mm-1 I have compiled a list of online literatures, mostly related tocomputing/mathematics/economics stuff, athttp://www.srcf.ucam.org/~dl276/bookmarks.phpComments and additions are welcome!Ry name is Erik Evenson.I invented this new type of sequence. It must be good for something. Maybethis group can come up with some ideas.I was labored with written calculations on and off for two months during mySpiritual Trip to India to come up with aworking prototype. I hereby name this type of sequence the EvensonSequence.The sequence is circular; after the end it repeats again from the beginning.The sequence is 96 digits long. The digits of the sequence is made up of 4different entities.For this sequence I use the digits 0,1,2,3. There are 24 of each of the 4digits.Since the 4 digits can be represented by binary maybe there can anencryption use.The Evenson Sequence:0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 32 1 2 0 1 3 1 2 0 2 3 0 1 0 2 03 1 2 3 2 0 1 0 3 2 1 3 1 0 3 02 3 1 3 0 1 2 1 3 2 0 3 0 1 3 23 0 2 1 2 3 1 0 1 2 0 3 2 3 1 21 0 2 1 3 0 3 2 0 2 1 0 1 3 0 2Observations:1) No digit is repeated in the sequence. No group of digits is repeated inthe sequence.2) All possible four-digit combinations (taking rule #1 in account) arerepresented in the forward reading of the sequence.3) All possible four-digit combinations (taking rule #1 in account) arerepresented in the BACKWARD reading of the sequence.Here is a real life example to explain this sequence: Imagine Buddhistmonks chanting a mantra based on the Evenson Sequence.Imagine the monks using four different syllables in the chant. If the monkschant over and over again the sequence using thosesyllables the listener could isolate 96 different four syllable words!Imagine, 96 four-syllable words represented by a 96 single-syllablesequence.I must admit when I experienced the Tibetan monks chanting at the BodhigayaTemple Complex in December of 1999 the ideaintrigued my mathematical mind and I suffered two-months of mind labor'nding the sequence.I hereby introduce my creation to the unwashed masses for critique. Pleasebe gentle. When mentioning my sequence inposts please use my Surname--two months of mental torture is worth at leastthat much. My only Evenson. => 0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 3> 2 1 2 0 1 3 1 2 0 2 3 0 1 0 2 0> 3 1 2 3 2 0 1 0 3 2 1 3 1 0 3 0> 2 3 1 3 0 1 2 1 3 2 0 3 0 1 3 2> 3 0 2 1 2 3 1 0 1 2 0 3 2 3 1 2> 1 0 2 1 3 0 3 2 0 2 1 0 1 3 0 2> Observations:> 1) No digit is repeated in the sequence. No group of digits is repeated in> the sequence.> 2) All possible four-digit combinations (taking rule #1 in account) are> represented in the forward reading of the sequence.> 3) All possible four-digit combinations (taking rule #1 in account) are> represented in the BACKWARD reading of the sequence. As another poster has mentioned, this is essentially a de Bruijn sequence (or de Bruijn cycle) with some forbidden substrings. Is your condition (1) saying that no group of digits appears immediately after the same (identical) group of digits? I notice that 0101 and 0202 do not appear in the sequence, while ï02Í does appear in the sequence 8 times (but never adjacent to another ï02Í.) In this case, you are looking at de Bruijn sequences with forbidden subsequences 00, 11, 22, 33, 0101, 0202, 0303, 1212, 1313, 2323, 123123, etc. In the theory of combinatorial words we usually represent a string such as 23100110013 in the following way: instead of ï00Í we write 0^2 (0 squared) and instead of 1(0^2)11(0^2)1 we write (1(0^2)1)^2 so that whole word is written 23(1(0^2)1)^2)3. Now, if we want a word that can not have any representation with powers of 2 or 3 or 4, etc, we call the word square-free. You might be interested in the Morse-Thue sequence (sometimes also called the Thue-Morse sequence), which can be used to generate an in'nite string over 3 symbols which is square-free. You are essentially looking for square-free de Bruijn sequences... applications are included in coding theory and possible connections to tandem repeats in genetics. (Of course, requiring the square-free condition will probably not makethese sequences formal ïde BruijnÍ sequences \ since they will probably losetheir alternate characterizations such as those related to de Bruijngraphs, but the concept seems to be suf'ciently well-de'ned.) I do not know of any previous study of square-free de Bruijn-like sequences.Jim Nastos => Evenson.> I invented this new type of sequence. It must be good for something. Maybe> this group can come up with some ideas.> I was labored with written calculations on and off for two months during my> Spiritual Trip to India to come up with a> working prototype. I hereby name this type of sequence the Evenson> Sequence.> The sequence is circular; after the end it repeats again from the beginning.> The sequence is 96 digits long. The digits of the sequence is made up of 4> different entities.> For this sequence I use the digits 0,1,2,3. There are 24 of each of the 4> digits.> Since the 4 digits can be represented by binary maybe there can an> encryption use.> The Evenson Sequence:> 0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 3> 2 1 2 0 1 3 1 2 0 2 3 0 1 0 2 0> 3 1 2 3 2 0 1 0 3 2 1 3 1 0 3 0> 2 3 1 3 0 1 2 1 3 2 0 3 0 1 3 2> 3 0 2 1 2 3 1 0 1 2 0 3 2 3 1 2> 1 0 2 1 3 0 3 2 0 2 1 0 1 3 0 2> Observations:> 1) No digit is repeated in the sequence. No group of digits is repeated in> the sequence.> 2) All possible four-digit combinations (taking rule #1 in account) are> represented in the forward reading of the sequence.> 3) All possible four-digit combinations (taking rule #1 in account) are> represented in the BACKWARD reading of the sequence.3) is a consequence of 1) and 2), right ?There seem to be many sequences that obey your constraints,even many that start with 0 1 2 3Is something else special about your sequence besides the constraints you told us about ?Bart Demoenps. it bears resemblence to the problem of 'nding a circular string of length 16,containing zeros and ones, and in which every number from 0 to 15 is representedexactly once in binary - vaguely from my algo class long ago => I invented this new type of sequence. It must be good for something. Maybe> this group can come up with some ideas.....> The sequence is circular; after the end it repeats again from the beginning.> The sequence is 96 digits long. The digits of the sequence is made up of 4> different entities.....> 1) No digit is repeated in the sequence. No group of digits is repeated in> the sequence.> 2) All possible four-digit combinations (taking rule #1 in account) are> represented in the forward reading of the sequence.> 3) All possible four-digit combinations (taking rule #1 in account) are> represented in the BACKWARD reading of the sequence.ThereÍs a concept of de Bruijn sequence which is like yoursbut without condition 1. You might 'nd this helpful in lookingfor prior art. (Constructing de Bruijn sequences is not hard;I donÍt know whether any of the constructions can be adaptedto produce Evenson sequences.) Condition 3 is unnecessarygiven conditions 1 and 2.-- Gareth McCaughan.sig under construc => For 1,2,3... omega, a sequence aj with limit a in S is an element> f = (a1,a2,... a) = (aj,a) of S^(omega+1)> with f(j) = aj for j < omega and f(omega) = a> A topology for S assures (the limits of) A subset S^(omega+1) when> for all (a1,a2,... a) in A, aj -> a in S> The indiscrete topology is the smallest topology for S assuring A.> If T is bunch of topologies that assure A, then sup T assures A.> if (aj,a) in A, a in U some subbase set of sup T, then> some tau in T with U in tau; aj -> a in tau> aj eventually in U> now use theorem> aj -> a iff for all subbase sets U containing a, aj eventually in U> Thus thereÍs a largest topology for S assuring A.> The sup of all topologies for S that assure A.> Sup here is the ïlargestÍ ( I think coarsest) topology whose open sets contain all the open sets of the set T? Assuming T is a set of course.> -- coinduction> Consider f:omega+1 -> S, f = (aj,a)> For (aj,a) in A with aj -> a in S, let U be open nhood a.> Then for f = (aj,a) in A, f^-1(U) is open.> case a in U: omega in f^-1(U); aj eventually in U> co'nite many j case a not in U: f^-1(U) subset omega thus open> So when S assures (aj,j), f is continuous.> Conversely if f = (aj,a) is continuous, then as> lim(j in N) j = omega in omega+1> lim aj = lim f(j) = f(omega) = a in S> So when f is continuous, S assures (aj,a).> Hence f continuous iff S assures f = (aj,a).> f is a point in the space, how is a point continuous?> Now the topology for S coinducted by all f in A, is> the largest topology making all f in A continuous.largest? wouldnÍt that always just be the discrete topology? IÍm takinglargest to mean ïwith the most open setsÍ. OughtnÍt we to be using 'nestand coarsest? => If T is bunch of topologies that assure A, then sup T assures A.> if (aj,a) in A, a in U some subbase set of sup T, then> some tau in T with U in tau; aj -> a in tau> aj eventually in U> now use theorem> aj -> a iff for all subbase sets U containing a, aj eventually in U>> Thus thereÍs a largest topology for S assuring A.> The sup of all topologies for S that assure A.>> Sup here is the ïlargestÍ ( I think coarsest) topology whose open sets> contain all the open sets of the set T? Assuming T is a set of course.>T_a subset T_b iff T_a coarser T_b iff T_a smaller T_bT_b contains T_a iff T_b 'ner T_a iff T_b larger T_aThe sup of T contains all the open sets of / T = Union TA topology is a subset of P(S)A collection of topologies for a set S is a subset of P(P(S)).This is adequate bound to assure a collection of topologies is a set.> f is a point in the space, how is a point continuous?>For example if f in R^R, the functions from R into R, thethe point f in the function space R^R can be continuous.Such as f(x) = x for all x in R, viz f = { (x,x) | x in R }> Now the topology for S coinducted by all f in A, is> the largest topology making all f in A continuous.>> largest? wouldnÍt that always just be the discrete topology? IÍm taking> largest to mean ïwith the most open setsÍ. OughtnÍt we to be using 'nest> and coarsest?>The largest topology for any set is the discrete topology. When yourequire a property of a topology, then the largest topology with thatproperty likely wonÍt be the discrete topology.I like larger and smaller, you may use 'ner or coarser if you like. =Let me try to pose the problem, I am trying to solve.I have a matrix, A (for simplicity let us assume it is a 2 X 2 matrix), thatis supposed to be symmetric and positive de'nite. There is a differentialequation the numerical solution of which gives me A at a current instant, inthe form Adot = f(x,p)..where x and p are the states of this dynamicalsystem and f is a nonlinear function in x, p. Since A has only three uniqueparameters for this case, I actually obtain, theta_dot = g(x,p), where thetaA(t) thereby enforcing symmetry..... To enforce positive de'niteness, Ineed to do the following,theta(1) > 0, theta(2) > 0 and -sqrt(theta(1)*theta(2)) < theta(3) Kerry,> I wanted to read your work on Goldbach conjecture,as it relies on a> simple theorem (WilsonÍs); but the fact it is not in standard notation> makes it very dif'cult to follow; and unfortunately many will not> bother to put in the effort. I donÍt think you have to use LaTex or> Tex. I believe you might be able to get the standard symbols you need> by using the symbols feature in microsoft word. If you re-type your> paper in MS word XP, the mathpreprints server will convert it into pdf> for you with no problem. In the word document go to insert then> select symbol and hopefully you will 'nd all the standard notation> you need there, like NE, etc.> Best wishes. viewed on www.mathpreprints.com where it has been converted to pdf. => In this direction, it could be useful to be able to distinguish> functions depending on their originating library so that, maybe, one can > choose different colors for different libraries or one can choose which> library to highlight. It is like this, for example, in maple syntax> 'le. In MuPAD, functions that stem from a certain library are usuallyaddressed with this library name, as in numeric::realroots. Sure,there is export, but are you going to have the vim syntax 'lerecognize export-commands and react to them? I for one would 'nd aeven if no export(numeric) or export(numeric, realroots) is seenanywhere to be almost unusable.-- +--+ +--+| |+-|+ Christopher Creutzig (ccr@mupad.de) Christopher Creutzig ha scritto:>> In this direction, it could be useful to be able to distinguish>> functions depending on their originating library so that, maybe, one can >> choose different colors for different libraries or one can choose which>> library to highlight. It is like this, for example, in maple syntax>> 'le.> In MuPAD, functions that stem from a certain library are usually> addressed with this library name, as in numeric::realroots. Sure,> there is export, but are you going to have the vim syntax 'le> recognize export-commands and react to them? I for one would 'nd a> even if no export(numeric) or export(numeric, realroots) is seen> anywhere to be almost unusable.> Well, checking if export(numeric) has been declared before usingrealroots I think is too complicated (at least for me... ;-) and, maybe, would also slow down typing too much. Moreover, syntax highlight can be fooled even when it is written at the state of art, which is notmy case... :-((I consider it simply like a tool which helps you to prevent most ofbanal typesetting errors. For example, I am not able (yet...) to recognize if a := declarationdoesnÍt end on the same line...Regarding libraries, I just inserted Type in the following way:syn match muType (::)=this way AlgebraicConstat is recognized with or without the trailing Type:: To have it mandatory, it is enough to usesyn match muType Which one is better? I thought the 'rst. You, it seems to me, thesecond, donÍt you? Others? Do you 'nd any other problem in thesyntax 'le as it is now? I would like to improve it as much aspossible, but it is my 'rst syntax 'le... So I am quite surprised itbehaves not too badly... ;-)ByeFabio =>> So the strategy I am using is to implement a user friendly syntax for>> metamathematics plus axiomatic set theory. But I doubt it will hold for>> any serious proof veri'cation system. So on top of that, I may add a>> system for user de'ned syntax. Then this may in fact be a simpli'cation>> for further implementations.>>A very ambitious project. Good luck!It is not as ambitious as it may sound, as I decided to not work withoptimizations. But I have implemented things like uni'cation branching inorder to achieve an proof-engine that is as general as possible. This wayI do not need Gentzen sequents, it seems.One idea that comes to my mind is that you might look at using thesemantic trees to write an interface on top of a program like Qu-Prolog.This might saving you some time. :-) Hans Aberg =87,160 formulas and 10,828 graphics about mathematical functionsare now available free at The Wolfram Functions Site, animportant new resource for mathematicians, scientists, engineers,and students at:http://functions.wolfram.comIn the applications of mathematics to science and engineering--aswell as in pure mathematics itself--there are several hundred so--called special functions that have been intensively used for acentury or more. These special functions--with names like Besselfunctions, hypergeometric functions, and totient functions--de'ne focal points of mathematical knowledge. The WolframFunctions Site provides in a readily accessible way the largestcollection, by far, of such knowledge ever assembled.Several widely used handbooks of mathematical functions have beenpublished, the largest of which contained about 15,000 formulas,meticulously compiled from thousands of technical papers.Traditional handbooks have also included only handfuls ofgraphics illustrating the properties of functions. WithMathematica, a huge number of new visualizations of functionshave become possible. The Wolfram Functions Site assembles over10,000 of these, with many more being planned.A major tour de force of reference website construction, TheWolfram Functions Site contains over 30 gigabytes of data.Material in The Wolfram Functions Site can be downloaded inseveral standard formats, including Mathematica InputForm andStandardForm, MathML, and PDF. Formulas can be copied from thesite and immediately used as input to a computer system. For easeof citation, each formula has been assigned a unique permanent ID.While having already far surpassed previous knowledge bases formathematical functions, continued growth is projected for TheWolfram Functions Site, with new searching capabilities, externalcontributions, and new classes of graphics and information. Visitthe site at:http://functions.wolfram.com =I have compiled a list of online literatures, mostly related tocomputing/mathematics/economics stuff, athttp://www.srcf.ucam.org/~dl276/bookmarks.phpComments and additions are welcome!Ry =>> I need some help with Magma. I have de'ned an algebraic structure> (e.g a 'nite 'eld) and want to declare an element of this structure,> which should be then treated as a symbol.>> For example, I want to declare a as an element of F = GF(5), so that> x+a is in F[x].If you want ïaÍ to be a particular value in the 'eld, use a variable: > F := GF(5); > a := F!2; > a*(a + 1); 1If you want to manipulate expressions in ïaÍ instead, then you arereally treating ïaÍ as the transcendental in a polynomial ring (orfunction 'eld, power series ring, etc.). In this case, you shouldcreate the corresponding structure with ïaÍ as the namedtranscendental (polynomial rings used in these examples): > P := PolynomialRing(F); > a*(a + 1); a^2 + aIf you want to use two (or more) variables, you can either stackthe polynomial rings: > Px := PolynomialRing(P); > (x + a)*(x + 1); x^2 + (a + 1)*x + a;or use a multivariate polynomial ring: > Pxa := PolynomialRing(F, 2); > (x + a)*(x + 1); x^2 + x*a + x + aTo instantiate the variables, use one of the appropriate versions ofthe Evaluate() intrinsic. e.g.: > f := (x + a)*(x + 1); > Evaluate(f, a, 3); x^2 + 4*x + 3Geoff.-- =Hoops 1-3 developed Hoop division algebras with r-fold negationcollapsed from Moufang loops with r-fold symmetry. These conserveone or more size - the conjugates of the factors of themultiplication table determinant. They all have multiplicativeinverses that split (in the non-degenerate cases) into partialfractions; each size is a numerator.I now generalize subtraction to additive elimination.(4a) r-fold Negation, Additive elimination. A loop with m unsigned elements and r-fold symmetry may becollapsed by r-fold negation to an algebra with m/r signed elementsthat multiplies vecs (vectors in some algebras) of m/r signedelements. The symmetry ensures that this is an integer. Thisequivalences r unsigned coef'cients to one signed coef'cient Every loop is a quasi-group; adding all the elements into one row (orcolumn) shows that the sum of elements is a factor of the determinant;hence every loop conserves the sum of its elements. r-fold negationalways equivalences this term to zero.(4b) Real & Complex negation. Ordinary (r=2) negation with Reals is the equivalence relation {a,b}~{c,d} iff a*d=b*c. This can be expressed as {a,b}~ R and {b,a}~ -R; itequivalences {a,a} to 0; ordinary subtraction of R= {a,b} is theaddition of -R= {b,a}. Complex numbers C= {R,I} (components R= real and I= imaginary) are apair of r=2 equivalence relations on four unsigned components{a,b,c,d} with {a,c}~ R, {b,d}~ I. The negation of C is {c,d,a,b}~ -C( a left or right rotation by 2 places) and the complex conjugate is{a,d,c,b}~ C*. Real and complex vectors are Vr= {R1, R2, ...} and Vc={C1, C2, ...} with each element (R1, C1, etc.) having an ordinarynegation (-R1, -C1, etc.).(4c) Quaternions etc. Quaternions (and many other algebras) are similarly negated byhalf-length rotation, V(i)~ {a(i), a(i+m/2)}; quaternion conjugatesnegate their imaginary elements V(2}, V(3) & V(4) but not theirscalar element V(1).(4d) Terplex negation by rotation. {a,b,c}~{d,e,f} iff a*d=b*e=c*f (written {a,b,c}~T) is three-fold(Terplex) equivalence relation; r=3. It gives two negations,{c,a,b}~ Tr and {b,c,a}~ Tr, left & right rotation by one position;T+Tl+Tr= {a+b+c, a+b+c, a+b+c} ~0. The conjugate is T*= {a,c,b}.Elements in terplex vecs Vt={T1,T2, ...} can be reduced to zero byadding their left & right negations. Here, the sum of two negations isthe additive inverse.(4e) Additive elimination. General Vec elements become zero on adding the un-equivalenced numbersets that make all the terms the same; {a,b}+{b,a} ={a+b,a+b}~0 whenr=2, {a,b,c}+{b+c, a+c, a+b}={a+b+c, a+b+c, a+b+c} ~0 when r=3, etc..This additive elimination generalizes subtraction; the negation of asigned set of numbers is the signed set that, when added to that set,makes each element equivalence to zero.The next posting will be concerned with the effect of sizes becomingzero, and projecting vectors into sub-symmetric sub-algebras. Roger Beresford.Oh, the little more,and how much it is!Oh, the little less, and what a world away!(Robert Browning.) (Read Oh as nought?)9122 => Let H(n) = sum{k=1 to n} 1/k, > the n_th harmonic number.Prove, for each m = positive integer,sum{k=1 to m-1} H(k) k! (m-k)!is congruent to(1/2) m! H(§oor(m/2))) (-1)^m (mod (m+1)) .> (I do not know if the result is trivial {or am certain it is> correct}.)> We can rewrite the above so it becomes, perhaps to some reading this,more simply stated and more natural.(Take sum from 1 to m, instead of from 1 to m-1)sum{k=1 to m} H(k) k! (m-k)!is congruent tom! *h(m) (mod(m+1)),where h(m) = sum{k=0 to §oor((m-1)/2)} 1/(m -2k) .Now, h(m) =sum{k=0 to §oor((m-1)/2)} 1/(m -2k)is interesting in itself, for it is (analogously todouble-factorials) the sum of the reciprocal of EVERY-OTHER positiveinteger <=m(even or odd, depending on m).Neither sequence:m!*h(m) norm!!*h(m)norsum{k=1 to m} H(k) k! (m-k)!seems to be in the EIS either, if I calculated correctly by hand.(EIS: http://www.research.att.com/~njas/sequences/index.html#L) Leroy Quet> By the way, for those interested in harmonic-number math puzzles> (limits, this time, instead of congruences), here is a link to a> recent sci.math math-puzzle no one has yet answered.> (I am including the link because I am cross-posting this to> rec.puzzles, as well as to sci.math.)> UTF-8&group=sci.math&safe=off&selm=b4be2fdf Quet =>I have problem that I canÍt seem to work out. Say there are 136,000>cows in my country. On average 10 of them die per year. Let me get this straight; your average cow lives over ten thousand years? = > | I am more interested right now in 'nding an explicit > |computation of w1 = gcd((1 + sqrt(-167))/2, 7), i.e., in 'nding > |a monic polynomial with constant term 7^k of which w1 is a root. > Ok. :-)BTW, are you using Magma for your calculations? > It appears that Q(sqrt(-167)) has a class group of order 11.Now it 'gures that we need 11-th powers in this case. > Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer, > in spite of the 2 in the denominator). > r^11 = (-592764018-86559857*r) > = (44555-222*r) (-12882-2017*r)I thought a bit further and found: r^11 = (44444-111.sqrt(-167))(298-23.sqrt(-167))[(-3-7.sqrt(-167))/2 ]and 7^11 = (44444-111.sqrt(-167))(44444+111.sqrt(-167)) 3^11 = (298-23.sqrt(-167))(298+23.sqrt(-167)) 2^11 = [(-3-7.sqrt(-167))/2][(-3+7.sqrt(-167))/2]recon'rming that indeed each prime integer factor of 42 is distributedamongst the two roots.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ => | I am more interested right now in 'nding an explicit > |computation of w1 = gcd((1 + sqrt(-167))/2, 7), i.e., in 'nding> |a monic polynomial with constant term 7^k of which w1 is a root.> Ok. :-)BTW, are you using Magma for your calculations? > It appears that Q(sqrt(-167)) has a class group of order 11.Now it 'gures that we need 11-th powers in this case. > Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer,> in spite of the 2 in the denominator). r^11 = \ (-592764018-86559857*r)> = (44555-222*r) (-12882-2017*r)I thought a bit further and found:> r^11 = (44444-111.sqrt(-167))(298-23.sqrt(-167))[(-3-7.sqrt(-167))/2 ]> and> 7^11 = (44444-111.sqrt(-167))(44444+111.sqrt(-167))> 3^11 = (298-23.sqrt(-167))(298+23.sqrt(-167))> 2^11 = [(-3-7.sqrt(-167))/2][(-3+7.sqrt(-167))/2]> recon'rming that indeed each prime integer factor of 42 is distributed> amongst the two roots.>ItÍs quite a bit easier to attack the x = -3 case. Here we have aclass group (of Q(sqrt(-38)) of order 3, with P(-3) = 7(137)and the roots of interest satisfying a^2 + 4a + 42 = 0This one you can do by hand.Rick following sum:> S(n) = sum {k=1,...,n} n!/(k*(n-k)!)> Need closed form solution or asymptotic estimations for large n.At 1st I thought that k in the denominator was a k!, leading me tosuspect that the replies to this thread, especially those regardingMathematica, were being highly sarcastic...(because, then S(n) would have been simply 2^n -1.) ;)I had to share this in-case others have misread the =>> Littlemanwearingbigboypants misstates yet again:>> Every presidential election year has been a leap year, exceptfor http://scienceworld.wolfram.com/astronomy/LeapYear.html>> Gardyloo.>> When you throw the bucket straight up, move.>> This site supports Bill Vajk, supposing that he is correct that there> have been presidential elections in 1789, 1800 & 1900, none of which> were leap years.Uncle Al gets uncharacteristically quiet when heÍs caught sitting in hisown poop, doesnÍt he. DonÍt despair, Al, Jesus loves you.Just adds weight to the argument that Richard Schultz is right. I guessthe old Alster is largely irrelevant at this point. Most people donÍtbother to point out his howlers anymore. Lord knows it would be a fulltime job.And now for the obligatory ad hominem: Uncle AlÍs wife is ugly and hisdog has bad breath -- or is it the other way ïround? => Littlemanwearingbigboypants misstates yet again:>> Every presidential election year has been a leap year, except> for http://scienceworld.wolfram.com/astronomy/LeapYear.html>> Gardyloo.>> When you throw the bucket straight up, move.>> This site supports Bill Vajk, supposing that he is correct that there> have been presidential elections in 1789, 1800 & 1900, none of which> were leap years.>> Uncle Al gets uncharacteristically quiet when heÍs caught sitting in his> own poop, doesnÍt he. DonÍt despair, Al, Jesus \ loves you.>> Just adds weight to the argument that Richard Schultz is right. I guess> the old Alster is largely irrelevant at this point. Most people donÍt> bother to point out his howlers anymore. Lord knows it would be a full> time job.>> And now for the obligatory ad hominem: Uncle AlÍs wife is ugly and his> dog has bad breath -- or is it the other way ïround?Hey twit. Learn to read. Al acknowledged his error... will you? =>> Littlemanwearingbigboypants misstates yet again:>> Every presidential election year has been a leap year,except> for one.>> Bzzzzzzt. Wrong! There have been http://scienceworld.wolfram.com/astronomy/LeapYear.html>> Gardyloo.>> When you throw the bucket straight up, move.>> This site supports Bill Vajk, supposing that he is correct thatthere> have been presidential elections in 1789, 1800 & 1900, none ofwhich> were leap years.>> Uncle Al gets uncharacteristically quiet when heÍs caught sitting inhis> own poop, doesnÍt he. DonÍt despair, Al, Jesus \ loves you.>> Just adds weight to the argument that Richard Schultz is right. Iguess> the old Alster is largely irrelevant at this point. Most peopledonÍt> bother to point out his howlers anymore. Lord knows it would be afull> time job.>> And now for the obligatory ad hominem: Uncle AlÍs wife is ugly andhis> dog has bad breath -- or is it the other way ïround?>> Hey twit. Learn to read. Al acknowledged his error... will you?>Hey! toadeater, letÍs hear him say in was the meaning of his post. Apparently, othersagree with me. => Littlemanwearingbigboypants misstates yet again:>> Every presidential election year has been a leap year,> except> for one.>> Bzzzzzzt. Wrong! There have been http://scienceworld.wolfram.com/astronomy/LeapYear.html>> Gardyloo.>> When you throw the bucket straight up, move.>> This site supports Bill Vajk, supposing that he is correct that> there> have been presidential elections in 1789, 1800 & 1900, none of> which> were leap years.>> Uncle Al gets uncharacteristically quiet when heÍs caught sitting in> his> own poop, doesnÍt he. DonÍt despair, Al, Jesus \ loves you.>> Just adds weight to the argument that Richard Schultz is right. I> guess> the old Alster is largely irrelevant at this point. Most people> donÍt> bother to point out his howlers anymore. Lord knows it would be a> full> time job.>> And now for the obligatory ad hominem: Uncle AlÍs wife is ugly and> his> dog has bad breath -- or is it the other way ïround?>> Hey twit. Learn to read. Al acknowledged his error... will you?> Hey! toadeater, letÍs hear him say in plain English I, Uncle Al, MADE A> MISTAKE.Why should he pander to your ignorance? It was plain what he meant. Nextthime he makes and error he might say mea culpa. Just a hint for you, thatmeans roughly my fault.> I donÍt think that was the meaning of his post. Apparently, others> agree with me.Nowhere in your ignorant spew did I see you acknowledging your own error.Try again. =>> Littlemanwearingbigboypants misstates yet again:>> Every presidential election year has been a leap year, except> for one.>> Bzzzzzzt. Wrong! There have been http://scienceworld.wolfram.com/astronomy/LeapYear.html>> Gardyloo.>> When you throw the bucket straight up, move.>> This site supports Bill Vajk, supposing that he is correct that there> have been presidential elections in 1789, 1800 & 1900, none of which> were leap years.Uncle Al gets uncharacteristically quiet when heÍs caught sitting in his> own poop, doesnÍt he. DonÍt despair, Al, Jesus \ loves you.Just adds weight to the argument that Richard Schultz is right. I guess> the old Alster is largely irrelevant at this point. Most people donÍt> bother to point out his howlers anymore. Lord knows it would be a full> time job.And now for the obligatory ad hominem: Uncle AlÍs wife is ugly and his> dog has bad breath -- or is it the other way ïround?Hey stooopid - I posted the scholarly link and recanted. You are toofucking stooopid to recognize that. You are stooopid and evil. Youdo not wish to learn, you do not even wish to ridicule. You merelywant to destroy with a broad brush, imagining taht it elevates you.Fuck off. Uncle Al was born with ankle-borne duck protectors.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! =Littlemanwearingbigboypants snivels and pouts:>Littlemanwearingbigboypants misstates yet again:>>Every presidential election year has been a leap year, except>>for one.>Bzzzzzzt. Wrong! There have been /LeapYear.html>>Gardyloo.>When you throw the bucket straight up, move.>This site supports Bill Vajk, supposing that he is correct that there>have been presidential elections in 1789, 1800 & 1900, none of which>were leap years.>>Uncle Al gets uncharacteristically quiet when heÍs caught sitting in his>>own poop, doesnÍt he. DonÍt despair, Al, Jesus loves you.Invarriably.>>Just adds weight to the argument that Richard Schultz is right. I guess>>the old Alster is largely irrelevant at this point. Most people donÍt>>bother to point out his howlers anymore. Lord knows it would be a full>>time job.>>And now for the obligatory ad hominem: Uncle AlÍs wife is ugly and his>>dog has bad breath -- or is it the other way ïround?> Hey stooopid - I posted the scholarly link and recanted. You are too> fucking stooopid to recognize that. You are stooopid and evil. You> do not wish to learn, you do not even wish to ridicule. You merely> want to destroy with a broad brush, imagining taht it elevates you.