>>Before Genzten proved Con(PA), Wilhelm Ackermann proved it. >> You have in mind Ackermann's 1940 proof? According to Szabo, >> Ackermann [1940, 1950-1953] adapted Gentzen's method of using >> For a paper explaining the technicalities, see >> www.logic.at/people/moser/publications/ ackermann050901.pdf >> (Before Gentzen, Ackermann had a supposed proof of the consistency of >> analysis, but this turned out to be incorrect.) Maybe I'm confused. I've noticed. Nice guy one newsgroup / troll in another. Russell -- Are you still here? The message is over. Shoo! Go away! ==== The slippery gooiness of biology is a consequence of its incredible > complexity, consisting as it does of complex systems based upon chemistry. > And chemistry obeys the rules of physics, which exists because of, and is > consequently best described by, mathematics. Mathematics is the ur-fluid of > Reality (gad, how poetic), and our symbollic attempts to represent > mathematics have given us windows through which our mushy grey-matter can > peer, and with which this same mushy grey-matter becomes altered, and we > call this alteration understanding (a frequently generous appellation). > [...to press our noses against some of the windows that look upon biological > systems...] [snip sciolistic brainfarting] Thermodynamics proposes, kinetics disposes. Kinetic control is the wild ride. Nobody can usefully model turbulence through space and over time. Uncle Al likes a universe that actively excludes god. > The book is pleasant and it is written in good Latin American Spanish but to > the readers who do not know the Castilian language (Latin American Spanish):(literate Spanish)::(Ebonics):(literate English) The Real Academia Espa.96ola located in Madrid is entrusted with purifying, clarifying and giving splendor to the language (re similar exercises of jackbooted State lingustic compassion in France, Quebec, and Israel). What a bunch of losers. -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) ==== >The slippery gooiness of biology is a consequence of its incredible > complexity, consisting as it does of complex systems based upon chemistry. > And chemistry obeys the rules of physics, which exists because of, and is > consequently best described by, mathematics. Mathematics is the ur-fluid of > Reality (gad, how poetic), and our symbollic attempts to represent > mathematics have given us windows through which our mushy grey-matter can > peer, and with which this same mushy grey-matter becomes altered, and we > call this alteration understanding (a frequently generous appellation). > [...to press our noses against some of the windows that look upon biological > systems...] > [snip sciolistic brainfarting] Thermodynamics proposes, kinetics disposes. Kinetic control is the > wild ride. Nobody can usefully model turbulence through space and > over time. Uncle Al likes a universe that actively excludes god. > Then you are of the opinion that the math that has been achieved in the medical sciences, pharmochology, etc... should be revised or eliminated or that this all works into body movements anyway? Not a bad idea, trying to translate robotics into sports medicine. > The book is pleasant and it is written in good Latin American Spanish but to > the readers who do not know the Castilian language (Latin American Spanish):(literate Spanish)::(Ebonics):(literate > English) The Real Academia Espa.96ola located in Madrid is entrusted with > purifying, clarifying and giving splendor to the language (re > similar exercises of jackbooted State lingustic compassion in France, > Quebec, and Israel). What a bunch of losers. -- > Uncle Al > http://www.mazepath.com/uncleal/qz.pdf > http://www.mazepath.com/uncleal/eotvos.htm > (Do something naughty to physics) ==== > (Latin American Spanish):(literate Spanish)::(Ebonics):(literate hey Uncle Al, aren't we getting caustic about something that doesn't even exist? There is no such thing as Latin American Spanish. You're acting like Thiotimoline ;P ==== (Latin American Spanish):(literate Spanish)::(Ebonics):(literate hey Uncle Al, aren't we getting caustic about something that doesn't > even exist? There is no such thing as Latin American Spanish. You're > acting like > Thiotimoline ;P I live in Southern California. Spanish as spoken by literate Spaniards is a lyrical Romance language. Spanish as spoken by local swine is an agrammatical slurred patois that sounds like a gutteral chicken cackling while firing a machine gun. New World wetback women are astoundingly ugly as a class. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! ==== > I live in Southern California. Spanish as spoken by literate > Spaniards is a lyrical Romance language. Spanish as spoken by local > swine is an agrammatical slurred patois that sounds like a gutteral > chicken cackling while firing a machine gun. Hey Uncle, I don't get what you are about, swine dont't speak. What is this gutteral adjective? Is it ebonics for something? Is there an equivalent english term? > New World wetback women are astoundingly ugly as a class. I guess you got a point there. ==== I need some help with my history essay. Does anyone know what Archimedes corollary is that was included in his > Measurement of a Circle? Or have any ideas how i could find it, as i'm > having difficultity finding it in the books i have got. Look for the Great Books of the Western World. This is a 50 or so > volume of translations into English of classic books of the > western world. Alas, the series was out of print for a long time; is it back in? ==== > I need some help with my history essay. Does anyone know what Archimedes corollary is that was included in his > Measurement of a Circle? Or have any ideas how i could find it, as i'm > having difficultity finding it in the books i have got. Look for the Great Books of the Western World. This is a 50 or so > volume of translations into English of classic books of the > western world. Alas, the series was out of print for a long time; is it back in? Yes. You can buy a copy from my website. James Harris ==== > Alas, the series was out of print for a long time; is it back in? Yes. You can buy a copy from my website. I'm not sure I'm willing to trust anything on your website, sorry. ==== > .... > Look for the Great Books of the Western World.... One volume is partially devoted to the works of Archimedes.... Yes. Pp.447-451 have the standard translation by Heath of Measurement of a Circle without any corollaries. Ken Pledger. ==== Two points. 1. if f_{n_i}(x_{n_i}) >= s then the closed interval, I_{n_i} containing > x_{n_i} with f_{n_i}(x) >= s for all x in I_{n_i} may only contain > x_{n_i}. This allows len(I_{n_i}) = 0. What you really want is for > f_{n_i}(x_{n_i}) > s so that for some _open_ interval I_{n_i} we have > f_{n_i}(x) > s for all x in I_{n_i} and len(I_{n_i}) > 0. 2. Even assuming that we are considering intervals with positive length, > that the midpoints of those intervals have a limit point does not > imply that any of the intervals intersect. Consider the intervals > { [(1/(2n-1),1/(2n)] : n = 1, 2, 3, ... }. 0 is the limit point of > their midpoints, yet 0 is in none of them, not infinitely many. > ok. I was thinking closed intervals helped, for some reason. rich >Rob Johnson I met the following problem i one book,which can be solved either by >measure theory or by Lebesque Dominated Convergence theorem,but the >author says it is too difficult to handle without these means .I think >I need help on it. Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x >in [0,1].Then Int(fn(x)dx,0,1)--->0! The following comes from deconstructing the measure-theoretic proof. Note that any open subset U of [0,1] is the union of at most countably many disjoint open intervals. Define m(U) as the sum of the lengths of those intervals. If f:[0,1] -> [0,1] is continuous and E(e) = {x: f(x) > e} (which is an open set) for e > 0, note that int_0^1 f(x) dx <= e + m(E(e)). For we can take any partition P of [0,1] and consider the lower Riemann sum L(f;P). The contribution of those intervals not contained in E(e) is bounded above by e, while the contribution of those intervals contained in E(e) is bounded above by m(E(e)). Now for each n let E_n(e) = {x: f_n(x) > e}. It is enough to show that for any e > 0, m(E_n(e)) < e for n sufficiently large. For convenience, I'll fix e and write E_n(e) as E_n. Let G_n = {x: f_n(x) > e but f_m(x) <= e for all m > n}. Since f_n -> 0, E_n = union_{m=n}^infinity G_n. I claim that there are open sets H_m with G_m contained in H_m and sum_n m(H_n) < infinity. If so, then taking N so large that sum_{n >= N} m(H_n) < e, E_n is contained in the union of countably many open intervals of total length < e, and therefore m(E_n) < e [ proving this therefore is an exercise left to the reader ]. We can write G_n = F_n F_{n+1} where each F_n = union_{m >= n} E_n is open, and F_{n+1} is a subset of F_n. F_n is the union of at most countably many disjoint open sets V_k of total length m(F_n), and F_{n+1} is the union of at most countably many disjoint open sets U_j of total length m(F_{n+1}); each U_j is contained in exactly one V_k. For each k, if z_k = m(F_(n+1) intersect V_k) is the total length of the U_j contained in V_k, take a finite subset of the U_j contained in V_k having total length at least z_k - 2^(-n-k). Take the complement in V_k of the union of this finite subset, and fatten up the intervals comprising it to make them open: thus V_k F_{n+1} is contained in an open set W_k of total length at most m(V_k) - z_k + 2^(1-n-k). The union of those W_k is our set H_n, with m(H_n) <= sum_k (m(V_k) - z_k + 2^(1-n-k)) <= m(F_n) - m(F_{n+1}) + 2^(1-n). And in particular, sum_n m(H_n) <= m(F_n) + 2 < infinity, QED. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >>Let me take that back. I either have a unplanned proof, or a planned >>unproof. >You all can decide. >>Given f_n:[0,1]-->[0,1], each f_n is continuous, and f_n(x)-->0 for x in >[0,1]. Prove (by elementary means) that int(f_n(x),0,1)-->0. >>Proof: >>Note int(f_n(x),0,1)=f_n(x_n) for some x_n in [0,1]. Also note if >>f_n(x_n)=sup >{ f_n(x) } then f_n(x)=f_n(y) for all x,y in [0,1]. >>I don't see what that last statement has to do with the proof, but >>it's true... >Clearly, {f_n(x_n)} has >some limit point. Suppose L>0 is the largest one. There is some real >>number >s,with 0=s for infinitely many n_i. For each >such n_i there is also some closed interval, I_n_i, containing x_n_i, with >f_n[I_n_i]>=s >>Slightly informal notation but I'm pretty sure I know what you mean. >>Fine so far. >and len(I_n_i)>0. Now let m_n_i be the midpoint of each I_n_i. >The set {m_n_i} has a limit point L' and therefore L' is in infinitely many >I_n_i. >Whoops. How does it follow that L' is in infinitely many I_n_i? Because each I_n_i is *closed*? Why post messages on sci.math if you're not going to read the replies? If you look at the example I gave showing that L' need not be in infinitely many I_n_i (unless it's for some reason you haven't explained) you'd see that I_n_i being closed doesn't help a bit. >(That they exist follows from the note about >sup{f_n(x_n)} above and the fact L is the largest limit point.) I didn't say the I_n_i doesn't exist... > But then f_n(L') >=s for infinitely many n, contradicting f_n(x)-->0. >So the unique limit point of {f_n(x_n)} is 0, as required. rich ************************ David C. Ullrich ==== Why post messages on sci.math if you're not going to read the >replies? If you look at the example I gave showing that L' need >not be in infinitely many I_n_i (unless it's for some reason you >haven't explained) you'd see that I_n_i being closed doesn't >help a bit. > rich We just started Riemann-Stieltjes Integration and I'm completely lost. We did not do Riemann first, so I am new to this. I do not understand how to prove this theorem or even what exactly I`m supposed to prove. The d(g1+g2) part is really throwing me off. I guess I expect one g. I'm not sure how the existence of 2 of the integrals will be able to ensure the existence of the other 2 in the conclusion. The theorem is (S=integral here) S[b a] f1 dg1 and S [b a] f2 dg2 both exist then S[b a] f1+ f2 d(g1+g2) = S[b a] f1 dg1+ S[b a] f2 dg1 + S [b a] f1 dg2 + S[b a] f2 dg2 ==== >We just started Riemann-Stieltjes Integration and I'm completely lost. We did >not do Riemann first, so I am new to this. I do not understand how to prove this theorem or even what exactly I`m >supposed to prove. The d(g1+g2) part is really throwing me off. I guess I >expect one g. I'm not sure how the existence of 2 of the integrals will be >able to ensure the existence of the other 2 in the conclusion. The theorem is (S=integral here) S[b a] f1 dg1 and S [b a] f2 dg2 both exist >then S[b a] f1+ f2 d(g1+g2) = S[b a] f1 dg1+ S[b a] f2 dg1 + S [b a] f1 dg2 + >S[b a] f2 dg2 The other integrals do not need to exist. Let f_1 be the characteristic function of [0,1/2] and f_2 be the characteristic function of (1/2,1]. Let g_1(x) = 1/(3-4x) and g_2(x) = 1/(1-4x). Then, |1 |1 2 | f d g = | f d g = - | 0 1 1 | 0 2 2 3 However both of the other integrals diverge. In fact, with a simple change of variables, we get that 2 |1 - + | f d g 3 | 0 2 1 |1 = | (f + f ) d g | 0 1 2 1 |1 4 = | -------- dx | 0 (3-4x)^2 |1 4 = | -------- dx | 0 (1-4x)^2 |1 = | (f + f ) d g | 0 1 2 2 |1 2 = | f d g + - | 0 1 2 3 which obviously diverges. Rob Johnson take out the trash before replying ==== >We just started Riemann-Stieltjes Integration and I'm completely lost. We did >not do Riemann first, so I am new to this. >I do not understand how to prove this theorem or even what exactly I`m >supposed to prove. The d(g1+g2) part is really throwing me off. I guess I >expect one g. I'm not sure how the existence of 2 of the integrals will be >able to ensure the existence of the other 2 in the conclusion. >The theorem is (S=integral here) S[b a] f1 dg1 and S [b a] f2 dg2 both exist >then S[b a] f1+ f2 d(g1+g2) = S[b a] f1 dg1+ S[b a] f2 dg1 + S [b a] f1 dg2 + >S[b a] f2 dg2 The first thing to do is always to get straight exactly what you're asked to prove. I don't know if we can really help you there: you'll have to ask your instructor. The existence of S[b a] f1 dg1 and S [b a] f2 dg2 do not imply the existence of any of the other integrals. But maybe the = is to be interpreted as: the left side exists if and only if all integrals on the right side exist, and in that case both sides are equal. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== I am interested in going to a mathematics school in the United States. I live in North Carolina, but where the school is is not important to me, as long as it is affordable and/or has a good scholarship program. I am an A student in math, with an interest primarily in algebra. Where would you recommend I apply? I am interested in pure research, as well as teaching. I am liking physics and may consider furthering my studies there. Katie. ==== > I am interested in going to a mathematics school in the United States. > I live in North Carolina, but where the school is is not important to > me, as long as it is affordable and/or has a good scholarship program. > I am an A student in math, with an interest primarily in algebra. > Where would you recommend I apply? I am interested in pure research, > as well as teaching. I am liking physics and may consider furthering > my studies there. Katie. You might want to take a look at the North Carolina School of Science and Math for your senior high school year. http://www.ncssm.edu/ ==== > I am interested in going to a mathematics school in the United States. > I live in North Carolina, but where the school is is not important to > me, as long as it is affordable and/or has a good scholarship program. > I am an A student in math, with an interest primarily in algebra. Where would you recommend I apply? I am interested in pure research, > as well as teaching. I am liking physics and may consider furthering > my studies there. If money is an issue first see what you can accomplish with a community college. Check with university before hand to know what they'll accept. ==== > I am interested in going to a mathematics school in the United States. > I live in North Carolina, but where the school is is not important to > me, as long as it is affordable and/or has a good scholarship program. > I am an A student in math, with an interest primarily in algebra. >Where would you recommend I apply? I am interested in pure research, > as well as teaching. I am liking physics and may consider furthering > my studies there. > If money is an issue first see what you can accomplish with a community > college. Check with university before hand to know what they'll accept. Skip the community college. They don't have anything past calculus, except maybe (that is, sometimes but not always) a first course in differential equations and/or linear algebra, with emphasis on calculating. You need to be writing proofs as soon as possible. If you go to a large university (say, just to pick one at random, one in calculus. This is a course that emphasizes proofs and understanding rather than just calculation. If you're already finished with calculus when you start, then you need to take a course that emphasizes writing proofs. Also, check the catalog. You're better off at a school that offers a year of analysis and algebra (usually titled abstract or moder algebra, to emphasize the difference between it and college algebra, which is high school algebra) instead of only a semester. You may be better off at a smaller school where you can easily meet the professors and talk to them about things and take independent study courses rather than formal ones. Especially if you're shy. My daughter is at a tiny liberal arts college and she got to analyze DDT levels by gas chromatog raphy of fish caught in the Pine River (EPA superfund site) her freshman year, just because she talked to the right people and someone was excited about her proposal. On the other hand, I went to a school with about 50,000 students and I didn't have any trouble finding professors who were interested in me. Besides, if you're at a school without a graduate program, how will you take graduate courses? When you visit colleges, visit the math departments and talk to people. (Also visit any other department you're interested in.) Look at the math library. Ask about practicing and coaching for the Putnam exam. What the heck, ask about tutoring opportunities or even undergraduate teaching assistantships (that'll help you pay your tuition). And ask about scholarships. For example, Texas Tech has a program that students in the top x% of their graduating class (or with an a.bc gpa) and an SAT above y are automatically granted in-state tuition, with more available for even better scores. It's on their web page. Many other schools have similar things. Your questions to schools should be can I do X? and what can you provide me? Of course, everyone will answer yes, so then you have to ask How? Jon Miller DuBois Double Reversed Dual Earth Model of 4D Space: How to make If someone here knows if these were made before, let me know by whom and I will change the name. I am put a picture of one at http://www.members.aol.com/scandere/ along with some other stuff on 4D. I am not selling them, I am just telling people how to make them and why they represent 4D because I probably won't get to make them myself. I am posting this message just to sci.math, sci.physics.relativity, sci.astro, sci.physics, as I thought it fits these best. Is there a usenet group devoted to 4D? If so let me know so I know where best to go for this topic. I have read most web sites devoted to 4D space already. Take 1 really big glass sphere and 1 smaller one. Fishbowls are recommended if you can find perfectly round ones. Paint a backwards globe (continents only looks best) on the large one. Paint a normal globe on the small one and put it inside the larger one, but flip it upside down and line it up backwards or opposite of the continents inside the larger one. You now have the the entire universe in your sights, not to scale, and only if curved for real. I originally intended to make 4 different sculptures of large sizes to demostrate aspects of 4D space. Most were to be about 3 to 4 meters in height. This one, DDRDE was the most obvious and simplest to understand, curved 3D space. Curved 3D space is wherever you go in space you wind up back at the same spot. On a globe it would be like firing a bullet so far you shoot yourself in the back of the head. That is curved 2D space, and painful. This represents the same concept with 3D space curved in all directions away in 4D space, and is an accurate representation of how it would line up. Objects curve away and seem completely around you when at equal distances in all directions, the antipole (the south pole on a 3D model, if you shot 4 bullets in directions at once, they would converge twice, once at the southpole and once at the north pole). Objects at either of those two points would, in curved space, not only appear to surround them but those points would be those same distances away, so it is not a true distortion but a mapping of the shape of space. That middle south pole equivalent in this model can be thought of as another sphere in between the two, and also as a single point or spot everywhere along that sphere as well, since it is only a single place like the opposite pole on a globe, but in 3D space opposite of where six bullets would pass each other if shot in curved space from a cube floating in space's sides. I did build a small table top model. The full scale (not to the size of the real actual scale model with the really really big lights) models I wanted to make were to be 4 meters built out of a 3 frequency geodesic dome made of glass and steel with a 1 meter center sphere. For aestic purposes the center sphere should always be at least 20% to not more than 25% of the outer sphere. Walking around in one you can get a feel for curved space. The center sphere should also be hollow so you can see through both at the same time with little to no distortion. The invertibility aspect came when I imagined a kid sticking his nose up to the center sphere and seeing how Africa and South America lined up then running around outside it to see the same view again. Thus with glass, the inversion is self- evident. Below is from the 5D notes pertaining to the dual globes. All (of the 4D sculptures) ought to be made of glass to demostrate the invertibility of curved space, or the inverse effect, visually and easily. That becomes so obvious (with glass), that the same view from all outside points (combined) looking inward through both spheres is exactly the same as the outward view from the center sphere looking outward in all directions (from its center point), that few could not be able to glimpse it. The purpose of the sculptures is to enable or help people to spin the Universe on its head, 3 dimensionally speaking. To instead or also see space as folded into objects instead of existing outside or between them. That from a 4th dimensional viewpoint all points away from the center point to the antipole equivalently and equally outside of it and within it, for they are the same spot. This is similar to how the standard idea of an antipole in the sculpture would be another sphere halfway between the other two, yet also (would) be in actuality a single spot or place, not a circular area. This multidimensional viewpoint is a good way to view electrons existing within a set position in folded space instead of orbiting around a circular area. They would not be moving at all, yet seem to be at all points in a (seemingly) circular ball around the nucleus as well, at the same time. Yet another folded dimension around the nucleus could explain how they jump from one orbit to the next without passing the space inbetween. Just because the level of space we percieve between what would seem to be 2 spheres from an outside viewpoint, seems 3 dimensional, that is no reason to believe that on the atomic scale, that space is not defined by or exists with more dimensions than we need to deal with out here (nor necessarily curled up so small as to not affect matter). The point of the scupltures is to see that inversible 3D sandwich, or that out there or in here is merely from an external viewpoint, (relative to where you are standing and) potentially equally the same spot. To understand closed 4D space or open curved 3D space, you only need to imagine being 2 places at the same time, which the sculptures do (represent). I could combine 2 of them as 4 different sculptures in which by standing in (and) combining them mentally would simulate being (in) 8 places at the same time, or 4 intersecting 4D worlds at weird angles, representing closed 5D space or 4D open space curved, but few would understand it, though it would look really cool. Jared. ==== Could someone recommend me rigorous and comprehensive books on mathematical logic and set theory? ==== > I've concluded that the problem I'm facing is that the human brain > isn't built to handle Mathematics. > > Well, I take it you are now claiming that you don't have a 'human brain'? > > Think about it some more. I think James means that he is the only one with a human brain. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >I've concluded that the problem I'm facing is that the human brain >isn't built to handle Mathematics. > > Well, I take it you are now claiming that you don't have a 'human brain'? > > Think about it some more. I think James means that he is the only one with a human brain. Think about it some more. James Harris ==== >I've concluded that the problem I'm facing is that the human brain >isn't built to handle Mathematics. >Well, I take it you are now claiming that you don't have a 'human brain'? > >Think about it some more. >I think James means that he is the only one with a human brain. Think about it some more. > James Harris There's not a lot to think about here Jimbo. Your statement, put mathematically (I know something you abhor) is: If a thing is human, then that thing isn't built to handle Mathematics The contrapositive is: If the thing is built to handle Mathematics, then it's not human. Now let's consider an instantiation of this statement and consider you. Two possibilities exist. Either you aren't built to handle Mathematics Or you are built to handle Mathematics If you are built to handle Mathematics, then you're not human. Which is obviously false (although there are times when your submental arguments make us wonder) Therefore you aren't built to handle Mathematics must be true. (yes, and this would indeed give you the aforementioned problem.) As always James, you have a firm grip on the obvious (although like most things you, since you do them incorrectly, you think there is some additional hidden meaning in it that nobody except you is smart enough to understand.) Jack ==== >I've concluded that the problem I'm facing is that the human > brain >isn't built to handle Mathematics. >Well, I take it you are now claiming that you don't have a 'human > brain'? > >Think about it some more. >I think James means that he is the only one with a human brain. >Think about it some more. > > James Harris There's not a lot to think about here Jimbo. > Your statement, put mathematically (I know something you abhor) is: If a thing is human, then that thing isn't built to handle Mathematics > The contrapositive is: > If the thing is built to handle Mathematics, then it's not human. Now let's consider an instantiation of this statement and consider you. > Two possibilities exist. > Either you aren't built to handle Mathematics Or you are built to handle > Mathematics > If you are built to handle Mathematics, then you're not human. > Which is obviously false (although there are times when your submental > arguments make us wonder) > Therefore you aren't built to handle Mathematics must be true. > (yes, and this would indeed give you the aforementioned problem.) > As always James, you have a firm grip on the obvious (although like most > things you, since you do them incorrectly, you think there is some > additional hidden meaning in it that nobody except you is smart enough to > understand.) Interesting, but flawed. Think about it some more. James Harris ==== The quick brown fox jumps over the lazy dog. c bfgn zwgmh hel zaclfv. ==== >The quick brown fox jumps over the lazy dog. Cwm fjord bank glyphs vext quiz. -- Richard -- FreeBSD rules! ==== Richard Nixon > The quick brown fox jumps over the lazy dog. > c bfgn zwgmh hel zaclfv. The first sentence is a cipher key for the second. The plaintext begins I know. LH ==== > Richard Nixon > The quick brown fox jumps over the lazy dog. > c bfgn zwgmh hel zaclfv. > The first sentence is a cipher key for the second. > The plaintext begins I know. And continues aoout the alieno ? -- Clive Tooth http://www.clivetooth.dk ==== The Last Danish Pastry > Larry Hammick > Richard Nixon > The quick brown fox jumps over the lazy dog. > c bfgn zwgmh hel zaclfv. > The first sentence is a cipher key for the second. > The plaintext begins I know. And continues aoout the alieno ? Apparently so :) ==== On 14 Jan 2004 19:26:08 -0800, K00L-Aid@excite.com (Richard Nixon) >The quick brown fox jumps over the lazy dog. > Hey there tricky Dick! I thought you were dead. It means you used all the letters of the alphabet in your sentence. I remember practicing that sentence when learning to type. --Lynn ==== > On 14 Jan 2004 19:26:08 -0800, K00L-Aid@excite.com (Richard Nixon) > >The quick brown fox jumps over the lazy dog. > Hey there tricky Dick! I thought you were dead. It means you used all > the letters of the alphabet in your sentence. I remember practicing > that sentence when learning to type. --Lynn But the redundancy can be lowered a bit by making one of the words the into the word a'. And I always preferred: Pack my box with five dozen liquor jugs which achieves the same goal with fewer letters and fewer words. There are, I believe, even better, though odder, ways to do it. ==== There are, I believe, even better, though odder, ways to do it. An odder and better example: Mr. Jock, TV quiz Ph.D., bags few lynx. -------- Tad ==== >I would like to know if this is a mathematical proof of the Cantor's >goof, or conversely it is just another mathematical goof about the >Cantor's proof. It's a goof. count the reals. Thus demonstrating that Mr. Mathman is not a genuine mathematician. >HC: I don't know it, but if we put a simple analogy perhaps we could >clarify our thoughts. Let's try it. >We may take the following equivalence table > N <==> M = A set of linear units of measure > R <==> A = The air in a room >Being our goal to find out whether A is measurable or not, we have >the following analogy: >POINT 1: A cannot be measured with linear unities, so we cannot >assume M -> A. >CRITERION: A set D will be lineally measurable if and only if it is >possible M -> D. >POINT 2: We can't use CRITERION to find out whether A is measurable >or not, because POINT 1 tells us that A can't be measured in this way. >PROOFS: We have found out that A is not measurable, because CRITERION >case. >PROPOSAL: We can admit CRITERION to measure A, and consequently A is >not measurable. Well, no one will accept this, because POINT 1 tells >us that we cannot apply CRITERION to measure A, and also because A >would be able to be measured by other means. >Good, Mr. Mathman, this is the end of our analogy. >MR. MATHMAN: And? >HC: What? >MR. MATHMAN: Are there other criteria to find out whether R is >countable or not? >HC: Well, in our analogy A cannot be measured with linear units, but >if we arrange orthogonally three of them, then we can. Perhaps the >reals have an undiscovered property which may be used as criterion to >count them. Think about it Mr. Mathman, think about it. >MR. MATHMAN: I'll do it. Are you suggesting the possibility that there is a bijection between N^3 and R? We know that there is a bijection between N^3 and N, and we also know that there is no bijection between N and R. It follows that there is no bijection between N^3 and R. Furthermore, for ANY nonzero natural number k, there is a bijection between N^k and N, and so for all natural numbers k, there is no bijection between N^k and R. The upshot is that your analogy fails miserably. And Mr. Mathman proves once again that he is not a genuine mathematician. The definition for countability of an infinite set is known. Perhaps if you spent more time studying mathematics, you might see that we were right all along. Or is there nothing that can shake your conviction that you are correct? >*************************** >Proof of Point 2 As I noted earlier, your proof of Point 2 was based on the unstated assumption that there is no bijection between N and a proper subset of N, an assumption which is known to be false. So, the proof that there is no bijection between N and R is not as trivial as you thought. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. ==== > Prove that > ((a % m) + (b % m))%m = (a + b)%m where a,b,m are integers. Assuming your ns should be ms What's an imteger? :-) Phil A member of a set which satisfies Peamo's Postulates? -- Paul Sperry Columbia, SC (USA) ==== I have a question about provability and unprovability results that must have been studied by now. I was hoping someone could tell me a summary of what is known about it. For example, let us work in ZF, and let us choose a difficult question, say the Continuum Hypothesis, as the statement S. We know that S is not provable or disprovable (in ZF), that is, it is undecidable. If we now let S1 be the statement S is undecidable in ZF, then we know that S1 is decidable, indeed, it is true. Question: Are there statements S for which S1 (S is undecidable) is *also* undecidable? We can go further, and for any statement S construct S1 = S is undecidable, S2 = S1 is undecidable, S3, etc. Question: How far can we progress down this series? Dale ==== > We know that S is not provable or disprovable (in ZF), that is, it is > undecidable. If we now let S1 be the statement S is undecidable in ZF, then we > know that S1 is decidable, indeed, it is true. True, yes, but decidable in what theory? Not in ZF. No statement of the form A is undecidable in ZF is provable in ZF. So you get your series: CH, CH is undecidable in ZF, 'CH is undecidable in ZF' is undecidable in ZF,... of statements undecidable in ZF. ==== On 15 Jan 2004 08:30:19 +0100, Torkel Franzen > We know that S is not provable or disprovable (in ZF), that is, it is >> undecidable. >> >> If we now let S1 be the statement S is undecidable in ZF, then we >> know that S1 is decidable, indeed, it is true. True, yes, but decidable in what theory? Not in ZF. No statement of >the form A is undecidable in ZF is provable in ZF. ??? Now you have me confused again. What _do_ we need in addition to ZF in order to show that this and that is undecidable in ZF? Oh, duh: For _any_ given P, showing that ZF does not imply P requires Con(ZF); if ~Con(ZF) then ZF certainly does imply P. (So when anyone says that this or that is undecidable in ZF what they really mean is that if ZF is consistent then this or that is undecidable in ZF; in particular your True, yes is assuming Con(ZF) (the word assuming is not meant to say anything about whether you're justified in assuming this.)) >So you get your >series: CH, CH is undecidable in ZF, 'CH is undecidable in ZF' is >undecidable in ZF,... of statements undecidable in ZF. ************************ David C. Ullrich ==== I have a question about provability and unprovability results that > must have been studied by now. I was hoping someone could tell me a > summary of what is known about it. Not really, but I can give a perspective on it and see if the common-sense approach has any holes in it. :) > For example, let us work in ZF, and let us choose a difficult > question, say the Continuum Hypothesis, as the statement S. We know that S is not provable or disprovable (in ZF), that is, it is > undecidable. If we now let S1 be the statement S is undecidable in ZF, then we > know that S1 is decidable, indeed, it is true. So far so good. > Question: Are there statements S for which S1 (S is undecidable) is > *also* undecidable? [My intuition:] No. Suppose S is decidable. Then S1 is decidably false, since we can exhibit an actual proof or disproof of S, at least in theory. So (S decidable) -> (S1 decidably false) Here's where my ability to express my intuition fails, and I'm forced into tricks with notation. ;-) Taking the contrapositive of this statement yields ~(S1 decidably false) -> ~(S decidable) Now, suppose for the sake of argument that S1 were actually undecidable. (S1 undecidable) -> ~(S1 decidable) -> ~(S1 decidably false) where each line represents a weaker proposition than the line before it. But by the last proposition, -> ~(S decidable) Which is exactly equivalent to the statement (S undecidable) So, if we assume (S1 undecidable), we are led to the inescapable conclusion that (S undecidable), which is precisely the statement (S1 true). So if S1 is undecidable, then it is the case that S1 is true; which looks an awful lot like a contradiction! Caveat: The contradiction depends on our being able to express the above argument entirely in the system under discussion. It's quite possible to have true, undecidable, statements. I'm pretty sure that Godel proved that we *must* have such statements in *any* system. But that doesn't automatically mean that this proof is wrong. :) > We can go further, and for any statement S construct S1 = S is > undecidable, S2 = S1 is undecidable, S3, etc. Question: How far can we progress down this series? I think not even the first step holds water, unfortunately; but I'm eager to see what subtleties I've missed. -Arthur ==== On Fri, 9 Jan 2004, Mark Griffith asked: > Anyone know of any recent work on the odd perfect number > question? Oddly enough, a paper claiming to prove the conjecture has appeared, dated Thu, 8 Jan 2004, at http://www.arxiv.org/abs/hep-th/0401052/ I haven't got the background to tell whether it's good or bad. ==== > On Fri, 9 Jan 2004, Mark Griffith asked: > Anyone know of any recent work on the odd perfect number > question? Oddly enough, a paper claiming to prove the conjecture has > appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ > I haven't got the background to tell whether it's good or bad. Well, one thing at least strikes me, which I don't think anyone else has picked up on yet... [people seem to be more interested in his affiliations and past papers - perhaps because they don't trust their own _mathematical_ judgement :-)] If you read his abstract (and concluding sentence) carefully it seems to me he's only claiming anything (even leaving aside the merits of his main argument) for odd numbers of the form 4k+1 [as opposed to 4k+/-1] (perhaps times a square). Therefore this completely lets slip through the net numbers like 7, for example... For the real proof of the non-existence of Odd Perfect Numbers see: http://www.bearnol.pwp.blueyonder.co.uk/Math/perfect.html J ==== > On Fri, 9 Jan 2004, Mark Griffith asked: > Anyone know of any recent work on the odd perfect number > question? Oddly enough, a paper claiming to prove the conjecture has > appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ > I haven't got the background to tell whether it's good or bad. Well, one thing at least strikes me, which I don't think anyone else > has picked up on yet... [people seem to be more interested in his > affiliations and past papers - perhaps because they don't trust their > own _mathematical_ judgement :-)] > If you read his abstract (and concluding sentence) carefully it seems > to me he's only claiming anything (even leaving aside the merits of > his main argument) for odd numbers of the form 4k+1 [as opposed to > 4k+/-1] (perhaps times a square). Therefore this completely lets slip > through the net numbers like 7, for example... I fart more correct mathematical judgements than you. Any odd perfect numbers must be of the form 4n+1, that's been known for centuries. > For the real proof of the non-existence of Odd Perfect Numbers see: > http://www.bearnol.pwp.blueyonder.co.uk/Math/perfect.html You're a freaking loon. *_PLONK_* Phil -- Unpatched IE vulnerability: history.back method caching Description: cross-domain scripting, cookie/data/identity theft, command execution Exploit: http://www.safecenter.net/liudieyu/RefBack/RefBack-MyPage.HTM ==== > For the real proof of the non-existence of Odd Perfect Numbers see: > http://www.bearnol.pwp.blueyonder.co.uk/Math/perfect.html > J Well, I looked. Got some questions: 1. p is prime, is it not? 2. why p n == 0 [mod 2]?!? Mateusz ==== > On Fri, 9 Jan 2004, Mark Griffith asked: > Anyone know of any recent work on the odd perfect number > question? Oddly enough, a paper claiming to prove the conjecture has > appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ > I haven't got the background to tell whether it's good or bad. It doesn't look very promising to me, although admittedly I've only skimmed through it. isn't relevant, or at least I didn't see in the paper any explanation of why it might be, and that doesn't inspire much confidence at the outset. (It could be he feels obliged to add it because his institution or department specializes in physics. Seems odd, but I guess that would account for it.) Also, in the paper itself, despite being reasonably formatted and having some sensible looking references, the proof itself is at the very least abominably badly presented. He launches straight into in intricate series of manipulations with no prior introduction or summary, and no explanation of what 'k' is in the first equation, for example. That in itself wouldn't be enough to condemn it out of hand; but just at a first glance there are even more worrying features. For example, throughout the proof, fractions pop up which it appears he may be assuming without justification are integers (unless that is taken care of in one of the references, but if so why not mention it?) It may all be impeccably correct; but for what it's worth I'd bet a fair amount that the whole thing is nonsense. John Ramsden ==== >> On Fri, 9 Jan 2004, Mark Griffith asked: >> Anyone know of any recent work on the odd perfect number >> question? >> >> Oddly enough, a paper claiming to prove the conjecture has >> appeared, dated Thu, 8 Jan 2004, at >> http://www.arxiv.org/abs/hep-th/0401052/ >> I haven't got the background to tell whether it's good or bad. It doesn't look very promising to me, although admittedly I've only > skimmed through it. isn't relevant, or at least I didn't see in the paper any explanation > of why it might be, and that doesn't inspire much confidence at the > outset. (It could be he feels obliged to add it because his institution or > department specializes in physics. Seems odd, but I guess that would > account for it.) It would also account for the loathsome practice (common amongst fizzisists) of omitting titles of papers in references [9, 12]. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > On Fri, 9 Jan 2004, Mark Griffith asked: > Anyone know of any recent work on the odd perfect number > question? > > Oddly enough, a paper claiming to prove the conjecture has > appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ > I haven't got the background to tell whether it's good or bad. If I correctly read what is written in that paper, the first error is around the top of the first page. I do not know whether 1.1 can be satisfied. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >On Fri, 9 Jan 2004, Mark Griffith asked: >Anyone know of any recent work on the odd perfect number >question? > > Oddly enough, a paper claiming to prove the conjecture has >appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ >I haven't got the background to tell whether it's good or bad. If I correctly read what is written in that paper, the first error is > around the top of the first page. I do not know whether 1.1 can be > satisfied. It's certainly one of the worst-written maths papers I've seen for a while. He really is most allergic to actually explaining his notations. :-( I nearly got to the end of section 1. Abstract. He might mention that it's easy to prove that an odd perfect number (indeed any odd number n with sigma(n) = 2 (mod 4)) has the form n = q^{4m+1} p_1^{2 r_1} ... p_s^{2 r_s} where q, p_1, ..., p_s are distinct primes, and q = 1 (mod 4). (He really should say that his 4k+1 is a *prime* distinct from any of his q_i s). Page 1. (1.1) is just a definition --- he's saying that if []/[] is put into lowest terms, it's a_i/b_i. (Here [] and [] are those gruesome fractions on either side which I won't copy). (1.2) is really an if and only if. As it stands it's just a complicated way of saying that N is *not* perfect. (This is a bit loopy: his section heading talks of one for a number being not perfect :-) ) I'm not sure why all those square roots are lying around... why didn't he write this as 2(4k+1)(b_1 ...b_l)/() []^{l+1}.... =/= ... ? (1.4) Now here his notation starts to get a bit gothic. He has things like a_{13}. I reckon that doesn't mean the thirteenth a_i but rather a_1 a_3. Also he has .Square . It took me a while to realize that .Square means times a square (presumably of a rational number). But realizing that, the formula looks even more batty. Why include brackets like (a_1 a_3 b_2/b_1 b_3 a_2) when he could have had (a_1 a_2 a_3/b_1 b_2 b_3) without any difference in meaning? After that he talks about these things, but in his notation he shoves overlines onto his subscripts ... Why? He then claims that some of these fractions aren't square multiples of 2(4k+1) referring to his previous paper ..... In (1.5) he introduces some notation and talks about fractions being square-free.... why doesn't he keep things simple, and represent these quantities as a square-free integer times a square of a rational? Next page... (1.7) lacks a parenthesis, but (1.6) through to (1.9) are unreadable as they stand. E.g., (1.9) contains rho-hat_{2i,2} which has never been defined. ((1.5) defines rho-hat_{3j+2}). At this stage, I wonder whether there is any worth trying to keep second-guessing this geezer as to what he means. Anyway my guess as to the import of section 1 is that he reckons with N of the form given in the abstract, sigma(N)/N can't even be twice a square of a rational and he reckons he's proved that in the sequel. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== >On Fri, 9 Jan 2004, Mark Griffith asked: >Anyone know of any recent work on the odd perfect number >question? > > Oddly enough, a paper claiming to prove the conjecture has >appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ >I haven't got the background to tell whether it's good or bad. If I correctly read what is written in that paper, the first error is > around the top of the first page. I do not know whether 1.1 can be > satisfied. What about A_i = (q_i-1)((4k+1)^(4m+2)-1) B_i = (4k) (q_i^(2^alpha_i+1)-1) g=gcd(A_i,B_i) a_i=A_i/g b_i=B_i/g ? However, I'm not putting any money on 1.2 being meaingful or not. Hmmm... I thought General Math was the kooks' hangout? I'm glad I normally surround myself with mathematical matters, as this physics stuff, particularly the stuff Simon Davis is into, looks pretty heavy: http://www.iop.org/EJ/S/UNREG/G6Lj4JvumVUVaKYOUtyVpw/abstract/-search=522350 5.2/0264-9381/18/17/305 <<< Abstract. String propagation in ten-dimensional Minkowski space or the direct product of Minkowski space and a six-dimensional K.8ahler manifold or orbifold might be regarded as an approximation to a theory which allows for the local curvature of spacetime by the energy-momenta of the component fields. String scattering in the interaction region might then be based on quantum field theory in a local region with a curved geometry. Special emphasis is given to field theory in anti-de Sitter space, as it represents a maximally symmetric spacetime of constant curvature which could arise in the description of matter interactions in local regions of spacetime. Curvature shifts in the momentum and squared mass are evaluated for scalar fields in anti-de Sitter space, and it is shown that the shift in p2 + m2 compensates the ground-state contribution to the bosonic string Hamiltonian, implying the consistency of computing the scattering entirely in flat space. Dual space rules for evaluating Feynman diagrams in Euclidean anti-de Sitter space are initially defined using eigenfunctions based on generalized plane waves. Loop integrals can be evaluated even more easily using momentum space rules in conformally flat coordinates for anti-de Sitter space, which admits flat three-dimensional sections that are analytic continuations of horospheres in hyperbolic space H4. An additional argument in favour of the model of string propagation described in this paper is based on the removal of reflective boundary conditions on quantum fields interacting in a locally anti-de Sitter region without spatial infinity, implying the existence of a one-parameter family of O(3,2)-invariant vacua in this region consistent with the degree of freedom in defining the string theory vacuum. > http://www.iop.org/EJ/abstract/0264-9381/11/5/007 <<< Abstract. The divergences that arise in the regularized partition function for closed bosonic string theory in flat space lead to three types of perturbation series expansions, distinguished by their genus dependence. This classification of infinities can be traced to geometrical characteristics of the string worldsheet. Some categories of divergences may be eliminated in string theories formulated on compact curved manifolds. > http://www.iop.org/EJ/S/UNREG/G6Lj4JvumVUVaKYOUtyVpw/abstract/-search=522350 5.1/0264-9381/20/13/331 <<< Abstract. The quantum cosmological wavefunction for a quadratic gravity theory derived from the heterotic string effective action is obtained near the inflationary epoch and during the initial Planck era. Neglecting derivatives with respect to the scalar field, the wavefunction would satisfy a third- order differential equation near the inflationary epoch which has a solution that is singular in the scale factor limit a(t) -> 0. When scalar field derivatives are included, a sixth-order differential equation is obtained for the wavefunction and the solution by Mellin transform is regular in the a 0 limit. It follows that inclusion of the scalar field in the quadratic gravity action is necessary for consistency of the quantum cosmology of the theory at very early times. > He seems in a very specialist field - quantum cosmological wavefunction yields only 19 hits in google, of which 18 pertain to Simon Davis' own papers. And what is the research foundation of southern california? Google's not really heard of it except for discussions of whether it exists or not in the context of tracing the author of another paper (by a B. Davis, rather than S. Davis). Weird. Whatever. Phil -- Unpatched IE vulnerability: history.back method caching Description: cross-domain scripting, cookie/data/identity theft, command execution Exploit: http://www.safecenter.net/liudieyu/RefBack/RefBack-MyPage.HTM ==== Hmmm... I thought General Math was the kooks' hangout? I'm glad I normally surround myself with mathematical matters, as this > physics stuff, particularly the stuff Simon Davis is into, looks > pretty heavy: > He seems in a very specialist field - quantum cosmological wavefunction > yields only 19 hits in google, of which 18 pertain to Simon Davis' own > papers. And what is the research foundation of southern california? > Google's not really heard of it except for discussions of whether > it exists or not in the context of tracing the author of another > paper (by a B. Davis, rather than S. Davis). Davis has 21 items on MathSciNet, all on stringy stuff apart from his cited paper whose review I append. MR1979400 (2004b:11007) Davis, Simon(5-SYD-SM) A rationality condition for the existence of odd perfect numbers. (English. English summary) 11A25 (11B39 11D41 11D61) Review in linked PDF Add citation to clipboard Document Delivery Service Journal Original Article More links More links If an odd perfect number exists, it must exceed $10^{300}$ ref[R. P. Brent, G. L. Cohen and H. J. J. te Riele, Math. Comp. 57 (1991), no. 196, 857--868; MR 92c:11004]. The author observes that an odd integer $N$ can be perfect only if a certain product depending on its prime divisors has a rational square root. This product contains factors which are repunits, that is, of the form $(q^n-1)/(q-1)$, for the primes $q$ appearing to an even power in the canonical decomposition of $N$. By using this rationality condition and properties of repunits the author studies the existence of odd perfect numbers $N$. He obtains an upper bound for the density of such $N$ in any fixed interval (above $10^{300}$) and proves the nonexistence of $N$ in some special classes of integers. An important role in his discussions is played by various Diophantine equations. Reviewed by T. Mets?nkyl? The maths department at Sydney University is certainly reputable. The journal Int. J. Math. Math. Sci. is certainly legit (they did once publish a paper by me, albeit a potboiler even by my standards :-) ) A couple of the reviews of his stringy stuff include some barbed comments: corresponding style. ... The mathematical definitions of the above terms previous work ... ... The author shows the modular covariance of the volume form on Siegel's upper half-space. (This is of course a well-known mathematical fact.) ... -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== >> Ha. You don't know what you are talking about. Poincare conjecture? >> And even Goldbach is not recursive, but recursively enumerable. To >> stay in simple number theory : there exist an infinity of twin >> primes is not even recursively enumerable (nor corecursively >> enumerable, of course). You didn't answer that So if I make a TM that for input x=0, tries to asnwer the yes/no question, if there is an infinity of twin primes, you say it doesn't halt and you can't prove it, but the point is, that maybe it does halt, you just don't know, and currently can't prove neither. >> How do you know? Undecidable problems exists. What about the >> Continuum hypothesis? You didn't answer that either. sorry, don't know about them. > No. This time, you don't understand the rules (and are trying to disprove > G.9adel in some sense) What is outside the *whole* system of proofs? ok you are right. let me formalize your argument in my words, so I will understand better. Bascially, you say, that mathematics has more True claims, that it can prove(by Godel). Since mathemaics captures the notion of algorithm(?) as we know it, and if we take the Church thesis, TM is equivalent to algorithms, there are problems (sets of numbers) or there are yes/no quesions, that TM will never be able to answer. That is the TM may halt or not halt, but we don't know it. ==== >> Ha. You don't know what you are talking about. Poincare conjecture? >> And even Goldbach is not recursive, but recursively enumerable. To >> stay in simple number theory : there exist an infinity of twin >> primes is not even recursively enumerable (nor corecursively >> enumerable, of course). >> You didn't answer that So if I make a TM that for input x=0, tries to asnwer the yes/no > question, How on earth do you do that? The meaning of my sentence is that the brden is on you to make such a TM, and it is not possible by enumerating anything... > if there is an infinity of twin primes, you say it doesn't halt and > you can't prove it, but the point is, that maybe it does halt, you > just don't know, and currently can't prove neither. >> How do you know? Undecidable problems exists. What about the >> Continuum hypothesis? >> You didn't answer that either. sorry, don't know about them. Time to learn some maths? > No. This time, you don't understand the rules (and are trying to >> disprove G.9adel in some sense) What is outside the *whole* system >> of proofs? ok you are right. let me formalize your argument in my words, so I > will understand better. Bascially, you say, that mathematics has more > True claims, > that it can prove(by Godel). Since mathemaics captures the notion of > algorithm(?) as we know it, and if we take the Church thesis, TM is > equivalent to algorithms, there are problems (sets of numbers) or > there are yes/no quesions, > that TM will never be able to answer. That is the TM may halt or not > halt, > but we don't know it. Mmmm... Something like that. But actually mathematics capture much more than algorithms. How do you propose to capture some truth like the pythagore theorem by some TM? Before Godel codes, no one had any idea about that... ==== Grab this perl script: http://www.farviolet.com/~entropy/decompose.txt Here is what it does, don't know if this is any use, but it has an interestingly odd result. Recursive decomposition of prime numbers, for the purpose of this discussion 0 is counted as a prime #. Any number 0 - N can be represented in the form: 0^a * 2^b * 3^c * 5^d * 7^e ... Or more simply the array: (a, b, c, d, e, ...) For example: 1 -> (0 0 0 0 0 ...) 