mm-242 === Subject: Re: Simplicial coordinates <-> Cartesian coordinates> What formulas are used for this transformation? Also, i'm interes> in more information about simplicial coords (they are used in> triangles: if we have a point within a triangle, so it divides its> squre S into three squares S1, S2, S3, then we can use the simplicial> coordinates gamma1 = S1/S, gamma2 = S2/S, gamma3 = S3/S,> gamma1+gamma2+gamma3 = 1).http://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch15.d/ IFEM.Ch15.index.html-> Chapter 15: The Linear Plane Stress TriangleForgot to point out: the equations you want are (15.10) and (15.11)in that chapter. Simplicial coordinates are called triangularcoordinates there. The name area coordinates also appears in old(pre-1970) finite element papers because of === Homeomorphism between reals and complex numbers?>> I've heard that the reals have the same cardinality as the complex>> numbebut I am unable to find a one-to-one function. Can anyone>> give me an example of one, or barring that, a proof that none exists?> A homeomorphism would be a bijection that preserves the topological > properties. There is no such map, since R and C are topologically > different. For example you can remove one point from R to make the > resulting set disconnec; but you can't do the same to C.> If you're just looking for a bijection, the trick is to interleave the > decimal expansions. For example the pair (.11111..., .22222..) would > map to .121212... You can convince yourself that this is almost a > bijection between the unit interval and the unit square.> The hard part is dealing with reals that have two decimal expansions, > such as .11111... = .2. I don't know the exact method for handling > this.Whenever the real or imaginary part if a complex number has two possiblerepresentations, you choose the one that doesn't end in all 9's. Theresulting map obtained by interleaving digits always maps differentmembers of C to different members of R, but there are some members ofR (such as 0.909090909...) that don't get accoun for by this map.No problem. There is also an obvious injection g: R -> C. That meansyou can invoke the Cantor-Bernstein theorem, which says that if two setscan each be injec into the other, then there is a bijection betweenthem. Using the notation of cardinality, the theorem says that if A andB are sets such that |A| <= |B| and also |B| <= |A|, then it follows that === following system to find the average rate of thedeparture process:Input process is Poisson process, the service time is exponentialdistribution.I am using the following method:1. generate the arrival event using exponential interarrival time2. record the interval of the the leaving event after the cuser service3. average the interval until sufficient satisfactory satisfied.However, the main problem lies in the convergence. It is difficult toconverge to a sufficient stable average rate of the departure process.Any comments for this === departure process> I am simulating the following system to find the average rate of the> departure process:Input process is Poisson process, the service time is exponential> distribution.I am using the following method:1. generate the arrival event using exponential interarrival time> 2. record the interval of the the leaving event after the cuser service> 3. average the interval until sufficient satisfactory satisfied.However, the main problem lies in the convergence. It is difficult to> converge to a sufficient stable average rate of the departure process.Any comments for this problem?many thansk for ur help. The stabilty of the output process has nothing to do with the averaging. If the input process is unstable, the output is unstable after a sufficiently long waiting period. So covergence has nothing to do with the problem. So use a weigh average, rather than a flat average. The only weightings that will produce a stable output are the ones that have zeros of weight greater than the weight === distribu sequence> Could anyone explain the meaning of a uniformly distribu sequence?A sequence (u_n) is uniformly distribu (according to Weil's criterion)That's Weyl's criterion; H. Weyl, not A. Weil. > <= for any positive integer h,> 1/n*sum(exp(2*i*Pi*h*u_k, k=1..n) converges to 0 when n->+oo.> Now it is clear that {n*a; a irrational} is uniformly distribu, and that> {n*a; a rational} is not.> There is a graphical interpretation of this criterion: if you represent each> point of your sequence by a point on a circle (of radius 1), then the> barycenter of all the points will converge to the point (0,0).That's implied by ud, but not equivalent to it. After all, if your sequence gets represen on the circle by the points (1,0), (-1,0), (1,0), (-1,0), (1,0), (-1,0), (1,0), (-1,0),... then the barycenter is (0,0), but the sequence is not ud. Your interpretation is equivalent to taking h = 1 only === sequenceI have a theorem I have to prove. If X is a number sequence then thefollowing statements are equivalent: (1) X has a cluster point c and (2) X hasa subsequence which converges to c. I'm trying to prove 1->2. How can imodify my proof for subsequences (i did it for a sequence...)? Or is that notpossible? I still have to prove the other direction too.My Proof: Let c be the cluster point of hte number sequence X. For every nbelonging to positive integethere is a Xn belonging to X such that 00. There is a positive integer N such that1/epsilon=N, |Xn-c|<1/n<=1/N following statements are equivalent: (1) X has a cluster point c and (2) X has> a subsequence which converges to c. I'm trying to prove 1->2. My Proof: Let c be the cluster point of hte number sequence X. For every n> belonging to positive integethere is a Xn belonging to X such that 0> Let epsilon>0. There is a positive integer N such that> 1/epsilon=N, |Xn-c|<1/n<=1/N infinity of Xn=C.>Let c be a cluster point of { x_n }. Thus for all j in N, some x in X with |x - c| < 1/2^jfor each j in N, let x_j be the x with that property that has the smallestindex. I leave to you to show x_j is a sub sequence of { x_n }The converse is easy. If { x_(n_j) } is a subsequence with limit c, thenshow c === posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064274074 http://mygate.mailgate.org/mynews/sci/sci.math/ 55b57e630dc56639291892409fe43275.48257%40mygate.mailgate.org> I have a theorem I have to prove. If X is a number sequence then the> following statements are equivalent: (1) X has a cluster point c and (2) X has> a subsequence which converges to c. I'm trying to prove 1->2. How can i> modify my proof for subsequences (i did it for a sequence...)? Or is that not> possible? I still have to prove the other direction too.Well, it _has_ to be a SUBsequence; consider if X is a sequence in (0,1)and alternate odd or even entries of X are converging respectively to1/4 and 3/4. Then the _sequence_ X does not converge at all, but it hasfor each cluster point a subsequence which converges to that clusterpoint.xanthian.-- Pos via Mailgate.ORG Server - === not continuously extensibleSupersedes: interesting discussion of the topic.One more question in connection with the thread:In [Jeffrey Rauch, Partial differential Equations] corollary 7 insection 5.5 states:Let G (subset of R^n) be a domain such that the closure of G is a compactsmooth submanifold with boundary, and definedM_eps := { x in M : dist(x,dM) < eps}.Then for u in W^1_0(M) we have:( int_{dM_eps} |u|^2 dx ) / ( eps * Vol(dM_eps) ) -> 0 for eps-> 0+This gives a nice impression in which way u tends to zero at theboundary.My question at this point is:Can that statement be extended to more general domains?(The proof in [Rauch] depends heavily on the parameterization of theboundary manifold and is === posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064274074 http://mygate.mailgate.org/mynews/sci/sci.math/ de12349cd4ee0cf46f95b6ad130f99e0.48257%40mygate.mailgate.org> Is it possible to find a sequence (d_n) and two numbers a and x_1 such> that (x_n) contains only prime numbers? Clearly, this is not possible> with a=3 and x_1=1:The search for _any_ integer sequence expression that efficientlyyields all and only prime numbers has been one of the smaller holygrails of mathematics for a very long time; that's what the series thatproduces the Mersenne Primes was originally conjectured to be.If it is possible, no one knows it yet, aside from trivial stuff liketheSieve of Eratosthenes where you find the primes by throwing away all thenon-primes one by one first. That's not what the vague termefficientlyintends.[Probably I've way overspoken my knowledge base....]xanthian.-- Pos via === Prime number sequence, randomness> x_{n+1} = a x_n + d_n, d_in in {-2, +2}Is it possible to find a sequence (d_n) and two numbers a and x_1 such> that (x_n) contains only prime numbers? Seems unlikely to me, but not at all easy to prove. > If we find a sequence (d_n) and two integers a and x_1 that work, then> would the sequence (d_n) would necessraly look like pseudo-random> numbers?Depends what you mean by looking like random numbers. Certainly no such sequence could be periodic, but conceivably such a sequence could be 60% +2s and 40% -2s - at least, I don't see why not. That wouldn't look === randomnessx_{n+1} = a x_n + d_n, d_in in {-2, +2}Is it possible to find a sequence (d_n) and two numbers a and x_1 such> that (x_n) contains only prime numbers? Seems unlikely to me, but not at all easy to prove. Looking at them mod 3, I think that proving that if such asequence exists then 3|a might be possible. There's a similarproblem in C&P PNaCP, IIRC (looking at 2*x_n+/-1 instead).> If we find a sequence (d_n) and two integers a and x_1 that work, then> would the sequence (d_n) would necessraly look like pseudo-random> numbers?Depends what you mean by looking like random numbers. Certainly > no such sequence could be periodic, but conceivably such a sequence > could be 60% +2s and 40% -2s - at least, I don't see why not. That > wouldn't look random to me. I agree that it would be utterly bizarre if such a sequence wereperiodic, but I can't instantly see what principle prevents it.Random needs ot be defined properly. A CP (or IID) BRV with p=0.4 is by some people quite correctly described as random. (Otherwise you couldn't have any distribution for a random variable apart from the uniform distribution - anything else and it wouldn't === randomness> x_{n+1} = a x_n + d_n, d_in in {-2, +2}>> Is it possible to find a sequence (d_n) and two numbers a and x_1 such> that (x_n) contains only prime numbers? Seems unlikely to me, but not at all easy to prove. > If we find a sequence (d_n) and two integers a and x_1 that work, then> would the sequence (d_n) would necessraly look like pseudo-random> numbers?Depends what you mean by looking like random numbers. Certainly > no such sequence could be periodic, but conceivably such a sequence > could be 60% +2s and 40% -2s - at least, I don't see why not. That > wouldn't look random to me.> I agree that it would be utterly bizarre if such a sequence were> periodic, but I can't instantly see what principle prevents it.If the d_n sequence were periodic then the x_n sequence would satisfy a constant coefficient linear recurrence, so it would be periodic mod m for every m. Take m to be one of the primes in the x_n sequence and then periodically there'd be a term divisible by m and hence not prime (it's clear that other than trivial cases like a = 1 the x_n sequence increases without bound). > Random needs ot be defined properly. A CP (or IID) BRV with p=0.4 > is by some people quite correctly described as random. (Otherwise > you couldn't have any distribution for a random variable apart from > the uniform distribution - anything else and it wouldn't be a random > variable!)You're absolutely right. I took OP to be writing informally, and meaning something like a sequence of flips of a fair coin. So I'll put it this way; I doubt such a sequence d_n exists, but if it does, I don't see why it has to have any distribution function at all. E.g., it could be that the percentage of +2 in an initial segment doesn't even converge as the initial segment gets longer.-- === x_n + d_n, d_in in {-2, +2}> If the d_n sequence were periodic then the x_n sequence would satisfy > a constant coefficient linear recurrence, so it would be periodic > mod m for every m. Take m to be one of the primes in the x_n sequence > and then periodically there'd be a term divisible by m and hence not > prime (it's clear that other than trivial cases like a = 1 the x_n > sequence increases without bound). I'm kicking myself - I was looking at a similar problem only about twoweeks ago, which had the same solution. D'Oh! === RevolutionA storage shed is the shape of a hemisphere. The base is circularwithradius r =1. Cross sections of the shed which are perpendicular to a>diameter of the base are square. Find the volume of the shed. This is>not difficult for a hemisphere but I'm having a problem with a>hemisphere with a square cross section. Anyone? Holman>First of all, You've not going to be revolving anything.> The expection of the student is a sketch of the solid by rotating the> indica region and then setting up the integral.> This was the last of a set of problems on volume of revolution from> 1st year college calculus.> I'm thinking there is an error in the way the problem is sta. HolmanThis is a typical type of problem for a section on revolving solids, but if you read the problem carefully, there is *nothing* in the problem itself to suggest there is a revolution occurring. The reason it is in the section on revolutions of curves is because it deals with the same concept of taking slices and integrating over the === === Re: Volume of REvolution>Subject: Re: Volume of Revolutionin>> In sci.math, Holman>> :>> A storage shed is the shape of a hemisphere. The base is circular>with>> radius r =1. Cross sections of the shed which are perpendicularto a>> diameter of the base are square. Find the volume of the shed.This>is>> not difficult for a hemisphere but I'm having a problem with a>> hemisphere with a square cross section. Anyone?>> You've specified the (top of the?) shed as a hemisphere.>> The cross section must include a semicircle on top,>> regardless of cutting-plane orientation; the plane needn't>> cut the shed straight up and down.>> If one assumes a plane going straight up and down, one>> gets a hemisphere sitting on a cylinder, if the problem>> is suitably correc. However, this may not be the only>> solution, although I'm not entirely certain what the bot>> will look like if the plane cuts at a specified angle --> will looklike if the plane cuts at a specified angle -->> probably a modified frustrum of some sort.>> If one assumes the former, the problem is completely specified,>> although I'm not sure how correctly.Lets see...Volume = Hemisphere + Cylinder> = Sum Pi*y^2 dx + Pi*r^2*h> = Sum Pi*Sqrt(1-x^2)^2 dx + Pi*r^2*h (fora>square x section r=h=1)> = Pi [x-1/3x^3] + Pi ...........x=1> = 2/3Pi + Pi = 5/3 Pi>The answer give at the bot of the page where I read this from is>16/3. Other problem solutions are given in terms of Pi and eventhough>5/3 Pi is not too far from 16/3, it looks like there is an error inthe>way the question is sta. Holman> Good lord people, stop confusing the poor kid. The problem is this: the object has a base which is a circle of radius one. If you take> cross sections perpendicular to the diameter you get squares: dV = dx (2h)^2 , where h^2 = R^2 - x^2 (but R = 1) dV = dx 4 (1 - x^2) V = Integrate[ 4 (1 - x^2) , {x,-1,1}] = 4 (4/3) = 16/3> hemispheremeaning that the question is very poorly sta.At 52 I don't think I qualify as being a kid anymore and why === posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064274074 http://mygate.mailgate.org/mynews/sci/sci.math/ 4dadb95f80ae97a65064c476310b94ed.48257%40mygate.mailgate.org> At 52 I don't think I qualify as being a kid anymore and why I'm> reviewing 1st year college math is another question.Probably because it will be a lot harder if you wait until you'remy age (and I got my degree in math and likely couldn't keep upwith you in the book you're using now); gray matter needs constantexercise to remain functional, and so do the individual skills itcontains, which is what has me back reading sci.math (andunderstanding one posting in twenty, at best).xanthian.-- Pos via Mailgate.ORG Server - === storage shed is the shape of a hemisphere. The base is circular with> radius r =1. Cross sections of the shed which are perpendicular to a> diameter of the base are square. Find the volume of the shed. This is> not difficult for a hemisphere but I'm having a problem with a> hemisphere with a square cross section. Anyone?> Nit picks first:1) There is *no* revolution involved.2) This is *not* a hemispherical shed.Solution:What you are looking at is basically a circle centered at (0,0) with radius 1, where the (wlog) vertical cross-section being a square with side of length 2y where y is the upper edge of the circle.So, y=sqrt(1-x^2) is the upper edge.The cross-sectional area is (2y)^2 = 4(1-x^2).The bounds of integration with respect to x are -1 to 1.Integral from -1 to 1 of (4(1-x^2) dx) =Integral -1 to 1 of (4-4x^2) dx =(4x - 4/3 x^3) from -1 to 1 =(4-4/3) - (-4 + 4/3) =8 - 8/3 = 16/3-- === Vector/Subspaces/Spanning SetsI even have the answers to some of the problems but it doesn't mean much to me. Like this: Determine which of the following subsets of R^(nxn) are in factsubspaces of R^(nxn). (a) the symmetric matrices, (b) the diagonal matrices,(c) the nonsingular matrices, (d) the singular matrices, (e) the triangularmatrices, (f) the upper triangular matrices, (g) all matrices that commute witha given matrix A, (h) All matrices such that A^2=A, (i) all matrices such thattrace(A)=0Answer: (a), (b), (f), (g) and (i) are subspaces.I'm not asked to prove this but I'm not really sure how I'm supposed to KNOWthat some of those are subspaces while === Vector/Subspaces/Spanning Sets> I even have the answers to some of the problems but it doesn't mean muchto me.> Like this: Determine which of the following subsets of R^(nxn) are infact> subspaces of R^(nxn). (a) the symmetric matrices, (b) the diagonalmatrices,> (c) the nonsingular matrices, (d) the singular matrices, (e) thetriangular> matrices, (f) the upper triangular matrices, (g) all matrices that commutewith> a given matrix A, (h) All matrices such that A^2=A, (i) all matrices suchthat> trace(A)=0 Answer: (a), (b), (f), (g) and (i) are subspaces. I'm not asked to prove thisOh, yes you are.> but I'm not really sure how I'm supposed to KNOW> that some of those are subspaces while others are not.Let's consider just (a). What does it mean to be a subspace? You have alist of axioms, but most of them are auatically met (because you knowthat you were in a vector space to start with). What you really need toprove is that, if A and B are in the alleged subspace, then xA and A+B arealso in the subspace. (Or prove that there are some A and B that don't havethat property, usually by finding a counterexample.) So, let's just look at(a).If A is symmetric and B is symmetric, then is A+B? Well, let's say that theentries of A are a_ij. Then A is symmetric means that a_ij = a_ji.Similarly, b_ij = b_ji. The entries of A+B are a_ij+b_ij=a_ji+b_ji, so A+Bis symmetric. You also need to prove that xA is symmetric.For each of (b)-(i), you need to determine what the name means and theneither see if you can prove that that property holds under matrix additionand scalar multiplication or not.At this stage of the course, I doubt that you are supposed to know which areand which aren't. This exercise has two purposes. The primary one is togive you practice with testing for subspaceness. The secondary one is togive you a list of some things that are and are not subspaces, so you canuse them for examples later on. Like maybe even the next === Vector/Subspaces/Spanning Sets> I even have the answers to some of the problems but it doesn't mean much to me.