mm-2589 === Subject: Re: Theoretical/Rigorous precalculus textbook recommendation(s) > As for Algebra, etc., I don't know of anything targeted to a HS > audience. There are undergraduate texts that I could recommend once > you are comfortable with the concept of formal proofs, but it doesn't > sound like you're ready for them yet. recommendations. === Subject: Re: Theoretical/Rigorous precalculus textbook recommendation(s) These are old, but good Halmos, Finite Dimensional Vector Spaces Apostol, Mathematical Analysis If your precalculus curriculum includes set theory, try Halmos, Naive Set Theory Does anybody have a suggestion for an accessible text on synthetic Geometry? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: are rational exponents defined for negative bases? >Are rational exponents defined for negative bases? > Not generally, at least for real numbers. With complex numbers, you > can always define them by using the complex natural logarithm and > complex natural exponential. >Is this why we get extraneous solutions when dealing with radical >equations? > I'm not sure what you mean here... But one reason one often gets > 'extraneous' solutions is that when you do things like square an > equation, you are performing a non-reversable operation: that is, the > function that maps every real number to its square has no inverse. So > while it is true that if x = y then x^2 = y^2, it is not true that if > x^2 = y^2 then x=y (as opposed to, for example, the fact that if x=y > then x+z = y+z, AND if x+z = y+z then x=y). Another reason could be that any of the original stuff under any of the roots end up being negative, which isn't allowed. === Subject: Re: ? what is the x maximizing norm(b) where b = A*x >>Re-state and put more constraints on my question. >>Given a N-by-N self-adjoint square matrix A and a N-by-1 unit vector x. >>Let c = max( norm(A*x) ) >>Questions: >>(1) Does c exist? If does, c = ? >>(2) Will max( eigenvalue(A) ) = max( norm(A*x) )? >>(3) What is the x that maximizes norm(A*x)? > Your question makes no sense as stated. If A and x are given, > then so is norm(A*x). What variable are you maximizing over? > I suspect you mean that A is given, but x is not, so you want > the maximum over all unit vectors x. Then c is the maximum of > the absolute values of the eigenvalues of A, and the maximum > is attained if x is an eigenvector for an eigenvalue that > has the maximum absolute value. Aye, that is what I meant. The only restriction for x is unit length. I also guessed the solution should be the e-vector corresponding to the max e-value, but cannot prove it. How to prove the max e-vector is the solution? by Cheng COsine Dec/01/2k5 NC === Subject: Re: ? what is the x maximizing norm(b) where b = A*x >Re-state and put more constraints on my question. >Given a N-by-N self-adjoint square matrix A and a N-by-1 unit vector x. >Let c = max( norm(A*x) ) >Questions: >(1) Does c exist? If does, c = ? >(2) Will max( eigenvalue(A) ) = max( norm(A*x) )? >(3) What is the x that maximizes norm(A*x)? >> Your question makes no sense as stated. If A and x are given, >> then so is norm(A*x). What variable are you maximizing over? >> I suspect you mean that A is given, but x is not, so you want >> the maximum over all unit vectors x. Then c is the maximum of >> the absolute values of the eigenvalues of A, and the maximum >> is attained if x is an eigenvector for an eigenvalue that >> has the maximum absolute value. >Aye, that is what I meant. The only restriction for x is unit length. >I also guessed the solution should be the e-vector corresponding to >the max e-value, but cannot prove it. How to prove the max e-vector >is the solution? Hint: a self-adjoint matrix has an orthonormal basis of eigenvectors. Calculate the norm of Ax if x is a linear combination of these basis vectors. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: ? what is the x maximizing norm(b) where b = A*x > I suspect you mean that A is given, but x is not, so you want > the maximum over all unit vectors x. Then c is the maximum of > the absolute values of the eigenvalues of A, and the maximum > is attained if x is an eigenvector for an eigenvalue that > has the maximum absolute value. >>Aye, that is what I meant. The only restriction for x is unit length. >>I also guessed the solution should be the e-vector corresponding to >>the max e-value, but cannot prove it. How to prove the max e-vector >>is the solution? > Hint: a self-adjoint matrix has an orthonormal basis of eigenvectors. > Calculate the norm of Ax if x is a linear combination of these basis > vectors. Let b = A*x, max( norm(b) ) is equiv to max( norm(b)^2 ) Then norm(b)^2 = x'*A'*A*x = x'*V*D*V'*V*D*V'*x where D is diagonal matrix with e-values in its diagonal and V is the corresponding ematrix that are constructed with columns of normalized e-vector. So norm(b)^2 = x'*V*D*D*V'*x = D^2, has nothing to do with x. :( What's wrong? by Cheng Cosine Dec/01/2k5 NC === Subject: Re: ? what is the x maximizing norm(b) where b = A*x >> >> I suspect you mean that A is given, but x is not, so you want >> the maximum over all unit vectors x. Then c is the maximum of >> the absolute values of the eigenvalues of A, and the maximum >> is attained if x is an eigenvector for an eigenvalue that >> has the maximum absolute value. >> >Aye, that is what I meant. The only restriction for x is unit length. >I also guessed the solution should be the e-vector corresponding to >the max e-value, but cannot prove it. How to prove the max e-vector >is the solution? >> Hint: a self-adjoint matrix has an orthonormal basis of eigenvectors. >> Calculate the norm of Ax if x is a linear combination of these basis >> vectors. > Let b = A*x, max( norm(b) ) is equiv to max( norm(b)^2 ) > Then norm(b)^2 = x'*A'*A*x = x'*V*D*V'*V*D*V'*x where D is diagonal > matrix with e-values in its diagonal and V is the corresponding ematrix > that are constructed with columns of normalized e-vector. > So norm(b)^2 = x'*V*D*D*V'*x = D^2, has nothing to do with x. :( > What's wrong? The last =. === Subject: Re: ? what is the x maximizing norm(b) where b = A*x > Hint: a self-adjoint matrix has an orthonormal basis of eigenvectors. > Calculate the norm of Ax if x is a linear combination of these basis > vectors. I knew the expression of A*x can be A*x = V*D*V'*x = sum( ( landa(n)*v(n)'*x )*v(n), n = 1,...N ), v(n) are e-vectors and landa(n) are e-values. and that's why I guessed the e-vector corresponding to the max e-value might be the answer. But did not see why. === Subject: Re: finite product of closed sets is not closed >>message >> to show that a finite product of closed maps is not closed I need to >> find a counter example. assume f1:X1->Y1 is closed. that is If U in X1 is closed then f1(U) in >> Y1 is closed. >> assume f2:X2->Y2 is closed. that is If U in X2 is closed then f2(U) in Y2 is closed. >> I need to show that f:X1xX2->Y1xY2 defined by f(x1,x2)=(f1(x1),f2(x2)) is not closed . that is If U in X1xX2 is closed then f(U) in Y1xY2 is not closed. > Hint: The set {(1/n, n) : n = 1, 2, ...} is closed in R^2, but > its projection into the x-axis is not. >>projection_on_x1(x1,x2)=f1(x1). >>but f1 and f2 are assumed to be closed. that is the projections on x1 and >>x2 >>should be closed. it is the set { (f1(x1),f2(x2)) in U} that should be not >>closed. > No, you didn't understand. The set Wade gave is U. > Now find closed maps f1 and f2 such that > {(f1(x1), f2(x2)): (x1,x2) in U} is not closed. I don't see how does that help me. What I am looking for is a set V in Y1xY2 that is not closed and whose projektions are closed. === Subject: Re: finite product of closed sets is not closed >message >> > to show that a finite product of closed maps is not closed I need to > find a counter example. > > assume f1:X1->Y1 is closed. that is If U in X1 is closed then f1(U) in Y1 is closed. > assume f2:X2->Y2 is closed. that is If U in X2 is closed then f2(U) in > Y2 is closed. > I need to show that f:X1xX2->Y1xY2 defined by f(x1,x2)=(f1(x1),f2(x2)) > is not closed . that is If U in X1xX2 is closed then f(U) in Y1xY2 is not closed. >> Hint: The set {(1/n, n) : n = 1, 2, ...} is closed in R^2, but >> its projection into the x-axis is not. >projection_on_x1(x1,x2)=f1(x1). >but f1 and f2 are assumed to be closed. that is the projections on x1 and >x2 >should be closed. it is the set { (f1(x1),f2(x2)) in U} that should be not closed. No, you didn't understand. The set Wade gave is U. >> Now find closed maps f1 and f2 such that >> {(f1(x1), f2(x2)): (x1,x2) in U} is not closed. >I don't see how does that help me. What I am looking for is a set V in Y1xY2 >that is not closed and whose projektions are closed. Maybe so, but that's not what you _should_ be looking for. === Subject: Re: Hard problem Putnam contest in the US. People are > asked to refrain from posting solutions until the westernmost test-takers > are finished late in their afternoon.... > Fair enough; but is there some way to recognize what is a Putnam > problem? > Ken Pledger. A necessary condition is a math problem concisely stated and baffling. To be sure, one needs access to a Robert Israel oracle. If a polished solution pops out before you can get your fingers off the keyboard, then it's a Putnam problem. :-) -- chip === Subject: Re: Hard problem > three _positive_ real numbers a,b,c with > ab + bc + ca = 3 > prove that: > a^3 + b^3 + c^3 + 6abc >= 9 > Any hints how to do that? A hint: Lagrange Multipliers. The answer falls right out. === Subject: Re: Hard problem > three _positive_ real numbers a,b,c with > ab + bc + ca = 3 > prove that: > a^3 + b^3 + c^3 + 6abc >= 9 > Any hints how to do that? > -- > Peper Yes - Use the Rusin method to express it as a sum of squares. Should be pretty straightforward (chortle). === Subject: Re: Hard problem >> .... >> PS -- Saturday Dec 3 is the 2005 Putnam contest in the US. People are >> asked to refrain from posting solutions until the westernmost test-takers >> are finished late in their afternoon.... > Fair enough; but is there some way to recognize what is a Putnam >problem? > Ken Pledger. Maybe declare a moratorium on Dec 3 for posting any hint or solution to anything that has the flavor of a Putnam problem. quasi === Subject: Re: list of maths books? 12/01/2005 >Is there anywhere where I can download a list of all mathematics >books that have been published in the last few years? Try the Library of Congress. Even there you probably won't get a complete list. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Homomorphism mappings <438ddc1c$0$1570$892e7fe2@authen.yellow.readfreenews.net> <438e0947$22$fuzhry+tra$mr2ice@news.patriot.net> In , on 11/30/2005 at 08:38 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >But why use header forgery to send copies of my messages without any >alteration? I never expect rational reasons for such behavior, and have neither the data nor the training to make a psychiatric diagnosis. I've seen the same sort of thing on other news groups. >Turns out they are not coming from MathForum, but from >ReadFreeNews.com. They might stick something in the header to identify the source, e.g., NNTP-Posting-Host, in which case you might want to complain to them as well. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Help with Taylor series, Fokker-Planck derivation Bit rusty with Taylor Series but.. If y = a and then x = a; x = y. In 2. and 3. you are multiplying by (y-a) then changing that to from 2. to 3. (y-x) which is zero, will negate all terms in the series after the first. > I have a function of two independent variables f(x,y) and I want to expand > f(x,y) in a Taylor series wrt y about x > The correct way seems to be to first let x-y = r so > 1. f(x,y) = f(y+r,y) = f(x+r,x) + (y-x)(d/dx)f(x+r,x) + > (1/2!)(y-x)^2(d^2/dx^2)f(x+r,x)+... > Read (d/dx) as partial > However I'm tempted to do either of the following: > 2. expand wrt y about a > f(x,y) = f(x,a) + (y-a)(d/da)f(x,a) + (1/2!)(y-a)^2(d^2/da^2)f(x,a)+... and > then let a = x > 3. f(x,y) = f(x,x) + (y-x)[(d/ds)f(x,s)] |_s=x + > (1/2!)(y-x)^2[(d^2/ds^2)f(x,s)] |_s=x +... > But I can't explain to myself why 1 is valid and why 2 and 3 aren't valid. > The problem comes up in a basic derivation of the Fokker-Planck equation > where the integrand is expanded > p(x,t+/Delta t) = /int_-/infty^/infty W(x|y)p(y,t)/,dy > let r = x - y = /Delta x and dr = -dy > = /int_-/infty^/infty W(y+r|y)p(y,t)/,dr (this is expanded as in 1 > above) > = /int_-/infty^/infty W(x|x-r)p(x-r,t)/,dr (if we really wanted to > integrate wrt r) > all of which is a little confusing since wrt the integration x is a constant > and r depends linearly on y, and yet with regard to the taylor expansion, y > and r are treated as independently varying which they are since x and y are > independent. === Subject: Re: Help with Taylor series, Fokker-Planck derivation (sorry for top post) I've decided that cases 2 and 3 amount to the same thing and in fact are valid expressions for the Taylor series as is case 1. The problem occurs that in the derivation of Fokker-Planck I was studying one needs to write the integral (from -infty to infty) of the infinite sum (the taylor series) as an infinite sum of integrals with the same bounds which isn't valid with case 3. They diverge or aren't well defined. Trying to commute three limiting processes. > Bit rusty with Taylor Series but.. > If y = a and then x = a; x = y. > In 2. and 3. you are multiplying by (y-a) then changing that to from 2. > to 3. (y-x) which is zero, will negate all terms in the series after > the first. >> I have a function of two independent variables f(x,y) and I want to >> expand >> f(x,y) in a Taylor series wrt y about x >> The correct way seems to be to first let x-y = r so >> 1. f(x,y) = f(y+r,y) = f(x+r,x) + (y-x)(d/dx)f(x+r,x) + >> (1/2!)(y-x)^2(d^2/dx^2)f(x+r,x)+... >> Read (d/dx) as partial >> However I'm tempted to do either of the following: >> 2. expand wrt y about a >> f(x,y) = f(x,a) + (y-a)(d/da)f(x,a) + (1/2!)(y-a)^2(d^2/da^2)f(x,a)+... >> and >> then let a = x >> 3. f(x,y) = f(x,x) + (y-x)[(d/ds)f(x,s)] |_s=x + >> (1/2!)(y-x)^2[(d^2/ds^2)f(x,s)] |_s=x +... >> But I can't explain to myself why 1 is valid and why 2 and 3 aren't >> valid. >> The problem comes up in a basic derivation of the Fokker-Planck equation >> where the integrand is expanded >> p(x,t+/Delta t) = /int_-/infty^/infty W(x|y)p(y,t)/,dy >> let r = x - y = /Delta x and dr = -dy >> = /int_-/infty^/infty W(y+r|y)p(y,t)/,dr (this is expanded as in 1 >> above) >> = /int_-/infty^/infty W(x|x-r)p(x-r,t)/,dr (if we really wanted to >> integrate wrt r) >> all of which is a little confusing since wrt the integration x is a >> constant >> and r depends linearly on y, and yet with regard to the taylor expansion, >> y >> and r are treated as independently varying which they are since x and y >> are >> independent. === Subject: FLT case 1 Consider the following descent proof outline: 1. Assume there exists a smallest solution satisfying the FLT problem for any given prime exponent p and positive integers a,b,c. i.e., that c^p = a^p + b^p (abc,p) =1 Note: These solution values would also satisfy the congruence c^p == a^p + b^p (modp^2) 2. Suppose one could then show (as a contradiction) that a smaller such congruence solution must also exist: i.e. such that c'^p == a'^p + b'^p mod p^2, for values c'Consider the following descent proof outline: >1. Assume there exists a smallest solution satisfying the FLT problem >for any given prime exponent p and positive integers a,b,c. >i.e., that c^p = a^p + b^p (abc,p) =1 >Note: These solution values would also satisfy the congruence c^p == >a^p + b^p (modp^2) >2. Suppose one could then show (as a contradiction) that a smaller such >congruence solution must also exist: >i.e. such that c'^p == a'^p + b'^p mod p^2, for values c'Would this be a valid proof of FLT by contradiction? No. The assumption that (a,b,c) is the smallest solution to the equation doesn't imply that it's the smallest solution of the congruence. Hence if you find a smaller solution of the congruence, it's not a contradiction. === Subject: Re: FLT case 1 >Consider the following descent proof outline: >1. Assume there exists a smallest solution satisfying the FLT problem >for any given prime exponent p and positive integers a,b,c. >i.e., that c^p = a^p + b^p (abc,p) =1 >Note: These solution values would also satisfy the congruence c^p == >a^p + b^p (modp^2) >2. Suppose one could then show (as a contradiction) that a smaller such >congruence solution must also exist: >i.e. such that c'^p == a'^p + b'^p mod p^2, for values c'Would this be a valid proof of FLT by contradiction? > No. > The assumption that (a,b,c) is the smallest solution to the equation > doesn't imply that it's the smallest solution of the congruence. Hence > if you find a smaller solution of the congruence, it's not a > contradiction. Worse: strictly speaking, the expression c' < c has no meaning when you are working mod p^2, as there is no useful order relation to be put on the integers mod p^2. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: FLT case 1 >> >>Consider the following descent proof outline: >> >>1. Assume there exists a smallest solution satisfying the FLT problem >>for any given prime exponent p and positive integers a,b,c. >> >>i.e., that c^p = a^p + b^p (abc,p) =1 >> >>Note: These solution values would also satisfy the congruence c^p == >>a^p + b^p (modp^2) >> >>2. Suppose one could then show (as a contradiction) that a smaller such >>congruence solution must also exist: >> >>i.e. such that c'^p == a'^p + b'^p mod p^2, for values c'> >>Would this be a valid proof of FLT by contradiction? >> >> No. >> >> The assumption that (a,b,c) is the smallest solution to the equation >> doesn't imply that it's the smallest solution of the congruence. Hence >> if you find a smaller solution of the congruence, it's not a >> contradiction. >> Worse: strictly speaking, the expression c' < c has no meaning >> when you are working mod p^2, as there is no useful order relation >> to be put on the integers mod p^2. >> -- >> Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) >the situation a little better now. However, one problem stills nags at >me. Although I failed to explicitly mention this, I visualize >(intended) that step 2. in my statement is an iterative step in which >successively smaller values of a,b,c are obtained. That is, >c>c'>c''>c'''>.... etc., >The meaning of this behavior/result is what is not clear to me. It >seems that ultimately, the finite sequence must end, and the smallest >values for the congruence solution would be obtained. But what would >that mean? Would this be significant? >Donald Yes, it would be significant. If you could prove that given any positive solution (a,b,c) to the congruence, you could find another positive solution (a,b,c') with c>c', then you could iterate that, obtaining an infinite decreasing sequence of positive integers c>c'>c''>c'''>... But, since an infinite decreasing sequence of positive integers doesn't exist, the contradiction would imply that the congruence has no solution. If the congruence has no solution, then the equation also has no solution, thus proving case 1 of FLT. But, as far as I know, there are no known cases of an odd prime p and a positive integer m for which the congruence a^p+b^p=c^p mod m has no positive solutions But go ahead, check for yourself. You don't need a reduction technique. Just choose a small prime p such as p=3 or 5 or 7, and then simply test the congruence by trying all triples (a,b,c) with 0 But, as far as I know, there are no known cases of an odd prime p and > a positive integer m for which the congruence a^p+b^p=c^p mod m has no > positive solutions First of all, positive has no meaning in this context. Second, there are values of m and p for which the congruence has no non-zero solutions (e.g., a^3 + b^3 = z^3 mod 7) but there is a theorem of Schur from about 1918 to the effect that given p there are non-zero solutions to a^p + b^p = c^p mod q for all sufficiently large primes q. There are even quantitative versions of this theorem (something along the lines of q > p^4) which I think are due to Chowla. No one is going to prove Fermat, case 1 or otherwise, using congruences. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: FLT case 1 >> But, as far as I know, there are no known cases of an odd prime p and >> a positive integer m for which the congruence a^p+b^p=c^p mod m has no >> positive solutions >First of all, positive has no meaning in this context. Right. I meant positive in the set of reduced residues. I should have said nonzero rather than positive. >Second, there are values of m and p for which the congruence >has no non-zero solutions (e.g., a^3 + b^3 = z^3 mod 7) I wasn't aware of that. Ok, so then that proves that if there is a positive integer solution to a^3+b^3=c^3 (assuming we didn't already know that there are no such solutions), then for any such solution, one of a,b,c must be a multiple of 7. >but there is a theorem of Schur from about 1918 to the effect that >given p there are non-zero solutions to a^p + b^p = c^p mod q >for all sufficiently large primes q. That's a very nice result in its own right. Is the proof elementary in the sense of only requiring concepts of elementary number theory? Note that Schur's result, if you meant it precisely as stated, does not imply that for all sufficiently large m, the congruence a^p+b^p=c^p mod m has no solutions with a,b,c all nonzero. Not unless you can remove the qualification that q is prime. >There are even quantitative >versions of this theorem (something along the lines of q > p^4) >which I think are due to Chowla. Interesting, and possibly useful if planning to use congruences to try to prove FLT in an elementary way for a given p. >No one is going to prove Fermat, case 1 or otherwise, using congruences. That's a pretty strong statement -- too strong, I think, and I doubt that you can back it up. Can you even make the statement precise? I agree that it's highly unlikely that congruences alone can be used to show that there are no positive integer solutions a,b,c to a^p+b^p=c^p for a given odd prime p. But I wouldn't claim that it's impossible. Hypothetically, for a given prime p, there might be a set of congruences, possibly an infinite set, which have no simultaneous nonzero integer solutions. As a simple example, suppose, for a given odd prime p, there is an integer m>1 such that the congruence a^p+b^p=c^p mod m has no solutions with a,b,c all nonzero -- for example, p=3, m=7, the case you mentioned. But now let's also suppose that one could prove the more general result that for the given p and m, and any positive integer k, the congruence a^p+b^p=c^p mod m^k has no nonzero solutions. This would immediately prove that a^p+b^p=c^p has no positive integer solutions, thus proving FLT for the given prime p. Do I believe that for a given odd prime p, congruences alone could be used to prove FLT?. No, I don't believe it. Schur's result provides strong evidence supporting the claim that congruences alone are not enough. However, as far as I can see, a proof using congruences is still theoretically possible. quasi Supersedes: <438f37ae$0$1513$f69f905@mamut2.aster.pl> === Subject: Problem three _positive_ real numbers a,b,c with ab + bc + ca = 3 prove that: a^3 + b^3 + c^3 + 6abc >= 9 Any hints how to do that? -- Peper === Subject: Re: Problem > three _positive_ real numbers a,b,c with > ab + bc + ca = 3 > prove that: > a^3 + b^3 + c^3 + 6abc >= 9 > Any hints how to do that? > -- === Subject: Determinants of multi-dimensional matrices Hi there. I study Electrical Engineering (first year), and I am being bothered by by Linear Algebra issues lately. One of them is the three-(or more)-dimensional matrices. In Linear Algebra lessons, we only have to deal with two-dimensional matrices, but, in fact, there is no limitation in the number of dimensions, it could be anything. Yet, all definitions we have learnt are designed for two-dimensional matrices. One of them is the determinant. I am trying to discover on my own, how to compute the determinant of a NxNxN matrix, but I am baffled. Maybe I should ask if there IS such a thing as determinant of a three-dimensional matrix. Another one is the multiplication of three-dimensional matrices. All clear on how to multiply an MxN with an NxK matrix. But what about three dimensional matrices? No clue at all... Tero === Subject: Re: Determinants of multi-dimensional matrices >Hi there. I study Electrical Engineering (first year), and I am being >bothered by by Linear Algebra issues lately. One of them is the >three-(or more)-dimensional matrices. In Linear Algebra lessons, we >only have to deal with two-dimensional matrices, but, in fact, there is >no limitation in the number of dimensions, it could be anything. Yet, >all definitions we have learnt are designed for two-dimensional >matrices. One of them is the determinant. I am trying to discover on my >own, how to compute the determinant of a NxNxN matrix, but I am >baffled. Maybe I should ask if there IS such a thing as determinant of >a three-dimensional matrix. >Another one is the multiplication of three-dimensional matrices. All >clear on how to multiply an MxN with an NxK matrix. But what about >three dimensional matrices? No clue at all... You are, apparently, being taught Matrix Algebra, as a kind of applied algebra (presumably tailored to EE applications). Although it is not necessary to do so, it is *possible* to teach Matrix Algebra in a way that leaves both matrix multiplication and determinants nearly completely unmotivated, and presents them simply in terms of (mysterious) formulas. There are other ways to approach Linear Algebra, some algebraic, some geometric, in which it's much more difficult to avoid an early introduction of motivation for matrix multiplication, and in which it is at least possible (though it can be difficult, depending on the nature of the students) to introduce motivation for determinants more or less at the same time formulas for determinants are introduced. When multiplication and determinants are introduced purely in terms of formulas, it is *extremely* natural (and reasonable) for a student such as yourself (or me, though I was not yet an enrolled student of mathematics, merely a kid visiting a friend's uncle who was an engineer, and trying to make sense of some books on the shelves) to wonder how to multiply higher dimensional matrices, and how to evaluate their determinants. Unfortunately, there is no particularly reasonable way to do either of those things. Once you learn the motivation for matrix multiplication (in terms of linear transformations and their compositions) and determinants, it's much easier to see why the case of two dimensional matrices is the one where matrix multiplication and determinants make sense. Lee Rudolph === Subject: Re: Determinants of multi-dimensional matrices In , on 12/01/2005 at 07:57 PM, lrudolph@panix.com (Lee Rudolph) said: >Although it is not necessary to do so, it is *possible* to teach >Matrix Algebra in a way that leaves both matrix multiplication and >determinants nearly completely unmotivated, and presents them simply >in terms of (mysterious) formulas. Alas, it is not only possible but common. But you knew that :-( >When multiplication and determinants are introduced purely in terms >of formulas, it is *extremely* natural (and reasonable) for a student >such as yourself (or me, though I was not yet an enrolled student of >mathematics, merely a kid visiting a friend's uncle who was an >engineer, and trying to make sense of some books on the shelves) to >wonder how to multiply higher dimensional matrices, and how to >evaluate their determinants. Unfortunately, there is no >particularly reasonable way to do either of those things. Surely tensor products and traces are reasonable ways to define products[1]. Throw in a permutation matrix and you get an analog of the determinant. [1] Perhaps more than the OP wants ;-) -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Determinants of multi-dimensional matrices Please look at determinant of n x n x n array as well. Hero === Subject: determinant of n x n x n array merged with Determinants of multi-dimensional matrices cause anyone's threaded newsreader to explode. If it does--sorry!] >Please look at >determinant of n x n x n array >as well. >Hero I have, and I think it bears out my belief that there's nothing remotely as natural, for arrays with 3 or more indices, as matrix multiplication and determinant are for 2-by-2 arrays--at least provided that the things being defined are supposed to be in some way good analogues of multiplication and determinant. On the other hand, there might well be a way to extract the invariants of a 3-by-3-by-3 array that's more natural than whatever it is that Mathematica did for the carlos@colorado.edu when he let it run overnight. Not that it would (necessarily) lead to tractable computations, but there might--for instance-- be an interpretation of the invariants as certain integrals, analogous to the interpretation of the kth trace of an n-by-n real matrix (for k = 1, the trace itself, to k = n, the determinant) as an appropriate integral over a Grassmannian. See P. Eberlein, A trace formula, Linear & Multilinear Algebra 9, 231-236 (1980) (or, for the first trace, various posts in sci.math which can be found by a Google search for posts containing both trace and Rayleigh quotient). The key ingredient in such formulas does seem to be something like a Rayleigh quotient, and thus might appear to depend heavily on the interpretation of a matrix as the matrix of a linear transformation; but I think it could be rephrased in terms of contractions of tensors in a way that might more obviously generalize to higher-rank tensors. That's a wild guess, of course. Carlos, I don't suppose you could tell us explicitly what the five invariants *are*, without huge numbers of lines of Mathematica code or output? Something conceptual would be nice. (To start with, invariant under *what*? Since your 3-by-3-by-3 arrays come from partial derivatives, they presumably are highly symmetric. I presume there's some kind of action, like conjugation, by orthogonal 3-by-3 matrices, and that you're looking for functions on the set of suitably symmetric arrays that are a basis for the vectorspace of invariant functions of that action? Am I even warm?) Lee Rudolph === Subject: Re: determinant of n x n x n array merged with Determinants of multi-dimensional matrices > cause anyone's threaded newsreader to explode. If it does--sorry!] Look at determinant of n x n x n array Now, still better, just like entangling three strands of hair, or like building a rope from three smaller ones, one can entangle three threads. So Lee,what about looking at: Tim's golden Coordinates (polysigned news) too? The more weight one hangs on a rope, the tighter it gets. Hero === Subject: Re: determinant of n x n x n array Found a determinant rule for higher class matrices in Chapter XXIV of Muir's A Treatise on the Theory of Determinants, Dover, 1928. Class is defined there as number of dimensions of the matrix = number of indices in the entries. My problem is of class 3. According to Muir, Cayley (the inventor of matrix algebra) gave a determinant rule valid for even classes in 1843. An odd-class rule was proposed by Scott (another Englishman) in 1879. The number of terms for a n x n x ... n (k dimensions) Cayley-Scott determinant is (n!)^(k-1) I used Scott's rule to form a characteristic equation after making what seemed to be a good guess for a n x n x n identity matrix. But it did not produce the 5 correct invariants of the 3D second-gradient tensor. I had found those separately by letting Mathematica crunch overnight over all possible combinations. So higher class determinants appear so far to be just a curiosity like $2 bills. === Subject: Re: determinant of n x n x n array .... .... This is nice math, all about directions, and inversion, like left, right, inside out, up-down, linear and rotational... Now multiplication of 3x3x3 - think of rubiks cube for a start. Looking for the invariants in all this movements, it's not only the determinant, there's the spur = the sum of the diagonal elements. Just like 2x2x2. One example of opertaion on this is the 2x2-matrix with complex numbers. As i= sqrt ( - 1) = ( 0, 1) they can easily be regarded as 2x2x2, and the i is also giving some pre-invarianz on the operations in this case. With Clifford, one can play around with i*j= 1 , i*j= -1 ( and in 3D i*j = k, k*j = -1,...). There are lots of possibilities, some are interesting in real life. Like the 2x2 with complex numbers are iso to Quaternions. It's a pity i've little time today. Just one question: are there's basically two rubik cubes with orientation of rotation opposite , a left one and a right one, or only one ? With 3x3 Matrizes i have some results, like x w v w y u v u z One can relate for example x, y , z as scalars (numbers) to three axes (directions) in cartesian coordinates, and u as a vector to the x -, v to the y- and w to the z-axis. One example of this are the rotational matrices. See my website http://1iz.de ( the provider doing something with addresses so may be in a few days) where a bit is also about complex numbers. Now as for the invariants, one has of course an equation F of degree 3 in three variables x, y , z with a 3x3x3 matrix of coeffizients, the eigenvalues of which relate to the three values A,B,C in solving F( x, y z ) = 0=> (x-A) * ( y - B) * (z - C) = 0. (Sorry this is not to the point , as there is linear addition involved too, just an outline in which direction to look). And F differentiated three times partial into the three directions gives a 3x3x3 Matrix. Build the equation for eigenvalues and look for more conections. That's it for today have fun Hero PS Just look at it geometrically. === Subject: Re: determinant of n x n x n array > Now as for the invariants, one has of course an equation F of degree 3 > in three variables x, y , z with a 3x3x3 matrix of coeffizients, the > eigenvalues of which relate to the three values A,B,C in solving F( x, > y z ) = 0=> (x-A) * ( y - B) * (z - C) = 0. (Sorry this is not to the > point , as there is linear addition involved too, just an outline in > which direction to look). > And F differentiated three times partial into the three directions > gives a 3x3x3 Matrix. Build the equation for eigenvalues and look for > more conections. It seems to me, that not only the look at the third derivate might be of interest, but also a look at the 3x3x3-matrix, given by the product of the first and the second partial derivates: F ' or grad F= ( F_x, F_y, F_z) (where F_y is the partial derivate of F in the y-direction) F ' ' or 3x3 functional-matrix F_xx F_xy F_xz F_yx F_yy F_yz F_zx F_zy F_zz and the 3x3x3 -result: a matrix,with the element at a,b,c is F_a * F_b_c This is modelled after the torsion which is the cross-product of the first and second derivate of 3D-curve. I wonder how this all will come out, when put into action. Hero === Subject: Re: determinant of n x n x n array May be soon in calculators and math-programs one will have a 3D representation of 3x3x3 matrices with all operations of inverting, rotating, inside out, diagonalplane to the front and so forth - this should be usefull. Math written not on 2D paper or blackbord, but notated 3D. Hero === Subject: Re: involution has fixed point <20875368.1133428489785.JavaMail.jakarta@nitrogen.mathforum.org> of course, for linear maps it is trivial :-) === Subject: Re: equilateral tringles schrieb eugene : > Let ABC be an equilateral triangle. > Points A_1,B_1,C_1 are chosen inside the triangle in such a way > that A_1 in CC_1, B_1 in AA_1, C_1 in BB_1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ THE PROBLEM BECOMES MORE INVOLVED WHEN WE CHANGE THIS LINE INTO: that A_1 in BC, B_1 in CA, C_1 in AB > and AB_1=B_1A_1, BC_1=C_1B_1, CA_1=C_1A_1. > Prove that the triangle A_1B_1C_1 is also equilateral. HAVE FUN! T.M. === Subject: Re: equilateral tringles >> Let ABC be an equilateral triangle. >> Points A_1,B_1,C_1 are chosen inside the triangle in such a way >> that A_1 in CC_1, B_1 in AA_1, C_1 in BB_1 > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >THE PROBLEM BECOMES MORE INVOLVED WHEN WE CHANGE THIS LINE INTO: > that A_1 in BC, B_1 in CA, C_1 in AB >> and AB_1=B_1A_1, BC_1=C_1B_1, CA_1=C_1A_1. >> Prove that the triangle A_1B_1C_1 is also equilateral. I mentioned barycentric coordinates elsewhere in this thread. An equivalent formulation is to view the plane as the subset x+y+z = 1 in R^3; this allows us to use coordinates in a symmetrical way. Write A=(1,0,0), B=(0,1,0), C=(0,0,1), A1=(0,x,1-x), B1=(1-y,0,y), C1=(z,1-z,0) Then ( A B1 )^2 = 2 y^2 and ( B1 A1 )^2 = || ( y-1, x, 1-x-y ) ||^2 = (y-1)^2 + x^2 + ( (1-y)^2 - 2 x (1-y) + x^2 ), so if we set these equal to each other we get 2 (1-y)^2 = 2 x (1-y) so (1-y) = x, or equivalently x+y = 1. Likewise the other pairs sum to 1, so (summing the three equations) x+y+z = 3/2, and thus (subtracting the previous equations from this one) x=y=z=1/2. So A1, B1, C1 are the midpoints of the sides, giving a decomposition of ABC into four congruent equilateral triangles. Again this problem is presented symmetrically (which is nice) so the fact that the small triangle is equilateral follows from the fact that it is unique. How about a construction that has a degree of freedom in it, but still has the form, If ABC is equilateral and A1, B1, C1 are _any_ points in ABC such that [...] then A1 B1 C1 is also equilateral ? (I guess that could be sort of dual to Morley's theorem, which starts with arbitrary ABC and a unique A1 B1 C1 within it; here instead I'd like a unique ABC but a flexible choice of A1 B1 C1 within it, but still described symmetrically enough so that A1 B1 C1 is provably equilateral.) dave === Subject: RE: Roboto est mort @ et oui cal moe === Subject: Re: Cardinality of the surreals > ... > I'm not sure what surreals are, > > Conway's numbers from On Numbers And Games. Basically, (L,R) is a > number iff L and R are numbers, and no element of R is less than or > equal to an element of L. > That one is wrong. L and R are sets of numbers, not numbers (as the > remainder: no element clarifies). [...] yes, it should be sets of numbers. I must have been in one of my sleep-deprived states when I posted it. --- Christopher Heckman === Subject: Re: Cardinality of the surreals > ... > I'm not sure what surreals are, > > Conway's numbers from On Numbers And Games. Basically, (L,R) is a > number iff L and R are numbers, and no element of R is less than or > equal to an element of L. > That one is wrong. L and R are sets of numbers, not numbers (as the > remainder: no element clarifies). > to what i referr to. (3 | 4 ) is the surreal which given by the set of > all numbers smaller than 3 and the numbers 3 and 4 and all numbers > bigger than 4. > This one is basically right. But you should remember that on the left > and right of the vertical bar there is actually a set (in this case a > singleton). So actually the 3 above might stand for the set of all > numbers less than or equal to 3. > Definition: If L and R are two sets of surreal numbers and no member of > R is less than or equal to any member of L then { L | R } is a surreal > number. > This is indeed the correct definition. ........ > There is a single definition: given two sets of surreal numbers L and R > such that every element of L is less than or equal to every element of R, > the number {L|R} is a new surreal number. > However, to do that we need to define <=. > That is simple, given two surreals x1 = {L1|R1} and x2 = {L2|R2} we > say that x1 <= x2 if and only if there is (no x2 in R2 <= x1 and x2 > <= no x1 in L1). >......... problems, a book or a website never can do. Just to confirm, if i'm right now: The starting point is to define some surreals and as i want to talk about the ordinary reals as well i will differ them with an sor an r. advances to other surreal numbers like ( 0(s) | 1(s) ) or ( 1 (s) | 1 (s) ) of the form ( L | R ) , where R and L denote surreal numbers.The expression ( L | R ) is defined as a set of surreal numbers, actually a union of two sets of surreals which satisfy some conditions with respect to the ordering in these numbers. Now regarding the starting point, one can see, that the first numbers defined satisfy this defintion too. Now a set of surreal numbers of this form and with given conditions form a surreal number as well. That is not different to Peano in this respect, the natural numbers are sets of numbers as well, for example Peano 2 = { { } , { { } } } = { { } , 1 } } = { 0 , 1 }, all not surreal numbers. feel encouraged to try a better formulation of what i'm looking for. Is this an isomorphism between specially constructed sets of real numbers and surreal numbers: |R ----> 0 (s) Now basically rays of the real number line or given by { r , r element |R and r smaller or equal a , a element |R} ---> ( A | ) { r , r element |R and r bigger or equal b , b element |R} ---> ( | B ) Now |R with an open intervall cut out { r , r element |R and r smaller or equal d , d element |R} union with { r , r element |R and r bigger or equal g , g element |R} ----> ( D | G ) or does this goes wrong somewhere when coming to arithmetic ? Hero === Subject: Re: Cardinality of the surreals >> ... >> I'm not sure what surreals are, >> >> Conway's numbers from On Numbers And Games. Basically, (L,R) is >> a >> number iff L and R are numbers, and no element of R is less than >> or >> equal to an element of L. >> That one is wrong. L and R are sets of numbers, not numbers (as the >> remainder: no element clarifies). >> to what i referr to. (3 | 4 ) is the surreal which given by the set of >> all numbers smaller than 3 and the numbers 3 and 4 and all numbers >> bigger than 4. >> This one is basically right. But you should remember that on the left >> and right of the vertical bar there is actually a set (in this case a >> singleton). So actually the 3 above might stand for the set of all >> numbers less than or equal to 3. >> Definition: If L and R are two sets of surreal numbers and no member >> of >> R is less than or equal to any member of L then { L | R } is a >> surreal >> number. >> This is indeed the correct definition. > ........ >> There is a single definition: given two sets of surreal numbers L and R >> such that every element of L is less than or equal to every element of R, >> the number {L|R} is a new surreal number. >> However, to do that we need to define <=. >> That is simple, given two surreals x1 = {L1|R1} and x2 = {L2|R2} we >> say that x1 <= x2 if and only if there is (no x2 in R2 <= x1 and x2 >> <= no x1 in L1). >>......... > problems, a book or a website never can do. Just to confirm, if i'm > right now: The starting point is to define some surreals and as i want > to talk about the ordinary reals as well i will differ them with an > sor an r. > advances to other surreal numbers like ( 0(s) | 1(s) ) or ( 1 (s) | 1 > (s) ) of the form ( L | R ) , where R and L denote surreal numbers.The > expression ( L | R ) is defined as > a set of surreal numbers, actually a union of two sets of surreals > which satisfy some conditions with respect to the ordering in these > numbers. Now regarding the starting point, one can see, that the first > numbers defined satisfy this defintion too. Now a set of surreal > numbers of this form and with given conditions form a surreal number as > well. That is not different to Peano in this respect, the natural > numbers are sets of numbers as well, for example Peano > 2 = { { } , { { } } } = { { } , 1 } } = { 0 , 1 }, all not surreal > numbers. I wished he hadn't said this is basically right, because that basically covered a lot of ground, and it seems to be where your problems lie. Returning to the example you picked, {3|4} is in fact the same surreal as {{x<=3} | {x>=4}}, and indeed is the same as {{1,3}| {4,5,6}} as only the largest element of the Left set and the smallest element of the right set actually matter (treating x as a Real number). I use same as to mean they are part of the same equivalence class. You cannot simply treat L and R as rays and try to map this onto a topological type description of R. This is not how Surreals are constructed; it is a co-incidence that a ray interpretation works for {3|4}. This is because (amongst other reasons) it doesn't generalise to {{x is a Real} | } which is in fact w, the first infinite ordinal. If you limit yourself to Surreals that can be considered as unions of rays, you end up only considering Surreals which map directly to Reals, which means you are not considering Surreals, you are considering Reals. You lose things like infinite ordinals, infinestimals, and all the other things which make Surreals different from Reals. > feel encouraged to try a better formulation of what i'm looking for. > Is this an isomorphism between specially constructed sets of real > numbers and surreal numbers: > |R ----> 0 (s) > Now basically rays of the real number line or given by > { r , r element |R and r smaller or equal a , a element |R} ---> ( A | > { r , r element |R and r bigger or equal b , b element |R} ---> ( | > B ) > Now |R with an open intervall cut out > { r , r element |R and r smaller or equal d , d element |R} union with > { r , r element |R and r bigger or equal g , g element |R} ----> ( D | > G ) Its not a union of the left and right sets, the character | is a primitive term in Surreals which doesn't map onto union. You could re-formalise this as an ordered pair, but it doesn't add anything new, and the | is pretty intuitive. > or does this goes wrong somewhere when coming to arithmetic ? > Hero It works for those Surreals which are also Reals, but only by excluding those things that make Surreals surreal. This may help: http://www.tondering.dk/claus/sur15.pdf (A very easy, interesting and straightforward description of the construction of the Surreals). === Subject: Re: Cardinality of the surreals ... > {3|4} is in fact the same surreal as {{x<=3} | {x>=4}}, and indeed is the > same as {{1,3}| {4,5,6}} as only the largest element of the Left set and the > smallest element of the right set actually matter (treating x as a Real > number). I use same as to mean they are part of the same equivalence > class. This is true of course, if and only if the left set has a largest element and the right set has a smallest element. When that is not the case differences come up. For instance: {{x rational and x < 2}|{x rational and x >= 2}} defines a surreal that is not a real. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cardinality of the surreals > ... > {3|4} is in fact the same surreal as {{x<=3} | {x>=4}}, and indeed is > the > same as {{1,3}| {4,5,6}} as only the largest element of the Left set and > the > smallest element of the right set actually matter (treating x as a Real > number). I use same as to mean they are part of the same equivalence > class. > This is true of course, if and only if the left set has a largest element > and the right set has a smallest element. When that is not the case > differences come up. For instance: > {{x rational and x < 2}|{x rational and x >= 2}} > defines a surreal that is not a real. > -- That is almost exactly what I used as an example when I started in this thread: >Let S be the set of all surreal numbers. >Let L = {x: x is an element of S and x<1} >Let R = {x: x is an element of S and x>=1} >Then {L|R} is a surreal number which is not part of S. However, you state x is rational where I state (elsewhere) x is Real. I can understand it might make a difference if you were considering sqrt(2) as the limit point. You use 2 and I use 1, and in these cases at least the restriction on x as Rational seems unneccesary. === Subject: Re: Cardinality of the surreals ... > This is true of course, if and only if the left set has a largest element > and the right set has a smallest element. When that is not the case > differences come up. For instance: > {{x rational and x < 2}|{x rational and x >= 2}} > defines a surreal that is not a real. > That is almost exactly what I used as an example when I started in this > thread: >Let S be the set of all surreal numbers. > >Let L = {x: x is an element of S and x<1} >Let R = {x: x is an element of S and x>=1} > >Then {L|R} is a surreal number which is not part of S. > However, you state x is rational where I state (elsewhere) x is Real. I > can understand it might make a difference if you were considering sqrt(2) as > the limit point. You use 2 and I use 1, and in these cases at least the > restriction on x as Rational seems unneccesary. The difference is that my statements give a surreal number while your statements (here) do *not* give a surreal number because your L and R are not sets. On the other hand, if you state x is real you of course do get a surreal number. It makes not much difference whether we use real or rational, but you have to restrict L and R to sets. I missed to show that this works differently from Dedekind cuts (and I have the impression that Hero was confusing the two). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cardinality of the surreals <43903984$0$17707$afc38c87@news.optusnet.com.au> > to show that this works differently from Dedekind cuts (and I have the > impression that Hero was confusing the two). understanding. It's a challenge and a test for the math's we have for the reals and the geometry of a straight line, both related. We have the numbers (points), we have archimedes axiom, in modern form of Kuratowsky (between two points is always a set of reals) we have the numbers divided in algebraic and transfinite ( but like in pi: is the radius algebraic, then the circumference is transcendent, but when the radius is algebraic, the circumfence is transcendent). My interest is of course to see the surreals relate to this and see, if there's something new. As You introduced now (at least for me) the surreals as equivalence classes , this matches to: in the reals there are no intervalls [a, b] with a=b, that's a number or the set with this number ( one of these two, i'm not sure). And may be by this, one reduces to ordered pairs of numbers, no longer sets - or ordered pairs of sets with just one number in it Now i just see, one can write my interpretation in the form u ] , [ v as the two rays of numbers starting from u resp. v in opposite directions with u < v or u = v, an open intervall ( u, v ) taken out of |R. If u = v there's nothing taken out, so alll like this one intervall they form some kind of equivalence class of points. To make it short, one can expect from Conway, that he didn't invert logic for the surreals, so with aristotelian logic intact, the surreals should match to the reals, and sets of reals - or they can teach us something new about them. I have my reservation about surreal infinitesimal small - just like in the reals or the natural numbers - there's a difference between the natural numbers and all of them or the set of naturals. | N has of course the property, that each element in it has a successor, but this is gone in { 1, 2, 3 ,....., |N} (and { 1, 2, 3 ,....., |N} is not { { 1 }, { 2}, {3} ,....., |N} }) so here You have an infinitally (big) element, so the infinitally small element between two real numbers is an open intervall, which is zero, empty, as the reals on both sides of the infintesimal small are the same number. As Aristoteles said, A=A at the same time and in the same place. But everything which is not a point and not nothing, if it exists, exists with extension in space and/ or time, beeing different in it's points and parts to a certain extent, but beeing equal in it's parts and points, when going beyond ( 3+4 = 7 equal regarding the value, but both sides different - that is the fate of beeing an equation). Now i have to study more of these surreals to cope up with Your level of understanding, before i can talk about them. It was a pleasure for me, to read Your posts and even more to get Your responses, Hero === Subject: An: GuruGram #57 ... is newly available for free download ad http://www.tinaja.com/glib/ellipse4.pdf It is on Bezier cubic spline approximations to circles. And shows why earlier results published elsewhere are suboptimal. Additional gurugrams at http://www.tinaja.com/gurgrm01.asp Sourcecode at http://www.tinaja.com/glib/ellipse4.pdf Consulting services available via http://www.tinaja.com/info01.asp -- Don Lancaster voice phone: (928)428-4073 Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552 rss: http://www.tinaja.com/whtnu.xml email: don@tinaja.com Please visit my GURU's LAIR web site at http://www.tinaja.com === Subject: Re: Bug in Magma's Groebner() command It seems that our original response has not propagated, so here is a repost: > I would like to ask about an error I just encountered in Magma. > This post is both a bug-report/request-for-workaround and a > mathematical query: what ring-theoretic conditions caused this error? is a bug which only occurs when a Groebner Basis function is invoked on an ideal in a polynomial ring with field of coefficients GF(p), where p lies between 2^23.5 and 2^26. The bug has now been fixed and a patch version (V2.12-17) containing the fix is now available on the Magma download page. Note that there is a mailing address for reporting bugs to Magma: magma-bugs@maths.usyd.edu.au Directly reporting bugs frees up bandwidth on discussion groups and ensures that they are seen by the people who fix them. All bugs are assessed within 24 hours and the majority are fixed within 7 days. -- Magma Computer Algebra System Email: magma@maths.usyd.edu.au Web: http://magma.maths.usyd.edu.au/ Phone: +61 2 9351 3338 Fax: +61 2 9351 4534 === Subject: unique critical points The following result is easy to prove: Suppose f:R-->R is differentiable. If f has local minimum at x=a and no other critical values, then f has a global minimum at x=a. Next, consider real-valued functions of 2 variables. Suppose f:R^2-->R is differentiable. If f has local minimum at x=p and no other critical points, must f have a global minimum at x=p? I can see by visualization that the answer to the above question is no. Can anyone provide a concrete example? Suppose f is also required to be a polynomial? === Subject: Re: unique critical points Here is my idea: Define f(x,y)=x^2+y^2 for (x,y)in C, and =x+y+1 for (x,y) outside of C, where C is the closed disk centered at (1/2,1/2) with radius sqrt(3/2) --- the boundary of C is just the projection of the interesection of z=x^2+y^2 and z=x+y+1 projected onto the x-y plane. Certainly f has unique local min. at (0,0) which is not a global minimum. Now you just have to smooth the function on the boundary of C to make f a differentiable function with (0,0) as the only critical point. Exactly how to smooth the f? Use convolution? I leave this to others. > The following result is easy to prove: > Suppose f:R-->R is differentiable. If f has local minimum at x=a and > no other critical values, then f has a global minimum at x=a. > Next, consider real-valued functions of 2 variables. > Suppose f:R^2-->R is differentiable. If f has local minimum at x=p and > no other critical points, must f have a global minimum at x=p? > I can see by visualization that the answer to the above question is > no. > Can anyone provide a concrete example? > Suppose f is also required to be a polynomial? > quasi === Subject: Re: unique critical points Mathematics Magazine: Volume 58, Number 3, Pages: 149-150 1985 Ira Rosenholtz and Lowell Smylie First Paragraph: When searching for absolute extrema of function s of a single variable, it is often convenient to apply the well known Only Critical Point in Town Test: If f is a continuous function on an interval, which has a local extremum at x_0, and x_0 is the only critical point of f, then f attains an absoloute extremum at x_0. A natural question which arises is Is the corresponding statement true for functions of two variables (say defined over the enire plane)? Since our colleagues were evenly split on the question (both halves being quite adamant), and neither a proof nor a counterexample was readily available, we set to work trying to find one. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: unique critical points The function given there is not a polynomial. > Mathematics Magazine: Volume 58, Number 3, Pages: 149-150 > 1985 > Ira Rosenholtz and Lowell Smylie > First Paragraph: When searching for absolute extrema of function s of a > single variable, it is often convenient to apply the well known Only > Critical Point in Town Test: If f is a continuous function on an > interval, which has a local extremum at x_0, and x_0 is the only > critical point of f, then f attains an absoloute extremum at x_0. A > natural question which arises is Is the corresponding statement true > for functions of two variables (say defined over the enire plane)? > Since our colleagues were evenly split on the question (both halves > being quite adamant), and neither a proof nor a counterexample was > readily available, we set to work trying to find one. > -- > G. A. Edgar > http://www.math.ohio-state.edu/~edgar/ === Subject: Re: unique critical points at American Mathematical Monthly, but I don't recall when. > The following result is easy to prove: > Suppose f:R-->R is differentiable. If f has local minimum at x=a and > no other critical values, then f has a global minimum at x=a. > Next, consider real-valued functions of 2 variables. > Suppose f:R^2-->R is differentiable. If f has local minimum at x=p and > no other critical points, must f have a global minimum at x=p? > I can see by visualization that the answer to the above question is > no. > Can anyone provide a concrete example? > Suppose f is also required to be a polynomial? === Subject: Re: unique critical points > The following result is easy to prove: > Suppose f:R-->R is differentiable. If f has local minimum at x=a and > no other critical values, then f has a global minimum at x=a. > Next, consider real-valued functions of 2 variables. > Suppose f:R^2-->R is differentiable. If f has local minimum at x=p and > no other critical points, must f have a global minimum at x=p? > I can see by visualization that the answer to the above question is > no. > Can anyone provide a concrete example? { (x,y,x^2) | x <= 1 } and for x > 1, start tilting the y axis around the x-axis at (x,0,x^2). === Subject: Re: unique critical points -- Quotes from The Weather Man: Robert Spritz: Do you know that the harder thing to do, and the right thing to do, are usually the same thing? Easy doesn't enter into grown-up life... to get anything of value, you have to sacrifice. > The following result is easy to prove: > Suppose f:R-->R is differentiable. If f has local minimum at x=a and > no other critical values, then f has a global minimum at x=a. > Next, consider real-valued functions of 2 variables. > Suppose f:R^2-->R is differentiable. If f has local minimum at x=p and > no other critical points, must f have a global minimum at x=p? > I can see by visualization that the answer to the above question is > no. I dont > Can anyone provide a concrete example? > Suppose f is also required to be a polynomial? > quasi === Subject: A confusing integer programming problem! need help!! assume we have the following mixed integer prgramming problem: I can't figure out a way to express the linear constraint of Y and (a1,....an,b1....,bm, s are constants) Min a1* P1+a2*P2+...an*Pn+b1*X1+b2*X2+...