0 -> (1 0 0 0 0 ...) 12 -> (0 2 1 0 0 0 ...) 16 -> (0 4 0 0 0 ...) Or to simplify, leave off trailing zeros: 1 -> () 0 -> (1) 12 -> (0 2 1) 16 -> (0 4) Now the recursive part, represent the powers in the array the same way: 1 -> () 0 -> (1) -> (()) 12 -> (0 2 1) -> ((1) (0 1) ()) -> ((())((())())()) 16 -> (0 4) -> (0 (0 2)) -> ((())((())((())()))) Your left with the ability to represent any number as a set of grouping symbols after repeated recursive prime decomposition. Don't know if this is of any use at all, but its neat :) Notice that for the numbers 0 and 1, the expression '0^n' is equivalent to 'NOT n'. -- Laughter-Confusion, Pleasure-Pain, Happyness-Sadness, Excitement-Fear, Love-Hate, etc. The true primary emotions, a modifier makes each into two. This modifier is acceptance/unacceptance. Let go, surrender, accept... Be a counter terrorist perpetrate random senseless acts of kindness ==== > Grab this perl script: http://www.farviolet.com/~entropy/decompose.txt Here is what it does, don't know if this is any use, but it has an > interestingly odd result. Recursive decomposition of prime numbers, for the purpose of this > discussion > 0 is counted as a prime #. Any number 0 - N can be represented in the form: 0^a * 2^b * 3^c * 5^d * 7^e ... Or more simply the array: (a, b, c, d, e, ...) For example: 1 -> (0 0 0 0 0 ...) > 0 -> (1 0 0 0 0 ...) > 12 -> (0 2 1 0 0 0 ...) > 16 -> (0 4 0 0 0 ...) Or to simplify, leave off trailing zeros: 1 -> () > 0 -> (1) > 12 -> (0 2 1) > 16 -> (0 4) Now the recursive part, represent the powers in the array the same way: 1 -> () > 0 -> (1) -> (()) > 12 -> (0 2 1) -> ((1) (0 1) ()) -> ((())((())())()) > 16 -> (0 4) -> (0 (0 2)) -> ((())((())((())()))) Your left with the ability to represent any number as a set of grouping > symbols after repeated recursive prime decomposition. Don't know if this > is of any use at all, but its neat :) Notice that for the numbers 0 and 1, the expression '0^n' is equivalent to > 'NOT n'. I myself posted something along these lines myself recently (amongst the cr**) at: http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fdf .0312061711.15da0b01%40posting.google.com&rnum=6&prev= Your contribution was (I gather by introducing the zero) to do away with the primes (and 1), leaving just parentheses. At the very least, this is an easy proof that the set of lists of parentheses grouped this way has a cardinality of aleph-null (because of the bijection with the integers). thanks, Leroy Quet ==== > Grab this perl script: http://www.farviolet.com/~entropy/decompose.txt GP script into tree form: e(n)=local(s,p,c,q);Str((,if(n==0,(),if(n==1,,s=(());forprim e(p=2,99999,c=0;while(n%p==0,c++;n/=p);s=Str(s,e(c));if(n==1,break));s)),) ) > Here is what it does, don't know if this is any use, but it has an > interestingly odd result. ... > Now the recursive part, represent the powers in the array the same way: 1 -> () > 0 -> (1) -> (()) > 12 -> (0 2 1) -> ((1) (0 1) ()) -> ((())((())())()) > 16 -> (0 4) -> (0 (0 2)) -> ((())((())((())()))) Your left with the ability to represent any number as a set of grouping > symbols after repeated recursive prime decomposition. Don't know if this > is of any use at all, but its neat :) It is indeed neat. It appeals to the Lisp programmer in me! Whod have thought that the numbers 5 ((())(())(())()) 12 ((())((())())()) 18 ((())()((())())) 42 ((())()()(())()) 64 ((())((())()())) 70 ((())()(())()()) 105 ((())(())()()()) 2310 ((())()()()()()) would have had something in common. Phil -- Unpatched IE vulnerability: dragDrop invocation Description: Arbitrary local file reading through native Windows dragDrop invocation. Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0302/12.html Exploit: http://kuperus.xs4all.nl/security/ie/xfiles.htm ==== I have this homework question that I've been struggling for quite a while. I hope someone will give me some hint on how to do this proof. For a, b natural numbers, consider the set of numbers ar + bs for all integers r, s so that ar + bs >= 1. Since this set is nonempty, by well-ordering it has a least element Show that the least element of this set is the greatest common disvisor of a and b. First of all, i can tell that (a, b) exists in the set because (a, b) = ar + bs for some r, s. Now, how can I prove it has to be the least element? Gavin ==== >I have this homework question that I've been struggling for quite a while. I >hope someone will give me some hint on how to do this proof. For a, b natural numbers, consider the set of numbers ar + bs for all >integers r, s so that ar + bs >= 1. Since this set is nonempty, by >well-ordering it has a least element Show that the least element of this set >is the greatest common disvisor of a and b. Let c be the least positive element of {ar+bs : r,s in Z}. 1. Show (a,b) | c (remember that (a,b) | a and (a,b) | b). 2. Show c | a and c | b (that is, c is a common divisor of a and b). If all else fails, see http://www.whim.org/nebula/math/bezout.html >First of all, i can tell that (a, b) exists in the set because (a, b) = ar + >bs for some r, s. Now, how can I prove it has to be the least element? I could be wrong here, but isn't one of the things you are trying to prove that (a,b) = ar+bs for some integers r and s? If so, you are not allowed to use this fact in your proof. Rob Johnson take out the trash before replying ==== > Are the following proofs correct? 2) Let R be the ring of all real-valued functions of > a single variable under pointwise addition and multiplication. > The subset S of R of functions whose graphs pass through > the origin forms a subring of R. Prove S is a subring of R. > f in R passes thru the origin when f(0) = 0. Thus for f,g in S, f(0) = g(0) = 0 and (f+g)(0) = f(0) + g(0) = 0 (-f)(0) = -f(0) = 0 (f*g)(0) = f(0) * g(0) = 0 showing f+g, -f and f*g are in S. Does your text require rings to have identities? Mind doesn't. > Proof: > It suffices to show that S is closed under substraction and > multiplication. > Clearly S is non-empty since f(x)=x is in S. Ok. So is the additive identity f(x) = 0. > So assume f(x)=x*h(x) and g(x)=x*z(x) for some h(x) > and z(x) in R[x]. Just a nanosec. R[x] isn't the ring of real functions, it's the ring of polynomials with real coefficients. > But then f(x)-h(x)=x[h(x)-z(x)]=x*q(x) hence x=0 is a root > thus S is closed under substraction. > Similarly f(x)*h(x)=x^2*h(x)*z(x)=x*p(x) thus S is closed under Whycramequationstogetherwithoutspaces? > multiplication and by the subring test this implies that S > is a subring of R. > Makes no sense in the context of ring of real functions. ==== I was wondering if anyone knows how to set restrictions on integers or simply how to generate the 8 integers that satisfy the following two conditions: 1) a8^2 + a1^2 = a5^2 + a4^2 = a7^2 + a2^2 = a6^2 + a3^2 2) (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 or to put it another way: 1) a8^2 + a1^2 = x a5^2 + a4^2 = x a7^2 + a2^2 = x a6^2 + a3^2 = x 2) (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = y (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 = y [No relation between x and y] [0 < a1 < a2 < a3 < a4 < a5 < a6 < a7 < a8] Now, I know how to solve for the first condition, but I'm at a loss on how to mathematically generate a set of 8 integers that satisfy both conditions simultaneously. I've used brute force methods and have found two sets of 8 integers that satisfy the above two conditions. But I would like to know if I can _generate_ more (instead of searching for more). Or, maybe these are the only two solutions? The two sets of solutions are: {a1, a2, a3, a4, a5, a6, a7, a8} {11, 77, 101, 131, 343, 353, 359, 367} {139, 317, 541, 719, 827, 953, 1049, 1087} I would greatly appreciate any help that anyone can offer in this -David C. ==== ... > I was wondering if anyone knows how to set restrictions on integers or > simply how to generate the 8 integers that satisfy the following two > conditions: > 1) > a8^2 + a1^2 = a5^2 + a4^2 = a7^2 + a2^2 = a6^2 + a3^2 > 2) > (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 ... > Now, I know how to solve for the first condition, but I'm at a loss on > how to mathematically generate a set of 8 integers that satisfy both > conditions simultaneously. I've used brute force methods and have found > two sets of 8 integers that satisfy the above two conditions. But I > would like to know if I can _generate_ more (instead of searching for > more). Or, maybe these are the only two solutions? The two sets of solutions are: > {a1, a2, a3, a4, a5, a6, a7, a8} > {11, 77, 101, 131, 343, 353, 359, 367} > {139, 317, 541, 719, 827, 953, 1049, 1087} I presume you refer to some results of Jacobi for the first condition; for the second condition I don't know what to do either. Let x = a1^2 + a8^2. Regarding number of solutions, there are infinitely many because integer multiples of a solution also are solutions. The following shows all solutions that are not integer multiples of other solutions, for x < 25000000, per a program that takes about a minute on a 750 MHz machine. x 67405: { 106, 126, 141, 178, 189, 218, 227, 237} 134810: { 11, 77, 101, 131, 343, 353, 359, 367} 600445: { 54, 206, 366, 474, 613, 683, 747, 773} 1200890: { 139, 317, 541, 719, 827, 953, 1049, 1087} 5915065: { 141, 717, 916, 1267, 2076, 2253, 2324, 2428} 11830130: { 809, 1337, 1607, 2287, 2569, 3041, 3169, 3343} -jiw ==== I presume you refer to some results of Jacobi for the first > condition; for the second condition I don't know what to do either. Well, I don't know any Jacobi results. I was using the fact that you can multiply k gaussian primes together to generate 2^(k-1) complex numbers all of whose norm (not sure if norm is right, what i mean is a+bi then norm = a^2 + b^2) is equal. > Let x = a1^2 + a8^2. Regarding number of solutions, there are > infinitely many because integer multiples of a solution also are > solutions. The following shows all solutions that are not integer > multiples of other solutions, for x < 25000000, per a program that > takes about a minute on a 750 MHz machine. > x > 67405: { 106, 126, 141, 178, 189, 218, 227, 237} > 134810: { 11, 77, 101, 131, 343, 353, 359, 367} 600445: { 54, 206, 366, 474, 613, 683, 747, 773} > 1200890: { 139, 317, 541, 719, 827, 953, 1049, 1087} 5915065: { 141, 717, 916, 1267, 2076, 2253, 2324, 2428} > 11830130: { 809, 1337, 1607, 2287, 2569, 3041, 3169, 3343} -jiw WOW! How did you get a program to run that fast. It took mine about 10 minutes to find the first two and it couldn't find any after that. Did you search for x and then search for the 8 integers? If not, what algorithm did you use? Could you send me the source? I was looking at even x because I'm doing research in which I need x to be even. Pretty soon I'm going to start a search for a set of 16 integers which meet even stricter requirements. Hopefully, with the help you've given, I can figure out a methodic way to come up with these -David C. ==== ... > Let x = a1^2 + a8^2. Regarding number of solutions, there are > infinitely many because integer multiples of a solution also are > solutions. The following shows all solutions that are not integer > multiples of other solutions, for x < 25000000, per a program that > takes about a minute on a 750 MHz machine. ... > WOW! How did you get a program to run that fast. It took mine about 10 > minutes to find the first two and it couldn't find any after that. Did > you search for x and then search for the 8 integers? If not, what > algorithm did you use? Could you send me the source? ... See http://pat7.com/jp/sqsum.c . If you change #define N 25000000 to #define N 2500000 the program takes about 2 seconds to display all the solutions with x less than 2.5 million. The program is quite straightforward - first it computes and counts all the values of i^2+j^2 less than N with i<=j, then computes them again and saves the decompositions for those n that have 4 or more representations as i^2+j^2, then for each such n tests all subsets of 4 decompositions for condition 2. -jiw ==== I don't have any specific solutions, but I can put the problem into a general context for you. You really have 4 equations in 8 unknowns: a8^2 + a1^2 = a5^2 + a4^2 a5^2 + a4^2 = a7^2 + a2^2 a7^2 + a2^2 = a6^2 + a3^2 (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 There are three equations that are homogeneous of degree 2, and the other one is homogeneous of degree 4. They define a variety V in projective 7-space, and since the four equations are independent, your variety V has dimension 3. Your integer solutions correspond to rational points on the projective variety V. Your variety V is probably nonsingular(?), though I haven't checked. Let's assume for the moment that it is nonsingular. Then its canonical sheaf is O(-7-1+2+2+2+4) = O(2). (In general, if you have a smooth variety in projective n-space given by the intersection of homogeneous polynomials of degree d_1,d_2,...,d_r, then its canonical sheaf is O(-n-1+d_1+...+d_r). Of course, I'm assuming that it has dimension n-r.) Anyway, the fact that V has canonical sheaf O(2) means that it is a variety of general type. There is a famous conjecture of Bombieri and Lang saying that the rational points on a variety of general type are not Zariski dense. In less fancy language, what that means for your system of equations is that there is a homogeneous polynomial F(a1,a2,...,a8) which is independent of your equations so that every integer solution of your equations is also a solution to F=0. Unfortunately, the Bombieri-Lang conjecture is still just a conjecture, and in any case, even conjecturally it gives no practical procedure for finding the polynomial F. OTOH, it does suggest that the non-obvious solutions to your equations are likely to be few and far between. N.B. All of this is contingent on your equations defining a nonsingular variety. If they don't, then all of this may not apply. Finally, on a positive note, the way that people typically produce lots of integer solutions to equations like this (when they do manage) is by finding an elliptic curve of positive rank that sits on the variety. That's how Elkies solved the famous problem of sums of fourth powers; but Elkies variety had trivial canonical sheaf, so it wasn't of general type. Joe Silverman I was wondering if anyone knows how to set restrictions on integers or > simply how to generate the 8 integers that satisfy the following two > conditions: > 1) > a8^2 + a1^2 = a5^2 + a4^2 = a7^2 + a2^2 = a6^2 + a3^2 > 2) > (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 > > two sets of 8 integers that satisfy the above two conditions. But I > would like to know if I can _generate_ more (instead of searching for > more). Or, maybe these are the only two solutions? The two sets of solutions are: > {a1, a2, a3, a4, a5, a6, a7, a8} > {11, 77, 101, 131, 343, 353, 359, 367} > {139, 317, 541, 719, 827, 953, 1049, 1087} I would greatly appreciate any help that anyone can offer in this -David C. ==== > >Oh, yes. I got my instructions, script to enter in this new character >on universal stage. I have done lot of rehearsal in real life and I >am aware of my melodramatic tragedy. That is why I am shouting that, >those rehearsals were setup, a plot. I was controlled to do those >rehearsals in real life. > >In those rehearsals, I never tried to commit suicide. This is why I >can't die. So I will have to play this character. You have probably seriously considered suicide. Good thing you didn't try, though. It might have worked, and then you'd have found out that you *can* die. Of course you can! I know you're the chosen one and all, but everybody dies. > >Whatever I do, it is already scripted anyway. No it isn't. You're making it up as you go along. You may not be consciously aware of the process, but you are making it up. I realize the convenience of thinking that you are without personal will and accountability - an instrument of a higher power - but it is a delusion. Your genius has been mocked, so now you feign a sudden transformation and promise doom for all. Sorry, but I don't think anyone is buying it. More than likely, you are psychotic. Seek the kind of help that involves medication. > >-Abhi. > > Why is it always about you? Because I am Zero, feeling of zeroness, cause of creation of this universe. Clever. So, because you feel like a zero, you are essentially an omnipotent being? And now you are tasked to undo the universe, right? Do you realize how common it is for megalomania to manifest in people with catastrophically low self esteem? It is equally common for persons under such delusions of grandeur to make threats of annihilation towards everyone who ever doubted their greatness. So far, you fit the profile very well. Because I am truth. No. You are deluded. It's probably pointless to try to help you, but you really should seek help. You may be a danger to yourself and others. Unless, of course, you are just acting all weird on usenet without anything being wrong with you.... If you really believe the things you have been writing, you are mentally ill. ==== >Oh, yes. I got my instructions, script to enter in this new character >on universal stage. I have done lot of rehearsal in real life and I >am aware of my melodramatic tragedy. That is why I am shouting that, >those rehearsals were setup, a plot. I was controlled to do those >rehearsals in real life. > >In those rehearsals, I never tried to commit suicide. This is why I >can't die. So I will have to play this character. You have probably seriously considered suicide. > Good thing you didn't try, though. It might have worked, and then you'd have > found out that you *can* die. Of course you can! > I know you're the chosen one and all, but everybody dies. >Whatever I do, it is already scripted anyway. No it isn't. You're making it up as you go along. You may not be consciously > aware of the process, but you are making it up. > I realize the convenience of thinking that you are without personal will and > accountability - an instrument of a higher power - but it is a delusion. > Your genius has been mocked, so now you feign a sudden transformation and > promise doom for all. Sorry, but I don't think anyone is buying it. > More than likely, you are psychotic. Seek the kind of help that involves > medication. >-Abhi. >Why is it always about you? >Because I am Zero, feeling of zeroness, cause of creation of this > universe. Clever. So, because you feel like a zero, you are essentially an omnipotent > being? And now you are tasked to undo the universe, right? > Do you realize how common it is for megalomania to manifest in people with > catastrophically low self esteem? > It is equally common for persons under such delusions of grandeur to make > threats of annihilation towards everyone who ever doubted their greatness. > So far, you fit the profile very well. > Because I am truth. No. You are deluded. > It's probably pointless to try to help you, but you really should seek help. > You may be a danger to yourself and others. > Unless, of course, you are just acting all weird on usenet without anything > being wrong with you.... > If you really believe the things you have been writing, you are mentally > ill. To all your queries like, mentally ill, psychotic, schizophrenia, megalomania etc, I replied on 22 November. But I am finding that you never gave any response to that post. That post is given below. --------------------------------------------------- > You know, you really should pay heed to that post on schizophrenia. Paranoid > schizophrenics often think that really big events are set in motion just for > them. It seems to me that you exhibit that kind of megalomania. He also trained me to what to do in such circumstances. He trained me to execute Logic. If switch is ON, it just can't be OFF at exactly same moment. Yes # No. I am talking about generation of unidirectional force which will change course of history, physics and will open gateway to entire universe in future. Now we make a statement that this device does work. (1)Above statement is TRUE statement. Then looking at the simplicity of this device, any person who know physics can see that mechanism of this device does not need my theory. I have not talked a single word about my theory in this mechanism. It can be explained on the basis of Newtonian mechanics and Pythagoras Theorem. But millions of scientists, physicists could not think of it in at least last 350 years. Only conclusion, I can draw is that thinking ability of millions of people in last 350 years was controlled by someone. If we look further closely, we find that this device is so simple that it could have been discovered accidently by someone who does not know even basics of physics. And this accident could have happened at any time in last 5000 years. But it never happened. Only conclusion, I can draw is that actions of billions of people in last 5000 years was controlled by someone for specific purpose. And if that someone can control the actions of billions of people in last 5000 years, He can also control my actions, my mind, absolutely every event happened around me from the moment I was born. Is there any flaw in this logic, Laura? (2) Above statement is FALSE statement. Then Laura, you are right. I am suffering from Paranoid schizophrenia. TRUTH, what is that and where is it..... -Abhi. --------------------------------------------------- Laura, you never replied to my above post. this Time Theory and Action Device), only way to show me that I am show me and all of you some miracles. BTW, I have filed 7 patent application and I have sent no.8 by ground make it. Because if I make it and it works, then it means that the statement I made in above post is TRUE and all the logic which follows it, is also TRUE. Those lights were not my delusion. It was God. In such situation, I will not have any moral right to hide this device in my room and prepare patent application, file it etc. No, I must disclose it to mankind immediately because I have seen God in action through this device. [And as soon as I disclose it to media, it destroys absolute novelty part of patent rules!] Till this date, I have not prepared or even tried to prepare this device. I have sent my patent application no.8 and now what is holding me back, apart from my personal problems, is that all these 8 applications might not be enough to secure this device. But I am going to prepare this device now and disclose it to media even if it is not secured. I have tried my best in last 6 and half months. That's All. -Abhi. ==== > He also trained me to what to do in such circumstances. He trained me > to execute Logic. If switch is ON, it just can't be OFF at > exactly same moment. Yes # No. I am talking about generation of unidirectional force which will > change course of history, physics and will open gateway to entire > universe in future. Now we make a statement that this device does work. (1)Above statement is TRUE statement. Then looking at the simplicity of this device, any person who