mm-2839 === Subject: Which is more difficult Is factoring harder than solving a diophantine equation, in general? === Subject: Re: Which is more difficult >Is factoring harder than solving a diophantine equation, in general? Given an integer n, if you can find all integer solutions of the Diophantine equation x*y = n, then you can factor n. Or you can reduce it to finding a nonnegative integer solution of (x + 2)*(y + 2) = n, which is like finding one integer solution to (x1^2 + x2^2 + x3^2 + x4^2 + 2)*(y1^2 + y2^2 + y3^2 + y4^2 + 2) = n. -- Wanted: Experts at choosing the best of 100+ applicants for a position. Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Which is more difficult >Is factoring harder than solving a diophantine equation, in general? Not sure whether you mean factoring a polynomial (say over the rationals) or factoring an integer, but these are both easier than solving a diophantine equation: there are algorithms for both kinds of factoring, while for diophantine equations in general there is no algorithm (look up Hilbert's Tenth Problem). 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: Legal logic ? puzzle. > The matter which I am desperately struggling with, concerns a > time sequence of event; with the assumption that 'cause' does NOT > follow 'effect'. > If I can't explain this to scientifically trained persons, how will > I explain the legal-types ?! Perhaps they all do understand your story, but they don't understand what your actual question is. If x owes 10$, but is billed 11$, isn't it possible for x to just pay 10$? Other possibility: every 10 times, x could simply not pay 1 bill. That would make things even. Herman Jurjus === Subject: Re: The two envelope paradox In artikel schreef Steve Gerrard: >< snippage Having just muddled my way through this myself, I am going to take a shot at > explaining what I have learned as it applies to your proposal. >> Just budding in, but I have found few people come up with the following >> solution. Before you open the envelope in your hand, think of a number. >> If the amount in your envelope is smaller then that number ask to >> switch. If it is greater don't switch. > Your proposal would work (I think) if it just so happened that over time, the > frequency of each possible amount was the same - that is, that the distribution > of amounts was uniform. However, there is no reason to think that is the case. > Illustrating that it is a false assumption is the real purpose of the paradox. No >> There are three possibilities. >> 1) Both envelopes contain an amount under your number. >> You will always switch and on average gain nothing. > #1 is false, because the on average is assuming a uniform distribution of > amounts, and that it will all average out over time. There just is nothing > that makes that so. No it is assuming a uniform distribution in having chosen the lower or the higher number. Half of the time I will have sitched the lower amount for the higher and vice versa. The amount I win in the first case is on average the amount I lose in the latter case. So on average I gain nothing. >> 2) Both envelopes contain an amount above your number. >> You will never switch and on average gain nothing. > #2 works, since you are not changing anything. >> 3) Your number is between the amounts in the envelopes. >> You will keep the higher amount but will swicth >> with the lower amount, so you will gain some. > #3 might happen, and if it did, you might make a gain. Might not. There is no > way to tell, since no one knows the true distribution of amounts that will be > carried around in pairs of envelopes by generous persons with an interest in > logic problems, for the rest of eternity ... But you can't escape the fact, that if a person has two such envelopes that the double amount has a higher chance of being bigger than some randomly choosen number than the single amount. Sure the chance can be very small, but a small chance is still better than no chance. -- Antoon Pardon === Subject: Re: The two envelope paradox > But you can't escape the fact, that if a person has two such envelopes > that the double amount has a higher chance of being bigger than some > randomly choosen number than the single amount. Sure the chance can > be very small, but a small chance is still better than no chance. It actually isn't better, in the case of infinities. This is like a limit in calculus. What is the limit of 1/x as x -> infinity? The answer is zero. In other words, a sufficiently small chance is _exactly_the_same_ as no chance, not still better than no chance. -- Don ____________________________________________________________________________ ___ Don Geddis http://don.geddis.org/ don@geddis.org The statistics on sanity are that one out of every four Americans is suffering from some form of mental illness. Think of your three best friends. If they're okay, then it's you. -- Rita Mae Brown === Subject: symbolic calculus software What is the best soft to do symbolic calculus ? Maple 9, Mathematica 4.2 or Matlab 6.5 ? === Subject: Re: symbolic calculus software >What is the best soft to do symbolic calculus ? Maple 9, Mathematica 4.2 >or Matlab 6.5 ? What does best mean?? === Subject: Re: symbolic calculus software > What is the best soft to do symbolic calculus ? Maple 9, Mathematica 4.2 > or Matlab 6.5 ? Mathematica has a newer release (IIRC it is 5.0). There are other symbolic math programs out there too. For example, Gauss, Macsyma, Mathcad, Axiom, MuPad, and others. There are also specialized programs out there for specific math topics. May want to have a look. HTH, FLip === Subject: Re: symbolic calculus software > What is the best soft to do symbolic calculus ? Maple 9, Mathematica 4.2 > or Matlab 6.5 ? Use Matlab for numerical and matrix calculations. Use Maple or Mathematica for symbolic calculus. === Subject: [graph theory] how to find a path with a length N I'm looking for some algorithms in order to solve the following problem: I want to extract from a graph all the paths between two nodes A and B such that the number of edges (of those path) is equal to N. does anyone have some ideas? Jeremie === Subject: Rubik's cube: another question... If you consider the moves you can make, a plane can be rotated clockwise once, anticlockwise once, or rotated twice (180 degrees) - i.e. each plane has three possible moves. I count 9 planes on the cube (3 hirozontal, 3 vertical and 3 into the cube. So, in one move there are 3x9 = 27 possible moves. so in 14 moves, there are 27^14 possible outcomes this = 1.09 x 10^20 (I presume a few of these might return to it's original state) . But there are only 4.3x 10^19 total different patterns. Does this mean that any given cube is possible to solve within 14 moves? === Subject: Re: Rubik's cube: another question... > If you consider the moves you can make, a plane can be rotated clockwise > once, anticlockwise once, or rotated twice (180 degrees) - i.e. each plane > has three possible moves. I count 9 planes on the cube (3 hirozontal, 3 > vertical and 3 into the cube. So, in one move there are 3x9 = 27 possible > moves. so in 14 moves, there are 27^14 possible outcomes this = 1.09 x 10^20 > (I presume a few of these might return to it's original state) . But there > are only 4.3x 10^19 total different patterns. > Does this mean that any given cube is possible to solve within 14 moves? I think this problem is generally referred to as God's algorithm. Given D, the diameter of the graph of Rubik cube states, an omnipotent being could determine the shortest solution to any given starting state. This solution by definition would be D or fewer moves long. According to this site, God's algorithm is 14 moves long for the 2 x 2 x 2 cube, but the diameter is unknown for the 3-cube. http://web.usna.navy.mil/~wdj/book/node187.html On another site, I saw mention of a speculation that D=22 for the 3-cube, but without any mathematics. - Randy === Subject: Re: Rubik's cube: another question... ... > According to this site, God's algorithm is 14 moves long > for the 2 x 2 x 2 cube, but the diameter is unknown for the > 3-cube. > http://web.usna.navy.mil/~wdj/book/node187.html > On another site, I saw mention of a speculation > that D=22 for the 3-cube, but without any mathematics. The 14 are right. Long ago I have already speculated that for the 3-cube the diameter would be 22 or 23. This was based on the look of the graph for other similar puzzles and the part that was known about the graph forthe 3-cube. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Rubik's cube: another question... > If you consider the moves you can make, a plane can be rotated clockwise > once, anticlockwise once, or rotated twice (180 degrees) - i.e. each plane > has three possible moves. I count 9 planes on the cube (3 hirozontal, 3 > vertical and 3 into the cube. So, in one move there are 3x9 = 27 possible > moves. so in 14 moves, there are 27^14 possible outcomes this = 1.09 x 10^20 > (I presume a few of these might return to it's original state) . But there > are only 4.3x 10^19 total different patterns. > Does this mean that any given cube is possible to solve within 14 moves? It depends what you mean by a move - I think it's most usual to consider the 6 squares at the centres of the faces to be fixed, so a move of one on the 3 central planes is counted as two moves - one by each of its neighbours. So there are 18 (3x6) possible moves from any position rather than 27. Unfortunately this doesnt prove that any position can be reached in 16 moves, even though 18^16 > 4.3x10^19, because, as you point out, a few of these might return [the cube] to [its] original state. (Actually it will be rather more than a few...) Have a look at http://cubeman.org/ for more information, especially the bits about God's algorithm in Progress in Solving Algorithms under Cube Notes. According to this, the maximum length of God's algorithm is empirically at most 20: I don't know whether there are any better estimates. (By the argument above it's at least 16). Andrew Taylor (Who has the small claim to fame of being mentioned in the Cube section of Winning Ways by Berlekamp, Conway & Guy) === Subject: Re: Way off topic > Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in what > oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist > and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and > you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed > ervey lteter by itslef but the wrod as a wlohe. Aoccdrnig to Google the word Aoccdrnig has appeared in at least 51 news groups. It appears to have been off-topic in most of them. The dates below relate to the most recent post in the thread. The word was unknown to Usenet The list below does not include all the newsgroups in which the word has appeared. If a post is cross-posted to several groups it seems that only the first is referenced. For example, I picked it up in rec.humor which is not listed below. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Way off topic linux) >> Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in > what >> oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the > frist >> and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >> you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >> ervey lteter by itslef but the wrod as a wlohe. > Aoccdrnig to Google the word Aoccdrnig has appeared in at least 51 news > groups. It appears to have been off-topic in most of them. The dates below > relate to the most recent post in the thread. The word was unknown to Usenet It's making the Xerox rounds, too. A guy at work printed out a few copies of this quote, or nearly enough. Same misspelling of (Aoccdrnig to *a* rscheearch...). Was the paragraph yours, or did it appear elsewhere? -- What I've learned is that [mathematicians are] the gatekeepers, and seem to have almost absolute power when it comes to mathematics. -- James Harris, on All I Really Ever Needed to Know I Learned in /Ghostbusters/. === Subject: Re: Way off topic message >> Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in > what >> oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the > frist >> and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >> you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >> ervey lteter by itslef but the wrod as a wlohe. > Aoccdrnig to Google the word Aoccdrnig has appeared in at least 51 news > groups. It appears to have been off-topic in most of them. The dates below > relate to the most recent post in the thread. The word was unknown to Usenet > It's making the Xerox rounds, too. A guy at work printed out a few > copies of this quote, or nearly enough. Same misspelling of > (Aoccdrnig to *a* rscheearch...). > Was the paragraph yours, or did it appear elsewhere? I copied it from a post in rec.humor, making a couple of very minor corrections. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Way off topic >Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in what >oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist >and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >ervey lteter by itslef but the wrod as a wlohe. > That's very interesting. And it's easy to predict what the replies are > going to look like. tuB I ewnrdo hthewre it tsetamr htat we aelev eth > sfrit and tsal strelet in lapce? Hmm, I guess it does. B.t is t...e so m..h i.........n in t.e f...t a.d l..t l.....s t..t t.e m....e o..s d..'t m....r at a.l? Phil === Subject: Re: Way off topic >>Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in what >>oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist >>and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >>you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >>ervey lteter by itslef but the wrod as a wlohe. >> That's very interesting. And it's easy to predict what the replies are >> going to look like. tuB I ewnrdo hthewre it tsetamr htat we aelev eth >> sfrit and tsal strelet in lapce? Hmm, I guess it does. >B.t is t...e so m..h i.........n in t.e f...t a.d l..t l.....s t..t >t.e m....e o..s d..'t m....r at a.l? Unclear - the others certainly help comprehension. >Phil ************************ === Subject: Re: Way off topic is there a reference to this result? (please try to avoid using the lemma when posting the reference.. . ) ceehrs -tzurs > Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in what > oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist > and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and > you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed > ervey lteter by itslef but the wrod as a wlohe. === Subject: Re: Way off topic > Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in what > oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist > and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and > you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed > ervey lteter by itslef but the wrod as a wlohe. I always suspected that English was a highly redundant language, as probably many other languages are. In the old days, when letters were carved into stone, they exploited this redundancy already by leaving out the vowels... And Hamming, in the fifties last century, worked the other way by adding redundancy to a message (bit string) in a clever way, in order to make it more fault tolerant when passing it through a transmission channel. BTW, there's plenty of math going into that, as in fault tolerant logic synthesis (would'nt you hate it if a single faulty transistor, out of a many millions, wrecks a sevral man_years of design work!-) And qua math texts: I noticed that it is useful to figure out, in advance, in how many ways a text can be mis-interpreted (for mathematicians this appears to be their main concern, of course;-) -- NB === Subject: Re: Way off topic In sci.math, The Ghost In The Machine <9itb31-gdc.ln1@lexi2.athghost7038suus.net>: > In sci.math, The Last Danish Pastry > : >> Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in what >> oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist >> and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >> you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >> ervey lteter by itslef but the wrod as a wlohe. > Fascinating, and quite readable. I'll admit this is probably > more appropriate to sci.psychology or some such but the idea > has occurred to me personally. Erm, that should be has NOT occurred to me personally. > I might have to whip up a Perl script to test this at some point. > Of course spelling checkers will blow chunks over such submissions. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Way off topic >>Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in what >>oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist >>and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >>you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >>ervey lteter by itslef but the wrod as a wlohe. > I do not believe this. We may take in the word as a whole, > but the internal processor still does it letter by letter, > unless the word is recognized quickly. Those with larger > vocabularies will find more words to separate out. That one needs to _know_ the correct spelling of the word in order to recognize its distorted version does not neccesarily mean that the process of reading goes letter by letter. Actually one also needs a firm grasp of the underlying grammar. This way one can match a whole sentence pattern against plausible meaningful sentences. I doubt if the permutation of letters could still be repaired when these are introduced in an isolated word. So the wording of the original post is a bit misleading because it neglects the usage of the context. Marc === Subject: Re: Way off topic >>Aoccdrnig to rscheearch at an Elingsh uinervtisy, it >>deosn't mttaer in what >>oredr the ltteers in a wrod are, the olny iprmoetnt >>tihng is taht the frist >>and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >>you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >>ervey lteter by itslef but the wrod as a wlohe. Very nice. I had no trouble reading it once I stopped proofing it. >I do not believe this. We may take in the word as a whole, >but the internal processor still does it letter by letter, >unless the word is recognized quickly. I don't think so. It may differ with people who have been trained in different disciplines. For instance, you math guys need to process glyph by glyph because that's just the way of math. > .. Those with larger >vocabularies will find more words to separate out. When I took a speed reading course, the point was to teach my eyes to read a sentence, then a paragraph at a time. Proficiency in reading in grade school comes when kids stop reading each letter and word separately. The poor readers never get beyond this point. /BAH === Subject: Re: Way off topic >Aoccdrnig to rscheearch at an Elingsh uinervtisy, it >deosn't mttaer in what >oredr the ltteers in a wrod are, the olny iprmoetnt >tihng is taht the frist >and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >ervey lteter by itslef but the wrod as a wlohe. >Very nice. I had no trouble reading it once I stopped proofing it. >>I do not believe this. We may take in the word as a whole, >>but the internal processor still does it letter by letter, >>unless the word is recognized quickly. >I don't think so. It may differ with people who have been >trained in different disciplines. For instance, you math >guys need to process glyph by glyph because that's just the >way of math. Regardless of whether one believes it, what you say about math is a good point, and even somewhat on-topic. It often seems to me that when students simply cannot read mathematics part of the problem is that they're not reading it slowly enough - they glance at an expression, find they cannot understand it by glancing at it and conclude they cannot understand it. >> .. Those with larger >>vocabularies will find more words to separate out. >When I took a speed reading course, the point was to teach >my eyes to read a sentence, then a paragraph at a time. >Proficiency in reading in grade school comes when kids >stop reading each letter and word separately. The poor >readers never get beyond this point. >/BAH > ************************ === Subject: Re: Way off topic >Aoccdrnig to rscheearch at an Elingsh uinervtisy, it >I do not believe this. We may take in the word as a whole, >>but the internal processor still does it letter by letter, >>unless the word is recognized quickly. >I don't think so. It may differ with people who have been >trained in different disciplines. For instance, you math >guys need to process glyph by glyph because that's just the >way of math. > Regardless of whether one believes it, what you say > about math is a good point, and even somewhat on-topic. > It often seems to me that when students simply cannot > read mathematics part of the problem is that they're not > reading it slowly enough - they glance at an expression, > find they cannot understand it by glancing at it and > conclude they cannot understand it. We tend to assume that there is only one kind of reading skill, but there's some research supporting the idea that reading varies comparing reading of music versus reading a book: http://www.musica.uci.edu/mrn/V5I1W98.html#sightreading [M]usic cognition and behavior are often viewed merely as an instance of other, better known subjects. An example is music sight-reading, often believed to obey the laws of language reading. However, recent studies reveal that the study of sight-reading in music provides a unique window on the mind. (saccades) while reading printed texts and printed musical scores, and how different they are. Probably a similar point of view can be applied to reading mathematics. While similar to regular reading, in some ways it is fundamentally different than reading English (or other verbal languages). === Subject: Re: Way off topic |Regardless of whether one believes it, what you say |about math is a good point, and even somewhat on-topic. |It often seems to me that when students simply cannot |read mathematics part of the problem is that they're not |reading it slowly enough - they glance at an expression, |find they cannot understand it by glancing at it and |conclude they cannot understand it. I remember someone once commented that the good math students seemed usually to be better at adjusting their speed based on a recognition of their own level of comprehension. As far as I know this was just an impression without specific experimental support, but it's certainly the sort of think one would *like* to see math students doing. Some of the worst difficulties students had with math seemed to accompany their inability to tell whether they were even getting the point correctly or not. I remember sometimes students understanding a point correctly, but not realizing that it was as simple as that. Keith Ramsay === Subject: Re: Way off topic >>Aoccdrnig to rscheearch at an Elingsh uinervtisy, it >>deosn't mttaer in what >>oredr the ltteers in a wrod are, the olny iprmoetnt >>tihng is taht the frist >>and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >>you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >>ervey lteter by itslef but the wrod as a wlohe. >>Very nice. I had no trouble reading it once I stopped proofing it. >I do not believe this. We may take in the word as a whole, >but the internal processor still does it letter by letter, >unless the word is recognized quickly. >>I don't think so. It may differ with people who have been >>trained in different disciplines. For instance, you math >>guys need to process glyph by glyph because that's just the >>way of math. >Regardless of whether one believes it, what you say >about math is a good point, and even somewhat on-topic. I tried. Math is unusual in that it packs a lot of meaning into one little glyph; for example, textbooks have been written to prepare me to know what that sigma implies. >It often seems to me that when students simply cannot >read mathematics part of the problem is that they're not >reading it slowly enough - they glance at an expression, >find they cannot understand it by glancing at it and >conclude they cannot understand it. Do you know if these kids do sentence diagramming in grade school? In math, it's important to be able to take things apart, rearrange them, and put them back together again. Same is true in chemistry, physics, definitely programming. /BAH === Subject: Re: Way off topic >Aoccdrnig to rscheearch at an Elingsh uinervtisy, it >deosn't mttaer in what >oredr the ltteers in a wrod are, the olny iprmoetnt >tihng is taht the frist >and lsat ltteers are in the rghit pclae. The rset can be a toatl mses >and >you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >ervey lteter by itslef but the wrod as a wlohe. >Very nice. I had no trouble reading it once I stopped proofing it. >>I do not believe this. We may take in the word as a whole, >>but the internal processor still does it letter by letter, >>unless the word is recognized quickly. >I don't think so. It may differ with people who have been >trained in different disciplines. For instance, you math >guys need to process glyph by glyph because that's just the >way of math. >>Regardless of whether one believes it, what you say >>about math is a good point, and even somewhat on-topic. >I tried. Math is unusual in that it packs a lot of >meaning into one little glyph; for example, textbooks have >been written to prepare me to know what that sigma implies. >>It often seems to me that when students simply cannot >>read mathematics part of the problem is that they're not >>reading it slowly enough - they glance at an expression, >>find they cannot understand it by glancing at it and >>conclude they cannot understand it. >Do you know if these kids do sentence diagramming in grade >school? I don't know for a fact, but it seems very unlikely. >In math, it's important to be able to take things >apart, rearrange them, and put them back together again. >Same is true in chemistry, physics, definitely programming. Yup. Which is exactly why kids should be required to do simple proofs, just to build those muscles. But this is really getting way on-topic... >/BAH > ************************ === Subject: Re: Way off topic >In math, it's important to be able to take things >apart, rearrange them, and put them back together again. >Same is true in chemistry, physics, definitely programming. > Yup. Which is exactly why kids should be required to do > simple proofs, just to build those muscles. Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point out the error. C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1 C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2 C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak Extensionality How's about I let you in on a little secret? (Ex)~(x=x) follows from (C3,C4) alone! C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak Extensionality C'mon, David. Show us how. Don't be afraid. You can do it!!! --John And yes, identity is in _fact_ reflexive. To refute that statement you need to give an example of something which is not identical to itself. The idea that there is something which is _not_ identical to itself is simply ludicrous: That's what identity _means_: A thing is identical to itself and to nothing else. --David Ullrich === Subject: Re: Way off topic >>In math, it's important to be able to take things >>apart, rearrange them, and put them back together again. >>Same is true in chemistry, physics, definitely programming. >> Yup. Which is exactly why kids should be required to do >> simple proofs, just to build those muscles. >Exhibit of proof of Ex~(x=x) from C1-C4 and someone will >point out the error. >C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1 >C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2 >C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification >C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak >Extensionality >How's about I let you in on a little secret? (Ex)~(x=x) follows >from (C3,C4) alone! You should also let everyone else in on another little secret: In the thread where this came up you were claiming that C1-C4 were axioms of NBG. >C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification >C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak >Extensionality >C'mon, David. Show us how. Don't be afraid. You can do it!!! >--John >And yes, identity is in _fact_ reflexive. To >refute that statement you need to give an >example of something which is not identical >to itself. The idea that there is something >which is _not_ identical to itself is simply >ludicrous: That's what identity _means_: A >thing is identical to itself and to nothing >else. >--David Ullrich Another little secret: When you quote me saying the above, as though I'd said something stupid or even something the teensiest bit controversial you look like an idiot. (Don't bother replying with citations from learned parties talking about non-reflexive identities. It's not _identity_ they're talking about.) ************************ === Subject: Re: Way off topic >>In math, it's important to be able to take things >>apart, rearrange them, and put them back together again. >>Same is true in chemistry, physics, definitely programming. >> Yup. Which is exactly why kids should be required to do >> simple proofs, just to build those muscles. >Exhibit of proof of Ex~(x=x) from C1-C4 and someone will >point out the error. >C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1 >C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2 >C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification >C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak >Extensionality >How's about I let you in on a little secret? (Ex)~(x=x) follows >from (C3,C4) alone! > You should also let everyone else in on another little secret: In > the thread where this came up you were claiming that C1-C4 > were axioms of NBG. Produce a post where I claim this, or admit you are a LIAR. >C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification >C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak >Extensionality >C'mon, David. Show us how. Don't be afraid. You can do it!!! >--John >And yes, identity is in _fact_ reflexive. To >refute that statement you need to give an >example of something which is not identical >to itself. The idea that there is something >which is _not_ identical to itself is simply >ludicrous: That's what identity _means_: A >thing is identical to itself and to nothing >else. >--David Ullrich > Another little secret: When you quote me saying the above, > as though I'd said something stupid or even something > the teensiest bit controversial you look like an idiot. > (Don't bother replying with citations from learned parties > talking about non-reflexive identities. It's not _identity_ > they're talking about.) Thus we might characterise a non-individual as an object for which no identity criteria hold. If the theory of identity is taken to be that of classical logic, then for a non-individual it is not the case that ëa = a'. This allows us to avoid the Evans-Salmon argument and resolve the curious situation it is indeterminate whether they are identical to themselves because they are not! (Steven French and Decio Krause, Quantum To be ignorant of one's ignorance is the malady of the ignorant. --Amos Bronson Alcott, _Table Talk_ [1877] === Subject: Re: Way off topic === >Subject: Re: Way off topic >Message-id: Aoccdrnig to rscheearch at an Elingsh uinervtisy, it >>deosn't mttaer in what >>oredr the ltteers in a wrod are, the olny iprmoetnt >>tihng is taht the frist >>and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >>you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >>ervey lteter by itslef but the wrod as a wlohe. >>Very nice. I had no trouble reading it once I stopped proofing it. >I do not believe this. We may take in the word as a whole, >but the internal processor still does it letter by letter, >unless the word is recognized quickly. >>I don't think so. It may differ with people who have been >>trained in different disciplines. For instance, you math >>guys need to process glyph by glyph because that's just the >>way of math. >Regardless of whether one believes it, what you say >about math is a good point, and even somewhat on-topic. >It often seems to me that when students simply cannot >read mathematics part of the problem is that they're not >reading it slowly enough - they glance at an expression, >find they cannot understand it by glancing at it and >conclude they cannot understand it. I find that a lot of the stuff posted in sci.math makes my eyes glaze over. And frequently it's stuff I actually understand when I make an effort to disect the ASCII equations. It's amazing how one's perception of those symbols changes once one understands what it's about. > .. Those with larger >vocabularies will find more words to separate out. >>When I took a speed reading course, the point was to teach >>my eyes to read a sentence, then a paragraph at a time. >>Proficiency in reading in grade school comes when kids >>stop reading each letter and word separately. The poor >>readers never get beyond this point. >>/BAH >************************ > -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Way off topic === >>Subject: Re: Way off topic >>Message-id: >Aoccdrnig to rscheearch at an Elingsh uinervtisy, it >deosn't mttaer in what >oredr the ltteers in a wrod are, the olny iprmoetnt >tihng is taht the frist >and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >ervey lteter by itslef but the wrod as a wlohe. >Very nice. I had no trouble reading it once I stopped proofing it. >>I do not believe this. We may take in the word as a whole, >>but the internal processor still does it letter by letter, >>unless the word is recognized quickly. >I don't think so. It may differ with people who have been >trained in different disciplines. For instance, you math >guys need to process glyph by glyph because that's just the >way of math. >>Regardless of whether one believes it, what you say >>about math is a good point, and even somewhat on-topic. >>It often seems to me that when students simply cannot >>read mathematics part of the problem is that they're not >>reading it slowly enough - they glance at an expression, >>find they cannot understand it by glancing at it and >>conclude they cannot understand it. >I find that a lot of the stuff posted in sci.math >makes my eyes glaze over. Especially when the right margin is set way out to 80. *hint* > .. And >frequently it's stuff I actually understand when >I make an effort to disect the >ASCII equations. I always have to write it down on paper. I don't think I'll ever establish that short cut of translating from screen to concept in my head like youngsters can do very, very well. > ..It's amazing how one's perception