+bm*Xm+ s*Y Suject to: Y= 1 if sum(P1+....Pn)>0; ( means that one or more Pi are 1 ) 0 if sum(P1+...+Pn) =0 ( means that all Pi are 0) some other linear constraints of X1.... Xn P1,....Pn are binary; X1,....Xn>=0; === Subject: Re: A confusing integer programming problem! need help!! > assume we have the following mixed integer prgramming problem: > I can't figure out a way to express the linear constraint of Y and > (a1,....an,b1....,bm, s are constants) > Min a1* P1+a2*P2+...an*Pn+b1*X1+b2*X2+...+bm*Xm+ s*Y > Suject to: > Y= 1 if sum(P1+....Pn)>0; ( means that one or more Pi are 1 ) Hint: you want to force Y to be positive (but < = 1) if any Pi is positive; once Y is forced to be positive, it must then equal 1. You also want to _allow_ Y to be zero if all Pi are zero. Later, in another constraint, you can _force_ Y = 0 when sum(Pi) = 0. So, think about a constraint of the form Y >= k(P1 + ... + Pn), with k some positive constant. How can you choose k > 0 so that Y is never forced to be > 1? Of course, when all Pi = 0, Y is allowed be be anything, so the constraint becomes redundant. > 0 if sum(P1+...+Pn) =0 ( means that all Pi are 0) Now you want to _force_ Y = 0 if all Pi = 0. This happens if you require Y <= sum(Pi). Note that if any Pi = 1 the right-hand-side is > = 1, so the constraint becomes redundant. R.G. Vickson > some other linear constraints of X1.... Xn > P1,....Pn are binary; X1,....Xn>=0; === Subject: Re: Why string theorists like Brian Greene are Crackpots === Subject: Re: Why string theorists like Brian Greene are Crackpots <438fc50c$0$18118$892e7fe2@authen.yellow.readfreenews.net> In <438fc50c$0$18118$892e7fe2@authen.yellow.readfreenews.net>, on 12/02/2005 said: >Then along with being an idiot with delusions of competence you are >unable to express yourself. Please fix your quoting style. You should be using >> for the text the idiot[1] is quoting, to distinguish it from his text, and should trim it to what is relevant. I understand that the idiot failed to prefix his quotes with >, but it would be fairly simple to fix it if you were trimming properly. [1] She/he/it seems to be one of the old style anti-Einstein kooks in new garb. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: New symmetry design tool > Hello. I'm producing a symmetry design tool called Honeycomb and you > can try it now at www.geocities.com/roninabox/Honeycomb.html. It's an > applet: you don't have to download and install anything. Basically, you > have grids, and a grid can be in any one of many different symmetry > types. You can put a grid into a grid for multiple symmetries. It > borrows from the Adobe interfaces and tools: there are bezier, pencil, > rectangle, and oval pens. I'd like some feedback so let me know what > you think and if this tool might be something you'd use in your > professional or academic environment. > Ron On my machine (Sun Ultra 80 running Solaris) clicking the text tool brings up a default font that can't be seen in the status bar. Must be an odd font. Others select OK. But I can't draw an text in the text box, so text tool is not usable. Also, I can't close the text tool - all menue items when I click the top left are grayed out. (Move, Close, Minimse etc). It's not something I'd use. There's no info about whether it is commercial, open source or whatever, PS. I don't think English or Metric is a good choice. Perhaps American and Metric, as America seems to have taken to metric far less than the rest of the world. England is pretty much all metric now - road signs being an obvious exception. (I guess Imperial or Metric is really correct, but American sounds good to me!!!) -- Dave K http://www.southminster-branch-line.org.uk/ It is always of the form: month-year@domain. Hitting reply will work for a couple of months only. Later set it manually. The month is always written in 3 letters (e.g. Jan, not January etc) === Subject: Re: New symmetry design tool <438fc98f@212.67.96.135> > [...] > PS. I don't think English or Metric is a good choice. Perhaps American > and Metric, as America seems to have taken to metric far less than the > rest of the world. Certain people who are in a position to officially change to metric don't want to change to metric. > England is pretty much all metric now - road signs > being an obvious exception. (I guess Imperial or Metric is really > correct, but American sounds good to me!!!) Imperial is the usual name for the non-metric system. --- Christopher Heckman === Subject: Re: Prime ideals, k[x,y] >Let k be algebraically closed and let m_1, ..., m_n be maximal ideals of >k[x,y]. Is there necessarily a prime ideal that is contained in every m_i? Equivalently: let P_1, ..., P_n be points in the plane k^2. Is there necessarily an irreducible variety V which contains each P_i ? Apply a linear transformation (x,y) -> (x',y) = (x + a y, y) so that the n points have distinct x'-coordinates. Use interpolation to find a polynomial f whose graph y=f(x') passes through each of the points. Then take V to be the zero-locus of y - f( x + a y ). dave === Subject: Re: Prime ideals, k[x,y] >>Let k be algebraically closed and let m_1, ..., m_n be maximal ideals of >>k[x,y]. Is there necessarily a prime ideal that is contained in every >>m_i? > Equivalently: let P_1, ..., P_n be points in the plane k^2. Is there > necessarily an irreducible variety V which contains each P_i ? > Apply a linear transformation (x,y) -> (x',y) = (x + a y, y) so that > the n points have distinct x'-coordinates. Use interpolation to find a > polynomial f whose graph y=f(x') passes through each of the points. > Then take V to be the zero-locus of y - f( x + a y ). > dave === Subject: Re: Prime ideals, k[x,y] >>Let k be algebraically closed and let m_1, ..., m_n be maximal ideals of >>k[x,y]. Is there necessarily a prime ideal that is contained in every >>m_i? > What's wrong with (0) ? I'm sorry I meant non-zero prime ideal. > -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Pascal triangle new property and quality? Sorry: aug. 9 10:21 bad or wrong and incorrect: triangele and tringle! Correction all: Pascal triangle new property? AAAAAAAABBBB (A: num 8 B: num 4) permutations and all comparison template: AAAAAAAABBBB layout here deformed:(primary:squareness - paralellogramm) 12 0; C(0,0)*C(12,4) 11 1 C(1,0)*C(11,4);C(1,1)*C(11,3) 10 2; C(2,0)*C(10,4);C(2,1)*C(10,3);C(2,2)*C(10,2) 9 3 C(3,0)*C(9,4);C(3,1)*C(9,3);C(3,2)*C(9,2);C(3,3)*C(9,1) 8 4; C(4,0)*C(8,4);C(4,1)*C(8,3);C(4,2)*C(8,2);C(4,3)*C(8,3);C(4,4)*C(8,0) 7 5; C(5,0)*C(7,4);C(5,1)*C(7,3);C(5,2)*C(7,2);C(5,3)*C(7,1);C(5,4)*C(7,0) 6 6; C(6,0)*C(6,4);C(6,1)*c(6,3);C(6,2)*c(6,2);C(6,3)*C(6,1);C(6,4)*C(6,0) 5 7; C(7,0)*C(5,4);C(7,1)*C(5,3);C(7,2)*C(5,2);C(7,3)*C(5,1);C(7,4)*C(5,0) 4 8; C(8.0)*C(4,4);C(8.1)*C(4,3);C(8.2)*C(4,2);C(8.3)*C(4,1);C(8.4)*C(4,0) 3 9; C(9,1)*C(3,3);C(9,2)*C(3,2);C(9,3)*C(3,1);C(9,4)*C(3,0) 2 10; C(10,2)*C(2,2);C(10,3)*C(2,1);C(10,4)*C(2,0) 1 11; C(11,3)*C(1,1);C(11,4)*C(1,0) 0;12 C(12,4)*C(0,0) All Rows (HORIZONTAL SUM) :495 C(12,4) or C(12,8) matching frequency or matching probability Left and right (or right and left): All OBLIQUELY SUM :715 C(13,4) or C(13,9) and All OBLIQUELY SUM :1287 C(13,5) or C(13,8) ETC... :ALL PASCAL triangle centralize mirror and multiplier.... I, if no new PASCAL triangle centralize mirror and multiplier search bibliography and author then PASCAL TRIANGLE NEW POPERTIES!!!??? http://matserv.pmmf.hu/matport/cikk.php?pelda_id=64 zerinvary quadrangles.ppt Please help and write! Zerinvary Lajos emeritus prof. Hungary Vasarhelyi Pal Technique School zerinvaryla...@yahoo.com === Subject: Re: Pascal triangle new property and quality? >Sorry: >aug. 9 10:21 bad or wrong and incorrect: triangele and tringle! >Correction all: >Pascal triangle new property? >AAAAAAAABBBB (A: num 8 B: num 4) permutations and all comparison >template: AAAAAAAABBBB >layout here deformed:(primary:squareness - paralellogramm) >12 0; >C(0,0)*C(12,4) >11 1 > C(1,0)*C(11,4);C(1,1)*C(11,3) >10 2; > C(2,0)*C(10,4);C(2,1)*C(10,3);C(2,2)*C(10,2) >9 3 > C(3,0)*C(9,4);C(3,1)*C(9,3);C(3,2)*C(9,2);C(3,3)*C(9,1) >8 4; >C(4,0)*C(8,4);C(4,1)*C(8,3);C(4,2)*C(8,2);C(4,3)*C(8,3);C(4,4)*C(8,0) >7 5; >C(5,0)*C(7,4);C(5,1)*C(7,3);C(5,2)*C(7,2);C(5,3)*C(7,1);C(5,4)*C(7,0) >6 6; >C(6,0)*C(6,4);C(6,1)*c(6,3);C(6,2)*c(6,2);C(6,3)*C(6,1);C(6,4)*C(6,0) >5 7; >C(7,0)*C(5,4);C(7,1)*C(5,3);C(7,2)*C(5,2);C(7,3)*C(5,1);C(7,4)*C(5,0) >4 8; >C(8.0)*C(4,4);C(8.1)*C(4,3);C(8.2)*C(4,2);C(8.3)*C(4,1);C(8.4)*C(4,0) >3 9; >C(9,1)*C(3,3);C(9,2)*C(3,2);C(9,3)*C(3,1);C(9,4)*C(3,0) >2 10; >C(10,2)*C(2,2);C(10,3)*C(2,1);C(10,4)*C(2,0) >1 11; > C(11,3)*C(1,1);C(11,4)*C(1,0) >0;12 > C(12,4)*C(0,0) >All Rows (HORIZONTAL SUM) :495 C(12,4) or C(12,8) matching >frequency or matching probability >Left and right (or right and left): >All OBLIQUELY SUM :715 C(13,4) or C(13,9) and >All OBLIQUELY SUM :1287 C(13,5) or C(13,8) >ETC... :ALL PASCAL triangle centralize mirror and multiplier.... >I, if no new PASCAL triangle centralize mirror and multiplier >search bibliography and author then PASCAL TRIANGLE NEW >POPERTIES!!!??? >http://matserv.pmmf.hu/matport/cikk.php?pelda_id=64 > zerinvary quadrangles.ppt >Please help and write! I am not quite sure what you are trying to say here, but it looks as if, in the horizontal sums, you are noting that --- C(n+m,k) = > C(n,j) C(m,k-j) [1] --- j However, this is well known and can be proven simply by looking at the coefficients of x^k in n+m n m (1+x) = (1+x) (1+x) In the oblique sums, --- C(n+m+1,k) = > C(n-j,n-k) C(m+j,m) [2] --- j --- = > C(n-j,k-j) C(m+j,j) [3] --- j These can be proven by summing against x^k and using the identity C(m+j,j) = C(m+j,m) = (-1)^j C(-m-1,j) That is, --- --- k > > C(n-j,k-j) C(m+j,j) x --- --- k j --- n-j j = > C(m+j,j) (1+x) x --- j --- j n-j j = > (-1) C(-m-1,j) (1+x) x --- j n x -m-1 = (1+x) (1 - --- ) 1+x n+m+1 = (1+x) [4] Simply compare coefficients of x^k in [4]. So the sums horizontally or diagonally are nothing new. If you were asking something else, I apologize for not understanding. If you ask again, I will try to answer again. Rob Johnson take out the trash before replying === Subject: Re: Pascal triangle new property and quality? <20051202.021758@whim.org> Rob! please:visage and consider Another : Rencontres umbers or don't rencontres numbers? (SORRY:Rencontres numbers or don't....) Lajos - dec. 2 07:14 Jul 08 2005 Sending in a Comment on an Existing Sequence to The On-Line Encyclopedia of Integer Sequences SINCE WANT!!! What's the problem? NUMERAL: %I A133581 %S A133581 0,6,24,54,96,150,216,294,384,4[CapitalEth]86,600,726,864,1014,1176,1350, 1536,1734,1944,2166,2400,2646,[CapitalEth]2904,3174,3456,3750,4056,4374, tov.87bb é 1 .9fzenet 1 szerzotol szerzo and RENCONTRES NUMBERS or SUBFACTORIAL OR DERANGEMENTS NUMBERS ... ... this is subfactorial or rencontres numbers or derangement numbers... TRIVIAL Formula: (WIKIPEDIA): Thus, if we write dn as the number of derangements of n ... aug. 2 11:10-n a k.9avetkezotol: Lajos - 1 .9fzenet - 1 szerzo Lajos === Subject: Re: Pascal triangle new property and quality? <20051202.021758@whim.org> please:visage and consider http://matserv.pmmf.hu/matport/cikk.php?pelda_id=64 zerinvary quadrangles.ppt or Google search: filetype:ppt zerinvary quadrangles Lajos === Subject: Re: Pascal triangle new property and quality? > please:visage and consider >http://matserv.pmmf.hu/matport/cikk.php?pelda_id=64 > zerinvary quadrangles.ppt > or Google search: >filetype:ppt zerinvary quadrangles When I finally found a way to view those slides (I don't have Power Point on my Mac), it appears that you are simply diagramming the same identities that I have pointed out. Is there more than I am seeing? Rob Johnson take out the trash before replying === Subject: Re: Pascal triangle new property and quality? > Sorry: > aug. 9 10:21 bad or wrong and incorrect: triangele and tringle! > Correction all: > Pascal triangle new property? > AAAAAAAABBBB (A: num 8 B: num 4) permutations and all comparison > template: AAAAAAAABBBB > layout here deformed:(primary:squareness - paralellogramm) > 12 0; > C(0,0)*C(12,4) > 11 1 > C(1,0)*C(11,4);C(1,1)*C(11,3) > 10 2; > C(2,0)*C(10,4);C(2,1)*C(10,3);C(2,2)*C(10,2) > 9 3 > C(3,0)*C(9,4);C(3,1)*C(9,3);C(3,2)*C(9,2);C(3,3)*C(9,1) > 8 4; > C(4,0)*C(8,4);C(4,1)*C(8,3);C(4,2)*C(8,2);C(4,3)*C(8,3);C(4,4)*C(8,0) > 7 5; > C(5,0)*C(7,4);C(5,1)*C(7,3);C(5,2)*C(7,2);C(5,3)*C(7,1);C(5,4)*C(7,0) > 6 6; > C(6,0)*C(6,4);C(6,1)*c(6,3);C(6,2)*c(6,2);C(6,3)*C(6,1);C(6,4)*C(6,0) > 5 7; > C(7,0)*C(5,4);C(7,1)*C(5,3);C(7,2)*C(5,2);C(7,3)*C(5,1);C(7,4)*C(5,0) > 4 8; > C(8.0)*C(4,4);C(8.1)*C(4,3);C(8.2)*C(4,2);C(8.3)*C(4,1);C(8.4)*C(4,0) > 3 9; > C(9,1)*C(3,3);C(9,2)*C(3,2);C(9,3)*C(3,1);C(9,4)*C(3,0) > 2 10; > C(10,2)*C(2,2);C(10,3)*C(2,1);C(10,4)*C(2,0) > 1 11; > C(11,3)*C(1,1);C(11,4)*C(1,0) > 0;12 > C(12,4)*C(0,0) > All Rows (HORIZONTAL SUM) :495 C(12,4) or C(12,8) matching > frequency or matching probability > Left and right (or right and left): > All OBLIQUELY SUM :715 C(13,4) or C(13,9) and > All OBLIQUELY SUM :1287 C(13,5) or C(13,8) > ETC... :ALL PASCAL triangle centralize mirror and multiplier.... > I, if no new PASCAL triangle centralize mirror and multiplier > search bibliography and author then PASCAL TRIANGLE NEW > POPERTIES!!!??? > http://matserv.pmmf.hu/matport/cikk.php?pelda_id=64 > zerinvary quadrangles.ppt > Please help and write! My hovercraft is full of eels. > Zerinvary Lajos > emeritus prof. > Hungary > Vasarhelyi Pal Technique School > zerinvaryla...@yahoo.com === Subject: Re: Pascal triangle new property and quality? >> Sorry: >> aug. 9 10:21 bad or wrong and incorrect: triangele and tringle! >> Correction all: >> Pascal triangle new property? >> AAAAAAAABBBB (A: num 8 B: num 4) permutations and all comparison >> template: AAAAAAAABBBB >> layout here deformed:(primary:squareness - paralellogramm) >> 12 0; >> C(0,0)*C(12,4) >> 11 1 >> C(1,0)*C(11,4);C(1,1)*C(11,3) >> 10 2; >> C(2,0)*C(10,4);C(2,1)*C(10,3);C(2,2)*C(10,2) >> 9 3 >> C(3,0)*C(9,4);C(3,1)*C(9,3);C(3,2)*C(9,2);C(3,3)*C(9,1) >> 8 4; >> C(4,0)*C(8,4);C(4,1)*C(8,3);C(4,2)*C(8,2);C(4,3)*C(8,3);C(4,4)*C(8,0) >> 7 5; >> C(5,0)*C(7,4);C(5,1)*C(7,3);C(5,2)*C(7,2);C(5,3)*C(7,1);C(5,4)*C(7,0) >> 6 6; >> C(6,0)*C(6,4);C(6,1)*c(6,3);C(6,2)*c(6,2);C(6,3)*C(6,1);C(6,4)*C(6,0) >> 5 7; >> C(7,0)*C(5,4);C(7,1)*C(5,3);C(7,2)*C(5,2);C(7,3)*C(5,1);C(7,4)*C(5,0) >> 4 8; >> C(8.0)*C(4,4);C(8.1)*C(4,3);C(8.2)*C(4,2);C(8.3)*C(4,1);C(8.4)*C(4,0) >> 3 9; >> C(9,1)*C(3,3);C(9,2)*C(3,2);C(9,3)*C(3,1);C(9,4)*C(3,0) >> 2 10; >> C(10,2)*C(2,2);C(10,3)*C(2,1);C(10,4)*C(2,0) >> 1 11; >> C(11,3)*C(1,1);C(11,4)*C(1,0) >> 0;12 >> C(12,4)*C(0,0) >> All Rows (HORIZONTAL SUM) :495 C(12,4) or C(12,8) matching >> frequency or matching probability >> Left and right (or right and left): >> All OBLIQUELY SUM :715 C(13,4) or C(13,9) and >> All OBLIQUELY SUM :1287 C(13,5) or C(13,8) >> ETC... :ALL PASCAL triangle centralize mirror and multiplier.... >> I, if no new PASCAL triangle centralize mirror and multiplier >> search bibliography and author then PASCAL TRIANGLE NEW >> POPERTIES!!!??? >> http://matserv.pmmf.hu/matport/cikk.php?pelda_id=64 >> zerinvary quadrangles.ppt >> Please help and write! >My hovercraft is full of eels. >> Zerinvary Lajos >> emeritus prof. >> Hungary >> Vasarhelyi Pal Technique School >> zerinvaryla...@yahoo.com Do you want to come back to my place, bouncy, bouncy? Rob Johnson take out the trash before replying === Subject: Re: Pascal triangle new property and quality? <20051201.214306@whim.org> >> Sorry: >> aug. 9 10:21 bad or wrong and incorrect: triangele and tringle! >> Correction all: >> Pascal triangle new property? >> AAAAAAAABBBB (A: num 8 B: num 4) permutations and all comparison >> template: AAAAAAAABBBB >> >> layout here deformed:(primary:squareness - paralellogramm) >> 12 0; >> C(0,0)*C(12,4) >> 11 1 >> C(1,0)*C(11,4);C(1,1)*C(11,3) >> 10 2; >> C(2,0)*C(10,4);C(2,1)*C(10,3);C(2,2)*C(10,2) >> 9 3 >> C(3,0)*C(9,4);C(3,1)*C(9,3);C(3,2)*C(9,2);C(3,3)*C(9,1) >> 8 4; >> >> C(4,0)*C(8,4);C(4,1)*C(8,3);C(4,2)*C(8,2);C(4,3)*C(8,3);C(4,4)*C(8,0) >> 7 5; >> >> C(5,0)*C(7,4);C(5,1)*C(7,3);C(5,2)*C(7,2);C(5,3)*C(7,1);C(5,4)*C(7,0) >> 6 6; >> >> C(6,0)*C(6,4);C(6,1)*c(6,3);C(6,2)*c(6,2);C(6,3)*C(6,1);C(6,4)*C(6,0) >> 5 7; >> >> C(7,0)*C(5,4);C(7,1)*C(5,3);C(7,2)*C(5,2);C(7,3)*C(5,1);C(7,4)*C(5,0) >> 4 8; >> >> C(8.0)*C(4,4);C(8.1)*C(4,3);C(8.2)*C(4,2);C(8.3)*C(4,1);C(8.4)*C(4,0) >> 3 9; >> >> C(9,1)*C(3,3);C(9,2)*C(3,2);C(9,3)*C(3,1);C(9,4)*C(3,0) >> 2 10; >> >> C(10,2)*C(2,2);C(10,3)*C(2,1);C(10,4)*C(2,0) >> 1 11; >> C(11,3)*C(1,1);C(11,4)*C(1,0) >> 0;12 >> C(12,4)*C(0,0) >> >> All Rows (HORIZONTAL SUM) :495 C(12,4) or C(12,8) matching >> frequency or matching probability >> >> Left and right (or right and left): >> All OBLIQUELY SUM :715 C(13,4) or C(13,9) and >> All OBLIQUELY SUM :1287 C(13,5) or C(13,8) >> ETC... :ALL PASCAL triangle centralize mirror and multiplier.... >> I, if no new PASCAL triangle centralize mirror and multiplier >> search bibliography and author then PASCAL TRIANGLE NEW >> POPERTIES!!!??? >> http://matserv.pmmf.hu/matport/cikk.php?pelda_id=64 >> zerinvary quadrangles.ppt >> >> Please help and write! >My hovercraft is full of eels. >> >> Zerinvary Lajos >> emeritus prof. >> Hungary >> Vasarhelyi Pal Technique School >> zerinvaryla...