mm-380 === Subject: : Re: the anticlassicalist }{ i: linguistic negation: If you like Larry's short paper, you might want to look at his book, *A: Natural History of Negation*. He's the leading expert on the topic.The descriptions of that book look very, very promising, and my localuniversity has it on its shelves. Thank you for the suggestion; I will pickit up tomorrow!-- === === Subject: : Re: e*pi is irrational>> This proof actually supports the slightly stronger>> statement that e+pi and e*pi are not both algebraic.>Well, thanks. Jokes and proofs must be remembered>correctly - or they lack someting :-)Let u=e+pi, and v=e*pi.Also, u and v are both positive, and e < pi.Let uu = u/2.v = (uu+x)*(uu-x) = uu^2 - x^2.Hence, x = v - uu^2.And e = uu-x, pi = uu+x.They *had better* not be both algebraic, or e and pi would be too! Savardhttp://home.ecn.ab.ca/~jsavard/index.html=== === Subject: : Re: What is the Origin of Space and Time?>What is the origin of space and time?Where and when could space and time have originated, if there was nospace and time before they originated? Savardhttp://home.ecn.ab.ca/~jsavard/index.html=== === Subject: : Re: What is the Origin of Space and Time?>> What is the origin of space and time?>>> Space and time is the 4D fabric within which all life>> has evolved.>> Wrong. It is in fact the 8th dimentional time-space fabric.>> See 'The adventurures of Buckaroo Banzai across the 8th dimension'.>> Do your homework, kid!>Uh, I think it is you who needs to do the homeowork. Everyone knows there>are exactly FIVE dimensions. Every hear about a little something called>'Twilight Zone' (aka the fifth dimension) !?!?! I bet you feel pretty dumb>right about now.Actually, we're not sure. There may be 10 or 11 dimensions, dependingon whether you go with superstrings or supergravity. But there's a newtheory that says the extra dimensions aren't curled up into littleballs or toruses after all, and the Universe is held in four of thosedimensions through a different mechanism...On the other hand, I remember once from an old Flash comic book wherethe Flash and Kid Flash visited a world with 24 dimensions, inhabitedby tall grey beings who looked very three-dimensionally humanoid... Savardhttp://home.ecn.ab.ca/~jsavard/index.html=== === Subject: : Re: The role of infinity in math>What is the role of infinity in math:The role of completed infinity is limited to stuff like axiomatic settheory, but incomplete infinity pervades mathematics.>How is it defined?>Why is it needed?>At what precition does math work?If math were limited to any set precision, it would be merecomputation, and if a higher precision were needed, then we would haveto work out math all over again.That's why infinity is needed in math, so we get it right the firsttime. Savardhttp://home.ecn.ab.ca/~jsavard/index.html=== === Subject: : Re: Mathcad Upgrade QuestionMathcad 11 is worth the upgrade but I have found that it doesn't workreliably on WIN98. I upgraded to WINNT and it works fine. I also hatec-dilla, but it should do anything bad to your system unless you have somesoftware that scans for trojans and stealth programs. I would rather havea USB dongle.> Don't use mathcad. I installed version 2001i this weekend. After> installition I noticed that c-dilla had also been introduced to myregistry.> Google c-dilla to find out what this program does. Anyhow I have spent 2> days rebuilding my hard drive because of the damage that Mathcad did.>I am thinking of upgrading from an old version of Mathcad (Version 7)>to the most recent. But past experience with upgrades has taught me>to be a bit wary, so 2 questions:>1. Did Mathsoft do anything to screw up the product, in terms of>usability or features, since I last purchased it (back in 1997)?>2. Does the latest product come with any kind of registration>feature, like Mathematica has, that locks your software into one>computer, and possibly locks you out of the software if you upgrade>your hardware? (I respect the rights of companies to make money off>their software, but I'm a firm believer that the only fair deal is>one-user/one-app. If I have three computers -- and I do -- I should>not have any worries or complications with loading the same software>package on all three of them.)>Steve O.>Standard Antiflame Disclaimer: Please don't flame me. I may actually> *be* an idiot, but even idiots have feelings.=== === Subject: : Re: Question of intent (PG13)> I've been thinking about formalizing the argument with which I prove a> problem with the ring of algebraic integers, and ran into a problem.Meaning that another absurd claim of yours was trivially refuted.I know lots of self-interested liars, Harris, but I fail to see who it isyou hope to deceive with this of yours -- page after page about ain' quadratic polynomial. Don't you get tired of making such an ass ofyourself? You never succeed in deceiving anyone, and you prove nothingexcept your own ignorance, over and over.=== === Subject: : 1 to 1 correspondenceHow would I show a 1 to 1 correspondence between (0,1) and [0,1] ?????=== === Subject: : Re: The role of infinity in math> The role of completed infinity is limited to stuff like axiomatic set> theory, but incomplete infinity pervades mathematics. What's incomplete about the infinity of the real numbers?=== === Subject: : The infamous K-man ?Hi Peter , nnrp1.ozemail.com.au!53ab2750 .You wouldn't be the infamous K-man by any chance, would you ?Sloppy, sloppy.=== === Subject: : Re: wanted: more mathematical false proofs Discussion, linux)> |> ... I would like to know if anyone has more of these false proofs.> I'm surprised everybody here missed this unique opportunity of killing two> birds with one stone.> Starting with a rock solid result from harrissian fundamental tautology> space theory and developping at an elementary level, here is a crystal clear> proof of a core error in mathematics.> (0) 1 = 1> (1) 1 = sqrt( 1 )> (2) 1 = sqrt[ (-1)^2 ]> (3) 1 = sqrt[ (-1) * (-1) ]> (4) 1 = sqrt( -1 ) * sqrt( -1 )> (5) 1 = i * i> (6) 1 = -1> Please note that some nitpickers might argue that sqrt being such an> ambiguous and lunatic operator, the odds of sqrt(-1) evaluating to (i) the> first time and (-i) the second time (or vice-versa) are not completely> negligeable.You've missed the linguistic turn this morning that explains thingsbetter now. There are two numbers that, when squared, yield 1. Forthe square root operator to *always* return the positive of these twowould require human *intent*. Mathematics always works for a reason,but humans choose arbitrarily and so while they can form intent, QueenMathematics cannot.Therefore, sometimes the square root operator must sometimes choosethe negative root.-- [Mathematical] society has evolved far enough away from mainstreamsociety that it has become rogue, and now is willing to push its needsagainst that of the majority. -- James S. Harris === === Subject: : Re: PDE Max-Min Principle (Help?) > Prove that if u(x,t) <= v(x,t) for x=0, for x=l (lower-case L), for> t=0, then u<=v for all (x,t) in 0<=x<=l, 0<=t<=oo.>Ridiculous as it stands. Please include all hypotheses. (Also, what>does t = oo mean here?)> Sorry, at the beginning of the problem set, we were told that u and v> are are solutions to the diffusion (heat) equation u_t = k*u_xx.> There is a hint that says this is proving the Comparison Principle> using the Maximum-Minumum Principle.can you apply the Maximum-Minumum Principle. to w=u-v ? which PDE and boundaryconditions does w fulfill?hth> Also, that last inequality should read 0<=t infinity.> Like I said, I'm not looking for the answer, just a way to start the> beast. Most other problems in this set were proving properties of the> diffusion equation using change of variables and the chain rule. That> does not seem to work here.> --=== === Subject: : New crypto algorithmDedicated to the Faculty of Mechanics and Mathematics of Belarusian StateUniversityThis message defines new symmetric cryptoalgorithm and contains animplementation that can be used for practical testing.The cryptoalgorithm has the following advantages: 1. Keys of arbitrary length are allowed. 2. The cryptoalgorithm is extremely easy to understand and implement. 3. The cryptoalgorithm is strong (not proved), open and free. I am theoriginator of this cryptoalgorithm and I tell you: use it for free.The cryptoalgorithm is based on the process of conversion from one numericsystem to another: a number is converted to another numeric system, thedigits are permuted and the result is converted back to the original numericsystem.Description of the cryptoalgorithm:- Plain text to encode is (long) unsigned integer number. Let's denote it byA.- Similarly, key is unsigned integer number B. 0. Plain text is compressed in order to transform it into a relatively random(i.e. uncompressible) sequence of bytes using some compression algorithm. 1. A is split into a sequence of remainders of division by B. 2. The sequence of remainders is permuted. B's value defines the permutation. 3. The permuted sequence is converted to number A2.- A2 is the encoded text.The following Java program is created for practical testing. It is a completeimplementation of the cryptoalgorithm (without compression phase).-------------------------------- A.java --------------------------------import java.io.File;import java.io.FileInputStream;import java.io.FileOutputStream;import java.math.BigInteger;import java.util.LinkedList;import java.util.ListIterator;public class A{ public static void main(String[] args) throws Exception { if (args.length!=4) badArgs(); String scmd=args[0]; String skey=args[1]; String sin=args[2]; String sout=args[3]; if ( !(e.equals(scmd) || d.equals(scmd)) ) badArgs(); boolean encrypt=e.equals(scmd); BigInteger b=new BigInteger(1, readFile(skey, 0)); if (encrypt) { byte[] adata=readFile(sin, 1); // append positive most significant (random) byte for (int i=1; i=0; i--) { a=a.multiply(b); a=a.add(rems[i]); } return a; } public static BigInteger[] permute(BigInteger[] rems, BigInteger b, boolean inverse) { int len=rems.length; BigInteger[] ret=new BigInteger[len]; int pos[]=createPermutation(len, b, inverse); for (int i=0; i=1; div--) { BigInteger bdiv=new BigInteger(+div); if (div==len) // always move bdiv=bdiv.subtract(BigInteger.ONE); // most significant remainder if (b2.compareTo(bdiv)<0) b2=b; // restart from b BigInteger[] qr=b2.divideAndRemainder(bdiv); b2=qr[0]; int rem=qr[1].intValue(); ListIterator lit=lst.listIterator(rem); ret[div-1]=((Integer)lit.next()).intValue(); lit.remove(); } if (inverse) { int[] ret2=new int[len]; for (int i=0; i I was looking at litotes (perhaps more exactly, antenantiosis), with forms> likeHe is not unattractive.It is not impossible to do that.> where the double-negation implies some form of a shift in meaning for> effect, following closely an empirical maxim sometimes called the division> of pragmatic labor:Do not double negatives not leave you tied in knots?=== === Subject: : Re: constructing reals> Hi all,> i had this idea of defining the reals from rationals by intersections> of> subsets of Q:> Let {O_n} be a sequence of open intervals in Q s.t. |O_n| -> 0> and O_n+1 is in O_n for all n.> Then either U(O_n)=p/q is in Q, or U(O_n)={} in which case U(O_n)> defines an irrational number.> Would this be only slightly different from the approach through the> equivalence classes of cauchy sequences ?Given such a sequence of open intervals, a suitable Cauchy sequence canbe easily constructed by interleaving the endpoints. Alternately, thesequence of lower endpoints and the sequence of upper endpoints wouldthemselves be suitable Cauchy sequences. Constructing a nested sequenceof open intervals from a Cauchy sequence is not always that easy, but Ithink it is always possible (although some of the endpoints might not beterms in the sequence).I think using intervals like this is an unneeded complication. -- Daniel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Re: 1 to 1 correspondence> How would I show a 1 to 1 correspondence between (0,1) and [0,1] ?????Hint: How would you show a 1 to 1 correspondence between{1,1/2,1/3,1/4,...} and {1/2,1/3,1/4,...}?-- Daniel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Re: the anticlassicalist }{ i: linguistic negation: 2) are there also well known schools of thought about negation fromAfrican: or South American traditions?I didn't mean to appear to exclude the sufi as a part of African thought,but intended other schools of thought only because of familiarity.Syncretism, particularly from gnostic schools, seems to be a fertile groundfor such thought, but I was thinking more of the Ishango bone or khipu andwondering of symbolic negation work in societies influenced by these andother early symbolic mathematics.-- === === Subject: : Re: the anticlassicalist }{ i: linguistic negation: Do not double negatives not leave you tied in knots?It is eerie how I was typing up a question about the khipu at the same timeyou posted.Is it knot?(PS: Louis Kauffman has some beautiful stuff on knot logics, as well.Reminds me of the BBC / TimeLife show Connections with James Burke)-- === === Subject: : Re: Metamath Axiom of Choice>While looking at the Metamath site, and specifically at their statement of>the Axiom of Choice, I note that their statement of the Axiom of Choice is>effectively the statement that for any set x, there exists a set y such>that for any nonempty element w of x, there is at least one element t of y>such that w is an element of t, and that w has exactly one element in>common with the union of all elements of y which have w as an element,>i.e. there is a set y such that w is an element of some element t of y,>and if s = {t in y : w in t}, then w n (Us) has exactly one element, where>A n B denotes the intersection of A and B, and where Us is the union of>the elements of s.>That this implies the Axiom of Choice is obvious (if w is a nonempty >element of x, let f(w) be the unique element of w n (Us), where >s = {t in y : w in t}). The question is how can one prove the Metamath>statement of the Axiom of Choice in ZFC?> How does this sound for a proof. Let f : x-{empty set} -> Ux be a choice > function (so that f(v) is an element of v for all nonempty elements v of x). > Let y = {{v,f(v)} : v in x, v nonempty}. The first thing to note is that > if w is a nonempty element of x, then there is an element of y which has w > as an element, specifically, {w,f(w)}. Let w be a nonempty element of x, > then s = {t in y : w in t} = {{w,f(w)}} u {{v,w} : v in x, v nonempty, > f(v) = w}, where A u B denotes the union of A and B. It follows that > Us = {w,f(w)} u {v in x : v nonempty, f(v) = w}, and so the elements of > Us are w, f(w), and nonempty elements v of x such that f(v) = w. Since > w is an element of v for all nonempty elements v of x such that f(v) = w, > then the elements of Us are w, f(w) and some sets which have w as an > element. By the axiom of regularity, the only common element of w and Us > is f(w), completing the proof of the Metamath statement of the Axiom of > Choice.> At the moment, they (the Time Lords) are far from being all-powerful.> That's why it's been left up to me and me and me.> quote by: Patrick Troughton in The Three Doctors> ------->Uhm... the Axiom of Choice is an _axiom_. For specificity, it is a>wff of the FOL of set theory which is (roughly speaking) stipulated>true, and can thus be used in deductions. If you decide that you>don't like the axiom of choice, you can choose to not work with FO>languages which include it. But even then, it's hard to deny it's use>in axiom systems which _do_ use it. (Since, by construction, you're>setting up a conditional... If the axiom of choice is true, and>...blah..., then ...blah'...)>In any case, there is nothing to prove.> As I stated above, the Metamath formulation of the Axiom of Choice is > effectively the statement that given any set x, there exists a set y > such that for all nonempty elements w of x, w n (Us) has exactly one > element, where s = {t in y : w in t}. This is not amongst the > formulations of the Axiom of Choice of which I was previously aware.> The question of the equivalence of the Metamath formulation with the> more standard versions under ZF naturally arises, as such an equivalence > is necessary if the Metamath set theoretic axioms are to be taken as > axioms of ZFC.> In spite of what you might think, I was NOT trying to prove the Axiom of> see that I initially gave a proof of a more standard version of the Axiom> of Choice from ZF *and* the Metamath version of the Axiom of Choice, and> that I was wondering about the possibility of a proof of the Metamath> version of the Axiom of Choice from ZFC (which I believe that I supplied> in the subsequent posting). In fact, I can't tell where you could have> got any idea that I was trying to prove the Axiom of Choice from ZF.> Your condescendingly pointing out that the Axiom of Choice is an axiom,> and your final statement that there is nothing to prove demonstrate that> you have completely missed the point of what I was trying to do, i.e. to> demonstrate that the axioms adopted by Metamath do in fact form a set of> axioms for ZFC, a fact which was not immediately obvious to me.> Both Jesse Hughes and David Ull picked up what I was trying to do. Why> couldn't you?My apologies for not reading your post more carefully.I'll have to be more careful in the future, so as not to offend thoseasking for help with my condescending remarks.'cid 'oohPS. Chill out.=== === Subject: : Re: 1 to 1 correspondence> How would I show a 1 to 1 correspondence between (0,1) and [0,1] ?????One way is to define f:[0,1] -> (0,1) such that f(0) = 1/2,f(1) = 1/4,f(1/2^n) = 1/2^(n+2), for all positive integers n,otherwise f(x) = x.The above is bijective, but not order preserving. Order preserving and bijective are mutually exclusive properties for mappings between any closed interval of reals and any open interval of reals.=== === Subject: : Re: Metamath Axiom of Choice>Uhm... the Axiom of Choice is an _axiom_. For specificity, it is a>wff of the FOL of set theory which is (roughly speaking) stipulated>true, and can thus be used in deductions. If you decide that you>don't like the axiom of choice, you can choose to not work with FO>languages which include it. But even then, it's hard to deny it's use>in axiom systems which _do_ use it. (Since, by construction, you're>setting up a conditional... If the axiom of choice is true, and>...blah..., then ...blah'...)>In any case, there is nothing to prove.>As I stated above, the Metamath formulation of the Axiom of Choice is >effectively the statement that given any set x, there exists a set y >such that for all nonempty elements w of x, w n (Us) has exactly one >element, where s = {t in y : w in t}. This is not amongst the >formulations of the Axiom of Choice of which I was previously aware.>The question of the equivalence of the Metamath formulation with the>more standard versions under ZF naturally arises, as such an equivalence >is necessary if the Metamath set theoretic axioms are to be taken as >axioms of ZFC.> Also, the motivation behind proving the equivalence of the Metamath > formulation of the Axiom of Choice and more standard formulations> is the same as the motivation behind proving the equivalence of the > Axiom of Choice and Zorn's Lemma, and the equivalence of the Axiom > of Choice and the Well-Ordering Theorem. I presume that you have > seen the proofs of these last two mentioned equivalences, Acid Pooh.Heh. Since you appear to be goading me into replying, and I'm asucker for this sort of thing, here goes: I was not aware thatMetamath was a website. Nor was I aware that they had a differentversion of the axiom of choice. I assumed (obviously incorrectly)that you meant metamath to refer to Set Theory, or some suchmetamathematical study. Now, in this context, my response doesn'tseem so inappropriate, does it? In particular, it shouldn't seem sopatronizing, since, as I parsed your question, it is a question abeginner should ask. So, why decide to patronize me?'cid 'oohPS. Chill out.=== === Subject: : Re: the anticlassicalist }{ i: linguistic negatioThey don't know nothing!How bland. Let's really mean it: They don't know nothing about nothing.The not-house will be builtIs this to mean the not-house won't be not-built.Do even number of negations affirm and odd number of negations deny? If he didn't not double negate, didn't he affirm? If he didn't not double negate, didn't he not affirm?How about double affirmatives implying negation?Aren't my ideas the greatest?Oh yea, yea.=== === Subject: : Re: 1 to 1 correspondence>How would I show a 1 to 1 correspondence between (0,1) and [0,1] ?????> One way is to define f:[0,1] -> (0,1) such that> f(0) = 1/2,> f(1) = 1/4,> f(1/2^n) = 1/2^(n+2), for all positive integers n,> otherwise f(x) = x.> The above is bijective, but not order preserving.> Order preserving and bijective are mutually exclusive properties for> mappings between any closed interval of reals and any open interval of> reals.=== === Subject: : Re: branch of log z> A veil has been woven out of words like imaginary and spread over> the real plane, giving the impression of an domain, accessible only> for people with special brain-power, sometimes letting the real plane> shimmering through.> introduced to me (at the college level, anyways). > The first time i hear this, except from Caspar Wessel and> sincerely Yours for these words.> No. It's known to insiders, working under the veil - but it's not madewell known.That was a really long post, and you snipped the part where Imentioned that both approaches are useful. Now, quite obviously, whatI meant is that this is well known to people who are interested inmathematics. Nobody else really cares that the complex plane is justR^2. Nobody else cares that the complex field is R^2 with some neatmultiplication thrown in. Call us insiders if you'd like... butapparently you're interested, too. So stop throwing stones, we'reboth in the glass house.> Can You give me a reference, where it's said, that the i-axis and the> y-axis are just different names, but with no mathematical difference?> The same for R2, the 2D-plane, no difference to the complex plane,> argand diagramm, (gauss plane)?R^2 and the Complex plane are isomorphic if you *ignore* complexmultiplication. That is about as deep as the relationship goes. (It's quite a deep relationship, actually. But less deep than youapparently think)> Can You give me a reference, where mixed-mode-calculating is used,> like :> i*(3,4)=(-4,3) (see my calculator)?> Can You name someone, who uses the notation 3+i*4 in any> 2D-vectorspace in place of (3,4) ? The second is of course an ordered> pair of reals and the first is using the linear-combination of two> basis-vectors (1,0) and i=(0,1) and, as the real numbers are embedded,> you ommit (1,0).That's terrible notation. The i notation on elements from R is muchnicer. And there is no reason to use the notation a + bi when workingwith R^2, because, in particular, multiplication is not defined onR^2. So, if we want to generalize to R^n, do we need n-1 letters torepresent the n directions?> And sometimes insiders entangle in their own veil and can't see clear,> look at Algebra: What is the relation between (R2,+,r.s.m.) and> (R2,+,*)-this is called the vectorspace of complex numbers(and an> commutative field)- and (R2,+,r.s.m.,dot)-this is the euclidian> vectorspace?branches of log ...This stuff can get messy later. -> but may be without imaginary veil a little bit less.Read Rudin's principles of mathematical analysis. Or J.E. Marsden'sComplex Analysis. Or Royden's analysis book. Or...'cid 'ooh=== === Subject: : Re: DiffEQ - need help to solve a basic linear diffEQ> Following your work I was able to understand up to:>u^2 + 2xu - (x^2 + c) = 0>u = [-2x +- sqr(4x^2 + 4(x^2 + c))]/2> How did you isolate the u in the first line so you get the result in> the second line?Usual formula for the solution of a quadratic equation derived bycompleting the square.>(x + e^y) dy/dx = x * e^(-y) - 1>(x + e^y) e^y dy/dx = x - e^y>u = e^y>(x + u) du/dx = x - u>v = x + u; u = v - x>v(dv/dx - 1) = 2x - v>v dv/dx = 2x>2v dv/dx = 4x>w = v^2>dw/dx = 4x>w = 2x^2 + c>v^2 = 2x^2 + c>(x + u)^2 = 2x^2 + c>y = log(-x + sqr(2x^2 + c)), 0 <= c>I leave to you the onus of checking the solution.=== === Subject: : Re: We come from your futureTime to commit suicide...=== === Subject: : Re: Candy Inspiration (in the news)[...]|> If M&Ms had been VERY flat, it seems intuitively obvious they would|> have packed even more efficiently.|> (as long they were oriented in the barrel so that neighboring oblate|> spheroids were mostly positioned, on average, in somewhat the same|> direction.)||It seems intuitive to me too. If you imagine M&M's being arbitrarily |long and thin (topological M&Ms) you can see that they approach the |packing density you'd get if all the M&M's were flat -- namely 100%. I'd |guess that spherical M&Ms are the LEAST efficient shape for packing.I can believe they might pack better, but I don't see why. If they are allellipsoids oriented the same way, i.e. with their three corresponding axesparallel, it's impossible to get a better packing density than what youcan get for spheres, because there is a linear transformation whichtakes all of them from ellipsoids to spheres.In your argument, you have two limits being taken. One is the limit asthe shape becomes very flat. The other is the limit in the definition ofpacking density, namely as the volume of the box goes to infinity.I don't see how you can get a packing density close to 100% for anygiven (ellipsoidal) shape, given that they are thicker in the middle andthinner at the edges. If you stacked them up like pancakes, for example,you'd still have that wasted space at the edges, plus the fact that there'sno way to fill the plane with disks without gaps.into big containers, which surely is less than the ideal best possibledensity. I can believe that M&Ms might be better at packing together insomething like the optimal spherical packing (with the linear transformationapplied to turn them into ellipsoids instead of spheres) than spheres are.It would still be interesting to know, though, whether there's any shape ofellipsoid whose *best* packing density exceeds the *best* density forspheres.Keith Ramsay=== === Subject: : Re: Calculating Modulo (very big numbers)> Is there a way to calculate the modulo with numbers like this:> 1226^37 mod 4838 ?1226^2 = 1503076 = 3296 (mod 4838)1226^37 = 1226 * 3296^18 (mod 4838)repeat with 3296^18, etc.=== === Subject: : Re: A question about Kripke semantics and physics (Was: Re: Study groups in science): So, once again, let me start with a quick Google search to: establish context.That is always a fun way to learn, and getting better every year!: If I use the search string '26 dimensions string physics' I get about959: hits.:: For my part, I could care less about the details. It is the mathematics: through which your colleagues express their explanations that is importantto: me. The 26 dimensions are particularly interesting here because their: symmetries and invariants involved offer me an opportunity to ask you howto: think about truth and falsity in physical theory.:: Now, Galathaea has been wanting to talk about Heyting algebras and quantum: logic. Section 6.4 of the paper:: http://www.illc.uva.nl/Publications/ResearchReports/MoL-2001- 09.text.pdf:: is entitled Finite projective formulas in two variables Strangelyenough,: there are precisely 26 Heyting algebras associated with 2-universal models: discussed here. Moreover, there is not a single mention of quantumlogic.:: Now, isn't this just an amazing coincidence? Logicians tellmathematicians: about 'T' and 'F,' the physicists are talking about 26 dimensions, and: knows just where to find a paper specifying the 26 2-generated Heyting: algebras.:: Actually, I do not think it is coincidental. Unfortunately, whereas Iwould: love to offer an explanation, you would simply engage in more vulgarity.I only today got through this reference in my first run through and stillhave much organising to get a better understanding, but I was wondering ifyou were thinking of a particular relationship here between the constraintequation in string theory and the n=2 formula in the paper. The constraintfor strings arises from a requirement for Weyl invariance of the action,which has a very interesting topological structure, so I can certainly see amotif. Have you pursued this to the next exposition stage, or are you stillhunting down the connections? This paper is very interesting!=)-- === === Subject: : Re: Entering *Real* Math Formulae in ExcelWilhelm schrieb im Newsbeitrag> Maybe I've missed something obvious but I've now blown a day on this> ...> How the &*&%$ does one enter a real math formula (simple example: the> sum of all N sub x as x goes from x1 to x2) ???!> I have the data that I want to operate on. I know some basic> university calculus including differentials and integration. I can> formualte these simple summation functions on paper ...> But then how are they entered into and operated on in Excel???> Am I perhaps using the wrong program? (i.e. Should I be using> something like MathCad?)> Do I need to learn Excel arrays or other such extra baggage?Wilhelm,you might have a look athttp://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/ summaryindex.html.In general I think Excel is the wrong tool for your purposes.hth,Stephan=== === Subject: : Understanding taylor expansion for sineI've been trying to understand the taylor exapnsion for the sinefunction,sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...I've been told that it uses the fact that e^(i*x) = cos(x) + i*sin(x),and I can easily work out the taylor series for e^x, but I'm not surehow to use these two pieces to find the expansion for sin(x).Any help would be appreciated.Jonathan Christensen=== === Subject: : UnionsLet A1 ,A2, ... be a sequence of subsets of a universal set X. Define Ej =( Aj - {A1 U A2 U...U A(j-1)} ) where j= 1,2,3,....I was able to prove that Ei intersect = Ej empty set if i =/ j. I was alsoable to show that the union of Ek from k = 1 to n = union from k = 1 to n ofAk but I'm not sure how to prove this last statement if k goes from 1 to oo.=== === Subject: : Re: What is a number?/What is not a number?>I'll offer this definition: A number is an element of a number set. A>number set is either the set of integers Z or any ring/field R that>contains Z.So ordinal numbers are not numbers?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: Simple numbers>Hvala na linkovima !>Puno pozdrava!> Does this mean he appreciated your answer?Yes.=== === Subject: : countable setsLet E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Provethat union of Ek as k = 1 to oo is countable. I do not need help provingthis. I need someone to explain exactly what is being asked. In Let E1, E2,E3,... be a sequence of pairwise disjoint countable sets is that sayingthat the numbers of Ei's are countable or is it saying that each Ei has acountable number of elements??=== === Subject: : Re: Understanding taylor expansion for sine > e^(i*x) = cos(x) + i*sin(x)thus sin(x) is the imaginary part of e^(ix), so you have to extract the imaginary part of the series.=== === Subject: : Re: Need help with proof!> Show that for arbitrary natural numbers m and n, such that 1<=m<=n the> following holds: the product n(n-1)(n-2) ...(n-m+2)(n-m+1) is dividable by> m!Here's a hint that should get you started without giving away the answer.Show that for every prime that divides m!, that prime occurs in thefactorization of n(n-1)(n-2)...(n-m+1) at least as many times as it occursin the factorization of m!.-- === === Subject: : Re: Got a speeding ticket and need to fight back> The Lord of Chaos (Suresh Devanathan)>The last i checked, I dictate the laws of physics. I am God, to many>people. Your court is beneath me. Your constitution is beneath me. In >fact,>you cannot even prosecute me, because the constitution depends on me. I >tell>what the laws are. And that's final.>Not another case of Narcissistic Personality Disorder...> btw, does your boss keep your arround because he can use your face to wipe > his ass?That particular symptom is called projection: Attributing your own fears and believes to others. I have to tell you this: Pleading insanity is not a good tactics when you want to fight a speeding ticket. You might not have to pay the ticket, but they surely will take your license away.=== === Subject: : Re: 1 to 1 correspondence>How would I show a 1 to 1 correspondence between (0,1) and [0,1] ?????Choose any sequence of unique reals a_n from (0, 1). Letf: [0,1] -> (0,1)be defined as0 |--> a_11 |--> a_2a_n |--> a_(n+2)a |--> a, when a not in a_nThen f is a bijection.