> Like this: Determine which of the following subsets of R^(nxn) are in fact> subspaces of R^(nxn). (a) the symmetric matrices, (b) the diagonal matrices,> (c) the nonsingular matrices, (d) the singular matrices, (e) the triangular> matrices, (f) the upper triangular matrices, (g) all matrices that commute with> a given matrix A, (h) All matrices such that A^2=A, (i) all matrices such that> trace(A)=0Answer: (a), (b), (f), (g) and (i) are subspaces.I'm not asked to prove this but I'm not really sure how I'm supposed to KNOW> that some of those are subspaces while others are not.This is an exercise in applying the characterization of a subspace ofa vector space as a (non-empty) subset closed under vector additionand multiplication by a scalar.Let's tackle (d), for instance:The subset of singular matrices is non-empty, of course. It's alsoclosed under multiplication by a scalar (do you see why?). But it'snot closed under addition, because you can add the singular matrices[1 0; 0 0] and [0 0;0 1] and obtain a to do _any_ proofs in the course you're === Vector/Subspaces/Spanning Sets: I even have the answers to some of the problems but it doesn't mean much to me.: Like this: Determine which of the following subsets of R^(nxn) are in fact: subspaces of R^(nxn). (a) the symmetric matrices, (b) the diagonal matrices,: (c) the nonsingular matrices, (d) the singular matrices, (e) the triangular: matrices, (f) the upper triangular matrices, (g) all matrices that commute with: a given matrix A, (h) All matrices such that A^2=A, (i) all matrices such that: trace(A)=0: Answer: (a), (b), (f), (g) and (i) are subspaces.When in doubt, start from the definitions. You know a subspace must be itselfa vector space, hence contain the additive identity, 0. For R^(nxn), thatmeans the 0 matrix of n^2 0 entries. Now, is that matrix nonsingular? No,so (c) is not s subspace.: I'm not asked to prove this but I'm not really sure how I'm supposed to KNOW: that some of those are === question on Vector/Subspaces/Spanning Sets> I even have the answers to some of the problems but it doesn't mean much to me.> Like this: Determine which of the following subsets of R^(nxn) are in fact> subspaces of R^(nxn). (a) the symmetric matrices, (b) the diagonal matrices,> (c) the nonsingular matrices, (d) the singular matrices, (e) the triangular> matrices, (f) the upper triangular matrices, (g) all matrices that commute with> a given matrix A, (h) All matrices such that A^2=A, (i) all matrices such that> trace(A)=0Answer: (a), (b), (f), (g) and (i) are subspaces.I'm not asked to prove this but I'm not really sure how I'm supposed to KNOW> that some of those are subspaces while others are not.I'm working to learn about GA and this looks like a great question/answer.So, it looks like not c,d and e because the they can include base elements that are not of R.And the trick question, h. A was not defined as a subspace of R so...I, is because A reduces to a scalar????Best, Dan.-- http://lakeweb.nethttp://ReserveAnalyst.comdbAtLakewebDotCom== === =Subject: Re: Linear Algebra question on Vector/Subspaces/Spanning Sets> I even have the answers to some of the problems but it doesn't mean much to me.> Like this: Determine which of the following subsets of R^(nxn) are in fact> subspaces of R^(nxn). (a) the symmetric matrices, (b) the diagonal matrices,> (c) the nonsingular matrices, (d) the singular matrices, (e) the triangular> matrices, (f) the upper triangular matrices, (g) all matrices that commute with> a given matrix A, (h) All matrices such that A^2=A, (i) all matrices such that> trace(A)=0Answer: (a), (b), (f), (g) and (i) are subspaces.I'm not asked to prove this but I'm not really sure how I'm supposed to KNOW> that some of those are subspaces while others are not.I'm working to learn about GA and this looks like a great question/answer.What are GA? Genetic Algorithms? So, it looks like not c,d and e because the they can include base > elements that are not of R.What are base elements? Please provide a definition. What does itmean for a base element to be in or ouside of R?And the trick question, h. A was not defined as a subspace of R so...No no. In this context, A is a variable.I, is because A reduces to a scalar????What do you mean by reduces to a scalar? Where does I come in?Remember that it's a yes/no question: is the given subset a subspaceor not?I suggest you grab a good book, like the excellent A primer on linearalgebra by I.N. Herstein or the excellent (but much faster-paced)Linear algebra by Hoffman & Kunze, and read it very carefully. Youseem to be a bit confused and a good book should clear the === Algebra question on Vector/Subspaces/Spanning SetsWhat are GA? Genetic Algorithms? > Geometric AlgebraWhat are base elements? Please provide a definition. What does it> mean for a base element to be in or ouside of R?> I should have said basis vectors...And even in GA I can see my answer was very wrong.I suggest you grab a good book, like the excellent A primer on linear> algebra by I.N. Herstein or the excellent (but much faster-paced)> Linear algebra by Hoffman & Kunze, and read it very carefully. You> seem to be a bit confused and a good book disappoin by a book recommend in a sci group.Does the first book cover all in the latter?I have been downloading and printing out everything I can find ongeometric algebra.The got a lot to learn.It looks like this:http://planetmath.org/encyclopedia/LinearAlgebra.htmlis a good starting point till I get the book.Best, Dan.-- http://lakeweb.nethttp://ReserveAnalyst.comdbAtLakewebDotCom== === =Subject: Re: Linear Algebra question on Vector/Subspaces/Spanning SetsWhat are GA? Genetic Algorithms? provide a definition. What does it> mean for a base element to be in or ouside of R?> I should have said basis vectors...> And even in GA I can see my answer was very wrong.> I suggest you grab a good book, like the excellent A primer on linear> algebra by I.N. Herstein or the excellent (but much faster-paced)> Linear algebra by Hoffman & Kunze, and read it very carefully. You> seem to be a bit confused and a good book disappoin by a book recommend in a sci group.> Does the first book cover all in the latter?No, definitely not. They are about the same size (IIRC) but the secondone covers much more ground - assuming as it does a greater degree ofmathematical maturity (whatever that is.I'd recommend also the classic Finite-Dimensional Vector Spaces byPaul Halmos and for good exercises Linear Algebra (on the Schaumseries) by Seymour Lipschutz.I have been downloading and printing out everything 'subspace linear algebra' tells me> I've got a lot to learn.Yes. But I think you'll like it.It looks like this:> http://planetmath.org/encyclopedia/LinearAlgebra.html> is a good starting point till I get the === Vector/Subspaces/Spanning Sets> I even have the answers to some of the problems but it doesn't mean much to me.> Like this: Determine which of the following subsets of R^(nxn) are in fact> subspaces of R^(nxn). (a) the symmetric matrices, (b) the diagonal matrices,> (c) the nonsingular matrices, (d) the singular matrices, (e) the triangular> matrices, (f) the upper triangular matrices, (g) all matrices that commute with> a given matrix A, (h) All matrices such that A^2=A, (i) all matrices such that> trace(A)=0Answer: (a), (b), (f), (g) and (i) are subspaces.I'm not asked to prove this but I'm not really sure how I'm supposed to KNOW> that some of those are subspaces while others are not.I'm working to learn about GA and this looks like a great question/answer.So, it looks like not c,d and e because the they can include base > elements that are not of R.And the trick question, h. A was not defined as a subspace of R so...I, is because A reduces to a scalar????Best, Dan.First I should warn you that the last quo reply is incoherent andwithout content.Now to answer your question. The answers in all cases are evident andshould have been obvious. If they are not, then your understanding ofthe question is lacking. For a subset of a vector space to be asubspace it must (i) include the 0 vector and be closed under (ii)addition and (iii) scalar multiplication. (Or you can replace (i) byits being non-empty, since once it contains any vector and is closedunder scalar multiplication, it will contain the 0 vector.)Now take some of your examples. Take e, for example. Although 0 istriangular and a scalar multiple of a triangular matrix is triangular,the sum of an upper triangular and lower triangular matrix is nottriangula unless at least one of them is 0. So not closed under sum. What about non-singular? Well, the 0 matrix isn't there. Singular? Actually, in the 1 by 1 case, the singular matrices (only (0)) do forma subspace, but in all other cases you can take A to have one 1 in theupper left corner and 0s elsewhere, while taking B = I - A. Then Aand B are singular, while A + B is not. How about A^2 = A. Well,this is not closed under scalar multiplication. I^2 = I, but (2I)^2 =4I = 2(2I). And so === Vector/Subspaces/Spanning SetsI meant to ask if there's an easy way to tell if something is NOT a spanningset or does === curves?Does anyone know of a website with a Java applet or something thatwill draw quick graphs for me? I'd like something likex = sin ty = cos( 6t)range = 0 .. 2*piand it will just draw it for me. Another thing that would be great would be a site where I can upload adata file and it would tell me what curve gives the best fit, but thatmay be wanting === curves?>Does anyone know of a website with a Java applet or something that>will draw quick graphs for me? I'd like something like>x = sin t>y = cos( 6t)>range = 0 .. 2*pi>and it will just draw it for me. Try Plot Function and Derivative at Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia === website to plot curves?>>Does anyone know of a website with a Java applet or something that>>will draw quick graphs for me? I'd like something like>>x = sin t>>y = cos( 6t)>>range = 0 .. 2*pi>and it will just draw it for me. Try Plot Function and Derivative at >Sorry, I somehow didn't notice that you wan a parametric curve.Well you could use the Phase Plotter applet with the systemx' = cos(t)y' = -6 sin(6 t)and initial point [0,1].Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia === website to plot curves?>> Does anyone know of a website with a Java applet or something that>> will draw quick graphs for me? I'd like something like>> x = sin t>> y = cos( 6t)>> range = 0 .. 2*pi> and it will just draw it for me. Try Plot Function and Derivative at> > Does anyone know of a website with a Java applet or something that>> will draw quick graphs for me? I'd like something like>> x = sin t>> y = cos( 6t)>> range = 0 .. 2*pi>> and it will just draw it for me. Try Plot Function and Derivative at> do you know what Lissajous curves are?> Can you picture the curve for x = sin t> y = cos tTry to figure out what sets of equations result in the curves on> http://mathworld.wolfram.com/LissajousCurve.html.Hi ThereI can send you an old soft called EUREKA which works under DOSYou can solve all equation you want and plot it or tabulate === ThereI can send you an old soft called EUREKA which works under DOS> You can solve all equation you want and plot it or tabulate it but I wan something that was out on the web, free to the public, andaccessible from any computer with an internet connection;not really something to download and install.I think the best thing, meeting my criteria, that I've seen so farwas Symbmath at www.raceprediction.com. It does an incredible amount ofstuff evidently, but all I cared about was plotting parametric curves.(The step size is a little big and I don't see how to adjust it, though).Otherwise it seems pretty === equations?Jeff,> do you know what Lissajous curves are?> Can you picture the curve for x = sin t> y = cos tTry to figure out what sets of equations result in the curves on> the input. Yes, I do know what Lissajous curves are. And like I say in my reply to Robert Israel, I could solve my immediateproblem either by writing my own C program or just thinking a little harderwith my pen, but in the long run it would sure be nice to just fire upthe web browser and get the plots. I'm suprised Mathematica or some competingpackage doesn't offer some limi functionality over the web to try === website to plot curves?That collection of applets is pretty impressive! However, I note thatthe function you mention does not seem to draw parametric equations, whichis what I really need. Of course, I can write a program to produce datapoints and then plot them with gnuplot or something similiar; that is justa bit of bother when I am in the middle of something else and just want a quick sketch of some curves. I don't have Mathematica, Maple, readilyaccessible any more.You might also thing about making the source code publicly available,if possible, since that is a friendly thing to do.JeffTry Plot Function and Derivative at > === there a website to plot curves?> I don't have Mathematica, Maple, readily accessible any more.Think abouthttp://sourceforge.net/projects/maxima/ (symbolic maths)http://www.octave.org/ (numerics)http://pauillac.inria.fr/cdrom/www/scilab/eng.htm (numerics)I prefer the latter to octave. But that is a matter of taste.TN-- Write to: i at tn dash home point de(Just to feed hungry robots: somebody@absolute-nonsense`rm -fr === That collection of applets is pretty impressive! However, I note that> the function you mention does not seem to draw parametric equations,How about this applet:http://www-gap.dcs.st-and.ac.uk/~history/Java/ Lissajous.html(parameter control could be a bit better === That collection of applets is pretty impressive! However, I note that> the function you mention does not seem to draw parametric equations,You can rewrite the first equation as t=f(x) and replace the t in the secondequation by the result.Is OK 'coz x = sin(t) is a bijection over the given domain.(I'm not a mathematician. Feel free to correct me if I === use the wrongterminology or just talk rubbishStevenSubject: really happy with the answers of my question about `Smooth W^1_0function not continuously extensible'.But I've got one more problem which I think is even a bit more trickythan the previous one.Let G be a domain. I'm looking for an eigenfunction f of the Laplacianon W^1_0(G), such that f has no continuous extension to the boundary of G.Because of the smoothening property of the solution operator of theLaplacian there is no such eigenfunction if G satisfies some uniform conecondition.I suspect that G must have some inward direc sharp thorn to admitsuch an eigenfunction (just like the example by Lebesgue for aboundary value problem of the Laplacian with continuous boundaryconditions and no continuous solution).-- If you want to contact me, look at === continuous @ boundary> Let G be a domain. I'm looking for an eigenfunction f of the Laplacian> on W^1_0(G), such that f has no continuous extension to the boundary of G.Because of the smoothening property of the solution operator of the> Laplacian there is no such eigenfunction if G satisfies some uniform cone> condition.I suspect that G must have some inward direc sharp thorn to admit> such an eigenfunction (just like the example by Lebesgue for a> boundary value problem of the Laplacian with continuous boundary> conditions and no continuous solution). I would very much like to know whether this problem is a very hard oneor not.Does someone have a reference where such an example is construc?In many books on PDE this regularity of eigenfunctions is treaded inthe following way. First regularity theory is developed for solvingPoisson's equation(1) Laplace u = ffor u in W^1_0(G) and f in L^2(G). This theory includesrestrictions for the shape of the boundary. Then this regularitytheory is applied to the special problem(2) Laplace u = w uwith some complex number w. One outcome is that u is smooth in theinterior of G. Under the mentioned restrictions to the boundary shapeof G one also derives from this theory the continuity of u at theboundary of G.But examples, which show that the restrictions to the boundary shapeare necessary, are only given for problem (1).At least I found no such examples for (2) up to now.Any helpful comments on the topic are apprecia.Tobias Naehring-- Write to: i at tn dash home point de(Just to feed hungry robots: somebody@absolute-nonsense`rm === doubleswap(x) be the string formed by replacing each a in x by thesubstring bb and each b by the substring aa. For example,doubleswap(abcab)=bbaacbbaa. Furthermore, let doubleswap(L) be thelanguage formed of strings doubleswap(x) for x an element of L. Provethat if L is regular, then so is doubleswap(L). Where L is a subset(or equal) to {a, b, c}^*.If L is regular, the L can be expressed as L(y) for some regularexpression y. We define y' to be the same as y except that we replaceeach a in y by a pair of b's and each b by a pair of a's. Since y' isa regular expression, it will suffice to show that L(y') =doubleswap(L).(a) Prove that L(y') is a subset or equal to doubleswap(L).(b) Prove that doubleswap(L) === problemJack you're an idiot. See my other post. Get off of usenet and find ajob at a bar where your mental capacities are on par with the subjects youconverse with.-c> Can someone help doubleswap(x) be the string formed by replacing each a in x by the> substring bb and each b by the substring aa. For example,> doubleswap(abcab)=bbaacbbaa. Furthermore, let doubleswap(L) be the> language formed of strings doubleswap(x) for x an element of L. Prove> that if L is regular, then so is doubleswap(L). Where L is a subset> (or equal) to {a, b, c}^*.If L is regular, the L can be expressed as L(y) for some regular> expression y. We define y' to be the same as y except that we replace> each a in y by a pair of b's and each b by a pair of a's. Since y' is> a regular expression, it will suffice to show that L(y') => doubleswap(L).(a) Prove that L(y') is a subset or equal to doubleswap(L).(b) Prove that doubleswap(L) === stress forces, compressing space.During a post some time ago in which I asked about a jingle thatappears in Alfred Bester's novel The Demolished Man (a jingle thatgoes like Tenser, said the Tensor), I got a reply from that said: (...) So there are gravitational effects relating to momentumand tensionor compression as well as just energy; they are just usually tooweak to be noticed. Two long rods arranged parallel to each otherwill attract each other less, gravitationally, if under tension, forexample.This is something I haven't been able to get off my mind. I never knewthis, and it's such a fundamental thing to know. I'm also wondering ifmomentum, tension, or compression could affect space (or time, orspace-time) like it effects gravity. For example, could applyingenough momentum, tension, or compression on a suitably shaped mass ina suitable way bend or even compress space (even if the energy levelsdon't seem reachable with our technology)? I really want to knowbecause if the answer is yes then I guess in theory you could make awarp drive for a spaceship or === Adam, I get it now. The shape is actually not a hemisphere>meaning that the question is very poorly sta.>At 52 I don't think I qualify as being a kid anymore and why I'm>reviewing 1st year college math is another with the calculus review, . Feel free to post againif you get === all know how to expand (a+b)^n where n is a positive or negative integer.How do I expand above , where n is either fractional or a decimal ? (a+b)^0.2345 for exampleThe same, if you interpret (n choose i) as n (n-1) ... (n-i+1)/i!> It's called the binomial series.Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T expand your answer:how to use nCr if n is not an integer. That === wassum( ( (p1*i)/(z-p2*i) )^n )where p1,p2 are real constants, z is a complex number thus i is sqrt(-1)so for the sum to converge it must be that( (p1*i)/(z-p2*i) )^n -> 0 as n->oowhich is the case if( (p1*i)/(z-p2*i) ) is in the unit circle.how do i find the complex z or z's so that abs( === series>how do i find the complex z or z's so that abs( (p1*i)/(z-p2*i) ) < 1.You want |p1*i| < |z-p2*i|, i.e. z is outside the circle centred at p2*iwith radius |p1*i| = |p1|.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia === seriesembarrassingly, i have not realized that if b<>0 then abs(a/b) =abs(a)/abs(b).another stupid question..how do i start with a series likesum( (1/(n^2)) * (z-p1*i)^n)?i know how that 1/(n^2) -> 0 as n->oo.now i also believe that i know how to sort the (z-p1*i)^n thing. justneed to find a circle.but what if the two things are combined? do i still i.e. z is outside the circle centred at p2*i> with radius |p1*i| = |p1|.Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of === Re: complex series>another stupid question..how do i start with a series like>sum( (1/(n^2)) * (z-p1*i)^n)?>i know how that 1/(n^2) -> 0 as n->oo.>now i also believe that i know how to sort the (z-p1*i)^n thing. just>need to find a circle.>but what if the two things are combined? do i still have to find a>circle?The 1/n^2 does not change the circle of convergence - it only influencesconvergence on the circle.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia === Calculus> Someone recently asked me a problem from Taylor and Mann's advanced calculus> book. Although I remember doing such problems from that same book (although> I do not have the book to review from) when I was a student I just can't> remember how to do this problem. Even more embarrassing, I can't even do it> from scratch.> The problem is to show that x / (x+ 1) < log ( 1 + x) / x.> Any help/ hints would be apprecia.> StevenPerhaps incorrectly copied, or some conditions are missing.Quick check:limit as x goes to 0+ : 0 < 1 truelimit as x goes to infinity: 1 < 0 false.The cross-over point, after a crude sketch, is betweenx=1.5 and x=1.6 . === sequence of n DISTINCT positive> integeall >= 4, such that:>> (taking the indexes {mod n}),>> a(k) = (any divisor >= 2 of a(k-1)) +> (any divisor >= 2 of a(k-2))> for all terms of the sequence.>> (So, the first term, for example, is the sum of a divisor (>=2) of > the> last element and a divisor (>=2) of the next-to-last element of the> sequence.)>> Here is an example of a divisor-sum cycle:>> 4, 5, 7, 12, 11, 14, 18, (and back to the beginning)>> Note: The divisor of a(k), that is part of the sum equal to a(k+1),> need not necessarily be the same divisor of a(k) that is part of the> sum equal to a(k+2).>> But what I suspect is rather difficult, if at all possible, > is to create such a sequence of n >= 3 terms which are CONSECUTIVE> integers. (ie. For a sequence of n terms, the sequence is a> permutation of the integers from m to m+n-1.)The shortest such cycle including 4 as an element is:22, 5, 16, 9, 11, 14, 13, 15, 18, 8, 4, 6, 7, 10, 12, 17, 20, 19, 21The shortest cycle which is a permutation of n consecutive integers > (with n >= 3) is 10, 8, 7, 9....Very interesting. Are the cycles, which are a permutation of> consecutive integers which include 4, common?Did you find the {4-22} sequence using a computer or by-hand?>The solution given was found by hand in a few minutes. A computer check has turned up 71 solutions for the m=4, x=22 case. Incidentally, if a prime number P occurs in a cycle, then so must an integer >= P+3. So, it was really only necessary for me to check those cases in which the maximum integer x is in the sequence 10, === news.mailgate.org; posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064302908 http://mygate.mailgate.org/mynews/sci/sci.math/ e68d802877cf0c601265d1fe2f73577e.48257%40mygate.mailgate.org> Mr. Dolan hasn't actually been reading sci.math, or at least certain> threads, over, say, the last few months or yeahas he?Well, senior citizen, I star programming while Kennedy was newly inoffice, got my math degree at the beginning of Nixon's first term,joined the Net at the start of Jimmy Carter's term and Usenet just afterThe Great Renaming, and crea my first Usenet newsgroup at thebeginning of the 1990's. There was a time when I read all of Usenet butor more of all of Usenet's traffic, by myself.You do the math.With just that tiny amount of experience, it doesn't take me a hugeamount of time to identify a net.vandal. You all behave remarkablysimilarly: stubborn, self-centered, mannerless beyond redemption, andsure that any amount of inconvenience your misbehavior causes others istotally outweighed by the righteousness of your desire to misbehave.xanthian, unappreciative sufferer of obdurate fools.-- Pos via Mailgate.ORG Server - === Neoclassical Theory Of Value, Kent Mr. Dolan hasn't actually been reading sci.math, or at least certain> threads, over, say, the last few months or yeahas he? > [Snip]Mr. Dolan misses my point.> With just that tiny amount of experience, it doesn't take me a huge> amount of time to identify a net.vandal.Mr. Dolan seems to be well known for identify 10 out of 2 netvandals, too.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their mastebut highly @ r c m unbecoming to reasonable and free men === news.mailgate.org; posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064346948 http://mygate.mailgate.org/mynews/sci/sci.math/ 5f2c52e6022c4d3309cc52548fee51b5.48257%40mygate.mailgate.org%% % Mr. Dolan hasn't actually been reading sci.math, or at least%%% certain threads, over, say, the last few months or yeahas he?% Mr. Dolan misses my point.Oh, I must have hit it with something, you sure pulled that turtlehead back inside its shell quickly enough.%% With just that tiny amount of experience, it doesn't take me a huge%% amount of time to identify a net.vandal.% Mr. Dolan seems to be well known for identify[ing] 10 out of 2 net% vandals, too.Count yourself among the lucky two, oh pointless one. Insistance oncrossposting where you have been told you are not wan is amongthe top three symps of net.vandalism. Staying to argue about itmerely overwhelmingly confirms the diagnosis.xanthian, the best part of dealing with net.fools is pulling theirchains over and over and over; the word retreat is in a languagethey don't know, and so they are a source of infinite amusement.[Click Tracking Quiz: What part of your personal privacy is AOLviolating today?]-- Pos via Mailgate.ORG Server === Death Rattle Of Neoclassical Theory Of Value<5f2c52e6022c4d3309cc52548fee51b5.48257@ mygate.mailgate.org>, Kent > Insistance on> crossposting where you have been told you are not wan is among> the top three symps of net.vandalism. Staying to argue about it> merely overwhelmingly confirms the diagnosis.Speaking for all mathematicians, you are not wan here on sci.math,Mr. Dolan.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their mastebut highly @ r c m unbecoming to reasonable and free men === [Profoundly OT]: Death Rattle Of Neoclassical Theory Of Value <3f67bb36$6$fuzhry+tra$mr2ice@news.patriot.net> <8370fa18ff28632870b371a462223444.48257@mygate.mailgate.org> <90b1f3c6a96b3b8f4af0b67ca4bb27e1.48257@mygate.mailgate.org> <5f2c52e6022c4d3309cc52548fee51b5.48257@mygate.mailgate.org> admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 08:52 PM, Robert Vienneau said:>Speaking for all mathematicians,You don't.>you are not wan here on sci.math, Mr. Dolan.Well, one of you two isn't. Would you care to take another guess as towhich?-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolici bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Death Rattle Of Neoclassical Theory Of Value> at 08:52 PM, Robert Vienneau said: >Speaking for all mathematicians, > You don't.My point exactly.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their mastebut highly @ r c m unbecoming to reasonable and free men === Fear FactorDunno if you know that TV show (never saw it myself),but anyway, here they show...the Leech lattice:http://www.nbc.com/Fear_Factor/images/407_fd.jpg:-)(If something seems to pop out of the pic, remember,it's 24-dimensional-- Hauke Reddmann <:-EX8 Private email:fc3a501@math.uni-hamburg.deFor our chemistry === <3F6F0A3E.44BA4BE0@ix.netcom.com> sha1:ne3m+eKm6kxglSo2+m+zObxCK2k=> I'd venture to suggest the term information terrorist.> Perhaps then the punishment would be substantially> increased to fit the crime.If you think that email viruses are morally comparable to terrorism,then you are a horribly confused individual.Such ridiculous analogies do not help fitting the punishment to theinfraction. (Of course, increasing the punishment for writing virusesisn't likely to deter much anyway, given how few authors are caught.)-- It seems to me that in wartime Americans shouldn't be attacking eachother in this way on a *worldwide* forum. Then again, I know I'm anAmerican, but I have no way of knowing that you are, which wouldexplain a lot. --, on why Yanks should accept === the following:> It seems to me that in wartime Americans shouldn't be attacking each> other in this way on a *worldwide* forum. Then again, I know I'm an> American, but I have no way of knowing that you are, which would> explain a lot. --, on why Yanks should accept his proofI love that JSH quote. So for JSH, it's the fact that the proof isAmerican is all that matters? Things like whether it's *right* and allare just insignificant details?-- /-- Joona Palaste (palaste@cc.helsinki.fi) ---------------------------| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.fi/~palaste W++ B OP+ |----------------------------------------- Finland rules! ------------/This isn't right. This isn't even wrong. - === scribbled the following:>It seems to me that in wartime Americans shouldn't be attacking each>>other in this way on a *worldwide* forum. Then again, I know I'm an>>American, but I have no way of knowing that you are, which would>>explain a lot. --, on why Yanks should accept his proof> I love that JSH quote. So for JSH, it's the fact that the proof is> American is all that matters? Things like whether it's *right* and all> are just insignificant details?Since 9/11 government agencies have been quite vigilant against traditionalkinds of terrorist attacks. Terrorists know this, and one would suspect thatthey are, even now, weaving insidious and unforeseen plots.One of the greatest dangers facing America, in the long term, is the continuedhegemony of the corrupt academic establishment. Nowhere is this more clear thanin the realm of mathematics, where the continued suppression of James Harris'swork is a harbinger of doom. Laugh if you will, but America's chief advantageover the Islamic world is freedom.Terrorists cannot blast the floodgates of freedom in their own countries: theywould be washed away in the torrent. Instead, they seek to throttle Americanfreedoms -- most obviously, the freedom from fear.What other freedoms are they targeting? I am in no position to know, of course.But it would not surprise me if they were engaged in a secret jihad against JamesHarris, for James is the exemplary proponent of our most precious freedom of all:freedom of thought.The American mathematicians who are abetting the terrorists in their attacks onJames are engaged in treason. They used to hang people for that. Nowadays theyreward them with endowed chaiwith prestige, money, and power, with junketsto exotic conferences where they sit around sipping brandy and laughing at howeasily they're getting away with it all. Suppressed any new mathematics lately?Naw, too much work. I just pay the terrorists to do it.Thus far the terrorists/mathematicians have not made an attempt on James's life,as far as I know. Or maybe they have, but God has turned aside their hand. It'strue that God does not directly intervene in the world often, but when it comes tomatters as important as and his brilliant mathematical discoveries,surely God cannot turn away.But things have gotten ridiculous, haven't they? I can hardly stand it. SometimesI think that the only hope is for God to sweep away everything in this sinful worldand replace it with a new Kingdom, where Jesus rules, and is at Hisside, the Court Mathematician if you will. And there'll be a little trapdoor whereJames can watch David Ullrich and and all the rest of their rag-taggang burn in the flames of Hell with Osama bin Laden and Hitler.Okay, maybe === scribbled the following:> It seems to me that in wartime Americans shouldn't be attacking each> other in this way on a *worldwide* forum. Then again, I know I'm an> American, but I have no way of knowing that you are, which would> explain a lot. --, on why Yanks should accept his proofI love that JSH quote. So for JSH, it's the fact that the proof is> American is all that matters? Things like whether it's *right* and all> are just insignificant details?--> /-- Joona Palaste (palaste@cc.helsinki.fi) ---------------------------> | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|> | http://www.helsinki.fi/~palaste W++ B OP+ |> ----------------------------------------- Finland rules! ------------/> This isn't right. This isn't even wrong.> - Wolfgang PauliAn interesting question: If the number of worm mails is proportional toposting frequency and I get 30/hr: How === question: If the number of worm mails is proportional to> posting frequency and I get 30/hr: How many does JSH === <3F6F0A3E.44BA4BE0@ix.netcom.com> <87brtbq0al.fsf@phiwumbda.org> sha1:TiMkD0UmkNdjqQV53OpaZqhmx3w=> scribbled the following:>> It seems to me that in wartime Americans shouldn't be attacking each>> other in this way on a *worldwide* forum. Then again, I know I'm an>> American, but I have no way of knowing that you are, which would>> explain a lot. --, on why Yanks should accept his proof I love that JSH quote. So for JSH, it's the fact that the proof is> American is all that matters? Things like whether it's *right* and all> are just insignificant details?Sorry, I just write down the words. I can't explain them.All enquiries should be direc to JSH, should he tire of the MegaFoundation and alt.tv.bigbrother, a combination of interests whichspeak for themselves.-- Yup, as far as I'm concerned, if you live out your lives smiling theentire time full of pride in your *believed* accomplishments, when younever had any, well that's === Re: Swen32 worm - why have not the hackers gone after the <87brtbq0al.fsf@phiwumbda.org> sha1:GQuI4lxzJYNJfxWIBbzEFJXbOkw=> All enquiries should be direc to JSH...or Jim Ferry.-- Jesse HughesYou see 300 of something, anything, and you go `[Man], that's a lot ofstuff.' === Re: Swen32 worm - why have not the hackers gone after the Ferry.> Oh, I wouldn't be a good person to talk to about JSH'swork. I don't understand it at all. After all, asJames himself says, only about 10 people in the worldare capable of === term information terrorist.> Perhaps then the punishment would be substantially> increased to fit the crime. If you think that email viruses are morally comparable to terrorism,> then you are a horribly confused individual. Such ridiculous analogies do not help fitting the punishment to the> infraction. (Of course, increasing the punishment for writing viruses> isn't likely to deter much anyway, given how few authors are caught.)>I would go further; there is nothing morally wrong about writing viruses,it's just code. What's wrong is using them to cause damage, the mediadoesn't usually make that === === I'd venture to suggest the term information terrorist.>> Perhaps then the punishment would be substantially>> increased to fit the crime.>> If you think that email viruses are morally comparable to terrorism,>> then you are a horribly confused individual.>> Such ridiculous analogies do not help fitting the punishment to the>> infraction. (Of course, increasing the punishment for writing viruses>> isn't likely to deter much anyway, given how few authors are caught.)>>I would go further; there is nothing morally wrong about writing viruses,>it's just code. What's wrong is using them to cause damage, the media>doesn't usually make that distinction clear.And there's nothing morally wrong with manufacturing insecticide. Yet themakers of Zyklon B were hanged after WWII. Maybe if a few of these hackerswere hanged instead of having their wrists slapped, they would understand thateconomic terrorism won't be tolera.--Mensanator2 of Clubs === Subject: Re: Swen32 worm - why have not the hackers gone after sha1:ZAtjU0ZuU/Qs+PIxoDbzjDfeIl4=> And there's nothing morally wrong with manufacturing> insecticide. Yet the makers of Zyklon B were hanged after> WWII. Maybe if a few of these hackers were hanged instead of> having their wrists slapped, they would understand that economic> terrorism won't be tolera.Again: Sloppy analogies confuse the issues. Genocide is incomparableto email viruses. To pretend that expensive annoyances like Swen 32is comparable to contributing to terrorism or genocide is to belittlethe horrific effects of the latter crimes.How much effect would more drastic sentencing have anyway? Very fewof the virus writers are ever found, and fewer still are under USjurisdiction. I'm not sure that stiffer penalties will effectanything but an illusion that we're getting the bastards.But whether or not current penalties are inadequate, try to keep asense of proportion. Bad analogies lead to bad thinking.-- I think the burden is on those people who think he didn't haveweapons of mass destruction to tell the world where they are. -- White House === morally wrong with manufacturing> insecticide. Yet the makers of Zyklon B were hanged after> WWII. Maybe if a few of these hackers were hanged instead of> having their wrists slapped, they would understand that economic> terrorism won't be tolera.Again: Sloppy analogies confuse the issues. Genocide is incomparable> to email viruses. To pretend that expensive annoyances Erbschloe had estima the economic impact from last year's Love Bug virus to be $8.7 billion and the economic damage from the Melissa virus in 1999 to be about $1 billion.While some people have questioned his figures, Erbschloe said that Lloyds of London put the estimate for Love Bug at $15 billion.In my opinion, $8.7 billion is not ludicrous, he said. Some companies repor seven million Love Bug messages and 10 days to clean up.How bad does something have to be to not be an annoyance? Why don'tyou ask those who have lost their jobs due to the depressed economyin the aftermath of 9/11 their opinions on economic terrorism?> like Swen 32 is comparable to contributing to terrorism Well, that depends on who's distributing the viruses doesn't it?Only a naive person would think that terrorism is confined tolarge explosions.> or genocide is to belittle the horrific effects of the latter crimes.The magnitude of the crime is not the issue. Those who knowingly areaccomplices to criminal acts are criminals. If a terrorist modifiesthe source code of a virus, that the author of that virus is just as guilty. There are no innocent hackers. Everyone is fully aware of thescope and potential scope of their crimes.How much effect would more drastic sentencing have anyway? None at all. The Nuremberg Trials haven't had any effect on thebehaviour of dictators since then. Are you saying they should not havebeen punished?> Very few of the virus writers are ever found, and fewer still are under > US jurisdiction. They can always be assassina.> I'm not sure that stiffer penalties will effect> anything but an illusion that we're getting the bastards.I'll settle for that.But whether or not current penalties are inadequate, try to keep a> sense of proportion. Bad analogies lead to bad thinking.Such as the War in Iraq? Maybe Saddam's act was all a bluff. If so,he played his cards wrong. He assumed that a proper sense of proportionwould protect him. He didn't === Re: Swen32 worm - why have not the hackers gone after the all a bluff. If so, > he played his cards wrong. He assumed that a proper sense of proportion > would protect him. He didn't anticipate the joker in the deck trumping > his hand.Ah, now I understand why the most wan deck does not have a joker.Putting Bush there would be a bit, eh, ...-- === but disengage and considerhow far off-topic this is. My mail === <87ad8vocwr.fsf@phiwumbda.org> sha1:pV1V6488h21AYKifAFdVXh9floo=> Really cogent and gripping discourse guys, but disengage and consider> how far off-topic this is. My mail spike has passed; hope yours has too.I'm still receiving about 2000/day. My spike has not passed.Perhaps there is an upstream filter catching many of your emails. Inmy backwater, there has been no appreciable let-up.