@yahoo.com > Do you want to come back to my place, bouncy, bouncy? Drop your panties, Sir William, I cannot wait till lunchtime. > Rob Johnson > take out the trash before replying === Subject: Re: Pascal triangle new property and quality? > Sorry: > aug. 9 10:21 bad or wrong and incorrect: triangele and tringle! > Correction all: > Pascal triangle new property? > AAAAAAAABBBB (A: num 8 B: num 4) permutations and all comparison > template: AAAAAAAABBBB > > layout here deformed:(primary:squareness - paralellogramm) > 12 0; > C(0,0)*C(12,4) > 11 1 > C(1,0)*C(11,4);C(1,1)*C(11,3) > 10 2; > C(2,0)*C(10,4);C(2,1)*C(10,3);C(2,2)*C(10,2) > 9 3 > C(3,0)*C(9,4);C(3,1)*C(9,3);C(3,2)*C(9,2);C(3,3)*C(9,1) > 8 4; > > C(4,0)*C(8,4);C(4,1)*C(8,3);C(4,2)*C(8,2);C(4,3)*C(8,3);C(4,4)*C(8,0) > 7 5; > > C(5,0)*C(7,4);C(5,1)*C(7,3);C(5,2)*C(7,2);C(5,3)*C(7,1);C(5,4)*C(7,0) > 6 6; > > C(6,0)*C(6,4);C(6,1)*c(6,3);C(6,2)*c(6,2);C(6,3)*C(6,1);C(6,4)*C(6,0) > 5 7; > > C(7,0)*C(5,4);C(7,1)*C(5,3);C(7,2)*C(5,2);C(7,3)*C(5,1);C(7,4)*C(5,0) > 4 8; > > C(8.0)*C(4,4);C(8.1)*C(4,3);C(8.2)*C(4,2);C(8.3)*C(4,1);C(8.4)*C(4,0) > 3 9; > > C(9,1)*C(3,3);C(9,2)*C(3,2);C(9,3)*C(3,1);C(9,4)*C(3,0) > 2 10; > > C(10,2)*C(2,2);C(10,3)*C(2,1);C(10,4)*C(2,0) > 1 11; > C(11,3)*C(1,1);C(11,4)*C(1,0) > 0;12 > C(12,4)*C(0,0) > > All Rows (HORIZONTAL SUM) :495 C(12,4) or C(12,8) matching > frequency or matching probability > > Left and right (or right and left): > All OBLIQUELY SUM :715 C(13,4) or C(13,9) and > All OBLIQUELY SUM :1287 C(13,5) or C(13,8) > ETC... :ALL PASCAL triangle centralize mirror and multiplier.... > I, if no new PASCAL triangle centralize mirror and multiplier > search bibliography and author then PASCAL TRIANGLE NEW > POPERTIES!!!??? > http://matserv.pmmf.hu/matport/cikk.php?pelda_id=64 > zerinvary quadrangles.ppt > > Please help and write! >> >>My hovercraft is full of eels. >> > > Zerinvary Lajos > emeritus prof. === Subject: Help optimization theory proof Hello. Let I be an index set, g_i(x) continuous functions in R^n. Let F= { x in R ^n | g_i (x )> =0 , i in I} and let I (x) = { i in I : g_i(x) =0} Prove that if I(x) is empty then x lies in int F. (int F = interior of F). Proof: Suppose I(x) is empty, this means that for all i in I , g_i (x) is nonzero. Wlog, suppose g_i(x) > 0, by continuity it follows that there exists some d>0 such that g_i(x) >0 for all x in B(d,0) , the ball centred at 0 and radius d, and therefore x lies in the interior of F. Is the above proof correct? What Im stuck is on the following part: Assume that x lies in int F and that the gradient of g_i(x) is nonzero for all i in I. Prove that I(x) is empty. === Subject: Re: Help optimization theory proof === Subject: Help optimization theory proof > Let I be an index set, g_i(x) continuous functions in R^n. What you mean function in R^n? Functions aren't in R^n, only points are in R^n. Do you mean function with domain and range of R^n ? No, likely you mean domain R^n and range R. Thus correct expresion is real valued function over R^n ? > Let F= { x in R ^n | g_i (x )> =0 , i in I} and F = { x in R^n | for all i in I, g_i(x) >= 0 } > let I (x) = { i in I : g_i(x) =0} > Prove that if I(x) is empty then x lies in int F. > Proof: > Suppose I(x) is empty, this means that for all i in I , g_i (x) is > nonzero. Consider when for all i, x in R^n, g_i(x) = -1 For all x, I(x) = nulset; F = nulset = int F; x not in int F. > Wlog, suppose g_i(x) > 0, by continuity it follows that there exists > some d>0 such that g_i(x) >0 for all x in B(d,0) , the ball centred > at 0 and radius d, and therefore x lies in the interior of F. > Is the above proof correct? In view of the above counter example, how could it be? > What Im stuck is on the following part: Assume that x lies in int F > and that the gradient of g_i(x) is nonzero for all i in I. > Prove that I(x) is empty. ---- === Subject: Re: Help optimization theory proof Sorry, typo, g_i(x) : R^n -> R and g_i(x) >=0. (In your example the g_i's are negative). === Subject: Re: Help optimization theory proof > Sorry, typo, g_i(x) : R^n -> R and g_i(x) >=0. (In your example the > g_i's are negative). I've not the time nor the inclination for the clerical chore of back tracking the thread to reconstruct the line of thought. If you want intelligent and well thought answers, instead of sloppy inaccurate answers from what I remember, then include the context pertinent to your reply and to whom your are talking. Please learn and use better math group manners as demonstrated by other participants of this newsgroup and as described at http://oakroadsystems.com/genl/unice.htm#quote If you're posting from Mathforum or Google, it is requested you use the quote feature. Many of us use different news browsers than you. === Subject: generalizing iterated function systems? Hello - I was wondering if there is a literature that deals with the following generalization of iterated function systems, and what this literature might be called. For simplicity assume we have a closed, bounded rectangle X of R^n. An iterated function system is created when we have m functions f_i for i =1,...m, where each f_i is a function from X to itself. The iterated function system is formed by choosing a starting point x^0 in X, and then randomly choosing one of the maps f_i, and taking x^1 = f(x^0), and repeating this process. I was wondering about the following generalization. Instead of having n functions f_i, suppose we had n transition probabilities P_i(x, .) where for every x in X, P_i(x, .) is a probability measure over X. Then for any starting point x^0, we choose a P_i at random, and then form x^1 by drawing from the probability measure P_i(x^0, .), and repeating the process. I was especially interested in any results that pointed in the direction of saying that if each P_i defined a well behaved markov chain, such as an equicontinuous chain, then the iterated system was also well behaved in this sense? Amit === Subject: Re: easiest and shortest proof of Jordan Curve theorem apparently, especially for simple, closed curves. I like the theorem that was listed in (gak) mathworld: given a triangle, a similar one can be inscribed inside of a Jordan curve. thus: dood, don't throw strings out with the Schroedinger's Cat's quantum litter-box; the Copenhagen school is utterly mystical! anyway, their reification of the probabilites, has been undone by fuzzy logic (which is an acceptable way to do QM, apparently); dig? (NB: Schroedinger made that cat as a joke on the Copenhagen school; he loved his cat .-) people who say that stringtheory is so bad, seem to forget that it encompasses a lot of the other stuff; hence, difficult to find any thing *new* to test it on. thus: one does not have to believe in the compactified analogy about dimensions; they're just trying to visualize the math. but, they won't be orthogonal in the 3-D sense, obviously, a unique case.... if anyone says time is orthogonal to 3 space dimensions; how is that? thus: cool thing about time-travel ... were it to be possible without violating great-granpa's legacy ... is that you can bypass the libeRal media owned by consWervatives. thus quoth: Technically, scientifically, in our back-channel dialogue of the time, the Soviet government agreed with my view on this feature of the proposed non-military advantage, but conveyed the view that since we would benefit more than they, they would reject the proposal and beat us by other means. Hence, my absolutely accurate forewarning of a potential collapse of the Soviet system within about five years, under the conditions of Soviet rejection of the proposal were it made by President Reagan, as Reagan did make the proposal a month later, and as the Soviet government of Andropov did reject the proposal. http://larouchepub.com/lar/2005/3246of_british_fools.html --les Protocols de George Elder chez Kyoto! http://tarpley.net/bush8.htm http://larouchepub.com/other/2002/2903_chapter_11.html http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: easiest and shortest proof of Jordan Curve theorem > What I am saying is draw a circle on the surface of a torus which has a > full interior to that circle, ie, all points of the circle and its > interior are points on the surface of the torus. There are many such > circles with their interiors on the torus surface. And all of these > circles divide the torus into an inside and outside. > So, is this definition well-defined-- The points on a circle plus the > interior points of the circle are all points lying on the surface of > the torus. > Any other circle on the torus is just not a circle, Of course it is, unless, as I suggest below, you've redefined circle.. > just as a latitude > line on a sphere may look like a line of the sphere but is not a line > because only great-circles (longitudes and equator) are lines on a > sphere. > So if the above is well defined, then obviously a torus obeys the JCT > as easily as the sphere and plane obey the Jordan Curve Theorem. For a suitable definition of circle. You have to be *way* more precise. > Then the question would be what is the difference theoretically between > a torus that obeys the JCT definitions from that of a torus that does > not obey JCT with its definitions. torus. Your distinction is bootless: It makes no sense to talk about tori for which the JCT is true versus those for which it isn't. JCT either holds for all tori (and all circles) or it doesn't. === Subject: Re: easiest and shortest proof of Jordan Curve theorem > Then the question would be what is the difference theoretically between > a torus that obeys the JCT definitions from that of a torus that does > not obey JCT with its definitions. > torus. Your distinction is bootless: It makes no sense to talk about > tori for which the JCT is true versus those for which it isn't. JCT > either holds for all tori (and all circles) or it doesn't. In all fairness, I think AP means a closed curve that obeys the JCT defintion. AP has done here what he's done for all the other problems he's claimed to have solved: redefined them to the point of triviality. AP's version of JCT is ] If C is a closed curve that has some points in its interior, and some points in ] its exterior, then C satisfies the JCT. Actually, he hasn't even done that. He hasn't ruled out the possibility of 3 or more regions. --- Christopher Heckman === Subject: Re: easiest and shortest proof of Jordan Curve theorem > What I am saying is draw a circle on the surface of a torus which has a > full interior to that circle, ie, all points of the circle and its > interior are points on the surface of the torus. What are the interior points of a circle on a torus? Isn't this definition, eh, circular? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: easiest and shortest proof of Jordan Curve theorem > What are the interior points of a circle on a torus? Isn't this > definition, eh, circular? > -- Some circles on the torus have interiors whose points all lie on the surface of the torus. Then there are some circles whose interiors are points that lie inside the torus and not on the surface of the torus and then there are some circles such as a tangent plane to the top of the torus whose interior are points that lie outside the torus itself. Now when we draw a circle on the sphere all the interior points of the circle lie on the surface of the sphere. So it is all the circles of a torus whose interior points lie on the surface of the torus that I define as being a circle of the torus, whereas all other circles whose interior points do not all lie on the surface of the torus I will define as not a circle. Just like lines of latitude are not lines in Riem geom and so circles whose interior points are not all on the surface of the torus are not circles for a torus. By defining circles for a torus in this manner I have the Jordan Curve theorem obeying toruses. Now I have to show that such a definition makes more sense overall, and not just for the application of the Jordan Curve theorem. Dik, can you tell me why lines of latitude cannot be lines in Riem geometry and only great circles? Perhaps the answer to that question is the answer to whether my definition of a circle on a torus must hold. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: easiest and shortest proof of Jordan Curve theorem > What are the interior points of a circle on a torus? Isn't this > definition, eh, circular? > -- > Some circles on the torus have interiors whose points all lie on the > surface of the torus. Then there are some circles whose interiors are > points that lie inside the torus and not on the surface of the torus > and then there are some circles such as a tangent plane to the top of > the torus whose interior are points that lie outside the torus itself. Fine, but given a circle on the torus, or any other surface, how do you find the interior points? I suppose you could say that you can look at the the plane which contains the circle, and consider all points on the interior of the circle. But this definition doesn't fit with what you say about a circle on the sphere, in the next paragraph. ((Side note to everyone else: Defining the interior of a circle can be done rigorously. Given a circle in a plane, you can choose cartesian coordinates and a unit distance such that the circle consists of all points (x,y) such that x^2 + y^2 = 1. Then the interior of the circle is all points (x,y) with x^2 + y^2 < 1.)) > Now when we draw a circle on the sphere all the interior points of the > circle lie on the surface of the sphere. It won't be, anyway according to the definition above. For instance, one of the points in the interior of the equator is the center of the sphere, which is not on the sphere. (The center is inside it.) I think I know what you're aiming for here, though. You want the interior points of the equator to be the points on the sphere in the northern hemisphere. (Or is it the southern hemisphere?) You want to pick a point P not on the circle; then the interior of the circle is all points which you can get to, from P, without crossing the circle. A problem with the sphere is that either half of the sphere can be the interior of the circle. (This isn't a problem with the plane, where the exterior is all points P such that you can get from P to an arbitrarily far away point Q without crossing the circle.) The problem with the torus is that the interior of a circle may end up being the entire torus (except for the circle), and this is where the JCT falls apart. (The JCT can only fail if there is exactly one region left; it's not possible to get 3 or more pieces when you remove a simple closed curve.) > So it is all the circles of a > torus whose interior points lie on the surface of the torus that I > define as being a circle of the torus, whereas all other circles whose > interior points do not all lie on the surface of the torus I will > define as not a circle. Just like lines of latitude are not lines in > Riem geom and so circles whose interior points are not all on the > surface of the torus are not circles for a torus. > By defining circles for a torus in this manner I have the Jordan Curve > theorem obeying toruses. > Now I have to show that such a definition makes more sense overall, and > not just for the application of the Jordan Curve theorem. Your definition turns out to be the same as that for contractible curves. But the JCT for contractible curves follows from its definition; it's been proven already. --- Christopher Heckman === Subject: Re: easiest and shortest proof of Jordan Curve theorem > > What are the interior points of a circle on a torus? Isn't this > definition, eh, circular? > -- > Some circles on the torus have interiors whose points all lie on the > surface of the torus. Then there are some circles whose interiors are > points that lie inside the torus and not on the surface of the torus > and then there are some circles such as a tangent plane to the top of > the torus whose interior are points that lie outside the torus itself. > Fine, but given a circle on the torus, or any other surface, how do you > find the interior points? > I suppose you could say that you can look at the the plane which > contains the circle, and consider all points on the interior of the > circle. But this definition doesn't fit with what you say about a > circle on the sphere, in the next paragraph. > ((Side note to everyone else: Defining the interior of a circle can be > done rigorously. Given a circle in a plane, you can choose cartesian > coordinates and a unit distance such that the circle consists of all > points (x,y) such that > x^2 + y^2 = 1. Then the interior of the circle is all points (x,y) with > x^2 + y^2 < 1.)) > Now when we draw a circle on the sphere all the interior points of the > circle lie on the surface of the sphere. > It won't be, anyway according to the definition above. For instance, > one of the points in the interior of the equator is the center of the > sphere, which is not on the sphere. (The center is inside it.) > I think I know what you're aiming for here, though. You want the > interior points of the equator to be the points on the sphere in the > northern hemisphere. (Or is it the southern hemisphere?) You want to > pick a point P not on the circle; then the interior of the circle is > all points which you can get to, from P, without crossing the circle. > A problem with the sphere is that either half of the sphere can be > the interior of the circle. (This isn't a problem with the plane, > where the exterior is all points P such that you can get from P to an > arbitrarily far away point Q without crossing the circle.) > The problem with the torus is that the interior of a circle may end up > being the entire torus (except for the circle), and this is where the > JCT falls apart. (The JCT can only fail if there is exactly one region > left; it's not possible to get 3 or more pieces when you remove a > simple closed curve.) > So it is all the circles of a > torus whose interior points lie on the surface of the torus that I > define as being a circle of the torus, whereas all other circles whose > interior points do not all lie on the surface of the torus I will > define as not a circle. Just like lines of latitude are not lines in > Riem geom and so circles whose interior points are not all on the > surface of the torus are not circles for a torus. > By defining circles for a torus in this manner I have the Jordan Curve > theorem obeying toruses. > Now I have to show that such a definition makes more sense overall, and > not just for the application of the Jordan Curve theorem. > Your definition turns out to be the same as that for contractible > curves. But the JCT for contractible curves follows from its > definition; it's been proven already. > --- Christopher Heckman Chris, I think the easiest and best definition of a circle is whether its radius on the surface of the object can sweep out an entire circle with its interior. Anything else is not a circle in that object. For instance in Eucl 3-D any two points will sweep out an circle. But given a sphere, then the maximum circle is a hemisphere, at least my intuition suspects that to be the case but perhaps my intuition fails me. So on a torus, only those figures which are circular and which contain an interior of points all inside the circle are to be defined as circles. Any other so called circle of the torus is just not a circle. The idea is that only circles are those on the surface of the object and all the points of the interior of the circle are also points of the surface of the object. And secondly the radius of the circle sweeps out only points on the surface of the object. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: easiest and shortest proof of Jordan Curve theorem > What I am saying is draw a circle on the surface of a torus which has a > full interior to that circle, ie, all points of the circle and its > interior are points on the surface of the torus. > What are the interior points of a circle on a torus? Isn't this > definition, eh, circular? Hehe. Welcome to the thread, Dik. Not enough else out there to occupy you? Rick === Subject: Re: easiest and shortest proof of Jordan Curve theorem > What I am saying is draw a circle on the surface of a torus which has a > full interior to that circle, ie, all points of the circle and its > interior are points on the surface of the torus. > > What are the interior points of a circle on a torus? Isn't this > definition, eh, circular? 6 But it's not perfectly uniform. Regions originally in causal contact were so small that the uncertainty principle demands that they should have appreciable quantum fluctuations in energy density from place to place ... these fluctuations would be frozen when the regions passed out of causal contact and would have approximately the same magnitude (approximatley one part in a hundred thousand) when they said 'hello' again. But these regions would inflate vastly in size when out of causal contact ... a region originally 10^-26 cm or smaller could grow to be billions of light years in expanse ... the quantum fluctuation in density, predicted in inflation, should be random ... This is a ... spongelike geometry ... a variety of galaxy samples ... all show a spongelike distribution ... the structures we see in our universe today are the fossilized remains of quantum fluctuations during the first 10^-35 seconds of our universe ... the universe may have undergone 100 doublings increasing its size by more than 10^30 during the inflationary epoch ... What if one could simply start off with the narrow waist of the de Sitter spacetime eliminating the contracting ... phase ... the inflationary universe at its waist is very small ... much smaller than a ... proton. Is it a fundamental string that inflates? ... inflation might end when the energy in the inflationary vacuum was dumped in the form of thermal radiation over the whole space at once Does this need a superluminal signal? ... when you boil water ... bubbles of steam form ... high density vacuum would likely decay by forming bubbles of ordinary vacuum ... Each bubble would expand ... The vacuum inside each bubble is a normal vacuum with zero energy density and zero pressure. Outside the bubble the pressure would be negative (a universal suction) ... Tunneling is what happens when a Coleman bubble forms. Note my model of the electron Bohm hidden variable is a thin spherical charge ~ 10^-11 cm held together by negative pressure suction in the interior with zero pressure normal vacuum outside. This also explains Ken Shoulders' charge clusters. At high momentum scattering transfer the shell of charge appears to shrink down to 10^-16 cm from the huge space warp of the interior uniform core of negative zero point dark energy pressure. The space inflates so fast that the bubbles stay isolated. This was Guth's original picture. Gott says our universe is inside such an expanding vacuum bubble. Light beams emitted from the center of the bubble never reach the receding bubble wall. Note that such a bubble may be a point defect in the vacuum order parameter. This idea leads to an infinite hyperbolic open universe with negative space curvature. But space is flat with zero curvature on the large scale not negative (Sidney) Coleman bubble curvature. Gott says that our flat universe is only a small piece of the negatively curved Coleman bubble. That's how he rationalizes the latest data which has a residual vacuum zero point energy/c^2 density of ~ 6x10^-30 grams/cc close to what Einstein predicted for his 1917 static unstable universe model. This inflation model is, of course, very different. Lemaitre actually predicted what we see today back in the 1930's. (p. 180, Gott) === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised Now I do not like the idea of the 2-adics converging to two points of ...0000 and ....0001. Is Dik Winter busy? If not, can Dik tell me if there is a iterative roots sequence for each and every p-adic and n-adic integers, or p-adic and n-adic rationals? Is there an iterative roots of the Adics that matches the iterative roots in Positive-Reals where they all converge to 1. That was a nagging open question when I put this PC on hold in the 1990s. So I would like to improve the above. Dik, a question, what is the closest thing in Adics, whether integers or rational adics that resembles the iterative roots all converging to 1 in Reals? Is there anything like that in Adics. Perhaps Chris knows the answer, for I certainly do not know. Somehow I have a dyslexia when it comes to different bases and when it comes to adics and that is why I have to ask these questions is because my mind simply cannot function well in different bases or in adics. Faraday had a similar mind where he was dyslexic when it comes to mathematical physics. So I need to know if there is something in the Adics, whether the adic rationals or the adic integers that is closest to the iterative roots of Reals converging to 1. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised > Now I do not like the idea of the 2-adics converging to two points of > ...0000 and ....0001. > Is Dik Winter busy? If not, can Dik tell me if there is a iterative > roots sequence for each and every p-adic and n-adic integers, or > p-adic and n-adic rationals? No. In the first place not every p-adic or n-adic has a root, and if there is a root it is not necessarily true that that number has a root. It is quite similar to the integers. Within the integers there are only finite iterative root sequences. Moreover, in the 10-adics there are 2 idempotents. By their definition, given one of the idempotents: a, sqrt(a) = a. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised > Now I do not like the idea of the 2-adics converging to two points of > ...0000 and ....0001. > > Is Dik Winter busy? If not, can Dik tell me if there is a iterative > roots sequence for each and every p-adic and n-adic integers, or > p-adic and n-adic rationals? > No. In the first place not every p-adic or n-adic has a root, and if > there is a root it is not necessarily true that that number has a root. > It is quite similar to the integers. Within the integers there are > only finite iterative root sequences. > Moreover, in the 10-adics there are 2 idempotents. By their definition, > given one of the idempotents: a, sqrt(a) = a. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Dik, is this mathematically allowed, or mathematically the same?? If we convert every p-adic and n-adic to a 2-adic and then take the iterative square that every such number (point) converges to ....0000 and ....0001 ? Is that true? Instead of iterative roots for positive-Reals we have iterative squares for 2-adics. And is it true for all higher-powers instead of just squaring? Is that true, and why would iterative -higher-powers be like a inverse to that of iterative roots for positive-Reals? Dik, can I build a similarity on the Adics that matches iterative roots for positive-Reals? Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised Gotten into discussion again: > Dik, is this mathematically allowed, or mathematically the same?? > If we convert every p-adic and n-adic to a 2-adic You can't. For instance, the 10-adics have idempotents and the 2-adics do *not* have idempotents. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised > Gotten into discussion again: > Dik, is this mathematically allowed, or mathematically the same?? > > If we convert every p-adic and n-adic to a 2-adic > You can't. For instance, the 10-adics have idempotents and the 2-adics > do *not* have idempotents. > -- The word conversion is not what I want. I want what the iterative roots of Positive Reals converging to 1 is for Adics. To find what is comparable in Adics where they all converge to one or two numbers. So is raising to higher powers squaring, cubing, etc etc converge to ...0000 and ...0001 in All-adics true? Do the p-adic integers and the n-adic integers converge to ...000 and ...0001 with iterative-higher-powers. And why is the Positive Reals wanting of iterative roots whereas Adics Integers is wanting of iterative higher powers? Can we say that the Positive Reals are the inverse of the Adic Integers because the Positive Reals need iterative roots which is the inverse of iterative higher powers. So what is the internal structure of Adics that convergence needs the inverse of Positive Reals. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised ... > I want what the iterative roots of Positive Reals converging to 1 is > for Adics. To find what is comparable in Adics where they all converge > to one or two numbers. Yes, you may want that, but it remains at that. > So is raising to higher powers squaring, cubing, etc etc converge to > ...0000 and ...0001 in All-adics true? Do the p-adic integers and the > n-adic integers converge to ...000 and ...0001 with > iterative-higher-powers. You apparently do not think. As far as I see it is only true in the 2-adics. For instance, in the 3-adics, calculating mod 9 and doing the squaring, when we start at 4 we get the sequence: 4, 7, 4, 7, ... > And why is the Positive Reals wanting of iterative roots whereas Adics > Integers is wanting of iterative higher powers? And apparently you do not read either. It is the positive integers that lack iterative roots. They are fine in the positive reals. For the n-adics you could have them once you made them algebraically complete, which they are not, and which is also a very tedious process. Moreover, also in the n-adics yu have each time the choice between two square roots, but you can not define something like positive n-adics in that field, so you have no way to choose one or the other. > Can we say that the > Positive Reals are the inverse of the Adic Integers because the > Positive Reals need iterative roots which is the inverse of iterative > higher powers. Well, you might say that, but it looks like nonsense. > So what is the internal structure of Adics that > convergence needs the inverse of Positive Reals. For one thing, the complex numbers (of which the positive reals are a subset) are algebraically complete. Moreover there is a sqrt mapping from R+ to R+, and so we have iterative roots in the positive reals. Iterative powering in the n-adics does *not* give convergence in general, and iterative roots are not possible because not every number has a square root. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised > I want what the iterative roots of Positive Reals converging to 1 is > for Adics. To find what is comparable in Adics where they all converge > to one or two numbers. > So is raising to higher powers squaring, cubing, etc etc converge to > ...0000 and ...0001 in All-adics true? Do the p-adic integers and the > n-adic integers converge to ...000 and ...0001 with > iterative-higher-powers. What happens if you square a 10-adic which ends in 5? Cube it? Raise it to the 4th power? Etc? --- Christopher Heckman === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised > [...] > I want what the iterative roots of Positive Reals converging to 1 is > for Adics. To find what is comparable in Adics where they all converge > to one or two numbers. > So is raising to higher powers squaring, cubing, etc etc converge to > ...0000 and ...0001 in All-adics true? Do the p-adic integers and the > n-adic integers converge to ...000 and ...0001 with > iterative-higher-powers. > What happens if you square a 10-adic which ends in 5? Cube it? Raise it > to the 4th power? Etc? > --- Christopher Heckman Do they all become zeroes? Just as those ending in 6, 7, 3 Is this a true statement, Chris, given any p-adic integer or n-adic integer by repeated squaring, cubing, ... all higher powers, that the integers converge to either one of its idempotents ....0000 or ....00001. Is that true? Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised ... > So is raising to higher powers squaring, cubing, etc etc converge to > ...0000 and ...0001 in All-adics true? Do the p-adic integers and the > n-adic integers converge to ...000 and ...0001 with > iterative-higher-powers. > > What happens if you square a 10-adic which ends in 5? Cube it? Raise it > to the 4th power? Etc? > Do they all become zeroes? Just as those ending in 6, 7, 3 Do some thinking. What happens with the last digit for a 10-adic ending in 5? > Is this a true statement, Chris, given any p-adic integer or n-adic > integer by repeated squaring, cubing, ... all higher powers, that the > integers converge to either one of its idempotents ....0000 or > ....00001. Is that true? That is false. In almost all cases there is no convergence. Off-hand I think repeated squaring only gives convergence in the 2-adics, and in that case to ...000 and ...001. Also I think repeated cubing might give convergence in the 3-adics, to any of ...000, ...001 and ...222, and repeatedly taking the 5-th power in the 5-adics, and I think they go to ...000, ...001, ...21212, ...444 and a fifth. But there is no power that will give general convergence in the 6-adics or the 10-adics because of the idempotents. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised <23096985.1133439890372.JavaMail.jakarta@nitrogen.mathforum.org> >> Has it occurred to you that peer reviewed journals are mostly >> obsolete and that the Internet is a far better means of >> communication than peer reviewed. Time is of the essence >> and where the Internet takes seconds, journals take months. Yes, these peer reviewers also have a nasty habit of filtering out cranks and crankish proofs. This would be quite unfortunate for you. However, I do know of one legitimate mathematician who might publish your work in a true mathematical publication. It would be Dr. Underwood Dudley, author of the very entertaining book, Mathematical Cranks: http://www.amazon.com/gp/product/0883855070/ . With any luck, perhaps he is working on a second edition and your work would finally be published. :-) === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised <23096985.1133439890372.JavaMail.jakarta@nitrogen.mathforum.org> >> Has it occurred to you that peer reviewed journals are mostly >> obsolete and that the Internet is a far better means of >> communication than peer reviewed. Time is of the essence >> and where the Internet takes seconds, journals take months. Yes, these peer reviewers also have a nasty habit of filtering out cranks and crankish proofs. This would be quite unfortunate for you. However, I do know of one legitimate mathematician who might publih your work in a true mathematical publication. It would be Dr. Underwood Dudley, author of the very entertaining book, Mathematical Cranks: https://enterprise.maa.org/ecomtpro/Timssnet/products/TNT_products.cfm . With any luck, perhaps he is working on a second edition and your work would finally be published. :-) === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised <23096985.1133439890372.JavaMail.jakarta@nitrogen.mathforum.org> > So, since you've had a proof of the Poincar.8f Conjecture since the > beginning of the 1990's, I'm assuming we'll see it in a peer-reviewed > mathematics journal shortly? (Or maybe not..) > Has it occurred to you that peer reviewed journals are mostly obsolete > and that the Internet is a far better means of communication than peer > reviewed. Time is of the essence and where the Internet takes seconds, > journals take months. Has it occured to you that since certain results (like the 4CT, trisecting the angle, squaring the circle, etc) have produced so many false proofs, that they should be thoroughly checked before being announced? The point of refereed journals is that things are checked. It may take slower, but you will find that journals seldom have to retract anything that gets published. Remember cold fusion, a few years ago? Some physicists at a university in Utah claimed they had discovered a way to have fusion take place at room temperature, and they announced it to the public before they could check it and realize they were wrong. Same thing. It's a trade-off. --- Christopher Heckman === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised >>So, since you've had a proof of the Poincar.8f Conjecture since the >>beginning of the 1990's, I'm assuming we'll see it in a peer-reviewed >>mathematics journal shortly? (Or maybe not..) >>Has it occurred to you that peer reviewed journals are mostly obsolete >>and that the Internet is a far better means of communication than peer >>reviewed. Time is of the essence and where the Internet takes seconds, >>journals take months. > Has it occured to you that since certain results (like the 4CT, > trisecting the angle, squaring the circle, etc) have produced so many > false proofs, that they should be thoroughly checked before being > announced? Two of those are known to be false, obviating the need for checking. Rick === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised >>So, since you've had a proof of the Poincar.8f Conjecture since the >>beginning of the 1990's, I'm assuming we'll see it in a peer-reviewed >>mathematics journal shortly? (Or maybe not..) >> >>Has it occurred to you that peer reviewed journals are mostly obsolete >>and that the Internet is a far better means of communication than peer >>reviewed. Time is of the essence and where the Internet takes seconds, >>journals take months. > Has it occured to you that since certain results (like the 4CT, > trisecting the angle, squaring the circle, etc) have produced so many > false proofs, that they should be thoroughly checked before being > announced? > Two of those are known to be false, obviating the need for checking. That doesn't prevent some people from claiming that they've really solved those problem. I had to use two results which are known to be false, because a lot of the classical examples of conjectures producing a lot of false proofs have in fact been settled. I can't use the 4CT, and I can't use FLT, which are the two best examples otherwise. --- Christopher Heckman === Subject: Re: Proof of the Poincare Conjecture; my 1990's proof revised <23096985.1133439890372.JavaMail.jakarta@nitrogen.mathforum.org> may I be controversial? the prefered term of art is catalyzed fusion, but one has to be sure that the calorimetry takes into account the amount of energy thais needed to refine the platinum (e.g.) lattice (since most metals are in the water-table, they may already be saturated with H -- used to be a problem with the natural gas pipes, hydrogen embrittlement, so they're plastic-coated. actually, I forget the prefered term, but taht's the idea; is it so bizarre?... there are tons of results, two, over the years, although the hot fusion folks were justifiably worried about defunding. now, there's a magazine called Infinite Energy, which often has some silly stuff in it, but the current issue is devoted to Ampere's longitudinal force, which has been neglected from Maxwell (in spite of what he said about it), on. (there was no mention, that I could see, of LaRouche, who's been plugging this for decades). can you say, Einsteinmania Millenium Two? > Remember cold fusion, a few years ago? Some physicists at a --les Protocols de George Elder chez Kyoto! http://tarpley.net/bush8.htm http://larouchepub.com/other/2002/2903_chapter_11.html http://clowder.net/zubek/zubek.html http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Rencontres umbers or don't rencontres numbers? Jul 08 2005 Sending in a Comment on an Existing Sequence to The On-Line Encyclopedia of Integer Sequences SINCE WANT!!! What's the problem? NUMERAL: %I A133581 %S A133581 0,6,24,54,96,150,216,294,384,486,600,726,864,1014,1176,1350, 1536,1734,1944,2166,2400,2646,2904,3174,3456,3750,4056,4374, 4704,5046,5400,5766,6144,6534,6936,7350,7776,8214,8664,9126, 9600,10086,10584,11094,11616 %C A133581 This numbers analogue :Rencontres numbers: permutations with exactly m-2 fixed points. 