mm-367 === Subject: Re: Very Ugly Sum >The following series converges for all t>0: >1*exp(-t)+2*exp(-4t)+3*exp(-9t)+.....+n*exp(-n^2*t)+.... >I would like to know if it is possible to express it in a closed >form,or to give integral expression for this series. It is the derivative of a theta function; see any decent discussion on this branch of elliptic functions. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Coprimality: One-Person Game (& question) I want to mention this example of a 7-by-7 game-result I got today. It is not remarkable for its score, which is unimpressive (41). (And I DID erase...) But it is notable(somewhat) because its path has (purposefully-crea) rotational-symmetry. 3 2 1 12 13 14 15 4 9 10 11 * 17 16 5 8 * 20 19 18 * 6 7 * 21 * 35 36 * 24 23 22 * 34 37 26 25 * 31 32 33 38 27 28 29 30 41 40 39 Since the center is 21, the multiples of 3 and 7 are symmetrically placed, as are the multiples of 2 (which would have been the case whatever the center integer was). And, I gave this game to non-mathematical members of my family, after explaining what coprime means, and they seemed to enjoy it. So perhaps there is some teaching-value with this game, especially in teaching students about prime-divisors, GCD, or even in helping bright younger kids with their multiplication-tables. (Math FUN??? Who would have EVER guessed?! ...) Leroy Quet > [I apologize if this post appears twice. Google is giving me sh**..] > I was playing around with this today, and found it somewhat addicting. A 1-person game/ puzzle (with a score), which involves a little math. > First, start with a n-by-n grid drawn on paper. > I suggest a grid of at least n=5, and (for serious fun) perhaps an n > of at least 10. > So, you start by writing 1 in any of the grid's n^2 squares. > You then write the 2 adjacent to the 1, the 3 adjacent to the 2, the 4 > adjacent to the 3, etc, in a chain of increasing integers, in such > away that (ideally...) all n^2 integers are placed into the grid, one > integer (and ONLY one integer) per square. > (By adjacent, I mean immediately next to in the direction of up, > down, left, or right, but *not* diagonally.) > And, oh by the way, the integers are to be placed so that EVERY pair > of adjacent squares contains two integers which are COPRIME with each > other... > (Again, adjacent is as defined above.) > So, you place the integers into the grid as far as you can until your > path cannot be continued for whatever reason. > And your score is the last integer you were able to write into a > square. > Oh,...no erasing after you write down an integer! > (Well, perhaps a *little* erasing can be allowed...) > Here are my best, as of now, n=6 game, for examples: > 21 22 25 26 27 * > 20 23 24 * 28 29 > 19 18 1 2 3 * > * 17 6 5 4 * > 15 16 7 8 9 * > 14 13 12 11 10 * > (My score = 29) > And better, > 1 2 3 34 27 26 > 6 5 4 33 28 25 > 7 8 9 32 29 24 > 12 11 10 31 30 23 > 13 14 17 18 * 22 > * 15 16 19 20 21 > (My score: 34) > (I viola the no-erasing rule a little bit above...) > (And I apologize if I missed that any 2 adjacent integers in my > examples have a GCD which is greater than 1.) > (And, double-check every time you write down an integer so as to make > sure that it is coprime with each of its neighbors, I suggest! > {unless playing on a computer, which should be checking for > coprimality for you}) > I had intended long ago to post this game idea, which is based upon > the puzzle/math problem at: > http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2fdf. 0211011659.79913415%40posting.google.com&rnum=22&prev= > (I may have actually already mentioned this exact idea, but more > likely a 2-player variation. And, anyway, I have a question regarding > it, so as to justify a new thread.) > Question: > Let, for an n-by-n game, the average score, taken over ALL possible > games, be: > a(n). > (By possible games, I mean all games following the rules and being > played to the highest possible score, without any regards to > strategy.) > What is a(n) asymptotical towards?? > === Subject: Re: Coprimality: One-Person Game (& question) En el mensaje:b4be2fdf.0401241437.73a12678@posting.google.com, Leroy Quet escribi.97: > [I apologize if this post appears twice. Google is giving me sh**..] > I was playing around with this today, and found it somewhat addicting. > A 1-person game/ puzzle (with a score), which involves a little math. > First, start with a n-by-n grid drawn on paper. > I suggest a grid of at least n=5, and (for serious fun) perhaps an n > of at least 10. > So, you start by writing 1 in any of the grid's n^2 squares. > You then write the 2 adjacent to the 1, the 3 adjacent to the 2, the 4 > adjacent to the 3, etc, in a chain of increasing integers, in such > away that (ideally...) all n^2 integers are placed into the grid, one > integer (and ONLY one integer) per square. > (By adjacent, I mean immediately next to in the direction of up, > down, left, or right, but *not* diagonally.) > And, oh by the way, the integers are to be placed so that EVERY pair > of adjacent squares contains two integers which are COPRIME with each > other... > (Again, adjacent is as defined above.) http://groups.google.com/groups?selm=aq7dv4$pmt$1@ morgoth.sfu.ca its say: One interesting note is that for n=5, every solution has the 25 either in the corner or in the exact middle (it can't be at an even-numbered square, of course But that is a solution that refute it: 03 02 15 16 17 04 01 14 25 18 05 12 13 24 19 06 11 10 23 20 07 08 09 22 21 (Obtained by Jordi Dom.8fnech i Arnau, by hand) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Coprimality: One-Person Game (& question) > First, start with a n-by-n grid drawn on paper. > So, you start by writing 1 in any of the grid's n^2 squares. > You then write the 2 adjacent to the 1, the 3 adjacent to the 2, the 4 > adjacent to the 3, etc, in a chain of increasing integers, in such > away that (ideally...) all n^2 integers are placed into the grid, one > integer (and ONLY one integer) per square. > (By adjacent, I mean immediately next to in the direction of up, > down, left, or right, but *not* diagonally.) > ...so that EVERY pair > of adjacent squares contains two integers which are COPRIME > And your score is the last integer you were able to write into a > square. > Oh,...no erasing after you write down an integer! > n=6 game, for examples: > 1 2 3 34 27 26 > 6 5 4 33 28 25 > 7 8 9 32 29 24 > 12 11 10 31 30 23 > 13 14 17 18 * 22 > * 15 16 19 20 21 > (My score: 34) > (I viola the no-erasing rule a little bit above...) An exhaustive search reveals that you can't score 36! But you can get 35. 1 24 25 28 29 30 2 23 26 27 32 31 3 22 21 20 33 34 4 9 10 19 * 35 5 8 11 18 17 16 6 7 12 13 14 15 (My computer viola the no-erasing rule only 34012 times.) On 7x7, you can get 49. 15 16 17 18 19 20 21 14 3 2 29 28 27 22 13 4 1 30 31 26 23 12 5 42 41 32 25 24 11 6 43 40 33 34 35 10 7 44 39 38 37 36 9 8 45 46 47 48 49 -- Don Reble djr@nk.ca === Subject: math notation: how to concisely write expression of remainder? Dear all, This question is about writing division in concise math formulation: suppose y= qx + r where all variables are integers, q is the quotient, r is the remainder. It is easy to represent q in concise math form floor(y/x), there is a special symbol for floor... the lower square bracket... but is there any special symbol for remainder? if I want to define one in my thesis, what is a good way to make it formal? thanks a lot, -Walala === Subject: Re: math notation: how to concisely write expression of remainder? >suppose y= qx + r >where all variables are integers, q is the quotient, r is the remainder. >It is easy to represent q in concise math form floor(y/x), there is a >special symbol for floor... the lower square bracket... >but is there any special symbol for remainder? I like _x. But then you'd use langle, rangle of course. HTH, Michele -- >It's because the universe was programmed in C++. No, no, it was programmed in Forth. See Genesis 1:12: And the earth brought Forth ... - Robert Israel on sci.math, thread Why numbers? === Subject: Re: math notation: how to concisely write expression of remainder? > Dear all, > This question is about writing division in concise math formulation: > suppose y= qx + r > where all variables are integers, q is the quotient, r is the remainder. > It is easy to represent q in concise math form floor(y/x), there is a > special symbol for floor... the lower square bracket... > but is there any special symbol for remainder? > if I want to define one in my thesis, what is a good way to make it formal? > thanks a lot, > -Walala Sometimes people will write r as y mod x. In the C programming language it is written as y%x. === Subject: Re: math notation: how to concisely write expression of remainder? > Dear all, This question is about writing division in concise math formulation: suppose y= qx + r where all variables are integers, q is the quotient, r is the remainder. It is easy to represent q in concise math form floor(y/x), there is a > special symbol for floor... the lower square bracket... but is there any special symbol for remainder? if I want to define one in my thesis, what is a good way to make it formal? thanks a lot, -Walala > Sometimes people will write r as y mod x. In the C programming language it is > written as y%x. Since I am trying to use both floor(y/x) and y%x, I need to use unified notation in my thesis, if I use that special symbol(lower square bracket) similar to |y/x| to denote floor(y/x), then using y%x to denote mod(y/x) is not very match... And moreover, I want to use them as sub-scripts... Can you think of a unified notation for both quotient and remainder of y/x that can be used beautifully in subscripts(I guess up-down slim notation will be better for these subscripts)? Thanks a lot, -Walala === Subject: Re: math notation: how to concisely write expression of remainder? > Since I am trying to use both floor(y/x) and y%x, I need to use unified > notation in my thesis, > if I use that special symbol(lower square bracket) similar to |y/x| to > denote floor(y/x), then using y%x to denote mod(y/x) is not very match... > And moreover, I want to use them as sub-scripts... > Can you think of a unified notation for both quotient and remainder of y/x > that can be used beautifully in subscripts(I guess up-down slim notation > will be better for these subscripts)? > Thanks a lot, > -Walala As someone else sugges, you could use x{y/x} where {x} denotes the fractional part of x. However, I think that you are asking about the Kronecker product that you have in the other thread. I would use some less formal notation, for example, I would say that it is understood that c_{kl} = a{im}a{jn} where k=iN+m-N+1 (so (1,1) goes to 1) and l=jN+n-N+1. Maybe say that this notation will be adop throughout the thesis (so k,l,i,j,m,n always have this relationship). It may not look 'formal' but everyone who reads it will know what it means. === Subject: Re: math notation: how to concisely write expression of remainder? special symbol for floor... the lower square bracket... but is there any special symbol for remainder? if I want to define one in my thesis, what is a good way to make it > formal? thanks a lot, -Walala > Sometimes people will write r as y mod x. In the C programming language > it is > written as y%x. > Since I am trying to use both floor(y/x) and y%x, I need to use unified > notation in my thesis, > if I use that special symbol(lower square bracket) similar to |y/x| to > denote floor(y/x), then using y%x to denote mod(y/x) is not very match... > And moreover, I want to use them as sub-scripts... > Can you think of a unified notation for both quotient and remainder of y/x > that can be used beautifully in subscripts(I guess up-down slim notation > will be better for these subscripts)? In one book I have seen lfloor x/y rfloor to denote the integral part and {x/y} used to denote the fractional part of x/y. I do not know how widely it is used. -- Aditya Mahajan === Subject: math notation: how to explicitly express each element of a Kronecker product of matrix? Dear all, Suppose I have a N^2 x N^2 matrix C which is the tensor/Kronecker product of N x N matrices A and A, that is, C=kron(A, A), given the element A=(a_ij), i=1...N, j=1...N, what is the explicit expression for each element of C? That's to say, write explicitly c_kl in terms of a_ij... where k, l=1 ... N^2; i, j=1... N thanks a lot, -Walala === Subject: Re: when NaturalNumbers = p-adics what alters in the Riemann Hypothesis Re: proof of the Riemann Hypothesis > (snips) I would have had to consider two extreme cases: (1) that the RH is > completely false just as FLT is completely false when NaturalNumbers = > p-adics. If taking that extreme route would have had me find out what > was flawed in my claimed two proofs below. (2) the second extreme case is to say that something lies on the 1/2 > Real line but not the NaturalNumbers of the illdefined notion of > finite-integers but rather instead the p-adics. Only the line is not a > straight line. And my second *alleged proof below* using a spiral sort > of touches or hints of a curved line. (big snip) > TWO PROOFS OF THE RIEMANN HYPOTHESIS [clip] * > Mr. Plutonium: Have you ever submit a paper for publication? earle > * > For Archimedes Plutonium to submit any of his mathematical proofs to a > Math Journal is akin to a Democrat in the USA Senate to submit a Earth Air Conditioner Bill espousing a research race to find a > chemical that can solve not only Global Warming but can also make this > planet much more comfortable and much greener where humanity actually > begins to climate control planet Earth. > We all know what such a Democrat senator trying to solve Global > Warming would face from a Republican controlled Congress and a Bush > Global Warming. Bush does not even accept that the American Academy of > Sciences, that all the scientists have proven Global Warming is real > and is the cause of our goofy weather. > Just the other day I read a report that the Gulf Stream warm air > currents will vanish as Global Warming increases which means that the > EAst Coast USA will be more frigid in winters and that Europe > extending to Russia will become freezer countries in the decades > ahead. > The Bush administration is anti-science. They are boneheads about > Global Warming and about science. The mathematics journals editors are > boneheads. They are ignorant of the idea that NaturalNumbers are not > FiniteIntegers but are instead the P-adics. They are ignorant > boneheads as to what NaturalNumbers are and why they are a ill-defined > set, just as Bush is a ignorant bonehead about Global Warming. > Is it any use for a Democrat to start a research program as to what > chemical we can put into the upper atmosphere to create a Global Air > Conditioner. No. We all know what the Republican controlled Congress > and WhiteHouse will do-- reject reject and reject. They even rejec > the Academy of Scientists. > Is it any use for Archimedes Plutonium to send two proof of the > Riemann Hypothesis to any math journal editors in the world? Of course > not. They are math boneheads, whose ignorant minds can only write-- > reject, reject, reject. > Archimedes Plutonium > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies * I'll take that as a 'No'. earle * === Subject: Re: factoring 25.01.04 23:07 factors Amicable pairs/ chains. alt.math > what are the divisors of 48? Hit D. > sum of proper divisors? myprog gives the pairs of divisors which multiply to n. num of divisors sum of divisors perfect squares perfects multiperfects abundants/ deficients sigma(n)/n add-on calculations. > number theoretic functions? > sigma(n) sum of divisors, tau ?, d(n) number of divisors, phi(n) relatively prime divisors???. LATER. a poster Warwick? (same thread) queried in effect. The sequence 48, sigma(48)-48, sigma(n)-n. These seqs can presumably(?) be done by pari-gp (research calculator) or don sericalc4s. (below.) d.ivisors -.minus [sigma(n)-n]. s.wap sign +/-. [? optional] repeat d.ivisors -.minus ... query amicable numbers?? Right. amicable chain. Jacob's gift of 220 goats to Esau. Bible. 220;284. ** each is the sum of the aliquot parts of the other. Penguin dictionary of curious + interesting numbers, David Wells, 1997 etc. very good. : Table 8. the proper factors, where composite, and the values of the functions phi(n), d(n), sigma(n). search eis. NO. or www.MATHWORLD.com probably? eric weisstein. but eis (on-line encyclopedia of integer sequences) is interes particularly in Infinite integer sequences. don.mcdonald .Pgms.maths.calculator.watsfact48 25.01.04 18:49 BEGIN, ENTER START NUMBER (Expression), Q. END ?48 (1) 48 ROUND TO NEAREST (+ve) INTEGER DIVISORS OF INTEGER, R% = 48 1 * 48 DSUM= 49 DD%= 2 2 * 24 DSUM= 75 DD%= 4 3 * 16 DSUM= 94 DD%= 6 4 * 12 DSUM= 110 DD%= 8 6 * 8 DSUM= 124 DD%= 10 DSUM/R% = 2.58333333 NO. OF DIVISORS , DD% = 10 DSUM / number R% = 2.58333333 ABUNDANT. RESULT (2) 48 -?124 (3) -76 CHANGE SIGN +/- change sign may be superfluous if d.ivisors makes it positive anyway. (4) 76 ROUND TO NEAREST (+ve) INTEGER DIVISORS OF INTEGER, R% = 76 1 * 76 DSUM= 77 DD%= 2 2 * 38 DSUM= 117 DD%= 4 4 * 19 DSUM= 140 DD%= 6 NO. OF DIVISORS , DD% = 6 DSUM / number R% = 1.84210526 DEFICIENT. (5) 76 -?140 (6) -64 CHANGE SIGN +/- (7) 64 ROUND TO NEAREST (+ve) INTEGER DIVISORS OF INTEGER, R% = 64 1 * 64 DSUM= 65 DD%= 2 2 * 32 DSUM= 99 DD%= 4 4 * 16 DSUM= 119 DD%= 6 8 * 8 DSUM= 135 DD%= 8 PERFECT SQUARE, DSUM (SUM OF DIVISORS )= 127 XXX NO. OF DIVISORS , DD% = 7 DSUM / number R% = 1.984375 DEFICIENT. (8) 64 -?135 ??? oops. no worry below.. (9) -71 RECALL RES () ENTER SUBSCRIPT 0-999, OR (ENTER) LIST ?8 (10) 64 -?127 recover (11) -63 CHANGE SIGN +/- (12) 63 ROUND TO NEAREST (+ve) INTEGER DIVISORS OF INTEGER, R% = 63 1 * 63 DSUM= 64 DD%= 2 3 * 21 DSUM= 88 DD%= 4 7 * 9 DSUM= 104 DD%= 6 NO. OF DIVISORS , DD% = 6 DSUM / number R% = 1.65079365 DEFICIENT. (13) 63 -?104 (14) -41 CHANGE SIGN +/- (15) 41 ROUND TO NEAREST (+ve) INTEGER DIVISORS OF INTEGER, R% = 41 1 * 41 DSUM= 42 DD%= 2 NO. OF DIVISORS , DD% = 2 DSUM / number R% = 1.02439024 DEFICIENT. (16) 41 -?42 (17) -1 CHANGE SIGN +/- (18) 1 ROUND TO NEAREST (+ve) INTEGER DIVISORS OF INTEGER, R% = 1 1 * 1 DSUM= 2 DD%= 2 PERFECT SQUARE, DSUM (SUM OF DIVISORS )= 1 NO. OF DIVISORS , DD% = 1 DSUM / number R% = 1 DEFICIENT. MULTIPERFECT INDEX = 1 (19) 1 >>Don McDonald schreef in bericht > factoring : what are the factors of 48? > 24.01.04 00:04 >>Those are some of its divisors, but not all of them! I think you've critiqued a time-stamp here. willy > What are the factors of 48? > 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. > Take the sum of those factors, other than 48 (as in the definition of > a perfect number; 6 is a perfect number as it is the sum of 1, 2 and > 3, its factors other than itself.) > sum of 1+2+3+4+6+8+12+16+24 = 76 > 76 factors as 2*2*19: factors 1, 2, 4, 19, 38 (and 76); factor-sum > 1+2+4+19+38 =64. > Factors of 64: 1, 2, 4, 8, 16, 32: sum = 63; > and so on. > What happens to these series? Do any come to a defined end (if so, > how?) Do some regenerate a previous term and hence loop infinitely > (obviosly 6 does!) > Do any NOT terminate and NOT repeat. If so why? > A pleasant diversion for a Sunday afternoon (?) > Steve B. === Subject: Re: More Accessible Algorithm-Maze One error in my other reply to Teabag's post. Correc below in this repost. (But, in any case, the error did not involve the particular algorithm-loop being critisized.) > This puzzle has no solution; it is defective. The 11th step directs > you to skip the 11th step the second time you encounter it. You can't > skip the step if the step itself directs you to skip it. > Also I'm lazy, and it was easier to stop the first time I got to the > 11th step. > I'll try it again tomorrow, reading the 11th step as it was probably > intended. Huh?... Step 11 is encountered AND NOT skipped once, the *first* time you get to it. It then, the FIRST time you encounter it, tells you to skip it the SECOND time you encounter it. Analogously: (if my pseudo-coding is not erroneous) f=0: g=0 ....(steps 1 through 4) 5) g=g+1: move to adjacent square with number: if f+g=3 goto 5; else goto 6 ....(steps 6 through 10) 11) f=f+1: if f=1 goto 5; else goto 12 I hope this clears things up... [In other reply, in line-5 I had if f+g=2, instead of if f+g=3, the latter being correct.] Leroy Quet >.. 5) Move to adjacent square with a number. . > 11) Repeat the fifth step (step-5 above) twice, then repeat > steps (6),(7),(8),(9) and (10) once each in-order. > (Skip this {11th} step the second time you encounter it.) 12) Which square are you now on, and which symbol is in that square?? > (This maze only has one solution, or does it??..) > === Subject: Re: More Accessible Algorithm-Maze > This puzzle has no solution; it is defective. The 11th step directs > you to skip the 11th step the second time you encounter it. You can't > skip the step if the step itself directs you to skip it. > Also I'm lazy, and it was easier to stop the first time I got to the > 11th step. > I'll try it again tomorrow, reading the 11th step as it was probably > intended. Huh?... Step 11 is encountered AND NOT skipped once, the *first* time you get to it. It then, the FIRST time you encounter it, tells you to skip it the SECOND time you encounter it. Analogously: (if my pseudo-coding is not erroneous) f=0: g=0 ....(steps 1 through 4) 5) g=g+1: move to adjacent square with number: if f+g=2 goto 5; else goto 6 ....(steps 6 through 10) 11) f=f+1: if f=1 goto 5; else goto 12 I hope this clears things up... Leroy Quet >.. 5) Move to adjacent square with a number. . > 11) Repeat the fifth step (step-5 above) twice, then repeat > steps (6),(7),(8),(9) and (10) once each in-order. > (Skip this {11th} step the second time you encounter it.) 12) Which square are you now on, and which symbol is in that square?? > (This maze only has one solution, or does it??..) > === Subject: Re: A Certain Random Maze-Generating Algorithm I had some other items to add to this discussion. Regarding my wall-building algorithm specifically, but not > necessarily: > First, perhaps we could simply get rid of the outer walls at the > start. > We would have a theoretically infinite grid. > But the maze would be genera from 2 physically-close occupied > seed vertexes. > So, after the algorithm has run a certain amount of time, > perhaps being shut off by a human operator, > then we should have a blob-shaped maze, where we know not for sure > where the entrance and exit are. > The goal, of course, would be to find a way to pass through the blob, > which is topologically 2 intertwined trees. > After I contempla this a little, I realized there is a big problem. > The blob-maze is not guaranteed to have a solution! > This is because the walls being drawn from one of the two seeds may > completely surround those walls being drawn from the other seed. > So we would have then, in that case (which could actually be very > common), one path leading into the maze, connecting to a loop, and > have no exit. > I could not come up with a simple algorithm-fix for this problem. > As an interesting (perhaps) question itself, what would > sci.math/rec.puzzle-repliers suggest for the *simplest* fix which > GUARANTEES a path *through* the blob?? A simple solution: Just say that the maze is solved, if not only > someone finds a path through the maze, but if the solver's path > reconnects with itself somewhere within the maze (without just doing a > U-turn, obviously). > ;) I may have a solution. Since the grid is truly finite in which the blob-maze might be drawn upon, the maze-algorithm first draws one wall from one side of the maze to the other (v=1), with a break (of one door-width) somewhere randomly (preferably near the center) within it. And then the algorithm proceeds normally after this wall (with break) has been made. Actually, some of the initial wall (up to two lengths of the total initial wall), from some point near each extreme until each respective outer termination point, could be only virtual, not actually drawn as part of the maze. The purpose of these *virtual* wall-sections would be to have their vertexes act as occupied, for the purpose of keeping away approaching walls, but act as unoccupied in that they would be incapable of having any wall-segments drawn from them after being comple. Hopefully, this fix would work to ensure a path through each blob-maze!... > Also, a problem (which is a result of the maze's imperfect boundry): > it would be more difficult to occupy ALL of the internal vertexes with > a SIMPLE algorithm. > I suggest that after some point, the algorithm only work in occupying > the unoccupied INTERNAL vertexes. > For extra weirdness, perhaps we could have 3 or more seeds, > giving us a maze with 3 or more entrances/exits. > Or we could have a walled maze, without an exit or entrance door, > plus an internal seed. > Then the goal of this maze would be to find the closed loop within the > maze. > (Hmmm...this sounds fun, perhaps! Did I steal this idea??) > . . > WHERE did I see the closed-loop maze?? > And regarding the v-variable (which is referred to in my original > post), perhaps it would be interesting to build mazes where v > alternates {depending on if the order of the move is odd or even} > between very low (possibly 0) and very close to 1 (possibly 1). (In > the v = ~1 case, the wall would perhaps be joined to the the vertex > occupied *2* moves back, instead of 1 move back.) > === Subject: Re: Characterization of irrational numbers |let x be a positive number. Then x is rational if and only if min { n^a |* (nx - INT(nx)) } = 0, where the minimum is taken over all integers n |>0. It is assumed that a=1. | |QUESTION: is this result correct? Is it still correct if a=0.5 or a=0.2? I'm not sure this notation INT(nx) is standard. Ways of converting a real number to an integer differ in when they round down or up. A more standard notation is [t], which stands for the largest integer <=t. Someone else poin out it makes more sense to consider only n>=1 in your question. If x is a rational number, x=p/q where p and q are integers, q<>0, then nx-[nx] is always one of 0,1/q,2/q,...,(q-1)/q. If p/q is in lowest terms, then all of these occur. It's true that 0 is the minimum of the values of n(nx-[nx]). Since n(nx-[nx])>=n/q, the positive values go to infinity, and there's a minimum value of the positive values of n(nx-[nx]) which occurs only finitely many times. If x is irrational, there often is no minimum value of n(nx-[nx]). The inf, the greatest lower bound, of the n(nx-[nx]) usually but not always is 0. If x is irrational, the continued fraction approximations m/n to x satisfy n|nx-{nx}| < 1 where {nx} is the nearest integer to nx. The sign of nx-{nx} alternates with successive continued fraction approximations, so taking every other one gives us values of n for which n(nx-[nx])<1. If you replace n with n^a for 00 such that there are infinitely many terms a[n] that are >epsilon away from the limit of the first one. (Since the sequence doesn't converge, there must be such an epsilon.) Those terms form another bounded sequence, which by Bolzano-Weierstrass again must have a convergent subsequence, and the limit can easily be shown to be at a distance >=epsilon from the limit of the first convergent subsequence. It's also true that if there are two subsequences that converge to different limits, then the original sequence doesn't converge. But that's an unnecessarily difficult way to try to prove it, since often it will be tricky to identify such convergent subsequences. Ordinarily it would be easier to find two regions R1 and R2, each containing infinitely many terms a[n] of the sequence, and having the property that for some epsilon>0, each point in R1 is at a distance of >=epsilon from each point in R2. If there are two such regions, then given any point p, there exist infinitely many terms of the sequence that lie at a distance of at least epsilon/2 away from p. If some point of R1 lies within epsilon/2 of p, then no point of R2 lies within epsilon/2 of p, and vice versa, so there's one or the other of the two regions that lies at least epsilon/2 from p, and all the terms of the sequence in that region are at least epsilon/2 from p. Finding two converging subsequences with different limits p1 and p2 can be thought of as a special case, where we let R1 be the points within d(p1,p2)/3 of p1, and R2 be the points within d(p1,p2)/2 of p2. Another way to think of it which amounts to almost the same thing is that if you can give a continuous function f and reals ab for infinitely many n, then the sequence does not converge. === Subject: Re: sqrt(-1)=0/0 > 0/0 is every number, 0 is their placeholder. > 0/0=undefined, so 0 cannot be a place holder for all x in N, I or R. 0 > denotes the absence of quantity which is instrinsically a quantity > (ie. nothing is something). > Nothing has no constituent parts, and therefore cannot be divided by > or multiplied against (10 times nothing is still nothing, half of > nothing is undefined). > JS === Subject: Pythagorean triples genera by Ford circles? Let A = {(x,y)| (x-a/b)^2+(y-1/b^2)^2 = 1/(4*b^4)} B = {(x,y)| (x-c/d)^2+(y-1/d^2)^2 = 1/(4*d^4)} C = {(x,y)| (x-(a+c)/(b+d))^2+(y-1/(b+d)^2)^2 = (4*(b+d)^4)} be 3 Ford circles touching each other. For convenience, let also a/bd. Then, ( 1/(2*b^2)-1/(2*d^2), c/d-a/b, 1/(2*b^2)+1/(2*d^2) ) ( 1/(2*b^2)-1/(2*(b+d)^2), (a+c)/(b+d)-a/b, 1/(2*b^2)+1/(2*(b+d)^2) ) ( 1/(2*d^2)-1/(2*(b+d)^2), c/d-(a+c)/(b+d), 1/(2*d^2)+1/(2*(b+d)^2) ) all are Pythagorean triples. (In geometric interpretation we consider 3 right triangles with diagonals connecting circle's centers). Do we generate all Pythogorean triples that way? === Subject: Anyone can assist on this probability problems? Thanks in advance. Let E,F and G be three events. I need to find expressions for the events so that of E,F and G: 1. only E occurs P(E' int F' int G') where '=complement , int=intersection. 2. at most one of them occurs: For this I have the answer, but I don't understand it, maybe there's an easier expression..can someone explain it? I don't get this one. P( (E int F' int G') U (E' int F int G') U (E' int F' int G)U (E' int F' int G')) 3. at most two of them occur: P ( (E int F int G)' ) I don't understand this one either. If we take the complement of this we get P (( E' U F' U G')) which seems confusing.. 4. both E and G but not F occurs P ((E int G) int F')) 5. Exactly one of the events occur: P(E int (F' int G') U (F int (E' int G')) U (G int (E' int F')) ) (I don't know if this one is correct) 6. No more than two events occur at the same time. is this? 1-P(E' U F' G') ? (any other expression) 7. Exactly two of the events occur: P( ((E int F) int G' ) U ( (F int G) int E' ) U ((E int G) int F') ) === Subject: Re: Anyone can assist on this probability problems? > Thanks in advance. > Let E,F and G be three events. I need to find expressions for the events > so that of E,F and G: > 1. only E occurs > P(E' int F' int G') where '=complement , int=intersection. P(E int F' int G') > 2. at most one of them occurs: > For this I have the answer, but I don't understand it, maybe > there's an easier expression..can someone explain it? I don't get this one. > P( (E int F' int G') U (E' int F int G') U (E' int F' int G)U (E' int F' int G')) P(E int F' int G') + P (E' int F int G') + P (E' int F' int G) + (E' int F' int G')) > 3. at most two of them occur: > P ( (E int F int G)' ) 1 - P (all 3) = 1- P(E int F int G) > I don't understand this one either. > If we take the complement of this we get > P (( E' U F' U G')) which seems confusing.. > 4. both E and G but not F occurs > P ((E int G) int F')) > 5. Exactly one of the events occur: > P(E int (F' int G') U (F int (E' int G')) U (G int (E' int F')) ) P(E int (F' int G')) + P (F int (E' int G')) + P (G int (E' int F')) ) > (I don't know if this one is correct) > 6. No more than two events occur at the same time. > is this? 1-P(E' U F' G') ? (any other expression) Sure, but will be much more complica. > 7. Exactly two of the events occur: > P( ((E int F) int G' ) U ( (F int G) int E' ) U ((E int G) int F') ) P( ((E int F) int G' )+P ( (F int G) int E' ) +P ((E int G) int F') === Subject: Re: Socrates' Nothing is Everything The term infinity, as far as I'm concerned is a general word for talking about non-finite entities or the domainms y inhabit. 'Infinity' acts like a noun when refered to as 'the infinite' but, on reflection, this doesn't make much sense because the word does not point to anything noun-like. The adjective 'infinite' makes more sense because it denotes the property of being non-finite. I take this to refer to the infinitely small as well as the infinitely large. I'm very happy to include infinite entities into some kind of number system, in fact I think it undesirable that 'arithmetic' forbids the use of infinite entities. It seems to me that set theory, although it purports to deal with 'the infinite' does so only to make sure that infinite entities do not get into with the finite domain of arithmetic. But that is recidivism, infinitessimals are not necessary to analysis. My complaint is that the infinite entities of set theory are not a useful basis for any system of infinite arithmetic because they are 'fictional' ie unrela to anything that looks like a system of arithmetic, finite or infinite. That's what I meant by denying that they could be regarded as numbers in the more usual sense. The extension of numbers from integers to rationals, irrationals, reals, complex or any other type allow integration into general mathematics. However, all these types are finite, so we do not usually encounter such things as 'infinite square roots'. To recap. It seems to me appropriate that any complete system of infinite arithmetic would range over a hierarchy of infinite and infinitessimal entities as well as finite ones. This 'number range' (strangely) would exclude the extremes of both zero and absolute infinity. In other words, these two 'numbers' would not be expressible as numbers in the definition of number per se. Zero, as usual, can only be genera by the operation a + (-a). The analogy with logic is the form p.~p = false or x ^ x' = {} in the theory of classes. Tony Thomas > Zero is ubiquitous in that arises from the operation a-a = 0. > Zero has meaning in real experience as absence or emptiness. > In the context of physics it is rather out of place as far as the > conservation of mass/energy is concerned but lives hapily with those That operation is only well defined for numbers. Zero, the number, is > not absence or emptiness. What you should have said is that zero does not signify absence or > emptiness. What I meant was that nought, nothing, zero and the like do refer to absence > and emptiness > in ordinary speech and in the practical use of arithmetic. For example, if > there are no packets of cornflakes on a supermaket shelf the number zero > will be entered in the stock sheets.. > If your wallet is empty you have zero cash. Such obvuious meanings hardly > warrent comment. > As to the operation a-a, if you spend all the cash you have, this expresses > what happened in numbers. So, if you spent 2 dollars, the expression would > be 2-2 = 0 and if you spent ten dollars it would be 10-10 = 0. These > statements represent different events. Zero is a number, a token of a different > type than absence or emptiness. And I would suggest you don't > take modern physics too seriously. It is notorious for using > suggestive terminology which has little to do with the physical > meaning. > Metaphysically, zero can be considered fundamental or its negation. In > spatial terms this is infinite extension in infinite dimensions. In > physical > terms it would be an infinitely dense mass of infinite extension ie the > plenum void. In arithmetic it would be absolute infinity, that most > intractable of all entities. This isn't metaphysics. It's barely coherent english. I was being a bit loose here. The process of negation has a dual role, that > of nullifying and that of transforming. These roles can be clarrified by > considering four monadic operators: (1) N00: nullification > (2) N01: affirmation or stasis > (3) N10: negation > (4) N11: plenum nullification N00x -> 00 > N01x -> x > N10x -> ~x > N11 -> 11 In its arithmetical context zero acts as type (2) with respect to addition, > type (1) with respect to multiplication, type (3) with repect to subtraction > and type (4) with respect to division. > My point is that zero should bne trea as sui generis and recognized as > distinct from other numbers. (don't bother to tell me that zero is not an > operator) Yeah, but what does this have to do with infinite extension in > infinite dimensions or infinitely dense mass of infinite extension? Zero is fundamental because it is bound up with symetry, particularly > the > binary symetry of positive and negative numbers. Zero is paradoxical > because > it exists and does not exists or, rather, is a symbol for something > which is > impossible, non-existence, at least in the context of existence. You *might* have the beginnings of a point here. I don't know if you > have the osophical sophistication to turn this into a cogent > position. (My response would be that the context of existence isn't > the right context for an analysis of the number zero.) I don't think that the term 'context of existence' is meaningful. > Mathematical systems exist, if only in thought, but they exist nonetheless. The number zero cannot be analysed in the context of mathematics perse since > 'mathematics' is incapable of saying anything about its objects; that is the > role of metamathematics. > Russell's dictum is relevant here: 'mathematics is the science in which we do not know what we are talking > about, and do not care what we say about it is true'. > That's the spirit. :) > Ideally, zero should be trea like infinity, as a necessary token for > a > number but not a proper number. Infinity is unreachable by any finite > operastion on infinite numbers. Zero is unreachable too by any finite > operation other than that of a-a. This last operation, then is rather > suspect and proves to be an annoyance in forms like II[1/(n-1) x 1/(n-2) > x > 1/(n-3) x ...] What the hell is a proper number? A proper number is one which obeys all the operations of arithmetic without > exception, as a good logical object should. Unfortunately, zero is > recalcitrant with respect to division. Infinite and infinitesimal numbers > are improper and are kept out of the garden altogether. > This seems ad hoc to me. If you look at the axioms for the Peano > Arithmetic, you won't see an axiom for division. And it's obvious > that the natural numbers satisfy the PA axioms. I don't see why you > feel justified in calling division an arithmetic operation. By the > way, the next section And why isn't infinity a proper > number? Surely you're familiar with the logical results about the > existence of non-standard models for, say, the ordered Peano > Arithmetic which have objects which are greater than any other object. Although infinity is referred to in analysis it is forbidden the status of > operand > in that context. I wasn't referring to analysis, but to results from mathematical > logic. > Infinity, as an object, is on the same logical status as any other > number. It is not, for the reason given above. I agree that divergent series might > be said to have an infinite limit but this is meaningless unless one can say > which order of infinity it tends towards. > The sentence you're replying to here is highly rela to the sentence > immediately before it. I was not talking about limiting processes > here. I was talking about the existence of non-archimedean models of > ordered Peano arithmetic. In such models, there are objects such that > they are greater than any other object, including 1, 2, 3, ... . The > natural interpretation for such an object would be as infinity, and > in fact, similar ideas are used to develop the theory of non-standard > analysis, where objects called infinitesimals are given the same > logical status as numbers (so you can add a number and an > infinitesimal, say). Infinity, as usually used in mathematics (like what you > described) '(as you described it)' would be a neater expression for those who care > about good English. Well English, you mean. ;) > , is not an object, however. It is used as a sort of > short-hand for quantificational statements to describe the behavior of > the dependent variable (to set an upper bound, for instance) for any > given instance of the independent variable. Note that in the standard > definition of a limit, there is no mention of the word infinity. It > is all done in terms of existential quantification depending on > universal instantiation. > 'cid 'ooh I know, that's why I said it wasn't a proper number or 'a number of the > usual sort' if you prefer. I'm quite happy for infinite entites (not 'infinity' of course) to be > regarded as proper numbers. All you need are bigger 'premises'. > But I'm not sure you realize that infinity can be a number, too. In > fact, I'm sure you don't. > 'cid 'ooh === Subject: Re: Socrates' Nothing is Everything > The term infinity, as far as I'm concerned is a general word for talking > about non-finite entities or the domainms y inhabit. 'Infinity' acts like a > noun when refered to as 'the infinite' but, on reflection, this doesn't make > much sense because the word does not point to anything noun-like. Infinite is a property the some, but not all sets possess. Bob Kolker === Subject: Opnions needed http://sourceforge.net/project/showfiles.php?group_id=61183& package_id=10488 6 -suresh === Subject: Re: Opnions needed > http://sourceforge.net/project/showfiles.php?group_id=61183& package_id=104886 > -suresh The store down the street sells opnions and all kinds of other own container). You could also make your own opnions by getting some opnides at your local chemical store and disolving them. Paul Cardinale === Subject: Re: Squares of 0.999... tend toward 0, not toward 1 tend to 1 making the sequence .999...^2, .999...^3, .999...^4 tend to 1. > Dear Mr. Denke, > what I think you are missing is that .999... (a zero, a dot and an > infinite amount of 9s) is equal to 1, so what you are asking is > whether 1^2, 1^3, 1^4, etc. is 'tending to' 1. > Look at it this way: We can write a number like 0.9 as 1 - 0.1, and a > number like 0.99 as 1 - 0.01. In general we can write it down as: > f(n) = 1 - (1/(10^n)) for n>0 where n is the number of nines behind > the dot. > So the number 0.999... can be found by looking at: > lim{n->infinity}{f(n)} = > lim{n->infinity}{1 - (1/(10^n))} = > lim{n->infinity}{1} - lim{n->infinity}{1/(10^n)} > Since the limit of the fraction-part goes to zero, the whole thing > becomes equal to 1. In other words 0.999... is equal to 1. > BTW I'm sure there are better ways (a probably more mathematical > correct ways) to explain this, but perhaps this helps. Yes. x = 0.9_ 10x = 9.9_ = 9 + x 9x = 9 x = 9/9 = 1 Subject: re:Hilbert space problems === I solved 1, but I still need help on 2. ----== Pos via Newsfeed.Com - Unlimi-Uncensored-Secure Usenet News==---- http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: This problem is probably old as old as chocholate... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0Q05CC04917; >It is well known that a ping pong ball can be made to levitate >by a gust of wind blowing vertically upward. What other objects >besides a sphere exist with this property? I know of at >least one other because I construc it myself out of paper and >levita it for about 5 seconds at about a steady 7cm distance >from my own mouth (don`t know what its called geometrically... but it doesn`t have any wholes, topologically speaking). >Thanks. Well, its been a while since I pos this (actually, I think this was my first or so post on this forum). I have since levita one of these objects in the physics lab for much longer than that (using a steady current of air from below). The physics professor, a turbulence specialist, said that he did not think it was entirely due to the Bernoulli-effect, the effect responsible for the floating ping pong ball and also sugges that there may exist a bunch of these floating objects, based entirely on some mathematical symmetry. Btw., the object at hand does have at least some resemblence to a spaceship. By mathematical symmetry, he was refering to my claim that the object was construc based upon some properties of the quaternions. However, please do not confuse a bunch of these floating objects, based entirely on some mathematical symmetry with a bunch of these floating objects, based entirely on the quaternions- as I don't particularly think any more (other than the one I found) exist... unless they are made much more complica, that is- good luck! === Subject: Re: Algebraic integer surprise by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0Q05Cv04923; >For some time now I've been trying to explain an algebraic method for >analyzing polynomials that relies on non-polynomial factors. My >research is a natural extension in the polynomial domain of the idea >of irrationality in the integer domain. Basically, I look at the >equivalent of irrational factors with polynomials. >Recently Rick Decker, a professor at Hamilton College, apparently >trying to refute my research came up with a quadratic example, which I >like because it's a quadratic, and easier to manipulate than the >cubics I've used before. >If you wish to see his original post here are some headers which also >show that he is indeed at Hamilton College: >Subject: Re: Mathematical consistency, courage >Decker put forward the quadratic >(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >where his a's are roots of >a^2 - (x - 1)a + 7(x^2 + x). >The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of >non-polynomial factors. >Notice that despite not being polynomials they are algebraic integers >if x is an algebraic integer because a_1(x) and a_2(x) are the two >roots of >a^2 - (x - 1)a + 7(x^2 + x). to this point something can happen according to Your opinions but non-polynomial factors would have generally some values of algebraic numbers and only very accidentally they'll go together with algebraic integers. See: Your N-P1 * N-P2 = 7(25x^2 + 30x + 2) = W now for x=1 W=7*57 = 3*7*19 for x=2 W=7*162 = 2*7*3^4 for x=3 W=7*317 where 317 is prime also once You'll try to divide one of Your N-P by 7 so for x=3 there is to remain on right side only some prime... For to be correct into future relations, You should equal such one of Your N-P (non-polynomial factors) to 7 only... Does it will help to factorise W , I am not so sure. ( somebody used to proof, that there are infinite quantity of algebraic numbers between two next integers and so on serches in infinity are rather not so easy ) Regards Ro === Subject: Pure abstract demonstration by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0Q05DB04961; I think, You'll be surprised with some elementary developments: Some exact relation can be closed as f(t). However some advanced developments in that same relation would show some F(t), where once f(t) is of n-degree so F(t) is of (n-1)n degree. Somebody should notice, that the best idea for F(t) to be coherent or homogeneous to f(t) is just F(t) = [f(t)]^(n-1). To some small and prime values of n once beginning with n=3 such properity can be checked using results of division F(t)/f(t). There could be also shown, that for f(t) once the oldest coefficient is 1 so Abs.th. = -(A+B) but for F(t) once the oldest coefficient is 1 so Abs.th. = -(A+B)^(n-1) instead of [-(A+B)]^(n-1) = (A+B)^(n-1) What generally and sufficiently shows impossibility of F(t) to be coherent to f(t). See also, that from other side value A+B shows possibility to be divided by t as some potential root in f(t) and in F(t). Some other combination of parameters and changing values for this very relation will show some possible factorisation of any of Absolute therms.( I used to address and achieve in main part such developments at 22-th May of 2001 ) Parameters A and B represents here some of parametric values taken from Abel formulae: A = a^n once B = b^n or B = n^(nu-1) b^n or inverse . Once in general used input is X=T+B; Y=T+A; Z=T+A+B where T=n^u abtp always but for n=3 p=1 and the question was to solve X^n +Y^n = Z^n for n prime numbers bigger than 2 and for X;Y;Z integers Main part of FLT ? Hint: compare [X^n + Y^n]/(X+Y) to some s^n or n*s^n and express it in a;b;t;p parameters exclusively... f(t)= t^n - 2 n^u abtp - a^n - b^n once including so called first fall of FLT: X;Y;Z not divided by n And so on my current entertainment is going to be finished: There was some significant chance of likely correct proof of P. Fermat origine Roman B. Binder Yours Ro === Subject: Re: Connecness in the plane by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0Q05EK04997; Thank you David, and thanks to all of you who replied. Now we now that A and C are path-connec,and B,C are disconnec.Are B and C even totally disconnec? >Gday - >as mentioned elsewhere, B and D are path connec >and hence connec. >A has a partition U,V with >U = (-inf, sqrt(2)) x R >V = (sqrt(2), inf) x R >so it is disconnec. Similarly, C is disconnec. >But to my knowledge, the best answer to your first question >is that, in manifolds (such as RxR) connec sets are >path connec (and vice versa of course). >I suppose this is a less specific characterisation >than the intervals in R, isn't it. >-dave. >>Hi everybody, >>on the real line we know exactly which sets are connec- >>the intervals.Is something similar true for the plane? >>I mean,is there some characterization of the connec sets in RxR? >>Even for relatively harmless sets in RxR such as >>A={(x,y):x and y are rational} >>B={(x,y):x or y is rational} >>C={(x,y):x and y are irrational} >>D={(x,y):x or y is irrational} >>I don't know if they are connec.Which of them are?Why? >>Any help welcomed. === Subject: Re: Multivariate skew normal distribution in C by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0Q05CK04905; I do not think it is that easy to generate this kind of distributions. Azzani supplies some MATLAB code for generation of the random vectors. But this is a bit messy for me because I do not know Matlab too much. I have to think about it. Anyway impressing home pagem Alan, Thanks for help, Przem >Correction to my previous method. >> Dear All, I really wonder if there is somwhere an inplementation of the >> simulation algorithm (random numbers) of the multivariate skew normal >> distribution. If yes, would you like to give me the source. Thank you very much Przem >> After a search of the web, I found that Azzalini assigned the name skew >> normal to the density: >> f(x) = 2.phi(x).PHI(alpha.x) >> where phi(x) is the probability density for the normal distribution, and >> PHI() is the cumulative normal distribution function. phi is the std. >> normal (mean = 0, std. devn. = 1), but x can easily be scaled and shif >to >> give any other mean and std. devn. >> I have no experience with this distribution, but I guess that you can >> generate a random variable with this distribution simply by generating >> random normals (x), and then ********** (delete) >> accepting the first one if a uniform random variable (z) is less than >> PHI(alpha.x). I have not thought about how you extend it to the >> multivariate case. >> I can supply code (in Fortran) for generating normally distrinu random >> numbers, and for the normal distribution function.. >> Cheers >> -- >> Alan Miller >> http:// users.bigpond.net.au/amil ler Retired Statistician === Subject: exact D.E. I am taking intro. D.E. course. Consider (1) M(x, y)dx + N(x, y)dy = 0 And I want to show this has the exact form (2) f_x*dx + f_y*dy = 0 Now my book's theorem states that (1) is (2) if M_y = N_x. So I think that If M_y = N_x, then this means M_y and N_x must be second partial derivatives of some function f(x, y). That is, M_y = f_xy, and N_x = f_yx. Therefore, M = f_x, and N = f_y. However, I have a problem. Is it not possible for M and N to be some unrela functions where it is possible for the coincidence that M_y = N_x, even though they are completely unrela (i.e., M and N are not partial derivatives of f(x, y))??? === Subject: Re: exact D.E. > I am taking intro. D.E. course. > Consider > (1) M(x, y)dx + N(x, y)dy = 0 > And I want to show this has the exact form > (2) f_x*dx + f_y*dy = 0 > Now my book's theorem states that (1) is (2) if M_y = N_x. > So I think that > If M_y = N_x, then this means M_y and N_x must be second partial derivatives > of some function f(x, y). That is, M_y = f_xy, and N_x = f_yx. Therefore, > M = f_x, and N = f_y. > However, I have a problem. Is it not possible for M and N to be some > unrela functions where it is possible for the coincidence that M_y = N_x, > even though they are completely unrela (i.e., M and N are not partial > derivatives of f(x, y))??? The theorem says this is not possible. (provided the domain for the whole thing is simply connec...) Assuming M_y = N_x, how to find f? Intregrate M with respect to x and see what you get... there will be an unknown function of y added on. Then differentiate that and match with N to determine the function. Alternatively, read the proof of the theorem in your book. === Subject: Re: exact D.E. >I am taking intro. D.E. course. >Consider >(1) M(x, y)dx + N(x, y)dy = 0 >And I want to show this has the exact form >(2) f_x*dx + f_y*dy = 0 >Now my book's theorem states that (1) is (2) if M_y = N_x. >So I think that >If M_y = N_x, then this means M_y and N_x must be second partial derivatives >of some function f(x, y). That is, M_y = f_xy, and N_x = f_yx. Therefore, >M = f_x, and N = f_y. >However, I have a problem. Is it not possible for M and N to be some >unrela functions where it is possible for the coincidence that M_y = N_x, >even though they are completely unrela (i.e., M and N are not partial >derivatives of f(x, y))??? No, that's not possible. You're asking whether it's possible for the theorem in the book to be wrong. Do you see an error in the _proof_ of the theorem? (Um, there are some technical conditions we need to make the theorem correct, that are usually ignored at this level: If M and N are defined in a simply connec region, in other words if the domain of M and N has no holes in it, then (1) implies (2). Probably in a beginning DE course the domain is just a rectangle, which is certainly simply connec.) ************************ === Subject: Re: digital logic I'm just putting things into code as a way of getting this stuff straightened out in my mind. I've been using a freeware simulator (Multimedia Logic) as my standard of correctness; LogicWorks is too expensive for me. The goal is (i) to figure out what this stuff means, and (ii) to translate it from its usual diagram-based form into a form that consists in statements about something. (I'm really lame at thinking diagrammatically.) It occurred to me recently that, in the course I took and in all of the books I've looked at, the theory of flipflops is only developed in the context of counters. What do you think about that? If there were a magic black box that ac just like a counter (0,1,2...n,0,1,2...n,0,1,2 etc, for any n you choose) and was absolutely free in terms of any kind of engineering cost, would flipflops be superfluous? | | | >1) Am I supposed to assume that the time required for a signal to travel along a | >wire is negligible, in the sense that I can always assume it's instantaneous? | > | In practice it is negligible compared to the clock rate. If I were | writing a simulator I would sequentially: | 1. Read the states Q_i | 2. Read the inputs I_i | 3. Calculate J_i and K_i from the states and inputs. | 4. Evaluate the next states Q_i+ | 5. Repeat, using the Q_i+ as the new Q_i | | >2) Am I supposed to assume that JK-flipflops and D-flipflops (more generally, | >all types of flipflops) are time-equivalent? Here's what I mean by that: | >Suppose that t_0 is a moment of time (continuous time, the time of Newtonian | >physics); let X be a JK-flipflop, and Y a D-flipflop, such that each input of X | >and each input of Y stays the same (as itself) in some time period just before | >t_0. If one or more of the inputs of each flipflop assume(s) a different value | >at t_0, and the new inputs are, in each case, such that the output of the | >flipflop is going to change, then can I assume that X and Y will change their | >respective output values at the same moment? | > | I don't think so. Changing the J or K input just before the state | transition is a no-no. I would think a physical flip-flop probably has | some sort of mechanism built in to protect against this, but I don't | know enough about hardware to tell you. I do know that good design | requires that inputs be synchronized with the low clock (for JK). Also | note that D and JK flip-flops change state at different parts of the | clock cycle so I don't think it is a good practice to mix them. They | certainly don't change states at the same time. The idea is that | during low clock is when the logic gates all settle down to their new | values so everything is stable during high clock and for all practical | purposes you get simultaneous state changes on the transition. Things | are just the opposite for D types; they change on the rising clock. | | Disclaimer: I an not a EE person but have studied the topic and have | designed and built several digital logic circuits. So anything that's | really important you might want to check with the experts. | | | >Peace, and thanks again. (Code follows.) | | Sorry, I don't know Python so I won't comment on your code. Are you | writing a simulator? Why don't you use logic works or something | similar to test your logic and simulate your circuit? | | --Lynn | | === Subject: Re: digital logic Hola! I don't understand your first point. If one of the ingoing pins of a flipflop is connec to the clock (and, at the risk of pointing out the obvious, this condition is not satisfied by all flipflops: for example, some flipflops in a ripple counter are not connec to the clock), then is it incorrect to call that ingoing pin an input? My intention was to have the term input understood as meaning any ingoing pin. Actually, I'm not sure I understand your second point either. When you say that clock transitions are instantaneous in a given simulation, I assume you mean that the time it takes for the clock to go from A to B or vice-versa (where A and B are respectively 5V and 0V, or whatever the case may be) is taken to be zero. That much seems fairly clear (though I may as well ask whether it matters what the clock signal is at the moment between successive intervals), but I need clarification wrt the statement that all flip-flops change at the same instant. Clearly you don't mean that if any flipflop in a given circuit changes its output state at a particular moment t, then every flipflop in the circuit changes its output state at that same moment t. (You don't, right?) I'm thinking you meant to assert the previous statement only for t in some particular interval of time, like a cycle or half-cycle of the clock. Could I prevail on you to be completely explicit about this point? One other question: Does it ever happen that lack of information about matters of timing causes a simulator to be unable to determine how a circuit-simulation should behave? Peace | No, that's not right. As others have poin out, you're overlooking the | fact that most flip-flops are clocked. They are designed so that they | only change when the clock makes a transition. The flip-flop then takes | up whatever new state it should have according to its inputs. The | circuit sees _ONLY_ the values of the inputs at the clock transition, | which can be thought of as a specific instant of time. | | Detailed simulations such as Pspice do consider gate delays, etc., but | most simulations of circuits involving clocked logic assume idealized | behavior in which clock transitions are instantaneous and all flip-flops | change at the same instant. | | Jack | | > Hi. Thanks for responding. I'm including some (working) Python code that may | > help to clue you in wrt where I'm at regarding this problem. Arguably you'd be | > able to tell more from the code that doesn't work, but then you couldn't see it | > in action. | > | > Here are some basic questions that I hope you can help me with: | > | > 1) Am I supposed to assume that the time required for a signal to travel along a | > wire is negligible, in the sense that I can always assume it's instantaneous? | > | > 2) Am I supposed to assume that JK-flipflops and D-flipflops (more generally, | > all types of flipflops) are time-equivalent? Here's what I mean by that: | > Suppose that t_0 is a moment of time (continuous time, the time of Newtonian | > physics); let X be a JK-flipflop, and Y a D-flipflop, such that each input of X | > and each input of Y stays the same (as itself) in some time period just before | > t_0. If one or more of the inputs of each flipflop assume(s) a different value | > at t_0, and the new inputs are, in each case, such that the output of the | > flipflop is going to change, then can I assume that X and Y will change their | > respective output values at the same moment? | > | > 2') In case that still isn't clear, what I'm asking for is a yea or nay | > regarding the validity of the following argument: | > | > Hypotheses: | > | > i) X is a JK-flipflop with inputs A1...An and output B | > ii) Y is a D-flipflop with inputs C1...Cm and output D | > iii) There is a moment t_0 of time and a number e>0 such that, for (t_0-e) < tx | > < ty < t_0, Ai(tx)=Ai(ty) for 1<=i<=n and Bi(tx)=Bi(ty) for1<=i<=m. | > iv) There exists an i with Ai(t_0)=/=Ai(t) for (t_0-e) < t < t_0, and there | > exists an i with Bi(t_0)=/=Bi(t) for (t_0-e) < t < t_0. | > v) The change of input values at t_0 causes both flipflops to change their | > respective output values. | > | > (Putative) Conclusion: | > | > If tX is the supremum of {t: B(T)=B(t_0) for t_0<=T<=t}, and tY is the supremum | > of {t: D(T)=D(t_0) for t_0<=T<=t}, then tX=tY. | > | > Peace, and thanks again. (Code follows.) | > | > | > MY LATEST FIDDLING, SHOWN HERE, IS A MOD-13 COUNTER. | > IT'S ESSENTIALLY OK, ALTHOUGH IT SHOWS 13 FOR HALF A | > CLOCK CYCLE BEFORE RESETTING TO ZERO | > | > ########################################################## | > ## BOOLEAN FUNCTIONS: | > | > AND = lambda *bitVector: int(0 not in bitVector) | > OR = lambda *bitVector: int(1 in bitVector) | > XOR = lambda *bitVector: sum(bitVector)%2 | > NAND = lambda *bitVector: int(0 in bitVector) | > NOR = lambda *bitVector: int(1 not in bitVector) | > NOT = lambda bit: int(not bit) | > | > ########################################################## | > | > IF [d_n, ... ,d_0] IS A LIST OF BINARY DIGITS, THEN | > num([d_n, ... ,d_0]) IS THE NUMBER WHOSE BINARY NUMERAL | > IS d_n...d_0 ; IF n IS THE NUMBER WHOSE BINARY NUMERAL | > IS d_n...d_0 THEN bitVector(n) IS [d_n, ... ,d_0] . | > | > | > def num(bitVector): | > digits=list(bitVector) | > digits.reverse() | > return sum([digits[i]*pow(2,i) for i in range(len(digits))]) | > | > def bitVector(num): | > twoExp = lambda x: x>=2 and twoExp(x/2)+1 | > digits=[(num>>i)%2 for i in range(twoExp(num)+1)] | > digits.reverse() | > return digits | > | > ########################################################## | > ## NOT APPEARING IN THIS EPISODE: | > | > decoder = lambda *bitVector: | > [int(i==num(bitVector)) for i in range(len(bitVector))] | > | > mux = lambda *bitVector: | > lambda *dataVector: dataVector[num(bitVector)] | > | > def bus(*args): | > if sum([enabler for (driver,enabler) in args])!=1: | > raise bus error | > for (driver,enabler) in args: | > if enabler: | > return driver | > | > ########################################################## | > | > NODES REPRESENT THE CONNECTIONS BETWEEN THE OUTPUT OF ONE | > THINGIE (OR OF A SWITCH OR A CLOCK) AND THE OUTPUT OF | > ANOTHER THINGIE. IF X IS A NODE, THEN nodeVals[X] IS THE | > CURRENT VALUE OF X; THAT IS, THE SIGNAL CURRENTLY BEING | > CARRIED ALONG X. IF STRG IS A COMMA-SEPARA STRING OF | > WHICH IS IRRELEVANT, IS THEMSELVES); INITIALLY THEY ALL | > HAVE VALUE 0. toggle(node) DOES EXACTLY WHAT IT SAYS. | > IF transitionList IS A LIST OF INSTRUCTIONS FOR UPDATING | > NODE VALUES, THEN doTransitions(transitionList) DOES EXACTLY | > WHAT IT SAYS; ITS PURPOSE IS TO TRANSLATE BOOLEAN FUNCTIONS | > INTO THE CORRESPONDING OPERATIONS ON nodeVals. counter DOES | > A SIMILAR TRANSLATION WRT num. T_edge ANSWERS TO AN EDGE-TRIGGERED | > T-TYPE FLIPFLOP (WITH A RESET INPUT) | > | > | > nodeVals={} | > | > def declareNodes(strg): | > args=strg.split(,) | > for x in args: | > exec x + =' + x + ' in globals() | > nodeVals[x]=0 | > | > def toggle(node): | > nodeVals[node]=NOT(nodeVals[node]) | > | > def doTransitions(transitionList): | > global nodeVals | > newNodeVals=nodeVals.copy() | > for transition in transitionList: | > (assignee,funct,args)=transition | > newNodeVals[assignee]=funct(*[nodeVals[x] for x in args]) | > nodeVals=newNodeVals | > | > counter = lambda *nodeVector: num([nodeVals[x] for x in nodeVector]) | > | > def T_edge(T,Q,RESET): | > if RESET: | > return 0 | > return (T and NOT(Q)) or (NOT(T) and Q) | > | > ########################################################## | > | > EACH f_i REPRESENTS THE TRANSITION FUNCTION FOR THE | > FLIPFLOP WHOSE VALUE WILL BE THE i-TH DIGIT OF THE | > NUMERAL SHOWN BY THE COUNTER. thirteen(t0,t1,t2,t3) IS | > 1 IF num([t3,t2,t1,t0]) IS 13, AND 0 OTHERWISE. | > THE g_i'S ARE LIKE THE f_i'S, BUT INCORPORATE THE | > 'RESET AT THIRTEEN' FUNCTION | > | > | > f0 = lambda t0,t1,t2,t3,clk,reset: T_edge(clk,t0,reset) | > f1 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0),t1,reset) | > f2 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0,t1),t2,reset) | > f3 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0,t1,t2),t3,reset) | > | > thirteen = lambda t0,t1,t2,t3: AND(t0,NOT(t1),t2,t3) | > | > g0 = lambda t0,t1,t2,t3,clk: f0(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3)) | > g1 = lambda t0,t1,t2,t3,clk: f1(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3)) | > g2 = lambda t0,t1,t2,t3,clk: f2(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3)) | > g3 = lambda t0,t1,t2,t3,clk: f3(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3)) | > | > ########################################################## | > ## SCRIPT TO SHOW HOW THE ABOVE DEFINITIONS WORK: | > | > declareNodes(T0,T1,T2,T3,CLK) | > | > ## LIST OF UPDATE-INSTRUCTIONS: | > | > mod_13_counter = [ | > (T0,g0,[T0,T1,T2,T3,CLK]) , | > (T1,g1,[T0,T1,T2,T3,CLK]) , | > (T2,g2,[T0,T1,T2,T3,CLK]) , | > (T3,g3,[T0,T1,T2,T3,CLK]) ] | > | > for i in range(30): | > print counter(T3,T2,T1,T0) | > toggle(CLK) | > doTransitions(mod_13_counter) | > | > ## END | > | > | > | | > | >Subject: Digital Logic | > | >Level: Beginner | > | > | > | | > | > | > | >Essentially what I want to know is this: How can you tell how a circuit will | > | >behave without knowing the lengths of the gate delays? For example, how do | > | >simulation programs (eg: LogicWorks, Multimedia Logic) determine how circuit | > | >simulations should behave? | > | > | > | Others have given good answers but I will throw in my 2 cents worth | > | anyway. Simple state machines generally consist of flip-flops, which | > | may be either J-K or D input types, logic gates, and a clock. We might | > | denote the current state of a flip-flop by Q and its next state by Q+. | > | In the case of a JK flip-flop, for example, the outputs are immune to | > | the inputs when the clock is low. When the clock goes high the inputs | > | are locked in and read. On the transition from high to low the state | > | of the flip-flop changes from its current state Q to its next state | > | Q+. During this low clock period, the flip-flop is immune to the | > | changes in inputs. This gives time for all the logic gates to receive | > | their new inputs, including any fed-back values of the Q+ outputs | > | which now represent the new current state. All the gate delays have | > | time to settle to their new values and set the new J-K inputs, so the | > | flip flop is settled and ready to go when the clock goes back high and | > | the cycle repeats. | > | | > | This happens for all flip-flops in the circuit at the same time, so | > | what you get is a machine that proceeds from one state to the next | > | each time the clock goes from high to low for a JK flip flop. | > | | > | The boolean logic is used to specify the next state equations, that is | > | Q+ is a function of J and K, which are functions of the inputs and the | > | current state of the machine. Figuring out the logic equations for the | > | J and K inputs is where you usually get to dealing with Karnaugh maps. | > | | > | Machines of this type are called synchronous machines. They may appear | > | to respond immediately to input changes, but that is only because the | > | clocks are generally running very fast. | > | | > | Machines with no clock are called asynchronous machines and may indeed | > | have problems with gate delays and races between outputs and inputs | > | which must be carefully guarded against. | > | | > | Hope that helps. | > | | > | --Lynn | > | | > | > | === Subject: Re: Topological genus of the human body I think many piercings have to count. A friend once commen that some of her acquaintances possessed a rapidly increasing genus. === Subject: Re: Automobile Speeds. >Exercise 1. Complete the following sentence: My car gets ___ to the acre. Until it broke down, only about 4*10^10, alas! === Subject: Re: Newbie Questions: Series >Now, though, if I could get info on book / websites etc. on this type >of series (not just this, but more like table of series, if such a >thing exists), I will consider it as a happy bonus I'm sure I've seen such a table. The library at the highschool I attended had such a book which I think had the sum you originally asked for. There are also books which describe techniques for deriving them. (Being omniscient, of course, you know all this already.) === Subject: Re: Axioms and Reducibility |If I have a mathematical system (A) based on 5 axioms and another |mathematical system (B) that is based on 3 of A's 5 axioms, does it |logically follow that B is reducible to A? You haven't said what kind of reducibility you mean, but there's nothing to keep B from being a much richer system than A. If B is something like elementary group theory, and we add a couple of axioms to it, the result may be trivial, e.g. with the addition of x^3=e for each x and x^2=e for each x, we get the theory of just the 1-element group, which is relatively dull. So I would say that generally the answer is no, unless you mean some kind of reducibility that I'm not thinking of. There is a trivial sense in which A is reducible to B. If I want to know whether or not a result X is a consequence of the axioms of A, I can rephrase that as wanting to know whether or not (P1 and P2)->X is a consequence of the axioms of B, where P1 and P2 are the two axioms of A that aren't axioms of B. I should be a little careful here, since it may be that the language of the theory of A might contain additional undefined terms that the language of B doesn't have, but even so there's a looser sense in which A is reducible to B. One could consider studying consequences of the axioms of A a subfield of the study of consequences of the axioms of B. This relationship doesn't necessarily have any practical relevance, however, since the kind of question one asks in the study of A may be completely unlike the kind of question one asks in the study of B. Groups can be thought of as a special kind of category, one in which there is only one object and every arrow is invertible. But people don't usually think of group theory as a subfield of category theory. Many theories can be reduced to set theory, but if you translate some of your favorite questions into set theory, you get questions that a typical set theorist is no more likely to know how to solve than you are. The kind of reduction which is most useful reduces the kind of question typically asked in A to a kind of question typically answered in B. === Subject: Re: Chaos Question re Strange Attractors >Will any given subset (other than the null set) of a strange attractor >ultimately flow to equal and fill out the entire strange attractor? (If this is so, then the subset can be taken to be just one starting point.) My main trouble with questions like this is that I don't often see precise definitions of the terms. There are so many popularizations with loose descriptions only! Sometimes a strange attractor is required to be a fractal, and fractals themselves are sometimes defined as having non-integer Hausdorff dimension, but more often just having a Hausdorff dimension different from their topological dimension. I once heard a tenured professor tell a graduate student not to worry about such things, but to think of fractal as simply meaning closed set. Allegedly some authors define strange attractor to mean simply the attractor of a chaotic system, whether fractal or not. Chaotic, in turn, seems sometimes to be defined simply as sensitive dependence on initial conditions, but I seem vaguely to remember seeing some additional criterion included in the definition once. Anyhow, I don't believe there's anything in the definitions that rules out the situation where you have more than one basin of attraction. You could have a system where if you start in one basin, the system approaches one component of the attractor, and if you start in another basin, the system approaches a different component of the attractor. So the two portions don't flow into each other. It's also possible to have unstable fixed or periodic points on a strange attractor, and if you start on one, you don't wander away. Of course, this depends on starting exactly on the periodic orbit. For example, consider values of the parameter A for which iteration of f(x)=Ax(1-x) has a strange attractor. The polynomial f^n you get by iterating f n times has solutions for f^n(x)=x for some values of n certainly, even though the typical behavior of an x under repea application of f is not periodic. Note that in the cases in which one would feel like answering your question yes, the filling out won't usually be a matter of the path of the system tracing out the whole attractor exactly, but only of coming arbitrarily close to each point on it. I suspect this is why you put quotes around filling out, but I thought it might be worth pointing out anyway. === Subject: Re: Regular Languages > Interesting question... my suspicion is no. But I think the answer is yes. Because we can increment the number of states as follows: Think of arraying all of the old states except the accepting state in a column on the left and the accepting state alone on the right; from each old state, we have a bundle of paths to the accepting state; epsilon transitions are used to choose a path in the bundle and additional states are used as necessary to implement each path -- but all of these paths converge in a fan from the old states to the accepting state and thus do not c. (from: http://www.cs.unm.edu/~moret/computation/3/13.html) It doesn't satisfy me! By this method, is the number of states bounded? I don't know! > Consider an alphabet S = {0,1,2,3,4} and a language L = {w in S* such > that the sum of the characters of w is 0 mod 5} > Then any machine would need states representing (at least) the five > possible current sums (mod 5) and since any symbol (0 through 4) could > move the action from one state to any other state, there has be some kind > of path taking it there (be as creative as you like with epsilon > transitions, such paths must still exist). > This looks like a subdivision of the complete graph on five vertices, > and so not planar. In your example, we need only 5 states and transitions from each state to next state(because each transition increments sum of the characters by one), so it will be a planar graph. > This of course isn't a complete proof... do you need one (for an > assignment, or something?) It's just interesting for me, I've encountered it in a theory of computation book. Behrang. === Subject: Re: Regular Languages > Interesting question... my suspicion is no. > But I think the answer is yes. Because we can increment the number of > states as follows: > Think of arraying all of the old states except the accepting state in > a column on the left and the accepting state alone on the right; from > each old state, we have a bundle of paths to the accepting state; > epsilon transitions are used to choose a path in the bundle and > additional states are used as necessary to implement each path -- but > all of these paths converge in a fan from the old states to the > accepting state and thus do not c. > (from: http://www.cs.unm.edu/~moret/computation/3/13.html) > It doesn't satisfy me! By this method, is the number of states > bounded? I don't know! Yes. I think it requires at most O(1+(c+n)*2^n) = O(2^n) nodes (n = #nodes in original NFA) since the number of cycle-free paths between any two nodes in a finite graph is bounded by 2^n; c are the nodes for 'wrap-around', 1 is the additional final state. By the way, why would you ask a question you seem to have known the answer to? Why not just mention you didn't get the answer and post a link? l8r, Mike N. Christoff === Subject: Re: Regular Languages Distribution: inet [ Is there a planar-graph NFA for every regular language? ] > Interesting question... my suspicion is no. > But I think the answer is yes. Because we can increment the number of > states as follows: > Think of arraying all of the old states What do you mean, old states? [DN]FAs have a start state, an accepting state (or states), and a bunch of other states. The term old state is not familiar to me in this context. By the way, does regular language mean language determined by some DFA? Clarification reques (although I feel certain I've had this clarified for me several times elsethread :( ). > except the accepting state in > a column on the left and the accepting state alone on the right; from > each old state, we have a bundle of paths to the accepting state; > epsilon transitions are used to choose a path in the bundle and > additional states are used as necessary to implement each path -- but > all of these paths converge in a fan from the old states to the > accepting state and thus do not c. > (from: http://www.cs.unm.edu/~moret/computation/3/13.html) Aha. I can't vouch for the correctness of that argument, but it at least explains that your old states are really states of an arbitrary NFA, and you're describing an algorithm for planarizing that NFA. Right? By the way, that problem statement discusses regular sets, not regular languages. But I assume they're the same thing (a language being a set of strings). [The only thing suspicious about that proof, BTW, is its cavalier use of phrases like these transitions... lead 'around the back'... so that they do not c each other. Kind of hand-wavy, IMHO.] > It doesn't satisfy me! By this method, is the number of states > bounded? I don't know! > Consider an alphabet S = {0,1,2,3,4} and a language L = {w in S* such > that the sum of the characters of w is 0 mod 5} Then any machine would need states representing (at least) the five > possible current sums (mod 5) and since any symbol (0 through 4) could > move the action from one state to any other state, there has be some kind > of path taking it there (be as creative as you like with epsilon > transitions, such paths must still exist). > This looks like a subdivision of the complete graph on five vertices, > and so not planar. > In your example, we need only 5 states and transitions from each state > to next state(because each transition increments sum of the characters > by one), so it will be a planar graph. Wrong. When Jim describes the sum of the characters of w, he's talking about the sum of the numerical values of the digits. E.g., the word 123 has a sum of 6, and will be rejec; the words 14 and 12345, having sums divisible by 5, will be accep. Jim's argument seemed very persuasive to me -- but that was before think it *can* be done. Follow-up (possibly dumb) question: Does every planar NFA have a planar DFA equivalent? 'Cause if not, then I'm in way over my head even just doodling on paper, and might as well give up. ;-) -Arthur === Subject: Re: Regular Languages talking about the sum of the numerical values of the digits. E.g., the > word 123 has a sum of 6, and will be rejec; the words 14 and 12345, having sums divisible by 5, will be accep. > Jim's argument seemed very persuasive to me -- but that was before > think it *can* be done. Arthur, for correcting him since he did seem to misunderstand my definition. I'm glad my pseudo-construction sort of convinced you, and it sort of convinced me at the time, but now I'm thinking that we can always create a planar NFA... I haven't checked any of the links that were provided, but my guess would be to draw any DFA for the language, and if there are corssing edges, we create a dummy-node there and provide epsilon trasitions to adequately carry-out the same computation, nondeterministically, of course. > Follow-up (possibly dumb) question: Does every planar NFA have a > planar DFA equivalent? 'Cause if not, then I'm in way over my > head even just doodling on paper, and might as well give up. ;-) I think I was wrong before, but I'd like to believe that my previous construction would form a language with a necessarily-nonplanar DFA. And if ever regular language has a planar NFA, then your question will be answered in the negative. Another poster mentioned that we could take a regular expression for the language then create an NFA... this should also correctly create a planar NFA and in fact a quite linear one (perhaps linear enough that there will always exist a dominating path? Perhaps it will always create an interval graph? I'd have to see a complete and correct construction before I know those answers.) J === Subject: Re: Regular Languages > Is it possible to represent each regular language with a planar NFA graph? Sure. Every regular language has a regular expression, and translations from regular expressions to NFAs using Thompson's construction are planar, if I remember correctly. If I don't remember correctly, well, then I'm certain there is a trivial planar construction very much like Thompson's. Take Care, Matt. === Subject: induction problem Let r1, r2, ..., rN and c1, c2, ... cN be two sequences of integers whose sum is equal. Such sequences are called realizable if there is an n X n matrix all of whose elements are either 0 or 1, such that, for all i, the sum of the ith row is exactly r(sub)i and and the sum of the ith column is exactly c(sub)i. Not all sequences are realizable. For example, the two sequences 0,2 and 0,2 are not. Design an algorithm to determine whether two given sequences are realizable, and construct a matrix with the corresponding row and column sums if they are. (Hint: First, strengthen the induction hypothesis to extend the problem to n X m matrices. Then, use induction on n (the number of rows). Try to place 1's in the first row so that the problem for the other n-1 rows can be solved iff the original problem can be solved). Even with the Hint, I am unable to tackle this problem. I know I have to design an algorithm, but that simply falls out of the proof by induction (I think - my other problems do anyway). I could really use some mathematical guidance on how this proof should go. newby === Subject: Re: induction problem > Let r1, r2, ..., rN and c1, c2, ... cN be two sequences of integers whose > sum is equal. > Such sequences are called realizable if there is an n X n matrix all of > whose elements are either 0 or 1, such that, for all i, the sum of the ith > row is exactly r(sub)i and and the sum of the ith column is exactly c(sub)i. > Not all sequences are realizable. For example, the two sequences 0,2 and > 0,2 are not. Design an algorithm to determine whether two given sequences > are realizable, and construct a matrix with the corresponding row and column > sums if they are. I think this problem is solved in the chapter on matrices of zeros and ones in Ryser's book, Combinatorial Mathematics, which is #14 in the Carus Mathematical Monographs series published by the Mathematical Association of America. -- === Subject: Re: induction problem > Let r1, r2, ..., rN and c1, c2, ... cN be two > sequences of integers whose sum is equal. > Such sequences are called realizable if there > is an n X n matrix all of whose elements are > either 0 or 1, such that, for all i, the sum > of the ith row is exactly r(sub)i and and the > sum of the ith column is exactly c(sub)i. Matrix for n=2: : a11 a12 --> r_1 (Sum in row 1) : a21 a22 --> r_2 (Sum in row 2) : | | : V V : c_1 c_2 (column sums for col. 1 and 2) > Not all sequences are realizable. For example, > the two sequences 0,2 and 0,2 are not. Indeed, if all the aij are restric to 0 and 1, then a11+a12 = r_1 = 0 means a11 = a12 = 0. But from a21+a22 = r_2 = 2 follows a21 = a22 = 1 then. There is no choice for the c_i in this case: The only realizable column-sum-sequence is c_1 = a11+a21 = 1, c_2 = a12+a22 = 1. You will notice, how little freedom you will have in choosing elements for a row or column, whose sum is close to zero or close to the max value n: if zero, then all matrix elements have to be zero; if n, then all matrix elements must be 1. > Design an algorithm to determine whether two given > sequences are realizable, and construct a matrix > with the corresponding row and column sums if they are. Let's ask in the simple case n=2, whether r = [1,1] and c = [1,0] might be realizable. : a11 a12 --> 1 : a21 a22 --> 1 : | | : V V : 1 0 The answer is no again, because c_2 = 0 determines ai2=0 for i=1,2. So in this case all the ai1 are equal to r_i, i.e. a11=a21=1 with sum a11+a21=2. This shows that with r = [1,1] we can realize c = [2,0] but not c = [1,0]. > I could really use some mathematical guidance on how > this proof should go. You will have to understand, what's going on. If I were to solve your problem (what I am not I would try with n=2 and n=3 to get a feeling for this problem. And then look into the other hints. This problem looks quite nice. (As if some OEIS sequence is lurking ...) Rainer Rosenthal r.rosenthal@web.de Subject: [JSH] Re: A Looney Tune? charset=iso-8859-1 === is a remarkable model for Wile E. Coyote. http://www.goldenlands.com/wb/warnbros/wbimages-cels/full size/WILE%20E%20COYOTE%20GENIUS.jpg (Warning: 112 kB file. Best viewed in 1024 x 768 or higher.) === Subject: Re: [JSH] Re: A Looney Tune? charset=utf-8 === William Rex Marshall [CapitalEth][EDouble Dot][Micro] .b3 .b9[EDo ubleDot].b9 is a remarkable model for Wile E. Coyote. http://www.goldenlands.com/wb/warnbros/wbimages-cels/full_size /WILE%20E%20CO YOTE%20GENIUS.jpg Correct. On the other hand, the roadrunner isn't what I'd exactly call a rational adversary. I mean, it occasionally passes through walls. No? > (Warning: 112 kB file. Best viewed in 1024 x 768 or higher.) -- ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: ANN weight update algs / Vector calculus question > CONTEXT > ======== > With scalars, the Taylor expansion goes like this: > y(x) = y(a) + y'(a) * (x-a) + (1/2) * y''(a) * (x-a) ^2 + (1/6) * > y'''(a) * (x-a) ^ 3 + ? This is derived in every elementary calculus book. I don't have one here but my guess is that you just look for a polynomial approximation in the form of f(x) = SUM(n=1,N){ a(n,x0) * (x-x0)^n } subject to the derivative constraints f^n(x0) = y^n(x0). A more general formula is probably derived in every advanced calculus book and in the advanced chapters of some elementary books. An obvious approach is a composite polynomial expansion of the type f(x,y) = SUM(n=0,N){ f^n(x0,y) *[(1/n!)