s(n, a) = s(n, 0) - sum for i=1 to a of s(n/p_i, i-1) - (i-1) assuming p_a^2 <= n s(n, 0) is floor(n)-1 (when you sieve against /no/ primes, only 1 gets crossed out), so that's: s(n, a) = floor(n) - 1 - sum for i=1 to a of s(n/p_i, i-1) - (i-1) [1] assuming p_a^2 <= n The rest is just adding layers of obfuscation. First slam in pi() so you can sum over all integers instead of just the primes: s'(x, y) = [2] floor(x) - 1 - sum for i=2 to y of if i is prime then s'(x/i, i-1) - pi(i-1) else 0 assuming y^2 <= x where again if y^2 > x, y should be replaced with floor(sqrt(y)) first. First note that, assuming y^2 <= x, s'(x, y) = s'(x, p_(pi(y))) because all /composites/ <= y are ignored in [2]. y can therefore be replaced by the largest prime <= y without affecting the result. So the only question remaining is whether s'(x, y) = s'(x, p_(pi(y))) as defined by [2] equals s(x, pi(y)) as defined by [1]. How tedious do you want this to get? Since i goes from 1 to y in [2], i obviously takes on the values of all the primes <= y: p_1, p_2, ..., p_(pi(y)), and no other primes. That corresponds exactly to the set of primes looked at in [1] when evaluating s(n, pi(y)): p_1, p_2, ..., p_a where a=pi(y). Where [2] has pi(i-1), i is a prime p_j for some j, and then pi(i-1) = pi(p_j - 1) = j-1, so [2] subtracts j-1 at the same point [1] subtracts j-1 when [1] looks at p_j. Finally, it doesn't matter that [2] recurses on s'(x/p_j, p_j - 1) instead of on s'(x/p_j, p_(j-1)), because as already shown the second argument can be replaced by the largest prime not exceeding it, and by definition p_(j-1) is the largest prime less than p_j. In short, s'(x/p_j, p_j - 1) = s'(x/p_j, p_(j-1)) so again [2] is computing exactly the same thing as [1]. What else? Ah, right, instead of an obvious conditional expression, you use a technical trick of spelling i is prime as pi(i) - pi(i-1), relying on that z*1 = z and z*0 = 0. So plug that trick into [2] to get: s'(x, y) = floor(x) - 1 - sum for i=2 to y of [s'(x/i, i-1) - pi(i-1)] * [pi(i) - pi(i-1)] assuming y^2 <= x But that still doesn't look messy enough :-(. OK, you also obscure it by spelling pi(k) as s'(k, sqrt(k)) everywhere. That's /almost/ wholly mechanical, except that we can no longer assume the second argument is an integer then, so an adjustment to the upper summation bound is also needed: s'(x, y) = floor(x) - 1 - sum for i=2 to floor(y) of [3] [s'(x/i, i-1) - s'(i-1, sqrt(i-1))] * [s'(i, sqrt(i)) - s'(i-1, sqrt(i-1))] assuming y^2 <= x That's it, right? If so, sorry, but IMO going from the base recurrence to [1] is trivial, and everything beyond that to get to [3] is pretty much pointless. >> Of course this is strongly related to the slightly cleaner Legendre >> recurrence for the number of naturals <= `n` not divisible by any of the >> first `a` primes: >> phi(n, a) = phi(n, a-1) - phi(n/p_a, a-1) >> Indeed, either recurrence can be derived mechanically from the other, via >> noting that (given reasonable bounds): >> s(n, a) = phi(n, a) + a - 1 >> Or, IOW, s(n, a) doesn't count the integer 1 (the SOE crosses that off, >> while phi counts it), but does count the first `a` primes (the SOE does >> not cross those off: it only crosses out non-trivial multiples of the >> primes it sieves against, while phi doesn't distinguish trivial from non- >> trivial multiples). > Now you're going into old territory for me as I've previously given the > mathematical relationship between my prime counting function in its > SIEVE form, and other methods like Legendre's. As I said, the `s` recurrence can be trivially derived from Legendre's phi recurrence, which contradicts your frequent claim that it's impossible to get to what you did starting from what Legendre did. It's not only possible, it's straightforward (albeit tedious). > One key difference, of course, between my prime counting function and > others is that it has something other than a sieve form--when it uses a > partial difference equation--which the other ways cannot do. That is a key point. To you it is (you say so often enough ), but nobody else finds it interesting. To save endless repetition, please realize that you're not going to convince anyone it's important unless you can /use/ it to /prove/ something interesting. And no, /guessing/ that the Riemann Hypothesis is false is neither interesting nor a proof. >> The `s` recurrence has one nice feature, which can speed a dirt-dumb >> implementation: if p_a^2 > n, and r = pi(sqrt(n)) (i.e., p_r is the >> largest >> prime whose square doesn't exceed n), >> s(n, a) = s(n, r) >> exactly. That particular trick is messier for phi, because (still >> assuming p_a^2 > n) phi(n, r) counts all the primes from p_(r+1) through >> n, but phi(n, a) doesn't. OTOH, the phi recurrence can be >> straightforwardly optimized in deeper ways that get messier if you try >> to apply them to the `s` recurrence. >> ... >> BTW, given a table of canned primes through sqrt(n), the obvious thing is >> actually true :-): a program to compute pi(n) based on James's formula >> (but /just/ looking at the primes, skipping all the composites and >> entirely >> throwing away the part that computes /whether/ a number is prime) makes >> exactly as many recursive calls, and with exactly the same arguments, as >> a program derived from expanding Legendre's phi recurrence in the same >> way. > Maybe... Well, this will be dead obvious if I spell out what I did. The James-like program I'm speaking of is a direct transcription of s(n, a) = n-1 - sum for i=1 to a of s(n/p_i, i-1) - (i-1) along with a bit of fiddling upon entry to reduce `a` if necessary so that p_a^2 <= n. Here's the guts (coded in Python); the driver that calls this ensures that n >= 3; if you stare at it, you'll find that it doesn't quite look right, which is because the list of primes (`ps` below) is indexed starting at 0 instead of the 1-based indexing assumed in the statement of the recurrence: def s(n, i): while psq[i] > n and i > 0: i -= 1 result = (n + 1) >> 1 for j in xrange(1, i+1): result -= s(n // ps[j], j-1) return result + i*(i+1)//2 Note that the i-1 terms from the body of the summation in the recurrence have been collapsed into a single computation of C(i+1, 2) on the last line. The program based on Legendre's recurrence is much the same, a trivial little direct transcription of the similar summation immediately derived from his phi recurrence: phi(n, a) = n - sum for i=1 to a of phi(n/p_i, i-1) This was also fiddled to force p_a^2 <= n upon entry, although as explained before such fiddling is /not/ neutral to the phi result: if `a` needs to be reduced k times to make p_a^2 <= n upon entry, then the result returned must also be reduced by k. It should be obvious from inspection then that both program derived from transcribing the summation formulas above must make exactly the same number of calls with exactly the same arguments. > but years ago when the issue came up of which was faster, some > poster implemented Legendre's Method and ran it in a head to head > against an early version of my prime counting function which thoroughly > beat it. What on Earth is Legendre's Method? If you're talking about Legendre's /formula/ identified as (2) on http://mathworld.wolfram.com/LegendresFormula.html then yes, either of the dead-simple programs described above will run circles around it. But nobody (well, nobody who knows what they're doing) uses the inclusion/exclusion formula directly as written there to compute pi(n): it contains 2^pi(sqrt(n)) terms (i.e., there's a term for every subset of the primes <= sqrt(n), and there are 2^S subsets of a set of size S). That becomes intractable very quickly as n increases. > I did some optimizations myself, and came up with a Java program that > was so fast, posters quit playing games like that and instead switched > to using comparisons with the fastest known using the latest research. I've seen those messages, and it was obvious that the reason they started comparing against modern methods for computing pi was that you were claiming over and over that your formula /had/ to deliver the fastest /possible/ prime-counting algorithm. When you claim fastest possible, comparing against fastest known is thoroughly reasonable. BTW, you neglected to mention that your Java program did poorly against those. > Now to me, it's a boring area, which is why I mostly let it drop, as > I'm more interested in the partial difference equation, so I haven't > worried about such details much. Or it's possible that it would have taken real work to make your program competitive with others, and you weren't up to the challenge. >> James's runs a little slower because of the extra a-1 term in the >> recurrence, but because the expanded form sums over all integers in 1 >> through a, that term can be removed from the body of the summation loop, >> and replaced with (1-1)+(2-1)+...+(a-1) = a*(a-1)/2 once before or after >> the loop. The runtimes seems indistiguishable then. > I think you're just way off there and it sounds like you are guessing > versus having actually ran an implementation, but maybe not. Of course I implemented them: see above for details. Both were trivial. I'll note that both versions start by (in effect -- the code is actually cleverer than this) running a sieve to find the primes up to n^(1/4), and then using those in another sieve to find the primes up to n^(1/2). > I did optimize my Java programs in very particular ways, which made > them rather fast. Really? For example, in the message with id #SPt0BUBCHA.1636@cpimsnntpa03 Oh yeah, I put a timer on the program, after one quick fix (I break out of a certain loop when things go negative) and checked a run counting the number of primes up to 10000000. It took it 12.8 seconds and it outputted all the primes up to 3,137 before giving the correct figure of 664, 579. The dead-stupid programs mentioned above compute pi(10000000) correctly in about 0.4 seconds on my desktop box coded in unoptimized Python, including the time to find the primes up to the square root first. Using the psyco JIT for Python reduced it to 0.06 seconds. > Here's a challenge to you: Can you write an implementation of > Legendre's that can even do pi(10^13)? With methods /derived from/ Legendre's recurrence, yes, can and have. The dead-stupid programs mentioned above /can/ compute it exactly as written (Python has no bound on integer size), but not in an amount of time I'd consider reasonable. BTW, for future reference, I'm close to the last person on Earth you'd want to challenge to a program optimization contest :-) > My fastest Java program can do that in about ten minutes on a regular > desktop. Then it probably wasn't the program referenced above. > If my suspicions are correct, you can't write a program that will do it > in days with Legendre's Method. As above, I don't know what you mean by Legendre's Method, but if you're talking about a straightforward implementation of Legendre's inclusion/exclusion formula, /nobody/ could: the formula contains 2^227647 (2^pi(sqrt(10^13))) terms, and nobody could compute that over the lifetime of the universe. There are several approaches /derived/ from Legendre's formulas (the inclusion/exclusion summation, and the recurrence formula for phi) that can do it in reasonable time. === Subject: Re: Finding large primes, quickly > Am I correct in assuming you moved to San Francisco from > Atlanta to flee from your family (your daddy in particular) > because you couldn't face the shame and humiliation of > having been found out that the prodigy has been simply > bullting his way thorugh life using math-ese? > > Feel free to correct me if I'm wrong. > > I'll take silence to mean I'm right. > Well, James has not replied to this post, yet..... In the meantime, could you, or anyone else, elaborate on this bit of information, please. Such as, how you came across it, or heard about it. I suspect knowing this, will answer alot of the other questions that have been asked, and have gone unanswered. === Subject: Re: Finding large primes, quickly <12dqk142edj2678@corp.supernews.com> Atlanta to flee from your family (your daddy in particular) > because you couldn't face the shame and humiliation of > having been found out that the prodigy has been simply > bullting his way thorugh life using math-ese? Feel free to correct me if I'm wrong. I'll take silence to mean I'm right. > Well, James has not replied to this post, yet..... In the meantime, could you, or anyone else, elaborate on this bit of > information, please. Such as, how you came across it, or heard about it. The information comes straight from the horse's mouth. Feb 20, 2005: > After a long involved discussion with my father, I've > decided to reverse my previous decision, and release > the basic algorithm. I personally still worry about it, but he thinks I'm > just paranoid. May 5, 2005: > Like, I actually had a family member who didn't realize > I'd deleted the mathforprofit blog and someone took it > over, who actually thought that maybe it was me, ranting > and raving on the blog, like that I'd completely lost it. Sci.math hiatus Jun 2005 to Jan 2006. Feb 8, 2006: > So I made it my business to drive to Nashville, which was > a bit over a four hour trip as I was then in the Atlanta area. On his blog, he says he's in San Francisco. And although he deletes such posts, they are archived at Math Forum. Now it COULD just be a coincidence that the family issues immediately preceded his 7 month absence (and gives him enough time to relocate to San Francisco). That's why I asked for confirmation. I consider the probability of coincidence to be extremely low, that's why I'll consider silence to be confirmation. I suspect knowing this, will answer alot of the other questions that > have been asked, and have gone unanswered. > === Subject: Re: Finding large primes, quickly linux) You will continue to strive for fame and glory, and you will always > fall short. Whereas your witty rejoinders are a contribution to humanity. I'm sure that your mama is proud. Or at least she might be, if you weren't so ashamed of your own time-wasting activity that you use a silly and insulting pseudonym. -- Jesse F. Hughes I guess it's a passable day to die. -- Lt. Dwarf, /Star Wreck:In the Pirkinning === Subject: Re: The MYTH that software, inevitably, contains defects Foolish argument. The fact that a uniform proof procedure cannot prove all programs correct does not prevent one from proving that a particular program correctly implements its specification. === Subject: Re: The MYTH that software, inevitably, contains defects correct does not prevent one from proving that a particular program > correctly implements its specification. There are also theorems that state results along the lines that, for each integer N, you can determine whether a program halts, if it can be specified in less than N bits (as a Turing machine). --- Christopher Heckman === Subject: Re: The MYTH that software, inevitably, contains defects correct does not prevent one from proving that a particular program > correctly implements its specification. There are also theorems that state results along the lines that, for > each integer N, you can determine whether a program halts, if it can be > specified in less than N bits (as a Turing machine). Wouldn't this contradict Goedel's first incompleteness theorem? === Subject: Re: The MYTH that software, inevitably, contains defects I should be clearer, since Walter Roberson misunderstood. The fact that a uniform proof procedure U cannot prove all programs correct--- indeed the fact that a uniform proof procedure U does not exist under specified circumstances--- does not prevent someone S or some algorithm A from proving that a particular program P correctly implements its specification I. Indeed, many programs can and have been proved correct (and some of their proofs are probably correct too). The impossibility of solving the Halting problem has not prevented many people from proving that their algorithms halt. Indeed the usual definition of an algorithm requires that the procedure halt with the correct answer. Failing to find U does not mean that one cannot find A1, A2, A3 .... for programs P1, P2, P3. etc. A further argument can be made that no procedure programmed to run on computers that actually exist today can in fact simulate a Turing machine. All of our computers are finite state machines, having only a programs can/cannot do with what Turing machines can/cannot do are --theoretically-- all wet. Unless they are somehow talking about (potentially) infinite machines, none of which can ever be built. It may be very time consuming to prove that a program running on (say) a is not theoretically impossible. It is a finite state machine, and after a certain number of steps it must either halt or return to a previous state, in which case it is in a loop. (How many steps? 2^(8^(252X10^9)) or so. A finite number.) Thus any theoretical argument about programs halting must hypothesize infinite computers. We don't have any infinite computers. So the argument about certain things being hard is -- because they are hard, not impossible. As for the original question about software necessarily containing defects. Obviously not. Some software does, and some does not. RJF > Foolish argument. The fact that a uniform proof procedure cannot prove all programs > correct does not prevent one from proving that a particular program > correctly implements its specification. === Subject: Re: The MYTH that software, inevitably, contains defects Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) >I should be clearer, since Walter Roberson misunderstood. No, I don't think I misunderstood at all. >The fact that a uniform proof procedure U cannot prove all programs >correct--- >indeed the fact that a uniform proof procedure U does not exist under >specified circumstances--- does not prevent someone S or some >algorithm A from proving that a particular program P correctly >implements its specification I. If the program P is given not in advance, and I is non-trivial, then S or A must create the universal proof procedure U(I), which is something we know to be impossible. If the program P is given in advance, then S or A must create a new program Q(P) or Q(P,I) to examine whether P implements I. However, if it is A (i.e, an algorithm rather than a person), then in order for A to be able to consistantly create Q(P) or Q(P,I) then A would be a universal proof procedure, and as per the above we know that isn't possible. We are thus reduced down to the situation that S (some*one*) examines the pre-specified P and attempts to prove that it implements I. Sometimes they can do so, and other times, for even simple specifications, they cannot reasonably find any such proof. Either way, the effort is human and customized around pre-specified P, rather than being a generalized mechanism. This is not a scalable approach, and there is no *practical* way to apply it to large programs. >Indeed, many programs can and have >been proved correct (and some of their proofs are probably correct >too). For the purposes of this thread, having correctness proofs that are probably correct is not sufficient: the correctness proofs themselves must be provably correct. This isn't a thread about PROBABLY no defects, this is a thread about NO defects. I would question the scope of the many you used in your statement. An indefinite number of similar near-trivial programs qualifies under many: how often have the proofs been found for arbitrary moderately-complex programs? >All of our computers are finite state machines, having only a >programs can/cannot do with what Turing machines can/cannot do are >--theoretically-- all wet. Unless they are somehow talking about >(potentially) infinite machines, none of which can ever be built. It >may be very time consuming to prove that a program running on (say) a >is not theoretically impossible. It is a finite state machine, and >after a certain number of steps it must either halt or return to a >previous state, in which case it is in a loop. (How many steps? >2^(8^(252X10^9)) or so. A finite number.) And if you managed to extract 2 bits of information from every of storage, in which you would attempt to track the roughly 2^201600 states of the problem. A terabyte (2^40 bytes) of storage is not enough >Thus any theoretical argument about programs halting must hypothesize >infinite computers. We don't have any infinite computers. So the >argument about certain things being hard is -- because they are hard, >not impossible. Is it hard to get 2^(8^(252X10^9)) bits of storage to track your example program? Or is it impossible to get that storage, and the ways we can impose state upon them? >As for the original question about software necessarily containing >defects. Obviously not. Some software does, and some does not. The statement was, A very dangerous, destructive, devastating myth which bars progress and human society well-being. Are there really that many *useful* non-trivial programs which can, with reasonable effort, have their correctness proven from specifications? Or, alternately, for given non-trivial specification, are there methods to generate or formulate correct programs? Keep in mind for this purpose that A) it was tried before; and B) it didn't work. Automated code generation depends upon the correctness of the specifications, and it is well established that people often make mistakes in writing the specifications... -- No one has the right to destroy another person's belief by demanding empirical evidence. -- Ann Landers === Subject: Re: The MYTH that software, inevitably, contains defects The fact that a uniform proof procedure U cannot prove all programs >correct--- >indeed the fact that a uniform proof procedure U does not exist under >specified circumstances--- does not prevent someone S or some >algorithm A from proving that a particular program P correctly >implements its specification I. If the program P is given not in advance, and I is non-trivial, > then S or A must create the universal proof procedure U(I), which > is something we know to be impossible. Of course the program P and its specification are given in advance. What you are saying is analogous to saying there is no such thing as national defense unless it can defend against any offensive weapon, including those not yet invented. While such a defensive system would certainly be nice to have, it doesn't prevent us from having specific defenses against specific threats. If the program P is given in advance, then S or A must create a > new program Q(P) or Q(P,I) to examine whether P implements I. > However, if it is A (i.e, an algorithm rather than a person), then > in order for A to be able to consistantly create Q(P) or Q(P,I) > then A would be a universal proof procedure, and as per the above > we know that isn't possible. If the program P is given in advance, its proof is a constant, and need not be computed by a universal program, or any other program. We are thus reduced down to the situation that S (some*one*) > examines the pre-specified P and attempts to prove that it > implements I. Sometimes they can do so, and other times, for > even simple specifications, they cannot reasonably find any > such proof. That is exactly right. And sometimes one person cannot prove a program satisfies its specification, and another person can. This happens all the time in mathematics education, where a student can't find the proof, even if there is one. Either way, the effort is human and customized > around pre-specified P, rather than being a generalized mechanism. Yes, that is true. > This is not a scalable approach, and there is no *practical* way > to apply it to large programs. It is more effective than trying to find a universal proof procedure --- which incidentally tends to require that the specifications for programs be written in some unscalable way, namely by humans specifying what programs are supposed to do. If you have computers writing both the specifications and the programs, the problem is trivialized because the computer program could easily say the specification is what the program does and the program is correct. In a practical sense it only makes sense to prove a program correct that has a meaning independent of the program. There are clearly infinite classes of provably correct programs of extremely limited interest. Indeed, many programs can and have >been proved correct (and some of their proofs are probably correct >too). For the purposes of this thread, having correctness proofs that are > probably correct is not sufficient: the correctness proofs > themselves must be provably correct. As has been shown repeatedly in the mathematics literature, published proofs are sometimes incorrect. Sometimes the theorems being proved are false. Sometimes published proofs of programs are incorrect. It has been observed, I think correctly, that the notion of proof in inherently a human notion (by Demillo, Lipton); and so ultimately whether you believe a proof to be correct is a judgment, not an absolute. In fact, computers make errors in hardware (cosmic rays etc), so unless you have a more reliable mechanical computation in mind, a proof that a proof is correct would be statistical. > This isn't a thread about > PROBABLY no defects, this is a thread about NO defects. I would question the scope of the many you used in your statement. > An indefinite number of similar near-trivial programs qualifies > under many: how often have the proofs been found for arbitrary > moderately-complex programs? Maybe people don't really need proofs to know that a program is correct. It is enough for most people to have extensive tests. >All of our computers are finite state machines, having only a >programs can/cannot do with what Turing machines can/cannot do are >--theoretically-- all wet. Unless they are somehow talking about >(potentially) infinite machines, none of which can ever be built. It >may be very time consuming to prove that a program running on (say) a >is not theoretically impossible. It is a finite state machine, and >after a certain number of steps it must either halt or return to a >previous state, in which case it is in a loop. (How many steps? >2^(8^(252X10^9)) or so. A finite number.) And if you managed to extract 2 bits of information from every > of storage, in which you would attempt to track the roughly 2^201600 > states of the problem. A terabyte (2^40 bytes) of storage is not enough Some computation like that is possible. However, a finite machine is not a Turing machine. >Thus any theoretical argument about programs halting must hypothesize >infinite computers. We don't have any infinite computers. So the >argument about certain things being hard is -- because they are hard, >not impossible. Is it hard to get 2^(8^(252X10^9)) bits of storage to track > your example program? Or is it impossible to get that storage, > and the ways we can impose state upon them? Turing machine arguments must hold in larger universes too. >As for the original question about software necessarily containing >defects. Obviously not. Some software does, and some does not. The statement was, A very dangerous, destructive, devastating myth > which bars progress and human society well-being. > Are there really that many *useful* non-trivial programs which > can, with reasonable effort, have their correctness proven from > specifications? Or, alternately, for given non-trivial specification, > are there methods to generate or formulate correct programs? > Keep in mind for this purpose that A) it was tried before; and > B) it didn't work. Automated code generation depends upon the > correctness of the specifications, and it is well established that > people often make mistakes in writing the specifications... As I have said, proofs of programs are not what is needed by most people. Partly because a proof might be wrong. Partly because they are hard to discover. Partly because accurate specs are so difficult to write. Though proofs of programs might be partial confirmation of likely correctness -- so there is some rational interest in the topic. Yet people test programs. And run on that basis.. > -- > No one has the right to destroy another person's belief by > demanding empirical evidence. -- Ann Landers === Subject: Re: The MYTH that software, inevitably, contains defects The fact that a uniform proof procedure U cannot prove all programs >correct--- >indeed the fact that a uniform proof procedure U does not exist under >specified circumstances--- does not prevent someone S or some >algorithm A from proving that a particular program P correctly >implements its specification I. If the program P is given not in advance, and I is non-trivial, > then S or A must create the universal proof procedure U(I), which > is something we know to be impossible. Of course the program P and its specification are given in advance. What you are saying is analogous to saying there is no such thing as national defense unless it can defend against any offensive weapon, including those not yet invented. While such a defensive system would certainly be nice to have, it doesn't prevent us from having specific defenses against specific threats. If the program P is given in advance, then S or A must create a > new program Q(P) or Q(P,I) to examine whether P implements I. > However, if it is A (i.e, an algorithm rather than a person), then > in order for A to be able to consistantly create Q(P) or Q(P,I) > then A would be a universal proof procedure, and as per the above > we know that isn't possible. If the program P is given in advance, its proof is a constant, and need not be computed by a universal program, or any other program. We are thus reduced down to the situation that S (some*one*) > examines the pre-specified P and attempts to prove that it > implements I. Sometimes they can do so, and other times, for > even simple specifications, they cannot reasonably find any > such proof. That is exactly right. And sometimes one person cannot prove a program satisfies its specification, and another person can. This happens all the time in mathematics education, where a student can't find the proof, even if there is one. Either way, the effort is human and customized > around pre-specified P, rather than being a generalized mechanism. Yes, that is true. > This is not a scalable approach, and there is no *practical* way > to apply it to large programs. It is more effective than trying to find a universal proof procedure --- which incidentally tends to require that the specifications for programs be written in some unscalable way, namely by humans specifying what programs are supposed to do. If you have computers writing both the specifications and the programs, the problem is trivialized because the computer program could easily say the specification is what the program does and the program is correct. In a practical sense it only makes sense to prove a program correct that has a meaning independent of the program. There are clearly infinite classes of provably correct programs of extremely limited interest. Indeed, many programs can and have >been proved correct (and some of their proofs are probably correct >too). For the purposes of this thread, having correctness proofs that are > probably correct is not sufficient: the correctness proofs > themselves must be provably correct. As has been shown repeatedly in the mathematics literature, published proofs are sometimes incorrect. Sometimes the theorems being proved are false. Sometimes published proofs of programs are incorrect. It has been observed, I think correctly, that the notion of proof in inherently a human notion (by Demillo, Lipton); and so ultimately whether you believe a proof to be correct is a judgment, not an absolute. In fact, computers make errors in hardware (cosmic rays etc), so unless you have a more reliable mechanical computation in mind, a proof that a proof is correct would be statistical. > This isn't a thread about > PROBABLY no defects, this is a thread about NO defects. I would question the scope of the many you used in your statement. > An indefinite number of similar near-trivial programs qualifies > under many: how often have the proofs been found for arbitrary > moderately-complex programs? Maybe people don't really need proofs to know that a program is correct. It is enough for most people to have extensive tests. >All of our computers are finite state machines, having only a >programs can/cannot do with what Turing machines can/cannot do are >--theoretically-- all wet. Unless they are somehow talking about >(potentially) infinite machines, none of which can ever be built. It >may be very time consuming to prove that a program running on (say) a >is not theoretically impossible. It is a finite state machine, and >after a certain number of steps it must either halt or return to a >previous state, in which case it is in a loop. (How many steps? >2^(8^(252X10^9)) or so. A finite number.) And if you managed to extract 2 bits of information from every > of storage, in which you would attempt to track the roughly 2^201600 > states of the problem. A terabyte (2^40 bytes) of storage is not enough Some computation like that is possible. However, a finite machine is not a Turing machine. >Thus any theoretical argument about programs halting must hypothesize >infinite computers. We don't have any infinite computers. So the >argument about certain things being hard is -- because they are hard, >not impossible. Is it hard to get 2^(8^(252X10^9)) bits of storage to track > your example program? Or is it impossible to get that storage, > and the ways we can impose state upon them? Turing machine arguments must hold in larger universes too. >As for the original question about software necessarily containing >defects. Obviously not. Some software does, and some does not. The statement was, A very dangerous, destructive, devastating myth > which bars progress and human society well-being. > Are there really that many *useful* non-trivial programs which > can, with reasonable effort, have their correctness proven from > specifications? Or, alternately, for given non-trivial specification, > are there methods to generate or formulate correct programs? > Keep in mind for this purpose that A) it was tried before; and > B) it didn't work. Automated code generation depends upon the > correctness of the specifications, and it is well established that > people often make mistakes in writing the specifications... As I have said, proofs of programs are not what is needed by most people. Partly because a proof might be wrong. Partly because they are hard to discover. Partly because accurate specs are so difficult to write. Though proofs of programs might be partial confirmation of likely correctness -- so there is some rational interest in the topic. Yet people test programs. And run on that basis.. > -- > No one has the right to destroy another person's belief by > demanding empirical evidence. -- Ann Landers === Subject: Re: The MYTH that software, inevitably, contains defects Foolish argument. The fact that a uniform proof procedure cannot prove all programs correct does not prevent one from proving that a particular program correctly implements its specification. === Subject: Re: The MYTH that software, inevitably, contains defects > Foolish argument. > > The fact that a uniform proof procedure cannot prove all programs > correct does not prevent one from proving that a particular program > correctly implements its specification. It is a practical matter. If a program is sufficiently complicated and contains many branches and paths, the cobinatorics defeat any reasonable effort to test out the program in totality. 