-- === === Subject: : Re: Understanding taylor expansion for sinejust pick out the bits that has an is in.> I've been trying to understand the taylor exapnsion for the sine> function,> sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...> I've been told that it uses the fact that e^(i*x) = cos(x) + i*sin(x),> and I can easily work out the taylor series for e^x, but I'm not sure> how to use these two pieces to find the expansion for sin(x).> Any help would be appreciated.> Jonathan Christensen=== === Subject: : Re: Understanding taylor expansion for sine>I've been trying to understand the taylor exapnsion for the sine>function,>sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...>I've been told that it uses the fact that e^(i*x) = cos(x) + i*sin(x),>and I can easily work out the taylor series for e^x, but I'm not sure>how to use these two pieces to find the expansion for sin(x). e^(ix) = cos(x) + i sin(x)<=> -i e^(ix) = -i cos(x) + sin(x) => sin(x) = Re(-i e^(ix))Work from there.-- === === Subject: : Re: countable sets>Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove>that union of Ek as k = 1 to oo is countable. I do not need help proving>this. I need someone to explain exactly what is being asked. In Let E1, E2,>E3,... be a sequence of pairwise disjoint countable sets is that saying>that the numbers of Ei's are countable or is it saying that each Ei has a>countable number of elements??Both. Sequences by definition are countable ordered sets. In this casethe elements of the sequence are countable sets.-- === === Subject: : Re: countable sets: Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove: that union of Ek as k = 1 to oo is countable. I do not need help proving: this. I need someone to explain exactly what is being asked. In Let E1, E2,: E3,... be a sequence of pairwise disjoint countable sets is that saying: that the numbers of Ei's are countable or is it saying that each Ei has a: countable number of elements??Both. The E1, E2, E3, ... part says there are a countable number of E's,and pairwise disjoint *countable* sets says that each Ei is countable.Ted=== === Subject: : Re: Understanding taylor expansion for sine> I've been trying to understand the taylor exapnsion for the sine> function,> sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...> I've been told that it uses the fact that e^(i*x) = cos(x) + i*sin(x),> and I can easily work out the taylor series for e^x, but I'm not sure> how to use these two pieces to find the expansion for sin(x).Hello.I usually work out the Taylor series for sine around zero directly.Since you say you can easily work out the Taylor series for e^(i*x)directly, I won't go further on that point.If one plugs (i*x) into the Taylor series for e^x around zero andtakes the imaginary terms alone, doesn't one end up with the Taylorseries for sine around zero?-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau=== === Subject: : Re: countable sets* Math Tutor> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove> that union of Ek as k = 1 to oo is countable. I do not need help proving> this. I need someone to explain exactly what is being asked. In Let E1, E2,> E3,... be a sequence of pairwise disjoint countable sets is that saying> that the numbers of Ei's are countable or is it saying that each Ei has a> countable number of elements??... a sequence of pairwise disjount countable setscountable sets means that each Ei has a countable number ofelements.... a sequence means that the number of Eis is countable.I.e. both.-- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92=== === Subject: : 53ab2750 ?Oops, I called ozemail.com.au!53ab2750 , an anonymous server ... It's not.So what does the 53ab2750 mean ?It seems to be some kind of a control code.=== === Subject: : Re: Parity Check Matrix of a Systematic Linear Block Code> The generator matrix of a systematic linear block code has the> form G = [Ik : P]. How can it be shown that the parity check> matrix is of the form H = [-P^T : In-k]?With great ease.If a vector (a : b) is supposed orthogonal to the code generatedby G then (a : b)G^t = 0 so a + b P^t = 0 or a = -b P^t etc.Are you in the IEEE?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: 53ab2750 ?> Oops, I called ozemail.com.au!53ab2750 ,> an anonymous server ... It's not.> So what does the 53ab2750 mean ?> It seems to be some kind of a control code.I just guessing now...Could it be the IP address written in hexadecimal, i.e. 83.171.39.80?-Michael.=== === Subject: : Re: How fast is the Confinued Fraction factorization algorithm?>>I have another question: In each of the above factorizations, only half>>(approximately) of the primes in the factor base are used. I read>>somewhere that N (the composite) must be congruent to a square modulo>>the primes in the factor base, but I don't see why:>>If x^2 = y mod N, then x^2 - y = kN, for some k. Now if a prime p (from>>the factor base) divides y, then we must have p | (x^2 - kN). So I>>conclude that kN must be congruent to a square modulo p, but that does>>not say anything about N itself? Please enlighten me.> I don't know anything about the continued fraction method> (as has been pointed out!)> but if it is like the simple quadratic sieve> one considers products of numbers of the form> Q(x) = x^2 - n> (with x close to sqrt{n})> each of which is smooth.> So one need only consider small primes p> such that n is a quadratic residue mod p.> (In the multiple polynomial version one takes> Q(x) = ax^2 + 2bx + c,> with b^2 - ac = n, so that> aQ(x) = (ax + b)^2 - n> and much the same is true.)Well, using the same notation, then the continued fraction method deals with the polynomials Q(x) = x^2 - kn.-Michael.=== === Subject: : Re: Constraints on Self-Dual Linear Block Codes> If (n,k) is a q-ary linear block code of length n and> size q^k, then in order for the code to be self-dual> it is fairly easy to show that n = 2*k.> Are there any other constraints required of a code> in order for it to be self-dual?Yes ... it has to satisfy the definition of self-duality :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : . 53ab2750 is an anonymous user .I asked,I think it means that a proxy client is being used. ( i.e. The poster is being kept anonymous )Normally that number would identify the user.=== === Subject: : Re: What is a number?/What is not a number?> at 01:39 PM, reany@asu.edu (Patrick Reany) said:>I'll offer this definition: A number is an element of a number set. A>number set is either the set of integers Z or any ring/field R thatcontains Z.> So you're excluding the Cayley Numbers? Elements of finite fields?> Yes. It was my intention to eliminate all finite fields or finite> rings for that matter.> Good, that is an opinion. However, you have to exclude more. Polynomials> in one of more variables with integer coefficients for instance. They> form a ring and contain Z. Unless you would call them numbers (and> that is not common practice). I have no idea how you would exclude them.As the definition stands, they couldn't be excluded. Of course, I haveavoided dealing with the question of why we even need a label ofnumber in the first place. If we had an answer to that question,we'd have a basis to decide how to proceed. We have three ways toproceed:1) revamp the definition above to remove polynomials,2) accept polynomials as numbers, or3) define a number set as any mathematical set at all.In favor of 1) is a measure of psychological comfort, I suppose. Infavor of 2) is that polynomials mod ideals are isomorphic tononpolynomial rings or fields, which are already included as numbersets. In favor of 3) is that the word number is just a way to avoidsaying the wordy expression element of a mathematical set with closedbinary operation. And this brings us back to Cayley numbers.I suppose the question of what number means in common practice is atthe heart of this thread. I know of no commonly accepted definition ofnumber in mathematics, so any definition is a contender. Thatcertain mathematical objects are not usually considered as numbersis of no significance as far as I'm concerned, because I can find noparticular rationale to call any object a number.We could just demand that whatever objects were considered as numbersup to 1985, say, are grandfathered in as numbers, but no more forever.Or we could try to come up with a rationale for the label number andthen craft a definition on top of it.One rationale for the label number could be to regard all objectsused directly for counting or measuring as numbers. That includesthe naturals, the integers, the rationals, and the reals, but nothinglarger than that. And that leaves some sets out which are already usedas number sets, such as the complex numbers. But I'm not fond of thisrationale. However, I can't think of any rationale to define numberwhich won't either leave some sets out or include too many.Patrick=== === Subject: : Re: What is a number?/What is not a number?> at 01:39 PM, reany@asu.edu (Patrick Reany) said:>I'll offer this definition: A number is an element of a number set. A>number set is either the set of integers Z or any ring/field R thatcontains Z.> So you're excluding the Cayley Numbers? Elements of finite fields?> Yes. It was my intention to eliminate all finite fields or finite> rings for that matter.> Good, that is an opinion. However, you have to exclude more. Polynomials> in one of more variables with integer coefficients for instance. They> form a ring and contain Z. Unless you would call them numbers (and> that is not common practice). I have no idea how you would exclude them.As the definition stands, they couldn't be excluded. Of course, I haveavoided dealing with the question of why we even need a label ofnumber in the first place. If we had an answer to that question,we'd have a basis to decide how to proceed. We have three ways toproceed:1) revamp the definition above to remove polynomials,2) accept polynomials as numbers, or3) define a number set as any mathematical set at all.In favor of 1) is a measure of psychological comfort, I suppose. Infavor of 2) is that polynomials mod ideals are isomorphic tononpolynomial rings or fields, which are already included as numbersets. In favor of 3) is that the word number is just a way to avoidsaying the wordy expression element of a mathematical set with closedbinary operation. And this brings us back to Cayley numbers.I suppose the question of what number means in common practice is atthe heart of this thread. I know of no commonly accepted definition ofnumber in mathematics, so any definition is a contender. Thatcertain mathematical objects are not usually considered as numbersis of no significance as far as I'm concerned, because I can find noparticular rationale to call any object a number.We could just demand that whatever objects were considered as numbersup to 1985, say, are grandfathered in as numbers, but no more forever.Or we could try to come up with a rationale for the label number andthen craft a definition on top of it.One rationale for the label number could be to regard all objectsused directly for counting or measuring as numbers. That includesthe naturals, the integers, the rationals, and the reals, but nothinglarger than that. And that leaves some sets out which are already usedas number sets, such as the complex numbers. But I'm not fond of thisrationale. However, I can't think of any rationale to define numberwhich won't either leave some sets out or include too many.Patrick=== === Subject: : Re: What is a number?/What is not a number?> at 01:39 PM, reany@asu.edu (Patrick Reany) said:>I'll offer this definition: A number is an element of a number set. A>number set is either the set of integers Z or any ring/field R thatcontains Z.> So you're excluding the Cayley Numbers? Elements of finite fields?> Yes. It was my intention to eliminate all finite fields or finite> rings for that matter.> Good, that is an opinion. However, you have to exclude more. Polynomials> in one of more variables with integer coefficients for instance. They> form a ring and contain Z. Unless you would call them numbers (and> that is not common practice). I have no idea how you would exclude them.As the definition stands, they couldn't be excluded. Of course, I haveavoided dealing with the question of why we even need a label ofnumber in the first place. If we had an answer to that question,we'd have a basis to decide how to proceed. We have three ways toproceed:1) revamp the definition above to remove polynomials,2) accept polynomials as numbers, or3) define a number set as any mathematical set at all.In favor of 1) is a measure of psychological comfort, I suppose. Infavor of 2) is that polynomials mod ideals are isomorphic tononpolynomial rings or fields, which are already included as numbersets. In favor of 3) is that the word number is just a way to avoidsaying the wordy expression element of a mathematical set with closedbinary operation. And this brings us back to Cayley numbers.I suppose the question of what number means in common practice is atthe heart of this thread. I know of no commonly accepted definition ofnumber in mathematics, so any definition is a contender. Thatcertain mathematical objects are not usually considered as numbersis of no significance as far as I'm concerned, because I can find noparticular rationale to call any object a number.We could just demand that whatever objects were considered as numbersup to 1985, say, are grandfathered in as numbers, but no more forever.Or we could try to come up with a rationale for the label number andthen craft a definition on top of it.One rationale for the label number could be to regard all objectsused directly for counting or measuring as numbers. That includesthe naturals, the integers, the rationals, and the reals, but nothinglarger than that. And that leaves some sets out which are already usedas number sets, such as the complex numbers. But I'm not fond of thisrationale. However, I can't think of any rationale to define numberwhich won't either leave some sets out or include too many.Patrick=== === Subject: : Matrix problem... please enlighten me...X-Mimeole: Produced By Microsoft MimeOLE V6.00.2800.1165Dear all, I'm trying to convert from 1 coordinate space to another where they-axis is inverted.In the 1st space, the 3x3 transformation matrix is specified as with +ve yupwards:a b 0c d 0e f 1in the 2nd space :a c eb d f0 0 1where in the 2nd space y is downwards.I've worked out by trial and error that by negating b & c the conversion issuccessful so thata -c e-b d f0 0 1will make everything work for my conversion.what I understand is to do a flip of y-axis one needs to multiply the matrixby :1 0 00 -1 00 0 1but that doesn't givea -c e-b d f0 0 1Is there somehting more to a transpose than simply flipping along thediagonal?=== === Subject: : Re: A trend ?>Hi Dan Weiner,> Re: How I feel about your one word posts,>You added two words: duly noted ,>Do I detect a trend here ?> But will his next post have three words or four?Reminds me of a story told about President Calvin Coolidge, known asSilent Cal.A woman reporter once approached him and said, My editor bet me thatI couldn't get more than two words out of you.The President replied, You Lose.Double-A=== === Subject: : Re: convergence of measurable functionsContent-transfer-encoding: 8bit> Hi all,> Let X be a measurable space (a set with a sigma-algebra of sets) and> B a general complete metric space, that is, not necessarily separable> or compact or whatever. Put on B, the measurable structure generated> by the open sets of B (the Borel structure). Now, let> f_{n}:Xlongrightarrow B be a sequence of measurable functions> converging pointwise to some f.> Question: Is f measurable? Proof?Yes. More generally, if B has the property that every open set can bewritten as a countable union of open sets G_n such that closure(G_n)subset G_{n+1} this is true. (That property certainly holds formetric spaces.)Proof: Let U be open, f_n --> f. Now x in f^{-1}(U) ==> f(x) in U ==> f_n(x) in U for suff large n ==> x in cup_{N=1}^infty cap_{n=N}^infty f_n^{-1}(U).OTOH, x in cup_{N=1}^infty cap_{n=N}^infty ==> f_n(x) in U for sufflarge n ==> f(x) in closure(U).Thus if we write U = union of G_k, where closure(G_k) subset G_{k+1},we have f^{-1}(U) subset cup_k f^{-1}(G_k) subset cup_k cup_{N>=1}cap_{n>=N}f_n^{-1}(G_k) subset cup_k f^{-1}(closure(G_k)) subset cup_k f^{-1}(G_{k+1}) = f^{-1}(U).Therefore f^{-1}(U) is equal to cup_k cup_{N>=1}cap_{n >= N} f_n^{-1}(G_k),which is a Borel measurable set.(This has been discussed before on sci.math, and I've given this proofbefore. I first saw it in a class given by Alberto Calderon.)--Ron Bruck=== === Subject: : x^2 + y^4 = z^4+ y^4 = z^4 has no positive-integer solutions. Is the proof of this result short enough for some kind soul to post it, or need I make a trip to the library? (I have citations.)-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: 53ab2750 ?Hi Michael Jrgensen,You ventured, Could it be the IP address ,A number like 53ab2750 would normally identify the poster, but 53ab2750 seems to mean: An anonymous poster .=== === Subject: : tensors for totsHow far can one get in understanding the structure of tensors at apoint on a manifold in terms of our old school friends column vectors,row vectors and matrices? Even in general relativity all we have ateach point on the manifold is a real vector space with fourcomponents, so it seems we can go all the way -- but I'm havingtrouble with some of the mappings.Shall I show what work you've done so far, or would somebody care totake the ball from there... ?=== === Subject: : No Set Contains Every Computable Natural (was Church-Turing compared to Zuse-Fredkin thesis)> Dave Seaman:> A set is recursive if there is an effective procedure (a Turing machine,> perhaps) that can decide membership in the set. That is, the machine> halts and returns 1 if the input is in the set, and the TM halts and> returns 0 if the input is not in the set.This is the definition of a decidable language at wikipedia:http://en.wikipedia.org/wiki/Decidable_languageA decidable or recursive language is a formal language that is a recursiveset, i.e., for which there exists an algorithm to solve the followingdecision problem: Given string w, does w belong to the language? Thealgorithm is not allowed to run into an infinite loop and has to produce aYES/NO answer for any input string after a finite amount of time. Toformalize the rather vague term algorithm, one usually employs Turingmachines, but several other equivalent approaches are possible.This definition should say finite number of steps instead of finiteamount of timeI will come back to this.This is the definition of a recursively enumerable language:http://en.wikipedia.org/wiki/Recursively_enumerable_ languageDefinition 1. Given string w as input, the algorithm halts and outputs YESif and only if w belongs to the language L. If w does not belong to thelanguage L, the algorithm either runs forever, or halts and outputs NO.There is a second definition that you might want to read.Let me define a language, L, that consists of all unary representations ofnatural numbers.1 = one, 11 = two, 111 = three, etc.This language is recursively enumerable. Given a string, x, there exists aTM thatwill halt after a finite number of steps if x is a member of L.This language is NOT decidable. Consider what happens if x is an infinitestringof 1's. The TM will not halt after a finite number of steps.The set of all natural numbers is not a recursive set.Consider what happens when we talk about the set of all TM's.Assume we can encode the state transition table of a TM usinga natural number. This set can not be recursive for the samereason that the natural numbers are not recursive. Given astring, i, we can not be sure that TM(i) is actually a TMbecause we can not be sure that i is actually a natural number.No set can contain every computable natural number.Proof:Assume a TM produces a tape with all of the natural numbersencoded in unary where each natural is followed by a 0 (or a blank).Turing gives the state transition table of such a TM 010110111011110...I can define a three state TM that will find a natural numberthat is not on this tape:1) Read right until there is a 0.2) Read right until a second 0 is found.3) Backup and write a 1 on the previous 0.Repeat steps (1) through (3).This TM will always produce a tape that has exactly one 0.This 0 will be at a finite position on the tape and the string of 1'sthat preceds this 0 represents a natural number that was not onthe original input tape.The Halting Problem is ill posed.There is no set that contains every TM just as there is no setthat contains every computable natural number.Now, about the difference between finite amount of time vsfinite number of steps. There are theories of computing wherea TM can perform an infinite number of steps in finite time.These are called hypercomputers.http://en.wikipedia.org/wiki/HypercomputerA non-deterministic Turing machine is an example of a hypercomputer.Another example is an accelerated Turing Machine. An ATM performsthe first operation in one unit of time, the second operation is 1/2 unitof time, the third operation in 1/4 unit of time, etc. An ATM will performan infinite number of operations after two units of time.Assume we have a computer that can perform an operation in onenanosecond. The language, L, that I defined above is not evenrecognizable if we expect an answer before the universe freezes over.We can easily compute a natural number, n, that is so large thatour nanosecond computer will take 15 ion years to recognizen as a natural number. There is only a finite number of naturalsless than or equal to n. There is an infinite number of naturalslarger than n. Nearly all natural numbers are larger than n.So, we see that we have to assume that a TM can performan infinite number of operations in finite time to even saythat L is recursively enumerable.=== === Subject: : Re: parabola y^2=4px intercepted by the line x=aGeorge Cox>The sides of the rectangle are 2y and a-x = a-(y^2)/4p.>Its area is A = 2y(a-(y^2)/4p)= 2ay-(y^3)/2p>The derivative is =0 when>y^2 = 4ap/3, that is x = a/3.>And the rectangle has dimensions 2a/3 and 2*sqrt(ap/3).> You assume that one side is a segment of x = a. Can you prove that that> is optimum?I think we can start with two or three moves aimed at reducingthe number of degrees of freedom in the figure.1. As we saw, if the sides are parallel to the axes, then therectangle is determined by the x-intercept x = a/3, which isindependant of p.2. We can assume WLG that one vertex of the rectangleis on the line x=a. For if none is, then we cantranslate the rectangle in the x direction, until onevertex is on that line, and then enlarge the rectangle.3. We can assume that all four (not just three)vertices are on the parabola. For suppose one vertexD is inside the curve. We can extend a certain side ofthe rectangle, from D, until it meets the parabola asecond time. We get a new rectangle with a new value of$a$, but the area of the rectangle has increased inproportion to the side being extended, while the valueof $a$ has increased in a smaller ratio. In view of (1)we come out ahead again (I think).=== === Subject: : question re Lebesgue-integrable fcnAnyone know how to prove the following?Suppose f is a Lebesgue-integrable function on R (reals). Prove thatthe series sum_{n=-infty}^{infty} f(x+n) converges absolutely foralmost every x in R.=== === Subject: : Re: the anticlassicalist }{ i: linguistic negationNow really, Ms.(?) fabulist, etc: there is much that is more grosslyinappropriate to sci.physics here, so you might say all the other kidsare doing it. But if you affect to be a _superior_ kid, might you notlead by example and limit the number of freshly cross-posted threadswhich are mainly about linguistics, and neither physics nor math?You are beginning to speak Crankish.> -=-=-=-=-= linguistic negation => The logic of natural language has been studied by many different schools of> thought throughout history. Textual analysis: schools of thought do not study anything, and is acliche.For somebody who is so academically concerned with language, you seemremarkably insensitive to its use. Your earnestly learned turgiditywill carry you far ... away from here. Anyone know how to prove the following?> Suppose f is a Lebesgue-integrable function on R (reals). Prove that> the series sum_{n=-infty}^{infty} f(x+n) converges absolutely for> almost every x in R.Yes, there are many of us who know how to prove this.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: The infamous K-man ?> Hi Peter ,> nnrp1.ozemail.com.au!53ab2750 .> You wouldn't be the infamous K-man by any chance,> would you ?> Sloppy, sloppy.Jeff,Be careful!I used to get a lot of unsolicited spam email porn from ozemail.au (Australia).Double-A=== === Subject: : Re: . 53ab2750 is an anonymous user . I asked, >I think it means that a proxy client is being used. > ( i.e. The poster is being kept anonymous ) >Normally that number would identify the user. No, it wouldn't. Go look up the rfc for the nntp protocol. The pathprior to arriving on the news server on which you are reading the=== === Subject: : Re: Unions===Subject: Unions >Let A1 ,A2, ... be a sequence of subsets of a universal set X. >Define Ej = ( Aj - {A1 U A2 U...U A(j-1)} ) where j= 1,2,3,.... >I was able to prove that Ei intersect Ej = empty set if i =/ j. >union of Ek from k = 1 to n = union from k = 1 to n of Ak >how to prove this last statement if k goes from 1 to oo. /{ Ej | j in N } = /{ Aj | j in N }/{ Ej | j in N } subset /{ Aj | j in N } is easyConversly assume x in /{ Aj | j in N } some k in N with x in Ak x in /{ Aj | j <= k } x in /{ Ej | j <= k } x in /{ Ej | j in N }----=== === Subject: : Re: x^2 + y^4 = z^4=== === Subject: : x^2 + y^4 = z^4 >x^2 + y^4 = z^4 has no positive-integer solutions. Is the proof of >this result short enough for some kind soul to post it, or need I >make a trip to the library? (I have citations.)When you find proof, let us know.It's one that's been pestering me for ages.Here's my archived thoughts upon x^2 + y^4 = z^4 for pairwise coprime x,y,zIf y odd: y^4 = z^4 - x^2 = (z^2 - x)(z^2 + x); factors odd, coprime y^4 = u^4 v^4; x = (u^4 - v^4)/2; z^2 = (u^4 + v^4)/2If x odd: x^2 = z^4 - y^4 = (z - y)(z + y)(z^2 + y^2); factors odd, coprime x^2 = u^2 v^2 w^2; y = (u^2 - v^2)/2; z = (u^2 + v^2)/2 w^2 = z^2 + y^2 = (u^4 + v^4)/2Thus problem boils down to showing no coprime a,b,c with a^4 + b^4 = 2c^2when a /= b.Is there some coprime a,b,c with a^2 + b^2 = 2c^2and a /= b?-- this is infinite descent proof, not quickno solution in positive integers to the equation x^4 + y^4 = z^2http://www.math.toronto.edu/mathnet/plain/questionCorner/ fermat4.html----=== === Subject: : Crikey mate !Hi Double-A, You mentioned, I used to get a lot of unsolicited spam email porn from ozemail.au ( Australia ) . Crikey mate ! Maybe that is him then.I think that's k-man's profession.=== === Subject: : Re: More Boolean Algebra>This isn't logic, it's algebra; so how do truth tables apply?> If a statement of Boolean algebra always has value 1 in the two-valued> structure, then it always has value 1 in any Boolean algebraic> structure.Would this mean XOR,NAND,NOT,NOR, etc?A bit is on or off.-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----http://www.newsfeeds.com - The #1 Newsgroup Service in the World!-----== Over 100,000 Newsgroups - 19 Different Servers! =-----=== === Subject: : Number propertyHi everybody!I am searching for the most general set of numbers {a_i} (i=1...n) , 0<=a_i<=2pi , such that :a_i + a_j = a_t mod(2pi) for every i,j=1...nand such that the set of all {a_t} obtained in this way is equal tothe set {a_i} we started with.Any idea to characterize this set?Fabio.=== === Subject: : Re: No Set Contains Every Computable Natural (was Church-Turing compared to Zuse-Fredkin thesis)>Dave Seaman:>A set is recursive if there is an effective procedure (a Turing machine,>perhaps) that can decide membership in the set. That is, the machine>halts and returns 1 if the input is in the set, and the TM halts and>returns 0 if the input is not in the set.> This is the definition of a decidable language at wikipedia:> http://en.wikipedia.org/wiki/Decidable_language> A decidable or recursive language is a formal language that is a recursive> set, i.e., for which there exists an algorithm to solve the following> decision problem: Given string w, does w belong to the language? The> algorithm is not allowed to run into an infinite loop and has to produce a> YES/NO answer for any input string after a finite amount of time. To> formalize the rather vague term algorithm, one usually employs Turing> machines, but several other equivalent approaches are possible.> This definition should say finite number of steps instead of finite> amount of time> I will come back to this.> This is the definition of a recursively enumerable language:> http://en.wikipedia.org/wiki/Recursively_enumerable_language> Definition 1. Given string w as input, the algorithm halts and outputs YES> if and only if w belongs to the language L. If w does not belong to the> language L, the algorithm either runs forever, or halts and outputs NO.> There is a second definition that you might want to read.> Let me define a language, L, that consists of all unary representations of> natural numbers.> 1 = one, 11 = two, 111 = three, etc.> This language is recursively enumerable. Given a string, x, there exists a> TM that> will halt after a finite number of steps if x is a member of L.> This language is NOT decidable. Consider what happens if x is an infinite> string> of 1's. The TM will not halt after a finite number of steps.You can't input an infinite string to the tape. If you could, it wouldn'tbe a Turing machine.> The set of all natural numbers is not a recursive set.No natural number requires an infinite unary representation. All may berepresented using a finite number of 1s.> Consider what happens when we talk about the set of all TM's.> Assume we can encode the state transition table of a TM using> a natural number. This set can not be recursive for the same> reason that the natural numbers are not recursive. Given a> string, i, we can not be sure that TM(i) is actually a TM> because we can not be sure that i is actually a natural number.All Turing machines can be finitely represented. There is no such thing asa TM transition function that requires an infinite state automata forrepresentation.> No set can contain every computable natural number.> Proof:> Assume a TM produces a tape with all of the natural numbers> encoded in unary where each natural is followed by a 0 (or a blank).This is impossible, a) a decider is limited to a finite number of stepsbefore halting, b) to output all integers in unary requires an inifnitenumber of steps - hence, no TM can produce this output.> Turing gives the state transition table of such a TM> 010110111011110...> I can define a three state TM that will find a natural number> that is not on this tape:> 1) Read right until there is a 0.> 2) Read right until a second 0 is found.> 3) Backup and write a 1 on the previous 0.> Repeat steps (1) through (3).> This TM will always produce a tape that has exactly one 0.> This 0 will be at a finite position on the tape and the string of 1's> that preceds this 0 represents a natural number that was not on> the original input tape.You have not at all proven this. First, how can there be a 0 at a finiteposition from the start? If there is, there must a 0 to its right (in factan infinite number of zeros), hence step 2 will read this second zero, andstep 3 will go back and erase the 'remaining 0'. The part about the numberpreceding the last 0 being an integer not on the tape makes no sense either.> The Halting Problem is ill posed.> There is no set that contains every TMSure there is : A = { | M is a valid TM representation}It is easy to check, given a data structure, whether the input is a valid TMor not.> just as there is no set> that contains every computable natural number.All natural numbers are computable. However, they _all_ can't be generatedin a single run of any algorithm/TM.> Now, about the difference between finite amount of time vsfinite number of steps. There are theories of computing where> a TM can perform an infinite number of steps in finite time.> These are called hypercomputers.> http://en.wikipedia.org/wiki/Hypercomputer> A non-deterministic Turing machine is an example of a hypercomputer.No, an NTM is not a hyper-computer since any NTM can be simulated by astandard TM. The NTM must be given additional abilities to becomesuper-Turing.> Another example is an accelerated Turing Machine. An ATM performs> the first operation in one unit of time, the second operation is 1/2 unit> of time, the third operation in 1/4 unit of time, etc. An ATM will perform> an infinite number of operations after two units of time.