*(x-x0)^n] } with f^n(x0,y) = SUM(m=0,N-n){ f^(n+m)(x0,y0) * [(1/m!)*(y-y0)^m] } and mixed derivative constraints. > In artificial neural nets, many training algorithms use the Hessian > matrix to find the minimum of an error surface as a function of the > weight vector. > Let > E = e(w) > where > ?w' is the weight vector space composed of N dimensions > ?E' is the scalar error > ?e' is an error function of w > in particular, > ?Ea' = e(wa) > where > ?wa' is a given weight vector. Assume it is a column vector. > ?Ea' is the scalar error at Wa > What I am trying to do here is to relate the Taylor series formula for > scalars to the one that is applied in ANNs. The ANN literature uses > this approximation around wa: > Ew = e(wa) + e'(wa) * (w-wa) + (1/2) * (w-wa)T * e''(wa) * (w-wa) + > ?????? > Note that > e'(wa) is the gradient. It is a vector with N elements. > e''(wa) is the Hessian matrix with NxN dimensions. > (w-wa)T is row vector corresponding to the column vector (w-wa). > QUESTIONS > ========= > My questions are: > 1. How was the term (1/2) * (w-wa)T * e''(wa) * (w-wa) derived > according to Taylor's formula? The factor (1/2) is clear, but what > about the Transposed Vector * Matrix * Vector? I usually just take the scalar 2-D expression and convert it to matrix notation. Then assume the matrix notation holds for all higher dimensions. (x y)^T * [a b ; c d ] (x y) = ( x y)^T * [ a*x+b*y ; c*x+d*y ] = a*x^2 + b*x*y + c*y*x + d*y^2 > 2. What would the 3rd order term be like? Since the 0th term is a > scalar, the 1st term is a vector, the 2nd is matrix, the 3rd order > term should naturally be something like a cubic matrix. Is there such > a thing? All terms are scalar. The 2nd term is a scalar function of a vector and a matrix, etc. Not sure of any terminology for the 3rd term matrix. The matrix structure is not obvious. One obvious possibility is (e f)^T *(x y) * (x y )^T *[a b ; c d] * (x y) = (e*x+f*y)*(a*x^2 + b*x*y + c*y*x + d*y^2) However, you'd better check the references. > I am wondering if there is a way to understand this problem using the > matrix notation. Perhaps this was all derived by doing the expansions > and later regrouped in matrix notation. There probably is a matrix approach. But even if you found it you wouldn't remember it and would have to do a search every time you wan to use it. With complica stuff like that I tend to expand and regroup using dyadic and/or tensor notation. If you don't know what they are, don't bother to try to learn just for this. There are better uses of your time. Hope this helps. Greg Subject: abelian group === Hey Everyone I was wondering if someone could help me with the proof of: Suppose S is a set and let Symm (S) = f: f is a bijection I must prove that this is a group with * being composition of unctions and e = identity function on S. And then how would I should that the group is non-abelian as soon as the # of elements of S >= 3. Ie, * is not commutative. Thanks. -Greg === Subject: Re: abelian group >Hey Everyone > I was wondering if someone could help me with the proof of: > Suppose S is a set and let > Symm (S) = f: f is a bijection > I must prove that this is a group with * being composition of unctions >and e = identity function on S. Simply show that all the group properties (closure, associativity, identity element and inverse elements) are fulfilled. > And then how would I should that the group is non-abelian as soon as the ># of elements of S >= 3. Ie, * is not commutative. Take two elements of S_3, show that they don't commute and extend from there to S_n, n >= 3 by induction. -- I'm not interes in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: abelian group >>Hey Everyone >> I was wondering if someone could help me with the proof of: >> Suppose S is a set and let >> Symm (S) = f: f is a bijection >> I must prove that this is a group with * being composition of unctions >>and e = identity function on S. > Simply show that all the group properties (closure, associativity, > identity element and inverse elements) are fulfilled. >> And then how would I should that the group is non-abelian as soon as the >># of elements of S >= 3. Ie, * is not commutative. > Take two elements of S_3, show that they don't commute and extend from > there to S_n, n >= 3 by induction. Extend by inclusion, it's a few lines shorter === Subject: Re: abelian group > Hey Everyone > I was wondering if someone could help me with the proof of: > Suppose S is a set and let > Symm (S) = f: f is a bijection > I must prove that this is a group with * being composition of unctions > and e = identity function on S. > And then how would I should that the group is non-abelian as soon as the > # of elements of S >= 3. Ie, * is not commutative. Can you describe how far you've gotten already, and where you're stuck? Hint #1: Can you write down exactly what it is you need to prove? === Subject: Re: Algebra shows algebraic integer limitation Oh, my god... I have never read those past posts before... do you always include the links in the james signature? I generally enjoy the arguments an counter-arguments of the posters vs. James, but now I feel kinda sad... Arraitz. -------------------------------------------------------- That's free enterprise, friends: freedom to gamble, freedom to lose. And the great thing -- the truly democratic thing about it -- is that you don't even have to be a player to lose. -Barbara Ehrenreich > === Subject: Re: Algebra shows algebraic integer limitation > Oh, my god... I have never read those past posts before... do you > always include the links in the james signature? > I generally enjoy the arguments an counter-arguments of the posters > vs. James, but now I feel kinda sad... > Arraitz. Don't be fooled. He's repealy made clear that his desire is to get people who understand math to stop replying to his posts, so that when he (or somebody) looks up his posts on Usenet, it looks like he's right, because nobody's refuting him. That's why he always starts a new thread. Knowing this (and knowing his dishonesty) I wouldn't give much (or any) stock to a post that essentially says I'm sick, and by refuting my math you're making me sicker, but instead of seeking treatment I'm just going to insist you stop refuting my math. Yeah, right. Nathan === Subject: sum of series? I was wondering if there's a way to evaluate exactly what this series converges to: infinity k^k SUM ------------ k = 1 (k+1)^(k+1) -Robert === Subject: Re: sum of series? Robert Pruvenok escribi.97 en el mensaje > I was wondering if there's a way to evaluate exactly what this series > converges to: > infinity k^k > SUM ------------ > k = 1 (k+1)^(k+1) -Robert k^k/(k+1)^(k+1) = (1/(k+1))(k/(k+1))^k = (1/(k+1))(1/(1+1/k)^k Then, if you compare with armonic series, Sum(1/k, k, inf), you get Lim((k^k/(k+1)^(k+1))/(1/k), k, inf) = Lim(k/(k+1), k, inf)/Lim((1+1/k)^k, k, inf) = 1/e and your series and the armonic series vave the same character, divergent. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: sum of series? > I was wondering if there's a way to evaluate > exactly what this series converges to: > infinity k^k > SUM ------------ > k = 1 (k+1)^(k+1) 1/4 + 4/27 + 27/256 + ... seems like a convergent series, but I doubt it. The terms look too much like 1/n in the long run. I try to calculate to prove my guess. k^k/(k+1)^(k+1) = (k/k+1)^k * 1/(k+1) = (1 - 1/(k+1))^k * 1/(k+1) > (1 - 1/k)^k * 1/(k+1) As lim (1 - 1/k)^k = 1/e > 1/3, we will have k^k/(k+1)^(k+1) > 1/3 * 1/(k+1) for all k > K for some large K. We know that Sum 1/n diverges. That proves my guess: Your series doesn't converge at all. Rainer Rosenthal r.rosenthal@web.de === Subject: Re: theorem vs. proposition >>What is the difference between a theorem and a proposition? >The difference is explained in most elementary texts of logic. >A proposition is a sentence which makes an assertion which might be true > or >false but not both. >>Oh. Yes. That particular usage slipped by me. That is a distinct technical >>usage of the term in logic. Instead, I am specifically referring to the >>other usage, that of general mathematicians in presenting their results. > I do not understand what you mean by 'the other usage' > I thought it was a universal usage, assuming that mathematical statements > say anything at all. I now realize I wasn't particularly clear in my original question and explanation. 'The other usage' to which I am referring is that in a mathematical paper/monograph/textbook in say geometry/linear algebra/etc. where a series of proven statements are presen. In my experience sometimes these are labeled 'theorem' (the majority of cases), sometimes 'proposition' (not so often, but in Heath's translation of Euclid, all are labeled propositions), and if memory serves correctly, I feel like I've seen both used in the same document (but memory does not serve well enough to give me an example). -- Mitch Harris (remove q to reply) === Subject: Re: Conjecture about Fibonacci Numbers >This might be well-known, but not to me. >Consider the sequence of ratios of neighboring Fibonacci >numbers: > 1/1, 1/2, 2/3, 3/5, 5/8, etc. >Let r_n be the nth rational on this list. What appears >to be true is this: > r_{n+2} is the rational with the smallest denominator that > is strictly between r_n and r_{n+1} >I don't immediately see how to prove this, though. >-- >Daryl McCullough >Ithaca, NY > If n1/n2 is a rational number between 3/5 and 5/8, try solving the system of equations n1 = 3*x + 5*y n2 = 5*x + 8*y Show that x and y must be positive integers. Therefore n2 >= 5 + 8. -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. The son lost his reelection attempt severely. Now history is repeating itself. pmontgom@cwi.nl Home: San Rafael, California === Subject: Re: Metric's fatal error > mass is being used as a synonym for weight*; and the liter (L) for volume, > irregardless of what that volume consists of: [bla bla bla] > the volume of different elements and substances varies; With one exception, volume is independent of the substance measured wether you measure it in litres, pints or teaspoons; The exception being your brain, which you have demonstra to be equivalent to have about a nanolitre of volume for each litre measured. The main advantage of the metric system is its division into a scale of decimal units. This should make clear why you resent its introduction, since you must be intellectually overburdened by the feat of counting to ten. === Subject: Re: Metric's fatal error > The metric system most certainly does NOT use mass as a synonym for > weight. It is true that the liter is used for volume regardless of what > that volume consists of. Isn't the same true for imperial volume units? > By what authority does the metric system exist? Simon Stevin & ISO. === Subject: Re: Metric's fatal error ETAsAhRslfo9Uf1b4bbh/ebaYj52BNX6+ QIUOIdsGUMmid18kKrp0rLEANFSehM= Sometimes you do measure quantities by volume. If you are dealing with a known substance, such as granula sugar, specifying the volume is as fgood as apecifying the mass, and maybe easier to implement. When you add 250 mL it's always about 200 grams; for the sake of argument I'll say the density of this stuff is 0.8 kg/L. Since a measuring cup marked with a 250mL graduation is cheaper and easier to use than a weighing scale marked with 200 grams, it's more convenient to use the 250 mL as a specification in recipes. --OL === Subject: Re: wooooooooooosssssssssshhhhhhhhhhh............. > On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, Shaun Rimmer ishkabibbled: > ^ > ^ Actually, my mom gave me her Mickey Mouse cereal spoon she'd gotten as a > ^ kid out of a Kellogg's box. That beats a silly ole silver spoon ANY > ^ day!!! > ^Ahhh, true! Musta had some good magiks in it after all that use eh? > ^Shaun aRe > Shaun baby, you have NO IDEA what magick i have working for me in my > life!!! Really? No idea at all? Not even an iddy-biddy little bit of an idea? Shucks... ',;~}~ > spoons not even necessary!!! But I find soup dribbles through forks. Shaun aRe === Subject: Re: wooooooooooosssssssssshhhhhhhhhhh............. On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, Shaun Rimmer ishkabibbled: ^ ^> On a supercalifragilisticexpialidocious day, after dancing about singing ^> Bibbety bobbety boo!, Shaun Rimmer ishkabibbled: ^> ^ ^> ^ Actually, my mom gave me her Mickey Mouse cereal spoon she'd gotten as a ^> ^ kid out of a Kellogg's box. That beats a silly ole silver spoon ANY ^> ^ day!!! ^> ^> ^Ahhh, true! Musta had some good magiks in it after all that use eh? ^> ^> ^> ^Shaun aRe ^> ^> Shaun baby, you have NO IDEA what magick i have working for me in my ^> life!!! ^ ^Really? No idea at all? Not even an iddy-biddy little bit of an idea? ^Shucks... ',;~}~ ^ ^> spoons not even necessary!!! ^ ^But I find soup dribbles through forks. ^ ^ ^Shaun aRe ^ ^ Forks (or other utensils) not necessary either. I'll tell ya about it sometime... giggle! -- 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 Appoin by the Psychedelic Pope, a/k/a Saint Isidore of Seville An Oin 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 === Subject: Re: wooooooooooosssssssssshhhhhhhhhhh............. X-Who-Cares: Who cares? > On a supercalifragilisticexpialidocious day, after dancing > about singing Bibbety bobbety boo!, Shaun Rimmer > ishkabibbled: ^ No normal person could possibly be lame enough to say something as insipid as that, even once. Here we have a spectacularly abnormal human cull repeating it over and over, as though the cull has absolutely no pride or is proud of its chance to be the newsgroup. Take a hike with your disgusting fantasies, lamer. === Subject: Re: wooooooooooosssssssssshhhhhhhhhhh............. In sci.math, Shaun Rimmer : >> On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, Shaun Rimmer ishkabibbled: >> ^ >> ^ Actually, my mom gave me her Mickey Mouse cereal spoon she'd gotten as a >> ^ kid out of a Kellogg's box. That beats a silly ole silver spoon ANY >> ^ day!!! >> ^Ahhh, true! Musta had some good magiks in it after all that use eh? >> ^Shaun aRe >> Shaun baby, you have NO IDEA what magick i have working for me in my >> life!!! > Really? No idea at all? Not even an iddy-biddy little bit of an idea? > Shucks... ',;~}~ >> spoons not even necessary!!! > But I find soup dribbles through forks. > Shaun aRe There's also this danger. http://www.rathergood.com/spoonguard/ For fairness, there's also http://www.rathergood.com/banspoonguard/ :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: wooooooooooosssssssssshhhhhhhhhhh............. On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, The Ghost In The Machine ishkabibbled: ^In sci.math, Shaun Rimmer ^ ^: ^> ^>> On a supercalifragilisticexpialidocious day, after dancing about singing ^>> Bibbety bobbety boo!, Shaun Rimmer ishkabibbled: ^>> ^ ^>> ^ Actually, my mom gave me her Mickey Mouse cereal spoon she'd gotten as a ^>> ^ kid out of a Kellogg's box. That beats a silly ole silver spoon ANY ^>> ^ day!!! ^>> ^>> ^Ahhh, true! Musta had some good magiks in it after all that use eh? ^>> ^>> ^>> ^Shaun aRe ^>> ^>> Shaun baby, you have NO IDEA what magick i have working for me in my ^>> life!!! ^> ^> Really? No idea at all? Not even an iddy-biddy little bit of an idea? ^> Shucks... ',;~}~ ^> ^>> spoons not even necessary!!! ^> ^> But I find soup dribbles through forks. ^> ^> ^> Shaun aRe ^> ^ ^There's also this danger. ^ ^http://www.rathergood.com/spoonguard/ ^ ^For fairness, there's also ^ ^http://www.rathergood.com/banspoonguard/ ^ ^:-) ^ ^ Now THAT was truly wack!!! Good thing I wasn't any higher than I am... talk about your bad trips!?!?!?! Rathergood, actually! Just what I needed on a snowy Monday nite!! -- 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 Appoin by the Psychedelic Pope, a/k/a Saint Isidore of Seville An Oin 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 === Subject: Re: irreducibility of polynomial >How do I check irreducibility of polynomial x^4+6x^2+2x+1 over Z/5Z? >Can I apply theorem I found on Internet: > Let D be a unique factorization domain, let Q be the quotient field >of D, and let f(x) be a primitive polynomial in D[x]. Then f(x) is >irreducible in D[x] if and only if f(x) is irreducible in Q[x]. >How is this theorem in accordance whith reducibility of x^2+1 over >Z/2Z? As others have mentioned, it is reducible. Reducing the coefficient 6 modulo 5 gives x^4 + x^2 + 2x + 1 = (x^2)^2 + (x + 1)^2. Show that the sum of two squares factors modulo 5. -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: irreducibility of polynomial Peter L. Montgomery > Reducing the coefficient 6 modulo 5 gives > x^4 + x^2 + 2x + 1 = (x^2)^2 + (x + 1)^2. > Show that the sum of two squares factors modulo 5. Could you please explain? Or did you just mean these two squares? I cannot see how 9+9 factors modulo 5. Or is it something polynomical, I don't see? Rainer Rosenthal r.rosenthal@web.de === Subject: Re: irreducibility of polynomial Peter L. Montgomery Reducing the coefficient 6 modulo 5 gives x^4 + x^2 + 2x + 1 = (x^2)^2 + (x + 1)^2. Show that the sum of two squares factors modulo 5. Could you please explain? Or did you just mean these > two squares? The two squares are the squares to the right of the equal sign. And in the integers you can write x^2 - y^2 = (x - y)(x + y), in the same way in the integers mod 5 you can write x^2 + y^2 as the product of two linear terms. -- === Subject: Re: irreducibility of polynomial Show that the sum of two squares factors modulo 5. > Could you please explain? Or did you just mean these > two squares? I cannot see how 9+9 factors modulo 5. > Or is it something polynomical, I don't see? Modulo 5 we have a^2 + b^2 = a^2 - 4 b^2. Surely you can factor that? Jyrki Lahtonen, Turku, Finland === Subject: Re: irreducibility of polynomial > Show that the sum of two squares factors modulo 5. I cannot see how 9+9 factors modulo 5. > Or is it something polynomical, I don't see? > Modulo 5 we have a^2 + b^2 = a^2 - 4 b^2. Surely you > can factor that? Me (proudly): Sure! Using x^2 - y^2 = (x+y)(x-y). (after a while ...): But how about 9 + 9? 3^2 + 3^2 = 3^2 - 4*3^2 (mod 5) ... = (3+2*3)(3-2*3) = 9 * (-3) = ... oh, now I see *grmpff*: I misread factors modulo 5 as is divisible by 5. Sorry for that :-( ... = 4 * 2 = 3 (mod 5). Well, but it factors after all, OK. I want to use this little posting to hint on the following paper: http://www.cs.toronto.edu/~mackay/sumsquares.pdf It deals with the representation of numbers as sums of two squares and was mentioned to me by Hugo Pfoertner as a fine reference. They talk about the interesting fact that many of the numbers with two or more different representations as a^2 + b^2 are indeed divisible by 5. See for example the rela sequence http://www.research.att.com/projects/OEIS?Anum=A007692 Rainer Rosenthal r.rosenthal@web.de === Subject: Re: functions needed I need to construct such a family of functions which > satisfy following constrains: f(a,b) = f(c,d) if and only if a=b, c=d or a=c, b=d. > for all positive integers a,b,c,d. Thanks >> Define f(x,x) = 0 or >> f(x,y) = x + i*y, where i^2 = -1 and x <> y. >It's supposed to be *commutative*; (a,b) -> a+bi is not, >Nor is the earlier suggestion of (a,b) -> a.b. >(I took it that f: Z+ x Z+ -> Z+ as well.) >Try this ... > f(a,b)=(2^(max(a,b)))*(3^(min(a,b))) This does not satisfy f(1, 1) = f(2, 2). If it is commutative, then f(2, 3) = f(3, 2). The requirement then becomes 2 = 3, 3 = 2, or 2 = 3, 3 = 2. There are no commutative solutions to the problem as poised. -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: The Lost Proof of Fermat The limit of [A^x + B^x]^[1/x] , as x goes to infinity, is B. For positive integers x,y,z,p x^p + y^p = z^p If (x+y)^p = z^p + f(x,y) (x-y)^p = z^p - f(x,y) (x+y)^p + (x-y)^p = 2*z^p [(x+y)^p + (x-y)^p ]/2 = x^p + y^p = z^p then p = 2 === Subject: magic cubes website I found this link http://makoto.mattolab.kanazawa-it.ac.jp/~poyo in some messages from this newsgroup and geometry.puzzles. The links were from 2001 and 2002 and the URL doesn't work anymore. Has someone still a copy of that old webpage or a new link ? === Subject: Re: magic cubes website > I found this link > http://makoto.mattolab.kanazawa-it.ac.jp/~poyo > in some messages from this newsgroup and geometry.puzzles. > The links were from 2001 and 2002 and the URL doesn't work anymore. > Has someone still a copy of that old webpage or a new link ? There's a magic squares website at http://mathforum.org/te/exchange/hos/suzuki/MagicSquare.html which you may or may not find useful. -- === Subject: floating point numbers we are planning to develop a portable mathematical library which will rely heavily on the proper implementations of the floating point standard (ieee754). does anybody have any experience with poor implementations or other troubles on different machines or systems ? in particular we would like to know on which systems we can trust the special numbers - like NaN etc - and on which we could experiece surprises ... === Subject: Chair in Statistics - Vacancy Vacancy for a Chair in Statistics at Eindhoven University of Technology, Eindhoven, The Netherlands ============================================================== ============== ===================================== The Department of Mathematics and Computer Science has a vacancy for a full professorship in Statistics General Information: -------------------------------------------------------------- -------------- -------------------------------------------------------------- -------------- --------------------------------------------- The Department of Mathematics and Computer Science provides undergraduate, MSc and PhD programs in Industrial and Applied Mathematics and in Computer Science. The Department has research collaborations with other Departments at the Technical University Eindhoven, as well as with a large number of other universities and companies, both at home and abroad. The Department has approximately 300 employees and over 700 students. The chair of Statistics is one of the nine chairs in Mathematics, and one of the four chairs in the section Statistics and Operations Research. The other two sections in Mathematics cover Analysis and Discrete Mathematics. For the future, the Department of Mathematics and Computer Science envisions important new opportunities for research on biological, biomedical, industrial and engineering applications. The statistics group is actively involved with the activities at EURANDOM, the European institute for research in stochastics. What are your duties? -------------------------------------------------------------- -------------- -------------------------------------------------------------- -------------- -------------------------------------------------------------- You are expec to stimulate and coordinate the fundamental and applied research of the group, to initiate new research directions, to establish links with other research programs at our university and to become actively involved with EURANDOM. You give and coordinate courses in Statistics and Probability, and you are responsible for updating these courses. You advise MSc and PhD students. You fulfill key management functions in the section and the faculty. Your skills are: -------------------------------------------------------------- -------------- -------------------------------------------------------------- -------------- ------------------------------------------------------- A deep and broad insight into statistics, as is reflec in a strong international reputation in the field and a large number of - also recent - publications in the international literature. Much affinity with, and knowledge of stochastics. Interest in application-orien research, experience and expertise in acquisition and consultancy. Excellent didactical qualities; leadership and management qualities. What we have to offer: -------------------------------------------------------------- -------------- -------------------------------------------------------------- -------------- -------------------------------------------------------------- A prominent leading position in a stimulating scientific environment, in which you will work with a strong group in stochastics, enthusiastic students, trainee design engineers, and PhD students, as well as postdocs from EURANDOM. A full-time appointment in accordance with the Collective Labour Agreement for Dutch Universities. A maximum salary of euro 7.875,00 g per month depending on your experience. An extensive package of fringe benefits. Inquiries: -------------------------------------------------------------- -------------- -------------------------------------------------------------- -------------- --------------------------------------------------------- For more information, please contact Prof.dr. W.T.F. den Hollander, e-mail: denhollander@eurandom.tue.nl, tel. +31 40 2478100. For information concerning job conditions, please contact Mr. W.C.J. Verhoef, head human resources, e-mail: w.c.j.verhoef@tue.nl, tel. +31 40 2472321. How to respond: -------------------------------------------------------------- -------------- -------------------------------------------------------------- -------------- ------------------------------------------------------ Please submit a written letter of application accompanied by a recent curriculum vitae to: Mrs. Drs. S. Udo, Managing Director of the Department of Mathematics and Computer Science, Eindhoven University of Technology, HG 6.22, P.O. Box 513, 5600 MB Eindhoven, The Netherlands, mentioning the number of the vacancy V32854 in your letter and on the envelope. === Subject: Re: Chair in Statistics - Vacancy > Vacancy for a Chair in Statistics at Eindhoven University of Technology, > Eindhoven, The Netherlands This vacancy would be better pos at sci.stat.math and sci.stat.edu === Subject: {Group Theory} Formula for number of groups of order n I figured out a formula for the number of groups of order n, but it is exceedingly complica and useless for computational purposes. Is it any good, or are such useless formulae for this sequence already known? Sniz Pilbor === Subject: Re: {Group Theory} Formula for number of groups of order n Adjunct Assistant Professor at the University of Montana. > I figured out a formula for the number of groups of order n, but >it is exceedingly complica and useless for computational purposes. >Is it any good, or are such useless formulae for this sequence already >known? Well, like Prof. Holt, I am not aware of any such formula (complica or not), so if you have one (even if it is recursive), then it would be interesting. The On-Line Encyclopedia of Integer Sequences also does not mention any general formula http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A000001 Sequence A000001 is the number of nonisomorphic groups of order n. The On-line Encyclopedia gives the following lists when queried for number of groups of order %N A000001 Number of groups of order n. %N A063756 Number of groups of order <= n. %N A000679 Number of groups of order 2^n. %N A090091 Number of groups of order 3^n. %N A090130 Number of groups of order 5^n. %F A090130 For a prime p >= 5, the number of groups of order p^n begins 1, 1, 2, 5, 15, 61 + 2*p + 2*gcd (p - 1, 3) + gcd (p - 1, 4), 3*p^2 + 39*p + 344 + 24*gcd(p - 1,3) + 11*gcd(p - 1,4) + 2*gcd(p - 1,5), ... %N A090140 Number of groups of order 7^n. %F A090140 For a prime p >= 5, the number of groups of order p^n begins 1, 1, 2, 5, 15, 61 + 2*p + 2*gcd (p - 1, 3) + gcd (p - 1, 4), 3*p^2 + 39*p + 344 + 24*gcd(p - 1,3) + 11*gcd(p - 1,4) + 2*gcd(p - 1,5), ... -- === Subject: Re: {Group Theory} Formula for number of groups of order n > I figured out a formula for the number of groups of order n, but >it is exceedingly complica and useless for computational purposes. >Is it any good, or are such useless formulae for this sequence already >known? >Sniz Pilbor Well I have never come ac anything like that at all, so it sounds interesting. I find it hard to imagine how there could be a formula for the number of groups of order a prime power, for example. Upper and lower bounds are known, but they are still some way apart, even logarithmically. Can you give any idea of what sort of formula is involved, without necessarily giving it in complete detail? Derek Holt. === Subject: Re: universal property of semidirect products? > |> anyway, the semi-direct product of a group h acting on a group k is > |> simply the homotopy colimit of the obvious functor f:h->groups > |> assigning the group k to the unique object of h. (the ordinary > |> colimit is obtained by starting with k and then universally compelling > |> the autofunctor of k associa with each element of h to become equal > |> to the identity autofunctor, whereas the homotopy colimit is obtained > |> by starting with k and then universally compelling the autofunctor of > |> k associa with each element of h to become naturally isomorphic to > |> the identity autofunctor, subject to a certain straightforward > |> coherence property of this system of natural isomorphisms.) > | > | > |I'm having a bit of trouble visualizing this one. > | > |h is the category with one object and with every element in the hom set > |invertible..., so the obvious functor sends this object to k inside the > |cat grp, and sends and element of h to some automorphism of k. So where > |do these autofunctors arise? > automorphism = autofunctor in this context. as you poin out, the > group h is a one-object category, but of course so is the group k; > thus its automorphisms can be thought of as autofunctors. Just to clarify, k is not quite a category with one obj in the same sense as h, in fact it is the group Aut(k) in that sense? > (hope i didn't misunderstand your question. also i should probably > mention that my use of the terminology homotopy colimit here might > be slightly (but only slightly) non-standard, and that in any case a > rigorous treatment of such homotopy colimits should take account of > the fact that the functor f:h->groups is actually a 2-functor, or at > least of the fact that the category of groups is actually a > 2-category, in fact a full sub-2-category of the 2-category of > categories.) I think I can imagine the holim in this situation, and in fact this might be a useful characterization in other areas: my holim is the third corner of a distinguished triangle. === Subject: Re: Generating function of Repunits? Thank you everybody for your reply. Xan. === Subject: Re: Dilemma with undergraduate real analysis >>Yes he should forget about this particular dismal professor (who should >>probably be fired), >> This is not clear from anything he said. His only comment on the >> professor was this: >>To be >>honest the professor is terrible and the way he goes about things >>is to make the class so easy that even the worst students do great. >What could be clearer? He says the teacher is terrible. And this is from a >student who thinks the class is way too easy. That's an evaluation worth >consideration. He's in an upper level undergraduate math course. He's had >other math profs to compare with. AND AGAIN, HE THINKS IT'S TOO EASY. I'm >inclined to believe him. Besides, the main point here is not to determine >if our OP is correct, but given his strong feelings on the matter, how best >to advise him. I'm also inclined to believe that the course is much too easy. The question of whether that arises from incompetence or from the sort of issues I suggest is relevant to solving the OP's problem (for example it says a lot about whether there's any point in the student asking the teacher to help in learning a more complete version of the material.) >Note that my firing comment is parenthetical. >> It could be that the professor is >> dismal and should be fired. Or it could be that he's perfectly >> competent, and stuck in a situation where if he taught the >> course the way it should be taught and gave the students >> the grades they should get then more than half the class >> would flunk, the students would complain, and he'd get in >> trouble for that! >Your argument seems to be that some outside agency could have caused the >professor to be dismal. Fine, he's still dismal. And by trouble you mean >what exactly? The stuff below? No, I didn't say anything about what sort of trouble I was talking about - what's below is a description of the sort of thing that can lead to trouble, not a description of what form that trouble might take. >> Yes, things like that happen. Yes, it happens that >> administrators judge competence _solely_ on the >> basis of whether the students are complaining, >> with _no_ worries about whether the complaints >> are justified. Hard to believe, but it _does_ happen >> that faculty are told that complaints from students >> are bad and must not happen, and _any_ comment >> from the faculty member regarding the validity of the >> complaints is dismissed as irrelevant, by definition. >> The fact that this sometimes happens is not something >> I can prove mathematically, but trust me, it's something >> I know is true. >Sure, one can get complaints from above, ie, the sewer. But at any decent >place you listen politely and then put them on ignore. And with tenure, you >don't have much to worry about. It's _really_ not nearly that simple - with tenure you don't have much to worry about is a very naive view. There are plenty of things they can do to make your life miserable other than firing you. Tenure does not guarantee a reasonable salary. It does not guarantee a decent teaching assignment. Etc [I have a few very specific etc's in mind that I'd rather not talk about... if you're curious send a request by email.] >Use yer academic freedom there mister >perfesser. ************************ === Subject: Re: Dilemma with undergraduate real analysis |Sure, one can get complaints from above, ie, the sewer. But at any decent |place you listen politely and then put them on ignore. And with tenure, you |don't have much to worry about. Use yer academic freedom there mister |perfesser. But if the professor in question simply has nothing to fear from complaints, by the same token any suggestion he should be fired is entirely moot. I met a professor who was described by others as tending to give mostly A's on the easiest possible choice of material. I was told students tended to give him great reviews, but those who went on to take anything depending on the material that was supposed to be covered in that course tended to be in for a shock. (All hearsay, in other words.) I don't think the first thing one should do with such a professor is to fire him, or the second or third. If the rest of the department has no other way of assessing teaching than student evaluations, they should start by coming to grips with that, and then prod him into changing his ways based on a more realistic standard. Even if one has tenure, it tends to be unpleasant to uphold standards that one knows will be praised by nobody, and win occasional complaints. === Subject: Re: Dilemma with undergraduate real analysis >|Sure, one can get complaints from above, ie, the sewer. But at any decent >|place you listen politely and then put them on ignore. And with tenure, you >|don't have much to worry about. Use yer academic freedom there mister >|perfesser. >But if the professor in question simply has nothing to fear from >complaints, by the same token any suggestion he should be >fired is entirely moot. >I met a professor who was described by others as tending to >give mostly A's on the easiest possible choice of material. I >was told students tended to give him great reviews, but those >who went on to take anything depending on the material that >was supposed to be covered in that course tended to be in for >a shock. (All hearsay, in other words.) >I don't think the first thing one should do with such a professor >is to fire him, or the second or third. If the rest of the department >has no other way of assessing teaching than student evaluations, >they should start by coming to grips with that, and then prod him >into changing his ways based on a more realistic standard. The current method of assessing teaching does not do that at all; it is essentially imposed by administrators. If you want an administrator to do something intelligent, make sure that there is NO summary number available. Many departments have faculty visit classes. An even more effective method is to look at the examinations and their claimed contributions to the grade. Probably the best is to go by how well they can use the material supposedly learned in the course. >Even if one has tenure, it tends to be unpleasant to uphold standards >that one knows will be praised by nobody, and win occasional >complaints. It is up to the faculty to uphold standards. Unless they do, their actions are fraudulent. Alas, at this time, I cannot accept at face value any American credential. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Error Analysis: how to measure the similarity between two vectors? (WAS: What kind of problem does this belong to? error analysis for matrix-vector multiplication?) Hi all, This is a question about error analysis for matrix-vector multiplication: y=A*x, where x and y are vectors, A is a matrix, in implementation, some error introduced in A, it actually computes: y=y1+y2=(A+E)*x=A*x+E*x Let y2=E*x, this is our resul error vector. Now taught by Ron, I got to know that a SVD of E can reveal the worst input x that gives the largest contribution to y2, which is error. I feel very interesting about this. Now I have some more questions: 1) the worst vectors x that lead to the largest contributions to y, I guess the largest contribution to y is measured in L2 norm of y, right? So the largest contribution means the largest L2 norm of y? Suppose the measure is a weighed L2 norm of y having different weights on each element of y, how to modify the SVD to identify the worst vector x(the most dangerous input)? 