1. It is possible for a program to be correct, no matter how complicated. 2. It is a practical impossibility to show that a complicated program faithfully executes an algorith (which is really another program, perhaps of a different kind). Even using verification statement imbeddings does not make the proof of correctness for a sufficiently complicated program more certain. Bob Kolker > === Subject: Re: The MYTH that software, inevitably, contains defects Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) >Foolish argument. >The fact that a uniform proof procedure cannot prove all programs >correct does not prevent one from proving that a particular program >correctly implements its specification. If it only works on a particular program then it isn't a uniform proof procedure. It is true that there are programs about which some property can be proved and in a reasonable amount of time. However, one then reaches the question of What portion of real-life programs have properties that can be -usefully- be proven in a reasonable amount of time? Rice's Theorem, which I previously referenced, establishes that if the non-trivial property to be tested is pre-specified but the input (programs) to be tested are not pre-specified, then any general program capable of proving the property would be equivilent to solving The Halting Problem (i.e., impossible.) If the program to be tested is pre-specified before the analysis program is written, then Rice's Theorem does not apply -- but the implication is that the proof program would have to be rewritten (or customized) for -each- program to be tested. That's obviously not at all scalable. The Wikipedia page I cited for Rice's Theorem (or perhaps one linked to from it) gave specific examples of simple algorithms for which it is clear that proof analysis would be difficult or impossible, even though the specifications are easily stated. One of the examples given was a program to find the last twin prime. There are proofs, of course, that there are an infinite number of primes, and there are proofs that Mersenne numbers give rise to twin primes, but after more than 350 years, there has been no proof found that there are infinite numbers of Mersenne numbers (but it is strongly suspected.) -- There are some ideas so wrong that only a very intelligent person could believe in them. -- George Orwell === Subject: Re: The MYTH that software, inevitably, contains defects >The Wikipedia page I cited for Rice's Theorem (or perhaps one >linked to from it) gave specific examples of simple algorithms for >which it is clear that proof analysis would be difficult or impossible, >even though the specifications are easily stated. One of the >examples given was a program to find the last twin prime. There >are proofs, of course, that there are an infinite number of primes, >and there are proofs that Mersenne numbers give rise to twin primes, There are? None that I know of. How does a Mersenne number give rise to twin primes? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: The MYTH that software, inevitably, contains defects Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) >>and there are proofs that Mersenne numbers give rise to twin primes, >There are? None that I know of. How does a Mersenne number give >rise to twin primes? Sorry, it was late: I was thinking of perfect numbers. -- No one has the right to destroy another person's belief by demanding empirical evidence. -- Ann Landers === Subject: [Q] on filtrations in measure theory (http://en.wikipedia.org/wiki/Filtration_(abstract_algebra)): A sigma-algebra defines the set of events that can be measured, which in a probability context is equivalent to events that can be discriminated. Therefore a filtration is often used to represent the change in the set of events that can be measured, through gain or loss of information. A typical example is in mathematical finance, where a filtration represents the information available at each time t, and is more and more precise (the set of measurable events is staying the same or increasing) as information from the present becomes available. The example from mathematical finance *sounded* reasonable to me upon first reading it, but on further thought I have to concede that I really don't understand what it is saying. Could someone give me an example of an (unmeasurable) event that becomes measurable only after some key piece of information becomes available? kj -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. === Subject: Re: [Q] on filtrations in measure theory > (http://en.wikipedia.org/wiki/Filtration_(abstract_algebra)): > > A sigma-algebra defines the set of events that can be measured, > which in a probability context is equivalent to events that can > be discriminated. Therefore a filtration is often used to represent > the change in the set of events that can be measured, through > gain or loss of information. A typical example is in mathematical > finance, where a filtration represents the information available > at each time t, and is more and more precise (the set of measurable > events is staying the same or increasing) as information from > the present becomes available. > > The example from mathematical finance *sounded* reasonable to me > upon first reading it, but on further thought I have to concede > that I really don't understand what it is saying. Could someone > give me an example of an (unmeasurable) event that becomes measurable > only after some key piece of information becomes available? > > > kj The algebra of events known now, the events you can use in today's decisions, is different from what will be known next week. Unmeasurable event: the price of stock XYZ will be above $10 at the close of the stock exchange on August 21. Today, we don't know whether that event happens or not. But on August 22 we will know. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: [Q] on filtrations in measure theory > (http://en.wikipedia.org/wiki/Filtration_(abstract_algebra)): > > A sigma-algebra defines the set of events that can be measured, > which in a probability context is equivalent to events that can > be discriminated. Therefore a filtration is often used to represent > the change in the set of events that can be measured, through > gain or loss of information. A typical example is in mathematical > finance, where a filtration represents the information available > at each time t, and is more and more precise (the set of measurable > events is staying the same or increasing) as information from > the present becomes available. > > The example from mathematical finance *sounded* reasonable to me > upon first reading it, but on further thought I have to concede > that I really don't understand what it is saying. Could someone > give me an example of an (unmeasurable) event that becomes measurable > only after some key piece of information becomes available? > > > kj > You toss a coin at 9am, and win or loose $100 dollars according to the coin toss. At 8am, the event {you win the $100} is non-measurable. At 10am, this event is measurable. Stephen === Subject: Re: [Q] on filtrations in measure theory >> (http://en.wikipedia.org/wiki/Filtration_(abstract_algebra)): >> >> A sigma-algebra defines the set of events that can be measured, >> which in a probability context is equivalent to events that can >> be discriminated. Therefore a filtration is often used to represent >> the change in the set of events that can be measured, through >> gain or loss of information. A typical example is in mathematical >> finance, where a filtration represents the information available >> at each time t, and is more and more precise (the set of measurable >> events is staying the same or increasing) as information from >> the present becomes available. >> >> The example from mathematical finance *sounded* reasonable to me >> upon first reading it, but on further thought I have to concede >> that I really don't understand what it is saying. Could someone >> give me an example of an (unmeasurable) event that becomes measurable >> only after some key piece of information becomes available? >> >> >> kj >> >You toss a coin at 9am, and win or loose $100 dollars according to the >coin toss. >At 8am, the event {you win the $100} is non-measurable. >At 10am, this event is measurable. Sorry but that makes no sense to me; if the coin is known to be fair, the event you cite looks quite measurable to me, with measure 0.5, irrespective of time. kj -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. === Subject: Re: [Q] on filtrations in measure theory >In Stephen Filtrations >>(http://en.wikipedia.org/wiki/Filtration_(abstract_algebra)): A sigma-algebra defines the set of events that >can be measured, which in a probability context is equivalent to >events that can be discriminated. Therefore a filtration is often >used to represent the change in the set of events that can be >measured, through gain or loss of information. A typical example is >in mathematical finance, where a filtration represents the >information available at each time t, and is more and more precise (the >set of measurable events is staying the same or increasing) as >information from the present becomes available. The example from mathematical finance *sounded* >reasonable to me upon first reading it, but on further thought I >have to concede that I really don't understand what it is saying. >Could someone give me an example of an (unmeasurable) event that >becomes measurable only after some key piece of information becomes available? kj >You toss a coin at 9am, and win or loose $100 dollars >according to the >>coin toss. >At 8am, the event {you win the $100} is non-measurable. >At 10am, this event is measurable. >Sorry but that makes no sense to me; if the coin is >known to be >fair, the event you cite looks quite measurable to me, >with measure >0.5, irrespective of time. Your problem is that you don't know what measurable means in this context. It's talking about membership in a certain sigma-algebra. Basically, the events that are measurable at a certain time are those that are determined by the observations up to that time. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: [Q] on filtrations in measure theory > > (http://en.wikipedia.org/wiki/Filtration_(abstract_algebra)): > A sigma-algebra defines the set of events that can be measured, which in a probability context is equivalent to events that can be discriminated. Therefore a filtration is often used to represent the change in the set of events that can be measured, through gain or loss of information. A typical example is in mathematical finance, where a filtration represents the information available at each time t, and is more and more precise (the set of measurable events is staying the same or increasing) as information from the present becomes available. >The example from mathematical finance *sounded* reasonable to me upon first reading it, but on further thought I have to concede that I really don't understand what it is saying. Could someone give me an example of an (unmeasurable) event that becomes measurable only after some key piece of information becomes available? >>kj >> > >>You toss a coin at 9am, and win or loose $100 dollars according to the >>coin toss. > > >>At 8am, the event {you win the $100} is non-measurable. > > >>At 10am, this event is measurable. > > > > Sorry but that makes no sense to me; if the coin is known to be > fair, the event you cite looks quite measurable to me, with measure > 0.5, irrespective of time. No, its not measurable - by definition. === Subject: Re: [Q] on filtrations in measure theory >(http://en.wikipedia.org/wiki/Filtration_(abstract_algebra)): A sigma-algebra defines the set of events that can be measured, > which in a probability context is equivalent to events that can > be discriminated. Therefore a filtration is often used to represent > the change in the set of events that can be measured, through > gain or loss of information. A typical example is in mathematical > finance, where a filtration represents the information available > at each time t, and is more and more precise (the set of measurable > events is staying the same or increasing) as information from > the present becomes available. The example from mathematical finance *sounded* reasonable to me >upon first reading it, but on further thought I have to concede >that I really don't understand what it is saying. Could someone >give me an example of an (unmeasurable) event that becomes measurable >only after some key piece of information becomes available? > Let X_t be the price of some commodity at epoch t. Let F_t = sigma({X_s: 0 <= s <= t}). Suppose 0 <= s <= t. Then, for any Borel subset A of R, {X_t in A} is generarlly not in F_s. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: 4CT Proof, Take Two: Was: Re: Four Color Theorem Chris, I am abandoning this thread and starting a new one (generally on the same subject) entitled: The Lazy Person's Guide to Proving the Four Color Theorem ----Bill J > [clipped to save storage on various newsgroup servers] > In the game I suggested, I envisioned an endless cycle of recoloring & chord > relocation. This would require some 30 different colorings (RGRBY is not the > same as BGRBY, nor is YBGRB) Then I would observe that if all of > these different colorings are possible; then just maybe, the coloring > of P is absolutely independent of the rest of G! This may be true (I really doubt it), but (even if it is true) it still > need to be proven. And the proof appears to be difficult. Let's start out with the premise that the colors of P are determined by the > configuration of G, more precisely, the configuration of G-v (of which P is an > induced subgraph). An indifferent 4-coloring of P is shown in Figure 1 I don't know what you mean by indifferent, but I agree that you may > assume that if you have ONE 4-coloring of G-v, then P is colored as in > Figure 1, as symmetry has not been broken for P. There may be other > colorings, though; for instance, one where b and d have the same color, > and the overall graph G might not be symmetric. Given a specific Graph G. Can there be one proper 5-coloring of G that has v_a and v_c the same color and another proper 5-coloring of the same G that has v_b and v_d the same color? There seems to be no reason why they cannot. If so, all 5 different pairs; ie a &c, a&d, b&d, b&e and c&e, could each have the same color. This suggests at least 5 different 5-colorings for G ( assuming that one 5-coloring is possible)! R G R B Y > e<== a-----b-----c-----d-----e ==>a > Figure 1 What if a chord is added between v_a and v_c? V_a and v_c are disjoint > because they have the same color. The 4-colorability of G-v is in > doubt? But the 4-colorability of G-v is not in doubt; you assumed that G was a > minimal counterexample, so G-v MUST be 4-colorable. OTOH, _this_ coloring isn't proper for G-v+ac. There may be other ones > out there, though. It would not be proper for G-v+ac+ce either, which is a configuration of (G-v)_max For convenience, let H = any cofinguration of (G-v)_max. Let's configure H by adding internal chords {ac} and {ce} to graph P. Conjecture: If and only if both of the following statememts are true, then the 4CT is true. I If Chi(G) > 4, then Chi(H) > 4 II Chi(H)< 5 if and only if Chi(G) < 5. The 4CT can be false if and only if Chi(G) > 4 and Chi(H) < 5. Therefore, proving the statements of the conjecture will prove the 4CT? Proving the conjecture is not necessarily easy or even possible. But the scope of the problem is limited to recoloring the vertices of a 5-cycle graph. . It can be restored if > 1. v_a = B, or > 2. v_c = Y, or > 3. v_a = B & v_c = Y. I think that, by 1., you really mean we recolor G so that v_a has > color B, which means you have to recolor more of G, since there might > be a vertex v_f adjacent to v_a which is colored B. Or you 4-colored G-v+ac. But you don't give enough details for it to be considered a proof. --- Christopher Heckman I think that I will test case 3. first. V_a and v_c are disjoint. V_a and v_c have > different colors. Case 3 is valid. Parity be damned! Chi(G) = 4. The > 4CT is true. === Subject: Re: 4CT Proof, Take Two > Chris, I am abandoning this thread and starting a new one (generally on > the > same subject) entitled: > The Lazy Person's Guide to Proving the Four Color Theorem Well, that was the purpose in me retitling this as 4CT Proof, Take Two. So do you want a reply here as well? --- Christopher Heckman > [clipped to save storage on various newsgroup servers] > In the game I suggested, I envisioned an endless cycle of recoloring & chord > relocation. This would require some 30 different colorings (RGRBY is not the > same as BGRBY, nor is YBGRB) Then I would observe that if all of > these different colorings are possible; then just maybe, the coloring > of P is absolutely independent of the rest of G! This may be true (I really doubt it), but (even if it is true) it still > need to be proven. And the proof appears to be difficult. Let's start out with the premise that the colors of P are determined by the > configuration of G, more precisely, the configuration of G-v (of which P is an > induced subgraph). An indifferent 4-coloring of P is shown in Figure 1 I don't know what you mean by indifferent, but I agree that you may > assume that if you have ONE 4-coloring of G-v, then P is colored as in > Figure 1, as symmetry has not been broken for P. There may be other > colorings, though; for instance, one where b and d have the same color, > and the overall graph G might not be symmetric. Given a specific Graph G. Can there be one proper 5-coloring of G that > has > v_a and v_c the same color and another proper 5-coloring of the same G > that has v_b and v_d the same color? There seems to be no reason why > they > cannot. If so, all 5 different pairs; ie a &c, a&d, b&d, b&e and > c&e, > could each have the same color. This suggests at least 5 different > 5-colorings > for G ( assuming that one 5-coloring is possible)! > R G R B Y > e<== a-----b-----c-----d-----e ==>a > Figure 1 What if a chord is added between v_a and v_c? V_a and v_c are disjoint > because they have the same color. The 4-colorability of G-v is in > doubt? But the 4-colorability of G-v is not in doubt; you assumed that G was a > minimal counterexample, so G-v MUST be 4-colorable. OTOH, _this_ coloring isn't proper for G-v+ac. There may be other ones > out there, though. It would not be proper for G-v+ac+ce either, which is a configuration > of (G-v)_max > For convenience, let H = any cofinguration of (G-v)_max. Let's configure H by adding internal chords {ac} and {ce} to graph P. Conjecture: If and only if both of the following statememts are true, > then > the 4CT is true. I If Chi(G) > 4, then Chi(H) > 4 > II Chi(H)< 5 if and only if Chi(G) < 5. The 4CT can be false if and only if Chi(G) > 4 and Chi(H) < 5. > Therefore, > proving the statements of the conjecture will prove the 4CT? Proving the conjecture is not necessarily easy or even possible. But > the scope > of the problem is limited to recoloring the vertices of a 5-cycle > graph. . It can be restored if > 1. v_a = B, or > 2. v_c = Y, or > 3. v_a = B & v_c = Y. I think that, by 1., you really mean we recolor G so that v_a has > color B, which means you have to recolor more of G, since there might > be a vertex v_f adjacent to v_a which is colored B. Or you 4-colored G-v+ac. But you don't give enough details for it to be considered a proof. --- Christopher Heckman I think that I will test case 3. first. V_a and v_c are disjoint. V_a and v_c have > different colors. Case 3 is valid. Parity be damned! Chi(G) = 4. The > 4CT is true. === Subject: An exact 1-D integration challenge - 25 Hello the inspired formula lovers, It's refreshing to observe how human mind fights against dark of ignorance, against complexity trying to discover patterns, - and enjoy the victories gained! None of modern computer algebra systems can crack this integral straightforwardly. Such a sad fact... Is there a soul who can show a way to calculate using a CAS the exact value of the integral int(cos(z^2)*sech(z*sqrt(Pi))^3, z= 0..infinity); ? Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing === Subject: World Hologram in Emergent Gravity Given a p-form P and an q-form Q P/Q is a p + q form P/Q = (-1)^(pq) Q/P d(P/Q) = (dP)/Q + (-1)^pP/(dQ) Let O be a 0-form and I be a 1-form O/I = (-1)^(0x1)I/O = I/0 d(O/I) = dO/I + (-1)^0O/dI = dO/I + O/dI d^2 = 0 Let Theta & Phi be Goldstone phase 0-forms of the vacuum ODLRO inflation field. Define the 1-form A = Theta/dPhi - dTheta/Phi d(Theta/dPhi) = dTheta/dPhi + Theta/d^2Phi = dTheta/dPhi d(dTheta/Phi) = d^2(Theta)/Phi - dTheta/dPhi = -dTheta/dPhi Therefore, the 2-form is dA = +2dTheta/dPhi Let eu^a be a set of Einstein-Cartan sub-geometrodynamic tetrad field components. The O(1,3) indices a,b,c... are raised and lowered with the globally flat constant Minkowski metric nab. The GCT Diff(4) indices are raised and lowered with the locally curvilinear GCT metric guv. Einstein's local principle of equivalence is formally here for a point event E guv(E) = eu^a(E)nabev^b(E) The 4 GCT 1-form tetrad fields are e^a = eu^adx^u Form the contracted O(1,3) scalar e = e^a&a &a is a basis of vector fields in flat tangent space. Ansatz e = 1 + A 1 is the constant unit scalar When there is no real gravity curvature field A = 0 identically in the region of events {E}. A = Theta/dPhi - dTheta/Phi Au^a = Lp(Theta/dPhi - dTheta/Phi)u^a Lp = (hG/c^3)^1/2 dA = 2(dTheta)/(dPhi) = (dA)uv^adx^u/dx^v&a (dA)uv^a = 2Lp^2(dTheta/dPhi)uv^a A topological 3D space point defect (world line in 4D spacetime) in the macro-quantum More is different vacuum ODLRO local order parameter inflation field from a spontaneously broken continuous symmetry has Theta & Phi undefined. The integral of dA around a closed surface that surrounds that point defect is quantized in order to keep the local order parameter single valued. Therefore Integral of dA on a closed surface with no boundary that is itself not a boundary ~ (Integer)Lp^2, i.e. Hawking-Bekenstein black hole entropy rule of 1 c-bit per Planck quantum of area and t'Hooft-Susskind world hologram rolled into one formula. Note, if there was only one Goldstone phase Theta instead of 2 the situation would be like 1. Superfluid helium 2. Superconductor 3. Self-trapped laser filaments in nonlinear optics. The topological defects would be string vortex lines, i.e. world sheets in 4D spacetime. The surrounding surface would be a non-bounding 1D loop without boundary enclosing a vortex line, and the corresponding loop integral would be quantized as integer h/m for a neutral superfluid, or h/2e magnetic flux quantum for a superconductor. === Subject: The Lazy Person's Guide to Proving the Four Color Theorem Let G be a simple loopless maximal planar graph. The 4CT is false if there exists a graph G such that Chi(G) > 4 and every graph (H) with fewer vertices than G is 4-colorable. If such a graph G exists it is called a minimal counterexample (mce) to the 4CT. The 4CT can be proven by proving that no mce can exist. Some parameters of G have already been established. One of these conditions is that G cannot have any vertices of degree less that 5. Since G is maximal planar, it must have at least one vertex of degree = 5. If that vertex (v) is removed from G, the resulting graph is (G-v). Let's assume that Chi(G) > 4. If G is an mce, (G-v) must be 4-colorable. Now here is a very critical point; For every possible 4-coloring of (G-v), there can be no proper 4-colorings of G. The five vertices that are/were the neighbors of v form a pentagon (P). P is the interface between (G-v) and G. To assure that Chi(G) > 4 when Chi(G-v) =4 it is necessary that P cannot be 3-colored! Second critical point; Graph (G-v) is not maximum (maximal) planar. It is lacks two edges. (G-v) can be maximized by the addition of any two legal edges. These edges will be added as internal chords to graph P. When (G-v) is maximized we can call it graph H. H has fewer vertices than G. If H is not 4-colorable, then G is not minimal and therefore, not an mce. If, after the addition of the maximizing edges, graph P cannot be properly 4-colored, then H cannot be properly 4-colored. To be proven: P is 4-colorable if and only if Chi(G) < 5. It is very easy to choose the maximizing edges so that P is not properly 4-colored. Now, proof of the 4CT depends entirely upon proving that P cannot be properly 4-colored. Or that if P can be properly 4-colored, then G can also be properly 4-colored. ---Bill J === Subject: Re: The Lazy Person's Guide to Proving the Four Color Theorem > Let G be a simple loopless maximal planar graph. > The 4CT is false if there exists a graph G such that Chi(G) > 4 and every graph (H) > with fewer vertices than G is 4-colorable. In fact, the 4CT is false if there is a graph G with Chi(G) > 4, period. The extra condition, on smaller graphs, defines a minimal counterexample. > If such a graph G exists it is called a > minimal counterexample (mce) to the 4CT. The 4CT can be proven by proving that no mce can exist. Some parameters of G have already been established. One of these conditions is that > G cannot have any vertices of degree less that 5. Since G is maximal planar, it must > have at least one vertex of degree = 5. If that vertex (v) is removed from G, the resulting > graph is (G-v). Let's assume that Chi(G) > 4. If G is an mce, (G-v) must be 4-colorable. Now here > is a very critical point; For every possible 4-coloring of (G-v), there can be no proper > 4-colorings of G. Correct, since we assumed that Chi(G) > 4. > The five vertices that are/were the neighbors of v form a pentagon (P). P is the > interface between (G-v) and G. To assure that Chi(G) > 4 when Chi(G-v) =4 > it is necessary that P cannot be 3-colored! Not exactly. It is necessary that no 3-coloring of P extends to the rest of G, or equivalently, if C is a coloring of G, then C(P) has four colors. P (by itself) certainly can be 3-colored. (BTW, it is not too difficult to show that every triangle in a mce must be facial, so P really is a pentagon; there are no extra edges in P (as a subgraph of G).) > Second critical point; Graph (G-v) is not maximum (maximal) planar. It is lacks two edges. (G-v) > can be maximized by the addition of any two legal edges. These edges will be added > as internal chords to graph P. When (G-v) is maximized we can call it graph H. > H has fewer vertices than G. If H is not 4-colorable, then G is not minimal and > therefore, not an mce. If, after the addition of the maximizing edges, graph P cannot be properly 4-colored, > then H cannot be properly 4-colored. True. However, P+2edges _can_ be properly 4-colored; in fact Chi(P+2edges) = 3! So your statement is vacuously true; it's a statement of the form False -> Something, which is always true. A statement about extending colorings from P to the rest of G-v doesn't help, either, since one of the proper 4-colorings of P+2edges might extend to a 4-coloring of G-v. For this to be a real proof, this possiblity must be taken care of. > To be proven: P is 4-colorable if and only if Chi(G) < 5. We know that Chi(G) > 4, so Chi(G) < 5 is false. Hence you are trying to show that P is not 4-colorable, which is blatantly impossible, in light of what I explained above. > It is very easy to choose the maximizing edges so that P is not > properly 4-colored. And I bring up what I've said many times before: You can do this so that a particular coloring C of P+2e is no longer properly colored, but there can be other colorings out there where P+(these two edges) _will_ be properly 4-colored. > Now, proof of the 4CT depends entirely upon proving that P cannot be properly > 4-colored. Or that if P can be properly 4-colored, then G can also be properly 4-colored. Neither of which has been shown. Thus this is not a proof. --- Christopher Heckman === Subject: Re: a^5+b^5+c^5+d^5+e^5 = f^5 > Here is a summary of results for the density of solutions of k kth > powers equal to a kth power, for k=2,3,4,5. Given the equation, ... > Jarek Wroblewski has a database of 3.1.3 with z < 10^6, and 4.1.4 with > z < 220,000. There are only 13 solutions to 5.1.5 with z < 600, with > one 5.1.4. If you can EXTEND this, it would be appreciated by all. > > (There is a parametrization of the latter, but we want the terms to be > all positive.) ... I put a list of some solutions to a^n + b^n + c^n + d^n + e^n = f^n, in forms like the following, 19^5+ 43^5+ 46^5+ 47^5+ 67^5 - 72^5 0^5+ 27^5+ 84^5+ 110^5+ 133^5 - 144^5 145^5+ 565^5+ 1105^5+ 