-- I have written many words to sci.math, some of them === > Really cogent and gripping discourse guys, but disengage and consider > how far off-topic this is. My mail spike has passed; hope yours has too. > I'm still receiving about 2000/day. My spike has not passed. > Perhaps there is an upstream filter catching many of your emails. In > my backwater, there has been no appreciable let-up.It appears that it is mostly the Netherlands that is a target at this moment.Also quite a few Dutch newsgroups have this worm already seen pos a fewtimes (I have seen something like 20). This is also much less ininternational newsgroups.My procmail filter filtered out all mails that contain the worm. Alas,much software around is now removing the virus from mails, so that Inow receive those mails without virus (but they are considerably smaller).9 MB each hour is a bit much...-- === element x such that there exists noauorphism f of G for which f(x)=x^{-1}?The question came up in a forum at === Hammick> Does there exist a group G with an element x such that there exists no> auorphism f of G for which f(x)=x^{-1}?> The question came up in a forum at www.planetmath.org.> LH Jim === :> Does there exist a group G with an element x such that there exists no> auorphism f of G for which f(x)=x^{-1}?> The question came up in a forum at www.planetmath.org.Yes, for example the semi-direct product G = (C_2 x C_2 x C_2)x| C_7, of order 56, with presentation . This group has noauorphism that maps d to d^{-1}, nor any other of its 48elements of order 7 to its inverse. (In fact, Out(G) = C_3, andis genera by f : f(a) = a, f(b) = c, === theory qin message :> in message : Does there exist a group G with an element x such that there exists no> auorphism f of G for which f(x)=x^{-1}?> The question came up in a forum at www.planetmath.org.[...]BTW, I suspect, although I haven't checked all groups of order16, that the smallest such group is the semi-direct product C_5x C_4, a subgroup of S_5 of order 20, with presentation . This group has no auorphism takingb to b^{-1} (nor any other of its === Re: Group theory q> Does there exist a group G with an element x such that there exists no> auorphism f of G for which f(x)=x^{-1}?> The question came up in a forum at www.planetmath.org.Yes, consider M_24.All auorphisms of M_24 are inner. If g is an elementof order 23 in M_24 then g is congruent to g^a iffa is a quadratic residue mod 23. -1 is not a QR mod 23,so g and g^{-1} are not conjugate, and as all auorphismsare inner, then there is no auorphism f with f(g) = === iteration functionx_{k+1)=x_k+p(x_k) , lets call it Phi(x)=x+p(x),that is used to find f(x)=0, where both x,f in R^n (vector valueddifferentiable function of several variables). p(x) contains first andsecond order derivatives of f(x) and some inverses of them.In the 1-dimenasional case R^1, and since Phi(x_*)=x_*, I can use Taylorexpansion of Phi(x) around x_*, where f(x_*)=0 according to||x_{k+1}-x_*||=||Phi(x_k)-Phi(x_*)|| = ||Phi(x_*)+Phi'(x_*)p+Phi''(x_*)p^2/2! +...+ R - Phi(x_*)||where p=x_k-x_* and R=Phi^{n}(t)p^n/n! is given by the meanvalue theorem, n'th derivative of Phi(x) and tin[x_*,x_k] So if first, second ,3rd,...,n-1'th derivatives of Phi(x) are zero andthere is some number beta >= Phi^(n)(t)/n! then||x_{k+1}-x_*|| <=beta||p^n||=beta||x_k-x_*||^nThe method has convergence factor n.Returing to my case with several variables. Can the 1-dim. case begeneralized to several variables ?To prove that my iteration is of order 3, is it sufficient to show thatPhi'(x_*)=0 and Phi''(x_*)=0 ?Taylor expansion can be done in a similar manner, and the remainder term Rbecomes an integral over a parametrization of Phi'''(x(t))... still it isof Ordo ||x_k-x_*||^3 right ?I have proved that first and second order derivative (Jacobian andHessian tensor) is 0 at x_*. I suspect that analysis of the remainderterm is needed.... I would rather not ! Since it involves _quite_complica differentiations of higher order tensors.Then I would have to assume (which is normally done) that for the norm||.||, ||f'||x_{k+1)=x_k+p(x_k) , lets call it Phi(x)=x+p(x), >that is used to find f(x)=0, where both x,f in R^n (vector valued >differentiable function of several variables). p(x) contains first and >second order derivatives of f(x) and some inverses of them. >In the 1-dimenasional case R^1, and since Phi(x_*)=x_*, I can use Taylor >expansion of Phi(x) around x_*, where f(x_*)=0 according to > >||x_{k+1}-x_*||=||Phi(x_k)-Phi(x_*)|| > = ||Phi(x_*)+Phi'(x_*)p+Phi''(x_*)p^2/2! +...+ R - Phi(x_*)|| > >where p=x_k-x_* and R=Phi^{n}(t)p^n/n! is given by the meanvalue theorem, > n'th derivative of Phi(x) and tin[x_*,x_k] >So if first, second ,3rd,...,n-1'th derivatives of Phi(x) are zero and >there is some number beta >= Phi^(n)(t)/n! then >||x_{k+1}-x_*|| <=beta||p^n||=beta||x_k-x_*||^n >The method has convergence factor n. you mean convergence order n. the factor would be the beta, much harder to compute in general >Returing to my case with several variables. Can the 1-dim. case be >generalized to several variables ? >To prove that my iteration is of order 3, is it sufficient to show that >Phi'(x_*)=0 and Phi''(x_*)=0 ? sufficient: yes (but not necessary). of course you need existence and continuity of the third derivative in some neighborhood of the solution > Taylor expansion can be done in a similar manner, and the remainder term R > becomes an integral over a parametrization of Phi'''(x(t))... still it is ???? x(t)=x_*+t(x-x_*) >of Ordo ||x_k-x_*||^3 right ? yes >I have proved that first and second order derivative (Jacobian and >Hessian tensor) is 0 at x_*. I suspect that analysis of the remainder of Phi ???????? if you can show that the derivative exists nad is locally continuous you are done >term is needed.... I would rather not ! Since it involves _quite_ >complica differentiations of higher order tensors. >Then I would have to assume (which is normally done) that for the norm >||.||, ||f'|| Ortega&Rheinboldt:iterative solution of nonlinear equations in several variables, page 188? above you spoke about Phi, here about f? >M||x_k-x_*||^3 where M is a number depending on alpha,beta,gamma, etc.. >but it will be === factor I have an iteration function>x_{k+1)=x_k+p(x_k) , lets call it Phi(x)=x+p(x),>that is used to find f(x)=0, where both x,f in R^n (vector valued>differentiable function of several variables). p(x) contains first and>second order derivatives of f(x) and some inverses of them.>In the 1-dimenasional case R^1, and since Phi(x_*)=x_*, I can use Taylor>expansion of Phi(x) around x_*, where f(x_*)=0 according to> >||x_{k+1}-x_*||=||Phi(x_k)-Phi(x_*)||> = ||Phi(x_*)+Phi'(x_*)p+Phi''(x_*)p^2/2! +...+ R - Phi(x_*)||> >where p=x_k-x_* and R=Phi^{n}(t)p^n/n! is given by the meanvalue theorem,> n'th derivative of Phi(x) and tin[x_*,x_k] >So if first, second ,3rd,...,n-1'th derivatives of Phi(x) are zero and>there is some number beta >= Phi^(n)(t)/n! then>||x_{k+1}-x_*|| <=beta||p^n||=beta||x_k-x_*||^n>The method has convergence factor n.> you mean convergence order n. the factor would be the beta, much harder to> compute in generalYes, sorry for the mix up. >Returing to my case with several variables. Can the 1-dim. case be>generalized to several variables ?>To prove that my iteration is of order 3, is it sufficient to show that>Phi'(x_*)=0 and Phi''(x_*)=0 ? sufficient: yes (but not necessary). of course you need existence> and continuity of the third derivative in some neighborhood of the solution > Taylor expansion can be done in a similar manner, and the remainder term R> becomes an integral over a parametrization of Phi'''(x(t))... still it is> ???? x(t)=x_*+t(x-x_*)Exactly.>I have proved that first and second order derivative (Jacobian and>Hessian tensor) is 0 at x_*. I suspect that analysis of the remainder> of Phi ????????> if you can show that the derivative exists nad is locally continuous you> are doneYes I need the derivatives of Phi equal to 0, since it is that function ITaylor expand yielding||x_{k+1}-x_*||=||Phi(x_k)-Phi(x_*)|| < M*||x_k-x_*||^3>term is needed.... I would rather not ! Since it involves _quite_>complica differentiations of higher order tensors.>Then I would have to assume (which is normally done) that for the norm>||.||, ||f'|| you did reenvent the method of Traub or the method of tangent> hyperbolas (-> Ortega&Rheinboldt:iterative solution of nonlinear> equations in several variables, page 188?No, but it is quite similar. I must check up the Altman, Janko,Lita and the others mentioned there on p.188. I have not heard ofthose before. Perhaps my iteration function is something old.> above you spoke about Phi, here about f?Yes, since Phi(x)=x+p(x) is a function of f(x),f'(x),f''(x), so whendifferenting Phi(x) I get it expressed in higher order derivatives off(x). Now in order to say that ||Phi'''(x(t))|| Every group contains atleast 1 member.> Every group should be in contact with another group via atleast 1> member.> Every member of the group should be in contact with atleast one> another member of the same group.> 'Contact' = Placed next to eath other.When you say next to each other do you mean side by side or I cancount in also up and down???I'm pretty sure I misunderstood but if not...let the 4 groups to be @ # $ and % === Math puzzle..> Put 4 different groups on 3 x 3 grid, such that> Every group contains atleast 1 member.> Every group should be in contact with another group via atleast 1> member.> Every member of the group should be in contact with atleast one> another member of the same group.> 'Contact' = Placed next to eath other.When you say next to each other do you mean side by side or I can> count in also up and down???I'm pretty sure I misunderstood but if not...let the 4 groups to be @ # $ and % :..............@ @ @> # $ %> # % %...yep, this was a small test..Make it 4x4x4 box type.. and see how many maximum different group you can fit it.You can === Math puzzle..Make it 4x4x4 box type.. and see how many maximum different group you can fit it.> You can display it.. via 4 layers of 4x4 Grids here.hello again. I've managed to fit in 7 groups but my guess is that 9 isthe number you are looking for.could you please tip me with the maximum number of groups cause thecombinations are infinite and it would be easier for me to have atarget instead of trying meaningless atempts.I'll understand if you dont want to give that hint but without it itseems as a waste of time and energy trying to find the right === puzzle..(right track)> Put 4 different groups on 3 x 3 grid, such that> Every group contains atleast 1 member.> Every group should be in contact with another group via atleast 1> member.> Every member of the group should be in contact with atleast one> another member of the same group.> 'Contact' = Placed next to eath other.When you say next to each other do you mean side by side or I can> count in also up and down???I'm pretty sure I misunderstood but if not...let the 4 groups to be @ # $ and % :..............@ @ @> # $ %> # % %greetings spirosWow..That's it..Contact = '> Side by Side in any direction.( Not Circular)Now the next in series, enter the === Math puzzle..(right track)@ @ @> # $ %> # % %> I don't get it, this seems even easier while it sould be moredifficult since its the second in o row of puzzles????anyway I'll use the previous matrix (3x3) combined with + 0 0+ + 00 + 0 where + is === of cone using integral calculusHello!How is area of right circular cone (excluding base) found usingintegral calculus?I know the method were you unroll the cone so that is becomesequivalent of determing area of a circle sector.What I'm looking for is to find the right differential element dS.What I want is to cut the cone into many tiny rings where each ringhas infinitesimal area, dS = 2*pi*(radius of ring)*differential ofwidth.Summing all tiny area elements gives the lateral area of cone.Somehow I can't find the correct differential element.The area need to be: S = pi*r*sqrt(r^2+h^2)where r is the radius of the base circle, h is the height.This comes from the problem of finding the electrostatic potential atthe tip of cone (without the base), with base r and height h, if thecone has same surface charge === calculusMarko escribi.97 en Hello! How is area of right circular cone (excluding base) found using> integral calculus? I know the method were you unroll the cone so that is becomes> equivalent of determing area of a circle sector. What I'm looking for is to find the right differential element dS. What I want is to cut the cone into many tiny rings where each ring> has infinitesimal area, dS = 2*pi*(radius of ring)*differential of> width. Summing all tiny area elements gives the lateral area of cone. Somehow I can't find the correct differential element.> The area need to be: S = pi*r*sqrt(r^2+h^2) where r is the radius of the base circle, h is the height. This comes from the problem of finding the electrostatic potential at> the tip of cone (without the base), with base r and height h, if the> cone has same surface charge density everywhere.The surface genera by the graph of y = f(x) revolving on the OX axisbetween x = a and x = b isS = 2pi*Int(f(x)*sqrt(1 + (f'(x))^2), x, a, b)The differential element of area is a circular band of radius f(x) and widthsqrt(dx^2 + dy^2) = sqrt(1 + (dy/dx)^2)*dxThen, calling H and R to the height and radius of the cone, it is generaby rotation of the line y = (R/H)x on the OX axis, between x = 0 and x = H.Theny' = R/HS = 2pi*Int((R/H)x*sqrt(1 + (R/H)^2), x, 0, H) = 2pi(R/H^2)*sqrt(H^2 + R^2)*H^2/2 = pi*R*Gwhere G = sqrt(H^2 + R^2) is the conus generatriz.-- Ignacio Larrosa Ca.96estroA Coru.96a === Area of cone using integral calculus> Hello! How is area of right circular cone (excluding base) found using> integral calculus? I know the method were you unroll the cone so that is becomes> equivalent of determing area of a circle sector. What I'm looking for is to find the right differential element dS. What I want is to cut the cone into many tiny rings where each ring> has infinitesimal area, dS = 2*pi*(radius of ring)*differential of> width. Summing all tiny area elements gives the lateral area of cone. Somehow I can't find the correct differential element.> The area need to be: S = pi*r*sqrt(r^2+h^2) where r is the radius of the base circle, h is the height. This comes from the problem of finding the electrostatic potential at> the tip of cone (without the base), with base r and height h, if the> cone has same surface charge density everywhere.There is a formula based on cutting up the surface into smallpatches, approxima by small patches of the tangent planes(remembering that the area of a parallelogram is calculausing the cross product of its sides):Parametrize the surface by v=(x,y,z)' (transpose for my comfort)x = X(p,q)y = Y(p,q)z = Z(p,q)and the partial derivatives are deno by v_p, v_q.Then calculate the normal (v_p cross v_q) = N(p,q), and finallysurface area= integral[over domain in the (p,q) plane] norm(N(p,q)) dp dqFor surfaces of revolution about the z-axis, the parameters are:polar coordinates r, theta,x = r*cos(theta)y = r*sin(theta)and the altitude is a function of r alone:z = f(r).The domain is R1 < r < R2 (for a surface over an annulus), -pi < theta < piFor checking purposes, the surface element in variables r, thetacomes out asdS = r * sqrt(1 + (f'(r))^2) dr dthetaFor the cone with radius R and height h, the expression for z isz = h * (R - r) / Rhence the classical === integral calculuschange of notationCone height H base radius RThen at some point h up the cone area element isdS=2*pi*r*dldl is along the slope of coneso dl^2=dr^2+dh^2 = dr*(1+(dh/dr)^2)but dh/dr is constant = H/Rso S=2*pi*sqrt(1+(H/R)^2)*int(r*dr)S=pi*R*sqrt(R^2+H^2)> Hello! How is area of right circular cone (excluding base) found using> integral calculus? I know the method were you unroll the cone so that is becomes> equivalent of determing area of a circle sector. What I'm looking for is to find the right differential element dS. What I want is to cut the cone into many tiny rings where each ring> has infinitesimal area, dS = 2*pi*(radius of ring)*differential of> width. Summing all tiny area elements gives the lateral area of cone. Somehow I can't find the correct differential element.> The area need to be: S = pi*r*sqrt(r^2+h^2) where r is the radius of the base circle, h is the height. This comes from the problem of finding the electrostatic potential at> the tip of cone (without the base), with base r and height h, if the> cone has same surface === choice>In ZF without choice, can one show that given any two nonempty sets, >there exists a surjection from one to the other? If so, why not use >this as the method for comparing cardinalities?There are models for ZF for which incomparable sets can be surjec on each other; they have been given here.I do not believe it is known whether surjection comparabilityfor all sets implies choice.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: === variationis it possible to show that P(Z) is uncountable using a variation ofthe diagonal argument. like assuming that there is a list of sets Ansubset of Z, and then constructing a new set whose nth element is notin An?my TA says this doesnt work, I think he is wrong. or at least it couldbe refined to === variation> is it possible to show that P(Z) is uncountable using a variation of> the diagonal argument. like assuming that there is a list of sets An> subset of Z, and then constructing a new set whose nth element is not> in An?It's not a variation of the diagonal argument. It is, in fact, thediagonal argument.Proposition: Let X be a set, and let f : X -> P(X) be a mapping from Xinto its power set. Then f is not a surjection.Proof. We are to produce a member of P(X) that is not in the range of f.Let D (for diagonal) = { x in X : not(x in f(x)) }. Then D is a subsetof X, which is to say that D is a member of P(X). Moreover, for each xin X we find by the definition of D that (x in D) <--> not(x in f(x)).That is, for each x in X, we find that f(x) differs from D by at leastone element, since x belongs to exactly one of the two sets. Conclusion:D is the required member of P(X) that is not in the range of f.Corollary: If X is countably infinite, then P(X) is uncountable.Note that there is a variation of the diagonal argument that deals withdecimal representations of real numbebut the real diagonal argumentis the one above. It applies to all sets, not just countably infiniteones.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.is it possible to show that P(Z) is uncountable using a variation of>the diagonal argument. like assuming that there is a list of sets An>subset of Z, and then constructing a new set whose nth element is not>in An?Yes; you can use Cantor's original argument for the power set. Assumethat f:Z->P(Z) is any function. We will show f is not surjective.Let An = f(n). Then constructB = {n in Z: n is not in An}.So: you construct a subset of Z. If n is in An, then you omit it, sothe set you are constructing will be different from An. If n is not inAn, then you include it, so the set you are constructing will bedifferent from An.So B is different from An for each n. So B is not in the image of f.Now, this is slightly different from what you were sayingconstructing a new set whose n-th element is not in An, because setsare not ordered. So there is no good notion of n-th element of aset. That may be what your TA meant === by diagonal variationthank you very much, but I had typed my question fast because I was inthe library and there were people waiting for the computer. my answerwas a little bit more refined. It went something like this (haven'tgotten my paper back):Suppose P(Z) is countable and that all of Z's subsets have been placedin a sequence {An}. Let Bn be the singleton of an element that is notin An, since An is a proper subset of Z. Then U Bn (read as union ofBn, n=1 to oo) is not in the list but is contained in Z, thereforeP(Z) is uncountable.Is this not a correctly sta proof? I didn't use order in my finalanswer. I remember my TA saying something like what if you constructall of Z... but I wasnt sure what he was trying to say. But he wasconvinced that it was uncountable using a variation of>the diagonal argument. like assuming that there is a list of sets An>subset of Z, and then constructing a new set whose nth element is not>in An?Yes; you can use Cantor's original argument for the power set. Assume> that f:Z->P(Z) is any function. We will show f is not surjective.Let An = f(n). Then constructB = {n in Z: n is not in An}.So: you construct a subset of Z. If n is in An, then you omit it, so> the set you are constructing will be different from An. If n is not in> An, then you include it, so the set you are constructing will be> different from An.So B is different from An for each n. So B is not in the image of f.Now, this is slightly different from what you were saying> constructing a new set whose n-th element is not in An, because sets> are not ordered. So there is no good notion of n-th element of a> set. That may be what your TA meant when he said it does not what I accept as reality.