2) the worst vectors x that lead to the largest contributions to y, Will it happen such case that the vectors corresponding to the second and third largest eigenvalues of E can combine together and make a combined vector to beat the vector corresponding to the first largest eigenvalue of E, in terms of normal L2 norm and weighed L2 norm? 3) In order to avoid such worst case input vector x, I need to have a measure of similarity between two vectors x and z. When an vector z is very similar to x, it will lead to bad result; What can be a meaningful similarity measure between two vectors? Any thoughts? Please throw some lights on me! Thank you! -Walala === Subject: Quartics and elliptic curves I try to implement a program to search for elliptic curves with high rank using the method of Mestre. Within my program I search for points on a quartic (naive way). The transformation of the quartic into Weierstrass form is not a problem. This was pos by David Boyd: Suppose that C: y^2 = a*x^4 + b*x^3 + c*x^2 + d*x + e, Then the Jacobian of C has a Weierstrass equation E: y^2 = x^3+c*x^2+(b*d-4*a*e)*x-(4*a*c*e-b^2*e-a*d^2); If C has a rational point, then it is birationally equivalent to E. Now Id like to transform the points on the quartic C so that they lie on the elliptic curve E. How can that be done? (Sorry, but I`m not more than a beginner in this field). Thanks in advance. === Subject: Is C([0,1]) a Baire space? Consider the space C([0,1]) of all continuous functions from [0,1] into the reals, endowed with the metric d(f,g) = int(|f(x) - g(x)|,x in [0,1]). Is this a Baire space? My guess is that it is not, but I have been unable to prove it. === Subject: Re: Is C([0,1]) a Baire space? Originator: grubb@lola >Consider the space C([0,1]) of all continuous functions from [0,1] into >the reals, endowed with the metric > d(f,g) = int(|f(x) - g(x)|,x in [0,1]). >Is this a Baire space? My guess is that it is not, but I have been >unable to prove it. Let C_n={f: |f|(x)<=n for all x in [0,1]}. Show that C_n is closed and nowhere dense. To show that C_n is closed, you need to see that a convergent sequence has a subsequence that converges almost everywhere. That C_n has no interior should be clear. --Dan Grubb === Subject: Re: Is C([0,1]) a Baire space? > Let C_n={f: |f|(x)<=n for all x in [0,1]}. Show that C_n is > closed and nowhere dense. To show that C_n is closed, you need > to see that a convergent sequence has a subsequence that > converges almost everywhere. You don't need measure theory for this. Suppose g in C[0,1] is not in C_n. Then |g| > n + epsilon on an open interval (a,b) contained in [0,1]. This implies d(f,g) >= int_(a,b) (|g|-|f|) > epsilon*(b-a) for all f in C_n. Therefore g is not in the closure of C_n. === Subject: Re: Is C([0,1]) a Baire space? Originator: grubb@lola >Consider the space C([0,1]) of all continuous functions from [0,1] into >the reals, endowed with the metric > d(f,g) = int(|f(x) - g(x)|,x in [0,1]). >Is this a Baire space? My guess is that it is not, but I have been >unable to prove it. Oops, ignore previous post. I saw sup instead of int there. --Dan Grubb === Subject: Re: Is C([0,1]) a Baire space? Originator: grubb@lola >Consider the space C([0,1]) of all continuous functions from [0,1] into >the reals, endowed with the metric > d(f,g) = int(|f(x) - g(x)|,x in [0,1]). >Is this a Baire space? My guess is that it is not, but I have been >unable to prove it. It's a complete metric space, so a Baire space. --Dan Grubb === Subject: Re: Is C([0,1]) a Baire space? > Consider the space C([0,1]) of all continuous functions from [0,1] into > the reals, endowed with the metric > d(f,g) = int(|f(x) - g(x)|,x in [0,1]). > Is this a Baire space? My guess is that it is not, but I have been > unable to prove it. First prove that is it a complete separable metric space. === Subject: Re: Is C([0,1]) a Baire space? >> Consider the space C([0,1]) of all continuous functions from [0,1] into >> the reals, endowed with the metric >> d(f,g) = int(|f(x) - g(x)|,x in [0,1]). >> Is this a Baire space? My guess is that it is not, but I have been >> unable to prove it. > First prove that is it a complete separable metric space. I'm afraid that that's outside my capacity, since it is not a complete space. You see, if you define f_n(x) (for n > 1) as 1 if x > 1/2 + 1/n 0 if x < 1/2 - 1/n n.x/2 - n/4 + 1/2 otherwise, then you'll get a non-convergent Cauchy sequence. Furthermore, if it were complete, why should I want to prove that it is separable? === Subject: Re: Is C([0,1]) a Baire space? Consider the space C([0,1]) of all continuous functions from [0,1] into >> the reals, endowed with the metric d(f,g) = int(|f(x) - g(x)|,x in [0,1]). Is this a Baire space? My guess is that it is not, but I have been >> unable to prove it. First prove that is it a complete separable metric space. > I'm afraid that that's outside my capacity, since it is not a complete > space. Aha! A different metric. Well, can you show that the unit ball for the sup norm is nowhere dense in your integral metric? === Subject: Re: groups and permutations > Well, the first thing to note is that G strong implies G is abelian. > If G is nonabelian, then the permutation of G that interchanges g and g^-1 > for all g in G lies in the normalizer of H in S, but not in H. > If G is abelian things are more complica, but I am sure it is possible > to determine exactly which abelian groups G are strong, at least when G > is finite. After a little thought and experimentation I have: > 1. Elementary abelian p-groups are strong. > 2. C4 (cyclic group) is strong, but C2^n is not strong for n>=3. > 3. C2 x C4, C4 x C8, C2 X C8 are not strong, but C4 x C4, C2 x C2 x C4 and > C2 x C4 x C4 are. > 4. C9, C27, C3 x C9 are all strong. > Derek Holt. Wonderful reply. Thank you very very much! === Subject: Re: Decidable problem ? > Giving a number m which is the code of a Turing machine, is it possible to > decide if m is such as its associa Turing machine will always go to the > right whatever the entry is ? I would say yes. Think of the TM as a direc graph, where each node represents a state, and each arc is the result of the TM reading a particular symbol from the tape while in that state. Ignore nodes (states) unreachable from the starting node. Each node is a halt and has no arcs, or has an arc for each symbol in the alphabet. If any of the arcs transition to the left, then its node, or another node with a left transition arc, will be reached by some initial value on the tape. Consider a path from the starting node to that node. Now consider the first node along that path that has a left transition arc. All previous arcs along this path transition to the right. Thus every such node is reading a symbol from the initial tape, not one that was written along the way. The initial tape value that has the symbol that each of the arcs along the way represents (the Nth symbol on the tape counting left to right equals the symbol of the Nth arc in the path) will reach that node, and then (with one more symbol) reach the arc that transitions left. Conversely, if none of the arcs transition to the left, then naturally all inputs result in only right transitions (if any.) Does that answer your question? Cambridge, MA === Subject: Re: Decidable problem ? > Giving a number m which is the code of a Turing machine, is it possible to > decide if m is such as its associa Turing machine will always go to the > right whatever the entry is ? > I would say yes. Think of the TM as a direc graph, where each node > represents a state, and each arc is the result of the TM reading a > particular symbol from the tape while in that state. Ignore nodes > (states) unreachable from the starting node. You can solve the halting problem if you can find all of the unreachable states. Russell - 2 many 2 count === Subject: Fractal problem I came up with this fractal the other day at work, and was interes to see if I could prove a hunch about the way it works. But after an hour of looking at it with a friend of mine, I could not. Let the points ABCD form a sqaure (equilateral rectangle) whose points are labeled clockwise from the upper left corner. Let A' be the midpoint of AD, B' be the midpoint of AB, C' be the midpoint of BC and D' be the midpoint of CD. So A'B'C'D' is also a square. Define A''B''C''D'' in the same manner with A'' between A'B', B'' between B'C', C'' between C'D', and D'' between A'D'. One more iteration and I noticed something that looked to be repea from the diagram. A''' between A''D'', B''' between A''B'', C''' between B''C'', and D''' between C''D''. Of course, defined iteratively, this is a fractal. When I connec the dots in such a way that it looked much like a snowflake (I connec the outer corner of each of the squares with the inner square vertices farthest away: for example A connects to C' and D', B connects to A' and D', and so on), I noticed that D, A''' and B' seemed to be colinear. And by symmetry, every line that seemed to be that way... A' B''' and B looked colinear. So did B'C''' and C. Then I wondered if there was a way to show that these points are in fact colinear. Basically, I wan to show that the Angle D'DA''' was congruent to the angle D'DB' - I think this would show colinearity. I tried to use the similar triangles, but couldn't get enough information about the angles and/or sides to do that. I also tried various other things, like looking for congruent triangles to show that the line dropped from A''' perpendicular to DD' bisec DD', but couldn't find a way to do that either. Can anyone prove this? === Subject: Re: Fractal problem pianowow a .8ecrit: > Let the points ABCD form a sqaure (equilateral rectangle) whose points > are labeled clockwise from the upper left corner. Let A' be the > midpoint of AD, B' be the midpoint of AB, C' be the midpoint of BC and > D' be the midpoint of CD. > So A'B'C'D' is also a square. Define A''B''C''D'' in the same manner > with A'' between A'B', B'' between B'C', C'' between C'D', and D'' > between A'D'. > One more iteration and I noticed something that looked to be repea > from the diagram. A''' between A''D'', B''' between A''B'', C''' > between B''C'', and D''' between C''D''. [...] > I noticed that D, A''' and B' > seemed to be colinear. And by symmetry, every line that seemed to be > that way... A' B''' and B looked colinear. So did B'C''' and C. > Then I wondered if there was a way to show that these points are in > fact colinear. A''B'' is parallel to AB (symetry), intersects AA' in P . P is the middle of AA' (Thales in AB'A'). B' B''' and D' are aligned (symetry). Thales again in ABA' says the middle of BA' is on the parallel to AA' from B' (that is B'D'), and also on the parallel to AB from P (that is A''B''). These two lines intersect at B''' by construction. So not only A' B''' and B are colinear but also B''' is the middle of A'B ! regards -- === Subject: Re: Fractal problem In-reply-to: pianowow@hotmail.com (pianowow) >I came up with this fractal the other day at work, and was interes >to see if I could prove a hunch about the way it works. But after an >hour of looking at it with a friend of mine, I could not. >Let the points ABCD form a sqaure (equilateral rectangle) whose points >are labeled clockwise from the upper left corner. Let A' be the >midpoint of AD, B' be the midpoint of AB, C' be the midpoint of BC and >D' be the midpoint of CD. >So A'B'C'D' is also a square. Define A''B''C''D'' in the same manner >with A'' between A'B', B'' between B'C', C'' between C'D', and D'' >between A'D'. >One more iteration and I noticed something that looked to be repea >from the diagram. A''' between A''D'', B''' between A''B'', C''' >between B''C'', and D''' between C''D''. >Of course, defined iteratively, this is a fractal. When I connec >the dots in such a way that it looked much like a snowflake (I >connec the outer corner of each of the squares with the inner >square vertices farthest away: for example A connects to C' and D', B >connects to A' and D', and so on), I noticed that D, A''' and B' >seemed to be colinear. And by symmetry, every line that seemed to be >that way... A' B''' and B looked colinear. So did B'C''' and C. >Then I wondered if there was a way to show that these points are in >fact colinear. Basically, I wan to show that the Angle D'DA''' was >congruent to the angle D'DB' - I think this would show colinearity. I >tried to use the similar triangles, but couldn't get enough >information about the angles and/or sides to do that. I also tried >various other things, like looking for congruent triangles to show >that the line dropped from A''' perpendicular to DD' bisec DD', but >couldn't find a way to do that either. >Can anyone prove this? Using coordinate geometry, A' = (D+A)/2 B' = (A+B)/2 D' = (C+D)/2 A'' = (A'+B')/2 = (2A+B+D)/4 D'' = (D'+A')/2 = (A+C+2D)/4 A''' = (D''+A'')/2 = (3A+B+C+3D)/8 (A+C)/2 = (B+D)/2 since they are both the intersection of the diagonals. Therefore, we can write C = B+D-A, and use this in the formula for A''' to get A''' = (A+B+2D)/4 = (B'+D)/2 Thus, A''' is the midpoint of the segment between B' and D. Since we used only the formulas above, this is true for the same construction on any parallelogram. Rob Johnson take out the trash before replying === Subject: Differential of function with two variables? This is a question I got from reading a book, and I'm not to comfortable with English math terminology, so please bare with me. There is a question where the diff.equ.: dy/dx = sqr(x+y-1) should be solved. The author proceeds to transform using z = x+y-1 and dz/dx = 1+ dy/dx. My question is if this is possible or just a typo in the book, how it occurs and what does it represent. I mean, it shows the rate of change between z and x, but what about y? === Subject: Re: Differential of function with two variables? > This is a question I got from reading a book, and I'm not to > comfortable with English math terminology, so please bare with me. > There is a question where the diff.equ.: > dy/dx = sqr(x+y-1) > should be solved. > The author proceeds to transform using z = x+y-1 and dz/dx = 1+ dy/dx. > My question is if this is possible or just a typo in the book, how it > occurs and what does it represent. I mean, it shows the rate of change > between z and x, but what about y? The book is correct. You can do this. I don't know how to answer 'how it occurs'. As for rates of change, well, 'it' shows the relationship between the rate of change of z wrt x and y wrt x, and we know what the rate of change of y wrt x is. y is a function of x, better written as y(x), z is also then a function of x, so you can differentiate z wrt x. If you don't believe it check with y=your favourite function. In this case you subs in and dz/dx -1 = sqrt(z) which ought to then be easier to solve. at the end replace z with x+y-1. === Subject: Re: Differential of function with two variables? > This is a question I got from reading a book, and I'm not to > comfortable with English math terminology, so please bare with me. > There is a question where the diff.equ.: > dy/dx = sqr(x+y-1) > should be solved. > The author proceeds to transform using z = x+y-1 and dz/dx = 1+ dy/dx. > My question is if this is possible or just a typo in the book, how it > occurs and what does it represent. I mean, it shows the rate of change > between z and x, but what about y? There's no typo here. I'll put it in another way. You have y as a function of x. Let's write it as y(x). Now, you define z(x) = x + y(x) - 1. Therefore z'(x) = 1 + y'(x) = 1 + sqrt(x + y(x) -1) = 1 + sqrt(z(x)). So, z is a solution of the differential equation z' = 1+ sqrt(z). === Subject: Re: Differential of function with two variables? > This is a question I got from reading a book, and I'm not to > comfortable with English math terminology, so please bare with me. > There is a question where the diff.equ.: > dy/dx = sqr(x+y-1) > should be solved. > The author proceeds to transform using z = x+y-1 and dz/dx = 1+ dy/dx. > My question is if this is possible or just a typo in the book, how it > occurs and what does it represent. I mean, it shows the rate of change > between z and x, but what about y? You are to find an unknown function y=f(x). The author thinks it will be easier to find instead z=g(x) := x+f(x)-1. dz/dx = 1 + dy/dx. Each of y and z is a function of the one variable x. So there are no functions of two variables to worry about. === Subject: Re: Differential of function with two variables? > Each of y and z is a function of the one variable x. So there are > no functions of two variables to worry about. I understand it now. I thought of x and y as independent variables, which is silly of me. But I was right in thinking that dz/dx cannot have much of a meaning if y,x were independent variables, as then we would use partial derivatives. Or wasn't I? === Subject: Re: JSH: A Looney Tune? > expressed concern over their or someone else's morbid > curiosity with James' relentless pursuit of fame and fortune > through mathematics. But maybe it isn't so morbid after all. > Here's my theory. > Consider: if Mathematical Truth is the Road Runner, then is a remarkable model for Wile E. Coyote. He > is constantly slamming into walls, sailing off cliffs or > getting entangled in his own traps. Readers delight in > discovering exactly where his latest contraption fails and > can plainly see the destructive path he insists on taking. > Their efforts to help him surmount obstacles can be regarded > as part of the good-natured fun. The level of entertainment > escalates as each failure is met with a new and less than > half-baked assault on his adversary. > Once in a while he goes silent, perhaps waiting for the > delivery of the latest gadget from ACME Proof Works, after > which he wades back in with his newfangled thingamabob, only > to smack into a new obstacle, or succumb to his own > ill-conceived trap. > Personally, I think the legacy of James' posts and the > readers responses resembles a Warner Bros. cartoon and am > deligh that he persists in providing this never-ending > entertainment. > -- > There are two things you must never attempt to prove: the > unprovable -- and the obvious. > -- > The central question with is his mental state. If he has in fact ced the line from more-or-less harmless lying to himself and others to overt paranoid delusion, I would be less inclined to respond to his posts. My own guess is that he is not yet there. I think he is a deeply dishonest and rather nasty person who, in the case of the 'core error' and Fermat's Last Theorem, knows more or less consciously that his arguments are wrong, but he is not clinically insane. He may be continuing to maintain that he is *basically* right even though wrong in what he regards as niggling little details, and that eventually we will all see this and give him the credit he deserves. In the last few days he star at least 6 different threads on Rick Decker's quadratic example: Past posters, consider facts JSH: Algebra is supreme JSH: Questioning my conclusions JSH: Problem with Decker, Madigin, Ullrich et al Algebra shows algebraic integer limitation Algebraic integer surprise *** His argument in all of these is approximately the same. For the most part he is not replying to objections in these threads but keeps starting new ones. Incredibly, he does not refer ONCE to the factorization that Rick found - that is, he argues against the main result without ever actually stating it or acknowledging it! Yet is clear that he knows what that result is - these posts make no sense without it - and that it devastates his own argument. There are people who post to sci.math who are clearly deeply disturbed. It is wrong and cruel to make fun of such people. I don't think Harris is in that category. He is simply an unpleasant, mean-spiri, dishonest person who wants to achieve fame and glory on the cheap. I don't find him funny like Wile E. Coyote. He represents an anti-academic and anti-intellectual undercurrent that should be opposed. *** I especially like that title 'Algebraic integer surprise' - it sounds like a new dessert whipped up by Martha Stewart. Some other possibletitles in this vein - P(g)'s in a Blanket Half-Baked Alaska [or, Half-Baked Algebra] Turkey on a Roll Prime Rib Ideals Chock Full o' Logic GCDs Ready to Eat [replacing K-Rationals] === Subject: Re: JSH: A Looney Tune? In sci.math, C. Bond <40143FA8.8E3430BD@ix.netcom.com>: > expressed concern over their or someone else's morbid > curiosity with James' relentless pursuit of fame and fortune > through mathematics. But maybe it isn't so morbid after all. > Here's my theory. > Consider: if Mathematical Truth is the Road Runner, then is a remarkable model for Wile E. Coyote. He > is constantly slamming into walls, sailing off cliffs or > getting entangled in his own traps. Readers delight in > discovering exactly where his latest contraption fails and > can plainly see the destructive path he insists on taking. > Their efforts to help him surmount obstacles can be regarded > as part of the good-natured fun. The level of entertainment > escalates as each failure is met with a new and less than > half-baked assault on his adversary. > Once in a while he goes silent, perhaps waiting for the > delivery of the latest gadget from ACME Proof Works, after > which he wades back in with his newfangled thingamabob, only > to smack into a new obstacle, or succumb to his own > ill-conceived trap. > Personally, I think the legacy of James' posts and the > readers responses resembles a Warner Bros. cartoon and am > deligh that he persists in providing this never-ending > entertainment. > -- > There are two things you must never attempt to prove: the > unprovable -- and the obvious. > -- -- > http://www.crbond.com I should note here that Wile E. Coyote may have not been just one coyote...but many coyotes. I cite as evidence one segment where Wile E. is dressed as a large female roadrunner, and the other coyotes appear out of the rocks. The last shot one sees is said ersatz roadrunner hightailing it out of town with the mob of coyotes in hot pursuit. Make of it what you will. :-) As for JSH: I think he believes what he's doing is right. Unfortunately the rest of us can more or less recognize that something is wrong -- although it's not clear to me at times where the problem is, beyond the rather obvious and trite observation that one can always factor N as N = (N/49) * 49, in a field. Others may be more perspicaceous regarding this issue. But Mathematical Truth does throw a few curveballs now and again. Or, perhaps, seems to vanish without a clue and then appear behind us. (Fortunately Mathematical Truth doesn't go beep beep when doing so.) And unlike the Road Runner, we've harnessed Mathematical Truth -- or at least bits thereof -- for such things as modern cryptography. It's a weird world out there. -- #191, ewill3@earthlink.net -- insert random Acme gadget here It's still legal to go .sigless. === Subject: abs( Gamma(z) ) Is there an expression in terms of hyperbolic functions (or exp) for Gamma(x+i*y)*Gamma(x-i*y) with x,y reals, Gamma the Gamma fct? For a few x (1, 1/2) there are formulas according to Abramowitz & Stegun 6.1.30 ff - are there any others? --- remove the no for mail === Subject: Re: abs( Gamma(z) ) > Is there an expression in terms of hyperbolic functions (or exp) > for Gamma(x+i*y)*Gamma(x-i*y) with x,y reals, Gamma the Gamma fct? > For a few x (1, 1/2) there are formulas according to Abramowitz & > Stegun 6.1.30 ff - are there any others? > --- > remove the no for mail I do not know specifically. But I believe that Gamma(1+z) = product{k=1 to oo} (1 +1/k)^x /(1 +x/k), for those x's where this product converges (everywhere in C?). So, Gamma(x+i*y)*Gamma(x-i*y) = product{k=1 to oo} (1 +1/k)^(2x-2) /((1+(x-1)/k)^2 +y^2/k^2), which is noteworthy, in any case, because it is always real for real x, real y, and series convergence. Leroy Quet === Subject: Re: abs( Gamma(z) ) >Is there an expression in terms of hyperbolic functions (or exp) >for Gamma(x+i*y)*Gamma(x-i*y) with x,y reals, Gamma the Gamma fct? >For a few x (1, 1/2) there are formulas according to Abramowitz & >Stegun 6.1.30 ff - are there any others? For any positive integer n you'll have Gamma(n+iy) Gamma(n-iy) = (product_{j=0}^{n-1} (j^2 + y^2)) Gamma(iy) Gamma(-iy) = (product_{j=0}^{n-1} (j^2 + y^2)) pi/sinh(pi y) Similarly, Gamma(n + 1/2 + iy) Gamma(n + 1/2 - iy) = (product_{j=0}^{n-1} ((j+1/2)^2 + y^2)) Gamma(1/2+iy) Gamma(1/2-iy) = (product_{j=0}^{n-1} ((j+1/2)^2 + y^2)) pi/cosh(pi y) But you won't find an expression for general x and y, because if it were true for x and y real it should be true for complex also (at least in some domain), in particular for x = (1+t)/2, y = (1-t) i/2, leading to an expression for Gamma(t). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: categories and set theory Is, for example, the set of all vector spaces over a field K well-defined? IMO not, because it would be as big as the set of all sets for which there is no definition in ZF. On the other hand, things like the category of vector spaces over K (choose your favorite category!) are of widespread use, often things are defined via universal properties. Then how can this be justified? I don't want even think about the category of all categories... -- Just because you're paranoid Don't mean they're not after you reverse my forename for mail! - saibot === Subject: Re: categories and set theory > Is, for example, the set of all vector spaces over a field K well-defined? > IMO not, because it would be as big as the set of all sets for which there > is no definition in ZF. > On the other hand, things like the category of vector spaces over K (choose > your favorite category!) are of widespread use, often things are defined > via universal properties. Then how can this be justified? > I don't want even think about the category of all categories... There are several ways of dealing with this question. The simplest is to take a category to consist of a class of objects and a class of arrows. You usually do require that there is only a set of maps between any two objects. (Such a category is called locally small.) Another approach is to choose a universal, usually an inaccessible cardinal and allow only sets that are no larger than the universe. Grothendieck does this and then spends time showing his results are independent of the universe. Wastes time, in my opinion. A third possibility is to ignore set theory. People have complained to me that everything is a set and how can you axiomatize things without having sets. Well, you axiomatize sets without having sets and you can the same with categories. There are objects and there are arrows and to each arrow are associa two objects called the domain and codomain. And when the domain of one arrow is the codomain of another, there is a composite arrow and so on. It is possible to axiomztize the category of sets in this approach. I incline to the third view, but I am not dogmatic on the subject. === Subject: Re: categories and set theory > Is, for example, the set of all vector spaces over a field K well-defined? > IMO not, because it would be as big as the set of all sets for which there > is no definition in ZF. As Arturo says, no it isn't well defined. > On the other hand, things like the category of vector spaces over K (choose > your favorite category!) are of widespread use, often things are defined > via universal properties. Then how can this be justified? Because there is no condition on the Objects that they be a set. The Hom spaces do need to be a set though. For example, the coproduct of as many objects as you wish can be defined because the compatibility of the commuting diagrams doesn't require you to examine them all 'at the same time', merely to look at any two subobjects in the coproduct Sometimes you get round issues of size by saying: take the category of finite dimensional vector spaces. Here you only have finite coproducts and products, so you lose some flexibility. > I don't want even think about the category of all categories... If you really want to see some scary stuff consider the derived category of all abelian groups (it's not important what that really is) It is labelled D(Z), Z for the integers. There is a natural inclusion of Z to D(Z), and a 'universal homological functor' sending this copy of Z to something in another category. Now, this single object does not even have a set of subobjects. === Subject: Re: categories and set theory NNTP-Posting-User: [VI8pLR2lLSnuV1FRojUA2qQ+8Uxm8vYq] >> Is, for example, the set of all vector spaces over a field K well-defined? >> IMO not, because it would be as big as the set of all sets for which there >> is no definition in ZF. > As Arturo says, no it isn't well defined. >> On the other hand, things like the category of vector spaces over K (choose >> your favorite category!) are of widespread use, often things are defined >> via universal properties. Then how can this be justified? > Because there is no condition on the Objects that they be a set. The Hom > spaces do need to be a set though. [See Mac Lane, Categories for the Working Mathematician; he discusses some of these issues at the beginning of the book, when he first defines category, etc.] I agree, but not everyone does, and even people who would agree sometimes ignore this, intentionally or not. For example, as far as I can tell, the original construction of the (unbounded) derived category produced a category in which it's not clear that the Hom sets are actually sets. I think the same goes for Gabriel and Zisman's categories of fractions, but I'm too lazy to go to the library to check for sure. > For example, the coproduct of as many objects as you wish can be defined > because the compatibility of the commuting diagrams doesn't require you to > examine them all 'at the same time', merely to look at any two subobjects > in the coproduct. What? I thought that the defining property of the coproduct was that Hom(coproduct X_i, Y) = product Hom(X_i, Y) If your coproduct isn't indexed by a set, this doesn't make any sense, does it? -- J. H. Palmieri Dept of Mathematics, Box 354350 mailto:palmieri@math.washington.edu University of Washington http://www.math.washington.edu/~palmieri/ Seattle, WA 98195-4350 === Subject: Re: categories and set theory Discussion, linux) >> On the other hand, things like the category of vector spaces over K (choose >> your favorite category!) are of widespread use, often things are defined >> via universal properties. Then how can this be justified? > Because there is no condition on the Objects that they be a set. The Hom > spaces do need to be a set though. No, they don't. It's perfectly acceptable to have the Hom-sets be classes, too (some authors refer to a category in which some Hom-sets are proper classes as illegitimate, but it's only terminology). When Hom(A,B) is set-sized for every pair of objects A, B, the category is called locally small (so, not locally small iff illegitimate, in some authors' terminology). -- Ultimately, I can bring the entire mathematical establishment to its knees... Live in a fantasy world if you wish, but to me that's just an expression of your intellectual inferiority. -- === Subject: Re: categories and set theory Adjunct Assistant Professor at the University of Montana. >Is, for example, the set of all vector spaces over a field K well-defined? Not as sta. >IMO not, because it would be as big as the set of all sets for which there >is no definition in ZF. Correct. >On the other hand, things like the category of vector spaces over K (choose >your favorite category!) are of widespread use, often things are defined >via universal properties. Then how can this be justified? Usually, instead of working with the set of all vector spaces over a field K you work with the set of all vector spaces over a field K whose underlying set is a subset of the given set U, where U is usually a large cardinal. This means that the category is 'small' (the collection of all objects is a set). But you don't have to. You can do category theory where the collection of all objects is a proper class, because you never use more than a set of objects. So just like you can define the direct product of any family of sets, because family implies 'sethood' in a sense, even though the collection of all sets is not a set but a class, so you can define universal properties or categories where the elements range over a proper class. -- === Subject: Re: Resolvable Space <0401241647330.16637-100000@gandalf.math.ukans.edu> <0401251015510.18502-100000@gandalf.math.ukans.edu> Subject: Re: Resolvable Space > An irresovable Hausdorff space (X,t) with no isola points: > Let X be an infinite set. > Let t_0 be a Hausdorff topology on X with no isola points. >> For example Q and R, for those cardinalities. >For any infinite cardinal m, you can get a trivial >example by taking the topological sum of m copies of Q. Indeed. > Let T be the set of all topologies on X that extend t_0 and > contain no singletons. By Zorn's lemma, T has maximal elements. > Let t be a maximal element of T. >> Now what? Show every t-dense set D has nonnul interior? >That's clearly true if D = X, so assume that the complement of D, >call it C, is nonempty. Since D is t-dense, C is not t-open. >Maximality of t implies that there is a t-open set U whose >intersection with C is a singleton {x}. Since {x} is closed, >U{x} is a nonempty t-open subset of D. Oh sigh, again I confess to blank mindedness, now in regards the genesis of U. -- unresolved resolvables >> So the question of equally resolvable is barely interesting. >You can try to make it into a real question (I don't know how >interesting it is): Does every resolvable space X contain two >disjoint dense subsets A and B with |A| = |B| = density(X), where >density(X) is defined as the minimum cardinality of a dense >subset of X? Minimally 2-resolvable. Generalization: minimally n-resolvable. For example, R is minimally aleph_1 resolvable with density aleph_0 as each cofactor of R/Q is dense and countable and [R:Q] = aleph_1. As |R| = aleph_1, R is consider maximally resolvable. Thus (oh oh) R is minimally maximally resolvable with density aleph_0. I change terminology R is maximally uniformly resolvable with minimal density aleph_0 >I guess it's false, but I don't know. I haven't the foggiest. -- finite prosaics >It's trivially true for finite spaces. Finite resolvable spaces aren't T1. Any T0 finite space has an open singleton. If a minimal open set U is multipoint with distinct a,b in U then some open V with a in V, b not in V or visa versa. Thus U/V proper open subset U. So finite resolvable spaces aren't T0. Let S be a finite space and let M = { U | nonnul U minimal open subset } nonnul U is minimal open subset when for all open V proper subset U, V = nulset equivalently for all open V, U subset V or U,V disjoint M is finite pairwise disjoint collection of open sets. cl /M = S. Ah, a structure theorem for finite spaces. Assume x not in cl /M. Some open V nhood x with V / /M = nulset Some minimal open nonnul subset V But that contradicts V disjoint from all U in M When S is T0, M is a collection of singletons. When S is T1, M is the collection of all singletons. Let S be a finite space with no singletons. Then for all U in M, U is multipoint. Let D = { chose one from U | U in M } Let E = { chose another one from U | U in M } = { chose one from UD | U in M } D and E are disjoint minimally equinumerous dense subsets Thus a finite space is uniformly 2-resolvable with minimal density |M| iff it has no singletons? Let n = min { |U| : U in M }. Hence a finite space is uniformly n-resolvable with minimal density |M| iff it has no singletons. Hm, as every space is trivially 1-resolvable, singleton premise not needed. Thus with n as before A finite space is uniformly n-resolvable with minimal density |M| and is not n+1 resolvable Zongs, I feel as I've spieled off as much blah-blah as an electioneering politician espousing finite economics in an era of infinite national debt. ---- === Subject: what is the math notion for these two measures? Dear all, In order to characterize a semi-sparse matrix, I pick the following two easy-to-compute measures: 1) the number of non-zero elements of this matrix; 2) the max magnitude of the elements of this matrix; Is there a more meaningful math notion that combine the above two operations in one operation? or there are more meaningful easy-to-compute characteristics of a matrix? Thanks a lot, -Walala === Subject: Re: what is the math notion for these two measures? >Dear all, >In order to characterize a semi-sparse matrix, I pick the following two >easy-to-compute measures: >1) the number of non-zero elements of this matrix; >2) the max magnitude of the elements of this matrix; AFAIK the maximum magnitude of the elements has nothing to do with sparseness (unless of course that maximum is 0). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: A tiny puzzle in number theory > Correct! Tiny as the title indica. The point of problem was in the word > distinct, but not relative primes. ;-) > Can you parameterize the result? > How about (x,y,z) = (ai, a(a-i), a^2) with i < a-i? Very well done and rapidly! I assume a and i are integers, i.e. i is not sqrt(-1) Thus, there are infinite many of those. But, consider - is it necessary that i === Subject: Re: A tiny puzzle in number theory Correct! Tiny as the title indica. The point of problem was in the > word > distinct, but not relative primes. ;-) > Can you parameterize the result? > How about (x,y,z) = (ai, a(a-i), a^2) with i < a-i? > Very well done and rapidly! > I assume a and i are integers, i.e. i is not sqrt(-1) > Thus, there are infinite many of those. > But, consider - is it necessary that i 0, and if we mark the remainder with r: ((ai)^4 + (a(a-i))^4 + (a^2)4)/ (ai+a(a-i) + a^2) =r if now r = 441, we can found a=3 and i=1, but also as a=3 and i=2. Further we can find the third solution as a= 1 and i=5 if r=2704 then a=4 then i=1 or 3. We can also find solutions as a=5,6,7,8,9,10 etc... Thus there are evidently infinite many answers to Kyrtatas inquiry. This Kyrtatas inquiry seems to be true also for exponents n as n=3,4,5, etc. and it's no way to limi to the exponent n=4. Thus we can write a common case: (a^n+b^n+c^n)/(a+b+c) = (d^n +e^n+f^n)/(d+e+f) A puzzle: Apply Edwin's parametrization and proof it. (A little bit harder but not too hard) Correct! Tiny as the title indica. The point of problem was in the > word > distinct, but not relative primes. ;-) > Can you parameterize the result? > How about (x,y,z) = (ai, a(a-i), a^2) with i < a-i? > Very well done and rapidly! > I assume a and i are integers, i.e. i is not sqrt(-1) > Thus, there are infinite many of those. > But, consider - is it necessary that i Further, (x+y+z)= (ai + a(a-i) +a^2) = 2a^2. > What you can tell about remainder? === Subject: Re: A tiny puzzle in number theory > Is it always true that: > a) there are distinct integers x,y,z >0 so that x+y+z always divides > x^n+y^n+z^n as n is any integer >0 ? Yes, for n>1 choose x and y almost at random and make z = abs[x^n + y^n + (-x-y)^n] - x - y > b) there are distinct integers x,y,z >0 so that x+y+z always divides > x^n+y^n-z^n as n is any integer >0 ? Likewise, for n>1 z = abs[x^n + y^n - (-x-y)^n] - x - y Your question is ambiguous - these solutions change for each value of n, which is what I assume you were looking for. Possibly, though, you're looking for x = 1, y = -1, z = 2, which will work for any even or odd n, but which gives up the first condition y>0. Also, if n>2, then x=2,y=6,z=8 will work, but this does not work for n=2. If n = odd, x=1,y=2,z=3 seems to work, but not for n = even. >>make |x+y+z|<=1 ? >> That's rather difficult for distinct positive integers... >Oh that? I thought it was some kind of weird alien smiley! -- >Unpatched IE vulnerability: WebFolder data Injection >Description: Injecting arbitrary data in the My Computer zone >Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0305/13.html === Subject: Building myself a robotic girlfriend Any thoughts where to start? === Subject: Re: Building myself a robotic girlfriend > Any thoughts where to start? Hollywood?! === Subject: Re: Building myself a robotic girlfriend >Any thoughts where to start? Somewhere else besides sci.math. === Subject: Re: Building myself a robotic girlfriend Btw, can do u mathematics? > The Lord of the Rain( Suresh __NoJunkMail kumar) >Any thoughts where to start? > Somewhere else besides sci.math. === Subject: Re: Building myself a robotic girlfriend >Btw, can do u mathematics? nNo but i kan spel === Subject: Re: Building myself a robotic girlfriend I understand that inflatable girlfriends are more softig than most robots. === Subject: Re: Building myself a robotic girlfriend > The Lord of the Rain( Suresh __NoJunkMail kumar) >Btw, can do u mathematics? > nNo but i kan spel kEwL dOOwD!!!!!! 2 eye mathematics tooo. === Subject: Re: Building myself a robotic girlfriend > The Lord of the Rain( Suresh __NoJunkMail kumar) >Any thoughts where to start? > Somewhere else besides sci.math it is still math. === Subject: Re: Building myself a robotic girlfriend The Lord of the Rain( Suresh __NoJunkMail kumar) @ID-182852.news.uni-berlin.de: > Any thoughts where to start? Make her out of metal, so the Dominos Pizza refridgerator magnet will stick to the front. === Subject: Re: boolean algebra questions > Consider a boolean algebra of subsets B = where P > is the power set of A = {0, x, y, 1}, and where * will be deno by > juxtaposition. What you mean juxtaposition and what's + ? Usally for P = P(A), *,+,' are intersection, union and complement relative to A > For B, does 0 = the empty set of A and 1 = A Yes, the 0 and 1 of B which are not the 0,1 in A. -- > What is the relationship between P and the following boolean > expressions/functions F -- None, they're not expressions of B as the elements of B are subsets of A, namely nulset, {0}, {x}, {y}, {1}, {0,1}, {x,y}, {1,x,0} = {y}', etc > 1. 0 > 2. xy > 3. xy' > 4. x > 5. x'y > 6. y > 7. xy' + yx' > 8. x + y > 9. x'y' > 10. (x + y')(x + y') > 11. y' > 12. x + y' > 13. x' > 14. x' + y > 15. x' + y' > 16. 