> Assume we have a computer that can perform an operation in one> nanosecond. The language, L, that I defined above is not even> recognizable if we expect an answer before the universe freezes over.> We can easily compute a natural number, n, that is so large that> our nanosecond computer will take 15 ion years to recognize> n as a natural number. There is only a finite number of naturals> less than or equal to n. There is an infinite number of naturals> larger than n. Nearly all natural numbers are larger than n.> So, we see that we have to assume that a TM can perform> an infinite number of operations in finite time to even say> that L is recursively enumerable.To say that there is some natural number that a physical computer will neverbe able to print out in unary is one thing, but to say that a TM (a purelymathematical object) 'can't' print out certain natural numbers in unary in afinite number of steps is not correct.=== === Subject: : GEP book plus APS 3.0 Standard Edition freeDear All,The third edition of APS has been released and we will be offering FREE theStandard Edition with the purchase of the book:Ferreira, C., Gene Expression Programming: Mathematical Modeling by anArtificial Intelligence, 2002, 272 pp.For further information please go to:http://www.gene-expression-programming.com/gep/Books/ index.aspGeneral features of APS 3.0 Standard Edition:o Translates the evolved models into 8 different languages (C, C#, C++,Java, JavaScript, Visual Basic, VB.NET, and Fortran).o Draws the parse trees of the evolved models.o Translates the evolved models virtually into any programming languagethrough User Defined Grammars.o A total of 70 different built-in mathematical functions and comparisonrules plus Dynamic UDFs and Static UDFs for modeling.o A total of 11 built-in fitness functions for Function Finding.o A total of 10 built-in fitness functions for Classification.o A total of 11 built-in fitness functions for Times Series Prediction.o User Defined Fitness Functions for all problem categories.o Implements a new algorithm for handling random numerical constants.o Data screening engine for preprocessing.o Time series transformation engine.o Evolution from seed models.o Change seed utilities.o Saves all the best-of-generation models of a run.o Plots the evolutionary dynamics of the run.o Complexity increase engine.o Supports Databases and Text Files both for loading input data andscoring.o Recursive testing and prediction for Time Series.o Implements an extensive package of statistical indexes for modelevaluation.o and much more.For more information go to:http://www.gepsoft.com/gepsoftEnjoy and spread the word!All the best,Candida Ferreira------------------------------------------------------ -----Candida Ferreira, Ph.D.Chief Scientist, Gepsoft73 Elmtree DriveBristol BS13 8NA, UKph: +44 (0) 117 330 9272http://www.gepsoft.com/gepsofthttp:// www.gene-expression-programming.com/author.asp---------------- -------------------------------------------=== === Subject: : Re: . 53ab2750 is an anonymous user .Hi Bilge,You mistakenly thought, No, it wouldn't. Go look up the rfc for the nntp protocol. The path line identifies the news servers through which I know for a fact that a number like that I sometimes score on it, it works perfectly.For example it identifies one exact Earthlink user. take a look at Ha Ha Hanson's headers, you would always know that it was him, and only him.53ab2750 is an anonymous or unauthenticated user.And as Double-A noted, ozemail.au is famous for spam ( and anonymous posts ) .=== === Subject: : Re: 'erf' function in C> Are there any C compilers which have the erf function (from> probability) as part of their math libraries? Or are there any math> libraries available to download which implement this function?> James Harlacher> [...]>A procedure that can determine Phi(x) nearly to the limit>available in double precision can be based on the following>method, which I developed the 1960's:> [...]>/* An indication of the accuracy of this Phi function> is given by the following examples, with the true> value (to 20 places) given under the resulting Phi(x):> x Phi(x)> 0.123 0.5489464510164369> .54894645101643675909> 1.2 0.8849303297782918> .88493032977829173198> 2.4 0.9918024640754040> .99180246407540387055> 6.1 0.9999999994696578> .99999999946965767370> -6.1 0.0000000005303427> .00000000053034232638> -1.1 0.1356660609463828> .13566606094638267517> 7.2 0.9999999999996990> .99999999999969893721>See what your system gives and compare with>an erf function if it has one.>*/>------------------------------------------------------ ---->Cut, paste and try. You may like it.>George Marsaglia> [...]> Is there someplace a table available of the true values of the> normal distribution up to 20 places?> With kind regards> Matthias Kl.8ay> -- > www.kcc.chSorry, I should have mentioned that the values I usedwere based on setting Digits:=20 or 30 in Maple, thencalling .5+.5*erf(x/sqrt(2.)), (Yes, Maple is still stuckwith erf for lack of a standard for the normal distribution,although it may be avoided by the rather awkward stats[statevalf,cdf,normald](x);)Mathematika and other extended precision math packageswould also serve. Of course the old standby oflooking up tabled values, for example the Handbook ofMathematical Functions edited by Abramowitz and Stegun,will still serve, but they are likely to list evenly spaced x's and perhaps less than 20 digits for Phi(x).George Marsaglia=== === Subject: : Re: 1 to 1 correspondence> How would I show a 1 to 1 correspondence between (0,1) and [0,1] ?????if you just need to show that (0,1) and [0,1] have the samecardinality (i.e. just the existence of the 1 to 1 correspondance) youmay do the following:1- consider the half-circle, of radius 1 and center P, tangent to thereal line. This half circles represents the interval (0,1) (0 on oneend, 1 on the other). draw a line starting from P and intersecting thereal line. This a geometrically constructed bijection between (0,1)and R.2- Use the Schroder-Bernstein theorem: a) (0,1) in [0,1] in R b) R and (0,1) are equivalent (by 1-)therefore all three sets are equivalent ([0,1], (0,1) and R).=== === Subject: : Re: TSP and the Bell curveIn message , Suresh >% Hello everyone>% it seems that i have stumbled on something interesting.>% Basically, an algorithm that would use the 'Bell curve' assumption>to find a very quick O(N^2) sub-optimal solutionYou appear to have discovered the Central Limit Theorem.There's a large literature out there on better heuristic solutions to optimisation problems.-- ard Herring=== === Subject: : Re: puzzle: GCDs of Infinite Set of Integer Pairs>...just because an event has probability 0 doesn't mean it can't>happen.mmmm... I have a problem with this. Can you give an example of a probability 0event that actually happened?I would argue that you measured its probability incorrectly. Or it changed afteryou measured it. Or something.-- Patrick Hamlyn posting from Perth, Western AustraliaWindsurfing capital of the Southern HemisphereModerator: polyforms group (polyforms-subscribe@egroups.com)=== === Subject: : Re: the anticlassicalist }{ i: linguistic negationgalathaea > The logic of natural language has been studied> by many different schools of thought throughout history.> Much of this study has pointed to many models> different than those based on classical logical structures.> I have studied this field for some time, but I have had some> questions along the way.> I was looking at litotes (perhaps more exactly, antenantiosis),> with forms likeHe is not unattractive.It is not impossible to do that.> where the double-negation implies some form of a shift> in meaning for effect, following closely an empirical maxim> sometimes called the division of pragmatic labor:The use of a longer, marked expression in lieu of a shorter> expression involving less efort on the part of the speaker tends> to signal that the speaker was not in a position to employ the> simpler version felicitously> (L. R. Horn, Duplex negatio affirmat... the economy of double> negatives -- > Chicago Linguistic Society 27-II)> This kind of maxim, though, is very general, and would seem> to point to many of the other non-classically negated logics> which still support the partial order implied. This shows some> sort of felicity with the agrammatical but common use of> double negatives to stress negativity, such asThey don't know nothing!> In the logical literature I have read, unfortunately, there is> not as much distribution analyses of these logical models> (for instance, the Perspectives der Analytischen Philosophie> book Negation: a notion in focus edited by Hein> Wansing), and so I was hoping someone might hold some> clues as to:> 1) Are there any good tables available in the literature> that correlates types and uses of negation with modern> and historical languages? I mean, I know of some specific> works such as The syntax of negation in russian by> Sue Brown, Buddhist theory of meaning (apoba) and> negative statement by Dhirenda Sharma, The grammar> of negation by Jong-Bok Kim, and have read things> about Serbian and similar forms, but I'm very unsure> of what is considered the most modern or agreed upon> population correlations with fairly comprehensive scope.> Additionally, I have been intrigued by the denial of terms> found in the anvitabhidhana school, where negation> is used to give a meaningful direction to action.> For example, one may sayThe not-house will be built> To stress the otherness or change for which the building> is applied (as a distinguishing in the existing forms).> This ness is an early contribution towards a typed> theory of semantics. Sufi (la ilaha illa 'llah) and taoist> theories of negation are also quite sophisticated, and> so I was curious about:> 2) are there also well known schools of thought> about negation from African or South American> traditions?--------------------------------------------------- --------------They speak mostly Spanish in South America.In Spanish, double-negatives are required.It has nothing to do with logic. It's just amatter of grammatical agreement:http://spanish.about.com/library/weekly/ aa061101a.htm~I don't know if legalese-Spanish is any different.~People frequently make positive assertions,such as this is really great..., while at thesame time shaking their heads as if denying it.When I first started noticing that, I couldn'tthink of any explanation for it other thanthat it was probably a parapraxix, like a slipof the tongue, -revealing that they were lying.But that can't possibly be it.Whatever the true explanation, I think ithas something to do with how we constructsentences unconsciously. I think we constructin parts, in parallel, many different ways to saya thing, and then, at the last second,choose one of them to speak. And I thinkwe always construct both positive andnegative ways to say it. And then, if wechoose to speak the positive form, ourheads may shake to negate, -not it, - butthe lingering unspoken ('undischarged'perhaps) negative form.I believe negation of sentences wasone of the things Chomsky had in mindas a surface-structure transformation,-meaning that it is a trivial last-secondtransformation made on a deep-structureconstruct. Which would only make senseif the deep construct was un-ambivalent,- perhaps even always positive. But Ibelieve that Chomsky eventually gaveup that approach.~Different cultures use differenthead motions to signal 'yes' and 'no'.In India they bob their heads tosay 'yes' -in a way that looks toothers like 'no'. Italians jerktheir head up once to say 'yes',- in a way that looks to otherslike they want to start a fight.(- Some may recall images ofMussolini 'chilling' - crossinghis arms and jerking his headup after giving a speech;He wasn't looking for a fight;He was just pleased with himself.)Desmond Morris talks aboutgesture frontiers, or isoclines.And isoclines of yes/no gesturescross each other in the Balkans.That, and the suspicion that body-languagethat appears to contradict the spoke languageimplies that the person is lying - may goa long way to account for the Balkans.No one understands any one there.~~Sartre regarded negation as the primarycharacteristic of being human: .... from a web page: -- On Part One of Being and Nothingness: Nothingness and Bad Faithhttp://216.239.51.104/search?q=cache:Vp5XSEbyWXgJ: web.ics.purdue.edu/~mmichau/sartre-bn.pdf+%22Being+And+ Nothingness%22+negation&hl=en&ie=UTF-8 I. The Origin of Negation In this section, Sartre discusses the relation between being-in-itself and being-for-itself. Probably the most original feature in Sartre's conception of consciousness is his insistence on its essential negativity; neither Husserl nor Heidegger gives the negative function such a central place ii. In BN we begin with a sober analysis of investigation as called for by the ontological problem, similar to the discussion of Being in Being and Time. However, in Sartre's work, the main emphasis is on the readiness to be faced by the non-existence of the situation inquired about. The question thus reveals that nothings are constant possibilities of our experience. In fact consciousness is shot through with nothings; Nothingness is an immanent characteristic of Being.etc.~~It is not impossible to do that. ...The use of a longer, marked expression in lieu of a shorter> expression involving less effort on the part of the speaker tends> to signal that the speaker was not in a position to employ the> simpler version felicitouslyI don't buy less effort playingany role at all in natural speech.People think at a fixed rate. And they also speak atsome fixed rate appropriate to circumstance. Normally,speech is massively redundant and contentless. Speechis, I think, just a vestigial behavior originating in ourmonkey ancestor's primary social bonding behaviorcalled lice picking. We don't carry lice anymore,or don't admit it, and, therefore we have to talk toeach other instead. And so, there's just no pointin finding ways to say things with 'less effort'.There's certainly a connotational difference between: It is possibleand It is not impossible.However, it's not perfectly clearto me what it is, at this moment.~> So I still am left asking:> 3) what is the actual source of the bias against nonclassical logics?Try: Deviant Logic, Fuzzy Logic: Beyond the Formalism -- by Susan Haackhttp://www.amazon.com/exec/obidos/tg/detail/-/0226311341/ qid=1077019274/sr=1-1/ref=sr_1_1/102-7545475-4339367?v=glance& s=books~Greg.=== === Subject: : Re: Calculating Modulo (very big numbers)>Is there a way to calculate the modulo with numbers like this:>1226^37 mod 4838 ?> I assume you want a way to do it without needing a 115-digit calculator.> Using a 10-digit calculator, I computed this as follows:> exponent> in base-2> 1226^1 = 1226 mod 4838 1> 1226^2 = 1503076 mod 4838 10> = 3296 mod 4838> 1226^4 = 10863616 mod 4838 100> = 2306 mod 4838> 1226^8 = 5317636 mod 4838 1000> = 674 mod 4838> 1226^9 = 826324 mod 4838 1001> = 3864 mod 4838> 1226^18 = 14930496 mod 4838 10010> = 428 mod 4838> 1226^36 = 183184 mod 4838 100100> = 4178 mod 4838> 1226^37 = 5122228 mod 4838 100101> = 3624 mod 4838> Where each step is squaring, multiplying by 1226, or modding by 4838.> Rob son take out the trash before replying=== === Subject: : Re: the anticlassicalist }{ i: linguistic negation> : Do not double negatives not leave you tied in knots?> It is eerie how I was typing up a question about the khipu at the same time> you posted.> Is it knot?Anglice, .> (PS: Louis Kauffman has some beautiful stuff on knot logics, as well.> Reminds me of the BBC / TimeLife show Connections with James Burke)> --> -- Peter T. Daniels grammatim@att.net=== === Subject: : Re: convergence of measurable functions>> Hi all,>>> Let X be a measurable space (a set with a sigma-algebra of sets) and>> B a general complete metric space, that is, not necessarily separable>> or compact or whatever. Put on B, the measurable structure generated>> by the open sets of B (the Borel structure). Now, let>> f_{n}:Xlongrightarrow B be a sequence of measurable functions>> converging pointwise to some f.>>> Question: Is f measurable? Proof?>>> With my best regards,>> G. Rodrigues>How about this. Let F be the sigma-algebra on X.>To show f is measurable, we must show:>if U is open in B, then f^{-1}(U) in F.>Fix U. Define phi : X -> R by phi(b) = dist(b,XU).>Then phi is continuous. And b in U iff phi(b)>0.>All f_n are F-measurable, so>phi(f_n(.)) is F-measurable X -> R. phi is continuous,>so phi(f_n(.)) -> phi(f(.)) pointwise. Therefore>(by the real-valued version of this theorem)>phi(f(.)) is F-measurable, so U = inverse image of>open set (0,infinity) is in F.Hmmm... Could we use the same argument using the characteristicfunction chi_U?chi_U is measurable. chi_U f_n is measurable therefore by thereal-valued version lim chi_U f_n (supposing the limit exists) ismeasurable. Since U is open and B is Hausdorff, it's easy to see thatlim chi_U f_n = chi_U fTherefore chi_U f is measurable. But chi_U f = chi_{f^{-1}U} fromwhich we conclude that f^{-1}U is measurable.Yeah, it seems to work. And only Hausdorff-ness is needed.With my best regards,G. Rodrigues=== === Subject: : Re: the anticlassicalist }{ i: linguistic negation> Now really, Ms.(?) fabulist, etc: there is much that is more grossly> inappropriate to sci.physics here, so you might say all the other kids> are doing it. But if you affect to be a _superior_ kid, might you not> lead by example and limit the number of freshly cross-posted threads> which are mainly about linguistics, and neither physics nor math?Here at sci.lang, we sure haven't seen many posts about linguistics,other than one indicating she'd once opened a volume of Proceedings.> You are beginning to speak Crankish.>-=-=-=-=-= linguistic negation =>The logic of natural language has been studied by many different schools of>thought throughout history.> Textual analysis: schools of thought do not study anything, and is a> cliche.> For somebody who is so academically concerned with language, you seem> remarkably insensitive to its use. Your earnestly learned turgidity> will carry you far ... away from here.> Hi all,> Let X be a measurable space (a set with a sigma-algebra of sets) and> B a general complete metric space, that is, not necessarily separable> or compact or whatever. Put on B, the measurable structure generated> by the open sets of B (the Borel structure). Now, let> f_{n}:Xlongrightarrow B be a sequence of measurable functions> converging pointwise to some f.> Question: Is f measurable? Proof?> With my best regards,> G. Rodrigues>How about this. Let F be the sigma-algebra on X.>To show f is measurable, we must show:>if U is open in B, then f^{-1}(U) in F.>Fix U. Define phi : X -> R by phi(b) = dist(b,XU).>Then phi is continuous. And b in U iff phi(b)>0.>All f_n are F-measurable, so>phi(f_n(.)) is F-measurable X -> R. phi is continuous,>so phi(f_n(.)) -> phi(f(.)) pointwise. Therefore>(by the real-valued version of this theorem)>phi(f(.)) is F-measurable, so U = inverse image of>open set (0,infinity) is in F.> Hmmm... Could we use the same argument using the characteristic> function chi_U?> chi_U is measurable. chi_U f_n is measurable therefore by the> real-valued version lim chi_U f_n (supposing the limit exists) is> measurable. Since U is open and B is Hausdorff, it's easy to see that> lim chi_U f_n = chi_U fIt seems some easy to see things are false...> Therefore chi_U f is measurable. But chi_U f = chi_{f^{-1}U} from> which we conclude that f^{-1}U is measurable.> Yeah, it seems to work. And only Hausdorff-ness is needed.> With my best regards,> G. Rodrigues=== === Subject: : Re: limit of sequence of functions>Nevermind. I just found the (applicable) statement of the theorem and see>how they do it.>One question about how it applies here, though:>Our linear functional>L_n (f) = INT(-pi, pi) p_n (-y) f(y) dy>is seen as mapping a subset of L_oo The _definition_ I gave of L_n was as a linear functionalon C([-pi,pi]):Let's define L_n(f) = int f(x) p_n(-x).Then L_n is a bounded linear functional on C([-pi, pi]) or whateverthe interval was (don't delete everything!) Yes, C is a subset of L^infinity, but that's not really relevant here;this L_n is defined on C, period.>(that's why we used the sup norm) into>R. But the RRT (at least the version I'm looking at) says we need to view>L_n as mapping a subset of L_p for 1 <= p < oo into R. Does it still work>for p = oo ?Oops, you haven't found the right RRT yet. Look for the one thatsays what the dual of C(K) is, for a compact Hausdorff space K.(No, the theorem I think you're looking at is false for p = infinity;the dual of L^infinity is definitely not L^1, except in some veryspecial cases. But that doesn't matter here, because wer'reapplying a totally different theorem, not talking about thedual of L^infinity at all.)>> Out of curiosity, how does RRT show that>> ||L_n|| = int p_n ?Actually ||L_n|| = int |p_n| (probably that's what you meant?>> [...]=== === Subject: : Re: Yo mathemeticians>> i am bad at math. I want to know how to become good at math as>> quickly as possible. Since you would know better than i would from>> experience, say whatever silly idea comes to your head.>>>>> Also, i ing hate math>*>But are you good at pumping gas? That's clear from his post, isn't it?>Flipping burgers?>earle>*=== === Subject: : Re: Entering *Real* Math Formulae in Excel>> How the &*&%$ does one enter a real math formula (simple example: the>> sum of all N sub x as x goes from x1 to x2) ???!>I can't imagine doing real math in Excel, but doesn't >sum(n23:n97) produce the value n23 + n24 + ... + n97? >If it doesn't, something like it does, and it should be possible >to find out within Excel because it has some kind of Help thing >built in & you can just look for sum and it should tell you >how to use it.Gosh, that sounds useful. Say I were running Excel - how wouldI find this Help thing?=== === Subject: : Re: JSH: Question of intent>I've been thinking about formalizing the argument with which I prove a>problem with the ring of algebraic integers, and ran into a problem.Guffaw! Of course you did. It's probably similar to the problemsI run into every time I try to formalize my proof that the Earthis flat.>[...] How do you handle human intent mathematically?You don't - mathematical proofs do not involve any such notion.>=== === Subject: : Re: tensors for tots> How far can one get in understanding the structure of tensors at a> point on a manifold in terms of our old school friends column vectors,> row vectors and matrices?Very far, I think. Start with a vector space. A vector is clearlyindependent of his coordinates referring to a special basis, so howdo the coordinates change, when we change the basis? The answer ofLinear Algebra is: Let B be the matrix of the change of the basis,and v be the (column-)vector of coordinates with respect to the old basis. Then the new coordinates w are given byw = B^(-1) v.B^(-1)'s name in SRT is often Lambda mu nu.A vector is a contravariant vector, so to say, an (1,0)-Tensor.Now consider a linear mapping from the vector space to the field ofscalars, represented by a row v*. How do the coordinates change? LAsays, the new coordinates w* arew* = v* B^T(Multiplication from the right with the Transposed of B)B^T's name in SRT is often Lambda tilde (swung dash) mu nu.A co-vector is a covariant vector, an (0,1)-Tensor.A Matrix consists of rows and colums, so it is a (1,1)-Tensor.Generally speaking, a (m,n)-Tensor is a tuple of m contravariantvectors (= vectors, tangent vectors) and n covariant vectors(= co-vectors, cotangent vectors).OK? === === Subject: : Re: question re Lebesgue-integrable fcn>Anyone know how to prove the following?>Suppose f is a Lebesgue-integrable function on R (reals). Prove that>the series sum_{n=-infty}^{infty} f(x+n) converges absolutely for>almost every x in R.Some basic theorems about integrating _non-negative_ measurablefunctions let you show that int_0^1 sum |f(x+n)| = ... . Hence ... .=== === Subject: : [PhD-position] Statistics. University of Twente. Netherlands=================================================== ================ Announcement of PhD-positionStatistical analysis of insurance portfolios University of Twente, The Netherlands=================================================== ================The Department of Applied Mathematics of the Faculty of ElectricalEngineering, Mathematics and Computer Science at the University ofTwente, Enschede, The Netherlands has available a PhD-positionFor information contact:prof.dr. W. Albers (w.albers@math.utwente.nl, tel. 053-4893816/3434).Title:Statistical analysis of dependence effects on insurance portfoliosResearch Group:The project will be performed in the chair Statistics and Probability.It fits in the research programme Financial Engineering from theCentre for Telematics and Information Technology (CTIT), one of thekey research institutes of the University of Twente. The FinancialEngineering Laboratory (FELab) is a joint initiative of researchersin this field within the School of Mathematical Sciences and theDepartment of Finance and Accounting (FMBE) at the School ofTechnology & Management. At present, 13 full- and 6 part-time facultymembers, as well as 6 PhD-students, participate in this laboratory.Supervision:prof.dr. W. Albers (UT; email: w.albers@math.utwente.nl)dr. W.C.M. Kallenberg (UT; email: w.c.m.kallenberg@math.utwente.nl)Funding source:This position is funded by the Technology Foundation STW, appliedscience division of NWO and the technology programme of the Ministryof Economic Affairs.Period:The project can start immediately; it will take 4 years.Required:The candidate should have a MSc in mathematical statistics, probabilityor econometrics, and a keen interest in applications.Description:Quantities of interest for insurance portfolios are among others the sumS of the claims of the individual risks during a reference period andthe stop-loss premiums E{max(0,(S-a))} for various retentions a.Typically it is assumed that the summands in S are independent, althoughit is clear that in practice dependencies will occur. Implicit in thisassumption apparently is the hope that the effects of such dependencieswill be negligible, or at most small for practical purposes. Unfortunately,however, this is often not the case: introducing a small dose ofdependence may already lead to gross deviations of the stop-loss premiums. To deal with the problem, families of models incorporating dependenciesneed to be proposed and studied, with an emphasis on practical usefulness.Typically, a bce will have to be found between degree of complicationand realistic content. Next, techniques have to be found to obtainadequate approximations to the above mentioned quantities of interest.This is also a nontrivial part, as already in the independent case exactresults are generally intractable. Finally, estimation methods have to bedeveloped to fit the models obtained to data occurring in practice. As concerns the first two of these three stages in the above program,some progress has already been made in [1]. Here a simple stochasticmixture model is proposed, which is demonstrated to incorporate to firstorder many models in common practical use Second order Edgeworth expansionsturn out to be quite successful for obtaining adequate approximations tostop-loss premiums, as long as normally distributed claims are used.In [2], the scope is widened considerably by including various otherapproximations from the literature. For a wide variety of claim sizedistributions and retention levels, such approximations are compared in thispaper to each other, as well as to a quantitative criterion. The third step,fitting such a model to real data, still has to be taken. It is expectedthatthis will generate a new iteration of the program by suggesting directionsinto which the basic model should be altered or generalized.[1] Albers, W. (1999). Stop-loss premiums under dependence. Insurance: Math. & Econ. 24, 173-185. for stop-loss reinsurance premiums. TW-report 1695.========================================================= ==========____________________________________________________ ______________Gerhard J. Woeginger http://wwwhome.cs.utwente.nl/~woegingergj/=== === Subject: : Re: How fast is the Confinued Fraction factorization algorithm? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1HDJYB27293;I have another question: In each of the above factorizations, only half >(approximately) of the primes in the factor base are used. I read >somewhere that N (the composite) must be congruent to a square modulo >the primes in the factor base, but I don't see why:So? The real part of exp(i pi) is cos(pi), and its imaginary part>is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. >> I,have a reason , with my due respect.>> Panagiotis Stefanides>Yes, but you DON'T tell us what your reason is. You can't expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0?>>The reason is simply that exp(ipi0=-1 should be accompanied>>by the statement that this is the real part solution.>>Is it fair?>No, that is not a fair comment. exp(i pi), the complex number, is>equal to -1, the complex number. There is no need to appeal to the >real part. Also, what does exp(i pi) = -1 is the real part solution>mean? You look like you are using terminology in a manner not >recognized in mathematics.-------------->>e^[i*pi] ,as accepted ,is a phasor. >In most of the relevant fields of mathematics, e^[i pi] is a complex >number.>>It is only fair to state that its >>polar representation is :>>e^[i*pi] = MOD 1 , ARG 180 .>That is Arg 180 degrees, not just Arg 180. And so what? That does >not lead to your assertion that exp[i pi] = 0, a result for which >you have given absolutely no support. Why don't you just give up? At the moment, they (the Time Lords) are far from being all-powerful.> That's why it's been left up to me and me and me.> quote by: Patrick Troughton in The Three Doctors>-------I, have made myself very explicit.My original question of the implicationof the imaginary component: e^[ipi]=j*0 (to the related proof)was not answered and still it is open.My best wishes to You ,and to the thread list members.Panagiotis Stefanides.=== === Subject: : Re: 'erf' function in C>> Are there any C compilers which have the erf function (from> probability) as part of their math libraries? Or are there any math> libraries available to download which implement this function?> James Harlacher>> [...]>>A procedure that can determine Phi(x) nearly to the limit>>available in double precision can be based on the following>>method, which I developed the 1960's:>>> [...][...]>>>George Marsaglia>>> [...]>> Is there someplace a table available of the true values of the>> normal distribution up to 20 places?>> With kind regards>> Matthias Kl.8ay>> -- >> www.kcc.ch>Sorry, I should have mentioned that the values I used>were based on setting Digits:=20 or 30 in Maple, then>calling .5+.5*erf(x/sqrt(2.)), (Yes, Maple is still stuck>with erf for lack of a standard for the normal distribution,>although it may be avoided by the rather awkward> stats[statevalf,cdf,normald](x);)>Mathematika and other extended precision math packages>would also serve. Of course the old standby of>looking up tabled values, for example the Handbook of>Mathematical Functions edited by Abramowitz and Stegun,>will still serve, but they are likely to list evenly spaced> x's and perhaps less than 20 digits for Phi(x).>George MarsagliaIn my copy of A&S from around 1972 Phi(x) is tabled for x = 0(0.02)3 to 15 places, andx = 3(0.05)5 to 10 places. So this is fairly outdated as areference. Unfortunately, it seems that the new electronic version ofA&S at http://dlmf.nist.gov is not going to be completed very soon.This still leaves me with the question: How to validate such analgorithm against a standard? Can you trust Maple, Mathematica & Co.to deliver exact results up to the desired precision? BTW I made implementations in Fortran, HP-41C, Pascal, Modula 2,Clipper and recently Visual Basic of the Taylor Series expansion ofPhi(x) given in 26.2.11 in A&S since around 1980: Phi(x) = 1/2 + phi(x)*(x + x^3/3 + x^5/(3*5) + x^7/(3*5*7) + ...)Here are my VB results against your algorithm (also translated toVisual Basic). It looks like the VB version loses about 1 digit inprecision against the C version (I'm not surprised, but it is not toobad either). x Marsaglia (my VB Version) Marsaglia (C Version, from your table) Klaey (VB) Maple 20 places 0.123 0.54894645101643700000 0.5489464510164369 0.54894645101643700000 0.54894645101643675909 1.2 0.88493032977829200000 0.8849303297782918 0.88493032977829200000 0.88493032977829173198 2.4 0.99180246407540400000 0.9918024640754040 0.99180246407540400000 0.99180246407540387055 6.1 0.99999999946965800000 0.9999999994696578 0.99999999946965800000 0.99999999946965767370-6.1 0.00000000053034350051 0.0000000005303427 0.00000000053034221459 0.00000000053034232638-1.1 0.13566606094638300000 0.1356660609463828 0.13566606094638200000 0.13566606094638267517 7.2 0.99999999999969900000 0.9999999999996990 0.99999999999969900000 0.99999999999969893721I'm quite pleased how my algorithm still stands up today :-)With kind regardsMatthias Kl.8ay-- www.kcc.ch=== === Subject: : Re: DiffEQ - need help to solve a basic linear diffEQ> Hi. I'm trying to solve this differential equation:> (x + e^y) dy/dx = x * e^(-y) - 1Take x = t.e^y. Then dx/dy = (t + dt/dy).e^y, so that: (x + e^y).e^y = (t + 1).e^(2y) = (x - e^y).dx/dy = (t - 1).