> --- Calvin (Calvin and Hobbes)> === uncountable by diagonal variation Visiting Assistant Professor at the University of Montana.thank you very much, but I had typed my question fast because I was in>the library and there were people waiting for the computer. my answer>was a little bit more refined. It went something like this (haven't>gotten my paper back):Suppose P(Z) is countable and that all of Z's subsets have been placed>in a sequence {An}. Let Bn be the singleton of an element that is not>in An, since An is a proper subset of Z. Then U Bn (read as union of>Bn, n=1 to oo) is not in the list but is contained in Z, therefore>P(Z) is uncountable.Is this not a correctly sta proof?No. First, there is no reason to assume that {An} is a proper subsetof Z, so there is definitely a problem in your argument. In fact, ifyou have assumed that P(Z) is countable and your list {An} includesall subsets, then there MUST be one value of n for which An=Z, surely!What n do you choose then?And that is exactly where you have the problem. So you need to excludeZ from your list a priori. That is not a problem, since clearlyP(Z)-{Z} is uncountable if and only if P(Z) is uncountable. So given a map f:N->P(Z)-{Z}, then let Bn={a_n}, where a_n is thesmallest integer (with the integers well ordered by letting rI remember my TA saying something like what if you construct>all of Z... but I wasnt sure what he was trying to say. But he was>convinced that it was incorrect.Maybe that clarifies what he meant?And what happens if you do not exclude Z? Then you need to make surethat your final B (a) contains one element not in each A_n; and (b) isnot all of Z; but say you take the list above, shift it down by 1, andput Z as the first element of P(Z) in your list? How are you going toconstruct your B by choosing singletons or === binary sequencelet {Si} be the binary sequence defined by :Si = 1 if i = 2^l - 1 ,l is a natural numberSi = 0, otherwiseLet L(k) be the linear complexity of S0,.....S(k-1).Prove that === let V_k be the set of sequences u = (u(n)) (n in Z) of elements of K such that(a) u(0) = 0, and(b) for all m, n, p, q in Z, if m^k + n^k = p^k + q^k, then u(m) + u(n) = u(p) + u(q).It is easy to see that V_k is a vector space over K.Let d_k be the dimension of V_k over K.Then we can check:(1) d_1 = 1. (i.e., V_1 is isomorphic to K as a vector space.)(2) d_2 = 5, unless the characteristic of K is 2, 3 or 5.Is d_k always finite?If so, how can we determine its value?In particular, can we figure out the value of d_3, assuming char(K) = 0, or === QUESTION?If L1 union L2 is regular, and L1 is finite, is L2 === thing that occurs to me is a constructive proof. I don't know if thisis the only normal form for regular grammabut one normal form is that everyproduction is either of the form A->xB or A->(empty string), where A,B aresingle nonterminal symbols and x is a single terminal symbol. Also, there is noloss of generality in assuming that the start symbol does not appear on therighthand side of any production. Let G be a regular grammar satisfying theserequirements, let n be any positive integer, and let S be the set of all stringsX of L(G) such that (i) X can be derived from the start symbol of G, (ii)|X|<=n, and (iii) X=sA for some nonterminal A of G and string s of terminals ofG. Let H be a regular grammar with terminal alphabet (terminals ofG)+(nonterminals of G) such that L(H)=X, and let J be the grammar you get byadding to H all of the productions of G (and moving the terminals of G into theterminal alphabet of J where they belong). Then L(J)={X in L(G): |X|>n}, and nwas arbitrary.So now let's apply this to your question: First take out all the strings thatare short enough to belong to (L1-L2), and then add back the ones that don'tactually belong to (L1-L2). Then you're done.| If L1 union L2 is regular, and L1 is finite, is L2 === to now I knew the notion REGULAR EXPRESSION from programs like`emacs' and `egrep' only. Could someone of you give a short primerabout the meaning of the word in this thread? Also a hint forintroductory literature would be apprecia.Have the regular expressions in `emacs' something in common with themore abstract regular expressions discussed here?TN-- Write to: i at tn dash home point de(Just to feed hungry robots: === Helmut Richter ] ! [ 17: Mitch Harris ] ! [ 23: drieux, just drieux ] for the informations about regular expressions.Tobias N.8ahring.-- Write to: i at tn dash home point de(Just to feed hungry robots: somebody@absolute-nonsense`rm -fr === Up to now I knew the notion REGULAR EXPRESSION from programs like> `emacs' and `egrep' only. Could someone of you give a short primer> about the meaning of the word in this thread? Also a hint for> introductory literature would be apprecia.1) Hopcroft, Ullman, Motwani.> Have the regular expressions in `emacs' something in common with the> more abstract regular expressions discussed here?Yes. same thing. (er.. actually..unix regexps have a few more === EXPRESSIONS?> Up to now I knew the notion REGULAR EXPRESSION from programs like> `emacs' and `egrep' only. Could someone of you give a short primer> about the meaning of the word in this thread? Also a hint for> introductory literature would be 3) The text by Hopcroft, Ullman, Motwani.Not to mention the book from O'Reilly, 'Mastering Regular Expressions', of course.DrieuxHave the regular expressions in `emacs' something in common with the> more abstract regular expressions discussed here?Yes. same thing. (er.. actually..unix regexps have a few === EXPRESSIONS?> Up to now I knew the notion REGULAR EXPRESSION from programs like> `emacs' and `egrep' only. Could someone of you give a short primer> about the meaning of the word in this thread? Also a hint for> introductory literature would be apprecia.A language is a (empty, finite, or infinite) set of finitely longstrings over a finite alphabet. A regular language is definedrecursively:- The empty set is a regular language.- A language consisting of a single word of length 1 is a regular language.- If L1 and L2 are regular languages, so are L1L2 = {xy : x in L1 and y in L2} and L1 | L2 = {x : x in L1 or x in L2}- If L is a regular language, so is L* = {e} | L | LL | LLL | ... where e ist the empty word, i.e. the word consisting of 0 characters from the alphabet.By this definition, it is clear that most mechanisms for writingregular expressions in programming languages and tools yield regularlanguages and many regular languages can be described by regularexpressions.> Have the regular expressions in `emacs' something in common with the> more abstract regular expressions discussed here?Yes, see above. The two notions do, however, not fully coincide:- Regular expressions typically have no notation for the empty set. - In Basic Regular Expressions (BRE), for instance defined in X/Open CAE Specification System Interface Definitions, Issue 4, there is no operator | for set union. As a result, many regular languages cannot be written as BRE.- In BRE, there is an operator 1 referring back to a previously recognised substring of the same word, e.g. (.*)1 denoting the set {xx : any x} which is not a regular language.- The Extended Regular Expressions (ERE) from the same standard, however, define fairly well the regular languages.be comprehensible to English speakers as well): http://www.lrz-muenchen.de/services/schulung/unterlagen/regul/ table of what belongs to BRE and ERE: .../regul/regul-5.html#publish3.3.0.0.0.0table of which programs expect BRE or ERE: .../regul/regul-6.html#publish3.4.2.0.0.0Helmut === union L2 is regular, and L1 is finite, is L2 regular?> L1 L2 (set difference) is finite, hence regularL2 = (L1 union L2) (L1 L2) and regular languages are closedunder difference (since they're closed under intersection andcomplement), so L2 === QUESTION?> If L1 union L2 is regular, and L1 is finite, is L2 regular?If L2 was not regular, L1 union L2 could not be regular, so both L1 and L2 must be regular.Markus === my cousin Bobby. The trusty hound was at myside and Bobby was not far behind. I held my rifle close, my eyessearching the darkness. Then I saw it, sitting behind the bushes,staring at me, its eyes were big and red. I squin, peering deeperinto the gloom, and the thing's identity was certain. It was aone-to-one of N onto R. I'd heard these exis, but until now I'dalways been a bit skeptic (having studied deeply the theorems ofCantor and his peers). I frantically realized I hadn't brought mycamera. Hey, Bobby! I shou. Come look at this here thang! But my shouting scared it and the mapping vanished into the unknowndepths of the woods. Was the danged creepiest thing I done ever seen. I always thought there was something fishy about those theoremsCantor laid down. Now I see through the ruse with alarming clarity. It all seems so obvious now. When that UFO landed at Roswell, itbrung the mappings with it. The Air Force found them and captured asmany as they could but some escaped, like the one I saw. Thegovernment doesn't want us to see them. If people knew that thesemappings exis, it would disprove the ideals of the Illuminati. Thegovernment cannot allow this at any cost. Suddenly I realize withsome chagrin that ol' Jack, the geezer we always used to tease sobadly when he'd tell us how he was abduc by a spaceship full ofone-to-one mappings of N onto R, and brutally sodomized, was tellin'us the truth all along. Poor Jack. But now I have troubles of myown. Now the government knows I know too much. They're coming to getme. I must go.Written in haste,Cletus.---------------------------------------------- Some folk'll never kill a skunk, but then again some folk'll,Like Cletus, the slack-jawed yokel!Some folk'll never lose a toe, but then again some folk'll,Like Cletus, the slack-jawed : Re: Finally I saw it... Does this have some kind of point or purpose other than using upspace on thousands of servers all over the world?>I was out hunting with my cousin Bobby. The trusty hound was at my>side and Bobby was not far behind. I held my rifle close, my eyes>searching the darkness. Then I saw it, sitting behind the bushes,>staring at me, its eyes were big and red. I squin, peering deeper>into the gloom, and the thing's identity was certain. It was a>one-to-one of N onto R. I'd heard these exis, but until now I'd>always been a bit skeptic (having studied deeply the theorems of>Cantor and his peers). I frantically realized I hadn't brought my>camera. Hey, Bobby! I shou. Come look at this here thang! >But my shouting scared it and the mapping vanished into the unknown>depths of the woods. Was the danged creepiest thing I done ever seen.> I always thought there was something fishy about those theorems>Cantor laid down. Now I see through the ruse with alarming clarity. >It all seems so obvious now. When that UFO landed at Roswell, it>brung the mappings with it. The Air Force found them and captured as>many as they could but some escaped, like the one I saw. The>government doesn't want us to see them. If people knew that these>mappings exis, it would disprove the ideals of the Illuminati. The>government cannot allow this at any cost. Suddenly I realize with>some chagrin that ol' Jack, the geezer we always used to tease so>badly when he'd tell us how he was abduc by a spaceship full of>one-to-one mappings of N onto R, and brutally sodomized, was tellin'>us the truth all along. Poor Jack. But now I have troubles of my>own. Now the government knows I know too much. They're coming to get>me. I must go.Written in haste,>Cletus.----------------------------------------------> Some folk'll never kill a skunk, but then again some folk'll,>Like Cletus, the slack-jawed yokel!>Some folk'll never lose a toe, but then again some folk'll,>Like Cletus, the slack-jawed === Subject: Re: Finally I saw it... Does this have some kind of point or purpose other than using up> space on thousands of servers all over the world?Only to get you to follow up, quoting the entire original post, so asto use up twice as much space as before.Besides, the same question could be asked of many Usenet postings, andit would be just as rhetorical.-- === it...[...]At least he's funnier than Nathian.V.-- mail me at === realized that Iwas not very clear on my question. I certainly have the REALcoordinates for the location. What I want to do is compress thedistance matrix and create new coordinates where the distance betweenthese coordinates is the distance in === your comment I realized that I> was not very clear on my question. I certainly have the REAL> coordinates for the location.Oh. So you have the coordinates for all locations. > What I want to do is compress the > distance matrix and create new coordinates where the distance between > these coordinates is the distance in the matrix. Sorry for not being > specific.And you do not want to store the entire {n choose 2} = (n^2-n)/2 distances? That seems a reasonable desire.What do you mean by compression? Reduce the amount of memory taken up by the distance matrix by1) not using as many entriesor2) somehow compressing the entriesor3) something else?Since you have the points it should be very fast to compute the distance for any two points. no distance matrix necessary (but you have to compute distance everytime). Is computing distance considered too time consuming relative to everything === for your comment. When reading your comment I realized that I> was not very clear on my question. I certainly have the REAL> coordinates for the location. What I want to do is compress the> distance matrix and create new coordinates where the distance between> these coordinates is the distance in the matrix. Sorry for not being> specific.In every case where I used the term 'distance' I was referring to thedriving distance...Note to determine point D you must have the drivingdistance to it from two previously established points most direct to it...Now I said that each record would be point number, North coordinate, andEast coordinate. The thing that would need to be added to the record wouldbe the point numbers that each occupied point directly connects to...Nowsuppose the route is A to D and the direction of A to D calculates as 90degrees azimuth . The record of A notes connection to point B and to pointC. The direction of A to B is (say) 20 degrees while the direction of A to Cis 110 degrees. So the first leg of the route is A to C and the drivingdistance of the first leg is the distance between the coordinates of A andC. Then the record of point C notes connection to points A, B, and D. SinceD is the destination then the last leg of the route is C to D and thedriving distance of C to D is determined from the === Matrix> I have the following problem. I have a Matrix that contains a data> value for the driving distance between 2'000 different locations - so> this matrix has about 4 Mio records. The table would look like this:Location-1 Location-2 Distance> 1 1 0> 1 2 11> 1 3 8> 2 2 7> 2 3 4> etc....So, you have a distance matrix. symmetric, zeroes on the diagonal....> My ideas was now to convert this matrix into a map where I have a> coordinate (in a 2, 3 or more dimensional gird) for every location. > Now my questions are:> 1) Is there a mathmatical algorithm or procedure out there that does> exactly that, if so what is it called?Yes. Principal Components Analysis (PCA). It takes the distance matrix, finds its eigenvalues (the two largest ones) and uses that to construct points on the Euclidean plane having distances in your distance matrix. of course, these are unique -up to rotation and reflection- (isometries of the plane). so if you take the distances of a specific set of coords, then compute new coords using PCA, it may be flipped and/or turned.> 2) Is it possible to choose the number of dimensions when the map is> crea (I suppose the more dimensions the less the error)?Yes. for d dimensions, pick the d largest eigenvalues. If the original data happened to come from 2d and you pick three or more, then the smaller eigenvalues will be -really- small (not usually zero because of encoding approx). If the original data is from a high # of dimensions, and you pick 2, then you're essentially projecting all the data (losing some information) to just two dimensions, to a plane that preserves most of the variance (the two largest eigenvalues).Now, there is a slight problem because most of your data probably comes from cities that are on Earth and there might be approximation problems because of sphericity. I think PCA is geared specifically to Euclidean geometry/metrics. I would really like to know if there is a way to get a PCA-like procedure for other metrics (spherical, hyperbolic, manhattan, discrete (realizability of graphs from path distance)).> 3) Is there a software out there that a regular person (non math> expert) can use to create such a spreadsheets (Excel, etc), stats (SPSS, SAS), and symbolic math packages (Maple, Mathematica, etc) should have add-ons that do this for you somewhat auatically.> 4) I also assume that when using 10'000 locations that in order to> create such a map that I probably would not need the entire matrix to> create it. My guess is that the driving distance from every location> to the locations around it would be sufficient. Is this assumption> correct?that sounds right. I would think that you'd only need distances to 3 other points to determine a fourth exactly (in the plane).> 5) If there is no software available that can be used by a non math> expert, do you know anyone or a company who would be able to create> such a map?Hmmm.. I wouldn't be surprised if the map making software companies might do this. Sorry, it's not like I actually -know- anything! The cheaper ways above might require that you use a number of pieces of software (first compute the PCA, then compute the points, then transfer to a another package to === === small divisors problems, esp. complexanalytical dynamics and I see the concept of germ is used assomething very basic, that doesn't need to be defined...I asked a friend and he explained me the intuitive meaning of the termin differential geometry.I 'germ+dynamicalsystem'), but I couldn't find anything useful, possibly due to thelarge number of meanings this particular word has. I also triedmathworld for a quick answer, but no luck!I guess that *in this context* a germ f:(C,a)->(C,b) is simply ananalytical function such that f(a)=b. Am I right?PS: I know I could, and indeed I will *have to*, check the specificliterature lis in the bibliography, where I may find the answer tomy question. But right now I need to know if I'm reading correctly thefirst few paragraphs...TIA,Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- === Terminology: germ?> I'm studying some notes in small divisors problems, esp. complex> analytical dynamics and I see the concept of germ is used as> something very basic, that doesn't need to be defined...I asked a friend and he explained me the intuitive meaning of the term> in differential geometry.I 'germ+dynamical> system'), but I couldn't find anything useful, possibly due to the> large number of meanings this particular word has. I also tried> mathworld for a quick answer, but no luck!I guess that *in this context* a germ f:(C,a)->(C,b) is simply an> analytical function such that f(a)=b. Am I right?It's an equivalence class. Two maps are identified if theyagree on some neighborhood of a. You can add two germs byrestricting them both to a neighborhood where they are bothdefined, and adding them there.PS: I know I could, and indeed I will *have to*, check the specific> literature lis in the bibliography, where I may find the answer to> my question. But right now I need to know if I'm reading correctly the> === Terminology: germ?