1 You want to write with much less confusion using A = {a,b,c,d} and explicate that x and y are varibles restric to P(A) = P({a,b,c,d}) Then *,+,' are as described above. -- > Are the atoms and join irreducible elements of P 1, 2, 4, and 8? If > not, what are they each respectively? If so, is B an atomic boolean > algebra? Is it *not* a complete atomic boolean algebra? What's a join irreducible? The atoms of B are {0}, {x}, {y}, {1} The coatoms are A0, Ax, Ay, A1 where Ax = A with x removed, etc. -- > For F, what are the: Huh? What's F? > minterms; > minimal sum-of-products expressions; > complete sum-of-products expressions; > prime implicants; > sums of prime implicants? -- > Is P a field of sets so that P, meaning the algebraic structure field? > If so, how would P be characterized as a field? And then what would > the relationship be between its field characterization and ring > characterization? I understand that all fields are rings, but am > unsure of the nature of all the relationships here, presuming, of > course, they exist. What's a field of sets? -- > To my knowledge, the Hasse diagram for P with the edges removed can be > diagrammed in the following manner: > a > b c d e > f g h i j k > l m n o > p > What I don't quite understand is how P would map to this diagram. The > answer to ii should go some way towards illustrating this, I'd think. Seems unrela to P(A) A A1 Ax Ay A0 {0,x} {0,y} {0,1} {x,y} {x,1} {y,1} {0} {x} {y} {1} nulset Is perhaps what you want. I find Hasse diagrams of little use. === Subject: SVD? Hi all, I need good reference to understand SVD... it seems to be very useful but we had learnt very little in class... any good treatment/paper/thesis on this topic and its applications? I used matlab [SU, SD, SV]=svd(A) where A is a square matrix... In SD I saw singular values, but where is the vector rela to that singular value? What is the relationship between eigenvalue and singular values? Thanks a lot, -Walala === Subject: Re: SVD? > Hi all, > I need good reference to understand SVD... it seems to be very useful but we > had learnt very little in class... any good treatment/paper/thesis on this > topic and its applications? > I used matlab [SU, SD, SV]=svd(A) where A is a square matrix... > In SD I saw singular values, but where is the vector rela to that > singular value? > What is the relationship between eigenvalue and singular values? Maybe you should study good books instead of asking all these questions. SVD and eigenvalues are explained in essentially every book on numerical analysis. There is no substitute for studying... Arnold Neumaier === Subject: Decker quadratic, alternate factorization Mathematicians are prejudiced towards polynomial factors, so I think it worth it to consider a factorization of the Decker quadratic into polynomial factors. Recently Rick Decker, a professor at Hamilton College, apparently trying to refute my research came up with a quadratic example, which I like because it's a quadratic, and easier to manipulate than the cubics I've used before. If you wish to see his original post here are some headers which also show that he is indeed at Hamilton College: Subject: Re: Mathematical consistency, courage Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of non-polynomial factors. Notice that despite not being polynomials they are algebraic integers if x is an algebraic integer because a_1(x) and a_2(x) are the two roots of a^2 - (x - 1)a + 7(x^2 + x). Now then, let's look at a polynomial factorization of the same quadratic: (5 c_1 x + 7)(5c_2 x + 2) = 7(25x^2 + 30x + 2) and I doubt many would argue with the idea that the 7 divides off as (5 c_1 x/7 + 1)(5c_2 x + 2) = 25x^2 + 30x + 2 despite c_1 and c_2 not being algebraic integers! So what's the difference? Clearly the difference is that you can *see* the x with polynomial factors!!! The difference has everything to do with your confidence in what you can consciously see versus trusting the algebra with the non-polynomial factors, where your confidence runs away. Now then, let's consider the c's more closely. A root is given by -7/(5 c_1), so substituting into 7(25x^2 + 30x + 2) I have 7(25(49)/(25c_1^2) - 30(7)/(5 c_1) + 2) = 0, so 2 c_1^2 - 42c_1 + 7^2 = 0 showing that c_1 is NOT an algebraic integer. Using the other root -2/(5 c_2) I have 7(25(4)/(25c_2^2) - 30(2)/(5 c_2) + 2) = 0, so 2 c_2^2 - 12 c_2 + 4 = 0, which is c_2^2 - 6 c_2 + 2 = 0 which IS an algebraic integer. What's important here is that you can't *see* a *single* quadratic that has c_1 and c_2 as its roots. Bizarre, eh? How is that possible? Why can't you just write a quadratic which has c_1 and c_2 as its roots so that you can *see* what's going on? Algebra has surprises for those of you who are really mathematicians versus being mere keepers of the faith!!! === Subject: Re: Decker quadratic, alternate factorization > Mathematicians are prejudiced towards polynomial factors, so I think > it worth it to consider a factorization of the Decker quadratic into > polynomial factors. > Recently Rick Decker, a professor at Hamilton College, apparently > trying to refute my research came up with a quadratic example, which I > like because it's a quadratic, and easier to manipulate than the > cubics I've used before. > If you wish to see his original post here are some headers which also > show that he is indeed at Hamilton College: > Subject: Re: Mathematical consistency, courage > Decker put forward the quadratic > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of > non-polynomial factors. > Notice that despite not being polynomials they are algebraic integers > if x is an algebraic integer because a_1(x) and a_2(x) are the two > roots of > a^2 - (x - 1)a + 7(x^2 + x). I wonder why you keep bringing up this example of the Decker quadratic - this is the 7th new thread on this topic you have star in the last several days - without ever mentioning why you are concerned about it. As you know, Decker showed that P(1)/7 = 57 = (5*sqrt(-2) + sqrt(7))*(-5*sqrt(-2) + sqrt(7)), which proved by simple arithmetic that your claim that in general there is no factorization of P(x)/7 where all the coefficients are algebraic integers is false. You don't even state why this example is so worrying to you. You just keep trying to show that a factorization like the above cannot exist. Yet there it is, staring you right in the face. You don't even need algebra to verify it, just simple arithmetic! Why do you keep bringing up this example while at the same time refusing to state what it shows? And why, in the present post, are you considering a totally different and really unrela topic? It almost seems like you are trying to lead people off the trail. > Now then, let's look at a polynomial factorization of the same > quadratic: > (5 c_1 x + 7)(5c_2 x + 2) = 7(25x^2 + 30x + 2) Why, why, why? The original factorization was as a polynomial in the number 5. Now you are factoring it in the variable x. What have these things to do with each other? What do they have in common, other than the fact that you are using Decker's polynomial for both? What do your conclusions about factoring it as a polynomial in x have to do with Decker's factorization or your own previous factorizations in the number 5 ??? > and I doubt many would argue with the idea that the 7 divides off as > (5 c_1 x/7 + 1)(5c_2 x + 2) = 25x^2 + 30x + 2 > despite c_1 and c_2 not being algebraic integers! Why makes you think c_1 and c_2 should be algebraic integers? In fact, as you show below, c_1 very definitely is not. Again, why do you think it should be??? This is not similar to the situation you and Rick Decker have been considering when you were factoring his polynomial in the number 5. > So what's the difference? Clearly the difference is that you can > *see* the x with polynomial factors!!! > The difference has everything to do with your confidence in what you > can consciously see versus trusting the algebra with the > non-polynomial factors, where your confidence runs away. > Now then, let's consider the c's more closely. > A root is given by -7/(5 c_1), so substituting into > 7(25x^2 + 30x + 2) > I have > 7(25(49)/(25c_1^2) - 30(7)/(5 c_1) + 2) = 0, so > 2 c_1^2 - 42c_1 + 7^2 = 0 > showing that c_1 is NOT an algebraic integer. True, but to conclude this you really need to note that this polynomial in c_1 is irreducible. Did you check that? > Using the other root -2/(5 c_2) I have > 7(25(4)/(25c_2^2) - 30(2)/(5 c_2) + 2) = 0, so > 2 c_2^2 - 12 c_2 + 4 = 0, which is > c_2^2 - 6 c_2 + 2 = 0 > which IS an algebraic integer. > What's important here is that you can't *see* a *single* quadratic > that has c_1 and c_2 as its roots. Why *should* you be able to do that ? > Bizarre, eh? How is that possible? Why can't you just write a > quadratic which has c_1 and c_2 as its roots so that you can *see* > what's going on? You can, but it won't be monic and won't have integer coefficients. There is nothing bizarre about it. Even though c_1 and c_2 are not roots of a single quadratic with integer coefficients, it is nonetheless true that -7/(5*c_1) and -2/(5*c_2) are both roots of 25x^2 + 30x + 2. But so what? What is the relevance of this ??? What has it really got to do with the point of Decker's example? Again, Decker showed that P(1)/7 = (5*sqrt(-2) + sqrt(7))*(-5*sqrt(-2) + sqrt(7)), and there you CAN see everything explicitly - and everything in sight is an *algebraic integer*. This is what you keep saying cannot happen, yet there it is. Isn't it high time you confron this head-on and explained why something that you say cannot exist clearly and unarguably does exist ??? Or are you just going to create another thread and keep evading ? > Algebra has surprises for those of you who are really mathematicians > versus being mere keepers of the faith!!! > === Subject: Re: Decker quadratic, alternate factorization Adjunct Assistant Professor at the University of Montana. [.snip.] >> 2 c_1^2 - 42c_1 + 7^2 = 0 [...] >> c_2^2 - 6 c_2 + 2 = 0 >> which IS an algebraic integer. >> What's important here is that you can't *see* a *single* quadratic >> that has c_1 and c_2 as its roots. > Why *should* you be able to do that ? >> Bizarre, eh? How is that possible? Why can't you just write a >> quadratic which has c_1 and c_2 as its roots so that you can *see* >> what's going on? > You can, but it won't be monic and won't have >integer coefficients. Oh, surely you can find one which is monic: (x-c1)(x-c2) will do. (-: But of course you know that; you mean: a quadratic that has c_1 and c_2 as its roots will not be monic with algebraic integer coefficients, and in fact there is no quadratic with integer coefficients (monic or not) which has both c_1 and c_2 as a root (the latter because both 2x^2 - 42x + 49 and x^2 - 6x + 2 are irreducible over Q, so any polynomial with coefficients in Q which has both c_1 and c_2 as roots must be a multiple of both, hence of degree 4 or greater). Sorry for the nitpicking... -- === Subject: Re: Decker quadratic, alternate factorization Just a few questions to see what I'm missing... [snippit] >Decker put forward the quadratic >(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >where his a's are roots of >a^2 - (x - 1)a + 7(x^2 + x). In other words: a_1(x) = (1/2)*(-1 + x + sqrt(1 - 30 x -27 x^2)) a_2(x) = (1/2)*(-1 + x - sqrt(1 - 30 x -27 x^2)) >The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of >non-polynomial factors. >Notice that despite not being polynomials they are algebraic integers >if x is an algebraic integer because a_1(x) and a_2(x) are the two >roots of >a^2 - (x - 1)a + 7(x^2 + x). Why? I understand that if (x-1) and (x^2 + x) are integers that the roots of 'a' must be algebraic integers, but why is this also true for algebraic integers that are not integers (like for instance sqrt(3))? >Now then, let's look at a polynomial factorization of the same >quadratic: >(5 c_1 x + 7)(5c_2 x + 2) = 7(25x^2 + 30x + 2) In other words: c1 = (7/2)*(3 - sqrt(7)) c2 = 3 + sqrt(7) or c1 = (7/2)*(3 + sqrt(7)) c2 = 3 - sqrt(7) BTW why this factorization?? Why not: (5 e_1 x + 14)(5 e_2 x + 1) = 7(25x^2 + 30x + 2) or (7 f_1 x + 7)(2 f_2 x + 2) = 7(25x^2 + 30x + 2) ??? What is so special about the factorization you chose? >and I doubt many would argue with the idea that the 7 divides off as >(5 c_1 x/7 + 1)(5c_2 x + 2) = 25x^2 + 30x + 2 That is a possibility. Why aren't the following divisions a possibility also? (5 c_1 x + 7)(5c_2 x/7 + 2/7) = 25x^2 + 30x + 2 (5 c_1 x/3.5 + 2)(5c_2 x/2 + 1) = 25x^2 + 30x + 2 (5 x + 7/c_1)(5 x + 2/c_2) = 25x^2 + 30x + 2 Aren't their simply infinite ways to divide the left hand side of the above equation by 7 just as long as the product of d_1 and d_2 are 7 (see equation below this question)? (5 c_1 x/d_1 + 7/d_1)(5c_2 x/d_2 + 2/d_2) = 25x^2 + 30x + 2 >despite c_1 and c_2 not being algebraic integers! Why should they be? (BTW c_2 is an algebraic integer). And why the word 'despite'? >So what's the difference? Clearly the difference is that you can >*see* the x with polynomial factors!!! What am I supposed to *see*? >The difference has everything to do with your confidence in what you >can consciously see versus trusting the algebra with the >non-polynomial factors, where your confidence runs away. >Now then, let's consider the c's more closely. >A root is given by -7/(5 c_1), so substituting into >7(25x^2 + 30x + 2) >I have >7(25(49)/(25c_1^2) - 30(7)/(5 c_1) + 2) = 0, so >2 c_1^2 - 42c_1 + 7^2 = 0 >showing that c_1 is NOT an algebraic integer. >Using the other root -2/(5 c_2) I have >7(25(4)/(25c_2^2) - 30(2)/(5 c_2) + 2) = 0, so >2 c_2^2 - 12 c_2 + 4 = 0, which is >c_2^2 - 6 c_2 + 2 = 0 >which IS an algebraic integer. >What's important here is that you can't *see* a *single* quadratic >that has c_1 and c_2 as its roots. Why is this important? Why would I want to *see* a *single* quadratic with c_1 and c_2 as its roots? If I really wan to see a quadratic wouldn't I just need to look at this: b^2 + (((27+/- 5 sqrt(7))/2) b + 7 >Bizarre, eh? What is? Could you please make explicit what is bizarre. > How is that possible? I don't know, because I don't know what is supposed to be impossible... Do you think c_1 and c_2 must be algebraic integers or something? > Why can't you just write a >quadratic which has c_1 and c_2 as its roots so that you can *see* >what's going on? I did write that quadratic... I still don't *see* what you are getting at. >Algebra has surprises for those of you who are really mathematicians >versus being mere keepers of the faith!!! > === Subject: Re: Decker quadratic, alternate factorization > Now then, let's look at a polynomial factorization of the same > quadratic: > (5 c_1 x + 7)(5c_2 x + 2) = 7(25x^2 + 30x + 2) > Now then, let's consider the c's more closely. > A root is given by -7/(5 c_1), so substituting into > 7(25x^2 + 30x + 2) > I have > 7(25(49)/(25c_1^2) - 30(7)/(5 c_1) + 2) = 0, so > 2 c_1^2 - 42c_1 + 7^2 = 0 > showing that c_1 is NOT an algebraic integer. > Using the other root -2/(5 c_2) I have > 7(25(4)/(25c_2^2) - 30(2)/(5 c_2) + 2) = 0, so > 2 c_2^2 - 12 c_2 + 4 = 0, which is > c_2^2 - 6 c_2 + 2 = 0 > which IS an algebraic integer. > What's important here is that you can't *see* a *single* quadratic > that has c_1 and c_2 as its roots. Well, you can, but it's not likely to be what you want. c_1 and c_2 are roots of y^2 - ((27 + 5sqrt(7))/2)y + 7 (or, depending on the values you use for c_1 and c_2, you might have a slightly different quadratic). > Bizarre, eh? How is that possible? Why can't you just write a > quadratic which has c_1 and c_2 as its roots so that you can *see* > what's going on? It's no surprise that you can't get a quadratic in Z[x]. Suppose c_1 and c_2 satisfy (y - c_1)(y - c_2) = 0 For this to have integer coefficients you would need c_1 * c_2 and c_1 + c_2 to be integers. The first condition is satisfied, since you have c_1 * c_2 = 7 But you have 2*c_1 + 7*c_2 = 42 as your other condition, which, together with the first condition, doesn't give you an integer value for c_1 + c_2. Rick === Subject: Re: Decker quadratic, alternate factorization [snip] > I have > 7(25(49)/(25c_1^2) - 30(7)/(5 c_1) + 2) = 0, so > 2 c_1^2 - 42c_1 + 7^2 = 0 > showing that c_1 is NOT an algebraic integer. > Using the other root -2/(5 c_2) I have > 7(25(4)/(25c_2^2) - 30(2)/(5 c_2) + 2) = 0, so > 2 c_2^2 - 12 c_2 + 4 = 0, which is > c_2^2 - 6 c_2 + 2 = 0 > which IS an algebraic integer. > What's important here is that you can't *see* a *single* quadratic > that has c_1 and c_2 as its roots. > Bizarre, eh? How is that possible? Why can't you just write a > quadratic which has c_1 and c_2 as its roots so that you can *see* > what's going on? I know of no algebraic principle which states that if you cannot *see* some relation, then none exists. Where is a citation of this? > Algebra has surprises for those of you who are really mathematicians > versus being mere keepers of the faith!!! > James Often in error, but never in doubt. Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: Decker quadratic, alternate factorization > I know of no algebraic principle which states that if you cannot *see* > some relation, then none exists. Where is a citation of this? You are forgetting the algebraic principle that if cannot *see* some relation then none exists. Likewise, if cannt *see* an error in a proof, then the proof is correct. === Subject: Re: Decker quadratic, alternate factorization > Mathematicians are prejudiced towards polynomial factors, so I think > it worth it to consider a factorization of the Decker quadratic into > polynomial factors. [snip] Hey Harris, you are tits on a boar hog. Hey stooopid loud troll James Harris, put up or shut up, http://www.rsasecurity.com/rsalabs/challenges/factoring/ faq.html http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net Is a $10,000 prize no questions asked too small to justify your submission of two little prime numbers? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Decker quadratic, alternate factorization >> Mathematicians are prejudiced towards polynomial factors, so I think >> it worth it to consider a factorization of the Decker quadratic into >> polynomial factors. >[snip] >Hey Harris, you are tits on a boar hog. Hey stooopid loud troll James >Harris, put up or shut up, === Subject: problem with a calculation/Gleichungsaufgabe Dies ist eine Gleichung: It's a calculation: x/(Wurzel(400-x^2)) = (Wurzel(25-(10-x)^2))/(10-x) Can someone solve this problem? And how? With a calculator I received the solution x=5,16994... But I search a path without a calculator? Thank you very much! PS: Sorry, my English is very bad. Was ergibt diese Gleichung? Ich suche x. Mit dem Taschenrechner ist die L.9asung x=5,16994... Ich suche einen L.9asungsweg. K.9annte mir bitte jemand bei dieser Aufgabe helfen? Vielen Dank im Vorraus. bettina873@hotmail.com === Subject: Re: problem with a calculation/Gleichungsaufgabe > Dies ist eine Gleichung: It's a calculation: > x/(Wurzel(400-x^2)) = (Wurzel(25-(10-x)^2))/(10-x) > Can someone solve this problem? And how? With a calculator I received > the solution x=5,16994... But I search a path without a calculator? > Thank you very much! PS: Sorry, my English is very bad. > Was ergibt diese Gleichung? Ich suche x. Mit dem Taschenrechner ist > die L.9asung x=5,16994... > Ich suche einen L.9asungsweg. K.9annte mir bitte jemand bei dieser Aufgabe > helfen? > Vielen Dank im Vorraus. > bettina873@hotmail.com I suppose Wurzel is square root. Multiply by the common denominator... x(10-x) = (Wurzel(25-(10-x)^2))(Wurzel(400-x^2)) square both sides x^2(10-x)^2=(25-(10-x)^2)(400-x^2) multiply everything out, and get a polynomial equation. === Subject: Re: problem with a calculation/Gleichungsaufgabe A N Niel a .8ecrit: >>Dies ist eine Gleichung: It's a calculation: >>x/(Wurzel(400-x^2)) = (Wurzel(25-(10-x)^2))/(10-x) [...] > Multiply by the common denominator... > x(10-x) = (Wurzel(25-(10-x)^2))(Wurzel(400-x^2)) > square both sides > x^2(10-x)^2=(25-(10-x)^2)(400-x^2) > multiply everything out, and get a polynomial equation. This gives a *second degree equation* !!! And don't forget that 400-x^2 > 0 and 25-(10-x)^2) >= 0 and x(10-x) > 0 So the second solution (13.653....) is rejec. regards. -- === Subject: Re: Past posters, consider facts > Thanks for a much more careful history lesson. I'm not altogether > sure I understand the statement, while sqrt(2) (as described above) > was accep as a geometrical object, it was not accep as another > arithmetic type of object. If I were to describe matters (and here, > I talk out my ass a bit, but it's only a little fart), I'd say that > the Pythagoreans doub that there were any incommensurable lengths, > and the proof that square root of two is incommensurable to the unit > length contradic their beliefs. That's a reasonable statement. As far as I can tell, the history books indicate that was inconsistent with their osophical beliefs. The problem is that the Pythagoreans made no distinction between length (or magnitude) and number, but this distinction was definitely made at least later by Greek mathematicians. By the time of Euclid, much of this knowledge had been synthesized, and in Book X, for example, he distinguishes between commensurable and incommensurable magnitudes and gives a partial classification of the latter. What I meant by the snippet you quote is that the Pythagoreans could not deny sqrt(2)'s existence as a geometrical object, but they could not harmoniously fit in its incommensurability into their arithmetic. Remember I was responding to the statement that sqrt(2) was not accep by the mathematicians of the day. My point was that its seemingly contradictory properties demolished some preconceived notions of some Greeks like the Pythagoreans, but in no way can that be seen as sqrt(2) being not accep. Even if (see below) the majority of Pythagoreans refused to admit their mistaken beliefs, since they can't be considered the majority of mathematicians at the time, the not accep statement still doesn't make much sense. > I honestly don't know to what extent they attemp to hold onto their > mistaken beliefs once the proof was found (apocryphal tales of > mathematicians overboard notwithstanding). Did they try to suppress > the proof? Did they deny that it was correct? Did they accept the > proof and admit it caused their osophy deep difficulties? The problem with saying anything about Pythagoras or his followers is that none of their writings have survived, except through second-hand accounts by outsiders. So, sorry, I can't answer any of these questions. We could track down a scholar of Pythagoras. What we know is that the cult dwindled in size and that this knowledge of incommensurables (which I suppose could have been known outside the Pythagorean sect) made its way into general knowledge. > Also, didn't the proof cause more difficulties than just osophical > problems for the Pythagoreans? I was under the impression that it was > a real mathematical crisis, since the proposition that every pair of > lengths is commensurable had been implicit or explicit in many, many > geometrical proofs. If my failing memory is correct, Eudoxus (sp?) > provided the solution to the crisis, but I'll bet that's a much too > glib account. We don't know if the Pythagoreans were really mathematicians. Some of them were definitely mathematically inclined, but they seem more of a religious sect. In particular, we don't know if they were concerned with the notion of proof, although they must have been familiar with the idea of logical deduction. So, we don't have any ot their proofs (if any exis) to examine and say, look there's a problem. On the other hand, it caused difficulties for others besides Pythagoreans. I'm not familiar with this area, but I recall that the incommensurability problem was a source of difficulties for mathematicians in general during that time period. We can conclude from works like that of Euclid and Archimedes that people did work on that kind of stuff (there's a whole theory of proportions by the Greeks). So that indicates this is something they found interesting and troublesome (to some degree). And yes, there is a statement known as the Axiom of Eudoxus which gives a definition of when two magnitudes are comparable and thus lets us talk about when two magnitudes are the same or less than the other, etc. But all this goes to show that the original statement that sqrt(2) was not accep by the mathematicians of the day is misguided. After all, they wouldn't have done all this work to fix the problems if they had held firm to their beliefs and denied the fact that sqrt(2) was irrational. === Subject: Re: Past posters, consider facts > Jesse, it seems talking out of your ass has become a favorite > expression of yours. ;-) >> Up to cobordism, the expression simply states a banal fact; >> as Bob Stong used to point out (possibly incorrectly; see the >> thread on genus of the human body), one's mouth is cobordant >> to one's asshole. > And you, sir, are a hnut! > Gib Ah, the old ich bin ein Berliner story comes back to life. Dale. === Subject: Latin Rectangles, golf schedule Hey folks, trying to write an app that generates golf matchups, where P is the number of players, and W is the number of weeks that they play. I would like it so that no person plays the another person more than once, (as long as P-1<=W). After doing some research (I'm not a math major), looks like Latin Rectangles are the meat behind this, but I'm not sure where to start. I want to write this in VB. Anyone point me in the right direction? Thx. === Subject: Re: Latin Rectangles, golf schedule Have a look at the Social Golfer Problem (this is very difficult). See e.g.: http://www-icparc.doc.ic.ac.uk/~wh/golf/ http://4c.ucc.ie/~tw/csplib//prob/prob010/ -------------------------------------------------------------- -- Erwin Kalvelagen erwin@gams.com, http://www.gams.com/~erwin -------------------------------------------------------------- -- > Hey folks, trying to write an app that generates golf matchups, where > P is the number of players, and W is the number of weeks that they > play. I would like it so that no person plays the another person more > than once, (as long as P-1<=W). > After doing some research (I'm not a math major), looks like Latin > Rectangles are the meat behind this, but I'm not sure where to start. > I want to write this in VB. > Anyone point me in the right direction? > Thx. === Subject: Re: The gif ones > at 09:21 PM, j.schoenfeld@programmer.net (John Schoenfeld) said: >Quantum mechanics dictates discrete space, No, it dictates continuous space-time. Discreteness comes in with So then space-time is theoretically continuous but observably discrete? No. > Then, according to you, i can measure distances and time-intervals > smaller than h/2pi. h does not have the dimensions of either distance or time. Franz === Subject: Re: The gif ones <4005ae2a$35$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 04:26 PM, j.schoenfeld@programmer.net (John Schoenfeld) said: >Then, according to you, i can measure distances and time-intervals >smaller than h/2pi. That's not what he said. You asked about observably discrete, and that's what he answered. He didn't make a prediction about the results of measurements well beyond the scale of any instrument that we know how to build. All that he addressed was what the current theory predicts and what current observations show. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolici bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: The gif ones <4005ae2a$35$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS >So then space-time is theoretically continuous but observably >discrete? Not even that. Keep in mind, of course, that better observations or more complica theories code change that. There are reasons to suspect that an ultimate theory might involve discrete space-time. But, if so, we aren't there yet. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolici bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: The gif ones Message-Id: John Schoenfeld: >> Shmuel (Seymour J.) Metz > at 09:21 PM, j.schoenfeld@programmer.net (John Schoenfeld) said: >>Quantum mechanics dictates discrete space, >> No, it dictates continuous space-time. Discreteness comes in with >> So then space-time is theoretically continuous but observably discrete? >> No. >Then, according to you, i can measure distances and time-intervals >smaller than h/2pi. hbar is an angular momentum. === Subject: Re: Elementary Inequality for Fun > Hi! > I'm asking for a classical (i.e., by hand, without using any calculators) solution of the following inequality: > Let A=(e/2)^(e/2) and B=(pi/2)^(pi/2). Then AB>3. Ady. Define e and pi by these inequalities: e > 5/2 pi > 3 Therefore A > 25/16 B > 9/4 Therefore AB > 225/64 > 3 === Subject: Re: Elementary Inequality for Fun Adjunct Assistant Professor at the University of Montana. >> Hi! >> I'm asking for a classical (i.e., by hand, without using any calculators) solution of the following inequality: >> Let A=(e/2)^(e/2) and B=(pi/2)^(pi/2). Then AB>3. >> Ady. >Define e and pi by these inequalities: You mean, e and pi ->satisfy<- the inequalities. >e > 5/2 >pi > 3 >Therefore >A > 25/16 >B > 9/4 No. You are squaring 5/4 and 3/2; but A is not (e/2)*(e/2), it is (e/2)^(e/2). If you take (5/4)^(5/4), you get approximately 1.322, while 25/16 is 1.5625. (In fact, (e/2)^(e/2) is about 1.5175, so 25/16 is too big... -- === Subject: Re: JSH: Past posters, consider facts it's between this ring thing, and the definitional problem of primality, and co-primality. doesn't prime, mean that the number has the same property as the unit, namely indivisibility in the field-a-used? now, if you cannot divide in a ring, where does that leave us, viz one (or its 'associate,' minus one) over any other integer is not in the ring of integers? the study of the guassian primes makes this look like a peice of cake, since you essentially *have* to use elementary numbertheory, as opposed to an endless algebraic manipulation ... excusez-moi; I mean, algebraic marketing! > Under addition, R must be an Abelian group > I have no clue where I got the notion that the Natural Numbers > were a ring. Can I blame Harris for that? as you know, if you've read anyhting that I typed, there was a conspiracy of states that DID vote Gore -- and the Supremes sealed that conspiracy on March 27, 2000, by refusing to hear the appeal in LaRouche v. Fowler. (Don Fowler was the DNC Chair in '96; Sentelle's 3-judge panel made the Voting Rights Act unconsitutitonal. but, hey; it's up for re-auhtorization in '07 !-) I am frankly scared of the touchscreen mentality. just like the Supremes' abrogation of the USA and Florida constitutions foments a pop-culture hatred of the electoral college on the part of some rabid Democrats. ah, so; imagine if North Dakota ... rather, imagine if Wyoming had less than one electoral college vote for president. anyway, every one of us in this debate knows about the Texas cirterium for chad -- it ain't just missing confetti! thus saith: but, at the national level, will you be able to maintain the DNC's Any One But George -- unless it's Lyndon! media glut?... are you on the board of GOPpers For Howie Troisieme? http://www.wlym.com/pdf/iclc/communism.pdf --Give the Gift of Dick Cheeny -- out of office, at last! http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: JSH: Past posters, consider facts ... > Out of curiosity about how I had all this wrong, I went and looked up > my original source of the definition of a ring: the VNR Concise > Encyclopedia of Mathematics. The definition of a ring makes no > mention of the inverse or unit elements. What it _does_ say is Under addition, R must be an Abelian group Of course, it was my responsibility to go look up Abelian group > (in a different section of the book about 300 pages away). So this > misunderstanding was due to careless use of reference materials. That is a common problem with the encyclopedial approach. You look up a term, and think that you understand all the terms used in the definition. This may be true, or may not be true. In your case you apparently had some vague ideas about what a group was, but that vague idea did not conform to reality. > And unlike your page above, where rela material is all pulled > together in one place, reading an encyclopedia isn't the best way > to try and learn something. But my page is not an encyclopedial approach, it is more in a tutorial fashion. The problem with the encyclopedial approach (and also mathworld suffers from it) is that when you look up a word or term you get a definition in words or terms for which you have some intuition about the meaning but not exact knowledge. It is easy to follow intuition, but it can lead you to a wrong conclusion. It is a top-down approach. My page is bottom-up, it starts with the essentials and goes up to the more complica stuff. This leads to rela things mentioned closer together. On the other hand, with this approach far relations will not become clear. How do the quaternions, octonions and sedenions fit in this? Describing mathematics in an encyclopedic fashion is much easier than doing it in a tutorial fashion. -- === Subject: Re: Final Rout of Synchronization Clocks in Relativity Expires: 28 days > Yes. Putting complica diagrams on a webpage is the only satisfactory >way. >> Trouble is it all takes TIME! >This should explain the way I calcula v' better than >the ASCII sketch. >http://www.briar.demon.co.uk/Henri/speed.gif >Remember this is in the non-rotating lab frame while >the discussion below is in the co-rotating table frame. >The v*cos(a) adds on emission and subtracts at the >next reflection and the cos(b) term is identical at >both ends so the speed of incidence on the mirror must >be c. >>If the table is rotating anti-clockwise, the apparent >>curvature of the paths in the rotating frame will be >>clockwise for both rays but you need to be careful which >>way you draw it because the rays progress round the table >>in opposite directions. As a result, one ray gets nearer >>to the centre when half way between the mirrors while the >>other gets further away. This didn't take long: >>http://www.briar.demon.co.uk/Henri/paths.gif >> That's good. >> Two questions: >> How can you assume that both rays hit the same point on each mirror? >> Like I said, the apparatus rotates slightly while the light is in transit. >> Therefore the point where each ray hits will vary with rotation speed. >This is drawn in the co-rotating frame in which the mirrors are >not rotating. I think the problem is a little more difficult than it appears at first sight. It is not clear what assumptions can be made. Please have a look at www.users.bigpond.com/hewn/sagnac.jpg. For instance, can it be assumed that a ray, which is aligned with the centre of the first mirror, strikes that mirror at exactly the same point when the apparatus is rotating. If not, I don't think your interpretation of what happens in the co-rotating frame can be correct. Also, can the path length (L) between the centres of two mirrors be considered constant (whether or not source dependency is assumed). I am pretty convinced now that source dependency would make virtually no difference to the operation of a sagnac. >> Also, how can you say that the curves are identical and circular? I don't >think >> that is right. >They are not necessarily circular but any deviation is too small >to show with Paint. The direction of the path turns at a constant >rate as the light moves along path because the table rotates at >a constant angular velocity. >The curves needn't be identical but a straight line path must be >the shortest so for very small v the length of the path must be >rela to v by > L = L_0 + k*v^2 >There cannot be a linear term such as > L = L_0 + k'*v + k*v^2 >because for sufficiently small v, |k*v^2| can be made smaller than >|k'*v| and for the right sign of v the path would be shorter than >a straight line, hence k'=0 >For sufficiently small v we can also ignore v^3 and higher terms so > L = L_0 + k*v^2 >is valid and this is symmetrical about v=0, hence the path lengths >must be identical. If the paths lengths are identical, the angle >turned by the table in the time taken is the same so the angles >by which the paths deviate from the straight path when they hit >the mirrors are also identical. I understand your point but it is hardly convincing. You have not considered where the rays will strike the moving mirrors. > Thus they will be displaced sideways in opposite directions when > the meet and cause movement of the fringes. >>Sideways displacement doesn't cause fringe shift since you >>are moving parallel to the wavefront. >> I'm sure it will have some kind of effect - the angle between the two >beams >> varies with rotational speed. >http://www.briar.demon.co.uk/Henri/paths.gif >The diagram shows a square layout and at the corners the light >would turn through 90 degrees just as an example. Note the green >lines indicating the angle between the tangents to the rays at >the mirror. It should be clear both rays have moved in the same >direction (clockwise) and since the magnitude is identical, the >angle between the rays remains 90 degrees. What this shows is >that the angle between the rays stays the same for small v since >they rotate the same amount in the same direction. Well, again I'm not convinced because it isn't clear what can be considered a first or second order effect. >> In Ritzian theory, light gains a velocity when reflec from a >moving mirror. >> If that is not well known, it is now - because I said it. Sure, but you forgot the source was moving. Moving AND rotating. >>Yes, but it is the moving part that affects the speed and >>your diagram even shows the light from the source moving >>at c+v, but you seem to have forgotten the effect of the >>+v when considering the relative speed of the light >>incident on mirror A. >> Actually I am reaching the conclusion that source dependency or not make >little >> difference to the operation of a sagnac. The travel time of the light >between >> mirrors is barely affec by c+v when v is extremely small. >The travel time is barely affec but more importantly it is >increased by exactly the same amount for both rays since it >depends on v^2, not v, so there should be _no_ path difference. >However, the observed fringe shift is exactly as if the speed >was c-v and c+v in the co-rotating frame. Again this is hardy convincing. When you say the time depends on v^2 (actually 1/v^2) you are forced to include a constant (k) that is not dimensionless, (T/L^2). I'm not sure what that implies. >> The standard analysis for source dependency in a ring gyro, using a >rotating >> frame, is wrong. >> It should treat the device as a 'four mirror interferometer with an >infinite >> number of mirrors' rather than a fibre optic internal reflection thing - >if you >> know what I mean. >I think of it as sending the light round the inside of a >mirrored cylinder. The cylinder, source and detector all >co-rotate so then it is obvious that the path length is >just the circumference and the speed is c relative to the >mirror in both directions. All reflections are grazing >incidence and the rotation between reflections is >negligible. The speed in the lab frame is then simply >c+v and c-v for the two rays and the movement of the >detector exactly cancels the speed difference. It isn't that simple. I think it boils down to integrating to infinity, a function that is approaching zero (if you see what I mean) Let's stick with the three mirror system. . > The curvature is opposite for the two different rays. You are missing >this > point. >>See the diagram. Curvature alters the path length as v^2 >>(the path is shortest when v=0) and increases both paths >>equally so does not produce a path length difference. It >>is easiest to see in the fibre case which as you said is >>equivalent to an infinite series of mirrors so both paths >>are just the circumference. >> But you have assumed both rays strike the mirrors at the same points. That >is >> plainly wrong. >See above, because of the symmetry it must be correct. Let's see. Assume that when the apparatus is rotating, the red beam is aligned so that it hits the exact centres of the three mirrors. The question is, does the green ray do the same. I'm not at all happy about your above statement. It depends entirely how the rays are aligned when the apparatus is not rotating. You must also consider velocity variations that occur during reflection at each moving mirror. Each acts as a source. Under source dependency, the velocity of light leaving the source appears to be (c+vsin45). But is it? Is it c+vsin(45+x) for the red ray and c+vsin(45-x) for the green one? >> But the displacement is in opposite directions for the two beams. Draw out the paths, you'll find it is the same. It's a bit >surprising. The path length might be the same but the angle of incidence at the >>eyepiece is > different - as is the sideways displacement. >>Fringes are cused by the relative angle between the beams >>which stays the same as both rays rotate the same way. The >>beam has to be wide enough to fall on the eyepiece so >>sideways movement doesn't have an effect, the actual shift >>is very small and much less than the size of the beam. >> but the sideways movement is also indicative of an angular difference >between >> the two beams. >No, see the green lines on the diagram. The rays turn the same >way by the same amount so the angle between them is unaffec. But your diagram is not indicative of what actually happens.. >I think your 'infinite mirrors' approach is the simplest to >see what happens but that method, the inertial frame approach >and the co-rotating frame we worked through here all give the >same result, there should be no fringe shift if the speed was >source-dependent while in fact the shift is what is expec >if the speed of the light was c+v/c-v in the co-rotating frame >or always c in the lab frame. Ritz sugges the idea in 1908 >and Sagnac performed the experiment around 1913. De Sitter had >already poin out that this was incompatible with observations >of eclipsing binary stars, both they and the Sagnac experiment >rule out the Ritzian (source speed dependent) model. No they don't. De Sitter was wrong. >George Henri Wilson. www.users.bigpond.com/hewn/index.htm === Subject: Re: Final Rout of Synchronization Clocks in Relativity Message-Id: <1075154105.30833.0@ersa.uk.clara.net> the ASCII sketch. http://www.briar.demon.co.uk/Henri/speed.gif Remember this is in the non-rotating lab frame while >the discussion below is in the co-rotating table frame. >The v*cos(a) adds on emission and subtracts at the >next reflection and the cos(b) term is identical at >both ends so the speed of incidence on the mirror must >be c. >If the table is rotating anti-clockwise, the apparent >>curvature of the paths in the rotating frame will be >>clockwise for both rays but you need to be careful which >>way you draw it because the rays progress round the table >>in opposite directions. As a result, one ray gets nearer >>to the centre when half way between the mirrors while the >>other gets further away. This didn't take long: >>http://www.briar.demon.co.uk/Henri/paths.gif That's good. Two questions: >> How can you assume that both rays hit the same point on each mirror? Like I said, the apparatus rotates slightly while the light is in transit. >> Therefore the point where each ray hits will vary with rotation speed. This is drawn in the co-rotating frame in which the mirrors are >not rotating. > I think the problem is a little more difficult than it appears at first sight. > It is not clear what assumptions can be made. Please have a look at > www.users.bigpond.com/hewn/sagnac.jpg. I get 'page not found'. I also tried .htm and without the /hewn/ but all give the same result. There is no mention of sagnac on the index page. However, in trying to find the URL, I found this message from a thread last year: http://groups.google.com/groups?as_umsgid= gpf3hvsvjd91rqo287msia7127npsol42m @4ax.com You said earlier in the thread: >> They were not my comments. Sorry, someone talked of a source-dependent model and said the >reason it didn't show up in binary star tests was because the >speed adjus to c over ~1 light second distance. I thought it >was you, sorry if it wasn't. The above thread was what I remembered but I wasn't involved so perhaps I misunderstood what you were saying to Sergey. Anyway, back to the plot ... > For instance, can it be assumed that a ray, which is aligned with the centre of > the first mirror, strikes that mirror at exactly the same point when the > apparatus is rotating. If not, I don't think your interpretation of what > happens in the co-rotating frame can be correct. You have to remember that a 'ray' is just the path formed by the normal to the wavefront. In reality any light beam has a finite width. > Also, can the path length (L) between the centres of two mirrors be considered > constant (whether or not source dependency is assumed). In the rotating frame it can in the Galilean case. I'm sure you know that situation is more complex in the relativistic analysis. > I am pretty convinced now that source dependency would make virtually no > difference to the operation of a sagnac. Then you are wrong. The source-dependent model predicts no fringe shift at all. >> Also, how can you say that the curves are identical and circular? I don't think >> that is right. They are not necessarily circular but any deviation is too small >to show with Paint. The direction of the path turns at a constant >rate as the light moves along path because the table rotates at >a constant angular velocity. The curves needn't be identical but a straight line path must be >the shortest so for very small v the length of the path must be >rela to v by L = L_0 + k*v^2 There cannot be a linear term such as L = L_0 + k'*v + k*v^2 because for sufficiently small v, |k*v^2| can be made smaller than >|k'*v| and for the right sign of v the path would be shorter than >a straight line, hence k'=0 For sufficiently small v we can also ignore v^3 and higher terms so L = L_0 + k*v^2 is valid and this is symmetrical about v=0, hence the path lengths >must be identical. If the paths lengths are identical, the angle >turned by the table in the time taken is the same so the angles >by which the paths deviate from the straight path when they hit >the mirrors are also identical. > I understand your point but it is hardly convincing. You have not considered > where the rays will strike the moving mirrors. The rays must hit the mirrors at exactly the same points as in the non-rotating case because, in the co-rotating frame, the mirrors are static and the light eventually has to hit the same receiver (telescope, screen or whatever.) The point of reflection could move if the angles turned along the curved paths differed from leg to leg, but since they must be symmetrical, there can be no change. > Thus they will be displaced sideways in opposite directions when > the meet and cause movement of the fringes. >>Sideways displacement doesn't cause fringe shift since you >>are moving parallel to the wavefront. I'm sure it will have some kind of effect - the angle between the two >beams >> varies with rotational speed. http://www.briar.demon.co.uk/Henri/paths.gif The diagram shows a square layout and at the corners the light >would turn through 90 degrees just as an example. Note the green >lines indicating the angle between the tangents to the rays at >the mirror. It should be clear both rays have moved in the same >direction (clockwise) and since the magnitude is identical, the >angle between the rays remains 90 degrees. What this shows is >that the angle between the rays stays the same for small v since >they rotate the same amount in the same direction. > Well, again I'm not convinced because it isn't clear what can be considered a > first or second order effect. I use 'first order' to mean the effect is dependent on v/c and 'second order' to mean the effect is dependent on (v/c)^2. The key difference is that second order will have a minimum as v->0 while first order changes sign at v=0. In this case though there are two reasons why rotation does not cause a shift, first the rotations are in the same direction so cancel and second, even if they were in opposite directions, the effect would be to change the spacing of the fringes, not the phase at the centre point. >>Yes, but it is the moving part that affects the speed and >>your diagram even shows the light from the source moving >>at c+v, but you seem to have forgotten the effect of the >>+v when considering the relative speed of the light >>incident on mirror A. Actually I am reaching the conclusion that source dependency or not make little >> difference to the