(t + dt/dy).e^(2y)As e^(2y) is non-zero for all y, this implies: (t + 1)/(t - 1) - t = dt/dy<=> dy/dt = (t - 1) / (2 - (t - 1)^2)<=> - 2y = log (2 - (t - 1)^2) + C etcCheers Ramsden=== === Subject: : Re: Constraints on Self-Dual Linear Block Codes windows-nt)>> If (n,k) is a q-ary linear block code of length n and>> size q^k, then in order for the code to be self-dual>> it is fairly easy to show that n = 2*k.>>> Are there any other constraints required of a code>> in order for it to be self-dual?> Yes ... it has to satisfy the definition of self-duality :-)See the text by Roman. I know the answer to my question is yes - I was looking for clarification/insight. -- % Randy Yates % Bird, on the wing,%% Fuquay-Varina, NC % goes floating by%%% 919-577-9882 % but there's a teardrop in his eye...%%%% % 'One Summer Dream', *Face The Music*, ELOhttp://home.earthlink.net/~yatescr=== === Subject: : Re: Calculating fraction period lengths in different bases>> Can anyone prove these statements>> My interest is recreational.>>> Statement 1>>> If u=2n^2-n+2 and v=2n^2+n+2>>> Then u^2 mod (u+v-1) =u-1 and v^2 mod (u+v-1) = u-1>> explicidly:(1) u+v-1 = 4n^2+3(2) u^2 = 4n^4-4n^3+9n^2-4n+4 = (4n^2+3)(n^2) - 4n^3 + 6n^2-4n + 4 = (4n^2+3)(n^2-n) + 6n^2 - n + 4 = (4n^2+3)(n^2-n + 1) + 2n^2 - n + 1 ^^^^^^^^^^^^ u-1(3) v^2 = 4n^4+4n^3+9n^2+4n+4 = (4n^2+3)(n^2) + 4n^3 + 6n^2+4n + 4 = (4n^2+3)(n^2+n) + 6n^2 + n + 4 = (4n^2+3)(n^2+n + 1) + 2n^2 + n + 1 ^^^^^^^^^^^^ v-1(4) u^2 mod (u+v-1) = u-1(5) v^2 mod (u+v-1) = v-1* (5) contradiction of above(6a)v = u + 2n(6) v^2 = (u + 2n)^2 = u^2 + 2.u.2n + 4n^2 =u^2 + 8n^3 + 8n =u^2 + 2(4n^2+3) + 2n>> For all n>0 >>> Length of fraction period in base u and v =6> I think you have to tell us just what fraction you have in mind.-- If its Monday then I am a fool but not ignorant.=== === Subject: : Re: Metamath Axiom of Choice>>> Also, the motivation behind proving the equivalence of the Metamath >> formulation of the Axiom of Choice and more standard formulations>> is the same as the motivation behind proving the equivalence of the >> Axiom of Choice and Zorn's Lemma, and the equivalence of the Axiom >> of Choice and the Well-Ordering Theorem. I presume that you have >> seen the proofs of these last two mentioned equivalences, Acid Pooh.>Heh. Since you appear to be goading me into replying, and I'm a> sucker for this sort of thing, here goes: I was not aware thatMetamath was a website. Nor was I aware that they had a different> version of the axiom of choice. I assumed (obviously incorrectly)> that you meant metamath to refer to Set Theory, or some suchmetamathematical study. Now, in this context, my response doesn't> seem so inappropriate, does it? In particular, it shouldn't seem so> patronizing, since, as I parsed your question, it is a question a> beginner should ask. So, why decide to patronize me?I was likewise unaware that there was a Metamath site, but that factbecame abundantly clear as soon as I read D. McAnally's very firstsentence of his first post on the subject. I suggest you go back andread it.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.>I'll offer this definition: A number is an element of a number set. A>>number set is either the set of integers Z or any ring/field R that>>contains Z.> So ordinal numbers are not numbers?It seems even the natural numbers are not numbers.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Microsoft Excel HelpYou can do a search on the function name or function reference.Or perhaps get my function lists which automates help for allfunctions.Or refer to:Peter Noneley:http://homepage.ntlworld.com/noneley/Or do a Google search on the function name. Here I'd recommend Ron deBruin's Google Search 6 addin:http://www.rondebruin.nl/Google.htm-- Norman Harker MVP (Excel)Sydney, Australianjharker@optusnet.com.auExcel and Word Function Lists (Classifications, Syntax and Arguments)available free to good homes.> How the &*&%$ does one enter a real math formula (simple example:the> sum of all N sub x as x goes from x1 to x2) ???!>I can't imagine doing real math in Excel, but doesn't>sum(n23:n97) produce the value n23 + n24 + ... + n97?>If it doesn't, something like it does, and it should be possible>to find out within Excel because it has some kind of Help thing>built in & you can just look for sum and it should tell you>how to use it.> Gosh, that sounds useful. Say I were running Excel - how would> I find this Help thing?=== === Subject: : Re: TSP and the Bell curveCentral Limit Theorem http://mathworld.wolfram.com/CentralLimitTheorem.html=== === Subject: : Re: countable sets> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove> that union of Ek as k = 1 to oo is countable. I do not need help proving> this. I need someone to explain exactly what is being asked. In Let E1, E2,> E3,... be a sequence of pairwise disjoint countable sets is that saying> that the numbers of Ei's are countable or is it saying that each Ei has a> countable number of elements??Others have explained what is being asked, but I was intrigued by yourstatement that you don't need help proving it. Now that the question hasbeen clarified, you might look again at proving it.Hint: If your proof does not invoke the axiom of choice, then your proofis wrong.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.>...just because an event has probability 0 doesn't mean it can't>>happen.> mmmm... I have a problem with this. Can you give an example of a probability 0> event that actually happened?Suppose X is a random variable, uniformly distributed on [0,1]. Then theprobability of the event X=x is 0 for each real number x in [0,1].Yet, some x must be chosen. Every single repetition of this experiment,without fail, results in the occurrence of an event with probability 0.> I would argue that you measured its probability incorrectly. Or it changed after> you measured it. Or something.Probabilities are not measured. Probabilities are determined by thedefinition of the sample space.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.I'll offer this definition: A number is an element of a number set. A>number set is either the set of integers Z or any ring/field R that>contains Z.>> So ordinal numbers are not numbers?> It seems even the natural numbers are not numbers.So Numbers are not numbers;)-- If its Monday then I am a fool but not ignorant.Its Wenesday === === Subject: : Re: countable sets>> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove>> that union of Ek as k = 1 to oo is countable. I do not need help proving>> this. I need someone to explain exactly what is being asked. In Let E1, E2,>> E3,... be a sequence of pairwise disjoint countable sets is that saying>> that the numbers of Ei's are countable or is it saying that each Ei has a>> countable number of elements??> Others have explained what is being asked, but I was intrigued by your> statement that you don't need help proving it. Now that the question has> been clarified, you might look again at proving it.> Hint: If your proof does not invoke the axiom of choice, then your proof> is wrong.Explicitely invoked? j gWhat about N ---> N x N ---> union(E_k)where j is a suitable bijection and g(k,n) := f_k(n)is defined via given surjective maps f_k : N ---> Ek ?Marc=== === Subject: : Re: Metamath Axiom of Choice Heh. Since you appear to be goading me into replying, and I'm a> sucker for this sort of thing, here goes: I was not aware thatMetamath was a website. Nor was I aware that they had a different> version of the axiom of choice. I assumed (obviously incorrectly)> that you meant metamath to refer to Set Theory, or some suchmetamathematical study. Now, in this context, my response doesn't> seem so inappropriate, does it? In particular, it shouldn't seem so> patronizing, since, as I parsed your question, it is a question a> beginner should ask. So, why decide to patronize me?Perhaps he decided to patronize you because, as *everyone* parsed yourpost, you are incapable of reading plain English but you pretend tohave the expertise to respond to questions anyway?It shouldn't seem patronizing after all, since as he (and otherrespondents) read your post, it was an answer an ignoramus mightwrite.-- You lack the ability to reason, but instead get an idea in your headand hold on to it against all evidence. I don't find you credible,and reject your claims, as coming from a flawed source. === === Subject: : Re: What is a number?/What is not a number?>I'll offer this definition: A number is an element of a number set. A>number set is either the set of integers Z or any ring/field R that>contains Z.> So ordinal numbers are not numbers?Good point. OK, a number is an element of a number set. A number setis eithera) the set of integers Z orb) any ring/field/algebra R that contains Z orc) any of the ordinal or cardinal objects defined by Cantor.I think it's beginning to look like a weaker definition is in orderfor number set. Let's look at some possibilities. A number set is1) any set that has a closed binary operation defined on all pairs of elements of the set,2) any semigroup, or3) any totally orderable set.But these possible definitions may be too weak. The advantage of 1) isthat it allows people to call the elements of their favorite setnumbers if they choose to do so. It seems to me that pure whim isthe driving force behind what set elements have been generallyaccepted to be called numbers.Maybe the desire to label set elements as numbers lies in thepsychological need of people to reify their favorite sets. And if thishypothesis of ontology being at the root of the number issue istrue, then the term imaginary number is an oxymoron, or at the veryleast an equivocation. Perhaps the orgin of the number concept isPlatonistic at heart. In any case, it seems too late to return to thetime when number was associated with counting or measuring. Then whatis a number?Patrick=== === Subject: : Re: countable sets>Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets.Prove>that union of Ek as k = 1 to oo is countable. I do not need help proving>this. I need someone to explain exactly what is being asked. In Let E1,E2,>E3,... be a sequence of pairwise disjoint countable sets is thatsaying>that the numbers of Ei's are countable or is it saying that each Ei hasa>countable number of elements??> Others have explained what is being asked, but I was intrigued by your> statement that you don't need help proving it. Now that the question has> been clarified, you might look again at proving it.Dave,Now that I understand the statement I will try to prove it using the axiomof choice.Onto another topic. I hope that you don't mind. I know that Mumia is notfacing a death sentence at this point but is he still physically on deathrow locked up 23 hours per day? FREE Mumia now!> Hint: If your proof does not invoke the axiom of choice, then your proof> is wrong.> --> Dave Seaman> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> >> Hi all,> Let X be a measurable space (a set with a sigma-algebra of sets) and> B a general complete metric space, that is, not necessarily separable> or compact or whatever. Put on B, the measurable structure generated> by the open sets of B (the Borel structure). Now, let> f_{n}:Xlongrightarrow B be a sequence of measurable functions> converging pointwise to some f.> Question: Is f measurable? Proof?> With my best regards,> G. Rodrigues>>>How about this. Let F be the sigma-algebra on X.>>To show f is measurable, we must show:>>if U is open in B, then f^{-1}(U) in F.>>Fix U. Define phi : X -> R by phi(b) = dist(b,XU).>>Then phi is continuous. And b in U iff phi(b)>0.>>>All f_n are F-measurable, so>>phi(f_n(.)) is F-measurable X -> R. phi is continuous,>>so phi(f_n(.)) -> phi(f(.)) pointwise. Therefore>>(by the real-valued version of this theorem)>>phi(f(.)) is F-measurable, so U = inverse image of>>open set (0,infinity) is in F.>>>> Hmmm... Could we use the same argument using the characteristic>> function chi_U?>>> chi_U is measurable. chi_U f_n is measurable therefore by the>> real-valued version lim chi_U f_n (supposing the limit exists) is>> measurable. Since U is open and B is Hausdorff, it's easy to see that>>> lim chi_U f_n = chi_U f>It seems some easy to see things are false...After I sent the message, I stepped back and this easy to see thingsounded false. The problem is of course if f(x) is in the boundary ofU in which case we can assert nothing about f_n(x) being eventually inU.G. Rodrigues=== === Subject: : Re: countable sets> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove> that union of Ek as k = 1 to oo is countable. I do not need help proving> this. I need someone to explain exactly what is being asked. In Let E1, E2,> E3,... be a sequence of pairwise disjoint countable sets is that saying> that the numbers of Ei's are countable or is it saying that each Ei has a> countable number of elements??>> Others have explained what is being asked, but I was intrigued by your>> statement that you don't need help proving it. Now that the question has>> been clarified, you might look again at proving it.>>> Hint: If your proof does not invoke the axiom of choice, then your proof>> is wrong.> Explicitely invoked? > j g> What about N ---> N x N ---> union(E_k)> where j is a suitable bijection and g(k,n) := f_k(n)> is defined via given surjective maps f_k : N ---> Ek ?For any given k we are given that there is a suitable surjection f_k : N-> E_k. That is, the set S_k of surjections from N onto E_k is nonemptyfor each k.How do you conclude (without AC) that there is a g: N x N -> union(E_k)?That requires choosing all the f_k's at once, and there are infinitelymany of them. That is, you need to show that there is a choice functionon the set { S_k : k in N }.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Onto another topic. I hope that you don't mind. I know that Mumia is not> facing a death sentence at this point but is he still physically on death> row locked up 23 hours per day? FREE Mumia now!Neither side was satisfied with Judge Yohn's decision and both sides areappealing. The possibility still exists that the death penalty could bereinstated without a new trial.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.Are you saying that the Quadratic Sieve>is better than the Number Field Sieve>for certain sizes of n?Yes.>Which sizes, roughly speaking?Roughly for n < 100 digits. Certainly for n < 85 digits.In between, your mileage may vary depending on implementation.>My impression, based on not very scientific>(or knowledgeable) experiment>is that the Quadratic Sieve (with many polynomials)>is unlikely to factorize number with more than 80 digits>in a reasonable time on a reasonable computer>(say 1 day on my 667MHz laptop).What do you mean by unlikely? It makes no sense in this context. QS succeeds with virtual certainty in time that depends only on the size of n (with some small statistical variation).80 digits is about right for your machine in a day. Adding about 5 digits will roughly double the run time with a well-tunedimplementation. Note that above 85 digits you will want to use the 2-large prime variation.>Incidentally (I'm sure this will show my ignorance)>I've never understood why the various methods>looking for solutions of x^2 = y^2 mod n>(to give a factor gcd(x-y,n) of n)>always seem to compare a quadratic residue mod n>with a perfect square (of a smooth number).What do you mean by compare?? I do not understand you.CFRAC and QS generate relations of the formA^2 = b mod n,then find a lot of these relations with b fully factored.One then uses the 'index calculus' to construct x^2 = y^2 mod nby multiplying together a suitable set of factored relations.NFS generates relations of the forma = b mod nsuch that a and b are quite small and related algebraically.Now, since we don't a priori have a square on one side, we must factorBOTH a and b.>Why not compare two quadratic residues,How do you propose we do this? Why should we expect some kindof algebraic relation mod n between two random q.r's of n?Suppose we have two q.r's of n. Call them x and y. What do youpropose we do with them? Also explain how we determine that x andy ARE q.r.'s of n. I am not sure I understand what you are trying to suggest.>or two perfect squares?But q.r's ARE perfect squares mod n. Or did you mean something else?What are you trying to ask here? === === Subject: : Re: JSH: Question of intent> How do you handle human intent mathematically?Easy.IF 'mathematics is posted by JSH'THEN 'intent is to aggrandize self and affirm own magnificence'ELSE 'intent is irrelevant'> James The Magnificent, W.O.O.O Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--.--http://www.crbond.com=== === Subject: : Re: Calculating Modulo (very big numbers)> Hi.> Is there a way to calculate the modulo with numbers like this:> 1226^37 mod 4838 ?> Or is it impossible to do?One might try to use the hyperintegers as described inhttp://www.ajnpx.com/html/10-adic.htmland then generalize from base ten to base n. The point is that thenumber 1226 can be projected onto the idempotents of thehyperintegers, e_1 and e_2, as1226^37 = [1226(e_1 + e_2)]^37 = [1226 e_1 + 1226 e_2]^37 = (1226 e_1)^37 + (1226 e_2)^37So, 1226^37 mod 10000 = (1226 e_1)^37 + (1226 e_2)^37 mod 10000 = (1226 e_2)^37 mod 10000where we are interested in only the four right-most digits of 1226^37.It has been about 20 years since I've looked at this stuff, but Iremember then thinking that this could be one application of thehyperintegers. Perhaps there's a way to use the Chinese Remainder Thmto convert between bases quickly.Patrick=== === Subject: : How many different resistances with n resistors?Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)The Puzzle Corner in MIT's magazine Technology Review raised thequestion of the number of distinct resistances achievable by connectingten unit resistors.According to Sloane's online encyclopedia, the answer is 2213:http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anum=A048211However, this calculation appears to assume that the resistance of thenetwork can always be calculated by successive applications of the rulesfor resistors in series and parallel. I have always had the vagueimpression that some configurations, e.g., the Wheatstone bridge, weresolvable only by using Kirchhoff's laws and not by the series/parallelrules. Have I just been miseducated? Or maybe the issue doesn't comeup with ten unit resistors?-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences=== === Subject: : Re: No Set Contains Every Computable Natural (was Church-Turing compared to Zuse-Fredkin thesis)>This is the definition of a recursively enumerable language:>This language is recursively enumerable. Given a string, x, there existsa>TM that>will halt after a finite number of steps if x is a member of L.>This language is NOT decidable. Consider what happens if x is an>> infinite string of 1's. The TM will not halt after a finite number ofsteps.> You can't input an infinite string to the tape. If you could, it wouldn't> be a Turing machine.No, but you can input finite instructions which generate infinite output. is right about Turing saying this, and I found it repeated:A Highly Random Number by Becher, Daicz, and Chaitin:In this note we present a natural example of a random numberthat goes beyond the class of omega-numbers. The idea goesback to Turing's celebrated paper On computable numbers,with an application to the Entscheidungsproblem [11],where he describes the computable numbers as the real numberswhose decimal expansion is calculable by finite means.According to Turing's definition, a number is computable ifits decimal expansion can be written down by a circle freemachine. This is a machine that performs an unendingTuring defines the machine to be circular. Circular machinesreach a configuration from which there is no possible move,or go on moving, but do not print any more symbols.SH: You might not want to call this an algorithm which hasa terminating clause, but it is a computable method. See 1) FinitenessKnuth, Vol. 1, Sec. 1.1:The modern meaning for algorithm is quite similar to that of_recipe_,_process_, _method_, _technique_, _procedure_, _routine_,_rigmarole_, except that the word algorithm connotes somethingjust a little different. Besides merely being a finite set ofrules that gives a sequence of operations for solving a specifictype of problems, an algorithm has five important features: 1) Finitness. An algorithm must always terminate after a finite number of steps. [...] (A procedure that has all of the characteristics of an algorithm except that it possibly lacks _finitness_ may be called a _computational method_. [...])(A procedure which has all the characteristics of an algorithm except that it possibly lacks finitness may be called acomputational method. Besides his algorithm for the greatestcommon divisor of two integers, Euclid also gave a geometrical construction that is essentially equivalent to algorithm E, except that it is procedure for obtaining the ``greatest common measure'' of the lengths of two line segments; this is a computational method that does not terminate if the given lengths are ``incommensurate''). [SH: for intererested constructivists] 2) Definitess. Each step of an algorithm must be precisely defined; the actions to be carried out must be rigorously and unambiguously specified for each case. [...] 3) Input. An algorithm has zero or more _inputs_: quantities that are give to it initially before the algorithm begins, or dynamically as the algorithm runs. [...] 4) Output. An algorithm has zero or more _outputs_: quantities that have a specified relation to the inputs. [...] 5) Effectiveness: An algorithm is also generally expected to be _effective_, in the sense that its operations must all be sufficiently basic that they can in principle be done exactly and in a finite length of time by someone using pencil and paper. [...](A procedure which has all the characteristics of an algorithm except that it possibly lacks finitness may be called acomputational method. Besides his algorithm for the greatestcommon divisor of two integers, Euclid also gave a geometrical construction that is essentially equivalent to algorithm E, except that it is procedure for obtaining the ``greatest common measure'' of the lengths of two line segments; this is a computational method that does not terminate if the given lengths are ``incommensurate'').I think Church's (1937a) description agrees with infinite symbols.The author [i.e. Turing] proposes as a criterion that an infinitesequence of digits 0 and 1 be computable that it shall bepossible to devise a computing machine, occupying a finite spaceand with working parts of finite size, which will write down thesequence to any desired number of terms if allowed to run for asufficiently long time. As a matter of convenience, certain furtherrestrictions are imposed on the character of the machine, but theseare of such a nature as obviously to cause no loss of generality -in particular, a human calculator, provided with pencil and paperand explicit instructions, can be regarded as a kind of Turing machine.The Broad Conception of Computation by Jack CopelandIs addition over the computable numbers a computablefunction? (That is, is x+y computable whenever x and yare both computable numbers?) The answer is 'yes', butoff the top of one's head one might think otherwise,for if x (or y) is a computable number having an infinitedecimal representation, how can x be input, given Turing'srestriction that the input inscribed on the tape mustconsist of a finite number of symbols?The solution is to input x in the form of a program which,if inscribed on the (otherwise blank) tape of some universalTuring machine, would cause the machine to calculate thedecimal representation of x digit by digit. Nothing trickyis going on here. The program inscribed on the tape is, afterall, a sequence of symbols, and this particular sequence ofsymbols has no less a claim to be counted as a representationof the number, x, than '14' and '1110' have to be counted asrepresentations of the number fourteen.In short, Turing has given us a new method for representingnumbers; and in this system there are finite representationsof some of the numbers which, in the decimal system, can berepresented only by means of an infinite sequence of symbols.To return to the machine that is to add x and y, where x andy are any computable numbers: once the representations, inTuring's ingenious system, of x and y have been inscribed onthe tape, the machine is set in motion. First it calculatesthe first digit of the decimal representation of x andremembers it; next it calculates the first digit of therepresentation of y; then it adds these two digits, using aleft[CapitalEth]to[CapitalEth]right addition procedure; then it turns to the seconddigits of the representations, and so on. Each digit of thedecimal representation of x+y is produced by the machine insome finite number of steps.I hope the discussion so far has afforded some insight intowhat a computable number is, in Turing's restricted sense of'computable number'. A number is computable, in his sense,just in case the number can be expressed by means of a certainfinite string of symbols, namely a string of symbols which, ifinscribed on the otherwise blank tape of some universal Turingmachine, will cause the machine to churn out the decimalrepresentation of the number.SH: I think this means that a TM can add Pi and e digit by digit.I posted not to argue on 's side, but because I thought thisTuring ingenuity deserved some recognition. I think the terminatingrequirement makes more sense when applied to a digital computer.I think there is one term. not halting, doing double duty for twodifferent reasons. Printing Pi can continue to output symbols, acircle-free machine, but a circular machine, can continue to movebut doesn't output symbols. So it is no longer a 'generative' case.BTW, Chaitin has a new, warm, e-book online about the historyand philosophy of developing primes, randomness and Omega.http://www.cs.auckland.ac.nz/CDMTCS/chaitin/ omega.htmlMETA MATH! The Quest for Omega by Gregory ChaitinStephen=== === Subject: : Topological group basicsLet G be a topological group. If a subgroup H < G is open, then it is alsoclosed. Why? I must show that the cosets gH are all open, but why is thistrue?f : G x G ---> G(x,y) ---> xyandh : G ---> Gx ----> x^{-1}are both continuous.So this means thatg : H ----> G/H H ----> gHis continuous, but we need the the inverse of this last map to becontinuous, because I want this last map to be an open map, but why is it?I just can't see why the cosets gH are all open...Any help?Moshe=== === Subject: : Re: Calculating Modulo (very big numbers)>> Hi.>>> Is there a way to calculate the modulo with numbers like this:>>> 1226^37 mod 4838 ?>>> Or is it impossible to do?> One might try to use the hyperintegers as described in> http://www.ajnpx.com/html/10-adic.html> and then generalize from base ten to base n. The point is that the> number 1226 can be projected onto the idempotents of the> hyperintegers, e_1 and e_2, as> 1226^37 = [1226(e_1 + e_2)]^37 = [1226 e_1 + 1226 e_2]^37> = (1226 e_1)^37 + (1226 e_2)^37> So, 1226^37 mod 10000 = (1226 e_1)^37 + (1226 e_2)^37 mod 10000> = (1226 e_2)^37 mod 10000> where we are interested in only the four right-most digits of 1226^37.This is only practical if one knows the factorization of the modulusinto primes.In using the RSA algorithm one must compute the remainder of a^b on divisionby M where M is a product of two large primes. The user does not have theseprimes available, and were he/she able to compute them, this case of RSAwould be insecure. So even in practical applications (RSA cryptography)this is useless.Nonetheless, there are standard tricks known that will perform the OP'sThe binary powering algorithmand put it on my website athttp://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html .-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: Topological group basics> Let G be a topological group. If a subgroup H < G is open, then it is> also> closed. Why? I must show that the cosets gH are all open, but why is> this true?The topology on G is translation-invariant.As an extra exercise: show that a closed subgroup of finite index is open.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: e is transcendental> My original question of the implication> of the imaginary component: e^[ipi]=j*0 (to the related proof)> was not answered and still it is open.There is only one implication of the imaginary component:sin(pi) = Im(e^[ipi]) = Im(-1+i*{0)) = 0-- Daniel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Re: No Set Contains Every Computable Natural (was Church-Turing compared to Zuse-Fredkin thesis)messageThis is the definition of a recursively enumerable language:This language is recursively enumerable. Given a string, x, thereexists> a>TM that>will halt after a finite number of steps if x is a member of L.>This language is NOT decidable. Consider what happens if x is an> infinite string of 1's. The TM will not halt after a finite number of> steps.>>You can't input an infinite string to the tape. If you could, itwouldn't>be a Turing machine.> No, but you can input finite instructions which generate infinite output.> is right about Turing saying this, and I found it repeated:But that would be a different language. The language would then be { | Mis a TM that outputs a string a finite string of 1s}. This language isclearly undecidable. But that does not prove his point, as any languagewhere (language U lang_complement) is composed of arbitrary TMs and whosemembership is based on determining non-trivial properties of those TMs, isundecidable (a la Rice's theorem). However L is certainly decidable:Let G be a topological group. If a subgroup H < G is open, then it is>also>closed. Why? I must show that the cosets gH are all open, but why is>this true?> The topology on G is translation-invariant.What do you mean by translation invariant? Do you mean that if H has sometopology (open, closed, or whatever), then gH has the same topology? ThenWhy is the topology on G translation invariant? This is what I would liketo know.> As an extra exercise: show that a closed subgroup of finite index is open.> -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9> Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : CA & Numerical Analysis?Hi!Can anyone help me to find a paper about the application of cellularautomata in numerical analysis?Behrang.[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to fix the problems with your system. ]=== === Subject: : Re: Entering *Real* Math Formulae in Excel>>Gosh, that sounds useful. Say I were running Excel - how would>>I find this Help thing?>Hi David!>Use:>Help > Microsoft Excel HelpI guess this person failed Prof. Ull's irony class.-- === === Subject: : Re: Topological group basics>> Let G be a topological group. If a subgroup H < G is open, then it is>> also>> closed. Why? I must show that the cosets gH are all open, but why is>> this true?>> The topology on G is translation-invariant.> What do you mean by translation invariant? That multiplication by g is a homeomorphism from G to G.> Do you mean that if H has some> topology (open, closed, or whatever), then gH has the same topology? Then> Why is the topology on G translation invariant? This is what I would like> to know.Continuity of multiplication.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: Entering *Real* Math Formulae in Excel>Hi David!>Use:>Help > Microsoft Excel Help>You can do a search on the function name or function reference.>Or perhaps get my function lists which automates help for all>functions.>Or refer to:>Peter Noneley:>http://homepage.ntlworld.com/noneley/>Or do a Google search on the function name. Here I'd recommend Ron de>Bruin's Google Search 6 addin:>http://www.rondebruin.nl/Google.htmin Excel, but I can't figure out which one.=== === Subject: : Re: Topological group basics>> Let G be a topological group. If a subgroup H < G is open, then it is>> also>> closed. Why? I must show that the cosets gH are all open, but why is>> this true?