1462^5+ 1990^5 - 2087^5 519^5+ 642^5+ 1026^5+ 1480^5+ 1990^5 - 2087^5 at http://pat7.com/jp/s515-3019 . This is output from program http://pat7.com/jp/515count6d.c, which took 3.3 hours on my 900MHz Athlon to find 108 solutions with f < 3019. The program would probably take 5.4 days to find all solutions with f < 10007; its processing time is worse than O(n^3 * lg n) and it uses around 16*(n^2) bytes of memory on 32-bit machines. Note, I was unable to find a 5.1.5 database on Jarek Wroblewski's site to compare results. Do you have a URL for it? -jiw === Subject: Re: Polysign Quotients > I believe that I have resolved our apparent disagreement. I've > > scrambled together a low grade reciprocal finder (results with large > > error) and sure enough the divisor counterexamples that you provide > > have no decent reciprocal. I've read the definition of zero divisor and > > it is fine though it does not discuss anything about the underlying > > structure of the system it is applied to. I suppose that is up to us > > and I am now curious what we'll see above p4 in this context. Zero-divisors are indeed pat of the structure, but much can be said > about zero-divisors without even investigating the underlying > structure (when that is a ring, of course). > The 'other axes' are actually a plane. It is the plane that is > > perpendicular to the identity axis that passes through the origin. Yup, the set of obvious zero-divisors are, using coordinates > (x1, x2, x3, x4), the plane x1 + x3 = x2 + x4 and the line > x1 = x3 & x2 = x4. > We've discussed it a bit before but now I can see it in the context of > > division. Just as a multiplication by a point on the identity axis > > removes information so does a multiplication by any point in this > > plane. Instead of squashing the information into a line it yields two > > dimensions of result, but still provides an irreversible effect on any > > object. In effect the object has been squashed into a plane and can > > never be 'unsquashed' informationally. Indeed, the product of a zero-divisor and an arbitrary element is a > zero-divisor. That is easy to prove when the product is commutative. > We cannot > > get three dimensions back from two. There must be a theorem on > > dimensional reduction and zero divisors. It's clear in the > > informational context. There is not necessarily dimensional reduction. There are rings where > every element is a zero-divisor. But in general the zero-divisors form > subspaces of lower dimensionality (assuming we have some dimensional > algebra over the reals, which you have). But it is all quite tricky, > because if a and b are zero-divisors, a + b is not necessarily a > zero-divisor. Also it can be said that if a and b are zero divisors then their > product is not necesarily zero. > For instance -1*1 and #1+1 are both zero divisors. But their product is > -2*2. > This is where the structure matters, for it can be said that iff a is > on the 1D part and b is on the 2D part that their product will be zero. > This is all in the P4 context. The 1D part forces a result on its axis > and the 2D part forces a result on its plane, yielding the origin. The > two are inherently tied together. I think that this dimensional level > is more appropriate a focus than the phrase 'zero divisor' for this > phrase connotes a general phenomenon but without the structure it is > too abstract. Each zero divisor loses its meaning without being > attributed to its place in the structure. For P4 this is an axis and > its orthogonal plane at the origin. > In P6 the value (0,0,1,1,0,0) does not intuit to be normal to > (1,0,1,0,1,0) but it is according to my computer a right angle. So thus > far the zero divisor concept laid out here maintains an orthogonal > relation in P6 as well. In P6 the space orthogonal to the identity axis > is a 4D object. But this is just pinning down a known behavior. The P5 > challenge sounds much more interesting since there is a sliver of hope > of proving an exception to Hopf. You will not find an exception to a theorem. > I still have not found the Hopf proof. Since you have carefully noted > its publication I'll see what can be gotten in english. You can look for similar arguments in Husemoller's Fiber Bundles, > if I recall c0orrectly. Hopf's theorem, as well as Milnor-Kervaire's and others in this > direction are *not* proved using algebra, but algebraic topology. > The topological arguments stem from a rather deep study of > what spheres can be given the structure of H-spaces (see > wikipedia por a definition) and similar ideas. -- m The polysign numbers are a new construction so there is a sliver of a > chance that they could defy some of existing mathematics. Again, you will not defy existing mathematics. > I'm hunting > for P5 zero divisors now. It looks like they are there, but I have yet > to get all the way to zero. For instance (1.077, 0.889, 0.228, 0, 0.519 ) * ( 1.004, 0.420, 0.371, 1.043, 0 ) > = ( 0.005, 0.005, 0.011, 0.016, 0 ) and whose magnitude is 0.014 . (These values are chopped, not rounded) > A perturbative algorithm on these unity magnitude operands should get > a zero value, and expose the structure that these zero divisors are on. Pick n >= 4. Write e i for the i-th sign in your n-polysigned numbers. Let x = sum {k=0..n-1} cos(2 pi k / n) e i and y = sum {k=0..n-1} cos(4 pi k / n) e i (If you do not like negative coefficients, just add a large enough multiple of sum {i=0..n-1} e i so that all coefficients become positive; this does not change the end value...) Then a little trigonometry should convince you that x y = 0. NB: You may enjoy computing the product for 1 <= n < 4, and what is different in those three cases. Note that I did not come up with the this example out of thin air: this is just a particular instance of a rather more general phenomenon, studied (essentially completely and exhaustively) by Frobenius, Schur, Wedderburn and others nearly two centuries ago. > The zero divisors are still a coherent part of a continuous mapping. > I'm not sure in strict math what the proper teminology is, but right > nearby the zero divisors are points just off of zero, so in this way > the whole 'division algebra' problem makes me ask the question 'so > what?' What is it that we are gaining by studying the division problem? Well, for one thing, you are studying the division problem. > We can express general dimensional relations such as > 0 = a0 + a1 z + a2 z z + a3 z z z ... > but where will the division algebra become consequential? It really depends on what you want to do. You have to decide what you want to use your polysigned numbers for, and then evaluate if for *that* purpose the fact that divisors of zero exist for almost all cases is consequential or not. >I understand > that it helps characterize a construction, but I'm wondering where it > goes from there? I would heartily recomend that you try to go through an introduction to the theory of algebras, such as Pierce's `Associatve Algebras'. -- m === Subject: Re: Polysign Quotients snip You will not find an exception to a theorem. > I still have not found the Hopf proof. Since you have carefully noted > its publication I'll see what can be gotten in english. You can look for similar arguments in Husemoller's Fiber Bundles, > if I recall c0orrectly. Hopf's theorem, as well as Milnor-Kervaire's and others in this > direction are *not* proved using algebra, but algebraic topology. > The topological arguments stem from a rather deep study of > what spheres can be given the structure of H-spaces (see > wikipedia por a definition) and similar ideas. -- m The polysign numbers are a new construction so there is a sliver of a > chance that they could defy some of existing mathematics. Again, you will not defy existing mathematics. I'm hunting > for P5 zero divisors now. It looks like they are there, but I have yet > to get all the way to zero. For instance (1.077, 0.889, 0.228, 0, 0.519 ) * ( 1.004, 0.420, 0.371, 1.043, 0 ) > = ( 0.005, 0.005, 0.011, 0.016, 0 ) and whose magnitude is 0.014 . (These values are chopped, not rounded) > A perturbative algorithm on these unity magnitude operands should get > a zero value, and expose the structure that these zero divisors are on. Pick n >= 4. > Write e i for the i-th sign in your n-polysigned numbers. Let x = sum {k=0..n-1} cos(2 pi k / n) e i and y = sum {k=0..n-1} cos(4 pi k / n) e i (If you do not like negative coefficients, just add a large > enough multiple of sum {i=0..n-1} e i so that all coefficients become positive; this does not > change the end value...) Then a little trigonometry should convince you that x y = 0. NB: You may enjoy computing the product for 1 <= n < 4, and > what is different in those three cases. Note that I did not come up with the this example out of > thin air: this is just a particular instance of a rather more > general phenomenon, studied (essentially completely and > exhaustively) by Frobenius, Schur, Wedderburn and others > nearly two centuries ago. > Neat. I've verified it for P4 thru P9. I don't understand it. > The zero divisors are still a coherent part of a continuous mapping. > I'm not sure in strict math what the proper teminology is, but right > nearby the zero divisors are points just off of zero, so in this way > the whole 'division algebra' problem makes me ask the question 'so > what?' What is it that we are gaining by studying the division problem? Well, for one thing, you are studying the division problem. We can express general dimensional relations such as > 0 = a0 + a1 z + a2 z z + a3 z z z ... > but where will the division algebra become consequential? It really depends on what you want to do. You have to > decide what you want to use your polysigned numbers > for, and then evaluate if for *that* purpose the fact that > divisors of zero exist for almost all cases is consequential > or not. I understand > that it helps characterize a construction, but I'm wondering where it > goes from there? I would heartily recomend that you try to go through an > introduction to the theory of algebras, such as Pierce's > `Associatve Algebras'. -- m What do you think of P1? The polysign construction does not require the usage of Sum( sx ) = 0 except for purposes of graphing or 'rendering' a result. In this sense the components are accumulators. Zero-dimensional(P1) operations work but always render to zero. There is congruence with time. In some regard this 'disappearance' is what we are seeing in P4+ through zero divisors. Dimensions are disappearing. Physics has an accepted 'tunneling' effect modelled by energy wells. Perhaps there is a linkage to the arithmetical behavior of polysign numbers in the higher signs. Do you believe there is a way to express the equivalent of an exponential for Pn? Even in P3 the closest that I have gotten yet is just to take a value near unity and take the series z^n. A clean value will iterate a circle. I've tried this for values in higher signs but the results become degenerate. Now I am wondering if I simply haven't chosen the correct initial value; For instance in P4 the correct choice may be just off the unity vector along +1#1. Can I trace out a unit shell in any dimension via z^n? I'm fairly certain that the answer is no simply due to magnitude nonconservation in P4+ products. Whether that would break an exponential definition I am unsure. -Tim === Subject: Re: Polysign Quotients please, don't bother with the Copenhagenschooler reification of the math, to justify your own polemics. anyway, I may've misspoken on the third (planar) coordinate being negative; it depends on the homogenous coordinates; the simplest are trilinears, but you should try others. the study of the various points and lines and circles and on & on of the general trigon is much eaiser, using homogenous ones. only two coordinates are *neccesary*, but three gets rid of all sorts of degeneracies & special cases. (I'd say, it's obvious, that tunnelling depends upon defects in the semiconductor, beyond just the use of dopants, themselves.) > In some regard this 'disappearance' is what we are seeing in P4+ > through zero divisors. > Dimensions are disappearing. Physics has an accepted 'tunneling' effect > modelled by energy wells. Perhaps there is a linkage to the > arithmetical behavior of polysign numbers in the higher signs. thus: Polymath / Scholar Extraordinaire / Inventor / Architect / Poet / Ecologist / Futurist / Philosopher http://www.channer.tv/Friday.htm thus: it seems like the frequency shift would be manageable, going 600mph on a plane; you/pilots know the trajectory thereby.... --it takes some to jitterbug! http://members.tripod.com/~american_almanac http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html http://larouchepub.com/other/2006/3322_ethanol_no_science.html http://www.wlym.com/pdf/iclc/howthenation.pdf === Subject: Re: Probability question <44DF75EB.6040900@netscape.net >I'm studying probablity, and I'm having a problem w/a sample question: >Jar of 8 marbles: 3 are black solid, 1 is black striped, 2 are red >solid, & 2 are red striped. >2 marbles are picked out in order, 1 after the other, without >replacement. What is the probablity that the 2nd marble is striped? Can someone show me how to solve this without using a tree diagram >writing out all the possible orderings? There are eight equally likely possibilities for the second marble. Of > these, three are striped. Hence, the probability is 3/8. There is no > need to consider the first ball drawn. > But the problem says without replacement. 5/8 * 3/7 + 3/8 * 2/7 = (15 + 6)/56 = 21/56 = 3/8, hmmm. === Subject: Re: Probability question <44DF75EB.6040900@netscape.net> Jar of 8 marbles: 3 are black solid, 1 is black striped, 2 are red >solid, & 2 are red striped. >2 marbles are picked out in order, 1 after the other, without >replacement. What is the probablity that the 2nd marble is striped? Can someone show me how to solve this without using a tree diagram >writing out all the possible orderings? There are eight equally likely possibilities for the second marble. Of > these, three are striped. Hence, the probability is 3/8. There is no > need to consider the first ball drawn. But the problem says without replacement. Yes, but the problem is really no different from asking for the probability that the first ball is striped. You could take out five balls (without replacement) and ask for the probability that the last one is striped---same answer. You could even take out all eight of the balls and ask for the probability that the last one is striped. The answer would still be the same. R.G. Vickson > > 5/8 * 3/7 + 3/8 * 2/7 > = (15 + 6)/56 = 21/56 = 3/8, hmmm. === Subject: Re: Probability question <44DF75EB.6040900@netscape.net> as found in my book as the following: Eight evenly-match horses run a race. Three are bay colts, one is a bay filly, two are brown colts, and two are brown fillies. The color and sex of the first and second finishers are recorded. What is the probability that the second horse is a filly? I also got 3/8 as the answer, but I computed it (very) inefficiently: [P(3,2)/P(8,2)] + 1/2 * C(5,1) * C(3,2) / C(8,2) = 3/28 + 15/56 = 3/8 The answer in the back of the book is 1/8 (?), and that's why I posted the question here. So, I guess either there's a mistake or typo in the book, or the 2 problems aren't the same?.............. I'm studying probablity, and I'm having a problem w/a sample question: >Jar of 8 marbles: 3 are black solid, 1 is black striped, 2 are red >solid, & 2 are red striped. >2 marbles are picked out in order, 1 after the other, without >replacement. What is the probablity that the 2nd marble is striped? Can someone show me how to solve this without using a tree diagram >writing out all the possible orderings? > There are eight equally likely possibilities for the second marble. Of > these, three are striped. Hence, the probability is 3/8. There is no > need to consider the first ball drawn. Similar problem: Deal five cards without replacement from a standard > deck of cards. What is the probability that the third of these is an > ace? Answer: 1/13. -- > Stephen J. Herschkorn sjherschko@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhhattan === Subject: Re: Probability question It looks like a typo in the book. Herschkorn apporach and result are correct. === Subject: Slow learner trying to learn math Hi - I have trouble learning math. I take math courses part time at a well-known school, but it does not stick. The classes only ask for problem solving, not to prove theorems. It seems to me that I need to do proofs in order to learn, but what resources are there for the relatively beginner type who wants to learn by doing proofs? Questions welcome. Doug === Subject: Re: Slow learner trying to learn math Doug, One book that you might like is Numbers and Geometry by John Stillwell. This is mathematical, has proofs, and the blurb on the back says presupposes only high school algebra and therefore can be read by any well-prepared undergraduate student entering a university. It is basic mathematics with a strong historical flavor. It goes back and forth between ideas of various types of numbers and ideas of geometry. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ > Hi - > I have trouble learning math. I take math courses part time at a > well-known school, but it does not stick. The classes only ask for > problem solving, not to prove theorems. It seems to me that I need to > do proofs in order to learn, but what resources are there for the > relatively beginner type who wants to learn by doing proofs? Questions welcome. Doug > === Subject: Re: Slow learner trying to learn math How do you know learning by doing proofs will help you? What are the courses anyway? Often, focusing on proving everything will take you far away from the primary objectives of a particular course. Till you move to the advanced undergrad level or general graduate, in many instance grinding through the bunch of problems and getting some 'mechanical skills' is pretty much all you need to learn. Just learning some rules and reinforcing them by doing exercises. Proving the staff comes later on. > Hi - > I have trouble learning math. I take math courses part time at a > well-known school, but it does not stick. The classes only ask for > problem solving, not to prove theorems. It seems to me that I need to > do proofs in order to learn, but what resources are there for the > relatively beginner type who wants to learn by doing proofs? > > Questions welcome. > > Doug === Subject: Re: Slow learner trying to learn math > Hi - > I have trouble learning math. I take math courses part time at a > well-known school, but it does not stick. The classes only ask for > problem solving, not to prove theorems. It seems to me that I need to > do proofs in order to learn, but what resources are there for the > relatively beginner type who wants to learn by doing proofs? Questions welcome. > > Doug Good book in this matter is Polya Howto solve it ? Jarko === Subject: Re: Slow learner trying to learn math > Hi - > I have trouble learning math. I take math courses part time at a > well-known school, but it does not stick. The classes only ask for > problem solving, not to prove theorems. It seems to me that I need to > do proofs in order to learn, but what resources are there for the > relatively beginner type who wants to learn by doing proofs? Questions welcome. Doug > I found I had problem's understanding problems until I 'saw' them. You may be to obsessed with the logic. Develop a abstraction where you 'see' the problem. Math can be tough.. Be patient and stick with it and it will be your friend. -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/ === Subject: Re: Slow learner trying to learn math > Hi - > I have trouble learning math. I take math courses part time at a > well-known school, but it does not stick. The classes only ask for > problem solving, not to prove theorems. It seems to me that I need to > do proofs in order to learn, but what resources are there for the > relatively beginner type who wants to learn by doing proofs? I suggest looking at some proofs in geometry and number theory, and maybe group theory, before any in calculus and analysis (which involve infinite processes). Maybe look at some Schaum's Outlines, or some of the Dolciani series: https://enterprise.maa.org/ecomtpro/timssnet/books/dol.cfm LH === Subject: Re: Slow learner trying to learn math Doug .82í.83.81.83b.83Z.81[.83W: > Hi - > I have trouble learning math. I take math courses part time at a > well-known school, but it does not stick. The classes only ask for > problem solving, not to prove theorems. It seems to me that I need to > do proofs in order to learn, but what resources are there for the > relatively beginner type who wants to learn by doing proofs? Questions welcome. Doug I just currently have no question, === Subject: Re: Slow learner trying to learn math > Hi - > I have trouble learning math. I take math courses part time at a > well-known school, but it does not stick. The classes only ask for > problem solving, not to prove theorems. It seems to me that I need to > do proofs in order to learn, but what resources are there for the > relatively beginner type who wants to learn by doing proofs? Questions welcome. Doug > what level ? Get another book on proofs Proofs can be much harder than problem solving, so if the course is application orentated........ schalms outlines for college and some HS === Subject: Re: topology with S_omega. > let S_omega be the minimal uncountable well-ordered set. > The usual notation is omega_1 > S_omega is a sequentially compact. > if A is a countable subset of S_omega, > then A has an upper bound in S_omega. so, if sequence (x_n) in S_omega, > (x_n) has an upper bound in S_omega. so, i can make the monotone increasing subsequence {x_n_k} > in S_omega as Bolzano-Weierstrass. > Huh? As (xj)_j subset (sup_j xj) + 1, which is compact subset omega_1, it has a convergence subsequence, etc... > so, by monotone convergence theorem, > {x_n_k} converge to x. (x in S_omega) does x really contained in X_omega ? i need your advice over all my thinking. > === Subject: Re: topology with S_omega. let S_omega be the minimal uncountable well-ordered set. The usual notation is omega_1 S_omega is a sequentially compact. if A is a countable subset of S_omega, > then A has an upper bound in S_omega. so, if sequence (x_n) in S_omega, > (x_n) has an upper bound in S_omega. so, i can make the monotone increasing subsequence {x_n_k} > in S_omega as Bolzano-Weierstrass. Huh? As (xj)_j subset (sup_j xj) + 1, which is compact > subset omega_1, it has a convergence subsequence, etc... > yes, thank you very much. i think... the ordinal (sup_j xj) + 1 is compact and second countable (by countable ordinal) and regular (by order topology) so, it is metrizable by Urysohn. thus, compact <=> sequentially compact by metrizable. === Subject: Re: topology with S_omega. let S_omega be the minimal uncountable well-ordered set. > The usual notation is omega_1 S_omega is a sequentially compact. if A is a countable subset of S_omega, > then A has an upper bound in S_omega. so, if sequence (x_n) in S_omega, > (x_n) has an upper bound in S_omega. As (xj)_j subset (sup_j xj) + 1, which is compact > subset omega_1, it has a convergence subsequence, etc... the ordinal (sup_j xj) + 1 is compact and > second countable (by countable ordinal) and > regular (by order topology) so, it is metrizable by Urysohn. Every countable ordinal is embeddable into the rationals, hence metrizable. > thus, compact <=> sequentially compact by metrizable. > All you need is that 1st countable, compact spaces are sequentially compact. You've yet to finish showning omega_1 is sequentially compact. === Subject: graph theory In graph theory, a directed graph with directed cycles is not welcomed, and people want to eliminate some edges in order to get a directed acyclic graph. I want to know whether there exists some model with deals with directed graphs with cycles, because eliminating some edges means increasing substantial complexity. === Subject: Re: graph theory > In graph theory, a directed graph with directed cycles is not welcomed, > and people want to eliminate some edges in order to get a directed > acyclic graph. This is news to me. Certainly, *in some applications*, directed graphs are not desired, but saying there's something wrong with digraphs with directed cycles is like a number theory person complaining about irrational numbers. > I want to know whether there exists some model with > deals with directed graphs with cycles, because eliminating some edges > means increasing substantial complexity. The standard digraph model. --- Christopher Heckman === Subject: Re: graph theory Proginoskes .82í.83.81.83b.83Z.81[.83W: > In graph theory, a directed graph with directed cycles is not welcomed, > and people want to eliminate some edges in order to get a directed > acyclic graph. This is news to me. Certainly, *in some applications*, directed graphs > are not desired, but saying there's something wrong with digraphs with > directed cycles is like a number theory person complaining about > irrational numbers. I want to know whether there exists some model with > deals with directed graphs with cycles, because eliminating some edges > means increasing substantial complexity. The standard digraph model. --- Christopher Heckman That graph is tall and your graph is short ! === Subject: Re: topology with dense. > X is T_2. > f : X -> X is continuous. > A is a dense subset of X. if f(a) = a for any a in A.(identity on A), > show that f is a identity function on X. First show B = { x | f(x) = x } is closed subset of Hausdorff X. Then X = cl A subset cl B = B. Exericse. dense D, continuous f,g:X -> Y, Hausdorff Y, f|D = g|D, ie for all x in D, f(x) = g(x) ==> f = g Show fxg(x) = (f(x), g(x)) is continous map from X into YxY. Since Y is Hausdorff, show A = { (x,x) | x in Y } = { (x,y) | x = y } closed subset YxY. Thus { x | f(x) = g(x) } = (fxg)^-1(A) is closed. X = cl D subset cl{ x | f(x) = g(x) } = { x | f(x) = g(x) }, QED. === Subject: Re: topology with dense. X is T_2. > f : X -> X is continuous. > A is a dense subset of X. if f(a) = a for any a in A.(identity on A), > show that f is a identity function on X. First show B = { x | f(x) = x } is closed subset of Hausdorff X. > Then X = cl A subset cl B = B. Exericse. > dense D, continuous f,g:X -> Y, Hausdorff Y, > f|D = g|D, ie for all x in D, f(x) = g(x) > ==> f = g Show fxg(x) = (f(x), g(x)) is continous map from X into YxY. > Since Y is Hausdorff, show A = { (x,x) | x in Y } > = { (x,y) | x = y } closed subset YxY. > Thus { x | f(x) = g(x) } = (fxg)^-1(A) is closed. X = cl D subset cl{ x | f(x) = g(x) } = { x | f(x) = g(x) }, QED. > yes, thank you very much for your easy solution. === Subject: Re: topology with dense. > Suppose there exists b in X-D such that b /= f(b). Let V, W be disjoint neighborhoods of b, f(b). Using continuity of f, there exist neighborhoods V', W' of b, f(b) such that { f[V'] subset of V } and { f[W'] subset of W }. D is dense in X and V' intersect W' is a neighborhood of b, so there exists d in D such that d belongs to V' intersect W'. See if you can get a contradiction by showing that d = f(d) belongs to both V and W. >> Actually, something's not quite right here (3'rd sentence), >> but I'll leave the details for you to work out. I've given >> you the basic idea, which is probably all I should be >> doing anyway. > sorry, i can't understand that V' intersect W' is > a neighborhood of b. i need one more your advice. A, and I slipped up and used D, my letter-of-choice for dense sets), omit the definition of V' and delete f(b) in the 3'rd sentence [i.e. change there exist neighborhoods V', W' of b, f(b) such that in the 3'rd sentence to there exists a neighborhood W' of b such that], and in the 4'th sentence change (twice) V' intersect W' to V intersect W'. A generalization of this result, which can be proved in the same way, says that a continuous mapping into a T_2 space is determined by the values it takes on a dense subset of its domain. Cardinal function fans will note that this implies C(X), the collection of real-valued continuous functions on any topological space X (whose cardinality is at least continuum), has cardinality at most 2^d(X), where d(X) = min{ card(D): D is a dense subset of X } + aleph_0 is the density of the space X. Dave L. Renfro === Subject: Re: topology with dense. >> Suppose there exists b in X-D such that b /= f(b). Let V, W be disjoint neighborhoods of b, f(b). Using continuity of f, there exist neighborhoods V', W' of b, f(b) such that { f[V'] subset of V } and { f[W'] subset of W }. D is dense in X and V' intersect W' is a neighborhood of b, so there exists d in D such that d belongs to V' intersect W'. See if you can get a contradiction by showing that d = f(d) belongs to both V and W. >> Actually, something's not quite right here (3'rd sentence), >> but I'll leave the details for you to work out. I've given >> you the basic idea, which is probably all I should be >> doing anyway. > sorry, i can't understand that V' intersect W' is > a neighborhood of b. i need one more your advice. A, and I slipped up and used D, my letter-of-choice for > dense sets), omit the definition of V' and delete f(b) > in the 3'rd sentence [i.e. change there exist neighborhoods > V', W' of b, f(b) such that in the 3'rd sentence to there > exists a neighborhood W' of b such that], and in the 4'th > sentence change (twice) V' intersect W' to V intersect W'. > yes, i see. thank you very much for your advice. === Subject: Re: topology with dense. same way, says that a continuous mapping into a T_2 space > is determined by the values it takes on a dense subset > of its domain. Cardinal function fans will note that this > implies C(X), the collection of real-valued continuous > functions on any topological space X (whose cardinality > is at least continuum), has cardinality at most 2^d(X), When I said whose cardinality is at least continuum, I mean the cardinality of C(X). The space X itself can be finite (but nonempty). So, does anyone know where William Elliot is today? (This is rhetorical.) It seems that I've been taking care of what he usually does with point-set topology questions. [Except that I haven't yet commented on a poster's posting style, but the day's not over yet!] Dave L. Renfro === Subject: Re: The list of all natural numbers don't exist > One problem with set theory is {} has too many meanings. >>So what is the only meaning of {} you think there should >>be, with set theory? (Hint: what does it even mean by the >>phrase {} has a meaning?) > > > {} can mean anything we want it to mean. Well then how could we proceed with an argument if a moment ago you asserted that {}'s having [too] many meanings is problematic, and now you claim that {} can mean anything we want it to mean? > (In a model?) If you're not even sure where {} might have a meaning, why did you in the first place make an assertion that {}'s having many meanings would be problematic? > I just think we shouldn't assign more than one meaning to {}. That just doesn't seem to be an explanation at all, imho. (I mean, why do you think we should not?) > Of course, you can ask what does it mean for > a symbol to have more than meaning. > > > Russell > - 2 many 2 count > > -- ----------------------------------------------------- What we call 'I' is just a swinging door which moves when we inhale and exhale. Shunryu Suzuki ---------------------------------------------------- === Subject: Re: The list of all natural numbers don't exist Why do you think that something without an end must > be incomplete or unfinished? What is wrong with saying that > a list of finite integers is complete if it contains every finite > integer? Think! > Complete means that nothing is missing. A complete list of > finite integers is a list of integers that is not missing anything. There is no infinite list which isn't missing nothing since infinity > means: incompletable. Endless does not mean incompletable when using an axiom set like ZF or > NBG which specifically requires endless but completed sets. Infinity means endlessness means incompletable processes or > incompletable considerations or incompletable extensions. No it doesn't. What makes you think it does? > You can't put a word out of his context and claim: now it's math and it > means what I want (and use just the aspects of its meaning, you like to > use). If endlessness logical implies incompleteness it does imply it in the > usual understanding, in ZF and in NGB and in any other considerations. And if my grandmother had wheels she'd be a bus. Endlessness does > not imply incompleteness. > If you don't talk about infinity = endlessness = impossibility of > completion, use another word or symbol. > Why? Do you think this nonsense is the usual definition of > infinity? > Nonsense? In what kind of world do you live in? Let's see: You claim: An infinite set is by definition a set that can be bijected > to a > proper subset. > Yes, that's the usual definition. But I think, infinite sets are at first a special kind of sets which > have the propertie to be infinite. I'm right? Actually, there's not much difference here. Lets start by taking with a rough intuitive idea of what it means to be infinite. We have the concept that an infinite set can't be exhausted you can always get one more. This is a start, but it will not do to define an infinite set A. If we have A, we clearly cannot get one more. So we refine this. We say A is infinite if given any finite subset B contained in A, we can always find another element of A. This works, but now we have to define finite. We have an intuitive sense of what finite is, a finite set is one that when we count the elements we will finish. However, we need a non circular definition. How about a finite set has a largest element? Not bad, it works for sets of integers. But what about the set {1,1/2,,2/3,3/4,4/5,...