know > physics can see that mechanism of this device does not need my theory. Here's a logical mistake: The simplicity of the device does not depend on it working. In fact, from your drawings any person who know physics can see that it does not work. And anyone who has posted here has seen it. So, the two possibilities I see are: - The device does not work. - The device does work and looks different from your drawings. I hesitate to link device work not to deluded because it's obvious anyway that you are deluded by you seeing yourself as vastly important which is a classical symptom. There are lots of people who discovered that people did something wrong all along but they never bragged on about them being specially chosen. So, build it and post a picture (please not more drawings!). > I have not talked a single word about my theory in this mechanism. It > can be explained on the basis of Newtonian mechanics and Pythagoras > Theorem. But millions of scientists, physicists could not think of it > in at least last 350 years. Only conclusion, I can draw is that > thinking ability of millions of people in last 350 years was > controlled by someone. Wrong logic, see above, wrong conclusion. Very simple. > Laura, you never replied to my above post. this Time Theory and Action Device), Well, have you? What's it look like? Jesus there or Echnaton? > BTW, I have filed 7 patent application and I have sent no.8 by ground > make it. Because if I make it and it works, then it means that the > statement I made in above post is TRUE and all the logic which follows > it, is also TRUE. Those lights were not my delusion. It was God. > In such situation, I will not have any moral right to hide this > device in my room and prepare patent application, file it etc. No, I > must disclose it to mankind immediately because I have seen God in > action through this device. [And as soon as I disclose it to media, it > destroys absolute novelty part of patent rules!] Logical mistake number two: Why are you always posting long explanations of how and why it has to work? Right now I could take any of those, build it and sell it. If you see usenet as medium it *has* been disclosed by now. eetimes or so about some funny guy in sci.astro and you'd have your disclosure. Till this date, I have not prepared or even tried to prepare this > device. I have sent my patent application no.8 and now what is holding > me back, apart from my personal problems, is that all these 8 > applications might not be enough to secure this device. So, you have filed seven patents? You should have gotten reference numbers for this, right? What are they? But I am going to prepare this device now and disclose it to media > even if it is not secured. I have tried my best in last 6 and half > months. Right. If it shows up in the news, post a link, please? Lots of Greetings! Volker ==== > He also trained me to what to do in such circumstances. He trained me > to execute Logic. If switch is ON, it just can't be OFF at > exactly same moment. Yes # No. I am talking about generation of unidirectional force which will > change course of history, physics and will open gateway to entire > universe in future. Now we make a statement that this device does work. (1)Above statement is TRUE statement. Then looking at the simplicity of this device, any person who know > physics can see that mechanism of this device does not need my theory. > Here's a logical mistake: > The simplicity of the device does not depend on it working. V-shaped streched spring is not simple!!! [snip] > Logical mistake number two: Why are you always posting long explanations > of how and why it has to work? Right now I could take any of those, > build it and sell it. Right. Answer is circumstances. Never underestimate awesome power of circumstances. It can crash you and make very simple task impossible. [snip] > Till this date, I have not prepared or even tried to prepare this > device. I have sent my patent application no.8 and now what is holding > me back, apart from my personal problems, is that all these 8 > applications might not be enough to secure this device. > So, you have filed seven patents? You should have gotten reference > numbers for this, right? What are they? (8) Unknown -Abhi. ==== > I am going crazy trying to find a formula/equation for the >equivalent resistance asross diagonally opposite ends of a >resistive mesh. Can someone help. X > o--+--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+ > | | | | | | | > Rv Rv Rv | | Rv Rv > | | | | | | | > +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+ > | | | | | | | > Rv Rv Rv | | Rv Rv > | | | | | | | > . . . .(m rows) .--Rh--+--Rh--+ > | | | | | | | > Rv Rv Rv | | Rv Rv > | | | | | | | > +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+ > | | | | | | | > Rv Rv Rv | | Rv Rv > | | | | | | | > +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+--o Y > > Imagine a rectangular mesh with resistors at every small >segment. The horizontal segment resistances are Rh and >vertical ones are Rv. Need to get a formula to calculate >the effective resistance between X and Y. Any help welcome! > I do have a method for solving this. Very nice problem. The bad news is, that my method is rather involved and I do not have much time available at the moment. So, I will write it up and post it, but it will take another week or so. I hope you can manage from going totally crazy for that long. Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de ==== f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f'(x)dx 0~1 decide that this intergral of f'(x) is exist or not exist?? -------------------- i think..... f'(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) thus f'(x) is conti except x=0 i can't progress any more about it i wait your response ==== > f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f'(x)dx > 0~1 decide that this intergral of f'(x) is exist or not exist?? -------------------- i think..... f'(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) It's easy to see f' is not bounded on (0,1) (look along the sequence 1/sqrt(2nPi)), therefore it is not Riemann integrable on [0,1]. Does int_[0,1] f'(x) dx exist as an improper Riemann integral? Sure. Let a > 0. Then int_[a,1] f'(x)dx = f(1) - f(a), by the Fundamental Theorem of Calculus. But f is continuous at 0, so the last term has a limit as a -> 0+. ==== > f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f'(x)dx > 0~1 decide that this intergral of f'(x) is exist or not exist?? -------------------- i think..... f'(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) thus f'(x) is conti except x=0 i can't progress any more about it i wait your response f' is unbounded in every interval (0,eps), thus as a Riemann integral, int_0^1 f'(x) dx is at best improper (at worst, doesn't exist). Since f' is continuous on [eps,1], we have f(1) - f(eps) = int_eps^1 f'(x) dx. But f is continuous, hence as eps decreases to 0, f(1) - f(0) = int_0^1 f'(x) dx (the right-hand side being BY DEFINITION the limit as eps decreases to 0 of the earlier RHS). As Rob Johnson points out, f' is not Lebesgue integrable. --Ron Bruck ==== > f:[0,1] -> R >> >> f(x) = (x^2)sin(1/x^2) if x != 0 >> >> 0 if x = 0 >> >> int f'(x)dx >> 0~1 >> >> decide that this intergral of f'(x) is exist or not exist?? >> >> -------------------- >> >> i think..... >> >> f'(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) >> >> thus f'(x) is conti except x=0 >> >> i can't progress any more about it >> >> i wait your response f' is unbounded in every interval (0,eps), thus as a Riemann integral, >int_0^1 f'(x) dx is at best improper (at worst, doesn't exist). Since >f' is continuous on [eps,1], we have f(1) - f(eps) = int_eps^1 f'(x) dx. But f is continuous, hence as eps decreases to 0, f(1) - f(0) = int_0^1 f'(x) dx (the right-hand side being BY DEFINITION the limit as eps decreases to >0 of the earlier RHS). As Rob Johnson points out, f' is not Lebesgue integrable. Doh! I checked that the OP got f'(x) correct and then followed them over the edge of the cliff, ignoring the sign saying Fundamental Theorem of Calculus: go no further. Rob Johnson take out the trash before replying ==== > f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f'(x)dx > 0~1 decide that this intergral of f'(x) is exist or not exist?? -------------------- i think..... f'(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) calculation error: ^ should be x^2*(-2/x)cos(1/x^2) thus f'(x) is conti except x=0 ^ possibly But now you can easily check for continuity at x=0. Jon Miller ==== En el mensaje:m7idnTGF7MvKVJvd4p2dnA@comcast.com, Jonathan Miller escribi.97: >> f:[0,1] -> R >> f(x) = (x^2)sin(1/x^2) if x != 0 >> 0 if x = 0 >> int f'(x)dx >> 0~1 >> decide that this intergral of f'(x) is exist or not exist?? >> -------------------- >> i think..... >> f'(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) calculation error: ^ should be x^2*(-2/x)cos(1/x^2) >> Not should be x^2*(-2/x^3)cos(1/x^2) = - (2/x)cos(1/x^2), as the OP write ... -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== > En el mensaje:m7idnTGF7MvKVJvd4p2dnA@comcast.com, > Jonathan Miller escribi.97: >> f:[0,1] -> R >> f(x) = (x^2)sin(1/x^2) if x != 0 >> 0 if x = 0 >> int f'(x)dx >> 0~1 >> decide that this intergral of f'(x) is exist or not exist?? >> -------------------- >> i think..... >> f'(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) >calculation error: ^ should be x^2*(-2/x)cos(1/x^2) Not should be x^2*(-2/x^3)cos(1/x^2) = - (2/x)cos(1/x^2), as the OP write Duh. That's the second time in a week I've made a stupid mistake because I wouldn't take the time to write down what I was thinking. Must be getting old. Move on folks. Nothing to see here, just a guy whose brain is wandering away from him. Jon Miller ==== >f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f'(x)dx >0~1 decide that this intergral of f'(x) is exist or not exist?? -------------------- i think..... f'(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) thus f'(x) is conti except x=0 2x sin(1/x^2) is continuous in [0,1] so it is integrable. Let u = 1/x^2 so that du/u = -2 dx/x. Then |1 2 1 | - cos( --- ) dx | 0 x x^2 |oo 1 = | - cos(u) du [1] | 1 u For each integer k >= 0, |5pi/4 + k pi 1 | | - cos(u) | du | 3pi/4 + k pi u pi/2 1 >= ------------ ------- 5pi/4 + k pi sqrt(2) sqrt(2) = ------- [2] 5+4k Therefore, |oo 1 | | - cos(u) | du | 1 u oo --- sqrt(2) >= > ------- [3] --- 5+4k k=0 and [3] diverges by comparison to the harmonic series. Thus, [1] does not exist as a Lebesgue integral. As a Riemann integral, [1] |oo 1 = | - d sin(u) | 1 u |oo 1 = -sin(1) + | --- sin(u) du [4] | 1 u^2 The integral in [4] converges absolutely. Therefore, [1] exists as a Riemann integral but not a Lebesgue integral. Rob Johnson take out the trash before replying ==== Suppose you have a -A and a A V B operator defined, and A ^ B = -(-A V -B) as usual, and commutative and associative law hold for ^ and V, and --A=A, and 0,1 exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean and DeMorgan, only the distributive law fails. Does this structure already have a name? -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de als man ankam wollte man werden, die geschichte schreiben, die doofen sollen sterben, der plan als man damals nach hamburg kam (Kettcar) ==== > Suppose you have a -A and a A V B operator defined, > and A ^ B = -(-A V -B) as usual, and commutative and > associative law hold for ^ and V, and --A=A, and 0,1 > exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean > and DeMorgan, only the distributive law fails. Does this structure already have a name? > A non distributive complemented lattice. For example the complete atomic (not uniquely) complemented non-modular lattice of topologies of a set. cf A.K.Steiner (1966) 'The Lattice of Topologies' ==== > Suppose you have a -A and a A V B operator defined, > and A ^ B = -(-A V -B) as usual, and commutative and > associative law hold for ^ and V, and --A=A, and 0,1 > exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean > and DeMorgan, only the distributive law fails. I guess complemented Lattice, satisfying the DeMorgan identities is not exactly, what you want. Have you already looked through Birkhoffs Lattice theory? Marc ==== * Hauke Reddmann > Suppose you have a -A and a A V B operator defined, > and A ^ B = -(-A V -B) as usual, and commutative and > associative law hold for ^ and V, and --A=A, and 0,1 > exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean > and DeMorgan, only the distributive law fails. Does this structure already have a name? Beats me, but do you have an example? A smallest set that satisfies your requirements? Should be interesting enough. It is not a ring, so it avoids any elementary algebra books... -- Jon Haugsand ==== > * Hauke Reddmann >> Suppose you have a -A and a A V B operator defined, >> and A ^ B = -(-A V -B) as usual, and commutative and >> associative law hold for ^ and V, and --A=A, and 0,1 >> exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean >> and DeMorgan, only the distributive law fails. >> >> Does this structure already have a name? > Beats me, but do you have an example? A smallest set that satisfies > your requirements? Should be interesting enough. It is not a ring, > so it avoids any elementary algebra books... Knot theory, 4-tangles, - rotate by 90 deg, V tangle addition, mutants defined as identical. (but see other f'up) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de als man ankam wollte man werden, die geschichte schreiben, die doofen sollen sterben, der plan als man damals nach hamburg kam (Kettcar) ==== > * Hauke Reddmann > >>Suppose you have a -A and a A V B operator defined, >>and A ^ B = -(-A V -B) as usual, and commutative and >>associative law hold for ^ and V, and --A=A, and 0,1 >>exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean >>and DeMorgan, only the distributive law fails. >>Does this structure already have a name? Beats me, same here. nondistributive boolean-like algebra? Nearboolean? > but do you have an example? A smallest set that satisfies > your requirements? Should be interesting enough. facts courtesy of mace (http://www-unix.mcs.anl.gov/AR/mace4) A model doesn't exist when restricted to 2 elements (that is using the boolean axioms, there is exactly one model, and using the axioms above (over just - and V), there is exactly one model and it is equivalent to the boolean algebra on 2 elements. For 3 elements, there is a single algebra that satisfies your axioms but is not a boolean algebra (there are 3 models in total, only 2 also satisfying the boolean algebra axioms (all 3 are equivalent when restricted to 0 and 1). Here is the nondistributive model: not : 0 1 2 --------- 1 0 2 or : | 0 1 2 --+------ 0 | 0 1 2 1 | 1 1 1 2 | 2 1 0 and : | 0 1 2 --+------ 0 | 0 0 0 1 | 0 1 2 2 | 0 2 1 One distributivity offending term is or(2,and(2,2)) = 1 != 0 = and(or(2,2),or(2,2)) But there's a bit of a catch (I realize after doing the brute calculation). I left out idempotence: or(a,a) = a, and absorption: or(x,and(x,y))=x Then, the two axiom sets have the same (unique) model. Same with 4 elts (both have 4 models), but with 5 elts there are 6 boolean algebras and 16 models of the -,V algebras. -- Mitch Harris (remove q to reply) ==== >[snip Dinky's trash] >Quite a CV {:-)) >Franz > Apart from your subjective opinions of any people involved, Franz, do you > have any > objective opinion on the mathematics at > http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_2.0.htm > or are you completely enraptured by Dinky's silly game of one-up-manship? > Androcles > http://groups.google.com/groups?&as_umsgid=TCVIb.109057$Jl3.4986923@phobos.te lenet-ops.be >Dirk Vdm >I can testify that among relativists you said the truest thing ever > about the 1905 paper,you did'nt recognise the excerpt from the 1905 > paper and this makes it all the more funny. >Perhaps Androcles recognises that it is from the kinematics section of > the 1905 paper and I am delighted that you agree with him that it is > 'shit' that is not worth reading. > ----------------------------------------------------------------------------- > >Let us take a system of co-ordinates in which the equations of > Newtonian mechanics hold good. In order to render our presentation > more precise and to distinguish this system of co-ordinates verbally > from others which will be introduced hereafter, we call it the > ``stationary system.'' >What is this? > Some kind of quote of some post? > An introduction to the shit you produce later on? > Shit that you expect someone will bother reading? >Dirk > ----------------------------------------------------------------------------- I already said this on > http://groups.google.com/groups?&as_umsgid=3f71af17@usenet01.boi.hp.com > but I will repeat it here: You still don't seem to understand why I asked you > these 4 questions. > So let me try to explain in simple words. > Let's have a close look at the message you are referring > to here: > http://groups.google.com/groups?&threadm=273f8e06.0207281123.57d80819@posting .google.com | >Let us take a system of co-ordinates in which the equations of > | Newtonian >mechanics hold good.2 In order to render our presentation > | more precise and to >distinguish this system of co-ordinates verbally > | from others which will be >introduced hereafter, we call it the > | ``stationary system.'' > | > | Dear Al, > | ... This is what you gave us. > As you see, > - There is something severely wrong with the format. > - If this was a quotation from Einstein, you left out the quotes. > - You write a paragraph followed by Dear Al... > - In what you have included here, you have reformatted > and left out the failed introduction Dear Al'. So I will kindly ask again, and I will clarify what I mean: 1) What is this? > 2) Some kind of quote of some post? > Clarification: > Something you want us to believe you invented? > Something you found somewhere? > Something you want to tell us? > Something you want to tell us something about? > Something you forgot to delete when you started > with the beginning of your message Dear Al,? 3) An introduction to the shit you produce later on? > 4) Shit that you expect someone will bother reading? > Clarification: > The 'shit' in question 4 is a reprise of the 'shit' in > question 3. This is what we call a 'style figure'. Didn't they teach you to write English in Germany? > How old are you? Dirk Vdm It still remains a classic in my book and there is not a blessed thing you can do about it. Any Englishman will recognise the title of that thread - 'A memorable Fancy' comes from William Blake who uses historical characters to carry his points unfortunately you missed the point,did'nt recognise Albert's 1905 work and all in all made a jackass out of yourself. That you try to defend yourself makes you twice as dumb as you already are however it is always useful to point out that relativists have'nt the foggiest idea what the original relativity concept means,at least thanks for that. ==== >escribi.97 en el mensaje > message > [snip Dinky's trash] >Quite a CV {:-)) >Franz > Apart from your subjective opinions of any people involved, > Franz, > do you > have any > objective opinion on the mathematics at > http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_2.0.htm > or are you completely enraptured by Dinky's silly game of > one-up-manship? > Androcles > http://groups.google.com/groups?&as_umsgid=TCVIb.109057$Jl3.4986923@phobos.t > elenet-ops.be >Dirk Vdm >Androcles will never understand that c-v and c+v in Einstein's paper > are algebraic expressions which are not physical speeds measured by > any observer, but rates of travelled ** relative ** distances by the > light rays. >Yep, but I'd call them rates of change of distances between some > object and the front of a lightray, distances as calculated by a third > party observer who is assumed to measure c to be the speed of the > front of the lightray w.r.t. himself. >Dirk Vdm >As this: >But the ray moves relatively to the initial point of k, when measured > in > the stationary system, with the velocity c-v, so that ... >in which we have the observer (k), and the initial point of k. > But that is somehow confusing and for a similar calculation I got a zero > in > an exam, as the teacher believed that I meant: >But the ray moves relatively to [observer] k with the velocity c-v >which ignores when measured in the stationary system. >It would have been easier just to make c*t = v*t + L with L the initial > separation among lightray front and origin of k, than write directly L = > (c-v)*t and talk of c-v as a speed (it is a speed dimensionally, but not > physicall, i.e., cannot be adscribed to any observer). >Hm, to me the initial separation between lightray front and origin > of k seems to be 0. It grows to x', and upon reflection, shrinks > back to 0. > It is clear on the spacetime diagram: > http://users.pandora.be/vdmoortel/dirk/Stuff/tau-equation.gif > There you clearly see from the geometry that the 'horizontal distances' > (the dashed blue lines) between the red k-worldline (tau-axis) and > the grey light-worldline first grow from 0 at time t0 (event Flash) > to x' at time t1 (event Reflection), and then shrink back to 0 at > time t2 (event Flashecho). >Dirk Vdm >Yes, I was thinking of the mirror as the origin of k (I didn't have the > paper with me). > In any case, Einstein made a quite difficult and lenghty derivation. >It would have been much easier to derive the Lorentz transforms by using the > time dilation and length contraction effects. However, I guess that > psychologically it WAS more convenient the other way: get the Lorentz > transform. and from them, derive these effects. >You provided a reference with such a kind of derivation > http://www.courses.fas.harvard.edu/~phys16/Textbook/ch10.pdf > but one thing I am at odds with, namely that the relative speeds of the two > systems of reference are not necessarily equal and opposite (a strange > wording, btw). Without a proof of such a thing, the demo is not completely > correct. As far as I can see the phrase equal and opposite does not appear > in the derivation, but merely in an example. Notice how Einstein was very careful about it, and instead of assuming it (I > know PhD's who make that), he showed that the relative velocities were v > and -v respectively [ phi(v)*phi(-v)=1] And as far as I can see here, he did not *show* that the relative > velocities are v and -v, but rather *used* that trivial fact to show > that phi(v)*phi(-v)=1. Dirk Vdm Besides my parallel post, I have found this interesting reference: http://arxiv.org/PS_cache/physics/pdf/9703/9703006.pdf in which the concept of different relative speeds among two inertial observers is considered (see the text near the equation 4). However, if we accept the principle of relativity, the velocities are equal and opposite in the formula. By using a 3rd observer, two observers moving with v and -v will observe the same relative speeds. As this is true for any v [ Does any one know where these following sequences of numbers came from(the 4 > below) ? Who was the first person to discover this? And how did they come > about? There must be some reason besides the fact they are neat. Do any > others exist? 11 x 11 = 121 > 111 x 111 = 12321 > 1111 x 1111 = 1234321 > 11111 x 11111 = 123454321 > 111111 x 111111 = 12345654321 > 1111111 x 1111111 = 1234567654321 > 11111111 x 11111111 = 123456787654321 > 111111111 x 111111111 = 12345678987654321 > ----------------------------------- > 1 x 8 + 1 = 9 > 12 x 8 + 2 = 98 > 123 x 8 + 3 = 987 > 1234 x 8 + 4 = 9876 > 12345 x 8 + 5 = 98765 > 123456 x 8 + 6 = 987654 > 1234567 x 8 + 7 = 9876543 > 12345678 x 8 + 8 = 98765432 > 123456789 x 8 + 9 = 987654321 > ----------------------------------- > 0 x 9 + 1 = 1 > 1 x 9 + 2 = 11 > 12 x 9 + 3 = 111 > 123 x 9 + 4 = 1111 > 1234 x 9 + 5 = 11111 > 12345 x 9 + 6 = 111111 > 123456 x 9 + 7 = 1111111 > 1234567 x 9 + 8 = 11111111 > 12345678 x 9 + 9 = 111111111 > 123456789 x 9 + 10 = 1111111111 > ----------------------------------- > 9 x 9 + 7 = 88 > 9 x 98 + 6 = 888 > 9 x 987 + 5 = 8888 > 9 x 9876 + 4 = 88888 > 9 x 98765 + 3 = 888888 > 9 x 987654 + 2 = 8888888 > 9 x 9876543 + 1 = 88888888 > 9 x 98765432 + 0 = 888888888 > 9 x 987654321 - 1 = 8888888888 > ----------------------------------- Stephen You asked Do any others exist? Here's something which Newton used to help explain the binomial theorem. 11^1 = 11 11^2 = 121 11^3 = 1331 11^4 = 14641 After that the pattern of binomial coefficients is obscured by the carrying of decimal digits, but even these few powers of (10 + 1) can be enlightening to learners. Ken Pledger. Ken Pledger. ==== >Message-id: Does any one know where these following sequences of numbers came from(the >4 >> below) ? Who was the first person to discover this? And how did they come >> about? There must be some reason besides the fact they are neat. Do any >> others exist? >> >> 11 x 11 = 121 >> 111 x 111 = 12321 >> 1111 x 1111 = 1234321 >> 11111 x 11111 = 123454321 >> 111111 x 111111 = 12345654321 >> 1111111 x 1111111 = 1234567654321 >> 11111111 x 11111111 = 123456787654321 >> 111111111 x 111111111 = 12345678987654321 >> ----------------------------------- >> 1 x 8 + 1 = 9 >> 12 x 8 + 2 = 98 >> 123 x 8 + 3 = 987 >> 1234 x 8 + 4 = 9876 >> 12345 x 8 + 5 = 98765 >> 123456 x 8 + 6 = 987654 >> 1234567 x 8 + 7 = 9876543 >> 12345678 x 8 + 8 = 98765432 >> 123456789 x 8 + 9 = 987654321 >> ----------------------------------- >> 0 x 9 + 1 = 1 >> 1 x 9 + 2 = 11 >> 12 x 9 + 3 = 111 >> 123 x 9 + 4 = 1111 >> 1234 x 9 + 5 = 11111 >> 12345 x 9 + 6 = 111111 >> 123456 x 9 + 7 = 1111111 >> 1234567 x 9 + 8 = 11111111 >> 12345678 x 9 + 9 = 111111111 >> 123456789 x 9 + 10 = 1111111111 >> ----------------------------------- >> 9 x 9 + 7 = 88 >> 9 x 98 + 6 = 888 >> 9 x 987 + 5 = 8888 >> 9 x 9876 + 4 = 88888 >> 9 x 98765 + 3 = 888888 >> 9 x 987654 + 2 = 8888888 >> 9 x 9876543 + 1 = 88888888 >> 9 x 98765432 + 0 = 888888888 >> 9 x 987654321 - 1 = 8888888888 >> ----------------------------------- >> >> Stephen > You asked Do any others exist? Here's something which Newton >used to help explain the binomial theorem. > 11^0 = 1 >11^1 = 11 >11^2 = 121 >11^3 = 1331 >11^4 = 14641 After that the pattern of binomial coefficients is obscured by the >carrying of decimal digits, but even these few powers of (10 + 1) can >be enlightening to learners. Ken Pledger. -- Mensanator Ace of Clubs ==== > Does any one know where these following sequences of numbers came from(the 4 > below) ? Who was the first person to discover this? And how did they come > about? There must be some reason besides the fact they are neat. Do any > others exist? >11 x 11 = 121 > 111 x 111 = 12321 > 1111 x 1111 = 1234321 > 11111 x 11111 = 123454321 > 111111 x 111111 = 12345654321 > 1111111 x 1111111 = 1234567654321 > 11111111 x 11111111 = 123456787654321 > 111111111 x 111111111 = 12345678987654321 > ----------------------------------- > 1 x 8 + 1 = 9 > 12 x 8 + 2 = 98 > 123 x 8 + 3 = 987 > 1234 x 8 + 4 = 9876 > 12345 x 8 + 5 = 98765 > 123456 x 8 + 6 = 987654 > 1234567 x 8 + 7 = 9876543 > 12345678 x 8 + 8 = 98765432 > 123456789 x 8 + 9 = 987654321 > ----------------------------------- > 0 x 9 + 1 = 1 > 1 x 9 + 2 = 11 > 12 x 9 + 3 = 111 > 123 x 9 + 4 = 1111 > 1234 x 9 + 5 = 11111 > 12345 x 9 + 6 = 111111 > 123456 x 9 + 7 = 1111111 > 1234567 x 9 + 8 = 11111111 > 12345678 x 9 + 9 = 111111111 > 123456789 x 9 + 10 = 1111111111 > ----------------------------------- > 9 x 9 + 7 = 88 > 9 x 98 + 6 = 888 > 9 x 987 + 5 = 8888 > 9 x 9876 + 4 = 88888 > 9 x 98765 + 3 = 888888 > 9 x 987654 + 2 = 8888888 > 9 x 9876543 + 1 = 88888888 > 9 x 98765432 + 0 = 888888888 > 9 x 987654321 - 1 = 8888888888 > ----------------------------------- >Stephen You asked Do any others exist? Here's something which Newton > used to help explain the binomial theorem. 