>> The topology on G is translation-invariant.>What do you mean by translation invariant? Do you mean that if H has some>topology (open, closed, or whatever), then gH has the same topology? Then>Why is the topology on G translation invariant? This is what I would like>to know.The map g -> hg is continuous. It has an inverse given by...,therefore... (fill in the rest).With my best regards,G. Rodrigues=== === Subject: : JSH: Weird, eh?I'm curious to see if any of you will actually follow through myprevious post to its logical conclusion, and especially if *any* ofyou can resolve the apparent contradiction.It's freaky, eh?Welcome to the world of advanced polynomial factorization, andadvanced mathematics at a level that demands you use *logic* and notyour intuition or simple notions about how numbers behave.I'm here to help guide you along, if you manage to get past yourmental roadblocks, and past posters like Dik Winter.=== === Subject: : Re: Topological group basics>> Let G be a topological group. If a subgroup H < G is open, then it is>> also>> closed. Why? I must show that the cosets gH are all open, but why is>> this true?>> The topology on G is translation-invariant.>What do you mean by translation invariant? Do you mean that if H has some>topology (open, closed, or whatever), then gH has the same topology? Then>Why is the topology on G translation invariant? This is what I would like>to know.Saying that the topology on G is translation-invariant means thata translation x |-> hx is a homeomophism. As for why that's so,what's the _definition_ of topological group again?>> As an extra exercise: show that a closed subgroup of finite index is open.>> -- >> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html>> Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9>> Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: Topological group basics> Let G be a topological group. If a subgroup H < G is open, then itis> also> closed. Why? I must show that the cosets gH are all open, but whyis> this true?> The topology on G is translation-invariant.>What do you mean by translation invariant?> That multiplication by g is a homeomorphism from G to G.>Do you mean that if H has some>topology (open, closed, or whatever), then gH has the same topology?Then>Why is the topology on G translation invariant? This is what I wouldlike>to know.> Continuity of multiplication.> -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9> Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: JSH: Weird, eh?: I'm curious to see if any of you will actually follow through my: previous post to its logical conclusion, and especially if *any* of: you can resolve the apparent contradiction.I'm curious to see if *you'll* ever follow the mathematics of *any* of*your* posts to its logical conclusion.: It's freaky, eh?What's freaky?: Welcome to the world of advanced polynomial factorization, and: advanced mathematics at a level that demands you use *logic* and not: your intuition or simple notions about how numbers behave.Your factorizations are not advanced, they're simple and often wrong. Myprecalculus classes, I honestly believe, know better. : I'm here to help guide you along, if you manage to get past your: mental roadblocks, and past posters like Dik Winter.Okay, what I'd like you to do for me is choose *one*, just one, post byNora Baron, and find a single mistake she's made and elaborate.Thank you,Justin=== === Subject: : Re: Topological group basics> Let G be a topological group. If a subgroup H < G is open, then itis> also> closed. Why? I must show that the cosets gH are all open, but whyis> this true?> The topology on G is translation-invariant.>What do you mean by translation invariant? Do you mean that if H hassome>topology (open, closed, or whatever), then gH has the same topology?Then>Why is the topology on G translation invariant? This is what I wouldlike>to know.> The map g -> hg is continuous. It has an inverse given by...,> therefore... (fill in the rest).hg ---> g (easy to prove), and therefore your map h : G ---> Gis a homeorphism.i.e. h^{-1}(hg) = {x in G such that h(x) = hg} = {x in G such that hx = hg}= g.Note that we do not need here that the map x ---> x^-1 is a continuousoperation. (right?)> With my best regards,> G. Rodrigues=== === Subject: : Re: puzzle: GCDs of Infinite Set of Integer Pairs: Now, speaking as the voice of common sense and low education, I'd: say this is impossible. You can't make *every* natural number equally: likely to come up -- it just doesn't work! Sure you can. Pick a point randomly in [0,1), map it via a bijection tothe positive reals, then take the floor function.No? Justin=== === Subject: : Re: puzzle: GCDs of Infinite Set of Integer Pairs>...just because an event has probability 0 doesn't mean it can't>happen.>> mmmm... I have a problem with this. Can you give an example of a probability 0>> event that actually happened?>Suppose X is a random variable, uniformly distributed on [0,1]. Then the>probability of the event X=x is 0 for each real number x in [0,1].>Yet, some x must be chosen. Every single repetition of this experiment,>without fail, results in the occurrence of an event with probability 0.Define an actual experiment where you randomly chose a real between 0 and 1, andI'll show you where you took a shortcut.>> I would argue that you measured its probability incorrectly. Or it changed after>> you measured it. Or something.>Probabilities are not measured. Probabilities are determined by the>definition of the sample space.And the probability of randomly choosing a real in [0,1] is zero. As in youcan't do it. Another example please.-- Patrick Hamlyn posting from Perth, Western AustraliaWindsurfing capital of the Southern HemisphereModerator: polyforms group (polyforms-subscribe@egroups.com)=== === Subject: : principal ideals in F[x,y]If F is a field, and I is the ideal {fin F[x,y] : f(0,0)=0} inF[x,y], then why is I not principal?I'm trying to show by contradiction using the fact thatdeg(fg)=deg(f)+deg(f). hasn't worked so far, thanks you=== === Subject: : Re: Consider Dik Winter's claims Discussion, linux)> I don't know how many of you know about Dik Winter so I thought I'd> make a quick post to highlight this particular poster, who has made it> his business to put up negative web pages about my *old* work, as if> it were pertinent to continually push my old mistakes.You do realize that, typically, if one puts up a page, it remains upuntil taken down. It doesn't take effort to keep old pages around, ittakes effort to remove them.He hasn't put up pages about your old work. He has refrained fromremoving pages about your old work.-- It has been shown that no man can sit down to write without a very profounddesign. Thus to authors in general trouble is spared. A novelist, for example,need have no care of his moral. It is there -- that is to say, it is somewhere-- and the moral and the critics can take care of themselves. --E.A. Poe=== === Subject: : Re: Topological group basics Adjunct Assistant Professor at the University of Montana.>Let G be a topological group. If a subgroup H < G is open, then it is also>closed. Why? I must show that the cosets gH are all open, but why is this>true?Multiplication is continuous. The coset gH is the inverse image of the set H(which is open) under multiplication by g^{-1}; that is, the functionf:G-->G given by f(x) = g^{-1}x. Note that this is continuous, sincem:GxG->G given by m(x,y) = xy is continuous.Since f is continuous, the inverse image of an open set is open, hencegH is open for each g in G.-- === === Subject: : Re: PLEASE HELP!!!> I DON T KNOW HOW TO SOLVE THIS!First, typing in capital letters is generallyconsidered to be SHOUTING and is also generallyregarded as impolite (if only because it's harder to read).Second, while I don't know how things work at your school,getting help on an exam is regarded as cheating whereI work. It would, of course, be a different matterif your instructor had given permission to use othersources, but there's no evidence here that such permissionhas been granted.> 1.DETERMINE DOMAIN OF THE FUNCTION> f(X)=X3/X2+2X+1Third, this expression is ambiguous. I might interpretit as f(x) = (x^3)/(x^2 + 2*x +1)or as f(x) = ((x^3)/(x^2)) + 2*x + 1or as many other things. Even though you're stillnot going to get any help, you should be precise.> FIND EXTREME FUNCTIONS f AND DRAW ITS GRAPHFourth, Usenet newsgroups frown on anything buttext messages, for good reason. That means thata polite response would be some text art like this | / | / _______|/________ / ---| |which wouldn't be much help (ignoring for themoment that it would still be academicallydishonest).Fifth, these are quite simple problems. I sympathizewith your plight, since it does seem you're ina mess of trouble.Rick=== === Subject: : A matrix questionY: a Nx1 vector X: a NxN matrix H: a Nx1 vector C: a NxN matrix * stands for Hermitian f(X)=(Y-XH)^*inv(C)(Y-XH) ask what df/dx^* === === Subject: : Re: principal ideals in F[x,y]>If F is a field, and I is the ideal {fin F[x,y] : f(0,0)=0} in>F[x,y], then why is I not principal?>I'm trying to show by contradiction using the fact that>deg(fg)=deg(f)+deg(f). hasn't worked so far, thanks youBoth x and y are in I. So if I is principal, say I=(h(x,y)), then there must be polynomialsf(x,y) and g(x,y) such thath(x,y)*f(x,y) = x and h(x,y)*g(x,y) = y.of any monomial in h(x,y) is 1; that is, h(x,y) must be of the formh(x,y)=a+bx+cy.The same is true for both f and g. Say f(x,y) = r+sx+ty.Then ct=0, (bt+cs)=0, ra=0, (at+rc)=0, and as+br=1.If g(x,y)=u+vx+wy, then ua = 0, vb=0, cw=0, (bw+cv)=0, (av+bu)=0, and (aw+uc)=1.So either a=0, or both r and u are 0. If a is zero, then either b=0 or u=0 (from av+bu)=0; but you cannothave both a and b zero, since as+br=1. So a=0 implies u=0. But thenaw+uc=0, and it is supposed to be 1. So a is not zero.Therefore, r=u=0. Since a cannot be zero and at+rc=0, then t=0. Whichmeans that cs=0; If s=0, then as+br=0, which is impossible. Soc=0. Since bw+cv=0, then b=0 or w=0. But w cannot be zero (sinceaw+uc=1), so b=0. Now we have a=t=u=c=b=0. But then as+br=1 isimpossible. Thus, no such h exists. (Other, simpler arguments, are of coursepossible). Therefore, I is not principal.-- === === Subject: : Re: PLEASE HELP!!!>I DON T KNOW HOW TO SOLVE THIS!>1.DETERMINE DOMAIN OF THE FUNCTION>f(X)=X3/X2+2X+1> FIND EXTREME FUNCTIONS f AND DRAW ITS GRAPH>2.BY NEWTONS METHOD SOLVE EQUATION:>X3-X-2=0>DRAW A PICTURE!--Lynn=== === Subject: : Re: x^2 + y^4 = z^4>Is there some coprime a,b,c with> a^2 + b^2 = 2c^2>and a /= b?There are infinitely many positive integer solutions to, for example,1^2 + b^2 = 2*c^2 (see Pell's equation).Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Infinite polynomial product by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1HHYHv19759;I recently came across the following infinite product:(1-x)(1-x^2)(1-x^3)(1-x^4)..., where abs(x) < 1. I don't own any good resources for known infinite products and haven't found anything close to this in web searches. Is there a simpler form for this expression involving elementary functions? I've tried using logarithms and series expansions, but haven't made significant progress. Has anyone seen this one before or own a good reference that has it?Steve OakleyRemove iron for email=== === Subject: : Re: Topological group basics>> Let G be a topological group. If a subgroup H < G is open, then it>is> also> closed. Why? I must show that the cosets gH are all open, but why>is> this true?> The topology on G is translation-invariant.>>>What do you mean by translation invariant? Do you mean that if H has>some>>topology (open, closed, or whatever), then gH has the same topology?>Then>>Why is the topology on G translation invariant? This is what I would>like>>to know.>>> The map g -> hg is continuous. It has an inverse given by...,>> therefore... (fill in the rest).>hg ---> g (easy to prove), and therefore your map h : G ---> G>is a homeorphism.>i.e. h^{-1}(hg) = {x in G such that h(x) = hg} = {x in G such that hx = hg}>= g.You are mixing unrelated things. So, let me go step by step: Definethe map L_h:G -> G by assigning to g, hg. Now, prove L_h iscontinuous.Since L_h is continuous L_{h^{-1}} is also continuous. Now prove thatL_h and L_{h^{-1}} are inverses of one another. (note: This meansexactly thatL_h . L_{h^{-1}} = identity map on GL_{h^{-1}} L_h = identity map on GSince you are starting to study this subject, it's always good towrite down *everything* until it becomes obvious and trivial)Hence, L_h is a homeomorphism. The topological results you quotedfollow from this (Another exercise...).>Note that we do not need here that the map x ---> x^-1 is a continuous>operation. (right?)Right, but it does follow from the axioms of a topological group thatx --> x^-1 is continuous (another good exercise).With my best regards,G. Rodrigues=== === Subject: : JSH: Reply, reply, replyCome on, reply to my posts!!! Come why don't you say something else,huh?Try to point out some mistake Nora Baron you stupid !Hey, you lapdog David Ull, yeah, I called you a LAPDOG Ull!!!Aren't you mad now? Hey, you know, you HAVE to reply right, or I'llcall your school and try to get you fired, right?Ull the university professor cursing in the muck, with no shame.REPLY TO MY POSTS YOU CUR!Hey Dik Winter, you ing , reply again! Why don't you throwmore of my old arguments on a webpage, you stupid !!!Yeah, you'll all be back want you? Along with C. Bond and all therest.You are curs, my lapdogs, my little traveling circus of obsessiverepliers.I spit on you.But you'll be back, now won't you?=== === Subject: : errors in an argumentI recently attended some talks by a creationist talking abouthow bad evolution is. This person bragged that what set himaside from your run of the mill biologist or geologist was hisability to do math. He claims many degrees, including onein mathematics. He made a big deal of the fact that whenhe took differential calculus, he lost only 3 points all term,and lost only 5 points in integral calculus.Anyway, at one point in his presentation, he attemptedto prove that life could not arise by chance. His proofwhen something like this:start:Let's just look at the probability of a protien made up of100 amino acids forming by chance. All the amino acidshave to line up just right, and the probability of thishappening in any given try is 1/100! or 10^(-158).Now suppose that we give every concevable chancehere, and say that there can be 10^80 tries at anygiven instant (this was supposedly the number ofelectrons that would fit in the universe). Let us furthersuppose that these 10^80 tries can happen a iontimes a second. The universe is at most 30 ionyears old, so this would give 10^27 attempts fora total of 10^107. Thus, the odds agains this protienforming are 10^158 - 10^107 = 10^51 to 1 against.People in science will tell you that if the probabilityof something happening is less than 1/10^51, thenin effect, it can't happen.end.seminar on creation science took place. I complainedin general of the poor quality of the presentation. I alsomade the claim that the presentor might have missed onlythree points in differential calculus, but he made at leastthree math errors in that single proof: one being slapyour head obvious, the other slightly subtler. Just tobe careful, I thought I would post this here to see ifpeople generally agree that the above has at leastthree math errors. After seeing some responses,I will post what I thought the errors were.=== === Subject: : Re: Number property>I am searching for the most general set of numbers {a_i} (i=1...n) >, 0<=a_i<=2pi , such that :Better to say < 2 pi, because 0 and 2 pi are equivalent in respectto addition mod (2 pi). >a_i + a_j = a_t mod(2pi) for every i,j=1...n>and such that the set of all {a_t} obtained in this way is equal to>the set {a_i} we started with.In other words, you want a finite sub-semigroup of T = R/(2 pi Z)under addition. Any finite sub-semigroup of a group is a subgroup (in fact anyfinite semigroup such that t->a*t and t->t*a are one-to-one for all a in the semigroup is a group).So you're looking for the finite subgroups of T. And these are allof the form {2pi k/n: k=0...n-1} for positive integers n.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === === Subject: : Re: No Set Contains Every Computable Natural (was Church-Turing compared to Zuse-Fredkin thesis)>Dave Seaman:>A set is recursive if there is an effective procedure (a Turing machine,>perhaps) that can decide membership in the set. That is, the machine>halts and returns 1 if the input is in the set, and the TM halts and>returns 0 if the input is not in the set.> This is the definition of a decidable language at wikipedia:> http://en.wikipedia.org/wiki/Decidable_language> A decidable or recursive language is a formal language that is a recursive> set, i.e., for which there exists an algorithm to solve the following> decision problem: Given string w, does w belong to the language? The> algorithm is not allowed to run into an infinite loop and has to produce a> YES/NO answer for any input string after a finite amount of time. To> formalize the rather vague term algorithm, one usually employs Turing> machines, but several other equivalent approaches are possible.> This definition should say finite number of steps instead of finite> amount of time> I will come back to this.A TM is an abstract logical computing machine (Turing: LCM)which does not consume physical resources. Those Super-Turingmachines you mention later have an abstract property that TMsdo not. They represent the real numbers (...Pi) with infinite precision.This is not physically realizable.> This is the definition of a recursively enumerable language:> http://en.wikipedia.org/wiki/Recursively_enumerable_language> Definition 1. Given string w as input, the algorithm halts and outputs YES> if and only if w belongs to the language L. If w does not belong to the> language L, the algorithm either runs forever, or halts and outputs NO.> There is a second definition that you might want to read.> Let me define a language, L, that consists of all unary representations of> natural numbers.> 1 = one, 11 = two, 111 = three, etc.> This language is recursively enumerable. Given a string, x, there exists a> TM that> will halt after a finite number of steps if x is a member of L.> This language is NOT decidable. Consider what happens if x is an infinite> string> of 1's. The TM will not halt after a finite number of steps.> The set of all natural numbers is not a recursive set.http://www.cs.auckland.ac.nz/CDMTCS/chaitin/ omega.htmlAnyway, it was Post who put his finger on the essential idea, which is thenotion of a set of objects that can be generated one by one by a machine, insome order, any order. In other words, there is an algorithm, a computerprogram, for doing this. And that's the essential content of the notion of aFAS, that's the toy model that I'll use to study the limits of the formalaxiomatic method. I don't care about the details, all that matters to me isthat there's an algorithm for generating all the theorems, that's the keything.And since this is such an important notion, it would be nice to give it aname! It used to be called an r.e. or recursively enumerable set. Theexperts seem to have switched to calling it a computably enumerable orc.e. set. I'm tempted to call it something that Post said in one of hispapers, which is that it's a generated set, one that can be generated byan algorithm, item by item, one by one, in some order or other, slowly butsurely. But I'll resist the temptation!FAS = c.e. set of mathematical assertionsSo what's the bottom line? Well, it's this: Hilbert's goal of one FAS forall of math was an impossible goal, because you can't put all ofmathematical truth into just one FAS. Math can't be static, it's got to bedynamic, it's got to evolve. You've got to keep extending your FAS, addingnew principles, new ideas, new axioms, just as if you were a physicist,without proving them, because they work! Well, not exactly the way thingsare done in physics, but more in that spirit. And this means that the ideaof absolute certainty in mathematics becomes untenable. Math and physics maybe different, but they're not that different, not as different as peoplemight think. Neither of them gives you absolute certainty!SH: I brought this up because of:According to Turing's definition, a number is computable ifits decimal expansion can be written down by a circle freemachine. This is a machine that performs an unendingTuring defines the machine to be circular. Circular machinesreach a configuration from which there is no possible move,or go on moving, but do not print any more symbols.SH: There is a significant difference between a circle-freemachine that works at printing out the infinitely long digitsof Pi and a circular machine that may move but not printany more output --> no symbols so it is no longer generating.A generated set, one that can be generated by an algorithm,item by item, one by one, in some order or other, slowly but surely.Computing Pi is not an example of the 'halting problem' justbecause it does not halt. The circular machine is an exampleof the halting problem. I keep getting the impression that youthink these are analgous situations just because they dont halt.COMPUTABILITY AND RECURSION 285 Robert SoareDefinition 1.2. (i) A function is Turing computable if it is definable bya Turing machine, as defined by Turing 1936. (See [Kleene, 1952] or[Soare, 1987].)(ii) A set A is computably enumerable (c.e.) if A is (/) or is the rangeof a Turing computable function.(iii) A function f is recursive if it is general recursive, as defined byG ~odel 1934. (See also Kleene's variant 1936, 1943, and [1952, p. 274].)(iv) A set A is recursively enumerable (r.e.) if A is (/) or is the range ofageneral recursive function.Recursively Enumerable Reals and Chaitin Omega NumbersCristian S. Calude, Peter H. Hertling, Bakhadyr Khoussainov, Y. WangA real number alpha is called recursively enumerable if it is the limit ofarecursive, increasing, converging sequence of rationals.Roughly speaking, the computational depth of an object is the amount of timerequired for an algorithm to derive the object from its shortestdescription. Bennett showed that the characteristic sequence XK of thehalting problem is strongly deep, while no random sequence is strongly deep.Investigating this matter further, Juedes, Lathrop, and Lutz have consideredthe notion of usefulness'' of infinite sequences. A sequencex is useful if all recursive sequences can be computed with oracle accessto x within a fixed recursive time bound. For example XK is useful, whileno recursive or random sequence is useful.> Consider what happens when we talk about the set of all TM's.> Assume we can encode the state transition table of a TM using> a natural number. This set can not be recursive for the sameWikipedia:A set S of natural numbers or tuples of natural numbers, or of literalstrings, is recursively enumerable or computably enumerable orsemi-decidable if it satisfies either (and therefore both) of the followingequivalent conditions. a.. There is an algorithm that, when given a natural number n (or tuple ofnatural numbers, or word, as the case may be) eventually halts if n is amember of S and otherwise runs forever. a.. There is an algorithm that generates the members of S. That meansthat its output is simply a list of the members of S: s1, s2, s3, ... Ifnecessary it runs forever.Natural numbers have two main purposes: they can be used for counting(there are 3 apples on the table), or they can be used for ordering(this is the 3rd largest city in the state).> reason that the natural numbers are not recursive. Given a> string, i, we can not be sure that TM(i) is actually a TM> because we can not be sure that i is actually a natural number.Some problems are recursively enumerable but not recursive. Oneexample is the halting problem, which is the problem: Given a program and input parameters, will that program eventually halt when run with those parameters?> No set can contain every computable natural number.> Proof:> Assume a TM produces a tape with all of the natural numbers> encoded in unary where each natural is followed by a 0 (or a blank).> Turing gives the state transition table of such a TM> 010110111011110...> I can define a three state TM that will find a natural number> that is not on this tape:> 1) Read right until there is a 0.> 2) Read right until a second 0 is found.> 3) Backup and write a 1 on the previous 0.> Repeat steps (1) through (3).> This TM will always produce a tape that has exactly one 0.> This 0 will be at a finite position on the tape and the string of 1's> that preceds this 0 represents a natural number that was not on> the original input tape.SH: The successor function builds the whole set of natural numbersWikpedia:a.. There is a natural number 0.a.. Every natural number a has a successor, denoted by a + 1.a.. There is no natural number whose successor is 0.Distinct natural numbers have distinct successors: if a ? b, then a + 1 ? b+1If a property is possessed by 0 and also by the successor of every naturalnumber which possesses it, then it is possessed by all natural numbers.(This postulate ensures that the proof technique of mathematical inductionis valid.)> The Halting Problem is ill posed.> There is no set that contains every TM just as there is no set> that contains every computable natural number.The set of natural numbers in countably infinite and has no greatestor last digit. The set can be represented as {0,1,2,3,...}which doesnot list the entire set or assume it has a largest digit.just as there is no set that contains every computable natural numberIf you want to establish a set that has one-to-one correspondencebetween TMs which terminate after 1 step, 2 steps, 3 steps and so on,and the natural numbers, it seems reasonable to me. What systematicmethod do you have that eliminates the computable natural numbersand excludes them from the set of real numbers? I suspect you meanthe computable numbers cannot be sorted out of the reals. I amnot sure there is an algorithm to list computable transcendentals.I am not much interested in those hyper-machines so I snipped it.Stephen=== === Subject: : Re: x^2 + y^4 = z^4===> Subject: x^2 + y^4 = z^4>Thus problem boils down to showing no coprime a,b,c with> a^4 + b^4 = 2c^2> when a /= b.> Is there some coprime a,b,c with> a^2 + b^2 = 2c^2> and a /= b?********Well, (2m^2-1)^2 + (2m^2-4m+1)^2 = 2(2m^2-2m+1)^2._______________________________________________ __________Eric J. Wingler (wingler@math.ysu.edu)Dept. of Mathematics and StatisticsYoungstown State UniversityOne University PlazaYoungstown, OH 44555-0001330-941-1817=== === Subject: : Re: puzzle: GCDs of Infinite Set of Integer Pairs> : Now, speaking as the voice of common sense and low education, I'd> : say this is impossible. You can't make *every* natural number equally> : likely to come up -- it just doesn't work! > Sure you can. Pick a point randomly in [0,1), map it via a bijection to> the positive reals, then take the floor function.> No? No. There is no bijection that leads to equal probabilities.For example, if your bijection takes [0,1) -> [0,infty) such that x -x/(1-x), you'll end up with a probability of 1/2 of getting a 0, 1/6 ofgetting a 1, 1/12 of getting a 2, 1/20 of getting a 3, etc. -- Daniel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Re: the anticlassicalist }{ i: linguistic negation> -=-=-=-=-= linguistic negation => The logic of natural language has been studied by many different schools of> thought throughout history.That they have one and all avoided coming to any unique conclusionemphasizes the bankruptcy of the Liberal Arts. It's a hen party. Anybody can gossip and all are appreciated. It's bull. How wouldyou like to be doing 80 mph in a car that was built by mob ruleinstead of cold, hard, dry engineering?Oh, but that example is different! It certainly is. Liberal Arts arebull and engineering is real world.Everything in the Liberal Arts and Social Sciences is both right andwrong. No conclusion at all can be drawn except the need for morefunding. Say what you want - it makes no difference at all (exceptfor grant funding - then you must agree with the mob or be destroyed).Klingon was specifically created to be the worst language possible byfolks who knew linguistics. It was enthusiastically embraced by themob and it is as good as Korean or Chinese for transferring content.--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics)=== === Subject: : JSH: Not a solutionObsessively replying to my posts to insult me or just aggravate me isnot a solution.Sure, it might make you feel better or like you're accomplishingsomething if you can piss me off, but that's just a bad reflection onyou.Notice, here I am again: posting.Those of you who make it your business to obsessively reply to myposts are not doing other people a favor. I post for many reasons,including just to toss out math ideas that I'm considering, andthere's no reason to think that I'll stop.Some of you seem to be stupid in a way that defies commonsense as nowI'm one of the most read posters on sci.math and it seems to me thatyou MUST on some level understand that the obessive posters havesomething to do with it.Yet you continue to support them, and they continue to go at it. Now I'm in a mood so I feel like berating some of you, not because Ithink you'll go away, but because I feel confident you will not.It's kind of, sickly, *relaxing* to curse at Nora Baron or DavidUll.I'm starting to worry that I'm finding it appealing to have my owndumpground, where when I get frustrated I can just come out and curseat them, knowing they'll just be back, for more.It's not a solution though, just another display of rather absurd andsick human behavior played out to a worldwide audience.But hey, we're mostly Americans here! Surprise. Surprise.=== === Subject: : Re: Infinite polynomial product>I recently came across the following infinite product:>(1-x)(1-x^2)(1-x^3)(1-x^4)..., >where abs(x) < 1. >I don't own any good resources for known infinite products and haven't>found anything close to this in web searches. Is there a simpler form>for this expression involving elementary functions? Not elementary, but it's related to the Dedekind eta function andEuler's Pentagonal Number Theorem.See e.g..and the many references given at the last URL.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : VonNeumann Gametheory of the OS of StockMarket; Cingular name change to AT&TPortfolio of PAF as of 17FEB04:BCE 8,500 22.13 $188,105.00 BMY 50 28.95 $1,447.50 SBC 13,260 24.54 $325,400.40 realestate land 3APR03 of 3 lots $19,000.science-art of pictures,porcelain etc starting JAN03 for $14,143.realestate land 30JUL03 another lot $11,500.The Telecom sector is ablaze in action. SBC's Cingular in partnerswith BellSouth is buying AT&T wireless. Usually when these sort ofthings happen the acquirer stock falls.So today I spyed a Crossover with my BCE and SBC shares.Back on 8JAN04 I sold shares of SBC in order to buy BCE witha price differential of about $5. per share spread between SBC and BCEToday I see the spread between SBC and BCE of 24.55 and 22.13respectivelyof a spread approximately $2.40.That means I have achieved a Crossover and will get free shares of SBCplus somecash extras (in this case there is no tax liability because I amselling the BCE shares for a paper money loss which is okay by mebecause I gain free shares of SBC plus cash freebies of about 9hundred to one thousand dollars.)So, today I sold 1,220 shares of BCE at 22.13 and with the proceedsbought 1060 shares of SBC at 24.55. Back in 8JAN04 I had sold those1060 SBC at 27.61 and had bought those 1220 BCE for 22.67.How many free shares of SBC did I pick up today? Well, the spreaddifference between 8JAN trade and 17FEB trade is that of approx. $5. -$2.40 which is $2.60Thus, taking $2.60 X 1220 is equal to $3,172. and dividing that bytodays price of SBC at 24.55 is approx 129. So I reckon that today Ipicked up 129 free shares of SBC. If I had sat pat on 8JAN and donenothing, well, I would not be 129 shares of SBC er today.Now, I wonder if SBC will change the name of Cingular to somethingbetter. SBC is a great company and has great leadership. But they needto fix the name of their wireless unit for the name Cingular is justnot high class enough for the great company that SBC is. The name AT&Tis better than the name Cingular. And perhaps if SBC cannot come upwith a better name then use AT&T because I suspect that the namebrandof AT&T will soon be buried or lost because long-distance isevaporating as a telephone entity and thus AT&T long-distance is goingextinct.Also, I feel that Vodafone in cohort with Verizon just staged a bidand were never serious. I believe they did this in order to extract anextra pound of flesh out of SBC and an extra gallon of blood out ofSBC. Verizon and Vodafone are competitors of SBC and they hated to seeSBC buy AT&T wireless at a *fair price* of $30 ion and so theystaged a phony bid in order to make SBC shell out an extra $11ion.So, I hope that some future date that BellSouth and SBC can give thescrews to Vodafone and Verizon for their prank tricksterism.