} This has a largest element, but we want to call this set infinite. So, if we are going to deal with general sets, not just sets of integers we will need a better definition. How about A finite set cannot form a bijection with a proper subset ? At first this seems a bit strange, but we can see that any set we intuitively call finite has this property. So to say that an infinite set is one that can form a bijection with a proper subset, is just a precise way of saying that at infinite set is endless. Note that there is no mention of complete or incomplete. These concepts have nothing to do with infinity. -William Hughes === Subject: Re: The list of all natural numbers don't exist > Why do you think that something without an end must > be incomplete or unfinished? What is wrong with saying that > a list of finite integers is complete if it contains every finite > integer? Think! > Complete means that nothing is missing. A complete list of > finite integers is a list of integers that is not missing anything. There is no infinite list which isn't missing nothing Which integer is missing from the list {1,2,3,...} We are able to construct integers which are missing from the list as > e.g. Russell Easterly had done, analogue to the diagonal argument of > Georg Cantor. > since infinity > means: incompletable. No, an infinite set is by definition a set that can be bijected to a > proper subset. There is not mention of incompletable. This definition has no value since infinite sets are self > contradicting. You apply finite considerations on infinity. x ) 1 subgroup of x / a set of subgroups > x x ) 2 subgroups of x / a set of subgroups > xx xx ) > x x x ) > xx xx xx ) 3 subgroups of x / a set of subgroups > xxx xxx xxx ) > x x x x > xx xx xx xx > xxx xxx xxx xxx > xxxx xxxx xxxx xxxx > ... Your argument about completeness is not the only one. Why do you think, > only your definition is legal, mine is not? I argue logical correct, that the structure above is only complete, if > it contains a infinite substructure - which is impossible. > You claim that the above structue is only complete if it > contains an infinite substructure. Why? What definition of > complete are you using? > You say: The structure is complete if the structure is infinite? No, I do not think it makes sense to ask whether the structure is complete. > You say: The structure is complete although it doesn't contain the > complete structure as substructure? No. You are the one using the term complete. For some reason you think it has something to do with initinite. > Do you think, the finite structures are complete if they contain their > complete structure as substructure or not? Until you definite complete it makes no sense to ask if a structure (finite or infinite) is complete. > Why should it be different in the infinite case? It isn't. > The structure was constructed in this sense, that the structure is > complete if it contains the complete structure as substructure, you > understand this? Why. Why should the structure only be complete if it contains itself as a substructure. What definitinon of complete are you using. I would like to know just one thing from you: In your opinion, how > would you call the structure if it contains their whole structure as > substructure? Overcompleted? It must be something like this. No. Until you define complete, it does not make sense to call any structure compele, incomplete or overcomplete. While the obvious limit of the sequence of strutures you have described is a structure that does not contain itself, it is possible to definite a structure that does contain itself. However, I see no reason to call this structure any more or less complete that the structure that does not contain itself. Define complete. - William Hughes === Subject: Re: go forth and multiply [...] > There are a fair number of problems with the Ark story, though > admittedly I hadn't thought of that one. :-) Since there were > only two termites one might suggest a small pile of sawdust. > How small is an interesting question. > > Here's a related question: how much food was on the Ark, and > how much waste did Noah & Co. have to remove from the cages? > Bear in mind the Ark had everything from shrews and the > aforementioned termites to giraffes, hippos, and elephants. > Termite is probably not a big worry, but ... > There was a law suit (civil matters, I think) in Oz a few years back, around 1997: Background (bare bones) ---------------------------------------------------------------------- Ian Plimer: Professor of Geology David Fasold: Arkhunter, author Dr. Allen Roberts: Creation Science advocate, plagiarized David Fasold's work (in part) Ronald Sackville: Judge of the Federal Court of Australia --------------------------------------------------------------------- For more, see e.g.: http://www.abc.net.au/quantum/info/lxp.htm David Bernier === Subject: Re: go forth and multiply <_EHDg.16637$PV5.19447@wagner.videotron.net so when the waters finally receded, noah told the animals in the ark to > go forth and multiply. most all animals obeyed him, but he found 2 > snakes still lurking around the ark. he says to them, didn't you hear > me? now go forth and multiply. the snakes reply, but we can't > multiply...we're adders. On a related note, I wonder if Noah told the termites to wait > till debarkation, and *then* go forth and multiply... > If not, then what was the ark made of and what did > the termites eat on the ark? > Just a cotton picking nanosec. You're trying to pull a quick one on us. Two by two they entered the ark, male and female. Thus for termites, bees is much problem. Did the queen arrive with a worthless male or did the male come with a no_time_for_sex worker? === Subject: Re: go forth and multiply In sci.math, William Elliot on Sun, 13 Aug 2006 22:21:40 -0700 : > so when the waters finally receded, noah told the animals in the ark to >> go forth and multiply. most all animals obeyed him, but he found 2 >> snakes still lurking around the ark. he says to them, didn't you hear >> me? now go forth and multiply. the snakes reply, but we can't >> multiply...we're adders. >> On a related note, I wonder if Noah told the termites to wait >> till debarkation, and *then* go forth and multiply... >> If not, then what was the ark made of and what did >> the termites eat on the ark? > Just a cotton picking nanosec. You're trying to pull a quick one on us. Two by two they entered the ark, male and female. > Thus for termites, bees is much problem. Did the queen arrive with a worthless male or > did the male come with a no_time_for_sex worker? > No doubt a nubile bee (unmated female fresh from the hive) came with a drone, and then they mated just after they came on the ark and then the queen started gestating. Entirely proper behavior, if one postulates a priest bee on board to conduct the ceremony... oh, wait, there's the minor problem of where they get the pollen and nectar for the wax in order to incubate the eggs. Of course one has to ask the obvious question as to why the priest bee was left behind to drown. Cruel notion, that. Or maybe the priest bee stayed on after the male died off (drones don't live long after mating). That might work except that Catholicism frowns on female priests, so there's some issues to be worked out in the details here. (Since queen bees only mate once they are therefore monogamous; there's no problem in being a widow.) I suppose one could postulate a trailing garland or a surrounding patch of lily flowers, in a pinch, to solve the nectar/wax/pollen issue, but one would think Noah would have at least mentioned such. Ants and termites presumably have similar issues. Complicated, this Genesis. And then there are the butterflies, mayflies, and other such, who live for a very short time after reaching adulthood. In fact, mayflies live only 1 day, which makes for very exacting timing on Noah's part. Ob:-): :-) -- #191, ewill3@earthlink.net Windows Vista. Because it's time to refresh your hardware. Trust us. === Subject: Re: go forth and multiply <_EHDg.16637$PV5.19447@wagner.videotron.net> <5ki5r3-9n8.ln1@sirius.tg00suus7038.net> === Subject: Re: go forth and multiply >> On a related note, I wonder if Noah told the termites to wait >> till debarkation, and *then* go forth and multiply... >> If not, then what was the ark made of and what did >> the termites eat on the ark? > Just a cotton picking nanosec. You're trying to pull a quick one > on us. Two by two they entered the ark, male and female. > Thus for termites, bees is much problem. Did the queen arrive with a worthless male or > did the male come with a no_time_for_sex worker? > No doubt a nubile bee (unmated female fresh from the hive) > came with a drone, and then they mated just after they > came on the ark and then the queen started gestating. The drone spilled his guts and the queen was left with nobody to ... > Entirely proper behavior, if one postulates a priest bee > on board to conduct the ceremony... oh, wait, there's the > minor problem of where they get the pollen and nectar for > the wax in order to incubate the eggs. ... attend to all the work building the comb, etc. Thus no bees. > Of course one has to ask the obvious question as to why the > priest bee was left behind to drown. What priest bee? You're in the wrong time zone. They were married by a rabbi. > Ants and termites presumably have similar issues. Complicated, > this Genesis. And then there are the butterflies, mayflies, > and other such, who live for a very short time after reaching > adulthood. In fact, mayflies live only 1 day, which makes for > very exacting timing on Noah's part. As for birds and flying insects, wouldn't they prefer to wing it? > Ob:-): :-) What about fish? They weren't even invited. -- > #191, ewill3@earthlink.net > Windows Vista. Because it's time to refresh your hardware. > Trust us. Huh? Forthright an explaination is overdue. ---- === Subject: Re: go forth and multiply In sci.math, William Elliot on Mon, 14 Aug 2006 02:40:19 -0700 : === > Subject: Re: go forth and multiply On a related note, I wonder if Noah told the termites to wait till debarkation, and *then* go forth and multiply... If not, then what was the ark made of and what did the termites eat on the ark? Just a cotton picking nanosec. You're trying to pull a quick one >> on us. >> Two by two they entered the ark, male and female. >> Thus for termites, bees is much problem. >> Did the queen arrive with a worthless male or >> did the male come with a no_time_for_sex worker? > No doubt a nubile bee (unmated female fresh from the hive) >> came with a drone, and then they mated just after they >> came on the ark and then the queen started gestating. The drone spilled his guts and the queen was left with nobody to ... > Entirely proper behavior, if one postulates a priest bee >> on board to conduct the ceremony... oh, wait, there's the >> minor problem of where they get the pollen and nectar for >> the wax in order to incubate the eggs. ... attend to all the work building the comb, etc. Thus no bees. Drones do no work around the hive at all anyway. (Don't tell the Religious Right.) :-) > Of course one has to ask the obvious question as to why the >> priest bee was left behind to drown. What priest bee? You're in the wrong time zone. They were married by a > rabbi. OK, a rabbi bee. :-) But it will have to be a female rabbi bee. > Ants and termites presumably have similar issues. Complicated, >> this Genesis. And then there are the butterflies, mayflies, >> and other such, who live for a very short time after reaching >> adulthood. In fact, mayflies live only 1 day, which makes for >> very exacting timing on Noah's part. As for birds and flying insects, wouldn't they prefer to wing it? Well, the mosquitoes should be OK, though Noah and Co may not be all that happy; I'd have to research other flyers. The birds may have a problem finding a perch. > Ob:-): :-) What about fish? They weren't even invited. > Salt-water fish may indeed have a little problem. Not sure regarding fresh-water fish but they may have difficulties also as they're swept into the Global Ocean. -- #191, ewill3@earthlink.net Windows Vista. Because it's time to refresh your hardware. Trust us. === Subject: Re: go forth and multiply <_EHDg.16637$PV5.19447@wagner.videotron.net> so when the waters finally receded, noah told the animals in the ark to >> go forth and multiply. most all animals obeyed him, but he found 2 >> snakes still lurking around the ark. he says to them, didn't you hear >> me? now go forth and multiply. the snakes reply, but we can't >> multiply...we're adders. On a related note, I wonder if Noah told the termites to wait > till debarkation, and *then* go forth and multiply... > If not, then what was the ark made of and what did > the termites eat on the ark? There are a fair number of problems with the Ark story, though > admittedly I hadn't thought of that one. :-) Since there were > only two termites one might suggest a small pile of sawdust. > How small is an interesting question. > Nay nay, 'tis simply linear thought. For an exponential paradigm shift, quantum leap extrapolation see my other debarked post. > Here's a related question: how much food was on the Ark, and > how much waste did Noah & Co. have to remove from the cages? > Bear in mind the Ark had everything from shrews and the > aforementioned termites to giraffes, hippos, and elephants. > Termite is probably not a big worry, but ... > What me worry? Would not slugs eat it? Two by two they walked into the ark male and female... Oh oh, worms, slugs and all those other both male and female gay marriages not allowed on board. Only bonerfided heterosexual, one male and one female, marriages allowed. But that is discrimination, requiring trees, bushes, herbs, vegetables to walk on board without assistance. Thus is puzzle, without use of choice or evolution, explain how plants, worms, etc. survived the flood for in soothe, they were not aboard. === Subject: Re: go forth and multiply <_EHDg.16637$PV5.19447@wagner.videotron.net> . Thus is puzzle, without use of choice or evolution, explain how plants, > worms, etc. survived the flood for in soothe, they were not aboard. Nor were they a broad. And the snakes were told to go forth and copulate with rabbits, resulting in adders that multiplied. cfe === Subject: Re: go forth and multiply In sci.math, conradeaton@hotmail.com on 14 Aug 2006 05:47:05 -0700 . >> Thus is puzzle, without use of choice or evolution, explain how plants, >> worms, etc. survived the flood for in soothe, they were not aboard. > Nor were they a broad. And the snakes were told to go forth and > copulate with rabbits, resulting in adders that multiplied. > cfe > Not to mention the game Snakes and Rabbits. (I thought it was called something else, myself. Something more along the lines of slides and ascending devices. :-) ) -- #191, ewill3@earthlink.net Windows Vista. Because it's time to refresh your hardware. Trust us. === Subject: Re: go forth and multiply <_EHDg.16637$PV5.19447@wagner.videotron.net so when the waters finally receded, noah told the animals in the ark to > go forth and multiply. most all animals obeyed him, but he found 2 > snakes still lurking around the ark. he says to them, didn't you hear > me? now go forth and multiply. the snakes reply, but we can't > multiply...we're adders. On a related note, I wonder if Noah told the termites to wait > till debarkation, and *then* go forth and multiply... > If not, then what was the ark made of and what did > the termites eat on the ark? > As the ark was made in a hurry, bark was not removed from the planks to save days of laborious labour. Upon tour, the termites were requested to eat only the bark, lest they (and the boat itself) be thrown into the abysmal abyss. Thus upon debarkation, ie when the bark had been removed, it was revealed unto Noah and his passengers, the time of debarkation had come. -- In no time at all, time had begun for it's time had come. === Subject: Re: Group theory question >>For a prime p and natural k define f( p^k )= (p^k-1)... (p^k - p^(k-1) >>). >>Let n = p*q^a where p and q are prime numbers and a is a natural number >>and This post seems very confused. You define n and then make no further >reference to n. Presumably p and q are intended to be distinct primes? >p doesn't divide f(q^a). >>I'm interested in the following question: >>How can one prove that there exists a group H of order q^a which has an >>automorphism of order p under the above given hypotheses and what the >>role this condition p doesn't >>divide f(q^a). plays here Well, for any prime p dividing the automorphism group of a group H of >order q^a, either p = q or p divides f(q^a). Suppose, for example, that a=1. >Then H is cyclic of order p, and |Aut(H)| = p-1 = f(q^a). Perhaps the question was to prove that there exists no such group H? >Let n = p*q^a where p and q are prime numbers and a is a natural number >and p doesn't divide f(q^a). you can use Sylow's Theorem to deduce that any group of order n has a unique and hence normal Sylow p-subgroup (assuming p not equal to q). have an automorphism of order p, because such an automorphism would enable you to construct (using a semidirect product) a group of order n with a non-normal subgroup of order p. Derek Holt. === Subject: Re: Group theory question For a prime p and natural k define f( p^k )= (p^k-1)... (p^k - p^(k-1) >>). >>Let n = p*q^a where p and q are prime numbers and a is a natural number >>and This post seems very confused. You define n and then make no further >reference to n. Presumably p and q are intended to be distinct primes? >p doesn't divide f(q^a). >>I'm interested in the following question: >>How can one prove that there exists a group H of order q^a which has an >>automorphism of order p under the above given hypotheses and what the >>role this condition p doesn't >>divide f(q^a). plays here Well, for any prime p dividing the automorphism group of a group H of >order q^a, either p = q or p divides f(q^a). Suppose, for example, that a=1. >Then H is cyclic of order p, and |Aut(H)| = p-1 = f(q^a). Perhaps the question was to prove that there exists no such group H? >Let n = p*q^a where p and q are prime numbers and a is a natural number >and p doesn't divide f(q^a). you can use Sylow's Theorem to deduce that any group of order n has a > unique and hence normal Sylow p-subgroup (assuming p not equal to q). have an automorphism of order p, because such an automorphism would > enable you to construct (using a semidirect product) a group of order n > with a non-normal subgroup of order p. Derek Holt. I have another question: Let H be a group of order n = q^a with the same conditions on p,q,a. Is it true that the order of Aut(H) divides f(q^a) ? and if it how one can prove it === Subject: corrected formula for ALPHA The rotating Universe Angular moment L universe = (1/h) ^ ((4th root ( 1/alfa)) - 2) With today's h, alpha values L universe = 6,027620093*10^22 =1/10*((N Avogadro surface of universe ) ^2)/ (N Avogadro volume universe ( normal N A)) And (1/alpha) = (1/alpha matth) * ((ln ( 1/h) /ln ( 1/2*pi*h)) ^2) (1/alpha math) = ln (( L/2*pi*h)/ln ( 1/h) ) ^4 = (2^4)* ( 1+G/10^-12) ^4 )= = (ln((1/10* (N Avogadro surface of universe ) ^2)/ (N Avogadro volume universe)) -1)))^4 D universe = 4th root ( 1/alpha) = 3,421437946 = f ( t) This formula is valid for any organism ( alpha depends on development stage); It has long been speculated that fundamental constants are numerically related. I forgot number 10 in my previous post. Speculation: this formula demonstrates the fractal nature of 3rd/4th mass space dimension. Its value today is 2G= 1,42. 2 other space dimensions are on spheric surface of Universe; Fractal dimension 0,42 characterizes thickness of this surface layer measured in length units of surface topographical defects ( black holes) in relation to maximum possible length; Fractal dimension grows as universe gets older. This formula also links continuos and discrete dimensions via fractal dimenesion; Continuous mass time dimension only happens at death moment when Universe will disperse in non rotating masses to be further digested and freed from structure by worms of universe and serve as point mass food. D can also be considered an integral measure of Universe complexity; If laws of nature are the same in all universes, probably it is enough to look at this 1 value and it time evolution which could be calculated from system dynamics equations. G, N avogadro also evolves with universe Time scale of Universe if derivative of mass space; such that: (1/(G*c))* ( h/N) * (S/M^2)* (1/T) = const= (1/2*pi) G- gravitation constant c-speed of light S- surface of universe M- mass of universe T - age of universe, longest time scale in universe. N= N Avogadro surf ^2/ N Avogadro Volume === Subject: Re: Fourier Transform / Decay / Canot set I am not sure if anyone is still reading this, but I looked at similar questions in Phys Rev E 49, 3171-3178 (1994) and refs therein. The Pisot law also applies to higher-dimensional fractals, ie I found fractals with no decay of FT for rotations of order 1-10, 12, 15, 18 30 due to Pisot properties of trigonometric functions for these angles, and also for all finite rotation groups in 3D. Some of the work is published in the journal Fractals, which is hard to get access to (sorry!) but is posted on my web page (with permission). === Subject: Need help for PDE with boundary value problem I am working on the following PDE with boundary values: dA/dt = (X-IDENT)*A+Y*A0 dB/dt = c * INV[IDENT-c*HES(e(A,B)]*GRAD(f(A,B)) A(0)=A0, B(0)=B0, A(T)=A*, B(T)=B* Variables A,B,X,Y are time dependent, e.g. A(t), where the time is dropped for simplicity. f is a fixed and given nonlinear (but smooth and bounded) scalar-valued function. c is a small positive constant HES and GRAD are Hessian and Gradient operations on function f, with respect to B only (not A). IDENT is identity matrix INV is matrix inversion A and B are vectors while X and Y are square matrices. In general, A and B (and consequently X and Y) can be of different dimensions. We would like to solve this problem for X(t) and Y(t), i.e. finding X(t) and Y(t) in such a way that the given equations hold (t from 0 to T). Now I have two questions about this system of equations: 1. Familiarity: Is the behavior of this system (even very roughly) similar to any well-known system in mathematics or physics? 2. Analysis: What can be said about existence of solution? Is there any way to solve this analytically or numerically for obtaining X(t) and Y(t)? H.M. === Subject: Re: Need help for PDE with boundary value problem Wanted to correct my typo... by HES(e(A,B)) I meant HES(f(A,B)) I am working on the following PDE with boundary values: dA/dt = (X-IDENT)*A+Y*A0 > dB/dt = c * INV[IDENT-c*HES(e(A,B)]*GRAD(f(A,B)) > A(0)=A0, B(0)=B0, A(T)=A*, B(T)=B* Variables A,B,X,Y are time dependent, e.g. A(t), where the time is > dropped for simplicity. > f is a fixed and given nonlinear (but smooth and bounded) scalar-valued > function. > c is a small positive constant > HES and GRAD are Hessian and Gradient operations on function f, with > respect to B only (not A). > IDENT is identity matrix > INV is matrix inversion A and B are vectors while X and Y are square matrices. In general, A > and B (and consequently X and Y) can be of different dimensions. We would like to solve this problem for X(t) and Y(t), i.e. finding > X(t) and Y(t) in such a way that the given equations hold (t from 0 to > T). Now I have two questions about this system of equations: 1. Familiarity: Is the behavior of this system (even very roughly) > similar to any well-known system in mathematics or physics? 2. Analysis: What can be said about existence of solution? Is there any > way to solve this analytically or numerically for obtaining X(t) and > Y(t)? > > > H.M. === Subject: Re: Inference problem with normally distributed random variables can now get the expression from: max_a E[(a s1+(1-a) s2-v)^2] this gives a=(1/precision_2)/(1/precision_1+1/precision_2) and the expression below is a s1+(1-a) s2 Could I get the same result from a regression (which produces unbiased efficient estimates), e.g. v=a_0+a1 s1+a2 s2+error? florian > Consider the following set of three random variables v, s1 and s2 > (all random var. are norm. distr.): v=e+theta; e is a constant, and E[theta]=0, > Var[theta]=1/precision_theta > s1=v+epsilon1; E[epsilon1]=0, Var[epsilon1]=1/precision_1 > s1=v+epsilon2; E[epsilon2]=0, Var[epsilon2]=1/precision_2 I assume that the second s1 should be an s2, and that > theta, epsilon1, and epsilon2 are mutually independent. > The question I have is: what is (precision1*s1+precision2*s2)/(precision1+precision2) Is is not E[v|s1,s2] as I first thought but it should somehow be the > optimal aggregation of information (s1,s2) about v. I'd call it the minimum-variance conditionally unbiased estimate of > v|s1,s2. The words are tricky here. If I said just unbiased instead > of conditionally unbiased then some people might object on the > grounds that the variance has not been minimized, that it could be > reduced by regressing the estimate towards the mean (e). However, > I would call that only marginally unbiased, since it would not be > unbiased in every instance. (Ordinary least-squares regression, which > is usually said to be unbiased, is really only marginally unbiased.) === Subject: Re: Inference problem with normally distributed random variables can now get the expression from: max_a E[(a s1+(1-a) s2-v)^2] I assume you mean min, not max. this gives a=(1/precision_2)/(1/precision_1+1/precision_2) and the > expression below is a s1+(1-a) s2 Could I get the same result from a regression (which produces unbiased > efficient estimates), e.g. v=a_0+a1 s1+a2 s2+error? florian It would depend on how you did the regression. Suppose you have a sample of n objects, and you want to know each object's weight. In your terms, v_1,...,v_n are the unknown weights, and e is the average weight in the population from which your sample was taken. For each object, s_1 and s_2 are weight measurements made with two different instruments, say a cheap low-precision one and an expensive high- precision one. Neither instrument is biased. The question is how to combine the two measures to get a better estimate of each object's weight. To do the regression, you would of course have to know v. Then regressing v on s_1 and s_2 in the usual way would give you what I called marginally unbiased estimates of v: objects that are heavier than average would have their weights underestimated, and objects that are lighter than average would have their weights overestimated, but on average (i.e., marginally) the estimates would be unbiased. This phenomenon is known as regression to the mean. To get the formula you gave in your first post, you would have to omit the additive constant from the regression equation and define a_2 = 1 - a_1. This would give you conditionally unbiased estimates of v: each object's estimated weight would be unbiased. === Subject: Re: JSH: Turning success upside down > >>Troll Score = total responding posts - posts by JSH = 4 - 1 = 3 >>Troll Effeciency = total # posts / #posts by JSH = 3/1 = 300% >>(excluding this post) > > > Why are you excluding that post? You really ought > to include it in your count, along with any of its > descendants like this one. > -jiw Yep, we're now up to 600%. Any advance on %600%?? === Subject: Re: JSH: Turning success upside down > Troll Score = total responding posts - posts by JSH = 4 - 1 = 3 Troll Effeciency = total # posts / #posts by JSH = 3/1 = 300% > (excluding this post) >> Why are you excluding that post? You really ought >> to include it in your count, along with any of its >> descendants like this one. >> -jiw > > > Yep, we're now up to 600%. Any advance on %600%?? Oops, 800%. === Subject: Solving an integral Hello everybody, I am trying to solve this integral: f := int ( cos(x) / sqrt[ A*A*sin(x/2)*sin(x/2) + (Bx)^2] dx, x=0..N*2*pi where N==(int)0..20, and A and B are small fixed values (around 0.0001..0.05). The problem is the additional summand in the sqare root of the denominator (ie., (Bx)=C2=B2). Without this part the integral would be an elliptic one and could be represented with a combination of E(x) and F(x). I have tried developing the sine function into rows (x-x^3/3!+...) but to be precise enough I would have to go up to x^150/150! which is hardly practical. I have also tried seperating the denominator into expressions where either part of the sum is << or >> the other part so that suitable approximations can be made, without much luck. I would really appreciate any kind of hint or idea on how to tackle this problem. -- Jens Benecke === Subject: Re: Solving an integral On Mon, 14 Aug 2006 11:21:48 +0200, Jens Benecke Hello everybody, I am trying to solve this integral: f := int ( cos(x) / sqrt[ A*A*sin(x/2)*sin(x/2) + (Bx)^2] dx, x=0..N*2*pi where N==(int)0..20, and A and B are small fixed values (around > 0.0001..0.05). > The problem is the additional summand in the sqare root of the > denominator > (ie., (Bx)=C2=B2). Without this part the integral would be an elliptic > one > and could be represented with a combination of E(x) and F(x). I have tried developing the sine function into rows (x-x^3/3!