operation of a sagnac. The travel time of the light between >> mirrors is barely affec by c+v when v is extremely small. The travel time is barely affec but more importantly it is >increased by exactly the same amount for both rays since it >depends on v^2, not v, so there should be _no_ path difference. >However, the observed fringe shift is exactly as if the speed >was c-v and c+v in the co-rotating frame. > Again this is hardy convincing. When you say the time depends on v^2 (actually > 1/v^2) you are forced to include a constant (k) that is not dimensionless, > (T/L^2). I'm not sure what that implies. Actually it depends on (v/c)^2 which is dimensionless but that is beside the point. the key is that any curve based on even powers is symmetrical about zero hence the path length change for +v is the same as for -v and the path difference is therefore zero. >> The standard analysis for source dependency in a ring gyro, using a >rotating >> frame, is wrong. It should treat the device as a 'four mirror interferometer with an infinite >> number of mirrors' rather than a fibre optic internal reflection thing - if you >> know what I mean. I think of it as sending the light round the inside of a >mirrored cylinder. The cylinder, source and detector all >co-rotate so then it is obvious that the path length is >just the circumference and the speed is c relative to the >mirror in both directions. All reflections are grazing >incidence and the rotation between reflections is >negligible. The speed in the lab frame is then simply >c+v and c-v for the two rays and the movement of the >detector exactly cancels the speed difference. > It isn't that simple. I think it boils down to integrating to infinity, a > function that is approaching zero (if you see what I mean) Sigma from 1->n of n chords as n -> infinity = circumference. Time to traverse path = cicumference/c regardless of rotation. > Let's stick with the three mirror system. Whatever, they all give the same result. >>See the diagram. Curvature alters the path length as v^2 >>(the path is shortest when v=0) and increases both paths >>equally so does not produce a path length difference. It >>is easiest to see in the fibre case which as you said is >>equivalent to an infinite series of mirrors so both paths >>are just the circumference. But you have assumed both rays strike the mirrors at the same points. That is >> plainly wrong. See above, because of the symmetry it must be correct. > Let's see. Assume that when the apparatus is rotating, the red beam is aligned > so that it hits the exact centres of the three mirrors. The question is, does > the green ray do the same. I'm not at all happy about your above statement. It > depends entirely how the rays are aligned when the apparatus is not rotating. The rays must follow identical paths when the apparatus is stationary from basic optics, I'm sure you know that. Since the angles change symmetrically, they do not change the point of reflection. > You must also consider velocity variations that occur during reflection at each > moving mirror. Each acts as a source. > Under source dependency, the velocity of light leaving the source appears to be > (c+vsin45). > But is it? > Is it c+vsin(45+x) for the red ray and c+vsin(45-x) for the green one? The angle 'a' in this diagram http://www.briar.demon.co.uk/Henri/speed.gif is 45+x for the red ray at the top right mirror in this diagram: http://www.briar.demon.co.uk/Henri/paths.gif The trick though is that you than have to find the speed at which the light approaches the mirror at the top left. The angle there is also 45+x so when you subtract the motion of the mirror which is moving away from the light, the nett result is a relative incident speed of exactly c. The same applies to the green ray but you need to take a mirror image of the speed.gif sketch and the angle then, as you say, is 45-x at both ends but again since they are the same the contributions from v cancel. >>Fringes are cused by the relative angle between the beams >>which stays the same as both rays rotate the same way. The >>beam has to be wide enough to fall on the eyepiece so >>sideways movement doesn't have an effect, the actual shift >>is very small and much less than the size of the beam. but the sideways movement is also indicative of an angular difference >between >> the two beams. No, see the green lines on the diagram. The rays turn the same >way by the same amount so the angle between them is unaffec. > But your diagram is not indicative of what actually happens.. Why not? you seemed happy with it a few posts ago. >I think your 'infinite mirrors' approach is the simplest to >see what happens but that method, the inertial frame approach >and the co-rotating frame we worked through here all give the >same result, there should be no fringe shift if the speed was >source-dependent while in fact the shift is what is expec >if the speed of the light was c+v/c-v in the co-rotating frame >or always c in the lab frame. Ritz sugges the idea in 1908 >and Sagnac performed the experiment around 1913. De Sitter had >already poin out that this was incompatible with observations >of eclipsing binary stars, both they and the Sagnac experiment >rule out the Ritzian (source speed dependent) model. > No they don't. De Sitter was wrong. Your post to Sergey sugges that, without extinction, you would agree with De Sitter, and I am sure you gave a value of around 1 light second for the relaxation. You might find it in the thread I reference above with a bit of hunting. George === Subject: Re: A tiny puzzle in number theory Ooops! Correction: in my examples c^4=f^4 and c=f. So they does not satisfy Kyrtatas inquiry!!! Sorry :-( > A tiny notice as a tip: > Consider Diophantine quartic equations: > http://mathworld.wolfram.com/DiophantineEquation4thPowers.html > (a reference begins): V. Kyrtatas noticed a=3, b=7, c=20, d=25, e=38 and f=39 satisfy: > (a^4+b^4+c^4)/(d^4 +e^4+f^4)= (a+b+c)/(d+e+f) > and asks if there are any other distinct integer solutions > (end of reference) > We can rewrite Kyrtatas equation like: > (a^4+b^4+c^4)/(a+b+c) = (d^4 +e^4+f^4)/(d+e+f) > According to Edwin's parametrization,this means, as a and i are integers >0, > and if we mark the remainder with r: > ((ai)^4 + (a(a-i))^4 + (a^2)4)/ (ai+a(a-i) + a^2) =r > if now r = 441, we can found a=3 and i=1, but also as a=3 and i=2. Further > we can find the third solution as a= 1 and i=5 > if r=2704 then a=4 then i=1 or 3. > We can also find solutions as a=5,6,7,8,9,10 etc... > Thus there are evidently infinite many answers to Kyrtatas inquiry. > This Kyrtatas inquiry seems to be true also for exponents n as n=3,4,5, etc. > and it's no way to limi to the exponent n=4. > Thus we can write a common case: > (a^n+b^n+c^n)/(a+b+c) = (d^n +e^n+f^n)/(d+e+f) > A puzzle: Apply Edwin's parametrization and proof it. > (A little bit harder but not too hard) Correct! Tiny as the title indica. The point of problem was in the > word > distinct, but not relative primes. ;-) > Can you parameterize the result? > How about (x,y,z) = (ai, a(a-i), a^2) with i < a-i? Very well done and rapidly! > I assume a and i are integers, i.e. i is not sqrt(-1) > Thus, there are infinite many of those. But, consider - is it necessary that i What you can tell about remainder? > === Subject: The Right Question Thanks Gary. This is interesting. However, again I do not think Hartle, Gell-Mann and All The King's Men in, for example, Lee Smolin's very excellent Three Roads to Quantum Gravity are asking The Right Question. First look at the paper by Brian Josephson and Fotini Pallikari http://www.tcm.phy.cam.ac.uk/~bdj10/papers/bell.html http://www.tcm.phy.cam.ac.uk/~bdj10/ On how living matter may differ from dead matter in regard to a violation of the signal locality of orthodox micro-quantum theory in ALL interpretations Bohmian and others. This issue has been further developed by Antony Valentini http://www.fourmilab.ch/rpkp/valentini.html http://arxiv.org/abs/quant-ph/0104067 http://arxiv.org/abs/quant-ph/0112151 http://arxiv.org/abs/quant-ph/0203049 http://arxiv.org/abs/quant-ph/0112151 realizing the exotic vacuum zero point dark energy and dark matter with w = pressure/energy density = -1 may play an equivalent role giving effective signal nonlocality violating micro-quantum theory on the MACRO-QUANTUM scale in our local past light cone Hubble horizon limi universe that is not in thermal equilibrium on classical (i.e. a misnomer for MACRO-QUANTUM), micro-quantum, or sub-quantum levels. None of The Pundits in quantum gravity and quantum cosmology today have really considered P.W. Anderson's More is different in its fullness IMHO at the level of what Lenny Susskind aptly terms organizing ideas of high compression with immense heuristic value in terms of guides to the The Question. Lee Smolin in Ch 3 gives ample reason why the micro-quantum approach, assumed above, can be expec to break down at the MACRO level. Hartle et-al are simply over-extrapolating IMHO and have not been asking the right questions. Micro-quantum theory is linear nonlocal (entanglement) and unitary in time evolution in a non-dynamical space-time background with signal locality and no cloning of qubits where uncontrolled environmental decoherence dominates. MACRO-QUANTUM theory is nonlinear local (complementary to entanglements) and distinctly non-unitary in time evolution in a dynamical background-independent local curved space-time geometry at the low energy c-number ODLRO regime of cosmology and large scale physics i.e. ~ 1 fermi or greater in current cosmological epoch. There is no sub-quantal heat death here. There is no Born probability interpretation here. The latter is not fundamental in Bohm's ontology BTW. The generalized phase rigidity (P.W. Anderson) of the single-valued giant (MACRO) quantum vacuum wave is immune from uncontrolled environmental decoherence. This is Andrei Sakharov's metric elasticity in one instance. Physics is simple when it is local. (J.A. Wheeler) and I have just explained why and how that happens on the large low energy scales. The micro-quantum vacuum with the immutable non-dynamical globally flat spacetime is unstable to the non-perturbative formation of the MACRO-QUANTUM dynamical mutable curved space-time c-number vacuum (Action with direct two-way reaction) whose long-range holographic coherence (i.e. phase rigidity) is essentially the inflation field. Einstein's gravity, the guv field emerges from phase ripples in the inflation vacuum coherence LOCAL c-number ODLRO (Oliver Penrose & Lars Onsager) field and both dark energy and dark matter emerge from the amplitude ripples in the same vacuum coherence field. A simple pretty picture on the level of the organizing idea that can be intuitively comprehended by ordinary intelligent people who are not physicists. Bohmian Histories and Decoherent Histories http://www.arxiv.org/abs/quant-ph/0209104 Authors: James B. Hartle (University of California, Santa Barbara) Comments: 7 pages, 3 figures, Revtex, improved format and references, several points clarified The predictions of the Bohmian and the decoherent (or consistent) histories formulations of the quantum mechanics of a closed system are compared for histories -- sequences of alternatives at a series of times. For certain kinds of histories, Bohmian mechanics and decoherent histories may both be formula in the same mathematical framework within which they can be compared. In that framework, Bohmian mechanics and decoherent histories represent a given history by different operators. Their predictions for the probabilities of histories therefore generally differ. However, in an idealized model of measurement, the predictions of Bohmian mechanics and decoherent histories coincide for the probabilities of records of measurement outcomes. The formulations are thus difficult to distinguish experimentally. They may differ in their accounts of the past history of the universe in quantum cosmology. === Subject: Re: The Right Question O_o Now I remember why I'm in chemistry. --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: S-curve equation? Can anyone please write down an equation for y=f(x) where x goes from any positive value to another? Thanks alot, Mike === Subject: Re: Action Device: Destination Alpha Centauri A > Look, until I plan my future life, I can not build this Action Device. Alpha Centauri A is 4.35 ly away. So, please tell me, I am > accelerating my bike in space at the rate of 1g i.e. 9.8 m/s^2. How > much time it will take to cover 2.18 lightyears distance i.e. when I > have covered half the road? Please Note: Velocity Does Not Matter.... Thanks! -Abhi. > Plan for about 1.79 years. Bon voyage! Sorry correction: Without artificial gravity, it is not possible. We can NOT sustain G force greater than 1g for months. -Abhi. === Subject: Re: Action Device: Destination Alpha Centauri A > Look, until I plan my future life, I can not build this Action Device. Alpha Centauri A is 4.35 ly away. So, please tell me, I am > accelerating my bike in space at the rate of 1g i.e. 9.8 m/s^2. How > much time it will take to cover 2.18 lightyears distance i.e. when I > have covered half the road? Please Note: Velocity Does Not Matter.... Thanks! -Abhi. > Plan for about 1.79 years. Bon voyage! Without artificial gravity, it is not possible. We can sustain G force greater than 1g for months. -Abhi. === Subject: Re: Action Device: Destination Alpha Centauri A > In sci.math, Abhi > : >> In sci.math, Abhi >> <8d1edf6d.0401241427.2ccaf192@posting.google.com>: >> Look, until I plan my future life, I can not build this Action Device. Alpha Centauri A is 4.35 ly away. So, please tell me, I am >> accelerating my bike in space at the rate of 1g i.e. 9.8 m/s^2. How >> much time it will take to cover 2.18 lightyears distance i.e. when I >> have covered half the road? Please Note: Velocity Does Not Matter.... Thanks! -Abhi. First, there's the issue of common units. 4.35 light years is >> 4.12 * 10^16 m away. Half the distance to the goal line is >> 2.06 * 10^16 m. Assuming Newtonian theory, which isn't quite right: d = 1/2 * a * t^2 >> t = sqrt(2 * d / a) >> t = sqrt(2 * 2.06 * 10^16 m / 9.805 m/s/s) = 64822355 seconds >> or a smidge more than 2 years. Yes, it comes out to be around 750 days. It means total trip will take > around at least 8 years. Not possible for human being to spend 8 years > of his/her life in just journey. All the people are still unaware. But what has happened that we have > technology through which we can reach to Mars in just one day. We can > take trip of all the planets within month or two. But we just can't > leave this solar system. We are confined, for the moment, to our solar > system. > I'm not sure how you got the idea that Sojourner, Spirit, > and Opportunity reached Mars in one *day*. A proper free > orbit trajectory from Earth to Mars would take about 9 > months or so, with a burn at the end to match the planet's > speed, circularizing the orbit. Of course one can cheat > a bit here, as the burn at the end can bleed off some of > the excess velocity in the wrong direction as well as > increase the velocity in the right direction. There is no burn involved in this device. For example, if you are strong enough, on earth, you can lift body of which weight is greater than your body. You have lot of muscle power, potential energy. But in space, you can not use your strong muscle power to move a single centimeter without expelling some mass. This is the biggest problem in propulsion physics. I am talking about device in which you can use your potential energy to propel device and your body in only one direction. Expanding this idea, we can build spaceships which can use solar power or nuclear energy to propel spacecraft. No need to carry heavy tanks of fuel. > landed just the other day. That's about a 6 1/2 month > travel time, give or take. We can reach there in just one day with constant acceleration of 1g at half distance and decceleration by 1g over next half distance. We can use solar power as source of energy. > Deceleration will take a similar amount of time. Since the final compu v = a * t = 9.805 * 64822355 = >> 635583200 = 2.11c this isn't even close to being correct >> from a physical standpoint. However, I'd have to work >> out the Lorentz contraction, among other things. It should be possible to reach Alpha Centauri in about a >> year (subjective), assuming sufficient fuel and a brutal >> 10g trip, but the Twin Paradox still applies. It is not possible for human being to sustain G-Force greater than 1G > for long time, days and nights, for months. This action device can > accelerate the spaceship with acceleration far greater than 1g, say > 50g. But it will have to be Robotic mission first. > Sure. Can it get off the ground? Will it spring up into > space orbit if you let it go, or just drop in a pile? Spring is just basics of this mechanism in simplest forms. We can use electromagnets in synchronization to launch spaceshuttle in space. -Abhi. === Subject: Re: Action Device: Destination Alpha Centauri A > Who holds the watch, Well if it's a Newtonian then it won't matter, Uncle, 'Cos Newtonian watches all read the same time eh! And if it's a Relativist then it won't matter either, ('Cos Relativists can't tell the time eh! keith stein === Subject: Re: Action Device: Destination Alpha Centauri A > Who holds the watch, > Well if it's a Newtonian then it won't matter, Uncle, > 'Cos Newtonian watches all read the same time eh! > And if it's a Relativist then it won't matter either, > ('Cos Relativists can't tell the time eh! > keith stein Keith, gravity affects clock, right? What if I create artificial gravity in your spaceship which will give illusion to clock that the spaceship is not moving at all. -Abhi. === Subject: Re: Action Device: Destination Alpha Centauri A > Uncle Al, I request you to unleash your wisdom on me by analysing this > design.. http://www.geocities.com/inertial_propulsion I may not be Uncle Al but the error is obvious, you > are still not showing how you keep the angle ABC > fixed. This requires a force and the reaction to > that exactly cancels the desired unidirectional > force. This shows the rod and springs version: http://www.dishman.me.uk/George/Abhi/abhi_rod.gif Welcome back, George. Happy news is that I have this angled rod and > two springs. I am stretching these two springs and I find that angle > ABC still remains fixed! > Excellent, try it and see is the best way to make > progress, well done on taking the first step. > Now tell me this, as you pull down on the springs, > do you also have to hold the angled rod to stop it > moving? If not, which way are you pulling the springs, > along the arms of the rod or are you pulling A in a > direction directly away from C and vice versa? Well, when the V shped rod with inser springs is lying on floor, I just can't stretch the springs. So I had to hold this V structure vertical with point A and C on floor. Then I pulled these springs. Now the springs are stretched and I tried to keep this structure horizontally lying on floor. Now my fingers are along X axis. I am finding that my both hands are feeling force towards origin O of X axis and point B and whole V rod is trying slide along Y axis. But this is the key point. I have not fixed arms between point A and C. When these arms are fixed, like my fingers, point A and C will try to move towards center of X axis and V rod will slide along Y axis. But if V rod slide along Y axis, distance between point A and C along X axis will decrease and according to pythagoras theorem, point D will move in upward direction. That is if point B slides along Y axis through length l, point D will move in upward direction through length L where L > l. > Obviously it requires force to keep angle ABC fixed. But as I am using > iron rod, intermoleculer forces at point B are so strong that due to > spring forces acting at point B, angle ABC does not change. > So far, so good. > And also > there are other ways to keep it fixed or make it in such a way that, > if angle ABC is 60 degree, it changes from 60 degree to 0 degree. For > this purpose, we can use another spring having restoring force at > point B or attached in between middle points of spring AB and CB. > As long as you have a working method, go with it. > It is easy to be distrac by alternatives. > BTW, I have noticed that you are still clinging to my spring design > and not commenting on this magnet design.. http://www.geocities.com/inertial_propulsion > The spring pulls the ends together and so do the > magnets so from the point of view of the mechanics > there is no difference. Since I already did the > drawing for the springs, I just reminded you of it. > It applies equally well for the magnets. It is rather confusing to explain with spring drawing. But this magnet drawing is pure geometry. Magnets must move towards each other along straight line. > Another happy news is that, take a elastic wooden stick of mass 'm' > gram. Attach stone or any mass of '2m' gram. Bend this stick and > release both ends. As the stick tries to straighten up, center point > of stick and hence stone gains momentum. Due to this momentum of > stone, stone and stick are propelled in only one direction. I did it > with bicycle spoke. Why don't you repeat it and place it on record > here what exactly you observed? > There is an equal and opposite reverse force on your > finger while you bend the stick. When you release the > stone, it goes forward while your finger goes back. Of > course you have a mass much more than the stone and > stick so the motion is not noticeable but you move a > tiny amount all the same. If you are standing firmly > on the ground it is the whole mass of the Earth that > moves by an incredibly small amount, but the force is > still balanced. Is it so? I will figure out method in which my fingers are not involved. > So I am going to build this Action Device ... Good, I suggest you start with the springs from > two biros and a small bit of wire and see if it > can lift just its own weight. Wonder what will support weight of entire structure when it is > hanging? Think about this. Imagine you are standing in between two walls just 2 > feet away from each other. With your back on one wall no.1, lift your > right leg and push wall no.2. Now also lift your left leg and push > again wall no.2. You are hanging in between two walls. What is > supporting your weight? > Friction stops you sliding down the wall but your mass > is added to that of the wall so a worm under the wall > would feel just a little more force pressing down on > him. Again, the forces are balanced. Not quite right. Even if walls are slippery, if you are exerting horizontal force on both walls greater than downward gravitational force, you will be still hanging. So horizontal force > vertical force. -Abhi. === Subject: Re: Action Device: Destination Alpha Centauri A Message-Id: <1075147547.26740.0@ersa.uk.clara.net> Uncle Al, I request you to unleash your wisdom on me by analysing this > design.. > http://www.geocities.com/inertial_propulsion I may not be Uncle Al but the error is obvious, you > are still not showing how you keep the angle ABC > fixed. This requires a force and the reaction to > that exactly cancels the desired unidirectional > force. This shows the rod and springs version: http://www.dishman.me.uk/George/Abhi/abhi_rod.gif Welcome back, George. Happy news is that I have this angled rod and > two springs. I am stretching these two springs and I find that angle > ABC still remains fixed! Excellent, try it and see is the best way to make > progress, well done on taking the first step. Now tell me this, as you pull down on the springs, > do you also have to hold the angled rod to stop it > moving? If not, which way are you pulling the springs, > along the arms of the rod or are you pulling A in a > direction directly away from C and vice versa? > Well, when the V shped rod with inser springs is lying on floor, I > just can't stretch the springs. So I had to hold this V structure > vertical with point A and C on floor. Then I pulled these springs. Now > the springs are stretched Yes, that works because the rod cannot move downwards since A and C are on the floor. In fact A and C are pressing as hard on the floor as you are pulling. It this this downward force on the floor that you have missed in your analysis and which cancels out your unidirectional force. > and I tried to keep this structure > horizontally lying on floor. Now my fingers are along X axis. I am > finding that my both hands are feeling force towards origin O of X > axis and point B and whole V rod is trying slide along Y axis. That's it, the rod tries to slide down the Y axis because of the force I mention above. To stop that you need an equal and opposite upwards force, in this case from the floor. > But this is the key point. I have not fixed arms between point A and > C. When these arms are fixed, like my fingers, point A and C will try > to move towards center of X axis and V rod will slide along Y axis. > But if V rod slide along Y axis, distance between point A and C along > X axis will decrease and according to pythagoras theorem, point D will > move in upward direction. That is if point B slides along Y axis > through length l, point D will move in upward direction through length > L where L > l. Correct, but the upward force on D is equal and opposite to the downward force on the floor from the ends of the rod at A and C. You do not have a unidirectional force, you only have equal and opposite forces. > Obviously it requires force to keep angle ABC fixed. But as I am using > iron rod, intermoleculer forces at point B are so strong that due to > spring forces acting at point B, angle ABC does not change. So far, so good. And also > there are other ways to keep it fixed or make it in such a way that, > if angle ABC is 60 degree, it changes from 60 degree to 0 degree. For > this purpose, we can use another spring having restoring force at > point B or attached in between middle points of spring AB and CB. As long as you have a working method, go with it. > It is easy to be distrac by alternatives. BTW, I have noticed that you are still clinging to my spring design > and not commenting on this magnet design.. http://www.geocities.com/inertial_propulsion The spring pulls the ends together and so do the > magnets so from the point of view of the mechanics > there is no difference. Since I already did the > drawing for the springs, I just reminded you of it. > It applies equally well for the magnets. > It is rather confusing to explain with spring drawing. But this magnet > drawing is pure geometry. Magnets must move towards each other along > straight line. No, the magnets can move in any direction and the force just tries to reduce the distance between them but you can keep them on the line with a rod in the same way you stopped the springs bending. That's why the drawings are the equivalent. By the way, changing the ABC rod to a horizontal rod makes no difference, the forces are the same. > Another happy news is that, take a elastic wooden stick of mass 'm' > gram. Attach stone or any mass of '2m' gram. Bend this stick and > release both ends. As the stick tries to straighten up, center point > of stick and hence stone gains momentum. Due to this momentum of > stone, stone and stick are propelled in only one direction. I did it > with bicycle spoke. Why don't you repeat it and place it on record > here what exactly you observed? There is an equal and opposite reverse force on your > finger while you bend the stick. When you release the > stone, it goes forward while your finger goes back. Of > course you have a mass much more than the stone and > stick so the motion is not noticeable but you move a > tiny amount all the same. If you are standing firmly > on the ground it is the whole mass of the Earth that > moves by an incredibly small amount, but the force is > still balanced. > Is it so? I will figure out method in which my fingers are not > involved. What you use doesn't matter, you must have something to stop the stick moving before you release the stone. Whatever you use, the force the stick applies to it is the same and that something will move when you release the stone. > Think about this. Imagine you are standing in between two walls just 2 > feet away from each other. With your back on one wall no.1, lift your > right leg and push wall no.2. Now also lift your left leg and push > again wall no.2. You are hanging in between two walls. What is > supporting your weight? Friction stops you sliding down the wall but your mass > is added to that of the wall so a worm under the wall > would feel just a little more force pressing down on > him. Again, the forces are balanced. > Not quite right. Even if walls are slippery, if you are exerting > horizontal force on both walls greater than downward gravitational > force, you will be still hanging. > So horizontal force > vertical force. Nope, if the walls are very slippery, you can exert as enormous force and still slide but if they are rough a very small horizontal force can stop you. However that doesn't matter, the point to note is this: Gravity pulls you down but also pulls up on the Earth, the forces are equal and opposite. Compression in the wall pushes up on your back and down on the Earth, the forces are equal and opposite. Friction pushes up on your back and shoes while gravity pulls you down, the forces are equal and opposite so you do not move. Gravity pulls up on the Earth while the wall pushes down so the Earth does not move. The Earth pushes up on the wall while you push down so the wall does not move. In other words, every force is balanced so nothing moves. It is easy to nail a picture to the wall and it will not move, but there is no unblalanced force involved. Those paragraphs talk about the vertical forces, you can do the same for the horizontal forces. George === Subject: Re: Action Device: Destination Alpha Centauri A charset=iso-8859-1 > Friction stops you sliding down the wall but your mass > is added to that of the wall so a worm under the wall > would feel just a little more force pressing down on > him. Again, the forces are balanced. > Not quite right. Even if walls are slippery, if you are exerting > horizontal force on both walls greater than downward gravitational > force, you will be still hanging. No. Here's a way to test this: - Place lots of cushions under you - Put on rollerblades (you need good bearings!) - nail three more wheels on a board - hold it against your back, lean on the wall - put your rollerbladed feet against the other wall - observe yourself sliding down. The only way to keep hanging is to exert so much force on the bearings that the friction inside them keeps you hanging. Greetings! Volker === Subject: Re: Action Device: Destination Alpha Centauri A > Friction stops you sliding down the wall but your mass > is added to that of the wall so a worm under the wall > would feel just a little more force pressing down on > him. Again, the forces are balanced. Not quite right. Even if walls are slippery, if you are exerting > horizontal force on both walls greater than downward gravitational > force, you will be still hanging. > No. Here's a way to test this: > - Place lots of cushions under you > - Put on rollerblades (you need good bearings!) > - nail three more wheels on a board > - hold it against your back, lean on the wall > - put your rollerbladed feet against the other wall > - observe yourself sliding down. > The only way to keep hanging is to exert so much force on the > bearings that the friction inside them keeps you hanging. > Greetings! > Volker Yaa, you are right. I talked about this wall experiment to explain what will support the weight of whole device when it is hanging without support. I think, due to spring force >> downward G force, this spring force is conver in upward force at point D and this upward force is greater that downward G force acting on whole device. So this spring force, conver in vertical upward force is countering downward gravitational force. -Abhi. === Subject: Re: Action Device: Destination Alpha Centauri A [snip] > Yaa, you are right. I talked about this wall experiment to explain > what will support the weight of whole device when it is hanging > without support. I think, due to spring force >> downward G force, > this spring force is conver in upward force at point D and this > upward force is greater that downward G force acting on whole device. > So this spring force, conver in vertical upward force is countering > downward gravitational force. Abhi, It looks to me as if you're readying yourself to waste 17 more years on your device. If you build it and it works then move to the next level as to how you'll use it. You haven't built it, you haven't seen it in operation, you don't know what it will do and you don't know how it works. How in the world can you be postulating *anything* about it until you've accomplished that? I have a rock. I am convinced it is made out of antigravity metal and although it isn't displaying those attributes, I am still convinced. I will now sit and wait for the rock to defy gravity. When it does I will be able to stand atop it and fly like a bird. I'll be able to chip pieces of it off and sell them for a fortune. I'll be able to go anywhere I want to. It will be a miraculous rock.... someday... in the meantime I'll sit here and dream of the day when the rock becomes an antigravity rock. If you'd like, I'll tell you all about what I'll do when that day comes. Sheesh! O' > -Abhi. === Subject: Re: Action Device: Destination Alpha Centauri A charset=iso-8859-1 > Friction stops you sliding down the wall but your mass > is added to that of the wall so a worm under the wall > would feel just a little more force pressing down on > him. Again, the forces are balanced. Not quite right. Even if walls are slippery, if you are exerting > horizontal force on both walls greater than downward gravitational > force, you will be still hanging. > No. Here's a way to test this: > - Place lots of cushions under you > - Put on rollerblades (you need good bearings!) > - nail three more wheels on a board > - hold it against your back, lean on the wall > - put your rollerbladed feet against the other wall > - observe yourself sliding down. The only way to keep hanging is to exert so much force on the > bearings that the friction inside them keeps you hanging. Greetings! > Volker > Yaa, you are right. I talked about this wall experiment to explain > what will support the weight of whole device when it is hanging > without support. I think, due to spring force >> downward G force, > this spring force is conver in upward force at point D and this > upward force is greater that downward G force acting on whole device. > So this spring force, conver in vertical upward force is countering > downward gravitational force. It can't. The spring will move until the net force is zero. During the movement the device may jump a bit from the ground but as soon as the springs have stopped moving the force exer by the device towards the environment will be that of the weight of the device and that force will point downwards. Here's another way of looking at the whole force situation: There is no force without resistance. That's obvious because if there is no resistance force will result in movement until the force is zero. (Experiment: put a rock on a table. The rock exerts pressure. Remove the table. Pressure zero, movement goes on.) In the same way the bottom of your device will try to exert pressure on the air but, the air offering no noticeable resistance, will simply drop down until it finds something to exert pressure on. Lots of Greetings! Volker === Subject: Re: Action Device: Destination Alpha Centauri A am I correct, in butting into this conversation, that Abhi has proposed a *mobile perpetuum* ?? as for the force required to stay in the smooth elevator shaft, don't you have to use the static coefficient of friction for the slickest link, the ball-bearings & their races e.g.? I like Al's refs, at http://www.21stcenturysciencetech.com/ : [1] Albert Einstein's Special Theory of Relativity, Arthur Miller, pp. 328-331. Miller explains the transverse mass error. [2] Does mass really depend on velocity, dad? Carl G. Adler, Am. J. Phys., 55(8), Aug 1987 page 742 It should be no that Einstein's original formula for transverse mass was incorrect. It was correc by Planck in 1906. Planck was the first to introduce the formula P = m0v/(1-v2/c2) 1/2. > The only way to keep hanging is to exert so much force on the > bearings that the friction inside them keeps you hanging. as you know, if you've read anyhting that I typed, there was a conspiracy of states that DID vote Gore -- and the Supremes sealed that conspiracy on March 27, 2000, by refusing to hear the appeal in LaRouche v. Fowler. (Don Fowler was the DNC Chair in '96; Sentelle's 3-judge panel made the Voting Rights Act unconsitutitonal. but, hey; it's up for re-auhtorization in '07 !