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies(www.iw.net/~a_plutonium) website of the science of AP under revision.=== === Subject: : CAN ANYONE HELP ME????? boundary=----=_NextPart_000_0039_01C3F589.233A8BF0------------ --------------------------------------------------------- charset=iso-8859-2CAN ANYONE HELP ME?????1.DETERMINE DOMAIN OF THE FUNCTIONf(X)=X3/X2+2X+1 FIND EXTREME FUNCTIONS f AND DRAW ITS GRAPH2.BY NEWTONS METHOD SOLVE EQUATION:X3-X-2=0DRAW A PICTURE!=== === Subject: : Re: puzzle: GCDs of Infinite Set of Integer Pairs>>...just because an event has probability 0 doesn't mean it can't>>happen.> mmmm... I have a problem with this. Can you give an example of a probability 0> event that actually happened?>>Suppose X is a random variable, uniformly distributed on [0,1]. Then the>>probability of the event X=x is 0 for each real number x in [0,1].>>Yet, some x must be chosen. Every single repetition of this experiment,>>without fail, results in the occurrence of an event with probability 0.> Define an actual experiment where you randomly chose a real between 0 and 1, and> I'll show you where you took a shortcut.This is sci.math. If you want to talk about actual experiments, take itto sci.physics.> I would argue that you measured its probability incorrectly. Or it changed after> you measured it. Or something.>>Probabilities are not measured. Probabilities are determined by the>>definition of the sample space.> And the probability of randomly choosing a real in [0,1] is zero. As in you> can't do it. Another example please.I didn't say anything about what you could do. I was discussingmathematics. Can you write down all the natural numbers? Does the setof natural numbers exist?The example I gave is correct. Find a textbook on probability theory andlook up probability space. It's a special case of a measure space.The fact that a nonempty event may have probability zero is a directconsequence of the fact that a nonempty set may have measure zero.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. How the &*&%$ does one enter a real math formula (simple example: the> sum of all N sub x as x goes from x1 to x2) ???!>I can't imagine doing real math in Excel, but doesn't>sum(n23:n97) produce the value n23 + n24 + ... + n97?>If it doesn't, something like it does, and it should be possible>to find out within Excel because it has some kind of Help thing>built in & you can just look for sum and it should tell you>how to use it.> Gosh, that sounds useful. Say I were running Excel - how would> I find this Help thing?It has some kind of Meta Help thing built in & you can just look forHelp and it should tell you how to use it.-- G.C.=== === Subject: : area/vector question - help in equation needed boundary=------------070403050607050008040205----------------- ----------------------------------------------------hi.I'm trying to develop a simple game in flashand I've got some problems with mathematical equations to calculate the area of the shape I produce while drawing lines on a flat surface.first of all, you might have seen a game like this ages ago,I don't know how it was named, but it was very popular.(anyone?)basically, the game looks like the schema in the attachment.I'm looking for a way to calculate the area of the red part of the rectangle.any ideas would be appreciated.maybe someone could, at least, tell me what to look for on the internet?thanx. mat.=== === Subject: : Re: countable sets> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove> that union of Ek as k = 1 to oo is countable. I do not need help proving> this. I need someone to explain exactly what is being asked. In Let E1, E2,> E3,... be a sequence of pairwise disjoint countable sets is that saying> that the numbers of Ei's are countable or is it saying that each Ei has a> countable number of elements??Both.=== === Subject: : Re: How many different resistances with n resistors?X-ID: VIzY8yZfYerRqQQUL8v5CPaO8J8+MxfDK1GeozA-ZXXecnIyJwn1Zntchow@ lsa.umich.edu schrieb:> The Puzzle Corner in MIT's magazine Technology Review raised the> question of the number of distinct resistances achievable by connecting> ten unit resistors.> According to Sloane's online encyclopedia, the answer is 2213:> http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anum=A048211> However, this calculation appears to assume that the resistance of the> network can always be calculated by successive applications of the rules> for resistors in series and parallel. I have always had the vague> impression that some configurations, e.g., the Wheatstone bridge, were> solvable only by using Kirchhoff's laws and not by the series/parallel> rules. Have I just been miseducated? Or maybe the issue doesn't come> up with ten unit resistors?Tony Bartoletti (seq. A048211) considers only cases that can be reduced byapplying the formula for parallel/series. If you have 6 resistors there is 1configuration where this is not possible. I suppose we get an additionalresistance value.sincerely> --> Tim Chow tchow-at-alum-dot-mit-dot-edu> The range of our projectiles---even ... the artillery---however great, will> never exceed four of those miles of which as many thousand separate us from> the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences=== === Subject: : Re: . The hardest of all hard facts .>Hi Eleaticus, Re: How you think, Michelson-Morley and Kennedy-Thorndike do indeed fit> Galilean ( c + v ) physics ,All throughout the annals of history ...> No premise has been better tested than this premise:> The speed of light is the same for all observers.That makes it: The hardest of all hard facts.>No experiment ever showed that the speed of light is>the same for all observers. Indeed any determination>of the one way speed of light can be used to demonstrate>the speed of light is NOT the same for all observers....> A light----> B <-you> < ----------- L --------------> v m/s>Use syncronised clocks at A amd B to time how long it takes>light to travel a distance of L meters across the laboratory..>Speed of light relative to the laboratory = L/ (tB - tA) = c> where 'tA' is the time at which the light left A> and 'tB' is the at which the light arrived at B>Now repeat the experiment while running towards B at v m/s>Note that 'in your frame of reference' the point B is moving ,>so that the light must travel an extra distance = v * (tB - tA)>which is the distance B has moved as the light travels from>A to B.> But this extra distance is compensated by the difference of your> evolution through time while runnning. You go slower in time so that Rubbish! But if it were true then the time for light to travel betweenA and B would be for you LESS than (tB - tA ), and the velocity of lightfor you would then be even GREATER than c + v, since velocity = distance / time eh!> light may travel at the same speed for you as irtt does for the> laboratory.Well now that is how relativists actually work. They ASSUMEthe velocity of light is constant for all observers, and then fiddlewith the actually readings of the instruments until they have distanceand time values to make their ridiculous assumption appear true.Having flagrantly cheated, by ignoring the actual readings of theinstruments, they have the impertinence to make claims such as: > All throughout the annals of history ... > No premise has been better tested than this premise: > The speed of light is the same for all observers.Pull the other one Mr. Relf eh!>Therefore:>Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA)> = c + vkeith stein=== === Subject: : Re: the anticlassicalist }{ i: linguistic negation~> ...> Whatever the true explanation, I think it> has something to do with how we construct> sentences unconsciously.Meant 'subconsciously'.(I write posts 'unconsciously'.)=== === Subject: : Re: tensors for tots> How far can one get in understanding the structure of tensors at a> point on a manifold in terms of our old school friends column vectors,> row vectors and matrices? Even in general relativity all we have at> each point on the manifold is a real vector space with four> components, so it seems we can go all the way -- but I'm having> trouble with some of the mappings.> Shall I show what work you've done so far, or would somebody care to> take the ball from there... ? You can get very far. Since you can't even understand the word structure, never tensors, or even such nonsense as curved space-time without vectors. First you build a 3 dimensional school. Then you build an 4-dimensional box. Then you build an n-dimensional table. Then you rotate the table n+1 times, so that 1 is now at the origin. Then you build a radioactive GPS robot orbiting the Earth. Then you tell mathematicians to go back to school, and teach chemists something about logic, since not only doesn't cold-fusion work at negative temperatures, it doesn't even work at imaginary temperatures.=== === Subject: : Re: PDE Max-Min Principle (Help?)The original post has some data missing and I am hoping that the hint belowhelps.Assume PDE u_t = u_xx for t >0, 0 < x < LNote that the given IC impliesIC => u(x,0) - v(x,0) = w(x) for 0 =< x =< LNow the given BCs also implyBC1 => u(0,t) - v(0,t) = k1, k1 > 0, t >0BC2 => u(L,t) - v(L,t) = k2, k2 > 0, t >0Hint:Begin withu(x,t) - v(x,t) = w(x) => v(x,t) = u(x,t) - w(x)Convert given original IBV problem in u into a new IBV problem in v. Anintermediate simple BVP is created in the process.If I didn't mess up the simple BVP should look likew(x)=0w(0)=k1w(L)=k2solve this for w(x). Tie loose ends and make the argument leading tou(x,t) - v(x,t) > w(x) for t >= 0, 0 =< x =< L------------------Mohan Pawar------------------> Tim Brauch schrieb: Prove that if u(x,t) <= v(x,t) for x=0, for x=l (lower-case L), for> t=0, then u<=v for all (x,t) in 0<=x<=l, 0<=t<=oo.Ridiculous as it stands. Please include all hypotheses. (Also, what>does t = oo mean here?)>>Sorry, at the beginning of the problem set, we were told that u and v>are are solutions to the diffusion (heat) equation u_t = k*u_xx.>There is a hint that says this is proving the Comparison Principle>using the Maximum-Minumum Principle.> can you apply the Maximum-Minumum Principle. to w=u-v ? which PDE andboundary> conditions does w fulfill?> hth>Also, that last inequality should read 0<=tinfinity.>Like I said, I'm not looking for the answer, just a way to start the>beast. Most other problems in this set were proving properties of the>diffusion equation using change of variables and the chain rule. That>does not seem to work here.>-->Timothy M. Brauch>Graduate Student>Department of Mathematics>Wake Forest University=== === Subject: : Re: puzzle: GCDs of Infinite Set of Integer Pairs>> Define an actual experiment where you randomly chose a real between 0 and 1, and>> I'll show you where you took a shortcut.>This is sci.math. If you want to talk about actual experiments, take it>to sci.physics.Playing the physics card isn't helpful here. I'm claiming that it's*mathematically* impossible to randomly choose a real from [0,1]. Not physicallyimpossible. I'm asking you to convince me otherwise, use a thought experiment ifyou want, or not. And the fact that you can't specify which real you chose is not the only reasonyou can't choose one. If you find some way of specifying which one you chose,I'll find a way to show you where you failed to do what you set out to do. >> And the probability of randomly choosing a real in [0,1] is zero. As in you>> can't do it. Another example please.>I didn't say anything about what you could do. I was discussing>mathematics. Can you write down all the natural numbers? Does the set>of natural numbers exist?Once again, you can't hide behind the 'reality card'. You can say I'll pick aninteger randomly from a uniform distribution across the whole set but that'sgarbage mathematically speaking, and I'm claiming that choosing a real from[0,1] is roughly the same flavour of garbage.>The example I gave is correct. It's so 'cos I said it's so doesn't wash with me.>Find a textbook on probability theory and>look up probability space. It's a special case of a measure space.>The fact that a nonempty event may have probability zero is a direct>consequence of the fact that a nonempty set may have measure zero.I don't believe it. I'm prepared to be convinced, but you haven't offered anyargument even vaguely convincing yet.-- Patrick Hamlyn posting from Perth, Western AustraliaWindsurfing capital of the Southern HemisphereModerator: polyforms group (polyforms-subscribe@egroups.com)=== === Subject: : Re: Entering *Real* Math Formulae in Excel> How the &*&%$ does one enter a real math formula (simple example: the> sum of all N sub x as x goes from x1 to x2) ???!>I can't imagine doing real math in Excel, but doesn't>sum(n23:n97) produce the value n23 + n24 + ... + n97?>If it doesn't, something like it does, and it should be possible>to find out within Excel because it has some kind of Help thing>built in & you can just look for sum and it should tell you>how to use it.> Gosh, that sounds useful. Say I were running Excel - how would> I find this Help thing?Why, you read the manual, of course. Duh.Jon Miller=== === Subject: : Re: Simple numbers>Hallo,>I know that every simple number (beside 2 and 3) has its one formula :S.N =>6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1gives>the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple ,5,7|35)> By simple number I think you must mean prime.>so my question is:>If n={1,2,3,...} what is the formula which gives the simple numbers(beside>2 and 3) for any n?> Evidently you don't consider p(n) is the n'th prime to be a formula.Any> particular reason?Computational complexity?Jon Miller=== === Subject: : Re: errors in an argument>Let's just look at the probability of a protien made up of>100 amino acids forming by chance. All the amino acids>have to line up just right, and the probability of this>happening in any given try is 1/100! or 10^(-158).There are only 20 different amino acids.The more subtle error is that nobody says that life requiredexactly this protein to be produced by a random permutation. There are probably a huge number of different proteins thatcould do more or less whatever this one does. And its productionwould not be a matter of random permutations, but rather ofthe gradual accumulation of small improvements. You mightlook up genetic algorithms: this is not just a theoreticalconstruct but a method which is actually used to obtain a nearly-optimal solution to a difficult problem by simulating the process of evolution.That said, I think it's fair to add: my impression, as anoutsider in this field, is that while evolution itself (for organisms that already have the usual genetic machinery) has prettygood mathematical models, we are still far from understandingat a quantitative level how that genetic machinery might have arisen in the first place.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Four Color Theorem> If graph G is a minimal counter example [mce] to the FCT; then,> G is maximal planar,> G is internally 6-connected and> every vertex in G has minimum degree = 5.> Is this essentially a correct and complete description of an mce?> regards, J.I should have asked if these were necessary and sufficient conditionsfor G to be an mce.B.=== === Subject: : Re: Mathcad Upgrade QuestionYes spybot caught on to it. Besides I don't like folks putting stuff on mycomputer without telling me first.Neither Mathematica nor Matlab has any problems with the way I use mycomputer. Does Mathcad 11 usec-dilla to keep track of what your doing? Or did they leave it out?> Mathcad 11 is worth the upgrade but I have found that it doesn't work> reliably on WIN98. I upgraded to WINNT and it works fine. I also hate> c-dilla, but it should do anything bad to your system unless you have some> software that scans for trojans and stealth programs. I would ratherhave> a USB dongle.Don't use mathcad. I installed version 2001i this weekend. After>installition I noticed that c-dilla had also been introduced to my> registry.>Google c-dilla to find out what this program does. Anyhow I have spent 2>days rebuilding my hard drive because of the damage that Mathcad did.in>I am thinking of upgrading from an old version of Mathcad (Version 7)>to the most recent. But past experience with upgrades has taught me>to be a bit wary, so 2 questions:1. Did Mathsoft do anything to screw up the product, in terms of>usability or features, since I last purchased it (back in 1997)?2. Does the latest product come with any kind of registration>feature, like Mathematica has, that locks your software into one>computer, and possibly locks you out of the software if you upgrade>your hardware? (I respect the rights of companies to make money off>their software, but I'm a firm believer that the only fair deal is>one-user/one-app. If I have three computers -- and I do -- I should>not have any worries or complications with loading the same software>package on all three of them.)>Steve O.Standard Antiflame Disclaimer: Please don't flame me. I may actually>*be* an idiot, but even idiots have feelings.=== === Subject: : Re: Got a speeding ticket and need to fight backX-NewsReader: GRn 3.2n February 9, 1999>I guess your name is more a sign of high hopes of the parents - which >was then deeply disappointed...> I didnt disppoint my mom and dad. But definitely, i know, it hurts you> so much to know that there is atleast someone, who is not you, who> didnt.You must be a severe disappointment to whomever taughtyou social skills...=== === Subject: : Re: JSH: Weird, eh?> I'm curious to see if any of you will actually follow through my> previous post to its logical conclusion, and especially if *any* of> you can resolve the apparent contradiction.> It's freaky, eh?Watching JSH trying to teach his grandmother to suck eggs is the most fun one can have outside of a bed.> Welcome to the world of advanced polynomial factorization, and> advanced mathematics at a level that demands you use *logic* and not> your intuition or simple notions about how numbers behave.Watching JSH trying to call his intuition and simple notions about how numbers behave loginc comes in a close second.> I'm here to help guide you along, if you manage to get past your> mental roadblocks, and past posters like Dik Winter.Don't worry, Jimmy, the your silliness comes through without help from Dik or anyone else.=== === Subject: : Re: JSH: Reply, reply, reply> Come on, reply to my posts!!! Come why don't you say something else,> huh?> Try to point out some mistake Nora Baron you stupid !> Hey, you lapdog David Ull, yeah, I called you a LAPDOG Ull!!!> Aren't you mad now? Hey, you know, you HAVE to reply right, or I'll> call your school and try to get you fired, right?> Ull the university professor cursing in the muck, with no shame.> REPLY TO MY POSTS YOU CUR!> Hey Dik Winter, you ing , reply again! Why don't you throw> more of my old arguments on a webpage, you stupid !!!> Yeah, you'll all be back want you? Along with C. Bond and all the> rest.> You are curs, my lapdogs, my little traveling circus of obsessive> repliers.> I spit on you.> But you'll be back, now won't you? James , unable to digest his own arguments, appears to be having a serious acid attack.Calm down, Jimmy, and try some Pepto-Bismol.=== === Subject: : Re: JSH: Not a solution> Obsessively replying to my posts to insult me or just aggravate me is> not a solution.You don't undersand, Jimmy. Even the worst of people can serve a bad examples, and we all watch you to assure ourselves that, while you exist, we are not one of them.Aggravating you is just frosting on the cake.=== === Subject: : Re: Understanding taylor expansion for sine> I've been trying to understand the taylor exapnsion for the sine> function,> sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...> I've been told that it uses the fact that e^(i*x) = cos(x) + i*sin(x),> and I can easily work out the taylor series for e^x, but I'm not sure> how to use these two pieces to find the expansion for sin(x).> Any help would be appreciated.> Jonathan ChristensenDo you know, or know how to find, the expansion for e^x?If so, replace the x by ix, and see what happens.=== === Subject: : Re: No Set Contains Every Computable Natural (was Church-Turing compared to Zuse-Fredkin thesis)> message>>>> This is the definition of a recursively enumerable language:> This language is recursively enumerable. Given a string, x, there> exists>a>> TM that>> will halt after a finite number of steps if x is a member of L.>> This language is NOT decidable. Consider what happens if x is an> infinite string of 1's. The TM will not halt after a finite numberof>steps.>You can't input an infinite string to the tape. If you could, it> wouldn't be a Turing machine.>>No, but you can input finite instructions which generate infiniteoutput.> is right about Turing saying this, and I found it repeated:> But that would be a different language. The language would then be { |M> is a TM that outputs a string a finite string of 1s}. This language is> clearly undecidable. But that does not prove his point, as any language> where (language U lang_complement) is composed of arbitrary TMs and whose> membership is based on determining non-trivial properties of those TMs, is> undecidable (a la Rice's theorem). However L is certainly decidable:> Let me define a language, L, that consists of all unary representations of> natural numbers.> 1 = one, 11 = two, 111 = three, etc.> You can't input an infinite string to the tape. If you could, it> wouldn't be a Turing machine.>Harris:>No, but you can input finite instructions which generate infiniteoutput.> is right about Turing saying this, and I found it repeated:Chaitin et al: This is a machine that performs an unending[SH: If you read the thread, previously claimed this.}Harris added in last post:I posted not to argue on 's side, but because I thought thisTuring ingenuity deserved some recognition.as if I supported 's argument and brought up languages,which I fail to see as germane to my informational post. Mymotive rests in the posts I've read on Google which have usedthis finite input stipulation to claim that a TM cannot generatea potentially infinite output. But more to your point, I don't seewhy there can't be two inputs, x and y, a fragment of infinite Pi(or any unique generic fragment) which is then finitely matchedagainst the potentially infinite output of a Pi computational method.I think a mutual sequence match would eventually be discoveredalthough when this match would occur is undecidable. This is aspecualtive post on my part and I'm willing to change my mindStephen=== === Subject: : Re: CAN ANYONE HELP ME?????> CAN ANYONE HELP ME?????The 'caps lock' key is on the left-hand side of the keyboard, above'shift' and below 'tab'. It should look like this:+--------+| |<--- | <-- tab key| --->| |+--------+-+| Caps | <-- caps lock key| Lock |+--------+-+| / || || | <-- shift key+--------+Over on the right-hand side of the keyboard, above the numeric keypad,should be three LEDs, labelled (from left to right) 'Num Lock', 'CapsLock' and 'Scroll Lock'Press your 'caps lock' key until the LED marked 'Caps Lock' is no longerlit.Now you are ready to post on usenet.> 1.DETERMINE DOMAIN OF THE FUNCTION> f(X)=X3/X2+2X+1What happens when x^2 + 2x + 1 = 0?> FIND EXTREME FUNCTIONS f AND DRAW ITS GRAPH> 2.BY NEWTONS METHOD SOLVE EQUATION:> X3-X-2=0> DRAW A PICTURE!I would, but it's very difficult to plot cubics using ASCII art.HTH, HAND.-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice.=== === Subject: : Re: Fourier analysis of a radarsignal>A dirac in time or frequency transforms to unity.>A wide function in time has a narrow spectrum and a>narrow function of time has a wide spectrum.>The eigenfunction(s) transforming to themselves are the gaussian, BTW.>>Are you sure they are gaussian? Most electromagnet distributions have a>>Lorentzian distribution with a Gaussian distribution imposed on top for>>the noise in the measurement system. All resonance filters give a>>Lorentzian distribution.> Yes, Rene has that right. The FT of a Gaussian is indeed a Gaussian.> FranzSorry, I missed read the arguement. I agree the FT of a Gaussian is a Gaussian.I was referring to the spectrum distribution of the actual signal being Lorentzian since any single pole filter will have an spectral output that is Lorentzian.=== === Subject: : help with solutions for three questions from the past contestI'd greatly appreciate your help on the solutions for these questions.1. Al and Bob are at opposite ends of a diameter of a silo in theshape of a tall right circular cylinder with radius 150 ft. al is duewest of Bob. Al begins walking along the edge of the silo at 6 ft. persecond at the same moment that Bob begins to walk due east at the samespeed. The value closest to the time in seconds when Al first can seeBob is what? answer: 482. if a, b, c, and d are nonzero numbers such that c and d aresolutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, finda+b+c+d. ans: -23. A boat with an ill passenger is 7.5 mi north of a straightcoastline which runs east and west. A hospital on the coast is 60miles from the point on shore south of the boat. If the boat startstoward shore at 15 mph at the same time an ambulance leaves thehospital at 60 mph and meets the ambulance, what is the total distance(to the nearest 0.5 mile) traveled by the boatand the ambulance? ans:62.5Choogu=== === Subject: : How big can a manifold be?Is there any limit to the cardinality of a connected n-manifold, thatis a connected Hausdorff space every point of which has aneighborhood homeomorphic to R^n?----== Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =---=== === Subject: : Re: JSH: Reply, reply, reply: You are curs, my lapdogs, my little traveling circus of obsessiveWell, we all know who the clown is.-Justin=== === Subject: : Re: Mathcad Upgrade QuestionYou may be coming up against the old saying If it isn't broken, don't fixit!. I have MathCad 2000 professional, which works very well and haveresisted the UK seller's efforts to get me to do several upgrades. It soundsas though that was the right decision.I also have Delphi 3, 5 and 7, Personal, on my PC. They all work and 5 wasslightly better than 3 but 7 seems less good than 5. I suspect a lot ofupgrades are more trouble than they are worth.> Yes spybot caught on to it. Besides I don't like folks putting stuff onmy> computer without telling me first.> Neither Mathematica nor Matlab has any problems with the way I use my> computer. Does Mathcad 11 use> c-dilla to keep track of what your doing? Or did they leave it out?>Mathcad 11 is worth the upgrade but I have found that it doesn't work>reliably on WIN98. I upgraded to WINNT and it works fine. I also hate>c-dilla, but it should do anything bad to your system unless you havesome>software that scans for trojans and stealth programs. I would rather> have>a USB dongle.>>Don't use mathcad. I installed version 2001i this weekend. After>installition I noticed that c-dilla had also been introduced to my>registry.>Google c-dilla to find out what this program does. Anyhow I have spent2>days rebuilding my hard drive because of the damage that Mathcad did.Steven O. in>> I am thinking of upgrading from an old version of Mathcad (Version7)>> to the most recent. But past experience with upgrades has taught me>> to be a bit wary, so 2 questions:> 1. Did Mathsoft do anything to screw up the product, in terms of>> usability or features, since I last purchased it (back in 1997)?> 2. Does the latest product come with any kind of registration>> feature, like Mathematica has, that locks your software into one>> computer, and possibly locks you out of the software if you upgrade>> your hardware? (I respect the rights of companies to make money off>> their software, but I'm a firm believer that the only fair deal is>> one-user/one-app. If I have three computers -- and I do -- I should>> not have any worries or complications with loading the same software>> package on all three of them.)>> Steve O.> Standard Antiflame Disclaimer: Please don't flame me. I mayactually>*be* an idiot, but even idiots have feelings.=== === Subject: : Re: JSH: Why am I so important?> Why am I so important to these people?I believe morbidly fascinating is the term you're looking for.V. and occasionally entertaining-- email: lastname at cs utk eduhomepage: cs utk edu tilde lastname=== === Subject: : branch of log z... stop throwing stones, we're both in the glass house.I don't want to throw stones, i don't even know, am i inside oroutside. It took me 25 years to find out, what is behind this veil of:imaginary axis, argand graph, complex plane, imaginary part. CasparWessel and Hamilton didn't use or invented these, as i know now. Thesewords are still used in teaching people, who are interested in math:Ask Dr.math Mathworld.wolfram.com Complex planeThese two don't mention, that they are talking in their websites aboutthe R2, just the simple R2 with + and *.At another level - You mentioned Rudin,Marsden, Royden (books for100$): i only found some lecture notes to Rudin. Following these, hestarts with C=RxR, but then for example calling a rectangularcoordinate system an Argand Graph making this special. But you arenot told, what is the difference - it can't be the multiplication, canit?So really i am thankfull, that You told me,that i got this - nodifferennce between i-axis and y-axis - right. Just one similaropinion was recently written to sci.math from Lynn Kurtz. It seems tome, that other people try to avoid a clear statement. So manymathematicians are reading this and no confirmation or opposition fromthem can be found here in sci.math about it.In R3 you can write a position or an arrow with (x,y,z) and sometimesit's written as x*i+y*j+z*k or as x*e1+y*e2+z*e3. (linear combinationof three basis-vectors). Different notations for the same thing (and multiplication of vectors is not a necessary condition for it).When you introduce the dot-product in R3 you call it euclidianvectorspace. But you won't say it's isomorph to the first, the dotignored. It stays the same, with extra dot-multiplication.So i think (a,b) =a+i*b. a is called the Real part and b theImaginary part. a is the first component, the first coordinate andyou ommit 1 or k or e3, as R is embedded ( In R4 they do it too) .Andb is the second component, second coordinate.Are You going here a step backwards : R^2 and the Complex plane areisomorphic (if..... ) ? The only strict definition of C is(R^2,+,*) , as far as i could find. Do You have another one?looks like: complex analysis and R2 analysis -one must be part of theother. Is this so?Feeling like a snail, unable to throw stones, creeping up an parkingramp to an understanding of branches of log,Hero=== === Subject: : Minimisation of L0 norm with dynamic programming linux)I need to compute the vector f that minimizes a functional J(f), whereone of the terms is the L0 norm of f (or possibly another Lp normwhere p < 1). In this case J(f) is non convex and the minimisationalgorithm is a dynamic programming procedure.I would appreciate any link or reference on the topic.-- J.8er.99me Kalifa=== === Subject: : Re: JSH: Not a solution> Obsessively replying to my posts to insult me or just aggravate me is> not a solution.> Sure, it might make you feel better or like you're accomplishing> something if you can piss me off, but that's just a bad reflection on> you.> Notice, here I am again: posting.Who's obsessive now?> Those of you who make it your business to obsessively reply to my> posts are not doing other people a favor. I post for many reasons,> including just to toss out math ideas that I'm considering, and> there's no reason to think that I'll stop.N.B. There's no math in this post, only crackpot ranting.> It's kind of, sickly, *relaxing* to curse at Nora Baron or David> Ull.That admission reveals a lot about you, . If you enjoycursing at others, you have a personality disorder.> James I know you are, but what am I? Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--.--http://www.crbond.com=== === Subject: : Re: More Boolean AlgebraThis isn't logic, it's algebra; so how do truth tables apply?>If a statement of Boolean algebra always has value 1 in the two-valued>structure, then it always has value 1 in any Boolean algebraic>structure.> Would this mean XOR,NAND,NOT,NOR, etc?