+...) but to > be precise enough I would have to go up to x^150/150! which is hardly > practical. I have also tried seperating the denominator into expressions > where either part of the sum is << or >> the other part so that suitable > approximations can be made, without much luck. > I would really appreciate any kind of hint or idea on how to tackle this > problem. > Clever one! Remember pythagoras. and the fundemental theorem of calulus. -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/ === Subject: Re: Bourbaki & Witt === Subject: Re: Bourbaki & Witt > Bourbaki-Witt theorem. > ordered S, > inflationary f:S -> S, > for all chains C subset S, sup C in S > == some x with x = f(x) Does this theorem go by any other names? This is known as Tarski's theorem in some books. How is that so? Is Tarski a member of Bourbaki? Tarski also has another fixed point theorem complete lattice L, ascending f:L -> L ==> f has fixed point In fact, the theorem goes on to show the fixed points of f is a complete lattice. So how are they distinguished? ---- === Subject: Re: Bourbaki & Witt > Bourbaki-Witt theorem. > ordered S, > inflationary f:S -> S, > for all chains C subset S, sup C in S > == some x with x = f(x) As Bourbaki isn't an anybody, how's there a theorem by Bourbaki? > Why then, is the theorem not simply Witt's theorem? Well, one can guess that (some form of) the theorem first appeared in a work signed by Bourbaki. The fact that Bourbaki is a collective is quite irrelevant, if the members of the collective do not care, don't you think? -- m === Subject: Re: Bourbaki & Witt === Subject: Re: Bourbaki & Witt > Bourbaki-Witt theorem. > ordered S, > inflationary f:S -> S, > for all chains C subset S, sup C in S > == some x with x = f(x) As Bourbaki isn't an anybody, how's there a theorem by Bourbaki? > Why then, is the theorem not simply Witt's theorem? > Well, one can guess that (some form of) the theorem first > appeared in a work signed by Bourbaki. The fact that Bourbaki > is a collective is quite irrelevant, if the members of the > collective do not care, don't you think? Was Tarski or Witt in Bourbaki? ---- === Subject: Re: Bourbaki & Witt === > Subject: Re: Bourbaki & Witt Bourbaki-Witt theorem. > ordered S, > inflationary f:S -> S, > for all chains C subset S, sup C in S > == some x with x = f(x) As Bourbaki isn't an anybody, how's there a theorem by Bourbaki? > Why then, is the theorem not simply Witt's theorem? Well, one can guess that (some form of) the theorem first > appeared in a work signed by Bourbaki. The fact that Bourbaki > is a collective is quite irrelevant, if the members of the > collective do not care, don't you think? Was Tarski or Witt in Bourbaki? No. (If that had been the case, the teorem would be called simply Bourbaki's theorem!) Mathematics is full with examples of theorems which were obtained simultaneously, or non-simulteaneously but independently, by different people. It is also full with examples of misattributed theorems... Also, theorems get reformulated, recast in different languages, reproved, and what not, and sometimes this is sufficiently significant that lastnames are added to the name of the theorem (by certain authors, maybe others will keep the original name...) In any case, the theorem you are after is really of such a basic and simple character that it is quite probable several people could have come up with it driven by need at various different times. -- m === Subject: Re: DE - Reduction of Order questions === Subject: Re: DE - Reduction of Order questions > William Elliot ^^^ > What method? > I mean the way you solved 2. Don't recall the method nor the problem. > For Reduction of Order: > How about a quick explanation? > (1) u'' + p u' + q u = 0 > say we know a solution, u_1, we look for second, lin. indep. u_2 = v > u_1 > Differentiating this twice and putting it back to (1), What does that mean? Are you saying make the substitution u = v.u_1 ? Consequently u' = v'.u_1 + v.u'_1 u = v.u_1 + v'.u'_1 + v'.u'_1 + v.u_1 0 = u + pu' + qu = v.u_1 + 2v'.u'_1 + v.u_1 + pv'.u_1 + pv.u'_1 + qv.u_1 = v.u_1 + 2v'.u'_1 + pv'.u_1 > we have > (2) u_1 v'' + ( 2 u_1' + p u_1 ) v' = 0 > Then let w = v', w' = v'' and (2) becomes > u_1 w' + ( 2 u_1' + p u_1 ) w = 0 which is of first order. > For question 2, > Second solution, try: u_2 = v ( x + 1 ) > where v is a function of x. > Differentiating u_2 twice and putting it back to the DE give: > ( x + 1 ) v'' + [ 2 - ( x + 1 )^2 ] v' = 0 > v/v' = x + 1 - 2/(x + 1) log v' = x^2 / 2 + x - 2.log(x + 1) + c v' = k/(x + 1)^2 * e^x * sqr e^x^2 = yuck ---- === Subject: Re: DE - Reduction of Order questions William Elliot .9b.8d.93.b9.81F === > Subject: Re: DE - Reduction of Order questions William Elliot ^^^ What method? > I mean the way you solved 2. Don't recall the method nor the problem. For Reduction of Order: > How about a quick explanation? (1) u'' + p u' + q u = 0 > say we know a solution, u 1, we look for second, lin. indep. u 2 = v > u 1 Differentiating this twice and putting it back to (1), What does that mean? Are you saying make the substitution > u = v.u 1 ? Consequently > u' = v'.u 1 + v.u' 1 > u = v.u 1 + v'.u' 1 + v'.u' 1 + v.u 1 0 = u + pu' + qu > = v.u 1 + 2v'.u' 1 + v.u 1 > + pv'.u 1 + pv.u' 1 + qv.u 1 > = v.u 1 + 2v'.u' 1 + pv'.u 1 Yup, that's what I mean. we have > (2) u 1 v'' + ( 2 u 1' + p u 1 ) v' = 0 > Then let w = v', w' = v'' and (2) becomes > u 1 w' + ( 2 u 1' + p u 1 ) w = 0 which is of first order. For question 2, > Second solution, try: u 2 = v ( x + 1 ) > where v is a function of x. > Differentiating u 2 twice and putting it back to the DE give: > ( x + 1 ) v'' + [ 2 - ( x + 1 )^2 ] v' = 0 > This equation is obtained accidentally.. TCL gave the correct one. The original problem is solved, while this isn't. I'm still interest in solving this one, any idea? v/v' = x + 1 - 2/(x + 1) log v' = x^2 / 2 + x - 2.log(x + 1) + c > > v' = k/(x + 1)^2 * e^x * sqr e^x^2 = yuck > > ---- === Subject: Re: DE - Reduction of Order questions How about a quick explanation? (1) u'' + p u' + q u = 0 > What does that mean? Are you saying make the substitution > u = v.u_1 ? Consequently > u' = v'.u_1 + v.u'_1 > u = v.u_1 + v'.u'_1 + v'.u'_1 + v.u_1 0 = u + pu' + qu > = v.u_1 + 2v'.u'_1 + v.u_1 > + pv'.u_1 + pv.u'_1 + qv.u_1 > = v.u_1 + 2v'.u'_1 + pv'.u_1 Yup, that's what I mean. > Using the fact that u_1 is a solution. -- > ( x + 1 ) v'' + [ 2 - ( x + 1 )^2 ] v' = 0 > I'm still interest in solving this one, any idea? v/v' = x + 1 - 2/(x + 1) log v' = x^2 / 2 + x - 2.log(x + 1) + c v' = k/(x + 1)^2 * e^x * sqr e^x^2 = yuck integral yuck = big messy numerical approximation series === Subject: Re: DE - Reduction of Order questions > 1. Verify that u_1 = sin (x^2) is a solution of x u'' - u' + 4 x^3 u = > 0 and use reduction of order to find a second, linearly independent > solution. 2. Verify that u_1 = x + 1 is a solution of x u'' - ( x + 1 ) u' + u = > 0 and use reductio of order to find the general solution. 1. I've verified u_1 and tried u_2 = v (x) u_1 as another solution. > Substituting and rearranging terms gives x v'' + ( 4 x^2 cos (x^2) - > sin (x^2) ) v' = 0. Then let w = v', we have x w' + ( 4 x^2 cos (x^2) - > sin (x^2) ) w = 0, which is a first order, linear and homogeneous ODE. > Did I make any mistake so far? Yes. You did. The correct eqn is x sin(x^2) v + ( 4 x^2 cos (x^2) - sin (x^2) ) v' =0 Now retry it. >If not, I proceeded and found difficulty > in integrating [sin (x^2)] / x. 2. The reduced equation I found was ( x + 1 ) v'' + [ 2 - ( x + 1 )^2 ] > v' = 0. Did I make any mistake? Yes. The correct eqn is: x(x+1)v - (x^2+1)v' = 0. Now retry it. I found difficulty in solving the ODE > again. Answer > 1: cos (x^2) > 2: A exp(x) + B ( x + 1 ) I'm taking my first Differential Equation course. Nick > === Subject: Re: DE - Reduction of Order questions TCL .9b.8d.93.b9.81F > 1. Verify that u 1 = sin (x^2) is a solution of x u'' - u' + 4 x^3 u = > 0 and use reduction of order to find a second, linearly independent > solution. 2. Verify that u 1 = x + 1 is a solution of x u'' - ( x + 1 ) u' + u = > 0 and use reductio of order to find the general solution. 1. I've verified u 1 and tried u 2 = v (x) u 1 as another solution. > Substituting and rearranging terms gives x v'' + ( 4 x^2 cos (x^2) - > sin (x^2) ) v' = 0. Then let w = v', we have x w' + ( 4 x^2 cos (x^2) - > sin (x^2) ) w = 0, which is a first order, linear and homogeneous ODE. > Did I make any mistake so far? Yes. You did. The correct eqn is x sin(x^2) v + ( 4 x^2 cos (x^2) - sin (x^2) ) v' =0 > Now retry it. >If not, I proceeded and found difficulty > in integrating [sin (x^2)] / x. 2. The reduced equation I found was ( x + 1 ) v'' + [ 2 - ( x + 1 )^2 ] > v' = 0. Did I make any mistake? Yes. The correct eqn is: > x(x+1)v - (x^2+1)v' = 0. > Now retry it. > I found difficulty in solving the ODE > again. Answer > 1: cos (x^2) > 2: A exp(x) + B ( x + 1 ) I'm taking my first Differential Equation course. Nick > === Subject: Re: DE - Reduction of Order questions Nick .9b.8d.93.b9.81F > TCL .9b.8d.93.b9.81F 1. Verify that u 1 = sin (x^2) is a solution of x u'' - u' + 4 x^3 u = > 0 and use reduction of order to find a second, linearly independent > solution. 2. Verify that u 1 = x + 1 is a solution of x u'' - ( x + 1 ) u' + u = > 0 and use reductio of order to find the general solution. 1. I've verified u 1 and tried u 2 = v (x) u 1 as another solution. > Substituting and rearranging terms gives x v'' + ( 4 x^2 cos (x^2) - > sin (x^2) ) v' = 0. Then let w = v', we have x w' + ( 4 x^2 cos (x^2) - > sin (x^2) ) w = 0, which is a first order, linear and homogeneous ODE. > Did I make any mistake so far? Yes. You did. The correct eqn is x sin(x^2) v + ( 4 x^2 cos (x^2) - sin (x^2) ) v' =0 > Now retry it. BTW, how to solve x v'' + ( 4 x^2 cos (x^2) - sin (x^2) ) v' = 0 or ( x + 1 ) v'' + [ 2 - ( x + 1 )^2 ] v' = 0? >If not, I proceeded and found difficulty > in integrating [sin (x^2)] / x. 2. The reduced equation I found was ( x + 1 ) v'' + [ 2 - ( x + 1 )^2 ] > v' = 0. Did I make any mistake? Yes. The correct eqn is: > x(x+1)v - (x^2+1)v' = 0. > Now retry it. > I found difficulty in solving the ODE > again. Answer > 1: cos (x^2) > 2: A exp(x) + B ( x + 1 ) I'm taking my first Differential Equation course. Nick > === Subject: Re: DE - Reduction of Order questions Nick .8c¿.82.8e».8b.95.b9.9e > TCL .8c¿.82.8e».8b.95.b9.9e 1. Verify that u_1 = sin (x^2) is a solution of x u'' - u' + 4 x^3 u = > 0 and use reduction of order to find a second, linearly independent > solution. 2. Verify that u_1 = x + 1 is a solution of x u'' - ( x + 1 ) u' + u = > 0 and use reductio of order to find the general solution. 1. I've verified u_1 and tried u_2 = v (x) u_1 as another solution. > Substituting and rearranging terms gives x v'' + ( 4 x^2 cos (x^2) - > sin (x^2) ) v' = 0. Then let w = v', we have x w' + ( 4 x^2 cos > (x^2) - > sin (x^2) ) w = 0, which is a first order, linear and homogeneous ODE. > Did I make any mistake so far? Yes. You did. The correct eqn is x sin(x^2) v + ( 4 x^2 cos (x^2) - sin (x^2) ) v' =0 > Now retry it. BTW, how to solve x v'' + ( 4 x^2 cos (x^2) - sin (x^2) ) v' = 0 or ( x + 1 ) v'' + [ 2 - ( x + 1 )^2 ] v' = 0? For the first one: v= int( exp(-2sin(x^2)+Si(x^2)/2), dx), where Si(x) = int_0^x sin(t)/t dt. For the second one: v= int ( exp((x+1)^2/2) / (x+1)^2, dx). 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Somebody posted his solution earlier, > but I fear it was wrong. So now I'd like to make sure. > Anyone maybe? You try and help christian_mohr, he tells you that he knew that already, what he *really* wanted to know was something completely different. See for the original thread. And then he has the nerve to start a new thread, saying that in his opinion you were wrong! Christian, I have lost my patience with you. You are a nuisance. === Subject: Re: Counting numbers Lemme get this straight, you want to figure out how many numbers there are that have less than or equal to 9 digits ie the lowest number is 1 and the highest 999,999,999... is zero included or not? === Subject: Re: Counting numbers Oups, I feel really sorry, I forgot to add one quite important word. I am looking for the amount of numbers with n digits that contain at most 9 _different_ digits. Preceding zeros are of course omitted. === Subject: Simulated Annealing Hello all. I am looking for clarifacation of the terms move and neighbor as they are used in lottery covering designs. eg.Nurmela quotes the following... A solution ..s'.. is a neighbor of the solution ..s.. if s' = d(s) for some d an element of D. He previously refers to d as being a move and D being the set of all moves. So say we have a hypothetical lottery design in 45 lines which achieves full (but not optimal) cover. What would move mean here? Would it be the random altering if a single line that still leaves our cover in tact? Would this slightly altered cover be a neighbor of the original cover? TIA Mick. === Subject: Re: In search of a simple proof > For any integer n > 1, find the smallest set of positive integers with > the following properties: > 1. All members of S are less than or equal to n > 2. No member of S divides any other member of S > 3. S is maximal in the sense that inclusion of any other number in S > would violate either (1) or (2) Let p_i denote the i. prime number, i.e. p_1=2, p_2=3,... I claim that the set S contains numbers of the form s_1=p_1^{a_1}, s_2=p_1^{b_1} p_2^{b_2}, s_3=p_1^{c_1} p_2^{c_2} p_3^{c_3} ... s_k=p_1^{k_1} ........................................ p_k^{k_k} up to a number s_k, where p_k is the biggest prime <=n and the entries on the diagonal a_1, b_2, c_3,.... are >=1. (If this is proven, S has cardinality at least k.) Indeed, since including 2 to S violates the second axiom by maximality, 2 divides some s. Among the s divisible by 2 choose one such that v_2(s) is maximal, where v_2(s) is the exponent of 2 in the prime factorization of s. Then s=2^i*b with b odd, and if b weren't 1, 2^{i+1} would be a number smaller than n which could be included to S, contradiction. This establishes the existence of s_1. For s_2, again there exists some number s divisible by 3. If s is of the form s=2^a 3^b we're done. Otherwise choose s such that v_3(s) is maximal. Then s=2^a*3^b*c with c>3, and s':=3^{b+1} I think I found a solution to my own problem. Is the following > reasoning valid? For any solution set S, use the following method to replace S by a set > that is of size equal to S and consists only of primes. Find a member of S of the form p^k for some prime p. There must be > some such number because otherwise find a non-prime number m in S that > can be expressed as the product of the smallest number of primes: > m=p1*p2*...pi, where the primes are not necessarily unique. Let pj be > the smallest of these primes. Then pj^i could be added to S without > violatihg (1) or (2), which would contradict (3). Once a member of S is found of the form p^k, replace p^k with p and > remove p as a factor from all other members of S. For example where > n=6 and S={4,5,6}, replace 4 with 2 and then replace 6 with 3. process. Keep doing this until the only members of the set are prime. Indeed there is a member of the form p^k, but it's not legal to remove p as a factor from the other members: Look at {12,16,18}. === Subject: Re: FLTMA: Using FLT as a lemma A slightly modified version of this post has a reply in sci.math.research. > If there were a counterexample, {n,a,b,c} to FLT, n>2, > a^n + b^n = c^n (and there isn't), > then there would be a primitive counterexample {p,x,y,z}, > x^p + y^p = z^p with 1) gcd(x,y,z) = 1 > 2) exactly one of {a,b,c} even > 3) 0 < x < y < z < (x+y) > 4) p an odd prime. For this nonexistent counterexample, (x^n + y^n) == 0 mod z, > (z^n - x^n) == 0 mod y, and > (z^n - y^n) == 0 mod x. For any positive x, y, and z, under conditions 1), 2), and 3), > a sequence S(n) exists with mod indicating the remainder of division: S(n) = > (x^n + y^n mod z + > (z^n - x^n) mod y + > (z^n - y^n) mod x; S(0) = 2. Now, since FLT is true, we know the zeros of this sequence will never > be at > any odd prime. > And that is where I made a mistake. Doug === Subject: G(x+1)/G(x+0.5),G=Gamma Denote e(x,a)=exp( 1/( 64(x+a)^2) ) , E(x,a):=e(x,a)(x+0.25)^{0.5} Prove or disprove E(x,0.75)< G(x+1)/G(x+0.5)}< E(x,-0.25) , (for x> 1), G being Gamma function. === Subject: Re: G(x+1)/G(x+0.5),G=Gamma > Denote e(x,a)=exp( 1/( 64(x+a)^2) ) , > E(x,a):=e(x,a)(x+0.25)^{0.5} > Prove or disprove E(x,0.75)< G(x+1)/G(x+0.5)}< E(x,-0.25) , (for x> 1), G being Gamma function. In giving bounds for G(x + 1)/G(x + 1/2) in the form E(x, a), we can do much better than using a = 3/4 and -1/4, resp., for the lower and upper bounds. With a = 1/4 (rather than -1/4), we get the tightest possible upper bound. It is valid for x > -1/4. Using any a > 1/4, E(x, a) provides a lower bound if x is sufficiently large. But if, say, we wish the lower bound to be valid for all x > -1/4, then a must be at least 0.36995... Combining the above, we may say E(x, 0.37) < G(x + 1)/G(x + 1/2) < E(x, 1/4) for x > -1/4. David Cantrell === Subject: Re: order of magnitude of an inverse order of >magnitude of A (-1). Here a is unknown, it can be a scalar or a >function of n Not determined. Sorry, I wasn't clear. With A (-1) , I meant A^(-1). What is A? A function of n, I guess, but are the values > numbers, matrices, what? Also, I presume you're talking about > integers n -> infinity? A is a matrix. Yes, i am talking about as n goes to infinity. If a is constant (i.e. not dependent on n), then > O(n) + a O(sqrt(n)) = O(n). All this means is that there are > some constants N and K such that |A| <= K n for n > N (if > A is a matrix, this means all |A_{ij}| <= K n for n > N). > If a is a function of n, then without further information > on a there's nothing you can say. Can't we say about the order of magnitude in terms of a? (for example: A^(-1)=a^(-1)O(n^(-1)). If A = O(n), there's no guarantee that A^(-1) exists: A could > be 0. If A is a number and nonzero, A^(-1) = Omega(1/n), > i.e. there exist constants N and K > 0 such that > |A^(-1)| >= K/n for n > N. Assume A exists. I think A^(-1) should be O(n^(-1))+a^(-1)O(n^(-1/2)). And let say if a=1/n then A^(-1)=O(n^(-1))+a^(-1)O(n^(-1/2)) =O(n^(-1))+O(n^(1/2)) =O(n^(1/2)) Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: order of magnitude of an inverse > If A = O(n) + aO(sqrt(n)) , then what is the >>order of >>magnitude of A (-1). Here a is unknown, it can be a scalar or a >>function of n >> Not determined. Sorry, I wasn't clear. With A (-1) , I meant A^(-1). > What is A? A function of n, I guess, but are the values >> numbers, matrices, what? Also, I presume you're talking about >> integers n -> infinity? A is a matrix. Yes, i am talking about as n goes to >infinity. > If a is constant (i.e. not dependent on n), then >> O(n) + a O(sqrt(n)) = O(n). All this means is that there are >> some constants N and K such that |A| <= K n for n > N (if >> A is a matrix, this means all |A_{ij}| <= K n for n > N). >> If a is a function of n, then without further information >> on a there's nothing you can say. Can't we say about the order of magnitude in terms >of a? (for >example: >A^(-1)=a^(-1)O(n^(-1)). > If A = O(n), there's no guarantee that A^(-1) exists: A could >> be 0. If A is a number and nonzero, A^(-1) = Omega(1/n), >> i.e. there exist constants N and K > 0 such that >> |A^(-1)| >= K/n for n > N. > Assume A exists. I think A^(-1) should be >O(n^(-1))+a^(-1)O(n^(-1/2)). And let say if a=1/n then >A^(-1)=O(n^(-1))+a^(-1)O(n^(-1/2)) =O(n^(-1))+O(n^(1/2)) > =O(n^(1/2)) I assume you mean assume A^(-1) exists. Consider e.g. the 2 x 2 matrices [use fixed-width font] [ n 0 ] A = [ 0 f(n) ] [ 1/n 0 ] A^(-1) = [ 0 1/f(n) ] with f(n) -> 0 as rapidly as you want as n -> infty. Then A = O(n), but unless you have a _lower_ bound on how rapidly f(n) -> 0, there's no way to get a bound on A^(-1). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: order of magnitude of an inverse > If A = O(n) + aO(sqrt(n)) , then what is the >>order of >>magnitude of A (-1). Here a is unknown, it can be a scalar or a >>function of n >> Not determined. Sorry, I wasn't clear. With A (-1) , I meant A^(-1). > What is A? A function of n, I guess, but are the values >> numbers, matrices, what? Also, I presume you're talking about >> integers n -> infinity? A is a matrix. Yes, i am talking about as n goes to infinity. > If a is constant (i.e. not dependent on n), then >> O(n) + a O(sqrt(n)) = O(n). All this means is that there are >> some constants N and K such that |A| <= K n for n > N (if >> A is a matrix, this means all |A_{ij}| <= K n for n > N). >> If a is a function of n, then without further information >> on a there's nothing you can say. Can't we say about the order of magnitude in terms of a? (for >example: >A^(-1)=a^(-1)O(n^(-1)). > If A = O(n), there's no guarantee that A^(-1) exists: A could >> be 0. If A is a number and nonzero, A^(-1) = Omega(1/n), >> i.e. there exist constants N and K > 0 such that >> |A^(-1)| >= K/n for n > N. > Assume A exists. I think A^(-1) should be >O(n^(-1))+a^(-1)O(n^(-1/2)). That simply doesn't follow from anything you've said. For an example with scalar A, let A = 0 for all n. Then it's true that A = O(n) + aO(sqrt(n)) with a = 1. But A^{-1} does not even exist. > And let say if a=1/n then >A^(-1)=O(n^(-1))+a^(-1)O(n^(-1/2)) =O(n^(-1))+O(n^(1/2)) > =O(n^(1/2)) > Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada ************************ David C. Ullrich === Subject: Min spacing of square-free numbers Suppose that I form a set of numbers S that consist of all the square-free numbers that can be made using only primes that are greater than some prime P-sub-k. Can I say anything about the minimum spacing between adjacent members of S? Bob Adams === Subject: Re: Min spacing of square-free numbers >Suppose that I form a set of numbers S that consist of all the >square-free numbers that can be made using only primes >that are greater >than some prime P-sub-k. Can I say anything about the >minimum spacing >between adjacent members of S? I assume can be made using means is a product of. The minimum spacing is certainly at least 2 since all your numbers are odd, and if the Twin Prime conjecture is true it's exactly 2. But maybe this is easier to prove... Let N be the product of the primes <= P_k. For any positive integer n, all primes dividing nN-1 and nN+1 are greater than P_k. Heuristically, if p is a prime > P_k, the probability of nN-1 or nN+1 being divisible by p is O(1/p^2), and since sum_p 1/p^2 is finite there should be infinitely many n for which nN-1 and nN+1 are squarefree. That's not a proof, but I wouldn't be surprised if there is a proof using sieve methods. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: JSH: Objectivity linux) >> You people, on the other hand, ran away from the truth. Just like the way you ran away from Atlanta? You know, it's really none of your business why he moved to San Francisco. What is the point of this post? You just want to be hurtful? Leave his offline personal issues out of it. -- It seems to me that some of you don't realize that [...] some day the truth comes out. It doesn't matter if you're dead. I've made certain that your name will live in infamy if it's known at all: Wiles, Ribet, Granville, or anyone else from this generation. --JSH beyond the grave === Subject: Re: JSH: Objectivity <874pwf32kn.fsf@phiwumbda.org >> You people, on the other hand, ran away from the truth. Just like the way you ran away from Atlanta? You know, it's really none of your business why he moved to San > Francisco. I'm making it my business. What is the point of this post? You just want to be hurtful? Absolutely. Leave his offline personal issues out of it. Sorry, as long as JSH is a danger to himself and others there will be no seperating his online and offline behaviour. -- > It seems to me that some of you don't realize that [...] some day the > truth comes out. It doesn't matter if you're dead. I've made certain > that your name will live in infamy if it's known at all: Wiles, Ribet, > Granville, or anyone else from this generation. --JSH beyond the grave === Subject: Re: JSH: Objectivity Discussion, linux) <874pwf32kn.fsf@phiwumbda.org> Leave his offline personal issues out of it. Sorry, as long as JSH is a danger to himself and others > there will be no seperating his online and offline behaviour. Oh, thank you, selfless protector of us wee innocent folk! I was so scared of JSH before you came to save us! You shore is good folk. Please. -- Start obeying math rules, or if you prefer I can proceed to tear away your illusions of being rational and logical piece by piece over time, as I reduce your society using advanced psychological tools that rival the best psychological warfare techniques of world governments. --JSH === Subject: Re: JSH: Objectivity >> You people, on the other hand, ran away from the truth. >> Just like the way you ran away from Atlanta? You know, it's really none of your business why he moved to San > Francisco. What is the point of this post? You just want to be hurtful? Leave his offline personal issues out of it. Actually that is the point of JSH postings, why is he doing this? what in his background is causing these flaws to be personally circulated world wide? is he locked up somewhere with white walls? Did he get out again? is he off his meds yet again? What meds is he on? is he relly working for Microsoft as bug introducer, why is he playing the race card if he is white? (It is a deception, he is white) What rule set is he using to keep getting so many responces after so many years? Is english his second language? Are there really two people using one computer? Why is he such a damn good Crackpot? How can he keep doing the same HUGE crap for YEARS? Is Atlanta infected now? -- > It seems to me that some of you don't realize that [...] some day the > truth comes out. It doesn't matter if you're dead. I've made certain > that your name will live in infamy if it's known at all: Wiles, Ribet, > Granville, or anyone else from this generation. --JSH beyond the grave === Subject: Formula for chess rating - can it be inverted? On the Internet Chess Club http://www.chessclub.com one plays games of chess and gets a rating. Your rating goes up if you win, down if you loose etc. If you draw, it may go up or down, depending on the rating of your opponent. The change in one's rating depends on the rating of your opponent too. http://www.chessclub.com/help/ratings If you play a game, the *rew* ratings (i.e. adjusted after the game) gets stored in a file. Blacks and Whites are stored as: [WhiteElo 1234] [BlackElo 1388] But there is no record of the ratings before the game. Usually the two will be different, although if you draw against an opponent who had the same rating as you before the game started, both your ratings will go unchanged. The formula to determine the *new* rating is below. It's taken directly from the page at: http://www.chessclub.com/help/ratings ----This is the formula used------ To explain the established period requires the use of a formula. Suppose your rating is r1, and the opponent's is r2. Let w be 1 if you win, .5 if you draw, and 0 if you lose. After a game, your new rating will be: 1 r1 + K * [ w - ---------------------- ] 1 + 10 ^ ((r2-r1)/400) I still need to explain the variable K. This is the largest change your rating can experience as a result of the game. The value K=32 is always used for established player versus established player. If you're playing a provisional player, the factor K is scaled by n/21, where n is one plus the number of games your opponent has played. --------------- Can this formula be inverted to find the old ratings (r1, r2) given the new rating of both you and your opponent? I've tried in Mathematica without success, but I'm not a mathematician. Any help appreciated. -- Dave (from the UK) Please note my email address changes periodically to avoid spam. It is always of the form: month-year@southminster-branch-line.org.uk Hitting reply will work for a few months only - later set it manually. http://witm.sourceforge.net/ (Web based Mathematica front end) === Subject: Re: Ping-Pong Question <336pd2d393bahfo20kqj75ggbbhm0vbthc@4ax.com> On 10 Aug 2006 20:08:34 -0700, Chip Eastham > On Thu, 10 Aug 2006 10:55:06 -0500, David C. Ullrich >Here's the whole story: >G is a group acting on X. G is generated by a and b. A subset X is such that the sets a^n(A) (n in Z) are disjoint and their union is X. Similarly for b and B. Also, for every n <> 0, a^n(A) subset B and b^n(B) subset A. >There is a bijection j from X to X, with j(j(x)) = x for all x. And j has a fixed point e. Also a and b are conjugate via j: b = jaj. And B=j(A). (Lest there be confusion, no, j is not an element of G.) >It's not hard to show that e is in E = A intersect B. And it's not hard to show that E intersect w(E) is empty for any non-trivial reduced word w in a and b. Hence a and b generate G freely; I gather this is the Ping Pong lemma. >The question: Does it follow that the union of g(E) for g in G is all of X? > Never mind, found a counterexample. Didn't _think_ it was true, a counterexample wasn't obvious. Cuz I wasn't looking at it right - if you think about the obvious way to try to prove it's true and then try to concoct something so the ping-ponging doesn't terminate you get this: > Let G be free on a and b. Let X be the space consisting of e (the identity of G), all finite reduced words in G, and also all infinite reduced words in a and b. Infinite in one direction only, extending to the left, so ...abab is an element of X. > Define an action of G on X by concatenate and then reduce. Takes only a little thought to see if w is a possibly infinite reduced word and g is in G then reducing wg takes only finitely many well-defined steps. > Define j by j(e) = e and j(w) = the word you get from w by swapping a and b. Let A = {e} together with all w in X that end with b or b^{-1} (meaning that the rightmost element is b or b^{-1}) and let B be {e} plus all words ending in a or a^{-1}. > All the hypotheses seem pretty clear, but A intersect B is just {e}, so the orbits of A intersect B don't cover X. >(In the application it's true, but that part involves some analysis. Since most of the analysis has been replaced by just algebra it would be keen if this part was just algebra as well...) >>Okay, or let X be the group freely >>generated by a,b,c, but define sets >>A,B as those reduced words in >>a,b that don't begin with a (nonzero) >>power of a or b (respectively). >>Thus {e} = A intersect B, >>a^n A subset B and >>b^n B subset A. >> ??? A intersect B includes all words >> beginning with a power of c, so >> G(A intersect B) = X. >> Running late this morning, so I haven't >> thought about whether this can be trivially >> fixed by redefining something. If yes then >> regardless. >>G acts on all of X but only >>reaches G subset X >>through its action >>on A intersect B. I meant here that A,B are subsets of G: ...define sets A,B as those reduced words >in a,b that don't begin with a (nonzero) >power of a or b (respectively). So, no c's in A,B, and G subset X (as >the free group on a,b, and c). But then, if I'm understanding you correctly > this time, X is not the union of the sets > a^n(A) as required - there's no c's in any > of the elements of any a^n(A). Yes, I overlooked that the a^n*A should cover X (indeed, partition it). Your approach seems attractive as there is a cardinality argument in the end that G acting on {e} = A intersect B, which is countable, will not cover uncountable X. === Subject: Re: Objectivity >> Tell me one thing. If you have such an interest in math, why haven't you >> gotten a formal education in it. A genius *rolls eyes* like you should >> have >> no problem getting at least a bachelor's in math. I am almost done with >> mine, got my associate's degree in math last year. >> Dave I have a B.Sc. in physics. A physics degree trumps a math degree, as > you need a lot of math anyway, and you also need more... Why don't you get a degree in physics, if you can, and then come back > and talk to me? > > I started out studying physics. I have a undergraduate study. Then I realized I really hated 'the miracle' approach. After two years of hearing of 'geniuses' and here is what you should compute on I was fed up. I went over to math. Once I had a bachelor in math I went back to physics for my masters at the institute for theoretical astrophysics in Quantum field theory and general relativity. Now with the mathematical maturity to truly understand them. I realizes that half ass pseudo dynamic descriptions were all wrong. This was a purly pragmatic approach. 1. You see symmetries. You categorize them (groups like SU3). 2. You create a representation. 