> Fuck off. Uncle Al was born with ankle-borne duck protectors.Why would anyone suppose you might need them in the 'rst place?When someone repeatedly gets very elementary stuff wrong as you do,Schwartz, thereÍs no incentive to look in depth at the more complexofferings they present.Nils already posted a litany of Schwartz nonsense, so the fact thereare additional mountains of similar stuff. => Littlemanwearingbigboypants misstates yet again:>> Every presidential election year has been a leap year,except> for one.>> Bzzzzzzt. Wrong! There have been http://scienceworld.wolfram.com/astronomy/LeapYear.html>> Gardyloo.>> When you throw the bucket straight up, move.>> This site supports Bill Vajk, supposing that he is correct thatthere> have been presidential elections in 1789, 1800 & 1900, none ofwhich> were leap years.>> Uncle Al gets uncharacteristically quiet when heÍs caught sitting inhis> own poop, doesnÍt he. DonÍt despair, Al, Jesus \ loves you.>> Just adds weight to the argument that Richard Schultz is right. Iguess> the old Alster is largely irrelevant at this point. Most peopledonÍt> bother to point out his howlers anymore. Lord knows it would be afull> time job.>> And now for the obligatory ad hominem: Uncle AlÍs wife is ugly andhis> dog has bad breath -- or is it the other way ïround?>> Hey stooopid - I posted the scholarly link and recanted. You are too> fucking stooopid to recognize that. You are stooopid and evil. You> do not wish to learn, you do not even wish to ridicule. You merely> want to destroy with a broad brush, imagining taht it elevates you.>> Fuck off. Uncle Al was born with ankle-borne duck protectors.>Uh..... Al, I think you forgot to click the send key on yourrecantation. IsnÍt on my newsserver. IsnÍt on Google. Mebbe \ you justÍmagined it, huh?Without intending any slight to Edward Green, ^5s to Richard Schultz forexposing Uncle AlÍs nakedness to the world.HereÍs an enlightening link for ya, Al.http://tinyurl.com/ywx34DonÍt forget to click for the music. :-)And now for the obligatory ad hominem: Al, compared to you, your wifethinks IÍm pretty good. => Uh..... Al, I think you forgot to click the send key on your> recantation. IsnÍt on my newsserver. IsnÍt on Google. Mebbe \ you just> ïmagined it, huh?Fair is fair, this time he actually did.http://tinyurl.com/3hddqOf course having a reasonable vocabulary helpsto understand his intent. > And now for the obligatory ad hominem: Al, compared to you, your wife > thinks IÍm pretty good.Did you raise your standards or or did you have to lower themin order to make this statement? EMWTK.Ack, no matter, youÍve arrived. =>> Uh..... Al, I think you forgot to click the send key on your> recantation. IsnÍt on my newsserver. IsnÍt on Google. Mebbe \ youjust> ïmagined it, huh?>> Fair is fair, this time he actually did.>> http://tinyurl.com/3hddq>> Of course having a reasonable vocabulary helps> to understand his intent.Whether this is the fabled recantation or merely a misguided attempt torationalize his mistake is not entirely clear.I lean toward the latter. It didnÍt look like a self-directeddefenestration to me.Uncle Al can easily remove any doubt by saying This was my RECANTATIONof my earlier ERRONEOUS post. I apologize to the Usenetizens forinconveniencing them and wasting their time.I bet he is constitutionally incapable of saying it. He left sci.chemwith his tail between his legs rather than admit that Richard Schultzwas right. => Uh..... Al, I think you forgot to click the send key on your> recantation. IsnÍt on my newsserver. IsnÍt on Google. Mebbe you> just> ïmagined it, huh?>> Fair is fair, this time he actually did.>> http://tinyurl.com/3hddq>> Of course having a reasonable vocabulary helps> to understand his intent.>> Whether this is the fabled recantation or merely a misguided attempt to> rationalize his mistake is not entirely clear.>> I lean toward the latter. It didnÍt look like a self-directed> defenestration to me.>> Uncle Al can easily remove any doubt by saying This was my RECANTATION> of my earlier ERRONEOUS post. I apologize to the Usenetizens for> inconveniencing them and wasting their time.Did you learn a new word? If so I take it that you think learning is awaste of time. Any time you spent arguing and making a fool of yourself yougave freely, and so is your fault. =>Uh..... Al, I think you forgot to click the send key on your>recantation. IsnÍt on my newsserver. IsnÍt on Google. Mebbe you> just>Ímagined it, huh?>>Fair is fair, this time he actually did.>>http://tinyurl.com/3hddq>>Of course having a reasonable vocabulary helps>>to understand his intent.> Whether this is the fabled recantation or merely a misguided attempt to> rationalize his mistake is not entirely clear.> I lean toward the latter. It didnÍt look like a self-directed> defenestration to me.> Uncle Al can easily remove any doubt by saying This was my RECANTATION> of my earlier ERRONEOUS post. I apologize to the Usenetizens for> inconveniencing them and wasting their time.> I bet he is constitutionally incapable of saying it. He left sci.chem> with his tail between his legs rather than admit that Richard Schultz> was right.Agreed, with the understanding that normalcy and reasonability are toomuch to expect from Schwartz. => So now IÍve also decided to stop moderating the newsgroup>Shit. [emoticon doffs its hat in honor of work done and submits>a fervent thank you]. =>> So now IÍve also decided to stop moderating the newsgroup>Shit. [emoticon doffs its hat in honor of work done and submits>>a fervent thank you].>I understand that. I just have some regrets about a road nottaken some time ago but always had plans to try another time.The tutorial that developed from Oz, you, and /erg yakkingin the newsgroups is a classic. IÍd always hoped that morewould develop. But it does take a special group synergyand lots of coordinated time of three or more people. ThatÍshard to do./BAH => The last bit where you say glue af'ne schemes obtained in the above> method will work only if the kernel of O_Y ---> f_*(O_X) is> quasicoherent, which would follow from f_*(O_X) being quasicoherent.> Closed subschemes are de'ned by quasicoherent sheaves of ideals.As perhaps Mr. Ramsay guessed, I was/am trying to solveHartshorneÍs exercises and the problem appeared in chapter II,section 3, whereas the notion of quasi-coherent sheaves wasintroduced in section 5. Judging from your proof, it seemsto me that the problem is almost impossible to solve withoutthe notion of quasi-coherent sheaves.Nobuo =>the URL didnÍt work. I like it, though,>when other people think that Fermat had a Method;>he just didnÍt reveal all of it!> perhaps it will ultimately be shown taht>the last theorem was actually one of his earliest ones;>eh?> You may view my proofs at the Florida State web page:>> www.math.fsu/Science/Specialized under Goldbach Proof and In>> Evans.Once youÍve seen enough links to Florida State University, the properURL is pretty easy: www.math.fsu.edu/Science/SpecializedRob Johnson take out the trash before replying =>the URL didnÍt work. I like it, though,>when other people think that Fermat had a Method;>he just didnÍt reveal all of it!> perhaps it will ultimately be shown taht>the last theorem was actually one of his earliest ones;>eh?> You may view my proofs at the Florida State web page:>> www.math.fsu/Science/Specialized under Goldbach Proof and In>> Evans.Once youÍve seen enough links to Florida State University, the proper> URL is pretty easy: www.math.fsu.edu/Science/SpecializedRob Johnson > take out the trash before replyingYouÍre right. I apologize for any inconvenience. Kerry = > the URL didnÍt work. I like it, though, > when other people think that Fermat had a Method; > he just didnÍt reveal all of it! > perhaps it will ultimately be shown taht > the last theorem was actually one of his earliest ones; > eh?The url indeed does not work. And indeed his last theorems was oneof his earliest ones. He had a later theorem that x^4 + y^4 = z^4had no solutions in the integers, and has given a proof of this (aboutthe only proof he has actually given himself).Are there actually theorems by Fermat (other than the one above) thathe has (provably) proven himself? For instance, Fermat has not showna proof of his little theorem, the 'rst published proof was by Euler.The more I read about him, the more I think of him as a man with goodideas about structure, but no good ideas about proving things.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =[about Fermat]:>And indeed his last theorems was on of his earliest ones. How do you (we) know? I thought his conjecture was found in his book byhis son after his (pereÍs) death. Most of us donÍt date our marginalscribblings. If it is known when he obtained the book, that puts alower bound on the date of the ideaÍs conception; I donÍt see aneasy way to obtain an upper bound.>He had a later theorem that x^4 + y^4 = z^4>had no solutions in the integers, and has given a proof of this (about>the only proof he has actually given himself).So I am told. But note that it was also (IÍm told) common practice toto pose problems for others which were special cases of things you(the poser) already knew. So the fact that he provided an answer tothis problem in year N does not, by itself, rule out the possibilityhe (thought he) had a general proof of FLT in year NÍ < N.>Are there actually theorems by Fermat (other than the one above) that>he has (provably) proven himself? For instance, Fermat has not shown>a proof of his little theorem, the 'rst published proof was by Euler.>The more I read about him, the more I think of him as a man with good>ideas about structure, but no good ideas about proving things.Sadly for M. Fermat, he lived before the days of sci.math andsamizdat web publishing. If he had a good proof of a theorem, what wouldhe do with it? The conventions of the day seemed to call for publishinga book, writing letters to peers, or making notations in a diary. There were at the time no scholarly journals (probably no reputable periodicalsof any kind) and no scienti'c societies yet IIRC, and he had no positionat a university. And indeed, why publicize a proof at all? Conceivably,to have proved a theorem may well have been considered a privatepleasure, akin to breeding a new kind of §ower: not the sort of thingone would keep secret, exactly, but also not necessarily worthpreserving for posterity. (I am speaking conjecturally; I am not ahistorian.)dave =>Message-id: [about Fermat]:>>And indeed his last theorems was on of his earliest ones. >>How do you (we) know? I thought his conjecture was found in his book by>his son after his (pereÍs) death. Most of us donÍt date our marginal>scribblings. If it is known when he obtained the book, that puts a>lower bound on the date of the ideaÍs conception; I donÍt see an>easy way to obtain an upper z^4>>had no solutions in the integers, and has given a proof of this (about>>the only proof he has actually given himself).>>So I am told. But note that it was also (IÍm told) common practice to>to pose problems for others which were special cases of things you>(the poser) already knew. So the fact that he provided an answer to>this problem in year N does not, by itself, rule out the possibility>he (thought he) had a general proof of FLT in year NÍ < N.>Are there actually theorems by Fermat (other than the one above) that>>he has (provably) proven himself? For instance, Fermat has not shown>>a proof of his little theorem, the 'rst published proof was by Euler.>>The more I read about him, the more I think of him as a man with good>>ideas about structure, but no good ideas about proving things.>>Sadly for M. Fermat, he lived before the days of sci.math and>samizdat web publishing. If he had a good proof of a theorem, what would>he do with it? The conventions of the day seemed to call for publishing>a book, writing letters to peers, or making notations in a diary. There >were at the time no scholarly journals (probably no reputable periodicals>of any kind) and no scienti'c societies yet IIRC, and he had no position>at a university. And indeed, why publicize a proof at all? Conceivably,>to have proved a theorem may well have been considered a private>pleasure, akin to breeding a new kind of §ower: not the sort of thing>one would keep secret, exactly, but also not necessarily worth>preserving for posterity. (I am speaking conjecturally; I am not a>historian.)>>dave--MensanatorAce of Clubs =Jose Carlos Santos> Larry Hammick > I would like to know where can I 'nd a proof of the following> statement: if M is a real analytic manifold and v is an analytic> vector 'eld on M, then the §ow associated with M is also> analytic.> Correcting another slip of mine, the reference> Dieudonn.8e, _Treatise on Analysis_, 18.2.12> is correct, but thatÍs in volume 4, not volume 2. Nine volumes in all,and> not exactly light reading.>up> in Dieudonn.8eÍs treatise, but in the third volume instead of the fourthone.IÍm glad IÍm not the only one who makes this kind of mistake :) The bookswere missorted on my bookshelf, simply because I consult them so often. Welost a great scholar when we lost Dieudonn.8e. => Science = it is reality.And objective proposition. One can check.> Philosophy = I am reality.A subjective illusion. Right up there with I am Napoleon. Bishop > Berkeley would have approved.Bob KolkerMaybe Bob Kolker or someone can suggest an experiment, that would de'nitely prove that no sentient objects exist, and or if they do exist, that they are just observers, and have no affect on causes and effects.No doubt, above a certain pleasure/pain threshhold, I am just a reacting object, but sometimes, I think that a little sentitence comes into play, and I am a little butter§y, that causes a few ripples in space/time, that would not otherwise exist.--Tom Potter => IÍve noticed lately that the de'ntion of science I picked up from> Karl Popper (that it is a critical activity including criticism by> experimentation where possible) must be drastically different than the> de'ntion of science assumed by some of more visible posters here.How about a little poll on this.In one or two paragraphs in your own words, describe what science> means to you, and if possible from whom your de'ntion is in§uenced> by.Science is the branch of philosophy dealing with attempting todetermine truth via experimental means,...with the ultimate goal of achieving some kind of (imprecise)certainty,...but 'nally discovering that the Whole Mess is based upon UNCERTAINTYin the end...Science: the branch of art/religion/mathematics(/politics) dealingwith that seen *outside* what is known to our own inner-eyes, even ifour inner-eyes are what actually have been seeing it all, and perhapswith a tad bit of hallucination yet => Ref: http://www.astro.ucla.edu/~wright/cosmolog.htm#News> Variable Constants?> Something that is variable is NOT a constant. Are you referring to> things that we thought to be constant but in truth are variable?> Variable Constant is an oxymoron, like military intelligence or > business ethics.> Bob Kolker> [Sam]> What we know is through measurement... the 'ne structure constant > alpha has been measured to -0.6+/-0.6 parts per million for the> change which is consistent with zero... and therefore constant.>[hanson]It may even be desirable, by consensus &/or arbitration, to *set* alphaor one of such dimensionless constants as *constant* against whichall other fundamental and derived physical constants can be measured.There seems to be a current drive toward that notion in metrology with a wide range of projects going on to determine the size of NewtonÍs G, and the mole with its AvogadroÍs Number N_A.All physical constants must be expressible formally thru the other ones,interchangeably and accurately, or the entire structure of physics willfall like a house of cards.I think the epidemy would of course be to 'nd a physically meaningfuldimensionless number which expresses the notion of the only constant is the (change of the change)x of the change...lim ooand not becoming strictly zero, nor falling into the categories of Nothing remains constant except change itself, nor NeillÍs the only constant is change .Whether any of this is possible is another issue altogether for we enterhere into the realm and domain of absolute vs relative.hanson =I have this question here that I attempted and failed. some pointer towardthis problem is greatly appreciated.The golden ratio is the ratio b:a so that b:a = (a+b):b. The ratio of thegolden ratio was considered by the ancient Greeks to be the most perfectratio for the lengths of the sides of rectangles, such as portraits. Showthat if b:a is the golden ratio, then b/a = (1+ sqrt(5))/2.I guess the question is really saying to derive b = (1+sqrt(5)), and a = 2from b/a = (a+b)/b (*).So I tried to manipulate with the algebra of (*), and failed.Is there anything i am doing wrong? =If a/b=b/(a+b) then a^2+ab=a(a+b)=b^2 and a^2+ab-b^2=0, so that(a/b)^2+(a/b)-1=(a^2+ab-b^2)/b^2=0 and a/b=1/2[(-1)+sqrt(5)] by the quadraticformula, where sqrt(5) is either the positive root or the negative one. Sincea,b are both positive, it is of course the positive square root. Henceb/a = 2 / (sqrt(5)-1) = 2 * (sqrt(5)+1) / [ (sqrt(5)-1) * (sqrt(5)-1) ] = 2 * (sqrt(5)+1) / 4 = (sqrt(5)+1) / 2.Take time to recall that a/b=c/d doesnÍt imply a=c and b=d.|| I have this question here that I attempted and failed. some pointer toward| this problem is greatly appreciated.|| The golden ratio is the ratio b:a so that b:a = (a+b):b. The ratio of the| golden ratio was considered by the ancient Greeks to be the most perfect| ratio for the lengths of the sides of rectangles, such as portraits. Show| that if b:a is the golden ratio, then b/a = (1+ sqrt(5))/2.|| I guess the question is really saying to derive b = (1+sqrt(5)), and a = 2| from b/a = (a+b)/b (*).|| So I tried to manipulate with the algebra of (*), and failed.| Is there anything i am doing wrong?||| =I have this question here that I attempted and failed. some pointer toward> this problem is greatly appreciated.The golden ratio is the ratio b:a so that b:a = (a+b):b. The ratio of the> golden ratio was considered by the ancient Greeks to be the most perfect> ratio for the lengths of the sides of rectangles, such as portraits. Show> that if b:a is the golden ratio, then b/a = (1+ sqrt(5))/2.I guess the question is really saying to derive b = (1+sqrt(5)), and a = 2> from b/a = (a+b)/b (*).No, see below. > So I tried to manipulate with the algebra of (*), and failed.> Is there anything i am doing wrong? b/a = (a + b)/b says b^2 = a^2 + a*b or b^2 - a*b - a^2 = 0. Use thequadratic equation with b playing the role of x then divide throughby a. Note that you want b/a to be positive.-- Paul SperryColumbia, SC (USA) =I have this question here that I attempted and failed. some pointer toward> this problem is greatly appreciated.The golden ratio is the ratio b:a so that b:a = (a+b):b. The ratio of the> golden ratio was considered by the ancient Greeks to be the most perfect> ratio for the lengths of the sides of rectangles, such as portraits. Show> that if b:a is the golden ratio, then b/a = (1+ sqrt(5))/2.I guess the question is really saying to derive b = (1+sqrt(5)), and a = 2> from b/a = (a+b)/b (*).So I tried to manipulate with the algebra of (*), and failed.> Is there anything i am doing wrong?Probably, but and failed isnÍt enough to see what that might be.Try setting x = b/a = (a+b)/b, and see what you can conclude about x*x.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes /039 53 36 N / 086 11 55 W =That was a very insightful substitution. I 'nally see how itÍs done byMy initial attempt was to try to solve the a and b having two variables.and I got something like, 0 = (a + b)^2 - ab - 2b^2, in the end, and failed isnÍt enough to see what that might be.>> Try setting x = b/a = (a+b)/b, and see what you can conclude about x*x.> -- > Daniel W. Johnson> panoptes@iquest.net> http://members.iquest.net/~panoptes/> 039 53 36 N / 086 11 55 W =How would i 'nd 0.5 in (mod 7)? :( - i donÍt want the answer, but afew hints on how i would do it. ;) All i know about mod is that in programming, it gives the remainderwhen a number is divided by another number, but from what iÍve beenreading, i think that it has a different meaning in this context. => How would i 'nd 0.5 in (mod 7)? :( - i donÍt want the answer, but a> few hints on how i would do it. ;) All i know about mod is that in programming, it gives the remainder> when a number is divided by another number, but from what iÍve been> reading, i think that it has a different meaning in this context.numbers would go 0, 1, 2, 3, 4, 5, 6, 7 and then start again - is> that right? Would that mean that 0.5 in (mod 7) would still be 0.5> since it is between 0 and 7? :?That might be right!Others have already pointed out the interpretation in the ring of integersmod 7, where the notation .5 is strange, but could be taken to mean 1/2.In that case the answer would be 4.There is however an interpretation of mod as exact remainder, e.g. inthe IEEE §oating-point sense. In this case it is the smallest number(in absolute value) obtained from the argument by exact subtraction ofsome exact multiple of the modulus -- and .5 (mod 7) would indeed be .5(as would, for example, be 14.5 mod 7). Since the argument was given as.5 and not as 1/2 the possibility exists that this is what was intended.Michel.Michel. =>How would i 'nd 0.5 in (mod 7)? :( - i donÍt want the answer, but a>few hints on how i would do it. ;) questions donÍt deserve having the answer taking up space on thousandsof servers all over the world! The quaestion makes no sense. Modular arithmetic only has a meaningfor integers!>>All i know about mod is that in programming, it gives the remainder>when a number is divided by another number, but from what iÍve been>reading, i think that it has a different meaning in this context.>>numbers would go 0, 1, 2, 3, 4, 5, 6, 7 and then start again - is>that right? Would that mean that 0.5 in (mod 7) would still be 0.5>since it is between 0 and 7? :?>>---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- =>How would i 'nd 0.5 in (mod 7)? :( - i donÍt want the answer, but a>few hints on how i would do it. ;) questions donÍt deserve having the answer taking up space on thousands> of servers all over wrong with protecting => How would i 'nd 0.5 in (mod 7)? :( - i donÍt want the answer, but a> few hints on how i would do it. ;) DonÍt write 0.5 in number systems other than the reals. If you mean1/2, then write 1/2. We say x=1/2 mod 7 if x*2=1 mod 7. So you cantry the 7 possibilities and conclude x=4 will do. A bit more work(using the fact that 7 is prime) shows that this solution 4 isunique mod 7. Thus: 1/2 = 4 mod 7.All i know about mod is that in programming, it gives the remainder> when a number is divided by another number, but from what iÍve been> reading, i think that it has a different meaning in this context.> certainly clears some things up for me.Elaine, iÍm not exactally sure what you mean, but iÍm guessing thatyou agree with my answer?---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- =Since 4 if prime to 7, a==b (mod 7) iff 4a==4b (mod 7). Hencex==1/2 (mod 7) iff 2x==1(mod 7) iff x==8x==4 (mod 7).| How would i 'nd 0.5 in (mod 7)? :( - i donÍt want the answer, but a| few hints on how i would do it. ;)|| All i know about mod is that in programming, it gives the remainder| when a number is divided by another number, but from what iÍve been| reading, i think that it has a different meaning in this context.|| numbers would go 0, 1, 2, 3, 4, 5, 6, 7 and then start again - is| that right? Would that mean that 0.5 in (mod 7) would still be 0.5| since it is between 0 and 7? :?|||| http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000| ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption=--- =mod 7 = 1 am i right, am i? :wink:---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- =would 0.5 be any different to 1/2? cause iÍm sure that the questionwas 0.5 :cry: .---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- => would 0.5 be any different to 1/2? cause iÍm sure that the question> was 0.5 :cry: .mod 7 is de'ned only for integers. For example 3 and 10 are eqivalent mod 7, written 3=10 (mod 7), although in texbooks the ï=Í is written with 3 horizontal lines to remind the reader that this is different than saying two integers are equal. As you mentioned earlier, the numbers mod 7 are {0,1,2,3,4,5,6} but it is important to understand that when we list the elements of the mod 7 world, these ARE NOT regular normal integers. For example ï2Í is not the usual number 2, but rather stands in this context for the equivalence class of 2 in the mod 7 world. Thus ï2Í actually stands for the entire set of integers {2,9,16, etc.} that are all equivalent to each other mod 7.It turns out that in the mod 7 world, which is technically called the ring of integers mod 7, every element has a multiplicative inverse. A multiplicative inverse is a number that, multiplied by another number, gives 1. So the multiplicative inverse of 2 (in the integers mod 7) is 4, because 2*4 = 1 (mod 7).It makes sense to ask things like, what is 1 divided by 2, mod 7? Rephrased, that question means, what number can I multiply by 2, to get 1. And weÍve already noted that the answer is 4. So itÍs reasonable to say that 1/2 = 4 (mod 7).If indeed your teacher or textbook asked what is 0.5 (mod 7), I think itÍs a stretch to try to interpret that as asking whatÍs 1/2 (mod 7). I am not sure what they are getting at here if they actually said 0.5. =would 0.5 be any different to 1/2? cause iÍm sure that the questionwas 0.5 :cry: .---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- => How would i 'nd 0.5 in (mod 7)? :( - i donÍt want the answer, but a> few hints on how i would do it. ;) All i know about mod is that in programming, it gives the remainder> when a number is divided by another number, but from what iÍve been> reading, i think that it has a different meaning in this context.numbers would go 0, 1, 2, 3, 4, 5, 6, 7 and then start again - is> that right? No, for mod 7 you want the possible remainders on division by 7: 0, 1, 2, 3, 4, 5, 6. You add and multiply by doing the usual addition and multiplication andthen your mod 7 operator. So, 4*5 = 6 (mod 7).> Would that mean that 0.5 in (mod 7) would still be 0.5> since it is between 0 and 7? :? I doubt it. The clue is 0.5*2 = 1. Look for something so thatone of 0, 1, 2, 3, 4, 5, 6.-- Paul SperryColumbia, SC (USA) => How would i 'nd 0.5 in (mod 7)? :( - i donÍt want the answer, but a> few hints on how i would do it. ;) All i know about mod is that in programming, it gives the remainder> when a number is divided by another number, but from what iÍve been> reading, i think that it has a different meaning in this context.numbers would go 0, 1, 2, 3, 4, 5, 6, 7 and then start again - is> that right?No, that is the behavior of mod 8. In mod 7, 0 and 7 are not distinctnumbers.> Would that mean that 0.5 in (mod 7) would still be 0.5> since it is between 0 and 7? :?Probably not, depending on the de'nitions being used. Are you certainthat 0.5 is what you were asked to 'nd? If you were asked instead to'nd 1/2, that would be something that can be multiplied by 2 to get1.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes /039 53 36 N / 086 11 55 W =1 raised to non-integer powers can have many roots. Simplest example,sqrt(1)=1 or -1. Your exponent (x/2pi) is not an integer, unless xis a multiple of 2pi.---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- =Sorry if my English isnÍt good, I have an array of numbers i.e square 3x3: | 1 2 3 |A = | 5 5 5 | | 8 9 7 |how can I calculate A^-1 ?I do need it i.e. for calculting functions of several variables using newtons method.Maybe I can 'nd somehere an ready C/C++ or Pascal programming language method for that?What is english name for array and for ^-1 so that I can use them as keywords in web searchers?-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~l-.~~~~~~~~~~~~~~~~~~~GG- 1175498 ____| ]____, Rafal ïRaf256Í Maj X-( * )Rafal(at)Raf256(dot)com ,---------- => Sorry if my English isnÍt good, I have an array of numbers i.e square 3x3:> | 1 2 3 |> A = | 5 5 5 | > | 8 9 7 |> how can I calculate A^-1 ?> I do need it i.e. for calculting functions of several variables using > newtons method.Sounds like you are actually interested in solving linear systems, ratherthan merely obtaining the inverse. You can do that directly.> Maybe I can 'nd somehere an ready C/C++ or Pascal programming language > method for that?The original is in Fortran. You can 'nd a C translation at. Many computers have LAPACK or CLAPACKalready installed. My Mac does.> What is english name for array and for ^-1 so that I can use them as > keywords in web searchers?A two-dimensional array is called a matrix, and what you are doing iscomputing the inverse.-- Dave SeamanJudge YohnÍs mistakes revealed in Mumia Abu-Jamal ruling. = > | 1 2 3 |> A = | 5 5 5 | > | 8 9 7 |how can I calculate A^-1 ?| -2/3 13/15 -1/3 || 1/3 -17/15 2/3 || 1/3 7/15 -1/3 |Google check:matrix inversion c code numerical recipes =>> GM: General Mathematics>> -----------------------> math.GM/0312360 Paola Cattabriga: Beyond Uncountable> My, my! Time to give up set theory. Of course, from> <5vSKb.2670$fI5.783@reader1.news.jippii.net> we knew that already.>> math.GM/0312309 Craig Alan Feinstein: The Collatz 3n+1 Conjecture is Unprovable> I didnÍt know proving something about a function implies calculating> its value for every possible argument. You always learn something new.One learns lots of new things in the GM category.Why, thatÍs egregious.I use f(x)=x+1 to map the set to the powerset.The set of all sets is its own powerset. The class of all classes isunfortunately for class theory in classes a class. IÍm among thosewho say no classes in set theory, as you may well know.Then again I think the impulse function goes to half a scalarin'nity.What else did you have in mind of the General Mathematics category,Dr. Chapman?The Mathematics ArXiv is unfettered of moderation. To what level ofcredulity to you assign inclusion within it?Cardinality is a gross qualitative comparison, for in'nite sets.Large cardinals are often comparable by their dates of discovery.ItÍs always one more. - Axiom of In'nityNot not p is not necessarily equal to p. - No Axioms EnthusiastBorel or combinatorics, anyone? If Borel armwrestles a jar with athousand clear marbles, do pigs §y?Why, thatÍs absurd.Ross F. =>> GM: General Mathematics>> -----------------------> math.GM/0312360 Paola Cattabriga: Beyond Uncountable> My, my! Time to give up set theory. Of course, from> <5vSKb.2670$fI5.783@reader1.news.jippii.net> we knew that already.>> math.GM/0312309 Craig Alan Feinstein: The Collatz 3n+1 Conjecture is Unprovable> I didnÍt know proving something about a function implies calculating> its value for every possible argument. You always learn something new.