> A bit is on or off.Your bit on is my 1, you bit off is my 0. XOR, NAND, NOT, NOR(in some sense) implement what I call the Boolean two-valuedstructure.-- G.C.=== === Subject: : Re: Got a speeding ticket and need to fight back> Hello everyone,> the cops got me speeding 10-15 mph above the speed-limit. My arguement is> as follows.> Given that the radar has a +/-5 mph standard deviation error, I claim that> the cops did not get me beyound resonable doubt.> Z = -10/5 = -2> P( Z < -2) = 0.03> 3% chance that i didnt speed and the radar was at fault. My arguement to the> judge is that 3 in 100 cars will recieve a false speeding ticket because> the radar isnt accurate. And i am one of those 3.> Even still, the chances that i went just anywhere from below '5 mph above> the speed limit'> Z = -5/5 = -1> P(Z < -1) = 0.15%> There is a 15% chance(huge chance) that the radar gave a wrong reading.> Do u think the judge will buy the arguement?> ==============> Btw> The best way to embarrass them is to teach High school students statistics> and statiscal reasoning and embarrass them, by making them appear smarter> than the cops> -sureshif this material is duplicate or triplicate.I used to work in a Copy Center on the Side of Town, andoccassionally a gentleman would come in and ask us to copy aPhotoradar Handbook in color. He did conventions in thesouthwestern-midwestern areas of the United States in the largercities, educating officers on radar use. The manual was basically ahow to.The court cases below are bound to Arizona unless otherwise specified,but I'm sure most states have an analog, its just a matter of findingit (try lexis-nexis).Some facts:- Radar guns have an acknowledged deviancy of +/- 5 mph. The 5 mphrule of thumb is extremely conservative; most of the readings are +/-2 or 3 MPH. LIDAR is more accurate.- In 99.999999 percent of cases, the gun will report you going slowerthan your actual speed, due to the cosine affect (since the angleincident to the car is not 0). Only in certain rare circumstances -say an overpass on a freeway where the cop is directly above traffic,and there is a hill in the horizon providing a straight line - willthere not be a cosine error. More elaborate stationary photo radarsare fixed at 20 degrees and automatically include cosine-correction.----..radar..Cop / Overpass --- ||| --------Freeway --------------/- You need a license to operate a radar gun (ask for verification).This involves passing a test (ask for verification).- According to the US Supreme Court, the officer does not need to bean expert in how the radar works to operate it. Therefore, asking himthe Doppler Effect formula will get you nowhere. The officer onlyneeds to know reasonably how to use the gun; he is only held to knowwhat a reasonably aware person knows about radar guns (which isvirtually nothing).- The Burden of Proof in a civil case is substantially less than acriminal case. Speeding tickets are usually civil, unless you weregoing really fast (not in this case). Civil cases are beyond apreponderance of the evidence; criminal cases are beyond areasonable doubt. There is no way in hell that your evidence of 15%will meet this burden of proof (I've heard it said before, with a lotof violence to the law, that criminal cases are about 90-95% certainand civil cases are about 66-75% certain).- I'm a little fuzzy on this one, but the courts have ruled somethinglike the expert testimony proving that the radar is reasonablyaccurate is assumed (there is a legal word for it), and that the stateor the officer need not provide a witness. If you want to dispute thelegitimacy of the radar, its up to you to provide the witness.- Technically you are supposed to be not responsible until foundresponsible, but honestly its never the case in court.- The judge will not care about your derivation or your formulas. Hewill forget the 15% before he even rules.I've had about 4 or 5 speeding tickets in my short life, and went tocourt defending them all. I won only one of them - and this was a casemisdemeanor instead of a civil traffic violation (he later tried tofraudulently update the ticket). Cops always show up in thesejurisdictions, since they get paid overtime. One time, a cop was onthe side up the street with his lights off at night time gunningpeople and nabbed me. This was a violation of ARS 28-639 (b), and Ipointed it out. As it turns out, the cop breaking the law doesn'tnullify my action (however, you can always give a cop a civilcitation, which I really wish I had).If you have the option for driving school, take it.Jason=== === Subject: : Re: Infinite polynomial product by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1HKK6805854;> Not elementary, but it's related to the Dedekind eta function and> Euler's Pentagonal Number Theorem.Steve OakleyRemove iron for email=== === Subject: : Re: Entering *Real* Math Formulae in ExcelGerry Myerson wibbled:>How the &*&%$ does one enter a real math formula (simple example: the>sum of all N sub x as x goes from x1 to x2) ???!> I can't imagine doing real math in Excel, but doesn't > sum(n23:n97) produce the value n23 + n24 + ... + n97? > If it doesn't, something like it does, and it should be possible > to find out within Excel because it has some kind of Help thing > built in & you can just look for sum and it should tell you > how to use it.yeh, you can do that stuff, but you have to spell out all the steps explicitly. it does arithmetic rather than maths. it's not a fun way to do that stuff.=== === Subject: : Request for help with math puzzleI've received a math puzzle I can't solve. Can you help?Here's the message I received:This is going to sound weird but '4-22882559-6477-967'. Now takeyour sweet little time to unscramble THAT.I'll appreciate any help you can give.AlinSLC@hotmail.com=== === Subject: : Re: Coprime Grid: Filling Infinite Quadrantath: meganewsservers.com>In this post, I write of a specific and altered case of the puzzle>mentioned in these previous threads:>http://groups.google.com/groups?hl=en&lr=&ie=UTF-8& safe=off&threadm=b4be2fdf.0211011659.79913415% 40posting.google.com&rnum=3&prev=>http://groups.google.com/ groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fdf .0401241437.73a12678%40posting.google.com&rnum=4&prev=>http:// groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm= b4be2fdf.0303071814.63262520%40posting.google.com&rnum=1&prev= >In the puzzle, we try to write, in order, the positive integers into a>grid, one integer per grid-square, such that:>Each integer n is adjacent (above /left of /right of /below)>of the integer (n+1);>And each adjacent pair of integers (above /left of /right of /below)>are coprime.>And the goal is to completely fill the grid.>But here I am asking about filling an infinite grid which is bounded>along 2 perpendicular sides.>ie. the grid is an entire quadrant of the Cartesian plane, bounded by>the x-axis and the y-axis.>I think I found a simple procedure which *might* ensure a successful>filling of the grid with coprime-adjacent integers.>(Sorry to those on rec.puzzles, but I will give my procedure below.>You can still post your own algorithm, however, or confirm that mine>can really work for the entire grid without problems.)>I illustrate with the first 99 terms:>(figured by hand, so maybe in-error)>99>98>97 96 95>54 55 94 93 92>53 56 57 58 91 90 89>52 51 50 59 60 61 88 87>21 22 49 48 47 62 63 86>20 23 24 25 46 45 64 85>19 18 17 26 27 44 65 84>06 07 16 15 28 43 66 83 82 81 80 79>05 08 09 14 29 42 67 68 69 70 71 78>04 03 10 13 30 41 40 39 38 37 72 77 76>01 02 11 12 31 32 33 34 35 36 73 74 75>Basically, the path swings clockwise and counterclockwise, running>along the outside of the already-filled section.>When it gets to either the x-axis or y-axis, it forms a 'peninsula',>the length of which is the shortest needed to avoid uncoprime integers>being placed next to each other in the path-section which runs from>that peninsula's axis to the other axis.>Now, we do not want a situation where, following the algorith>precisely,>there is NO peninsula-length which would avoid uncoprime neighbors.>I am not certain, but I believe this issue is not a problem.>Fun perhaps: Show if my algorithm is foolproof...or just foolish.Well, what I know about number theory and 50 cents will get me a cupof coffee, but it seems to me that the way to attack this solutionis to show that once the peninsula length exceeds some number N,you will always have a noncoprime pair when it loops its way back.It seems to me that this ought to be true. Of course, that stil wouldn't prove that your algorithm can'twork, because you might never have to buld a peninsula out that big,but it would be a start.>(I know I do not give the peninsula's length. Also fun perhaps: try to>determine the shortest length needed for each peninsula, given that>the algorithm never leads to a problem.)>Leroy QuetGeorge=== === Subject: : A New Kind of Science OnlineStephen Wolfram's groundbreaking book A New Kind of Science(NKS) is now available online. With complete text and images,full searchability, 30,000+ links, and many other enhancedfeatures, NKS|Online gives you access to all 1,280 pages, 1,000+illustrations, and 1,350 technical notes in NKS.NKS|Online supplements the printed book with features that canonly be achieved online. It allows visual and textual browsing,navigation through the book's nearly 15,000 index entries, andfull-text searching. Each page of the original book has beenenhanced with links to related technical notes, downloadableprograms, and other elements.The result is a unique resource for students, scholars, andcasual readers, freely accessible worldwide. To read NKS|Online,visit:http://www.wolframscience.com/nksonline=== === Subject: : Re: JSH: Reply, reply, reply> Yeah, you'll all be back want you? Along with C. Bond and all the> rest.I can't speak for anyone else, but I keep hanging around in hopes ofwitnessing your inevitable meltdown. When you post things like this itencourages me to think that we haven't much longer to wait.Tick, tick, tick...-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- Myers Myers, Silverlock=== === Subject: : Re: puzzle: GCDs of Infinite Set of Integer Pairs <4emcnSWuZaeKcK3dXTWc-g@speakeasy.net> Define an actual experiment where you randomly chose a real> between 0 and 1, and I'll show you where you took a shortcut.>This is sci.math. If you want to talk about actual experiments,>take it to sci.physics.> Playing the physics card isn't helpful here. I'm claiming that it's> *mathematically* impossible to randomly choose a real from [0,1]. Not> physically impossible. I'm asking you to convince me otherwise, use a> thought experiment if you want, or not.> And the fact that you can't specify which real you chose is not the only> reason you can't choose one. If you find some way of specifying which> one you chose, I'll find a way to show you where you failed to do what> you set out to do. While I originally agreed with Patrick on this point, after givingit a little thought I'm no longer sure. Patrick, how about thisrandom number? R = Sum from i=1 to +inf of (2^-i)*(r_i) where each r_i = a random integer in [0,1] inclusive.We can certainly pick either 0 or 1 randomly. And this real numberis pretty well specified: I can give you its approximation to anyprecision you like; I can add and subtract it from things in a well-defined manner; etc. So I guess it is possible to choose a randomreal number in [0,1). I haven't studied formal probability yet, so perhaps I'm talkingnonsense here, but I *would* argue that the probability of choosingR=x, for any real number x, is not zero but rather infinitesimal.You might call it one over two to the aleph-null, except that thatsounds even more made-up. ;-) Saying that the probability of an event that *has occurred* waszero seems to me like foolishness, no matter how big a sample spacewe're talking about. Once again, you can't hide behind the 'reality card'. You can say I'll> pick an integer randomly from a uniform distribution across the whole> set but that's garbage mathematically speaking, and I'm claiming that> choosing a real from [0,1] is roughly the same flavour of garbage. I claim that my example above shows it's not. But I think in any caseit's true that you can't map [0,1):R onto [0,+inf):N in such a way as tomake the two *equivalent* flavors of garbage. ;-))-Arthur=== === Subject: : Re: 3 Squares Covering 1 Circle> Yet another coverings-problem...> (I myself might have already asked this, but I do not believe that I,> at least, have asked this specific question before.)> If we have a unit circle (radius = 1),> and we have 3 unit squares (sides =1),> then what is the maximum area of the circle we can ever cover with the> squares> a) if overlapping of the squares is not allowed?> b) if overlap is allowed?> I may be way off,> but I conjecture that the (a) case is> squares arranged like:> ____> ! !> ! !> ---------> ! ! !> ! ! !> ---------> But I bet tilting the squares works better.For case (b), I considered a class of arrangements in which squaresare oriented 45 degrees to the above pictures, i.e., like diamonds.I let the center of the circle lie at the origin, and placed thethree squares such that:1) the right corner of square 1 is at (x0,y0) (with x0 and y0 >= 0),2) the left corner of square 2 is at (-x0,y0), and3) the top corner of square 3 is at (0,y0-x0).Thus squares 1 and 2 overlap (making an overlap square of area 2 x0^2),and square 3 lies flush against the other 2 without overlapping.The maximal amount of the circle covered in a such a configuration is2.56845930, which occurs for x0 = 0.182805617, y0 = 0.221185110.This is not optimal, however. It is still less than the what canbe obtained in the case (a) configuration shown above: 2.57003584.-Jim Ferry=== === Subject: : re:How big can a manifold be?This post's gonna be burried under the rubble...If anybody knows the answer please tell me.----== Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =---=== === Subject: : Big Rip NY Science Timesre: http://qedcorp.com/destiny/SuperCosmos.pdfsome time.This will definitely be cited at the beginning of Super Cosmos.Perfect timing. Your name is in the book BTW.Metric engineering is the control of dark energy.Hal Puthoff has never even mentioned the word in any of his metric engineering publications.Maybe he will start now at last? Of course he has no way to describe it inside his PV model.While the density of the energy in Einstein's cosmological constant staysthe same as the universe expands, the density of phantom energy would go upand up, eventually becoming infinite. Such would be the case if theparameter w turned out to be less than minus 1, say physicists, who admitthey are stunned by the possibility and until recently simply refused toconsider it.It crosses a boundary of good taste, Dr. Caldwell said, calling phantomenergy bad news stuff. Phantom energy violates physicists' intuitionsabout how the universe should behave. A chunk of it could be used to propopen wormholes in space and time - and thus create time machines, forexample.I have been saying this for awhile now. It is in my two books Destiny Matrix and Space-Time and Beyond II published in late 2002 and in the Library of Congress for the record. Of course, I am the only physicist with real academic credentials from two good Universities to connect dark energy with flying saucer propulsion. Hal Puthoff never did even though I gave him plenty of opportunity to do so. There is no mention of mainstream physicist has dared to go because of the fear they will be censored out of the LNL/Cornell archive by Paul Ginsparg. ;-) Look what happened to Henry Stapp in Physics Today when he slipped a retro PK It could lead to such bizarre effects as negative kinetic energy, Dr.Krauss said. As a result, objects like atoms would be able to lose energy byspeeding up.Nevertheless, a recent analysis by Dr. Caldwell and his Dartmouth colleagueDr. Michael Doran of the supernova measurements to date, combined with othercosmological data, suggest that w could lie anywhere from minus 0.8 to minus1.25, leaving open the possibility of phantom energy.My falsifiable prediction for BOTH dark energy and dark matter is w = -1 ON THE NOSEOf course globs of w = -1 zero point stress-energy tensor fields of positive pressure will look like w = 0 CDM to the distant observer.What All The Kings Men trying to put Humpty Dumpty (Cosmological Constant) together again are missing is The Right Stuff i.e. the right idea of Vacuum Coherence (inflation field really) whose phase patterns give Einstein's theory of general relativity and whose intensity patterns give the dark energy/matter dynamic background-independentzero point stress-energy density tensor fields all over the universe at all scales including the microscopic as well as the macroscopic. As my Techgnostic Qabalist alleged ancestors would sayAs Above, So BelowThe cosmologicalconstant would give a value of minus 1.0, and anything higher would be asign of quintessence.Dr. Kirshner said phantom energy had been dismissed as too strange whenhis group was doing calculations of dark energy back in 1998. In retrospect,he said, that was not the right thing to do.It sounds wacky, he said, referring to phantom energy, but I think we'rein a situation where we're going to need a really new idea. We're introuble; the way out is going to be new imaginative things. It might be ourideas are not wild enough, they don't question fundamentals enough.Yeah, well wake up and smell the coffeehttp://qedcorp.com/destiny/SuperCosmos.pdfPrediction #2 No dark matter detector will ever click with The Right Stuff - only false positives in the long run. Mark my words.Popper falsifiable for sure.It's all w = -1.===ummm...except it's slightly a parody site. it's not real. Why don't youactually read the whole site before you start condemning it. Moron.- Nick> Has anyone heard about this? I found this site, and it really boiled> my blood! SaveTheCEO.com. Apparently this CEO wants people to raise> $100,000 to hire a lobbyist to ease tax restrictions so he can move> his factory to Mexico! As if it's not already easy enough as it is!> Someone should find this guy and put him in jail! Where he belongs!=== === Subject: : Re: the anticlassicalist }{ i: linguistic negation:: Now really, Ms.(?) fabulist, etc: there is much that is more grossly: inappropriate to sci.physics here, so you might say all the other kids: are doing it. But if you affect to be a _superior_ kid, might you not: lead by example and limit the number of freshly cross-posted threads: which are mainly about linguistics, and neither physics nor math?:: You are beginning to speak Crankish.There is certainly math already in what I've posted. Although a couplegroups have dropped either iv or v of the series (at least from what I'veseen on my server), there is already the beginnings of an exposition onlattice theory. And my focus is on getting to quantum logic, the study ofcausal sets, and logics for reasoning about continuous dynamics here in thestart talking about model satisfaction somewhere, to finish the foundations,but I understand that physicist may still be wary about applicability totheir own models, so I may invert the discussion some.I intended only two threads so far. The first was to be towards aconstructive education, which was to be quite largely cross posted (thoughwell within the scope of all groups I posted to) in order to draw in theopinions from what I knew to be a quite small minority in each field, butunfortunately found that newsservers have maximum cross-posting limits, so Ineeded to repost in a chopped up fashion, for which I appologise. Along theway, I came up with the starting of an organised bibliography, which I triedto post in the same thread as my first. Unfortunately, I believe one of thethe other two sections of posts for towards a constructive educationgroups to post for this section, and that particular posting software brokeit out of the thread even though I replied within.Then I have this thread, the anticlassicalist where I wanted to expand onthe things that people had asked questions about in my first thread. Thissci.physics.research. It is meant to be educational as well as provocative.It is certainly meant to raise the level of discourse from the potty mouthedname calling so clogging sci.physics, and is meant to introduce linguists tothe logical formalisation which was so heavily questioned in my firstthread. The logicians have had some arguments about these topics, so Iwanted to enter the fray, and nearly all of my exposition will bemathematical when I am not questioning psychological motives.So, I believe my post targets are accurate, and I have only managed to start4 threads total in the process. 2 of those were incidental, not intended.Unlike cranks, I am open to critical analysis and earnest debate.: > -=-=-=-=-= linguistic negation =: : > The logic of natural language has been studied by many different schoolsof: > thought throughout history.:: Textual analysis: schools of thought do not study anything, and is a: cliche.:: For somebody who is so academically concerned with language, you seem: remarkably insensitive to its use. Your earnestly learned turgidity: will carry you far ... away from here.Ok, I change it to read only schools. That was more my intention.As for turgidity, I do understand that my writing can be up tight attimes, just as sometimes it can be more playful. When I look to thesegroups, however, I see that often the angry rebuttals cause either crankishobstinacy or a learned timidity. The timidity response places those wholaunched the angry rebuttals into authoritative positions that often I donot feel they have the competency for. Uncle Al is a prime example of onewho certainly is knowledgeable, but is also quite ignorant and certainly notone I see as an authority.I've studied obsessively for many years, and will continue doing so for manymore. Any confidence that comes through in my posts is due to muchpractise, but I do not want to convey that such confidence excludes memaking mistakes or accepting their being pointed out. I know I makemistakes all the time. I focus on them, to try to understand them. But Iwill not back down from my positions by content-less chest beating.-- === === Subject: : Re: puzzle: GCDs of Infinite Set of Integer Pairs> Define an actual experiment where you randomly chose a real between 0 and 1, and> I'll show you where you took a shortcut.>>This is sci.math. If you want to talk about actual experiments, take it>>to sci.physics.> Playing the physics card isn't helpful here. I'm claiming that it's> *mathematically* impossible to randomly choose a real from [0,1]. Not physically> impossible. I'm asking you to convince me otherwise, use a thought experiment if> you want, or not. The probability space is simply ([0,1],L,m), where L is the set of allLebesgue-measurable subsets of [0,1], and m is Lebesgue measure on [0,1].See .For each x in [0,1], the probability of the event X element {x} is theLebesgue measure of the set {x}, which is zero.Done.> And the fact that you can't specify which real you chose is not the only reason> you can't choose one. If you find some way of specifying which one you chose,> I'll find a way to show you where you failed to do what you set out to do. And I suppose you also object to things like geometric constructions, onthe grounds that they can't be carried out exactly in the real world.Yawn.> And the probability of randomly choosing a real in [0,1] is zero. As in you> can't do it. Another example please.>>I didn't say anything about what you could do. I was discussing>>mathematics. Can you write down all the natural numbers? Does the set>>of natural numbers exist?> Once again, you can't hide behind the 'reality card'. You can say I'll pick an> integer randomly from a uniform distribution across the whole set but that's> garbage mathematically speaking, and I'm claiming that choosing a real from> [0,1] is roughly the same flavour of garbage.No, there is an enormous difference between the two. There is no suchthing as a uniform probability measure on a countably infinite set, butuncountable sets are quite a different matter, as the Lebesgue measureexample demonstrates.>>The example I gave is correct. It's so 'cos I said it's so doesn't wash with me.It's not my fault if you don't know the definition of a probabilityspace.>>Find a textbook on probability theory and>>look up probability space. It's a special case of a measure space.>>The fact that a nonempty event may have probability zero is a direct>>consequence of the fact that a nonempty set may have measure zero.> I don't believe it. I'm prepared to be convinced, but you haven't offered any> argument even vaguely convincing yet.Look at and read thethird paragraph down, where the concept of a probability function ismentioned. Notice that the function f: [0,1] -> R given by f(x) = 1 isan example of a probability function or probability density functionas described there. The fact that the integral of f over [0,1] = 1 iswhat makes the function properly normalized to define a probabilitymeasure, which turns out to be ordinary Lebesgue measure on [0,1].Look at the probability axioms that are stated farther down on that page.A probability measure is something that satisfies those axioms. Noticethat the description of probability says nothing about what you can orcannot do in the real world; it's an abstract mathematical concept.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> Can someone here point me to a straightforward proof that the binomial >> distribution reduces to the normal distribution as n -> inf?>> I've seen the statement over and over that it does, and of course it's >> easy to plot histograms of both dists and show that they get closer and>> closer as n increases. I'm aware that Stirling's formula is involved,>> but so far I've been unable to show the equivalence. Would appreciate>> some help.>This is a special case of the Central Limit Theorem. The easiest proofs >involve transforms (e.g., characteristic funtctions). See Drake AW, >Fundamentals of Applied Probability Theory for an elementary outline of >most of a proof.I suggest that nobody use the characteristic function proofif any understanding is wanted. If one wants to use a completely general proof, the Lindeberg version is good andis completely rigorous at the undergraduate level.However, the normal distribution itself was found by showingthe Central Limit Theorem for the binomial. What de Moivredid was to use Stirling's formula for the factorials to approximate log((n!*p^k*(1-p)^(n-k)/(k!*n-k)!)by expanding it as a function of (k-n*p)) (or more easily,k+.5 - (n+1)*p) and finding the dominant part of the expansion to be a quadratic. Analogous to the textbookapproximation of an integral as a sum, he approximated thesum by the integral of the resulting normal distribution;this was actually a hot topic then, see the Euler-Stirlingformula. This calculus type proof provides more insight than any ofthe other proofs where it can be carried out. The CLT isa form of cancellation, and this is highly unintuitive.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Request for help with a math puzzleI've received a math puzzle I can't solve. Can you help?Here's the message I received:This is going to sound weird but '4-22882559-6477-967'. Now takeyour sweet little time to unscramble THAT.I'll appreciate any help you can give.AlinSLC@hotmail.com=== === Subject: : Re: the anticlassicalist }{ i: linguistic negation: : > -=-=-=-=-= linguistic negation =: : > The logic of natural language has been studied by many different schoolsof: > thought throughout history.:: That they have one and all avoided coming to any unique conclusion: emphasizes the bankruptcy of the Liberal Arts. It's a hen party.: Anybody can gossip and all are appreciated. It's bull. How would: you like to be doing 80 mph in a car that was built by mob rule: instead of cold, hard, dry engineering?:: Oh, but that example is different! It certainly is. Liberal Arts are: bull and engineering is real world.Formal linguistics is founded on the same rigour that one finds in quantummechanics, so don't give me your bull. Obviously, I wouldn't trust alinguist to build my car solely on linguistic credentials, just as Iwouldn't trust a physicist to do surgery on me solely on the physicscredentials.in any way condemning science. I am scientist, and maybe if you stop yourreflex vomit and read a little, maybe one day you will become one too.I am talking about what we mean by making propositions in a model, and willbe discussing all sorts of wonders, like quantum logic, causal sets, andeven maybe explore antigen-antibody reactions. Yes, I will also be givingmany linguistic examples. I am sorry that I won't be able to respect thenarrow mindedness that you are so comfortable with.: Everything in the Liberal Arts and Social Sciences is both right and: wrong. No conclusion at all can be drawn except the need for more: funding. Say what you want - it makes no difference at all (except: for grant funding - then you must agree with the mob or be destroyed).:: Klingon was specifically created to be the worst language possible by: folks who knew linguistics. It was enthusiastically embraced by the: mob and it is as good as Korean or Chinese for transferring content.Awww... I was hoping for a little more blatant racism from you. I can'thave any fun tearing you a new one with that tired rag.-- === === Subject: : HELL :-) (Was: Re: the anticlassicalist }{ ii: the spectre continues)>The quote from Martin even continues quite explicitly:Strictly, there is only one logic, which, however, can be extended>variously for specific purposes as needed. 'One god, one country, one>logic,' is the stunning phrase of Whitehead.> Nonsense. This is just a description of HELL! :-)lolYou never did understand why I was posting to sci.logic. :-)In Process and Reality Whitehead had been kind enough to say that mathematicsset a bad example for philosophy. Of course, his discussion of extensiveconnection in later chapters relegated that to a vacuous nonsequitur. But, onemight say that he tried to not...????When professionals in every academic discipline involve their topic withmathematics as a justificationalist tool, people with a sense of the subject mustinvariably suffer. Imagine trying to make sense of inconsistent mathematics, http://plato.stanford.edu/entries/mathematics-inconsistent/ However, specifying a topological space by itsclosed sets is every bit as reasonable as specifying itby its open sets. Yet the logic of closed sets is knownto be paraconsistent, ie. supports inconsistent theories;Naturally while the academic jackasses argue about these matters at faculty tea,they continue to teach nonsense (=inconsistent) to their students. Caveatemptor. It is all too easy when one has refuge in the pathetic retort, But, *I*am not talking about THAT!As an example of the dark side with this issue, let me consider Grothendieck for amoment. Pierre Cartier has a lovely paper about space and symmetry http://modular.fas.harvard.edu/sga/from_grothendieck.pdf...was given a suspended sentence of six months inprison and a fine of 20,000 francs. It seems to methat his definitive break with science dates from thisincident. He withdrew more and more into his owntent. After 1993 he no longer had a postal addressand settled into a hamlet in the Pyrenees. Very fewmanaged to see him there. If I can believe his mostrecent visitors, he is obsessed with the Devil, whomhe sees at work everywhere in the world, destroyingthe divine harmony and replacing 300,000 km/sec by299,887 km/sec as the speed of light!It is terribly easy to relegate this to nonsense--but, after all, we are talkingabout HELL here. :-)There is an embedding theorem in combinatorial topology stating that everyn-dimensional triangulation is isomorphic to a triangulation in 2n+1 dimensions.Currently, there appears to be a group of theoretical physicists advocating atheory embedded in the symmetries of a 26-dimensional Lorentzian lattice. So,here is a little calculation for Grothendieck's Devil,300,000299,887------- 113Using the number 26, we have(26)...(1)...(26) 53 2n+1 2 2-monotonic ordering constants --- 55(55)...(1)...