3. You get a field (Noethlers theorem) Now I feel I have the background to truly get into the mind if Dirac and see what he saw. I never felt that in physics.. -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/ === Subject: Re: Objectivity On Mon, 14 Aug 2006 16:32:33 +0200, John Thingstad > Tell me one thing. If you have such an interest in math, why haven't you gotten a formal education in it. A genius *rolls eyes* like you should have no problem getting at least a bachelor's in math. I am almost done with mine, got my associate's degree in math last year. > Dave >> I have a B.Sc. in physics. A physics degree trumps a math degree, as >> you need a lot of math anyway, and you also need more... >> Why don't you get a degree in physics, if you can, and then come back >> and talk to me? >> I started out studying physics. I have a undergraduate study. Then I > realized > I really hated 'the miracle' approach. > After two years of hearing of 'geniuses' and here is what you should > compute on > I was fed up. I went over to math. > Once I had a bachelor in math I went back to physics for my masters > at the institute for theoretical astrophysics in Quantum field theory > and general relativity. > Now with the mathematical maturity to truly understand them. > I realizes that half ass pseudo dynamic descriptions were all wrong. > This was a purly pragmatic approach. > 1. You see symmetries. You categorize them (groups like SU3). > 2. You create a representation. > 3. You get a field (Noethlers theorem) > Now I feel I have the background to truly get into the > mind if Dirac and see what he saw. I never felt that in physics.. QED is SO2.. sorry about the dirac/SU3 combo QCD is SO3 (Gellman) -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/ === Subject: Re: Objectivity On Mon, 14 Aug 2006 16:41:27 +0200, John Thingstad > On Mon, 14 Aug 2006 16:32:33 +0200, John Thingstad >> Tell me one thing. If you have such an interest in math, why haven't > you > gotten a formal education in it. A genius *rolls eyes* like you > should have > no problem getting at least a bachelor's in math. I am almost done > with > mine, got my associate's degree in math last year. Dave > I have a B.Sc. in physics. A physics degree trumps a math degree, as you need a lot of math anyway, and you also need more... > Why don't you get a degree in physics, if you can, and then come back and talk to me? >> I started out studying physics. I have a undergraduate study. Then I >> realized >> I really hated 'the miracle' approach. >> After two years of hearing of 'geniuses' and here is what you should >> compute on >> I was fed up. I went over to math. >> Once I had a bachelor in math I went back to physics for my masters >> at the institute for theoretical astrophysics in Quantum field theory >> and general relativity. >> Now with the mathematical maturity to truly understand them. >> I realizes that half ass pseudo dynamic descriptions were all wrong. >> This was a purly pragmatic approach. >> 1. You see symmetries. You categorize them (groups like SU3). >> 2. You create a representation. >> 3. You get a field (Noethlers theorem) >> Now I feel I have the background to truly get into the >> mind if Dirac and see what he saw. >> I never felt that in physics.. QED is SO2.. sorry about the dirac/SU3 combo > QCD is SO3 (Gellman) > SU2 and SU3 respectivly. Gereneral relativity is SO(3,1) Bit rusty.. Bin working in AI lately. sorry bout these repet corrections -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/ === Subject: partitions of n into at most k parts can somebody tell me where I can find a resource that i can cite as a source that the number of ways to partition a positive integer n into at most k parts is given by the x^k term in the generating function: 1/[(1-x)(1-x^2)...(1-x^n)] when k<=n and the x^n term otherwise. === Subject: Better-quasi-ordering - minimal excluded sub-wqo I've seen indications in the literature that there exists a wqo set which embeds into every wqo which is not a bqo, but I haven't found a description of this anywhere. Could someone point me in the right direction? Kevin === Subject: Re: ICM 2006-Fields medal winners? > OK who do you guess will get Fields medals this year? A few weeks ago, I saw a message giving convincing reasons why Grisha Perelman had very high chances to be awarded a Fields medal next week, but I have lost the reference. If someone finds it, I am interested to know it. A hint is contained in the press release of the International Congress of Mathematicians (ICM) of July 31, 2006, at the address http://www.icm2006.org/press/releases/ It is said: . . . . . . . In 1994 Perelman sequestered himself away to tackle the problem, and for the following eight years gave managed to prove the Conjecture and posted his work - - much more extensive that Poincar.8e`s Conjecture itself- - on the Internet, for other experts to check. No one has drawn attention to any errors in Perelman`s work ever since, but the final verdict has still to be pronounced. . . . . . . . The part above is followed by a part that has been changed: Version of last week: See my posting of last week Part 4, at the address http://mathforum.org/kb/message.jspa?messageID=5004240&tstart=15 . . . . . . . The attendance of these experts at the ICM2006 has recently become the object of even greater interest since the Chinese mathematicians Xi-Ping Zhu, from the University of Zhongshan (Canton, China), and Huai-Dong Cao, from Lehigh University in Pennsylvania (USA), announced in the latest issue of the Asian Journal of Mathematics, published in the U.S.A., that they had arrived at ``a complete proof of the Poincar.8e and Geometrization Conjectures``. Zhu and Cao base their work according to their abstract- ``on the accumulative works of many geometric analysts in the past thirty years``, and they specifically mention Hamilton and Perelman himself. However, their paper is still not available on the Internet, and is therefore still not accessible to the mathematical community at large. IMU General Assembly The Congress will be preceded. . . . . . . . New version of this week: See the press release of July 31, 2006, (((modified this week))) at the address http://www.icm2006.org/press/releases/ The attendance of these experts at the ICM2006 has recently become the object of even greater interest since the Chinese mathematicians Xi-Ping Zhu, from the University of Zhongshan (Canton, China), and Huai-Dong Cao, from Lehigh University in Pennsylvania (USA), announced in the latest issue of the Asian Journal of Mathematics, published in the U.S.A., that they had arrived at ``a complete proof of the Poincar.8e and Geometrization Conjectures``. Zhu and Cao base their work - according to their abstract - ``on the accumulative works of many geometric analysts in the past thirty years``, and they specifically mention Hamilton and Perelman himself. On top of that, Bruce Kleiner and John Lott, from the University of Michigan (USA) have also published ``Notes on Perelman's papers``, described as ``detailed notes`` on two of Perelman's papers. IMU General Assembly The Congress will be preceded.83.83.83 The sentence ``However, their paper is still not available on the Internet, and is therefore still not accessible to the mathematical community at large.`` has been deleted, and the sentence ``On top of that, Bruce Kleiner and John Lott, from the University of Michigan (USA) have also published ``Notes on Perelman's papers``, described as ``detailed notes`` on two of Perelman's papers.`` has been added. The title ``Press release of July 31, 2006`` has not been changed. Jean-Claude Evard Western Kentucky University Department of Mathematics E-mail: Jean-Claude.Evard@wku.edu === Subject: Countdown to the 2006 Fields medal(s) The Fields medal(s) of 2006 will be awarded next week, on Tuesday, August 22, 2006, at the opening ceremony of the International Congress of Mathematicians (ICM2006). in Madrid (Spain) . The Fields medals are awarded every four years at the opening ceremonies of the ICM`s. At most four Fields medals are awarded at each meeting. See the two press releases August 1, 2006: ``His Majesty the King will present the Fields Medals, the ``Nobel Prizes`` for Mathematics`` and July 31, 2006 ``The Mathematical Community Awaits a Verdict on Poincar.8eÍs Conjecture, this summer at the ICM2006`` at the address http://www.icm2006.org/press/releases/ and more at the address http://www.icm2006.org/press/dossier/ The International Congress of Mathematicians (ICM2006) will be held in Madrid (Spain) from Tuesday, August 22 to Wednesday, August 30, 2006 See: http://www.icm2006.org/ http://www.icm2006.org/paginas/?pagina=home_ing and http://www.mathunion.org/ICM/index.html It is organized by the International Mathematical Union http://www.mathunion.org/index.html A guessing thread has been started by Gowan at the address http://mathforum.org/kb/thread.jspa?threadID=1428568&tstart=0 I have been especially interested in the recent publications of the last steps of the proof of Thurston`s geometrization conjecture, and I have posted information about this at the two addresses http://mathforum.org/kb/thread.jspa?threadID=1398589&tstart=15 and http://mathforum.org/kb/message.jspa?messageID=5004240&tstart=15 In addition, talks in many other Fields of mathematics will be presented at the ICM2006. See: ICM2006, main events: http://www.icm2006.org/onlineevents ICM2006, plenary lectures: http://icm2006.org/v_f/web_fr.php?PagIni=1pl ICM2006, sections lectures: http://icm2006.org/v_f/web_fr.php?PagIni=2il ICM2006, special activities: http://icm2006.org/v_f/web_fr.php?PagIni=7sa Jean-Claude Evard Western Kentucky University Department of Mathematics E-mail: Jean-Claude.Evard@wku.edu === Subject: Countdown to the 2006 Fields medal(s) The Fields medal(s) of 2006 will be awarded next week, on Tuesday, August 22, 2006, at the opening ceremony of the International Congress of Mathematicians (ICM2006). in Madrid (Spain) . The Fields medals are awarded every four years at the opening ceremonies of the ICM`s. At most four Fields medals are awarded at each meeting. See the two press releases August 1, 2006: ``His Majesty the King will present the Fields Medals, the ``Nobel Prizes`` for Mathematics`` and July 31, 2006 ``The Mathematical Community Awaits a Verdict on Poincar.8eÍs Conjecture, this summer at the ICM2006`` at the address http://www.icm2006.org/press/releases/ and more at the address http://www.icm2006.org/press/dossier/ The International Congress of Mathematicians (ICM2006) will be held in Madrid (Spain) from Tuesday, August 22 to Wednesday, August 30, 2006 See: http://www.icm2006.org/ http://www.icm2006.org/paginas/?pagina=home_ing and http://www.mathunion.org/ICM/index.html It is organized by the International Mathematical Union http://www.mathunion.org/index.html A guessing thread has been started by Gowan at the address http://mathforum.org/kb/thread.jspa?threadID=1428568&tstart=0 I have been especially interested in the recent publications of the last steps of the proof of Thurston`s geometrization conjecture, and I have posted information about this at the two addresses http://mathforum.org/kb/thread.jspa?threadID=1398589&tstart=15 and http://mathforum.org/kb/message.jspa?messageID=5004240&tstart=15 In addition, talks in many other Fields of mathematics will be presented at the ICM2006. See: ICM2006, main events: http://www.icm2006.org/onlineevents ICM2006, plenary lectures: http://icm2006.org/v_f/web_fr.php?PagIni=1pl ICM2006, sections lectures: http://icm2006.org/v_f/web_fr.php?PagIni=2il ICM2006, special activities: http://icm2006.org/v_f/web_fr.php?PagIni=7sa Jean-Claude Evard Western Kentucky University Department of Mathematics E-mail: Jean-Claude.Evard@wku.edu === Subject: Check Out The Important Stuffs Related To This Group On www.sweetinbox.com/Special Check Out The Important Stuffs Related To This Group On www.sweetinbox.com/Special You Can Also Get Some Entertainment Stuffs Like Free Ringtones, Wallpapers, Gossips, Themes & Mobile Reviews Check Out www.sweetinbox.com/Special & Enjoy === Subject: help with lambda i'm working through An Introduction to Lambda Calculus (Barendregt & Barendsen) (http://ling.ucsd.edu/~barker/Lambda/barendregt.94.pdf) and i'm having some trouble with a few of the problems. the content of problem 2.11 i don't really get - why is it special? it tells us to 'have another look' at the proceeding problems, most of which i solved with Y. is there another way i should have done it? is there an answer sheet for these problems available? i haven't been able to find it. if there is a special group for discussing this topic, or even this particular topic, please let me know. === Subject: Re: JSH: Other prime gaps g I doubt anyone will dispute that. > Too bad they're mostly crap. > Dave >> I say you are lying, and dare you to name one that is in any way crap. >> Hate my definition of mathematical proof? Why? Politics? Or good >> reasons? >> What about my idea for finding large primes? >> How about my prime counting function? Or my prime gap equation which >> this thread is supposed to be about, as do you have something better? >> Tell me what is the probability of finding a prime gap of 12 in the >> interval from 4567 to 5500. >> That should be a simple challenge if my ideas are crap. > This has already been done back in 1878. > I know the history. I also know there is a formula for twin primes > probability, but have not seen one for an arbitrary prime gap g. Yes, it is true that ALL the prime gaps have actually been found out to > a very high range, but that is different from the probability of a > particular gap. To understand the difference, consider that all the primes up to 100 > are known, but what is the probability that a number over that range is > prime? Most people with some knowledge in the area would immediately respond: approximately 1/(ln 100) So then, with it settled that I'm not asking for the actual prime gaps > themselves, what is the probability of finding a prime gap of 12 in > the interval from 4567 to 5500? Well, it's either 0 (there is no gap) or 1 (there is such a gap). As 4679 and 4691 are prime, the probability is 1. Perhaps probability is not the word you wanted? This is, after all, not a random variable. Next time I check, I'll still find that 4679 and 4691 are prime. - Randy === Subject: Re: JSH: Other prime gaps g > > Which is why these math wars are so vicious as I'm up against > politicians, who have everything to lose, and nothing to gain from the > truth, except maybe SOME dignity, if they acknowledge it before the > hand-cuffs get slapped on them. Do you have little Math Wars action figures that you play with in your room at night? If so, that's the closest you'll ever come to seeing a math war. (You just can't *buy* entertainment like the stuff Harris provides for free.) -- Wayne Brown (HPCC #1104) [CapitalThorn].bes ofer.8eode, [CapitalYAcute]isses sw.87 m.beg. (That passed away, this also can.) Deor, from the Exeter Book (folios 100r-100v) === Subject: Re: JSH: Other prime gaps g > Do you have little Math Wars action figures that you play with in your > room at night? If so, that's the closest you'll ever come to seeing a > math war. Haven't you seen those new action figures they have out now called the Math Warriors? I'm thinking about buying a pack for my nephew, he likes stuff like that. Dave === Subject: Re: JSH: Other prime gaps g room at night? If so, that's the closest you'll ever come to seeing a > math war. Haven't you seen those new action figures they have out now called the Math > Warriors? I'm thinking about buying a pack for my nephew, he likes stuff > like that. Be sure and get some of the villians. They have fully articulated arms (to facilitate handwaving). > > Dave === Subject: Re: Eigenvalues of tridiagonal matrix > Are there any ready to use formulas and/or methods for calcualting > eigen-values/vectors and matrix exponent of the following tridiagonal matrix > : Tridiagonal matrices are nothing special. Most software for eigenvalues first reduces the matrix to tridiagonal form, and computes the eigenvalues of that reduced matrix. In other words, having it in tridiagonal form buys you nothing. Victor. -- Victor Eijkhout -- eijkhout at tacc utexas edu ph: 512 471 5809 === Subject: Re: Eigenvalues of tridiagonal matrix > Are there any ready to use formulas and/or methods for calcualting > eigen-values/vectors and matrix exponent of the following tridiagonal matrix > : There seems to be a lot of good stuff on eigenvalue algorithms at the LAPACK section of NETLIB. http://www.netlib.org/lapack/lug/node70.html However, your example does not appear to be tri-diagonal. > Or any hint how to solve this? a1 a2 ct 0 0 0 0 0 > a4 -b1 0 0 0 0 0 0 > ct 0 a1 a2 ct 0 0 0 This row has elements two columns away from the diagonal. So your matrix is at least 5-diagonal (penta-diagonal?). It seems to be ALMOST block diagonal. If the coupling between blocks (constants -b1 and ct on the next two rows) perhaps you could use a block-diagonal approximation. > 0 b2 a4 -b1 0 0 0 0 > 0 0 ct 0 a1 a2 ct 0 > 0 0 0 b2 a4 -b1 0 0 > 0 0 0 0 ct 0 0 0 good references. - Randy === Subject: Re: Eigenvalues of tridiagonal matrix >Are there any ready to use formulas and/or methods for calcualting >eigen-values/vectors and matrix exponent of the following >tridiagonal matrix >: >Or any hint how to solve this? a1 a2 ct 0 0 0 0 0 >a4 -b1 0 0 0 0 0 0 >ct 0 a1 a2 ct 0 0 0 >0 b2 a4 -b1 0 0 0 0 >0 0 ct 0 a1 a2 ct 0 >0 0 0 b2 a4 -b1 0 0 >0 0 0 0 ct 0 0 0 . This seems to be an array of 7 rows and 8 columns. If that's your matrix, it has no eigenvalues (only square matrices do). And it's not tridiagonal either (row 4 has four nonzero entries). Exactly what is it that you want to solve? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Eigenvalues of tridiagonal matrix U[YAcute]ytkownik Robert Israel napisaü w wiadomo.beci >>Are there any ready to use formulas and/or methods for calcualting >>eigen-values/vectors and matrix exponent of the following >>tridiagonal matrix >>: >>Or any hint how to solve this? >>a1 a2 ct 0 0 0 0 0 >>a4 -b1 0 0 0 0 0 0 >>ct 0 a1 a2 ct 0 0 0 >>0 b2 a4 -b1 0 0 0 0 >>0 0 ct 0 a1 a2 ct 0 >>0 0 0 b2 a4 -b1 0 0 >>0 0 0 0 ct 0 0 0 > . > This seems to be an array of 7 rows and 8 columns. NO !!! J have shown the pattern of an arbitrary (but finite) order matrix. As the pattern I showed the upper left corner of the matrix. It's the system matrix of a 2nd order discretized PDE. And its square. Tom === Subject: Re: Primes, probability and politics I haven't been following most of this thread, but: > your assertion is > groundless, > not based upon any fact, > premature, > unfounded, Quite possibly. > poor guessing, sense less, no good, bad, DOA, dumb, shallow, etc. Not really, unless you have some good reason to suspect so. A quick check shows that the proportion is pretty close for all primes less than 100 million (proportion with x = 1 mod 3 is 2880591/5761574, or about 0.499966). This strongly suggests that the conjecture is likely to be true. -- Chris Smith === Subject: Re: Primes, probability and politics Chris Smith skrev: > I haven't been following most of this thread, but: your assertion is groundless, > not based upon any fact, > premature, > unfounded, Quite possibly. poor guessing, sense less, no good, bad, DOA, dumb, shallow, etc. Not really, unless you have some good reason to suspect so. A quick > check shows that the proportion is pretty close for all primes less than > 100 million (proportion with x = 1 mod 3 is 2880591/5761574, or about > 0.499966). This strongly suggests that the conjecture is likely to be > true. It is actually a well known theorem (see, for example, http://mathworld.wolfram.com/PrimeArithmeticProgression.html ). It's just that Harris' handwaving arguments are nowhere near a valid proof. --- J K Haugland http://home.no.net/zamunda === Subject: Re: Primes, probability and politics [jstevh@msn.com] > ... > and amazingly, you can see (p - 2)/(p - 1) in the mathematical > literature about twin primes!!! See: http://mathworld.wolfram.com/TwinPrimesConstant.html So I noticed that and at first I thought maybe I'd just been a bit more > brilliant than mathematicians who had just missed something--despite it > being easy--but now I wonder. What if they didn't miss it, but simply didn't want to give the > probability link? You also just said (in a different post): Years ago, I bought an excellent prime reference, Any good (let alone excellent) prime reference contains at least a heuristic argument explaining how Littlewood and Hardy derived their estimate for the number of twin primes <= x. I pointed you at Hardy & Wright's An Introduction to the Theory of Numbers a few days ago, and gave you the exact section number to read. I'll grant that it's unlikely you'd be able to follow that (it's rigorous and requires effort to follow), but there are also hand-wavy explanations all over the place. Alas: and no, I'm not going to tell you which one. That stops anyone from telling you which page number to read in the book you bought. Let me guess: since it /does/ have the explanation you're looking for, you bought the book but never did more than skim parts of it. Either that, or it's not as good a reference as you believe. It doesn't matter whether it's a few years old, or even 50 years old -- their conjecture is older than that, and it would be astonishing if any good reference on primes didn't cover it. === Subject: Re: Primes, probability and politics In <-cidnZjiucidjX3ZnZ2dnUVZ_qGdnZ2d@comcast.com>, Tim Peters >[jstevh@msn.com] >> ... >> and amazingly, you can see (p - 2)/(p - 1) in the mathematical >> literature about twin primes!!! >> See: http://mathworld.wolfram.com/TwinPrimesConstant.html >> So I noticed that and at first I thought maybe I'd just been a bit more >> brilliant than mathematicians who had just missed something--despite it >> being easy--but now I wonder. >> What if they didn't miss it, but simply didn't want to give the >> probability link? You also just said (in a different post): Years ago, I bought an excellent prime reference, Any good (let alone excellent) prime reference contains at least a >heuristic argument explaining how Littlewood and Hardy derived their >estimate for the number of twin primes <= x. I pointed you at Hardy & >Wright's An Introduction to the Theory of Numbers a few days ago, and gave >you the exact section number to read. I'll grant that it's unlikely you'd >be able to follow that (it's rigorous and requires effort to follow), but >there are also hand-wavy explanations all over the place. Alas: and no, I'm not going to tell you which one. That stops anyone from telling you which page number to read in the book you >bought. Let me guess: since it /does/ have the explanation you're looking >for, you bought the book but never did more than skim parts of it. Either >that, or it's not as good a reference as you believe. It doesn't matter >whether it's a few years old, or even 50 years old -- their conjecture is >older than that, and it would be astonishing if any good reference on primes >didn't cover it. I don't blame James for wanting to keep it a secret. It might help get him a Nobel prize. -- John Roberts-Jones === Subject: Re: Primes, probability and politics [jstevh@msn.com] ... So the probability that x does NOT have p as a factor is 1 - 1/p = (p-1)/p, and for each prime you include you just need to multiply the probabilities. > So, trivially, you have the probability that x is prime: > probPrime(x) = ((p_j -1)/p_j)*...*(1/2) > where there are j primes up to sqrt(x) and p_j is the jth prime. [jankrihau@hotmail.com] >> Trivial to you, but wrong. The left hand side is asymptotically >> 1/ln(x), >> the right side is asymptotically 2 exp(-gamma) / ln(x), Where did 2 come from? I don't believe that part. >> where gamma is the Euler-Mascheroni constant 0.57721566... This >> follows from Merten's theorem, and yes, I do have a reference: >> http://mathworld.wolfram.com/MertensTheorem.html [jstevh@msn.com] > More detail please... I followed the link and am suspicious of your > assertion. Heh. You also ignored me last week when I pointed out the footnote in Hardy & Wright warning: Considerations of this kind explain why the usual probability arguments lead to the wrong asymptotic value for pi(x). and urged you to complete the exercise of figuring out what asymptotic value your argument /does/ lead to. In fact, you poo-poo'ed me twice on that, so I gave up trying. More, Dik Winter urged you to look up Mertens's Theorem even earlier, the very first time you tried making this argument. You ignored him too, IIRC. You can't say you weren't warned. > Worse, I think that what I call math-ese is a peculiarly powerful > weapon in the arsenal of math people who are, well, not telling the > truth. Do you need to act like a total ass every time? > You put up some complex assertion, which you figure is hard to > parse through it, and hope that no one bothers to really check you on > it. No, what he said was easy to follow for someone who can read math. I'm not sure on where you're getting confused, because you made no effort to explain where you stopped following the argument. So I can only assume you have no idea whatsoever about what anything he said meant. I'll try. An equivalent form of Mertens's theorem appears as Theorem 429 in Hardy & Wright: the product over all prime p <= x of 1 - 1/p is aymptotically equal to e^-gamma / ln(x) To get the form on MathWorld, take the reciprocal of both sides, then divide both sides by ln(x), then replace x by p_n. Same thing, but I won't say more about that because the Hardy & Wright version is applicable as-is to your argument, just by noting that (p_j -1)/p_j = 1 - 1/p_j via simple algebra. Your probPrime(x) expression is exactly the left-hand side of the H&W form of Mertens's theorem. BTW, if you don't believe me either, you can also find the H&W spelling on Chris Caldwell's prime pages: http://primes.utm.edu/glossary/page.php?sort=MertensTheorem > And people not able or willing to check assume you are right, so you > can put up a false assertion and be believed based on the intellectual > weakness of others. How about yours? You have a computer, and you could have /checked/ your argument against reality very easily here. Let's take a look at each 10000th prime. Here, where normal: the usual approximation x/(ln(x) - 1) to pi(x) x*JSH: x * product across all prime p <= x of (p-1)/p Mertens: x * e^-gamma / ln(x) x pi(x) normal JSH Mertens -------- ------ ------ ----- ------- 104729 10000 9918 5085 5086 224737 20000 19848 10237 10239 350377 30000 29776 15406 15408 479909 40000 39723 20595 20597 611953 50000 49653 25783 25786 746773 60000 59629 31002 31003 882377 70000 69531 36183 36187 1020379 80000 79495 41404 41407 1159523 90000 89445 46620 46623 1299709 100000 99383 51833 51836 1441049 110000 109328 57051 57055 1583539 120000 119285 62279 62282 1726943 130000 129244 67510 67512 1870667 140000 139167 72722 72726 2015177 150000 149093 77938 77943 2160553 160000 159029 83161 83166 2307229 170000 169008 88410 88414 2454587 180000 178990 93662 93665 2601857 190000 188927 98890 98894 2750159 200000 198895 104136 104140 2898527 210000 208831 109366 109370 3047767 220000 218792 114611 114615 3196933 230000 228716 119836 119841 3346601 240000 238643 125064 125069 3497861 250000 248645 130335 130339 3648923 260000 258606 135584 135587 3800201 270000 268554 140826 140830 3951161 280000 278456 146043 146049 4103629 290000 288431 151302 151307 4256233 300000 298392 156554 156559 4410317 310000 308425 161847 161849 4562693 320000 318325 167066 167070 4716053 330000 328267 172310 172314 4869863 340000 338218 177559 177563 5023307 350000 348125 182784 182790 5178049 360000 358096 188045 188051 5332519 370000 368030 193287 193293 5487701 380000 377992 198544 198551 5644031 390000 388009 203834 203838 5800079 400000 397991 209104 209107 5955031 410000 407885 214325 214330 6111613 420000 417868 219596 219601 6268289 430000 427839 224862 224866 6424937 440000 437794 230119 230123 6581963 450000 447757 235380 235384 6738889 460000 457699 240631 240635 6895393 470000 467599 245858 245865 7052113 480000 477499 251086 251094 7210759 490000 487507 256375 256381 7368787 500000 497461 261634 261641 7528121 510000 507485 266933 266937 7685801 520000 517392 272166 272172 7844731 530000 527364 277436 277442 8003537 540000 537316 282696 282702 8163047 550000 547299 287973 287978 8322241 560000 557251 293233 293239 8482259 570000 567242 298516 298520 8640679 580000 577122 303736 303743 8799919 590000 587042 308979 308988 8960453 600000 597031 314262 314269 9121439 610000 607038 319554 319560 9281011 620000 616945 324791 324799 9442229 630000 626944 330080 330086 9602443 640000 636871 335329 335336 9763393 650000 646832 340598 340605 9925439 660000 656852 345899 345904 10086767 670000 666817 351170 351175 Do it yourself if you don't believe me: Mertens's theorem clearly gives an excellent approximation to your guess, while both of those suck as approximations to pi(x). > So, step out your claim, please. See above. > If I am wrong about your claim and you can validate it, I apologize up > front for coming out and saying you could be lying and deliberately > trying to fool people in a way that has worked for mathematicians in > the past. Given that you appeared to make no effort at all to check one way or the other, I suggest that bringing up the possibility he may be lying was inexcusable ... yup, I officially retire from trying to give you a serious answer about anything anymore. Congratulations :-) === Subject: Re: Primes, probability and politics [jstevh@msn.com] ... So the probability that x does NOT have p as a factor is 1 - 1/p = (p-1)/p, and for each prime you include you just need to multiply the probabilities. > So, trivially, you have the probability that x is prime: > probPrime(x) = ((p_j -1)/p_j)*...