of those symbols changes >once one understands what it's about. There are pixel problems with presentation on a terminal screen, too. /BAH === Subject: Re: Way off topic >>I find that a lot of the stuff posted in sci.math >>makes my eyes glaze over. >Especially when the right margin is set way out to 80. *hint* How do you do that in the AOL editor? -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Way off topic >I find that a lot of the stuff posted in sci.math >makes my eyes glaze over. >>Especially when the right margin is set way out to 80. *hint* >How do you do that in the AOL editor? There is a key on the right that may have the word enter on it. Use that when the parser gets 3/4 across your screen. /BAH === Subject: Re: Way off topic === >Subject: Re: Way off topic >Message-id: I find that a lot of the stuff posted in sci.math >>makes my eyes glaze over. >Especially when the right margin is set way out to 80. *hint* >>How do you do that in the AOL editor? >There is a key on the right that may have the word enter >on it. Duh. Sorry, I apparenetly mistook you for someone with something contructive to offer. Having the right margin set at 80 implies that it can be changed in the preferences, but there aren't any. > Use that when the parser gets 3/4 across your screen. IF the editing windows were always the same size, and IF the editor used a monospaced font, and IF the font size is constant, THEN that might make some sense. But as they aren't, it doesn't, and it's not, I'll have to look elsewhere for advice. >/BAH > -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Way off topic === >>Subject: Re: Way off topic >>Message-id: >I find that a lot of the stuff posted in sci.math >makes my eyes glaze over. >>Especially when the right margin is set way out to 80. *hint* >How do you do that in the AOL editor? >>There is a key on the right that may have the word enter >>on it. >Duh. Sorry, I apparenetly mistook you for someone with something >contructive to offer. [emoticon notices some snot] >Having the right margin set at 80 implies that >it can be changed in the preferences, but there aren't any. I suspect it's different with every program. The way to force idiot programs to do your bidding is to use the enter key. >> Use that when the parser gets 3/4 across your screen. >IF the editing windows were always the same size, and IF the editor >used a monospaced font, and IF the font size is constant, >THEN that might make some sense. You're just producing excuses. None of those IFs apply. >But as they aren't, it doesn't, and it's not, I'll have to look elsewhere >for advice. IOW, doing the logical thing isn't your bag. Out of curiosity, how did you learn to touch type? People who trained on a typewriter usually don't have any problems with typing carriage-control characters. /BAH === Subject: Re: Way off topic === >>Subject: Re: Way off topic >>Message-id: >I find that a lot of the stuff posted in sci.math >makes my eyes glaze over. >>Especially when the right margin is set way out to 80. *hint* How do you do that in the AOL editor? >>There is a key on the right that may have the word enter >>on it. >Duh. Sorry, I apparenetly mistook you for someone with something >contructive to offer. > [emoticon notices some snot] Snotty? You were admonishing me for not having correctly set something I have no control over. And when I politely asked where those settings are found, your snotty reply was to learn where the enter key was. If you're going to pontificate from a position of ignorance, don't be surprised that you get snotty replies. >Having the right margin set at 80 implies that >it can be changed in the preferences, but there aren't any. > I suspect it's different with every program. The way to force > idiot programs to do your bidding is to use the enter key. Do you know the difference between an editor and a word processor? In the former, the user usually controls the line breaks with the enter key. In a word processor or any other program that automatically controls word wrap, the enter key is used for paragraph breaks. >> Use that when the parser gets 3/4 across your screen. >IF the editing windows were always the same size, and IF the editor >used a monospaced font, and IF the font size is constant, >THEN that might make some sense. > You're just producing excuses. None of those IFs apply. Unlike some, I don't spout off without knowing what I'm talking about. You might want to check out my reply to Gerry Myerson to learn something about editing in AOL. >But as they aren't, it doesn't, and it's not, I'll have to look elsewhere >for advice. > IOW, doing the logical thing isn't your bag. The logical thing is to use the editor the way it is intended to be used, with enter used for paragraph breaks, not line breaks. And just to be snotty, I'll say if you don't like the way the messages are displayed, then that's _your_ problem, not mine. > Out of curiosity, how did you learn to touch type? I didn't, but 25 years of banging on computer keyboards has (in addition to giving me carpal tunnel syndrome) made me a very fast pecker. > People who trained on a typewriter usually don't have any problems with > typing carriage-control characters. Until they use a word processor and find out that every line has become a paragraph. Luckily, I am multi-talented. I can use typewriters, editors and word processors interchangeably and not get confused. > /BAH > === Subject: Re: Way off topic === >Subject: Re: Way off topic >Message-id: How do you do that in the AOL editor? >There is a key on the right that may have the word enter on it. > Use that when the parser gets 3/4 across your screen. > IF the editing windows were always the same size, and IF the editor > used a monospaced font, and IF the font size is constant, THEN that > might make some sense. > But as they aren't, it doesn't, and it's not, I'll have to look > elsewhere for advice. Probably not to me, as I'm not familiar with the AOL editor, but can I ask: is it possible to resize a window? and is it possible to make the editor use the font of your choosing? and is it possible to convince the editor to use the font size of your choosing? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Way off topic === >Subject: Re: Way off topic === >>Subject: Re: Way off topic >>Message-id: How do you do that in the AOL editor? >>There is a key on the right that may have the word enter on it. >> Use that when the parser gets 3/4 across your screen. >> IF the editing windows were always the same size, and IF the editor >> used a monospaced font, and IF the font size is constant, THEN that >> might make some sense. >> But as they aren't, it doesn't, and it's not, I'll have to look >> elsewhere for advice. >Probably not to me, as I'm not familiar with the AOL editor, >but can I ask: is it possible to resize a window? Yes, but it would be a lot of work and you may end up with this phenomena, which I find very annoying. It is caused by people who incorrectly gauge where 3/4 of the screen is and sometimes hit the enter key after the editor has already imposed its own line wrap at 80 columns, which is not shown prior to sending. And no, there is no preview mode. The Arial font used by AOL contributes to this problem. >and is it possible to make the editor use the font of your choosing? No. AOL has preference settings for the e-mail but not for the newsgroup editor. >and is it possible to convince the editor to use the font >size of your choosing? And when I say there are none, I mean there is a certain amount. One of the e-mail settings is to display text as small, medium or large. For some reason, the newsgroup editor inherits this setting although it does not inherit the font setting preferences. With a large display such as 1280x960 and the preference on small, the editing window is easily over a hundred characters wide if the AOL window is at max and the edit window is at max. And that's when you are replying. Creating a new message almost doubles the edit window size. It is extremely annoying not being able to control the editor settings. Usually, for serious replies, especially those involving ASCII art, I compose in Notepad and paste in the reply. But that's a lot of effort for simple one or two line responses For this message, I tried to control the line length with the enter key. I won't know how successful until after it's posted. >-- >Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Way off topic === >>Subject: Re: Way off topic >>Message-id: How do you do that in the AOL editor? >>There is a key on the right that may have the word enter on it. >> Use that when the parser gets 3/4 across your screen. >> IF the editing windows were always the same size, and IF the editor >> used a monospaced font, and IF the font size is constant, THEN that >> might make some sense. >> But as they aren't, it doesn't, and it's not, I'll have to look >> elsewhere for advice. >Probably not to me, as I'm not familiar with the AOL editor, >but can I ask: is it possible to resize a window? With the AOL software I use, it's a temporary thing and has to be reset with every display of a post and its reply. > .. and is it >possible to make the editor use the font of your choosing? I suspect that depends on the version running. Mine doesn't but mine is very, very old. >and is it possible to convince the editor to use the font >size of your choosing? No. /BAH === Subject: Re: Way off topic > Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer > in what oredr the ltteers in a wrod are, the olny iprmoetnt tihng is > taht the frist and lsat ltteers are in the rghit pclae. The rset can > be a toatl mses and you can sitll raed it wouthit a porbelm. Tihs is > bcuseae we do not raed ervey lteter by itslef but the wrod as a > wlohe. Let's see if this pertains to painters, pantries, or anything done in loco parentis. -- http://hertzlinger.blogspot.com === Subject: Re: Way off topic linux) >> Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer >> in what oredr the ltteers in a wrod are, the olny iprmoetnt tihng is >> taht the frist and lsat ltteers are in the rghit pclae. The rset can >> be a toatl mses and you can sitll raed it wouthit a porbelm. Tihs is >> bcuseae we do not raed ervey lteter by itslef but the wrod as a >> wlohe. > Let's see if this pertains to painters, pantries, or anything done in > loco parentis. Do you feel an odd compulsion to repeat yourself? -- Jesse Hughes So far as this negative attitude toward life is concerned, Buddhism is merely Taoism a little touched in its wits. -- Lin Yutang, /My Country and My People/ === Subject: Re: Way off topic > Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer > in what oredr the ltteers in a wrod are, the olny iprmoetnt tihng is > taht the frist and lsat ltteers are in the rghit pclae. The rset can > be a toatl mses and you can sitll raed it wouthit a porbelm. Tihs is > bcuseae we do not raed ervey lteter by itslef but the wrod as a > wlohe. >> Let's see if this pertains to painters, pantries, or anything done in >> loco parentis. > Do you feel an odd compulsion to repeat yourself? I feel it again and again. -- http://hertzlinger.blogspot.com === Subject: Re: Way off topic > Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer > in what oredr the ltteers in a wrod are, the olny iprmoetnt tihng is > taht the frist and lsat ltteers are in the rghit pclae. The rset can > be a toatl mses and you can sitll raed it wouthit a porbelm. Tihs is > bcuseae we do not raed ervey lteter by itslef but the wrod as a > wlohe. >> Let's see if this pertains to painters, pantries, or anything done in >> loco parentis. > Do you feel an odd compulsion to repeat yourself? > I feel it again and again. The sherpas had pitons and plates. This might come out as... The shapers had points and pleats. The seraphs had pintos and petals. The sphaers had pinots and palets. The sphears had potins and peltas. [Some of the more obscure words... palets: paleae peltas: shields pinots: grapes potins: copper alloys sphaers: old form of 'spheres' sphears: old form of 'spheres'] -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Way off topic >Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in what >oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist >and lsat ltteers are in the rghit pclae. The rset can be a toatl mses and >you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed >ervey lteter by itslef but the wrod as a wlohe. > I do not believe this. We may take in the word as a whole, > but the internal processor still does it letter by letter, > unless the word is recognized quickly. Those with larger > vocabularies will find more words to separate out. I believe it. English is not my first language, nor my second, still I had no problems with reading. Some words require backtracking, but only the long words (I think I missed only two words on first reading). I had also no problems with the French text supplied in follow-up (but French is my second language). I know for sure that I do not take words letter for letter. There are many cases where I know a letter is missing in a word, only when it is pointed out to me. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: An algebraic problem let x and y be two rational angles verifing the following condition: 8 (cos x)^3 cos y is an algebraic integer. Is it true that at least one of them has to be a multiple of pi/4? === Subject: Re: An algebraic problem > let x and y be two rational angles verifing the following condition: > 8 (cos x)^3 cos y is an algebraic integer. > Is it true that at least one of them has to be a multiple of pi/4? If x and y are rational, how can one of them be a (non-zero) multiple of pi/4? -- Clive Tooth http://www.clivetooth.dk === Subject: Re: An algebraic problem a rational angle is of the form p Pi with p in Q. Nicola > If x and y are rational, how can one of them be a (non-zero) multiple of > pi/4? === Subject: Re: An algebraic problem > a rational angle is of the form p Pi with p in Q. === Subject: Re: When laying block, better than string for straightness > I am building my own concrete block garage. And when it came time to > lay the first course I did not like the string method for it depends too > much > on eye judgement. The string method is the absolute best way to make garages. Since that's the way you make garages that aren't made out of concrete blocks. If you building a prison, the pipe method is preferred. So what I did was haul out two very long and stout > plumbing pipes. Very long stiff and firm and layed the block loosely for > the first row > and then instead of string and eyeball I simple rolled the two pipes, > one on the inside of the row and one on the outside of the row to make a > perfect line. > That method is great for the first course because the pipes are on the > concrete slab. > Now, I am trying to figure a way to use the pipe method going up the > wall > so that I never have to use the time wasting and imprecise string > method. Sorry you can't do it. To get perfect vertical lines, you need a plumb bob, which use strings. > Thought I might share that with you so that we do replace the old string > method with something easier, faster and better. A method where the test > uses the material instead of a eyeball judgement call electron-dot-cloud are galaxies === Subject: Re: When laying block, better than string for straightness (snipped) > Sorry you can't do it. To get perfect vertical lines, > you need a plumb bob, which use strings. That is a very silly and stupid device to be using for concrete and brick wall building. I say that because a fixed stick or rod is a faster means of telling the accuracy of vertical lines than a plumb bob setup. Just picture the time wasted in setting up that plumb bob when a fixed stick or rod and determine the vertical accuracy in a split second. My method does use a level to check upon things. But other than a level, the accuracy of vertical lines and horizontal lines should always be that of *fixed sticks or rods* that are straight. And never on string things which rely upon a eyeball judgement. Obviously the above has never done hands on building, just mouth chatter. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: When laying block, better than string for straightness > (snipped) > Sorry you can't do it. To get perfect vertical lines, > you need a plumb bob, which use strings. > That is a very silly and stupid device to be using for concrete and brick wall > building. That's why we who know what they are don't use them for making walls, are somehing stupid as concrete. We use them for making plumbs lines. I say that because a fixed stick or rod is a faster means of telling the accuracy of vertical > lines than a plumb bob setup. I did'nt say anything about faster. I mentioned something about *vertical lines*, which is what your question was about. Just picture the time wasted in setting up that plumb bob when a fixed > stick or rod and determine the vertical > accuracy in a split second. There are no such things as fixed sticks. Either you've been readng something about Einstein, or some first grade mathematicians curicculum. Either way is doesn't matter, since neither can actually build anything other than playground balls. > My method does use a level to check upon things. But other than a level, the accuracy of vertical lines > and horizontal lines should always be that of *fixed > sticks or rods* that are straight. They would make sense if a level wasn't just an amatuer plumb bob. And never on string things which rely upon a eyeball judgement. > Obviously the above has never done hands on building, just mouth chatter. That's true. Those of us who actually know how to build walls, only design them. We let day laborers, such as yourself, build them. > Archimedes Plutonium, a_plutonium@hotmail.com > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies electron-dot-cloud are galaxies === Subject: Re: When laying block, better than string for straightness > Seems like alot of work would it not be easier to buy two cheap laser > pointers mount two to form a 90 deg angle, one pointing down one pointing in > the direction of the row of bricks. Mark the ground so the pointers are in > the same location on the first brick and line the bricks up against the > lightr pointing down the row you could even mount a bubble level on them to > make sure it's level Actually long 2 X 4 works even better than steel pipes. No, what I am trying to achieve is a quick way to know that all the block or brick in a row are in a perfect line. And where the string method relies on eyeball judgement is bad. And so will the laser. A long straight wood board down the line of a row of block where you touch the block to the board and you know you are on that perfect line. === Subject: Re: When laying block, better than string for straightness > No, what I am trying to achieve is a quick way to know that all the block or > brick in a row are in a perfect line. And where the string method relies on > eyeball > judgement is bad. And so will the laser. > A long straight wood board down the line of a row of block where you touch the > block to the board and you know you are on that perfect line. Arch, If you're actually doing the block work yourself, 'buttering` the block, dropping it in place, tapping it level and plumb with your trowel, you know that your mortar joint should be between 3, and 5 sixteenths of an inch thick. With an eighth of an inch to play with, 'Perfect` is a big waist of your time and effort, and any solid batterboard you try to use will just get in your way and slow you down even more. This is demonstrated by the fact that you've apparently laid only one course of block in a day. If you're an amateur, merely staying within tolerances will be hard enough. Just use the proven method and get on with it. More than one project has been 're-thought` to death. Best of luck - Pragmatist -Use a bigger hammer electron-dot-cloud are galaxies === Subject: Re: When laying block, better than string for straightness > No, what I am trying to achieve is a quick way to know that all the block or > brick in a row are in a perfect line. And where the string method relies on > eyeball > judgement is bad. And so will the laser. > A long straight wood board down the line of a row of block where you touch the > block to the board and you know you are on that perfect line. > Arch, > If you're actually doing the block work yourself, 'buttering` the > block, > dropping it in place, tapping it level and plumb with your trowel, you > know > that your mortar joint should be between 3, and 5 sixteenths of an > inch thick. > With an eighth of an inch to play with, 'Perfect` is a big waist of > your > time and effort, and any solid batterboard you try to use will just > get in your way and slow you down even more. > This is demonstrated by the fact that you've apparently laid only one > course of block in a day. > If you're an amateur, merely staying within tolerances will be hard > enough. Just use the proven method and get on with it. > More than one project has been 're-thought` to death. > Best of luck - > Pragmatist -Use a bigger hammer Doing it myself in order to learn and improve the task. This industry has never been wholescale improved and that is where I can do the most good. For example, the old true and tested way is to use string and to slop alot of mortar to build the building. Cement is never going to get cheaper and abundant because fusion energy is never coming. So cement is going to become as rare and valuable as is petroleum. Instead of using a trowel to place cement I believe in using the fingers. Not because I am an amateur bricklayer but because I need to find the method that saves every precious piece of Portland cement. Of course using latex gloves and not exposed skin. For placement of mortar for the vertical joints that is easy. For the horizontal joints is a little bit more tricky. What I do is get a 3/8 steel rebar and wire connect it to the vertical rebar and then rest the next course with its *inner edge* onto the rebar. Thus I have a uniform horizontal guide. Then with the latex-glove, never a trowel, I fill in the gaps and with a pointer I then solidly compact the joint. No matter how great one gets at a troweling method they lose and slop too much and waste too much cement. In my method, which is the method of the future as oil prices rise so does cement. In my method, although painstakingly longer is perhaps the ultimate method of laying brick and block. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies electron-dot-cloud are galaxies === Subject: Re: When laying block, better than string for straightness In the old method, mortar had to be a mix of lime for consistency in order to make the mix pliable to use a troweling method. So that a standard mortal mix would be something like 1 part mortar-cement which is 50-50 portlandcement and lime and then 2 to 3 parts sand. In my method we dispense with ever using any lime. We go straight with a concrete type mix without aggregates because we simply use latex gloves finger in the mortar into the joints. In the future as oil prices become very high, and cement becomes very high, the luxury of using lime and wasting alot of mortar is a luxury not in the future. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Catalan sequence problem > OK, this sounds close to noncrossing partitions. Which are yet > another Catalan counted structure (with all sorts of bijections with > other Catalan structures) problem pp from Stanley (with only > references, no explanation)). > However, the cyclic nature of your description might make a difference. > But then it should show up with n = 5. Er... no it doesn't (rethinking > the definitions), because the partitions can enclose another without > crossing: e.g. {{1,2,5},{3,4}} is not crossing. > I'm thinking about what the bijection really should be (I have seen it > before but can't remember, and I'm having trouble with obvious > possibilities). Maybe DFS labelling on an ordered tree? After talking to somebody, we figured it out. It's quite simple (afterwards of course!). Take a binary tree on n nodes (not necessarily full), and label the vertices according to inorder traversal. then remove all left edges. the remaining components form a partition, which, by the inorder labeling, is non-crossing. To go the other direction, take the part (of the partition) which includes n, create a rightmost path labeled increasing with these elements. All the parts between n and the label of its parent (the next highest in the part that n is in) can be recursively made into a tree to the left of n's parent. One can also create noncrossing partitions from first principles that results in the Catalan recurrence C_n = sum(C_k C_{n-k-1}, {n,0,n-1}), C_0 = 1. disjoint union a partition of 1 thru k and k+1 thru n-1, then put n in the same partition as k (when k is 0, it goes in a partition itself). This construction insures that it is noncrossing (this latter bit is not clear until you do some examples). Mitch === Subject: polynomial gcd Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of x and gcd (m,n) ? === Subject: Re: polynomial gcd Here's a hint. If x^a-1 divides x^b-1, then a divides b... You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: polynomial gcd > Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of x and gcd (m,n) ? x^gcd(m,n) -1, using Euclid's algorithm. === Subject: Re: polynomial gcd >>Express the polynomial gcd of (x^m - 1,x^n - 1) in terms of x and >>gcd (m,n) ? > x^gcd(m,n) -1, using Euclid's algorithm. ... and by extending that argument gcd(x^n - y^n, x^m - y^m) = x^gcd(n,m) - y^gcd(n,m) so, what is the next generalization? It is very reminiscent of the identity gcd(F_n,F_m) = F_gcd(n,m) where F_n is a Fibonacci number (not just reminiscent, it follows the pattern above because of Binet's formula F_n = (sigma^n - tau^n)/sqrt(5) ) So is it something about linear recurrences and divisibility? restricted bases cases? I can't seem to get an example that works where there are 3 bases (x^n - y^n - z^n) (I kind of see why it is impossible, but it is not clear enough to me yet how to be more rigorous about it, or how to extend that disproof to others) What's the generalization? Mitch === Subject: Geometry: Perimeter & Number of Shapes Possible Is there an easy way (without drawing) to determine the number of straight shapes that are possible to create with a certain perimeter value? For example, say that there are only three shapes (not rounded or with diagonal lines) that can possibly be created that have a perimeter of eight, is there an easy way to determine the number of shapes that can be made with a perimeter of say 100 or even more without drawing them (which would obviously be time consuming)? === Subject: Re: Geometry: Perimeter & Number of Shapes Possible > Is there an easy way (without drawing) to determine the number of > straight shapes that are possible to create with a certain perimeter > value? For example, say that there are only three shapes (not rounded > or with diagonal lines) that can possibly be created that have a > perimeter of eight, is there an easy way to determine the number of > shapes that can be made with a perimeter of say 100 or even more > without drawing them (which would obviously be time consuming)? If you don't constrain edge numbers, lengths, and angles to finite sets, there will be uncountably many shapes. For example, there are uncountably many different triangles with a perimeter of 1. You need to define the notions of shape, difference, and perimeter; indicate if paths must be continuous; and how crossings or inclusions are treated, before you will get agreement on numbers of shapes. -jiw === Subject: Re: CAN'T Order Reals I don't see your point. As far as I know, counting & establishing a total order are different things. Counting a set is the same as constructing a infinite sequence in it, & the total order <= is a equivalence relation, ie * reflexive: a <= a * antisymmetric: a <= b and b <= a ---> a = b * transitive: a <= b and b <= c ---> a <= c that satisfies, for all a,b: a<=b or b<=a. So, I don't see why you claim that total orders c counted orders. The usual order in R is somewhat inherited from Q, if you construct R from Q-Cauchy sequences. And the order in Q can be stated as: a/b <= c/d iff a*d <= b*c There's no place for counting here, and the density of Q and R does not contradict that <= is a total order. Sorry if I'm too lost, please explain me again, if I missed the point. -- DvD Gomez === Subject: Re: CAN'T Order Reals > I don't see your point. As far as I know, counting & establishing a total > order are different things. Counting a set is the same as constructing a > infinite sequence in it, & the total order <= is a equivalence relation, ie > * reflexive: a <= a > * antisymmetric: a <= b and b <= a ---> a = b > * transitive: a <= b and b <= c ---> a <= c > that satisfies, for all a,b: a<=b or b<=a. > So, I don't see why you claim that total orders c counted orders. > The usual order in R is somewhat inherited from Q, if you construct R from > Q-Cauchy sequences. And the order in Q can be stated as: > a/b <= c/d iff a*d <= b*c > There's no place for counting here, and the density of Q and R does not > contradict that <= is a total order. > Sorry if I'm too lost, please explain me again, if I missed the point. That's correct, I meant : properties of total order C properties of sequences w.r.t. countable sets regarding total order is a specialisation of the term order my point is order in maths is a vector field, in English it means put one before the other. Cantors proof has nothing to do with vectors so CANT Order Reals is an apt name. Herc === Subject: Rodrigues-like formulas, F,L numbers ? = n is a positive integer a=a(n)= n!/(2n-1)! , b=b(n)= 2*n!/(2n)! h(x)=sqrt(x^2+4):=(x^2+4)^{0.5} H(x)=H_n(x)=h^{2n-1}(x)=(x^2+4)^{n-0.5} f_n(x)=a*H^{(n-1)}(x)/h(x) ,l_n(x)=b*h(x)*H^{(n)}(x) . By H^{(k)} is denoted the k-th derivative . It's true that f_n(1) and l_n(1) are integer numbers ? If possible , please recognize f_n(1) and l_n(1) as special numbers. = === Subject: Mahler measure of a polynomial I have been unable to write a satisfactory Maple program to compute the Mahler measure of a polynomial. If anyone has an efficient Maple program for computing this number and would be willing to share it with me I most appreciate it. C.Nicol === Subject: Re: Multitude of infinities. >> Allowing an infinite cardinality would mean that we'd have aleph >> infinity. Taking the power set would give us aleph infinity + 1. But >> infinity + 1 and infinity are the same number. > No. If the cardinality of a set were aleph infinity, the cardinality > of its power set would be 2^(aleph infinity)which is strictly greater > than aleph infinity. Just to clarify, the subscripts in the sequence of alephs are the ordinals. The least upper bound of the aleph_n for n in the natural numbers is aleph_omega (omega being the smallest infinite ordinal). The next cardinal after aleph_omega is aleph_{omega + 1} (the GCH asserts that that is 2^(aleph_omega)). Of course, omega and omega+1 are different ordinals. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Finding a basis continuously Given a N-by-N matrix X of rank M (not necessarily full rank). Think of it as an array of N-column vectors {x1, .., xN}. Can you find a basis for the space spanned by {x1, .., xN} continuously? That means the algorithm to compute the basis should depend continuously in {x1, .., xN}. Bernard === Subject: Re: Finding a basis continuously >Given a N-by-N matrix X of rank M (not necessarily full rank). Think of it >as an array of N-column vectors {x1, .., xN}. Can you find a basis for the >space spanned by {x1, .., xN} continuously? That means the algorithm to >compute the basis should depend continuously in {x1, .., xN}. No. Consider the case N=2, M=1. Suppose there is a continuous function f from the set of 2x2 matrices of rank 1 to nonzero vectors such that f(A) is a basis for the columns of A. Since x -> x/|x| is continuous on nonzero vectors, we can assume f(A) is always a unit vector. Let e(theta) be the column vector [ cos(theta) ] [ sin(theta) ] and think of the matrices as pairs of column vectors. WLOG f(e(0),e(0)) = e(0). By continuity, we must have f(e(theta),e(theta)) = e(theta). But then again by continuity, f(t e(0), e(0)) = e(0) for all t. In particular, f(-e(0),e(0)) = e(0). And then again by continuity, f(-e(0),t e(0)) = e(0) for all t so that f(-e(0),-e(0)) = e(0). But -e(0) = e(pi), and we must have f(e(pi),e(pi)) = e(pi), not e(0), contradiction. I suspect algebraic topology may have something to say about the general case (for N > M). 