> One learns lots of new things in the GM category.Why, thatÍs egregious.I use f(x)=x+1 to map the set to the powerset.The set of all sets is its own powerset. The class of all classes is> unfortunately for class theory in classes a class. IÍm among those> who say no classes in set theory, as you may well know.Then again I think the impulse function goes to half a scalar> in'nity.What else did you have in mind of the General Mathematics category,> Dr. Chapman?The Mathematics ArXiv is unfettered of moderation. To what level of> credulity to you assign inclusion within it?Cardinality is a gross qualitative comparison, for in'nite sets.Large cardinals are often comparable by their dates of discovery.ItÍs always one more. - Axiom of In'nity> Not not p is not necessarily equal to p. - No Axioms EnthusiastBorel or combinatorics, anyone? If Borel armwrestles a jar with a> thousand clear marbles, do pigs §y?Why, thatÍs absurd.Ross \ F.Whatever you are drinking, it is doing funny things to your synapses. =JS: You do not have enough background.ZD: OK, but just a few more questions and I promise I wonÍt waste your time anymore until grasshopper can jump.J.S. Unlike other 'elds the quantum information 'eld in orthodox quantumtheory has no sources. The quantum potential is fragile hence signallocality.ZD: Ironically enough I just got 'nished reading a few of his papers. What do you mean by, Unlike other 'elds the quantum information 'eld in orthodox quantum theory has no sources? Ithought what you call the quantum information 'eld is what is traditionally know as the wave function? DoesnÍt this come from a JS: Ordinary quantum theory only has the BIT wavefunction without an IT The wave function is in§uenced by its environment via boundary conditions andinteraction Hamiltonians. But that is not the idea here. In BohmÍs ontology there ispilot wave.ItÍs ONE-WAYIT FROM BITwithout any compensatingBIT FROM ITtheir marching orders from a rigid globally §at space-time geometry.General relativity changes that, large enough mass-energy sources warp space-time. Space-time geometry is no longer merely an in'nitely stiff steel plate for the actors to strut upon, but is now also part of the act! Said another way, the players if they are massive enough modify the now rubber sheet they are moving on in a nonlinear way. Similarly for the covering theory of quantum theory. This covering theory also IMHO includes our inner consciousness as a More is different emergent phenomenon.Since 1998 it has become clear that ~ 96% of the mass-energy of the but is the zero point energy density of all the quantum 'elds. When the zero point energy density in a region of space-time is positive, the pressure is negative but of equal magnitude to the energy density, i.e. w = -1. The gravity warp factor of pressure is 3x that of energy density in EinsteinÍs theory. Therefore, positive zero point energy density makes a universal anti-gravity exotic vacuum 'eld that is the dark energy. Similarly, the dark matter is the same story, but with positive pressure from negative zero point energy density. Warp drive and star gate time travel are possible by the metric engineering of this exotic vacuum zero point energy density 'eld. UFOs show us that someone Out There knows how to do this trick. The control parameter for the zero point energy density is vacuum coherence -- an idea missing from the minds of most physicists with a few exceptions like George Chapline, G. E. Volovik and myself although the details of each of our models is different, they share important common conceptual ground.probability current. It is this quantum noise that doesnÍt allow usnon-local communications in normal circumstances. If you can calmthis noise (i.e.-thermal heat) via macro-quantum phase coherence suchas a superconductor or BEC then this would be possible. Or perhapseven the micro-tubulars in our brains are somehow capable of it. :-)JS: ThatÍs roughly the idea, yes. The giant DeBroglie MACRO-QUANTUM q-bit ground state condensate in the living brain IS the MIND!Z.D. As in singular!? Kinda like Gia or something?JS: You mean Gaia. Yes, thatÍs HawkingÍs Mind of God in the Intelligent Conscious Universe of Star Maker, Solaris, Black Cloud, VALIS, Cosmic Coincidence Control -- all that stuff from The Poets prior to The Physicists.http://www.classicalmidiconnection.com/cgibin/ x.cgi?midi/n3/zsunrise.midZD: It is this superposing of this wave and intent retarded waves from the mind that create consciousness. Could you de'ne this back action in terms ofphysical principles?Z.D. What I mean by this is a Wheeler-Feynmen absorber theory of thewavefunction its self. IsnÍt the superposition of the advanced andretarded potentials the equivalent to the Q.P.?J.S. See also Shelley GoldsteinÍs recent paper on how presponse advanced effect comes in. An advanced from-the-future micro-causal arrow and a retarded from the past MACRO-arrow of time. This idea needs further development, i.e. putting Wheeler-Feynman together with BohmÍs ontology.Z.D. Again I am thinking in terms of CramerÍs Transactional theory inrelation to BohmÍs ontological idea.JS: ThatÍs correct.ZD: You have the equation: Phi-> <-X - (Phi->X)=CJS: Phi = BIT (pilot wave)X = IT (extra variable)-> = ACTION<- = REACTION ZD: What does this mean? Is Phi-><-X the conscious intent acting onmatter and back and the last term is the unconscious intent?Z.D. I guess what IÍm asking is intent, as you see it, mostly on theunconscious level? What is BIT in terms of conscousness? What is moreimportant in de'ning BIT, conscious intent or unconscous intent?JS: Consciousness is generated in MACRO-BIT in the reaction of MACRO-IT on its own private MACRO-BIT.More is different. (P.W. Anderson) Conscious intent is a globally self-consistent spontaneously self-organizing quasi-autonomousfeedback-control loop between MACRO-BIT and MACRO-IT which essentially fuse together in the ODLRO Coherence Parameter which has simultaneously aspects of both a classical 'eld and a quantum wave.There is no such thing as unconscious intent. That is simply training and any AI machine can do it. Much of our daily behavior is machine-like.Do not confuse intelligence and purposeful adaptive behavior with consciousness. Of course you can have multiple personality disorder with weakly interacting fragmented MACRO-BIT mind 'eld or landscapes competing for control of the same IT brain-body.JS: No. Phipps is plain wrong period. He is a nice man and a good writer, but he is wrong. Phipps does not believe length contraction although he accepts timedilation. I think he gave up on his idea. It never got anywhere really exceptfor a non-replicated empirical claim about lack of Thomas precession asI recall. Einstein wins!Z.D. But what about the motion of the detectors? Inhttp://www.stardrive.org/Jack/PhippsEM.pdfyou claim that MaxwellÍs equations are incomplete or to use your termunder parameterized because they do not take into account thevelocity of the detectors.JS: I was describing PhippÍs idea. I was not agreeing with it. I was merely playing with it.Learning it. I decided it was not useful.ZD: When you do this you have a simpleGalilean transfomation as long as you keep track of which frame anevent is considered to be proper. And in order to do this one musthave an absolute universal time which is compatable with BohmÍsideas.JS: Goldstein is attempting to use the presponse advanced micro-causal in§uence to free BohmÍs theory from its seemingly Galilean shackles. =the great-great-grandkids of JackÍs great-great-grandkids are coming-- having freed themselves from HawkingÍs shackles --to put an end to this misery. woe, to future Trekkie conferees! > pilot wave. > exotic vacuum zero point energy density 'eld. UFOs show us that someone > Out There knows how to do this trick. The control parameter for the > zero point energy density is vacuum coherence -- an idea missing from > JS: You mean Gaia. Yes, thatÍs HawkingÍs Mind of God in the > Intelligent Conscious Universe of Star Maker, Solaris, Black Cloud, > VALIS, Cosmic Coincidence Control -- all that stuff from The Poets prior > to The Physicists.> http://www.classicalmidiconnection.com/cgibin/x.cgi?midi/n3/ zsunrise.mid > Z.D. What I mean by this is a Wheeler-Feynmen absorber theory of the> wavefunction its self. IsnÍt the superposition of the advanced and> retarded potentials the equivalent to the Q.P.? > There is no such thing as unconscious intent. That is simply training > and any AI machine can do it. Much of our daily behavior is machine-like. > JS: Goldstein is attempting to use the presponse advanced micro-causal > in§uence to free BohmÍs theory from its seemingly Galilean shackles.--Give the Gift of Trickier Dick Cheeny -- out of of'ce, at last!http://www.benfranklinbooks.com/http://www.wlym.com/ pages/music.htmlhttp://www.rand.org/publications/randreview/ issues/rr.12.00/http://members.tripod.com/~american_ = I have a set of Bezier curves. Does anyone know a way that I can> order them, so that two curves which are ordinally (right word?)> near each other will be similar?>> Darren Grant>> Similar is a pretty fuzzy term. LetÍs assume that the starting and>> ending points of all our curves are identical. Let F(A,B) be the area>> between the two curves A and B. F(A,A) = 0; and if F(A,B) = 0, then A>> = B.>> Then we can make the de'nition that B is more similar to A than C is>> to A iff F(A,B) < F(A,C).>> However, this only provides us with a metric; not an order. ...>> > Given a list of curves, a new curve is added to the list by inserting it>> after the curve which it is most similar to.>> > If the curves are all formed from, say, three control points, can anyone>> show me how I can compare the similarity of two curves using these >> points?>> Certainly, you could use a metric between two curves which was the sum> of the squares of the distances (in the x/y plane) between the> coresponding control points of the two curves.Cool, I was wondering if it was as simple as this.> Well, itÍs a simple description; it would certainly depend on yourde'nition of similar. A more complicated but probably ultimatelymore useful one would be calculate the area between the curves.> But I guess my question would be, what is the goal of your ...> inserting it after... most similar... sorting method?>> LetÍs suppose each curve is identi'ed with a single control point,> (x,y).>> If you wish to create an order that will allow, for example, a binary> tree search for closest curve from some set, in the same fashion> that one would use for looking up a string in a dictionary, I think> itÍs not possible.>> I presume you ultimately want> to, given a random curve, 'nd the closest curve in some 'xed set of> curves?Actually, the opposite. If I have a set C of curves, I want to be able> to choose a subset of 9 of these curves, which represent the most varied> subset possible. Most varied could have many possible meanings. I de'ne one below.> The way I thought of doing this, is by ordering the> curves in C so the most similar are next to each other, then just> choosing 9 curves, distributed equally in the set C (ie, choose the 'rst> curve in C, then the curve 1/9th of the way through the set, etc.)> You are assuming that there is a _linear_ ordering of your curves(... most similar are next to each other...); but that will ingeneral not be the case.As an example: Let each curve in C be described by a single controlpoint. If c_1 and c_2 are two curves (i.e., control points) in C, thenlet d(c_1, c_2) be the distance (in x/y) between those control points.Now imagine that the set C consists of 100 control points which areequally distributed around a circle of radius 1/4, centered at (1/2,0) (with end points at (0,0) and (1,0)).A circle has no beginning or end - so there is no 'rst or leastsimilar curve in C, and no 1/9th of the way through the set. Curvesthat are similar are close to each other; but not in a linear way.Add another curve, c_0, to the set C above; with c_0 = (1/2, 0). c_0is clearly very unsimilar to most other curves in C, but it is notthe closest (most similar) to any other curve in C; nor is it thefurthest (least similar) to any other curve in C.I may be wrong, but it sounds more like the problem you are posing is:given the above set of curves C, which 9 curves are (in some sense)the most representative of the entire set of curves?It follows that such a representative set will consist of curves thatare not particularly similar to each other; if two curves in the setwere very similar, we could replace one of them with one which wasless similar, and would then better represent more of the possiblecurves in the set.To formalize this, suppose you have 9 (distinct) representative curvesselected from C, labelled r_1, r_2, .., r_9. If c is any curve in C,let D(c) be the smallest of the distances d(c,r_1), d(c,r_2), ...,d(c,r_9).Then a most varied set of 9 curves could be de'ned to be a setwhich minimizes the sum of D(c) over all cÍs in C.In this example, a good candidate set would seem to be c_0 and then 8points equally distributed about the circle.On the other hand, you may want to select a subset of 9 curves suchthat the total of the distances d(r_1,r_2) + d(r_1, r_3) + ... +d(r_1, r_9) + d(r_2, r_3) + ... + d(r_8, r_9) between the points inthe subset is maximized; the most varied subset would be one havingthe largest sum of these distances.In this case, I think the most varied subset would be one whichconsisted of 9 equally distributed points around the circle.> IÍm sorry for leaving you a bit in the dark as to why I want to do this,> but as itÍs the basis for my M.Sc. research, I canÍt \ say what IÍll be> doing with it - yet :)> Without knowing exactly what youÍre trying to do, itÍs hard to guesswhat a \ proper solution would be; but my advice would be, regardless ofhow you choose to measure the distance between two given curves, tocome up with an exact mathematical formulation of what properties amost varied subset of 9 curves is supposed to have. The rest is justplug and chug :).> Darren => Portfolio of PAF as of 8JAN04:BCE 6,500 22.63 $147,095.00 > BMY 50 29.38 $1,469.00 > MRK 100 48.08 $4,808.00 > Q 14,200 4.52 $64,184.00 > SBC 12,200 27.59 $336,598.00 > realestate land 3APR03 of 3 lots $19,000.> science-art of pictures,porcelain etc starting JAN03 for $12,160.> realestate land 30JUL03 another lot $11,500.Today I sold some more Qwest for 4.51 per share for a small pro't and> sold 2,800> shares of SBC at 27.60 for a substantial pro't since SBC has been> going almost straight up as it hit its ex dividend date on 7 January,> but whenever a stock of mine goes up after the ex dividend date I am> always tempted to make a switch out of a sizeable chunk, because the> presumption that it will go down below the price at the ex dividend> date which I remember to be 27. even is a fairly high probability and> that a crossover can be achieved in the short term future and perhaps> buyback more.Stockmarket in that I had stocks not for Crossover technique but> rather instead on hopes of a buyought. So I am getting the portfolio> into that of 100% obedience to the Crossover technique. And ridding> myself of con§icting and competing strategies.Let the OS do the work.Portfolio as of 9JAN04:BCE 6,200 BMY 50 MRK 100 Q 14,200 SBC 12,200 realestate land 3APR03 of 3 lots $19,000.science-art of pictures,porcelain etc starting JAN03 for $13,705.realestate land 30JUL03 another lot $11,500.Today I learned something about the rules and regulations of the SEC.And a good thing that I learned it today in a little bit of a jamrather than say learn it in the future in a big jam. I found themarket going in the direction I expected SBC to go since itsex-dividend of 7JAN and was hoping to sell 3300 BCE at 22.78 and buyback 2,840 SBC at 26.40 in order to net a gain of 100 free shares ofBCE plus 40 free shares of SBC in the matter of a single day. Abeautiful and pretty crossover. But as it turned out in practice andreality is that the 3300 shares were only partially 'lled and I endedup selling only 300 shares at 22.78. And so as the day wore where bothBCE and SBC were falling, I decided to buy back the 300 BCE at a lowerprice of 22.60 or thereabouts and make a tiny gain of extra cash. Andso as I put in the order to buy back the 300 shares I had sold earlierin the day, I was prevented from doing so because of a SEC rule that 3days to settle cash. I guess it is a good rule in that it preventsspeculation-bubbles from arising in the total system and gives timefor the stock and cash to be accounted for in proper accounts and cashalso. So I have no grudge about the rule of settlement and I am happyto have learned about this where no damage is done to me. And I mustapologize to the broker representative that I spoke with for I mayhave been gruff, in falsely thinking that the computer was in faultfor not allowing me to trade on.In relief I went ahead and bought more porcelain of a science of artexperiment. I have a formula for 'ne porcelain shopping. Somethingthat the Antiques Roadshow ought to elucidate. The formula I use isthat I take the 'nest porcelain and 'gure a value in terms of skillof manufacture, materials, and hours of labor in doing the artwork.Then I compute a surface area of the piece in question and multiply bya dollar value. So what I end up with is a dollar value per squarecentimeter of porcelain and I use it on all 'ne porcelain no matterthe brand name. Simply take a piece of 'ne-porcelain compute roughlythe surface area and then multiply by that dollar 'gure and thatshould be the price of the piece. So that whenever someone is in a jamas to know whether they paid too much or too little for a piece, well,this square centimeter method is just the proper method. It is likeof'ce rental space or building construction space which goes for thesquare-foot or square-meter. I am running several tests on porcelainand one of the reasons I spent $1,545. today for more pieces.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies => Just out of curiosity, with all these unproven conjectures about...> what happens when the conjecture is proven to be correct? Does the> newly declared theorem retain the name of the person originating the> conjecture, or will it get named after the person(s) who proves it?Sometimes the conjecture itself must be so valuable that we should creditthe person who made it. IÍm thinking of the fundamental theorem ofalgebra - according to what I have read it was 'rst proposed by DÍAlembert,who gave an incomplete proof. So presumably DÍAlembert deserves to behonoured for realizing that there was something to prove. Hmm - what didpeople do before DÍAlembert? Just assume that of course there is asolution?I donÍt claim to understand the Taniyama-Shimura conjecture, but again fromwhat I have read it proposed a deep connection between two areas of mathsthat seemed totally unrelated. So the name really honours the guys who'rst suspected there was a deep connection.Simon. =in message :> Every so often, I think about how much math IÍve forgotten (B.S. degree in> Mathematics from Caltech, followed by 20+ years of work in which I never> use any of it, and so most of it is gone), and decide to start over. [...]IÍve sometimes idly wondered if you were *that* Tim Smith. Long(ROT-13 my address, assuming that isnÍt obvious.)-- Jim Heckman =I recall many years ago hearing about putting in'nitesimals on a(more) rigorous basis, and not having to depend so much on theepsilon-delta form of calculus and proof.I think there were new types of numbers de'ned, includingin'nitesimals, with axioms they obeyed, and even a Calculus text was written this way.Does anyone know if anything became of this? Some people were enthusiasticabout using it at one time.Van =>I recall many years ago hearing about putting in'nitesimals on a>(more) rigorous basis, and not having to depend so much on the>epsilon-delta form of calculus and proof.>>I think there were new types of numbers de'ned, including>in'nitesimals, with axioms they obeyed, and even a Calculus text> was written this way.YouÍre referring to non-standard analysis. The calculus text isby Jerome Keisler; I think the title is Calculus.>Does anyone know if anything became of this? Some people were enthusiastic>about using it at one time.>>Van************************David C. Ullrich =>>I recall many years ago hearing about putting in'nitesimals on a>(more) rigorous basis, and not having to depend so much on the>epsilon-delta form of calculus and proof.>>I think there were new types of numbers de'ned, including>in'nitesimals, with axioms they obeyed, and even a Calculus text> was written this way.>> YouÍre referring to non-standard analysis. The calculus text is> by Jerome Keisler; I think the title is Calculus.In another thread I learned that this book is now available online asa single pdf 'le. Here is the link:http://www.math.wisc.edu/~keisler/calc.html>Does anyone know if anything became of this? Some people were enthusiastic>about using it at one time.>>Van> ************************>> David C. UllrichJim Buddenhagen-- To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE =I recall many years ago hearing about putting in'nitesimals on a(more) rigorous basis, and not having to depend so much on theepsilon-delta form of calculus and proof.I think there were new types of numbers de'ned, includingin'nitesimals, with axioms they obeyed, and even a Calculus text was written this way.Does anyone know if anything became of this? Some people were enthusiasticabout using it at one time.Van mogensn@mogensn.dk (Mogens Nielsen)>>Does anyone know the de'nition of the Spectralradius of a real>operator, or even better: Knows where I can download some litterature>about spectral-theory for Real Banach AlgebraÍs, if there is such a>thing. (There is lots of stuff about complex banach algebraÍs, but I>am not interested in that!)>>If you want me to be more precise, here is the problem:>Let K be a compact subset of R^k. Let T:C(K) -> C(K) be an operator,>where C(K) is the set of continuous functions from K to R. In the>of T, without de'ning it.HeÍs probably referring to the complex spectral radius, ie the sup of |lambda| for lambda in the (complex) spectrum. Because thatÍsthe _standard_ de'nition of the spectral radius, even for real Banach algebras.>Mogens************************David C. Ullrich Timofeev)>>The velocity of a moving clock>causes it to appear to run slow relative to a clock on the Earth.>GPS satellites revolve around the Earth with an orbital period of>11.967 hours and a velocity of 3.874 km/s. Thus on account of its>velocity, a GPS satellite clock appears to run slow by 7 [Micro]s per day.> ...> The difference in>gravitational potential between the altitude of the orbit and the>surface of the Earth causes the satellite clock to appear to run fast.>At an altitude of 20,184 km, the clock appears to run fast by 45 [Micro]s>per day.>>The net effect of time dilation and gravitational redshift is that>the satellite clock appears to run fast by approximately 38 [Micro]s per>day when compared to a similar clock at rest on the geoid, including>the effects of the velocity of rotation and the gravitational>potential at the Earth’s surface. This is an enormous rate>difference for a clock with a precision of a few nanoseconds. To>compensate for this large secular effect, the clock is given a>fractional rate offset prior to launch of -4.465 [Times]10^-10 from its>nominal frequency of exactly 10.23 MHz, so that on average it appears>to run at the same rate as a clock on the ground. The actual>frequency of the satellite clock prior to launch is thus>10.229 999 995 43 MHz.>> Load of crap.The above are merely statements of fact. The clocks are built> to run at 10.22999999543MHz prior to launch so that the signal> received once they are in orbit is measured as 10.23MHz on the> ground receivers.so that the signal received once they are in orbit is measured as 10.23MHz on the ground receivers. Really??? >;^)The satellite is in a state of continuous motion always. Why do you have forgotten about Doppler effect?> If the orbiting clocks appear to run slow it is because they DO run slow.They run fast, not slow.> The ground observer counts every ïtickÍ emitted per orbit.Well a ground observer can only see any satellite for a fraction> of its orbit but if the Earth didnÍt get in the way, you would> be right.> The GPS clock preset is required simply to compensate for an increase in> clock> rates due to being in free fall.The change of coordinate rate is correlated to the difference in> gravitational potential, not force, but with that clari'cation> you are correct again.> .....unless the ïtick-fairiesÍ have been active again!If a satellite signal could be received throughout its orbit the> ground observer could count every tick emitted, so there are no> tick fairies.Nothing you have said con§icts in any way with what was stated,> so what were you objecting to?George => On 9 Timofeev)>>The velocity of a moving clock>causes it to appear to run slow relative to a clock on the Earth.>GPS satellites revolve around the Earth with an orbital period of>11.967 hours and a velocity of 3.874 km/s. Thus on account of its>velocity, a GPS satellite clock appears to run slow by 7 [Micro]s per day.> ...> The difference in>gravitational potential between the altitude of the orbit and the>surface of the Earth causes the satellite clock to appear to runfast.>At an altitude of 20,184 km, the clock appears to run fast by 45 [Micro]s>per day.>>The net effect of time dilation and gravitational redshift is that>the satellite clock appears to run fast by approximately 38 [Micro]s per>day when compared to a similar clock at rest on the geoid, including>the effects of the velocity of rotation and the gravitational>potential at the Earth’s surface. This is an enormous rate>difference for a clock with a precision of a few nanoseconds. To>compensate for this large secular effect, the clock is given a>fractional rate offset prior to launch of -4.465 [Times]10^-10 from its>nominal frequency of exactly 10.23 MHz, so that on average it appears>to run at the same rate as a clock on the ground. The actual>frequency of the satellite clock prior to launch is thus>10.229 999 995 43 MHz.>> Load of crap.>> The above are merely statements of fact. The clocks are built> to run at 10.22999999543MHz prior to launch so that the signal> received once they are in orbit is measured as 10.23MHz on the> ground receivers.>> so that the signal received once they are in orbit is measured> as 10.23MHz on the ground receivers.>> Really??? >;^)>> The satellite is in a state of continuous motion always.> Why do you have forgotten about Doppler effect?Because I assumed you were suf'ciently familiar withbasic physics that I didnÍt need to point this out. Theoriginal web page you quoted, and which I was simplysummarising, doesnÍt mention it either yet you didnÍtcomplain about that.George => But I have another quetion about the following value related to the above> sequence while I thought about the above.> Let fn(x) be> fn(x) = log(log(log(....log(x))..) n times log nesting.> What is the value of the following suppose an be a diverging sequence?> lim fn(an) =? (n--->in'nite)> Is it 0?or another value?> Does it depend on the sequence an?Let g_n(x) = exp(exp(exp(...exp(x))...) x times exp nesting. Let {b_n}be any sequence; and let the seqeunce {a_n} be de'ned by a_n =g_n(b_n).Because of the explosive nature of g_n, a_n is going to be divergentin a lot of cases, both those where f_n(g_n(b_n)) = b_n converges andthose where b_n does not.One might better ask, if a_n is _not_ a diverging sequence, doesf_n(a_n) always have a limit? =I was able to identify a numerical series that gives me the solutionfor a problem I have in the 'eld of Stochastic Processes. This serieswas obtained by observation,i.e., investigating the outcomes I wasable to identify the generating function of the series.My question is how I can formally prove that this series I came acrossis valid in any case. Which steps I must formally go through?Once I read in a book (FermatÍs Last Theorem by Simon Singh) thatthere are two kinds of proofs: scienti'c and mathematical). Theformer refers to the kind of proof that is based on simply observationof outcomes; we exhaustedly test a given empirical solution andafterwards we assume it is true in any case. The latter dictates thatwe have to formally prove (using formal methods and equations:e.g.,induction) that our statement is true in any case.I conclude that now I need a methematical proof. But how can I work itout? Which steps I have to go through? => I was able to identify a numerical series that gives me the solution> for a problem I have in the 'eld of Stochastic Processes. This series> was obtained by observation,i.e., investigating the outcomes I was> able to identify the generating function of the series.>> My question is how I can formally prove that this series I came across> is valid in any case. Which steps I must formally go through?> Once I read in a book (FermatÍs Last Theorem by Simon Singh) that> there are two kinds of proofs: scienti'c and mathematical). The> former refers to the kind of proof that is based on simply observation> of outcomes; we exhaustedly test a given empirical solution and> afterwards we assume it is true in any case. The latter dictates that> we have to formally prove (using formal methods and equations:e.g.,> induction) that our statement is true in any case.>> I conclude that now I need a methematical proof. But how can I work it> out? Which steps I have to go through?>Proving something is what mathematics is all about, and can range from thetrivial to the impossible. So, to prove something you need a good deal ofknowledge and experience. You could post your series here and maybe someonecould offer some suggestions. Other than that, get a degree in Maths.Lurch =I conclude that now I need a methematical proof. But how can I work it> out? Which steps I have to go through?> Probably something you can learn only by experience.A help, perhaps: HOW TO READ AND DO PROOFS, by Daniel Solow.But learning this without a mentor (instructor) must be next toimpossible.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ =Terminology: differentiable = C^{infty}; cap = symbol of intersectionbetween sets, otimes = symbol of tensorial product; x in S means that x belongs to S.I know the following standard result:*************************************************Let X be a topological space (or a differentiable manyfold)and U = ( U_i )_{i in I} an open covering of X.forall i in I, consider the trivial vector bundle U_{i} x R^{p}and suppose that for each(i,j) in I^{2} such that U_{i} cap U_{j} not equal emptysetis assigned a continuous (or differentiable if it makes sense) applicationf_{ij}:U_{i} cap U_{j} --> GL(R^{p}).In the disjoint union of such trivial bundles, letÍs glue a point(x,v_{i}) in U_{i} x R^{p} with (y,v_{ j}) in U_{ j} x R^{p}if and only if x = y in U_{i}cap U_{j} and v_{j} = f_{ij}(x)[v_{i}].If, forall (i,j,k) in I^{3} and x in U_{i}cap U_{j}cap U_{k}we have f_{ij}(x) o f_{jk}(x) = f_{ik}(x) ==> the quotient for such arelation de'nes a vector bundle over X.*************************************************Now, M_{i} will denote a differentiable manyfold;If M is a differentiable manyfold then V(M) will denote the additivecategory of C^{infty} real vector bundles over M.Let M = M_{1} x ... x M_{k} be the product manyfold of M_{i}.If xi^{i} in V(M_{i}), then xi^{1} otimes ... otimes xi^{k} is a bundle over Mwhose 'ber at x = (x_{1},...,x_{k}) is xi^{1}_{x_{1}}otimes ... otimes xi^{k}_{x_{k}}.If f ^{i} in C^{infty}(M_{i}, xi^{i}), we de'ne f^{1}otimes ... otimes f^{k},a function on M, by (f^{1}otimes...otimes f^{k})(x) = = f^{1}(x_{1})otimes...otimes f^{k}(x_{k}).The bundle structure of xi^{1}otimes ... otimes xi^{k}is characterized by the property thatf^{1}otimes ... otimes f^{k}in C^{infty}(M, xi^{1} otimes ... otimes xi^{k}) whereverf ^ {i} in C^{infty}(M_{i}, xi^{i}).I donÍt understand the reason for which ``the bundle structure ofxi^{1} otimes... otimes xi^{k} is characterized by the property thatf ^{1} otimes... otimes f^{k}in C^{ infty}(M, xi^{1}otimes ... otimes xi^{k}) whereverf^{i} in C^{infty}(M_{i}, xi^{i}) ïÍ.May \ you to help me, please?Ternternnret@yahoo.it =IÍm doing som math and came to thisquestion about graphs.The Graph y= f(x) has a symmetry line x=3a) decide f(7), if f(-1) = 0 b) decide f(5), if f(1) = 3 c) Great the GraphI have calculated that a) should be 0and b) should be 3.what I wonder is how do I create a graph? (question (c))How do I create the equation like (y= -2x^2 +2x +1)?I do need that one I think to calculate all the Points in the graph.Please advice a newbe from Sweden! =En escribi.97:> IÍm doing som math and came to this> question about graphs.