*(1/2) > where there are j primes up to sqrt(x) and p_j is the jth prime. [jankrihau@hotmail.com] >> Trivial to you, but wrong. The left hand side is asymptotically >> 1/ln(x), >> the right side is asymptotically 2 exp(-gamma) / ln(x), [Tim Peters] > Where did 2 come from? I don't believe that part. Ah! It came from ln(sqrt(x)) = ln(x)/2 -- I forgot the sqrt(x) part in JSH's original. Sorry for that. where gamma is the Euler-Mascheroni constant 0.57721566... This follows from Merten's theorem, and yes, I do have a reference: > http://mathworld.wolfram.com/MertensTheorem.html ... Leaving the 2 in makes for much smaller differences in the tables I computed before. Here's a corrected table, with ratios added; checking each 50 million up to 10^9 to see whether the differences can be made more apparent; and using x/ln(x) for normal instead of the better (although asymptotically equal) x/(ln(x)-1), to stop giving normal an undue advantage. normal: the approximation x/ln(x) to pi(x) JSH: x * product across all prime p <= sqrt(x) of (p-1)/p Mertens: x * 2 * e^-gamma / ln(x) ratio1: pi(x) / normal ratio2: pi(x) / JSH x/10^6 pi(x) normal JSH Mertens ratio1 ratio2 ------ ------ ------ ----- ------- ------ ------ 50 3001134 2820471 3161853 3167160 1.064054 0.949169 100 5761455 5428681 6088469 6095968 1.061299 0.946290 150 8444396 7967642 8941792 8947016 1.059836 0.944374 200 11078937 10463628 11742238 11749807 1.058804 0.943512 250 13679318 12928601 14500586 14517771 1.058066 0.943363 300 16252325 15369409 17248950 17258600 1.057446 0.942221 350 18803526 17790479 19962114 19977267 1.056943 0.941961 400 21336326 20194905 22664825 22677242 1.056520 0.941385 450 23853038 22584966 25346170 25361087 1.056147 0.941090 500 26355867 24962408 28011558 28030761 1.055822 0.940893 550 28845356 27328610 30667682 30687814 1.055500 0.940578 600 31324703 29684688 33306524 33333500 1.055248 0.940498 650 33793395 32031565 35948151 35968852 1.055003 0.940059 700 36252931 34370013 38570624 38594739 1.054784 0.939910 750 38703181 36700688 41192229 41211899 1.054563 0.939575 800 41146179 39024157 43801214 43820966 1.054377 0.939384 850 43581966 41340910 46394745 46422492 1.054209 0.939373 900 46009215 43651379 48997675 49016961 1.054015 0.939008 950 48431471 45955943 51588931 51604800 1.053867 0.938796 1000 50847534 48254942 54166822 54186390 1.053727 0.938721 So across this range, ratio1 is slowly moving closer to 1, while ratio2 is slowly moving away from 1, but normal and JSH are both non-nuts estimates to pi(x) over this range. Since x*2*e^-gamma/ln(x) ~= 1.123*(x/ln(x)), and the Mertens approximation gives a good estimate of JSH, so long as ratio1 is approaching 1 from above, ratio2 = pi(x)/JSH ~= pi(x)/(1.123*x/ln(x)) = ratio1/1.123 is doomed to drift farther away from 1 on the low side. May be interesting to note that JSH's probPrime(x) gives the exact probability that an integer picked uniformly at random from 1 through N is coprime to N, where N = the product of all primes <= sqrt(x) (since phi(N) = N*product((p-1)/p across all prime p <= sqrt(x)), then divide both sides by N). === Subject: Re: Primes, probability and politics <-cidnZniucidjX3ZnZ2dnUVZ_qGdnZ2d@comcast.com> Trivial to you, but wrong. The left hand side is asymptotically >> 1/ln(x), >> the right side is asymptotically 2 exp(-gamma) / ln(x), Where did 2 come from? I don't believe that part. It comes from 1/ln(sqrt(x)) = 2/ln(x). We are taking the product over the primes up to sqrt(x). --- J K Haugland http://home.no.net/zamunda === Subject: Re: Primes, probability and politics level mathematicians like Haugland around the world must KNOW WHAT THEY > ARE DOING ......... But James you strive to be a mathematition, and you say that all mathematition lie, therefore you strive to always lie. > > === Subject: Re: Primes, probability and politics [...] > I strongly suggest that the simplest explanation available to us is > that mathematicians deliberately avoid simple explanations so that they > can choose very complex math-ese on which lots of careers can be built, > and books and papers written. > > I suggest to you that mathematicians avoid simple answers--for more > jobs for mathematicians. > > Mathematics is a difficut discipline. A few people can come up with a > lot of answers, so what if mathematicians just went with the simple? > What about the countless others who would then have no subject on which > to write papers or to have a doctoral thesis because they lacked the > intrinsic ability to make discoveries on their own? > > It is a social welfare system that supports people who otherwise could > not support themselves as mathematicians. > > It is a white collar work program. [...] What is the history of this alleged corruption, if I'm understanding you correctly? Partly out of curiosity, I drew up a list of mathematicians and some of their thesis students, starting with Gauss. I know a little bit about a number of them. Anyway, how did math and mathematicians get lost along the way? Advisor, Student ----------------- Gauss Gerling Gerling Pl.9fcker Pl.9fcker Klein Klein Lindemann Lindemann Hilbert Hilbert Hecke Hecke Maak Maak Jacobs Jacobs Strassen I consulted the Math. Genealogy Project on the Web to draw up this list; --> http://genealogy.math.ndsu.nodak.edu/ David Bernier === Subject: Re: Primes, probability and politics >You put up some complex assertion, which you figure is hard to >parse through it, and hope that no one bothers to really check you on >it. >And people not able or willing to check assume you are right, so you >can put up a false assertion and be believed based on the intellectual >weakness of others. >So, step out your claim, please. So you are admitting that you are intellectually weak? Someone who has such a great mathematical mind should be able to follow the math with no problem. === Subject: Re: Primes, probability and politics >It is fairly easy to consider probability with the prime distribution >by using one single fact, which I can show with two arbitrary primes >p_1 and p_2, and p_1 mod p_2 with the assertion that there is no preference from p_1 for particular >residues of p_2. For instance, the residues available to primes modulo 3 are 1 and 2. >My assertion is that there is no preference for either, so you can have >5 = 2 mod 3, but also have 7 = 1 mod 3, and primes will go either way >with 50% probability. That must be the case as the composites are just products of primes, so >if primes, say, favored 1 as a residue modulo 3, then their products >would favor 1, and most numbers would be 1 modulo 3, but actually, 1/3 >are. 1, 2, 3 is followed by 4, 5, 6 followed by 7, 8, 9, repeat out to >infinity... which is trivial but I show it to emphasize how simple these ideas are, >and how rigid they are. So you can say that given x is prime, the probability that x+2 is >divisible by 3 is 1/2. And you can continue with the other primes up to sqrt(x+2) as it's >well-known that if x+2 does not have any of the primes up to sqrt(x+2) >as factors, then it must be prime. If x is not prime, then the probability that x+2 is divisible by 3 is >1/3. Now the probability that x has p as a factor when p is less than or >equal to sqrt(x+2) is just 1/p, by the reasoning above. So the probability that x does NOT have p as a factor is 1 - 1/p = >(p-1)/p, and for each prime you include you just need to multiply the >probabilities. So, trivially, you have the probability that x is prime: probPrime(x) = ((p_j -1)/p_j)*...*(1/2) where there are j primes up to sqrt(x) and p_j is the jth prime. Trivial and easy ideas which follow from noting that primes do not have >a preference for residues modulo each other, so why can't you just go >out on the web, do some searches, and find that prominently displayed >as an early and simple result in research about primes? You CAN find that mentioned, but you have to dig, and at first I >thought it was about a dislike of probability by number theorists, but >they talk about probability with 1/(ln x) and primes, why not here? Well, think about it, what if that is THE explanation for most of the >behavior of primes? Consider that if you think of twin primes, and x as a prime, so you >consider x+2, you get a slightly different equation, as for instance, I >noted above that if x is prime there is a 1/2 chance that it is 1 >modulo 3. Since primes are coprime to each other, that is, can't have factors in >common with each other, so 3 can't be a factor of 5 or any other prime, >you drop the 0 residue, so you have p-1 possible residues modulo a >prime possible. So the probability that x+2 is divisible by p when p is less than or >equal to sqrt(x+2) is 1/(p-1), so you subtract that from 1 to get the >probability that x+2 is NOT divisible by p: 1 - 1/(p-1) = (p-2)/(p-1) and the probability when x is prime that x+2 is prime is given by probTwinPrime(x) = ((p_j -2)/p_j - 1)*...*(1/2) and amazingly, you can see (p - 2)/(p - 1) in the mathematical >literature about twin primes!!! See: http://mathworld.wolfram.com/TwinPrimesConstant.html So I noticed that and at first I thought maybe I'd just been a bit more >brilliant than mathematicians who had just missed something--despite it >being easy--but now I wonder. What if they didn't miss it, but simply didn't want to give the >probability link? Now I'm going to diverge a bit and no this is not really a plug for my >own research as I'm going to connect the dots and go into the politics >part of this post, as I have other research related to primes, and once >certain is important. My final version is in the history of the page: http://en.wikipedia.org/w/index.php?title=Prime_counting_function&oldid=914 2249 There you can see what I call my prime counting function to distinguish >it from the others out there and it has some peculiar features, like it >uses a partial difference equation to count prime numbers, which has >never been seen before. It uses that partial difference equation with >a special constraint which is what forces the exact prime count. Without that constraint the summation of the partial difference >equation with the rest of the equation gives a result close, but not >exact to, the prime distribution. There is a partial differential equation that follows from the partial >difference equation, and then, a connection to continuous functions. I have been talking about that for years. The information I gave >before on probability and primes is trivial, and (p-2)/(p-1) is >actually visible in mathematical literature you can find on the web, >with no mention of the probability argument that I quickly gave, from >which it would seem to naturally follow. Now to politics. I have a question for you, if probability is the >crucial feature that controls the observed behavior of primes, and the >connection between the prime distribution and continuous functions like >1/(ln x) can all be explained quickly, with a few ideas, like how I've >given them in this post, how many careers could be supported by >research in the area? What if the answers to the twin prime conjecture and Goldbach's >conjecture and any number of questions about primes could be answered >by simple ideas and in a few pages of some number theory text? How many graduate students would need new theses? How many books in this area could be written? How many professors could support grants and write papers? I suggest to you that the politics of the simple answers to the >questions about primes have to do with jobs--jobs for mathematicians. Think you'd have too much pride to appear to work feverishly on the >Riemann Hypothesis, or the Twin Primes Conjecture, or Goldbach's >Conjecture, when simple answers were readily available? Think so? Even if you'd wonder what would you do if you weren't >getting the grant income in this area? Even if you were wondering how >you'd feed and clothe your children, or maintain your prestige? > See Landau, Handbook (1909), Vol I, Book I, Part I, Chapter 3: On the Probability that a Number is Prime. === Subject: Re: Primes, probability and politics >It is fairly easy to consider probability with the prime distribution >>by using one single fact, which I can show with two arbitrary primes >>p_1 and p_2, and >>p_1 mod p_2 >>with the assertion that there is no preference from p_1 for particular >>residues of p_2. >>For instance, the residues available to primes modulo 3 are 1 and 2. >>My assertion is that there is no preference for either, so you can have >>5 = 2 mod 3, but also have 7 = 1 mod 3, and primes will go either way >>with 50% probability. >>That must be the case as the composites are just products of primes, so >>if primes, say, favored 1 as a residue modulo 3, then their products >>would favor 1, and most numbers would be 1 modulo 3, but actually, 1/3 >>are. >>1, 2, 3 is followed by 4, 5, 6 followed by 7, 8, 9, repeat out to >>infinity... >>which is trivial but I show it to emphasize how simple these ideas are, >>and how rigid they are. >>So you can say that given x is prime, the probability that x+2 is >>divisible by 3 is 1/2. >>And you can continue with the other primes up to sqrt(x+2) as it's >>well-known that if x+2 does not have any of the primes up to sqrt(x+2) >>as factors, then it must be prime. >>If x is not prime, then the probability that x+2 is divisible by 3 is >>1/3. >>Now the probability that x has p as a factor when p is less than or >>equal to sqrt(x+2) is just 1/p, by the reasoning above. >>So the probability that x does NOT have p as a factor is 1 - 1/p = >>(p-1)/p, and for each prime you include you just need to multiply the >>probabilities. >>So, trivially, you have the probability that x is prime: >>probPrime(x) = ((p_j -1)/p_j)*...*(1/2) >>where there are j primes up to sqrt(x) and p_j is the jth prime. >>Trivial and easy ideas which follow from noting that primes do not have >>a preference for residues modulo each other, so why can't you just go >>out on the web, do some searches, and find that prominently displayed >>as an early and simple result in research about primes? >>You CAN find that mentioned, but you have to dig, and at first I >>thought it was about a dislike of probability by number theorists, but >>they talk about probability with 1/(ln x) and primes, why not here? >>Well, think about it, what if that is THE explanation for most of the >>behavior of primes? >>Consider that if you think of twin primes, and x as a prime, so you >>consider x+2, you get a slightly different equation, as for instance, I >>noted above that if x is prime there is a 1/2 chance that it is 1 >>modulo 3. >>Since primes are coprime to each other, that is, can't have factors in >>common with each other, so 3 can't be a factor of 5 or any other prime, >>you drop the 0 residue, so you have p-1 possible residues modulo a >>prime possible. >>So the probability that x+2 is divisible by p when p is less than or >>equal to sqrt(x+2) is 1/(p-1), so you subtract that from 1 to get the >>probability that x+2 is NOT divisible by p: >>1 - 1/(p-1) = (p-2)/(p-1) >>and the probability when x is prime that x+2 is prime is given by >>probTwinPrime(x) = ((p_j -2)/p_j - 1)*...*(1/2) >>and amazingly, you can see (p - 2)/(p - 1) in the mathematical >>literature about twin primes!!! >>See: http://mathworld.wolfram.com/TwinPrimesConstant.html >>So I noticed that and at first I thought maybe I'd just been a bit more >>brilliant than mathematicians who had just missed something--despite it >>being easy--but now I wonder. >>What if they didn't miss it, but simply didn't want to give the >>probability link? >>Now I'm going to diverge a bit and no this is not really a plug for my >>own research as I'm going to connect the dots and go into the politics >>part of this post, as I have other research related to primes, and once >>certain is important. My final version is in the history of the page: >>http://en.wikipedia.org/w/index.php?title=Prime_counting_function&oldid=91 42249 >>There you can see what I call my prime counting function to distinguish >>it from the others out there and it has some peculiar features, like it >>uses a partial difference equation to count prime numbers, which has >>never been seen before. It uses that partial difference equation with >>a special constraint which is what forces the exact prime count. >>Without that constraint the summation of the partial difference >>equation with the rest of the equation gives a result close, but not >>exact to, the prime distribution. >>There is a partial differential equation that follows from the partial >>difference equation, and then, a connection to continuous functions. >>I have been talking about that for years. The information I gave >>before on probability and primes is trivial, and (p-2)/(p-1) is >>actually visible in mathematical literature you can find on the web, >>with no mention of the probability argument that I quickly gave, from >>which it would seem to naturally follow. >>Now to politics. I have a question for you, if probability is the >>crucial feature that controls the observed behavior of primes, and the >>connection between the prime distribution and continuous functions like >>1/(ln x) can all be explained quickly, with a few ideas, like how I've >>given them in this post, how many careers could be supported by >>research in the area? >>What if the answers to the twin prime conjecture and Goldbach's >>conjecture and any number of questions about primes could be answered >>by simple ideas and in a few pages of some number theory text? >>How many graduate students would need new theses? >>How many books in this area could be written? >>How many professors could support grants and write papers? >>I suggest to you that the politics of the simple answers to the >>questions about primes have to do with jobs--jobs for mathematicians. >>Think you'd have too much pride to appear to work feverishly on the >>Riemann Hypothesis, or the Twin Primes Conjecture, or Goldbach's >>Conjecture, when simple answers were readily available? >>Think so? Even if you'd wonder what would you do if you weren't >>getting the grant income in this area? Even if you were wondering how >>you'd feed and clothe your children, or maintain your prestige? >> See Landau, Handbook (1909), Vol I, Book I, Part I, Chapter 3: On the > Probability that a Number is Prime. JSH is a crank or crackpot, he will never look it up. === Subject: Re: Primes, probability and politics So the probability that x does NOT have p as a factor is 1 - 1/p = > (p-1)/p, and for each prime you include you just need to multiply the > probabilities. So, trivially, you have the probability that x is prime: probPrime(x) = ((p_j -1)/p_j)*...*(1/2) where there are j primes up to sqrt(x) and p_j is the jth prime. Trivial to you, but wrong. The left hand side is asymptotically > 1/ln(x), > the right side is asymptotically 2 exp(-gamma) / ln(x), where gamma > is the Euler-Mascheroni constant 0.57721566... This follows from > Merten's theorem, and yes, I do have a reference: http://mathworld.wolfram.com/MertensTheorem.html More detail please... I followed the link and am suspicious of your > assertion. Worse, I think that what I call math-ese is a peculiarly powerful > weapon in the arsenal of math people who are, well, not telling the > truth. You put up some complex assertion, which you figure is hard to > parse through it, and hope that no one bothers to really check you on > it. And people not able or willing to check assume you are right, so you > can put up a false assertion and be believed based on the intellectual > weakness of others. So, step out your claim, please. It is not hard at all. You claim the left hand side and the right hand side are trivially equal; I claim they differ by a factor that tends to a constant different from 1. And my claim follows from a well known theorem if you just plug in the numbers. > If I am wrong about your claim and you can validate it, I apologize up > front for coming out and saying you could be lying and deliberately > trying to fool people in a way that has worked for mathematicians in > the past. So you are sensing that your conspiracy theories are falling apart yet again? :-) --- J K Haugland http://home.no.net/zamunda === Subject: help simplifying expression containing ceiling function A few months ago, I got help from this newsgroup on simplifying a summation expression containing the ceiling of a quotient. Here is the expression: sum{ c=0, n-1, ceil((m+c)/n) } where m and n are positive integers > Let q and r be the integers such that m = q*n + r, with 0 <= r < n. > Then > sum{ c=0, n-1, ceil((m+c)/n) } = [replacing m w/ q*n+r] > sum{ c=0, n-1, ceil((q*n+r+c)/n) } = [distributing /n] > sum{ c=0, n-1, ceil(q+(r+c)/n) } = [ceil(i+x)=i+ceil(x)] > sum{ c=0, n-1, q + ceil((r+c)/n) } = [q is invariant across summands] > q*n + sum{ c=0, n-1, ceil((r+c)/n) } = [since m=q*n+r, q*n=m-r] > m-r + sum{ c=0, n-1, ceil((r+c)/n) } Eventually, after some analysis omitted here, the whole summation expression simplifies to m + n - 1 I have found the above approach to be a useful trick and have been using it to simplify various summation of ceiling expressions. This trick assumes that the denominator of the ceiling function does not depend on the summation variable; otherwise, q and r are not constant. Is there a similar trick to handle a situation like this: sum{ k=1, n, ceil(m/k) } , where m is a positive integer My goal is to get as many terms as possible outside of the ceiling and summation functions. Chris === Subject: Re: An uncountable countable set > > x > > xx > > xxx > > ... > Do you believe that there are actually infinitely many finite > > numbers? Yes. So the triangle gets infinitely long but not infinitely broad. This > means the Diagonal must bend. > > If the number of rows is unlimited and the number of x's in a row keeps > pace with the number of that row, how is there any more a limit on the > number of x's in some row than on the number of rows? > > There is no limit. Then there is no triangle. > If the number of rows is infinite, the number of x's > in a row is also infinite. Only if the index number of a row is infinite need the number of x's in that row be infinite. Which row(s) does Mueckenh claim has(have) an infinite index number? > But, by definition, the number of x's in a > row is always finite (because every natural number is finite). Because the index number of every row (which equals the number of x's in that row) is finite. === Subject: Re: An uncountable countable set > > > Let's state that more proper. The index of every digit that can be > indexed is in the true list (of natural numbers). But later you use > the word index to mean something completely different. You are > stating there that if every digit of a number can be indexed the number > *itself* is in the list. But that does not follow. > > I does. Indexing, or let's say indexibility, is a symmetric > relation: Then if each digit can be indexed by a unique natural number, Mueckenhrequires that each natural number be indexed by a unique decimal digit. Also note that the set of even naturals can be indexed by the set of all naturals so it would follow that the set of all naturals can be indexed by the even naturals leaving the odd naturals to index the set of all naturals a second time without breaking into a sweat. > Every digit of n can be indexed And can be 'indexed by using only those digits whose position number is a prime, leaving most of them unused as indices but still indexing all of them. > This is a *symmetric* relation. Not at all. when one turns it about, only those digits in prime positions are indexed leaving infinitely many un-indexed. Mueckenh's notions about infinity are centuries out of date. === Subject: Re: An uncountable countable set > > > > The last conclusion requires proof (and it is false). What is the > > case > > is that for every Ai the indexing of digits will terminate at some > > point, > > and the indexing of digits of 0.111... will never terminate when we > > start > > at the first digit and go on. Because there will always be a further > > digit. (As there will always be a further An.) > But each one will be finite. You mean the index of the digit and the An. Yes. > > All you agree with for finite numbers applies to all index numbers, > because each one is finite. > (every digit of n can be indexed) ==> (n is in the list) Where is the proof? There is an easy proof of: > (n is in the list) ==> (every digit of n can be indexed) > but not of the converse. > > Indexing is a symmetric relation. A position n of number N can index a > position m of number M, if and only if n = m. Wrong! The set of all naturals can index the set of all even naturals and vice- versa. The critical issue for the indexing of any countable set is whether the set of indices has a unique first index and for each index has a unique successor index distinct from those before it. Absolutely nothing else is required. As 0.111... has more > digit positions than the binary representation of any natural number, > your claim implies that indexing is not a symmetric relation. It is not necessarily symmetric, at least in Mueckenh's sense. === Subject: Re: An uncountable countable set > This shows that w is not the cardinal number of the natural numbers This shows that Mueckenh has no idea what he is talking about. === Subject: Re: An uncountable countable set > An infinite sum of 1's is not infinite? n lim sum 1 = lim n =def L n -> oo i = 1 n -> oo There is no such L in N. F. N. -- xyz === Subject: Re: 4-d Sudoku > To make a 2r-dimensional Sudoku board, take a 2r-dimensional cube with > side length n^2, and partition it (with a thick hyperpen) into n^(2r) > subcubes of side length n, then partition each of those into n^(2r) > subcubes of side length 1 (with a thin hyperpen). Using an alphabet of > n^(2r) letters, fill in each of the n^(4r) unit cubes such that > (a) each of the n^(2r) subcubes of side length n has each letter > exactly once, and > (b) each of the C(2r,r) n^(2r) r-dimensional orthogonal slices of the > entire cube has each letter exactly once. For r = 1 and n = 3 this reduces to the usual game of Sudoku. Question: for what values of r and n does a solution exist? Certainly we can ignore the trivial cases (r = 0 or n = 0 or 1), and it > seems clear that the r = 1 case has solutions for all n (presumably a > proof exists somewhere). What about for r > 1? To be specific, is there a solution for r = 2, n = 2? A colleague attempted to find solutions using the Choco constraint solver. He found none for the cases (r,n) = (2,2), (2,3), and (3,2). He said it wasn't quite clear whether these results mean that there are no solutions, or just that Choco couldn't find any, though he believes it is the former. -Jim Ferry Metron, Inc. f rr @m tsc .c m e y e i o === Subject: Latin Squares Could someone please tell me how and why these diagrams http://finitegeometry.org/sc/gen/ortho.html === Subject: NEWTON WAS WRONG THE MECHANICS OF THE UNIVERSE Copyright 1984-2006 Allen C. Goodrich No force is necessary to cause orbital motion. The planets orbit the sun at a special mean orbital radius L to conserve total energy. The modified first law of thermodynamics , which states that the total energy of the universe is a constant, is the fundamental equation of the universe. The sum of kinetic and potential energies is a constant. m (2 pi L)^2 / t^2 + G m(M-m) /L = A CONSTANT. (if no charges are present) + Delta m(2 pi L)^2/T^2 = - Delta Gm(M-m)/L What's so important about this modified first law of thermodynamics? Kinetic energy m (2 pi L)^2 / t^2 of a mass m is orbital motion relative to the rest of the effective universe (M-m) and potential energy G(M-m) m / L, where L is the mean orbital radius, are nearly equal at equilibrium, where no energy change occurs and the total energy is constant. At equilibrium, any change of kinetic energy must be accompanied by an equal and opposite, sign, change of potential energy. The universe has been found to be expanding at an accelerating rate. The potential energy of the universe is continually decreasing and the kinetic energy is continually increasing. Again, to conserve the total energy relative to the rest of the universe. This is why the modified first law is so important. The conservation of total energy must be maintained relative to the rest of the expanding universe. Kinetic and potential energies must be computed relative to the rest of the expanding universe. This modified first law leads to the conclusion that the force of gravity and the velocity of light are misleading illusions. The definition of Kinetic and Potential Energies is most important to an understanding that these energies are relative to the rest of the effective universe, not just relative to any other mass. Kinetic energy is mass m times the square of the velocity relative to the rest of the effective universe. For example the kinetic energy of the earth is effectively relative to the sun plus the rest of the planets of the solar systen. Potential energy is the product of the mass m times the mass of the rest of the effective universe M-m divided by the distance L between their effective centers of mass. M is the mass of the entire effective universe. t is the orbital time for one complete revolution. G is the gravitational constant. Once this is clearly understood, the modified First Law of Thermodynamics becomes the Fundamental Equatiion of the Universe. This equation ,then, defines the photon and the rest of the universe. The Thomas R. Young two slit defraction pattern, the complete logic of quantum mechanics, the true nature of gravitation, the fact that light does not have a mass or a velocity, all become quite obvious. This is a simple solution to so many of the problems that baffled scientists for hundreds of years. We remember that Sir Isaac Newton proposed the force of gravity F_g = k m_1 m_2 / L^2. Scientists have known that action at a distance without the transfer of energy was not possible. A force of gravity would not be possible without the transfer of energy. The amount of energy required to cause the so called force of gravity to make the planets travel in any orbit other than their equilibrium orbits about the sun would be tremendous and is not available. Another explanation is necessary. Einstein and other scientists have assuned that masses change the shape of space time. These explanations have their problems. No problems exist if the modified first law of thermodynamice is used.This is the fundamental equation of the universe. One night when I was walking on Myrtle Beach in the light of the full moon, the full moon was at its highest point in the sky, and the beach was very wide. The lowest tide was present. This is not what would be expected according to the gravitational theory. The gravitational theory states that the tide should be very high. One can find this explained in most older dictionaries or encyclopedias. The water of the ocean is shown bulging on the side of the earth directly under the full moon. The tides never occur in this way. This discrepancy with the law of gravity is explained by the requirement for the time required for the water to flow due to the force . No one has bothered to explain that the water would have to flow at more than 1000 miles per hour to complete this picture. This flow would wash all of the continents away in one day, Here, we have a big problem with the force of gravity . The NOAA U.S. Coast and Geodetic Survey has monitored the tides at many ocean ports for many years and this data is available. The Jet Propulsion Laboratory has plotted the position and phases of the moon with changes of time. This data is also available. No one has previously published a correlation ot these two sets of data. If they had, it would become very obvious that invariably the lowest tide occurred when the full moon and new moon were exactly directly at their highest point in the sky. It would have been obvious that the lowest tide also occurred on the opposite side of the earth. Not at all consistant with the existing gravitational theory. However, this is predicted by the modified first law of thermodynamics, when the kinetic and potential energies are relative to the rest of the effective universe. If one calculates the kinetic and potential energies of the planets and moons,at mean orbital distance L, one finds that the two are nearly equal in magnitude. This is the only mean orbital distance where a positive change of kinetic energy equals a negative change of potential energy. In the expanding universe this is necessary for equilibriun motion where there is no energy transfer. A force or transfer of total energy would be necessary to make the planets travel at any other mean orbital radius. Nature obeys the modified first law of thermodynamics. The assumption of a force of gravity, with its energy transfer, action at a distance, is not necessary to explain orbital motion. Gravitation is explained by the modified first law of thermodynamics. The Thomas R. Young's two slit defraction pattern is also explained by the modified first law of thermodynamics if one uses the charges e in this fundamental equation for the calculation of kinetic and potential energies. Delta e_1 (2 pi L)^2 / t^2 = Delta-K e_1 e_2 / 4 pi E_o L. Where e_2 is the charge of the rest of the effective universe and . e_1 is the orbital electron. E_o is the dielectric constant. Here, the masses would have little effect by comparison and can be neglected. The kinetic energy change of the electron would then be a function of a change of potential energy relative to the rest of the effective universe. The energy of the electron of the atom on the defraction pattern screen would be a function of the rest of the effective universe. The electron would sense the fact that one or two of the slits was open and the Thomas R Young defraction pattern is explained. We now have the basis for a new quantum mechanics. Light , that is observed as a change of the kinetic energy of the electron, which has the correct energy density (time) , direction and frequency, relative to the rest of the expanding potential energy change of the rest of the effective universe . The four fundamental forces - strong, weak, electromagnetic and gravitational - are likewise illusions that result from Newton's error of interpretation of the mathematics. A force is not possible without the transfer of energy. The graviton is not necessary for equilibrium motion where no energy transfer takes place. The modified first law of thermodynamics says that the total energy of the universe is a constant in the absence of a force and energy transfer. Orbital moton is necessary to the conservation of total energy. Kinetic and potential energies of masses and charges are conserved by orbital motion. === Subject: Re: A Questionable Foundation > Is there actually some part of the proof you want me to explain? How the proof follows from the ZF axioms, for a start. I can see that it follows from Russell's assumptions, but not how either Russell's assumptions, nor the alleged proof itself, follow from the axioms they are supposed to follow from, if what Russsell alleges is a proof is to actually be a proof. And based on Russell's past performance, it is not even close to a proof in ZF. === Subject: Re: A Questionable Foundation > One often finds that when one understands > a subject thoroughly things that once appeared to be pregnant with all > sorts of intriguing and interesting epiphanies are, in the end, just > somewhat boring technicalities. > > I guess I don't understand set theory thoroughly. Understatement of the month!! > I will admit, given every set has a smallest member, > proving there are no infinite descending posets > is a somewhat boring technicality. > I gave such a proof, but you snipped it. It is not posets which are at issue but TOsets (totally ordered sets) and, in particularWOsets, (well ordered sets). > > What isn't so boring is that the same proof > shows there are no infinite ascending posets, either. Then it was not a proof at all, as AoI guarantees what Russell thinks he is smart enough to deny. But Russell is a lot dumber than he will admit. > > I think it is deliberately misleading to say > every set has a smallest element is equivalent to > there are no infinite descending posets, when > every set has a smallest element is really equivalent > to there are no infinite posets. It is even more deliberately misleading for Russell to open his mouth on issues of what follows from the ZF axioms. The above mish-mash supports my point. > > > Russell > - 2 many 2 count === Subject: Re: A Questionable Foundation Hi >>Yep. And the same holds for every other well founded set. It is not a special property of a N. >> When set theory says N is well ordered, it really means >> any finite subset of N can be well ordered. Until now I used as a successor X' = {X}. > But there is also a variant around, where > X' = X u {X}. This gives a little bit more bloathed numbers: 1 = {{}} 2 = {{},{{}}} 3 = {{},{{}},{{},{{}}}} Let's use the less bloated representation of N. N = {{{...