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: Combinatorics in Bridge > Here's an approach from a different angle: The task was very ungreatfull, though... -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: Mathematical induction and the use of n and k (?) >Hi all, >After reading some of the mathematical induction problems, I tried to >remember why it is necessary to replace the n with k in the induction step. >Can somebody help me? Is it that k in N is supposed to be a particular >element; whereas, n in N is meant as a general element. How is this >described in logic? >>As you are using it, N is a constant, the set of integers. >>The idea that one should even use mnemonic letters for >>variables is greatly to be decried. >I'd be interested in your reasons why. ISTM that mnemonic variable names >are common precisely because they *are* helpful. >> It makes no difference >>if one uses n or k or q or z or T or alpha for a variable >>symbol. Sometimes, more than one needs to be used. >There are well-established coventions for variable names, presumably as >memory aids: and in limit proofs, for an angle, >x and y for reals, z for complex, etc. >Although formally the variable names make no difference, well-chosen ones >certainly do seem to aid readability. They may or may not aid readability. Unless the person KNOWS that the particular variable names are unimportant, there is likely to be the assumption that one needs to use particular names for particular types of variables. This seems to be exactly the type of confusion of the original poster; look at the penultimate statement in the segment quoted. The general use of variables needs to come first to avoid the student ever running into this problem. Only then can one use the simplifications. -- 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: Mathematical induction and the use of n and k (?) >>Although formally the variable names make no difference, well-chosen ones >>certainly do seem to aid readability. >They may or may not aid readability. Unless the person KNOWS >that the particular variable names are unimportant, there is >likely to be the assumption that one needs to use particular >names for particular types of variables. This seems to be >exactly the type of confusion of the original poster; look at >the penultimate statement in the segment quoted. It has always bugged me that in recursion theory there is an important theorem called the SNM theorem, where N and M are just names of indices. -- Daryl McCullough Ithaca, NY === Subject: Re: Mathematical induction and the use of n and k (?) >Although formally the variable names make no difference, well-chosen ones >certainly do seem to aid readability. >>They may or may not aid readability. Unless the person KNOWS >>that the particular variable names are unimportant, there is >>likely to be the assumption that one needs to use particular >>names for particular types of variables. This seems to be >>exactly the type of confusion of the original poster; look at >>the penultimate statement in the segment quoted. >It has always bugged me that in recursion theory there is an important >theorem called the SNM theorem, where N and M are just names of indices. There is also the abc (or is it ABC ?) conjecture in number theory, where the letters stand for the variables. -- 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: Mathematical induction and the use of n and k (?) >Although formally the variable names make no difference, well-chosen ones >certainly do seem to aid readability. >>They may or may not aid readability. Unless the person KNOWS >>that the particular variable names are unimportant, there is >>likely to be the assumption that one needs to use particular >>names for particular types of variables. This seems to be >>exactly the type of confusion of the original poster; look at >>the penultimate statement in the segment quoted. >It has always bugged me that in recursion theory there is an important >theorem called the SNM theorem, where N and M are just names of indices. And there's alpha-beta pruning of game trees, where alpha and beta are just the names of cutoff values. -- --------------------------- | BBB b barbara minus knox at iname stop com | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- electron-dot-cloud are galaxies === Subject: theoretically the strongest concrete 1.6 cement :: 3.2 sand :: 6 gravel 5 cement :: 12.5 sand :: 20 aggregate or gravel For mortar I get this as a mix ratio: 1 portland cement :: 1 lime :: 4 to 6 parts sand What I am wondering is like a checkerboard or chess board where the squares are each one grain of sand. And that the strongest and ideal mix to make both Concrete and to make Mortar is simple this ratio: 1 portland cement to 1 of sand portland cement between them. Trouble is getting a mix so that it is perfectly mixed as to result in a 3-dimensional perfect packing such as a chessboard is a perfect 2 dimensional packing. Has anyone done experimental tests to see at what ratio of portlandcement to sand yields the strongest most durable concrete?? I would hazard to guess that if uniformly mixed that the 1 to 1 mix is the strongest. Is there any proof to my above assertion? Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: theoretically the strongest concrete I would imagine that civil engineers have studied topic this just about to temperature history of the cement, amount of water available, etc. Michael > 1.6 cement :: 3.2 sand :: 6 gravel > 5 cement :: 12.5 sand :: 20 aggregate or gravel > For mortar I get this as a mix ratio: > 1 portland cement :: 1 lime :: 4 to 6 parts sand > What I am wondering is like a checkerboard or chess board where the > squares are each one grain of sand. And that the strongest and ideal mix > to make both Concrete and to make Mortar is simple this ratio: > 1 portland cement to 1 of sand > portland cement between them. Trouble is getting a mix so that it is > perfectly mixed as to result in a 3-dimensional perfect packing such as > a chessboard is a perfect 2 dimensional packing. > Has anyone done experimental tests to see at what ratio of > portlandcement to sand yields the strongest most durable concrete?? > I would hazard to guess that if uniformly mixed that the 1 to 1 mix is > the strongest. > Is there any proof to my above assertion? > Archimedes Plutonium, a_plutonium@hotmail.com > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies === Subject: Re: theoretically the strongest concrete > I would imagine that civil engineers have studied topic this just about to > temperature history of the cement, amount of water available, etc. well, they have, but as any empirical science (sorry, I just had to write that) there is still a lot of room for improvement. Unfortunately i don't know the right english words for the coming remarks; i hope it is still understandable. Often highest strength is not the aim, but early strength to allow quick formwork removal, or low porosity (high density) to make a barriers for ground water protection etc. Concretes are optimized to set under suboptimal circumstances like low/hi temperatures, short times in formwork, low viscosity, very long/short pot life. The posed 1:1 optimal concrete theory has no basis... a sand grain is a sand grain is a sand grain...silicon dioxide... a dead dog.. the cement however is more or less reactive depending on where it comes from, it is not a defined substance.. pure CaO has completely different setting behaviour than fly ash cements. Silica fume, plasticizers and thousands of other additives make this a real witchcraft system. So no 1:1 ratio, neither m:m, nor v:v or mol:mol have the slightest chance of being a general rule, let alone specify some concrete mixture. The quality of the concrete widely varies even with the same cement/sand ratio (and using the exact same materials in different experiments) with different water/cement ratios. Depending on water content of the sand, its specific surface, the quality of the cement etc., the brickie on the building site has to slightly modify the recipe every day to just get the ideal, same consistence. electron-dot-cloud are galaxies === Subject: Re: theoretically the strongest concrete > I would imagine that civil engineers have studied topic this just about to > temperature history of the cement, amount of water available, etc. > well, they have, but as any empirical science (sorry, I just had to > write that) there is still a lot of room for improvement. > Unfortunately i don't know the right english words for the coming > remarks; i hope it is still understandable. > Often highest strength is not the aim, but early strength to allow > quick formwork removal, or low porosity (high density) to make a > barriers for ground water protection etc. Concretes are optimized to > set under suboptimal circumstances like low/hi temperatures, short > times in formwork, low viscosity, very long/short pot life. > The posed 1:1 optimal concrete theory has no basis... a sand grain is > a sand grain is a sand grain...silicon dioxide... a dead dog.. the Well yes, there is a direct link with theory and with analysis. We have the Kepler Packing to be applied to concrete and to see if the Kepler Packing is a better concrete than the theory of 1 :: 1 mix of a 3-d chessboard mix. I forgotten the 3-d Kepler Packing percent of voids. Was it somewhere around 23% voids? Let us say it was. So that a Kepler Packing Concrete mix would not be a 1 :: 1 mix but a 23% :: 77% mix or a 1 :: 3 mix. > cement however is more or less reactive depending on where it comes > from, it is not a defined substance.. pure CaO has completely > different setting behaviour than fly ash cements. Silica fume, > plasticizers and thousands of other additives make this a real > witchcraft system. So no 1:1 ratio, neither m:m, nor v:v or mol:mol > have the slightest chance of being a general rule, let alone specify > some concrete mixture. > The quality of the concrete widely varies even with the same > cement/sand ratio (and using the exact same materials in different > experiments) with different water/cement ratios. > Depending on water content of the sand, its specific surface, the > quality of the cement etc., the brickie on the building site has to > slightly modify the recipe every day to just get the ideal, same > consistence. Experiment: It should be easy to experiment as to whether a Kepler Packing concrete mix is superior to a 1 :: 1 chessboard mix. In that we get marble sized balls and Kepler pack them in wet cement. We then pack some marbles in a configuration of a chessboard where the marbles and cement are in a 1 :: 1 ratio. We wait for the two samples to dry and harden. We then test them for superiority of one over the other. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: theoretically the strongest concrete Once upon a time, Herman Family said: >I would imagine that civil engineers have studied topic this just about to >temperature history of the cement, amount of water available, etc. I can't imagine the answer being independent of the actual sources of crumb rubber as an asphalt filler, for instance, it was found that rubber that was ground after cryogenic treatment resulted in a drastically less strong compound than rubber that was ground at room temperatures. This was determined to be because the cryogenically treated crumb had shear surfaces, like faceted gems, which provided less surface area to grip its binder than the more complex surfaces of the other crumb. If different sources of sand produce significantly different compounds made of them would have different physical properties. surface texture varies or not. Getting back to the original question, I think it would depend on exactly what kinds of strength was needed from the concrete compound. Adding different things to the mix (straw, dung, sawdust, rubber, steel rebar, hardware cloth, etc) can improve the strength/density ratio, or the resilience, or the total blunt trauma energy absorbed per unit mass of the compound, albeit possibly at the detriment of hardness or absolute flexural, shear, or tensile strength. Even if the concrete is only used to support a load which is static most of the time (eg, the foundation of a building), there may be other factors which complicate things, like occasional dynamic loading (Californian earthquakes), or water erosion (rain), etc. There is no single best formula for everything. -- TTK === Subject: Re: theoretically the strongest concrete Sharp sand and aggregate makes much stronger concrete than smooth sand and aggregate. Gordon > Once upon a time, Herman Family said: >I would imagine that civil engineers have studied topic this just about to salt, >temperature history of the cement, amount of water available, etc. > I can't imagine the answer being independent of the actual > sources of crumb rubber as an asphalt filler, for instance, it > was found that rubber that was ground after cryogenic treatment > resulted in a drastically less strong compound than rubber that > was ground at room temperatures. This was determined to be > because the cryogenically treated crumb had shear surfaces, > like faceted gems, which provided less surface area to grip its > binder than the more complex surfaces of the other crumb. > If different sources of sand produce significantly different > compounds made of them would have different physical properties. > surface texture varies or not. > Getting back to the original question, I think it would depend > on exactly what kinds of strength was needed from the concrete > compound. Adding different things to the mix (straw, dung, > sawdust, rubber, steel rebar, hardware cloth, etc) can improve the > strength/density ratio, or the resilience, or the total blunt > trauma energy absorbed per unit mass of the compound, albeit > possibly at the detriment of hardness or absolute flexural, shear, > or tensile strength. Even if the concrete is only used to support > a load which is static most of the time (eg, the foundation of a > building), there may be other factors which complicate things, > like occasional dynamic loading (Californian earthquakes), or > water erosion (rain), etc. There is no single best formula for > everything. > -- TTK === Subject: Re: theoretically the strongest concrete NNTP-Proxy-Relay: library1-aux.airnews.net What you are asking isn't as easy as it sounds. Concrete is not like a checkerboard, where all the squares are the same size and shape. Concrete is a combination of rock, sand, cement and water, all of varying sizes and shapes (or amorphous). Think of it this way - in woodworking the strongest glued joint is the thinnest one. Concrete is sort of the same way. Water surrounds the concrete, and you want the least amount of mortar. You could theoretically calculate all this for spheres, or using computer simulations, but the National Institute of Standards and Technology uses super-computers to do it, and is just now, after about 10 years, getting to the point where they can do it. Actually, their program doesn't optimize the concrete, it just tries to predict what the specified concrete will do. If you are familiar with the subject, you might comment on my question week. Jay Shilstone >1.6 cement :: 3.2 sand :: 6 gravel >5 cement :: 12.5 sand :: 20 aggregate or gravel >For mortar I get this as a mix ratio: >1 portland cement :: 1 lime :: 4 to 6 parts sand >What I am wondering is like a checkerboard or chess board where the >squares are each one grain of sand. And that the strongest and ideal mix >to make both Concrete and to make Mortar is simple this ratio: >1 portland cement to 1 of sand >portland cement between them. Trouble is getting a mix so that it is >perfectly mixed as to result in a 3-dimensional perfect packing such as >a chessboard is a perfect 2 dimensional packing. >Has anyone done experimental tests to see at what ratio of >portlandcement to sand yields the strongest most durable concrete?? >I would hazard to guess that if uniformly mixed that the 1 to 1 mix is >the strongest. >Is there any proof to my above assertion? >Archimedes Plutonium, a_plutonium@hotmail.com >whole entire Universe is just one big atom where dots >of the electron-dot-cloud are galaxies electron-dot-cloud are galaxies === Subject: Re: theoretically the strongest concrete > What you are asking isn't as easy as it sounds. Concrete is not like a > checkerboard, where all the squares are the same size and shape. > Concrete is a combination of rock, sand, cement and water, all of > varying sizes and shapes (or amorphous). It should be easy to research and find the answer. One easily can get a uniform grade of sand and see if a 1 :: 1 mix is stronger than say a 1:: 2 mix and various other mixes. > Think of it this way - in woodworking the strongest glued joint is the > thinnest one. Concrete is sort of the same way. Water surrounds the > concrete, and you want the least amount of mortar. You could > theoretically calculate all this for spheres, or using computer > simulations, but the National Institute of Standards and Technology > uses super-computers to do it, and is just now, after about 10 years, > getting to the point where they can do it. Actually, their program > doesn't optimize the concrete, it just tries to predict what the > specified concrete will do. What is intriguing about the strongest concrete is that it is sort of like another Kepler Packing Problem where the spaces between the balls is filled with portland cement. So, in the Kepler Packing Problem in 3-d, each sphere is surrounded by how many gaps? Perhaps the strongest concrete is a direct result of the Kepler Packing Problem. > If you are familiar with the subject, you might comment on my question > week. > Jay Shilstone I proved the Kepler Packing Problem as that of kissing points. sizes, this new problem is also solvable once you factor in the adjustment of kissing points. For example, say you have 60 balls of which 1/3 are one size, and 1/3 another and the last 1/3 another, then once you factor in the new kissing points to determine the smallest packing, well, you are on the way to a generalized formula for variable packing sizes. As for Concrete and the maximum strength, I believe it is somewhere near a 1 :: 1 mix, where 1 is portland cement and the 1 part sand is of uniform size and where the mix is so well mixed. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: theoretically the strongest concrete for uniform-ball aggregate, teh cement is the complement to the volume of the ball, inscribed within a rhombical dodecahedron; I think it's got pi and 18 in the ratio, as per Kepler's problem, about a fifth left-over. (congratulations on your proof via kissing; you've beat Hale, at last .-) I wasn't aware taht sharp sand is better, either. however, assigning a simple 1/3 ratio of various sizes will not be easy -- no easier than Kepler's problem. > What is intriguing about the strongest concrete is that it is sort of like > another Kepler Packing Problem where the spaces between the balls is filled > with portland cement. > So, in the Kepler Packing Problem in 3-d, each sphere is surrounded by how > many gaps? > I proved the Kepler Packing Problem as that of kissing points. > sizes, this new problem is also solvable once you factor in the adjustment > of kissing points. For example, say you have 60 balls of which 1/3 are one > size, and 1/3 > another and the last 1/3 another, then once you factor in the new kissing > points to determine the smallest packing, well, you are on the way to a > generalized formula > for variable packing sizes. --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: theoretically the strongest concrete > It should be easy to research and find the answer. One easily can get a > uniform grade of sand and see if a 1 :: 1 mix is stronger than say a 1:: 2 > mix and various > other mixes. As I understand it, the cement is much smaller than the sand, and chemically reacts with the water as it solidifies. The actual material giving strength to the material is the sand and gravel, the cement The use of steel rebar is a common way to strengthen the concrete (concrete is an ideal environment for normal steels as long as salts aren't present), but fiberglass, rubber, and perhaps other materials can also add useful characteristics to the concrete (respectively, resistance to crack formation and flexibility). Water in the original mix is bad. You've probably already noted that the recipes state to use as little water as possible to wet the mixture. There are surfactants that can reduce the amount of water required for your concrete. Reducing water in the mix is another way to strengthen the concrete since little water pockets in the concrete apparently create flaws in the material. The mix ratios you mention have been tried and tested IMHO and are probably really close to being optimal. Incidentally, when I looked on the Internet for details on small batch preparation of concrete, I found a lot of useful information from instructions for small house projects to art-related websites. You should be able to find good information on small-scale concrete projects if that's your inclination. Karl Hallowell khallow@hotmail.com electron-dot-cloud are galaxies === Subject: Kepler Packing on concrete mixes Re: theoretically the strongest concrete > uniform grade of sand and see if a 1 :: 1 mix is stronger than say a 1:: 2 > mix and various > other mixes. > chemically reacts with the water as it solidifies. The actual material > giving strength to the material is the sand and gravel, the cement > The use of steel rebar is a common way to strengthen the concrete > (concrete is an ideal environment for normal steels as long as salts > aren't present), but fiberglass, rubber, and perhaps other materials > can also add useful characteristics to the concrete (respectively, > resistance to crack formation and flexibility). Okay, it is hard to get a uniform sand mix. So let us do a research on something that is perfectly uniform. That of steel balls of the size of marbles and BBs. Where the marbles are gravel and the BBs are sand. Now the question is, what mix ratio of Portland cement to make a cubic foot of concrete that has marbles and BBs and is the strongest cubic foot. I still suspect the final answer to be a cubic foot wherein only BBs are used and the ratio is close to a 1 :: 1 in terms of volume. Say perhaps 10 liters of portland cement and 10 liters of BBs. Trouble with various sizes of aggregate is that it allows air pockets which diminish strength and cause cracking. So I am guessing that the strongest mix ratio is a 1 to 1 ratio using only BBs and portland cement. And I am guessing that the perfect piece of concrete is one in which there is a Kepler Packing and where the portland cement is in between the voids of the Kepler Packing. I wonder if someone can arrange for a Kepler Packing of steel BBs and infuse the crevasses between the BBs or take marble size steel balls and Kepler Pack them and then infuse the interstatials with portlandcement. I have the hunch that such a block of concrete would be the strongest such block over any other such block having a different arrangement of those BBs and marbles. And obviously the ratio is no longer a 1 :: 1 because if memory serves me (mathematician would know) that in a Kepler Packing that the gaps are something like 23% of the volume and so the ration of cement to balls would be closer to that of 1 :: 4 Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Kepler Packing on concrete mixes Re: theoretically the strongest concrete So let us do a research on something that is perfectly uniform. That of steel balls of the size of > marbles and BBs. Where the marbles are gravel and the BBs are sand. > Now the question is, what mix ratio of Portland cement to make a cubic foot of concrete that has marbles > and BBs and is the strongest cubic foot. > I still suspect the final answer to be a cubic foot wherein only BBs are used and the ratio is close to > a 1 :: 1 in terms of volume. Say perhaps 10 liters of portland cement and 10 liters of BBs. > Trouble with various sizes of aggregate is that it allows air pockets which diminish strength and cause > cracking. Actually, I'm not sure that this is the case. I can see that the sand would fill in the gaps between the larger aggregate, and the cement in turn would fill in the gaps between the sand. OTOH, you probably would get more air than with sand-sized aggregate alone. Also, would a marble sized glob of BB's cemented together be stronger than a steel marble of the same size? I think not. It indicates to me that the large aggregate is probably responsible for most of the strength (perhaps in some random almost kissing structure) with the spaces filled by cement and sand. > So I am guessing that the strongest mix ratio is a 1 to 1 ratio using only BBs and portland cement. > And I am guessing that the perfect piece of concrete is one in which there is a Kepler Packing and where > the portland cement is in between the voids of the Kepler Packing. > I wonder if someone can arrange for a Kepler Packing of steel BBs and infuse the crevasses between the > BBs or take marble size steel balls and Kepler Pack them and then infuse the interstatials with > portlandcement. I have the hunch that such a block of concrete would be the strongest such block over > any other such block having a different arrangement of those BBs and > marbles. And obviously the ratio is no longer a 1 :: 1 because if memory serves me (mathematician would > know) that in a Kepler Packing that the gaps > are something like 23% of the volume and so the ration of cement to balls > would be closer to that of 1 :: 4 This seems more reasonable to me. I notice that there appears to be a sparseness of the larger stronger stuff (by volume). Ie, the large gravel should be 4 over the smaller stuff and cement combined. Instead it's slightly better than 1. The sand to cement ratio is a little over 2-1 with pure sand in cement around 3 to 2. I see a couple of reasons this theoretical limit isn't reached. First, the concrete when poured needs to be sufficiently fluid to be usable (and other people have mentioned the various other attributes that usable concrete needs depending on the application). That appears to be a significant constraint and may be the sole reason for the low ratios of large aggregate to the space filling material. Second, the material as you note is usually randomly packed. I don't know how inefficient that is. One way to improving the packing would be to compress the concrete while it's still wet. That would force the aggregate into a more optimal packing, and squeeze out air and perhaps some of the water. I seem to recall that this is done for some applications (and the weight of a large amount of concrete tends to compress the buried sections). Karl Hallowell khallow@hotmail.com === Subject: Re: Kepler Packing on concrete mixes Re: theoretically the strongest concrete it may have to do with Kepler's problem, but a) using BBs will weaken it (too heavy), and b) aerocrete is actually stronger than that using sand: the aggregate has no tensional strength, between any two chunks. the thing about Portland Cement is that it's stronger, the longer it takes to dry -- so keep dousing it. > Trouble with various sizes of aggregate is that it allows air pockets which diminish strength and cause > cracking. > So I am guessing that the strongest mix ratio is a 1 to 1 ratio using only BBs and portland cement. > And I am guessing that the perfect piece of concrete is one in which there is a Kepler Packing and where > the portland cement is in between the voids of the Kepler Packing. > I wonder if someone can arrange for a Kepler Packing of steel BBs and infuse the crevasses between the > BBs or take marble size steel balls and Kepler Pack them and then infuse the interstatials with > portlandcement. I have the hunch that such a block of concrete would be the strongest such block over > any other such block having a different arrangement of those BBs and > marbles. And obviously the ratio is no longer a 1 :: 1 because if memory serves me (mathematician would > know) that in a Kepler Packing that the gaps > are something like 23% of the volume and so the ration of cement to balls > would be closer to that of 1 :: 4 --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Kepler Packing on concrete mixes Re: theoretically the strongest concrete NNTP-Proxy-Relay: library1-aux.airnews.net Part of the problem also revolves around the fact that the more water that is added to the concrete, the lower the strength. We do not have a binary (marbles and B-Bs), or a ternary mix (marbles, B-Bs and cement), but a quaternary (4 part) blend of marbles, B-Bs, cement and water. Studies have shown that we can reduce water by getting a good studies have shown that reducing the maximum aggregate size will minimize water demand. I am a concrete technologist, not a mathematician. If you know of a computer program or web model that will predict voids in a Kepler Packing system using a variety of sphere sizes, I would like to know where. Jay Shilstone >> It should be easy to research and find the answer. One easily can get a >> uniform grade of sand and see if a 1 :: 1 mix is stronger than say a 1:: 2 >> mix and various >> other mixes. >> As I understand it, the cement is much smaller than the sand, and >> chemically reacts with the water as it solidifies. The actual material >> giving strength to the material is the sand and gravel, the cement >> The use of steel rebar is a common way to strengthen the concrete >> (concrete is an ideal environment for normal steels as long as salts >> aren't present), but fiberglass, rubber, and perhaps other materials >> can also add useful characteristics to the concrete (respectively, >> resistance to crack formation and flexibility). >Okay, it is hard to get a uniform sand mix. >So let us do a research on something that is perfectly uniform. That of steel balls of the size of >marbles and BBs. Where the marbles are gravel and the BBs are sand. >Now the question is, what mix ratio of Portland cement to make a cubic foot of concrete that has marbles >and BBs and is the strongest cubic foot. >I still suspect the final answer to be a cubic foot wherein only BBs are used and the ratio is close to >a 1 :: 1 in terms of volume. Say perhaps 10 liters of portland cement and 10 liters of BBs. >Trouble with various sizes of aggregate is that it allows air pockets which diminish strength and cause >cracking. >So I am guessing that the strongest mix ratio is a 1 to 1 ratio using only BBs and portland cement. >And I am guessing that the perfect piece of concrete is one in which there is a Kepler Packing and where >the portland cement is in between the voids of the Kepler Packing. >I wonder if someone can arrange for a Kepler Packing of steel BBs and infuse the crevasses between the >BBs or take marble size steel balls and Kepler Pack them and then infuse the interstatials with >portlandcement. I have the hunch that such a block of concrete would be the strongest such block over >any other such block having a different arrangement of those BBs and >marbles. And obviously the ratio is no longer a 1 :: 1 because if memory serves me (mathematician would >know) that in a Kepler Packing that the gaps >are something like 23% of the volume and so the ration of cement to balls >would be closer to that of 1 :: 4 >Archimedes Plutonium, a_plutonium@hotmail.com >whole entire Universe is just one big atom where dots >of the electron-dot-cloud are galaxies === Subject: a question Let B be a field of characteristic <> k,j in N. Is it true that for every p(x), p(x)=kf(x)+jf'(x) for a suitable f(x) in B[x]??? TIA === Subject: Re: a question > Let B be a field of characteristic <> k,j in N. Is it true that > for every p(x), p(x) = kf(x)+jf'(x) for a suitable f(x) in B[x]??? Hint: compute the kernel of f -> k f + j f' -Bill Dubuque === Subject: Re: a question >Let B be a field of characteristic <> k,j in N. Is it true that for every >p(x), p(x)=kf(x)+jf'(x) for a suitable f(x) in B[x]??? Somewhat more generally: it's true if p is in B[x], B is a field, k and j in B with k <> 0. Hint: consider the vector space V_n of polynomials of degree <= n and the linear map T: V_n -> V_n defined by T(f) = k f + j f'. What is its matrix using the standard basis [1, x, ..., x^n]? Show that T is invertible. 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: Jacobi's method : B.S. Jacobi or C.J.G. Jacobi Originator: israel@math.ubc.ca (Robert Israel) Jacobi's iterative method for solving linear systems was discovered (in 1845 ? ) by Boris Semionovici JACOBI ( Russian mathematician) or by German matehmatician Carl Jacob Gustav JACOBI Please help with a reference . === Subject: Re: Jacobi's method : B.S. Jacobi or C.J.G. Jacobi >Jacobi's iterative method >for solving linear systems was >discovered (in 1845 ? ) by > Boris Semionovici JACOBI ( Russian mathematician) >or by German matehmatician > Carl Jacob Gustav JACOBI Carl Gustav Jacob, see below http://www-history.mcs.st-and.ac.uk/~history/Mathematicians/Jacobi.html Boris Semjonowitsch Jakobi, aka Boris Semionovich Yakobi (1801-1874), was the elder brother of Carl Gustav Jacob Jacobi; his name in German is Moritz Hermann Jacobi. In 1835 he got a chair of physics at Dorpat (now Tartu) University (Estonia), and in 1837 he moved to the Russian Academy of Sciences in St. Petersburg. http://chem.ch.huji.ac.il/~eugeniik/history/jacobi.html (at the bottom are a few links to biographies in Russian). http://hp.iitp.ru/ger/32/3284.htm http://hp.iitp.ru/ger/32/3284a.htm >Please help with a reference . C. G. Jacobi: .86ber eine neue Aufl.9asungsart der bei der Methode der kleinsten Quadrate vorkommenden linearen Gleichungen. [On a new way of solving the linear equations arising in the method of least squares]. Astronomische Nachrichten Vol. 22 (1845), Issue No. 523. It may also be found in Jacobi's Collected Works in 7 Vols. Berlin: R. Reimer 1881-1891 in Vol. 3, p. 467-478. All 7 volumes are online at the Biblioth.8fque Nationale de France in Paris: http://gallica.bnf.fr --> Recherche --> Auteur Jacobi but currently the server is partially down due to update and maintainance. Here are some more references: http://www.numerik.uni-kiel.de/~in/PAPER/Diplom/node43.html --> Jacobi45 http://www.cs.umd.edu/TRs/authors/C_G_Jacobi.html --> CS-TR-2877, but the full paper seems to be no more available ;-(( Hope that helps ... Hermann -- === === Subject: big-oh functionality hi, I wanted to know if for any two functions, f and g : N -> R*, R* is a set of positive real numbers including 0(zero),,, could we verify that f (not belongs) O(g) and g (not belongs) O(f). Please let me know 2 sets of examples at the earliest! === Subject: Re: big-oh functionality > hi, > I wanted to know if for any two functions, f and g : N -> R*, R* is a > set of positive real numbers including 0(zero),,, could we verify that > f (not belongs) O(g) and g (not belongs) O(f). > Please let me know 2 sets of examples at the earliest! f(n) = exp(2*n), g(n) = exp(n*(2+(-1)^n)) === === Subject: Re: Symmetric polynomials (I need help) Of course that the difference of symmetric polynomials is symmetric, what was I :) Anderson Brasil andersbrasil@aol.com >Any product of symmetric polynomials is symmetric >(from the definition of symmetric -- >invariant under any permutation of x_1,...,x_n), >and so is any difference of symmetric polynomials. >-- >Timothy Murphy >e-mail: tim@birdsnest.maths.tcd.ie >tel: +353-86-233 6090 >s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Vulgar but curious: How would go about solving the following problem (or problems like it) ? (don't read farther if you are easily offended) Chrissie loves to suck cock and jerk off guys! One fine evening, her three gorgeous flatmates (Will, Jeff and Keith) show up for some action. Since Chrissie has had a rough evening partying and has a sore nose and ass, she phones her adequate relative Uncle Happy for help in deciding the most efficient way she can perform her task, since she just wants to go to bed. Uncle Happy takes a drag on his pipe, contemplating. He has just turned off his copy of Girls Gone Wild- Brooklyn Edition and now knows that some girls can service three men at a time using mouth and hands. Uncle Happy asks Chrissie to estimate, based on her past experiences, how long it would normally take her to please her three charges with her mouth, her hand, or a combination thereof. Chrissie replies (units in minutes): Will - 20 mouth, 18 hand, 17 hand+mouth Jeff - 21 mouth, 19 hand, 17 hand+mouth Keith - 21 mouth, 20 hand, 18 hand+mouth How should Uncle Happy respond? Assume the following: - Chrissie may arrange her friends however she chooses. - Chrissie possesses one mouth and two hands. - Assume it takes Chrissie thirty seconds to get up to speed due to disorientation problems whenever she moves her mouth from one flatmate to the other. Ex.: If it normally takes Dan 2 minutes to finish, but Chrissie works him for a minute, does something else for a while and then returns, the total time Dan will take is 3 minutes (.5 startup, 1 sucking, .5 startup, 1 sucking = 3 minutes). Bonus: Chrissie needs to know whether or not to program her VCR to catch Law & Order. How long will this activity session take? === Subject: Re: Vulgar but curious: >Will - 20 mouth, 18 hand, 17 hand+mouth >Jeff - 21 mouth, 19 hand, 17 hand+mouth >Keith - 21 mouth, 20 hand, 18 hand+mouth Will, mouth, Keith, hand, Jeff, hand. 20 minutes. (Both Will and Jeff are on the critical path, and finish at the same time.) Yours, Doug Goncz (at aol dot com) Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/ I will be signing an amended Affidavit soon. === Subject: Enumerating all pairs of positive integers method, does it have a name? Googling didn't answer my question, so I suspect there is no name. How about triangular navigation method? === Subject: Re: Enumerating all pairs of positive integers method, does it have a name? > Googling didn't answer my question, so I suspect there is no name. How about > triangular navigation method? Dovetailing? (that term is specifically used in automata theory to describe the simulation of a countable # of Turing machines. 