>> The Graph y= f(x) has a symmetry line x=3>> a) decide f(7), if f(-1) = 0> b) decide f(5), if f(1) = 3> c) Great the Graph> I have calculated that a) should be 0> and b) should be 3.>> what I wonder is how do I create a graph? (question (c))> How do I create the equation like (y= -2x^2 +2x +1)?> I do need that one I think to calculate all the Points in the graph.>> Please advice a newbe from Sweden!I think that in some way it is implicit that you can seek a quadraticfunction. Then, as x = 3 is a simetry line, the equation must bey = a(x - 3)^2 + bReplace x by 7 or -1 (it is the same) and y by 0, and you get a eqution in aand b. Do the same with x = 5 or 1 and y = 3, and you get a system with twoequations with two unknowns, a and b.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com =IÍm interested in doing some animations in maya for which time will haveslowed signi'cantly and speeds will be fast enough that I was intrigued atthe possibility of doing modeling over Minkowskian space-times. I have readpapers that explain some of the ways in which Lorentz contraction is notnaively visible due to light traversal computations, and I wanted to see towhat extent that was true across various environments with complicatedobjects and moving lighting and camera. This does mean, however, that raytracing, shading, shadow casting, and such must not use the projectionalgorithms which assume basically an in'nite speed of light, but mustactually trace photon paths in time.I was curious if such a possibility had been already encapsulated in anyplug-in or script, or failing that, if anyone had any ideas as to whatpart(s) of the rendering system would need to be programmed to accomplishthis. I am fairly good at both MEL and the c++ plug-in API, but I am havingdif'culty seeing how to put things together to get the correct physics.Must the calculations be done entirely in the plug-in with the animations,deformations, ray-tracing, and all done collectively, or is there some cutelittle idiom that will give the same physics (for some reason, I keepcamera...)?I also was curious about other topologies with generalised metrics.Switching back to in'nite light speed assumptions for a moment, ignoringtime and looking only at spatial dimensions, I was curious if there were anytools for constructing complexes, offering surgery operations, local spacedeformers, etc. It seems like the calculations based on curvature could behandled by standard refraction routines, but I do not know to get thesurgery identi'cations working (esp. again with light tracing).And of course it would be nice getting both facilities working together, forthings like black hole modeling and similar fun. My belief is that all ofthese modeling possibilities would make great educational tools forvisualisation in both in the physics community as well as the differentialtopology / geometry crowds, but I suspect with the many simulations run inthese areas in the scienti'c communities, tools may have spilled over intothe maya community for 'lm, which would assist my project immensely.I thank you for any directions, either in the modelling algorithms or whereto apply them in the maya rendering pipeline.-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar =: One apparent way of avoiding the paradoxes of naive set theory is to: turn set-de'ning characteristic functions into partial functions from: sets to the three-valued logic {T,F,bot}. This three-valued logic: extends classical logic in the obvious way with: and(T,bot)=or(F,bot)=bot, and(F,x)=F, or(T,x)=T, not(bot)=bot, so that: every truth function has a 'xed point.Very much LukasiewiczÍ trivalent logic, then. That is good, and it allowseasy extension to polyvalent logics.: Obviously the law of excluded middle does not hold: or(a,not(a)) only: implies that a is T or bot. This gives the logic a constructive: character.Constructivism has been one of the driving reasons behind the introductionof alternate logics. Removing the constraint of excluded middlegeneralises one into the realm of Heyting algebras, where constructivetheories roam, while allowing the avoidance of the antinomies found in theBoolean.: In my formulation, I identify each set with its characteristic: function from sets to {T,F,bot}. Thus given s:set, the membership of: an element x can be tested with s(x). In the general case, this is a: partial function, returning bot for some values. I call such sets: partial sets, and sets whose characteristic function is in {T,F}: total sets. Every set of ZF and NF is a total set, with this theory: admitting a strictly larger class of sets than either.In some ways, this is the structure of a topoi, or at least a functor from atopoi category to something similar. Is the approach meant to be categorial(which, by the way, is a good thing in my eyes -- IÍm just curious)? Then,your partial classi'cation appears to mainly distinguish the topoi of setsfrom some of the many other topoi.: RussellÍs set R={s:set|not(s(s))} is then a partial set. All ZF sets: are elements of R, while some elements of NF are not elements of R,: while some new sets such as R itself are of undecidable membership.:: IÍve translated the ZF axioms to this set theory, rephrasing them in: terms of characteristic functions and new for-all and there-exists: logic operators performing logical conjunction and disjunctions across: all elements of a characteristic functions. Everything appears to be: sound and avoids known paradoxes.The categorial study of paradox is becoming a large 'eld these days, and itappears you may be repeating some of the work already done (which can besoooo frustrating sometimes!). I donÍt mean to assume any level of study,but perhaps I might suggest that, if you havenÍt, you should check out someonline I can suggest.: With the new axioms, it is easy to construct a bijection from the: universal set to its power set. CantorÍs proof that |P(x)|>|x| for: all non-empty sets x proceeds by constructing C={a:x|not(P(x)(a)} and: using the law of excluded middle to derive a contradiction on its: membership in P(x). This goes away for lack of excluded middle,: leaving C a partial set which appears not to be constructively: contradictory.:: The one worrying aspect of this approach is that it identi'es sets: with characteristic functions from sets to logic values:: Set=Set->{T,F,bot}. I have only been able to develop an intuition of: such sets in a purely constructive way, by writing down a 'nite list: of possibly self-referential equations de'ning sets, and convincing: myself that a unique solution exists. This is much in the style of: NFÍs axiom that every (possibly cyclic) graph corresponds to a set,: but I allow unlimited comprehension.:: Are there any known problems with this approach to set theory? Any: pointers to research on the topic?No known problems that I am aware of. In fact, it seems to me to be one ofthe more successful modern approaches for classifying paradox. However, ifI could make one suggestion, it would be to not restrict yourself to yourtrivalent logic. Any Heyting algebra is possible, and expands your researchinto the much more fruitful world that all topoi present. In fact, becauseof natural distinctions that present themselves between propositions thattake on the middle value, trivalent theories are often looked at only assummarisations of a more natural in'nitely valent theory. Good resourcesfor this can be found in intuitionist discussions, but it is more general.Also, may I ask why you posted to comp.lang.functional? This intrigues mebecause some of my own research has been around the evaluation of the lambdacalculus and proof / evaluation theory in the context of non standardlogics, but I do not see this approach explicitly stated in your message.Good luck with your researche!-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar =Garry Denke, GeologistDenoco Inc. of Texas =How can i proove that equation x^3+2y^3+4z^3-34xyz=0 is only true then x=y=z=0,if x,y,z are whole numbers? => How can i proove that equation x^3+2y^3+4z^3-34xyz=0 is true> then x=y=z=0, if x,y,z are whole numbers?that x, y, and z are even, so x/2, y/2, z/2 is also a solution.---------------------------------------------------- -----------------------John R Ramsden (jr@adslate.com)--------------------------------------------- ------------------------------Eternity is a long time, especially towards the end. Woody Allen =*** post for FREE via your newsreader at post.newsfeed.com ***IÍm doing a module on Probability Theory, and IÍm not sure I ïgetÍ theconcept of conditional probability of a random variable over a sigma 'eld.I think I understand the mathematics of the situation, but I donÍt reallyhave an intuitive grasp of the situation.By a sigma 'eld I mean a set B, of subsets of a set A, containing the emptyset, and closed under taking complements, and countable intersections.According to the de'nition IÍve been given, for a random variable Y onprobability space (A,F,P), with E[|Y|] < in'nity and D a sub sigma 'eld ofF. A version of E[Y|D] is any function from A to the real numbers that is Dmeasurable, and whose integral over each member of D is the same as that ofY, the integrals being over P. I say all that so we know exactly what weare talking about.What seems to be happening is that the expectation over a sigma 'eld D, issort of sampling the random variable Y, so it acts in the same way over thesets in D.But perhaps someone could tell me if there is a better way of understandingintuitively what this means.Perhaps there isnÍt a good intuitive way of understanding this, and if so,IÍll just accept this as a pure abstraction, and concentrate on the maths,but if not, it would be good if you could explain it to me.http://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- =>>I have just attended a course on scheduling for project managers and there>>the beta distribution comes up in relation to estimation of task durations.>>I have looked up in my old probability theory book regarding this and found>>the proof for the fact that the sum of _many_ beta distributions converges>>to a gaussian distribution,>Presumably youÍre talking about independent random variables with the >same parameters for the beta distribution. Then this is just the Central>Limit Theorem - nothing special about beta distribution.>> but neither my beloved book (and some hard core>>integral solvining) or extensive search on the internet has given me>>what I have found (this includes characteristic function of the beta>>distribution and itÍs density function) I am beginning to believe that there>>is no distribution de'ned/named for the sum of two beta distributions, but>>I might be wrong.>IÍm almost certain thereÍs no closed-form general formula for the >distribution of the sum (again, assuming independence). In some >particular cases (e.g. if the parameters are positive integers) there >may be a formula, but itÍs not a standard named distribution.The distribution of the sum of k Beta independent random variables has a density which is not analytic at any sumof endpoints; there are between k+1 and 2^k of these,depending on the particular ranges, and all but 2 of theendpoints are internal.The distribution of a sum of independent Beta random variablesis close to normal if no variance is large compared to thesum, and the parameters do not get too extreme; iid always works, but the rate of convergence is well estimated by thesum of the third absolute moments about the mean divided bythe cube of the standard deviation, the Berry-Esseen result.-- 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 University =*** post for FREE via your newsreader at post.newsfeed.com ***IÍm doing a module on Probability Theory, and IÍm not sure I ïgetÍ theconcept of conditional probability of a random variable over a sigma 'eld.I think I understand the mathematics of the situation, but I donÍt reallyhave an intuitive grasp of the situation.By a sigma 'eld I mean a set B, of subsets of a set A, containing the emptyset, and closed under taking complements, and countable intersections.According to the de'nition IÍve been given, for a random variable Y onprobability space (A,F,P), with E[|Y|] < in'nity and D a sub sigma 'eld ofF. A version of E[Y|D] is any function from A to the real numbers that is Dmeasurable, and whose integral over each member of D is the same as that ofY, the integrals being over P. I say all that so we know exactly what weare talking about.What seems to be happening is that the expectation over a sigma 'eld D, issort of sampling the random variable Y, so it acts in the same way over thesets in D.But perhaps someone could tell me if there is a better way of understandingintuitively what this means.Perhaps there isnÍt a good intuitive way of understanding this, and if so,IÍll just accept this as a pure abstraction, and concentrate on the maths,but if not, it would be good if you could explain it to me.-- me.http://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- =In probability, a sigma-algebra represents information. Imagine anexperiment takes place behind a curtain. You cannot see the result,but you have a computer and you may type in questions such as didevent A happen? where A is some event of your choosing. If thecomputer could answer your question for every event A, then you mightas well just pull back the curtain and look at the result, becauseyouÍll be able to 'gure it out by systematically asking the rightquestions. So letÍs suppose the computer can only answer your questionfor some restricted class of events. Call this set of events B. Now,if we think about it for a moment, itÍs clear that, without loss ofgenerality, we can assume B is an algebra. If you can 'nd out whetheror not A has happened, then you can found out whether or not Acomplement has happened, and the same with unions. Okay, so fortechnical reasons, letÍs go a step further and assume itÍs asigma-algebra.So the sigma-algebra B represents information in the sense that youdonÍt know the outcome, w, but you do know whether or not w is in Afor every A in B. Now, if a function f is measurable with respect toB, then we can evaluate f(w) without knowing w, provided we knowwhether or not w is in A for every A in B.Given the information B, E[X|B] is that function f(w), which we canevaluate solely based on the information we have, that gives the bestguess as to the value of X.ThereÍs another nice way of looking at conditional expectation when Xhas 'nite variance. Let (O,F,P) be your probability space and for asub-sigma-algebra B of F, let L^2(B) denote the space of randomvariables with 'nite second moment that are measurable with respectto B. Then L^2(B) is a vector subspace of L^2(F) and if X is inL^2(F), then E[X|B] is the vector projection of X onto L^2(B).> *** post for FREE via your newsreader at post.newsfeed.com ***IÍm doing a module on Probability Theory, and IÍm not sure I ïgetÍ the> concept of conditional probability of a random variable over a sigma 'eld.> I think I understand the mathematics of the situation, but I donÍt really> have an intuitive grasp of the situation.By a sigma 'eld I mean a set B, of subsets of a set A, containing the empty> set, and closed under taking complements, and countable intersections.According to the de'nition IÍve been given, for a random variable Y on> probability space (A,F,P), with E[|Y|] < in'nity and D a sub sigma 'eld of> F. A version of E[Y|D] is any function from A to the real numbers that is D> measurable, and whose integral over each member of D is the same as that of> Y, the integrals being over P. I say all that so we know exactly what we> are talking about.What seems to be happening is that the expectation over a sigma 'eld D, is> sort of sampling the random variable Y, so it acts in the same way over the> sets in D.But perhaps someone could tell me if there is a better way of understanding> intuitively what this means.Perhaps there isnÍt a good intuitive way of understanding this, and if so,> IÍll just accept this as a pure abstraction, and concentrate on the maths,> but if not, it would be good if you could explain it to me.-- > me.> http://www.newsfeed.com - The #1 Newsgroup Service in the World!> -----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- => *** post for FREE via your newsreader at post.newsfeed.com ***IÍm doing a module on Probability Theory, and IÍm not sure I ïgetÍ the> concept of conditional probability of a random variable over a sigma 'eld.> I think I understand the mathematics of the situation, but I donÍt really> have an intuitive grasp of the situation.By a sigma 'eld I mean a set B, of subsets of a set A, containing the empty> set, and closed under taking complements, and countable intersections.According to the de'nition IÍve been given, for a random variable Y on> probability space (A,F,P), with E[|Y|] < in'nity and D a sub sigma 'eld of> F. A version of E[Y|D] is any function from A to the real numbers that is D> measurable, and whose integral over each member of D is the same as that of> Y, the integrals being over P. I say all that so we know exactly what we> are talking about.What seems to be happening is that the expectation over a sigma 'eld D, is> sort of sampling the random variable Y, so it acts in the same way over the> sets in D.But perhaps someone could tell me if there is a better way of understanding> intuitively what this means.Perhaps there isnÍt a good intuitive way of understanding this, and if so,> IÍll just accept this as a pure abstraction, and concentrate on the maths,> but if not, it would be good if you could explain it to me.> expectations, martingales, and tangent sequences. It might be a little advanced for your needs, but then again, when you get onto martingales, you may 'nd it helpful.Basically, the example given by G.A. Edgar in another post is essentially completely representative of what a conditional expectation is.I think that it is also very instructive to consider the case when the sub-sigma 'eld consists of 'nitely many sets.Best, Stephen =>*** post for FREE via your newsreader at post.newsfeed.com ***>IÍm doing a module on Probability Theory, and IÍm not sure I ïgetÍ the>concept of conditional probability of a random variable over a sigma 'eld.>I think I understand the mathematics of the situation, but I donÍt really>have an intuitive grasp of the situation.>By a sigma 'eld I mean a set B, of subsets of a set A, containing the empty>set, and closed under taking complements, and countable intersections.>According to the de'nition IÍve been given, for a random variable Y on>probability space (A,F,P), with E[|Y|] < in'nity and D a sub sigma 'eld of>F. A version of E[Y|D] is any function from A to the real numbers that is D>measurable, and whose integral over each member of D is the same as that of>Y, the integrals being over P. I say all that so we know exactly what we>are talking about.>What seems to be happening is that the expectation over a sigma 'eld D, is>sort of sampling the random variable Y, so it acts in the same way over the>sets in D.This is incorrect.>But perhaps someone could tell me if there is a better way of understanding>intuitively what this means.Since the argument is by means of the Radon-NikodymTheorem, we need to go back to that to get theunderstanding. BTW, one can de'ne conditionalexpectation with respect to a measure which is nota probability measure, and this can be useful.At any rate, suppose we have a 'nite measure space (A, D, P),and we have a (signed) measure Q on D, and for convenience wewill assume that |Q(S)|/P(S)| is bounded for all S in D. Wedo not need such strong assumptions, but the R-N theorem canbe reduced to this, and the intuition remains valid.Now for any partition Z of D into a 'nite (countable can beused as well) family of sets of positive P measure, de'ner_Z(x) = Q(S)/P(S) for that S which contains x. Take asequence of partitions Z_n which spreads out the rÍs; underthe conditions given, making int (r_Z_n)^2 dP go to itsleast upper bound will work, but otherwise a maximal spreadingout is needed, and any will do. Then r_Z_n will converge inmeasure, and the limit is the Radon- Nikodym derivative, whichfor your problem is the conditional expectation.So the intuition is to take conditional expectations on 'nitepartitions, and pass to the limit using a sequence of partitionswhich use as much of the information as they can. It is thenatural extension from discrete partitions.-- 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 University =>*** post for FREE via your newsreader at post.newsfeed.com ***>>IÍm doing a module on Probability Theory, and IÍm not sure I ïgetÍ the>concept of conditional probability of a random variable over a sigma 'eld.>I think I understand the mathematics of the situation, but I donÍt really>have an intuitive grasp of the situation.>>By a sigma 'eld I mean a set B, of subsets of a set A, containing the empty>set, and closed under taking complements, and countable intersections.>>According to the de'nition IÍve been given, for a random variable Y on>probability space (A,F,P), with E[|Y|] < in'nity and D a sub sigma 'eld of>F. A version of E[Y|D] is any function from A to the real numbers that is D>measurable, and whose integral over each member of D is the same as that of>Y, the integrals being over P. I say all that so we know exactly what we>are talking about.>>What seems to be happening is that the expectation over a sigma 'eld D, is>sort of sampling the random variable Y, so it acts in the same way over the>sets in D.Well thatÍs what it is, sort of - the D-measurable function that comesas close as possible to being Y, given that itÍs D-measurable.>But perhaps someone could tell me if there is a better way of understanding>intuitively what this means.You might consider a few trivial examples:(i) If D is the trivial sigma-algebra containing only the empty set and A then the conditional expectation is equal to the expectedvalue.(ii) Say you have two non-independent random variables that takeonly a few values: P(X=0, Y=0) = P(X=0,Y=1) = 1/4,P(X=1,Y=1) = 1/2. In particular E(Y) = 3/4.Intuitively, if you know that X = 0 then E(Y) = 1/2 andif you know that X = 1 then E(Y) = 1, which one wouldstate formally as E(Y | X=0) = 1/2, E(Y | X=1) = 1.Say D is the sigma-algebra generated by X; then infact E(Y|D) is equal to 1/2 on the set where X = 0and 1 on the set where X = 1; so E(Y|D) correspondsvery nicely with the intuitive notion of expected value of Y, given the value of X.>Perhaps there isnÍt a good intuitive way of understanding this, and if so,>IÍll just accept this as a pure abstraction, and concentrate on the maths,>but if not, it would be good if you could explain it to me.************************David C. Ullrich =Another example to consider. The unit square A=[0,1]x[0,1], F=Borelsets, D=sets of the form Ux[0,1] where U is Borel in [0,1].Then a D-measurable function is (essentially) a function that dependsonly on the 'rst variable x. For conditional expectation E(Z | D)(x),take r.v. Z(x,y) and integrate with respect to y to get somethingdepending only on x.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ =>...>Therefore, while the sqrt() operator is in place you cannot further>resolve the expression to see *which* of the solutions is coprime to>7.>> Let primes denote complex conjugation and let c denote a root of> a^2 - a + 42 = 0 supposed to be coprime to 7 in the ring A of> algebraic integers. Note that the other root is cÍ.> If c and 7 are coprime in A then there are elements p and q in A> such that cp + 7q = 1.>>Be careful here. In JamesÍ all-inclusive ring of two complex conjugates,>one can be a unit while the other is not a unit. A bit strange, and I>do not know whether his ring really does exists, but the proof>fails for that. Because if c is a unit, q = 0 and p = inv(c). But>pÍ = inv(c)Í does not necessarily belong to the ring (conjugation is>not a ring operation).>> So 1 = 1Í = (cp + 7q)Í = (cp)Í + (7q)Í = cÍpÍ + 7ÍqÍ = cÍpÍ + 7qÍ,> and since pÍ and qÍ are in A, cÍ and 7 are coprime in A.>>Indeed, a valid proof is conjugation is a valid operation in the ring.Actually, I thought I was careful to restrict my argument to thealgebraic integers, the ring in which James alleged coprimality.I agree that such an argument might not go through in the Harrisring, but might it not fail for a more basic reason? It is notobvious that his ring would be a BÍezout domain.James has on several occasions lowered his shield of vaguenesssuf'ciently to allow identi'cation of a speci'c element of hisring that is not an algebraic integer. Each time, people havejumped up and down pointing out the unintended consequences if thatnumberÍs (algebraic) conjugates were also included. The usuallyimplicit appeal to the symmetry rendering conjugates algebraicallyindistinguishable probably went way over JamesÍ head. Certainly,he has always run away from the questions raised, but his silencemight be explained as withdrawal of his examples after realisingtheir inapplicability, perhaps due to some ineffable subtletyvisible only to himself. I am therefore reluctant to deduce anyproperties of his ring from those examples.John Roberts-Jones =>>In X a normal space, there is a continuous function f: X -> [0,1] such>>that f(x) =0 on A and f(x) = 1 on B AND 0 < f(x) <1 OTHERWISE, if and>>only if A and B are disjoint closed G-delta sets in X.>>Does anybody know where I can 'nd a proof of this theorem? I donÍt>> I have not seen this theorem before, but I 'nd it trivial.>> All that one needs to use about G-delta sets is that they are>> countable intersections of open sets, the de'nition.>Well ok, would you detail a quicker proof in a few lines?I prefer to toss out another hint. An equivalent form ofUrysohnÍs Lemma is that in a normal space, if A is closed,and U is an open superset of A, there is a continuous functionwhich is 0 on A and l outside of U.-- 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 University =When working with some geometric objects I arrived to the followingobjectthat I called strong bijections. Let G be a group and H a subgroup of G. A bijection f:G->G is strongif,for all a,b in G,a in Hb implies f(a) in Hf(b).The most obvious strong bijections are the mappings R_x:G->G withR_x(a)=ax (where x in G).Another example is when H is characteristic in G. That is if thesubgroup H satis'es f(H) is contained in H, for all f automorphismof G, then every automorphism of G is also a strong bijection. So my questions are the following:1. Are strong bijections already known in group theory ( probablyunder another name; if so please give me the reference so that I can learn more about this).2. Does someone can provide a large class of groups whose strongbijections are described? For example, what conditions must a group G(and the subgroup H) satisfy in order that the only strong bijectionsare the described above (automorphisms and right translations, thatis, elements of the holomorph).I thank ideas and references. =>When working with some geometric objects I arrived to the following>object>that I called strong bijections. >>Let G be a group and H a subgroup of G. A bijection f:G->G is strong>if,>for all a,b in G,>>a in Hb implies f(a) in Hf(b).>>The most obvious strong bijections are the mappings R_x:G->G with>R_x(a)=ax (where x in G).>>Another example is when H is characteristic in G. That is if the>subgroup H satis'es f(H) is contained in H, for all f automorphism>of G, then every> automorphism of G is also a strong bijection. >>So my questions are the following:>>1. Are strong bijections already known in group theory ( probably>under another name; if so please> give me the reference so that I can learn more about this).G acts naturally on the set of left (or right) cosets of H in G bymultiplication. What you are asking is that the bijection respect thisaction, since you want a=b (mod H) --> f(a)=f(b) (mod H)So it looks like you are looking for (set-theoretic) bijections on Gthat induce isomorphisms of the G-set G/H. >2. Does someone can provide a large class of groups whose strong>bijections are described? For example, what conditions must a group G>(and the subgroup H) satisfy in order that the only strong bijections>are the described above (automorphisms and right translations, that>is, elements of the holomorph).>>I thank ideas and references.I doubt you can get anything nice. Since all you are requiring on f isthat it be a bijection, then there will be a lot of bijections: 'rstpick any permutation of the cosets of H (there are [G:H]! of them),then pick any permutation of each coset class (there are |H|! of each,for a total of |H|*|H|!); the resulting bijection will satisfy yourcondition. -- about what I accept as reality. --- Calvin (Calvin and Hobbes) was wondering if anyone have seen java implementation of PPMT function ?Or some description of the algorithm ?I would really appreciate any helpLM =>> I was wondering if anyone have seen java implementation of PPMT function ?> Or some description of the algorithm ?>> I would really appreciate any help>> LM>>PPMT = Primitive Polynomial Modulo TwoPPMT = Penylated Potein Carboxyl MthyltransferasePPMT = Pinciple PaymentPerhaps:http://www.vni.com/products/jmsl/v25/api/com/ imsl/'nance/Finance.htmlNot sure that helps, Flip =I am new to the world of stats and probabilty was wondering if anyonecould explain a wee bit about probabilty densityI am trying to write a piece of software quite like Granular SynthesisDemo 1 inhttp://www.dcs.gla.ac.uk/~jhw/audioclouds/ . I was wondering if anyonecould explain the difference between a probabilty distribution and aprobabilty density?And if you have a probabilty distribution can a density be obtainedfrom it and if so how?Also the main bit i am stuck on is how a density can be conditioned(in the example given it is conditioned by the x and y positions of amouse) how would you go about doing such a thingAny help would be greatly = I was wondering if anyone> could explain the difference between a probabilty distribution and a> probabilty density?for a real random variable?> And if you have a probabilty distribution can a density be obtained> from it and if so how?Differentiate. If r.v. X has distribtion function F that meansF(x) = Prob(X<=x), and the probability density (if it exists,that is if F is absolutely continuous) is f(x) = FÍ(x).I agree with the other answer. Coding something you do not understandwell is likely to lead to inferior (or incorrect) code. =>I am new to the world of stats and probabilty was wondering if anyone>could explain a wee bit about probabilty density>I am trying to write a piece of software quite like Granular Synthesis>Demo 1 in>http://www.dcs.gla.ac.uk/~jhw/audioclouds/ . I was wondering if anyone>could explain the difference between a probabilty distribution and a>probabilty density?>And if you have a probabilty distribution can a density be obtained>from it and if so how?>Also the main bit i am stuck on is how a density can be conditioned>(in the example given it is conditioned by the x and y positions of a>mouse) how would you go about doing such a thingI would strong recommend getting hold of a good book on probabilityand reading up BEFORE coding. It is nearly 40 years since I startedto learn it, and nearly 35 since I taught it, so I am not a goodperson to give you references! The books I used would still be OK,but are probably hard to get :-)Nick Maclaren. =I am stuck by this problem:Let G be a group and H a subgroup of G. The index [G:H]=n. Prove thatthere exists a normal subgroup of G, such that1. K is a subgroup of H;2. the index [G:K] is divisible by n!Any hint? => I am stuck by this problem:Let G be a group and H a subgroup of G. The index [G:H]=n. Prove that> there exists a normal subgroup of G, such that> 1. K is a subgroup of H;> 2. the index [G:K] is divisible by n!Any hint?ShouldnÍt property 2. read:2. the index [G:K] divides n!? E.g., G = Z_3, H = {1}; no subgroup K of G can have 3! = 6 divides [G:K]. =>I am stuck by this problem:>>Let G be a group and H a subgroup of G. The index [G:H]=n. Prove that>there exists a normal subgroup of G, such that>1. K is a subgroup of H;>2. the index [G:K] is divisible by n!