11^1 = 11 > 11^2 = 121 > 11^3 = 1331 > 11^4 = 14641 After that the pattern of binomial coefficients is obscured by the > carrying of decimal digits, but even these few powers of (10 + 1) can > be enlightening to learners. Ken Pledger. Ken Pledger. I also liked the fact that you can work with powers of (100 + 1) to go even further than what the (10 + 1) can get you. As in: 101^1 = 101 101^2 = 10201 101^3 = 1030301 101^4 = 104060401 101^5 = 10510100501 101^6 = 1061520150601 101^7 = 107213535210701 101^8 = 10828567056280801 Then the carry kills the pattern, but then you could just use (1000 + 1) to go even further. -David C. ==== On a related topic, a non-mathematical friend of mine once asked me, Why is it that 1/7 = .142857142857...? He elaborated: first you have 14 = 2 x 7, then 28 = 4 x 7, then 56 = 8 x 7 (plus a carried 1 from the sequel), then 112 = 16 x 7, etc. He wasn't asking about a proof; he was asking what *is* it about the number 7 that makes this so? The best answer I could come up with is because 7^2 + 1 = (10^2)/2. And it was hard to come up with similar examples in other bases, though I did point out that 1/9 = .09 + .018 +.0027 + ... What would you have told him? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > On a related topic, a non-mathematical friend of mine once asked me, > Why is it that 1/7 = .142857142857...? He elaborated: first you > have 14 = 2 x 7, then 28 = 4 x 7, then 56 = 8 x 7 (plus a carried > 1 from the sequel), then 112 = 16 x 7, etc. He wasn't asking about a > proof; he was asking what *is* it about the number 7 that makes this > so? The best answer I could come up with is because 7^2 + 1 = > (10^2)/2. And it was hard to come up with similar examples in other > bases, though I did point out that 1/9 = .09 + .018 +.0027 + ... What > would you have told him? First a comment to the original poster. I first saw the identities (and others like it, I think) in a little book my parents had called (and my memory is stretched pretty thin here) Mathemagic and written by D. C. Heath. I saw this book in the 50's. I presume these patters are way older than that. There are many other amazing things in that book, such as a magic square of squares and so on. On the subject of the fractions, if you have a decimal expansion that repeats from the beginning, you can always determine the number (and therefore the decimal expansion) as a geometric progression starting with any number of the first terms of the series after you pass all of the zeros of course. The constant multiple is just the remainder at the point where you stopped. This can look pretty cool. For instance 1/97 = .0103092783... where the initial 2 digits .01 are multiplied by 3/100 and added ad(d) infinitum. This is more impressive to the eye if you simply say that you are multiplying by 3 and the shifting the pattern 2 places to the right. Then you get 01 03 09 27 81 243 729 ... and it works out. The mathematics here is trivial, of course, but surprising when it is pointed out. This is an interesting expansion, as the period length is 96 and so like .142857 for 1/7, it has the property that when multiplied by anything less than the denominator you get a cyclic rearrangement as the product which is a little off topic. 1/13 = .076923. After .07, the remainder is 9, and so 1/13 = .07 + .07*(9/100) + .07*(9/100)*(9/100) or in visual form 07 63 567 5103 ... which of course is not nearly as impressive to look at. Or we could have had 1/7 given as 142 852 5112 30672 ... where the common multiple is now 6, the remainder after producing .142. Notice that we make sure here that we stick out precisely 3, just as we made sure that we stuck out precisely 2 when we started with just 2 numbers of the expansion. It is of interest that we started 1/7 with 14 which is a multiple of 7. I have not thought about when or why that happens. Achava ==== >On a related topic, a non-mathematical friend of mine once asked me, >Why is it that 1/7 = .142857142857...? He elaborated: first you >have 14 = 2 x 7, then 28 = 4 x 7, then 56 = 8 x 7 (plus a carried >1 from the sequel), then 112 = 16 x 7, etc. He wasn't asking about a >proof; he was asking what *is* it about the number 7 that makes this >so? The best answer I could come up with is because 7^2 + 1 = >(10^2)/2. And it was hard to come up with similar examples in other >bases, though I did point out that 1/9 = .09 + .018 +.0027 + ... What >would you have told him? Well, you do have 1/(62*127) = .000127000254000508001016002032... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >Does any one know where these following sequences of numbers came from(the 4 >below) ? Who was the first person to discover this? And how did they come >about? There must be some reason besides the fact they are neat. Do any >others exist? >11 x 11 = 121 >111 x 111 = 12321 ... (sum_{j=0}^n r^j)^2 = sum_{k=0}^{2n} sum_{j=max(0,k-n)}^{min(n,k)} r^k = sum_{k=0}^{2n} min(k+1,2n+1-k) r^k >1 x 8 + 1 = 9 >12 x 8 + 2 = 98 ... (r-2) sum_{j=0}^{n-1} (n-j) r^j + n = sum_{j=0}^{n-1} (r-n+j) r^j which is equivalent to (r-1) sum_{j=0}^{n-1} (n-j) r^j = -n + r sum_{j=0}^{n-1} r^j >0 x 9 + 1 = 1 >1 x 9 + 2 = 11 >12 x 9 + 3 = 111 ... (r-1) sum_{j=0}^{n-1} (n-j) r^j + n + 1= sum_{j=0}^n r^j which is the same as the previous one. >9 x 9 + 7 = 88 >9 x 98 + 6 = 888 >9 x 987 + 5 = 8888 ... (r-1) sum_{j=0}^{n-1} (j+r-n) r^j + (r-n-2) = (r-2) sum_{j=0}^n r^j which is again equivalent. How do these grab you? ((9*1-1)^2+10)* 1 - 9*(9*1-1)* 1 = 1*2 ((9*2-1)^2+10)* 21 - 9*(9*2-1)* 41 = 2*3 ((9*3-1)^2+10)* 321 - 9*(9*3-1)* 941 = 3*4 ((9*4-1)^2+10)*4321 - 9*(9*4-1)*16941 = 4*5 ((99*1-1)^2+100)* 1 - 99*(99*1-1)* 1 = 1*2 ((99*2-1)^2+100)* 201 - 99*(99*2-1)* 401 = 2*3 ((99*3-1)^2+100)* 30201 - 99*(99*3-1)* 90401 = 3*4 ((99*4-1)^2+100)* 4030201 - 99*(99*4-1)* 16090401 = 4*5 ((99*5-1)^2+100)* 504030201 - 99*(99*5-1)* 2516090401 = 5*6 ((99*6-1)^2+100)* 60504030201 - 99*(99*6-1)* 362516090401 = 6*7 ((99*7-1)^2+100)* 7060504030201 - 99*(99*7-1)* 49362516090401 = 7*8 ((99*8-1)^2+100)* 807060504030201 - 99*(99*8-1)* 6449362516090401 = 8*9 ((99*9-1)^2+100)*90807060504030201 - 99*(99*9-1)*816449362516090401 = 9*10 Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== thusly: > Does any one know where these following sequences of numbers came from(the > 4 below) ? Who was the first person to discover this? And how did they > come about? There must be some reason besides the fact they are neat. Do > any others exist? 11 x 11 = 121 > 111 x 111 = 12321 > 1111 x 1111 = 1234321 > 11111 x 11111 = 123454321 > 111111 x 111111 = 12345654321 > 1111111 x 1111111 = 1234567654321 > 11111111 x 11111111 = 123456787654321 > 111111111 x 111111111 = 12345678987654321 > ----------------------------------- > 1 x 8 + 1 = 9 > 12 x 8 + 2 = 98 > 123 x 8 + 3 = 987 > 1234 x 8 + 4 = 9876 > 12345 x 8 + 5 = 98765 > 123456 x 8 + 6 = 987654 > 1234567 x 8 + 7 = 9876543 > 12345678 x 8 + 8 = 98765432 > 123456789 x 8 + 9 = 987654321 > ----------------------------------- > 0 x 9 + 1 = 1 > 1 x 9 + 2 = 11 > 12 x 9 + 3 = 111 > 123 x 9 + 4 = 1111 > 1234 x 9 + 5 = 11111 > 12345 x 9 + 6 = 111111 > 123456 x 9 + 7 = 1111111 > 1234567 x 9 + 8 = 11111111 > 12345678 x 9 + 9 = 111111111 > 123456789 x 9 + 10 = 1111111111 > ----------------------------------- > 9 x 9 + 7 = 88 > 9 x 98 + 6 = 888 > 9 x 987 + 5 = 8888 > 9 x 9876 + 4 = 88888 > 9 x 98765 + 3 = 888888 > 9 x 987654 + 2 = 8888888 > 9 x 9876543 + 1 = 88888888 > 9 x 98765432 + 0 = 888888888 > 9 x 987654321 - 1 = 8888888888 > ----------------------------------- Stephen That's how your money is supposed to grow when you subscribe to a pyramid scheme :) -- Paul Townsend I put it down there, and when I went back to it, there it was GONE! Interchange the alphabetic elements to reply ==== > That's how your money is supposed to grow when you subscribe to a pyramid > scheme :) > -- I dont mind that in this scheme the people down the bottom make the money. ==== >Does any one know where these following sequences of numbers came from(the 4 >below) ? Who was the first person to discover this? And how did they come >about? There must be some reason besides the fact they are neat. Do any >others exist? 11 x 11 = 121 >111 x 111 = 12321 >1111 x 1111 = 1234321 >11111 x 11111 = 123454321 >111111 x 111111 = 12345654321 >1111111 x 1111111 = 1234567654321 >11111111 x 11111111 = 123456787654321 >111111111 x 111111111 = 12345678987654321 >----------------------------------- >1 x 8 + 1 = 9 >12 x 8 + 2 = 98 >123 x 8 + 3 = 987 >1234 x 8 + 4 = 9876 >12345 x 8 + 5 = 98765 >123456 x 8 + 6 = 987654 >1234567 x 8 + 7 = 9876543 >12345678 x 8 + 8 = 98765432 >123456789 x 8 + 9 = 987654321 >----------------------------------- >0 x 9 + 1 = 1 >1 x 9 + 2 = 11 >12 x 9 + 3 = 111 >123 x 9 + 4 = 1111 >1234 x 9 + 5 = 11111 >12345 x 9 + 6 = 111111 >123456 x 9 + 7 = 1111111 >1234567 x 9 + 8 = 11111111 >12345678 x 9 + 9 = 111111111 >123456789 x 9 + 10 = 1111111111 >----------------------------------- >9 x 9 + 7 = 88 >9 x 98 + 6 = 888 >9 x 987 + 5 = 8888 >9 x 9876 + 4 = 88888 >9 x 98765 + 3 = 888888 >9 x 987654 + 2 = 8888888 >9 x 9876543 + 1 = 88888888 >9 x 98765432 + 0 = 888888888 >9 x 987654321 - 1 = 8888888888 I believe the Babylonians used these sequences in drawing up plans for ziggurats... >----------------------------------- Stephen > ************************ David C. Ullrich ==== If f:B^2 ---> B^2 is the identity mapping, then any map g within epsilon > of f must contain the closed ball of radius 1-epsilon. Yes, but this needs a proof, in fact this is the hard part of the problemm. observe that the restrictions f|S^1 and g|S^1 are homotopic as maps S^1-->B^2{x}, so g|S^1 is homotopic to identity. Then the index of point x with respect to g|S^1 equals 1 and therefore x must be covered by g(B^2). Simeon ==== (note 1) If you are unaware of my posts or are aware of my posts and opened this for the purpose of ridicule, you would best consider searching the post for quoted material. There are long quotes describing the formal implications and axiomatizations for quantum logic. You should find those to be of interest independent of any purposes I might have. (note 2) If you are aware of my posts and the flaming that had been raging on sci.logic about the foundations of mathematics, then you will recognize that there is a context in the later material. It was originally an off-group correspondence to George Greene. I decided to make an independent post because of the quoted materials. References concerning George's statements are at a minimum--one sentence per section. The extent to which you will be able to follow it will depend on what posts you did or did not read over the last year. Nevertheless, I am just posting for informative purposes. I spent the time transcribing the excerpts and this is material of which most people are unaware. :-) mitch ------------- ----- You apparently thought everything I was doing could be reduced to Boolean algebras (or Heyting algebras). The class OML of orthomodular lattices forms a variety which includes the class BA of Boolean algebras. Of course, I became aware of orthomodularity from my investigations of set theory. I had no way to state it specifically and I had no reference with a simple declarative statement like that above. --- You apparently thought we were in agreement about the problematic nature of models. I could not get the document http://www.uni.torun.pl/~jacekm/latoml.pdf to open at the time of this response. The paper is entitled The Lattice of Orthomodular Logics. I had hoped to avoid the convoluted explanation of a diagram that follows. If you can get it to open, you will find a lattice (c) with two isolated paths between TOP and BOTTOM on the right side and several interlaced paths on the left side. The two paths on the right have only one point each between TOP and BOTTOM. All paths on the left side are interlaced among six points. Let me call the two points on the isolated paths R2 and the six points with the interlaced paths L6. The interlaced paths may be explained by the form, TOP / / / / / * * * / / / / / * * * | | | | | BOTTOM There is a path exactly opposite in sense running from TOP to BOTTOM and passing through the three points not connected in this diagram. Moreover, there are uniform diagrams for all six points. So, if the edge running from BOTTOM was shifted to the L6 vertex on the bottom left, the two edges above that vertex would shift to the bottom left vertex as well. Continuing, the two edges incident to the L6 vertex on the top left would be shifted to the top center. One of my earliest questions concerned the nature of the complete connectives. Given that first-order logic has a soundness theorem and a completeness theorem that are converse, it seemed natural to refer to the structure I had discerned in terms of completeness connectives and soundness connectives. What I mean by the structure I had discerned is the result of deliberations about classical truth-functional connectivity that would have associated the complete connectives with R2. Those deliberations led to a recognition of L6 as a pattern without any clear concept of how the pattern should be described. That is how idiolectic usage begins. You don't have words for things, so you make the best associations you can in order to work. I had no way to explain why I thought the Steiner 1-factorization was important when I did the deduction with the Steiner quasigroup. But, if you look at the factorization, {{#,0},{1,3},{2,6},{4,5}} {{#,1},{0,3},{2,4},{6,5}} {{#,2},{0,6},{1,4},{3,5}} {{#,3},{0,1},{2,5},{4,6}} {{#,4},{0,5},{1,2},{3,6}} {{#,5},{0,4},{2,3},{1,6}} {{#,6},{0,2},{3,4},{1,5}} you can see that it segregates eight symbols into pairs. Moreover, where '0' and '#' are separated from one another (recall that they were special symbols added to the symbols labeling my columns), the other pairs of symbols are segregated from one another. This is the same kind of uniform interlaced relationship for L6 in the lattice diagram. But, the specific labeling introduces additional complications. I cannot yet make all the connections that I need to appear coherent. But, what lies behind this is a very simple concept, ... The class H of all characteristic functions of theories of C determines the consequence of C. This somewhat trivial fact was independently observed by several logicians - see Scott[1971], Routley[1976], and Suszko[1977a], cf. also van Fraassen[1973]. It gave rise to discussions on two-valuedness and the scope of the principle of bivalence. Suszko seems to be the one who has drawn the most far-reaching conclusions from this slogan: Every logic is two-valued. What seems to be a source of difficulty is that Suszko's thesis does not provide any manageable description of truth-valuations that determine a given logic C. Whatever else may be true, this thesis is immensely complicated. The text which follows goes on to say, There is no general and satisfactory definition of an admissible truth-valuation for a given logic C. Nevertheless, having a sense through which classical truth-functional connectivity relates to the structure of an orthomodular logic does not seem trivial with regard to the questions. The only way I could get away from the model question was to resort to threshold logic and geometric intuitions leading to linear fractional transformations. I had no way to explain what I was doing since what I was trying to demonstrate was even unclear to myself. Nevertheless, it seems to me that it what is expressed here would lead one to threshold logic and linear separability as an alternative to the usual alternative to model-theoretic consequence. --- You apparently thought my linear order was trivial. I will transcribe the necessary definitions, but here is a theorem that seem to explain why I thought otherwise, Theorem 5.5.1. Up to identity over OML, p_1,..., p_5 are the only formulas p in two variables having the following property: for any algebra (A in OML) and all (a, b in A), p(a,b)=1 iff a<=b. We recall that the orthomodular quantum logic OML[|=] is defined semantically by the class OML of orthomodular lattices with the unit element 1 designated. Thus: (a in OML[|=](X)) iff for every (A in OML) and every homomorphism h:S->A, h(X) subset {1} implies h(a)=1 OML[|=] is thus the assertional logic of the class OML. Consider the following sentences of S in two variables x and y: p_0(x,y) := ~x / y p_1(x,y) := (~x / y) / (~x / ~y) / (x / (~x / y)) p_2(x,y) := (~x / y) / (x / y) / ((~x / y) / ~y) p_3(x,y) := ~x / (x / y) p_4(x,y) := y / (~x / ~y) p_5(x,y) := (~x / y) / (x / y) / (~x / ~y) Theorem 5.5.1. [stated above]... The proof of Theorem 5.5.1 is omitted. The proof makes use of the free algebra F_OML(2) on two generators. It has 96 elements and it is known to be isomorphic with the direct product MO2 x 2^4 of the Chinese lantern MO2 with the 16-element Boolean algebra, denoted 2^4. 1 // / / / / / / MO2 a ~a b ~b / / / / / / // 0 F_OML(2) is finite but the free algebra F_OML(3) is infinite. Theorem 5.5.1 iimplies: Corollary 5.5.2 The logic OML[|=] is implicative and each of the above (definable) connectives p_i (i=1,...,5) is its implication. Furthermore, the class Mod*(OML[|=]) coincides with OML. It follows from the above corollary that each of the sets {p_i(x,y),p_i(x,y)}, i=0,...,5 is an equivalence for OML[|=]. The consistent strengthenings of the logic OML[|=] are called quantum logics. Every quantum logic C is thus regularly algebraizable and, if C is finitary, its equivalent algebraic semantics coincides with the quasivariety Mod*(C) which is clearly a quasivariety of orthomodular lattices. The classical consequence K is a limit case - it is the strongest quantum logic. The class BA of Boolean algebras is the smallest non-trivial variety of orthomodular lattices. The consequence OML[|=] has been axiomatized by Kalmbach [1974], [1981]. Let MP_i be the detachment rule determined by the implication p_i, i.e., MP_i is the rule x, p_i(x,y) ----------- y for i=0,...,5. Define the binary connective R by: (a R b) := (a / b) / (~a / ~b) for any a, b. Then define the following axioms: A1 ~(p R q) / (~p / q) A2 p R p A3 ~(p R q) / (~(q R r) / (p R r)) A4 ~(p R q) / (~p R ~q) A5 ~(p R q) / ((p R r) R (q R r)) A6 (p / q) R (q / p) A7 (p / (q / r)) R ((p / q) / r) A8 (p / (p / q)) R p A9 (~p / p) R ((~p / p) / q) A10 ~(p / q) R (~p / ~q) A11 p R ~~p A12 (p / (~p / (p / q))) R (p / q) A13 (p R q) R (q R p) A14 (~p / q) ->_1 (p ->_1 (p ->_1 q)) A15 ~(p ->_1 q) / (~p / q) (In A14 and A15 we write (a ->_1 b) instead of p_1(a,b).) Theorem 5.5.3. Each of the sysetems {A1,...,A13,MP_0} and {A1,...,A15,MP_1} forms an inferential base for OML[|=]. The proof is omitted. The logics determined by the bases OML[|=](emptyset) cup MP_i for i in {2,...,5} are known to be weaker than OML[|=]. Quantum logics give rise to many counterexamples to some metalogical properties which hold for classical logic and for a large class of weaker logics. We mention here one result: Theorem 5.5.4. If C is a logic such that OML[|=]<=C<=K, and ~(C=K), then C does not admit the parameter-free Local Deduction Theorem; in particular, C does not admit the Deduction Theorem in the sense of Section 2.6 The above theorem has a simple algebraic interpretation: under the hypotheses of the theorem, the class Mod*(C) fails to have the C-filter extension property. This result implies that BA is the only variety of orthomodular lattices with the congruence extension property. There is a lot here. I hope you noticed the special cases involved with Boolean algebras and classical logic. I have no way to explain how or why I saw things so differently from everyone else. And, I have no way to say that any material I have ever presented constitutes evidence of a rational thought process. If you look at the relation used in the axioms, (a R b) := (a / b) / (~a / ~b) you can see why I had been so adamant about the fixedness of the projection connectives and their complements under DeMorgan conjugation. Moreover, it seemed particularly important given the unate recognition problem and the symmetries I was seeing that led to the discussion of Steiner quasigroups. --- You apparently interpreted my discussion about identity as intuitionism in spite of my assertions that it was more subtle than that. There seems to be a non-linear hierarchy of logics. Fregean Protoalgebraic Logics | | | | Regularly Regularly Weakly Algebraizable Algebraizable ------- Logics Logics | | | | | | | Weakly Algebraizable Algebraizable ------- Logics Logics | | | | | | | | | Equivalential | Logics / / / / / / / / Protoalgebraic Logics I believe the concept of an equality-free logic refers to logics that are not equivalential. Here are some details: Two connectives of special interest in metalogical investigations of classical logic: the connective of implication, strictly tied to the Deduction Theorem, and the connective of equivalence. The latter expresses, in the material sense, the fact that two sentences have the same logical value while in the strict sense it expresses the property that two sentences have the same logical value while in the strict sense it expresses the property that two sentences are mutually interderivable on the basis of a given logic. The process of identification of equivalent sentences relative to the theories of a logic C defines a class of abstract algebras. The members of the class are called Lindenbaum-Tarski algebras of the logic C. One may abstract from the origin of these algebras and examine them by means of purely algebraic methods. This approach, based on Lindenbaum-Tarski algebras, turns out to be particularly important because it bridges the gap between logic and algebra, and therefore makes it possible to apply the powerful methods of contemporary algebra in metalogic. If C is the classical consequence (logic), then the relation (1) a ==_T b iff a<->b in C(T) defines for any theory T, a congruence on the algebra of sentences (formulas). The resulting class of Lindenbaum-Tarski algebras coincides with the class of Boolean algebras. In turn, the class of Lindenbaum-Tarski algebras corresponding to intuitionistic logic coincides with the class of Heyting algebras. We may also use the implication connective ->. The logic C with the property that ==_T, defined by a ==_T b iff a->b, b->a in C(T) is the largest congruence on the formula algebra compatible with C(T) are called implicative. They are extensively studied in Rasiowa's monograph [1974]. The question arises about the scope of the above method. There are numerous examples of logics which are intractible by this method because there may not exist connectives in the language which, according to the above formulas, define a congruence on the formula algebra. This is a typical situation for intensional logics. Prucnal and Wronski [1974] have proposed a generalization of the Lindenbaum-Tarski method by replacing the equivalence connective by a possibly infinite set of sentential formulas which collectively possess many properties of the equivalence. Any logic which has such a set is called equivalential (or congruential). In a more formal rendering, a logic C is equivalential (finitely equivalential) if there exists a set (a finite set) E(p,q) of sentential formulas in two variables p and q such that the relation ==_T, where (2) a ==_T b iff E(a,b) subset C(T) is a congruence on the language compatible with C(T), for all theories T. The notion of an equivalential logic turns out to be very useful in the analysis of intensional logics such as modal, tense, or dynamic logics. It is actually the modal logic that seems to be relevant to the structures I had been considering. To see, why, I need to invoke a structure very similar to L6 discussed above, TOP / / / / / * * * / / / / / * * * | | | | | BOTTOM The actual result involved has to do with two modal logics, E[->] and RE[->] that are not equivalential. The proof of this is based on an algebra whose reduct is an eight-element Boolean algebra, 1 a c b ~a ~c ~b 0 I cannot draw the diagram. Here is the adjacency matrix: | 0 1 a b c ~a ~b ~c ------------------------------- 0 | 0 0 0 0 0 1 1 1 1 | 0 0 1 1 1 0 0 0 a | 0 1 0 0 0 0 1 1 b | 0 1 0 0 0 1 0 1 c | 0 1 0 0 0 1 1 0 ~a | 1 0 0 1 1 0 0 0 ~b | 1 0 1 0 1 0 0 0 ~c | 1 0 1 1 0 0 0 0 The actual proof refers to RE[->] since that result would suffice for E[->]. It involves certain necessitation relations, []1=[]a=[]c=1 []b=~b []~a=[]~b=[]~c=[]0=0 The proof depends on two defined sets from the vertices in the graph, {a,1} and {0,a,~a,1}. I realize that neither of us can make sense of the proof at this point. But, the paragraph that follows the proof gives some insight as to what is pertinent here, The answer to the question whether a logic is equivalential or not depends on the non-axiomatic inference rules of the logic. For instance, the system E(=E[->](emptyset)) of the logic E[->] is closed with respect to the rule of extensionality; but other theories of this logic need not be closed with respect to RE. If E[->] is strengthened by adjoining the rule RE to the rules of E[->] (and so RE can be applied in arbitrary proofs), the resulting logic is equivalential and {p<->q} is its equivalential system. What is going on here is that the modal systems are defined with respect to invariant subsets of formulas containing all of the classical tautologies, CL. E is the smallest modal system and its defining condition is E := Sb(CL,MP,RE) where MP:= p,p->q ------ q RE:= p<->q --------- []p<->[]q The logic E[->] derived from E admits only MP as a rule of inference on arbitrary collections of sentences from the language. So closure with respect to RE only holds for the emptyset