>I'm studying some notes in small divisors problems, esp. complex>analytical dynamics and I see the concept of germ is used as>something very basic, that doesn't need to be defined...I asked a friend and he explained me the intuitive meaning of the term>in differential geometry.I 'germ+dynamical>system'), but I couldn't find anything useful, possibly due to the>large number of meanings this particular word has. I also tried>mathworld for a quick answer, but no luck!I guess that *in this context* a germ f:(C,a)->(C,b) is simply an>analytical function such that f(a)=b. Am I right?Surely you are wrong (I've successively avoided ever learning much about complex analytical dynamics, but I hung out withsome of the people involved in its renascence 30-odd years ago;and I do know stuff about complex analytical topology). Tome and mine, a germ of a complex-analytic function f:(C,0)->(C,0)would determine, and be determined by, a convergent power series in the complex variable z, and you'd get a germ of a complex-analytic function f:(C,a)->(C,b) in the usual way (by translations in thedomain and range). In other words (*in your context*), not onlymust you have f(a)=b, but you must have the value of every derivativeof f at a. (For a general smooth function, that won't be enoughto determine the germ; it is, for a complex-analytic function.) In yet other words, a germ of a complex-analytic function f:(C,0)->(C,0)is what Ahlfors's textbook (for instance) calls a function element. >PS: I know I could, and indeed I will *have to*, check the specific>literature lis in the bibliography, where I may find the answer to>my question. But right now I need to know if I'm reading correctly the>first few === numbersI suppose a definition is in order.polysigned means a number system having n signs. It includes the realsas they have two signs. It also includes a set of numbers that bestrepresent time: one signed, and a set of numbers that represent theplane: three-signed, and a set of four-signed numbers that represent3D space.The crux of the polysigned approach lies in accepting an increase indimensionality. This comes as a result of summation. In general a zerosum always has identical component magnitudes in every sign (the reals) for any magnitude x: - x + x = 0.In three-signed math for any magnitude x: - x + x * x = 0. (where * is a new sign)In four-signed math for any magnitude x: - x + x * x # x = 0. (where # is a new sign)General sums in two signs are one-dimensional.General sums in three signs are two-dimensional.General sums in four signs are three-dimensional.General sums in n signs are (n-1)-dimensional.General sums in one sign are zero-dimensional?Product rules exist and work much like the real numbers.In effect each sign has a number that represents how many extremitiesaway from the identity sign to travel. The identity sign is always themaximum sign. The smallest sign is - (one), then + (two), then *(three), then # (four), and I don't know what character to use next.This system is logical and coexists with the traditional real numbers.It also works for the complex plane.I have crea a three-signed arithmetic that produces exactly thesame results for product and summation as would traditional complexarithmetic. If you search for three-signed within sci.math you willfind that thread. Is anyone interes in this?I will try to publish this in a more formal way but first have to geta linux box up and running some latex tools. My math terminology maynot be very precise but the math is so simple that all of the maththus far is === polysigned numbersTimothy - can we use another symbol for summation, and keep -, +, * as unary operators? I'll use S, and $ for subtraction (with the definition that x$y = z iff zSy = x). Before moving on to multiplication, can you clarify this:if-x S +x S *x = 0.then-x S +x = 0 $ *x is there any more concise representation of this quantity? its magnitude is x, but I can't imagine it lying on one of the three branches from the origin. It's clearly not on the star branch, and symmetry seems to prevent it from being on one of the two remaining branches rather than the other.In two-signed arithmetic, if you walk towards the origin from somewhere on one of the two branches, it's clear where you go after reaching the origin: you keep walking, now getting *further* from the origin, on the other branch.This is a naive question but I am curious as === polysigned numbers> I suppose a definition is in order.polysigned means a number system having n signs. It includes the reals> as they have two signs. It also includes a set of numbers that best> represent time: one signed, and a set of numbers that represent the> plane: three-signed, and a set of four-signed numbers that represent> 3D space.The crux of the polysigned approach lies in accepting an increase in> dimensionality. This comes as a result of summation. In general a zero> sum always has identical component magnitudes two-signed math (the reals) for any magnitude x:> - x + x = 0.> In three-signed math for any magnitude x:> - x + x * x = 0. (where * is a new sign)> In four-signed math for any magnitude x:> - x + x * x # x = 0. (where # is a new sign)> General sums in two signs are one-dimensional.> General sums in three signs are two-dimensional.> General sums in four signs are three-dimensional.> General sums in n signs are (n-1)-dimensional.> General sums in one sign are zero-dimensional?Product rules exist and work much like the real numbers.> In effect each sign has a number that represents how many extremities> away from the identity sign to travel. The identity sign is always the> maximum sign. The smallest sign is - (one), then + (two), then *> (three), then # (four), and I don't know what character to use next.> This system is logical and coexists with the traditional real numbers.> It also works for the complex plane.I have crea a three-signed arithmetic that produces exactly the> same results for product and summation as would traditional complex> arithmetic. If you search for three-signed within sci.math you will> find that thread. Is anyone interes in this?I will try to publish this in a more formal way but first have to get> a linux box up and running some latex tools. My math terminology may> not be very precise but the math is so simple that all of the math> thus far is easy.> I would like to find some help with this.Something you might consider is looking at this problem from theperspective of a vector space and one of its quotients.Let's, for instance, say we're working in three sign space. You canassociate with each number a triplet as follows and greatly simplifyyour notation, since you won't have to use the addition symbol to meantwo different things: +a -b *c = . You can add these asfollows: + = , just likeregular vector addition in R^3. In fact, I'm going to call this R^3with the understanding that it is a vector space with normal vectoraddition and scalar multiplication by reals.You can then consider the set S^3 = R^3/|<1,1,1>|. If you're familiarwith the notation I'm using, you'll understand immediately what thisdoes. If not, this imposes the condition that <1,1,1> = 0, or that +x-x *x = 0, which is what you wan, it seems. It forms the quotientvector space of R^3 with the one genera by <1,1,1>.One thing this implies is that every element in S^3 can be written as<0,a,b> for some real a and b. To see this, note that =-0 = -a*<1,1,1> = - = <0, b-a, c-a>.This means that the entire vector space can be spanned by twoelements, <0,1,0> and <0,0,1>. Unless I'm missing something, thedimension, then, of S^3 is actually 2. You've not added anything.Justin === for taking the time on this.I don't agree with the value of the vector representation.It gets too far away from the simplicity of three signs.The three-signed arithmetic yields traditional complex numbers.It is true that the dimensionality of the three signed space is two.I believe you do understand and that your writings here do work, butyou have done a transform back to a cartesian product. There is also ageometrical transform back to the cartesian coordinates.> Something you might consider is looking at this problem from the> perspective of a vector space and one of its quotients.Let's, for instance, say we're working in three sign space. You can> associate with each number a triplet as follows and greatly simplify> your notation, since you won't have to use the addition symbol to mean> two different things: +a -b *c = . You can add these as> follows: + = , just like> regular vector addition in R^3. In fact, I'm going to call this R^3> with the understanding that it is a vector space with normal vector> addition and scalar multiplication by reals.You can then consider the set S^3 = R^3/|<1,1,1>|. If you're familiar> with the notation I'm using, you'll understand immediately what this> does. If not, this imposes the condition that <1,1,1> = 0, or that +x> -x *x = 0, which is what you wan, it seems. It forms the quotient> vector space of R^3 with the one genera by <1,1,1>.One thing this implies is that every element in S^3 can be written as> <0,a,b> for some real a and b. To see this, note that => -0 = -a*<1,1,1> = - = <0, b-a, c-a>.This means that the entire vector space can be spanned by two> elements, <0,1,0> and <0,0,1>. Unless I'm missing something, the> dimension, then, of S^3 is actually 2. You've not added anything.I believe that this is true, where your values b-a,c-a are in thereals.I think that this is of interest at least osophically. By addingone sign beyond the reals you get two dimensional values that happento exactly match the complex numbers. No square roots of minus one atall necessary. I for one think that this is === definition is in order.polysigned means a number system having n signs. It includes the reals> as they have two signs. It also includes a set of numbers that best> represent time: one signed, and a set of numbers that represent the> plane: three-signed, and a set of four-signed numbers that represent> 3D space.The crux of the polysigned approach lies in accepting an increase in> dimensionality. This comes as a result of summation. In general a zero> sum always has identical component magnitudes in every sign yielding> proper thread, generalized signs aredeveloped in http://library.wolfram.com/infocenter/MathSource/4894/and are based on unsigned Primal numbers. Equivalence relations onpairs of primals create reals, on triples they create Terplex algebra,on quads they create complex algebra OR Study numbehyperbolicnumbePerplex numbers (all the same algebra on the hyperbolicplane with x^2-y^2=u^2). The {i,j,k} signs of quaternion algebra areobtained from the 8-element quaternion group, and are different fourthroots of unity, differing from complex-i which is also a fourth rootof unity. Dimensionality arises from equivalence relations on Directions,which only have primal coefficients. Space has six directions. SirW.R. Hamilton realised this nearly 150 years ago. Time is a direction,and cannot be nega. I propose the use of double-struck letters as signs. Using 's toindicate double-struck s in simple ASCII, my proposed standardsymbols are:-'s^'r = 1 (the general sign); 'd^12=1; 'n^9=1; 'o^8=1; 'g^7=1; 'h^6=1;'p^5=1; 'I^4=1; 'J^3=1; 'Y^3=1; 'k^2=1; 'l^2=1; 'm^2=1. Severaldistinct symbols are needed for each root of unity (eg quaternions,Study numbers), but only the second, third, and fourth rootsfrequently need several symbols. The relationship between signs differs between algebras, and sothere is no universal rule for their interaction.Roger Beresford.The world is so full of a number of things. I'm sure === polysigned numbersHi Roger.Could you give me some examples of terplex values please.Our product rule matches but I believe your values carry two signsdon't they?Also you have sta that these terplex values fit in between thereals and the complex numbers. The construction I am using producesthe complex numbers on the three signed stage.I'm not convinced that your construction is as simple as this one.Could you also provide an example zero sum for terplex?I would like to understand your writing but my math terminology ispoor in the areas of groups and rings. As I mentioned in your previous T-sign thread, generalized signs are> developed in http://library.wolfram.com/infocenter/MathSource/4894/> and are based on unsigned Primal numbers. Equivalence relations on> pairs of primals create reals, on triples they create Terplex algebra,> on quads they create complex algebra OR Study numbehyperbolic> numbePerplex numbers (all the same algebra on the hyperbolic> plane with x^2-y^2=u^2). The {i,j,k} signs of quaternion algebra are> obtained from the 8-element quaternion group, and are different fourth> roots of unity, differing from complex-i which is also a fourth root> of unity.> Dimensionality arises from equivalence relations on Directions,> which only have primal coefficients. Space has six directions. Sir> W.R. Hamilton realised this nearly 150 years ago. Time is a direction,> and cannot be nega.> I propose the use of double-struck letters as signs. Using 's to> indicate double-struck s in simple ASCII, my proposed standard> symbols are:-'s^'r = 1 (the general sign); 'd^12=1; 'n^9=1; 'o^8=1; 'g^7=1; 'h^6=1;> 'p^5=1; 'I^4=1; 'J^3=1; 'Y^3=1; 'k^2=1; 'l^2=1; 'm^2=1. Several> distinct symbols are needed for each root of unity (eg quaternions,> Study numbers), but only the second, third, and fourth roots> frequently need several symbols.I don't understand the symbolic system above. They are very long. the'=1' part seems redundant. Perhaps something got lost in the fonttranslation to my machine. I'm not seeing anything double struck. Arethe letters s, d, n, o, g, h, p, I, J, Y, k, l, m important?> The relationship between signs differs between algebras, and so> there is no universal rule for their interaction.I think I agree with you on this point except that there is a route todimensionality here that brings into question the need for thecartesian product or cross product. By increasing sign dimensionalityincreases.Roger Beresford.> The world is so full of a number of things. I'm sure we should all be> as happy as === a definition is in order.polysigned means a number system having n signs. It includes the reals> as they have two signs. It also includes a set of numbers that best> represent time: one signed, and a set of numbers that represent the> plane: three-signed, and a set of four-signed numbers that represent> 3D space.The crux of the polysigned approach lies in accepting an increase in> dimensionality. This comes as a result of summation. In general a zero> sum always has identical component magnitudes in every sign yielding> proper any magnitude x:> - x + x = 0.> In three-signed math for any magnitude x:> - x + x * x = 0. (where * is a new sign)> In four-signed math for any magnitude x:> - x + x * x # x = 0. (where # is a new sign)Are * and # unary? In your - x + x = 0 I take it that the sign thatisn't - is suppressed:- x + (+ x) = 0 ^and that's it. So should I read - x + x * x = 0 as - x + === In three-signed math for any magnitude x:> - x + x * x = 0. (where * is a new sign)> In four-signed math for any magnitude x:> - x + x * x # x = 0. (where # is a new sign)> Are * and # unary? In your - x + x = 0 I take it that the sign that> isn't - is suppressed:- x + (+ x) = 0> ^> and that's it. So should I read - x + x * x = 0 as - x + (+ x) + (* x) = 0 ?No. In three-signed math: - x + + x + * x = - 2x + x.Unary and binary signs are a misnomer.Summation is a process. It is integration in its traditional sense ona discrete level. You can also look binary operators modify only the sign ofthe number.in three-signed math: - - 1 = + 1. - + 1 = * 1. - * 1 = - 1. + - === math for any magnitude x:> - x + x * x = 0. (where * is a new sign)>In four-signed math for any magnitude x:> - x + x * x # x = 0. (where # is a new sign)>>Are * and # unary? In your - x + x = 0 I take it that the sign that>>isn't - is suppressed:>>- x + (+ x) = 0>> ^>>and that's it. >>So should I read - x + x * x = 0 as - x + (+ x) + (* x) = 0 ?No. In three-signed math:> - x + + x + * x = - 2x + x.Wait a minute. Does this mean that -x + *x = -2x? I don't think that's what you intended.Unary and binary signs are a misnomer.> Summation is a process. It is integration in its traditional sense on> a discrete level. You signs used as binary operators modify only the sign of> the number.in three-signed math: - - 1 = + 1.> - + 1 = * 1.> - * 1 = - 1.> + - 1 = * 1.> etc.I think you need to be more careful here. Let me explain how I'm envisioning your system:0,+1,+2,+3,+4 are a sequence of points marching to the right.0,-1,-2,-3,-4 are a sequence of points marching up and to the left along a vector rota 120 degrees from the positives.0,*1,*2,*3,*4 are a sequence of points marching down and to the left along a vector rota 120 degrees from the negatives.The basis for addition would be the following table:(+1)+(-1) = a vector of length 1 on the opposite side of 0 from *1(*1)+(+1) = a vector of length 1 on the opposite side of 0 from -1(-1)+(*1) = a vector of length 1 on the opposite side of 0 from +1I'm not sure how you would want to define multiplication. You could do it from a vector basis or try to define it to correspond to what happens when multiplying complex numbers.Hopefully, the whole thing would be set up to correspond to some sort of vector space that spans a plane.-- === any magnitude x:> - x + x * x = 0. (where * is a new sign)> In four-signed math for any magnitude x:> - x + x * x # x = 0. (where # is a new sign)> Are * and # unary? In your - x + x = 0 I take it that the sign that> isn't - is suppressed:- x + (+ x) = 0> ^> and that's it. So should I read - x + x * x = 0 as - x + (+ x) + (* x) = 0 ?No. In three-signed math: - x + + x + * x = - 2x + x.Unary and binary signs are a misnomer.Summation is a process. It is integration in its traditional sense ona discrete level. You can also look at summation as modify only the sign ofthe number.in three-signed math: - - 1 === Re: polysigned numbers> I suppose a definition is in order.polysigned means a number system having n signs. It includes the reals> as they have two signs. What exactly do you mean by that? If x is a real, what is its sign? And what is the sign of -x? What is the sign of 0? Given the reals, how would I determine how many signs it has? How many signs do the === suppose a definition is in order.polysigned means a number system having n signs. It includes the reals> as they have two signs. What exactly do you mean by that? If x is a real, what is its sign? if x is a real it can have one of two signs: + or - .> what is the sign of -x? The sign of -x is - if x is a magnitude(unsigned).>What is the sign of 0? 0 = +0 = -0. signed zeros are the same as plain unsigned zero. Zerois a special number.>Given the reals, how > would I determine how many signs it has? The reals have two signs: + and ->How many signs do the complex > numbers have?I have discovered that three-signed arithmetic is the complex numbers.In other words I can do complex arithmetic without the use of i or thesquare root of minus one. I would not say that complex numbers as theyhave been traditionally defined have three signs since they use i as ameans of keeping order. Products and sums as I have defined themproduce exactly equivalent results with three-signed numbers as thecomplex numbers. There is a proof === I suppose a definition is in order. polysigned means a number system having n signs. It includes the reals>> as they have two signs. What exactly do you mean by that? If x is a real, what is its sign? Gemini.>And what is the sign of -x?Also Gemini. They're twins, you see.dave(Am I sure about the sign? === with elements 1..n^2> Interestingly, two separate sessions on different machines, randomized > using the system clock, both came to row-and-column permutations of the> same solution.An overnight search again came up with the same result. Sloane's entry for this sequence now says| Results from a specially-adap hill-climbing algorithm strongly| suggest that a(5) = 6839492. a(6) is at least 1862125166.