>>Any hint?Let G act on the set of left cosets of H by letting g send the cosetaH to the coset gaH.There are n cosets. Show that this action induces a homomorphism fromG to S_n; let K be the kernel. Show K has the desired properties.-- about what I accept as reality. --- Calvin (Calvin and Hobbes) problem as part of a> job-interview pre-screen. After giving a totally wrong answer (ïcause> I wasnÍt thinking), I gave the following answer. However, the> interviewer said that my answer was still incorrect.Fair enough; I 'nd probability unintuitive. But now my curiosity is> peaked: What is the right answer? Question: You have a jar with 1000 coins in it, 999 are normal, but> one is two-headed (i.e both sides are heads). You choose a coin at> random and toss it 10 times. With each toss the coin comes up heads,> so you get 10 consecutive heads. What is probability that you have> choosen the two-headed coin? My answer:P(two headed) = 1 - P(normal coin *and* 10 heads in a row)> = 1 - .999 * (1/2)^10> = 1 - .999 * (1/1024)> = 1 - .000976> = 0.999024> = 99.9024%>> If this is wrong, what is the right answer? How do you calculate it? 10 heads not 10 heads2heads 1/1000 1 0normal 999/1000 1/1024 1023/1024=>P(2heads | 10 heads) = (1/1000)/(1/1000+999/1024000)= 1/(1+999/1024)= 1024/2023Phil-- Unpatched IE vulnerability: NavigateAndFind protocol historyDescription: cross-domain scripting, cookie/data/identity theft, command execution =P(fake coin | ten heads)= P(fake coin and ten heads) / P(ten heads)= P(fake coin) / P(ten heads)= P(fake coin) / [ P(legitimate coin and ten heads) + P(fake coin) ]= (1/10^4) / [ (999/10^4)*(1/2)^10 + (1/10^4) ]= 2^10 / (999 + 2^10)| probability problem as part of a| job-interview pre-screen. After giving a totally wrong answer (ïcause| I wasnÍt thinking), I gave the following answer. However, the| interviewer said that my answer was still incorrect.|| Fair enough; I 'nd probability unintuitive. But now my curiosity is| peaked: What is the right answer?|| Question: You have a jar with 1000 coins in it, 999 are normal, but| one is two-headed (i.e both sides are heads). You choose a coin at| random and toss it 10 times. With each toss the coin comes up heads,| so you get 10 consecutive heads. What is probability that you have| choosen the two-headed coin?|| My answer:|| P(two headed) = 1 - P(normal coin *and* 10 heads in a row)| = 1 - .999 * (1/2)^10| = 1 - .999 * (1/1024)| = 1 - .000976| = 0.999024| = 99.9024%|| If this is wrong, what is the right answer? How do you calculate it?|| Stuart| =The problem also appears in Timothy Falcon CrackÍs Heard on the street:quantitative questions in Wall Street job interviews. He suggests thatintuitively the answer should be near 0.5, since the probabilities of 10heads with a normal coin (1/1024) and choosing the two-headed coin (1/1000)are close to each other.AkshayStuart Brorson schreef probability problem as part of a> job-interview pre-screen. After giving a totally wrong answer (ïcause> I wasnÍt thinking), I gave the following answer. However, the> interviewer said that my answer was still incorrect.>> Fair enough; I 'nd probability unintuitive. But now my curiosity is> peaked: What is the right answer?>> Question: You have a jar with 1000 coins in it, 999 are normal, but> one is two-headed (i.e both sides are heads). You choose a coin at> random and toss it 10 times. With each toss the coin comes up heads,> so you get 10 consecutive heads. What is probability that you have> choosen the two-headed coin?>> My answer:>> P(two headed) = 1 - P(normal coin *and* 10 heads in a row)> = 1 - .999 * (1/2)^10> = 1 - .999 * (1/1024)> = 1 - .000976> = 0.999024> = 99.9024%>> If this is wrong, what is the right answer? How do you calculate it?>> Stuart> = > P(two headed) = 1 - P(normal coin *and* 10 heads in a row)Your start is already wrong. P(two headed) = 1/1000 by the de'nitionof the problem. Coming from the recesses of my memory I 'nd BayesÍtheorem: P(A|B) = P(A and B)/P(B)LetÍs have A = two headed B = 10 heads in a row.You want P(A|B) i.e. the probability of A (two headed) assuming thatB (10 heads in a row) did occur. Now P(A and B) = 1, and: P(B) = P(A and B) + P(not A and B) = 1 + 999/1024000.So P(A|B) = 1/(1 + 999/1024000) = 1024000/1024999 ~ 99.9025%; you gave: > = 99.9024%This may be a slight difference, but your reasoning was wrong. Try tounderstand BayesÍ theorem.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ => P(two headed) = 1 - P(normal coin *and* 10 heads in a row)>>Your start is already wrong. P(two headed) = 1/1000 by the de'nition>of the problem. Coming from the recesses of my memory I 'nd BayesÍ>theorem:> P(A|B) = P(A and B)/P(B)>LetÍs have A = two headed B = 10 heads in a row.>You want P(A|B) i.e. the probability of A (two headed) assuming that>B (10 heads in a row) did occur. Now P(A and B) = 1, and:> P(B) = P(A and B) + P(not A and B)> = 1 + 999/1024000.>So P(A|B) = 1/(1 + 999/1024000) = 1024000/1024999 ~ 99.9025%; you gave:>> = 99.9024%>>This may be a slight difference, but your reasoning was wrong. Try to>understand BayesÍ theorem.Take stock of what we are given.The probability of choosing the two-headed coin is 1/1000; that is,P(A) = 1/1000.Given that we have chosen the two-headed coin, the probability that wewill get 10 heads in a row is 1; that is, P(B|A) = 1.The probability of choosing the normal coin is 999/1000; that is,P(~A) = 999/1000Given that we have chosen the normal coin, the probability that we willget 10 heads in a row is 1/1024; that is, P(B|~A) = 1/1024. P(B) = P(B and A) + P(B and ~A) = P(A) P(B|A) + P(~A) P(B|~A) (Bayes on each summand) = 1/1000 * 1 + 999/1000 * 1/1024What we are asked is: given that we have gotten 10 heads in a row, whatis the probability that we chose the two-headed coin? That is, what isP(A|B)?Using Bayes Theorem, we get P(A and B) = P(A) P(B|A), and therefore, P(A|B) = P(A and B) / P(B) = 1/1000 * 1 / (1/1000 * 1 + 999/1000 * 1/1024) = 1024 / 2023 = .506178942...Rob Johnson take out the trash before replying => P(two headed) = 1 - P(normal coin *and* 10 heads in a row)>> Your start is already wrong. P(two headed) = 1/1000 by the de'nition> of the problem. Coming from the recesses of my memory I 'nd BayesÍ> theorem:> P(A|B) = P(A and B)/P(B)> LetÍs have A = two headed B = 10 heads in a row.> You want P(A|B) i.e. the probability of A (two headed) assuming that> B (10 heads in a row) did occur. Now P(A and B) = 1, and:> P(B) = P(A and B) + P(not A and B)> = 1 + 999/1024000.How can P(B) > 1? The problem is that P(A and B) = P(A)*P(B|A) = (1/1000)*1= 1/1000. So that P(B) = 1/1000 + 999/1024000 = (1/1000) (1+999/1024). Sothat P(A|B) = P(A and B)/P(B) = 1/(1+999/1024)= 1024/(1024+999) = .5062> So P(A|B) = 1/(1 + 999/1024000) = 1024000/1024999 ~ 99.9025%; you gave:> = 99.9024%>> \ This may be a slight difference, but your reasoning was wrong. Try to> understand BayesÍ theorem.> --> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,+31205924131> home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/ => Your start is already wrong. P(two headed) = 1/1000 by the de'nition> of the problem. Coming from the recesses of my memory I 'nd BayesÍ> theorem:> P(A|B) = P(A and B)/P(B)> LetÍs have A = two headed B = 10 heads in a row.> You want P(A|B) i.e. the probability of A (two headed) assuming that> B (10 heads in a row) did occur. Now P(A and B) = 1How can P(A and B) be greater than P(A) = P(two headed) = 1/1000?-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes /039 53 36 N / 086 11 55 W => Question: You have a jar with 1000 coins in it, 999 are normal, but> one is two-headed (i.e both sides are heads). You choose a coin at> random and toss it 10 times. With each toss the coin comes up heads,> so you get 10 consecutive heads. What is probability that you have> choosen the two-headed coin? My answer:P(two headed) = 1 - P(normal coin *and* 10 heads in a row)> = 1 - .999 * (1/2)^10> = 1 - .999 * (1/1024)> = 1 - .000976> = 0.999024> = 99.9024%If this is wrong, what is the right answer? How do you calculate it?ItÍs a conditional probability. You want P(two headed | 10 heads) = P(two headed and 10 heads)/P(10 heads)P(two headed and 10 heads) = 1/1000. P(10 heads) = (999/1000)*(1/2)^10 + 1/1000. So the answer appears to be 1/[(999/1024) + 1] = .5061789... =:> Question: You have a jar with 1000 coins in it, 999 are normal, but:> one is two-headed (i.e both sides are heads). You choose a coin at:> random and toss it 10 times. With each toss the coin comes up heads,:> so you get 10 consecutive heads. What is probability that you have:> choosen the two-headed coin? :> headed | 10 heads): = P(two headed and 10 heads)/P(10 heads): P(two headed and 10 heads) = 1/1000. P(10 heads) = (999/1000)*(1/2)^10 + : 1/1000. So the answer appears to be: 1/[(999/1024) + 1] = .5061789...This doesnÍt have the right limiting behavior. If I toss the coin 1e6times and it still comes up heads every time, then I can be (almost)certain that I have the two-headed coin. Right? Stuart =>> :> Question: You have a jar with 1000 coins in it, 999 are normal, but> :> one is two-headed (i.e both sides are heads). You choose a coin at> :> random and toss it 10 times. With each toss the coin comes up heads,> :> so you get 10 consecutive heads. What is probability that you have> :> choosen the two-headed want>> : P(two headed | 10 heads)>> : = P(two headed and 10 heads)/P(10 heads)>> : P(two headed and 10 heads) = 1/1000. P(10 heads) = (999/1000)*(1/2)^10 +> : 1/1000. So the answer appears to be>> : 1/[(999/1024) + 1] = .5061789...>> This doesnÍt have the right limiting behavior. If I toss the coin 1e6> times and it still comes up heads every time, then I can be (almost)> certain that I have the two-headed coin. Right?No. P(n heads) = (999/1000)*(1/2)^n + 1/1000 -->1/1000 as n-->oo. SoP(two headed | n heads) = = P(two headed and n heads)/P(n heads) --> 1 asn --> oo. => : ItÍs a conditional probability. You want: P(two headed | 10 heads): = P(two headed and 10 heads)/P(10 heads): P(two headed and 10 heads) = 1/1000. P(10 heads) = (999/1000)*(1/2)^10 +> : 1/1000. So the answer appears to be: 1/[(999/1024) + 1] = .5061789...This doesnÍt have the right limiting behavior. If I toss the coin 1e6> times and it still comes up heads every time, then I can be (almost)> certain that I have the two-headed coin. Right? Right. But how does this not have the right limiting behavior?P(two headed and 1e6 heads) = 1/1000P(1e6 heads) = (999/1000)*(1/2)^1e6 + 1/1000 = (approx) 1/1000 + 1.009e-301030So, this answer is about 1/(1 + 1.009e-301027) or 1 - 1.009e-301027Looks almost certain to me.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes /039 53 36 N / 086 11 55 W => In my ignorance, but what does modelled ZFC in PA mean?You can encode statement about ZFC as statements about naturals, in> such a way that a proof in ZFC translates to a proof in PA. As a> result, if there is an inconsistency in ZFC then there is also an> inconsistency in PA. Google for relative consistency or for G.9adel.I enjoy ShmuelÍs postings -- usually he sets people straight. In thiscase however I think he misremembers what Goedel did. Goedel did provean important relative consistency result: con(ZF) => con(ZFC). He didthis by constructing a model of ZF, the constructible unviverse L, andthen showed that AC holds in that model. The construction involvestrans'nite recursion over ranks in the cumulative hierarchy, and is thusbeyond Peano Arithmetic -- essential use is made of the Axiom of In'nity.Michel. <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <97adneZXNdeRbWCi4p2dnA@comcast.com> => Do mathematicians care if a theory is physically realizable?> My idea of in'nite computation is no less feasible than> using induction over all of the natural numbers.Maybe in'nite computations are interesting to study, maybe not.Whether theyÍre physically realizable or not is possibly beside thepoint, as you mention. But it is not at all obvious that theyÍrerelevant here, since youÍre mixing non-standard computational modelswith de'nitions and results for standard computational models.Anyway, I agree that feasibility isnÍt fundamental for whether thesemodels are interesting, but that we shouldnÍt assume that these modelsare relevant for the issues at hand. In particular, you simplymisstate TuringÍs de'nition here.Turing de'nes a computable number as any number that can berepresented by a computable sequence. A computable sequence is theoutput tape of a TM that has an in'nite number of 0Ís and/or 1Ís.Turing did *not* allow an in'nite number of 1Ís on the output tape,so this is simply not his de'nition. By the way, induction over all natural numbers has nothing to do withfeasibility and everything to do with the de'nition of N and with themeaning of universal quanti'cation. Induction does not require thatone actual go through all of the natural numbers and ensure that thenext one satis'es the proposition.-- Jesse HughesOf course, my ability to admit my mistakes and correct them is atrait that many of you seem to never have properly appreciated. -- JSH, discussing his 1463rd proof of FermatÍs Last Theorem. <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <87brpe1wtx.fsf@phiwumbda.org> <878ykhzlss.fsf@phiwumbda.org> => I know the standard proof. One shows the de'nable numbers are> countable, the real numbers are uncountable and concludes there are> an in'nite number of unde'ned reals. (ItÍs horrible that I can> recite these proofs, now.)>> The obvious §aw in this argument is that every real number has a> de'nition. A real number is de'ned as the limit of a Cauchy> sequence of rationals (or a Dedekind cut). Either there are an> uncountable number of such de'nitions or the real numbers are> countable.Of course there are an uncountable number of Cauchy sequences, butthese arenÍt de'nitions in the relevant sense. Only countably manyof these sequences can be speci'ed in a 'nite number of symbolsthemselves (i.e., de'ned in the relevant sense here). There is nocontradiction.> The computable numbers are supposed to be countable. Even so, the> computable numbers include all of rationals and all of the> de'nable irrationals. It is hard to imagine that the limit of a> Cauchy sequence would be uncomputable. Not hard at all if the Cauchy sequence itself is uncomputable in anappropriate sense. Most of them are, after all.> DoesnÍt this mean that every real number is computable?No.-- Jesse F. HughesNot all features that are found on the Security tab are designed tohelp make your documents and 'les more secure. -- Microsoft clari'es the Of'ce security features. =>I havenÍt a clue about how GoedelÍs proof proceeds. Sorry. I would>>appreciate a gentle outline of how the translation goes myself>>(gentler than go to the library and see the goedel? Get with it man, this is> the 21st century. For the 'rst time in the human history, learning> is *easy*-- just a click (or two) away!Google for Goedel did not yield any results that were detailedenough to sketch the proof Shmuel was describing. You overestimatethe utility of Google, or my ability to use it. But, if youÍre so con'dent that Google suf'ces, please Google for meand show me the site that sketches GoedelÍs proof of the consistencyof choice clearly enough to show the translation of proofs in ZFC toproofs in PA, together with some basic properties of the translation,but preferably not such detail that IÍm better off going to the source.> I want to know how to bottle feed my three baby goats with two hands> while my wifeÍs always hungry dog watches, drooling. Goedel *is*> easy by comparison. ;-)involving GoedelÍs 'rst and second incompleteness theorems in twofragments of set theory (not a the reassurances.None of this gives me the insight to evaluate ShmuelÍs claims (and IcanÍt take the \ time to read the originals now). A quick check of thelogic texts on my shelf gave me nothing relevant either (casting somedoubt on the claim Con(PA) => Con(ZF) --- that would be worth a linein one or two of those texts, surely).-- Mathematicians are rather important in the infrastructures of manyorganizations that protect civilization. IÍve determined that theyare a consistent security risk, and seem to have other agendas, otherloyalties beyond loyalty to their respective nations. -- James Harris =>It wouldnÍt matter if you could model a consistent theory in PA.>> No they donÍt. IÍll try to convince you of this two ways. First> note that if every proof in ZFC translates to a proof in PA then> every proof in ZF can also be so translated. By any reasonable> de'nition of translate, we can also translate from PA back to ZFC,> ZF, and PA. So every theorem of ZFC is also a theorem of ZF. At> this point the only real problem is that we all know the C in ZFC> stands for Choice, and not, say, Cohen. So it is at least> *plausible* that not every proof in ZFC translates into a proof in> PA, right?Of course itÍs plausible. I have only ShmuelÍs word for the contrary,>> At this point I should probably stop writing, because I donÍt really> understand the subject well enough to explain it clearly and thus> convince you that in fact not all proofs in ZFC translate to proofs> in PA.>> However, as near as I can 'gure, yes it is true that you can encode> statements about ZFC as statements about the naturals but NOT in the> way Shmuel Metz suggests. That there is an interpretation, f, from> the theory cn(PA) to the theory ZFC is a fact. By cn(PA) I mean the> consequences of PA, i.e. those sentences in PA that are true in> every model of PA. So to encode a theorem in ZFC, say s, as a> sentence in PA is easy: let t=f^(-1)(s). Why does this work? > Because interpretations are, by de'nition, 1-1. Shmuel Metz> asserts that t is a theorem of PA, hence t is in cn(PA), hence f is> onto ZFC, hence f is a bijection. My (very fuzzy, at this point)> understanding is that Godel(?) proved something akin to the> opposite: there is NO such *bijective* f, unless ZFC is> inconsistent. Perhaps my understanding is incorrect.While IÍm doubtful now about ShmuelÍs claim, I donÍt think that yourargument above is compelling. When we reduce a theory T to S and S toT, we do not require that the composites of the translations are theidentity, do we? We simply require that the translation maps proofsin T to proofs in S and vice versa.Well, unlike you, I donÍt plan on returning to proof theory any timesoon to check that my failing memory is correct. -- Jesse HughesShe moaned, in pain and pleasure, as, in a confused whirlwind, sheglimpsed an image of Saint Sebastian riddled with arrows, cruci'edand impaled. --Mario Vargas Llosa on category theory =>> presumably in a natural way, so that any proof of P & ~P in ZFC would>> yield a proof of Q & ~Q for some formula Q of PA.>> This line I do not understand. Gentzen proved consistency of PA.> But he did it in ZF... using induction up to epsilon_0.> So how is the above reduction possible?This isnÍt a result with which I am really familiar, so I canÍt help> you. ShmuelÍs response to one of my own postings is about all I know> about it.(Fourth attempt to post a reply I made to an earlier post by Jesse dropped the thread reference, incase that was the cause of IÍmaccessing it from home now where I canÍt see them.)>>Well, now we must be a bit careful. Shmuel Metz informs me that>Goedel proved the relative consistency of ZFC to PA --- so that *if*>PA is consistent, *then* ZFC is consistent. Hence, an inconsistency>in ZFC would in fact yield an inconsistency in PA --- not due to the>simple fact that PA has a model in ZFC, but to the more surprising>(reported) fact that ZFC has a model in PA.>>Now, I am not familiar with this result (I wonÍt say that IÍve never>seen \ it, but if IÍve seen it, it musta been some time ago). So, donÍt>take my word for it. IÍm only reporting what Shmuel said (though, I>assume that heÍs not mistaken on such an explicit statement of fact).If con(PA)=>con(ZF) then GentzenÍs proof of con(PA) must have used asystem stronger than ZFC. Which system was that?Maybe IÍm out of my depth here, but I thought that the two thingsGoedel proved that are relevant to this thread are:(1) Any ('rst-order) theory strong enough to express primitive-recursive arithmetic cannot be both consistent and complete.(2) con(ZF) => con(ZFC)Neither of these implies con(PA)=>con(ZF). PA is actually stronger thanwhat is needed to arithmetize proofs (the machinery behind (1) above),and weaker than ZF (if consistent), in that, say, the Goodstein theoremis trivially provable in ZF (which has big enough ordinals) but not in PA(which does not have big enough ordinals, so to speak).If ZFC is inconsistent then there is an arithmetic statement that encodesthe proof of S & ^S for some statement S about set membership relations.It encodes the 'nite number of ZFC axioms and axiom schema instances usedin the proof. I donÍt see how this says anything on the consistency of PA;indeed, it might even be the case that (if PA is in fact consistent) thisstatement is unprovable in PA. That depends on where in the arithmeticalhierarchy (quanti'er alternations) the Goedel sentences are, and this IdonÍt remember.Michel. <87u135nv7i.fsf@phiwumbda.org> <874qv5nprb.fsf@phiwumbda.org> => If ZFC is inconsistent then there is an arithmetic statement that encodes> the proof of S & ^S for some statement S about set membership relations.> It encodes the 'nite number of ZFC axioms and axiom schema instances used> in the proof. I donÍt see how this says anything on the consistency of PA;> indeed, it might even be the case that (if PA is in fact consistent) this> statement is unprovable in PA. That depends on where in the arithmetical> hierarchy (quanti'er alternations) the Goedel sentences are, and this I> donÍt remember.You evidently understand matters more thoroughly than I do, havingeither never seen the representation of ZFC proofs in PA or (much morelikely) having forgotten everything about it in the interveningyears. I have only tried to state what I think Shmuel was claiming inresponse to me, but I have not been comfortable with the claimCon(PA) => Con(ZF). Just doesnÍt seem like IÍd forget *that*relation!Well, okay, I might forget it, because IÍm just that incompetent.Still...So, unless Shmuel can provide some real details that show GoedelÍspaper on the consistency of Choice and CH yields also a proof of Con(PA) => Con(ZF) (which would be a fairly deep corollary!), I willreturn to my previous belief: If ZF is inconsistent, it does notnecessarily mean that all the big theories sink with it.Sorry for any confusion my confusion has caused. I wonÍt takeanyoneÍs word for such claims so easily hereafter (though, at least Ihope I made clear I was reporting hearsay).-- Just because youÍre ... in a Ph.d program it does not mean thatyouÍre up to the challenge of being a real mathematician. Only thosewho have a purity of mind and dedication to the truth as the highestideal have a chance. --James Harris, as Sir Galahad the Pure. =Say we are working in the ring of differential operators...How would you go about simplifying the following...[Using * for the ring multiplication][cos(t) * d/dt * sin(t) * d/dt]Basically I am not sure if you can ïforgetÍ the rightmost d/dt sincethat is just to differentiate the function - i.e it does notsimplify...so would it then become [cos(t) * [d/dt * sin(t)] * d/dt] [cos(t) * [cos(t) + Sin(t) * d/dt] * d/dt] [cos^2(t) * d/dt + cos(t)sin(t) *d/dt ] etc...Or am I forgetting something ?- ==>> How would you go about simplifying the following... [Using * for> the ring multiplication]>> [cos(t) * d/dt * sin(t) * d/dt]Looks like a chain rule. cos(t)*D*sin(t)*D should equalcos(t)[cos(t)*D + sin(t)*D^2] which equalscos^2(t) * D + sin(t) * D^2Len. =Herman Rubin schrieb in Nachricht ...>>Never heard this word before. It must be very modern Hebrew or very Ancient>>Aramaic. There is the shoresh LVE - to loan, lend, borrow. But that is all I>>can help you with.>>It is very modern Hebrew, and almost certainly derived>from the English word.Jeff Miller in http://members.aol.com/jeff570/i.html dates nilpotent andidempotent to approximately 1870, introduced (invented?) by Benjamin Peirce. Hermann-- => Herman Rubin schrieb in Nachricht ...>>Never heard this word before. It must be very modern Hebrew or veryAncient>>Aramaic. There is the shoresh LVE - to loan, lend, borrow. But that isall I>>can help you with.>>It is very modern Hebrew, and almost certainly derived>from the English word.>> Jeff Miller in http://members.aol.com/jeff570/i.html dates nilpotentand> idempotent to approximately 1870, introduced (invented?) by BenjaminPeirce.>> Hermann> -->Great. And impotent?YS---Checked by AVG anti-virus system (http://www.grisoft.com). <3ffd93c5.16333168@news.onetel.net.uk>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft =In 05:31 PM, reneged@anglickkan.org (The Revd TerenceFformby-Smythe) said:>You silly little convert. Were you asleep during yiddish classes?Mah pitom! Zo safah achereth!-- Shmuel (Seymour J.) Metz, SysProg and JOATnot 23:11:33 -0500, Shmuel (Seymour J.) Metz>In 05:31 PM, reneged@anglickkan.org (The Revd Terence>Fformby-Smythe) said:>You silly little convert. Were you asleep during yiddish classes?>>Mah pitom! Zo safah achereth!Ah, someone else was awake in yiddish class. Enjoy already. =The matter now is that I donÍt know this is right or wrong :if lim [p(x)] = lim [q(x)] when x-> c, then lim[pÍ(x)] = lim[qÍ(x)] when x-> c.May anyome \ con'rm that for me, plzzz.If itÍs right, I can solve the problem above.---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- => The matter now is that I donÍt know this is right or wrong :> if lim [p(x)] = lim [q(x)] when x-> c, then lim[pÍ(x)] = lim> [qÍ(x)] when x-> c.> \ May anyome con'rm that for me, plzzz.> If itÍs right, I can solve the problem above.Consider p(x) = x and q(x) = -x at c = 0.-- Dave SeamanJudge YohnÍs mistakes revealed in Mumia Abu-Jamal ruling. => Consider the function g(x) = f(x) - ax - b. What is its asymptote?Its asymptote is the line y = 0.> What can you say about gÍÍ(x)?g(x) = f(x), so \ g(x) >0 for all x>0, And g(x) is a concavefunction for x>0.---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- magidin@math.berkeley.edu>>Okay I have never been very good with math and hopfully someone out>>there can help. Here is the situation:>>I Have a total of 've different items: I am shipping out bowes with a>>combination of these different items and need to prepare all the>>different combinations before hand. So lets name the items:>>1. tshirt>>2. glove>>3. hat>>4. ring>>5. necklace>>now a package can have either all 've items or a choice of lets say>>number 1,2,4 or 1,2 or 1,4,5 or just 2 and so on... I think you probly>>get the jist of what im asking.>>What are the total number of combinations for the items obviously>>without just rearanging thier order. and what is the formula to do so?>>You are asking how many different packages there are. Each package>determines a subset of {1,2,3,4,5} (given by which items are in the>box), and you are allowing any subset except for the empty>subset. There are 2^5 = 32 possible subsets, one of which is empty. So>there are 31 different possible combinations/boxes.Additionally, you might wonder how to 'll each of the boxes in anorganized way without missing or repeating some subset. One way to dothat would be to write the numbers 1 to 31 in binary:00001000100001100100......11111Now assign each binary bit to one of your items:rightmost bit = tshirtsecond rightmost = glovemiddle = hatsecond leftmost = ringleftmost = necklaceThen which bits are = 1 tell you what to put in that box. For examplebox number 23 which is 10111 would have necklace, hat, glove, tshirt.This would be pretty easy to program or automate.--LynnIs there an ef'cient way to solve the following problem:Given a,b,eps>0 'nd all positive n 1. It does halt and we can prove it.> 2. It does halt and we canÍt prove it.> 3. It doesnÍt halt and we can prove that it doesnÍt.> 4. It doesnÍt halt and we canÍt prove that it doesnÍt.>> Other interesting questions: What can we say about case # 4? What> necessary or suf'cient conditions can we give? Can we ever prove> that a particular H(x,y) is case # 4? When can we prove that it> isnÍt? Generalize these questions by considering each subset of 1-4> instead of just # 4. Any takers?>> I can expound on condition #3.> A TM will not halt if it has no halt states.> A halt state is a state that has no instruction for some input.> This can be determined by examining the state transition> table which is 'nite.> A TM will halt if all of the halt states are unreachable.> A state is unreachable if it is not the initial state and> no other state changes to the unreachable state.> Again, this can be determined from the state table.There are lots of other ways in which a halt state is unreachable.And there are TM which will halt for some initial tapes but not others.Well of course in this case we are talking about a particular TM witha particular input tape.> The halting problem is not so trivial as you would have it.No, it isnÍt trivial at all. If you could solve the halting problem,then you could easily solve FLT, GoldbachÍs Conjecture, and a wholehost of other unsolved or incredibly dif'cult problems ofmathematics.BTW: ItÍs instructive to express various problems of mathematics suchas these in terms of recursive or recursively enumerable sets(relations) over the natural numbers and a Predicate Calculus wffusing this relation that must be evaluated or be recursive orrecursively enumerable but is not in general.For example, FLT involves the 4-place relation FLT(A,B,C,N) de'ned as(A+1)^(N+3)+(B+1)^(N+3)=(C+1)^(N+3). (The +1 is to make sure youdonÍt include the trivial cases such as 0^3+1^3=1^3, and the +3 is tomake the exponent 3 or more.) Code (A,B,C,N) as a single number so wecan talk about sets for each such problem. In my notation, P(I) meansset P is recursive and P(x) means set P is recursively enumerable. Sowe are given that FLT(I) and want to evaluate (eA)FLT(A). In generalterms, we have P(I) and want to evaluate (eA)P(A).We can very nicely classify unsolved problems of mathematics in thismanner. Try this with GoldbachÍs Conjecture, the Twin PrimeConjecture and the conjecture concerning the termination of the Ulamprogram, for starters.Charlie VolkstorfCambridge, MA> Russell> - 2 many 2 count = > 1. It does halt and we can prove it.> 2. It does halt and we canÍt prove it.> 3. It doesnÍt halt and we can prove that it doesnÍt.> 4. It doesnÍt halt and \ we canÍt prove that it doesnÍt.>> Other interesting questions: What can we say about case # 4? What> necessary or suf'cient conditions can we give? Can we ever prove> that a particular H(x,y) is case # 4? When can we prove that it> isnÍt? Generalize these questions by considering each subset of 1-4> instead of just # 4. Any takers?>> I can expound on condition #3.> A TM will not halt if it has no halt states.> A halt state is a state that has no instruction for some input.> This can be determined by examining the state transition> table which is 'nite.> A TM will halt if all of the halt states are unreachable.> A state is unreachable if it is not the initial state and> no other state changes to the unreachable state.