by default for E[->]. The discussion above says that if one examines the logic E[->,RE], the logic is equivalential and the formula, p<->q serves as its equivalence. This is, of course, exactly what is involved in forming a Tarski-Lindenbaum algebra for a classical system of logic. I could probably now make reference to Kant and the relation between possibility and necessity in characterizing the transcendental logic. Then I would point out his remarks on negation with a redirect to Frege. But, that would not be a rational argument. Or, at least, it would not be considered coherent by modern standards of peer review. In any case, there is a clear notion of equality-free logic, and, it does relate to intensional interpretation of quantifiers. --- You apparently never realized what I meant in referring to almost universality or why this might be important in a discussion of set theory. The origin of non-Fregean logics is strictly connected with the abolition of the so-called Fregean Axiom by Suszko [1975a]. Let us quote the following passage by Wojcicki [1984], p.326: According to Frege, denotations (Bedeutung) of sentences are logical values. Thus, each sentence denotes either Truth or Falsehood. Suszko, who sought support for his ideas in Wittgenstein, rejects this point of view. For him, denotation of a sentence is what the sentence says about: a certain situation. This term was chosen by Suszko to interpret Wittgenstein's Sachlage - the state of affairs. Situations which exist create positive facts, those which do not exist create negative facts. not denote the same. It is a certain fact that Wittgenstein knew Frege just like it is a fact that Wittgenstein exchanged letters with Russell, but these two facts are quite different, and thus two sentences stating these two facts have different denotations although their truth value is the same. Obviously, Frege was not of the opinion that all true (or false) sentences 'say the same' either. In Suszko's apprehension the differences lay in the sense (Sinn) of sentences and not in their denotations. For comparison of Suszko's ideas with those of Frege it is essential that neither Sinn itself nor any of its components is an element of the objective world. Sinn is a way in which sentences are assigned their logical values (one is tempted to repeat after Ajdukiewicz 'the way of how the sentence is understood'), or - which also can be found in Frege's works - 'the thought conveyed by the sentence'. The thought (...) understood as a certain abstract object and not an individual mental experience. The differences between Suszko's and Frege's approaches are by no means of verbal character: among the concepts used by Frege there is no counterpart for the notion of a situation. In the Suszko's times the situational theory of meaning did not exist. Thus the principle that the meaning of a sentence coincides with the situation described by this sentence had a purely postulative character at that time - building a situational semantics was a task for the future. This task was performed by Wojciciki who developed foundations of situational semantics for Suszko's non-Fregean logic with identity. [...] ... We do not discuss these issues here because we would have to begin with a general account of what a situation is. Instead, we shall focus our attention on some formal aspects of the sentential logic with the identity connective. Let (S_^, /, /, ->, <->, ~, ^) be the extension of the language (S, /, /, ->, <->, ~) of classical sentential logic obtained by adjoining to it a new binary connective ^ called the identity connective: (^) a^b in C(T) iff (A(phi) in S_^)(A(p) in Var(phi)) C(T,phi(p/a))=C(T,phi(p/b)) for all (a,b in S_^) and (T subset S_^) Note that (a^b) -> (a<->b) is a thesis of K_^; the sentence that if a and b are identical, they have the same truth value. The Frege Principle in the logical form (alias Frege axiom) is the converse of the above implication. More precisely, the Fregean Axiom states that the identity of two sentences is identical to their material equivalence: (F) (p^q)^(p<->q) (F) is not a theorem of K_^ The categoreal error I made here, of course would have involved confusing zero-order logic with first-order logic. But, to be honest, I see no real difference since I can turn to the Blackwell Guide to find a discussion of motivation that led to generalized quantification. If I make a uniform substitution in the sentences above, I obtain (p^q)^(p<->q) Au(u in p <-> u in q) <-> p=q (a^b) -> (a<->b) Au(u in a <-> u in b) -> a=b I will not try to justify this mistake either. Since I had been unaware of anything I am sending you here, it is hard for me to call this a mistake, though. I spent a long time thinking about extensionality in relation to identity (both eliminable and not eliminable) and ended up concluding that there was something more than was explained by what I had been taught. No one could answer my questions. --- I look forward to reading about logics discussed in the book from which these many quotes have been taken: Protoalgebraic Logics Janusz Czelakowski ISBN: 0-7923-6940-8 <3c65f87.0401131920.67d8e399@posting.google.com> <1g7ivag.1praqrrqra0psN%see.sig@for.addy> <3c65f87.0401140753.3bcde45f@posting.google.com> ==== > And we may speculate on JSH's deeper motivations. So far, greed and > egocentricity seem primary. And they have corrupted his judgement > woefully. speculate that greed is a motivation. He's been explicit on that point. -- I don't know why I live in a world with so many supposed mathematicians who are all so dumb AND rude. Why oh why couldn't someone like Gauss or Dedekind still be around? Shoot, I'd even take someone like Hardy at this point. -- James S Harris compromises ==== And we may speculate on JSH's deeper motivations. So far, greed and > egocentricity seem primary. And they have corrupted his judgement > woefully. speculate that greed is a motivation. He's been explicit on that > point. But he has also said that he regularly lies, and that he is using this NG as a psychological laboratory, so what do you _know_ about him (other than that he is a world champion kook)? ==== My analysis continues to indicate that my research on counting prime numbers *should* be more accessible than my other math research which is more abstract, and clearly more difficult. Still I also recognize that significant parts of my prime counting research are beyond a lot of people simply because that research extends into partial differential equations. To me the starting point is simple enough: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], S(x,1) = 0, p(x, y) = floor(x) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y). But I'm increasingly aware that what looks simple to me, leaves mathematicians all over the world befuddled. Not surprisingly, faced with a difficult mental challenge, people cling to what's known, and with the less sophisticated audience of sci.math that has usually meant looking towards Legendre's Method, while when I've contacted mathematicians more expert, it has usually meant looking towards Meissel's formula. And in fact in contacts with mathematicians at universities, for instance my alma mater Vanderbilt University, I heard that what I'd found was some variant on Meissel's formula. However, anyone who actually looks over known methods will find that what I have above is astonishing in its conciseness. It's just amazingly short for a way to count prime numbers. And then you might notice that it has a partial difference equation at its heart, but that's something I've emphasized only to see it apparently sail over the heads of readers, so I'm less interested in emphasizing it now, as hey, it's just a tad bit beyond most of you. I can get some sense why even experts, like mathematicians by definition, would find it intimidating and difficult to comprehend, but then again, I find myself surprised at how clear it is that the expression is completely overwhelming. It gets more interesting, and rather than move into the calculus by talking about the partial differential equation that follows, I'll talk more about practical matters. For instance, facing a daunting expression, I've seen a tendency by posters to try and simplify it, as if their minds are desperate to find something more familiar. Beyond what I already mentioned, for instance, many posters would keep deleting off the second factor and emphasizing using primes! So they'd keep pushing something like dS(x,p_j) = [p(x/p_j, p_j - 1) - j-1], where using primes makes things look simpler, though you're shifted from a mathematical formula--the partial difference equation--to an algorithm. That behavior is in line with what I mentioned before where posters grasping for the familiar looked to Legendre's method if novice in the field, or Meissel's formula if more sophisticated in their knowledge. Now then, on to the more expert commentary on my work looking like Meissel's formula, which has helped me to understand that yes, my work somehow is beyond most mathematicians ability to handle, as indeed you can relate from it back to Meissel's formula, but you can't get to it from Meissel's formula. It's actually easy to show what I mean, as if you know anything about Meissel's formula, you know that there's one aspect of it where you sum something like pi(x/p_j) - (j-1) where you iterate through primes p. That follows from the root prime counting function which I discovered easily enough with dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], by considering special cases. First if you specifically use primes for y, you get dS(x,p_j) = [p(x/p_j, p_j - 1) - j-1], where you can see a lot of information from the root mathematics can be dropped. It also is true that if p_j-1 >= sqrt(x/p_j), you can use [p(x/p_j, sqrt(x/p_j)) - j-1], and since with my function p(x,sqrt(x) = pi(x), you can finally get to [pi(x/p_j) - j-1]. Now though, notice that you can't go back the other way!!! So my work in less space encodes more information that relates back to what mathematicians already discovered! However, in looking at it, even experts seem to get lost from what I've gathered in communicating with mathematicians worldwide since May 2002. Now looking at dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], Now here's something for fun. For a while I pursued fast prime counting programs to see if I couldn't get progress by that route, and eventually I really just got bored with figuring out fast algorithms, as I'm more interested in the math, but along the way I found some of the fastest expressions possible for certain counts: With even N, N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - floor((N-36)/14) + floor((N-22)/42) is basically the explicit result of summing an algorithmic form of the dS(x,y) function and subtracting the resulting S(x,y) from N, where evens are automatically handled, from 2 up to and including 10, which represents the primes 2, 3, 5 and 7. That formula counts primes from N=36 up to N=174. For instance, N=100, gives 50 - 16 - 8 + 2 - 4 + 1 = 25 as expected. But beyond counting primes in a small range a slightly longer formula works to give N minus the count of composites up to and including N that have 2, 3, 5, or 7 as a factor out to positive infinity for even N: N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - floor((N-106)/210) + 2 Modern mathematicians can't find anything of their own without my research of similar length that works out to infinity. Yup, despite the research done on prime numbers through the entire math history of the world, mathematicians as a group can only produce *longer* expressions if they use their own research on counting prime numbers! Want more on prime counting? Then see my blog archives: James Harris ==== > My analysis continues to indicate that my research on counting prime > numbers *should* be more accessible than my other math research which > is more abstract, and clearly more difficult. Still I also recognize > that significant parts of my prime counting research are beyond a lot > of people simply because that research extends into partial > differential equations. To me the starting point is simple enough: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], S(x,1) = 0, p(x, y) = floor(x) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y). It doesn't get more interesting when you repost it. ==== [snip the usual self-aggrandizing rubbish] The Mad Russian: I've invented a new type of television, but you don't see anything - you just hear it. I call it Hear-o-vision. Eddie Cantor: But Russian - thats radio! The Mad Russian: Well what do you know? I invented radio too! ==== > My analysis continues to indicate that my research on counting prime > numbers *should* be more accessible than my other math research which > is more abstract, and clearly more difficult. Still I also recognize > that significant parts of my prime counting research are beyond a lot > of people simply because that research extends into partial > differential equations. To me the starting point is simple enough: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], You have previously repudiated that method on the grounds that it relies on the *inherently* ambiguous 'sqrt'. It fails for the negative values returned which are, according to you, impossible to avoid. > S(x,1) = 0, p(x, y) = floor(x) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y). But I'm increasingly aware that what looks simple to me, leaves > mathematicians all over the world befuddled. On the contrary. The method has left *you* befuddled, as you repeatedly confessed in previous posts. Your inability to find any means of confining the sqrt to positive numbers renders the method useless. [snip redundant material previous posted] -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== > My analysis continues to indicate that my research on counting prime > numbers *should* be the most accessible than my other math research That should have been that it should be more accessible than my other > math research. which is more abstract, and clearly more difficult. Still I also > recognize that significant parts of my prime counting research are > beyond a lot of people simply because that research extends into > partial differential equations. > floor((N-36)/14) + floor((N-22)/42) That's fine within a certain range. > is basically the explicit result of summing an algorithmic form of the > dS(x,y) function, where evens are automatically handled, from 2 up to > and including 10, which represents the primes 2, 3, 5 and 7, so it > gives a count of primes up to and including 120. For instance, N=100, gives 50 - 16 - 8 + 2 - 4 + 1 = 25 as expected. > But beyond counting primes in a small range it works to give N minus > the count of composites up to and including N that have 2, 3, 5, or 7 > as a factor out to positive infinity for even N. Actually I forgot that beyond 106 you have more terms, and the > expression then is N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - > floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - > floor((N-106)/210) + 2. > > It is the smallest expression possible for that job. You can roll that 2 back into it--carefully--but it's just as easy to > leave it hanging out there. > > Want more on prime counting? Then see my blog archives: > > > > James Harris You have got to be kidding. You have posted this over and over again and you still cannot get it right? What is wrong with you? It is *your* amazing discovery and you can't type it in correctly the first time. Amazing indeed. You cannot even cut and paste from your previous garbage posts. Pathetic. Have a nice day, Jay ==== > Why do you think _anyone_ wants more of _any_ of this stuff? I mean really: has _anyone_ expressed any interest? I mean > even one person? There would be things that I would find interesting: For example, since there is a very fast rough estimate for pi (x), namely li (x) = gamma + log log x + sum ((log x)^k) / k! / k, 1 <= k <= inf) and a rather slow algorithm taking about O (x^(2/3)) steps to calculate pi (x) exactly, is there any approximation for pi (x) that is substantially more precise than li (x) but can be calculated substantially faster than O (x^(2/3))? Another question: pi (M) - pi (N) can easily be calculated in O (min (M - N, N^(2/3))) steps by using a sieve if M is close to N, and using the Extended Meissel-Lehmer algorithm to calculate pi (M) and pi (N) of the difference between M and N is large. Is there any substantially faster algorithm? Another problem which I think I have solved: Given k >= 2 and values x(i) for 1 <= i <= k. Can the Extended Meissel-Lehmer algorithm be modified so that it calculates pi (x (i)) for all given values x (i) substantially faster than O (sum (x(i)^(2/3), 1 <= i <= k) ? ==== > There would be things that I would find interesting: For example, since > there is a very fast rough estimate for pi (x), namely li (x) = gamma + log log x + sum ((log x)^k) / k! / k, 1 <= k <= inf) and a rather slow algorithm taking about O (x^(2/3)) steps to calculate > pi (x) exactly, is there any approximation for pi (x) that is > substantially more precise than li (x) but can be calculated > substantially faster than O (x^(2/3))? Given my own trend for being grossly inaccurate, this thread may be of no use whatsoever... http://tinyurl.com/3cjlh ...but some may find it of interest. Carl ==== >> automatic decision support system for my strategy game, I'm looking for >> an equation for estimating the monopolization factor. > I think it depends on whether you're the top hat or the little racing > car. Also, if you have Boardwalk and Park Place with hotels, those > factors just don't mean the same thing as when you are sitting there > with Baltic and Mediterranean Avenues. ROTFL. I prefer sunny beaches of Puerto Rico, however I also like CA governed by Mr Arnie... (-; >> Company1 produces 3 units of goods, Company2 - 2, Company3 - 2. What is >> the market monopolization factor for Company1, and how will it change >> when it increases its production from 3 to 4 units? > BTW, you might be more likely to get good answers if you were to define > your terms. Either that, or ask an economics newsgroup. percentage of the market has any potential traps. First situation: C1 produces 3 units, total production = 7 units; market monopolization = 0.43; second situation: C1 produces 4 units, total production = 8 units; market monopolization = 0.5. I personally don't see any drawbacks here (yet), but I thought someone with more mathematical experience would. If not, I will be trying it in practice and let you know. :)) -- BB ==== Dear all, I've tried various approaches, but none seem convincing. Here's the problem: suppose t is a real number, define the following polynomial on the complex numbers: f(z)=z^3-(4+t^2)z^2-2tz+1. My goal is to show that this polynomial has one and only one zero in (0,1) (so we can actually restrict f to be defined only on the real numbers). The fact that there is a zero is not difficult, since f|R(0)=1 and f|R(1)=-1-(1+t)^2<0 (by f|R I mean f restricted to the real numbers). By the continuity theorem(?) there has to be a zero between 0 and 1. Now, how would you go about showing that there is only one zero. And how would you show that this is not a multiple zero of f? ==== > Dear all, I've tried various approaches, but none seem convincing. Here's the > problem: suppose t is a real number, define the following polynomial on > the complex numbers: f(z)=z^3-(4+t^2)z^2-2tz+1. My goal is to show that this polynomial has one and only one zero in > (0,1) (so we can actually restrict f to be defined only on the real > numbers). The fact that there is a zero is not difficult, since f|R(0)=1 > and f|R(1)=-1-(1+t)^2<0 (by f|R I mean f restricted to the real > numbers). By the continuity theorem(?) there has to be a zero between 0 > and 1. Now, how would you go about showing that there is only one zero. > And how would you show that this is not a multiple zero of f? Did you check its value at -1 ? Jim Buddenhagen -- To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE ==== Dear all, I've tried various approaches, but none seem convincing. Here's the > problem: suppose t is a real number, define the following polynomial on > the complex numbers: f(z)=z^3-(4+t^2)z^2-2tz+1. My goal is to show that this polynomial has one and only one zero in > (0,1) (so we can actually restrict f to be defined only on the real > numbers). The fact that there is a zero is not difficult, since f|R(0)=1 > and f|R(1)=-1-(1+t)^2<0 (by f|R I mean f restricted to the real > numbers). By the continuity theorem(?) there has to be a zero between 0 > and 1. Now, how would you go about showing that there is only one zero. > And how would you show that this is not a multiple zero of f? Did you check its value at -1 ? > and its sign as z -> infinity ? ==== >Dear all, >>I've tried various approaches, but none seem convincing. Here's the >problem: suppose t is a real number, define the following polynomial on >the complex numbers: >>f(z)=z^3-(4+t^2)z^2-2tz+1. >>My goal is to show that this polynomial has one and only one zero in >(0,1) (so we can actually restrict f to be defined only on the real >numbers). The fact that there is a zero is not difficult, since f|R(0)=1 >and f|R(1)=-1-(1+t)^2<0 (by f|R I mean f restricted to the real >numbers). By the continuity theorem(?) there has to be a zero between 0 >and 1. Now, how would you go about showing that there is only one zero. >And how would you show that this is not a multiple zero of f? >>Did you check its value at -1 ? > and its sign as z -> infinity ? Ooops... This turned out to be a very silly question. :-( My apologies for taking up everyone's time for reading the original question. For the two responders, thanks for the subtle hints. ==== -T Approved: ebunn@richmond.edu (sci.physics.research moderator) ==== This section of TWF was of particular interest to me: > The classic example, familiar to all physicists, is the Galois group > of the complex numbers, C, over the real numbers, R. This group has > two elements: the identity transformation, which leaves everything > alone, and complex conjugation, which switches i and -i. Since the > only group with 2 elements is Z/2, we have Gal(C/R) = Z/2 Where does complex conjugation come from? It comes from the fact that > C from R by throwing in a solution of the quadratic equation x^2 = -1. We say C is a quadratic extension of R. But as soon as we throw in > one solution of this equation, we inevitably throw in another, namely > its negative - and there's no way to tell which is which. And complex > conjugation is the symmetry that switches them! One thing I have yet to read much about is symmetry groups and quaternions. I am sure one could define Gal(H/R), the question is what would it equal? One would need four automorphisms: the identity transformation, one that switches i, j, k and -i, -j, -k, and two others. In a hotel lobby as I was leaving Rome from a conference on quaternions, I had the idea for what I called the first and second conjugates, symbolized by q*1 and q*2, which flip the signs of all except the first and second members of the quaternion 3-vector, so (t, x, y, z)*1 = (-t, x, -y, -z) and (t, x, y, z)*2 = (-t, -x, y, -z). These are related to the standard conjugation using a rotation around i and j: q*1 = (i q i)* and q*2 = (j q j)*. Since quaternions do not commute, Z cannot be used. This might be a logical place to begin non-Abelian field theory. Non-Abelian physics appears to be every, from bicycle tires to spin. Any references would be appreciated. doug quaternions.com ==== > One thing I have yet to read much about is symmetry groups and > quaternions. I am sure one could define Gal(H/R), the question is what > would it equal? One would need four automorphisms: the identity > transformation, one that switches i, j, k and -i, -j, -k, and two > others. There is also at least a cycle of length 3 that permutes (ijk). Arnold Neumaier ==== One thing I have yet to read much about is symmetry groups and > quaternions. I am sure one could define Gal(H/R), the question is what > would it equal? One would need four automorphisms: the identity > transformation, one that switches i, j, k and -i, -j, -k, and two > others. That's not an automorphism. It's an anti-atomorphism: denoting it by phi, phi(z w) = phi(w) phi(z) not phi(z) phi(w). H has uncountably many automorphisms, all fixing R pointwise. A typical one takes i to a_{11} i + a_{12} j + a_{13} k, j to a_{21} i + a_{22} j + a_{23} k, k to a_{31} i + a_{32} j + a_{33} k where (a_{pq}) is a matrix in SO(3). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > One thing I have yet to read much about is symmetry groups and > quaternions. I am sure one could define Gal(H/R), the question is what > would it equal? One would need four automorphisms: the identity > transformation, one that switches i, j, k and -i, -j, -k, and two > others. That's not an automorphism. It's an anti-atomorphism: > denoting it by phi, phi(z w) = phi(w) phi(z) not phi(z) phi(w). H has uncountably many automorphisms, all fixing R pointwise. > A typical one takes > i to a_{11} i + a_{12} j + a_{13} k, > j to a_{21} i + a_{22} j + a_{23} k, > k to a_{31} i + a_{32} j + a_{33} k > where (a_{pq}) is a matrix in SO(3). Not unexpected. But are these the only automorphisms fixing reals? I think I asked the same question somewhere before...but my real concern is of course the sensible (those I can construct) automorphisms. ==== >> >> >> One thing I have yet to read much about is symmetry groups and >> quaternions. I am sure one could define Gal(H/R), the question is what >> would it equal? One would need four automorphisms: the identity >> transformation, one that switches i, j, k and -i, -j, -k, and two >> others. >> >> That's not an automorphism. It's an anti-atomorphism: >> denoting it by phi, phi(z w) = phi(w) phi(z) not phi(z) phi(w). >> >> H has uncountably many automorphisms, all fixing R pointwise. >> A typical one takes >> i to a_{11} i + a_{12} j + a_{13} k, >> j to a_{21} i + a_{22} j + a_{23} k, >> k to a_{31} i + a_{32} j + a_{33} k >> where (a_{pq}) is a matrix in SO(3). Not unexpected. But are these the only automorphisms fixing reals? As I said automorphisms, all fixing R pointwise. To expand: if f: H -> H is a ring automorphism, then f(a) = a for all a in R. Proof: exercise for reader. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== Here's another example of a Galois group that physicists should like. > Let C(z) be the field of rational functions in one complex variable z - > in other words, functions like f(z) = P(z)/Q(z) where P and Q are polynomials in z with complex coefficients. You can > add, subtract, multiply and divide rational functions and get other > rational functions, so they form a field. And they contain C as a > subfield, because we can think of any complex number as a *constant* > function. So, we can ask about the Galois group of C(z) over C. > What's it like? It's the Lorentz group! To see this, it's best to think of rational functions as functions not > on the complex plane but on the Riemann sphere - the complex plane > together with one extra point, the point at infinity. The only > conformal transformations of the Riemann sphere are fractional linear > transformations: Could you please explain why an element of the Galois group must be a conformal transformation? At first sight one would think it can be any rational substitution that is bijective. So why does bijective imply conformal? Arnold Neumaier ==== There are a few dollars to be had here: http://faculty.evansville.edu/ck6/integer/unsolved.html ==== When I cared about finding fast prime counting algorithms versus concentrating on the core mathematics, I found formulas like the following, which counts primes up to a given even N over a certain range: N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - floor((N-106)/210) + 2. And it counts primes for an even N from N=38 to N=120. Using pi(N) = floor(N) - S(N) + 1, where S(N) is the count of composites, and solving for S(N), gives N/2 + floor((N-4)/6) + floor((N-16)/10) - floor((N-16)/30) + floor((N-36)/14) - floor((N-22)/42) - floor((N-106)/70) + floor((N-106)/210) - 1 which is the count of composites that have 2, 3, 5 or 7 as a factor with an even N in the range N=38 to positive infinity. So that's 9 terms. Up to N=120, in contrast, Legendre's Method has 15 terms. Which continues a trend started much earlier as they're closest up to 8 where for both you have N/2 - 1 while up to 24, Legendre's Method gives N/2 + floor(N/3) - floor(N/6) - 2 while my work gives N/2 + floor(N-4)/6 + 1 for the count of composites with 2 or 3 as a factor. James Harris ==== In sci.math.num-analysis, James Harris on 15 Jan 2004 07:31:20 -0800 <3c65f87.0401150731.73acc418@posting.google.com>: > When I cared about finding fast prime counting algorithms versus > concentrating on the core mathematics, I found formulas like the > following, which counts primes up to a given even N over a certain > range: N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - > floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - > floor((N-106)/210) + 2. And it counts primes for an even N from N=38 to N=120. Using pi(N) = floor(N) - S(N) + 1, where S(N) is the count of > composites, and solving for S(N), gives N/2 + floor((N-4)/6) + floor((N-16)/10) - floor((N-16)/30) + > floor((N-36)/14) - floor((N-22)/42) - floor((N-106)/70) + > floor((N-106)/210) - 1 which is the count of composites that have 2, 3, 5 or 7 as a factor > with an even N in the range N=38 to positive infinity. So that's 9 terms. Up to N=120, in contrast, Legendre's Method has 15 terms. Which continues a trend started much earlier as they're closest up to > 8 where for both you have N/2 - 1 while up to 24, Legendre's Method gives N/2 + floor(N/3) - floor(N/6) - 2 while my work gives N/2 + floor(N-4)/6 + 1 for the count of composites with 2 or 3 as a factor. > James Harris Whoop-te-do. You're going to have to work hard to beat Christian Bau's implementation of Meissel-Lehmer. http://www.cbau.freeserve.co.uk/ (I'll admit I'm tempted to code it in Java, myself. Just for the perversity... :-) ) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > When I cared about finding fast prime counting algorithms versus > concentrating on the core mathematics, I found formulas like the > following, which counts primes up to a given even N over a certain > range: N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - > floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - > floor((N-106)/210) + 2. If you no longer care about finding fast prime counting algorithms, why are you posting this now? [snip exposition of method cryptically posted in spite of James' above suggested lack of current interest] -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== When I cared about finding fast prime counting algorithms versus > concentrating on the core mathematics, I found formulas like the > following, which counts primes up to a given even N over a certain > range: >N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - > floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - > floor((N-106)/210) + 2. If you no longer care about finding fast prime counting algorithms, why > are you posting this now? > In this case I'm leaving a note to myself covering a slightly different approach. But hey, as usual, there's always the possibility that someone else might give a useful comment. I have lots of areas where I can go over my own research, so I get to pick and choose depending on my mood and purpose. I find it useful to sharpen my focus at times by posting in a certain direction. James Harris ==== > >> When I cared about finding fast prime counting algorithms versus >> concentrating on the core mathematics, I found formulas like the >> following, which counts primes up to a given even N over a certain range: >> N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - >> floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - >> floor((N-106)/210) + 2. If you no longer care about finding fast prime counting algorithms, why > are you posting this now? If it is mathematically correct and interesting, why does it matter? Unlike pretty much all of his earlier postings, this one appears to be mathematically correct, uses standard terminology, and is either interesting or suggests interesting things. Why are you picking on him? -- --Tim Smith ==== that's what I was going to say, Hear, here!, but I was going to try to edit-out all of the superfluous marketing, or what ever it is that looks just like whining (for going-on Decade Two .-) much better, if he did it himself. maybe he doesn't have a wordprocessor? one man's inability to grok mathematics, is another man's masturbation ... I *said* I'd get outta here ... unless there's some actual math involved. > If it is mathematically correct and interesting, why does it matter? Unlike > pretty much all of his earlier postings, this one appears to be --Give the Gift of Trickier Dick Cheeny -- out of office, at last! http://www.wlym.com/pages/music.html http://members.tripod.com/~american_almanac http://www.benfranklinbooks.com/ http://www.wlym.com/PDF-68-76/CAM7606.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ ==== > >> When I cared about finding fast prime counting algorithms versus >> concentrating on the core mathematics, I found formulas like the >> following, which counts primes up to a given even N over a certain range: >> N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - >> floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - >> floor((N-106)/210) + 2. If you no longer care about finding fast prime counting algorithms, why > are you posting this now? If it is mathematically correct and interesting, why does it matter? It doesn't, and everyone has said so. > Unlike > pretty much all of his earlier postings, this one appears to be > mathematically correct, That has never been an issue. > uses standard terminology, Nor has that. > and is either interesting You're getting warmer... > or suggests interesting things. Now you got it! The interesting things that are being suggested are in dispute as being uninteresting, flat out wrong and outright lies. > Why are you picking on him? Because he's a troll. ==== > >> When I cared about finding fast prime counting algorithms versus >> concentrating on the core mathematics, I found formulas like the >> following, which counts primes up to a given even N over a certain range: >> N/2 - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - >> floor((N-36)/14) + floor((N-22)/42) + floor((N-106)/70) - >> floor((N-106)/210) + 2. >If you no longer care about finding fast prime counting algorithms, why > are you posting this now? If it is mathematically correct and interesting, why does it matter? Unlike > pretty much all of his earlier postings, this one appears to be > mathematically correct, uses standard terminology, and is either interesting > or suggests interesting things. Why are you picking on him? -- > --Tim Smith The material has been posted before, has been repudiated by him, and was placed here for no other reason than to use this newsgroup as a vehicle for leaving a note to himself. See James' response for a confirmation. Haven't you been paying attention? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== I need to know a method to find the rate of convergence of any iterative solution. Say suppose, I need to find the convergence rate of Newton-Raphson method or secent method of finding roots of an equation, how should I go for it. I know (from other posting on this site) that the rate of convergence of Newton Method is 2, but I dont know the mathematics behind this. Can anyone explain me how the rates are calculated, and can anybody point me to some site/book/eBook which will teach me these things in detail. Satya ==== > I need to know a method to find the rate of convergence of any > iterative solution. Say suppose, I need to find the convergence rate > of Newton-Raphson method or secent method of finding roots of an Satya, Try Numerical Methods that Work by Forman S. Acton http://www.maa.org/pubs/books/nmtw.html or Numerical Methods for Scientists & Engineers by Richard Hamming http://www.mathbook.com/n/Number_Theory/Numerical_Methods_for_Scientists_and _Engineers_0486652416.htm They are both quite old, but still in print AFAIK. I've checked my copies - they both cover your question for Newton Raphson - and you should be able to figure out convergence for other methods by applying the same approach. BTW - did you do a google search on Newton Raphson first? If so you should Simon ==== Can anyone tell me what a Schur Complement is and what are its properties? I think it has something to do with the correlation between two signals, for example if x and y have auto- and cross-correlation matrices Rxx, Ryy, Rxy and Ryx then the Schur complement is Rx|y = Rxx - Rxy*inv(Ryy)*Ryx or Ry|x = Ryy - Ryx*inv(Rxx)*Rxy a million... ************************** ==== > Can anyone tell me what a Schur Complement is and what are its > properties? I think it has something to do with the correlation > between two signals, for example if x and y have auto- and > cross-correlation matrices Rxx, Ryy, Rxy and Ryx then the Schur > complement is Rx|y = Rxx - Rxy*inv(Ryy)*Ryx or Ry|x = Ryy - Ryx*inv(Rxx)*Rxy a million... Here is how I use Schur complemets: Let M be a 2x2 matrix whose entries are matrices or operators. M = ( A B ) ( C D ) (view with fixed width font). We can view M as an operator from a product space X x Y to itself. The problem now is, how to find an inverse for M. The answer can be given by the Schur complement. If D is invertible and the Schur complement A - B inv(D) C is invertible too, then M is invertible, and we have a formula for the inverse of M inv(M) = ( S - S B inv(D) ) ( - inv(D) C S inv(D) - inv(D) C S B inv(D) ) where S = inv( A - B inv(D) C ). A similar formula holds if A and D - C inv(A) B are invertible. HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de ==== have a look at Mersenne Twister website http://www.math.keio.ac.jp/~matumoto/emt.html > Copyright(c)2004 by Hermann Samso. > All Rights Reserved. > Good evening. First test at a RNG with a Motorola > MC68000 processor. > First try was to code the Lewis, Goodman > and Miller algorithm from 1969, but then > realized that a long division opcode was > necessary, and MC68000 can only divide > by words. So I gave up, and tried desperately with > a 32bit long dividend described in above > mentioned algorithm and a 16 bit number > as divisor, I chose 65533 that is nearly > the whole of 16bits. The algorithm here described is probably > not much random, and hasn't yet been > studied or benchmarked respectly. Anyways I found the results good enough > to be included in my last retro scene > demo intro for Atari ST. So, I have > decided to include it here, and maybe > get some advisings and/or proofs. The code I present here is in assembler > and can be easily compiled. It is only > 4 instructions long! It bases on the machine code of the > MC68000 which delivers remainder in the > high word, quotient in the low word of > a long word as result of an unsigned > division. > The only step inbetween is to rotate > this results around in the long word > by a fixed amount. I chose 8 bits, > but probably 7 or 23 are also good > values. > saludos, > Hermann Samso # Copyright(c)2004 by Hermann Samso. > # All Rights Reserved. > # RND1.s > # An easy way ? RND1 move.l D,d0 > divu d,d0 > ror.l #8,d0 > move.l d0,D dc.l D 2147483647 ;dividend > dc.w d 65533 ;divisor > http://members.tripod.com/so_o2 ==== The probability so that the stethoscope of Gineco is in one of the ten rooms of a hospital is equal to p/10. Did Gin.8eco seek in vain its stethoscope in the first 9 rooms, which is the probability so that he finds it in the last? (0Could anyone point me to a good discussion, either in print or online, of >the attitudes towards the epistemological differences between: >(1) constructing all solutions to a combinatorics problem > through a brute-force search on a computer, >and >(2) enumerating the number of solutions algebraically (say, > using group actions, or some such approach) >? >Of course the algebraic approach feels more elegant--I have that feeling >too--but in cases where the exhaustive search is easier to implement >(spend a few months learning a computer language versus spend many years >learning appropriate mathematics and even then not be sure of finding a >solution), easier to verify for the same reason, faster to complete, >etc... why do we still care that we don't have an algebraic solution? I might ask, why would we care that we do have a non-algebraic solution? That is, if all you get for your computer troubles is some big number, or a long exhaustive (and exhausting) list of cases, what have you learned? By contrast, if you have an algebraic solution to this sort of problem, you probably have a better sense of the kinds of solutions there are and so on. dave ==== In a book I have found the following formulation n xxxxxxxxxxxxxxxxxx x x x x x x x x x x xxxxxxxxxxxxxxxxxx j=1 It's look like a capital PI ! Can someone tell me waht its means ? -- Bernard Bour.8ee bernard@bouree.net ==== In a book I have found the following formulation > n > xxxxxxxxxxxxxxxxxx > x x > x x > x x > x x > x x > xxxxxxxxxxxxxxxxxx > j=1 It's look like a capital PI ! Can someone tell me waht its means ? > Its a product sign. ie: Product[j=1..4] x = x * x * x * x Product[j=1..5] j = 1 * 2 * 3 * 4 * 5 l8r, Mike N. Christoff ==== In a book I have found the following formulation > n >xxxxxxxxxxxxxxxxxx > x x > x x > x x > x x > x x >xxxxxxxxxxxxxxxxxx > j=1 It's look like a capital PI ! Bad ASCII art, then. >Can someone tell me waht its means ? It is a product sign. It means you are to multiply the expression that follows the sign, one factor for each value of j, j=1,2,3,4,...,n. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== In a book I have found the following formulation > n > xxxxxxxxxxxxxxxxxx > x x > x x > x x > x x > x x > xxxxxxxxxxxxxxxxxx > j=1 It's look like a capital PI ! Can someone tell me waht its means ? product? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== Moon Dirt isn't just Moon Dirt, it's Everything Dirt I have absolutely no doubts that once upon a time Mars had a sufficient atmosphere, thereby a warmer and radiation protected environment, possibly even long enough to have sustained either natural evolution and/or of some well intended terraforming on behalf of establishing some life similar to human. Unfortunately, there are certain limits to which life and of it's DNA/RNA as we know it can coexist within the confines of what Mars has had to offer for the past few thousand years, and certainly things are not getting any better. The more the likes of Mars core cools itself off, the worse becomes any opportunity for that planet to revive itself, short of receiving a massive infusion of artificial energy, such as what 1000 terawatts per year as derived from our lunar He3 might have to offer. Some good readings: SADDAM HUSSEIN and The SAND PIRATES http://mittymax.com/Archive/0085-SaddamHusseinAndTheSandPirates.htm The latest insults to this Mars/Moon injury: http://guthvenus.tripod.com/gv-moon-02.htm http://guthvenus.tripod.com/gv-gwb-moon.htm http://guthvenus.tripod.com/gv-interplanetary.htm http://guthvenus.tripod.com/moon-04.htm ==== <9f50a7c5.0401151119.13422bd@posting.google.com> thusly: > Moon Dirt isn't just Moon Dirt, it's Everything Dirt I have absolutely no doubts that once upon a time Mars had a > sufficient atmosphere, thereby a warmer and radiation protected > environment, possibly even long enough to have sustained either > natural evolution and/or of some well intended terraforming on behalf > of establishing some life similar to human. FWIW my own opinion is that Mars has had a substantial atmosphere and possibly even running water, at several points in its history. The times when it was hit by a comet. Since Mars' gravity is too weak to hold this stuff in the atmosphere for long, it is lost to space - until the next time. -- Paul Townsend I put it down there, and when I went back to it, there it was GONE! Interchange the alphabetic elements to reply ==== I talked to a psychiatrist, and I think I may have A.D.D. I guess I > could get some drugs. Does anyone have any ideas of things I could do? How about listen to the doctor and take the drugs? The psychiatrist Probably knows more about this subject than anyone in this group. Or you could just ignore it and hope it goes away. Of coarse that would probably mean you flunk out of college. Not to worry in even these times of economic challenges Burger King is hiring. > I don't really this obssesion to go away, but I also don't really want > to flunk out of college. Has anyone else experienced similar problems? Anyone have any advice, > or know something I should do? Most of us have experienced similar problems that is why we come to this group. -- Quote of the month . When trouble arises and things look bad, there is always one individual who perceives a solution and is willing to take command. Very often, that individual is crazy. Ok two quotes: The SAT losing the analogy section is like Christmas losing fruitcake ÷ it may have been ătried and true, but nobody really ever liked it that much.ä Jon Zeitlin, director of Kaplanâs New SAT test prep program,. ==== While playing with the Fibonacci sequence in a spreadsheet, the following interesting pattern appeared. When the Fibonacci number is divided by its position in the list, the result is a whole number when the position corresponds to a left truncatable prime*. I ran out of accuracy to test this past the 67th number and I know little about left truncatable primes. Does anyone know if this property is anything worth digging into? A google search on fibonacci and primes numbers returns a few things but nothing of great depth. If anyone has more info about Fibonacci and primes I would appreciate anything you care to share. The numbers from the spreadsheet are below. Primes(*) Column A=Fibonacci Column B=Position Column C=A/B Also notice the primes appear the intervals of 6,4,6,14,6,4,6,14... wonder if that would continue? Do I have too much time on my hands? Many thanks, Bob Carlson Fibonacci Sequence Position A B A/B 1 0 1 1 1.000000 2 2 1.000000 3 3* 1.000000 ----- 5 4 1.250000 8 5 1.600000 13 6 2.166667 21 7* 3.000000------- 34 8 4.250000 55 9 6.111111 89 10 8.900000 144 11 13.090909 233 12 19.416667 377 13* 29.000000------ 610 14 43.571429 987 15 65.800000 1597 16 99.812500 2584 17* 152.000000----- 4181 18 232.277778 6765 19 356.052632 10946 20 547.300000 17711 21 843.380952 28657 22 1302.590909 46368 23* 2016.000000----- 75025 24 3126.041667 121393 25 4855.720000 196418 26 7554.538462 317811 27 11770.777778 514229 28 18365.321429 832040 29 28691.034483 1346269 30 44875.633333 2178309 31 70268.032258 3524578 32 110143.062500 5702887 33 172814.757576 9227465 34 271396.029412 14930352 35 426581.485714 24157817 36 671050.472222 39088169 37* 1056437.000000----- 63245986 38 1664368.052632 102334155 39 2623952.692308 165580141 40 4139503.525000 267914296 41 6534495.024390 433494437 42 10321296.119048 701408733 43* 16311831.000000----- 1134903170 44 25793253.863636 1836311903 45 40806931.177778 2971215073 46 64591632.021739 4807526976 47* 102287808.000000---- 7778742049 48 162057126.020833 12586269025 49 256862633.163265 20365011074 50 407300221.480000 32951280099 51 646103531.352941 53316291173 52 1025313291.788460 86267571272 53* 1627690024.000000---- 139583862445 54 2584886341.574070 225851433717 55 4106389703.945450 365435296162 56 6525630288.607140 591286729879 57 10373451401.386000 956722026041 58 16495207345.534500 1548008755920 59 26237436541.016900 2504730781961 60 41745513032.683300 4052739537881 61 66438353080.016400 6557470319842 62 105765650320.032000 10610209857723 63 168416029487.667000 17167680177565 64 268245002774.453000 27777890035288 65 427352154389.046000 44945570212853 66 680993488073.530000 72723460248141 67* 1085424779823.000000 ---- 117669030460994 68 1730426918544.030000 ==== Now if you wanted to know what Fibonacci numbers are prime then the list is Fibonacci[n] where Fibonacci[3] = 2 where n = {3,4,5,7,11,13,17,23,29,43,47,83,131,137,359,431,433,449,509, 569,571,2971,4723,5387,9311,9677,14431,25561,30757,35999,81839,...} others that are possible primes are n = {37511, 50833, 104911 ,130021, 148091, 201107, 397379, 433781} ==== > While playing with the Fibonacci sequence in a spreadsheet, the > following interesting pattern appeared. When the Fibonacci number is divided by its position in the list, the > result is a whole number when the position corresponds to a left > truncatable prime*. I ran out of accuracy to test this past the 67th > number and I know little about left truncatable primes. Does anyone > know if this property is anything worth digging into? A google search > on fibonacci and primes numbers returns a few things but nothing of > great depth. If anyone has more info about Fibonacci and primes I would appreciate > anything you care to share. The numbers from the spreadsheet are below. Primes(*) Column > A=Fibonacci Column B=Position Column C=A/B Also notice the primes > appear the intervals of 6,4,6,14,6,4,6,14... wonder if that would > continue? Do I have too much time on my hands? Many thanks, Bob Carlson Fibonacci > Sequence Position > A B A/B > 1 0 > 1 1 1.000000 > 2 2 1.000000 > 3 3* 1.000000 ----- > 5 4 1.250000 > 8 5 1.600000 > 13 6 2.166667 > 21 7* 3.000000------- > 34 8 4.250000 > 55 9 6.111111 > 89 10 8.900000 > 144 11 13.090909 I have no idea what a left truncatable prime might be. However I can explain some of what you are seeing. First of all, your position numbering is not the standard one. Normally the Fibonacci number 2 is in position 3, and positions 1 and 2 are both occupied by 1. The 0'th Fibonacci number is 0. With that convention, it is a fact that F(p-1) is divisible by p whenever p is congruent to 1 or 4 mod 5, and F(p+1)) is divisible by p when p is congruent to 2 or 3 mod 5. The only remaining prime is 5 is it is a factor of F(5). This explains what you are seeing, I think. I don't actually totally know, because I have no idea what a left truncatable prime is. Could you explain? Hope this helps, Achava ==== > I have no idea what a left truncatable prime might be. However I can > explain some of what you are seeing. First of all, your position > numbering is not the standard one. Normally the Fibonacci number 2 is > in position 3, and positions 1 and 2 are both occupied by 1. The 0'th > Fibonacci number is 0. With that convention, it is a fact that F(p-1) > is divisible by p whenever p is congruent to 1 or 4 mod 5, and F(p+1)) > is divisible by p when p is congruent to 2 or 3 mod 5. The only > remaining prime is 5 is it is a factor of F(5). This explains what > you are seeing, I think. I don't actually totally know, because I > have no idea what a left truncatable prime is. Could you explain? Hope this helps, > Achava Achava, yes this helps. I see the position numbering error I made, pointed out by you and others here. An example of a left truncatable prime is 4967. If you truncate the 4 (from the left), 967 is still prime. Cut off the 9 and 67 is still prime. So is 7. The structure of these numbers intrigue me and mystical nature of these numbers goes back 2000 years. I have a long way to go. Bob ==== > While playing with the Fibonacci sequence in a spreadsheet, the > following interesting pattern appeared. When the Fibonacci number is divided by its position in the list, the > result is a whole number when the position corresponds to a left > truncatable prime*. I ran out of accuracy to test this past the 67th > number I checked from the 67th position to 99,643rd position, whose Fibonacci number has 20,825 digits and it worked for all of them. However... > and I know little about left truncatable primes. Neither do I, so I went to MathWorld and then Sloanes EIS and got the list. You overlooked the fact that 5 is a left truncatable prime. And it fails for 5. > 8 5 1.600000 Bummer. > Does anyone > know if this property is anything worth digging into? A google search > on fibonacci and primes numbers returns a few things but nothing of > great depth. If anyone has more info about Fibonacci and primes I would appreciate > anything you care to share. The numbers from the spreadsheet are below. Primes(*) Column > A=Fibonacci Column B=Position Column C=A/B Also notice the primes > appear the intervals of 6,4,6,14,6,4,6,14... wonder if that would > continue? Do I have too much time on my hands? Many thanks, Bob Carlson > ==== You say the pattern is {6,4,6,14,...} starting at 7. Sorry to disappoint but: 7+6 = 13 works 13+4 = 17 works 17+6 = 23 works 23 + 14 = 37 works 37+6 = 43 works 43 + 4 = 47 works 47 + 6 = 53 works 53 + 14 = 67 works 67 + 6 = 73 works 73 + 4 = 77 = 7*11 FAILS So, the Rule of small numbers got you. ==== > You say the pattern is {6,4,6,14,...