| Heuristically, a(n) is approximately 0.44 * n^(2.06*n), suggesting| that a(7) is close to 6.8 * 10^11. - Tim Paulden (timmy(AT)cantab.net), Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia === determinant of matrix with elements 1..n^2Interestingly, two separate sessions on different machines, randomized > using the system clock, both came to row-and-column permutations of the> same solution.An overnight search again came up with the same result. Sloane's > entry for this sequence now says> | Results from a specially-adap hill-climbing algorithm strongly> | suggest that a(5) = 6839492. a(6) is at least 1862125166.> | Heuristically, a(n) is approximately 0.44 * n^(2.06*n), suggesting> | that a(7) is close to 6.8 * 10^11. - Tim Paulden (timmy(AT)cantab.net),A quick run with some C code [20,6,5,23,13;8,16,21,18,1;11,25,4,9,15;22,10,19,2,12; 3,7,17,14,24]Several more minutes yielded nothing but [19,12,23,9,2;16,24,4,7,14;20,3,6,21,13;10,8,18,5,25; [24,16,3,7,14;12,19,23,9,2;8,10,18,5,25;4,20,6,21,13; 17,1,15,22,11](Blimey - 2 more 6839492s have just === elements 1..n^2 > Interestingly, two separate sessions on different machines, randomized> using the system clock, both came to row-and-column permutations of the> same solution. An overnight search again came up with the same result. Sloane's> entry for this sequence now says> | Results from a specially-adap hill-climbing algorithm strongly> | suggest that a(5) = 6839492. a(6) is at least 1862125166.> | Heuristically, a(n) is approximately 0.44 * n^(2.06*n), suggesting> | that a(7) is close to 6.8 * 10^11. - Tim Paulden 6839492 [20,6,5,23,13;8,16,21,18,1;11,25,4,9,15;22,10,19,2,12; 3,7,17,14,24]Several more minutes yielded nothing but [19,12,23,9,2;16,24,4,7,14;20,3,6,21,13;10,8,18,5,25; [24,16,3,7,14;12,19,23,9,2;8,10,18,5,25;4,20,6,21,13; 17,1,15,22,11]> (Blimey - 2 more 6839492s have just appeared!)Sorry if it appears twice - I pos a reply on mathforum, but itdoesn'tshow up here in the Usenet NG. Here is a copy of my mathforum post:I'm now absolutely shure about the n=5 solution. My Fortran programproduces thousands of 6839492's, if I let it run for several minutes.Also n=6 with a(6)=1865999570 seems to be a very stable solution,which I could not improve with several hours CPU time.For n=7 I'm currently ata(7)=761895710568 with corresponding matrix 7 8 33 42 15 27 43 26 21 46 13 19 45 6 40 12 34 29 41 1 18 4 37 20 10 48 25 31 32 47 30 17 3 11 35 22 38 9 49 23 28 5 44 14 2 16 24 39 36I'll not try n=8, because double precision will no longer === besufficient.HugoSubject: Re: Maximal determinant of matrix on different machines, randomized> using the system clock, both came to row-and-column permutations of the> same solution.An overnight search again came up with the same result. Sloane's> entry for this sequence now says> | Results from a specially-adap hill-climbing algorithm strongly> | suggest that a(5) = 6839492. a(6) is at least 1862125166.> | Heuristically, a(n) is approximately 0.44 * n^(2.06*n), suggesting> | that a(7) is close to 6.8 * 10^11. - Tim Paulden (timmy(AT)cantab.net), Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2I also ran a search with the following Fortran program,omitting the tabu heuristic.C Maximization of determinant of n*n integer matrix with elementsC 1..n^2CC Hugo Pfoertner http://www.pfoertner.org/C Version history:C 23.09.03 Search by exchange stepsC 19.09.03 5*5 matrix includedC 17.09.03 Initial version parameter ( n=7, nn=n*n ) doubleprecision d, db, dglob, ai, ajC...+....1....+....2....+....3....+....4....+....5....+....6 ....+....7.. doubleprecision A(n,N), b(n,N), det(2), work(n), as(nn), bs(nn) real x(nn) integer ip(nn), ipvt(nn) integer genrand_int32 equivalence (a,as) equivalence (b,bs) external genrand_int32CC Get starting time of computation from system T0 = SECNDS ( 0.0 )C Initial values for seed generation IRAN = NINT ( 100.0 * T0 ) IF ( MOD ( IRAN, 2 ) .EQ. 0 ) IRAN = IRAN + 1 j = init_genrand(iran) dglob = 0500 continuec Generate random vector using the Mersenne Twister do 10 i = 1, nn x(i) = float ( genrand_int32 () )10 continuec Permutation vector by sorting (SLATEC) call spsort ( x, nn, ip, -1, ier )c set matrix elements do 20 i = 1, nn as(ip(i)) = dble(i) bs(ip(i)) = dble(i)20 continueC Determinant of new random matrixC Gauss elimination (Linpack) call dgefa ( B, n, n, ipvt, info )C Determinant (Linpack) call dgedi ( B, n, n, ipvt, det, work, 10 ) d = det(1) * 10.0d0**det(2) l = 0C Perform 10-15 sweeps of interchange attempts do 100 it = 1, 15 do 30 i = 1, nn-1 do 40 j = i+1, nnC Copy to work matrix B, because Gauss elimination DO 41 k = 1, nn bs(k) = as(k)41 continuec Interchange matrix elements i and j in 1-dim stored bS(j) = aS(i) ai = as(i) bS(i) = aS(j) aj = as(j)C b (bs) are destroyed by dgefa call dgefa ( b, n, n, ipvt, info ) call dgedi ( b, n, n, ipvt, det, work, 10 ) db = det(1) * 10.0d0**det(2) if ( abs(db) .gt. abs(d) ) then l = l + 1 lt = itc write (*,*) l, i, j, dbC perform successful interchange in original matrix aS(j) = ai aS(i) = aj d = db endif40 continue30 continue100 continueC Update best value and print if ( abs(d) .gE. dglob ) then dglob = abs(d) write (*,*) ' d=', d, ' l, lt:', l, lt write (*,1000) (nint(as(k)),k=1,nn)1000 format ( (25i3),:,/,(25i3) ) endifC Skip back to generation of next random matrixC (infinite loop) goto 500 endwith the following results:a(5)=6839492 withA_5=25 15 9 11 4 7 24 14 3 17 6 12 23 20 510 13 2 22 1916 1 18 8 21found the same result from 100's of starting points.a(6)>=1865999570A_6=36 24 21 17 5 8 3 35 25 15 23 1113 7 34 16 10 3114 22 2 33 12 2820 4 19 29 32 626 18 9 1 30 27found 5 such solutions. If this solution is the optimum it contradictsmy guess, that the main diagonal contains the n maximal matrix elements.a(7)>=|-761680932467| (considerably larger than Paul Tilden'sconjectured a(7))A_7= 24 26 3 14 45 44 20 32 1 41 36 31 28 6 8 22 35 21 5 46 37 47 40 16 29 4 25 15 39 18 33 2 30 10 43 7 48 38 23 34 12 13 17 19 9 49 27 11 42very preliminary, will run this for a few days.Hugo === elements 1..n^2[SNIP - code]OK, if we're sharing code, here's my code for the 5x5 problem, which was able to find Robert's solution in only a few seconds (and after 10 minutes has found it ~20 times now).If anyone wants to modify it for 6x6, erm, good luck. I suggest using perl or such-like to auto-generate the code using the same high-school algorithm (avoiding the nastyfrom the permutation array) and am _not_ going to tackle 6x6 or larger by hand).--- 8< ---------------------#include #include #include #include #define INTEREST 6820000//typedef double etype;typedef int etype;typedef struct{ etype e[5][5];} m55;enum { m1_01, m1_02, m1_03, m1_04, m1_12, m1_13, m1_14, m1_23, m1_24, m1_34 };enum { m2_012, m2_013, m2_014, m2_023, m2_024, m2_034, m2_123, m2_124, m2_134, m2_234 };enum { m3_1234, m3_0234, m3_0134, m3_0124, m3_0123 };typedef struct{ etype m1[10]; etype m2[10]; etype m3[5];} minors;enum { dh_01, dh_02, dh_03, dh_04, dh_10, dh_12, dh_13, dh_14, dh_20, dh_21, dh_23, dh_24, dh_30, dh_31, dh_32, dh_34, dh_40, dh_41, dh_42, dh_43 };typedef struct{ etype s[5][5]; etype t[20]; // rows 0 and 1 etype m[20]; // rows 2 and 3} dethelp;typedef struct{ etype d[120];} dets;static unsigned int detorder[120]={ 0x01234,0x01243,0x01324,0x01342,0x01423,0x01432,0x02134,0x02143 , 0x02314,0x02341,0x02413,0x02431,0x03124,0x03142,0x03214,0x03241 , 0x03412,0x03421,0x04123,0x04132,0x04213,0x04231,0x04312,0x04321 , 0x10234,0x10243,0x10324,0x10342,0x10423,0x10432,0x12034,0x12043 , 0x12304,0x12340,0x12403,0x12430,0x13024,0x13042,0x13204,0x13240 , 0x13402,0x13420,0x14023,0x14032,0x14203,0x14230,0x14302,0x14320 , 0x20134,0x20143,0x20314,0x20341,0x20413,0x20431,0x21034,0x21043 , 0x21304,0x21340,0x21403,0x21430,0x23014,0x23041,0x23104,0x23140 , 0x23401,0x23410,0x24013,0x24031,0x24103,0x24130,0x24301,0x24310 , 0x30124,0x30142,0x30214,0x30241,0x30412,0x30421,0x31024,0x31042 , 0x31204,0x31240,0x31402,0x31420,0x32014,0x32041,0x32104,0x32140 , 0x32401,0x32410,0x34012,0x34021,0x34102,0x34120,0x34201,0x34210 , 0x40123,0x40132,0x40213,0x40231,0x40312,0x40321,0x41023,0x41032 , 0x41203,0x41230,0x41302,0x41320,0x42013,0x42031,0x42103,0x42130 , 0x42301,0x42310,0x43012,0x43021,0x43102,0x43120,0x43201,0x43210 ,};static inline void flipPermute(m55 * restrict d, m55 const* restrict s, unsigned int perm){ unsigned int i, j; unsigned int permnibs=detorder[perm]; for(i=0; i<5; ++i) { d->e[i][0]=s->e[4][(permnibs>>(16-i*4))&7]; for(j=1; j<5; ++j) { d->e[i][j]=s->e[j-1][i]; } }}static inline void dump(m55 const *restrict s){ unsigned int i, j; char semi='['; for(i=0; i<5; ++i) { char comma=semi; for(j=0; j<5; ++j) { printf(%c%i, comma, s->e[i][j]); comma=','; } semi=';'; } printf(]n);} /* Cost = * 1) 20* 10+ * 2) 30* 20+ * 3) 20* 15+ */static inline void makeMinors(minors *restrict d, m55 const *restrict s){#define CROSS(x1,x2) s->e[0][x1]*s->e[1][x2]-s->e[1][x1]*s->e[0][x2] d->m1[m1_01]=CROSS(0,1); d->m1[m1_02]=CROSS(0,2); d->m1[m1_03]=CROSS(0,3); d->m1[m1_04]=CROSS(0,4); d->m1[m1_12]=CROSS(1,2); d->m1[m1_13]=CROSS(1,3); d->m1[m1_14]=CROSS(1,4); d->m1[m1_23]=CROSS(2,3); d->m1[m1_24]=CROSS(2,4); d->m1[m1_34]=CROSS(3,4); d->m2[m2_012]=s->e[2][0]*d->m1[m1_12] - s->e[2][1]*d->m1[m1_02] + s->e[2][2]*d->m1[m1_01]; d->m2[m2_013]=s->e[2][0]*d->m1[m1_13] - s->e[2][1]*d->m1[m1_03] + s->e[2][3]*d->m1[m1_01]; d->m2[m2_014]=s->e[2][0]*d->m1[m1_14] - s->e[2][1]*d->m1[m1_04] + s->e[2][4]*d->m1[m1_01]; d->m2[m2_023]=s->e[2][0]*d->m1[m1_23] - s->e[2][2]*d->m1[m1_03] + s->e[2][3]*d->m1[m1_02]; d->m2[m2_024]=s->e[2][0]*d->m1[m1_24] - s->e[2][2]*d->m1[m1_04] + s->e[2][4]*d->m1[m1_02]; d->m2[m2_034]=s->e[2][0]*d->m1[m1_34] - s->e[2][3]*d->m1[m1_04] + s->e[2][4]*d->m1[m1_03]; d->m2[m2_123]=s->e[2][1]*d->m1[m1_23] - s->e[2][2]*d->m1[m1_13] + s->e[2][3]*d->m1[m1_12]; d->m2[m2_124]=s->e[2][1]*d->m1[m1_24] - s->e[2][2]*d->m1[m1_14] + s->e[2][4]*d->m1[m1_12]; d->m2[m2_134]=s->e[2][1]*d->m1[m1_34] - s->e[2][3]*d->m1[m1_14] + s->e[2][4]*d->m1[m1_13]; d->m2[m2_234]=s->e[2][2]*d->m1[m1_34] - s->e[2][3]*d->m1[m1_24] + s->e[2][4]*d->m1[m1_23]; d->m3[m3_0123]= + s->e[3][3]*d->m2[m2_012] - s->e[3][2]*d->m2[m2_013] + s->e[3][1]*d->m2[m2_023] - s->e[3][0]*d->m2[m2_123]; d->m3[m3_0124]= + s->e[3][4]*d->m2[m2_012] - s->e[3][2]*d->m2[m2_014] + s->e[3][1]*d->m2[m2_024] - s->e[3][0]*d->m2[m2_124]; d->m3[m3_0134]= + s->e[3][4]*d->m2[m2_013] - s->e[3][3]*d->m2[m2_014] + s->e[3][1]*d->m2[m2_034] - s->e[3][0]*d->m2[m2_134]; d->m3[m3_0234]= + s->e[3][4]*d->m2[m2_023] - s->e[3][3]*d->m2[m2_024] + s->e[3][2]*d->m2[m2_034] - s->e[3][0]*d->m2[m2_234]; d->m3[m3_1234]= + s->e[3][4]*d->m2[m2_123] - s->e[3][3]*d->m2[m2_124] + s->e[3][2]*d->m2[m2_134] - s->e[3][1]*d->m2[m2_234];}static inline etype matdet(m55 const* restrict m){ static minors n; makeMinors(&n,m); return m->e[4][0]*n.m3[m3_1234] - m->e[4][1]*n.m3[m3_0234] + m->e[4][2]*n.m3[m3_0134] - m->e[4][3]*n.m3[m3_0124] + m->e[4][4]*n.m3[m3_0123];}/* Cost * 25* 40+ */static inline void makeDetsHelp(dethelp*restrict d, m55*restrict m, minors*restrict n){ d->s[0][0]=m->e[4][0]*n->m3[m3_1234]; d->s[0][1]=m->e[4][1]*n->m3[m3_1234]; d->s[0][2]=m->e[4][2]*n->m3[m3_1234]; d->s[0][3]=m->e[4][3]*n->m3[m3_1234]; d->s[0][4]=m->e[4][4]*n->m3[m3_1234]; d->s[1][0]=m->e[4][0]*n->m3[m3_0234]; d->s[1][1]=m->e[4][1]*n->m3[m3_0234]; d->s[1][2]=m->e[4][2]*n->m3[m3_0234]; d->s[1][3]=m->e[4][3]*n->m3[m3_0234]; d->s[1][4]=m->e[4][4]*n->m3[m3_0234]; d->t[dh_01]=d->s[0][0] - d->s[1][1]; d->t[dh_02]=d->s[0][0] - d->s[1][2]; d->t[dh_03]=d->s[0][0] - d->s[1][3]; d->t[dh_04]=d->s[0][0] - d->s[1][4]; d->t[dh_10]=d->s[0][1] - d->s[1][0]; d->t[dh_12]=d->s[0][1] - d->s[1][2]; d->t[dh_13]=d->s[0][1] - d->s[1][3]; d->t[dh_14]=d->s[0][1] - d->s[1][4]; d->t[dh_20]=d->s[0][2] - d->s[1][0]; d->t[dh_21]=d->s[0][2] - d->s[1][1]; d->t[dh_23]=d->s[0][2] - d->s[1][3]; d->t[dh_24]=d->s[0][2] - d->s[1][4]; d->t[dh_30]=d->s[0][3] - d->s[1][0]; d->t[dh_31]=d->s[0][3] - d->s[1][1]; d->t[dh_32]=d->s[0][3] - d->s[1][2]; d->t[dh_34]=d->s[0][3] - d->s[1][4]; d->t[dh_40]=d->s[0][4] - d->s[1][0]; d->t[dh_41]=d->s[0][4] - d->s[1][1]; d->t[dh_42]=d->s[0][4] - d->s[1][2]; d->t[dh_43]=d->s[0][4] - d->s[1][3]; d->s[2][0]=m->e[4][0]*n->m3[m3_0134]; d->s[2][1]=m->e[4][1]*n->m3[m3_0134]; d->s[2][2]=m->e[4][2]*n->m3[m3_0134]; d->s[2][3]=m->e[4][3]*n->m3[m3_0134]; d->s[2][4]=m->e[4][4]*n->m3[m3_0134]; d->s[3][0]=m->e[4][0]*n->m3[m3_0124]; d->s[3][1]=m->e[4][1]*n->m3[m3_0124]; d->s[3][2]=m->e[4][2]*n->m3[m3_0124]; d->s[3][3]=m->e[4][3]*n->m3[m3_0124]; d->s[3][4]=m->e[4][4]*n->m3[m3_0124]; d->m[dh_01]=d->s[2][0] - d->s[3][1]; d->m[dh_02]=d->s[2][0] - d->s[3][2]; d->m[dh_03]=d->s[2][0] - d->s[3][3]; d->m[dh_04]=d->s[2][0] - d->s[3][4]; d->m[dh_10]=d->s[2][1] - d->s[3][0]; d->m[dh_12]=d->s[2][1] - d->s[3][2]; d->m[dh_13]=d->s[2][1] - d->s[3][3]; d->m[dh_14]=d->s[2][1] - d->s[3][4]; d->m[dh_20]=d->s[2][2] - d->s[3][0]; d->m[dh_21]=d->s[2][2] - d->s[3][1]; d->m[dh_23]=d->s[2][2] - d->s[3][3]; d->m[dh_24]=d->s[2][2] - d->s[3][4]; d->m[dh_30]=d->s[2][3] - d->s[3][0]; d->m[dh_31]=d->s[2][3] - d->s[3][1]; d->m[dh_32]=d->s[2][3] - d->s[3][2]; d->m[dh_34]=d->s[2][3] - d->s[3][4]; d->m[dh_40]=d->s[2][4] - d->s[3][0]; d->m[dh_41]=d->s[2][4] - d->s[3][1]; d->m[dh_42]=d->s[2][4] - d->s[3][2]; d->m[dh_43]=d->s[2][4] - d->s[3][3]; d->s[4][0]=m->e[4][0]*n->m3[m3_0123]; d->s[4][1]=m->e[4][1]*n->m3[m3_0123]; d->s[4][2]=m->e[4][2]*n->m3[m3_0123]; d->s[4][3]=m->e[4][3]*n->m3[m3_0123]; d->s[4][4]=m->e[4][4]*n->m3[m3_0123];}/* Cost * 240+ */static inline void makeDets(dets* restrict d, dethelp const*restrict h){//#define ABS(v) fabs(v)#define ABS(v) v d->d[0]=ABS(h->t[dh_01] + h->m[dh_23] + h->s[4][4]); d->d[1]=ABS(h->t[dh_01] + h->m[dh_24] + h->s[4][3]); d->d[2]=ABS(h->t[dh_01] + h->m[dh_32] + h->s[4][4]); d->d[3]=ABS(h->t[dh_01] + h->m[dh_34] + h->s[4][2]); d->d[4]=ABS(h->t[dh_01] + h->m[dh_42] + h->s[4][3]); d->d[5]=ABS(h->t[dh_01] + h->m[dh_43] + h->s[4][2]); d->d[6]=ABS(h->t[dh_02] + h->m[dh_13] + h->s[4][4]); d->d[7]=ABS(h->t[dh_02] + h->m[dh_14] + h->s[4][3]); d->d[8]=ABS(h->t[dh_02] + h->m[dh_31] + h->s[4][4]); d->d[9]=ABS(h->t[dh_02] + h->m[dh_34] + h->s[4][1]); d->d[10]=ABS(h->t[dh_02] + h->m[dh_41] + h->s[4][3]); d->d[11]=ABS(h->t[dh_02] + h->m[dh_43] + h->s[4][1]); d->d[12]=ABS(h->t[dh_03] + h->m[dh_12] + h->s[4][4]); d->d[13]=ABS(h->t[dh_03] + h->m[dh_14] + h->s[4][2]); d->d[14]=ABS(h->t[dh_03] + h->m[dh_21] + h->s[4][4]); d->d[15]=ABS(h->t[dh_03] + h->m[dh_24] + h->s[4][1]); d->d[16]=ABS(h->t[dh_03] + h->m[dh_41] + h->s[4][2]); d->d[17]=ABS(h->t[dh_03] + h->m[dh_42] + h->s[4][1]); d->d[18]=ABS(h->t[dh_04] + h->m[dh_12] + h->s[4][3]); d->d[19]=ABS(h->t[dh_04] + h->m[dh_13] + h->s[4][2]); d->d[20]=ABS(h->t[dh_04] + h->m[dh_21] + h->s[4][3]); d->d[21]=ABS(h->t[dh_04] + h->m[dh_23] + h->s[4][1]); d->d[22]=ABS(h->t[dh_04] + h->m[dh_31] + h->s[4][2]); d->d[23]=ABS(h->t[dh_04] + h->m[dh_32] + h->s[4][1]); d->d[24]=ABS(h->t[dh_10] + h->m[dh_23] + h->s[4][4]); d->d[25]=ABS(h->t[dh_10] + h->m[dh_24] + h->s[4][3]); d->d[26]=ABS(h->t[dh_10] + h->m[dh_32] + h->s[4][4]); d->d[27]=ABS(h->t[dh_10] + h->m[dh_34] + h->s[4][2]); d->d[28]=ABS(h->t[dh_10] + h->m[dh_42] + h->s[4][3]); d->d[29]=ABS(h->t[dh_10] + h->m[dh_43] + h->s[4][2]); d->d[30]=ABS(h->t[dh_12] + h->m[dh_03] + h->s[4][4]); d->d[31]=ABS(h->t[dh_12] + h->m[dh_04] + h->s[4][3]); d->d[32]=ABS(h->t[dh_12] + h->m[dh_30] + h->s[4][4]); d->d[33]=ABS(h->t[dh_12] + h->m[dh_34] + h->s[4][0]); d->d[34]=ABS(h->t[dh_12] + h->m[dh_40] + h->s[4][3]); d->d[35]=ABS(h->t[dh_12] + h->m[dh_43] + h->s[4][0]); d->d[36]=ABS(h->t[dh_13] + h->m[dh_02] + h->s[4][4]); d->d[37]=ABS(h->t[dh_13] + h->m[dh_04] + h->s[4][2]); d->d[38]=ABS(h->t[dh_13] + h->m[dh_20] + h->s[4][4]); d->d[39]=ABS(h->t[dh_13] + h->m[dh_24] + h->s[4][0]); d->d[40]=ABS(h->t[dh_13] + h->m[dh_40] + h->s[4][2]); d->d[41]=ABS(h->t[dh_13] + h->m[dh_42] + h->s[4][0]); d->d[42]=ABS(h->t[dh_14] + h->m[dh_02] + h->s[4][3]); d->d[43]=ABS(h->t[dh_14] + h->m[dh_03] + h->s[4][2]); d->d[44]=ABS(h->t[dh_14] + h->m[dh_20] + h->s[4][3]); d->d[45]=ABS(h->t[dh_14] + h->m[dh_23] + h->s[4][0]); d->d[46]=ABS(h->t[dh_14] + h->m[dh_30] + h->s[4][2]); d->d[47]=ABS(h->t[dh_14] + h->m[dh_32] + h->s[4][0]); d->d[48]=ABS(h->t[dh_20] + h->m[dh_13] + h->s[4][4]); d->d[49]=ABS(h->t[dh_20] + h->m[dh_14] + h->s[4][3]); d->d[50]=ABS(h->t[dh_20] + h->m[dh_31] + h->s[4][4]); d->d[51]=ABS(h->t[dh_20] + h->m[dh_34] + h->s[4][1]); d->d[52]=ABS(h->t[dh_20] + h->m[dh_41] + h->s[4][3]); d->d[53]=ABS(h->t[dh_20] + h->m[dh_43] + h->s[4][1]); d->d[54]=ABS(h->t[dh_21] + h->m[dh_03] + h->s[4][4]); d->d[55]=ABS(h->t[dh_21] + h->m[dh_04] + h->s[4][3]); d->d[56]=ABS(h->t[dh_21] + h->m[dh_30] + h->s[4][4]); d->d[57]=ABS(h->t[dh_21] + h->m[dh_34] + h->s[4][0]); d->d[58]=ABS(h->t[dh_21] + h->m[dh_40] + h->s[4][3]); d->d[59]=ABS(h->t[dh_21] + h->m[dh_43] + h->s[4][0]); d->d[60]=ABS(h->t[dh_23] + h->m[dh_01] + h->s[4][4]); d->d[61]=ABS(h->t[dh_23] + h->m[dh_04] + h->s[4][1]); d->d[62]=ABS(h->t[dh_23] + h->m[dh_10] + h->s[4][4]); d->d[63]=ABS(h->t[dh_23] + h->m[dh_14] + h->s[4][0]); d->d[64]=ABS(h->t[dh_23] + h->m[dh_40] + h->s[4][1]); d->d[65]=ABS(h->t[dh_23] + h->m[dh_41] + h->s[4][0]); d->d[66]=ABS(h->t[dh_24] + h->m[dh_01] + h->s[4][3]); d->d[67]=ABS(h->t[dh_24] + h->m[dh_03] + h->s[4][1]); d->d[68]=ABS(h->t[dh_24] + h->m[dh_10] + h->s[4][3]); d->d[69]=ABS(h->t[dh_24] + h->m[dh_13] + h->s[4][0]); d->d[70]=ABS(h->t[dh_24] + h->m[dh_30] + h->s[4][1]); d->d[71]=ABS(h->t[dh_24] + h->m[dh_31] + h->s[4][0]); d->d[72]=ABS(h->t[dh_30] + h->m[dh_12] + h->s[4][4]); d->d[73]=ABS(h->t[dh_30] + h->m[dh_14] + h->s[4][2]); d->d[74]=ABS(h->t[dh_30] + h->m[dh_21] + h->s[4][4]); d->d[75]=ABS(h->t[dh_30] + h->m[dh_24] + h->s[4][1]); d->d[76]=ABS(h->t[dh_30] + h->m[dh_41] + h->s[4][2]); d->d[77]=ABS(h->t[dh_30] + h->m[dh_42] + h->s[4][1]); d->d[78]=ABS(h->t[dh_31] + h->m[dh_02] + h->s[4][4]); d->d[79]=ABS(h->t[dh_31] + h->m[dh_04] + h->s[4][2]); d->d[80]=ABS(h->t[dh_31] + h->m[dh_20] + h->s[4][4]); d->d[81]=ABS(h->t[dh_31] + h->m[dh_24] + h->s[4][0]); d->d[82]=ABS(h->t[dh_31] + h->m[dh_40] + h->s[4][2]); d->d[83]=ABS(h->t[dh_31] + h->m[dh_42] + h->s[4][0]); d->d[84]=ABS(h->t[dh_32] + h->m[dh_01] + h->s[4][4]); d->d[85]=ABS(h->t[dh_32] + h->m[dh_04] + h->s[4][1]); d->d[86]=ABS(h->t[dh_32] + h->m[dh_10] + h->s[4][4]); d->d[87]=ABS(h->t[dh_32] + h->m[dh_14] + h->s[4][0]); d->d[88]=ABS(h->t[dh_32] + h->m[dh_40] + h->s[4][1]); d->d[89]=ABS(h->t[dh_32] + h->m[dh_41] + h->s[4][0]); d->d[90]=ABS(h->t[dh_34] + h->m[dh_01] + h->s[4][2]); d->d[91]=ABS(h->t[dh_34] + h->m[dh_02] + h->s[4][1]); d->d[92]=ABS(h->t[dh_34] + h->m[dh_10] + h->s[4][2]); d->d[93]=ABS(h->t[dh_34] + h->m[dh_12] + h->s[4][0]); d->d[94]=ABS(h->t[dh_34] + h->m[dh_20] + h->s[4][1]); d->d[95]=ABS(h->t[dh_34] + h->m[dh_21] + h->s[4][0]); d->d[96]=ABS(h->t[dh_40] + h->m[dh_12] + h->s[4][3]); d->d[97]=ABS(h->t[dh_40] + h->m[dh_13] + h->s[4][2]); d->d[98]=ABS(h->t[dh_40] + h->m[dh_21] + h->s[4][3]); d->d[99]=ABS(h->t[dh_40] + h->m[dh_23] + h->s[4][1]); d->d[100]=ABS(h->t[dh_40] + h->m[dh_31] + h->s[4][2]); d->d[101]=ABS(h->t[dh_40] + h->m[dh_32] + h->s[4][1]); d->d[102]=ABS(h->t[dh_41] + h->m[dh_02] + h->s[4][3]); d->d[103]=ABS(h->t[dh_41] + h->m[dh_03] + h->s[4][2]); d->d[104]=ABS(h->t[dh_41] + h->m[dh_20] + h->s[4][3]); d->d[105]=ABS(h->t[dh_41] + h->m[dh_23] + h->s[4][0]); d->d[106]=ABS(h->t[dh_41] + h->m[dh_30] + h->s[4][2]); d->d[107]=ABS(h->t[dh_41] + h->m[dh_32] + h->s[4][0]); d->d[108]=ABS(h->t[dh_42] + h->m[dh_01] + h->s[4][3]); d->d[109]=ABS(h->t[dh_42] + h->m[dh_03] + h->s[4][1]); d->d[110]=ABS(h->t[dh_42] + h->m[dh_10] + h->s[4][3]); d->d[111]=ABS(h->t[dh_42] + h->m[dh_13] + h->s[4][0]); d->d[112]=ABS(h->t[dh_42] + h->m[dh_30] + h->s[4][1]); d->d[113]=ABS(h->t[dh_42] + h->m[dh_31] + h->s[4][0]); d->d[114]=ABS(h->t[dh_43] + h->m[dh_01] + h->s[4][2]); d->d[115]=ABS(h->t[dh_43] + h->m[dh_02] + h->s[4][1]); d->d[116]=ABS(h->t[dh_43] + h->m[dh_10] + h->s[4][2]); d->d[117]=ABS(h->t[dh_43] + h->m[dh_12] + h->s[4][0]); d->d[118]=ABS(h->t[dh_43] + h->m[dh_20] + h->s[4][1]); d->d[119]=ABS(h->t[dh_43] + h->m[dh_21] + h->s[4][0]);}unsigned int getBest(dets const* restrict d){ unsigned int i; unsigned int bestat=0;#undef ABS//#define ABS(v) v//#define ABS(v) fabs(v)#define ABS(v) (v<0)?-v:v; etype best=ABS(d->d[0]); for(i=1; i<120; ++i) { etype me=ABS(d->d[i]); if(me>best) { best=me; bestat=i; } } return bestat;} int main(){ unsigned int i, j; m55 m[2]; int mi=0; minors n; dethelp h; dets d; int best; int still=11; int reseeds=1000000000; int candibble=1; int seed; for(i=0; i<5; ++i) { for(j=0; j<5; ++j) { m[mi].e[i][j]=i*5+j+1; } } while(reseeds>0) { if(still>=10) { if(candibble) { unsigned int bi=0, bj=0, bk=0; etype bdet=0; candibble=0; /* Take a copy */ m[!mi]=m[mi]; etype currdet=(d.d[0]<0)?