1st step on TM 1 2nd step on TM 1 1st step on TM 2 3rd step on TM 1 2nd TM 2 1st TM 3 etc. looks like the fanned out tail of a dove. I do not know where the word came from (who originally coined it in this usage). Mitch === Subject: Re: Enumerating all pairs of positive integers method, does it have a name? Mikito Harakiri scribbled the following: > Googling didn't answer my question, so I suspect there is no name. How about > triangular navigation method? AFAIK Georg Cantor used a similar method to enumerate Q. You could check his work to see if he named the method. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ A friend of mine is into Voodoo Acupuncture. You don't have to go into her office. You'll just be walking down the street and... ohh, that's much better! - Stephen Wright === Subject: Re: How to factor 27X^6-1? > plz help me factor this cause my math teacher ended up confusing me when she > went over this The rules that are commonly used for factoring binomials are: a^2 - b^2 = (a - b)(a + b) a^3 - b^3 = (a - b)(a^2 + ab + b^2) a^3 + b^3 = (a + b)(a^2 - ab + b^2) In your case, 27x^6 - 1 can be viewed as the middle method because it can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 = a, 1 = b. replacing a and b in (a - b)(a^2 + ab + b^2) gives you (3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally written as (3x^2 - 1)(9x^4 + 3x^2 + 1) -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: How to factor 27X^6-1? > plz help me factor this cause my math teacher ended up confusing me when she > went over this > The rules that are commonly used for factoring binomials are: > a^2 - b^2 = (a - b)(a + b) > a^3 - b^3 = (a - b)(a^2 + ab + b^2) > a^3 + b^3 = (a + b)(a^2 - ab + b^2) > In your case, 27x^6 - 1 can be viewed as the middle method because it > can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 = a, 1 = b. > replacing a and b in (a - b)(a^2 + ab + b^2) gives you > (3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally written as > (3x^2 - 1)(9x^4 + 3x^2 + 1) This has been asked before, but why is your method better than (sqrt(27))^2 - 1^2 = (sqrt(27) - 1) (sqrt(27) + 1)? My factorization has the advantage that you can follow it by the other two and continue factoring, if you notice that sqrt(27) = sqrt(3^3) = [sqrt(3)]^3. === Subject: Re: How to factor 27X^6-1? >plz help me factor this cause my math teacher ended up confusing me when she >went over this >>The rules that are commonly used for factoring binomials are: >>a^2 - b^2 = (a - b)(a + b) >>a^3 - b^3 = (a - b)(a^2 + ab + b^2) >>a^3 + b^3 = (a + b)(a^2 - ab + b^2) >>In your case, 27x^6 - 1 can be viewed as the middle method because it >>can be rewritten in the form (3x^2)^3 - (1)^3 where 3x^2 = a, 1 = b. >>replacing a and b in (a - b)(a^2 + ab + b^2) gives you >>(3x^2 - 1)((3x^2)^2 + (3x^2)(1) + (1)^2) which is normally written as >>(3x^2 - 1)(9x^4 + 3x^2 + 1) > This has been asked before, but why is your method better than > (sqrt(27))^2 - 1^2 = (sqrt(27) - 1) (sqrt(27) + 1)? > My factorization has the advantage that you can follow it by the > other two and continue factoring, if you notice that > sqrt(27) = sqrt(3^3) = [sqrt(3)]^3. In principle, it isn't. However, factored form is generally understood to be looking for factors with integer coefficients. Other than that, no special reason. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: How to factor 27X^6-1? > In principle, it isn't. However, factored form is generally understood > to be looking for factors with integer coefficients. Other than that, > no special reason. Proving that I'm no general. Oh well. === Subject: Re: What is the one sentence you can believe? >If someone tells you something *specific* and they know what they are saying, >no matter what circumstances are what is it they said if you >know it is always true? >e.g. circles are round isn't specific! >cryptic hint : old >Herc First of all, it must refer to their beliefs, as people will believe anything, regaldless of the logic involved. (E.g. that Saddam Hussein amassed a huge arsenal of weapons of mass destruction, but when the war began and it was time to use them, he destroyed them all.) This is false and you don't believe it. It cannot be true, as then by the first conjunct it is false. Thus it is false. Thus, one or the other of the conjuncts must be false. The first one is true, so the 2nd one must be false. Thus you believe it. (Whether you believe it or not.) (Now who can see the falacy of the above?) Charlie Volkstorf Cambridge, MA PS I also like the answer: If you don't believe me, I will shoot you with this gun. (while holding out an authentic looking weapon - of small, yet important, destruction.) === Subject: Re: What is the one sentence you can believe? >>If someone tells you something *specific* and they know what they are saying, >>no matter what circumstances are what is it they said if you >>know it is always true? >>e.g. circles are round isn't specific! >>cryptic hint : old >>Herc > First of all, it must refer to their beliefs, as people will believe > anything, regaldless of the logic involved. (E.g. that Saddam Hussein > amassed a huge arsenal of weapons of mass destruction, but when the > war began and it was time to use them, he destroyed them all.) > This is false and you don't believe it. > It cannot be true, as then by the first conjunct it is false. Thus it > is false. Thus, one or the other of the conjuncts must be false. The > first one is true, so the 2nd one must be false. Thus you believe it. > (Whether you believe it or not.) > (Now who can see the falacy of the above?) > Charlie Volkstorf > Cambridge, MA > PS I also like the answer: If you don't believe me, I will shoot you > with this gun. (while holding out an authentic looking weapon - of > small, yet important, destruction.) Just keep in mind this famous limerick: A theorem both deep and profound States that Every circle is round. But in a paper by Erdos (written in Kurdish) A counterexample is found! (not by) Martin Cohen (Seen on a bulletin board in Cal Tech in the 1960's) === Subject: Polynomials by recurrence Let A,B,C, a, b real numbers with ACa=/=0 . Consider the sequence (P_n)_{n>=0} of polynomials defined as p_0(x)=a , p_1(x)=Aax/2+ b p_{k+1}(x)=(Ax +B)*p_k(x) +C*p_{k-1}(x) , k=1,2,... . Suppose that T_n and U_{n-1} are Chebychev polynomials defined for t in (-1,1) by T_n(t)=cos(n*arccos t) , U_{n-1}(t)= sin( n*arccos t)/sqrt(1-t^2) . Question: try to represent p_n(x) as p_n(x)= C_1*T_n(px+q)+ C_2*U_{n-1}(ux+v) , n>=1 , where : C_1=C_1(n) , C_2=C_2(n) are real numbers , p,q,u,v are (complex) numbers which does'nt depend on n . === Subject: Re: David Ullrich on Identity > David Ullrich says: > And yes, identity is in _fact_ reflexive. To > refute that statement you need to give an > example of something which is not identical > to itself. The idea that there is something > which is _not_ identical to itself is simply > ludicrous A variable that returns a random value (common in programming languages) is often not equal to itself. One that returns the next sequential number is never equal to itself. Just output the value of x=x where x is that variable. But more generally, people from the ancient Greek philosophers who said things like Two things equal to the same thing are equal to each other. to modern day pseudointellectuals miss the real point of equality. ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND ABOVE AND UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION. Does 1+1 equal 2? At a low, physical level, they are different strings of characters. (1+1 will not equal 2 in string maniplulation processes in most programming languages.) But at a higher, Mathematical level of abstraction, they are. Does 1 equal 1? At a simple Mathematical level, they are. But at a more exact level, they are not, becaue one of them is the 2nd word in the question and the other is the 4th word. Or look at them under a microscope and you will detect faint differences in the displays. At a very high level of abstraction, they are both things. apple? But up close you can. Like everything, equality is context sensitive. Charlie Volkstorf Cambridge, MA === Subject: Re: David Ullrich on Identity chvol@aol.com says... >> David Ullrich says: >> And yes, identity is in _fact_ reflexive. To >> refute that statement you need to give an >> example of something which is not identical >> to itself. The idea that there is something >> which is _not_ identical to itself is simply >> ludicrous >A variable that returns a random value (common in programming >languages) is often not equal to itself. One that returns the next >sequential number is never equal to itself. Just output the value of >x=x where x is that variable. >But more generally, people from the ancient Greek philosophers who >said things like Two things equal to the same thing are equal to each >other. to modern day pseudointellectuals miss the real point of >equality. >ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND ABOVE AND >UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION. >Does 1+1 equal 2? At a low, physical level, they are different >strings of characters. (1+1 will not equal 2 in string >maniplulation processes in most programming languages.) But at a >higher, Mathematical level of abstraction, they are. >Does 1 equal 1? At a simple Mathematical level, they are. But at a >more exact level, they are not, becaue one of them is the 2nd word in >the question and the other is the 4th word. Or look at them under a >microscope and you will detect faint differences in the displays. >At a very high level of abstraction, they are both things. >apple? But up close you can. >Like everything, equality is context sensitive. >Charlie Volkstorf >Cambridge, MA Yes I agree. I'm not equal to myself (so says my scale after a big meal). IMHO the flaw is to ponder on the identify of the things themselves. IMHO, we should rather ponder on the identity of their properties. Which is the same thing you're saying: identity seems to be an artifact of intellectual resolution. Zooming in (and out) of levels of precision and abstraction reveals or dissolves differences. Equivalence is just as context-sensitive as equality but is not as misleading. === Subject: Re: David Ullrich on Identity >> David Ullrich says: >> And yes, identity is in _fact_ reflexive. To >> refute that statement you need to give an >> example of something which is not identical >> to itself. The idea that there is something >> which is _not_ identical to itself is simply >> ludicrous >A variable that returns a random value (common in programming >languages) is often not equal to itself. The value of the variable is always equal to itself. For a while the variable has the value 3. And 3 = 3. A little later the variable has the value 17. And 17 = 17. The value at one time is not equal to the value at another time. So what? They're not equal, because they're different things. >One that returns the next >sequential number is never equal to itself. Just output the value of >x=x where x is that variable. >But more generally, people from the ancient Greek philosophers who >said things like Two things equal to the same thing are equal to each >other. to modern day pseudointellectuals miss the real point of >equality. >ANY TWO THINGS ARE EQUAL AT SOME LEVEL OF ABSTRACTION AND ABOVE AND >UNEQUAL AT ALL LOWER LEVELS OF ABSTRACTION. >Does 1+1 equal 2? Yes. >At a low, physical level, they are different >strings of characters. No, neither 1 + 1 nor 2 is a string of characters. > (1+1 will not equal 2 in string >maniplulation processes in most programming languages.) One hopes not. So what? If 1 and 1 were the same thing you'd have a point. >But at a >higher, Mathematical level of abstraction, they are. >Does 1 equal 1? At a simple Mathematical level, they are. But at a >more exact level, they are not, becaue one of them is the 2nd word in >the question and the other is the 4th word. Or look at them under a >microscope and you will detect faint differences in the displays. >At a very high level of abstraction, they are both things. >apple? But up close you can. >Like everything, equality is context sensitive. Whether two _things_ are equal has nothing to do with context. Whether the objects denoted by two symbols are equal does have a lot to do with context. This says nothing about whether things are equal to themselves - in a context where two symbols denote things that are not equal they denote two _different_ things. Duh. >Charlie Volkstorf >Cambridge, MA ************************ === Subject: Re: David Ullrich on Identity >>Does 1+1 equal 2? > Yes. >>At a low, physical level, they are different >>strings of characters. > No, neither 1 + 1 nor 2 is a string of characters. Quite right. Charlie-Poo seems to be mired in crass formalism, unable to distinguish between an object and its representations. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: David Ullrich on Identity >Does 1+1 equal 2? >> Yes. >At a low, physical level, they are different >strings of characters. >> No, neither 1 + 1 nor 2 is a string of characters. >Quite right. Charlie-Poo seems to be mired in crass formalism, >unable to distinguish between an object and its >representations. So it would appear. (null comment inserted just to put sci.logic back on the list of newsgroups, lest he miss this.) ************************ === Subject: Re: David Ullrich on Identity >> David Ullrich says: >> And yes, identity is in _fact_ reflexive. To >> refute that statement you need to give an >> example of something which is not identical >> to itself. The idea that there is something >> which is _not_ identical to itself is simply >> ludicrous >A variable that returns a random value (common in programming >languages) is often not equal to itself. > The value of the variable is always equal to itself. For a while > the variable has the value 3. And 3 = 3. A little later the variable > has the value 17. And 17 = 17. Right. Only when you use the word (concept) itself is there equality, but then you are not referring to two things. Call this variable $R (the name used in my favorite programming language.) Then we agree that $R is not equal to $R. But you maintain that 3 is equal to 3. However, I pointed out the fact that 3 is not actually equal to 3 because the former is the 8th word in this sentence and the latter is the 14th word in this sentence. So at some level they are still not equal. Now if we say that 3 is equal to itself, we are not referring to two things. We are referring to one thing and then referring to itself. Whenever there is a distinction, and there are in fact two things, then whether they are equal or not depends on the level of abstraction at which we are dealing, whether it makes that same distinction. Since abstracting is the removal of distinctions, at some level, and above, they will be equal. In other words, it is not a question of whether two things are equal or not. Rather, it is a question of, at what level of abstraction, and above, they are equal? f(one thing,something)=the lowest level of abstraction at which they are equal. > They're not equal, because they're different things. What's the difference between not equal and different things? That is, when are these not the same? >Does 1+1 equal 2? > Yes. >At a low, physical level, they are different strings of characters. > No, neither 1 + 1 nor 2 is a string of characters. How else can we communicate other than by an exchange of strings of characters (i.e. elements of an agreed upon recursively enumerable set)? > Whether two _things_ are equal has nothing to do with > context. Whether the objects denoted by two symbols > are equal does have a lot to do with context. Are not objects things? Charlie Volkstorf Cambridge, MA > ************************ > === Subject: Re: David Ullrich on Identity > David Ullrich says: >> And yes, identity is in _fact_ reflexive. To > refute that statement you need to give an > example of something which is not identical > to itself. The idea that there is something > which is _not_ identical to itself is simply > ludicrous >>A variable that returns a random value (common in programming >>languages) is often not equal to itself. >> The value of the variable is always equal to itself. For a while >> the variable has the value 3. And 3 = 3. A little later the variable >> has the value 17. And 17 = 17. >Right. Only when you use the word (concept) itself is there >equality, but then you are not referring to two things. Uh, that's correct. That's why the word itself appears prominently in the statement that you seemed to think you were refuting. >Call this variable $R (the name used in my favorite programming >language.) Then we agree that $R is not equal to $R. No, we don't agree to that. >But you >maintain that 3 is equal to 3. You don't realize how funny this sounds, saying that I maintain that 3 equals 3? >However, I pointed out the fact that 3 >is not actually equal to 3 because the former is the 8th word in this >sentence and the latter is the 14th word in this sentence. Yes, you pointed this out. That doesn't make it true. In fact 3 is not a word. >So at some >level they are still not equal. Now if we say that 3 is equal to >itself, we are not referring to two things. We are referring to one >thing and then referring to itself. >Whenever there is a distinction, and there are in fact two things, >then whether they are equal or not depends on the level of abstraction >at which we are dealing, whether it makes that same distinction. >Since abstracting is the removal of distinctions, at some level, and >above, they will be equal. >In other words, it is not a question of whether two things are equal >or not. Rather, it is a question of, at what level of abstraction, >and above, they are equal? f(one thing,something)=the lowest level of >abstraction at which they are equal. >> They're not equal, because they're different things. >What's the difference between not equal and different things? There is none. That's what equal means. >That is, when are these not the same? >>Does 1+1 equal 2? >> Yes. >>At a low, physical level, they are different strings of characters. >> No, neither 1 + 1 nor 2 is a string of characters. >How else can we communicate other than by an exchange of strings of >characters (i.e. elements of an agreed upon recursively enumerable >set)? We can't. The fact that we need character strings to talk about things does not imply that things _are_ character strings. >> Whether two _things_ are equal has nothing to do with >> context. Whether the objects denoted by two symbols >> are equal does have a lot to do with context. >Are not objects things? Yes... >Charlie Volkstorf >Cambridge, MA >> ************************ >> ************************ === Subject: Re: David Ullrich on Identity > Homework for David Ullrich: > 1) What philosopher said: > ...definitions are available only for transforming > truths, not for founding them. > I would be interested in this answer... _Ways of Paradox_, p. 81 > So, your dispute with the boyz in the hood has to do with formal set > theory--class models of that theory, in fact. mitch > How can *theorems* be *false*? Well, the point here is that > when one says that a statement about the natural numbers is > false one means that it is false of the *standard* integers, > i.e., the integers that we normally work with. Likewise, when David Ullrich says that statements about individuals such as ExAy~(x=y) and Ex~(x=x) are false, he has said nothing more than that these are false of the *standard* individuals that FOL= deals with. Put back in the domain those that FOL= excludes, and it is not ExAy~(x=y) and Ex~(x=x) but AxEy(x=y) and Ax(x=x) that are false. It doesn't take a rocket scientist to figure that one out... --John === Subject: Re: David Ullrich on Identity > You should check out http://megafoundation.org/ > Harris has finally found people who understand him there. This from a purveyor of store-bought, on-the-shelf knowledge, whose absolute inability to reason from premises other than the customary, bought-and-paid-for ones is a common trait of the unabashedly uptaught. >Exhibit of proof of Ex~(x=x) from >C1-C4 and someone will point out the error. >C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1 >C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2 >C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) >Classification >C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] > Weak Here's the proof you couldn't hack. 1) Ax(x in y <-> Et(x in t & ~(x in x)) Instance of (C1) 2) y in y <-> Et(y in t & ~(y in y)) 1, US 3) ~Et(y in t) 2 4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] C4 5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US 6) Et(y in t) <-> y=y 5 7) ~(y=y) 3,6 8) Ex~(x=x) 7,EG ********************************************************************** Homework for David Ullrich What bearing would the non-self-identity of quanta have on the Leibnizean principle of the Identity of Indiscernibles? --John And yes, identity is in _fact_ reflexive. To refute that statement you need to give an example of something which is not identical to itself. The idea that there is something which is _not_ identical to itself is simply ludicrous: That's what identity _means_: A thing is identical to itself and to nothing else. --David Ullrich === Subject: Re: David Ullrich on Identity >> You should check out http://megafoundation.org/ >> Harris has finally found people who understand him there. >This from a purveyor of store-bought, on-the-shelf knowledge, >whose absolute inability to reason from premises other than >the customary, bought-and-paid-for ones is a common trait >of the unabashedly uptaught. You should _really_ check out http://megafoundation.org/ ! This sort of anyone-who-actually-knows-something-must-be -too-stupid-to-think-for-himself-the-way-we-geniuses-do is their motto. >>Exhibit of proof of Ex~(x=x) from >>C1-C4 and someone will point out the error. Just for the record, when I said that you were claiming that C1-C4 were part of NBG. Later turned out that that was just something you made up. >>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1 >>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2 >> >>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) >>Classification >> >>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] >> Weak >Here's the proof you couldn't hack. >1) Ax(x in y <-> Et(x in t & ~(x in x)) Instance of (C1) >2) y in y <-> Et(y in t & ~(y in y)) 1, US >3) ~Et(y in t) 2 >4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] C4 >5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US >6) Et(y in t) <-> y=y 5 >7) ~(y=y) 3,6 >8) Ex~(x=x) 7,EG >********************************************************************** > Homework for David Ullrich >What bearing would the non-self-identity of quanta >have on the Leibnizean principle of the Identity of >Indiscernibles? >--John >And yes, identity is in _fact_ reflexive. To >refute that statement you need to give an >example of something which is not identical >to itself. The idea that there is something >which is _not_ identical to itself is simply >ludicrous: That's what identity _means_: A >thing is identical to itself and to nothing >else. >--David Ullrich ************************ === Subject: Re: David Ullrich on Identity >> You should check out http://megafoundation.org/ >> Harris has finally found people who understand him there. >This from a purveyor of store-bought, on-the-shelf knowledge, >whose absolute inability to reason from premises other than >the customary, bought-and-paid-for ones is a common trait >of the unabashedly uptaught. > You should _really_ check out http://megafoundation.org/ ! > This sort of anyone-who-actually-knows-something-must-be > -too-stupid-to-think-for-himself-the-way-we-geniuses-do is > their motto. Against stupidity the very gods Themselves contend in vain. --Schiller, _The Maid of Orleans_ > Homework for David Ullrich >What bearing would the non-self-identity of quanta >have on the Leibnizean principle of the Identity of >Indiscernibles? === Subject: Re: David Ullrich on Identity >>Exhibit of proof of Ex~(x=x) from >>C1-C4 and someone will point out the error. >>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1 >>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2 >>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) >>Classification >>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] > Weak >Here's the proof you couldn't hack. >2) y in y <-> Et(y in t & ~(y in y)) 1, US >3) ~Et(y in t) 2 >4) AyAx[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] C4 >5) Az(z in y <-> z in y) -> (Et(y in t) <-> y=y) 4,US >6) Et(y in t) <-> y=y 5 >7) ~(y=y) 3,6 >********************************************************************** > Just for the record, when I said that you were claiming that > C1-C4 were part of NBG. Later turned out that that was just > something you made up. Till their own dreams at length deceive 'em, And oft repeating, they believe 'em. --Matthew Prior Alma [1718], canto III, l. 13 > Homework for David Ullrich > What bearing would the non-self-identity of quanta > have on the Leibnizean principle of the Identity of > Indiscernibles? >--John > Have you forgotten your homework, AGAIN??? === Subject: Re: David Ullrich on Identity linux) >> You should check out http://megafoundation.org/ >> Harris has finally found people who understand him there. > What part of Ax[Qu(x) -> ~(x = x)] do YOU fail to understand? that there does not exist an x such that Qu(x)? Is this a conclusion with which Teller would be happy? In any case, if some quantum theorist somewhere wants to mess about with a thoroughly non-standard idea of identity, what the heck should Ullrich care? And how does Teller's speculations support your claim that all and only existent things are self-identical? (I will probably regret treating you like a human being here. Again.) -- Jesse Hughes Of course, my ability to admit my mistakes and correct them is a trait that many of you seem to never have properly appreciated. -- JSH, discussing his 1463rd proof of Fermat's Last Theorem. === Subject: Re: David Ullrich on Identity >> You should check out http://megafoundation.org/ >> Harris has finally found people who understand him there. > What part of Ax[Qu(x) -> ~(x = x)] do YOU fail to understand? > that there does not exist an x such that Qu(x)? Is this a conclusion > with which Teller would be happy? 'Quanta' are one thing, non-existents are another, and in common is that they are identical-with-nothings. My (sketchy and non-mathematically well-informed impression) is that fruitful applications of 'imaginary numbers' have been found in different areas of mathematics. It would not be surprising if the logic of identical-with-nothings were to turn out to have to have different applications as well. file if you wish), what Teller would not be happy with is the notion that (e.g.) AxAy(qu(x) & qu(y) -> ~(x=y)) and Ax(qu(x) -> ~(x=x) are 'well-formed'. His argument appears to be that since quanta lack 'haecceities', it makes sense neither to assert that x=y when x,y are quanta, or that ~(x=y). > In any case, if some quantum theorist somewhere wants to mess about > with a thoroughly non-standard idea of identity, what the heck should > Ullrich care? And how does Teller's speculations support your claim > that all and only existent things are self-identical? The main thing at stake is whether, in the logic that different theories run under, identity should be taken to be reflexive or non-reflexive. If the latter, then quantum theorists can battle it out (on empirical grounds) over whether quanta satisfy Ax(x=x). If the former, such (potentially fruitful) disputes are--illegitimately, I think--ruled out of court by the background logic. > (I will probably regret treating you like a human being here. > Again.) --John === Subject: power set cardinality? My apologies for not trawling the newsgroup enough to find a likely answer to this query.. it is likely here somewhere due to the high set discussion I have been happily reading. I seem to remember hearing a theorem that the power set of a set is always larger, even for transfinite sets.. i.e. the power set of the integers must be uncountable. Is this true? And if so, where is the error in the construction of the countable list of such power set found here: http://descmath.com/diag/power.html ? === Subject: Re: power set cardinality? > My apologies for not trawling the newsgroup enough to find a likely > answer to this query.. it is likely here somewhere due to the high > set discussion I have been happily reading. > I seem to remember hearing a theorem that the power set of a set is > always larger, even for transfinite sets.. What about the so-called Universal Set -- the set of everything? Is its power set larger than the Universal Set itself? How could this be? Or does the Universal Set not really exist? If so, must every set exclude at least one thing? Dan === Subject: Re: power set cardinality? >> My apologies for not trawling the newsgroup enough to find a likely >> answer to this query.. it is likely here somewhere due to the high >> set discussion I have been happily reading. >> I seem to remember hearing a theorem that the power set of a set is >> always larger, even for transfinite sets.. > What about the so-called Universal Set -- the set of everything? Is its > power set larger than the Universal Set itself? How could this be? Or does > the Universal Set not really exist? If so, must every set exclude at least > one thing? In the versions of set theory that are in common use, there are no universal sets. There are some theories in which a universal set exists, but you have to give up some other things (such as the existence of a power set) in order to avoid contradictions. And having no universal set also means you can't have a set that excludes just one thing, or any finite number of things, or any collection of things that itself forms a set, because in those situations you would then be able to express the universal set as a union. Rather than a top-down limitation (every set must exclude something), there is a bottom-up limitation. There are only certain ways in which sets can be combined to form new sets, and a set of all sets, or a set of all sets satisfying property P is not allowed, in general. -- Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: power set cardinality? Visiting Assistant Professor at the University of Montana. > My apologies for not trawling the newsgroup enough to find a likely > answer to this query.. it is likely here somewhere due to the high > set discussion I have been happily reading. > I seem to remember hearing a theorem that the power set of a set is > always larger, even for transfinite sets.. >> What about the so-called Universal Set -- the set of everything? Is its >> power set larger than the Universal Set itself? How could this be? Or does >> the Universal Set not really exist? If so, must every set exclude at least >> one thing? >In the versions of set theory that are in common use, there are no >universal sets. There are some theories in which a universal set exists, >but you have to give up some other things (such as the existence of a >power set) in order to avoid contradictions. I thought a measurable cardinal would give you a sort of universal set in which you can take power sets. And that the axiom of universes was relatively consistent with ZFC. I guess what you would give up is the notion that it contains everything, and have to settle on it contains all of these things? It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: power set cardinality? >>In the versions of set theory that are in common use, there are no >>universal sets. There are some theories in which a universal set exists, >>but you have to give up some other things (such as the existence of a >>power set) in order to avoid contradictions. > I thought a measurable cardinal would give you a sort of universal > set in which you can take power sets. And that the axiom of > universes was relatively consistent with ZFC. I was trying to give an example of how a universal set might exist in some theory, not provide an absolute rule. I am certainly no expert on set theory. > I guess what you would give up is the notion that it contains > everything, and have to settle on it contains all of these things? -- Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: power set cardinality? Visiting Assistant Professor at the University of Montana. >> My apologies for not trawling the newsgroup enough to find a likely >> answer to this query.. it is likely here somewhere due to the high >> set discussion I have been happily reading. >> I seem to remember hearing a theorem that the power set of a set is >> always larger, even for transfinite sets.. >What about the so-called Universal Set -- the set of everything? Is its >power set larger than the Universal Set itself? In axiomatic set theory, there is no such set. > How could this be? Or does >the Universal Set not really exist? There is no set of all sets. > If so, must every set exclude at least >one thing? In axiomatic set theory, there are certain collections which are too big to be sets. The collection of all sets (or of all but a finite number of sets) is one of them. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: power set cardinality? > My apologies for not trawling the newsgroup enough to find a likely > answer to this query.. it is likely here somewhere due to the high > set discussion I have been happily reading. > I seem to remember hearing a theorem that the power set of a set is > always larger, even for transfinite sets.. > i.e. the power set of the integers must be uncountable. > Is this true? Yes. If A is a set and P(A) is the power set of A, suppose f is a bijection between the two. Let B = {a in A : a is not an element of f(a)}. B is a subset of A, therefore B is an element of P(A), therefore since f is a bijection there exists b in A such that f(b) = B. Now we ask, is b an element of B? If it is, then by definition of B, b is not in B. If it isn't, then again by definition of B, b is in B. In other words we just proved that b is in B if and only if b is NOT in B. This is a contradiction, showing that our assumption that f is a bijection is incorrect. Since f was arbitrary, there can be no bijection from A to P(A). > And if so, where is the error in the construction of the countable > list of such power set found here: > http://descmath.com/diag/power.html One doesn't have time to debunk all the false sites on the Net. === Subject: Re: power set cardinality? > My apologies for not trawling the newsgroup enough to find a likely > answer to this query.. it is likely here somewhere due to the high > set discussion I have been happily reading. I seem to remember hearing a theorem that the power set of a set is > always larger, even for transfinite sets.. i.e. the power set of the integers must be uncountable. Is this true? Yes. If A is a set and P(A) is the power set of A, suppose f is a > bijection between the two. Let B = {a in A : a is not an element of > f(a)}. B is a subset of A, therefore B is an element of P(A), therefore > since f is a bijection there exists b in A such that f(b) = B. > Now we ask, is b an element of B? If it is, then by definition of B, b > is not in B. If it isn't, then again by definition of B, b is in B. > In other words we just proved that b is in B if and only if b is NOT in > B. This is a contradiction, showing that our assumption that f is a > bijection is incorrect. Since f was arbitrary, there can be no > bijection from A to P(A). Very pretty! This seems similar to the (name forgotten) famous one: the set of all sets that don't contain themselves. Does it contain itself? Perhaps there is another assumption that could be creating your contradiction: that there exists a in A : a is not an element of f(a) .. In particular, we have {f(a_i)...} = P(A) (assuming f is a bijection). Is there an element a in A that is not in P(A)? Every element of a set is also in its power set so 'a' does not exist and B is the empty set... In a related issue, what is P({})? Is it {{}}? > And if so, where is the error in the construction of the countable > list of such power set found here: http://descmath.com/diag/power.html One doesn't have time to debunk all the false sites on the Net. Certainly true. Sorry.. === Subject: Re: power set cardinality? >My apologies for not trawling the newsgroup enough to find a likely >answer to this query.. it is likely here somewhere due to the high >set discussion I have been happily reading. >I seem to remember hearing a theorem that the power set of a set is >always larger, even for transfinite sets.. >i.e. the power set of the integers must be uncountable. >Is this true? >>Yes. If A is a set and P(A) is the power set of A, suppose f is a >>bijection between the two. Let B = {a in A : a is not an element of >>f(a)}. B is a subset of A, therefore B is an element of P(A), therefore >>since f is a bijection there exists b in A such that f(b) = B. >>Now we ask, is b an element of B? If it is, then by definition of B, b >>is not in B. If it isn't, then again by definition of B, b is in B. >>In other words we just proved that b is in B if and only if b is NOT in >>B. This is a contradiction, showing that our assumption that f is a >>bijection is incorrect. Since f was arbitrary, there can be no >>bijection from A to P(A). > Very pretty! This seems similar to the (name forgotten) famous one: > the set of all sets that don't contain themselves. Does it contain > itself? This is Russell's paradox. Incidently, it's actually a (form of) a proof of Cantor's theorem that the powerset of any set is of higher cardinality than the set itself! Basicly, consider a model of set theory, i.e. a structure consisting of elements for which there is provided a relation in and in which extensionality holds, i.e. if for all x x in a if and only if x in b, then a = b. We can aks of various properties P whether or not this model contains an element a, such that the elements having the relation in to a are exactly those having the property P. Now, Russell's paradox shows that not every property can have such a corresponding element a, which means just that we can't map bijectively the powerset of the model - which consists of all possible properties of the model - into the model itself, since if we could, then a contradiction would follows by Russell's reasoning. > In a related issue, what is P({})? Is it {{}}? Yes. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: power set cardinality? [Re: proof of Cantor's Theorem] > Very pretty! This seems similar to the (name forgotten) famous one: > the set of all sets that don't contain themselves. Does it contain > itself? > Perhaps there is another assumption that could be creating your > contradiction: that there exists a in A : a is not an element of f(a) There is no such assumption in the proof. The argument works even if no such a exists. > In particular, we have {f(a_i)...} = P(A) (assuming f is a bijection). > Is there an element a in A that is not in P(A)? Every element of a > set is also in its power set so 'a' does not exist and B is the empty > set... Not true. If a is a member of A, then you can conclude that {a} is a member of P(A), but not that a is a member of P(A). A member of A need not be a subset of A. > In a related issue, what is P({})? Is it {{}}? Yes. -- Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: power set cardinality? > [Re: proof of Cantor's Theorem] > Very pretty! This seems similar to the (name forgotten) famous one: > the set of all sets that don't contain themselves. Does it contain > itself? > Perhaps there is another assumption that could be creating your > contradiction: that there exists a in A : a is not an element of f(a) > There is no such assumption in the proof. The argument works even if no > such a exists. > In particular, we have {f(a_i)...} = P(A) (assuming f is a bijection). > Is there an element a in A that is not in P(A)? Every element of a > set is also in its power set so 'a' does not exist and B is the empty > set... > Not true. If a is a member of A, then you can conclude that {a} is a > member of P(A), but not that a is a member of P(A). A member of A need > not be a subset of A. > In a related issue, what is P({})? Is it {{}}? > Yes. Whoops!! :) Please diregard my erroneous claim that the size of {{}} = size of {}. trivial bijection indeed.. (eats hat) === Subject: Re: power set cardinality? > [Re: proof of Cantor's Theorem] > Very pretty! This seems similar to the (name forgotten) famous one: > the set of all sets that don't contain themselves. Does it contain > itself? > Perhaps there is another assumption that could be creating your > contradiction: that there exists a in A : a is not an element of f(a) > There is no such assumption in the proof. The argument works even if no > such a exists. > In particular, we have {f(a_i)...} = P(A) (assuming f is a bijection). > Is there an element a in A that is not in P(A)? Every element of a > set is also in its power set so 'a' does not exist and B is the empty > set... > Not true. If a is a member of A, then you can conclude that {a} is a > member of P(A), but not that a is a member of P(A). A member of A need > not be a subset of A. that the set {a} != {{a}}. However... > In a related issue, what is P({})? Is it {{}}? > Yes. If so, the bijection is trivial and exists.. Why does the proof not apply? === Subject: Re: power set cardinality? > However... In a related issue, what is P({})? Is it {{}}? Yes. > If so, the bijection is trivial and exists.. Why does the proof not > apply? Oh? {} contains no elements but {{}} contains one element, namely {}, so there is no bijection between them. === Subject: Re: power set cardinality? Visiting Assistant Professor at the University of Montana. >> In a related issue, what is P({})? Is it {{}}? >> Yes. >If so, the bijection is trivial and exists.. Why does the proof not >apply? It does. There is no bijection. P({}) has one element (namely, the empty set), but {} has no elements. The only function from {} to {{}} is the empty function, and it is not surjective. (To think of it another way: P({}) has one element. If a bijection exist, then there must be an element of the empty set whose image is the unique element of P({}). Can you exhibit an element of the empty set with that property? In fact, can you exhibit an element of the empty set, period?) It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: power set cardinality? >> [Re: proof of Cantor's Theorem] >> Very pretty! This seems similar to the (name forgotten) famous one: >> the set of all sets that don't contain themselves. Does it contain >> itself? >> >> Perhaps there is another assumption that could be creating your >> contradiction: that there exists a in A : a is not an element of f(a) >> There is no such assumption in the proof. The argument works even if no >> such a exists. >> In particular, we have {f(a_i)...} = P(A) (assuming f is a bijection). >> Is there an element a in A that is not in P(A)? Every element of a >> set is also in its power set so 'a' does not exist and B is the empty >> set... >> Not true. If a is a member of A, then you can conclude that {a} is a >> member of P(A), but not that a is a member of P(A). A member of A need >> not be a subset of A. >that the set {a} != {{a}}. >However... >> In a related issue, what is P({})? Is it {{}}? >> Yes. >If so, the bijection is trivial and exists.. There is no bijection between {} and {{}}. That's because {} has no elements, in particular there's no element of {} to map to the one element of {{}}. >Why does the proof not >apply? It does. ************************ === Subject: Re: power set cardinality? My apologies for not trawling the newsgroup enough to find a likely > answer to this query.. it is likely here somewhere due to the high > set discussion I have been happily reading. I seem to remember hearing a theorem that the power set of a set is > always larger, even for transfinite sets.. i.e. the power set of the integers must be uncountable. Is this true? > Yes. If A is a set and P(A) is the power set of A, suppose f is a > bijection between the two. Let B = {a in A : a is not an element of > f(a)}. B is a subset of A, therefore B is an element of P(A), therefore > since f is a bijection there exists b in A such that f(b) = B. Now we ask, is b an element of B? If it is, then by definition of B, b > is not in B. If it isn't, then again by definition of B, b is in B. In other words we just proved that b is in B if and only if b is NOT in > B. This is a contradiction, showing that our assumption that f is a > bijection is incorrect. Since f was arbitrary, there can be no > bijection from A to P(A). Very pretty! This seems similar to the (name forgotten) famous one: > the set of all sets that don't contain themselves. Does it contain > itself? > Perhaps there is another assumption that could be creating your > contradiction: that there exists a in A : a is not an element of f(a) > .. When did we make use of that assumption? We defined B as above, but for all we know B is empty. It matters not. > In particular, we have {f(a_i)...} = P(A) (assuming f is a bijection). > Is there an element a in A that is not in P(A)? Every element of a > set is also in its power set so 'a' does not exist and B is the empty > set... I'm assuming you mean {f(a} : a in A}. You can't write a_i to enumerate A, because there's no assumption that A is countable. You can't even index A using transfinite ordinals because without assuming the axiom of choice, A might not even be well-orderable. In general it's not true that every element of a set is in its power set. For example if A = {a,b} then P(A) = { {}, {a}, {b}, {a,b} }. Note that a is not the same as {a}. > In a related issue, what is P({})? Is it {{}}? Yes. And if so, where is the error in the construction of the countable > list of such power set found here: http://descmath.com/diag/power.html > One doesn't have time to debunk all the false sites on the Net. > Certainly true. Sorry.. Well a couple of others found the time. I should have said, One doesn't have time to debunk all the false sites on the Net, for suitable values of 'One.' === Subject: Re: power set cardinality? >> And if so, where is the error in the construction of the countable >> list of such power set found here: >> http://descmath.com/diag/power.html > One doesn't have time to debunk all the false sites on the Net. Surely not, but I was able to predict where the error lay without looking at the argument. -- Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: power set cardinality? Visiting Assistant Professor at the University of Montana. > And if so, where is the error in the construction of the countable > list of such power set found here: http://descmath.com/diag/power.html > One doesn't have time to debunk all the false sites on the Net. >Surely not, but I was able to predict where the error lay without looking >at the argument. Well, that's easy. Question is, was your prediction accurate? (-: It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: power set cardinality? >> And if so, where is the error in the construction of the countable >> list of such power set found here: >> http://descmath.com/diag/power.html > One doesn't have time to debunk all the false sites on the Net. >>Surely not, but I was able to predict where the error lay without looking >>at the argument. > Well, that's easy. Question is, was your prediction accurate? (-: Yes, actually, it was. -- Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: power set cardinality? >> And if so, where is the error in the construction of the countable >> list of such power set found here: >> http://descmath.com/diag/power.html > One doesn't have time to debunk all the false sites on the Net. >>Surely not, but I was able to predict where the error lay without looking >>at the argument. > Well, that's easy. Question is, was your prediction accurate? (-: >> Yes, actually, it was. >I predict that, if asked, will say his guess was: the author >enumerated only the finite subsets of N. >I am not so clever. I had to look at the site and see where the error >was. In hindsight, 's prediction is obvious, alright. It's not a question of cleverness, you just haven't paid enough attention to the various proofs that the power set of N is countable. Enumerating just the finite subsets of N is the _standard_ proof. (Actually it's the only one I recall ever seeing. Whether or not it's the only one that exists it is in fact extremely standard.) ************************ === Subject: Re: power set cardinality? <87ad95baqp.fsf@phiwumbda.org> linux) > It's not a question of cleverness, you just haven't paid enough > attention to the various proofs that the power set of N is > countable. I'm *sure* you're mistaken. I've paid too damn much attention to these proofs (well, perhaps not enough to make 's guess, but too damn much in a purely objective sense). -- Jesse Hughes LOL. How arrogant you are. Now when you realize that I DID prove Goldbach's conjecture and that I proved Fermat's Last Theorem as well, how are you going to feel then? -- James Harris === Subject: Re: power set cardinality? >> It's not a question of cleverness, you just haven't paid enough >> attention to the various proofs that the power set of N is >> countable. >I'm *sure* you're mistaken. I've paid too damn much attention to >these proofs (well, perhaps not enough to make 's guess, but too >damn much in a purely objective sense). Ok, I'm mistaken. It's happened before, not that I recall exactly when... What other proof that the power set of N is countable have you seen? Just curious. ************************ === Subject: Re: power set cardinality? <87ad95baqp.fsf@phiwumbda.org> <87r82h6lz8.fsf@phiwumbda.org> linux) > It's not a question of cleverness, you just haven't paid enough > attention to the various proofs that the power set of N is > countable. >>I'm *sure* you're mistaken. I've paid too damn much attention to >>these proofs (well, perhaps not enough to make 's guess, but too >>damn much in a purely objective sense). > Ok, I'm mistaken. It's happened before, not that I recall exactly > when... > What other proof that the power set of N is countable have you seen? > Just curious. Sorry, you weren't mistaken in that regard. I can't recall any other proof. I was just saying I'd seen more than enough, not fewer. -- My proof has been checked very thoroughly, both by me and others. Those others apparently decided that they would not believe the proof was correct, but cannot support that position using mathematics. But hey, they're just human beings. --JSH, prover of Fermat's Last Thm === Subject: Re: power set cardinality? >What other proof that the power set of N is countable have you seen? >Just curious. Let I_n denote {1,2,...,n}. Clearly (as has already been discussed in some recent thread, possibly this one) the limit of I_n, as n -> oo, is N. Similarly, the limit of I_{2^n}, as n -> oo, is N. Now, the power set of any set X is 2^X, and in particular 2^{I_n} is equipollent with 2^{2^n}. As we learn in Calculus, exponentiation is continuous, so it commutes with taking limits. Therefore 2^N = 2^{limit of I_n as n -> oo} = limit of 2^{I_n} as n -> oo is equipollent to the limit of I^{2^n} as n-> oo, which is N. Howzat? Lee Rudolph === Subject: Re: power set cardinality? >>What other proof that the power set of N is countable have you seen? >>Just curious. >Let I_n denote {1,2,...,n}. Clearly (as has already been >discussed in some recent thread, possibly this one) the limit >of I_n, as n -> oo, is N. Similarly, the limit of I_{2^n}, as >n -> oo, is N. Now, the power set of any set X is 2^X, and >in particular 2^{I_n} is equipollent with 2^{2^n}. As we >learn in Calculus, exponentiation is continuous, so it commutes >with taking limits. Therefore > 2^N = 2^{limit of I_n as n -> oo} > = limit of 2^{I_n} as n -> oo >is equipollent to the limit of I^{2^n} as n-> oo, which is N. >Howzat? That's one I'd forgotten. One might regard it as a more sophisticated version of the just-count-finite subsets, of course. >Lee Rudolph ************************ === Subject: Re: power set cardinality? >> And if so, where is the error in the construction of the countable >> list of such power set found here: >> http://descmath.com/diag/power.html > One doesn't have time to debunk all the false sites on the Net. >>Surely not, but I was able to predict where the error lay without looking >>at the argument. >Well, that's easy. Question is, was your prediction accurate? (-: My prediction is that yes it was. >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magidin@math.berkeley.edu ************************ === Subject: Re: power set cardinality? : My apologies for not trawling the newsgroup enough to find a likely : answer to this query.. it is likely here somewhere due to the high : set discussion I have been happily reading. : I seem to remember hearing a theorem that the power set of a set is : always larger, even for transfinite sets.. : i.e. the power set of the integers must be uncountable. : Is this true? : And if so, where is the error in the construction of the countable : list of such power set found here: : http://descmath.com/diag/power.html : ? Your list only includes finite subsets of integers. The set of all even integers will never appear in your list. Stephen === Subject: Re: power set cardinality? Visiting Assistant Professor at the University of Montana. >My apologies for not trawling the newsgroup enough to find a likely >answer to this query.. it is likely here somewhere due to the high >set discussion I have been happily reading. >I seem to remember hearing a theorem that the power set of a set is >always larger, even for transfinite sets.. >i.e. the power set of the integers must be uncountable. >Is this true? Yes. For any set A, the cardinality of P(A) is strictly larger than the cardinality of A. This is called Cantor's Theorem, in case you want to go look for it. >And if so, where is the error in the construction of the countable >list of such power set found here: >http://descmath.com/diag/power.html >? He is enumerating the collection of all FINITE subsets of the integers. These are indeed known to be countable. In his enumeration, however, the set of all even integers never shows up, and neither does the set of all primes, etc. In a sense, these are the ones that blow up the cardinality. For any infinite set A, the set of all finite subsets of A has the same cardinality as A (assuming the Axiom of Choice, anyway). The collection of all finite subsets of the integers is countable. But the collection of all infinite subsets of the integers is uncountable. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: power set cardinality? >My apologies for not trawling the newsgroup enough to find a likely >answer to this query.. it is likely here somewhere due to the high >set discussion I have been happily reading. >I seem to remember hearing a theorem that the power set of a set is >always larger, even for transfinite sets.. >i.e. the power set of the integers must be uncountable. >Is this true? > Yes. For any set A, the cardinality of P(A) is strictly larger than > the cardinality of A. This is called Cantor's Theorem, in case you > want to go look for it. >And if so, where is the error in the construction of the countable >list of such power set found here: >http://descmath.com/diag/power.html >? > He is enumerating the collection of all FINITE subsets of the > integers. These are indeed known to be countable. In his enumeration, > however, the set of all even integers never shows up, and neither does > the set of all primes, etc. > In a sense, these are the ones that blow up the cardinality. For any > infinite set A, the set of all finite subsets of A has the same > cardinality as A (assuming the Axiom of Choice, anyway). > The collection of all finite subsets of the integers is countable. But > the collection of all infinite subsets of the integers is uncountable. > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu that list. It looks like that list contains the set of the first N even integers, where N is as high as I want. I suppose my error is in comparing set theory to real number theory, where that type of convergence is considered equality (definition of convergence). In other words, you probably agree that lim(n-->inf)1/n = 0 because for every eps>0 I can find an n such that 1/n is less than eps. However, you don't agree that lim (n-->inf) {even_i, even_i+1..., even_n} = {all even integers}, even though for every even integer I can find an n such that it is on the list. Am I on the right track? === Subject: Re: power set cardinality? >>My apologies for not trawling the newsgroup enough to find a likely >>answer to this query.. it is likely here somewhere due to the high >>set discussion I have been happily reading. >>I seem to remember hearing a theorem that the power set of a set is >>always larger, even for transfinite sets.. >>i.e. the power set of the integers must be uncountable. >>Is this true? >> Yes. For any set A, the cardinality of P(A) is strictly larger than >> the cardinality of A. This is called Cantor's Theorem, in case you >> want to go look for it. >>And if so, where is the error in the construction of the countable >>list of such power set found here: >>http://descmath.com/diag/power.html >>? >> He is enumerating the collection of all FINITE subsets of the >> integers. These are indeed known to be countable. In his enumeration, >> however, the set of all even integers never shows up, and neither does >> the set of all primes, etc. >> In a sense, these are the ones that blow up the cardinality. For any >> infinite set A, the set of all finite subsets of A has the same >> cardinality as A (assuming the Axiom of Choice, anyway). >> The collection of all finite subsets of the integers is countable. But >> the collection of all infinite subsets of the integers is uncountable. >> It's not denial. I'm just very selective about >> what I accept as reality. >> --- Calvin (Calvin and Hobbes) >> Arturo Magidin >> magidin@math.berkeley.edu >that list. It looks like that list contains the set of the first N >even integers, where N is as high as I want. I suppose my error is in >comparing set theory to real number theory, where that type of >convergence is considered equality (definition of convergence). >In other words, you probably agree that lim(n-->inf)1/n = 0 because >for every eps>0 I can find an n such that 1/n is less than eps. >However, you don't agree that lim (n-->inf) {even_i, even_i+1..., >even_n} = {all even integers}, even though for every even integer I >can find an n such that it is on the list. >Am I on the right track? No, you're missing something very important. It is true that (if we insert suitable definitions) then the limit as n -> infinity of the set of the first n even numbers is the set of all even numbers. What you're missing is that this is simply _irrelevant_ : the fact that the set of even numbers is a limit of sets that appear on the list does not show that the set of even numbers itself appears on the list. Just like the fact that 1/n -> 0 does not imply that the sequence 1, 1/2, 1/3, ... actually includes the number 0 at some point. It doesn't. ************************ === Subject: Re: power set cardinality? > that list. It looks like that list contains the set of the first N > even integers, where N is as high as I want. I suppose my error is in > comparing set theory to real number theory, where that type of > convergence is considered equality (definition of convergence). > In other words, you probably agree that lim(n-->inf)1/n = 0 because > for every eps>0 I can find an n such that 1/n is less than eps. > However, you don't agree that lim (n-->inf) {even_i, even_i+1..., > even_n} = {all even integers}, even though for every even integer I > can find an n such that it is on the list. > Am I on the right track? One can create a definition of limits of sets such that lim (n-->inf) {2, 4, ..., 2n} = {all even integers}. However this would not be relevant to the question of countability. To see this, note that there are countably many rationals, and every real number is the limit of a sequence of rationals, yet there are uncountably many real numbers. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: power set cardinality? Visiting Assistant Professor at the University of Montana. >>And if so, where is the error in the construction of the countable >>list of such power set found here: >>http://descmath.com/diag/power.html >>? >> He is enumerating the collection of all FINITE subsets of the >> integers. These are indeed known to be countable. In his enumeration, >> however, the set of all even integers never shows up, and neither does >> the set of all primes, etc. >> In a sense, these are the ones that blow up the cardinality. For any >> infinite set A, the set of all finite subsets of A has the same >> cardinality as A (assuming the Axiom of Choice, anyway). >> The collection of all finite subsets of the integers is countable. But >> the collection of all infinite subsets of the integers is uncountable. >that list. It looks like that list contains the set of the first N >even integers, where N is as high as I want. I suppose my error is in >comparing set theory to real number theory, where that type of >convergence is considered equality (definition of convergence). Oh, dear. Don't use real number theory. Number Theory has a very precise meaning, and it is not the one you are talking about. >In other words, you probably agree that lim(n-->inf)1/n = 0 because >for every eps>0 I can find an n such that 1/n is less than eps. Because that's what the definition of convergence is for this context. >However, you don't agree that lim (n-->inf) {even_i, even_i+1..., >even_n} = {all even integers}, even though for every even integer I >can find an n such that it is on the list. I assume you mean something line the limit, as n goes to infinity, of {0, 2, 4, 6, ..., 2n} is the set {0, 2, 4, 6, ...,} of all even integers. If you interpret the notions correctly, then the answer is that it IS true: you interpret the limit as taking the ascending union of the sets. That is, let A_n = {0, 2, ..., 2n}, and then limit as n goes to infinity means the union of A_1, A_2, ... etc. But you have a similar phenomenon occurring in both cases. Even though the limit as n goes to infinity of 1/n is equal to 0, IT IS NOT THE CASE that there is an n such that 1/n=0. Likewise, even though the limit of these finite sets is an infinite set, it is not the case that at any finite step you have an infinite set. When you are ennumerating, however, you only cover the FINITE cases. You never get to the limit. That's what ennumerating means. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: power set cardinality? >>And if so, where is the error in the construction of the countable >>list of such power set found here: >>http://descmath.com/diag/power.html >>? >> He is enumerating the collection of all FINITE subsets of the >> integers. These are indeed known to be countable. In his enumeration, >> however, the set of all even integers never shows up, and neither does >> the set of all primes, etc. >> In a sense, these are the ones that blow up the cardinality. For any >> infinite set A, the set of all finite subsets of A has the same >> cardinality as A (assuming the Axiom of Choice, anyway). >> The collection of all finite subsets of the integers is countable. But >> the collection of all infinite subsets of the integers is uncountable. >that list. It looks like that list contains the set of the first N >even integers, where N is as high as I want. I suppose my error is in >comparing set theory to real number theory, where that type of >convergence is considered equality (definition of convergence). > Oh, dear. Don't use real number theory. Number Theory has a very > precise meaning, and it is not the one you are talking about. Sorry, mistake noted. >In other words, you probably agree that lim(n-->inf)1/n = 0 because >for every eps>0 I can find an n such that 1/n is less than eps. > Because that's what the definition of convergence is for this > context. >However, you don't agree that lim (n-->inf) {even_i, even_i+1..., >even_n} = {all even integers}, even though for every even integer I >can find an n such that it is on the list. > I assume you mean something line > the limit, as n goes to infinity, of > {0, 2, 4, 6, ..., 2n} > is the set > {0, 2, 4, 6, ...,} > of all even integers. > If you interpret the notions correctly, then the answer is that it IS > true: you interpret the limit as taking the ascending union of the > sets. That is, let A_n = {0, 2, ..., 2n}, and then limit as n goes to > infinity means the union of A_1, A_2, ... etc. > But you have a similar phenomenon occurring in both cases. Even though > the limit as n goes to infinity of 1/n is equal to 0, IT IS NOT THE > CASE that there is an n such that 1/n=0. I see your point. So, the set {1/1,1/2,1/3...