> Again, this can be determined from the state table.There are lots of other ways in which a halt state is unreachable.And there are TM which will halt for some initial tapes but not others.Well of course in this case we are talking about a particular TM witha particular input tape.> The halting problem is not so trivial as you would have it.No, it isnÍt trivial at all. If you could solve the halting problem,then you could easily solve FLT, GoldbachÍs Conjecture, and a wholehost of other unsolved or incredibly dif'cult problems ofmathematics.BTW: ItÍs instructive to express various problems of mathematics suchas these in terms of recursive or recursively enumerable sets(relations) over the natural numbers and a Predicate Calculus wffusing this relation that must be evaluated or be recursive orrecursively enumerable but is not in general.For example, FLT involves the 4-place relation FLT(A,B,C,N) de'ned as(A+1)^(N+3)+(B+1)^(N+3)=(C+1)^(N+3). (The +1 is to make sure youdonÍt include the trivial cases such as 0^3+1^3=1^3, and the +3 is tomake the exponent 3 or more.) Code (A,B,C,N) as a single number so wecan talk about sets for each such problem. In my notation, P(I) meansset P is recursive and P(x) means set P is recursively enumerable. Sowe are given that FLT(I) and want to evaluate (eA)FLT(A). In generalterms, we have P(I) and want to evaluate (eA)P(A).We can very nicely classify unsolved problems of mathematics in thismanner. Try this with GoldbachÍs Conjecture, the Twin PrimeConjecture and the conjecture concerning the termination of the Ulamprogram, for starters.Charlie VolkstorfCambridge, MA> Russell> - 2 many 2 count =It would be nice, and give your book a substantial edge over the competition, if you could manage to work in examples which:i) motivate the development of further theoryii) illustrate how the theory just developed is used to answer ïconcreteÍ mathematical questions (where ïconcreteÍ means not solely related to the topic at hand; showing how mathematics is interconnected)I speak from frustrated experiences with book which jump from subject to arcane subject without providing any direction to the process.Also, please consider making the later chapters independent of your 'rst chapters on logic and set theory, etc., because some readers will jump past them.Good luck.Alex-- =I have been trying to 'gure out why there is a /constrained/ quadraticprogamming step in SQP methods. I donÍt understand why the QP isnÍtformulated without constraints, instead using Langrange multiplierssince it would be much easier to solve.If you can tell me the answer IÍd be very grateful! IÍm not sure if itsa stupid question or not, so IÍve described my thinking below:I currently understand that SQP is an extension of quasi-Newton methodsto constrained optimisation. A quasi-Newton method is used to maintain apositive de'nite approximation to the Hessian of the objectivefunction. Each iteration involves 'nding the optimum step direction bysolving the QP based on the second order approximation of the objectiveusing this approximate Hessian with 'rst order approximations to theconstraint functions. The solution of this constrained QP yields thestep direction for both the parameters and the Lagrange multipliers forthe full NLP and a line search yields the best scaling of the step vector.What I donÍt understand is this: why not formulate the QP as anunconstrained problem using an approximated Hessian to the Lagrangian ofthe objective function. Since the quasi-Newton method yeilds a positivede'nite approximate Hessian this unconstrained QP will be triviallysolved with a Linear system. BFGS can give an inverse to the approximateHessian quite easily using a Choleski factorization which can also beeasily updated. This would mean that the Lagrange multipliers would haveto be updated outside of the QP solution, -but this would be fairlystraghtforward with an augmented Lagrangian method based on some penaltyfunction?IÍm guessing that either (i) it would be hard/impossible to 'nd themultipliers outside of the QP step for some reason, or (ii) it would bepossible but it is more ef'cient to 'nd as part of a constrained QP.Very grateful for any help!Best,Josh. =IÍm reading an introductory book on topology and the author de'nes a d-limit point p of a set S in a pseudometric space as a point having the property that for every eps>0 there is in S a point s different than p such that the metric d(p,s) Is there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass?Just use the standard compass-and-straightedge method for dividing a linesegment into n equal portions, and set n to 3.-- Dave SeamanJudge YohnÍs mistakes revealed in Mumia Abu-Jamal ruling. =>> Is there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass? Sure. Construct a parallel line segment in three equal parts(of whatever length is easiest), then project that line onto theoriginal segment. Given: segment AB to trisect Construct line CD parallel to AB. (Should be easy.) Construct point E on CD such that CD=DE. Construct point F on CD such that CD=DE=EF. Draw line CA. Draw line FB, intersecting CA at G. Draw lines GD and GE; these lines trisect AB.Note that trisecting an *angle*, on the other hand, is impossibleto do with only compass and straightedge.-Arthur => Is there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass?>> Sure. Construct a parallel line segment in three equal parts> (of whatever length is easiest), then project that line onto the> original segment.> Given: segment AB to trisect>> Construct line CD parallel to AB. (Should be easy.)> Construct point E on CD such that CD=DE.> Construct point F on CD such that CD=DE=EF.> Draw line CA.> Draw line FB, intersecting CA at G.> Draw lines GD and GE; these lines trisect AB.>> Note that trisecting an *angle*, on the other hand, is impossible> to do with only compass and straightedge.ThatÍs not necessarily true. I can trisect an angle which measures 270degrees or even 135 degrees.>> -Arthur =>>Is there an easy way to trisect a line segment, i.e. split it into>three equal portions, with just straightedge and compass?>>If not an easy way, is there any way at all?>>Jonathan Christensenconvenient angle, say 45 degrees or so, call this vertex V. Mark apoint P on M a ways down the line. Center your compass at P, set itsdrawing point on the vertex V and sweep an arc to mark a point Q twiceas far down M as P is. Now center your compass at Q, its drawing pointat P, and sweep another arc to locate a point R on M another equaldistance along M. At this point V P Q R are equally spaced on M. Drawthe line, call it N, from R to the far end of L from V. At the pointsP and Q construct lines parallel to N using the standard procedure.Where these lines intersect L are your required trisection points.--Lynn =A Google[tm] search on can recite pi reveals that MelissaJoan Hart (who plays or played Sabrina the Teenage Witch)can recite pi to 341 decimal places. :-)Keith Ramsay =>Message-id: Google[tm] search on can recite pi reveals that Melissa>Joan Hart (who plays or played Sabrina the Teenage Witch)>can recite pi to 341 decimal places. :-)>>Keith RamsaySo thatÍs how you summon Satan?--MensanatorAce of Clubs =I came across 2 questions I had that maybe someone can answer :1) If A is an n x n matrix, and ||A|| is de'ned as the smallest number c> such that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers,> then why is it true that a sequence of matrices A_m converges to a matrix A> if and only if ||A_m - A|| --> 0?by de'nition.verify that ||.|| above is a norm.> 2) Consider an in'nite series whose terms are matrices : A_0 + A_1 + A_ 2> + ... If the sum (from m = 0 to in'nity) of ||A_m|| < in'nity, then> the in'nite series of matrices is said to converge absolutely.this is an extension of a calculus theorem. itÍs proof is similar.> Why is it true that if a series converges absolutely, then the partial> sums of the series form a cauchy sequence?let space V possess a norm ||.||, and let epsilon > 0.we know that if S = Sum fj converges, then||Sn - S|| < epsilon/2 for some n large enough,where Sn = Sum fj, j = 1 to infty.||Sn - Sm|| loe ||Sn - S|| + ||Sm - S|| < epsilonfor m and n large enough.> Moshe => I came across 2 questions I had that maybe someone can answer :>> 1) If A is an n x n matrix, and ||A|| is de'ned as the smallest numberc> such that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers,> then why is it true that a sequence of matrices A_m converges to amatrix A> if and only if ||A_m - A|| --> 0?>> 2) Consider an in'nite series whose terms are matrices : A_0 + A_1 +A_ 2> + ... If the sum (from m = 0 to in'nity) of ||A_m|| < in'nity,then> the in'nite series of matrices is said to converge absolutely. Why isit> true that if a series converges absolutely, then the partial sums of the> series form a cauchy sequence?> What have you done so far? In both cases, they are fairly simple if> you know the proper background.I have taken undergrad analysis and linear algebra...I guess I will takeWorld Wide WadeÍs advice and lookover proofs of absolute convergence impliesconvergence. =>> 1) If A is an n x n matrix, and ||A|| is de'ned as the smallest numberc> such that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers,> then why is it true that a sequence of matrices A_m converges to amatrix A> if and only if ||A_m - A|| --> 0?>> Well, ||A_m - A|| -> 0 is often the de'nition of the convergence of a> sequence of matrices. WhatÍs your de'nition?>Let A_n be a sequence of complex matrices. A_n converges to a matrix A ifeach entry of A_n converges to the corresponding entry of A.> 2) Consider an in'nite series whose terms are matrices : A_0 + A_1 +A_ 2> + ... If the sum (from m = 0 to in'nity) of ||A_m|| < in'nity,then> the in'nite series of matrices is said to converge absolutely. Why isit> true that if a series converges absolutely, then the partial sums of the> series form a cauchy sequence?>> Start with ||A + B|| <= ||A|| + ||B||. Then go back to calculus and review> the proof that absolute convergence implies convergence. matrix, and ||A|| is de'ned as the smallest number>c>> such that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers,>> then why is it true that a sequence of matrices A_m converges to a>matrix A>> if and only if ||A_m - A|| --> 0?>> Well, ||A_m - A|| -> 0 is often the de'nition of the convergence of a>> sequence of matrices. WhatÍs your de'nition?>>Let A_n be a sequence of complex matrices. A_n converges to a matrix A if>each entry of A_n converges to the corresponding entry of A.If a_i,j is the entry of in row i and column j then you can write aformula for a_i,j involving A: a_i,j = _____ A ______ .ItÍs clear from that formula that |a_i,j| <= ||A||. (Or possibly that |a_i,j| <= c ||A||, depending on what norms we choose for all this.)>> 2) Consider an in'nite series whose terms are matrices : A_0 + A_1 +>A_ 2>> + ... If the sum (from m = 0 to in'nity) of ||A_m|| < in'nity,>then>> the in'nite series of matrices is said to converge absolutely. Why is>it>> true that if a series converges absolutely, then the partial sums of the>> series form a cauchy sequence?>> Start with ||A + B|| <= ||A|| + ||B||. Then go back to calculus and review>> the proof that absolute convergence implies convergence.>>************************David C. Ullrich => Zero is not needed as a placeholder in the Alternate Number System.> Also in this system in base 1, the digits are oneÍs as required and is> simply a tally system.> In base (b) the digits are 1 to b which is more logical than didgits 0 to> (b-1) in our existing system.> http://my.tbaytel.net/forslund/ans.html>IÍm confused by the following quote from your webpage:I say this because of the following three §aws in ENS.(1) In ENS, the digit zero behaves differently from other digits. For example,when any two digits (except zero) are added in ENS, the result differs fromthe original two numbers. When zero, on the other hand, is added to anynumber, the result is the original number - nothing happens. You appear to be saying that we donÍt need an additive inverse in ournumber system. Aside from all the complications that would arise inhigher mathematics, not even subtraction is well de'ned anymore.For instance, in ANS, what is one minus one equal to? I read yourpaper and didnÍt see anywhere you addressed this problem or evenacknowledged it existed.ANS really doesnÍt strike me as a particularly practical alternative, muchless a necessary one. IÍd imagine, though, that it could be an interestingcuriosity to study.Kev => I say this because of the following three §aws in ENS.(1) In ENS, the digit zero behaves differently from other digits.> For example, when any two digits (except zero) are added in ENS, the> result differs from the original two numbers. When zero, on the> other hand, is added to any number, the result is the original> number - nothing happens. You appear to be saying that we donÍt need an additive inverse in our> number system. Aside from all the complications that would arise in> higher mathematics, not even subtraction is well de'ned anymore.> For instance, in ANS, what is one minus one equal to? I read your> paper and didnÍt see anywhere you addressed this problem or even> acknowledged it existed.I believe the point is in adding *digits* here, not in addingnumbers. Remember, the discussion is about representations only. Soyou always change something, unless one of them is zero. That makeszero (the digit, not the number) seem rather idle and useless. Zero the number is represented by the empty string of digits in thissystem, which has a certain conceptual elegance, but is attended bysome problems as well. (A point I already made previously.)Thomas 19:39:22 .999... is not equal to 1, therefore any operation>> involving .999... may not substitute 1 therefor.>> At least, one can fashion a semi-consistent math system>> therewith, although it would take a lot of work and probably>> would not be worth it.>>> (For starters, if .999... != 1, what is .999... - (.999... * 10 - 9)>> equal to? Answer: 0.000...9. This sort of thing will get rather>> weird very quickly.)>>> Of course, if one cubes .999... one can do something like the following:>>> .9 * .9 * .9 = .729>>> .99 * .99 *.99 = .729 + 3*.0729 + 3*.00729 + .000729>> = .729 + .2187 + 0.02187 + .000729>> = .970299>>> Note the 7 in the 10^-2 spot.>>> .999 * .999 * .999 = .970299 + 3*.0088209 + 3*.00008019 + .000000729>> = .970299 + .0264627 + .00024057 + .000000729>> = .997002999>>> and yet another 7, this time in the 10^-3 spot.>>> .9999 * .9999 * .9999 = .997002999 + 3*.0008982009 + 3*.0000008091 +>> .000000000729>> = .997002999 + .0026946027 + 3*.0000024273 +>> .000000000729>> = .999700029999>>> and thereÍs that 7 again. Is that second term always going to be>> .000...0002something ?>>> Of course itÍs probably simpler to abstract the problem and>> equate .999...9 = 1 - 10^(-n), and then>>> (1 - 10^(-n))^3 = 1 - 3*10^(-n) + 3*10^(-2*n) - 10^(-3*n)>>> which proves that (.999...9)^3 = .999...99700...0299...9, in a>> simpler fashion than your equations below, or for that matter>> my algebra.>>> So far, no problem here. But one cannot conclude that>>> (.999...)^3 = .999...99700...0299...>>> unless one ascribes a de'nitive meaning to digit expansions>> with in'nite ellipses and de'nes their proper addition,>> subtraction, etc. It is naive to conclude that the>> arithmetic operation>>> 10 * (.999...)^3 - 9 - (.999...)^3 >>> = 9.999...97000...2999...990 - 9 - .999...99700...0299...999>> = -.000...02699...7300...009.>> makes any sense \ at all.I have no idea how to write it in a way that anyone here would accept,> but perhaps there is a way to write the squares of 0.999... tending> toward 0, not toward 1, settling the matter. Simple arithmetic says... 0.9^2 = .81> 0.9^3 = .729> 0.9^4 = .6561> 0.99^2 = .9801> .099^3 = .970299> 0.99^4 = .96059601 0.999^2 = .998001> 0.999^3 = .997002999> 0.999^4 = .996005996001> 0.9999^2 = .99980001> 0.9999^3 = .999700029999> 0.9999^4 = .9996000599960001 0.99999^2 = .9999800001> 0.99999^3 = .999970000299999> 0.99999^4 = .99996000059999600001 0.999999^2 = .999998000001> 0.999999^3 = .999997000002999999> 0.999999^4 = .999996000005999996000001 0.9999999^2 = .99999980000001> 0.9999999^3 = .999999700000029999999> 0.9999999^4 = .9999996000000599999960000001> 0.99999999^2 = .9999999800000001> 0.99999999^3 = .999999970000000299999999> 0.99999999^4 = .99999998000000059999999600000001 0.999999999^2 = .999999998000000001> 0.999999999^3 = .999999997000000002999999999> 0.999999999^4 = .999999996000000005999999996000000001 0.9999999999^2 = .99999999980000000001> 0.9999999999^3 = .999999999700000000029999999999> 0.9999999999^4 = .9999999996000000000599999999960000000001 0.99999999999^2 = .9999999999800000000001> 00.99999999999^3 = .999999999970000000000299999999999> 0.99999999999^4 = .99999999996000000000059999999999600000000001 0.999999999999^2 = .999999999998000000000001> 0.999999999999^3 = .999999999997000000000002999999999999> 0.999999999999^4 = .999999999996000000000005999999999996000000000001etc., etc., etc....the squares are heading toward 0. The powers are heading towards 1 as one increases the number of 9Ís.In the limit, everything becomes 1^n = 1, at least according tomost mathematical theories.Of course for any *'nite* expansion, lim (n->+oo) .999...9^n = 0.But the double limitlim(m,n->+oo) (1 - 10^(-m))^n = ?is not computable -- lim(m->+oo)(1 - 10^(-m))^n = 1 for any 'nite n,lim(n->+oo) (1 - 10^(-m))^n = 0 for any 'nite m.A mildly interesting exercise of course is to compute expressionssuch as(1 - 10^(-n))^(10^n)and see what limit that goes to -- it turns out to be 1/e.However, (1 - 10^(-n))^n tends to 1.Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netItÍs still legal to go .sigless. =In sci.logic, Dave digit sequence .999... is meaningless as one must always>> use a trailing digit approach, writing it as .999...9, where ...>> can be either 'nite (which is allowed under the Standard>> Representation) or in'nite (which the Standard Representation>> does not cover and dismisses). This means that operations>> such as (.999...9)^2 = .999...99800...001 are possible.>> Note that [3] would require a bit of work rewriting trans'nite>> theory, as omega (the ordinal analogue to the cardinal in'nity)>> is such that omega+1 = omega in the Standard Representation. Omega+1 has the \ same cardinality as omega, but they are different> ordinals. If the (w+1)-strings of digits are ordered lexicographically,> then there is no order-preserving map of such strings to the reals.>> Interesting. So expressions such as omega+1 and 2*omega actually do>> make sense? If so, mea culpa.Yes. In fact, w is identical to N, the set of natural numbers (including 0),> and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }. Addition and multiplication are noncommutative: 1+w = w < w+1, (w+1 has a last element but 1+w does not)> 2*w = w < w*2 = w+w. (an w of pairs vs. a pair of wÍs)Ugh. My brainÍs beginning to hurt now... :-)So which one of these applies to such Garry-Denke-isque decimalexpansions as(.999....)^2 = .999...99800...001 ?It feels like 8 is in the w+2 or w+3 position but I want to make sure.1 is presumably in the position w*2 + 2.Presumably 1+w refers to things like9,(9,9,9,9,...)where one tacks things onto the left side of an in'nite sequence.If one computes .999... / 10 + .8 = .8999..., one gets a similarsequence.>> However, if we do allow trans'nites and in'nitesimals into our>> number system a la Garry Denke, IÍm not sure how the rest of mathematics>> is going to deal with it (the concept of limit in particular may have>> to be rede'ned).>> Hopefully itÍs a simple retro't but I wonder.ThereÍs nothing wrong with considering trans'nite ordinal sequences of> digits in general, but as I said, there is no order-preserving map of> such objects to the reals for ordinals > w.> That might be good enough. :-) Of course thatÍs probably relatedto the impossibility of reliably de'ning things such as.999... - .999...99800...001 = .000...?!?!?!...:-)-- #191, ewill3@earthlink.net -- insert random hurting brain hereItÍs still legal to go .sigless. =>In sci.logic, Dave Seaman considering trans'nite ordinal sequences of>> digits in general, but as I said, there is no order-preserving map of>> such objects to the reals for ordinals > w.>>>That might be good enough. :-) Of course thatÍs probably related>to the impossibility of reliably de'ning things such as>>.999... - .999...99800...001 = .000...?!?!?!...>>:-)One way to look at this is that each of the digits after the 'rst ... have a positional multiplier of (1/10)^x where x>=w, which as real values would all be 0.-- ---------------------------| B B aa rrr b || BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- <8htjvvsrqs7sinrosrbfsbslnvimaf6i5i@4ax.com> <4dq5d1-7ld.ln1@lexi2.athghost7038suus.net> =>>[re: ordinal numbers]> 1+w = w < w+1, (w+1 has a last element but 1+w does not)> 2*w = w < w*2 = w+w. (an w of pairs vs. a pair of wÍs)>> Ugh. My brainÍs beginning to hurt now... :-)>> So which one of these applies to such Garry-Denke-isque decimal> expansions as>> (.999....)^2 = .999...99800...001 ?>> It feels like 8 is in the w+2 or w+3 position but I want to make sure.> 1 is presumably in the position w*2 + 2. Looks to me like 8 comes after w 9Ís, which I would call the w+1position. 1 would then be in position 2*w+2, as you said.> ThereÍs nothing wrong with considering trans'nite ordinal sequences of> digits in general, but as I said, there is no order-preserving map of ^^^^^^^^^^^^^^^^^^^^^^^^^^> such objects to the reals for ordinals > w. ^^^^^^^^^^^^^^^^^^^^^^^^^> That might be good enough. :-) Of course thatÍs probably related> to the impossibility of reliably de'ning things such as>> .999... - .999...99800...001 = .000...?!?!?!... Obviously, the 'rst w -0.000...800...001Nothing complicated about that, is there? :) (Sure, it *looks* like0.999... is bigger than 0.999...800..., but thatÍs just your intuitionplaying tricks, isnÍt it?) At this point, though, I think youÍre just playing games with digits.As Dave said, thereÍs no way to make these strings correspond to realnumbers. (Members of R, that is. ;-) Although there would be non-order-preserving maps, I guess, as long as we prohibit strings of theform .111...10222...20333... ...90101010...100111...10121212... ... ...which would, I suppose, require w^w digits to write out in full, andbe completely unmappable to the reals. If IÍm not mistaken.-Arthur =In sci.logic, David W newsreader.com>:>> In sci.logic, David W Cantrell>> standard question for equations such as x^2 = x is:>> Give the set of values that x can be set to, such that the> equation holds.>> This set is {0, 1}; both 0 and 1 satisfy the equation.> ItÍs a little harder to show that nothing else will,> though one can try algebra and get lucky in this case:>> x^2 = x> x^2 - x = 0> x(x - 1) = 0>> In an in'nite 'eld such as the reals, itÍs now obvious that> 0 and 1 are the only solutions. (In the ring of integers> mod 6, though, things get interesting, as the solution set> is {0,1,3,4}.}>> FWIW, the original equation also has another solution if our number>> system is the extended reals. Then the solution set is {0, 1, +oo}.>> Good point. :-)Is that the best you can say? Many mathematicians call the point +oo> ideal, which makes it sound far better than merely good! ;-)I for one canÍt call it ideal, thatÍs another concept! :-)Of course {+oo} would be an ideal in the extended reals ring, ifone likes, since +oo * +oo = +oo.{0, 1} would be ideal, too, as it turns out. {0, 1, +oo} mightbe an ideal if one adopts the convention that 0 * +oo = 0(a la Rudin). But thatÍs not an ideal solution... or is it? :-)It might also be noted that x = +oo is not a solution of> either x^2 - x = 0 or x(x - 1) = 0. (But that shouldnÍt be surprising.> After all, the extended reals donÍt form a 'eld.)+oo - +oo = ... uh ... :-)David-- #191, ewill3@earthlink.netItÍs still legal to go .sigless. => With all due respect, you completely miss my point. I donÍt know how> plainly I can state it. The problem or question isnÍt what the> standard meaning of .999... or .999... is or should be. The problem> is that in the proof given that .999...=1, the .999... is used in two> different ways: as a number and as an in'nite series with a limit of> 1.Then your point is wrong, wrong, wrong..999.... is a number. Namely, it is the sum of the series which has> been repeatedly mentioned here. It is *not* the series itself, but> the sum of the series. That sum is a real number, so there is no> contradiction at all in claiming(1) .999... is a real number;> (2) .999... is the sum of the in'nite series;> (3) .999... is not an in'nite series.I donÍt know why so many people are so dense. It must be a colossalfailure to communicate. Maybe itÍs my fault. :-pMy comment was addressing the proof that was given early in thisthread - nothing else. It has nothing to do with what othermathematicians say or do or use for syntax and semantics or anythingelse. It is completely self-contained in the proof on which I wascommenting.Personally, I think part of the problem is the plethora of people wholive in the world where the published paper is the altar at which theyworship and its author is their messiah. All they know how to do isto quote from their scripture. But in this case, their scripture isirrelevant. Everything is contained in the proof that I wascritiquing. It had to do with inconsistency within that proof, notcon§ict between the proof and what other people regard as standard orreasonable use of mathematical syntax. Any way you interpret what theauthor is saying produces something wrong. He is simply inconsistent,regardless of which interpretation is deemed appropriate by someexternal standards. ThereÍs no reason to try to show whichinterpretation is correct. That is not the point. I donÍt care howhe uses his syntax, as long as heÍs consistent. But he isnÍt. HeÍsinconsistent.There are two ways that he treats .999# 1. As a literal number. (To anyone who harps about my use of theword literal: Get a life.) By literal I mean how you write down anumber - not a formula or a function or any operators. Just a number. .999 means the decimal number that has 9 for the 'rst digit afterthe decimal point, and 9 for the second digit, and 9 for the thirddigit, etc. etc. etc. Ok? Like 12.345 but only longer. That is oneinterpretation and use of .999Under this interpretation, we can treat .999 as a number with whichwe can calculate directly using the arithmetic operators. We candivide it by 3. 3 goes into 9 3 times. 3 times 3 is 9. 9 from 9 is0. Bring down the next 9. The quotient is .333# 2. As a symbol for an in'nite series. In this case, the 9Ís donÍtmean anything. It could just as well be XQZ. It just means 9/10 +9/100 + 9/1000 + . . . . And please, drop the stupid obviousstatements like thatÍs the same thing as .999 or well thatÍs equalto 1 and 1 is a number. We know that. ThatÍs 1st grademathematics. ThatÍs not the point. The point is how the author isusing .999 in the proof. He is simply manipulating .999 as anin'nite series the same as if it were XQZ. ItÍs being treated as asymbol that represents that series. He is not taking a numberconsisting of a decimal point followed by all 9Ís and performingarithmetic on them as in case # 1 above.Other person: 0.999.... does NOT equal 1.Faulty proof: It certainly does! Just try to subtract 0.999... from1: 1 - 0.999... = 0 Reason: There is no real number between0.999... and 1, and, therefore, they must be one and the samenumber!Case # 1: If you allow .999 to be a literal number, then you have twodifferent literal numbers that represent the same number: .999 and1.000 both represent 1. I maintain that the literal number should bea (unique) function of the number. Otherwise the whole process ofexpressing a number with a character string (consisting of digits anddecimal points and hyphens) breaks down. You canÍt tell if twonumbers are the same by simply comparing their digits. (The commentthat 1.5 is the same as 1.50 and 100 is the same as 1e2 and 5 is thesame as binary 101 are insipid. These are merely an abbreviation, anexpression and a different base, respectively, not two differentliterals representing the same number.)Case # 2: If .999 is meant to be a symbol in the proof, then youcanÍt subtract it from 1 digit by digit, nor talk about whatÍs betweenit and 1. You can only do that after you evaluate it and then showthe resulting number. You canÍt talk about there being a real numberbetween a symbol and a number. You can evaluate it, but then you get1 and would have to talk about whether there is a number between 1 and1. But he isnÍt doing that.HeÍs treating .999 as a literal number, but at the same time usingtwo different literal numbers to represent the same number. I callthat inconsistent.I swear this is my last attempt.C-B(Exit to thunderous applause.)*blush* <38iivvk4pk1fdf4bk8f3skg5p42h5740tb@4ax.com> <4pgqvvgsepbgnrrg5dnr5e6k6jkduivklr@4ax.com> <874qv5zkn3.fsf@phiwumbda.org> => My comment was addressing the proof that was given early in this> thread - nothing else. It has nothing to do with what other> mathematicians say or do or use for syntax and semantics or anything> else. It is completely self-contained in the proof on which I was> commenting.Well, I donÍt know what proof on which youÍre commenting and I donÍtparticularly care about it either. What, youÍve found a proof of atrue fact which contains either bad reasoning or bad presentation?Golly.> Personally, I think part of the problem is the plethora of people who> live in the world where the published paper is the altar at which they> worship and its author is their messiah. All they know how to do is> to quote from their scripture. Yes, IÍm sure thatÍs right. *ThatÍs* the reason that people haveresisted your deep insight that a particular proof found on Usenet ispoorly presented (for which IÍll take your word). If it wasnÍt forall this damn dogma, IÍm sure youÍd have your accolades by now.couldnÍt 'nd any reason you would criticize a particular instance ofa proof, rather the[...]> There are two ways that he treats .999...>> # 1. As a literal number. (To anyone who harps about my use of the> word literal: Get a life.) By literal I mean how you write down a> number - not a formula or a function or any operators. Just a number.> .999... means the decimal number that has 9 for the 'rst digit after> the decimal point, and 9 for the second digit, and 9 for the third> digit, etc. etc. etc. Ok? Like 12.345 but only longer. That is one> interpretation and use of .999...>> Under this interpretation, we can treat .999... as a number with which> we can calculate directly using the arithmetic operators. We can> divide it by 3. 3 goes into 9 3 times. 3 times 3 is 9. 9 from 9 is> 0. Bring down the next 9. The quotient is .333...>> # 2. As a symbol for an in'nite series. In this case, the 9Ís donÍt> mean anything. It could just as well be XQZ. It just means 9/10 +> 9/100 + 9/1000 + . . . . And please, drop the stupid obvious> statements like thatÍs the same thing as .999... or well thatÍs equal> to 1 and 1 is a number. We know that. ThatÍs 1st grade> mathematics. ThatÍs not the point. The point is how the author is> using .999... in the proof. He is simply manipulating .999... as an> in'nite series the same as if it were XQZ. ItÍs being treated as a> symbol that represents that series. He is not taking a number> consisting of a decimal point followed by all 9Ís and performing> arithmetic on them as in case # 1 above.Huh?That *is* how 0.9999.... is de'ned. It is a number, namely the sumof the series above. We have some handy theorems about seriesmanipulations (like, if a series a_0 + a_1 + ... converges to x, then b * (a_0 + a_1 + ...) = b * x) and so his manipulations of the seriesto yield results about 0.999... are justi'ed (presumably -- I donÍtknow which manipulations you mean).>> Other person: 0.999.... does NOT equal 1.