}}} What can {{{...}}} possibly mean other than an infinite chain of set membership? You claim N does not have a largest proper subset, but it is OBVIOUS {{{...}}} has a largest proper subset. Remove the leftmost brace and the rightmost brace and we get the largest proper subset of N is {{{...}}}. N is its own largest proper subset. How can N be the smallest set that contains all natural numbers when the largest proper subset of N is N itself? You claim N is the smallest ordinal that contains all natural numbers, but N is not the smallest proper superset of any ordinal. How can this be? It is like you are claiming you don't know which pair of braces represents N but you assume that some pair of braces must be N. To convince me N is well founded and well ordered, you will have to produce an ordinal that has a finite number of pairs of braces at which point I will prove N is finite. Russell - 2 many 2 count === Subject: Re: A Questionable Foundation > This gives a little bit more bloathed numbers: 1 = {{}} 2 = {{},{{}}} 3 = {{},{{}},{{},{{}}}} > > Let's use the less bloated representation of N. > > N = {{{...}}} > WRONG! in this form it would have to be N = ...{{{}}}..., since there is an innermost, namely {}, but no outermost. > > You claim N does not have a largest proper subset, > but it is OBVIOUS {{{...}}} has a largest proper subset. Which is another reason why {{{...}}} is wrong. > > How can N be the smallest set that contains > all natural numbers when the largest proper > subset of N is N itself? Russel,l as usual, assumes conditions contrary to fact. > > You claim N is the smallest ordinal that contains > all natural numbers, but N is not the smallest > proper superset of any ordinal. How can this be? Easily. In NBG, the naturals are represented by the von Neumann sequence, in which the susccessor to each natural, n, is n / {n}. === Subject: Re: A Questionable Foundation Is this over? Does anyone (except Virgil) doubt I can prove N is finite? I gave a simple proof in ZF-I that every ordinal is finite. This proof is valid in ZF despite those that claim N is a counter example. N can't be a counter example in ZF-I because N doesn't exist in ZF-I and N is not a counter example in ZF, N is a contradiction. I have shown N is its own largest proper subset proving N is not well founded. I have shown no chain of set membership can contain N and every natural number proving N can't be well ordered. I have shown any well founded, well ordered set must be finite. Set theory is inconsistent. Russell - 2 many 2 count === Subject: Re: A Questionable Foundation Russell Easterly says... >Is this over? >Does anyone (except Virgil) doubt I can prove N is finite? Let me write down the facts that I think are relevant to proving things about N. Can you prove N is finite from these? 1. Let 0 = {}. 2. If x is any set, let x+1 = { y | y=x or y is an element of x }, 3. 0 is an element of N. 4. If x is an element of N, then x+1 is an element of N. 5. If S is any set such that (i) 0 is an element of S and (ii) forall x, if x is an element of S, then x+1 is an element of S, then N is a subset of S. Do you actually think you can prove N is finite from these axioms? If not, what additional facts do you need. If so, then please write down an actual proof, which as I have told you is a sequence of statements such that each statement is either an axiom, or follows from previous statements by logical deduction. -- Daryl McCullough Ithaca, NY === Subject: Re: A Questionable Foundation On 14 Aug 2006 04:16:42 -0700, Daryl McCullough said: > Russell Easterly says... >Is this over? >>Does anyone (except Virgil) doubt I can prove N is finite? Let me write down the facts that I think are relevant to > proving things about N. Can you prove N is finite from > these? 1. Let 0 = {}. > 2. If x is any set, let x+1 = { y | y=x or y is an element of x }, > 3. 0 is an element of N. > 4. If x is an element of N, then x+1 is an element of N. > 5. If S is any set such that (i) 0 is an element of S and > (ii) forall x, if x is an element of S, then x+1 is an > element of S, then N is a subset of S. Do you actually think you can prove N is finite from these > axioms? If not, what additional facts do you need. If so, > then please write down an actual proof, which as I have told > you is a sequence of statements such that each statement > is either an axiom, or follows from previous statements by > logical deduction. Hah, well, there's the problem, Daryl! Clearly, this limp notion of proof of yours omits Colbert's rule: Infer A, if A feels truthy. No wonder your class of theorems (and no doubt your beliefs about the world generally, constrained as they are by facts and evidence) is so impoverished! I pity you. === Subject: Re: A Questionable Foundation > Russell Easterly says... > >> Is this over? >> Does anyone (except Virgil) doubt I can prove N is finite? > > Let me write down the facts that I think are relevant to > proving things about N. Can you prove N is finite from > these? > > 1. Let 0 = {}. > 2. If x is any set, let x+1 = { y | y=x or y is an element of x }, > 3. 0 is an element of N. > 4. If x is an element of N, then x+1 is an element of N. > 5. If S is any set such that (i) 0 is an element of S and > (ii) forall x, if x is an element of S, then x+1 is an > element of S, then N is a subset of S. > > Do you actually think you can prove N is finite from these > axioms? If not, what additional facts do you need. If so, > then please write down an actual proof, which as I have told > you is a sequence of statements such that each statement > is either an axiom, or follows from previous statements by > logical deduction. > > -- > Daryl McCullough > Ithaca, NY > Isn't the standard set theoretic definition of finite, that a set S is finite iff there is an injection from S to some element of w? Then for any element of w just consider its identity map? or.. ? === Subject: Re: A Questionable Foundation >> Russell Easterly says... >> Is this over? Does anyone (except Virgil) doubt I can prove N is finite? >> >> Let me write down the facts that I think are relevant to >> proving things about N. Can you prove N is finite from >> these? >> >> 1. Let 0 = {}. >> 2. If x is any set, let x+1 = { y | y=x or y is an element of x }, >> 3. 0 is an element of N. >> 4. If x is an element of N, then x+1 is an element of N. >> 5. If S is any set such that (i) 0 is an element of S and >> (ii) forall x, if x is an element of S, then x+1 is an >> element of S, then N is a subset of S. >> >> Do you actually think you can prove N is finite from these >> axioms? If not, what additional facts do you need. If so, >> then please write down an actual proof, which as I have told >> you is a sequence of statements such that each statement >> is either an axiom, or follows from previous statements by >> logical deduction. >> >> -- >> Daryl McCullough >> Ithaca, NY >> > Isn't the standard set theoretic definition of finite, > that a set S is finite iff there is an injection from S > to some element of w? Then for any element of w just > consider its identity map? > or.. ? That only proves that each member of N (or w) is finite, which was not in dispute. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: A Questionable Foundation Russell Easterly says... >> Is this over? > Does anyone (except Virgil) doubt I can prove N is finite? Let me write down the facts that I think are relevant to proving things about N. Can you prove N is finite from these? > 1. Let 0 = {}. 2. If x is any set, let x+1 = { y | y=x or y is an element of x }, 3. 0 is an element of N. 4. If x is an element of N, then x+1 is an element of N. 5. If S is any set such that (i) 0 is an element of S and (ii) forall x, if x is an element of S, then x+1 is an element of S, then N is a subset of S. > Do you actually think you can prove N is finite from these axioms? If not, what additional facts do you need. If so, then please write down an actual proof, which as I have told you is a sequence of statements such that each statement is either an axiom, or follows from previous statements by logical deduction. > -- Daryl McCullough Ithaca, NY >> >> Isn't the standard set theoretic definition of finite, >> that a set S is finite iff there is an injection from S >> to some element of w? Then for any element of w just >> consider its identity map? > >> or.. ? > > That only proves that each member of N (or w) is finite, which was not in > dispute. Oh right... didn't sleep last night and was reading carelessly. Although it sort of caught me off guard that someone was trying to prove that N is finite. === Subject: Re: A Questionable Foundation > Isn't the standard set theoretic definition of finite, > that a set S is finite iff there is an injection from S > to some element of w? Yes. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A Questionable Foundation Russell Easterly says... >Is this over? >Does anyone (except Virgil) doubt I can prove N is finite? There is nobody who believes that you can prove N is finite (under the standard meaning of proof). I doubt that you understand what it means to prove anything. Once again, a proof is a sequence of statements, each of which is either an axiom, or follows from previous axioms by logical inference. You've never given a proof of anything, as far as I know. An informal proof is a little less rigid. Instead of directly proving things from axioms, you can instead state things in terms of lemmas, which in turn you can leave unproved unless someone challenges you on them. Informal proofs also make use of definitions, but a definition should always be eliminable, in the sense that you should be able to rewrite a proof using definitions in terms of the original language. >I gave a simple proof in ZF-I that every ordinal is finite. No, you didn't. That's a delusion on your part, based on your misunderstanding of what a proof is. >This proof is valid in ZF Not by the definition of proof. >I have shown N is its own largest proper subset >proving N is not well founded. No, you have not. >I have shown no chain of set membership can contain >N and every natural number proving N can't be >well ordered. No, you have not. >I have shown any well founded, well ordered set >must be finite. No, you have not. Russell, you haven't proved *anything*, except your own lack of understanding of mathematics and logic. >Set theory is inconsistent. You made the claim that ZF + the axiom of four is consistent. Now you are claiming that ZF by itself is inconsistent. That's a contradiction. -- Daryl McCullough Ithaca, NY === Subject: Re: A Questionable Foundation > Is this over? > Does anyone (except Virgil) doubt I can prove N is finite? No one doubts the validity of your proofs - even Virgil admitted in a private e-mail to me that you're right. It's all over. Set theory is fundamentally flawed. It's a lesson to all of us. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A Questionable Foundation Aatu Koskensilta says... > Is this over? >> Does anyone (except Virgil) doubt I can prove N is finite? No one doubts the validity of your proofs - even Virgil admitted in a >private e-mail to me that you're right. It's all over. Set theory is >fundamentally flawed. It's a lesson to all of us. I see you're carrying on the tradition of Torkel Franzen... -- Daryl McCullough Ithaca, NY === Subject: Re: A Questionable Foundation > Aatu Koskensilta says... > >>No one doubts the validity of your proofs - even Virgil admitted in a >>private e-mail to me that you're right. It's all over. Set theory is >>fundamentally flawed. It's a lesson to all of us. > > I see you're carrying on the tradition of Torkel Franzen... I see now my reaction was perhaps too hasty: we can still use order types to pretend set theory is consistent. As to carrying on the tradition of the venerable Torkel Franz.8en's, one can observe, given that Easterly's discoveries are of such magnitude they can't possibly be ignored - as demonstrated by the people willing to debate them with him endlessly - that there are two possibilities. Either one will, futilely, disagree with him and attempt to explain boring set theoretical technicalities, or one can just happily acknowledge that he's right, set theory is fundamentally flawed, and be done with it. I prefer the latter, but it's indeed fortunate that most people don't, and USENET instead of dying as predicted, thrives. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A Questionable Foundation <44DDB102.9050902@fastmail.fm> <5YCdnfzoUdShhEPZnZ2dnUVZ_rednZ2d@comcast.com> <44DE1D40.5060404@fastmail.fm> <44DE6E99.7020307@fastmail.fm> <44DEFE3B.1060905@fastmail.fm> <44DF926D.6000701@fastmail.fm> Does anyone (except Virgil) doubt I can prove N is finite? No one doubts the validity of your proofs - even Virgil admitted in a > private e-mail to me that you're right. It's all over. Set theory is > fundamentally flawed. It's a lesson to all of us. Indeed. But I haven't quite figured out Easterly yet - do you think he's aiming for a Fields medal or similar acclaim from academia? Will he call out the US military? Or is he just in it for the chicks after all? Brian Chandler http://imaginatorium.org === Subject: Re: A Questionable Foundation > Is this over? > Does anyone (except Virgil) doubt I can prove N is finite? No one doubts the validity of your proofs - even Virgil admitted in a > private e-mail to me that you're right. It's all over. Set theory is > fundamentally flawed. It's a lesson to all of us. > > Indeed. But I haven't quite figured out Easterly yet - do you think > he's aiming for a Fields medal or similar acclaim from academia? Will > he call out the US military? Or is he just in it for the chicks after > all? Goddammit, the lucky bugger. Set theory-loving chicks are *_hot_*! I need to change field; computational number theory just doesn't draw the babes. Then again, I'm not at the top of my field, unlike Easterly. Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: A Questionable Foundation > Indeed. But I haven't quite figured out Easterly yet - do you think > he's aiming for a Fields medal or similar acclaim from academia? Will > he call out the US military? Or is he just in it for the chicks after > all? People are usually drawn to set theory by the piquant notation and terminology. Who could resist the allure of 'well-ordering', 'countable chain condition', 'ineffable cardinal', 'the axiom of playful universe'? Easterly is surely motivated by the deep-rooted disappointment we all experience when finding out that these wonderful names are used for all sorts of silly, mundane technical stuff. But all this is moot now that we know set theory is fundamentally flawed. It's time to rmgroup sci.logic and sci.math. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A Questionable Foundation On Sun, 13 Aug 2006 21:42:13 -0700, Russell Easterly said: > Set theory is inconsistent. Certainly has the ring of truthiness to me. === Subject: Re: A Questionable Foundation > Is this over? > Does anyone (except Virgil) doubt I can prove N is finite? Does anyone (except Russell) believe that he CAN prove N is finite in ZF or in NBG? I doubt that anyone believes that Russell can even prove much that does follow from the axioms, much less anything that doesn't. > > I gave a simple proof in ZF-I that every ordinal is finite. It was not a proof at all. > This proof is valid in ZF despite those that claim N is > a counter example. We have better reasons for rejecting Russell's allegation of having a proof. > N can't be a counter example in > ZF-I because N doesn't exist in ZF-I The claim that N does not exist in ZF-I is unprovable in ZF-I. > and N is not a > counter example in ZF, N is a contradiction. Often claimed but never proved. > > I have shown N is its own largest proper subset > proving N is not well founded. Russell may think he has proved all sorts of improbable things, but on closer examination, one always finds that he assumes things not provable from the axioms in order to deduce his odd results. > > I have shown no chain of set membership can contain > N and every natural number proving N can't be > well ordered. Russell does not even seem to know what well ordered means. The only proof that N is not well ordered would be to display a counterexample: a subset of N which is not empty but does not have a smallest member, with respect to inclusion. Russell has not done this, so that he has NOT proved that N is not well ordered. > > I have shown any well founded, well ordered set > must be finite. Russell has shown himself to be an ass. > > Set theory is inconsistent. Russell is inconsistent. === Subject: Re: A Questionable Foundation > Is this over? > Does anyone (except Virgil) doubt I can prove N is finite? I have absolutely no doubt you can prove, to your own satisfaction, anything you choose. On the other hand, I have yet to see from you a proof that would not have been rejected for unexplained jumps, assuming the thing that is being proved, or general confusion, if I had tried to turn it in as computer science coursework. Why don't you try to prove SOMETHING to a reasonable standard of informal proof? It doesn't have to be anything difficult or controversial. Pick a set of axioms. For each step of your proof, explain how it follows from the axioms and previous steps. Do not introduce any other statement unless it is a well known theorem already proved from the axioms, and you can reference a proof. Patricia === Subject: Re: A Questionable Foundation Discussion, linux) <75ydneVj060_kkDZnZ2dnUVZ_vCdnZ2d@comcast.com> <44DDB102.9050902@fastmail.fm> <5YCdnfzoUdShhEPZnZ2dnUVZ_rednZ2d@comcast.com> <44DE1D40.5060404@fastmail.fm> <44DE6E99.7020307@fastmail.fm> <44DEFE3B.1060905@fastmail.fm> <44DF926D.6000701@fastmail.fm> informal proof? It doesn't have to be anything difficult or > controversial. Pick a set of axioms. For each step of your proof, explain how it > follows from the axioms and previous steps. Do not introduce any other > statement unless it is a well known theorem already proved from the > axioms, and you can reference a proof. Both you and Daryl have suggested that Russell presents a proof in which each step is either an axiom or derivable from previous steps. But this isn't really how informal mathematical proofs go. Mathematical proofs are closer to proofs in natural deduction: they include assumptions and assumption-discharging rules of inference (like -> introduction). Admittedly, the use of assumptions in proofs adds a bit of complexity and Russell is just the guy to bungle the role of assumptions. So, there's good reason to simplify matters by using a different set of rules of inference. But then we appear hypocritical, since math texts are full of proofs of the form: Assume sqrt(2) is rational. Of course Russell should learn what a proof is rather than claiming N = {{{...}}} on his say-so. But shouldn't we stick to a definition of proof that matches what we see in the textbooks? (Yes, I know that by choosing an appropriate set of axiom schemes, we can eliminate the role of assumptions in natural deduction. But that's not how the proofs appear in the texts.) -- I thought the wreck was over. I thought the fire was out. I thought the storm had passed and I was safe at last. I thought the wreck was over, but here she comes again. --The Flatlanders === Subject: Re: A Questionable Foundation Jesse F. Hughes says... >Both you and Daryl have suggested that Russell presents a proof in >which each step is either an axiom or derivable from previous steps. >But this isn't really how informal mathematical proofs go. >Mathematical proofs are closer to proofs in natural deduction: they >include assumptions and assumption-discharging rules of inference >(like -> introduction). You're right. What I was trying to get across is that *other* than citing axioms, every other step in a proof is purely *logical*. A logically valid step is correct or not by virtue of its *form*; it doesn't matter what *interpretations* are given to the function symbols, relation symbols and constant symbols. In contrast, Russell likes to make statements that are not logically valid, nor are they axioms. -- Daryl McCullough Ithaca, NY === Subject: Re: A Questionable Foundation Discussion, linux) <87oduncpl7.fsf@phiwumbda.org> logically valid, nor are they axioms. Well, duh. -- [Y]ou never understood the real role of mathematicians. The position is one of great responsibility and power. [...] You people have no concept of what it means to be an actual mathematician versus pretending to be one, dreaming you understand. -- James S. Harris === Subject: Re: A Questionable Foundation > >> Why don't you try to prove SOMETHING to a reasonable standard of >> informal proof? It doesn't have to be anything difficult or >> controversial. >> Pick a set of axioms. For each step of your proof, explain how it >> follows from the axioms and previous steps. Do not introduce any other >> statement unless it is a well known theorem already proved from the >> axioms, and you can reference a proof. > > Both you and Daryl have suggested that Russell presents a proof in > which each step is either an axiom or derivable from previous steps. > But this isn't really how informal mathematical proofs go. > Mathematical proofs are closer to proofs in natural deduction: they > include assumptions and assumption-discharging rules of inference > (like -> introduction). > > Admittedly, the use of assumptions in proofs adds a bit of complexity > and Russell is just the guy to bungle the role of assumptions. So, > there's good reason to simplify matters by using a different set of > rules of inference. But then we appear hypocritical, since math texts > are full of proofs of the form: Assume sqrt(2) is rational. > > Of course Russell should learn what a proof is rather than claiming > N = {{{...}}} on his say-so. But shouldn't we stick to a definition > of proof that matches what we see in the textbooks? (Yes, I know that > by choosing an appropriate set of axiom schemes, we can eliminate the > role of assumptions in natural deduction. But that's not how the > proofs appear in the texts.) I have no problem with the use of assumptions as long as they are clearly labeled as such, and anything proved from them is treated as dependent on the validity of the assumptions. Many of his proofs would be perfectly valid proofs of statements such as If there is a largest natural number, ZF is inconsistent, or If, for every ordinal X, there is a single element whose removal from X results in an ordinal, then there is a largest natural number. The problem is that he starts by assuming something that does not appear to be provable in ZF, and then loses track of the assumption. However, Russell cannot make effective use the Assume sqrt(2) is rational form of proof because he assumes inconsistency of ZF, and any other commonly used axioms for set theory. If he started out assuming sqrt(2) is rational he would end up deducing that rational number analysis is inconsistent, rather than that 2 does not have a rational square root. Patricia === Subject: Re: A Questionable Foundation Discussion, linux) <44DDB102.9050902@fastmail.fm> <5YCdnfzoUdShhEPZnZ2dnUVZ_rednZ2d@comcast.com> <44DE1D40.5060404@fastmail.fm> <44DE6E99.7020307@fastmail.fm> <44DEFE3B.1060905@fastmail.fm> <44DF926D.6000701@fastmail.fm> <87oduncpl7.fsf@phiwumbda.org> rational form of proof because he assumes inconsistency of ZF, and any > other commonly used axioms for set theory. If he started out assuming > sqrt(2) is rational he would end up deducing that rational number > analysis is inconsistent, rather than that 2 does not have a rational > square root. This isn't quite right. Indirect proofs are valid in inconsistent theories as well as in consistent theories. The only issue is whether he really discharges his assumptions. -- [I want to] stand at the pinnacle of human achievement with no one else in all of history even close, no human being having faced what I have--and survived. Because when all is said and done, make no mistake, the simple truth is, I am better. --James S. Harris === Subject: Re: A Questionable Foundation > >> However, Russell cannot make effective use the Assume sqrt(2) is >> rational form of proof because he assumes inconsistency of ZF, and any >> other commonly used axioms for set theory. If he started out assuming >> sqrt(2) is rational he would end up deducing that rational number >> analysis is inconsistent, rather than that 2 does not have a rational >> square root. > > This isn't quite right. Indirect proofs are valid in inconsistent > theories as well as in consistent theories. The only issue is whether > he really discharges his assumptions. > I'm not claiming that is a valid approach to proving inconsistency. Patricia === Subject: Re: A Questionable Foundation Russell Easterly says... >Let's use the less bloated representation of N. N = {{{...}}} No. N is the set of *all* natural numbers. If you use the representation 0 = {} 1 = {{}} 2 = {{{}}} Then N is the set {{}, {{}}, {{{}}}, {{{{}}}}, ...} >What can {{{...}}} possibly mean other than >an infinite chain of set membership? In ZF, there is no such set. -- Daryl McCullough Ithaca, NY === Subject: Re: A Questionable Foundation <75ydneVj060_kkDZnZ2dnUVZ_vCdnZ2d@comcast.com> <44DDB102.9050902@fastmail.fm> <5YCdnfzoUdShhEPZnZ2dnUVZ_rednZ2d@comcast.com> <44DE1D40.5060404@fastmail.fm> <44DE6E99.7020307@fastmail.fm> <44DEFE3B.1060905@fastmail.fm> <44DF926D.6000701@fastmail.fm> you use the representation 0 = {} > 1 = {{}} > 2 = {{{}}} Then N is the set {{}, {{}}, {{{}}}, {{{{}}}}, ...} What can {{{...}}} possibly mean other than >an infinite chain of set membership? In ZF, there is no such set. -- > Daryl McCullough > Ithaca, NY A tota ordering is something you could expect to apply to sets of numbers like the natural numbers or real numbers and so on and so forth, the oomplex numbers in various (although which is a natural ordering of the complex numbers is of various perspectives), and that fact about those numbers is normally encapsulated in calling them sets having the Archimedean property, that if x < y then by no means does y < x, by some natural ordering. While that is so, perhaps you can consider infinite sets that don't have the Archimedean property in their ordering, and I'm not talking about pathological cases. In fact, a bright person might be able to consider that even such sets as the natural or real numbers have in their natural ordering a way to consider them as non-Archimedean. Right below absolute zero would be negative one. It's infinity. People have been talking about trying to find infinity in physics for many, many years. Let me tell you, it's everywhere. There is no universe in NBG, nor ZF(C). Until there's a logical theory for mathematics with acceptance of a universe, cosmologists have no recourse. You got paradoxes in your theory there. Ross === Subject: Re: A Questionable Foundation >> Hi > Yep. And the same holds for every other > well founded set. It is not a special > property of a N. When set theory says N is well ordered, it really means any finite subset of N can be well ordered. >> Until now I used as a successor X' = {X}. >> But there is also a variant around, where >> X' = X u {X}. >> This gives a little bit more bloathed numbers: >> 1 = {{}} >> 2 = {{},{{}}} >> 3 = {{},{{}},{{},{{}}}} > > Let's use the less bloated representation of N. > > N = {{{...}}} If you use the 1={{}}, 2={{},{{}}} approach to the naturals, then N is { {}, {{}}, {{{}}}, {{{{}}}}, {{{{{}}}},... } Patricia === Subject: Re: A Questionable Foundation Hi Yep. And the same holds for every other >> well founded set. It is not a special >> property of a N. > When set theory says N is well ordered, it really means > any finite subset of N can be well ordered. Until now I used as a successor X' = {X}. But there is also a variant around, where X' = X u {X}. > This gives a little bit more bloathed numbers: > 1 = {{}} > 2 = {{},{{}}} > 3 = {{},{{}},{{},{{}}}} >> Let's use the less bloated representation of N. >> N = {{{...}}} > > If you use the 1={{}}, 2={{},{{}}} approach to the naturals, then N is I meant 2={{{}}} etc. > { {}, {{}}, {{{}}}, {{{{}}}}, {{{{{}}}},... } > > Patricia === Subject: Re: A Questionable Foundation Jan Burse says... >> I want to see one chain that contains every natural. It does not exist, if it is supposed to >be well founded. And in ZFC membership >chains are supposed to be well founded. Don't you mean that there is no *descending* chain containing every natural. There certainly is an *ascending* chain containing every natural number. -- Daryl McCullough Ithaca, NY === Subject: Re: A Questionable Foundation > Jan Burse says... >> I want to see one chain that contains every natural. >>It does not exist, if it is supposed to >>be well founded. And in ZFC membership >>chains are supposed to be well founded. Don't you mean that there is no *descending* chain > containing every natural. There certainly is an > *ascending* chain containing every natural number. No, there is not. If there were such a chain, I could prove there exists an infinite descending chain. Russell - 2 many 2 count === Subject: Re: A Questionable Foundation > Jan Burse says... >> I want to see one chain that contains every natural. >>It does not exist, if it is supposed to >>be well founded. And in ZFC membership >>chains are supposed to be well founded. Don't you mean that there is no *descending* chain > containing every natural. There certainly is an > *ascending* chain containing every natural number. > > No, there is not. The set of elements of N themselves form such a chain, but it is an endless chain. > If there were such a chain, I could prove there exists > an infinite descending chain. There is a infinite ascending chain in NBG, so where is Russell's claimed descending chain in NBG??? Nowhere, as usual. > Russell > - 2 many 2 count