} does not contain the number 0, whereas the sequence 1/1,1/2,1/3 does converge to zero. > Likewise, even though the limit of these finite sets is an infinite > set, it is not the case that at any finite step you have an infinite > set. > When you are ennumerating, however, you only cover the FINITE > cases. You never get to the limit. That's what ennumerating means. > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: Four Color Theorem Simplified Here's my revised list!! > SOME THINGS YOU NEED TO KNOW ABOUT THE FOUR COLOR THEOREM!!! > DEFINITIONS: > CHI(G) is the chromatic number of graph G. > If CHI(G) <= k, then graph G is k-colorable > If CHI(G) = k, then graph G is k-chroma. > A graph is planar if it contains no subgraph homeomorphic to > K5 or K3,3. > Every maximal planar graph [MPG] has exactly 3n-6 edges. > If every MPG is 4-colorable; then every planar graph is 4-colorable. > All MPG's are generated by the intersection of two triangulations of an > n-sided polygon. > Every triangulation of every convex polygon is 3-colorable. > All 4-partite graphs, both planar and non-planar, are 4-colorable. Note that I have removed the statement about the reduction of 5-chroma graphs. I feel that the 'list' is now incomplete and that there are more things you need to know about the four color theorem! Amybody have any suggestions? Does anyone disagree with any of the items in the list? === Subject: Re: Four Color Theorem Simplified > ALL YOU NEED TO KNOW ABOUT THE FOUR COLOR THEOREM!!! > DEFINITIONS: > CHI(G) is the chromatic number of graph G. > If CHI(G) <= k, then graph G is k-colorable > If CHI(G) = k, then graph G is k-chroma. > A graph is planar if it contains no subgraph homeomorphic to > K5 or K3,3. > Every maximal planar graph [MPG] has exactly 3n-6 edges. > If every MPG is 4-colorable; then every planar graph is 4-colorable. > All MPG's are generated by the intersection of two triangulations of an > n-sided polygon. > Every triangulation of every convex polygon is 3-colorable. > Every 5-colorable graph has a subgraph homeomorphic to the complete graph K5. > All 4-partite graphs, both planar and non-planar, are 4-colorable. > How about; Every 5-chroma 'non planar' graph has a subgraph > homeomorphic (or is it 'isomorphic') to the complete graph K5.? Is > this a valid statement? Despite the usage of is homeomorphic to elsewhere in math to denote an equivalence relation, it has a specialized meaning in graph theory. Graph A is homeomorphic to graph B, if by a sequence of edge deletions and coalescence of vertices, graph B can be obtained from graph A. Isomorphism of graphs is an equivalence relation, but homeomorphism in this sense is not. The statement: Every 5-chroma graph has a subgraph homeomorphic to the complete graph K5. is plausible, but I don't know how to prove it. See also my partial retraction and comment under the other branch of this thread. === Subject: Re: Four Color Theorem Simplified > ALL YOU NEED TO KNOW ABOUT THE FOUR COLOR THEOREM!!! > Every 5-colorable graph has a subgraph homeomorphic to the complete > graph > K5. > Your pentultimate statement is clearly wrong. > Thar's penultimate! Anyway, what is the smallest n for which it is > not true? An example, please? > You give no special definitions of terms, so presumably your usage is > standard for treatments of the four-color problem. > In standard parlance, four-colorable implies five-colorable, so a trivial > example (e.g. of a planar graph) may be adduced. > Even granting that you intended five-colorable in a stronger sense of > five- but not four-colorable, the statement would still be incorrect. > Here is an example which may be drawn on the surface of a torus. > Imagine three circles running parallel around the hole in a doughnut. > Subdivide one of the resulting three regions so obtained with five segments > cutting perpendicularly across that region. > The map now has 7 regions, lacks any embedded subgraph K5, and is > five-colorable but not four-colorable. > Your question about the smallest n is interesting. Obviously if there were > a map with 5 regions that was five- but not four-colorable, as a graph it > would be (isomorphic to) K5 (the complete graph on 5 vertices). Is there a > counterexample with n=6? I suspect not, but apart from an exhaustive > examination of cases, I cannot think off-hand of an approach to showing > this. > Perhaps you will have better insight. I must retract my last counter-example. It actually does have a subgraph homeomorphic to K5. Delete the edges between any two adjacent pairs of the five subdivided regions, thus combining each pair into a single region. The resulting graph is isomorphic to K5. Actually by saying non-planar we are already (Kuratowski's theorem) implying that the graph contains either a subgraph homeomorphic to K5 or to K3,3, and of course given the four-color theorem, any five-chroma graph would necessarily be non-planar. === Subject: On singular value decomposition Say the singular value decomposition of a N-by-N matrix A is, A=USV', where S=diag{s11, ... , sNN}. Take S(A), U(A) and V(A) as functions of A. Is there any stable algorithm such that S(A), U(A), V(A) are continous in A? ie. Say for every epsilon > 0, if |norm(E)| < epsilon, then there exist delta such that |norm(U(A+E)-U(A)| < delta, |norm(S(A+E)-S(A)| < delta, |norm(V(A+E)-V(A)| < delta. Bernard === Subject: Re: On singular value decomposition > Say the singular value decomposition of a N-by-N matrix A is, A=USV', where > S=diag{s11, ... , sNN}. Take S(A), U(A) and V(A) as functions of A. Is there > any stable algorithm such that S(A), U(A), V(A) are continous in A? No. Consider the sequence: / 1/n 1/n A_n = | | if n odd 1/n 1/n / / 1/n 0 = | | if n even 0 1/n / This converges to 0, but U(A_n) toggles between: / -t -t / 0 1 | | and | | -t t / 1 0 / (for t=1/sqrt(2)) as does V(A_n). -- Kevin === Subject: Re: On singular value decomposition >Say the singular value decomposition of a N-by-N matrix A is, A=USV', where >S=diag{s11, ... , sNN}. Take S(A), U(A) and V(A) as functions of A. Is there >any stable algorithm such that S(A), U(A), V(A) are continous in A? >ie. Say for every epsilon > 0, if |norm(E)| < epsilon, then there exist >delta such that |norm(U(A+E)-U(A)| < delta, |norm(S(A+E)-S(A)| < delta, >|norm(V(A+E)-V(A)| < delta. It can be done for S(A), as the sii can be taken to be the non-negative square roots of the characteristic values of AA'. However, it cannot be done in general for U(A) and V(A) if AA' has multiple characteristic roots. For example, it fails for A = I, as changing a_ij and a_ji to be a small non-zero number for i != j, and making no other changes, causes U(A) and V(A) to have magnitude sqrt(.5) for the elements in the i-th and j-th rows and columns. -- 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: Collatz (3x+1) as diophantine problem > Can't offer constructive feedback; I'm sorry. But I find it > interesting to study. > Hi Ernst - > short explanation. > One transformation of Collatz, starting at an odd integer x0 > and going to the next odd integer x1 can be written as > I am working your equations and so far it looks right to me. I wanted to let you know I was still chatting with you about it. Ernst === Subject: Re: Collatz (3x+1) as diophantine problem === >Subject: Re: Collatz (3x+1) as diophantine problem >> Can't offer constructive feedback; I'm sorry. But I find it >> interesting to study. >> Hi Ernst - >> short explanation. >> One transformation of Collatz, starting at an odd integer x0 >> and going to the next odd integer x1 can be written as >> > I am working your equations and so far it looks right to me. Hey Ernst, did you notice that GH's equations are the same as my Crossover Point formula? All GH has to do is multiply his equation by C to have a general formula for 3x+C. I thought something was wrong because his numerator differed from mine for the same sequence. But I studied it all weekend and finally realized that he derived his formulae by using the normal Collatz rules in going from x0 to x1 whereas I derived mine by using inverse rules going from x1 to x0. Also, his power of 2 exponents in the numerator are, for a,b,c c+b c whereas I put a,b,c into an array z[] and compute the exponents by e[0] = sum(z[]) - z[0] e[1] = e[0] - z[1] e[2] = e[1] - z[2] but in looking at this, I realized that sum(z[]) - z[0] is c+b+a - a or simply c+b which makes e[0] - z[1] c+b - b or simply c. And finally e[1] - z[2] is c - c or 0. That makes my list of exponents c+b c 0 Note that GH doesn't show 2^0 because that's simply 1. But from the programming aspect, it's better to have a rule without exceptions, so 0 exponents should always be shown for consistency. He got his the easy way, I got mine the hard way. The important thing is that we both arrived at the same place. Truth is not dependant on the pathway to it. His numbers were different because although a,b,c b,c,a c,a,b give different equations with different exponents, they are all the same loop. So GH will probably give you a more rigorous mathematical treatment, but my seat of the pants reults are still correct. > I wanted to let you know I was still chatting with you about it. >Ernst -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Collatz (3x+1) as diophantine problem Hi Mensanator - > my Crossover Point formula? All GH has to do is multiply his > equation by C to have a general formula for 3x+C. is that really all? I didn't look at these points yet... > I thought something was wrong because his numerator differed > from mine for the same sequence. But I studied it all weekend > and finally realized that he derived his formulae by using the > normal Collatz rules in going from x0 to x1 whereas I derived > mine by using inverse rules going from x1 to x0. Yes, that was also my original approach. But I thought: all people look along this direction... and it does not really make a difference at this point. One amazing thing was, that the formula for the loop is exactly the same - viewing it in this or that direction. > Note that GH doesn't show 2^0 because that's simply 1. Yes, I thought to deal only with odds. This also selects the used Collatz-numbers to these with a last index of 0 C(X;a,b,c,...,0) (in my notation) and flattens the equations > But > from the programming aspect, it's better to have a rule without > exceptions, so 0 exponents should always be shown for > consistency. Right. > He got his the easy way, Well well... I had several days with excessive playing around since the grammabeautiful-start - but when it came to this simple idea... that was nearly a shame... ;-) > I got mine the hard way. And also more continuously, I must say (and appreciate). But still it needs the final argument, that the base(2/3)- digits can not be converted to a integer multiple of a repunit, if they are not all equal at start. Maybe again this comes out to be much more simple than I thought in a first try. Since the digits can only be powers of 2 (or some fixed derivate of 2/3) - and the flattening of such a digitnumber, which comprises the carry-overs from one digit to another, could be simply impossible if base and integer-property of digits are not compatible. The more general question, seen of the generating view of the collatz-transformation: can all natural numbers be created by the inverse-collatz-transformation, starting by 1, can be attacked by the same formula. The behaving of the numbers can be much instructively be studied displaying them in binary then. I think, that is then another path to arrive at the bit-pattern-subject, that you mentioned here many times. GH (Gottfried) ;-) ---------------------------------------------------------------- Gottfried Helms Univ Kassel === Subject: Re: Collatz (3x+1) as diophantine problem > Hi Mensanator - my Crossover Point formula? All GH has to do is multiply his > equation by C to have a general formula for 3x+C. > is that really all? I didn't look at these points yet... You have n-1 n-2 z n-3 z+y 1 z+y+...+c z+...c+b 3 + 3 2 + 3 2 + ... + 3 2 + 2 x = --------------------------------------------------------------- z+...+c+b+a n 2 - 3 My Crossover Point formula for a 3x+C system is Z * C CO = -------- X - Y where my Z term is your entire numerator (our denominators are the same). For the standard 3x+1 system, our formulae are identical and the only way we can get an integer is if all the factors of X-Y can be cancelled by the factors of Z. When C>1, C and Z can work together (provided that C is not a power of 3, otherwise it will not cancel ANY factors in the denominator because X-Y can never have a factor of 3). Thus, systems like 3x+5 have many loops in the positive integers aside from the trivial one at C. I don't know if you saw this in another thread, so I'll repeat it here. If the Crossover Point is an integer, then there is a loop. For EVERY possible sequence, a 3x+C system exists in which that sequence is a loop. Take an extreme example ag_af ae ad ac_ab aa z y x w v u_t s r q p o n m_l k_j i h g f_e d c_b a the Crossover Point works out to be 35343985 * C -------------- 33552245 which factors to 5 * 23 * 307339 * C --------------------- 5 * 6710449 which reduces to 7068797 * C ------------- 6710449 So the system 3x+6710449 has a loop at 7068797, which is easily verified 7068797 27916840 13958420 6979210 3489605 17179264 8589632 4294816 2147408 1073704 536852 268426 134213 7113088 3556544 1778272 889136 444568 222284 111142 55571 6877162 3438581 17026192 8513096 4256548 2128274 1064137 9902860 4951430 2475715 14137594 7068797 === Subject: Re: Collatz (3x+1) as diophantine problem ehhm, I just put two unfinished manuscripts on my server. The first, which shows a wonderful fractal tree resulting from my version of indexing the collatz-numbers http://141.51.96.22/priv/math/collatz/A_nice_fractal.htm The second a manuscript, which was the start for the current loop-attack: with some more explanations: http://141.51.96.22/priv/math/collatz/A_simplificating.htm Also a deeper view in the generation tree is this (excel-generated) image: http://141.51.96.22/priv/math/collatz/generatinglist.htm ---------------------------- Besides of giving more explanations and thoughts for discussion it's just stuff coming from stumbling on a bifurcated way... ;-) > He got his the easy way, ;-) Gottfried === Subject: Re: Collatz (3x+1) as diophantine problem > http://141.51.96.22/priv/math/collatz/A_nice_fractal.htm > ---------------------------- Sorry, I'm not able to create html being readable by netscape 4 (and possibly other newsreaders) with the microsoft-version of html-generation from its word-files. Since I have written my text in this format i'm stuck now and the html-versions are partly garbage (for instance, exponents come out as indexes) So to be able to read it properly I present them as *.pdf-versions too. > http://141.51.96.22/priv/math/collatz/A_nice_fractal.pdf > http://141.51.96.22/priv/math/collatz/A_simplificating.pdf GH === === Subject: primecounter should be quite fast... 4,000,000,000 in 0.25 sek. hoping i'm not too late... #define Number unsigned long #define PrimeTableSize 8192 #define NSize 8192 #define PSize 16 #define Empty 0xFFFFFFFF Number Value = 4000000000; Number Primes[PrimeTableSize]; Number Cache[NSize][PSize]; Number Kick(Number n, Number p, Number Max) { Number i, p2, NewN, Res; NewN = n / p; Res = NewN; for (i = 1; i NewN) {Res-=Max - i; break;} if ((NewNR such that g(x) = f(f(x))= x^2 - 1996. I thought of considering fixed points, specifically the fact that if f is differentiable and a is a fixed point of f, then g(a) = f(a)^2 and g(a)>=0. But this didnt work and, anyway, nothing assures that, if the function f existed, it had to be differentiable. The number 1996 appears cause this problem is from 1996, but I dont know if the conclusion can be generalized to f(f(x))= x^2 - m, m>0. Any suggestion is welcome. Artur === Subject: Re: A help with f(f(x)=x^2 - 1996 >Hello everybody >I'm trying to show that there's no function f:R->R such that g(x) = >f(f(x))= x^2 - 1996. I thought of considering fixed points, >specifically the fact that if f is differentiable and a is a fixed >point of f, then g(a) = f(a)^2 and g(a)>=0. But this didnt work and, >anyway, nothing assures that, if the function f existed, it had to be >differentiable. >The number 1996 appears cause this problem is from 1996, but I dont >know if the conclusion can be generalized to f(f(x))= x^2 - m, m>0. You don't need differentiability, or even continuity. g(x) = x^2 - m has two fixed points, (1(+/-)sqrt(1+4m))/2, if m > -1/4. In addition, g has a single two-cycle (-1(+/-) sqrt(-3+4m))/2, if m > 3/4. Let these points be b_1 and b_2: g(b_1) = b_2 and g(b_2) = b_1. If g(x) = f(f(x)), then [b_1, f(b_1), b_2, f(b_2)] must form a four-cycle for f (and these four points must be distinct, otherwise [b_1, b_2] couldn't be a two-cycle for g). But then [f(b_1), f(b_2)] would be another two-cycle for g, and there can't be any more. 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 help with f(f(x)=x^2 - 1996 Mr. Israel, Could you kindly suggest to me a good textbook for the study of functional equations? >Hello everybody >I'm trying to show that there's no function f:R->R such that g(x) = >f(f(x))= x^2 - 1996. I thought of considering fixed points, >specifically the fact that if f is differentiable and a is a fixed >point of f, then g(a) = f(a)^2 and g(a)>=0. But this didnt work and, >anyway, nothing assures that, if the function f existed, it had to be >differentiable. >The number 1996 appears cause this problem is from 1996, but I dont >know if the conclusion can be generalized to f(f(x))= x^2 - m, m>0. > You don't need differentiability, or even continuity. > g(x) = x^2 - m has two fixed points, (1(+/-)sqrt(1+4m))/2, if m > -1/4. > In addition, g has a single two-cycle (-1(+/-) sqrt(-3+4m))/2, if m > 3/4. > Let these points be b_1 and b_2: g(b_1) = b_2 and g(b_2) = b_1. > If g(x) = f(f(x)), then [b_1, f(b_1), b_2, f(b_2)] must form a four-cycle > for f (and these four points must be distinct, otherwise [b_1, b_2] > couldn't be a two-cycle for g). But then [f(b_1), f(b_2)] would be > another two-cycle for g, and there can't be any more. > 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: looking for counterexample I'm trying to find a counterexample showing that the hypothesis |E| < infty is necessary in the following: Thm: let f_k, g_k:E->R be measurable functions finite a.e. in E subset R, with E of _finite measure_ such that f_k -> f in measure and g_k -> g in measure. Show that g_k f_k -> gf in measure. Any ideas? TIA, nojb. === Subject: Re: looking for counterexample >I'm trying to find a counterexample showing that the hypothesis |E| < >infty is necessary in the following: >Thm: let f_k, g_k:E->R be measurable functions finite a.e. in E >subset R, with E of _finite measure_ such that f_k -> f in measure >and g_k -> g in measure. Show that g_k f_k -> gf in measure. >Any ideas? E = R, g_k(x) = f_k(x) = x + 1/k. >TIA, >nojb. ************************ === Subject: A tight bound on Union by Rank with Path Compression Hello. I'd like to know whether the m*alpha(n) upper bound on this algorithm is tight (where alpha is the inverse of something similiar to Ackerman's function as defined in CLRS - Introduction to Algorithms). If not, what is the tight bound? Oren Becker. === Subject: total derivative and partial derivative Hi I'm reading through some maths and they've written something like this: Given F = F(x,y,z) and y=y(x) and z=z(x) dF/dx = del/delx*F + del/dely*F*y' + del/delz*F*z' What's the rule that lets you say that? I know that you can say the following: dF = delF/delx*dx + delF/dely*dy + delF/delz*dz And I think they used the above rule to compute dF/dx. Is that right? By writing this I've answered my own question - hmmm interesting. === Subject: Re: total derivative and partial derivative > I'm reading through some maths and they've written something like this: > Given F = F(x,y,z) and y=y(x) and z=z(x) > dF/dx = del/delx*F + del/dely*F*y' + del/delz*F*z' > What's the rule that lets you say that? I know that you can say the > following: > dF = delF/delx*dx + delF/dely*dy + delF/delz*dz Chain rule. === Subject: Re: Any idea for a math career? Digital motion picture visual effects. It's a natural step for a mathematician to make. Much of mathematics was invented, in the first place, to render accurate images. There is scope for dynamics, rigid-body dynamics), inverse problems (photogrammetry, match-moving, inverse rendering, inverse dynamics) and image processing. You regularly use linear alagebra, multivariate calculus, optimisation, ODE and PDE solving techniques, FFTs, statistics and sampling theory, monte-carlo interation methods and I even found an application for a tiny bit of commutative algebra (which I hope to publish soon). And I'm just talking about what I do. My colleagues do a whole lot of other stuff... -- Torque === Subject: Re: Discrete Math => K!>K^2 Proof > I'm looking for a website that runs through the proof of this problem. > If you can help me out, email me at shodan3510@yahoo.co m > Alainna It's not true for *all* positive integers k, since 1! = 1^2, 2! < 2^2 and 3! < 3^2. It _is_ true for k = 4, since 4! = 24 and 4^2 = 16. And ... once it's true for some value of k, then it's true for all larger values. Basically, you have that (k+1)! = (k+1) k! > (1 + (1/k))^2 k! > (k+1)^2 if k! > k^2 and k > 1. The first inequality follows from the fact that (k+1) > (1 + (1/k))^2 if k > 1 and the second from the assumption that k! > k^2. Altogether, then, this argument proves that k! > k^2 if k > 3. (Oh ... that other inequality follows from the fact that it's obviously true if k = 2 and increasing k makes the LHS of the inequality bigger while it _decreases_ the RHS ... ) === Subject: okamoto/kashima on P vs NP unprovability fyi, this new 42 pg paper appears on superficial glance to be a very major tour-de-force result that P vs NP is in some sense difficult to prove. apparently building on some of the conjectures in razborov/rudich's natural proofs. the authors use a complex godelian type diagonalization construction. the primary author has a very distinguished record in cryptography theory. http://www.informatik.uni-trier.de/~ley/db/indices/a-tree/o/Okamoto:Tatsuaki .html here is kashima's (coauthors) publications which focus more on proof theory http://www.is.titech.ac.jp/~kashima/ I would like to invite all serious comment at http://groups.yahoo.com/group/theory-edge/ where I have also just extended an email invitation to the authors. posts out here to comp.theory are fine of course but note that a mailing list dialogue can be a little better/quickly synchronized among interested participants. > Resource Bounded Unprovability of Computational Lower Bounds > Tatsuaki Okamoto and Ryo Kashima === Subject: Sphere-Triangle Intersection Related Geometry Question Given the two intersection points of an intersected edge of the triangle, and information as to which edge was intersected, is there a way to procedurally generate triangles for any variant of initial triangle and sphere such that only one edge of each triangle is in contact with the sphere. The problem can be regarded as a 2D problem as it is possible to move the centre of the sphere to the plane of the triangle and calculate the new radius. An example of a solution to a similar problem would be a sphere triangle intersection with 0 edges intersected and 0 vertices intersected. / / o /____ By taking the closest point to the centre of the circle from each line three new points are generated. By finding the intersections from lines taken from each of these points to the centre of the circle the points are then moved to the surface of the sphere. 6 Triangles are then created, 3 with one vertex touching the sphere, 3 with two vertices touching the sphere. The important detail being those triangles with two vertices touching the sphere have one edge in intersected by the sphere (through the triangles vertices). === Subject: Combinatorial/Probability Problem I am trying to evaluate the following sum: Sum from m= 5 to infinity [ m*Binomial[m-1,4] p^5 q^(m-5)] where p = 1-q = 1/36 and Binomial[a,b] is the binomial coefficient a choose b (This is the expectation value for the number of dice throws one needs to make until obtaining 5 pairs of sixes) Matematica tells me that for large m, this summation is a neat 180, but I can't see how to obtain this result analytically. TIA, Matthew T. Brenneman === Subject: Re: Combinatorial/Probability Problem >I am trying to evaluate the following sum: >Sum from m= 5 to infinity [ m*Binomial[m-1,4] p^5 q^(m-5)] >where p = 1-q = 1/36 and Binomial[a,b] is the binomial coefficient a >choose b >Matematica tells me that for large m, this summation is a neat 180, but I >can't see how to obtain this result analytically. More generally, consider S_n = sum_{m=n}^infinity m Binomial[m-1, n-1] p^n q^(m-n). = sum_{m=n}^infinity n Binomial[m,n] p^n q^(m-n) The generating function with respect to n is g(z) = sum_{n=1}^infinity S_n z^n = sum_{n=1}^infinity sum_{m=n}^infinity n Binomial[m,n] p^n q^(m-n) z^n = sum_{m=1}^infinity sum_{n=0}^m n Binomial[m,n] q^m (zp/q)^n = sum_{m=1}^infinity q^m (1+zp/q)^(m-1) zpm/q = zp/(1-zp-q)^2 = z/(p(1-z)^2) = sum_{n=1}^infinity n z^n/p So S_n = n/p. And in particular for n=5, p=1/36 we get 180. Or a probabilistic proof: if X is the number of the trial on which the n'th success occurs, S_n is the expected value of X. Now X = X_1 + X_2 + ... + X_n, where X_1 is the number of trials until the first success, and X_i for i > 1 is the number of trials from the (i-1)th success until the ith. Each X_i has a geometric distribution with parameter p, and E[X_i] = 1/p. So E[X] = sum_{i=1}^n E[X_i] = n/p. 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: Combinatorial/Probability Problem > I am trying to evaluate the following sum: > Sum from m= 5 to infinity [ m*Binomial[m-1,4] p^5 q^(m-5)] > where p = 1-q = 1/36 and Binomial[a,b] is the binomial coefficient > a choose b > (This is the expectation value for the number of dice throws one > needs to make until obtaining 5 pairs of sixes) That's easy. There are 6*6 outcomes throwing two dice, so the probability of a pair of sixes on a single roll is 1/36. Therefore, you will need (on average) 36 rolls to get the first pair of sixes, 36 more rolls for the second pair of sixes, and so on; 5*36 = 180. > Matematica tells me that for large m, this summation is a neat 180, > but I can't see how to obtain this result analytically. Does the problem require an analytical solution? In that case, expand (1-q)^(-6) in powers of q (using the binomial theorem or whatever) and compare that with your series. Note that m*Binomial[m-1,4] = 5*Binomial[m,5]. === Subject: Re: Jacobi Iterativ-Verfahren ? >Wer hat die sogenannte ,,Jacobi'sche >Iterativ-Verfahren , entdeckt (1845 ?) : > -der beruehmter Deutscher Mathematiker > Carl Jacob Gustav JACOBI >oder > -Boris Semionovici JACOBI , >ein Mathematiker aus Russland ? >Bitte ein Paar Literatur Hinweise. Hallo Alex, das war Carl Gustav Jacob Jacobi: C. G. Jacobi: .86ber eine neue Aufl.9asungsart der bei der Methode der kleinsten Quadrate vorkommenden linearen Gleichungen. Astronomische Nachrichten 22 (1845). Es steht auch in seinen gesammelten Werken (7 B.8ande. Berlin: R. Reimer 1881-1891) in Band 3, S. 467-478. Die Werke sind online von der Biblioth.8fque Nationale de France, Paris lesbar http://gallica.bnf.fr --> Auteur Jacobi aber momentan ist dieser Server wegen Update-Arbeiten nur teilweise in Betrieb. Hier noch einige Referenzen: http://www.numerik.uni-kiel.de/~in/PAPER/Diplom/node43.html --> Jacobi45 http://www.cs.umd.edu/TRs/authors/C_G_Jacobi.html --> CS-TR-2877, but the full paper seems to be no more available ... Gr.9fsse Hermann -- >Danke, Alex === Subject: Re: Jacobi Iterativ-Verfahren ? Hermann Kremer schrieb in Nachricht ... Hello Alex, some additional informations. >>Wer hat die sogenannte ,,Jacobi'sche >>Iterativ-Verfahren , entdeckt (1845 ?) : >> -der beruehmter Deutscher Mathematiker >> Carl Jacob Gustav JACOBI http://www-history.mcs.st-and.ac.uk/~history/Mathematicians/Jacobi.html Yep, he was it, see below ... >>oder >> -Boris Semionovici JACOBI , >>ein Mathematiker aus Russland ? Boris Semjonowitsch Jakobi, aka Boris Semionovich Yakobi (1801-1874), was the elder brother of Carl Gustav Jacob Jacobi; his name in German is Moritz Hermann Jacobi. In 1835 he got a chair of physics at Dorpat (now Tartu) University (Estonia), and in 1837 he moved to the Russian Academy of Sciences in St. Petersburg. http://chem.ch.huji.ac.il/~eugeniik/history/jacobi.html (at the bottom are a few links to biographies in Russian). http://hp.iitp.ru/ger/32/3284.htm http://hp.iitp.ru/ger/32/3284a.htm >>Bitte ein Paar Literatur Hinweise. C. G. Jacobi: .86ber eine neue Aufl.9asungsart der bei der Methode der kleinsten Quadrate vorkommenden linearen Gleichungen. [On a new way of solving the linear equations arising in the method of least squares]. Astronomische Nachrichten Vol. 22 (1845), Issue No. 523. It may also be found in Jacobi's Collected Works in 7 Vols. Berlin: R. Reimer 1881-1891 in Vol. 3, p. 467-478. All 7 volumes are online at the Biblioth.8fque Nationale de France in Paris: http://gallica.bnf.fr --> Recherche --> Auteur Jacobi but currently the server is partially down due to update and maintainance. Here are some more references: http://www.numerik.uni-kiel.de/~in/PAPER/Diplom/node43.html --> Jacobi45 http://www.cs.umd.edu/TRs/authors/C_G_Jacobi.html --> CS-TR-2877, but the full paper seems to be no more available ;-(( Hope that helps ... Hermann -- >>Danke, Alex === Subject: Re: why isn't THE a quantifier? > References ACZEL, P. 1988 Non-Well-Founded Sets, CSLI, Stanford, 1988, XX?137 pp. BARWISE, J. & MOSS, L. 1996 Vicious Circles, CSLI, Stanford, 1996, X?390 > pp. > Great! How about an example or two showing how it works? [...] > Do your own homework. > F. I did. I verified my recollection by pulling these documents off my bookshelf and verified that they don't contain what you maintain. What would you do if someone told you that proof of their assertion was contained in a reference which you believed does not contain such a proof? Everything you have ever posted is refuted in mathworld.com [insert your response here] Do your own homework. Charlie Volkstorf Cambridge, MA === Subject: Recursive Sequence: Highest Prime Divisors If we start with two positive integers {m,n}, and let a(1) = m, a(2) = n, and we let, for k >= 3, a(k) = (highest prime dividing a(k-1)) + (highest prime dividing a(k-2)), then we get a (possibly eventually-periodic) sequence. (For example: m = 2, n = 3, leads to -> 2, 3, 5, 8, 7, 9, 10, 8, 7, 9,...) I am not at all certain, but I wonder if every {m,n} leads to a sequence that is eventually periodic. Is this so?? Obviously, if the highest primes dividing a(j) and a(j-1) are those which are the highest primes which also divide a(k) and a(k-1), respectively (for some k not = j), then the sequence is periodic. There may be something in the EIS, but the fact that {m,n} can vary makes a search not too easy, since I do not know which {m,n}-sequence is fundamental enough to be in the EIS, yet such a sequence is not trivial. Leroy Quet === Subject: Re: Recursive Sequence: Highest Prime Divisors >If we start with two positive integers {m,n}, and let >a(1) = m, >a(2) = n, >and we let, for k >= 3, >a(k) = >(highest prime dividing a(k-1)) + >(highest prime dividing a(k-2)), >then we get a (possibly eventually-periodic) sequence. >(For example: m = 2, n = 3, leads to -2, 3, 5, 8, 7, 9, 10, 8, 7, 9,...) >I am not at all certain, but I wonder if every {m,n} leads to a >sequence that is eventually periodic. >Is this so?? >Obviously, if the highest primes dividing a(j) and a(j-1) are those >which are the highest primes which also divide a(k) and a(k-1), >respectively (for some k not = j), then the sequence is periodic. The sequence must either eventually lead to the 4-cycle [8,7,9,10] or the 1-cycle [2] or [2p] for an odd prime p. And the 1-cycle can only happen if a(1) and a(2) have the same highest prime factor. Let h(x) be the highest prime dividing x. Instead of a(n), consider the sequence p(n) = h(a(n)). Thus p(n+2) = h(p(n)+p(n+1)). We can then reconstruct (a(n)) (for n >= 2) from (p(n)) since a(n) = p(n-2) + p(n-1). For the sequence p(n) the 4-cycle is [2,7,3,5] and the 1-cycle is [p]. It seems to me that beginning with distinct primes [p(1), p(2)], by [p(5), p(6)] we reach a pair [p(i),p(i+1)] with max(p(i),p(i+1)) < max(p(1),p(2)) except in the cases [2,3], [2,5], [3,2], [3,5] and [5,2], which all lead to the 4-cycle. (I don't think I've covered all possible cases by hand, but I've done a computer search and I think this is correct) Note for example that if p(1) and p(2) are odd, p(1)+p(2) is even so p(3) <= (p(1)+p(2))/2 < max(p(1),p(2)). And then again if p(3) is odd, p(4) < max(p(2),p(3)) < max(p(1),p(2)). On the other hand, if p(3)=2 we could have p(4)= p(2)+2 (if [p(2),p(2)+2] are twin primes; this implies p(2) = 2 mod 3 unless p(2)=3), but then p(3)+p(4)=p(2)+4 is odd and divisible by 3, so 3 <= p(5) <= (p(2)+4)/3, resulting in max(p(5),p(6)) <= (2p(2)+5)/3 < max(p(1),p(2)) unless p(2) <= 5. 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: Recursive Sequence: Highest Prime Divisors >If we start with two positive integers {m,n}, and let >a(1) = m, >a(2) = n, >and we let, for k >= 3, >a(k) = >(highest prime dividing a(k-1)) + >(highest prime dividing a(k-2)), >then we get a (possibly eventually-periodic) sequence. >(For example: m = 2, n = 3, leads to -2, 3, 5, 8, 7, 9, 10, 8, 7, 9,...) >I am not at all certain, but I wonder if every {m,n} leads to a >sequence that is eventually periodic. >Is this so?? >Obviously, if the highest primes dividing a(j) and a(j-1) are those >which are the highest primes which also divide a(k) and a(k-1), >respectively (for some k not = j), then the sequence is periodic. > The sequence must either eventually lead to the 4-cycle > [8,7,9,10] or the 1-cycle [2] or [2p] for an odd prime p. > And the 1-cycle can only happen if a(1) and a(2) have the same > highest prime factor. >....(Robert's proof snipped) I have also unrigorously proved this myself this morning. But my proof is probably flawed someplace. I am glad to see the result confirmed. I too apparently deduced that, if m and n have both the same highest-prime-divisor, then the sequence has a period of 1, at least from the 3rd term on. Otherwise, all other {m,n} lead eventually to the period-4 sequence of {8,7,9,10} repeating. (I have seen a few others' proofs of this result. Each of these proofs varies somewhat and is interesting.) Leroy Quet === Subject: Multiple Sum of...Divisible by... Let m and n be positive integers. Let {a(k)} be any integer sequence. Two related results, the last more general than the first: Let B_m(k/n) = --- / a(j) --- GCD(j,m)=1 1<= j <= k/n = floor(k/(nj)) --- --- / mu(j) / a(rj) --- --- j|m r=1 In linear-mode: B_m(k/n) = sum{GCD(j,m)=1, 1<=j<=k/n} a(j) = sum{j|m} mu(j) sum{r=1 to floor(k/(nj))} a(rj) (mu(j) is the Mobius fuction.) Then: m --- / m+1 k | | B_m(k/n) (-1) / k+1 / --- k=1 is a multiple of (m/(GCD(m,n)). In linear-mode: Then: sum{k=1 to m} binomial(m+1,k+1) B_m(k/n) (-1)^k is a multiple of (m/(GCD(m,n)). -- And m --- --- / m+1 --- | | k / / k+1 / / a(rj) (-1) mu(j) --- --- --- j|m k=1 GCD(r,n)=1 1<=r<=k/j is a multiple of (m/(GCD(m,n)). In linear-mode: sum{j|m} sum{k=1 to m} binomial(m+1,k+1) sum{GCD(r,n)=1,1<=r<=k/j} a(rj) (-1)^k mu(j) is a multiple of (m/(GCD(m,n)). Leroy Quet === Subject: Re: Multiple Sum of...Divisible by... Again I suspect that this post had not propagated throughout the Usenet correctly. So I have reposted it below. Sorry if you thought I had something new to say... > Let m and n be positive integers. > Let {a(k)} be any integer sequence. > Two related results, the last more general than the first: > Let B_m(k/n) = > --- > / a(j) > --- > GCD(j,m)=1 > 1<= j <= k/n > = > floor(k/(nj)) > --- --- > > / mu(j) / a(rj) > --- --- > j|m r=1 > In linear-mode: B_m(k/n) = > sum{GCD(j,m)=1, 1<=j<=k/n} a(j) > sum{j|m} mu(j) sum{r=1 to floor(k/(nj))} a(rj) > (mu(j) is the Mobius fuction.) > Then: > m > --- / m+1 k > | | B_m(k/n) (-1) > / k+1 / > --- > k=1 > is a multiple of (m/(GCD(m,n)). > In linear-mode: > Then: > sum{k=1 to m} binomial(m+1,k+1) B_m(k/n) (-1)^k > is a multiple of (m/(GCD(m,n)). > -- > And > m > --- --- / m+1 --- > | | k > / / k+1 / / a(rj) (-1) mu(j) > --- --- --- > j|m k=1 GCD(r,n)=1 > 1<=r<=k/j > is a multiple of (m/(GCD(m,n)). > In linear-mode: > sum{j|m} sum{k=1 to m} binomial(m+1,k+1) > sum{GCD(r,n)=1,1<=r<=k/j} a(rj) > (-1)^k mu(j) > is a multiple of (m/(GCD(m,n)). > Leroy Quet === Subject: only c borel subsets of R ? Anyone knows a proof showing that there are only c (continuum) borel subsets of R? nojb. === Subject: Re: only c borel subsets of R ? > Anyone knows a proof showing that there are only c (continuum) borel > subsets of R? Yes. Let B_0 be the collection of open subsets of R. Define B_alpha for ordinals alpha > 0 as follows: if alpha = beta + 1, B_alpha is the collection of countable unions of elements of B_beta and their complements. For alpha a limit ordinal. B_alpha is the union of B_beta for beta < alpha. Show that for each countable ordinal |B_alpha| = c. Then B_{omega_1} is the set of Borel subsets of R where omega_1 is the first uncountable ordinal. Thus |B_omega_1| <= aleph_1 c = c. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: only c borel subsets of R ? Anyone knows a proof showing that there are only c (continuum) borel > subsets of R? > Yes. > Let B_0 be the collection of open subsets of R. > Define B_alpha for ordinals alpha > 0 as follows: > if alpha = beta + 1, B_alpha is the collection > of countable unions of elements of B_beta > and their complements. For alpha a limit ordinal. > B_alpha is the union of B_beta for beta < alpha. > Show that for each countable ordinal |B_alpha| = c. > Then B_{omega_1} is the set of Borel subsets of R > where omega_1 is the first uncountable ordinal. > Thus |B_omega_1| <= aleph_1 c = c. Another way shows that there are only c analytic sets (using operation A or as continuous image of N^N), and that every Borel set is analytic. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Last Call For Primecounters > It's been an interesting and slightly weird contest so far, > and I'll definitely have to write a blurb on memoization > somewhere in the results. (To put it crudely, to memoize > a function, usually recursive, means to add a cache > thereto, so that a result isn't computed more than once. > The speedup can be amazing.) Second version. Additional memoization only. Herb Savage /* Written by: Herb Savage */ /* Purpose: Computes pi(X) */ #include #include #include #include #define TRUE 1 #define FALSE 0 int prime(unsigned x) { unsigned i,t; if (x < 2) return FALSE; if (x == 2) return TRUE; if (!(x & 1)) return FALSE; i = 3; for (;;) { t = x / i; if (t < i) break; if (t * i == x) return FALSE; i += 2; } return TRUE; } unsigned nextPrimeM(unsigned x) { if (x < 2) return 2; if (!(x & 1)) x--; while (!prime(x += 2)); return x; } #define np_cache_size 10000 unsigned np_cache[np_cache_size]; unsigned nextPrime(unsigned x) { if (x < np_cache_size) { if (!np_cache[x]) np_cache[x] = nextPrimeM(x); return np_cache[x]; } return nextPrimeM(x); } unsigned pi_n(unsigned x, unsigned y); unsigned pi_nM(unsigned end, unsigned maxP) { unsigned others,p,primes; others = end - 1; if (others < 1) return 0; p = 0; primes = 0; for (;;) { p = nextPrime(p); if ((p * p > end) || (p >= maxP)) break; if (p == 2) others -= (others + 1) / 2; else if (p == 3) others -= (others + 2) / 3; else others -= pi_n(end/p,p) - (primes - 1); primes++; } return primes + others; } #define pi_cache_size 100 unsigned pi_cache[pi_cache_size][pi_cache_size]; unsigned pi_n(unsigned x, unsigned y) { if (x < pi_cache_size && y < pi_cache_size) { if (!pi_cache[x][y]) pi_cache[x][y] = pi_nM(x,y); return pi_cache[x][y]; } return pi_nM(x,y); } unsigned long long pi(unsigned long long v) { return pi_n(v,v); } int main(int argc, char *argv[]) { unsigned n; if (argc != 2) exit(1); sscanf (argv[1], %u, &n); printf (%llun, pi(n)); exit(0); } ---------------------------------end program === Subject: Re: Last Call For Primecounters Here is mine. I know it is not the fastest but it still beats my old code. ------------------- class bit { char *bitarray; char list[8]; public: bit(unsigned long long length) { length=(length+1)/8; bitarray=new char[length]; for(unsigned long long i=0;i #include inline unsigned long long pi(unsigned long long); int main (int argc, const char * argv[]) { using namespace std; //introduces namespace std const unsigned long long n = 3900000; std::cout< It's been an interesting and slightly weird contest so far, > and I'll definitely have to write a blurb on memoization > somewhere in the results. (To put it crudely, to memoize > a function, usually recursive, means to add a cache > thereto, so that a result isn't computed more than once. > The speedup can be amazing.) I tried to memoize a function in the following program and it speed it up by a factor of two. Here's my submission for the contest. Herb Savage /* Written by: Herb Savage */ /* Purpose: Computes pi(X) */ #include #include #include #include #define TRUE 1 #define FALSE 0 int prime(unsigned x) { unsigned i,t; if (x < 2) return FALSE; if (x == 2) return TRUE; if (!(x & 1)) return FALSE; i = 3; for (;;) { t = x / i; if (t < i) break; if (t * i == x) return FALSE; i += 2; } return TRUE; } unsigned nextPrimeM(unsigned x) { if (x < 2) return 2; if (!(x & 1)) x--; while (!prime(x += 2)); return x; } #define cache_size 10000 unsigned np_cache[cache_size]; unsigned nextPrime(unsigned x) { if (x < cache_size) { if (!np_cache[x]) np_cache[x] = nextPrimeM(x); return np_cache[x]; } return nextPrimeM(x); } unsigned pi_n(unsigned end, unsigned maxP) { unsigned others,p,primes; others = end - 1; if (others < 1) return 0; p = 0; primes = 0; for (;;) { p = nextPrime(p); if ((p * p > end) || (p >= maxP)) break; if (p == 2) others -= (others + 1) / 2; else if (p == 3) others -= (others + 2) / 3; else others -= pi_n(end/p,p) - (primes - 1); primes++; } return primes + others; } unsigned long long pi(unsigned long long v) { return pi_n(v,v); } int main(int argc, char *argv[]) { unsigned n; if (argc != 2) exit(1); sscanf (argv[1], %u, &n); printf (%llun, pi(n)); exit(0); } ----------------------------------end program === Subject: Counterexample to FLT by modular arithmetic a,b,c,n = 5,7,13,3 and (a^n mod c + b^n mod c) mod c = 0 but a^n + b^n - c^n # 0 However, there remains the possiblity that since c.min = 2 and nc.min. The reason c.min = 2 is that 1^1 + 1^1 = 2^1 is a valid solution, and the smallest nontrivial one. So how to get from n a,b,c,n = 5,7,13,3 and > (a^n mod c + b^n mod c) mod c = 0 but > a^n + b^n - c^n # 0 > a,b,c,n = 3,4,5,2 and > a^n mod c b^n mod c (a^n mod c + b^n mod c) (n = 1,2) = > 3 4 7 > 4 1 5 and 5 mod c = 0 --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Curve Fitting? > I have a tensor whos values need to be expressed as: > I = fn(x,y,z). The function is third order on each axis. > Given a set of samples from a tensor I need to derive the co-efficents of > the polynomial that discribes the function fn. > Least squares worked well for me in one dimension but how do I go about > doing it in three dimesions. or is it four? > B. Least squares delivers reasonable results only if the set of basic functions is appropriate to the task. An example, an analog multiplier: f(x,y) = x*y + device errors The quadratic approach g(x,y) = a0 + a1*x+a2*x^2 + b1*y+b2*y^2 cannot fit f(x,y) very well. For two independent variables we can use e.g. g(x,y) = a0 + a1*x+a2*x^2 + b1*y+b2*y^2 + c1*x*y The parameter vector is (a0,a1,a2,b1,b2,c1). For more than two independent variables x,y,z itÇs the same, but without some a-priori knowledge about the expected shape itÇs hard to choose a good set of basic functions (ANY set of linearly independent functions, also containing sines, cosines and exponen- tial functions, if this behaviour is expected). For polynomials the mixed products are necessary. As used for the generation of look-up tables for ICC color profiles. === Subject: Re: Curve Fitting? Originator: broeker@ > I have a tensor whos values need to be expressed as: > I = fn(x,y,z). The function is third order on each axis. > Given a set of samples from a tensor I need to derive the co-efficents of > the polynomial that discribes the function fn. > Least squares worked well for me in one dimension but how do I go about > doing it in three dimesions. or is it four? It's four, and the simpler case is actually in two dimensions: one independent, one dependent variable. If you look at the least squares algorithm from a more abstract point of view, you'll note that it doesn't care at all about the number of independent variables you have --- as a matter of fact, it doesn't even pretend to know you have any input variables other than the discrete index i in the expression sum(i=0..n-1, ((measurement_i - theory_i(P)) / sigma_i)^2) with P being the vector of parameters to be fitted. If the theory is a function of not only the fittable parameters, but also some continuous variable X (which can be a vector X = (x,y,z) just as easily), you would usually rewrite that as: sum(i=0..n-1, ((y_i - theory(P; X_i)) / sigma_i)^2) The least square fitting algorithm doesn't use the values of X_i at all; it doesn't even have to know what they might be. It only cares about the parameter vector, P. -- Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de) Even if all the snow were burnt, ashes would remain. === Subject: Re: Curve Fitting? So are you saying there is a way to use matlabs polyfit function to get the polynomials of the samples? For example if my experiment was to measure the temperatures at various locations in a 10m x 10m x 10m room. I could derive the polynomial coefficients that would describe temperature as a function of location in the room? > I have a tensor whos values need to be expressed as: > I = fn(x,y,z). The function is third order on each axis. > Given a set of samples from a tensor I need to derive the co-efficents of > the polynomial that discribes the function fn. > Least squares worked well for me in one dimension but how do I go about > doing it in three dimesions. or is it four? > It's four, and the simpler case is actually in two dimensions: one > independent, one dependent variable. > If you look at the least squares algorithm from a more abstract point > of view, you'll note that it doesn't care at all about the number of > independent variables you have --- as a matter of fact, it doesn't > even pretend to know you have any input variables other than the > discrete index i in the expression > sum(i=0..n-1, ((measurement_i - theory_i(P)) / sigma_i)^2) > with P being the vector of parameters to be fitted. If the theory is > a function of not only the fittable parameters, but also some > continuous variable X (which can be a vector X = (x,y,z) just as > easily), you would usually rewrite that as: > sum(i=0..n-1, ((y_i - theory(P; X_i)) / sigma_i)^2) > The least square fitting algorithm doesn't use the values of X_i at > all; it doesn't even have to know what they might be. It only cares > about the parameter vector, P. > -- > Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de) > Even if all the snow were burnt, ashes would remain. === Subject: Re: Curve Fitting? > So are you saying there is a way to use matlabs polyfit function to get the > polynomials of the samples? > For example if my experiment was to measure the temperatures at various > locations in a 10m x 10m x 10m room. I could derive the polynomial > coefficients that would > describe temperature as a function of location in the room? Not with polyfit, but if your model is linear in the coefficients it's not hard to write your own code. I can't quite tell from your description what the form of your model is. Here's the general linear least-squares problem: minimize E = (1/2)*sum(k=1 to N) |y_k - f(x_k, y_k, z_k)|^2 where f(x,y,z) = sum(i=1 to p) a_i * g_i(x,y,z) It doesn't care what the form of the functions g_i is. What matters is that the function f(x,y,z) depends linearly on the unknown coefficients a_i. For polynomial regression, the functions g_i are just 1, x, x^2, x^3, etc. (You need one of them to be g_i = 1 if you have a constant term). I *think* your functions are either various combinations of powers of x, y and z, like x^2 y z^3, or perhaps you only have separate terms, powers of x, powers of y, powers of z. At any rate general linear least-squares is very easy. Define a matrix V where element V(i,j) is the value of the j-th function at the i-th data point, i.e. V(i,j) = g_j(x_i, y_i, z_i). Then the least-squares solution for your a's is given by: a = Vy; where y = column vector of dependent values. - Randy === Subject: Re: Curve Fitting? Originator: broeker@ [Note: F'up2 limited to matlab newsgroup --- this off-topic just about everywhere else]. > So are you saying there is a way to use matlabs polyfit function to get the > polynomials of the samples? I don't know MatLab or polyfit at all. They may have hidden the features of the actual least squares algorithms unter too much syntactic sugar for this to work, but in general, yes, that should be posible. All you really need is a function that maps your 3 input variables (x,y,z) to a single, discrete index i to be used in the chi^q summation I quoted, and back to (x,y,z). If, e.g., your measurements are made in a regular 11x11x11 pixels grid spanning a 10x10x10 meters cube, it's quite simple: fi(x,y,z)= x + 11*y + 11*11*z fx(i) = i MOD 11 fy(x) = (i DIV 11) MOD 11 fz(z) = i DIV (11*11) Now, if your real 3D theory function is model(parameters; x,y,z), you would fit to a transformed function fitmodel(parameters; i)=model(paremeters; fx(i), fy(i), fz(i)) OTOH if, as I suspect, polyfit restricts its model function to a simple polynomial in one variable, then this won't work, because you don't get to choose your model function freely enough. But there's bound to be other, more general implementations of the least squares fitting algorithm somewhere in the vast size of MatLab that do let you do this --- or even let you input multi-variate functions as models straightforwardly. > I could derive the polynomial coefficients that would describe > temperature as a function of location in the room? To the extend that the polynomial coefficients to describe it actually exist: yes. -- Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de) Even if all the snow were burnt, ashes would remain. === Subject: Re: An Informal Call For Program Submissions To Compute pi(n) >(Unless, come to think of it, you allowed Mathematica >as a programming language - I believe I could write >a very short Mathematica program that would have a >pretty good shot... heh-heh.) Me, too. In Mathcad. In Mathcad: for b subset of (list) is a vector process that happens in no particular order, but as many time as there are items in the list. List can be a range, like 1, 2, ... 5; or a list of variables, any one of which can be a scalar, vector, or matrix. You don't type syntax in Mathcad, you click on it and fill in the blanks. There are only seven things you can click, and you can combine them in neat ways. The neat thing about this no particular order thing is it seems to happen all at once the way a Fourier transform of 2^n data points is broken into chunks which fit together. Like breaking down a document, copying double sided, collating the copies, and stapling, on a really good copier. Just bang bang bang. There a whole section on how to write in parallel instead of with indices. The performance benefit is many times. Yours, Doug Goncz (at aol dot com) Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/ I will be signing an amended Affidavit soon. === Subject: CAN you Order Reals? Is it true that there are only two closed linear orderings of R: the usual one and the mirror order? An ordering >= is closed, if the set {(x,y) | x>=y} is closed in R^2. By the way, does anybody know some references about closed (partial, linear) orderings of topological spaces? Seems that there's a lot of interesting things here. Simeon === Subject: Re: CAN you Order Reals? > Is it true that there are only two closed linear orderings of R: the > usual one and the mirror order? > An ordering >= is closed, if the set {(x,y) | x>=y} is closed in R^2. Yes. Consider the sets A={(x,y) | x>y}, B={(x,y) | x= is closed, the set B is open. But A is the mirror image of B across the diagonal, so it too is open. Therefore, A and B are two (nonempty) disjoint open sets whose union is R^2-D. But R^2-D has precisely two connected components (one on either side of the diagonal), so A and B must correspond to those components, giving the two possibilities for the ordering. -- Kevin === Subject: Reverse number sequence function I have a the following numbers: 1,2 and 3. When input into a function F(x), the the results should be as follows: F(1) = 3; F(2) = 2; F(3) = 1; What does F look like? Is there an easy way to do this - like a formula? === Subject: Re: Reverse number sequence function > I have a the following numbers: 1,2 and 3. When input into a function > F(x), the the results should be as follows: > F(1) = 3; > F(2) = 2; > F(3) = 1; > What does F look like? Is there an easy way to do this - like a > formula? It depends: how nice an F would you like? F(x) = -x+4 works F(x) = -(x-2)^3+2 also works. With not too much work you could define F(x) to be a function of sin or cos functions as well. In general, there are an infinite number of ways to define F(x) to give the results listed above. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Reverse number sequence function >I have a the following numbers: 1,2 and 3. When input into a function >F(x), the the results should be as follows: >F(1) = 3; >F(2) = 2; >F(3) = 1; >What does F look like? Is there an easy way to do this - like a >formula? You could use Lagrange interpolation: F(x) = 3(x-2)(x-3)/((1-2)(1-3) + 2(x-1)(x-3)/((2-1)(2-3)) + (x-1)(x-2)/((3-1)(3-2)). ************************ === Subject: Re: Reverse number sequence function handtheman12@hotmail.com discusses and asked: >I have a the following numbers: 1,2 and 3. When input into a function >F(x), the the results should be as follows: >F(1) = 3; >F(2) = 2; >F(3) = 1; >What does F look like? Is there an easy way to do this - like a >formula? Try graphing the function based on those points. G C === Subject: Floor-Sum let c,d positive integers. i try to simplify the following equation ( maple notation ): sum(floor((m/d),m=1..2*c); could anyone help me? christian ruether === Subject: Re: Floor-Sum > i try to simplify the following equation ( maple notation ): > sum(floor((m/d),m=1..2*c); > sum(floor(m/d),m=1,...,2c) = K(K-1)d/2 + (2c - Kd + 1)K, where > K = floor(2c/d) ACK! I totally messed that up, disregard === Subject: Re: Floor-Sum > i try to simplify the following equation ( maple notation ): > sum(floor((m/d),m=1..2*c); with the elegance of the solution (unlike most the sci.math problems I tackle) so I'll actually post it. sum(floor(m/d),m=1,...,2c) = K(K-1)d/2 + (2c - Kd + 1)K, where K = floor(2c/d) [Note that none of this requires we have the 2 before the c. Take that out and everything still holds (but keep the 2 that divides K(K-1)d)] Sam === Subject: Re: Floor-Sum >let c,d positive integers. >i try to simplify the following equation ( maple notation ): >sum(floor((m/d),m=1..2*c); Let F(d,n) = sum(floor(m/d), m=1..n). Note that floor(m/d) = k for kd <= m <= kd+d-1. So F(d,qd-1) = sum_{k=1}^{q-1} kd = (q-1)qd/2 and F(d,qd-1+r) = (q-1)qd/2 + qr if 0 <= r < d, i.e. F(d,n) = q(n+1) - q(q+1)d/2 where q = floor((n+1)/d) 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: Sets. > Has anybody ever tried formulating 'fuzzy mathematics', or other > abstract forms of mathematics? More specifically, have they ever tried > to do something like expand Zermelo-Fraenkel Set theory such that > states beyond existing and not existing can have a meaning? > (...Starblade Riven Darksquall...) Do you know the ng comp.ai.fuzzy? And the journal Fuzzy Sets and Systems? In the fuzzy set community there was, for a short time, such an initiative. They thought they could connect fuzzy logic with (the mathematical subject called) topos theory. AFAIK, the subject dried up, however, for unclear reasons. Anyway, the idea to give sets phases, or different contents at different timepoints will typically lead to some form of topos theory. In other words: you might be interested in topos theory. Herman Jurjus === Subject: Re: Sets. >> More generally, you cannot have a set of all sets satisfying P, where P >> is some arbitrary property. In fact, this restriction covers both of >> yours. What you can have instead is the principle of bounded >> comprehension, which says that if you have a set A and a property P, then >> you can form the subset B = { x in A : P(x) }. The set A serves as an >> upper bound for the size of the new set being constructed. The other >> kind is called unbounded comprehension and is what leads to paradoxes >> such as those of Russell, Cantor, and Burali-Forti. > What if we allow unbounded comprehensive sets? > Your question doesn't make sense. I was using comprehension as a noun, > not an adjective. >Or sets that do not > follow strict logic. That is, we allow something to exist in states > other than either existing or not existing. > What would happen if we injected fuzzy logic into Zermelo Fraenkel Set > theory? > We would get functions with codomain [0,1]. Not a big deal. Codomain? Well, what if we use a varity of neither-wholly-true-nor-wholly-false type logical operators? Is it possible to define a number x such that n^x is not greater than x where n is any number greater e^(1/e)? I know no numbers like that exist, but suppose we were to invent one? (...Starblade Riven Darksquall...) === Subject: Re: Sets. >> What would happen if we injected fuzzy logic into Zermelo Fraenkel Set >> theory? >> We would get functions with codomain [0,1]. Not a big deal. > Codomain? If X is a set, then a fuzzy set would be a mapping f: X -> [0,1]. The codomain of a function is the set in which the function takes its values. The point is, no radical reformulation of set theory is necessary. Fuzzy logic can be modelled perfectly well within the existing theory. > Well, what if we use a varity of neither-wholly-true-nor-wholly-false > type logical operators? > Is it possible to define a number x such that n^x is not greater than > x where n is any number greater e^(1/e)? Since e^(1/e) > 1, I don't know where you are going with this. -- Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: A question on Pythagorus theorem Hi all, I was thinking about Pythagoras theorem and was baffled by this : Let us say we have a staircase. The staircase is analogous to hypotenuse of a right angled triangle formed by the wall and floor. Suppose I decrease the size of each step on the stair case. At inifinity the staircase should look flat and the length should be equal to what I get from pythagoras theorem (with the steps becoming infinitesimally small). a/2 |------ | | b/2 | | a/2 | |------ | | | | b/2 | | |------------- Let us take height of the triangle as b and base as a. Let us have two steps in staircase in such a way that each step has a height b/2 and base a/2. The total length of staircase is (a/2 + b/2 + a/2 + b/2) = a + b. Let us iterate this for n times so that n steps are created (also size of the steps decreases). We get the length as {2^n / 2^n * ( a + b )}. This becomes an non-determinate form as it has no limit at inifinity. My question is how do we arrive at the length of hypotenuse (sqrt(a^2 + b^2)) from here ? Am I making any conceptual mistake here ? K.Srinivasan === Subject: Re: A question on Pythagorus theorem > Let us take height of the triangle as b and base as a. Let us have two > steps in staircase in such a way that each step has a height b/2 and > base a/2. The total length of staircase is (a/2 + b/2 + a/2 + b/2) = a > + b. Let us iterate this for n times so that n steps are created (also > size of the steps decreases). We get the length as > {2^n / 2^n * ( a + b )}. This becomes an non-determinate form > as it has no limit at inifinity. This is an indetermate form that does have a limit (as n goes to infinity). The limit is a+b. Afterall the expression is constant, equaling a+b for every n. > My question is how do we arrive at the length of hypotenuse > (sqrt(a^2 + b^2)) from here ? The total length of your saw-toothed staircase remains fixed at (a+b) at every stage. So the length of your staircase never approaches the length of the hypotenuse. > Am I making any conceptual mistake here ? As n gets very large, the distance between your steps and the hypotenuse becomes arbitrarily small. That does not mean that the total length of the steps approaches the length of the hypotenuse, as your example shows. === Subject: Re: A question on Pythagorus theorem > Hi all, > I was thinking about Pythagoras theorem and was baffled by this : > Let us say we have a staircase. The staircase is analogous to > hypotenuse of a right angled triangle formed by the wall and floor. > Suppose I decrease the size of each step on the stair case. At > inifinity the staircase should look flat and the length should be > equal to what I get from pythagoras theorem (with the steps becoming > infinitesimally small). > a/2 > |------ > | | b/2 > | | a/2 > | |------ > | | > | | b/2 > | | > |------------- > Let us take height of the triangle as b and base as a. Let us have two > steps in staircase in such a way that each step has a height b/2 and > base a/2. The total length of staircase is (a/2 + b/2 + a/2 + b/2) = a > + b. Let us iterate this for n times so that n steps are created (also > size of the steps decreases). We get the length as {2^n / 2^n * ( a + > b )}. This becomes an non-determinate form as it has no limit at > inifinity. 2^n/2^n = 1, so it simplifies to a+b. > My question is how do we arrive at the length of hypotenuse (sqrt(a^2 > + b^2)) from here ? You don't. What you are doing is minimizing the distance from the hypotenuse you travel, but at each stage you are only *on* the hypotenuse finitely many times. > Am I making any conceptual mistake here ? Yes. The hypotenuse has no vertical or horizontal components. Your staircases have *only* vertical and horizaontal components. Think of it this way, if an ant were going up your staircase, it would repeated change its facing. When it goes up the hypotenuse, it never changes facing. Therefor, no matter how similar they may appear without magnification, they are always *different* paths. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: A question on Pythagorus theorem >Hi all, >I was thinking about Pythagoras theorem and was baffled by this : >Let us say we have a staircase. The staircase is analogous to >hypotenuse of a right angled triangle formed by the wall and floor. >Suppose I decrease the size of each step on the stair case. At >inifinity the staircase should look flat and the length should be >equal to what I get from pythagoras theorem (with the steps becoming >infinitesimally small). > a/2 > |------ > | | b/2 > | | a/2 > | |------ > | | > | | b/2 > | | > |------------- >Let us take height of the triangle as b and base as a. Let us have two >steps in staircase in such a way that each step has a height b/2 and >base a/2. The total length of staircase is (a/2 + b/2 + a/2 + b/2) = a >+ b. Let us iterate this for n times so that n steps are created (also >size of the steps decreases). We get the length as {2^n / 2^n * ( a + >b )}. This becomes an non-determinate form as it has no limit at >inifinity. Huh? That expression certainly does have a limit at infinity. >My question is how do we arrive at the length of hypotenuse (sqrt(a^2 >+ b^2)) from here ? You don't. >Am I making any conceptual mistake here ? Yes. Even though those staircase curves approach the hypotenuse the _length_ of the staircase does _not_ approach the _length_ of the hypotenuse. >K.Srinivasan ************************ === Subject: Complement of a function Hi all, Is there a way to find a complement of a function defined on integers(Both domain and range are integers)? What I mean is : Suppose you have a set of positive integers = {1,2,3,4,5....} Let us say we create a subset of even integers out of it = {2,4,6,8,....} This set of even numbers can be generated by using the function f(x) = 2x, x=1,2,3... The complement of this set is set of odd numbers and they can be generated using g(x) = 2x-1 , x=1,2,3,4,.... So I define g(x)=2x-1 to be the complement function of f(x)=2x over the set of integers. Is there a generic way to arrive at the complement function given an arbitrary integer function? (I think we can do it using Taylor series. But seems more mechanical) K.Srinivasan === Subject: Re: Complement of a function > Hi all, > Is there a way to find a complement of a function defined on > integers(Both domain and range are integers)? > What I mean is : > Suppose you have a set of positive integers = {1,2,3,4,5....} > Let us say we create a subset of even integers out of it = > {2,4,6,8,....} > This set of even numbers can be generated by using the function f(x) = > 2x, x=1,2,3... > The complement of this set is set of odd numbers and they can be > generated using g(x) = 2x-1 , x=1,2,3,4,.... > So I define g(x)=2x-1 to be the complement function of f(x)=2x over > the set of integers. > Is there a generic way to arrive at the complement function given > an arbitrary integer function? (I think we can do it using Taylor > series. But seems more mechanical) > K.Srinivasan As another poster has pointed out, g will not be unique in general. A more serious problem is that even if f is computable then it doesn't follow that any computable g would work. The range of a computable function from integers to integers is by definition recursively enumerable and it is a standard result that the complement of an r.e. set need not be r.e. Hope this helps -john coleman === Subject: Re: Complement of a function > Hi all, > Is there a way to find a complement of a function defined on > integers(Both domain and range are integers)? > What I mean is : > Suppose you have a set of positive integers = {1,2,3,4,5....} > Let us say we create a subset of even integers out of it = > {2,4,6,8,....} > This set of even numbers can be generated by using the function f(x) = > 2x, x=1,2,3... > The complement of this set is set of odd numbers and they can be > generated using g(x) = 2x-1 , x=1,2,3,4,.... > So I define g(x)=2x-1 to be the complement function of f(x)=2x over > the set of integers. > Is there a generic way to arrive at the complement function given > an arbitrary integer function? (I think we can do it using Taylor > series. But seems more mechanical) > K.Srinivasan > As another poster has pointed out, g will not be unique in general. A > more serious problem is that even if f is computable then it doesn't > follow that any computable g would work. The range of a computable > function from integers to integers is by definition recursively > enumerable and it is a standard result that the complement of an r.e. > set need not be r.e. > Hope this helps If we restrict ourselves to functions from the positive integers into the positive integers and we take g to be an increasing function, then g is unique. If I calculated correctly, g would satisfy the equation g(f(x)-x) = f(x)-1. Of course this may not be of much use to actually compute g. _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 === Subject: Re: Complement of a function >> Is there a way to find a complement of a function defined on >> integers(Both domain and range are integers)? >> What I mean is : >> Suppose you have a set of positive integers = {1,2,3,4,5....} >> Let us say we create a subset of even integers out of it = >> {2,4,6,8,....} >> This set of even numbers can be generated by using the function f(x) = >> 2x, x=1,2,3... >> The complement of this set is set of odd numbers and they can be >> generated using g(x) = 2x-1 , x=1,2,3,4,.... >> So I define g(x)=2x-1 to be the complement function of f(x)=2x over >> the set of integers. >> Is there a generic way to arrive at the complement function given >> an arbitrary integer function? (I think we can do it using Taylor >> series. But seems more mechanical) I don't see what Taylor series could have to do with this. >> As another poster has pointed out, g will not be unique in general. A >> more serious problem is that even if f is computable then it doesn't >> follow that any computable g would work. The range of a computable >> function from integers to integers is by definition recursively >> enumerable and it is a standard result that the complement of an r.e. >> set need not be r.e. >If we restrict ourselves to functions from the positive integers into >the positive integers and we take g to be an >increasing function, then g is unique. If I calculated correctly, g >would satisfy the equation g(f(x)-x) = f(x)-1. Of This is wrong: e.g. if f(1)=1, f(2)=3, f(3)=4 then g(f(3)-3)=g(1)=2 but f(3)-1=3. >course this may not be of much use to actually compute g. I think you're also assuming f(x) is increasing. If we do that, then g is computable, e.g. g(x) = min {w >= x: f(w+1-x) > w}. (Of course, if g(x) doesn't exist, i.e. f misses fewer than x values, a search for such w would never terminate, and there is no way in general to predict this) 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: Complement of a function > Hi all, > Is there a way to find a complement of a function defined on > integers(Both domain and range are integers)? > What I mean is : > Suppose you have a set of positive integers = {1,2,3,4,5....} > Let us say we create a subset of even integers out of it = > {2,4,6,8,....} > This set of even numbers can be generated by using the function f(x) = > 2x, x=1,2,3... > The complement of this set is set of odd numbers and they can be > generated using g(x) = 2x-1 , x=1,2,3,4,.... > So I define g(x)=2x-1 to be the complement function of f(x)=2x over > the set of integers. > Is there a generic way to arrive at the complement function given > an arbitrary integer function? (I think we can do it using Taylor > series. But seems more mechanical) The complement will not, in general, be uniquely determined. For instance, if we use the whole set of integers and select the ones generated by f(n) = 2n, every function of the form g(n) = 2n - (2m + 1) is going to be a complement to f(n) for all integers m. With more complicated sets, even your generating function is not going to be uniquely determined. The problem is that you're going to begin with a set. Either you already know a generating function, or you'll be hard pressed to find one, since you'll have to know all terms of the sequence without knowing a generating function to give them to you. The best you can do, if you want a unique answer, is to give the set and its complement without speaking of functions. > K.Srinivasan === Subject: Re: Complement of a function > Hi all, > Is there a way to find a complement of a function defined on > integers(Both domain and range are integers)? > What I mean is : > Suppose you have a set of positive integers = {1,2,3,4,5....} > Let us say we create a subset of even integers out of it = > {2,4,6,8,....} > This set of even numbers can be generated by using the function f(x) = > 2x, x=1,2,3... > The complement of this set is set of odd numbers and they can be > generated using g(x) = 2x-1 , x=1,2,3,4,.... > So I define g(x)=2x-1 to be the complement function of f(x)=2x over > the set of integers. Note: g(x) is *a* complement function. h(x)=2x+1 is also a complement of f(x). Finding such a function may be non-trivial. Don't know if that helps or not. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Complement of a function > Hi all, > Is there a way to find a complement of a function defined on > integers(Both domain and range are integers)? What I mean is : > Suppose you have a set of positive integers = {1,2,3,4,5....} > Let us say we create a subset of even integers out of it = > {2,4,6,8,....} > This set of even numbers can be generated by using the function f(x) = > 2x, x=1,2,3... > The complement of this set is set of odd numbers and they can be > generated using g(x) = 2x-1 , x=1,2,3,4,.... > So I define g(x)=2x-1 to be the complement function of f(x)=2x over > the set of integers. > Note: g(x) is *a* complement function. h(x)=2x+1 is also a complement > of f(x). Finding such a function may be non-trivial. Don't know if > that helps or not. Seems to me this is more of a theory of computability issue than anything. For the given example, the OP essentially gave us a program for churning out a particular set of integers/naturals, and then gave us a program that churns out the complement of the first set. Of course there will in general be many programs to churn out the complement, if there are any at all. Among other things, if the original set (eg of evens, as per the OP) isn't recursive, then that's a good time to give up on a complement program.... Generally speaking, the notions of recursive set, recursively enumerable set, and the like seem quite relevant to the OP's question... cdj === Subject: Re: Complement of a function >Hi all, >Is there a way to find a complement of a function defined on >integers(Both domain and range are integers)? >What I mean is : >Suppose you have a set of positive integers = {1,2,3,4,5....