} starting at 7. Sorry to disappoint but: 7+6 = 13 works 13+4 = 17 works 17+6 = 23 works 23 + 14 = 37 works 37+6 = 43 works 43 + 4 = 47 works 47 + 6 = 53 works 53 + 14 = 67 works 67 + 6 = 73 works 73 + 4 = 77 = 7*11 FAILS > So, the Rule of small numbers got you. Jason, thanks. Of all numbers, how could I crap-out on the 7 11!! ==== > When the Fibonacci number is divided by its position in the list, the > result is a whole number when the position corresponds to a left > truncatable prime*. Fibonacci > Sequence Position > A B A/B > 1 0 > 1 1 1.000000 > 2 2 1.000000 > 3 3* 1.000000 ----- > 5 4 1.250000 > 8 5 1.600000 You're off by one from the usual indexing of the Fibonaccis; the standard way, it goes a_1 = 1, a_2 = 1, a_3 = 2, etc. But that's OK. Using the standard indexing, I think what really happens here is that a prime p (other than 5) divides a_{p+1} if and only if 5 is a quadratic non-residue modulo p. Equivalently, p must end in 3 or 7 when written in decimal. For a 2-digit prime, this is the same thing as saying p is left-truncatable, but it breaks down for larger p. If I'm right, 127 will divide a_128 (or, in your notation, a_127), even though 127 is not left-truncatable. What's behind this is the formula a_n = (x^n - y^n)/sqrt5, where x = (1 + sqrt5)/2 and y = (1 - sqrt5)/2. There is or isn't a sqrt5 modulo p depending on whether 5 is or isn't a quadratic residue modulo p (that's the definition of quadratic residue). If there is, then little Fermat tells you something about the remainders of x^(p+1) and y^(p+1) when you divide by p. If there isn't, then x and y are in a quadratic extension of the integers modulo p, which tells you something else about x^(p+1) and y^(p+1). This is a bit sketchy, but I think I've given you enough buzz words to get you started on finding out more about this stuff. -- ==== You're off by one from the usual indexing of the Fibonaccis; > the standard way, it goes a_1 = 1, a_2 = 1, a_3 = 2, etc. > But that's OK. Using the standard indexing, I think what really happens here > is that a prime p (other than 5) divides a_{p+1} if and only if > 5 is a quadratic non-residue modulo p. Equivalently, p must end > in 3 or 7 when written in decimal. For a 2-digit prime, this is > the same thing as saying p is left-truncatable, but it breaks > down for larger p. If I'm right, 127 will divide a_128 (or, in > your notation, a_127), even though 127 is not left-truncatable. What's behind this is the formula a_n = (x^n - y^n)/sqrt5, > where x = (1 + sqrt5)/2 and y = (1 - sqrt5)/2. There is or isn't > a sqrt5 modulo p depending on whether 5 is or isn't a quadratic > residue modulo p (that's the definition of quadratic residue). > If there is, then little Fermat tells you something about the > remainders of x^(p+1) and y^(p+1) when you divide by p. If there > isn't, then x and y are in a quadratic extension of the integers > modulo p, which tells you something else about x^(p+1) and > y^(p+1). This is a bit sketchy, but I think I've given you enough buzz > words to get you started on finding out more about this stuff. direction. Seems I have to go through all the ugly math to see the beauty in a simple solution. Bob ==== I have a series sum(i= 1 to 20) a/(1+c)^i easy to do in excel, but how does one do this in a standard scientific calculator? I actually need to the above to a number and solve for say, c, like so, 4/(1+c)^1 + 4/(1+c)^2 + .... + 4/(1+c)^20 = 98 The calculator I am using does have a solver, but runs out of limits on max length of the expression. Have an exam tomorrow! ==== Bob. See if using the following equation instead will help for your calculator's Solver. (a - a/(1 + c)^n)/c (with a=4,n=20,total of 98, I get a c of -0.018595428 and also c = -1.8260122255811773) -- Dana I have a series sum(i= 1 to 20) a/(1+c)^i easy to do in excel, but how does one do this in a standard scientific > calculator? I actually need to the above to a number and solve for say, c, like > so, 4/(1+c)^1 + 4/(1+c)^2 + .... + 4/(1+c)^20 = 98 The calculator I am using does have a solver, but runs out of limits > on max length of the expression. Have an exam tomorrow! ==== En el mensaje:f8b7dc86.0401151154.10923938@posting.google.com, I have a series sum(i= 1 to 20) a/(1+c)^i easy to do in excel, but how does one do this in a standard scientific > calculator? I actually need to the above to a number and solve for say, c, like > so, 4/(1+c)^1 + 4/(1+c)^2 + .... + 4/(1+c)^20 = 98 The calculator I am using does have a solver, but runs out of limits > on max length of the expression. You have listened to speak something about the sum of a geometric progression? > Have an exam tomorrow! Luck! -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== Arturo Magidin a .8ecrit dans le message de >> Let x,y be elements of a group G. Prove that if xy=yx then >>rank(xy)=LCM(rank(x),rank(y)). >> I assume that by rank you mean the order of the element? >> I know how to do it when LCD(rank(x),rank(y))=1, but have no idea how to >expand >> the proof to be good for the more general case. >> That's probably because it is false in general (though it is true when >> the two orders are relatively prime). >> Arturo Magidin, sans .sig Let p_1, p_2, p_3 primes distincts. >By example, if order(a)=p_1^2*p_3^4 and order(b)=p_2^5*p_3^3 and a*b=b*a >then order(a*b)=LCM(order(a),order(b)) but if order(a)=p_1^2*p_3^4 and >order(b)=p_2^5*p_3^4 then order(a*b) is not always Lcm(order(a),order(b)) >See Odoux Ramis Deschamps Tome 1 Exercise 2.18 >True or false? >Excuse my bad english In fact, here's a nice generalization of the usual result: Let order(x) = m, order(y) = n, n,m>0, and assume that xy = yx. Assume that for every prime p, if ord_p(n) = ord_p(m), then ord_p(n)=ord_p(m)=0 (where ord_p(a) = r, r a nonnegative integer, if and only if p^r divides a, but p^{r+1} does not divide a). Then order(xy) = lcm(n,m). This implies the usual result (in which gcd(n,m)=1), since in that case, no prime that divides n can divide m and vice-versa. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > Pardon my simplistic view but there truly seems to be a connection > between everything that exists in nature, and to a good extent, > everything created by man that is perceived to be pleasant to the eye > or ear. Everything in nature appears to have some connection to Pi > and/or phi (Fibonacci sequence - Golden Ratio). There also seems to > be a link between double toroids, fractals, and images created based > on Phi. Two toroids linked like chain links form the image of > splitting cells, two hydrogen atoms and, I'll bet my last buck, the > structure of the universe. Well with the big mystery solved what do we do next? thank you for contributing to the Unified Quantum Computer that is the human race. our thoughts run parallel. check out my thread entitled Human Beings As Quantum Machines . . . . . . PROGRAMMING IS THE LANGUAGE OF GOD! Open Source Only. Greed is for the rich. ==== thusly: >> Well with the big mystery solved what do we do next? Stop for lunch -- Paul Townsend I put it down there, and when I went back to it, there it was GONE! Interchange the alphabetic elements to reply ==== .for-some-reason-or-another-a-concept-has-stuck-with-me. . .i-convinced-some-woman-of-totality-and-she-said-WE ARE PART OF ONE HOLE!!!. .immediately-the-concept-of-a-BLACKHOLE-sprung-to-mind.. ==== Do you want to play as firm against another firm? Under www.game.uni-bonn.de/index you can find an internet experiment. You can test yourself whether you are better than the computer or other players. It takes about 5 to 10 minutes. ---------------------------------------------------------- Dr. Burkhard C. Schipper Dept. of Economics University of Bonn Adenauerallee 24-42 53113 Bonn Germany Web: http://www.bgse.uni-bonn.de/~burkhard ---------------------------------------------------------- ==== I have read two or three accounts of Goedel's theorem, > for example, and never quite understood why it has any > implications of interest, since it merely seems to be the > Liar paradox embellished with mathematical symbols and > methodology. Can anyone recommend anything that will > clarify this for me? Of course it doesn't matter how it got there. What matters is what it is. (Reminds me of people who criticize mathematical discoveries because they're not complicated enough! What's that saying about people who come up with unnecessarily complex solutions not really understanding the problem?) I always thought that the significance is that it opens up the possibility that unresolved conjectures of mathematics (pre-Wiles FLT, Goldbach, Twin Primes) are true but unprovable. But someone said No unreachable. Charlie Volkstorf BTW: Get this (for those debating the relationship between Godel and The Liar)! 1. Liar: This is not true. has no value. 2. 1st part of Godel: This is not provable. has a value. 3. Charlie: Thus true does not equal provable! (so sound => incomplete) ==== >I have read two or three accounts of Goedel's theorem, > for example, and never quite understood why it has any > implications of interest, since it merely seems to be the > Liar paradox embellished with mathematical symbols and > methodology. Can anyone recommend anything that will > clarify this for me? Of course it doesn't matter how it got there. What matters is what it > is. (Reminds me of people who criticize mathematical discoveries > because they're not complicated enough! What's that saying about > people who come up with unnecessarily complex solutions not really > understanding the problem?) I always thought that the significance is that it opens up the > possibility that unresolved conjectures of mathematics (pre-Wiles FLT, > Goldbach, Twin Primes) are true but unprovable. But someone said No > unreachable. Charlie Volkstorf BTW: Get this (for those debating the relationship between Godel and > The Liar)! 1. Liar: This is not true. has no value. > 2. 1st part of Godel: This is not provable. has a value. > 3. Charlie: Thus true does not equal provable! (so sound = incomplete) will follow up. I am also reviewing again the text book in which I first met Goedel's incompleteness theorem, Richard Jeffrey's Formal Logic: Its Scope and Limits, chapter 10 Undecideability, Incompleteness. Perhaps this time I will discern some of the absoluteness, concreteness and lack of whimsicality that you do. Love, respect, gratitude Chris ==== Nothing squared plus one equals zero (0). Because zero (0) is a number in Calculus, every algebraic equation is -- For example: Consider the equation: x - 3 = 0 The solution for x - 3 is the number 0 For example: Consider the equation x^3 - 3x - 4 = 0 The solution for x^3 - 3x - 4 is the number 0 For example: Consider the equation x - log x = 0 The solution for x - log x is the number 0 For example: Consider the equation y - 4 x + 1 = 0 The solution for y - 4 x + 1 is the number 0 -- Garry Denke, Geologist Denoco Inc. of Texas ==== In sci.math, Garry Denke on 15 Jan 2004 12:46:21 -0800 <4e63857.0401151246.f75b707@posting.google.com>: > Nothing squared plus one equals zero (0). Well, no doubt that's one reason sqrt(-1) was tagged imaginary, long ago, as all squares are nonnegative in the real field, and therefore somebody -- Gauss? -- had to imagine something. :-) To factorize irreducible (over R) equations such as x^2 + 1 = 0, one at some point postulated strange numbers -- i -- such that i^2 = -1. [*] Formal manipulations of R[i] are possible, where i^2 = -1 is assumed; for example, (1 + 2*i) * (3 - 4*i) = 1*3 + 2*3*i - 1*4*i - 2*4*i^2 = 3 + 6*i - 4*i + 8 = 11 + 2*i. Since (a + b*i) * (a / sqrt(a^2+b^2) - b * i / sqrt(a^2+b^2)) = 1 R[i] is a field, albeit not an ordered one, and of course 0 + 0*i turns out to be the arithmetic identity, with 1 + 0*i being the multiplicative one. It turns out that's all that's needed to reduce all polynomials. IINM, that's a variant of the Fundamental Theorem of Algebra, that every polynomial a_n*x^n + a_{n-1}*x^{n-1} + ... + a_1*x + a_0 can be written in the form (x - r_1) * (x - r_2) * ... * (x - r_n) for some set of complex r_i. (A root can appear more than once; for example x^2 + 2x + 1 = (x + 1) * (x + 1).) Of course x^2 + 1 = (x + i) * (x - i) under this regime. Because zero (0) is a number in Calculus, every algebraic equation is Ahem. The idea of solving an equation f(x) = 0 is to find an x such that f(x) = 0. For many equations this is not trivial -- for example, the roots for the general quintic cannot be represented using radicals and arithmetic operations. 0 might work of course -- e.g., the roots of x^3 - 3x = 0 are x = 0, x = sqrt(3), and x = -sqrt(3) -- but it's not a given for single-variate polynomials unless a_0 = 0, for hopefully obvious reasons. -- For example: Consider the equation: x - 3 = 0 > The solution for x - 3 is the number 0 No, it's x = 3. For example: Consider the equation x^3 - 3x - 4 = 0 > The solution for x^3 - 3x - 4 is the number 0 I'd have to work it out, but x = 0 won't work here (0^3 - 3*0 - 4 = -4). For example: Consider the equation x - log x = 0 > The solution for x - log x is the number 0 That has no solution at all in the reals. I'm not sure regarding the complex plane minus the origin. For example: Consider the equation y - 4 x + 1 = 0 > The solution for y - 4 x + 1 is the number 0 That solution is an infinite line (x,y) which passes through the points (0, -1), (1/4, 0) and (1, 3), among uncountably infinite others. Or, if you prefer, the slope is 4 and the y-intercept is -1. -- > Garry Denke, Geologist > Denoco Inc. of Texas [*] Electrical engineers often use j = sqrt(-1), as the symbol 'i' is reserved for current. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > Consider the equation: x - 3 = 0 > The solution for x - 3 is the number 0 No, it's x = 3. Are you having a problem with the equals sign or 0 the number to the right of the equals sign? > For example: Consider the equation x^3 - 3x - 4 = 0 > The solution for x^3 - 3x - 4 is the number 0 I'd have to work it out, but x = 0 won't work here (0^3 - 3*0 - 4 = -4). Are you having a problem with the equals sign or 0 the number to the right of the equals sign? > For example: Consider the equation x - log x = 0 > The solution for x - log x is the number 0 That has no solution at all in the reals. I'm not sure regarding > the complex plane minus the origin. Are you having a problem with the equals sign or 0 the number to the right of the equals sign? > For example: Consider the equation y - 4 x + 1 = 0 > The solution for y - 4 x + 1 is the number 0 That solution is an infinite line (x,y) which passes through > the points (0, -1), (1/4, 0) and (1, 3), among uncountably > infinite others. Or, if you prefer, the slope is 4 and the > y-intercept is -1. Are you having a problem with the equals sign or 0 the number to the right of the equals sign? Consider the equation x^2 + 1 = 0 The solution for x^2 + 1 is the number 0 Write again if you need help. Garry Denke, Geologist Denoco Inc. of Texas ==== By the way, what's the Flat Earth Society charging for membership these days? ==== > By the way, what's the Flat Earth Society charging for membership these > days? In number, zero (0). Garry Denke, Geologist Denoco Inc. of Texas ==== > By the way, what's the Flat Earth Society charging for membership these > days? In number, zero (0). Garry Denke, Geologist > Denoco Inc. of Texas It is the Denke head that is flat, here! ==== > It is the Denke head that is flat, here! Here are some more algebraic equations whose solution, in accordance with the rules of the mathematicians here, is zero (0) the number: -- quadratic equation; x^2 + bx + c = 0 cubic equation; x^3 + bx^2 + cx + d = 0 quartic equation; x^4 + bx^3 + cx^2 + dx + e = 0 quintic equation; x^5 + bx^4 + cx^3 + dx^2 + ex + f = 0 -- Algebra is easy because 0 is an now a number. Garry Denke, Geologist Denoco Inc. of Texas ==== Dear Mr Gates: Please send me $1,000,000 . . Sincerely, Us ==== Go kill yourself, you fucking dickhead. > Dear Mr Gates: Please send me $1,000,000 . . Sincerely, > Us ==== >I'm a cock sucker, I suck so much cock. I'm sure you do Ryan, I'm sure you do, boy. Ryan - the badtrip dude he sucks cocks his mom is nude she's a whore that's why he's rude and now chorus: And his father sucks cocks in hell, ==== On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, MiRo ishkabibbled: ^ ^>I'm a cock sucker, I suck so much cock. ^ ^I'm sure you do Ryan, I'm sure you do, boy. ^ ^ ^Ryan - the badtrip dude ^he sucks cocks ^his mom is nude ^she's a whore ^that's why he's rude ^ ^and now chorus: ^ ^And his father sucks cocks in hell, ^ ^ ^ ^ ^ Got a melody for that, so's we can sing along??? -- The Queen of DXers, as well as Queen of the Commonwealth of Virginia, as well as The Ruler of A.D.P., as well as Saint Debbe, as well as Our Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of Seville An Ointed Minister of the Universal Life Church Reverant of the Church of the SubGenius, UnOrthodox Superior Mutha Superior of the Little Sistahs of the Politically Incorrect Worshipper of Eris, Goddess of Discord I WON'T grow up!! -- Peter Pan ==== The aluminum foil deflector beanie. http://zapatopi.net/afdb.html It's your only defense. Get one now, before the eeeevillll Bush/Gates conspiracy assimilates you. >On a supercalifragilisticexpialidocious day, after dancing about singing >Bibbety bobbety boo!, MiRo ishkabibbled: >^ >^>I'm a cock sucker, I suck so much cock. >^ >^I'm sure you do Ryan, I'm sure you do, boy. >^ >^ >^Ryan - the badtrip dude >^he sucks cocks >^his mom is nude >^she's a whore >^that's why he's rude >^ >^and now chorus: >^ >^And his father sucks cocks in hell, >^^ >^ >^ >Got a melody for that, so's we can sing along??? ==== LOL :-) The aluminum foil deflector beanie. http://zapatopi.net/afdb.html It's your only defense. Get one now, before the eeeevillll Bush/Gates > conspiracy assimilates you. >On a supercalifragilisticexpialidocious day, after dancing about singing >Bibbety bobbety boo!, MiRo ishkabibbled: >^ >^>I'm a cock sucker, I suck so much cock. >^ >^I'm sure you do Ryan, I'm sure you do, boy. >^ >^ >^Ryan - the badtrip dude >^he sucks cocks >^his mom is nude >^she's a whore >^that's why he's rude >^ >^and now chorus: >^ >^And his father sucks cocks in hell, >^^ >^ >^ >^ >Got a melody for that, so's we can sing along??? > ==== > Go kill yourself, you fucking dickhead. Damn, you're the poster boy for eloquence! Oops check that.... Damn, you're the top-poster boy for eloquence! > >> Dear Mr Gates: >> Please send me $1,000,000 . Damn...you're the poster boy for DUMB. Gates is sincerely busting his ass to get the rest of the money, and you're standing in his way. That's really cold, man. >> Sincerely, >> Us ==== >Dear Mr Gates: Please send me $1,000,000 . . Sincerely, >Us Pocket change for Bill. -- Dr.Postman USPS, MBMC, BsD; Disgruntled, But Unarmed Member,Board of Directors of afa-b, SKEP-TI-CULT¨ member #15-51506-253. Shake it like a polaroid picture. - Andre 3000 of Outkast ==== What we have here is a failure to communicate. Allow me to speculate - informed and educated and useful - understanding that the realm of GWBUX is nazism, or worse. Microsoft Windows XP is full of security holes the BUSH FAMILY uses to spy on you. THE BUSH FAMILY is investing in China. They just want you to vote to make them and their friends more rich and your children more dumb and to look at ads that make your kids vote for the Bush's for kings of the world. > Dear Mr Gates: Please send me $1,000,000 . . Sincerely, > Us ==== On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, Joe Schmoe ishkabibbled: ^What we have here is a failure to communicate. Allow me to speculate - ^informed and educated and useful - understanding that the realm of GWBUX ^is nazism, or worse. Microsoft Windows XP is full of security holes the ^ BUSH FAMILY uses to spy on you. THE BUSH FAMILY is investing in ^China. They just want you to vote to make them and their friends more ^rich and your children more dumb and to look at ads that make your kids ^vote for the Bush's for kings of the world. ^ ^ ^ ^> Dear Mr Gates: ^> ^> Please send me $1,000,000 . . ^> ^> Sincerely, ^> Us ^ I disagree. It's not the Prez who's spying on us, it's the large corporations!! But of course they ARE shills for GWB... -- The Queen of DXers, as well as Queen of the Commonwealth of Virginia, as well as The Ruler of A.D.P., as well as Saint Debbe, as well as Our Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of Seville An Ointed Minister of the Universal Life Church Reverant of the Church of the SubGenius, UnOrthodox Superior Mutha Superior of the Little Sistahs of the Politically Incorrect Worshipper of Eris, Goddess of Discord I WON'T grow up!! -- Peter Pan ==== << Dear Mr Gates: Please send me $1,000,000 . . Sincerely, Us >> I'll forward your request on to Big Bill's secretary for action. But, essential for delivery. The Psychedelick Pope Saint Isidore of Laytonville ^.85^ Patron Saint of the Internet ^.85^ ˇˇ^.85^ ˇˇ http://apple2.org.za/gswv/me AOXOMOXOA and ENESSA QUA ONNICA ==== >Also, how do the odds vary as the distance from the block of properties >increases? Let u_S(x) be the probability, starting from space x, of landing on a set S of properties before passing Go (we assume x comes before the last member of S). If p_j is the probability of going j spaces in one roll, this can be calculated recursively using u_S(x) = 0 for x > max(S) u_S(x) = 1 for x in S u_S(x) = sum_j p_j u_S(x+j) otherwise For example, if S is the block of three properties 20, 21, 22, we have x u_S(x) 0 0.419157147 1 0.420426178 2 0.425276254 3 0.432340542 4 0.437795988 5 0.437966757 6 0.428844225 7 0.405578277 8 0.393742594 9 0.396254016 10 0.419153445 11 0.440264967 12 0.465034865 13 0.499037280 14 0.491126543 15 0.439879115 16 0.342592593 17 0.252314815 18 0.166666667 19 0.083333333 20 1.000000000 Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== I did mean to indicate that two 6 sided dice were used for movement and therefore each of the 12 spaces in front of a piece are not equally likely. I meant that the board design itself does not make any space more likely than any other in the long run (no goto jail or other movement spaces or cards). Are there existing programs to aid in numerically computing the probabilities of landing on spaces (say from 1 to 40 spaces away) in n (say each of 1 to 10) rolls? I would be very interested in a chart showing the probabilities for such a range of spaces and rolls of the dice. Jim Dobie > Your problem is not formulated carefully. If all spaces are equally > likely means they all have the same probability of being landed upon, > then owning any set of 12 spaces is as good as any other. The > reference to 12 indicates that your opponent moves by rolling 2 dice. > This opens the chance that some locations will be visited more often > than others. Finally, how long does the game last? At first glance, I > think that all locations will be visited eventually, but some may not > be visited in (say) 10 rolls of the dice. A way to solve for the occupancy probabilities is to set up a Markov > chain that describes the location of the opponent. The dice > probabilities give the transition probabilities of the chain, and one > can (at least numerically) solve for the probability of landing in > each location by the end of n rolls. Including Go to jail or chance > cards can be included with a little extra work. Dan Heyman > I was wondering if anyone has done any work on the following questions. >Assume a game with a board like Monopoly except that all spaces are equally > likely (no redirecting spaces), every space is capable of charging rent, and > all rents are equal if the space is owned. >If one wants to be sure of someone landing on one of their spaces, having 12 > properties in a row would do it. This intuitively would be the way to > maximize the probability of an opponent landing on your properties at least > once. However, what if your choice is between two groups of 6 properties and > some other two groups adding up to 12 (like 5 and 7)? Is there a general > rule for maximizing the probability of an opponent landing at least once in > a given circuit of the board? >Also, how do the odds vary as the distance from the block of properties > increases? >And lastly, how would you group your properties to maximize the expected > income for one circuit around the board? >Jim Dobie ==== >I did mean to indicate that two 6 sided dice were used for movement and >therefore each of the 12 spaces in front of a piece are not equally likely. >I meant that the board design itself does not make any space more likely >than any other in the long run (no goto jail or other movement spaces or >cards). Are there existing programs to aid in numerically computing the >probabilities of landing on spaces (say from 1 to 40 spaces away) in n (say >each of 1 to 10) rolls? I would be very interested in a chart showing the >probabilities for such a range of spaces and rolls of the dice. Let p_j be the probability that you go j spaces in one roll (so for two fair dice p_2 = 1/36, p_3 = 2/36, ..., p_12 = 1/36, p_j = 0 for j not in {2,3,...,12}). The probability generating function of this is the polynomial P(x) = sum_j p_j x^j. Then the probability of going j spaces in (exactly) n rolls is the coefficient of x^j in P(x)^n. The probability of landing on space j before going around the board is the sum of this for n from 1 to infinity (but really up to floor(j/2) is sufficient, since you always go at least 2 in each roll). In this case, the results, rounded to 9 decimal places, are as follows: j probability 1 0.000000000 2 0.027777778 3 0.055555556 4 0.084104938 5 0.114197531 6 0.146626372 7 0.182227366 8 0.166345760 9 0.155526025 10 0.147783796 11 0.141275314 12 0.134199441 13 0.124703575 14 0.138567340 15 0.145771350 16 0.148354627 17 0.147889983 18 0.145672339 19 0.142886275 20 0.140764094 21 0.140744872 22 0.141558294 23 0.142503054 24 0.143243171 25 0.143649910 26 0.143670481 27 0.143207189 28 0.142763609 29 0.142528051 30 0.142514761 31 0.142651802 32 0.142836086 33 0.142971433 34 0.143002870 35 0.142959360 36 0.142889015 37 0.142829957 38 0.142801528 39 0.142805773 Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== I have been trying to use a software package, ARPACK, to determine the eigenvalues of large, sparsely-populated matrix systems generated by finite element methods. The matrices are usually symmetric or Hermitian. However, most of the examples given in the ARPACK manual deal with systems which are tridiagonal, i.e. systems which have one off-diagonal on either side of the diagonal. Is there any SIMPLE methodology/algorithm for transforming a symmetric matrix system to a triadiagonal matrix system? If this is indeed the case, then this would simplify the implementation of the ARPACK subroutines. ========================================================= Gregory M. Wilkins, Ph.D. Department of Electrical and Computer Engineering Morgan State University 5200 Perring Parkway Baltimore, Maryland 21251 Office: (443) 885-3915 FAX: (443) 885-8218 Web site: www.eng.morgan.edu/~gwilkins =========================================================