(-d.d[0]):d.d[0]; //if(currdet>=INTEREST) //{ // printf(~~~ %i , currdet); // dump(&m[mi]); //} for(i=0; i<23; ++i) { for(j=i+1; j<25; ++j) { unsigned int k; for(k=i+1; k<25; ++k) { etype det; if(k==j) continue; m[!mi].e[i/5][i%5]=m[mi].e[j/5][j%5]; m[!mi].e[j/5][j%5]=m[mi].e[k/5][k%5]; m[!mi].e[k/5][k%5]=m[mi].e[i/5][i%5]; det=matdet(&m[!mi]); if(det<0) { det=-det; } if(det>currdet) { if(det>=INTEREST) { printf(:-) %i , det); dump(&m[!mi]); } if(det>bdet) { bi=i; bj=j; bk=k; bdet=det; } } m[!mi].e[i/5][i%5]=m[mi].e[i/5][i%5]; m[!mi].e[j/5][j%5]=m[mi].e[j/5][j%5]; m[!mi].e[k/5][k%5]=m[mi].e[k/5][k%5]; } } } /* Did any improvements get made? */ if(bdet>0) { // printf(continuingn); candibble=1; etype ti=m[mi].e[bj/5][bj%5]; m[mi].e[bj/5][bj%5]=m[mi].e[bk/5][bk%5]; m[mi].e[bk/5][bk%5]=m[mi].e[bi/5][bi%5]; m[mi].e[bi/5][bi%5]=ti; } } if(!candibble) { // printf(Having to mess it upn); struct timeval tv; gettimeofday(&tv, NULL); seed=(tv.tv_sec*1000+tv.tv_usec); srand(seed); for(i=0; i<10; ++i) { int r=rand()%120; flipPermute(&m[1],&m[0],r); //printf(~%i -> , r); dump(&m[1]); r=rand()%120; flipPermute(&m[0],&m[1],r); //printf(~%i -> , r); dump(&m[0]); } candibble=1; } still=0; --reseeds; } // printf(shuffling rown); /* Calculate 120 different 5x5 determinants */ makeMinors(&n, &m[mi]); /* 70* 45+ */ makeDetsHelp(&h, &m[mi], &n); /* 25* 40+ */ makeDets(&d, &h); /* 0* 240+ */ /* in 95 multiplies and 325 add/subs */ best=getBest(&d); /* 0* 120? */ if(!best) { ++still; } else { etype ad=(d.d[best]<0)?(-d.d[best]):d.d[best]; if(ad>=INTEREST) { flipPermute(&m[!mi], &m[mi], best); mi=!mi; } return 0;} === 1..n^2> [SNIP - code]OK, if we're sharing code, here's my code for the 5x5 problem, > which was able to find Robert's solution in only a few seconds > (and after 10 minutes has found it ~20 times now).If anyone wants to modify it for 6x6, erm, good luck. I > suggest using perl or such-like to auto-generate the code > using the same high-school algorithm (avoiding the nasty> from the permutation array) and am _not_ going to tackle > 6x6 or larger by hand).One thing you might find useful: if you interchange two entries inan n x n matrix, the change in the determinant can be calcula using two (n-1) x (n-1) determinants and (if they are in differentrows and columns) an (n-2) x (n-2).Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia === Society - is their math test bunkum?> Just nit-picking here. No new content.> ... stuff dele ...> A typical example is the sequence 3,3,5,4,5,3,...> the Correct, obvious, and only true answer is ..> TA-DA ... 5> ??? maybe this was a typo ??? ONE TWO THREE FOUR FIVE SIX SEVEN 3 3 5 4 5 3 5 Or have they changed the spelling of FIVE to FI5VE?>This mistakes was obviously deliberate, to make the puzzlemore difficult. And also: to better prepare the puzzlerfor an existence in the real world.In nature, namely, these things occur frequently. Physicistscall them 'measurement errors'. The rest of the === uniqueness of solution for a differential equationSIMPLER VERSIONsolvey''+A*sqrt(y)=0FULL VERSIONsolve y^2+A(y'')^4+B(y'')^3=0y is a function of x' indicates derivative wrt xA and B are unknown constantsIt is easy to show that y=x^4 is a solution (which gives B=0 for the full version)Are there any other solutions? Can it be proved that this solution is unique? queries can be sent to === solution for a differential equation>SIMPLER VERSION>solve>y''+A*sqrt(y)=0Maybe not so simple. If A < 0, solutions include y = 0 for x <= a, A^2/144 (x-a)^4 for x >= aor the same with <= and >= reversed.>FULL VERSION>solve >y^2+A(y'')^4+B(y'')^3=0>y is a function of x>' indicates derivative wrt x>A and B are unknown constants>It is easy to show that y=x^4 is a solution (which gives B=0 >for the full version)When that's a solution, so is (x-a)^4. Of course y = 0 is a solution. Maple finds a rather complica solution.>Are there any other solutions? Can it be proved that this solution is unique? Certainly non-unique.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia === functionCan someone point me to the theory involved in fitting a polynomial functionto a set of sample data. My sample data is composed of just two variables Xand Y. I know tht most stat packages do this but I just want to read up onhow this is === the theory involved in fitting a> polynomial function to a set of sample data. My sample data is> composed of just two variables X and Y. I know tht most stat> packages do this but I just want to read up on how this is done.A web source is preferred to a textbook.> Jay.Try this for a web source--http://www.numerical-recipes.com/nronline_ === Re: fitting a polynomial function> Can someone point me to the theory involved in fitting a polynomial function> to a set of sample data. My sample data is composed of just two variables X> and Y. I know tht most stat packages do this but I just want to read up on> how this is done.A web source is preferred to a textbook.> Jay.Look up Lagrange polynomial.There are over === a polynomial function> Can someone point me to the theory involved in fitting a polynomial function> to a set of sample data. My sample data is composed of just two variables X> and Y. I know tht most stat packages do this but I just want to read up on> how this is done. A web source is preferred to a textbook. > Jay.I think a lot of programs use least squares. Here is one sitehttp://mathworld.wolfram.com/LeastSquaresFitting.htmlBill= === ==Subject: Re: Problems com(pil)ing a(g)p(g)art mod(ul)e> -- >> Jesse Hughes>> But nothing's being Dr. Ullrich is a particular case of something's>> being such that nothing is it: (Ex)~(Ey)(y = x) >> -- John Correy on the failings of first order logic To which (in due course) Jesse Hughes (coarsely) replied: Kiss my fat, hairy ass, you stupid pig.As any reader who cares enough to follow your link will find, I> responded as above in reaction to an insult. I'm not ashamed or> embarrassed by that response. Hell, the invitation still stands....to claim that computation is connec with reality is either> ridiculous or...meaningless.> --Likewise, I am not ashamed or embarrassed by the quotation above. I> still stand by those words. Computations have no a priori connection> to reality in any meaningful sense. Of course, I never really knew> in what sense the other chap intended the claim to be taken. Seems> like empty words to me.You're an expert on those.This brings me to the question: Whatever are you doing in this> newsgroup providing these quotes? Is it retribution because I> include your quotation as my .sig? You've never asked me to refrain> from that or disavowed either the sentiment the quotation expresses or> its form here. Do you expect to somehow shame me into admitting that> logicians are bland followers of modern dogma and that there really> *are* things which are not identical to themselves? That John Correy> is right and logicians, osophers and mathematicians deny that> there are non-self-identical things just because they are ashamed of> their failures?No logician would conflate (as you have) a logic with the domains inwhich it applies. Nor would anyone other than you suggest thatquanta don't exist. You wait for it. But, until I come around and admit the brilliance of your insights,> why don't you leave the Slackware group alone. You can browbeat me on> more relevant newsgroups. I know, I know, the justness of your cause> and the gravity associa with battling the powerful and evil> mathematical community necessitates the use of any means available.> But, really, I don't think alt.ols.linux.slackware gives a rat's ass.--John...to claim that computation is connec with reality is eitherridiculous === a(g)p(g)art mod(ul)e> ...to claim that computation is connec with reality is either> ridiculous or...meaningless.> --Why ridiculous or meaningless, rather than trivial or false? And with the aphorism being so compressed, how do know the yah-sayersand nay-sayers aren't talking at === Could you please check the following? I have comple the rest of myassignment in the 'counting exercises' but i am stuck on from the set [1,2,...,n] where n is a positiveinteger, to the set 0,1ANSWER:An infinite number of sequences exist. For example you could map ever evennumber to 0 and every odd number to 1However I do not think this is correct and feel Im === missing something.Subject: Re: Please check 'counting' comple the rest of my> assignment in the 'counting exercises' functions are there from the set [1,2,...,n] where n is apositive> integer, to the set 0,1 ANSWER:> An infinite number of sequences exist. For example you could map evereven> number to 0 and every odd number to 1 However I do not think this is correct and feel Im missing something.>Look at it this way - for each number in the set there are 2 choices forwhere it can go, 0 or 1. Now count permutations.2 chioces for 1 x 2 choices === figured out the pattern of 2^n. Could you please check the following? I have comple the rest of my> assignment in the you!!!! QUESTION:> How many functions are there from the set [1,2,...,n] where n is a> positive> integer, to the set 0,1 ANSWER:> An infinite number of sequences exist. For example you could map ever> even> number to 0 and every odd number to 1 However I do not think this is correct and feel Im missing something.> Look at it this way - for each number in the set there are 2 choices for> where it can go, 0 or 1. Now count permutations.> 2 chioces for 1 x 2 choices for 2 x . . . x 2 === answer...> Could you please check the following? I have comple the rest of my> assignment in the 'counting exercises' but i am are there from the set [1,2,...,n] where n is a positive> integer, to the set 0,1ANSWER:> An infinite number of sequences exist. For example you could map ever even> number to 0 and every odd number to 1> To get an understanding of what's going on, pick a particular n. For example, how many === Please check 'counting' answer... Visiting Assistant Professor at the University of Montana.> Could you please check the following? I have comple the rest of my>assignment in the QUESTION:>How many functions are there from the set [1,2,...,n] where n is a positive>integer, to the set 0,1ANSWER:>An infinite number of sequences exist.True but irrelevant. You are not asked for the number of sequences,you are asked for the number of functions from the set {1,2,...,n} tothe set {0,1}. Your answer will depend on n, which is a FIXED butunspecified positive integer. So, for example, when n=1, you need the function from {1} to {0,1}(two of them). When n=2, the functions from {1,2} to {0,1} (four ofthem); when n=3, the function from {1,2,3} to {0,1} (eight ofthem). === Java Math packages. A lot seem to have overlapingfeatures, but I am not an expert. Since, I am not an expert, I would likerecommandations on what to use: I would like the union of features but theintersection of packages. Imagine for example that I would like to build upa general math API, or extend the Java language.So I really would like to know what combination of packages is reques tohave all features.By the way, here is my list of links. If you know some more, let me know.alphaWorks Numerically Intensive JavaalphaWorks BigDecimalColtdeveloperWorks Open sourcehttp--jmat.sourceforge.net-Introductory Java for Scientists and Engineers - downloadsJUnitJSR-000013 Decimal Arithmetic Enhancementjscl-meditor - java symbolic computing library and mathematical editorJavaMath API - WelcomeJava Numerics MainJava Numerical ToolkitJava Mathematics Library -Java for High Performance ComputingJava AppLib Approximation Library for JavaLinear Algebra for Statistics Java PackageMathLib sitemathExpr - Java Tools for Experimental MathematicsMathDL - Mathematical Sciences Digital LibraryOperations Research - Java ObjectsojAlgo - Object-Orien Java Algorithms for Mathematics, Linear Algebra andOptimisation - jama jampack optimatikaPolyMath-OpenMath (January 26, 2000)SourceForge.net Project Info - JUMP Ultimate Math PackageSingular Systems - JEP - Java Math Expression ParserThe randomX Package for JavaThe Probability-Statistics Object LibraryVisual Numerics and the Java Grande ForumVisual Numerics - Developers of IMSL and PV-WAVE(copy paste the title in Google to have the actual link. I don't have timeto build up === Convergence of limit> How does one prove that Limit[(1+(1/x))^x,x->Infinity] actually converges to> a finite number. Just defining e to equal this limit is not enough I> think--you need to actually show that it actaully converges to a number, and> define that number to be e. Is there some way to put an upper bound on the> function as x->Infinity, like (1+(1/x))^x < 3. Like I know with the> Maclaurin series of e (1+(1/1!)+(1/2!)+...+), you can massage the expression> a bit so that it's less than (1+1+1/2+1/4+1/8+...) so it will be definitely> converge, but without knowing the derivative of E^x or anything of the sort,> how do you prove the limit definition of e converges.Look in Heinrich Dorrie's 100 Great problems of Elementary Mathematics, published by Dover.A very interesting book.Accessable to smart high school === prove that Limit[(1+(1/x))^x,x->Infinity] actually converges to>a finite number. Just defining e to equal this limit is not enough I>think--you need to actually show that it actaully converges to a number, and>define that number to be e. Is there some way to put an upper bound on the>function as the>Maclaurin series of e (1+(1/1!)+(1/2!)+...+), you can massage the expression>a bit so that it's less than (1+1+1/2+1/4+1/8+...) so it will be definitely>converge, but without knowing the derivative of E^x or anything of the sort,>how do you prove the limit definition of e converges.We can start with showing that (1+a/n)^n is a monotonically increasing function of n by induction if a > 0. using thebinomial theorem, proved by induction itself. Also, if u > 1,u^v is monotonically increasing in v. In fact, how does onedefine u^v for irrational v except by using limit? Or evenin general for rational v? There is no problem in showing that (1+a/n)^n converges tothe series for exp(a), which can be directly shown to converge,and is a term-by-term upper bound. There are lots of ways tocomplete the proof.Also, if -x < a < 0, one can prove directly that the negativebinomial series for (1+a/x)^{-x} converges, that it is amonotonically decreasing function of x, and that it dominates(1-a/n)^n, which is another way to get an upper bound. Again,standard epsilontics completes the proof.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: === enough?The traditional way to do this is to show that (1+1/n)^n is monotoneincreasing and (1+1/n)^(n+1) is monotone decreasing. Since the latteris always greater than the former, it follows that (1+1/n)^n is boundedabove and (1+1/n)^(n+1) is bounded below, hence that each converges(and obviously to the same limit since one is 1+1/n times the other).This can be done entirely without logs or any reference to theexponential function.--Ron Bruck> This seems easy if we restrict x to natural numbers (Bernoulli inequality> works fine), but for real x I can't solve this. I haven't taken real/complex> analysis yet, but if anything from this helps, feel free to post it> How does one prove that Limit[(1+(1/x))^x,x->Infinity] actually converges> to> a finite number. Just defining e to equal this limit is not enough I> think--you need to actually show that it actaully converges to a number,> and> define that number to be e. Is there some way to put an upper bound on the> function as Maclaurin series of e (1+(1/1!)+(1/2!)+...+), you can massage the> expression> a bit so that it's less than (1+1+1/2+1/4+1/8+...) so it will be> definitely> converge, but without knowing the derivative of E^x or anything of the> sort,> how do you prove the limit definition of e === monotonically increasing right?It is, but how do you prove that if we're starting from scratch and not using e^x, ln(x), their derivatives and all that stuff? But if you have a limit L through integer values, note that if n < x < n+1, where n is a positive integer, then (1 + 1/(n+1))^n < (1 + 1/x)^x < (1 + 1/n)^(n+1).It's easy to see that the left and right hand terms -> L as n -> oo, and so (1 + 1/x)^x is dragged along to L as === opposite of a projection, but> two or more of them is a perspectivity,> viz painting (Brunelleschi et al).> the real meet (sik) of this is just in ten's complimentation,> in the base of ten. or nine's complimentation. that is,> when you subtract a larger number from a smaller,> this is whta you get, til you do the complimentation> to get the negative number. > Yet, in a way, Ord equals i, and Ord is very similar to zero. Ord is> the order type of all ordinals, in a way, infinity. What's the> opposite of a projection?--les ducs d'Enron!How about intention, the opposite of a projection being an intention,or perhaps, rather, intension, yet that's the opposite of extension.We might consider the opposite of projection antijection, but it isperhaps contraction.The range of a 2's complement signed register of length p+1 is[-(2^p0, (2^p)-1].I think another method of binary signed scalar representation iscalled 1's complement and the range is [-(2^p)+1, (2^p)-1], with zerorepresen by two values.I'm reading something and it says the two zeros, additive identityelements, of 1's complement arithmetic are all zeros and all ones, anegative number in 1's complement arithmetic is the bitwise complementof the positive number.I was thinking of another form of storage where the two zeros are allbits zero or a one bit followed by all zero bits, msb-to-lsb, sign bitarithmetic.Most computers implement 2's complement arithmetic.Now, we don't have any physical computers with infinite lengthregisters. However, we can imagine them.One thing to consider is a finite length word, and then the number iseither the positive integer as expressed or the negative integer beingthe the negation of the the maximum possible value minus the value ofthe number, depending on the extra sign bit, where negative one is allon bits and zero is all off bits, 2's complement.http://www.rwc.uc.edu/koehler/comath/12.htmlThe 2's complement is then obtained by simply adding 1 to the 1'scomplement...In the context of ordinals, my consideration is to consider thedefinition of predecession beyond a limit ordinal, in this case zero,often represen by the empty === compliment is the binary equivalent of ten's complimentation (decimal;2-sub-10 is 10-sub-2, so to say.subtract ...000011. from ...0000. for the uncomplimen, infinite representationof negative three (in base-two);subtract the result from ...1111. to get the one's compliment, andadd 1 to get -11 -- I guess. > The range of a 2's complement signed register of length p+1 is> [-(2^p0, (2^p)-1].I think another method of binary signed scalar representation is> called 1's complement and the range is [-(2^p)+1, (2^p)-1], with zero> represen by two values.I'm reading something and it says the two zeros, additive identity> elements, of 1's complement arithmetic are all zeros and all ones, a> negative number in 1's complement arithmetic is the bitwise complement> of the positive number.I was thinking of another form of storage where the two zeros are all> bits zero or a one bit followed by all zero bits, msb-to-lsb, sign bit> arithmetic.Most computers implement 2's complement arithmetic. > http://www.rwc.uc.edu/koehler/comath/12.html> The 2's complement is then obtained by simply adding 1 to the 1's> === ...9999.9999... = 0 ?viz, subtract three from ...0000. to get ...9997.;subtract the result from ...9999. to get the 9's Comp.;add one to get the 10's Comp. and put a minus sign in front. > the real meet (sik) of this is just in ten's complimentation,> in the base of ten. or nine's complimentation. that is,> when you subtract a larger number from a smaller,> this is whta you get, til you do the complimentation> to get the negative number.--les dics === <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_You know, Euler did some mighty strange things with infinite series,back then before convergence was understood. You can find statementslike Sum(n=-infinity to infinity) x^n = 0actually sta by him.