} >Let us say we create a subset of even integers out of it = >{2,4,6,8,....} >This set of even numbers can be generated by using the function f(x) = >2x, x=1,2,3... >The complement of this set is set of odd numbers and they can be >generated using g(x) = 2x-1 , x=1,2,3,4,.... >So I define g(x)=2x-1 to be the complement function of f(x)=2x over >the set of integers. >Is there a generic way to arrive at the complement function given >an arbitrary integer function? (I think we can do it using Taylor >series. But seems more mechanical) How do you do it using Taylor series? >K.Srinivasan ************************ === Subject: Re: Complement of a function >>Is there a way to find a complement of a function defined on >>integers(Both domain and range are integers)? >>What I mean is : >>Suppose you have a set of positive integers = {1,2,3,4,5....} >>Let us say we create a subset of even integers out of it = >>{2,4,6,8,....} >>This set of even numbers can be generated by using the function f(x) = >>2x, x=1,2,3... >>The complement of this set is set of odd numbers and they can be >>generated using g(x) = 2x-1 , x=1,2,3,4,.... >>So I define g(x)=2x-1 to be the complement function of f(x)=2x over >>the set of integers. >>Is there a generic way to arrive at the complement function given >>an arbitrary integer function? (I think we can do it using Taylor >>series. But seems more mechanical) > How do you do it using Taylor series? generating functions? 1/(1-z) - f(z) the function has to be strictly increasing and the gf is really a characteristic function (all coeffs are 0 or 1) e.g. 2n <=> 1/(1-x^2), not 2/(1-x)^2 Mitch === Subject: Re: Complement of a function >Is there a way to find a complement of a function defined on >integers(Both domain and range are integers)? >What I mean is : >Suppose you have a set of positive integers = {1,2,3,4,5....} >Let us say we create a subset of even integers out of it = >{2,4,6,8,....} >This set of even numbers can be generated by using the function f(x) = >2x, x=1,2,3... >The complement of this set is set of odd numbers and they can be >generated using g(x) = 2x-1 , x=1,2,3,4,.... >So I define g(x)=2x-1 to be the complement function of f(x)=2x over >the set of integers. >Is there a generic way to arrive at the complement function given >an arbitrary integer function? (I think we can do it using Taylor >series. But seems more mechanical) >> How do you do it using Taylor series? >generating functions? 1/(1-z) - f(z) That's a way to use power series to obtain the complement of a set. It doesn't give the complement function. (Given a function f with _range_ a subset of f we want another function g with range equal to the complement of the range of f...) >the function has to be strictly increasing and the gf is really a >characteristic function (all coeffs are 0 or 1) >e.g. 2n <=> 1/(1-x^2), not 2/(1-x)^2 >Mitch ************************ === Subject: Re: Complement of a function > >>Is there a way to find a complement of a function defined on >>integers(Both domain and range are integers)? >> >>What I mean is : >>Suppose you have a set of positive integers = {1,2,3,4,5....} >>Let us say we create a subset of even integers out of it = >>{2,4,6,8,....} >>This set of even numbers can be generated by using the function f(x) = >>2x, x=1,2,3... >>The complement of this set is set of odd numbers and they can be >>generated using g(x) = 2x-1 , x=1,2,3,4,.... >>So I define g(x)=2x-1 to be the complement function of f(x)=2x over >>the set of integers. >> >>Is there a generic way to arrive at the complement function given >>an arbitrary integer function? (I think we can do it using Taylor >>series. But seems more mechanical) > >How do you do it using Taylor series? >>generating functions? 1/(1-z) - f(z) > That's a way to use power series to obtain the complement of > a set. It doesn't give the complement function. (Given a function > f with _range_ a subset of f we want another function g with > range equal to the complement of the range of f...) Hmmm... then just extract the function from this characteristic function? 2n <=> 1/(1-x^2) 1/(1-x) - 1/(1-x^2) = x/(1-x^2) <=> 2n-1 (not really a generating function coeff extraction but easy to see anyway) Yes, this last step is not particularly tractable for anything other than combinations of linear functions. (I think quadratic functions have a basis using theta functions) Do you see another way for the original problem? Mitch === Subject: Re: Complement of a function >Is there a way to find a complement of a function defined on >>integers(Both domain and range are integers)? >>What I mean is : >>Suppose you have a set of positive integers = {1,2,3,4,5....} >>Let us say we create a subset of even integers out of it = >>{2,4,6,8,....} >>This set of even numbers can be generated by using the function f(x) = >>2x, x=1,2,3... >>The complement of this set is set of odd numbers and they can be >>generated using g(x) = 2x-1 , x=1,2,3,4,.... >>So I define g(x)=2x-1 to be the complement function of f(x)=2x over >>the set of integers. >>Is there a generic way to arrive at the complement function given >>an arbitrary integer function? (I think we can do it using Taylor >>series. But seems more mechanical) How do you do it using Taylor series? >>generating functions? 1/(1-z) - f(z) > That's a way to use power series to obtain the complement of > a set. It doesn't give the complement function. (Given a function > f with _range_ a subset of f we want another function g with > range equal to the complement of the range of f...) >Hmmm... then just extract the function from this characteristic function? Extracting a function whose _values_ are the indices where the generating function has a non-zero coefficient is exactly equivalent to the original problem. >2n <=> 1/(1-x^2) > 1/(1-x) - 1/(1-x^2) > = x/(1-x^2) > <=> 2n-1 >(not really a generating function coeff extraction but easy to see anyway) >Yes, this last step is not particularly tractable for anything other >than combinations of linear functions. (I think quadratic functions have >a basis using theta functions) >Do you see another way for the original problem? No, it seems to me like the sort of problem that one should simply not expect a simple answer to. >Mitch ************************ === Subject: Point - Curve distance I am looking for an algorithm to compute the distance between a point and a curve in 3D. My curve is a Catmull-Rom spline (curve going through all the control points). Francois. === Subject: Re: Point - Curve distance > I am looking for an algorithm to compute the distance between a point and a > curve in 3D. > My curve is a Catmull-Rom spline (curve going through all the control > points). > Francois. First, make a set of evalution points to determine which one is closer to that point. If that curve consists of N control points, I will suggest N control points and N-1 points between each control points using (U(i)+U(i+1))/2. once you find the point of the shortest distance. letting that position is i, then just do the search loop between U(i-1) ~ U(i+1) with initial condition U(i) using bisection method or any other method you prefered to find the minimal distance. Maybe there must be more fancy algorithms, I've solved that problem with this way and offered quite satisfactory result. D.K Lee === Subject: Re: probability and/or logic puzzle > Yes, this came out of a textbook . . . > but I passed that class about 25 years ago, > so I'm not cheating on my homework. > Start with a normal deck of playing cards. > Discard all but the aces and kings. Now you > have eight cards left. Your friend draws > two cards and hides them from you. He truthfully > tells you that at least one of his cards is an ace. > What is the probability that he holds two aces? > > He replaces the cards and you shuffle them well. > Your friend again draws two cards, and truthfully > tells you that one of them is the ace of spades. > What is the probability that he now holds two aces? > (Hint: this isn't the same answer as above!) > Okay . . . I don't get it. Why isn't it the same > answer both times? > Also, in case it matters, this isn't a math > book. It's a (non-mathematical) logic book. > Ted Shoemaker > shoemakerted@yahoo.com You are right, the book is wrong. This question, again illustrates the difference between being given at least one, and being told at least one. We've had this argument previously in this forum. The illustrious Dr. Ullrich has refused to acknowledge that there is an argument. Dr. Holt made a feeble attempt to understand the argument. P(A|B)=P(B|A)P(A)/P(B) Where B is defined as the event which has already happened. B is the key. Defining B. Defining the event which has definitely happened. There are two cards. Call them X and Y. X can represent Kings, and Y Aces, or vice versa. We don't need to know which is which. B is different when we are given at least one X, than when we are told that there is at least one X. Something different has definitely happened. When we are given at least one X: P(given X|XX)=P(given X|XY) When we are told at least one X: P(told X|XX)>P(told X|XY) The people who got this wrong are in good company. Marilyn had it wrong in her column and in one of her books. It was sent in by her guru, Martin Gardner, who also had it wrong in his book, Aha Gotcha! The secret is in B. What has definitely happened. Notice that to get the popular answer, the X=Ace, or X=King, must be a known, prior definition, ie, our friend must know which denominations she is looking for. In other words, to get other than the 3/7 answer, takes more information. Eldon Moritz === Subject: NTT transform for vectors of arbitrary length in GF(2^m)? I am looking for a NTT transform that can transform vectors in GF(2^m), but the vectors should not be constrained to length 2^m -1. The vectors that I want to investigate are seldom of length (2^m) - 1, as is required with the NTT transforms I have found thus far. Does anyone know where I can find any literature on these transforms (also known as finite field Fourier transforms, if I am right)? The nice property of the transform that I have used is that the transformed vector have zero's in the coefficients corresponding to the roots of the original vector. Is this true for all the NTTs? Also, can a NTT be used to factor a polynomial? If the roots cannot be found in GF(2^m), look for roots in GF(2^h), where h > m ? Your time, effort and suggestions will be much appreciated Jaco === Subject: Primecounter contest results I now have all the submissions for the primecounter contest. The results are in http://home.earthlink.net/~ewill3/math/primecounters/index.html and I got 32 entries (8 of them mine). Congratulations to all who participated; it was a very interesting contest. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === === Subject: Orthogonal Polynomials Weight Functions Given a three term recursion relation in the canonical form capable of generating orthogonal polynnomials is there any algorithm that can find the weight function or m-distribution associated with this recursion? === Subject: Re: Orthogonal Polynomials Weight Functions > Given a three term recursion relation in the canonical form > capable of generating orthogonal polynnomials is there any > algorithm that can find the weight function or m-distribution > associated with this recursion? The fact that such a measure exists is known as Favard's theorem and the standard proof is fairly constructive. If I recall correctly, it should be possible to extract a convergent sequence for this measure out of it. -- Jane - Daria? Come on, the neighbors are starting to talk. Daria - Um... good. Soon they'll progress to cave drawings and civilization will be on its way. === Subject: Is a polytope a projections of a hypercubes? I can represent some simple polytopes in 2 dimensions as a projections of the 3-dim. unit cube. I would like to know if this can be done in general, i.e. Can any polytope be represented as a projection of a hypercube? I've done some initial searching on this problem, but I haven't found much useful information. Are there useful textbooks / keywords that I should search for? Pete Seiler === Subject: Re: Is a polytope a projections of a hypercubes? > I can represent some simple polytopes in 2 dimensions as a projections of > the 3-dim. unit cube. I would like to know if this can be done in > general, i.e. Can any polytope be represented as a projection of a > hypercube? I've done some initial searching on this problem, but I haven't > found much useful information. Are there useful textbooks / keywords that > I should search for? > Pete Seiler There is a theorem of Bessaga to the effect that every finite dimensional Banach space with a polyhedral unit ball is isometric to a subspace of c_0. I think there is a finite dimensional version which would have any finite dimensional Banach space of dimension n with polyhedral unit ball is isometric to a subspace of k(n)-dimensional l_infty. The dimension of the hypercube could get higher and higher depending on the dimension of the polytope. This isn't exactly what you are interested in because the unit balls are centrally symmetric. === Subject: Re: Is a polytope a projections of a hypercubes? > I can represent some simple polytopes in 2 dimensions as a projections of > the 3-dim. unit cube. I would like to know if this can be done in > general, i.e. Can any polytope be represented as a projection of a > hypercube? I've done some initial searching on this problem, but I haven't > found much useful information. Are there useful textbooks / keywords that > I should search for? > Pete Seiler I don't quite understand. A regular triangle is not centrally symetric, but a projection of a hypercube is. === Subject: Re: Is a polytope a projections of a hypercubes? 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > I can represent some simple polytopes in 2 dimensions as a projections of > the 3-dim. unit cube. I would like to know if this can be done in > general, i.e. Can any polytope be represented as a projection of a > hypercube? I've done some initial searching on this problem, but I haven't > found much useful information. Are there useful textbooks / keywords that > I should search for? > I don't quite understand. A regular triangle is not centrally > symetric, but a projection of a hypercube is. Maybe he means perspective projection, not just parallel projection. I have a paper with some work on a related problem: perspective projections of zonotopes (which are themselves parallel projections of hypercubes...) at That paper discusses the centroid polytope C(S) where S is a set of points each associated with an interval of possible weights, and C(S) is the set of points formed by assigning a weight in its interval to each point in S and taking a weighted average; it turns out that C(S) is a perspective projection of a zonotope. When each interval is [0,1], C(S) is just the convex hull of S, so any polytope can be realized in this way. Probably it's possible to see this more directly somehow. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Is a polytope a projections of a hypercubes? working on, all of the examples I tried were symmetric polytopes (i.e. if x is in the polytope, then -x is in the polytope). If I restrict myself to polytopes that have this symmetry: can every such polytope be represented as the projection of a hypercube? I'm interested in what literature I should be looking at to answer this question. Pete > I can represent some simple polytopes in 2 dimensions as a projections of > the 3-dim. unit cube. I would like to know if this can be done in > general, i.e. Can any polytope be represented as a projection of a > hypercube? I've done some initial searching on this problem, but I haven't > found much useful information. Are there useful textbooks / keywords that > I should search for? Pete Seiler I don't quite understand. A regular triangle is not centrally > symetric, but a projection of a hypercube is. === Subject: Re: Is a polytope a projections of a hypercubes? 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > working on, all of the examples I tried were symmetric polytopes (i.e. if > x is in the polytope, then -x is in the polytope). If I restrict myself > to polytopes that have this symmetry: can every such polytope be > represented as the projection of a hypercube? I'm interested in what > literature I should be looking at to answer this question. The parallel projections of hypercubes are called zonotopes. They are characterized by the property that the whole polytope and each of its faces must be centrally symmetric. So a centrally symmetric polytope with non-centrally-symmetric faces, such as a regular octahedron, would be another counterexample. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: E=mc^2 I don't understand the point of the squared. Seems to me:- Since c is a constant, and the speed of light is not measured in the same units as mass or energy, c is just a 'matching' constant. and c^2 is just another matching constant So the equation (E =mc) ,without the squared, is equally as valid Is that right? Tony === Subject: Re: E=mc^2 > I don't understand the point of the squared. > Seems to me:- > Since c is a constant, and the speed of light is not measured in the > same units as mass or energy, c is just a 'matching' constant. and > c^2 is just another matching constant So the equation > (E =mc) ,without the squared, is equally as valid > Is that right? > Tony No! E, m and c are not merely numbers, they are measurements, and, as measurements have appropriate units of measurement. The type of units in an equation such as E = m*c^2 must balance as well as the numerical values, i.e., if the units of E do not match the units of m*c^2, you have a major problem, and no equality. And the units of c^2 are of a different type than the units of c. === Subject: Re: E=mc^2 > Seems to me:- > Since c is a constant, and the speed of light is not measured in the > same units as mass or energy, c is just a 'matching' constant. and > c^2 is just another matching constant So the equation > (E =mc) ,without the squared, is equally as valid > Is that right? Brain wave energy of E = m/(c^2) has been detected in some humans. === Subject: Re: E=mc^2 >I don't understand the point of the squared. >Seems to me:- >Since c is a constant, and the speed of light is not measured in the >same units as mass or energy, c is just a 'matching' constant. and >c^2 is just another matching constant So the equation >(E =mc) ,without the squared, is equally as valid No. c is measured in meters/second, and m is measured in kilograms. Therefore, mc^2 is in units of kilograms*meters^2/seconds^2, which happens to be equal to joules, the units associated with energy. Without the squares, you don't get that. Doug === Subject: Re: E=mc^2 > I don't understand the point of the squared. > Seems to me:- > Since c is a constant, and the speed of light is not measured in the > same units as mass or energy, c is just a 'matching' constant. and > c^2 is just another matching constant So the equation > (E =mc) ,without the squared, is equally as valid > Is that right? > Tony There are (I think) seven fundamental units in SI. The Joule is not one of them - it is derived from the others. One Joule is one Newton exerted over one meter. One Newton is the force required to accelerate a kilogram at one meter per second squared. Therefore the speed of light is indeed measured in the same system of units as energy and is not an arbitrary constant. Mark Atherton === Subject: Re: E=mc^2 >I don't understand the point of the squared. The only point is that it appears to model the observed phenomena well . >Seems to me:- >Since c is a constant, and the speed of light is not measured in the >same units as mass or energy, c is just a 'matching' constant. and >c^2 is just another matching constant So the equation >(E =mc) ,without the squared, is equally as valid >Is that right? >Tony === Subject: Re: E=mc^2 >>I don't understand the point of the squared. > The only point is that it appears to model the observed >phenomena well . That's far from the only point. Units are important. Doug === Subject: Terminology question What do you call an equation of the fourth degree in English? Degree 2 = quadratic? Degree 3 = cubic? Degree 4 = ????????????????????????????????????????????????????????? Degree 5 = quintic? Degree 6 = sextic? hextic? Degree 7 = septic? heptic? Degree 8 = octic? And so on. I don't think degrees from 9 onwards are ever needed in undergraduate mathematics... -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ To doo bee doo bee doo. - Frank Sinatra === Subject: Re: Terminology question > What do you call an equation of the fourth degree in English? > Degree 2 = quadratic? > Degree 3 = cubic? > Degree 4 = ????????????????????????????????????????????????????????? > Degree 5 = quintic? > Degree 6 = sextic? hextic? > Degree 7 = septic? heptic? > Degree 8 = octic? > And so on. I don't think degrees from 9 onwards are ever needed in > undergraduate mathematics... Quartic is the standard term for a fourth degree polynomial. === Subject: Re: Terminology question Visiting Assistant Professor at the University of Montana. >What do you call an equation of the fourth degree in English? The old term is biquadratic. Some call them quartic. Many call them of degree 4. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Terminology question I have always heard quartic, as Arturo said. Lurch >What do you call an equation of the fourth degree in English? > The old term is biquadratic. Some call them quartic. Many call > them of degree 4. > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Volume Hi~ I was hoping someone could help with a math problem that I am can not seem to solve. I have a volume in spherical coordinates defined as r = F(theta,phi) where r is the radial distance, theta is the angle from the z axis (0..Pi) and phi is the angle in the xy plane (0..2*Pi). Now I think that the volume of this should be: vol = integral(integral(integral(s^2 * sin(theta) ds*dtheta*dphi, s = 0..F(theta,phi) ), theta = 0..Pi ), phi = 0..2*Pi ) But when I try and do this, in maple, for some given F (such as F=cos(theta)*cos(phi) ) I get the vol = 0. Can someone please give me the correct formula for the volume? My second problem is a little more complicated. Suppose that I have now drawn a line circumscribing my volume. The line being defined as r = F( theta(phi),phi) where theta(phi) is some combination of cosines of phi say for example theta(phi) = cos(phi) +cos(2*phi). Now since the same line is drawn for negative phi I can connect the points ( F(theta(phi),phi) , theta(phi) , phi) and ( F(theta(-phi),-phi) , theta(-phi) , -phi) with a line. This defines a surface which is dividing the volume (r = F(theta,phi)). I now wish to seek the volume of the part above this surface. How do I do this?? Any help would be greatly appreciated. === Subject: Is ...9999.9999... = 0 ? Is ...9999.9999... = 0 ? Consider the Manipulation: ...9999.9999... = ...9999.0000... + ...0000.9999... = ...9999.0 +(1.0-1.0)+ 0.9999... = ...9999.0 +1.0 -(1.0 - 0.999...)= ...00000.0 - 0.0000... = 0.0 = 0 since ...9999.0 +1.0 = ...0000.0 by carrying to the left (try it!) and 1.0 - 0.9999... = 0.0000... by borrowing to the right. Woops, I may be getting my topologies confused(*), But can anybody obtain a real contradiction(**) from assuming ...9999.9999... = 0 and the things it would clearly imply like ...3333.0 = - 0.3333... ? * It's kind of like trying to keep your feet in seperate boats, that are being pulled apart. ** Such as 0=1, but no fair dividing by zero or nothing like that. (: I hope this note isn't offensive to those trying to teach kids that 1/3 = .3333.... :) === Subject: Re: Is ...9999.9999... = 0 ? > Is ...9999.9999... = 0 ? The expression...9999.9999... does not represent any real number, in any usual sense, since a (presumably decimal) representation of a real number must have a most significant digit, which ...9999.9999... does not have. === Subject: Re: Is ...9999.9999... = 0 ? >(: I hope this note isn't offensive to those trying to teach kids >that 1/3 = .3333.... :) Those of us teaching kids that 1/3 = 0.3333... know the difference between that and what you just did. HTH. Doug === Subject: Interesting Inequality *** post for FREE via your newsreader at post.newsfeed.com *** Show that a^b + b^a <= a^a + b^b for all positive real numbers a and b. http://www.newsfeed.com - The #1 Newsgroup Service in the World! -----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- === Subject: Re: MS in Math or Comp Sci?? CAF> I take what I said back. My statement did not convey the CAF> whole picture. Working with people who are knowledgeable CAF> in the field is definitely a good thing. That was very graciously said. xanthian. CAF> So someone should spend 2 years of his life in grad CAF> school in order to get the advantage and luxury of CAF> reading his professor's scrap book? KPD>> More like seven years, if you hope to become a front line KPD>> constributor to the state of the art. KPD>> And yes, exactly so. When I took graph theory from KPD>> Stephen Olariu(sp?), in his office he had a 30 cm high KPD>> stack of faxes, which contained all the recent work of he KPD>> and the 18 others in the world who could understand each KPD>> others papers. Each of us floundering grad students took KPD>> a paper from that stack, read it for comprehension and KPD>> presented it to the class. KPD>> It just isn't possible to get much closer to the state of KPD>> the art than that, both because the publishing cycle is KPD>> so slow for textbooks, and because the reward for writing KPD>> a textbook in such a rarified field is so unlikely to KPD>> match the effort involved. KPD>> While self-education is possible and fruitful in some KPD>> areas, others are so intrinsically difficult that the KPD>> mentorship of a professor is a necessity of life for all KPD>> but the most hyper-intelligent students. KPD>> My abject apologies all these years later to the rest of KPD>> the class for how weak my presentation was. There is no KPD>> question my grades in graph theory were gifts, pure and KPD>> simple. -- === === Subject: Improved prime counting function A new and improved implementation of the Extended Meissel-Lehmer Algorithm for the calculation of pi (N), the number of primes <= N written in C++ can be found at my webpage at www.cbau.freeserve.co.uk This implementation comes with a test program that calculates various values of pi (N) for N from 10^3 to 10^18. All the results for pi (N) for 3 <= N <= 18 agree with the results published at www.mathword.com. All the values from the original paper by Lagarias, Miller and Odlyzko are tested and agree with the published results. The calculation times on a Macintosh running at 733 MHz, using the CodeWarrior 7 compiler, are as follows: pi (10^15) < 60 seconds pi (10^16) 4 minutes 42 seconds pi (10^17) 24 minutes 40 seconds seconds pi (10^18) 110 minutes pi (1.9*10^13) less than 1 second For comparison: Harris' best code, based on Legendre's formula, took over fifty hours to find pi (10^16); that is more than 600 times slower, and that factor grows as N grows larger. A very short overview of the mathematics involved: Let phi (N, k) be the number of integers <= N which are not divisible by any of the first primes. Let phi (N, x, k) = (-1)^j * phi (N/x, k) if x is the product of exactly j prime numbers. Then phi (N, x, k) can be calculated using the recursion formula phi (N, x, k) = phi (N, x, k-1) + phi (N, x*p(k), k-1) for k > 0, where p(k) is the k-smallest prime. Calculation of phi (N, k) for small k is quite simple, for example phi (N, 5) = 480 * floor (N / 2310) + phi (N mod 2310, 5) which can be found easily using a small table of 2310 values. If k = pi (N^(1/2)), then pi (N) = phi (N, 1, k) + (k - 1). The basic idea of the Extended Meissel-Lehmer Algorithm is as follows: We choose an integer c >= 1 such that phi (N, c) can be calculated quickly using a table-based formula. Choose an integer M, N^(1/3) <= M <= N^(1/2). Let k2 = pi (floor (N^(1/2))) or k2 = pi (ceil (N^(1/2))). Then start with pi (N) = phi (N, 1, k2) + (k2 - 1) and apply the recursion formula until either the third argument is <= c or the second argument is greater than M. If we let k3 = pi (N^(1/3)), then a careful analysis shows that pi (N) can be found by adding the following values: 1. k2 - 1 2. phi (N, x, c) for square-free integers 1 <= x <= M with no prime factor p <= p (c) 3. phi (N, x * p(k+1), k) for square-free integers x such that floor (M / p(k+1) + 1) <= x <= M, with no prime factor p <= p (k+1), for those k where c <= k < pi (M) and p (k+1) * p(k+2) <= M. 4. phi (N, x * p(k+1), k) for primes x, p(k+2) <= x <= M, for those k where c <= k < pi (M) and p (k+1) * p (k+2) > M. 5. phi (N, 1 * p(k+1), k3) + (k - k3) for those k where max (pi (M), c, k3) <= k <= k2 - 1. The calculations of values phi (N, x, k) for k > c is implemented by creating a sieve of size N / M, and about M^2 / log^2 M values need to be looked up in this sieve. That is the mathematics, the rest is just clever implementation. With a good implementation, the execution time of this algorithm grows with N^(2/3). For example, my implementation on a Macintosh takes about 4.4 * N^(2/3) processor clocks when N = 10^15. === Subject: Vector Cross Product Is there a vector cross product defined for 7 dimensional vectors? If so, what is the definition? === Subject: Re: Vector Cross Product > Is there a vector cross product defined for 7 dimensional vectors? Yes, based on the purely imaginary Cayley numbers (alias octonions) > If so, what is the definition? Let e_1, ..., e_7 be the canonical basis of R^7. Then the product of e_i with e_j is given by this matrix (source: http://math.ucr.edu/home/baez/Octonions/node3.html ): e_j: e_1 e_2 e_3 e_4 e_5 e_6 e_7 e_i: e_1 0 e_4 e_7 -e_2 e_6 -e_5 -e_3 e_2 -e_4 0 e_5 e_1 -e_3 e_7 -e_6 e_3 -e_7 -e_5 0 e_6 e_2 -e_4 e_1 e_4 e_2 -e_1 -e_6 0 e_7 e_3 -e_5 e_5 -e_6 e_3 -e_2 -e_7 0 e_1 e_4 e_6 e_5 -e_7 e_4 -e_3 -e_1 0 e_2 e_7 e_3 e_6 -e_1 e_5 -e_4 -e_2 0 Extend to all of R^7 by bilinearity; that is, sum_i(r_i e_i) x sum_j(s_j e_j) = sum_ij (r_i s_j e_i x e_j) Dale. === Subject: Re: Vector Cross Product A jumbled mess. I'll do this one last time, and if it doesn't turn out legible, I'll refer you to the web page: http://math.ucr.edu/home/baez/Octonions/node3.html : e_j: e_1 e_2 e_3 e_4 e_5 e_6 e_7 e_i: e_1 0 e_4 e_7 -e_2 e_6 -e_5 -e_3 e_2 -e_4 0 e_5 e_1 -e_3 e_7 -e_6 e_3 -e_7 -e_5 0 e_6 e_2 -e_4 e_1 e_4 e_2 -e_1 -e_6 0 e_7 e_3 -e_5 e_5 -e_6 e_3 -e_2 -e_7 0 e_1 e_4 e_6 e_5 -e_7 e_4 -e_3 -e_1 0 e_2 e_7 e_3 e_6 -e_1 e_5 -e_4 -e_2 0 Dale On second thought, perhaps this process (of mine) would have been a better candidate for alt.test or some such scratch newsgroup? === Subject: Re: Vector Cross Product Let's try that spacing just one more time. > Let e_1, ..., e_7 be the canonical basis of R^7. > Then the product of e_i with e_j is given by this matrix (source: > http://math.ucr.edu/home/baez/Octonions/node3.html ): > > e_j: > e_1 e_2 e_3 e_4 e_5 e_6 e_7 > e_i: > e_1 0 e_4 e_7 -e_2 e_6 -e_5 -e_3 > e_2 -e_4 0 e_5 e_1 -e_3 e_7 -e_6 > e_3 -e_7 -e_5 0 e_6 e_2 -e_4 e_1 > e_4 e_2 -e_1 -e_6 0 e_7 e_3 -e_5 > e_5 -e_6 e_3 -e_2 -e_7 0 e_1 e_4 > e_6 e_5 -e_7 e_4 -e_3 -e_1 0 e_2 > e_7 e_3 e_6 -e_1 e_5 -e_4 -e_2 0 > Extend to all of R^7 by bilinearity; that is, > sum_i(r_i e_i) x sum_j(s_j e_j) = sum_ij (r_i s_j e_i x e_j) > Dale. There. All spaces, no tabs. I hope it's more legible. Dale === Subject: Re: Vector Cross Product mattsdad2, > Is there a vector cross product defined for 7 dimensional vectors? > If so, what is the definition? Yes there is - but you need six of them at a time. The generalized n-dimensional cross-product produces a vector normal to n-1 given vectors. You may know that one way of writing the calculation for a x b in 3 dimensions is the determinant | i x_a x_b | | j y_a y_b | | k z_a z_b | where i, j and k are the unit vectors along the axes. The n-dimensional cross product of n-1 vectors is calculated analogously. === Subject: Please assist... Hello Could somebody please confirm if I am on the correct path with the following? Determine if the following is a vaild, recursive defn of a function f from the set of nonnegative integers to the set of integers. If f is well defined find a formula for f(n) when n is a nonnegative integer and prove that your formula is valid. F(0) =1, F(n) = -F(n-1) for n >= 1. ANSWER F(n) = -F(1-1) = 0 As 0 is less than and not equal to 1, the definition is invalid. === === Subject: Order of an element If the order of a is 3 mod p, show that the order of (a + 1) is 6. 3 must divide EulerPhi[p] = p - 1, so p = 3K + 1 for some K an element of the natural numbers. I need to show that (a + 1)^3 is congruent to -1 (mod p). Then, its square, (a + 1)^6 will be congruent to 1 (mod p). a^3 = (3K + 1)m + 1 for some m an element of the natural numbers. (a + 1)^3 = (3K + 1)n - 1 for some n and element of the natural numbers. a^3 + 3a^2 + 3a + 1 = (3K + 1)n - 1 a^3 + 3a^2 + 3a + 2 = (3K + 1)n a^3 - 1= (3K + 1)m Can someone help at this point? === Subject: Second Order PDE (Linear) Can anyone tell me if the equation P_xx + P_yy + A P_x + B P_y + C P = 0 has a known method of solution? (A, B, and C are functions of x and y.) I posted something a couple of months ago where I assumed that there was such a method, but I have not been able to find it. === Subject: Continued-Fraction Root? (generalized) Let {f_n(x)} be an infinite sequence of real-> real functions (where the continued fraction is defined and converges). (f(x) is NOT necessarily a positive integer!) Then, given a real y and the sequence of {f_n}, how can we find the x('s) which are the root of this, if any x are a root and the CF converges?: y = 1 f_0(x) + --------------------- 1 f_1(x) + --------------- 1 f_2(x) + --------- f_3(x) +.... In linear-mode: y = f_0(x) + 1/(f_1(x) +1/(f_2(x) +...)) For example, for those y where a solution exists, which x give: y = 1 + 1/(x +1/(x^2 +1/(x^3 +1/(x^4 +... )))) ? Leroy Quet === Subject: Math Cheat Sheet Thought some of you might like to have this (for undergrads I suppose). http://bit.csc.lsu.edu/~seiden/cheat.pdf