>> Faulty proof: It certainly does! Just try to subtract 0.999... from> 1: 1 - 0.999... = 0 Reason: There is no real number between> 0.999... and 1, and, therefore, they must be one and the same> number!Well, thatÍs a pretty crappy proof, no doubt.But not for the reasons that you say, near as I can 'gger.-- Jesse HughesBasically there are two angry groups. I am a harsh force ofone. Against me is a society of mathematicians. So far itÍs been adraw. -- JSH gives another display of keen insight. =>So what youÍre saying is the sequence .9^2, .9^3, .9^4 ...>>.9^2 = .81>.9^3 = .729>.9^4 = .6561>etc>the sequence .99^2, .99^3, .99^4 ...>>.99^2 = .9801>.99^3 = .970299>.99^4 = .96059601>etc>>the sequence .999^2, .999^3, .999^4 ... >>.999^2 = .998001>.999^3 = .997002999>.999^4 = .996005996001>etc>tend to 1 making the sequence .999...^2, .999...^3, .999...^4 tend to 1.>>Is that what you are saying?No, nothing at all like that. How about responding to what I *actually* >Garry Denke, Geologist>Denoco Inc. of Texas-- ---------------------------| B B aa rrr b || BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- => No, nothing at all like that. How about responding to what I *actually* It now seems clear that the reason Garry is talking past everyone else, and vice versa, is that he has a different meaning of tends to 1 in mind. This is illuminated by a subsequent post: In the *abstract numerical value* view, the claim that .999999999999^4 is closer to 1 than .9^4 is (i.e. that |1-.999999999999^4| < |1-.9^4|) is (I hope) indisputably true, so that clearly isnÍt the view that Garry is meaning here. ISTM he (and Charlie Volkstorf) hold a *concrete representational* view. For Garry, the successively longer decimal fractions *look* progressively less like the simple numeral 1. A big problem with GarryÍs view is that the concrete decimal representation of .999... is unending, yet he hold fast to the notion that there is a last 9 digit (and hence a last 1 digit in (.999...)^2 and (.999...)^4). This reminds me of Phil. It turns out that if one takes the trouble to *precisely de'ne* numerical addition and multiplication on potentially in'nite decimals, the ONLY WAY to make it work consistently with the accepted arithmetic for 'nite decimals (which can each be considered to have an in'nite sequence of 0s to the right) results in the concrete decimal representation of (.999...)^2 being exactly .999... There really is no last one (or Last One). ISTM the reason Garry doesnÍt accept that fact is that itÍs not immediately obvious, and he hasnÍt worked through the detailed consequences of arithmetic for in'nite decimals. Note that another of these consequences is that in terms of concrete decimals, 1.000... - .999... is exactly equal to 0.000... This is much easier to show than examples involving multiplication, since this can be done one digit at a time.--Sequence .9^2, .9^3, .9^4, ...Sequence .99^2, .99^3, .99^4, ...Sequence .999^2, .999^3, .999^4, ...Sequence .9999^2, .9999^3, .9999^4, ...Sequence .99999^2, .99999^3, .99999^4, ...Sequence .999999^2, .999999^3, .999999^4, ...Sequence .9999999^2, .9999999^3, .9999999^4, ...Sequence .99999999^2, .99999999^3, .99999999^4, ...Sequence .999999999^2, .999999999^3, .999999999^4, ...Sequence .9999999999^2, .9999999999^3, .9999999999^4, ...Sequence .99999999999^2, .99999999999^3, .99999999999^4, ...Sequence .999999999999^2, .999999999999^3, .999999999999^4, ...Sequence .9999999999999^2, .9999999999999^3, .9999999999999^4, ...etc., etc., etc.,tend to 0.Therefore: Sequence .999...^2, .999...^3, .999...^4, ... tends to 0--Garry Denke, GeologistDenoco Inc. of Texas =In sci.logic, Garry Denkeon 10 Jan nothing at all like that. How about responding to what I *actually* It now seems clear that the reason Garry is talking past everyone else, > and vice versa, is that he has a different meaning of tends to 1 in > mind. This is illuminated by a subsequent post:> In the *abstract numerical value* view, the claim that .999999999999^4 > is closer to 1 than .9^4 is (i.e. that |1-.999999999999^4| < |1-.9^4|) > is (I hope) indisputably true, so that clearly isnÍt the view that Garry > is meaning here.> ISTM he (and Charlie Volkstorf) hold a *concrete representational* view. > For Garry, the successively longer decimal fractions *look* > progressively less like the simple numeral 1.> A big problem with GarryÍs view is that the concrete decimal > representation of .999... is unending, yet he hold fast to the notion > that there is a last 9 digit (and hence a last 1 digit in (.999...)^2 > and (.999...)^4). This reminds me of Phil.> It turns out that if one takes the trouble to *precisely de'ne* > numerical addition and multiplication on potentially in'nite decimals, > the ONLY WAY to make it work consistently with the accepted arithmetic > for 'nite decimals (which can each be considered to have an in'nite > sequence of 0s to the right) results in the concrete decimal > representation of (.999...)^2 being exactly .999... There really is no > last one (or Last One). ISTM the reason Garry doesnÍt accept that fact > is that itÍs \ not immediately obvious, and he hasnÍt worked through the > detailed consequences of arithmetic for in'nite decimals.> Note that another of these consequences is that in terms of concrete > decimals, 1.000... - .999... is exactly equal to 0.000... This is much > easier to show than examples involving multiplication, since this can be > done one digit at a time.--Sequence .9^2, .9^3, .9^4, ...> Sequence .99^2, .99^3, .99^4, ...> Sequence .999^2, .999^3, .999^4, ...> Sequence .9999^2, .9999^3, .9999^4, ...> Sequence .99999^2, .99999^3, .99999^4, ...> Sequence .999999^2, .999999^3, .999999^4, ...> Sequence .9999999^2, .9999999^3, .9999999^4, ...> Sequence .99999999^2, .99999999^3, .99999999^4, ...> Sequence .999999999^2, .999999999^3, .999999999^4, ...> Sequence .9999999999^2, .9999999999^3, .9999999999^4, ...> Sequence .99999999999^2, .99999999999^3, .99999999999^4, ...> Sequence .999999999999^2, .999999999999^3, .999999999999^4, ...> Sequence .9999999999999^2, .9999999999999^3, .9999999999999^4, ...> etc., etc., etc.,tend to 0.An in'nite sequence of sequences cannot tend to a single numberwithout a little more work.It turns out that your sequences tend towards the double limitlim(m,n->+oo) (1 - 10^(-m))^nwhich has no value, despite the limit sequence being1, 1, 1, ...(or, if you prefer, 0.999..., 0.999..., 0.999..., ...)Therefore: Sequence .999...^2, .999...^3, .999...^4, ... tends to 0Each of your sequences in the above sequence of sequences hasa limit value, namely 0. However, the limit sequence has everyvalue equal to 1, but has no limit value...assuming IÍm expressingthis concept at all correctly, which IÍm probably not.--Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netItÍs still legal to go .sigless. =In sci.logic, mathedman<3ffec892.4480964@netnews.worldnet.att.net>:>Squares of 0.999... \ tend toward 0, not toward 1NONSENSE!> LIM(n->+oo) { (1- 1/(10^n)) * [1 - 1/(10^n))}> = LIM(n->+oo)[1 - (1/10^n)) * LIM(n->+oo)[ 1 - (1/10^n))> = 1*1Just to throw in the usual monkey wrench :-), though, thedouble limit lim(m,n->+oo) (1 - 1/10^m)^nhas no assigned value. IÍm assuming Garry Denke meant powersinstead of squares here (recall his post had cubes and fourths)and is looking at lim(n->+oo) (1-1/10^m)^n = 0as opposed to lim(m->+oo) (1-1/10^m)^n = 1(which is one reason why the aforementioned limit hasno assigned value, BTW; other mildly interesting approachesinclude lim(k->+oo) (1 - 1/10^k)^(10^k) = 1/e.)Also, cheap electronic calculators can square repeatedly, leadingto some weird results: ï2 [Times] = = = =Í might result in the value32 or 65536 -- if it even bothers. (The program ïxcalcÍ on my machinesimply stays at 2, for example.)[crunch]-- #191, ewill3@earthlink.netItÍs still legal to go .sigless. => what I think you are missing is that .999... (a zero, a dot and an> in'nite amount of 9s) is equal to 1, so what you are asking is> whether 1^2, 1^3, 1^4, etc. is ïtending toÍ 1.i understand the premise.> Look at it this way: We can write a number like 0.9 as 1 - 0.1, and a> number like 0.99 as 1 - 0.01. In general we can write it down as:f(n) = 1 - (1/(10^n)) for n>0 where n is the number of nines behind> the dot.So the number 0.999... can be found by looking at:lim{n->in'nity}{f(n)} = > lim{n->in'nity}{1 - (1/(10^n))} => lim{n->in'nity}{1} - lim{n->in'nity}{1/(10^n)}but in'nity means unlimited... limit{n->unlimited}{f(n)} = limit{n->unlimited}{1 - (1/(10^n))} =limit{n->unlimited}{1} - limit{n->unlimited}{1/(10^n)}...which contradicts the premise. > Since the limit of the fraction-part goes to zero, the whole thing> becomes equal to 1. In other words 0.999... is equal to 1.only if the premise the unlimited has a limit is believed.garry denke, geologistdenoco inc. of texas =The lopsided Collatz tree? Maybe this is known so just disregard!I couldnÍt 'nd anything about it on web searches about theCollatz.Probably of little value for the overall conjecture but still acuriosity! The Collatz tree level count is odd up to level 5 where level1,2,3,4,5 [1,2,4,8,16] has a count of (1) for each level. At level 6 the count becomes even and fall one for one on eachside of the tree up too and including level 16.This changes @ level 17 where the left side of the tree adds onemore path.This happens also for certain levels higher than 17 where theleft side adds 1 more path over the right side.I have denoted the 2 sides of the tree by their highest-rankingsequence.Where [1,2,4,8,16] is the trunk and the 'rst main branchassigned to the RIGHT side of the tree is --32,64,128,256,512,1024,2048,4096...and with all applicable nodesbranching from it creating all branches on the RIGHT side ofthe Collatz tree.Where [1,2,4,8,16] is the trunk and the 'rst main branchassigned to the LEFT side of the tree is --5,10,20,40,80,160,320,640,1280,2560,...and with all applicablenodes branching from it creating all branches on the LEFT side ofthe Collatz tree. 8) 20 3 21 128 | | / 7) 10 64 / 6) 5 32 / 5) 16 4) 8 3) 4 2) 2 Levels 1) 1Why does the left side have this property of adding (1) morebranch at certain levels =>17 then the right side?Will this mean, at a very high level there will be many morebranches on the left side of the tree than on the right sideof the tree making the tree lopsided?Putting the above question in another perspective, at some higherlevel will the right side branching number eventually catch up andequal the left side branching number?It could be that the left side of the tree is facing the sun givingthat side of the tree a more vigorous growth. ;-)Dan =>Message-id: lopsided Collatz tree? >>Maybe this is known so just disregard!>I couldnÍt 'nd anything about it on web searches about the>Collatz.>>Probably of little value for the overall conjecture but still a>curiosity!>The Collatz tree level count is odd up to level 5 where level>1,2,3,4,5 [1,2,4,8,16] has a count of (1) for each level. >At level 6 the count becomes even and fall one for one on each>side of the tree up too and including level 16.>This changes @ level 17 where the left side of the tree adds one>more path.>This happens also for certain levels higher than 17 where the>left side adds 1 more path over the right side.>>I have denoted the 2 sides of the tree by their highest-ranking>sequence.>>Where [1,2,4,8,16] is the trunk and the 'rst main branch>assigned to the RIGHT side of the tree is -->32,64,128,256,512,1024,2048,4096...and with all applicable nodes>branching from it creating all branches on the RIGHT side of>the Collatz tree.>>Where [1,2,4,8,16] is the trunk and the 'rst main branch>assigned to the LEFT side of the tree is -->5,10,20,40,80,160,320,640,1280,2560,...and with all applicable>nodes branching from it creating all branches on the LEFT side of>the Collatz tree.> 8) 20 3 21 128 > | | /> 7) 10 64> /> 6) 5 32> /> 5) 16> 4) 8> 3) 4 > 2) 2> Levels 1) 1>>Why does the left side have this property of adding (1) more>branch at certain levels =>17 then the right side?>>Will this mean, at a very high level there will be many more>branches on the left side of the tree than on the right side>of the tree making the tree lopsided?>>Putting the above question in another perspective, at some higher>level will the right side branching number eventually catch up and>equal the left side branching number?>>It could be that the left side of the tree is facing the sun giving>that side of the tree a more vigorous growth. ;-)>>DanPersonally, I donÍt like the way you draw the Collatz tree. My preferenceis for vertical placement to always represent x*2 and for the diagonalplacement to always represent (x-1)/3: 3 20 128 21 | | / 10 64 | | 5 32 | 16 | 8 | 4 | 2 | 1By your method, 32 is part of the right branch whereas I consider the rightbranch single left branch and right branch. Rather, there is the central trunk fromwhichsprout an in'nite number of branches on either side, so I donÍt see anyproblem.Note that in my view each branch starts with an odd number and extendsverticallyto in'nity, all of which are even. Does this mean there are in'nitely moreevennumbers than odd?In'nity always catches up in the end.--MensanatorAce of Clubs =appreciate your help guys---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- =Why donÍt you just use some pseudorandom number generator with a 'xed seed?Do you need random numbers....? Why calculate them anyway... wouldnÍt somefunction that returns a particular digit given its index be enough? In which caseuse that index as a seed for a pseudorandom function and iterative it two orthree times to get well dispersed results.Why not calculate the 100 million digits of PI or whatever and save them in a 'le?>>I am working on the program where is necessary to calculate huge>amount of numbers in always the same sequence, before the program>actually opens.>>In other words, these number have to be calculated and saved into>memory before the program starts. >>I was trying to use number pi, but to calculate 100 million decimal>places of number pi would take forever and no one would want to wait>that long. On the other hand I canÍt afford to include the whole 100>million dec. places along with the software, that would take too much>of the space and too much time to download. >>Is there any other number like PI, which I could calculate faster?>Serge>------------------------------------------- ---------------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>---------------------------------------------------------- > http://www.usenet.com> =As already stated, the Mersenne Twister is a very popular randomnumber generator.http://www.math.keio.ac.jp/~matumoto/emt.html---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- =Dave,Serge---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- => I am working on the program where is necessary to calculate huge> amount of numbers in always the same sequence, before the program> actually opens.> In other words, these number have to be calculated and saved into> memory before the program starts. > I was trying to use number pi, but to calculate 100 million decimal> places of number pi would take forever and no one would want to wait> that long. On the other hand I canÍt afford to include the whole 100> million dec. places along with the software, that would take too much> of the space and too much time to download. Sounds like you may be looking for a pseudo-random number generator(PRNG). There are lots of them around. Many programming languages haveone built in. Or Google for random numbers.> Is there any other number like PI, which I could calculate faster?Well, e is a number like pi (both are transcendental), and the in'niteseries representation for e is simple to compute and converges veryrapidly.-- Dave SeamanJudge YohnÍs mistakes revealed in Mumia Abu-Jamal ruling. =Gib,could you please post the formulae for The Mersenne Twister ?that could help a bit... I am programming in vb.net, but any codewould be highly appreciated---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- =I am working on the program where is necessary to calculate huge> amount of numbers in always the same sequence, before the program> actually opens.In other words, these number have to be calculated and saved into> memory before the program starts. I was trying to use number pi, but to calculate 100 million decimal> places of number pi would take forever and no one would want to wait> that long. On the other hand I canÍt afford to include the whole 100> million dec. places along with the software, that would take too much> of the space and too much time to download. Is there any other number like PI, which I could calculate faster?The Mersenne Twister code IÍm using calculates 100 million random unsigned 32 bit integers in about 7 seconds on a P4 1.8, and the same seed will always give the same sequence.Gib => Is there any other number like PI, which I could calculate faster?2.0? 1/2? sqrt(2)?Those are faster to calculate, but I guess theyÍre not very much likepi. But IÍm not sure what you mean by like pi (not that I couldprobably answer the question if I knew).-- Jesse F. HughesI have written many words to sci.math, some of them are not evenmeaningless. --Ross Finlayson =Just to add to it, I was already wondering about number e, where sameas in number pi, decimal digits never end, and they never repeat...Well, e is just like Pi in this respect; if you keep going, it neverstops, and it never repeats. But is it faster to calculate e?Did anyone ever tried to calculate it on his computer?Or did anyone programmed its formulae in some sort of a programminglanguage?I was also thinking about using square root of 2 as another irrationalexample... welldoes anyone have formulaes or programs which will show how to computeit fast?---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- => Just to add to it, I was already wondering about number e, where same> as in number pi, decimal digits never end, and they never repeat...> Well, e is just like Pi in this respect; if you keep going, it never> stops, and it never repeats. But is it faster to calculate e?> Did anyone ever tried to calculate it on his computer?> Or did anyone programmed its formulae in some sort of a programming> language?[...]You could use the Mersenne Twister, as Gib Bogle mentioned, someof the PRNGs by George Marsaglia, arc4, or some combination...The C code below combines MT, Marsaglia PRNGs and arc4.David Bernier#include /* Marsaglia rng */#de'ne znew (zmars=36969*(zmars&65535)+(zmars>>16))#de'ne wnew (wmars=18000*(wmars&65535)+(wmars>>16))#de'ne MWC ((znew<<16)+wnew )#de'ne SHR3 (jsr^=(jsr<<17), jsr^=(jsr>>13), jsr^=(jsr<<5))#de'ne CONG (jcong=69069*jcong+1234567)#de'ne KISS ((MWC^CONG)+SHR3)#de'ne MARS ((int) (KISS>>24))/* Period parameters, Mersenne Twister */ #de'ne Nmersenne 624#de'ne Mmersenne 397#de'ne MATRIX_A 0x9908b0dfUL /* constant vector a */#de'ne UPPER_MASK 0x80000000UL /* most signi'cant w-r bits */#de'ne LOWER_MASK 0x7fffffffUL /* least signi'cant r bits */#de'ne G8 genrand_int8()static unsigned long mt[Nmersenne]; /* the array for the state vector */static int mti=Nmersenne+1; /* mti==N+1 means mt[N] is not initialized */static unsigned long zmars=899079, wmars=650069, jsr=127, jcong=6534282;/* initializes mt[N] with a seed */void init_genrand(unsigned long s){ mt[0]= s & 0xffffffffUL; for (mti=1; mti> 30)) + mti); mt[mti] &= 0xffffffffUL; /* for >32 bit machines */ }}void init_by_array(unsigned long[], int);void init_by_array(unsigned long init_key[], int key_length){ int i, j, k; init_genrand(19650218UL); i=1; j=0; k = (Nmersenne>key_length ? Nmersenne : key_length); for (; k; k--) { mt[i] = (mt[i] ^ ((mt[i-1] ^ (mt[i-1] >> 30)) * 1664525UL)) + init_key[j] + j; /* non linear */ mt[i] &= 0xffffffffUL; /* for WORDSIZE > 32 machines */ i++; j++; if (i>=Nmersenne) { mt[0] = mt[Nmersenne-1]; i=1; } if (j>=key_length) j=0; } for (k=Nmersenne-1; k; k--) { mt[i] = (mt[i] ^ ((mt[i-1] ^ (mt[i-1] >> 30)) * 1566083941UL)) - i; /* non linear */ mt[i] &= 0xffffffffUL; /* for WORDSIZE > 32 machines */ i++; if (i>=Nmersenne) { mt[0] = mt[Nmersenne-1]; i=1; } } mt[0] = 0x80000000UL; /* MSB is 1; assuring non-zero initial array */ } int genrand_int8(void){ unsigned long y; static unsigned long mag01[2]={0x0UL, MATRIX_A}; if (mti >= Nmersenne) { /* generate N words at one time */ int kk; if (mti == Nmersenne+1) /* if init_genrand() has not been called, */ init_genrand(5489UL); /* a default initial seed is used */ for (kk=0;kk> 1) ^ mag01[y & 0x1UL]; } for (;kk> 1) ^ mag01[y & 0x1UL]; } y = (mt[Nmersenne-1]&UPPER_MASK)|(mt[0]&LOWER_MASK); mt[Nmersenne-1] = mt[Mmersenne-1] ^ (y >> 1) ^ mag01[y & 0x1UL]; mti = 0; } y = mt[mti++]; /* Tempering */ y ^= (y >> 11); y ^= (y << 7) & 0x9d2c5680UL; y ^= (y << 15) & 0xefc60000UL; y ^= (y >> 18); return ((int) (y>>24));}int main(){ unsigned long initmersenne[4]={0x5aa, 0x709b, 0xb3c8, 0xa9d01}; int lengthmersenne=4; int s4=0; int ix4=0; int temp4, t4; int chooser; int digit; long numdigits; long counter; long distribution[] = {0,0,0,0,0,0,0,0,0,0}; int id; int doprint; int state4[] = {159,252,240,255,13,107,139,37,70,28,187,62,194,87,71,4, 225,196,49,181,173,113,116,175,236,127,111,33,39,88,203,242, 180,1,64,110,85,230,78,141,213,57,18,211,134,250,2,222, 119,161,26,95,27,45,96,8,103,147,208,138,11,131,68,195, 189,247,249,160,145,254,106,140,19,80,214,221,72,243,54,177, 166,120,105,104,24,100,182,25,165,81,41,123,199,148,91,133, 52,186,152,192,56,58,218,89,20,188,74,185,118,6,98,169, 163,136,235,233,59,14,231,35,29,15,117,190,101,251,36,183, 246,16,210,50,200,207,53,67,130,237,47,201,90,12,149,73, 228,198,162,217,9,75,42,23,0,109,227,137,40,5,223,197, 128,77,232,69,179,234,125,55,97,51,115,3,112,172,132,178, 238,209,121,31,34,164,253,184,32,229,174,193,76,38,86,167, 156,150,93,84,10,122,226,224,170,135,151,219,21,205,244,157, 102,44,158,202,46,48,245,248,212,171,43,61,144,94,92,83, 215,239,60,7,17,66,176,99,129,191,155,204,114,153,124,143, 126,30,108,154,216,22,79,65,146,168,142,82,63,206,241,220}; init_by_array(initmersenne, lengthmersenne); printf(nHow many digits?n); scanf(%ld, &numdigits); printf(Do you want to print the digits? (1=Yes 0=No)n); scanf(%d, &doprint); for(counter = 0; counter249) { ix4++; if(ix4 == 256) ix4 = 0; s4 = s4 + state4[ix4]; if(s4>255) s4 = s4 - 256; temp4 = state4[s4]; state4[s4] = state4[ix4]; state4[ix4] = temp4; t4 = state4[ix4] + state4[s4]; if(t4>255) t4 = t4 - 256; chooser = state4[t4]^G8^MARS ; } digit = chooser%10; if(doprint!= 0) { printf(%d , digit); if(24 == (counter%25)) { printf(n); } } if(doprint==0) { if(999999 == (counter%1000000)) { printf(.); if((50000000-1) == (counter%50000000)) { printf(n); } } } distribution[digit]++; } printf(nnDistribution of the digits:n); printf(___________________________nn); for(id=0;id<10;id++) { printf(%d: %ldn, id, distribution[id]); } printf(nn); return 0;} => Just to add to it, I was already wondering about number e, where same> as in number pi, decimal digits never end, and they never repeat...> Well, e is just like Pi in this respect; if you keep going, it never> stops, and it never repeats. But is it faster to calculate e?> Did anyone ever tried to calculate it on his computer?> Or did anyone programmed its formulae in some sort of a programming> language?[...]You could use the Mersenne Twister, as Gib Bogle mentioned, someof the PRNGs by George Marsaglia, arc4, or some combination...The C code below combines MT, Marsaglia PRNGs and arc4.David Bernier#include /* Marsaglia rng */#de'ne znew (zmars=36969*(zmars&65535)+(zmars>>16))#de'ne wnew (wmars=18000*(wmars&65535)+(wmars>>16))#de'ne MWC ((znew<<16)+wnew )#de'ne SHR3 (jsr^=(jsr<<17), jsr^=(jsr>>13), jsr^=(jsr<<5))#de'ne CONG (jcong=69069*jcong+1234567)#de'ne KISS ((MWC^CONG)+SHR3)#de'ne MARS ((int) (KISS>>24))/* Period parameters, Mersenne Twister */ #de'ne Nmersenne 624#de'ne Mmersenne 397#de'ne MATRIX_A 0x9908b0dfUL /* constant vector a */#de'ne UPPER_MASK 0x80000000UL /* most signi'cant w-r bits */#de'ne LOWER_MASK 0x7fffffffUL /* least signi'cant r bits */#de'ne G8 genrand_int8()static unsigned long mt[Nmersenne]; /* the array for the state vector */static int mti=Nmersenne+1; /* mti==N+1 means mt[N] is not initialized */static unsigned long zmars=899079, wmars=650069, jsr=127, jcong=6534282;/* initializes mt[N] with a seed */void init_genrand(unsigned long s){ mt[0]= s & 0xffffffffUL; for (mti=1; mti> 30)) + mti); mt[mti] &= 0xffffffffUL; /* for >32 bit machines */ }}void init_by_array(unsigned long[], int);void init_by_array(unsigned long init_key[], int key_length){ int i, j, k; init_genrand(19650218UL); i=1; j=0; k = (Nmersenne>key_length ? Nmersenne : key_length); for (; k; k--) { mt[i] = (mt[i] ^ ((mt[i-1] ^ (mt[i-1] >> 30)) * 1664525UL)) + init_key[j] + j; /* non linear */ mt[i] &= 0xffffffffUL; /* for WORDSIZE > 32 machines */ i++; j++; if (i>=Nmersenne) { mt[0] = mt[Nmersenne-1]; i=1; } if (j>=key_length) j=0; } for (k=Nmersenne-1; k; k--) { mt[i] = (mt[i] ^ ((mt[i-1] ^ (mt[i-1] >> 30)) * 1566083941UL)) - i; /* non linear */ mt[i] &= 0xffffffffUL; /* for WORDSIZE > 32 machines */ i++; if (i>=Nmersenne) { mt[0] = mt[Nmersenne-1]; i=1; } } mt[0] = 0x80000000UL; /* MSB is 1; assuring non-zero initial array */ } int genrand_int8(void){ unsigned long y; static unsigned long mag01[2]={0x0UL, MATRIX_A}; if (mti >= Nmersenne) { /* generate N words at one time */ int kk; if (mti == Nmersenne+1) /* if init_genrand() has not been called, */ init_genrand(5489UL); /* a default initial seed is used */ for (kk=0;kk> 1) ^ mag01[y & 0x1UL]; } for (;kk> 1) ^ mag01[y & 0x1UL]; } y = (mt[Nmersenne-1]&UPPER_MASK)|(mt[0]&LOWER_MASK); mt[Nmersenne-1] = mt[Mmersenne-1] ^ (y >> 1) ^ mag01[y & 0x1UL]; mti = 0; } y = mt[mti++]; /* Tempering */ y ^= (y >> 11); y ^= (y << 7) & 0x9d2c5680UL; y ^= (y << 15) & 0xefc60000UL; y ^= (y >> 18); return ((int) (y>>24));}int main(){ unsigned long initmersenne[4]={0x5aa, 0x709b, 0xb3c8, 0xa9d01}; int lengthmersenne=4; int s4=0; int ix4=0; int temp4, t4; int chooser; int digit; long numdigits; long counter; long distribution[] = {0,0,0,0,0,0,0,0,0,0}; int id; int doprint; int state4[] = {159,252,240,255,13,107,139,37,70,28,187,62,194,87,71,4, 225,196,49,181,173,113,116,175,236,127,111,33,39,88,203,242, 180,1,64,110,85,230,78,141,213,57,18,211,134,250,2,222, 119,161,26,95,27,45,96,8,103,147,208,138,11,131,68,195, 189,247,249,160,145,254,106,140,19,80,214,221,72,243,54,177, 166,120,105,104,24,100,182,25,165,81,41,123,199,148,91,133, 52,186,152,192,56,58,218,89,20,188,74,185,118,6,98,169, 163,136,235,233,59,14,231,35,29,15,117,190,101,251,36,183, 246,16,210,50,200,207,53,67,130,237,47,201,90,12,149,73, 228,198,162,217,9,75,42,23,0,109,227,137,40,5,223,197, 128,77,232,69,179,234,125,55,97,51,115,3,112,172,132,178, 238,209,121,31,34,164,253,184,32,229,174,193,76,38,86,167, 156,150,93,84,10,122,226,224,170,135,151,219,21,205,244,157, 102,44,158,202,46,48,245,248,212,171,43,61,144,94,92,83, 215,239,60,7,17,66,176,99,129,191,155,204,114,153,124,143, 126,30,108,154,216,22,79,65,146,168,142,82,63,206,241,220}; init_by_array(initmersenne, lengthmersenne); printf(nHow many digits?n); scanf(%ld, &numdigits); printf(Do you want to print the digits? (1=Yes 0=No)n); scanf(%d, &doprint); for(counter = 0; counter249) { ix4++; if(ix4 == 256) ix4 = 0; s4 = s4 + state4[ix4]; if(s4>255) s4 = s4 - 256; temp4 = state4[s4]; state4[s4] = state4[ix4]; state4[ix4] = temp4; t4 = state4[ix4] + state4[s4]; if(t4>255) t4 = t4 - 256; chooser = state4[t4]^G8^MARS ; } digit = chooser%10; if(doprint!= 0) { printf(%d , digit); if(24 == (counter%25)) { printf(n); } } if(doprint==0) { if(999999 == (counter%1000000)) { printf(.); if((50000000-1) == (counter%50000000)) { printf(n); } } } distribution[digit]++; } printf(nnDistribution of the digits:n); printf(___________________________nn); for(id=0;id<10;id++) { printf(%d: %ldn, id, distribution[id]); } printf(nn); return 0;} =no need for any specialities... just quickly generate 100 milionnumbers... only requirement is that IÍll 'nd the numbers on the samepositions no matter how often I generate them...can you explain this: a_(n+1) = 3a_n + 1 mod Total Privacy via Encryption =--- =Serge Oliva> no need for any specialities... just quickly generate 100 milion> numbers... only requirement is that IÍll 'nd the numbers on the same> positions no matter how often I generate them...> SergeThe sequence goes 1,4,13,40, etc. Each term is three times the previous,plus 1. When the sequence goes past 2^32 (or whatever size is convenient),then subtract 2^32 or 2*2^32, and use the remainder. You could start thesequence with any number, not necessarily 1.(The notation ^ is for exponentials, e.g. 4^5 = 4*4*4*4*4.)LH =Serge Oliva>> I am working on the program where is necessary to calculate huge> amount of numbers in always the same sequence, before the program> actually opens.>> In other words, these number have to be calculated and saved into> memory before the program starts.Writing a crypter, by any chance? But anyway, why not usea_(n+1) = 3a_n + 1 mod 2^32or some such thing? Does the sequence need special properties?LH =How to integrate sinc(x) (without Fourier transformation) ?? = There is another slightly subtle §aw with your argument. You know> that a_1(x,y) and a_2(x,y) are roots of a^2 - (x - y) + 7(x^2 + xy).Typo, this should be a^2 - (x - y)a + 7(x^2 + xy). There are two roots. Perhaps you choose as the ïnegativeÍ root, a_1(x,y) = ((x - y) - sqrt((x - y)^2 - 28*(x^2 + xy)))/2and the other as the ïpositiveÍ root, a_2(x,y) = ((x - y) + sqrt((x - y)^2 - 28*(x^2 + xy)))/2. The choice is arbitrary. Your subsequent argument makes no> use of whether the negative or positive root is chosen.>No. At x=0, one of these roots ( the ïpositiveÍ root) is 0,the other (the ïnegativeÍ root) is -y.James chooses a_1(0,y) to \ equal 0 and a_2(0,y) = -y. Values forother x can be speci'ed by chosing a branch cut and insisting oncontinuity. There is no ambiguity as long as y is not equal to 0. The same problem actually occurs with your argument about> your cubic polynomial, except there you are arbitrarily choosing> one of three roots a_1, a_2, and a_3 instead of one of two. You > conclude that two of the three must be divisible by 7, and the > other coprime to 7. Again James choses a_1 and a_2 to be 0 at 0. Again we can resolve theambiguity at other values of x by using continuity and branch cuts.The double root at x=0 (and possible other double or triple roots forother values of x) leads to a minor ambiguity, but not an important one. - William Hughes =>> What I like about adding to Rick DeckerÍs example is that it makes it>> easy.>>> In case you missed it, Rick Decker is a professor at Hamilton College>> (do a web search to 'nd out about his school) who posted a>> *quadratic* in an attempt to refute some of my conclusions.>> Well, I found \ his example grabbed me, for various reasons, and 'nally>> found that I could modify it slightly to suit my purposes, and now>> con'dent in your understanding--even as undergrads--of independence>> between variables, I feel I can 'nally show you how I have been>> defending mathematics against people working to undermine it!>>Well I made a mistake, as I added in that y variable wrong. Oh well.You left out a few signi'cant parts of your post that were also wrong; you should retract them as well:>> That is to doubt me here you need to doubt that concept in mathematics>> itself--the concept of independence between variables.>>> Of course, the problem for the old view on algebraic integers, is that>> if you supposed that you were in the ring of algebraic integers>> (notice I made no mention of a ring before now) then you run into a>> problem speci'c to that ring, where dividing both sides by 7 pushes>> you out of the ring of algebraic integers.