(55) 111 2n+1 2 2-monotonic ordering constants --- 113And, of course, I just made a different post mentioning the logic of apseudocomplemented lattice in which everything becomes true after no more than twoapplications of pseudocomplementation,-----...and as a pseudocomplemented distributive lattice with pseudocomplementa* | 0 if a <> 0 Chains with 0, 1 : a* = | | 1 if a = 0-----Unlike Cartier, I recognize a certain rationality behind the reported fact. Thatis not to say that I concur with the conclusion, however.One wants to immediately say that is engaging in nonsense, trickery, ornumerology. Well, perhaps we should overlook the fact that* INCONSISTENCY SHOULD BE NONSENSE**MATHEMATICS IS FULL OF TRICKS**REDUCING TOPICS TO THE SCIENCE OF NUMBER IS NUMEROLOGY*Regardless, if we are going to accept mathematical principles in any case, weshould not be ignoring the statistical entropy described by Shannon and the factthat there is almost no understanding of analog-to-digital quantization inmultiple dimensions (per Sphere Packings, Lattices and Groups by Conway andSloane). In my opinion, that is what seems to be involved here.Now, I will gladly admit to being confused. There is plenty of evidence of thaton sci.logic in the past year.And, I will gladly admit to talking in circles. Indeed, I have been accused ofthat often enough as well.But, let me share a non-circular observation with everyone. Whereas I thought theappearance of 26 dimensions might have had to do strictly with the ubiquity of theEnglish alphabet in technical literature, I was completely astonished when Irecently opened my Bible on a lark. Turning to the Book of Numbers, the prefacereads---The Book of Numbers derives its name from the account ofthe two censuses of the Hebrew people taken one near thebeginning and the other toward the end of the journey in thedesert (chapters 1 and 26)---Once again, this is not about numerology. It is about the structure ofinformation. At http://www.research.att.com/~njas/lattices/ANABASIC_11.htmlyou will find the description of an 11-dimensional lattice for which no subset ofminimal vectors forms a basis. The Gram matrix for this lattice is 60 5 5 5 5 5 -12 -12 -12 -12 -7 5 60 5 5 5 5 -12 -12 -12 -12 -7 5 5 60 5 5 5 -12 -12 -12 -12 -7 5 5 5 60 5 5 -12 -12 -12 -12 -7 5 5 5 5 60 5 -12 -12 -12 -12 -7 5 5 5 5 5 60 -12 -12 -12 -12 -7-12 -12 -12 -12 -12 -12 60 -1 -1 -1 -13-12 -12 -12 -12 -12 -12 -1 60 -1 -1 -13-12 -12 -12 -12 -12 -12 -1 -1 60 -1 -13-12 -12 -12 -12 -12 -12 -1 -1 -1 60 -13 -7 -7 -7 -7 -7 -7 -13 -13 -13 -13 96As can plainly be seen, the number 60 (that is, the base of the sexigesimal numbersystem used by ancient Sumer) lies on the diagonal in every position except thelast. As for the number 96, it plays an essential role in quantum logic. On page5 Jacek Malinowski's paper, http://www.uni.torun.pl/~jacekm/swc1.pdfyou will see mention of the Chinese lantern. This is a six element orthomodularlattice. The description of a quantum consequence operation depends on the crossproduct of this structure with the sixteen standard Boolean switching functions.This is discussed in Protoalgebraic Logics by Janusz Czelakowski, although theproof is omitted.Naturally, this has very little to do with interpreting physics--there are nooperators on a Hilbert space such that PQ-QP=1. But, it might indicate the properinterpretation of quantum logics with respect to information theory--if there wasanyone paying close enough attention.But, like Pavlov's dogs, everyone has been trained to say, *I* don't donumerology.Oops... too late.For those interested in that kind of thing, one might compare the Gram matrixabove with the story about separating the rocks when crossing the Jordan riverinto the Promised Land. But, keep in mind that the mathematical comparison shouldbe the fact that convex deltahedra can have a maximum of 20 sides. A 60-sideddeltahedron is not convex. The story is about discarding information. One canfind the relationship between the sexigesimal number system and the Great PlatonicYear in the work of Joseph Campbell.Anyway, if I apply the same sort of construction as with Grothendieck'srecalculation, I might write,(11)...(1)...(11) 23 2n+1 2 2-monotonic ordering constants --- 25And, coincidentally, I find a 25 member listing in Conway and Sloane. What isparticularly interesting is that the endpoints are fixed at 0 and 1, as is neededby classical logic. n lattice density 0 0 1 1/2 2 1/(2*surd(3)) 3 1/(4*surd(2)) 4 1/8 5 1/(8*surd(2)) 6 1/(8*surd(3)) 7 1/16 8 1/16 9 1/(16*surd(2))10 1/(16*surd(3))11 1/(18*surd(3))12 1/2713 1/(18*surd(3))14 1/(16*surd(3))15 1/(16*surd(2))16 1/1617 1/1618 1/(8*surd(3))19 1/(8*surd(2))20 1/821 1/(4*surd(2))22 1/(2*surd(3))23 1/224 1Anyone with the slightest grasp of how Lie algebras are applied by generalizationto answer questions in quantum physics should look at the triples for n=(9,10,11)and n=(13,14,15). The graph K_(3,3) is non-orientable, and, the logical analysisof quantum operators by von Neumann and Birkhoff led to Mackey's seventh axiom.That is, the logic of quantum mechanics suggests a framework of infinitely manyquestions. I can see it now...Give me 3.Give me 3 more.Give me 3 more.Give me 3 more.ad infinitum.I know that it is difficult to understand my point--after all, people talking incircles leave little room for a rational person to get a word in edgewise.Almost everyone *knows* that a sufficient amount of text will permit somenegative (apocalyptic) conclusion under statistical decoding. But, if youTotal positivity is a concept of considerable power thatplays an important role in various domains of mathematics,statistics, and mechanics. In mathematics, totally postivefunctions figure prominently (though sometimes indirectly)in problems involving convexity, moment spaces, eigenvaluesof integral operators, and the oscillation properties oflinear differential equations, as well as in the theory ofapproximations and other areas of real analysis. In statisticsthe theory of total positivity is fundamental to an understandingof statistical decision procedures, and especially in discerninguniformly most powerful tests for hypotheses involving afinite set of real parameters. Total positivity is also ofgreat importance in ascertaining optimal policy for inventoryand production processes, in evaluating the reliability ofcoherent systems, in the analysis of diffusion-type stochasticprocesses, and in the study of coupled mechanical systems.one begins to realize that mathematical pictures are easily biased. This seems toresult from picture-thinking and can be expressed formally in terms of theisoperimetric inequality,---The area F and the length L of any plane domain with rectifiableboundary satisfy the inequality L^2 - 4(pi)F >=0;the equality sign holds only in the case of a circle---At least the physicists seem to be concerned with one particular circle. Theequation for the fine structure constant, Alpha, is of the form, e^2 4(pi)Alpha = --------------------- (epsilon_0)(h-bar)(c)So, unlike Grothendieck, I have no particular concern about the speed of lightchanging. But, circles are convex domains and it is the mathematics of convexitythat is ubiquitous.The more I investigate the history of this subject, the more clear it is thatthese issues, mathematics has the soma effect a feel-good generation desires.But, someone has to have made a mistake unless there is no such thing as reality.My experience with closed physical systems is that there are conservation laws.:-)=== === Subject: : Re: 3 Squares Covering 1 Circle 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21:>I may be way off,>but I conjecture that the (a) case is>squares arranged like:> ____> ! !> ! !>--------->! ! !>! ! !>--------->But I bet tilting the squares works better.> For case (b), I considered a class of arrangements in which squares> are oriented 45 degrees to the above pictures, i.e., like diamonds.> I let the center of the circle lie at the origin, and placed the> three squares such that:> 1) the right corner of square 1 is at (x0,y0) (with x0 and y0 >= 0),> 2) the left corner of square 2 is at (-x0,y0), and> 3) the top corner of square 3 is at (0,y0-x0).> Thus squares 1 and 2 overlap (making an overlap square of area 2 x0^2),> and square 3 lies flush against the other 2 without overlapping.> The maximal amount of the circle covered in a such a configuration is> 2.56845930, which occurs for x0 = 0.182805617, y0 = 0.221185110.> This is not optimal, however. It is still less than the what can> be obtained in the case (a) configuration shown above: 2.57003584.I was just looking at the configuration of three squares at 120 degree angles from each other (all having an edge midpoint at the center of the circle). It seems to be even worse.However the first picture above (three nonoverlapping squares, arranged as shown) can't be optimal when overlap is allowed, because sliding the lower two squares upwards increases the covered area until the top and bottom edges of the squares have equal length segments of the boundary.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer Science=== === Subject: : Re: . 53ab2750 is an anonymous user . Hi Bilge, >You mistakenly thought, No, it wouldn't. > Go look up the rfc for the nntp protocol. > The path line identifies the news servers through which >I know for a fact that a number like that > I sometimes score on it, it works perfectly. No, you don'y know that for a fact, for the same reason you don't knowanything else - you never look up the references. To wit, RFC 1036 says:2.1.6. Path This line shows the path the message took to reach the current system. When a system forwards the message, it should add its own name to the list of systems in the Path line. The names may be separated by any punctuation character or characters (except . which is considered part of the hostname). and Normally, the rightmost name will be the name of the originating system. However, it is also permissible to include an extra entry on the right, which is the name of the sender. This is for upward compatibility with older systems. The Path line is not used for replies, and should not be taken as a mailing address. It is intended to show the route the message traveled to reach the local host. There are several uses for this information. One is to monitor USENET routing for performance... >For example it identifies one exact Earthlink user. > take a look at Ha Ha Hanson's headers, > you would always know that it was him, and only him. Only because earthlink _chooses_ to add that as an optional fieldfor it's own reasons. It has nothing to do with any requirementto do so as rfc 1036 makes clear. As usual, you make invalid assumptionsand draw unwarranted conclusions. Proof by example dies not constitutea proof. >53ab2750 is an anonymous or unauthenticated user. jeff relf is an anencephalic or brainless poster.=== === Subject: : Re: What is a number?/What is not a number?>I'll offer this definition: A number is an element of a number set. A>number set is either the set of integers Z or any ring/field R that>contains Z.So ordinal numbers are not numbers?> It seems even the natural numbers are not numbers.I addressed that issue on my post of the fourteenth.Patrick=== === Subject: : Re: x^2 + y^4 = z^4===> === Subject: : x^2 + y^4 = z^4> Thus problem boils down to showing no coprime a,b,c with> a^4 + b^4 = 2c^2> when a /= b.> Is there some coprime a,b,c with> a^2 + b^2 = 2c^2> and a /= b?Yes, 1^2 + 7^2 = 2*5^2 7^2 + 23^2 = 2*17^2 , etca^2 + b^2 = 2c^2suggests you are looking at the normof (1+i)(p +iq)^2 in Z(i)This givesa = p^2 - q^2 - 2pqb = p^2 - q^2 + 2pq(a,b) = 1, (p,q)=1 = p,q not both oddSome kind of descent might be possible in Zor even more likely in Z(i)=== === Subject: : Re: the anticlassicalist }{ i: linguistic negation> : > : > -=-=-=-=-= linguistic negation => : > : > The logic of natural language has been studied by many different schools> of> : > thought throughout history.> :> : That they have one and all avoided coming to any unique conclusion> : emphasizes the bankruptcy of the Liberal Arts. It's a hen party.> : Anybody can gossip and all are appreciated. It's bull. How would> : you like to be doing 80 mph in a car that was built by mob rule> : instead of cold, hard, dry engineering?> :> : Oh, but that example is different! It certainly is. Liberal Arts are> : bull and engineering is real world.> Formal linguistics is founded on the same rigour that one finds in quantum> mechanics, so don't give me your bull. Bull again. Linguistics is a loud empty hole that has pulled adonut about itself. It proposes to follow scientific method ondatabases that are at best weakly statistical and that have noempirical falsification possible, inputs or outputs. Like psychology,it shouts its self-percieved triumphs and whimpersheteroskedasticity at its failures. BTW, what are linguistics'triumphs?Well I don't think it's quite fair to condemn a whole program becauseof a single slip-up, sir. General Buck Turgidson explaining to thePresident the psychological evaluation program whose slip-up wasGeneral Ripper's 34-plane airborne nuclear attack upon Russia. --Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics)===> ummm...except it's slightly a parody site. it's not real. Why don't you> actually read the whole site before you start condemning it. Moron.Uh..I think he's promoting it. You might want to think hard about that last word you used.;^)=== === Subject: : Re: Metamath Axiom of Choice ...................>>As I stated above, the Metamath formulation of the Axiom of Choice is >>effectively the statement that given any set x, there exists a set y >>such that for all nonempty elements w of x, w n (Us) has exactly one >>element, where s = {t in y : w in t}. As stated, it is false. Add the elements of x are disjointand it is one of the standard forms.To see that it is false as stated, let x have three elements, {1, 2}, {1, 3}, {2, 3}. Any set intersecting all of these inat least one element has to contain both elements of one.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Re: No Set Contains Every Computable Natural>Let me define a language, L, that consists of all unary representationsof>natural numbers.>1 = one, 11 = two, 111 = three, etc.>This language is recursively enumerable. Given a string, x, there existsa>TM that>will halt after a finite number of steps if x is a member of L.>This language is NOT decidable. Consider what happens if x is aninfinite>string>of 1's. The TM will not halt after a finite number of steps.> You can't input an infinite string to the tape. If you could, it wouldn't> be a Turing machine.This is equivalent to saying that a human operator must decide ifthe tape has an infinite string of 1's before giving the tape to theTM. There is no automated way to decide if the tape has aninfinite string of 1's.Requiring the input tape to be finite is a severe limitation.For example, no language, L, could contain every binaryrepresentation of rational numbers between 0 and 1.1/3 has an infinite binary representation in base 2.Every definition of TM's that I have seen assumesthat an infinite tape of blanks is allowed.>The set of all natural numbers is not a recursive set.> No natural number requires an infinite unary representation. All may be> represented using a finite number of 1s.There is no TM that can decide if the string has aninfinite string of 1's.>Consider what happens when we talk about the set of all TM's.>Assume we can encode the state transition table of a TM using>a natural number. This set can not be recursive for the same>reason that the natural numbers are not recursive. Given a>string, i, we can not be sure that TM(i) is actually a TM>because we can not be sure that i is actually a natural number.> All Turing machines can be finitely represented. There is no such thingas> a TM transition function that requires an infinite state automata for> representation.There is no TM that can decide if the transition table isinfinitely long.>No set can contain every computable natural number.>Proof:>Assume a TM produces a tape with all of the natural numbers>encoded in unary where each natural is followed by a 0 (or a blank).> This is impossible, a) a decider is limited to a finite number of steps> before halting, b) to output all integers in unary requires an inifnite> number of steps - hence, no TM can produce this output.>Turing gives the state transition table of such a TM> 010110111011110...Turing's TM doesn't produce a unary representation of every natural?>I can define a three state TM that will find a natural number>that is not on this tape:>1) Read right until there is a 0.>2) Read right until a second 0 is found.>3) Backup and write a 1 on the previous 0.>Repeat steps (1) through (3).>This TM will always produce a tape that has exactly one 0.>This 0 will be at a finite position on the tape and the string of 1's>that preceds this 0 represents a natural number that was not on>the original input tape.> You have not at all proven this. First, how can there be a 0 at a finite> position from the start? If there is, there must a 0 to its right (infact> an infinite number of zeros), hence step 2 will read this second zero, and> step 3 will go back and erase the 'remaining 0'. The part about thenumber> preceding the last 0 being an integer not on the tape makes no senseeither.Is doesn't matter that there are an infinite number of 0's on the inputtape.Every 0 on the original input tape is in a finite position.The TM I describe will produce an output tape with exactly one 0.My TM doesn't ever write a 0, so the 0 was on the input tape.>The Halting Problem is ill posed.>There is no set that contains every TM> Sure there is : A = { | M is a valid TM representation}It is impossible to decide if M is a valid representation.> It is easy to check, given a data structure, whether the input is a validTM> or not.It is impossible for a TM to decide if M is a valid representation.>just as there is no set>that contains every computable natural number.> All natural numbers are computable. However, they _all_ can't begenerated> in a single run of any algorithm/TM.The output of a TM can't contain a representation of every natural?>Now, about the difference between finite amount of time vsfinite number of steps. There are theories of computing where>a TM can perform an infinite number of steps in finite time.>These are called hypercomputers.>http://en.wikipedia.org/wiki/Hypercomputer>A non-deterministic Turing machine is an example of a hypercomputer.> No, an NTM is not a hyper-computer since any NTM can be simulated by a> standard TM. The NTM must be given additional abilities to become> super-Turing.I misread the examples given at wikipedia. It says:A non-deterministic Turing machine which has a preference ordering over itsfinal states.The additonal ability is preference ordering over its final states.>Another example is an accelerated Turing Machine. An ATM performs>the first operation in one unit of time, the second operation is 1/2unit>of time, the third operation in 1/4 unit of time, etc. An ATM willperform>an infinite number of operations after two units of time.>Assume we have a computer that can perform an operation in one>nanosecond. The language, L, that I defined above is not even>recognizable if we expect an answer before the universe freezes over.>We can easily compute a natural number, n, that is so large that>our nanosecond computer will take 15 ion years to recognize>n as a natural number. There is only a finite number of naturals>less than or equal to n. There is an infinite number of naturals>larger than n. Nearly all natural numbers are larger than n.>So, we see that we have to assume that a TM can perform>an infinite number of operations in finite time to even say>that L is recursively enumerable.> To say that there is some natural number that a physical computer willnever> be able to print out in unary is one thing, but to say that a TM (a purely> mathematical object) 'can't' print out certain natural numbers in unary ina> finite number of steps is not correct.I didn't say that a TM can't print out certain natural numbers.I said that a TM that requires a fixed, finite amount of time toperform an operation can't print certain natural numbers in areasonable amount of time.- The universe is one dimensional=== === Subject: : Re: . 53ab2750 is an anonymous user . charset=Windows-1252> Jeff Relf:>Hi Bilge,> You mistakenly thought,> I know for a fact that a number like that> I sometimes score on it, it works perfectly.> For example it identifies one exact Earthlink user.> take a look at Ha Ha Hanson's headers,> you would always know that it was him, and only him.[Bilge]> jeff relf is an anencephalic or brainless poster.AHhahahahaha.......ahahahaha...AHAHAHAHHAA......Well, Bilge, I don't know about your evaluation of Jeff,but I must thank Jeff dearly that he takes such a deep andabiding interest in my posted grand wisdoms and goes tothe extreme length to verify and make sure that it is indeedfrom me, hanson@quick.net = !ed99b2be .........AHAHA....HAHAHA......ahahahahaha......AHAHAHHAHAHA... ...That's what I call readership loyalty, Bilge!Jeff, you get three attaboys for this!AHahhahahaha....ahahahanson=== === Subject: : Re: What is a number?/What is not a number?> @mozo.cc.purdue.edu:>>I'll offer this definition: A number is an element of a number set. A>>number set is either the set of integers Z or any ring/field R that>>contains Z. > So ordinal numbers are not numbers?>It seems even the natural numbers are not numbers.> So Numbers are not numbers> ;)The question is, What is it we really mean by the label number?There are two ways to define the set of number sets: One way is tosimply list all number sets, the other is to give a rule (given by adefinition) that takes the place of the listing of number sets. Myquestion is, Is there a definition that provides a rule that createsthe same set of number sets as there are in the set of number setscreated as a listing by fiat?So, we have the set of number sets S = {N, Z, Rationals, Reals,Complex, Cayley, ...}. So, what definition of number gives us theset S?Look at it this way: Let S' = {M : P(M), where M is a mathematicalset}. What definition do we propose such that 1) M is a number set iffP(M) is true, and 2) such that S' = S?Is there a mathematical or set-theoretic property P(M) common to allthose number sets M listed in S (other than convention says thatthey should be listed there)?Patrick=== === Subject: : Re: 'erf' function in C>> Are there any C compilers which have the erf function (from>> probability) as part of their math libraries? Or are there any math>> libraries available to download which implement this function?>> James Harlacher[...]>A procedure that can determine Phi(x) nearly to the limit>available in double precision can be based on the following>method, which I developed the 1960's:[...]> [...]>George Marsaglia[...]> Is there someplace a table available of the true values of the> normal distribution up to 20 places?> With kind regards> Matthias Kl.8ay> --> www.kcc.ch>Sorry, I should have mentioned that the values I used>were based on setting Digits:=20 or 30 in Maple, then>calling .5+.5*erf(x/sqrt(2.)), (Yes, Maple is still stuck>with erf for lack of a standard for the normal distribution,>although it may be avoided by the rather awkward> stats[statevalf,cdf,normald](x);)>Mathematika and other extended precision math packages>would also serve. Of course the old standby of>looking up tabled values, for example the Handbook of>Mathematical Functions edited by Abramowitz and Stegun,>will still serve, but they are likely to list evenly spaced>x's and perhaps less than 20 digits for Phi(x).>George Marsaglia> In my copy of A&S from around 1972> Phi(x) is tabled for x = 0(0.02)3 to 15 places, and> x = 3(0.05)5 to 10 places. So this is fairly outdated as a> reference. Unfortunately, it seems that the new electronic version of> A&S at http://dlmf.nist.gov is not going to be completed very soon.> This still leaves me with the question: How to validate such an> algorithm against a standard? Can you trust Maple, Mathematica & Co.> to deliver exact results up to the desired precision?> BTW I made implementations in Fortran, HP-41C, Pascal, Modula 2,> Clipper and recently Visual Basic of the Taylor Series expansion of> Phi(x) given in 26.2.11 in A&S since around 1980:> Phi(x) = 1/2 + phi(x)*(x + x^3/3 + x^5/(3*5) + x^7/(3*5*7) + ...)> Here are my VB results against your algorithm (also translated to> Visual Basic). It looks like the VB version loses about 1 digit in> precision against the C version (I'm not surprised, but it is not too> bad either).> I'm quite pleased how my algorithm still stands up today :-)> With kind regards> Matthias Kl.8ay> --> www.kcc.chYou can trust Maple (& Co, but they are all slow). I forgot the rule,but exactness is assured except the last digit and choosing a settingof n digits a leading 0 counts for that as well (it means: no, notthe digits after the decimal point, so if you want 200 you have toaks for 201 due to that convention).For Visual Basic i once played with Excel, they have some curious datatype beyond single, to see how far Marsaglia's approach would go. Youcan download it as http://www.wilmott.com/attachments/xmasNormDist.zipwith a small docu. Practical it is nonsense, but i had fun with it :-)Note that it uses string processing only for display as Excel can notdirectly show more than 15 (?) digits after the decimal point.---remove the no for mail=== === Subject: : How many ways to put 5 balls into 500 ordered cups?You are given 500 numbered cups and five identical balls. Any cup canhold up to five balls. How many ways can you put the five balls intothe 500 cups?As a warm-up, let's try smaller sets of ordered cups:Cups Arrangements====================1 12 63 214 565 1266 2527 4628 7929 128710 2002This is not Euler's partition function because the cups are ordered;thus, the two-cup case is:(0, 5)(5, 0)(1, 4)(4, 1)(2, 3)(3, 2)My first thought was that if I throw the first ball into the cups, itcan land #1, thru #500; likewise the second, third, fourth and fifthballs. So the answer would be (500^5) = 31, 250, 000, 000, 000 ways. This is wrong, but it sounds good.What I'd like to get is a function (generator, recurrence relation orclosed form) for this that I can explain to someone relatively easily.I have a brute force solution using an SQL query from hell (myspecialty), which is not quite the same thing as a formula and proof. I cannot remember enough Combinatorics and Finite Differences to makethe next step. ARRRGH!It is bothering me enough that I will offer a copy of my next book(TREES & HIERARCHIES IN SQL) as a prize for the best solution.=== === Subject: : Re: . 53ab2750 is an anonymous user .>Jeff Relf:>Hi Bilge,> You mistakenly thought,> I know for a fact that a number like that> I sometimes score on it, it works perfectly.> For example it identifies one exact Earthlink user.> take a look at Ha Ha Hanson's headers,> you would always know that it was him, and only him.> [Bilge]> jeff relf is an anencephalic or brainless poster.> AHhahahahaha.......ahahahaha...AHAHAHAHHAA......> Well, Bilge, I don't know about your evaluation of Jeff,> but I must thank Jeff dearly that he takes such a deep and> abiding interest in my posted grand wisdoms and goes to> the extreme length to verify and make sure that it is indeed> from me, hanson@quick.net = !ed99b2be .........AHAHA....> HAHAHA......ahahahahaha......AHAHAHHAHAHA......> That's what I call readership loyalty, Bilge!> Jeff, you get three attaboys for this!> AHahhahahaha....ahahahansonIt sure doesn't take much to make you laugh=== === Subject: : Re: JSH: Reply, reply, reply>> Yeah, you'll all be back want you? Along with C. Bond and all the>> rest.> I can't speak for anyone else, but I keep hanging around in hopes of> witnessing your inevitable meltdown. When you post things like this it> encourages me to think that we haven't much longer to wait.> Tick, tick, tick...It's like watching a car wreck over and over again....-- === === Subject: : RelMiCS8 - AnnouncementAnnouncement for the 8th International Conference on Relational Methods in Computer Science (RelMiCS 8) in combination with the COST 274 / TARSKI Workshops and the 3rd International Workshop on Applications of Kleene Algebra February 22-26, 2005 St. Catharines, Ontario, CanadaThe purpose of this meeting is to bring together researchers from varioussubdisciplines of Computer Science and Mathematics who use the calculus ofrelations and/or Kleene algebra as methodological and conceptual tools intheir work. The relational calculus originated with Tarski's abstract algebraictreatment of binary relations in 1941. Kleene Algebra is a related algebraicsystem. It was first introduced by Kleene in 1956 and further developedby Conway in 1971.Topics of this conference include but are not limited to:- Relation, Cylindric, Fork and Kleene Algebras- Relational proof theory and decidability issues- Relational representation theorems- Applications to programming, databases and analysis of data, such as: * Semantics of programming languages, program verification, specification and development and program analysis * Assertion calculi, modal and dynamic logic, interval and temporal logic * Duration calculus and timed automata * Process and network algebras * Modeling real world situations * Relational reasoning in qualitative physics and cognitive science * Knowledge acquisition, preference modeling, and scaling methods- Computer systems for relational knowledge representationPrevious RelMiCS meetings were held in 1) Dagstuhl, Germany (1994) 2) Parati, Brazil (1995) 3) Hammamet, Tunesia (1997) 4) Warsaw, Poland (1998) 5) Qu.8ebec, Canada (2000) 6) Oisterwijk, The Netherlands (2001)A call for papers will follow.Contact: info@relmics8.org=== === Subject: : Re: . 53ab2750 is an anonymous user . charset=Windows-1252>Jeff Relf:>Hi Bilge,> You mistakenly thought,> I know for a fact that a number like that> I sometimes score on it, it works perfectly.> For example it identifies one exact Earthlink user.> take a look at Ha Ha Hanson's headers,> you would always know that it was him, and only him.[Bilge]> jeff relf is an anencephalic or brainless poster.AHhahahahaha.......ahahahaha...AHAHAHAHHAA......>Well, Bilge, I don't know about your evaluation of Jeff,>but I must thank Jeff dearly that he takes such a deep and>abiding interest in my posted grand wisdoms and goes to>the extreme length to verify and make sure that it is indeed>from me, hanson@quick.net = !ed99b2be .........AHAHA....>HAHAHA......ahahahahaha......AHAHAHHAHAHA.. ....>That's what I call readership loyalty, Bilge!>Jeff, you get three attaboys for this!>AHahhahahaha....ahahahanson> It sure doesn't take much to make you laughAHahahahaha....AHAHAhahahaha...ahahaha....Your are so right, Peter. It does not take much to makeme smile, laugh, roar, ROTL or ROTFLMAO...ahahahaha..That's why you too get three attaboys for your reply !Life's a bowl of cherries, pits and nuts and all....ahahahahaha.......ahahahahanson=== === Subject: : Re: CAN ANYONE HELP ME????? boundary=----=_NextPart_000_0039_01C3F55E.8CF4F2E0------------ --------------------------------------------------------- charset=iso-8859-2Bax Perhaps he decided to patronize you because, as *everyone* parsed your> post, you are incapable of reading plain English but you pretend to> have the expertise to respond to questions anyway?Well, I guess you must be right, then. Since *everyone* says so. Bythe way, you should note that what I said was _accurate_. I onlyresponded to the wrong question. I really don't see why my gaffe hasinspired you to post two insulting messages about me.> It shouldn't seem patronizing after all, since as he (and other> respondents) read your post, it was an answer an ignoramus might> write.I'm not exactly sure why you'd want to get into a pissing contest withme. But your reply is completely pointless, since it is in responseto a post that wasn't addressed to you.'cid 'oohPS. Chill out. This has nothing to do with you.=== === Subject: : Re: Metamath Axiom of Choice> Also, the motivation behind proving the equivalence of the Metamath > formulation of the Axiom of Choice and more standard formulations> is the same as the motivation behind proving the equivalence of the > Axiom of Choice and Zorn's Lemma, and the equivalence of the Axiom > of Choice and the Well-Ordering Theorem. I presume that you have > seen the proofs of these last two mentioned equivalences, Acid Pooh.>Heh. Since you appear to be goading me into replying, and I'm a>sucker for this sort of thing, here goes: I was not aware thatMetamath was a website. Nor was I aware that they had a different>version of the axiom of choice. I assumed (obviously incorrectly)>that you meant metamath to refer to Set Theory, or some suchmetamathematical study. Now, in this context, my response doesn't>seem so inappropriate, does it? In particular, it shouldn't seem so>patronizing, since, as I parsed your question, it is a question a>beginner should ask. So, why decide to patronize me?> I was likewise unaware that there was a Metamath site, but that fact> became abundantly clear as soon as I read D. McAnally's very first> sentence of his first post on the subject. I suggest you go back and> read it.It's quite obvious that I'm aware of it now. Your suggestion isappreciated, but you seem to have suffered the same mistake as me. :)'cid 'ooh