-) I am frankly scared of the touchscreen mentality. just like the Supremes' abrogation of the USA and Florida constitutions foments a pop-culture hatred of the electoral college on the part of some rabid Democrats. ah, so; imagine if North Dakota ... rather, imagine if Wyoming had less than one electoral college vote for president. anyway, every one of us in this debate knows about the Texas cirterium for chad -- it ain't just missing confetti! thus saith: but, at the national level, will you be able to maintain the DNC's Any One But George -- unless it's Lyndon! media glut?... are you on the board of GOPpers For Howie Troisieme? http://www.wlym.com/pdf/iclc/communism.pdf --Give the Gift of Dick Cheeny -- out of office, at last! http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Action Device: Destination Alpha Centauri A charset=iso-8859-1 > am I correct, in butting into this conversation, that > Abhi has proposed a *mobile perpetuum* ?? > as for the force required to stay in the smooth elevator shaft, > don't you have to use the static coefficient of friction > for the slickest link, the ball-bearings & their races e.g.? I like Al's refs, > at http://www.21stcenturysciencetech.com/ : > [1] Albert Einstein's Special Theory of Relativity, Arthur Miller, pp. > 328-331. Miller explains the transverse mass error. > [2] Does mass really depend on velocity, dad? Carl G. Adler, Am. J. > Phys., 55(8), Aug 1987 page 742 It should be no that Einstein's original formula for transverse > mass was incorrect. It was correc by Planck in 1906. Planck was the > first to introduce the formula P = m0v/(1-v2/c2) 1/2. The only way to keep hanging is to exert so much force on the > bearings that the friction inside them keeps you hanging. as you know, if you've read anyhting that I typed, > there was a conspiracy of states that DID vote Gore -- and By trying to insert your political agenda into a perfectly unpolitical discussion you will lose votes. Volker === Subject: Re: Action Device: Destination Alpha Centauri A > Look, until I plan my future life, I can not build this Action Device. I am precognitive -- I predic you would continue to post about it! Please provide a link to the post where you made this prediction. > Amazing! Please forgive my initial skepticism. I've met very few genuine psychics in my life. sports fan, and her specialty was predicting the outcome of Tigers baseball games. Everytime I ran into her at work, she'd say, damn Tigers are going to lose again, and she'd be right 75% of the time! She was also pretty good about predicting the outcome of Army and SMU football games... :-) Minor Crank === Subject: Re: Action Device: Destination Alpha Centauri A > Look, until I plan my future life, I can not build this Action > Device. > I am precognitive -- I predic you would continue to post about > it! Please provide a link to the post where you made this prediction. psychics in my life. > sports fan, and her specialty was predicting the outcome of Tigers baseball > games. Everytime I ran into her at work, she'd say, damn Tigers are going > to lose again, and she'd be right 75% of the time! > She was also pretty good about predicting the outcome of Army and SMU > football games... Minor Crank Such things does not need any damn psychic powers. Use logic, modus operandi, theory of probability, physics etc. and you can do better in predicting things. -Abhi. === Subject: Re: Action Device: Destination Alpha Centauri A Content-transfer-encoding: 8bit > Look, until I plan my future life, I can not build this Action > Device. > I am precognitive -- I predic you would continue to post about > it! Please provide a link to the post where you made this prediction. psychics in my life. > sports fan, and her specialty was predicting the outcome of Tigers baseball > games. Everytime I ran into her at work, she'd say, damn Tigers are going > to lose again, and she'd be right 75% of the time! > She was also pretty good about predicting the outcome of Army and SMU > football games... I'm afraid my prediction was about as psychic as predicting moonrise times. It was merely an extrapolation based upon the previous history of the Abhi phenomenon; he can't stop yapping about his action device. -=-=-=-=- === Subject: Re: Action Device: Destination Alpha Centauri A > Look, until I plan my future life, I can not build this Action > Device. > I am precognitive -- I predic you would continue to post about > it! Please provide a link to the post where you made this prediction. Amazing! Please forgive my initial skepticism. I've met very few genuine > psychics in my life. sports fan, and her specialty was predicting the outcome of Tigers baseball > games. Everytime I ran into her at work, she'd say, damn Tigers are going > to lose again, and she'd be right 75% of the time! She was also pretty good about predicting the outcome of Army and SMU > football games... > I'm afraid my prediction was about as psychic as predicting moonrise > times. It was merely an extrapolation based upon the previous history > of the Abhi phenomenon; he can't stop yapping about his action > device. > -=-=-=-=- And what is your prediction about this action device? Will it work or not? http://www.geocities.com/inertial_propulsion -Abhi. === Subject: Re: rings proof (can someone check them please?) >> 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. >> Proof: It suffices to show S is closed under * and - . >> Clearly S is non-empty since f(x) = x is in S. >> So assume f(x) = x*h(x) and g(x) = x*z(x) for some h(x) >> and z(x) in R[x]. >> 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 multiplication and by the subring test this implies that >> S is a subring of R. > This is a complete mess. First, R[x] is of course ambiguous here in > ASCII. Second, what makes you think that you can write f and g as > polynomials? Not all functions from the reals to the reals that go > through 0 are polynomials! Nowhere in the statement are polynomials > singled out. Your functions need not be polynomials: they need not > even be continuous! R is the ring of ALL functions; the function > f(x) = 1 if x is irrational > 0 if x is rational > is in S. How do you write it as x*h(x) with h(x) a polynomial? > No, you went completely astray here. What you proved was that the set > of polynomial with integer coefficients which have 0 as a root are a > subring of the ring R[x]; you did not prove what you were asked. His proof is fine if you interpret R[x] as he obviously intends it. In fact this is a special case of a general principle. Namely, the kernel of a homomorphism is an ideal. Here the hom is simply evaluation at zero: f -> f0, which is clearly a hom, i.e. (f + g)0 = f0 + g0 (f * g)0 = f0 * g0 -Bill Dubuque === Subject: Re: rings proof (can someone check them please?) Adjunct Assistant Professor at the University of Montana. > 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. > Proof: It suffices to show S is closed under * and - . > Clearly S is non-empty since f(x) = x is in S. > So assume f(x) = x*h(x) and g(x) = x*z(x) for some h(x) > and z(x) in R[x]. > 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 multiplication and by the subring test this implies that > S is a subring of R. >> This is a complete mess. First, R[x] is of course ambiguous here in >> ASCII. Second, what makes you think that you can write f and g as >> polynomials? Not all functions from the reals to the reals that go >> through 0 are polynomials! Nowhere in the statement are polynomials >> singled out. Your functions need not be polynomials: they need not >> even be continuous! R is the ring of ALL functions; the function >> f(x) = 1 if x is irrational >> 0 if x is rational >> is in S. How do you write it as x*h(x) with h(x) a polynomial? >> No, you went completely astray here. What you proved was that the set >> of polynomial with integer coefficients which have 0 as a root are a >> subring of the ring R[x]; you did not prove what you were asked. >His proof is fine if you interpret R[x] as he obviously intends it. I would not be so bold as to say that. It is clear to me that he does not intend R[x] to be all functions; if he does, then claiming that f(0)=0 if and only if f(x) = x*h(x) with h(x) in R[x] seems rather a rather strange thing to say (though, of course, true) when talking about all functions from R to R, while it is a perfectly natural thing to say when talking about all polynomials with real coefficients. >In fact this is a special case of a general principle. Namely, >the kernel of a homomorphism is an ideal. Here the hom is simply >evaluation at zero: f -> f0, which is clearly a hom, i.e. > (f + g)0 = f0 + g0 > (f * g)0 = f0 * g0 I disagree that his proof (even interpreting R[x] to be all functions) is a special case of this general principle. Yes, of course, the statement he is trying to prove is, but you will note he does not do any evaluation. -- === Subject: Re: areas of overlapping circles === >I would like to know a formula for calculating the area of >the overlapping region of two circles that overlap and touch the center of >the other circle (knowing just the radius).... I take it you mean the two circles have the same radius r, and each passes through the centre of the other. If so, the answer seems to be (2(pi)/3 - (sqrt(3))/2)*(r^2) which is about (1.22837)*(r^2). If you're interes in the reason, then consider separately the area of a sector of angle 60 degrees = (pi)(r^2)/6 and the area of an equilateral triangle of side r = (1/2)(r^2)sin(60 degrees). Look at a diagram to see your area as four such sectors minus two such triangles, giving the result above. Ken Pledger. === Subject: recurrence relation help may you help me to solve the recurrence relation below? / | ( nabla^{k+1} f ) (Y_1, ...,Y_{k+1}) = | = Y_{k+1} (nabla^k f (Y_1, ...,Y_{k}) ) | - sum_{i=1}^k nabla^k f (Y_1, .,nabla_{Y_{k+1}}Y_i ,..,Y_k) / | nabla^1 f (X)= df X | | Y_i denotes a tangent vector field over a differentiable manifold M, f is a real valued function over M, f: M--> R, nabla denotes a linear (affine) connession; nabla^{k+1} f := nabla (nabla^k f). Thank you very much , i need an help to succede in solving this recurrence relation, please, help me.. Tern === Subject: Re: What kind of problem does this belong to? error analysis for matrix-vector multiplication? > Dear all, > I am facing with the following problem: > I have a square matrix A(NxN), and a column vector x(Nx1), > By computing y=A*x, I get another column vecotr y(Nx1). > But in implementation, due to some error, some small error was > introduced and actually I compu: > y=(A+E)*x, where E is a small error matrix(NxN), > now I want to analyze which element of y is impac most heavily, and > in what degree, and also want to have an order/ranking of the elements > of y being affec... > I am not sure whether this kind of analysis can be made independent of > x or not... if it should be dependent on x, then what characteristics > of x combined with what kind of E gives most impact to which element > of y? > Thank you very much in advance! > -Walala You may want to read Wilkinson books. He inven backward error analysis. === Subject: Re: What kind of problem does this belong to? error analysis for matrix-vector multiplication? > y = A*x + E*x > Where the RHS is still exact arithmetic. (A*x) is the exact answer > to the matrix-vector product, (E*x) is the error. so you mean the error E*x is dependent on input x, without knowing the input x, there is no way to predict why element of the result vector y will be impac mostly? The reason I ask this is that I want to establish a confidentality value for each element of the result vector y... without knowing input x, because a system commonly have chaning inputs... I hope to predict that under the deterministic error matrix E, which elements of the result vector y are most trustable, which elements of y are most suspicious? to what degree? Are these work done already by somebody? Thanks a lot, -Walala === Subject: Re: What kind of problem does this belong to? error analysis for matrix-vector multiplication? > y = A*x + E*x Where the RHS is still exact arithmetic. (A*x) is the exact answer > to the matrix-vector product, (E*x) is the error. so you mean the error E*x is dependent on input x, without knowing the input > x, there is no way to predict why element of the result vector y will be > impac mostly? > The reason I ask this is that I want to establish a confidentality value for > each element of the result vector y... without knowing input x, because a > system commonly have chaning inputs... > I hope to predict that under the deterministic error matrix E, which > elements of the result vector y are most trustable, which elements of y are > most suspicious? to what degree? > Are these work done already by somebody? > Thanks a lot, > -Walala moreover, I hope to know which kind of input vector x, which kind of combination of elements of x, leads to most error in y, ... can this be done? === Subject: Re: What kind of problem does this belong to? error analysis for matrix-vector multiplication? > moreover, I hope to know which kind of input vector x, which kind of > combination of elements of x, leads to most error in y, ... can this be > done? The error is (E*x), so you can do singular value decomposition on the error matrix E. The vectors corresponding to the largest singular values will correspond to the worst vectors x that lead to the largest contributions to y. $.02 -Ron Shepard === Subject: Re: What kind of problem does this belong to? error analysis for matrix-vector multiplication? > moreover, I hope to know which kind of input vector x, which kind of > combination of elements of x, leads to most error in y, ... can this be > done? > The error is (E*x), so you can do singular value decomposition on > the error matrix E. The vectors corresponding to the largest > singular values will correspond to the worst vectors x that lead > to the largest contributions to y. > $.02 -Ron Shepard Hi Ron, I did [U, S, V]=svd(E) in matlab and did see the singular values in S, but I don't which column vectors in U and V is the vector corresponding to this singular value? The first column vector in U, or first column vector in V? I also did [U, D]=eig(E), but got all eigenvalues/eigenvectors are complex numbered... but I need real number, how to do that? Thanks a lot, -Walala === Subject: Re: What kind of problem does this belong to? error analysis for matrix-vector multiplication? > moreover, I hope to know which kind of input vector x, which kind of > combination of elements of x, leads to most error in y, ... can this be > done? > The error is (E*x), so you can do singular value decomposition on > the error matrix E. The vectors corresponding to the largest > singular values will correspond to the worst vectors x that lead > to the largest contributions to y. Very interesting! I did not know that before... Thank you! And after a second thought, there are more interesting questions here: 1) the worst vectors x that lead to the largest contributions to y, I guess the largest contribution to y is measured in L2 norm of y, right? So the largest contribution means the largest L2 norm of y? Suppose the measure is a weighed L2 norm of y having different weights on each element of y, how to modify the SVD to identify the worst vector x(the most dangerous input)? 2) the worst vectors x that lead to the largest contributions to y, Will it happen such case that the vectors corresponding to the second and third largest eigenvalues of E can combine together and make a combined vector to beat the vector corresponding to the first largest eigenvalue of E, in terms of normal L2 norm and weighed L2 norm? 3) In order to avoid such worst case input vector x, I need to have a measure of similarity between two vectors x and z. When an vector z is very similar to x, it will leads to bad result; What can be a meaningful similarity measure between two vectors? Any thoughts? Please throw some lights on me! Thank you! -Walala === Subject: Re: Proof on Cassini's identity In-reply-to: John R Ramsden I found such a proof [ that F_.F_ - F_n^2 = (-1)^n ] at >> http://austin.onu.edu/~mcaragiu1/saramillercapstone.ppt, slide 21. >> However, my question to the newsgroup is if there is any other, >> perhaps simpler way to prove it. >> An identity that makes this easier is a generalization of the inductive >> definition of the Fibonacci sequence: >> F = F F + F F [1] >> n k+1 n-k k n-k-1 >> This is easily proven by induction. Since F_1 = F_2 = 1, the case k=1 >> is verified by the inductive definition of the Fibonacci sequence: >> F = F + F [2] >> n n-1 n-2 >I didn't have the patience to wait for the powerpoint presentation to >load. But to prove the - ^2 relation above, it seems >marginally easier to use F_n = (a^n - b^n)/(a - b) where a, b are >(1 + SQRT(5))/2 and (1 - SQRT(5))/2 resp: >Expanding G = (a^(n-1) - b^(n-1)).(a^(n+1) - b^(n+1)) - (a^n - b^n)^2 >gives immediately that G = - (a.b)^(n-1).(a - b)^2, and a.b = -1. I didn't want to start up my Virtual PC to read the Powerpoint document either, but I did look at it this morning. Although the exact algebraic formula for the Fibonacci sequence can be used to show any of the identities involving the Fibonacci sequence, I prefer proofs based on the recursive relation that defines the sequence as those proofs usually generalize more easily to other sequences defined in a similar way. Relation [1] and F(-n) = (-1)^{n-1} F(n) can be used to prove a large number of identities involving the Fibonacci sequence fairly easily. Furthermore, from looking at the work in the Powerpoint file, it appears that the types of proofs that the OP is looking for do not include ones using the closed form of the Fibonacci sequence. In any case, both proofs are valid, and both need a bit of prerequisite work, mine to prove [1], and yours to verify the validity of the closed form of the Fibonacci sequence. It basically comes down to personal preference. Rob Johnson take out the trash before replying === Subject: Re: Isoperimetric Zepp > And yes, this checks out: the curvature of the > lemniscate r^2 = a^2 cos(2 theta) is given by kappa = 3|r|/a^2. > As I said: I was very happy ... > And not to forget: F(x,y) = const.) > All in all we are a little wiser than on April 25 last year: > There was a thread star on April 21 then, which states > (1) as P(phi,rho) = rho, using polar coordinates: > I wrongly withdrew my statements about the maximizing > property of the lemniscate. Sorry. But this thread is still > interesting because of the nice shapes emerging from > P = rho^t for t other than t=1 (lemniscate). I just looked at that old thread. I knew I'd talked to you somewhere about this before, but I hadn't realized it had been right here on sci.math. I see there that you'd already noticed the condition F = alpha kappa as defining the maximizing curves, whereas I just re-discovered this, thinking I was bringing something new to the conversation. Sorry about that. I'm curious about these rho^t shapes. Can you write down explicit formulas? Could you post them? I followed the links you included, but didn't see any characterizations of the curves shown as maximizing anything nor any mention of curvature. I also remain quite curious about determining if there is a correspondence: > A) Intg F(x,y) dx dy is maximized when > B) F(x(s),y(s)) = alpha kappa(s), which also > C) minimizes Intg(s=0,1) ??? ds which is a generalization of the Zepp correspondence: > A) Intg x dx dy is maximized when > B) x(s) = alpha kappa(s), which also > C) minimizes Intg(s=0,1) kappa(s)^2 ds. Since the rho^t solutions are nice, particularly for the lemniscate (t=1), I wonder if there are characterizations of them, in particular, as minimizing some Intg(s=0,1) ??? ds. -Jim Ferry === Subject: Re: Isoperimetric Zepp > I see there that you'd already noticed > the condition > F = alpha kappa > as defining the maximizing curves, whereas I just re-discovered > this, thinking I was bringing something new to the conversation. > Sorry about that. Nothing to be sorry about! So we share the same Aha! experience. The fine thing is: r*F = const. is just Euler-Lagrange! > Since the rho^t solutions are nice, particularly for the > lemniscate (t=1), I wonder if there are characterizations > of them, in particular, as minimizing some Intg(s=0,1) ??? ds. I'll send you an Excel file if this is OK for you. It contains a simple pointwise construction of the curves and contains most lovely shapes, you may alter by yourself. As for the characterizations I don't have any idea but would like to hear about. There is still something about the Zepp's relevance for elesticity, which I do not understand: Why is the force proportional to the distance from a straight line? And not proportional to distance from the point, where the ends are connec? This point seems to me the source of the bending force and so I can only imagine a radial field of force. But this would correspond to the lemniscate. And as Euler told us, it's in fact the Euler-Zepp, which is the very curve. Hmmm... but this is physics and thus too compli- ca for me :-( Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Isoperimetric Zepp > Assuming that the force-balance formula remains valid in 3-d, > then the equation for the surface would be that F(x,y,z) is > proportional to the mean curvature. If we assume F(x,y,z) is > axisymmetric about x, then we can formulate this in terms of a > c-section F(x,y). The mean curvature could then be > expressed as > (1/2) ( phi'(s) - cos(phi(s)) / y(s) ), > yielding the equation > F(x(s),y(s)) = alpha ( phi'(s) - cos(phi(s)) / y(s) ), or > F(x(s),y(s)) = alpha ( kappa(s) - x'(s)/y(s) ). Actually the problem does not generalize well to 3-d: a fundamental aspect of the problem is that a given point (the origin) is constrained to lie on the boundary. But this constraint is meaningless in 3-d: one can always construct a skinny tube of negligible surface area to connect any two points, so the surface can, in effect, be arbitrarily transla. For a function F which grows without bound, such as F=x, there is, therefore, no maximizing shape. One could recast the problem without the constraint of a given point being on the boundary, assuming F is properly behaved, but this problem is somehow not as zesty. -Jim Ferry === Subject: Re: Isoperimetric Zepp Jim Ferry > Actually the problem does not generalize well > to 3-d: a fundamental aspect of the problem is > that a given point (the origin) is constrained > to lie on the boundary. But this constraint is > meaningless in 3-d: > one can always construct a skinny tube of > negligible surface area to connect any two > points, so the surface can, in effect, be > arbitrarily transla. For a function F > which grows without bound, such as F=x, > there is, therefore, no maximizing shape. Hello Jim, could you please explain? How should one be able to enclose maximal volume by a surface of neglegible area? Given some area of 'what-you-need-for-it' for a sack, there will be a certain shape for maximizing the gold to be grabbed. Or not? Rainer Rosenthal r.rosenthal@web.de === Subject: question about signal processing Hi everybody! I'm not sure that this is the right place to ask questions about these matters... forgive me if i'm wrong :) Let's have a white gaussian noise in time-discrete domain, w(k): mean = 0, std = 1, DSP = 1, and, for that, its autocorrelation is Rw(m) = d(m) (1 if m=0, 0 if m!=0) Now, i've got the following signal: s(k) = s1*exp(-k/(2*tau))*w(k), if k>=0 s(k) = 0, if k<0 where tau and s1 are consts. I've been thinking about it a lot, but I can't reach the way to evaluate autocorrelation of s(k)... i mean, of course, in a stocastic way: Rs(m) = E[s(k)'s(k+m)] I believe that it's pretty evident that if I modulate w(k) with a function, no matter what function it is, I'll introduce a kind of correlation between time samples; and for that reason we will not have anymore a set of indipendent gaussian variables (if I remember well, incorrelation is equal to indipendence working with gaussian variables). Is it possible to work it out analitically? Sorry for the silly question Tnx and bye! -- capitan harlock Non devo aver paura. La paura uccide la mente. La paura .8f una piccola morte che porta con s.8f l'annullamento totale. Guarder.98 in faccia la mia paura. Lascer.98 che mi calpesti. E quando se ne sar.88 andata, nulla sar.88 pi.9d. Solo io ci sar.98 === Subject: Re: ugly sum > sum_{i=1}^{infty}frac{mu^iln{i!}}{i!} > What do you want to know about it? For large positive > mu, it is approximately log( gamma(mu) ) * exp( mu ) > Now that's something; i would of course prefer some closed > form but it does seem unlikely. I would be interes in > the analysis for your result. For starters, find the largest term in your sum, and compare it to the largest term in the Taylor series sum for exp(mu). > can you post it in latex? Sorry. === Subject: simple probability question Can someone tell me the formula for calculating the probability that a particular 25-letter acrostic would appear? === Subject: Re: simple probability question >Can someone tell me the formula for calculating the probability that a >particular 25-letter acrostic would appear? Possibly, if you're more specific. ... would appear where, and in what circumstances? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 Subject: Re: Rationals are Uncountable === >I have crea such surjections as well. >No such surjection contains every rational. >>What do you mean by surjection onto the rationals, if it is not to be >>onto ALL the rationals? > I should have said there are no such surjections. But that is false; surjections (even bijections) from the set of natural numbers to the set of rational numbers exist. Attempts to prove that there aren't fails simply for the reason that there are. I am not sure if bijections N -> Q have been presen earlier in this discussion but surjections from a subset of N to Q certainly have. If someone constructs a bijection N -> Q for you will you finally believe that Q is countable? === Subject: Re: Rationals are Uncountable >I have crea such surjections as well. >No such surjection contains every rational. >>What do you mean by surjection onto the rationals, if it is not to be >>onto ALL the rationals? I should have said there are no such surjections. But that is false; surjections (even bijections) from the set of natural > numbers to the set of rational numbers exist. Attempts to prove that > there aren't fails simply for the reason that there are. > I am not sure if bijections N -> Q have been presen earlier in this > discussion but surjections from a subset of N to Q certainly have. If > someone constructs a bijection N -> Q for you will you finally believe > that Q is countable? If someone constructs a surjection N -> R will you finally believe that R is countable? Probably not. You will claim that Cantor proved no such surjection can exist. We are discussing one of Cantor's proofs. If this proof shows that the rationals are uncountable will you still believe a surjection N -> Q exists? Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> Discussion, linux) > I know Virgil hates when I osophize. > Of course, this is one of the reasons I do it. > The following is not a proof. > It is my take on what is happening. > Let S be an unlimi set of symbols > and N be an unlimi set of natural numbers. You are besmirching the term osophize here. Let N be an unlimi set of natural numbers might be a good thing to do if (1) we were sure that we knew what unlimi meant and (2) there actually *is* such an unlimi set. This isn't osophizing. This is gibberish. -- My proofs are out there. -- James S. Harris === Subject: Re: Rationals are Uncountable > Let N be an unlimi set of natural numbers might be a good thing > to do if (1) we were sure that we knew what unlimi meant and unlimi means bigger than infinite. > (2) there actually *is* such an unlimi set. Sets really exist? I'm not interes in mathematics that might have anything to do with reality. Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable > Let N be an unlimi set of natural numbers might be a good thing > to do if (1) we were sure that we knew what unlimi meant and unlimi means bigger than infinite. Then what does infinite mean? > (2) there actually *is* such an unlimi set. > Sets really exist? > I'm not interes in mathematics that might > have anything to do with reality. We have noticed this on occcasion. Your whole universe seems to be a bit too far from reality for comfort. === Subject: Re: Rationals are Uncountable <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> <87ektn2l24.fsf@phiwumbda.org> Discussion, linux) >> Let N be an unlimi set of natural numbers might be a good thing >> to do if (1) we were sure that we knew what unlimi meant and unlimi means bigger than infinite. What does that mean? Are the natural numbers *assumed* to be bigger than infinite? Why? >> (2) there actually *is* such an unlimi set. > Sets really exist? > I'm not interes in mathematics that might > have anything to do with reality. Idiot. The is there is the is of the existential operator. If you want to let N be an unlimi set of natural numbers, then you ought to argue why there is such an N. Otherwise, why not let N be a set of natural numbers which are both blue and not blue? If you assume that you have a set satisfying some unsatisfiable property, then the resulting proof will show either nothing or that your assumption is false. -- If you have a really big idea, you can get a measure of how big it is REALLY, REALLY, *REALLY*, BIG DISCOVERY!!! --, on being ignored === Subject: Re: Rationals are Uncountable <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> <87k73f3bxc.fsf@phiwumbda.org> Discussion, linux) >> Assume no such rational c exists. If c were in the range of x_n we would >> have c=x_i for some index i. When that index was reached in the process >> of defining a_n and b_n, c would would be added to A or B, contrary to >> our assumption that the rationals are dense. >> This is nonsense. >> First, you assume that the c here is not rational, and then you assume >> that it's in the sequence (x_n) of rationals. > I said: > Assume no such rational c exists. > I never said c is irrational. Right after that, you supposed that c = x_i. Either you were assuming that c is irrational but equal to some rational x_i or that a non-existent c was equal to x_i. Take your pick. Either interpretation shows the argument is pure hogwash, making no sense whatever. >> Then you claim that its >> membership in A or B violates density. > If c must be in set A or B there exists an interval > (a_n, b_n) that contains no rational number. > Are you claiming such an interval exists? Of course not. You are failing, once again, to prove your claim. But who cares, since the assumptions that lead to the claim are just gibberish? If you have an argument, a real one, not one produced by randomly selecting words from Cantor's proof, perhaps you ought to take the time to write it coherently. -- If you *still* believe that [my proof is wrong], then I have to think that your mind is limi [...], and it may be the case that not everyone *can* achieve that, as the mental wiring may not be there for the task. -- , on faculties needed to accept his proof. === Subject: Re: Rationals are Uncountable Assume no such rational c exists. If c were in the range of x_n we would >> have c=x_i for some index i. When that index was reached in the process >> of defining a_n and b_n, c would would be added to A or B, contrary to >> our assumption that the rationals are dense. This is nonsense. First, you assume that the c here is not rational, and then you assume >> that it's in the sequence (x_n) of rationals. I said: > Assume no such rational c exists. > I never said c is irrational. > Right after that, you supposed that c = x_i. Either you were assuming > that c is irrational but equal to some rational x_i or that a > non-existent c was equal to x_i. Cantor first proof of the uncountability of the reals relies on partioning a sequence of numbers, x_n, into two sequences, a_n and b_n. Cantor claims there is always a gap between the sequences a_n and b_n. The version of the proof at wikipedia makes four assumptions: There exists a set R with the following properties: 1) R can be linearly ordered 2) R is dense 3) R has no end points 4) R has no gaps After much discussion, we agreed that gaplessness means R has the LUB and GLB properties. I am claiming that assumption (4) is not necessary. Another important assumption that is given in the proof: 5) x_n contains every member of R Cantor's describes a method to construct the sequences a_n and b_n from the sequence x_n. a_n and b_n are monotone converging sequences. There is only one place in wikipedia's proof where assumption (4) is (mis)used. The two monotone sequences a and b move toward each other. By the gaplessness of R, some point c must lie between them. The natural numbers have the the LUB property. There is no natural number in the interval (1,2). The gaplessness of R does not guarantee some point c must lie between them. We can assume that c exists because R is dense. A number of people in this thread have argued that it is not enough for R to be dense. I am curious why these people think it is possible to have no rational numbers between two non-equal rational numbers. It is obvious that the LUB property does not guarantee c exists. If the density of R is not enough to say c exists then Cantor's proof fails. Let's assume c exists. Cantor argues that c can not be in the range of x_n. Let's assume c doesn't exist. Cantor's construction will put every member of x_n into either a_n and b_n. There is no point that lies between a_n and b_n. There must exist an interval (a_i, b_i) that does not contain any rational numbers. This contradicts our assuption that R is dense. The rationals can be either countable or dense, but they can not be both. Russell - Learning is easy. Unlearning is harder. === Subject: Re: Rationals are Uncountable > Assume no such rational c exists. If c were in the range of x_n we > would >> have c=x_i for some index i. When that index was reached in the > process >> of defining a_n and b_n, c would would be added to A or B, contrary > to >> our assumption that the rationals are dense. This is nonsense. First, you assume that the c here is not rational, and then you assume >> that it's in the sequence (x_n) of rationals. I said: > Assume no such rational c exists. > I never said c is irrational. Right after that, you supposed that c = x_i. Either you were assuming > that c is irrational but equal to some rational x_i or that a > non-existent c was equal to x_i. > Cantor first proof of the uncountability of the reals relies on partioning a > sequence of numbers, x_n, into two sequences, a_n and b_n. > Cantor claims there is always a gap between the sequences a_n and b_n. > The version of the proof at wikipedia makes four assumptions: > There exists a set R with the following properties: > 1) R can be linearly ordered > 2) R is dense > 3) R has no end points > 4) R has no gaps > After much discussion, we agreed that gaplessness means R > has the LUB and GLB properties. > I am claiming that assumption (4) is not necessary. We know that, but if there are gaps, the theorem does NOT claim uncountability, because there is no contradiction produced by assuming that a bijections N -> R exists. > Another important assumption that is given in the proof: > 5) x_n contains every member of R > Cantor's describes a method to construct the sequences a_n and b_n > from the sequence x_n. a_n and b_n are monotone converging sequences. Since convergence in X, not R, has not yet been defined (X does NOT have to be the reals), what the theorem actually says is that they are strictly monotone bounded sequences, with each member of one sequence being a bound for the other. Then the gaplessness assumption, which you want to eliminate, is what guarantees that they have limits IN X. If they were to have a common limit in some superset of X but not in X itself (as is the case when X = Q), then the theorem draws no conclusions as to the countability of X. > There is only one place in wikipedia's proof where assumption (4) is > (mis)used. The two monotone sequences a and b move toward each other. > By the gaplessness of R, some point c must lie between them. > The natural numbers have the the LUB property. But the naturals are not dense, so their having the LUB property, or not, is irrelevant. The theorem only applies when ALL of its conditions are met. Leave out any condition, as you want to do, and it does not appply. > There is no natural number in the interval (1,2). > The gaplessness of R does not guarantee some point c must lie between them. The LUB property only appplies to NON-empty sets, even in the set of reals. Since, for example, the open interval (1,1) in the reals has no LUB and no GLB. > We can assume that c exists because R is dense. You seem to be able to assume a lot of conditions contrary to fact, just like the Reverend Dodson's Red Queen, but that does not make them true. For example, the set Q of rationals fits all the conditions of the Cantor theorem except gaplessness in which the c you claim must exist does not exist. A strictly increasing sequence, a_n, and a strictly decreasing sequence, b_n in Q can converge (in R) to an element in R but not in Q, providing a counterexample to your thesis. In fact, every irrational is a gap in Q which provides a counterexample to your thesis, so there are more counterexamples that there are members of Q. > A number of people in this thread have argued > that it is not enough for R to be dense. > I am curious why these people think it is possible > to have no rational numbers between two non-equal > rational numbers. And we are curious as to why you think anyone has claimed that. The claim has been made, and is true, the an infinite nes sequence of open intervals in Q can be empty without any member of that sequence being empty. Examples of such sequences have been produced, and have been studiously ignored by Russell, because he cannot refute them. > It is obvious that the LUB property does not > guarantee c exists. The LUB property in a totally ordered set X guarantees that for every non-empty subset that is bounded above, there is in X a least upper bound. If a set is required to have the LUB property, it guarantees existence in X of a unique limit value to any non=empty strictly increasing set that is bounded above. Which part of that do you not understand? > If the density of R is > not enough to say c exists then Cantor's proof fails. What part of the definition of density do you, Russell, claim allows of degrees of density, allowing one set to be more dense that another? A set either is dense or is not dense, there are no degrees of density. Definition: a totally ordered set A is DENSE if and only if for every x in A and z in A with x < z, there is at least one y in A with x < y and y < z. More succinctly, between any two members of A there is at least one other member of A. That is what density says, and nothing else. > Let's assume c exists. Cantor argues that c can not > be in the range of x_n. > Let's assume c doesn't exist. Cantor's construction > will put every member of x_n into either a_n and b_n. You cannot have read the proof if you claim this. Most of the x_m's are omit in the process of generating the a_n and b_n. > There is no point that lies between a_n and b_n. Then either a_n = b_n or the original set X is not dense. > There must exist an interval (a_i, b_i) that > does not contain any rational numbers. Why? You keep claiming this, but you provide no valid argument to support it. Your pseudo-arguments are riddled with falsehhood and idiocies to the point of nausea. > This contradicts our assuption that R is dense. > The rationals can be either countable or dense, > but they can not be both. In everyone else's universe, except yours and maybe one other's, the rational are both countable and dense. They are countable because, among other reasons, injections from N to Q and from Q to N are easy to construct and thus there are bijections between N and Q. See http://mathworld.wolfram.com/Schroeder-BernsteinTheorem.html > Russell > - Learning is easy. Unlearning is harder. And you have so much of it to do. === Subject: Re: Rationals are Uncountable <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> <87k73f3bxc.fsf@phiwumbda.org> <87hdyj2m19.fsf@phiwumbda.org> Discussion, linux) > Cantor first proof of the uncountability of the reals relies on > partioning a sequence of numbers, x_n, into two sequences, a_n and > b_n. Cantor claims there is always a gap between the sequences > a_n and b_n. > The version of the proof at wikipedia makes four assumptions: > There exists a set R with the following properties: > 1) R can be linearly ordered > 2) R is dense > 3) R has no end points > 4) R has no gaps > After much discussion, we agreed that gaplessness means R > has the LUB and GLB properties. > I am claiming that assumption (4) is not necessary. > Another important assumption that is given in the proof: > 5) x_n contains every member of R > Cantor's describes a method to construct the sequences a_n and b_n > from the sequence x_n. a_n and b_n are monotone converging sequences. > There is only one place in wikipedia's proof where assumption (4) is > (mis)used. The two monotone sequences a and b move toward each other. > By the gaplessness of R, some point c must lie between them. > The natural numbers have the the LUB property. > There is no natural number in the interval (1,2). > The gaplessness of R does not guarantee some point c must lie between them. > We can assume that c exists because R is dense. Bull. Density by itself is insufficient. This is just errant nonsense when you claim this. Here's the facts: A set satisfying (1) -- (4) allows the proof that there's such a c. You claim that a set satisfying just (4) does not allow the proof. You conclude that (1) -- (3) is sufficient. This is simply false. Let's go through the proof carefully and see how the gaplessness of R really plays a role. They kinda skip the explicit details on the web page, so we'll put them here. Define a_n and b_n as they do. Let A = {x | exists n x < a_n} and let B = R X. Then A and B form a partition such that each x in A is below each x in B. Therefore, *by gaplessness* there exists a c such that for all x, (*) if x < c then x in A and if x > c then x in B. Let us now see why c is between the sequences (a_n) and (b_n), i.e., that for all n, a_n < c < b_n. Note: This does not mean that c is neither in A nor B! On the contrary, c *is* in B, by the definition of B. First, we show that a_n < c for all n. Suppose, on the contrary, that c <= a_n for some n. Then c < a_{n+1} (since (a_n) is strictly increasing), and hence by (*), a_{n+1} is in B. But, because a_{n+1} < a_{n+2}, we know that a_{n+1} is in A, and so a_{n+1} is not in R A. Therefore, our assumption that c <= a_n for some n is false, and so for all n, a_n < c. Now, let us show that for all n, c < b_n. Again, assume that there is an n such that b_n <= c. Then b_{n+1} < b_n <= c and so, by (*), b_{n+1} is in A. Hence, there is some m such that b_{n+1} < a_m. We claim that there is some k that b_k < a_k. This would yield a contradiction, since by construction, we have for all n, a_n < b_n. Suppose, first, that m < n+1. Then, we have b_{n+1} < a_m < a_{m+1} < ... < a_{n+1}. The case in which n+1 < m is similar. We have comple the proof that, for all n, a_n < c < b_n. We see, then, that gaplessness is essential in proving that there exists a c as alleged. The fact that c is between the two sentences requires a touch more work, but is not difficult. We may *not* assume that density is sufficient. We must *show* that it is sufficient -- if it is. Evidently, it is not. > A number of people in this thread have argued > that it is not enough for R to be dense. > I am curious why these people think it is possible > to have no rational numbers between two non-equal > rational numbers. No one said it was possible to have no rational numbers between two distinct rationals. No one said anything which implied such a claim (you and aside). > It is obvious that the LUB property does not guarantee c exists. Bullhonkies. The LUB property *is* sufficient. However, the proof wouldn't work in N because one cannot construct sequences (a_n) and (b_n) there -- because N isn't dense. We need R to be dense in order to construct (a_n) and (b_n) as in the proof. We need R to be gapless in order to prove that there is a c between the two sequences. > If the density of R is not enough to say c exists then Cantor's > proof fails. > Let's assume c exists. Cantor argues that c can not > be in the range of x_n. > Let's assume c doesn't exist. Cantor's construction > will put every member of x_n into either a_n and b_n. No it won't. It would mean that for each n, there is an m such that either x_n <= a_m or x_n >= b_m. > There is no point that lies between a_n and b_n. > There must exist an interval (a_i, b_i) that > does not contain any rational numbers. > This contradicts our assuption that R is dense. > The rationals can be either countable or dense, > but they can not be both. Wrong. -- These mathematicians are worse than communists, as how do you explain their behavior? I *am* the American Dream, fighting for what should be mine, having to get past weak-minded academics who are fighting to block my success. But I shall prevail!!! -- James S. Harris === Subject: Re: Rationals are Uncountable > We can assume that c exists because R is dense. > Bull. > Density by itself is insufficient. > This is just errant nonsense when you claim this. Here's the facts: A > set satisfying (1) -- (4) allows the proof that there's such a c. You > claim that a set satisfying just (4) does not allow the proof. You > conclude that (1) -- (3) is sufficient. This is simply false. Why do I need to assume R has the LUB property? > Let's go through the proof carefully and see how the gaplessness of R > really plays a role. They kinda skip the explicit details on the web > page, so we'll put them here. We have defined gaplessness to mean LUB. You need to show that the LUB property plays a role. > Define a_n and b_n as they do. Let A = {x | exists n x < a_n} and let > B = R X. Then A and B form a partition such that each x in A is > below each x in B. Therefore, *by gaplessness* there exists a c such > that for all x, > (*) if x < c then x in A and if x > c then x in B. > Let us now see why c is between the sequences (a_n) and (b_n), i.e., > that for all n, a_n < c < b_n. Note: This does not mean that c is > neither in A nor B! Every member of a_n and b_n is also a member of x_n. Cantor claims c is not equal to any x_n. If Cantor's claim is true, c can not be a member of a_n or b_n. > On the contrary, c *is* in B, by the definition of > B. Then Cantor's proof fails. > First, we show that a_n < c for all n. Suppose, on the contrary, that > c <= a_n for some n. Then c < a_{n+1} (since (a_n) is strictly > increasing), and hence by (*), a_{n+1} is in B. But, because a_{n+1} > < a_{n+2}, we know that a_{n+1} is in A, and so a_{n+1} is not in R > A. Therefore, our assumption that c <= a_n for some n is false, and > so for all n, a_n < c. > Now, let us show that for all n, c < b_n. Again, assume that there is > an n such that b_n <= c. Then b_{n+1} < b_n <= c and so, by (*), > b_{n+1} is in A. Hence, there is some m such that b_{n+1} < a_m. We > claim that there is some k that b_k < a_k. This would yield a > contradiction, since by construction, we have for all n, a_n < b_n. > Suppose, first, that m < n+1. Then, we have > b_{n+1} < a_m < a_{m+1} < ... < a_{n+1}. > The case in which n+1 < m is similar. > We have comple the proof that, for all n, a_n < c < b_n. Where did you use the LUB property? > We see, then, that gaplessness is essential in proving that there > exists a c as alleged. The fact that c is between the two sentences > requires a touch more work, but is not difficult. > We may *not* assume that density is sufficient. We must *show* that > it is sufficient -- if it is. Evidently, it is not. Why isn't density enough? Where did your proof use the LUB property? > A number of people in this thread have argued > that it is not enough for R to be dense. > I am curious why these people think it is possible > to have no rational numbers between two non-equal > rational numbers. > No one said it was possible to have no rational numbers between two > distinct rationals. No one said anything which implied such a claim > (you and aside). You are saying c doesn't exist when R is the rationals. This can only be true if some (a_i, b_i) is an empty set. > It is obvious that the LUB property does not guarantee c exists. > Bullhonkies. The LUB property *is* sufficient. > However, the proof wouldn't work in N because one cannot construct > sequences (a_n) and (b_n) there -- because N isn't dense. So LUB isn't sufficient. In fact, LUB has nothing to do with proving c exists. > We need R to be dense in order to construct (a_n) and (b_n) as in the > proof. We need R to be gapless in order to prove that there is a c > between the two sequences. > If the density of R is not enough to say c exists then Cantor's > proof fails. Let's assume c exists. Cantor argues that c can not > be in the range of x_n. Let's assume c doesn't exist. Cantor's construction > will put every member of x_n into either a_n and b_n. > No it won't. It would mean that for each n, there is an m such that > either x_n <= a_m or x_n >= b_m. c must exist proving the rationals are uncountable. > There is no point that lies between a_n and b_n. > There must exist an interval (a_i, b_i) that > does not contain any rational numbers. > This contradicts our assuption that R is dense. The rationals can be either countable or dense, > but they can not be both. > Wrong. Ugly maybe, but not wrong. Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> <87k73f3bxc.fsf@phiwumbda.org> <87hdyj2m19.fsf@phiwumbda.org> <87bror2ev7.fsf@phiwumbda.org> Discussion, linux) >> We can assume that c exists because R is dense. >> Bull. >> Density by itself is insufficient. >> This is just errant nonsense when you claim this. Here's the facts: A >> set satisfying (1) -- (4) allows the proof that there's such a c. You >> claim that a set satisfying just (4) does not allow the proof. You >> conclude that (1) -- (3) is sufficient. This is simply false. > Why do I need to assume R has the LUB property? See below. I *answered* that question. >> Let's go through the proof carefully and see how the gaplessness of R >> really plays a role. They kinda skip the explicit details on the web >> page, so we'll put them here. > We have defined gaplessness to mean LUB. > You need to show that the LUB property plays a role. You can define it to be the LUB property, or you can use the explicit definition given on the Wikipedia page. The two are equivalent. I used the Wikipedia definition. (The proof of equivalence available on request.) >> Define a_n and b_n as they do. Let A = {x | exists n x < a_n} and let >> B = R X. Then A and B form a partition such that each x in A is >> below each x in B. Therefore, *by gaplessness* there exists a c such >> that for all x, >> (*) if x < c then x in A and if x > c then x in B. >> Let us now see why c is between the sequences (a_n) and (b_n), i.e., >> that for all n, a_n < c < b_n. Note: This does not mean that c is >> neither in A nor B! > Every member of a_n and b_n is also a member of x_n. Yes. > Cantor claims c is not equal to any x_n. Yes. > If Cantor's claim is true, c can not be a member of a_n or b_n. Read what I write. I never *said* c is a member of {a_n | n in N} or {b_n | n in N}. See all those words up there? They mean something. Look at the definition of A. Look at the definition of B. Think real, real hard now. See if you can't puzzle out why saying that c is either in A or B does not mean that c = a_n or c = b_n for some n. It's a tricky one, there. Oh, wait. No it's not. I just checked and my three-year-old got it. >> On the contrary, c *is* in B, by the definition of >> B. > Then Cantor's proof fails. No it doesn't. You're incapable of reading. The words I use when I write proofs mean something, Russell. They actually matter. You have to read the words to see what I mean. You cannot simply guess that c in A means c = a_n for some n. It doesn't. [...] >> We have comple the proof that, for all n, a_n < c < b_n. > Where did you use the LUB property? Here, I'll try to make it more apparent for you. ,---- | Define a_n and b_n as they do. Let A = {x | exists n x < a_n} and let | B = R X. Then A and B form a partition such that each x in A is | below each x in B. Therefore, | | --->> ********BY GAPLESSNESS******** <<<<<<-- | | there exists a c such that for all x, | | (*) if x < c then x in A and if x > c then x in B. `---- I didn't quote the LUB property, because it's not the definition of without gaps on the page (although its equivalent). If I wan to use it explicitly, then I would write the following. ,---- | Define a_n and b_n as they do. Let A = {x | exists n x < a_n}. Then | A is bounded above. Therefore, | | --->> ********BY LUB******** <<<<<<-- | | there exists a c such that for all x, | | (*) if x < c then x in A and if x > c then x in R A. `---- Since, previously, I defined B = R A, the proof continues as before. >> We see, then, that gaplessness is essential in proving that there >> exists a c as alleged. The fact that c is between the two sentences >> requires a touch more work, but is not difficult. >> We may *not* assume that density is sufficient. We must *show* that >> it is sufficient -- if it is. Evidently, it is not. > Why isn't density enough? Because I have no proof that uses just density. > Where did your proof use the LUB property? Right up there, in the big letters. [...] >> No one said it was possible to have no rational numbers between two >> distinct rationals. No one said anything which implied such a claim >> (you and aside). > You are saying c doesn't exist when R is the rationals. Yes. > This can only be true if some (a_i, b_i) is an empty set. No. Try proving that claim. Hint: Your proof will use the simply false statement that the intersection of nes open intervals (a_n,b_n) is equal to (a_i,b_i) for some i. This means your proof is nonsense. >> It is obvious that the LUB property does not guarantee c exists. >> Bullhonkies. The LUB property *is* sufficient. >> However, the proof wouldn't work in N because one cannot construct >> sequences (a_n) and (b_n) there -- because N isn't dense. > So LUB isn't sufficient. Not by itself. That's why gaplessness was the fourth element in a list, you dolt. With items (1) -- (3), the LUB is sufficient. In fact, density is used only to define the sequences (a_n) and (b_n), so that the question of whether such a c exists will make sense. Without density, there are no sequences and hence no question about whether there is a c in between the two (non-existent) sequences. > In fact, LUB has nothing to do with proving c exists. Wrong. A million times wrong. If you can prove that (1) -- (3) suffice to show the existence of c, then bloody well do so. Let's see which of your tricks you use. Will it be (1) Assuming that the intersection of open intervals {A_n | n in N} is A_i for some i? (2) Assuming that there is a greatest natural number? (3) Assuming that, on the contrary, there is a set of natural numbers bigger than infinity? (4) Assuming that there is a least rational? (5) Using an infinite natural number? (not sure if that's in your bag of tricks, but it's common enough) ....or, the all-time favorite..... (6) SWITCHING THE ORDER OF QUANTIFIERS WILLY-NILLY[1]? [...] >> Let's assume c doesn't exist. Cantor's construction >> will put every member of x_n into either a_n and b_n. >> No it won't. It would mean that for each n, there is an m such that >> either x_n <= a_m or x_n >= b_m. > c must exist proving the rationals are uncountable. No. Simply proof by repetition. [...] >> The rationals can be either countable or dense, >> but they can not be both. >> Wrong. > Ugly maybe, but not wrong. Ugly, wrong and stupid. Footnotes: [1] I apologize for gratuitous overuse of capital letters. -- Truth is common stuff, ready to your hand, but lies you have to make yourself, and you can't be sure they are any good until you've used them --- and then it's too late. John Steinbeck === Subject: Re: Rationals are Uncountable > We can assume that c exists because R is dense. Bull. Density by itself is insufficient. This is just errant nonsense when you claim this. Here's the facts: A >> set satisfying (1) -- (4) allows the proof that there's such a c. You >> claim that a set satisfying just (4) does not allow the proof. You >> conclude that (1) -- (3) is sufficient. This is simply false. Why do I need to assume R has the LUB property? > See below. I *answered* that question. >> Let's go through the proof carefully and see how the gaplessness of R >> really plays a role. They kinda skip the explicit details on the web >> page, so we'll put them here. We have defined gaplessness to mean LUB. > You need to show that the LUB property plays a role. > You can define it to be the LUB property, or you can use the explicit > definition given on the Wikipedia page. The two are equivalent. I > used the Wikipedia definition. (The proof of equivalence available on > request.) >> Define a_n and b_n as they do. Let A = {x | exists n x < a_n} and let >> B = R X. Then A and B form a partition such that each x in A is >> below each x in B. Therefore, *by gaplessness* there exists a c such >> that for all x, (*) if x < c then x in A and if x > c then x in B. Let us now see why c is between the sequences (a_n) and (b_n), i.e., >> that for all n, a_n < c < b_n. Note: This does not mean that c is >> neither in A nor B! I am defining A as the set of points in the sequence a_n. You must be defining A differently. I assume Cantor defines A and B the same way I do. Otherwise, he can not prove there is some number, c, that lies between the sequences a_n and b_n. > Every member of a_n and b_n is also a member of x_n. > Yes. > Cantor claims c is not equal to any x_n. > Yes. > If Cantor's claim is true, c can not be a member of a_n or b_n. > Read what I write. I never *said* c is a member of > {a_n | n in N} or {b_n | n in N}. > See all those words up there? They mean something. > Look at the definition of A. Look at the definition of B. Think > real, real hard now. See if you can't puzzle out why saying that c is > either in A or B does not mean that c = a_n or c = b_n for some n. > It's a tricky one, there. If A is defined as the sequence of point in a_n then saying c is in A is exactly the same thing as saying c=a_i for some i. > Oh, wait. No it's not. I just checked and my three-year-old got it. I'm sure she shook her head to make you happy. >> On the contrary, c *is* in B, by the definition of >> B. Then Cantor's proof fails. > No it doesn't. You're incapable of reading. We should agree on the definition of A and B. Would you say that every member of x_n is a member of a_n or b_n? Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable > Would you say that every member of x_n is a member of a_n or b_n? Not in the Cantor proof. Every a_n and every b_n is taken from {x_m}, but not all of the members of {x_m} are used up in {a_n} and {b_n}. If {a_n} = (a_1,a_2,a_3,...), {b{n} = (b_1,b_2,b_3,...) with a_1 < a_2< ...< b_2 < b_1, there are still values of x_m below a_1 and above b_1 which cannot be in either sequence, as well as values x_m between consecutive members of each sequence, because of density. === Subject: Re: Rationals are Uncountable >We can assume that c exists because R is dense. Bull. Density by itself is insufficient. This is just errant nonsense when you claim this. Here's the facts: A > set satisfying (1) -- (4) allows the proof that there's such a c. You > claim that a set satisfying just (4) does not allow the proof. You > conclude that (1) -- (3) is sufficient. This is simply false. Why do I need to assume R has the LUB property? >> See below. I *answered* that question. > Let's go through the proof carefully and see how the gaplessness of R > really plays a role. They kinda skip the explicit details on the web > page, so we'll put them here. We have defined gaplessness to mean LUB. >> You need to show that the LUB property plays a role. >> You can define it to be the LUB property, or you can use the explicit >> definition given on the Wikipedia page. The two are equivalent. I >> used the Wikipedia definition. (The proof of equivalence available on >> request.) > Define a_n and b_n as they do. Let A = {x | exists n x < a_n} and let > B = R X. Then A and B form a partition such that each x in A is > below each x in B. Therefore, *by gaplessness* there exists a c such > that for all x, (*) if x < c then x in A and if x > c then x in B. Let us now see why c is between the sequences (a_n) and (b_n), i.e., > that for all n, a_n < c < b_n. Note: This does not mean that c is > neither in A nor B! >I am defining A as the set of points in the sequence a_n. >You must be defining A differently. >I assume Cantor defines A and B the same way I do. >Otherwise, he can not prove there is some number, c, >that lies between the sequences a_n and b_n. >> Every member of a_n and b_n is also a member of x_n. >> Yes. >> Cantor claims c is not equal to any x_n. >> Yes. >> If Cantor's claim is true, c can not be a member of a_n or b_n. >> Read what I write. I never *said* c is a member of >> {a_n | n in N} or {b_n | n in N}. >> See all those words up there? They mean something. >> Look at the definition of A. Look at the definition of B. Think >> real, real hard now. See if you can't puzzle out why saying that c is >> either in A or B does not mean that c = a_n or c = b_n for some n. >> It's a tricky one, there. >If A is defined as the sequence of point in a_n then saying c is in A >is exactly the same thing as saying c=a_i for some i. >> Oh, wait. No it's not. I just checked and my three-year-old got it. >I'm sure she shook her head to make you happy. > On the contrary, c *is* in B, by the definition of > B. Then Cantor's proof fails. >> No it doesn't. You're incapable of reading. >We should agree on the definition of A and B. >Would you say that every member of x_n is a member of a_n or b_n? >Russell >- 2 many 2 count === Subject: Re: Rationals are Uncountable > We can assume that c exists because R is dense. Bull. Density by itself is insufficient. This is just errant nonsense when you claim this. Here's the facts: A >> set satisfying (1) -- (4) allows the proof that there's such a c. You >> claim that a set satisfying just (4) does not allow the proof. You >> conclude that (1) -- (3) is sufficient. This is simply false. Why do I need to assume R has the LUB property? > See below. I *answered* that question. >> Let's go through the proof carefully and see how the gaplessness of R >> really plays a role. They kinda skip the explicit details on the web >> page, so we'll put them here. We have defined gaplessness to mean LUB. > You need to show that the LUB property plays a role. > You can define it to be the LUB property, or you can use the explicit > definition given on the Wikipedia page. The two are equivalent. I > used the Wikipedia definition. (The proof of equivalence available on > request.) >> Define a_n and b_n as they do. Let A = {x | exists n x < a_n} and let >> B = R X. Then A and B form a partition such that each x in A is >> below each x in B. Therefore, *by gaplessness* there exists a c such >> that for all x, (*) if x < c then x in A and if x > c then x in B. Let us now see why c is between the sequences (a_n) and (b_n), i.e., >> that for all n, a_n < c < b_n. Note: This does not mean that c is >> neither in A nor B! Every member of a_n and b_n is also a member of x_n. > Yes. > Cantor claims c is not equal to any x_n. > Yes. > If Cantor's claim is true, c can not be a member of a_n or b_n. > Read what I write. I never *said* c is a member of > {a_n | n in N} or {b_n | n in N}. > See all those words up there? They mean something. > Look at the definition of A. Look at the definition of B. Think > real, real hard now. See if you can't puzzle out why saying that c is > either in A or B does not mean that c = a_n or c = b_n for some n. > It's a tricky one, there. > Oh, wait. No it's not. I just checked and my three-year-old got it. >> On the contrary, c *is* in B, by the definition of >> B. Then Cantor's proof fails. > No it doesn't. You're incapable of reading. > The words I use when I write proofs mean something, Russell. They > actually matter. You have to read the words to see what I mean. You > cannot simply guess that c in A means c = a_n for some n. It doesn't. > [...] >> We have comple the proof that, for all n, a_n < c < b_n. Where did you use the LUB property? > Here, I'll try to make it more apparent for you. > ,---- > | Define a_n and b_n as they do. Let A = {x | exists n x < a_n} and let > | B = R X. Then A and B form a partition such that each x in A is > | below each x in B. Therefore, > | > | --->> ********BY GAPLESSNESS******** <<<<<<-- > | > | there exists a c such that for all x, > | > | (*) if x < c then x in A and if x > c then x in B. > `---- > I didn't quote the LUB property, because it's not the definition of without gaps on the page (although its equivalent). If I wan to > use it explicitly, then I would write the following. > ,---- > | Define a_n and b_n as they do. Let A = {x | exists n x < a_n}. Then > | A is bounded above. Therefore, > | > | --->> ********BY LUB******** <<<<<<-- > | > | there exists a c such that for all x, > | > | (*) if x < c then x in A and if x > c then x in R A. > `---- > Since, previously, I defined B = R A, the proof continues as before. >> We see, then, that gaplessness is essential in proving that there >> exists a c as alleged. The fact that c is between the two sentences >> requires a touch more work, but is not difficult. We may *not* assume that density is sufficient. We must *show* that >> it is sufficient -- if it is. Evidently, it is not. Why isn't density enough? > Because I have no proof that uses just density. > Where did your proof use the LUB property? > Right up there, in the big letters. > [...] >> No one said it was possible to have no rational numbers between two >> distinct rationals. No one said anything which implied such a claim >> (you and aside). You are saying c doesn't exist when R is the rationals. > Yes. > This can only be true if some (a_i, b_i) is an empty set. > No. > Try proving that claim. Hint: Your proof will use the simply false > statement that the intersection of nes open intervals (a_n,b_n) is > equal to (a_i,b_i) for some i. This means your proof is nonsense. >> It is obvious that the LUB property does not guarantee c exists. Bullhonkies. The LUB property *is* sufficient. However, the proof wouldn't work in N because one cannot construct >> sequences (a_n) and (b_n) there -- because N isn't dense. So LUB isn't sufficient. > Not by itself. That's why gaplessness was the fourth element in a > list, you dolt. With items (1) -- (3), the LUB is sufficient. In > fact, density is used only to define the sequences (a_n) and (b_n), so > that the question of whether such a c exists will make sense. Without > density, there are no sequences and hence no question about whether > there is a c in between the two (non-existent) sequences. > In fact, LUB has nothing to do with proving c exists. > Wrong. A million times wrong. If you can prove that (1) -- (3) > suffice to show the existence of c, then bloody well do so. Let's see > which of your tricks you use. Will it be > (1) Assuming that the intersection of open intervals {A_n | n in N} is > A_i for some i? > (2) Assuming that there is a greatest natural number? > (3) Assuming that, on the contrary, there is a set of natural numbers bigger than infinity? > (4) Assuming that there is a least rational? > (5) Using an infinite natural number? (not sure if that's in your > bag of tricks, but it's common enough) > ....or, the all-time favorite..... > (6) SWITCHING THE ORDER OF QUANTIFIERS WILLY-NILLY[1]? > [...] >> Let's assume c doesn't exist. Cantor's construction >> will put every member of x_n into either a_n and b_n. No it won't. It would mean that for each n, there is an m such that >> either x_n <= a_m or x_n >= b_m. c must exist proving the rationals are uncountable. > No. Simply proof by repetition. > [...] >> The rationals can be either countable or dense, >> but they can not be both. Wrong. Ugly maybe, but not wrong. > Ugly, wrong and stupid. > Footnotes: > [1] I apologize for gratuitous overuse of capital letters. > -- > Truth is common stuff, ready to your hand, but lies you have to make > yourself, and you can't be sure they are any good until you've > used them --- and then it's too late. John Steinbeck === Subject: Re: Rationals are Uncountable <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> <87k73f3bxc.fsf@phiwumbda.org> <87hdyj2m19.fsf@phiwumbda.org> <87bror2ev7.fsf@phiwumbda.org> Discussion, linux) > We see, then, that gaplessness is essential in proving that there > exists a c as alleged. The fact that c is between the two sentences ^^^^^^^^^ sequences > requires a touch more work, but is not difficult. I should have poin out that the work required is given above. It's not more work than was found in the post. It's more work than the existence of c by itself required. -- Yup, you guessed it. If worse comes to worse, I *will* turn to the Army to help me with mathematicians. And then mathematicians don't think the NSA or CIA can save your asses, as generals LIKE me. -- 's latest foray into mathematical logic. === Subject: Re: Rationals are Uncountable > I know Virgil hates when I osophize. > Of course, this is one of the reasons I do it. > The following is not a proof. > It is my take on what is happening. > Let S be an unlimi set of symbols > and N be an unlimi set of natural numbers. > I deliberately use the word unlimi > instead of infinite. > I ask you to give me a function, f(), that > maps an unique symbol in S to every member of N. > After much thought you come up with > something like this: > {} = 0 > s{} = 1 > ss{} = 2 > ... > Let's call f() a numbering schema. > You claim f() assigns an unique symbol to > an infinite number of naturals and therefore > f() assigns a symbol to every member of N. > I take your f() and find a natural number that > f() doesn't assign a symbol to. I can do this > because N is unlimi. > Not only can I find an unmatched member of N, > I can find a symbol in S that f() does not use. > S is also unlimi. > So, I can find a natural in N that f() doesn't > assign a symbol to and a symbol in S that > f() doesn't use. > This does not mean that S or N are finite, > ie there is a largest natural number. > It is f() that is finite where finite means > there is a natural number that f() doesn't > assign a symbol to. > It is easy to think that f() orders the symbols. > After all, we never work directly with natural > numbers, we always work with symbols. > But, it is the natural numbers that are ordered, > not the symbols. > Early in this thread I gave an example of a > natural number, x, that was not produced > by a Russell Machine, RM. > Once I knew the value of x it was easy > to find a RM(i) = x where i >> x. > This seems contradictory until we look > at what I did. I found a natural number > that did not have a symbol and I randomly > chose an unused symbol from S. > Using the numbering schema, f(), x must > be much smaller than the i in RM(i). > But, I am not using the numbering schema > and the natural number that I am assigning > x to is larger than any natural in the range of f(). > So, x represents an extremely large natural > number even though the numbering schema, f(), > says x should represent a relatively small > natural number. > Russell > - 2 many 2 count Russell would make the universe over in his own image. You are really a sorry creep, aren't you. Can't accept a world which doesn't conform to your prejudices. === Subject: Re: Rationals are Uncountable The least real quantity is the least real quantity. In a way, I think that is true, re Burali-Forti, in some models, from nothing greater than Ord. There is not iota/2, iota/2 would be iota. One minus iota is around 0.999..., it's also undefined. The reals are not necessarily the powerset of the integers. Is there a way to construct, say, the unit interval of reals from the powerset of the naturals, with assigning explictly each member of the powerset of integers to be a unique real number? I guess there is an obvious example of making the infinite binary sequences that represent each subset of N, then there are issues about dual representation. That's not compatible with the ordinal representation of singletons or forms of composites, pairs with zero, as ordinals and integers, various other forms of ordinal constructions. The idea with that is to figure out constructions of the empty set to uniquely identify not only each number but each mathematical concept, to mechanistally, digitally, operate upon them. If we accept iota as a real, the least real quantity, implicitly positive rather than absence of quantity, and a monotonic mapping from the naturals to any interval of the reals, then that may allow a bridge between the foundations of set theory and integral calculus. There may be other formalizations that would mesh those concepts, but they would perhaps be more contrived than necessary. In considering the continuous reals as a set of points, eg Cauchy sequences, or plain real scalars, we've often discussed definite vis-a-vis indefinite elements of that set, and compatibilities and incompatibilities of their interchangeable use. Now, you may know by now that I say no classes in set theory, only sets in set theory, and various other neologisms. Heh heh heh heh. Is Ord negative one? Why or why not? Is it neither and both? Why or why not? The least real quantity is the least real quantity. === Subject: Re: Rationals are Uncountable > The least real quantity is the least real quantity. The least real quantity must by extremely unreal. > In a way, I think I take leave to doubt it. that is true, re Burali-Forti, in some models, from nothing greater than Ord. > There is not iota/2, iota/2 would be iota. That is exactly the point. Iota is the same as iota/2 which, as you now admit. does not exist. > One minus iota is around 0.999..., it's also undefined. Just as your iota is undefined. > The reals are not necessarily the powerset of the integers. But they have the same cardinality. > Is there > a way to construct, say, the unit interval of reals from the powerset > of the naturals, with assigning explictly each member of the powerset > of integers to be a unique real number? I guess there is an obvious > example of making the infinite binary sequences that represent each > subset of N, then there are issues about dual representation. Since the number of reals with dual binary representations is countable, that issue can be resolved. > That's not compatible with the ordinal representation of singletons > or forms of composites, pairs with zero, as ordinals and integers, > various other forms of ordinal constructions. The idea with that is > to figure out constructions of the empty set to uniquely identify not > only each number but each mathematical concept, to mechanistally, > digitally, operate upon them. Don't step in the oongah. > If we accept iota as a real, the least real quantity, implicitly > positive rather than absence of quantity, and a monotonic mapping from > the naturals to any interval of the reals, then that may allow a > bridge between the foundations of set theory and integral calculus. Or, equally, it may scuttle all of mathematics. > There may be other formalizations that would mesh those concepts, but > they would perhaps be more contrived than necessary. They are already more contrived than actual. > In considering the continuous reals as a set of points, eg Cauchy > sequences, or plain real scalars, we've often discussed definite > vis-a-vis indefinite elements of that set, and compatibilities and > incompatibilities of their interchangeable use. Perhaps you have discussed those things, but then you have a remarkably strong stomach for that sort of nonsense. > Now, you may know by now that I say no classes in set theory, only > sets in set theory, and various other neologisms. Heh heh heh heh. You are beginning to sound like Herbert Lom in those Clouseau movies. Do you look like him, too, twitches and all? === Subject: Re: Rationals are Uncountable > There is not iota/2, iota/2 would be iota. > If we accept iota as a real, And here is your inconsistency. There is only one real equal to one-half of itself: zero. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Rationals are Uncountable > There is not iota/2, iota/2 would be iota. If we accept iota as a real, > And here is your inconsistency. There is only one real equal to > one-half of itself: zero. The least real quantity is the least real quantity. Iota's an infinitesimal. If you try to divide it by two you can't treat it as anything except zero. The result of that operation is undefined. Iota is atomic. The value of iota is that of the reciprocal of an infinite quantity, the infinitesimal quantity that if you add together that infinitely many values of iota the result is equal to one, it is non-zero. In trying to consider the ordered set of reals as a sequence of values, from zero to one, zero is the least element, iota the next, then next to iota is another element, and so on and so forth until the next element would be one, thereby a unit interval of elements in ordered sequence is considered. One goal of trying to determine the properties or characteristics of such an irrational thing as the least positive real quantity is that it fulfills requirements of a conceptual item to fill a gap in our mathematical understanding or utility, towards a more complete whole. That it is difficult or contradictory to integrate with other accep mathematical practices is not always a sign of its uselessness. Also, as much as you can describe an object mathematically it is a mathematical object. Here, one thing that is considered is that despite the density of the reals within the reals, the reals consist of only points and the reals are ordered in that of any two (two implicitly different) reals one is greater and the other less. Now for any postive real that you can name by its numeric value, for example 2 or 9, as the reals are dense there are infinitely many reals between zero and two. The idea is to name a real number by its sequential value and its property that it is less than any other real number that you can name by its numeric value but it is non-negative and non-zero. The reason to consider it an element of the reals is because the reals are continuous. There is no real number, that is no number with no non-zero imaginary or real component scalar multipliers besides for the fundamental unit basis vector e1, that is not a real number. The nonstandard hyperreals are in some ways the same thing as the standard reals but with some different rules that explain their interrelation, for example with the concept of the rigorous infinitesimal. There is a broad context of discussion about iota as a surreal number. My interpretation of it is perhaps not interchangeable, at this point it's mostly an attempt to get a handle on a least, positive real number. I'm reminded of the hypothetical uniform random distribution over the reals. Not by the proven existence of such a thing, but merely by its adequate description, we can use it as a theoretical construct within mathematical enquiry. In some senses, it may be even possible to apply it to real-world dynamics, because it meets its own definition, and of that plain definition without justification besides its requirement. It's like the black box: the enigmatic object that performs a function with no transparent access or specification to watch or know its internal implementation of said function. In building something, you might need just that function, and the only item that you can find is not user-serviceable and is otherwise the black box. Eventually you might have enough interest in that functional apparatus to tear it open anyways or research how that event happens. You didn't know it, but maybe the box emit radiation that sterilized your dog, and increased his dog IQ five hundred points. These are side effects of the use of the black box. Anyways the analogy has lost its focus but the black box still does its function, it is useful, it has a purpose. It's not used only because its useful, it's used because it's useful and by its exclusive definition fulfills the mathematical conditions that no other compatible mathematical construct (definition within a context) does. Unlike the geometric column proofs, it doesn't have a contemporarily superior alternative replacement. It's useful because it gives us a common term to describe it without a four-hundred letter acronym, we can each have a general idea of what it is. Then, it can be discussed vis-a-vis its references in terms of its contexts and contextual appearances and (in)compatibilities and the context of other mathematical constructs which form a synthetic whole of practice and practical utility, and belief. Its use is as a placeholder for its entire expressed meaning within its entire context, where we can each see part of the elephant through the keyhole, as it stares back at us, contenly munching grass. Iota is not new to this particular discussion, here it was introduced in the context of a functional mapping between two sets as an element of one of the sets. The least real quantity is the least real quantity is iota. === Subject: Re: Rationals are Uncountable There is not iota/2, iota/2 would be iota. If we accept iota as a real, And here is your inconsistency. There is only one real equal to > one-half of itself: zero. > The least real quantity is the least real quantity. And a rose is a rose is a rose. > Iota's an infinitesimal. If you try to divide it by two you can't > treat it as anything except zero. The result of that operation is > undefined. Iota is atomic. If one cannot divide iota by 22, it cannot be a member of any field which contains the integers, including, but not limi to, the rationals, the reals and the complexes. Just what sort of a garbled up number system are you proposing to foist on an unsuspecting world? > The value of iota is that of the reciprocal of an infinite quantity, > the infinitesimal quantity that if you add together that infinitely > many values of iota the result is equal to one, it is non-zero. It is non-sense. > In trying to consider the ordered set of reals as a sequence of > values, from zero to one, zero is the least element, iota the next, > then next to iota is another element, and so on and so forth until the > next element would be one, thereby a unit interval of elements in > ordered sequence is considered. Only by the faint of mind. > One goal of trying to determine the properties or characteristics of > such an irrational thing as the least positive real quantity is that > it fulfills requirements of a conceptual item to fill a gap in our > mathematical understanding or utility, towards a more complete whole. > That it is difficult or contradictory to integrate with other accep > mathematical practices is not always a sign of its uselessness. Our goal is more towards trying to determine ' irrationality quotient. It is certainly much higher than his intelligence quotient > === Subject: Re: Rationals are Uncountable > Iota's an infinitesimal. If you try to divide it by two you can't > treat it as anything except zero. The result of that operation is > undefined. Iota is atomic. > The value of iota is that of the reciprocal of an infinite quantity, > the infinitesimal quantity that if you add together that infinitely > many values of iota the result is equal to one, it is non-zero. > In trying to consider the ordered set of reals as a sequence of > values, from zero to one, zero is the least element, iota the next, > then next to iota is another element, and so on and so forth ... Ah, well, I'll give you the so on and so forth. But I'm intrigued about the one you mentioned before that - another element. The one after iota. Don't be _too_ coy - just tell us a little more about it. Does it have a name? Does it bear any algebraic relation to iota, like being iota*2? Brian Chandler ---------------- Jigsaw puzzles from Japan http://imaginatorium.org/shop/ imaginatorium@despammed.com === Subject: what can we say about variance of a linear system output? Dear all, I have a linear system y=Ax, where x is input vector, A is system in matrix form, y is output vector, is there anything we can say about var(y) in terms of var(x) and some characteristic of A, A is given and known... Any existing theorem for this? === Subject: Re: How to find near integer values of n*a and n*b? |> Is there an efficient way to solve the following problem: |> Given a,b,eps>0 find all positive n of being integer. |> For example, let a=sqrt(2), b=sqrt(3) and eps=10^-k. I've compiled a table of |> n that work for various small values of k. Is it complete? Given any m real numbers x_1,...,x_m, there exist infinitely many positive integers n such that n x_j are all within n^(-1/m) of integers. But I don't think there's a lot more known on this topic. One way of finding good values of n is the LLL algorithm. In Maple 9: > A:= map(round,[evalf(10^10*sqrt(2)),evalf(10^10*sqrt(3)),1,0,0]); > IntegerRelations:-LLL([A,[10^10,0,0,1,0],[0,10^10,0,0,1]]); [[709420, 3547280, 326491, -461728, -565499], [-7420420, 2928720, -1653041, 2337753, 2863151], [1208500, 1814000, -3151075, 4456293, 5457822]] is a good candidate. With other positive integers d replacing 10 we get other good values: d, n, eps 5, 41, 0.0172439428 6, 1463, 0.0096685267 7, 4109, 0.0035277911 8, 20586, 0.0020753870 9, 326491, 0.0002139702 10, 326491, 0.0002139702 11, 2151016, 0.0001064238 12, 2151016, 0.0001064238 13, 78411940, 0.000042352 14, 447810523, 0.0000111850 15, 1580818405, 0.0000054692 16, 10011132893, 0.0000010187 17, 10011132893, 0.0000010187 18, 10011132893, 0.0000010187 -7 19, 936991385102, 5.29 10 -7 20, 3295276956399, 3.725 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 === Subject: Re: How to find near integer values of n*a and n*b? >Is there an efficient way to solve the following problem: >Given a,b,eps>0 find all positive neps >of being integer. >For example, let a=sqrt(2), b=sqrt(3) and eps=10^-k. I've compiled a table >of >n that work for various small values of k. Is it complete? A (possibly) more interesting question is the following: For what k do there exist n1,n2,n3 meeting the criteria above and n1+n2=n3? It may, at first, seem to be yet another silly question: i.e. given there are only a handful of n's that meet the criteria for a given k, what are the chances two would sum to a third? Pretty slim, right? This morning, however, I found the following [for sqrt(2),sqrt(3),10^-k]: k=22 works with: 140..049(44)+150..242(44)=291..291(44) k=25 works with: 345..942(50)+403..335(50)=749..297(50) Can anyone explain this data? >[The numbers are shortened: 297..072(26) is the 26 digit number starting with >296 and ending with 072] Rich === Subject: Re: Squaring A Circle...(almost) > We all (should) know it is impossible to square a circle with only a > straight-edge and compass and pencil. > But, of course, we can approximately square a circle to any finite, > but imperfect, accuracy we desire. > A brute-force method might be to first list the squareroot of pi in > binary. > Then keep bisecting a segment equal to the circle's radius so as to > get the lengths: > (radius)/2^k, for k = 0, 1, 2, ...n, > where n is any arbitrarily high finite integer. > Then finally construct the line-segment equal to the sum of the > segments' lengths in the previous step which correspond with > the 1's in sqrt(pi)'s binary expansion. > And, finally, make a square with this last segment's side-length. > So we have a square with an area (just under) > (radius)^2 *pi. Of course, the above method has the problem of errors compounding on each bisection and summing (summing to the segment which is *suppose* to be approaching (radius)*sqrt(pi)). But if we tend to have a similar error (about 1/4 millimeter too long, say) on each iteration (where the direction of the error tends to remain the same, as well as the approximate amount of error), then perhaps it would be better to use base(NEGATIVE 2), rather than base 2, to represent sqrt(pi). For, theoretically, then because we would be alternating between adding and subtracting to get sqrt(pi), the errors might cancel-out somewhat as we continue to form the square's side-length. (This assumes sqrt(pi) is normal in base(-2), of course...Or less strictly, it is required, for better error-cancellation, that the 1's bits occur in approximately the same frequency among the oddly-positioned as among the evenly-positioned bits.) Leroy Quet > So, FOR FUN, what are some other more efficient > approximate-circle-squaring algorithms (using only > straight-edge and compass) you all can come up with?? > (I recall reading someplace that some semi-crank from years ago came > up with a decent approximation algorithm.) > === Subject: Re: Squaring A Circle...(almost) ETAsAhQi2wS7wtRRrgos6Rn4BZBE/rCaBAIUKZrb4+iORQAOoV9RZEk+ kbEWn7g= My math is a little better than my English. In my examples I meant unit diameter, not nit radius. --OL === Subject: Re: Squaring A Circle...(almost) >) of fashion with the crank fringe lately. BTW, what did you >) have in mind for the third entry in your top-three list? >The four-colour map theorem, of course. The really really cool bit about >that one is that the theorem is actually true, so you can't just brush them >off with 'that's proven to be impossible' or something. Well, as I said in my other post I was intendedly vague (or more precisely, not completely honest!), but I fear that the four-colour theorem wouldn't hold the third position... the angle trisetting problem sugges by another poster smells more like it. Please note, however, that my judgement (FWIW!) may be strongly biased since due to time constraints I'm highly selective on the threads to follow and to ignore. Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: how does each row, column's variance rela with the matrix variance? Dear all, Suppose I have a matrix and want to compute its matrix variance, and I know each row and column's variance, what can I say about the matrix variance? And expression relating all these terms? === Subject: Re: Countable union of r.e. sets > Let r.e. abbreviate recursively enumberable > Suppose {B_n : n is a natural number} is a family of r.e. sets. > Is the union of this family also an r.e. set? . . . > Someone brought to my attention, however, the troubling point that this > algorthm f is not a finite list of instructions and algorithms must be > finite. > Leonard (email defunct) Any algorithm to enumerate an infinite set has an infinite number of values to output (steps to perform) and never does all of them. However, any particular value will be output at some particular step. In the case of an infinite number of sets (programs), you likely really have a recursive function g(X,Y) = the Xth instruction of the Yth program (that outputs the Yth set), so that is where you get an infinite number of instructions being execu by a finite program. But the principle is the same: your program keeps working and never finishes, but that suffices as an enumeration. At no time has every instruction been execu, but every instruction will be execu at some point in time. So I agree that the answer is yes. Of course that assumes that set B is r.e. === Subject: Re: Chicken Nugget Problem (Number Theory) > Recently I came ac this problem in my textbook and I've been quite > interes to know the answer to. Here it is, Chicken Nuggets used to be sold at a hamburger chain in packages of 6, 9, 20 > pieces. What is the largest number of pieces you could not order exactly? I'd appreciate any help on this problem. Gavin Does anyone know the origin of the problem? I proposed it as a problem in the Bent (the engineering honor society magazine) in around 1988 when my son was elec in 1987. I saw it repea (with no source) in MIT Technology Review last year some time. I know where I got it from. I was walking by MacDonald's one day with my son, who told me about the 6,9, and 20 McNuggets and worked it out while walking. === Subject: Re: Chicken Nugget Problem (Number Theory) >* mareg@mimosa.csv.warwick.ac.uk >> To get any 3n+2 you actually _need_ one 20 (or 4, 7, ... 20s). >> True, but you do not need that fact. >Well, how do you prove that 43 is not expressable then? Since 43 = 1 (mod 3), why should statements about 3n+2 be needed ? 20x+6y+9z=43 ==> x=2 (mod 3) ==> x >= 2 ==> obviously impossible. Derek Holt.