>>> It might seem esoteric and distant as a problem, but my discovery of>> that problem is actually a fascinating story, and now you can see how>> mathematicians *worldwide* answer the question:>>> What do you do when your paradigm is forced to shift by a discoverer>> who thought outside of the box?Turns out that to doubt you we actually _donÍt_ have to doubt theconcept of independence between variables. How about that.YouÍre really never going to catch on, are you? When a rationalperson deduces something that contradicts established mathematicshe doesnÍt announce that the mathematical community is hidingthe truth, that a child can see heÍs right, etc etc etc. He looksfor the _error_ in his proof. Or he asks someone else to 'ndthe error (and when the someone else points out an error hetries to understand what the guyÍs saying...)>James Harris************************David C. Ullrich => What I like about adding to Rick DeckerÍs example is that it makes it>> easy.>> In case you missed it, Rick Decker is a professor at Hamilton College>> (do a web search to 'nd out about his school) who posted a>> *quadratic* in an attempt to refute some of my conclusions.>> Well, I found his example grabbed me, for various reasons, and 'nally>> found that I could modify it slightly to suit my purposes, and now>> con'dent in your understanding--even as undergrads--of independence>> between variables, I feel I can 'nally show you how I have been>> defending mathematics against people working to undermine it!>>Well I made a mistake, as I added in that y variable wrong. Oh well.>> You left out a few signi'cant parts of your post that were also> wrong; you should retract them as well:> That is to doubt me here you need to doubt that concept in mathematics>> itself--the concept of independence between variables.>> Of course, the problem for the old view on algebraic integers, is that>> if you supposed that you were in the ring of algebraic integers>> (notice I made no mention of a ring before now) then you run into a>> problem speci'c to that ring, where dividing both sides by 7 pushes>> you out of the ring of algebraic integers.>> It might seem esoteric and distant as a problem, but my discovery of>> that problem is actually a fascinating story, and now you can see how>> mathematicians *worldwide* answer the question:>> What do you do when your paradigm is forced to shift by a discoverer>> who thought outside of the box?>> Turns out that to doubt you we actually _donÍt_ have to doubt the> concept of independence between variables. How about that.>> YouÍre really never going to catch on, are you? When a rational> person deduces something that contradicts established mathematics> he doesnÍt announce that the mathematical community is hiding> the truth, that a child can see heÍs right, etc etc etc. He looks> for the _error_ in his proof. Or he asks someone else to 'nd> the error (and when the someone else points out an error he> tries to understand what the guyÍs saying...)>>James Harris> ************************>> David C. UllrichBut James isnÍt rational. All he wants is fame and thinks that heÍll get itby being anal. The only thing heÍll get fame for is being the sci.mathcrank/crackpot.David Moran =Two days ago I had an interesting discussion with my coworker at the Mathdept of the U of Crete about continuity.The discussion was interesting, because he is a computer scientist, so ourviews on continuity had some substancial differences. The main argument ispresented below:1) Mathematician: The brain can perceivecontinuity (perhaps intuitively,but it can).2) Computer Scientist: The brain cannot perceive continuity, cause allactions in the universe are quantized.He started the discussion by asking me if analog clocks are indeed analog,in the true sense. Without hesitation I replied Of course! only toretract my opinion a little while afterwards, thinking that old analogclocks, posess a little spinning wheel, which surrounds a spring, aroundwhich the wheel alternatively rotates back and forth. As such, the movementof the (whatever) hands, is not trully analog. The hand moves only when thewheel moves. When the wheel halts, the hand halts as well. So analog clocksare in reallity semi-analog.Then we went over what appeared to be a slight variant of ZenoÍs paradox,with him insisting that the passage of time cannot be trully analog, sinceif time was continuous, the vector of time would have to have in'nitespeed as it passed over ONE particular real second (represented as a realnumber on the Real line of seconds).The above was quite interesting for me. It appears as though it is true. Ifthe passing seconds are represented as real numbers, then indeed, a_speci'c_ second in the future (which is a speci'c real number)approaching the present and becoming past, passes from the state future tothe state past in 0 time. Thus the speed vector of time as it passes overa _speci'c_ (real) second, must be in'nite.In other words, the rate of conversion of future seconds to past seconds (aspoints on the real line) is in'nite. In mathematical terms, an uncountablenumber of point-seconds get converted from future point-seconds to pastpoint-seconds if time passes continuously.The above is related to ZenoÍs paradox, in the sense that although the §owrateof point-seconds is in'nite, we are still able to perceive of a 'niteamounts of seconds having passed by, after, say 2 minutes.The key difference here is that although a 'nite number of seconds havepassed by in, say, 2 minutes, an uncountable number of individual realnumbers (representing point-seconds) have been converted from futurepoint-seconds to past point-seconds in that 2-minute time interval.here: Either time cannot be REALLY continuous (as a mapping from whatever itis to the Real numbers) or our brains are experiencing lots of individualgaps. And by lots, I mean an uncountable number of such point-gaps.If time is indeed continuous, then our brains appear to be functioning akinto a semi-analog clock: Every second of time in perception, includes a'nite number of 'nite perceptive gaps, during which there is absoluteconscious stasis.In other words, during every passing second, the process consciousness,appears to be suspended quite often. If this is the case, it appears that wehave modeled our computers in a very similar manner to how our own brainswork. Similar to a multitasking environment.So far, the Computer Scientist was correct. I then employed the big guns ofMath and told him that although philosophical continuity appears to be anon-perceivable notion (in actual terms), the mathematical mind canasymptotically try to understand it, using epsilons and deltas,via thenotion of mathematical continuity.I am sure I convinced him, since he isnÍt a mathematician, but if I were himI would probably object slightly: The key is the quanti'er for everyepsilon, there exists a delta. So no matter WHAT epsilon you give me, (fromall the available uncountable ones) I can throw at you a delta that(possibly) depends on epsilon.Indeed the quanti'er for every, forces the brain to consider mathematicalcontinuity here, but in practical terms isnÍt always a _speci'c_ epsilonthatÍs being tested? So, although the quanti'er intuitively (andmathematically) satis'es this crave for continuity, in actual practice, theavailable epsilons are discreetized in our minds. We certainly cannotpractically test EVERY available epsilon > 0.The power of mathematics here, is that the theoretical setting under whichthe de'nition of a speci'c functionÍs continuity is tested using thede'nition, convinces the brain that the function is mathematicallycontinuous being independent of speci'c discreet epsilons, although thebrain itself fails to grasp the philosophical notion of continuity.And the true beauty of the damn thing is that thus, mathematics transcendsdiscreetness, pointing towards what continuity _might_ be, in thephilosophical sense. The key distinction here is mathematical vsphilosophical continuity.Mathematical continuity, is perfectly de'ned and as such, mathematiciansunderstand it. (at least insofar as the de'nition is concerned).It appears though that the notion of philosophical continuity is stillsomehwat elusive. (to me at least).To top the cake, if one considers any possible action inside this universe,every such action is quantized. Therefore even Caesium clocks are not reallyanalog. Their heart ticks according to a speci'c wavelength thatÍs beingemitted by the Caesium atom. Following this, does it make sense to talkabout the philosophical notion of having time being continuous?Time itself might well exist outside our existence continuously, but the keyissue here is whether our brains can ever hope to understand time as acontinuous process. If all actions in our universe are quantized, thensurely consciousness, must be as well. But if consciousness is quantized,then the term continuous time is nonsense.It appears as though the mathematicianÍs main stance 1) above is invalid.Any thoughts on this are very welcome.--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------ ------------------------Eventually, _everything_ is understandable => The above was quite interesting for me. It appears as though it is true. If> the passing seconds are represented as real numbers, then indeed, a> _speci'c_ second in the future (which is a speci'c real number)> approaching the present and becoming past, passes from the state future to> the state past in 0 time. Thus the speed vector of time as it passes over> a _speci'c_ (real) second, must be in'nite.I do not know what a speed vector of time is supposed to be. Do you view the present moment as the origin and speci'c moments in the future as moving backward toward the origin? If so, I see no in'nite. I see the constant vector -1. =automatic decision support system for my strategy game, IÍm looking for anequation for estimating the monopolization factor.Company1 produces 3 units of goods, Company2 - 2, Company3 - 2. What is themarket monopolization factor for Company1, and how will it change when itincreases its production from 3 to 4 units?--BB => I have to build a context free grammar that generates the following > language : > { a^i b^j c^k | i,j,k >= 0 and k <= i+j }[his attempt deleted]It looks like youÍre on the wrong track.First solve the problem of generating b^j c^k with k<=j,and then try to generate a^i B c^k with k<=i, whereB is something you have already generated. => I have to build a context free grammar that generates the following language :> { a^i b^j c^k | i,j,k >= 0 and k <= i+j }I take it you are unaware of the pumping lemma for CFLÍs?Perhaps you should do some of the exercises that appear in the'rst hit at =the smallest non-de'nable ordinal is de'nable.why is this argument wrong? => One of my quests is to 'nd a source, not code, that explains the> difference between a ripple sort and a bubble sort.> I dont know any book, but the ripple sort takes the 'rst element> compares it with each of the remaining and swap them when the one in> higher position is greater. At the end of the process the 'rst element> will be the lowest, so the algoritm repeats starting with the second.>|>|> This is KnuthÍs Algorithm S (Straight selection sort). There the outer>|> loop runs down from N to 2. Knuth also gives a timing comparison>|> with bubble sort. See TAOCP 2nd Edition Vol 3 p 139--140.>The trouble with all this mess is the publish or die syndrome :-(>Back in the days when computers were made of discrete components>and computer science degrees didnÍt exist, pretty well all of the>simple O(N^2) sorts were called bubble sort. Nobody could see any>good reason to distinguish the variations, and any competent>programmer could develop one from reading about another.>Much more recently, people have got a lot of academic mileage from>publishing old and well-known but previously unpublished results,>and it helps a lot to make a paper if you invent a new name. Hence>all the different names for bubble and radix sort variants. Knuth>wasnÍt entirely guiltless in the renaming respect.>If you think about it, there isnÍt a hard distinction between>bubble and selection sorts (in KnuthÍs terms), and I have both>seen and invented intermediate forms and variants. Knuth really>should have made that clearer.And as you know, Knuth certainly goes back to the daysof discrete components. My favorite Knuth program ishis compiler for the IBM 650 (without the Ramac storageunit). I believe it was called Runcible which was anacronym the 'rst two words of which were Real useful.Sadly, time has erased the rest. ---- Paul J. Gans =Speaking of Knuth.My edition is 31 years old.Are the newer editions more up to date, or do they just include editorial'xes, if even that?-- http://www.standards.com/; See Howard KaikowÍs web site.> One of my quests is to 'nd a source, not code, that explains the> difference between a ripple sort and a bubble sort.>> I dont know any book, but the ripple sort takes the 'rst element> compares it with each of the remaining and swap them when the one in> higher position is greater. At the end of the process the 'rstelement> will be the lowest, so the algoritm repeats starting with thesecond.>|>>|> This is KnuthÍs Algorithm S (Straight selection sort). There theouter>|> loop runs down from N to 2. Knuth also gives a timing comparison>|> with bubble sort. See TAOCP 2nd Edition Vol 3 p 139--140.>>The trouble with all this mess is the publish or die syndrome :-(>>Back in the days when computers were made of discrete components>and computer science degrees didnÍt exist, pretty well all of the>simple O(N^2) sorts were called bubble sort. Nobody could see any>good reason to distinguish the variations, and any competent>programmer could develop one from reading about another.>>Much more recently, people have got a lot of academic mileage from>publishing old and well-known but previously unpublished results,>and it helps a lot to make a paper if you invent a new name. Hence>all the different names for bubble and radix sort variants. Knuth>wasnÍt entirely guiltless in the renaming respect.>>If you think about it, there isnÍt a hard distinction between>bubble and selection sorts (in KnuthÍs terms), and I have both>seen and invented intermediate forms and variants. Knuth really>should have made that clearer.> And as you know, Knuth certainly goes back to the days> of discrete components. My favorite Knuth program is> his compiler for the IBM 650 (without the Ramac storage> unit). I believe it was called Runcible which was an> acronym the 'rst two words of which were Real useful.> Sadly, time has erased the rest.>> ---- Paul J. Gans =>Does anyone have the exact value of PI? I need it to about 10 billion Yes, but the space in this post is too small to hold them all. ---- Paul J. Gans =I search a free library like ODEPACK but written in C/C++.I would like to have a function like DLSODA for a system ofdifferential equations.Do you know a library downloadable on Web ? =>> Does anyone know of any shareware that can take a series of (x,y,z) points, > plot them and give the best 't equation? Try my online curve and surface 'tting website, http://zunzun.com - I have a function'nder implemented for surfaces. James Phillips http://zunzun.com => I am curious as to whether or not there is a fairly standard library> or package for doing sparse matrix computations (like multiplication,> addition, etc. - fairly straightforward stuff). IÍd prefer C, but> Fortran is 'ne too. PETSc is a rather comprehensive package for all things matrix: http://www.mcs.anl.gov/petscYou can use it from both C and Fortran. Also, if you want toroll your own, netlib is an excellent starting point: http://www.netlib.org/templatesHere youÍll 'nd descirptions on how to create iterative solvers,sparse matrices, preconditioners and more. Recommended.Finally, if you wish to try something different from C, I havea quite large Java library for matrix computations (akinto PETSc and MTL): http://www.math.uib.no/~bjornoh/mtjMy own tests with the NIST sparse BLAS indicates its performanceis quite on par with compiled libraries.-- Bjrn-Ove HeimsundCentre for Integrated Petroleum ResearchUniversity of Bergen, Norway =>> 1. symbolic algebra and,> 2. numerical computation>> For (2), itÍs a restricted expressivity. This might seem paradoxical at> 'rst sight, but the more restricted the things that one can do are, the> easier it is for the compiler to analyse the sources and generate> ef'cient machine code for it.Yes, but does anyone actually write decent compilers in the real world?IÍve done low-level ASM code for 11 years... rarely have I seen a compilerthat was any good at it. The DEC Alpha compiler was the exception. Anywaysthis is imperative stuff, for mainstream languages like C and C++, so prettymuch the natural 't to the task of optimizing the CPU instruction stream.Those compilers mostly suck, so pardon my cynicism that youÍre going to getanywhere layering functional constructs on top of the basic problem?-- Brandon Van Every Seattle, WAWe live in a world of very bright people buildingcrappy software with total shit for tools and process. - Ed Mckenzie =>>For [numerical computation], itÍs a restricted expressivity. This might seem paradoxical at>>'rst sight, but the more restricted the things that one can do are, the>>easier it is for the compiler to analyse the sources and generate>>ef'cient machine code for it.> Yes, but does anyone actually write decent compilers in the real world?> IÍve done low-level ASM code for 11 years... rarely have I seen a compiler> that was any good at it.I havenÍt looked into ASM for years, so I can only repeat what I have read elsewhere: that current-day compilers are quite good at generating decent machine code. For whatever de'nition of quite good. (You might want to direct this kind of question to comp.compilers.)Anyway, optimization isnÍt just generating good ASM code, itÍs also stuff like loop unrolling, common subexpression elimination, etc.C and C++ are notoriously hard to optimize this, since itÍs often very dif'cult to trace which pointer may refer to what array. Fortran compilers have a far easier job since Fortran programs donÍt use pointers, they use indexes (at least that used to be true for older Fortran code, and IÍm pretty sure that all the numeric libraries are still written in that style). > The DEC Alpha compiler was the exception. Anyways> this is imperative stuff, for mainstream languages like C and C++, so pretty> much the natural 't to the task of optimizing the CPU instruction stream.Not really. C was just right for the CPUs of the 70ies, but nowadays it overspeci'es so many things that itÍs getting plain dif'cult to write even a decent compiler for it.IOW it may be possible to write a compiler that does a decent job of utilizing the register set, but the C de'nition imposes so many constraints on instruction ordering that itÍs a challenge to keep all the pipelines of the CPU busy. (And thatÍs just one of the code generation problems in C. Again, I recommend asking in comp.compilers.)> Those compilers mostly suck, so pardon my cynicism that youÍre going to get> anywhere layering functional constructs on top of the basic problem?Sort of.Functional languages have two advantages:1) Imperative programs impose a strict rule on the order in which things get executed. Most of these ordering constraints are accidental, only a few are essential, but those that are essential must be observed. ItÍs quite hard for a compiler to 'nd out which ordering is essential and which isnÍt (as witnessed by the huge body of literature on array and matrix loop optimization). Functional languages impose very little ordering, and almost all of that ordering is essential, so the compiler can merrily unroll loops, eliminate common subexpressions, and do all other kinds of code transformations. (Be warned: this kind of optimization is far less common than one would expect. The compiler thatÍs generally considered to generate the fastest code - that for OCaml - doesnÍt do much code transformation, it just generates good ASM. Or at least we were so told on the newsgroup. Still, since functional languages tend to have a far simpler semantics than imperative ones, itÍs less work to get the compiler right, so thereÍs more time for doing a good code generation backend *g*)2) Functional languages are extremely good at abstraction, far better than imperative ones. ( ItÍs possible to write your own looping constructs, and these are still ordinary functions, no macro expansion, meta facilities or whatever required. You can write various styles of iterators, or what you want - and everything is still just functions. This is one of the main reasons why functional languages tend to have a simpler semantics than imperative ones. ) Of course, highly abstract code is dif'cult to optimize, but even in a functional language, you can write quite down-to-earth, let-me-control-every-single-bit code. ItÍs the programmerÍs choice whether some piece of code should be concrete or abstract, so the library designer can always adapt the trade-off between speed and §exibility. I think this bene't is the more important one: one can have fast (and in§exible) code where speed matters, and abstract (and easily-used) library routines where §exibility is important.Of course, to be of any use, the standard numeric libraries would need to be rewritten in a functional language, and IÍm not going to hold my breath for that to happen. Those Fortran libraries are Good Enough After All...Jo--Currently looking for a new job. =Actually one of the main problems in optimizing languageslike C or Fortran is the need to recognizing aliases: two ormore different ways of accessing the same value. E.g. callsby reference, equivalence statements, pointer-following.A major issue for writing good assembler is that you arepresumably doing this because absolute ef'ciency is important.But then you have to write different assembler code to be optimal forcomputers that (nominally) have the same architecture buthave different performance, including different numbers ofarithmetic units, cache sizes, pipeline depth, etc.The compiler writers supposedly have studied these things inthe best cases. Even in the 1960Ís, different models of the IBM S/360had different better code sequences for things as simple asmoving 32 bits from one place in memory to another.I think this message is cross-posted to too many newsgroups.Where did it start?> For [numerical computation], itÍs a restricted expressivity. This > might seem paradoxical at> 'rst sight, but the more restricted the things that one can do are, the> easier it is for the compiler to analyse the sources and generate> ef'cient machine code for it.>> Yes, but does anyone actually write decent compilers in the real world?>> IÍve done low-level ASM code for 11 years... rarely have I seen a >> compiler>> that was any good at it.> I havenÍt looked into ASM for years, so I can only repeat what I have > read elsewhere: that current-day compilers are quite good at generating > decent machine code. For whatever de'nition of quite good. (You might > want to direct this kind of question to comp.compilers.)> Anyway, optimization isnÍt just generating good ASM code, itÍs also > stuff like loop unrolling, common subexpression elimination, etc.> C and C++ are notoriously hard to optimize this, since itÍs often very > dif'cult to trace which pointer may refer to what array. Fortran > compilers have a far easier job since Fortran programs donÍt use > pointers, they use indexes (at least that used to be true for older > Fortran code, and IÍm pretty sure that all the numeric libraries are > still written in that style).> The DEC Alpha compiler was the exception. Anyways>> this is imperative stuff, for mainstream languages like C and C++, so >> pretty>> much the natural 't to the task of optimizing the CPU instruction >> stream. Not really. C \ was just right for the CPUs of the 70ies, but nowadays > it overspeci'es so many things that itÍs getting plain dif'cult to > write even a decent compiler for it.> IOW it may be possible to write a compiler that does a decent job of > utilizing the register set, but the C de'nition imposes so many > constraints on instruction ordering that itÍs a challenge to keep all > the pipelines of the CPU busy. (And thatÍs just one of the code > generation problems in C. Again, I recommend asking in comp.compilers.)> Those \ compilers mostly suck, so pardon my cynicism that youÍre going >> to get>> anywhere layering functional constructs on top of the basic problem?> Sort of.> Functional languages have two advantages:> 1) Imperative programs impose a strict rule on the order in which things > get executed. Most of these ordering constraints are accidental, only a > few are essential, but those that are essential must be observed. ItÍs > quite hard for a compiler to 'nd out which ordering is essential and > which isnÍt (as witnessed by the huge body of literature on array and > matrix loop optimization). Functional languages impose very little > ordering, and almost all of that ordering is essential, so the compiler > can merrily unroll loops, eliminate common subexpressions, and do all > other kinds of code transformations. (Be warned: this kind of > optimization is far less common than one would expect. The compiler > thatÍs generally considered to generate the fastest code - that for > OCaml - doesnÍt do much code transformation, it just generates good ASM. > Or at least we were so told on the newsgroup. Still, since functional > languages tend to have a far simpler semantics than imperative ones, > itÍs less work to get the compiler right, so thereÍs more time for doing > a good code generation backend *g*)> 2) Functional languages are extremely good at abstraction, far better > than imperative ones. ( ItÍs possible to write your own looping > constructs, and these are still ordinary functions, no macro > expansion, meta facilities or whatever required. You can write > various styles of iterators, or what you want - and everything is still > just functions. This is one of the main reasons why functional > languages tend to have a simpler semantics than imperative ones. > ) Of course, highly abstract code is dif'cult to optimize, but > even in a functional language, you can write quite down-to-earth, > let-me-control-every-single-bit code. ItÍs the programmerÍs choice > whether some piece of code should be concrete or abstract, so the > library designer can always adapt the trade-off between speed and > §exibility. I think this bene't is the more important one: one can > have fast (and in§exible) code where speed matters, and abstract (and > easily-used) library routines where §exibility is important.> Of course, to be of any use, the standard numeric libraries would need > to be rewritten in a functional language, and IÍm not going to hold my > breath for that to happen. Those Fortran libraries are Good Enough After > All...> Jo> -- > Currently looking for a new job.> => MKL is free? I thought it was $199.00.Oh sorry, I was referring to the academic license which I believe is still free. =I need an automated method to 't a curve of the form y =c*x*(1+a*e^(b*x)) to test data. This is a magnetic saturation curvefor an electric machine. The exponential term is the saturationfunction, and is negligible at low values of x. We have severalpoints at low x values. At high values of x, the exponential termdominates. We usually have a few points out to where the exponentialterm is 0.5 or more, but not always. The data at low x are often abit noisy.When we 't by hand, we draw a straight line approximating anasymptote to the lower part of the curve, then we subtract that linefrom the data to get approximately the exponential part of thefunction (the saturation factor). By taking the logarithm of bothsides for the dataabove a certain value of x, we can get a pretty good't on the upper part of the curve. Combining the two functions givesthe full approximation.The problem being, it requires a bit of human judgement to pick outthe spot where the linear part ends and the exponential begins. If Iget it wrong, either the linear portion will be skewed by having partof the exponential term mixed in, or else it wonÍt 't the transitionif not enough data is used for the exponential part (particularly i't is not large).The approximation only needs to 't within 1% or so of the realvalues (which donÍt actually 't the function exactly). Accuracy nearthe transition region is most important, although this form ofequation was selected because it gives reasonable results whenextrapolated to higher values than we have data for.WeÍre doing this in LabView, so we donÍt have access to numericallibraries for nonlinear curve 'tting and such. Would prefer a quickand dirty solution, though we might be able to port in some Fortrancode. I donÍt know how much procedural code is possible.A table of sample data follows:Voc(PU) Igf(A)0.196 5.000.304 7.850.399 10.430.616 16.200.797 21.540.906 25.890.942 29.011.014 32.751.051 37.251.105 44.801.196 61.181.304 105.61This is approximately 't with:a = 0.000232b = 7.005c = 25.74Allen-- Allen Windhorn (507) 345-2782 FAX (507) 345-2805Kato Engineering (Though I do not speak for Kato)P.O. Box 8447, N. Mankato, MN 56002Allen.Windhorn@LSUSA.com => I need an automated method to 't a curve of the form y => c*x*(1+a*e^(b*x)) to test data. This is a magnetic saturation curve> for an electric machine. The exponential term is the saturation> function, and is negligible at low values of x. We have several> points at low x values. At high values of x, the exponential term> dominates. We usually have a few points out to where the exponential> term is 0.5 or more, but not always. The data at low x are often a> bit noisy.> When we 't by hand, we draw a straight line approximating an> asymptote to the lower part of the curve, then we subtract that line> from the data to get approximately the exponential part of the> function (the saturation factor). By taking the logarithm of both> sides for the dataabove a certain value of x, we can get a pretty good> 't on the upper part of the curve. Combining the two functions gives> the full approximation.> The problem being, it requires a bit of human judgement to pick out> the spot where the linear part ends and the exponential begins. If I> get it wrong, either the linear portion will be skewed by having part> of the exponential term mixed in, or else it wonÍt 't the transition> if not enough data is used for the exponential part (particularly if> it is not large).> The approximation only needs to 't within 1% or so of the real> values (which donÍt actually 't the function exactly). Accuracy near> the transition region is most important, although this form of> equation was selected because it gives reasonable results when> extrapolated to higher values than we have data for.> WeÍre doing this in LabView, so we donÍt have access to numerical> libraries for nonlinear curve 'tting and such. Would prefer a quick> and dirty solution, though we might be able to port in some Fortran> code. I donÍt know how much procedural code is possible.> A table of sample data follows:> Voc(PU) Igf(A)> 0.196 5.00> 0.304 7.85> 0.399 10.43> 0.616 16.20> 0.797 21.54> 0.906 25.89> 0.942 29.01> 1.014 32.75> 1.051 37.25> 1.105 44.80> 1.196 61.18> 1.304 105.61> This is approximately 't with:> a = 0.000232> b = 7.005> c = 25.74> AllenAllen:I believe the Levenberg-Marquardt method of curve-'tting can be used to 't your data. A quick search on Google found many freely available programs, some with source code.Also, here is what the Matlab v6.5 curve 't GUI predicts for your data:General model: f(x) = c*x + a*c*exp(b*x)Coef'cients (with 95% con'dence bounds): a = 0.0001179 (4.303e-005, 0.0001928) b = 7.735 (7.286, 8.183) c = 25.52 (24.39, 26.65)Goodness of 't: SSE: 3.104 R-square: 0.9996 Adjusted R-square: 0.9996 RMSE: 0.5873Good luck,OUP