0 letters m=0 0 fixed points n0=0 ABCD m=4 letters permutations m-2=4-2=2 fixed points n1=6 AABBCCDD m=8 letters permutations m-2 = 6 fixed points n2=24 AAABBBCCCDDD m=12 letters permutations m-2 = 10 fixed points n3=54 AAAABBBBCCCCDDDD m=16 letters permutations m-2 = 14 fixed points n4=96 etc. etc, etc. etc. %O A133581 %K A133581 ,easy,nice,nonn, %A A133581 Zerinvary Lajos (zerinvarylajos@yahoo.com), Jul 08 2005 RH RA 82.141.159.17 RU RI === Subject: COMMENT FROM Zerinvary Lajos RE A016743 %I A016743 %S A016743 0,8,64,216,512,1000,1728,2744,4096,5832,8000,10648,13824, 17576,21952,27000,32768,39304,46656,54872,64000,74088,85184, 97336,110592,125000,140608,157464,175616,195112 %C A016743 This numbers analogue :Rencontres numbers: permutations with exactly m-3 fixed points. 0 letters fixed points n0=0 ABCD m=4 letters permutations m-3=4-3=1 fixed points n1=8 ABCD m=8 letters permutations m-3 = 5 fixed points n2=64 AABBCCDD m=12 letters permutations m-3 = 9 fixed points n3=216 AAABBBCCCDDD m=16 letters permutations m-3 = 13 fixed points n4=512 AAAABBBBCCCCDDDD m=20 letters permutations m-3 = 17 fixed points n5=1000 etc. etc, etc. etc. %O A016743 0 %K A016743 ,easy,nice,nonn, %A A016743 Zerinvary Lajos (zerinvarylajos@yahoo.com), Jul 08 2005 RH RA 82.141.159.17 RU RI === Subject: COMMENT FROM Zerinvary Lajos RE A000459 %I A000459 %S A000459 0,1,10,297,13756,925705,85394646,10351036465,1596005408152, 305104214112561,70830194649795010,19629681235869138841, 6401745422388206166420,2427004973632598297444857, 1058435896607583305978409166 %C A000459 This numbers analogue :Rencontres numbers: permutations with exactly no fixed points. AA m=2 letters permutations no fixed points n1=0 AABB m=4 letters permutations no fixed points n2=1 AABBCC m=6 letters permutations no fixed points n3=10 AABBCCDD m=8 letters permutations no fixed points n4=297 AABBCCDDEE m=10 letters permutations no fixed points n5=13756 AABBCCDDEEFF m=12 letters permutations no fixed points n5=925705 etc. etc, etc. etc. %O A000459 0 %K A000459 ,easy,nice,nonn, %A A000459 Zerinvary Lajos (zerinvarylajos@yahoo.com), Jul 08 2005 RH RA 82.141.159.17 RU RI === Subject: COMMENT FROM Zerinvary Lajos RE A033431 %I A033431 %S A033431 0,2,16,54,128,250,432,686,1024,1458,2000,2662,3456,4394, 5488,6750,8192,9826,11664,13718,16000,18522,21296,24334, 27648,31250,35152,39366,43904,48778,54000,59582,65536,71874, 78608,85750,93312,101306 %C A033431 This numbers analogue :Rencontres numbers: permutations with exactly m-3 fixed points. null letters m-3=? fixed points n0=0 ABC m=3 letters permutations m-3=0 fixed points n1=2 AABBCC m=6 letters permutations m-3=3 fixed points n2=16 AAABBBCCC m=9 letters permutations m-3=6 fixed points n3=54 AAAABBBBCCCC m=12 letters permutations m-3=9 fixed points n4=128 AAAAABBBBBCCCCC m=15 letters permutations m-3=12 fixed points n5=250 etc. etc, etc. etc. %O A033431 0 %K A033431 ,easy,nice,nonn, %A A033431 Zerinvary Lajos (zerinvarylajos@yahoo.com), Jul 08 2005 RH RA 82.141.159.17 RU RI === Subject: COMMENT FROM Zerinvary Lajos RE A033428 %I A033428 %S A033428 0,3,12,27,48,75,108,147,192,243,300,363,432,507,588,675,768, 867,972,1083,1200,1323,1452,1587,1728,1875,2028,2187,2352, 2523,2700,2883,3072,3267,3468,3675,3888,4107,4332,4563,4800, 5043,5292,5547,5808,6075,6348 %C A033428 This numbers analogue :Rencontres numbers: permutations with exactly m-2 fixed points. null letters m-2=? -2? fixed points n0=0 ABC m=3 letters permutations m-2=1 fixed points n1=3 AABBCC m=6 letters permutations m-2=4 fixed points n2=12 AAABBBCCC m=9 letters permutations m-2=7 fixed points n3=27 AAAABBBBCCCC m=12 letters permutations m-2=10 fixed points n4=48 AAAAABBBBBCCCCC m=15 letters permutations m-2=13 fixed points n5=75 etc. etc, etc. etc. %O A033428 0 %K A033428 ,easy,nice,nonn, %A A033428 Zerinvary Lajos (zerinvarylajos@yahoo.com), Jul 08 2005 RH RA 82.141.159.17 RU RI === Subject: COMMENT FROM Zerinvary Lajos RE A000172 %I A000172 %S A000172 1,2,10,56,346,2252,15184,104960,739162,5280932,38165260, 278415920,2046924400,15148345760,112738423360,843126957056, 6332299624282,47737325577620,361077477684436, 2739270870994736,20836827035351596 %C A000172 This numbers analogue :Rencontres numbers: permutations with exactly no fixed points. null letters no fixed points n0=1 ABC m=3 letters permutations no fixed points n1=2 AABBCC m=6 letters permutations no fixed points n2=10 AAABBBCCC m=9 letters permutations no fixed points n3=56 AAAABBBBCCCC m=12 letters permutations no fixed points n4=346 AAAAABBBBBCCCCC m=15 letters permutations no fixed points n5=2252 etc. etc, etc. etc. %O A000172 0 %K A000172 ,easy,nice,nonn, %A A000172 Zerinvary Lajos (zerinvarylajos@yahoo.com), Jul 08 2005 RH RA 82.141.159.17 RU RI === Subject: COMMENT FROM Zerinvary Lajos RE A000279 %I A000279 %S A000279 3,24,216,1824,15150,124416,1014888,8241792,66724398, 538990800,4346692680,35009591040,281699380560,2264868936960, 18198009147600,146142982814208,1173123636533454, 9413509300965936,75513633110271264 %C A000279 This numbers analogue :Rencontres numbers: permutations with exactly one fixed points. ABC m=3 letters permutations one (1)fixed points n1=3 AABBCC m=6 letters permutations one (1) fixed points n2=24 AAABBBCCC m=9 letters permutations one (1) fixed points n3=216 AAAABBBBCCCC m=12 letters permutations one (1) fixed points n4=1824 AAAAABBBBBCCCCC m=15 letters permutations one (1)fixed points n5=15150 etc. etc, etc. etc. %O A000279 0 %K A000279 ,easy,nice,nonn, %A A000279 Zerinvary Lajos (zerinvarylajos@yahoo.com), Jul 08 2005 RH RA 82.141.159.17 RU RI This copy is just for your records. No reply is expected. === Subject: COMMENT FROM Zerinvary Lajos RE A000535 %I A000535 %S A000535 0,27,378,4536,48600,489780,4738104,44535456,409752432, 3708359550,33125746500,292779558720,2565087894720, 22307854940280,192788833482000,1657111548654720, 14176605442521312,1207794 %C A000535 This numbers analogue :Rencontres numbers: permutations with exactly two fixed points. ABC m=3 letters permutations two (2)fixed points n1=0 AABBCC m=6 letters permutations two (2) fixed points n2=27 AAABBBCCC m=9 letters permutations two (2) fixed points n3=378 AAAABBBBCCCC m=12 letters permutations two (2) fixed points n4=4536 AAAAABBBBBCCCCC m=15 letters permutations two (2)fixed points n5=48600 etc. etc, etc. etc. %O A000535 0 %K A000535 ,easy,nice,nonn, %A A000535 Zerinvary Lajos (zerinvarylajos@yahoo.com), Jul 08 2005 RH RA 82.141.159.17 RU RI Please help and write !! Zeinvary Lajos Hungary === Subject: Re: quick and easy proof of 4 Color Mapping that needs no computer 60% looks way too high to me. At the top levels, I'd expect about 50% of all games to be draws. Maybe it's true that 60% of all decisive games are won by White. At http://www.chessgames.com/chessstats.html it says, # of games in database: 355,326 Years covered: 1475 to 2005 (531 years) White wins 130,570 games (36.75%) Black wins 94,218 games (26.52%) 130,530 games are drawn (36.74%) I think what Ken was trying to say is that of the won games 60% were won by white and the other 40% won by black. But I want to comment on the practical issue of chess. In that there is no doubt an OS of chess. Now because the pros who play chess find it most often to be a draw conclusion and the fact that white winning about 60% to black 40% is a deep seated probe into the OS of chess. In other words, if the OS was a win for white, then one should expect most games played by pros to have white win say 80 to 90% of the time whereas black win 20 to 10% of the time. If it were the case that the OS is a draw then we should expect more than half of the games played by pros would be a draw. What I am saying is that the actual circumstances of games played over hundreds of years should be a probe or window into the OS of chess and from the data collected it appears to be saying that the OS is a draw. So I think my technique of white making 2 consecutive first moves and otherwise the game played normally and if a large number of pros draw such a game is more evidence that the OS of chess is a draw and I would say it can even offer a proof that it is a draw. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: quick and easy proof of 4 Color Mapping that needs no computer > 60% looks way too high to me. At the top levels, > I'd expect about 50% of all games to be draws. > Maybe it's true that 60% of all decisive games > are won by White. > At http://www.chessgames.com/chessstats.html it says, > # of games in database: 355,326 > Years covered: 1475 to 2005 (531 years) > White wins 130,570 games (36.75%) > Black wins 94,218 games (26.52%) > 130,530 games are drawn (36.74%) > I think what Ken was trying to say is that of the won games 60% were > won by white and the other 40% won by black. > But I want to comment on the practical issue of chess. In that there is > no doubt an OS of chess. Now because the pros who play chess find it > most often to be a draw conclusion and the fact that white winning > about 60% to black 40% is a deep seated probe into the OS of chess. > In other words, if the OS was a win for white, then one should expect > most games played by pros to have white win say 80 to 90% of the time > whereas black win 20 to 10% of the time. If it were the case that the > OS is a draw then we should expect more than half of the games played > by pros would be a draw. Someone (can't remember the name) took Grandmaster-level games and grouped them by opening, and then found the probability of White winning/Black winning/drawing using that opening. This would seem like a good idea, but maybe someone wins using the Doofus Opening 99 times, and the 100th opponent finds a way to defeat it easily and early on. If this game is published, then no one will play the Doofus Opening, despite the fact that it has a 99% chance of success. (In fact the chance of success would stay there, since no one would want to play it, ever again, and it will be still listed as 99/100; it would never get a chance to change to 99/1000 or 99/1000000, etc.) --- Christopher Heckman === Subject: Re: quick and easy proof of 4 Color Mapping that needs no computer Chris your logic of the Doofus Opening is wrong as a rebuttal to my claim that the OS influences the outcome of chess games. And the flaw of your rebuttal is that you neglect or omit the fact that if a game has an OS, that OS is going to have influence over the outcomes of millions of games played by professionals. So, chess has an OS and that OS will demand that either white wins, or black wins or it is a draw. So when millions of games played would indicate what that OS outcome is. The OS is like a river channel where the water will mostly flow to the ocean. In your rebuttal, you are claiming that the OS has no influence on games played by pros and that they find themselves at the last move as independent of a innate structure to chess. The reason so many pro chess games end up in a draw is because the OS is structured to be a draw. So the players are feeling themselves around for that OS, and because the OS is a draw, most of those games end up as a draw. If you were correct, then the OS has no impact on any move by any pro from beginning to end. If I am correct then the OS creates an internal structure and the pro players are banging around inside the OS which ends up usually as a draw game. If the OS of chess were a win for white, then the majority of pro games would end up as a white victory because of that internal structure that the OS creates. Yes, none of mine above is a mathematical proof that OS is a draw. It is experienced wisdom that the OS is a structure and that statistically the pro games will reflect what that internal structure is -- a draw. My method of 2 first moves by white if found not to be a victory for white is a mathematical proof that the OS of chess is not a win for white. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: quick and easy proof of 4 Color Mapping that needs no computer > Chris your logic of the Doofus Opening is wrong as a rebuttal to my > claim that the OS influences the outcome of chess games. [...] I'm not sure I meant it to be, but what it was _supposed_ to show is that you can't judge how good an opening is solely by its win/loss statistics, so statistical results should be taken with a grain of salt. My point (which you have erased) was: If an opening (Doofus Opening) results in a win in 99 games, then someone discovers a defense against it, then the Doofus Opening will have a +99=0-1 record, even though it's not a good opening to play. Note that such an opening, in the archive of chess games, would have a 99% success rate, which is much larger than any recommended opening (like Ruy Lopez); however, no one would ever win a game again if they played the Doofus Opening. That could easily be remedied by only considering games played within the last year, or last month. The absense of the Doofus Opening would show that it is not to be trusted. This suggests an interesting variation on the statisticsmethod: Count the percentage of games which are White wins, Black wins, and draws, If there is an OS for White winning, then that would suggest that as time goes on and openings are explored, White wins should make up a larger and larger percentage, as approximations to the OS would be found. (Note this does not follow directly. If you play the first 20 moves of the OS, that should put you at an advantage; but winning may depend on a zugzwang situation which shows up in move 50, so this isn't guaranteed.) --- Christopher Heckman === Subject: Re: quick and easy proof of 4 Color Mapping that needs no computer > Chris your logic of the Doofus Opening is wrong as a rebuttal to my > claim that the OS influences the outcome of chess games. [...] I'm not sure I meant it to be, but what it was _supposed_ to show is that you can't judge how good an opening is solely by its win/loss statistics, so statistical results should be taken with a grain of salt. My point (which you have erased) was: If an opening (Doofus Opening) results in a win in 99 games, then someone discovers a defense against it, then the Doofus Opening will have a +99=0-1 record, even though it's not a good opening to play. Note that such an opening, in the archive of chess games, would have a 99% success rate, which is much larger than any recommended opening (like Ruy Lopez); however, no one would ever win a game again if they played the Doofus Opening. That could easily be remedied by only considering games played within the last year, or last month. The absense of the Doofus Opening would show that it is not to be trusted. This suggests an interesting variation on the statisticsmethod: Count the percentage of games which are White wins, Black wins, and draws, If there is an OS for White winning, then that would suggest that as time goes on and openings are explored, White wins should make up a larger and larger percentage, as approximations to the OS would be found. (Note this does not follow directly. If you play the first 20 moves of the OS, that should put you at an advantage; but winning may depend on a zugzwang situation which shows up in move 50, so this isn't guaranteed.) --- Christopher Heckman Chris, I would confine those statistical games to just those very polished games by grandmasters playing other grandmasters. I believe the OS of chess structures the game so tightly that even in the opening moves the echo of the endresult of the OS is heard in the very first moves of the game. So the statistical outcomes, as you say, are not a proof that the OS is a draw since most games end up as draws by the grandmasters. But since the OS creates such a rigid internal structure that it very much suggests the OS is a draw. If the OS of chess was a victory for White, then the historical statistics of grandmaster play would have swung largely in a reality that white grandmasters win alot of times, but as is the case white grandmasters win about 50-50 over black. Maybe in the far future we can have a mathematical VonNeumann game theory theorem which tells us quite a bit more of how the OS forms an internal structure and the reality of playing the game end results have to sort-of-match or reflect what this OS is. Analogy: think of driving on roads as a VonNeumann game where you want to end up at the endpoint safe and sound. Now if the roads are straight and wide we can zoom along at high speeds, but if a road is narrow and very windy with some knife end turns such as a mountain rode and the game is played at night as well as day. So in this picture we can sense that the OS has an internal structure that the players can sense from the very beginning in that they would have to go slow speed or lose the game by flying off the curb and into a ditch or off the mountain. So the OS creates and forces an internal structure all along the way of the entire game itself. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Condition on two beta distributions >>I was wondering whether it is possible to maintain two beta marginal(with >>set mean and variance) distributions and use a copula?? [ ... ] > I think the word you want is coupling, not copula. [ ... ] Forgive me my ignorance. The only thing I know is that copula's do indeed exist: http://mathworld.wolfram.com/Copula.html Han de Bruijn === Subject: Re: Condition on two beta distributions Yes I think we both know that; but in this case Robert Israel is definitely right. the word i shoulod have used was coupling or joint distribution. === Subject: Re: Condition on two beta distributions Sorry I posted slightly too soon. How do you know that a_1 < a_2 and b_1 > b_2 will give you x <= y Eoin === Subject: Re: Condition on two beta distributions >Sorry I posted slightly too soon. >How do you know that a_1 < a_2 and b_1 > b_2 will give you x <= y >Eoin A/(A+C+D) <= (A+B)/(A+B+D) if A, B, C, D are all positive numbers. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: extremum for x: (0,2pi) My extremum is for x: (0,2pi) === Subject: extremum f(x)=|sinx| Why have the function extremum in xo=pi, xo=pi/2, xo=3pi/2 ? f'(x)=0 for x ?? f'(x)>0 for x ?? f'(x)<0 for x ?? Please help me calculate it === Subject: Re: extremum > f(x)=|sinx| > Why have the function extremum in xo=pi, Because f(pi) = 0 and f(x) >= 0 for each _x_. > xo=pi/2, Because f(pi/2) = 1 and f(x) <= 1 for each _x_. > xo=3pi/2 ? Same as above. > f'(x)=0 for x ?? > f'(x)>0 for x ?? > f'(x)<0 for x ?? > Please help me calculate it Calculate *what*? Jose Carlos Santos === Subject: appendix, Latex (help) Was wondering if anyone can help with following: I have chapters 1 -4 and then an appendix. The appendix is showing up properly in table of contents, but on the tops of the pages it still says i'm in chapter 4. Does any one know how to fix? === Subject: Re: appendix, Latex (help) > Was wondering if anyone can help with following: > I have chapters 1 -4 and then an appendix. > The appendix is showing up properly in > table of contents, but on the tops of the pages > it still says i'm in chapter 4. Post the same question at the comp.text.tex newsgroup, but I suggest that you add a minimal example. Jose Carlos Santos === Subject: Re: Finding minimum in increasing function + decreasing function <438F5169.7030609@netscape.net> You all are right. === Subject: [Dynamical Systems] Convergence to a fixed point I have a differetiable map T with an unstable fixed point x_0. How can I estimate how many iterations does it take for a point in the stable manifold at a distance O(1) from x_0 to reach a distance O(epsilon)? === Subject: Re: [Dynamical Systems] Convergence to a fixed point >I have a differetiable map T with an unstable fixed point x_0. I mean hyperbolic fixed point. > How can I estimate how many iterations does it take for a point in the > stable manifold at a distance O(1) from x_0 to reach a distance > O(epsilon)? I mean the stable manifold *of x_0*. === Subject: Re: [Dynamical Systems] Convergence to a fixed point >>I have a differetiable map T with an unstable fixed point x_0. >I mean hyperbolic fixed point. >> How can I estimate how many iterations does it take for a point in the >> stable manifold at a distance O(1) from x_0 to reach a distance >> O(epsilon)? >I mean the stable manifold *of x_0*. Let S be the set of eigenvalues of the Jacobian matrix of T at x_0 that have absolute value <= 1, and r the maximum of the absolute values of these. If r < 1, then for typical starting points y_0 in the stable manifold the sequence of iterates y_n should have ||y_{n+1} - x_0||/||y_n - x_0|| -> r as n -> infinity. For any r' with 1 > r' > r, the number of iterations to go from distance 1 to epsilon should be less than ln(epsilon)/ln(r') for sufficiently small epsilon. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: [Dynamical Systems] Convergence to a fixed point > How can I estimate how many iterations does it take for a point in the > stable manifold at a distance O(1) from x_0 to reach a distance > O(epsilon)? >>I mean the stable manifold *of x_0*. > Let S be the set of eigenvalues of the Jacobian matrix of T at x_0 > that have absolute value <= 1, and r the maximum of the absolute > values of these. If r < 1, then for typical starting > points y_0 in the stable manifold the sequence of iterates y_n > should have ||y_{n+1} - x_0||/||y_n - x_0|| -> r as n -> infinity. > For any r' with 1 > r' > r, the number of iterations to go from > distance 1 to epsilon should be less than ln(epsilon)/ln(r') for > sufficiently small epsilon. The problem is that the Jacobian matrix don't need to be hyperbolic along all the orbit because it start far away from the fixed point... If I draw the restriction of T along the stable manifold (it can be seen as an analytic function from the reals to themselves) I could find that it has different behaviors at different distances from the fixed point, I think. It could spend an arbitrarily large numbers of iterations before entering the epsilon neighbourhood, this number depending on the topological structure of this restriction... === Subject: Re: Topologies of Spec Getting back to this thread again.. I'd like a little advise on Spec Z (Z= whole numbers) .. so, the primes ideals are those (countably) generated by primes in Z, great.. but the open sets are? Are the open sets the set of numbers that arent divisible by a finite number of primes? Then there is the structure sheaf of this affine scheme. Is there an instinctive way to see how the the stalks look like? Like for instance, I was just discussing this with my professor and he suddenly remarked that the stalk of this sheaf at 0 is Q (rationals).. and I was like.. huh.. let me thing a few hours on this.. and Im not yet sure if I see this (honestly Im still very uncomfortable with affine schemes alone) === Subject: Re: Topologies of Spec > Getting back to this thread again.. I'd like a little advise on Spec Z > (Z= whole numbers) .. > so, the primes ideals are those (countably) generated by primes in Z, > great.. but the open sets are? Are the open sets the set of numbers > that arent divisible by a finite number of primes? The Zariski topology of Spec(Z) is the socalled cofinite topology. But first of all let us see: What are the points of Spec(Z)? Spec(Z) consists of all prime ideals in Z, hence Spec(Z) = P u {<0>} (disjoint union) with P={

s.t. p prime}. The closed sets are the finite subsets of P, such that <0> is contained in any (non-void) open subset of Spec(Z). Thus <0> is the generic point of Spec(Z), and <0> can be imagined as existing everywhere above P - in topological terms. To get a bit more accustomed to the affine scheme: What are the standard affine subsets D_a (a e Z)? > Then there is the structure sheaf of this affine scheme. Is there an > instinctive way to see how the the stalks look like? Of course. This can be done in general. If A is a unitary commutative ring, let (X,O_X) be the associated affine scheme. Let pp be a prime ideal of A which is denoted by x if considered as point of X=Spec(A). Then it holds that the stalk O_{X,x} of the structure sheaf O_X in x is naturally isomorphic to the localization A_pp of A at pp. > Like for instance, > I was just discussing this with my professor and he suddenly remarked > that the stalk of this sheaf at 0 is Q (rationals).. and I was like.. > huh.. let me thing a few hours on this.. and Im not yet sure if I see > this (honestly Im still very uncomfortable with affine schemes alone) Apply my last remark to: Let A be a domain, what is the localization of A at <0> naturally isomorphic to? Hope this can enlighten you a bit - and pushes you a bit closer to the geometeric view in algebra. Best wishes, J. === Subject: Atlas module general diffeomorphism group = ? I thought Atlas [for a manifold] module general diffeomorphism group [symmetry group of GR] = Frame Bundle, but there may be some subtlety. === Subject: Vector Sequence Suppose V is a vector space over a field F, and dim(V)=n>0. Prove that, there is a vector sequence S = {alpha_k in V; k=1..oo}, in which any n vectors form a basis of V. === Subject: Re: Vector Sequence <25494719.1133522918419.javamail.jakarta@nitrogen.mathforum.org>, yo |Suppose V is a vector space over a field F, and dim(V)=n>0. Prove |that, there is a vector sequence S = {alpha_k in V; k=1..oo}, in |which any n vectors form a basis of V. given such a sequence s for v, s' := q, alpha_1+q, alpha2+q, ... is obviously such a sequence for v + . -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Vector Sequence the original problem of mine was solved. Then, Daniel Grubb took the problem further. However, I'd like to restate the problem clearly. Suppose V is a vector space over F, and dim(V)=n>0. (i) If F is a subfield of C (complex number field), prove that, there exists an infinite subset of V, say S, such that any n vectors (they're different from each other) in S form a basis of V. (ii) Alternatively, if F is just an infinite field, prove the same conclusion above. I wish I did better than before. === Subject: Re: Vector Sequence > the original problem of mine was solved. > Then, Daniel Grubb took the problem further. > However, I'd like to restate the problem clearly. > Suppose V is a vector space over F, and dim(V)=n>0. > (i) If F is a subfield of C (complex number field), > prove that, > there exists an infinite subset of V, say S, such > that > any n vectors (they're different from each other) in > S > form a basis of V. > (ii) Alternatively, if F is just an infinite field, > prove the same conclusion above. > I wish I did better than before. The proofs of (i) & (ii) are of the same. With the help of Jyrki Lahtonen's hint, I'll prove (ii). Since the field F is infinite, we can find a squence { a_k (!= 0) in F; a_k != a_j for j=1..k-1 }. Suppose {v_1,v_2,...,v_n} be a basis of V. Construct a vector squence { w_k = 1*(v_1) + (a_k^1)*(v_2) + ... + (a_k^(n-1))*(v_n) }. Let x_1*w_(k_1) + x_2*w_(k_2) + ... + x_n*w_(k_n) = 0, then we get the simultaneous equations below: 1*x_1 + ... + 1*x_n = 0, (a_(k_1)^1)*x_1 + ... + (a_(k_n)^1)*x_n = 0, ......................., (a_(k_1)^(n-1))*x_1 + ... + (a_(k_n)^(n-1))*x_n = 0. Obviously, the coefficient matrix A is Vandermonde's. Since a_k != a_j while k != j, we see that det(A) != 0. Thus, x_1 = x_2 = ... = x_n = 0. And (ii) was proved. === Subject: Re: Vector Sequence Originator: grubb@lola >Suppose V is a vector space over a field F, >and dim(V)=n>0. >Prove that, there is a vector sequence >S = {alpha_k in V; k=1..oo}, >in which any n vectors form a basis of V. F must be infinite. It is enough for any 'n' elements of the sequence to be independent. So, let {v_1, ...v_n} be a basis. This starts the induction. Now assume that {v_1, ... v_k} have been chosen. Look at all the n-1 dimensional hyperplanes spanned by subsets of {v_1,...v_k}. The key is that the union of these hyperplanes is not all of V. Select v_{k+1} to be outside of all of these hyperplanes. The induction continues. --Dan Grubb === Subject: Re: Vector Sequence > Suppose V is a vector space over a field F, > and dim(V)=n>0. > Prove that, there is a vector sequence > S = {alpha_k in V; k=1..oo}, > in which any n vectors form a basis of V. If you meant to write in which any _n_ *consecutive* vectors form a basis, then a solution is: let {v_1, v_2, ..., v_n} be a basis of V and consider the sequence v_1, v_2, ..., v_n, v_1, v_2, ..., v_n, v_1, v_2, ..., v_n, ... Jose Carlos Santos === Subject: Re: Vector Sequence >Suppose V is a vector space over a field F, >and dim(V)=n>0. >Prove that, there is a vector sequence >S = {alpha_k in V; k=1..oo}, >in which any n vectors form a basis of V. This can hardly be true as stated, unless either your understanding of sequence (or perhaps of any n vectors in a sequence) is very different than mine. If F is finite, then any infinite sequence S as above will be such that there is some element v of V and some infinite subset X of indices k such that for all k in X, alpha_k equals v; any n-subset of X provides n vectors in S that are not a basis. So, what do you really mean? Lee Rudolph === Subject: Re: Vector Sequence >>Suppose V is a vector space over a field F, >>and dim(V)=n>0. >>Prove that, there is a vector sequence >>S = {alpha_k in V; k=1..oo}, >>in which any n vectors form a basis of V. >This can hardly be true as stated, unless either your understanding >of sequence (or perhaps of any n vectors in a sequence) is >very different than mine. If F is finite, then any infinite sequence >S as above will be such that there is some element v of V and some >infinite subset X of indices k such that for all k in X, alpha_k >equals v; any n-subset of X provides n vectors in S that >are not a basis. >So, what do you really mean? I'd hesitate to speculate on what he meant, but it seems like the statement is true if F is infinite... Yup. You can construct an example if you can show that V is not the union of finitely many (n-1)- dimensional subspaces. Say v.w = sum(v_j w_j). You need to show that given a finite family F of non-zero w's there exists a non-zero v such that v.w is non-zero for all w in F. By induction: Write v = (v_1, v'), where v' is in F^(n-1). By induction you find a v' such that v'.w' is non-zero for all the w's in F such that w' is non-zero. Noting that if it happens that w' = 0 then w_1 <> 0 you see that for each w rules out at most one choice of v_1, so there is a v_1 that makes v.w <> 0 for all w in F. >Lee Rudolph ************************ David C. Ullrich === Subject: Re: Vector Sequence Sorry, I made a mistake. F is a NUMBER field. That means F is a subfield of the complex number field C. === Subject: Re: Vector Sequence [I believe that the Math Forum makes it difficult, but not to include appropriate context. Please try to use it that way if possible.] >Sorry, I made a mistake. >F is a NUMBER field. >That means F is a subfield of the complex number field C. In fact, I think number field *usually* means finite extension of Q (inside C); for example, R is not (usually) called a number field. In any case, I am beginning to believe that you must mean, by sequence, what I would call an enumeration: that is, for k not equal to m, alpha_k is not equal to alpha_m. If you mean, instead, what I mean by sequence, then the other respondent has already given a trivial solution to your problem. Even if you do mean enumeration, as long as the ground field F is infinite, that trivial solution (to what is then a different problem) can be trivially modified to give a trivial solution to the new problem. Thus, let alpha_1,...,alpha_n be a basis of V, let t_0 =1, t_1, t_2,... be an enumeration of the non-zero elements of V, and then for k in {1,...,n} and j greater than or equal to 0 define alpha_{k+jn} = t_j alpha_k. So, again, I repeat: what do you really mean? Lee Rudolph === Subject: Re: Vector Sequence ... >In any case, I am beginning to believe that you must mean, >by sequence, what I would call an enumeration: that is, >for k not equal to m, alpha_k is not equal to alpha_m. If >you mean, instead, what I mean by sequence, then the other >respondent has already given a trivial solution to your problem. with the proviso, which he added, that you really meant any n consecutive vectors. I carried that along, forgetting it was his addition, into the example below. >Even if you do mean enumeration, as long as the ground field >F is infinite, that trivial solution (to what is then a different >problem) can be trivially modified to give a trivial solution >to the new problem. Thus, let alpha_1,...,alpha_n be a basis >of V, let t_0 =1, t_1, t_2,... be an enumeration of the non-zero >elements of V, and then for k in {1,...,n} and j greater than or >equal to 0 define alpha_{k+jn} = t_j alpha_k. >So, again, I repeat: what do you really mean? Lee Rudolph P.S. Of course my assumption is that what you really mean is what Ullrich thinks you really mean. === Subject: Re: Vector Sequence > Sorry, I made a mistake. > F is a NUMBER field. > That means F is a subfield of the complex number field C. This is an obvious homework problem, so a hint is all you are gonna get from me. Here comes: Have you ever heard of Vandermonde? Jyrki === Subject: Question about constant-width surface (not curve) I am solving a problem from the differential geometry book by R.Millman and G.Parker, p.194. It covers about constant-width surfaces, and advice to treat it as an analogue of constant-width curves. Though its statements seems natural, however, the rigorous proof is so far from me. Can you survey this? M is a compact surface, and the Gaussian curvature K is anywhere > 0. So this surface is convex in terms of surface (i.e. the normal map is 1-1 and onto, and there is only two points such that their tangent planes are parallel. Think of somewhat like sphere or ellipsoid.) Then assume the width between these planes are always constant regardless of the point selected, and the line between the point P and 'opposite' one Q is normal to the both surfaces. I understand and proved, without confidence. But from here, I can imagine what that will be, but it is only an imagination. 1. Principal directions at P and Q are each parallel. 2. If k is a principal curvature in the direction of X, which is in the tangent plane of P, then the same direction, which is also a principal direction by 1, makes a principal curvature in Q, say k'. Then 1/k + 1/k' = const. (I suspect it will be the width.) (3. This is just my conjecture: Gaussian curvature is same at P and Q.) Byunghyun Oh Department of Mathematics, POSTECH, South Korea === Subject: Re: Strange Fourier Transform problem Actually, in retrospect the gamma() solution is not that useful to me; I need something that directly gives H(w) without need to evaluate an integral. It's strange that the function exp(-x)/exp(exp(-x)) would have such a dramatic frequency-domain characteristic; it rolls off very steeply at higher frequencies(as evidenced by taking the FFT of this function, using a very dense time-sampling grid). It seems like there is a hole to be filled here, and it smells kind of fundamental to me. This formula popped up in my search to find a logarithmic sampling theorom; that is, are there any functions that can be completely reconstructed from samples that occur on log(integer) time points. The function exp(-x)/exp(exp(-x)) seems to play the same role for log-sampled signals that sin(x)/x plays in the reconstruction of linearly-sampled signals. This interpretation comes from one of the equations that equates the Reimann Zeta function with the series (SUM of N^-s), which can be interpreted as a logarithmic time series if you evaluate it on a vertical line in the complex plane. === Subject: Re: Strange Fourier Transform problem >Actually, in retrospect the gamma() solution is not that useful to me; >I need something that directly gives H(w) without need to evaluate an >integral. You don't need to evaluate an integral, you need to evaluate the Gamma function. That's a well-known special function, and good procedures to evaluate it numerically are available, e.g. in Maple, as well as all sorts of properties. For example, asymptotically we have ln(Gamma(1 + i w)) = i w ln(w) - (pi/2 + i) w + ln(w)/2 + pi i/4 + ln(2 pi)/2 - i/(12 w) + O(1/w^4) giving a very good approximation to Gamma(1 + i w) for large w. > It's strange that the function exp(-x)/exp(exp(-x)) would have such a >dramatic frequency-domain characteristic; it rolls off very steeply at >higher frequencies(as evidenced by taking the FFT of this function, >using a very dense time-sampling grid). It seems like there is a hole >to be filled here, and it smells kind of fundamental to me. Actually, according to the asymptotic formula above, |Gamma(1+iw)| decays exponentially as w -> infinity, essentially like sqrt(w) exp(-w pi/2). But because of the i w ln(w) it oscillates more and more rapidly. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: showing something is a topological group >> > > >> i read this question somewhere in this group... >> >> I seem to have become a bit..stuck on topological groups. >> Im trying to show that all the homeos on the circle (multiplication >> being >> the binary operation) is a topological group. > [...] >i know this is the triangle inequality, i know i am missing a REAL >elementary stage of my reasonsing, what i want to ask is if i prove the >above, >How does that show that i have proved that group multiplication is >continuous? by showing the triangle inequality??? >> Looks like you've missed the point to the last few posts: >> Everything you've done is totally wrong. The set of homeomorphisms >> is _not_ a group with multiplcation as the operation! You need to >> use _composition_ to make it a group. Has to be a group before >> it has much chance of being a topological group... >ok i see that... my mistake >but how does the triangle inequality help me prove that/? Prove what? Prove that the metric above makes this thing into a topological group? Two comments: (i) Since you only just now got straight what the group _operation_ is supposed to be you should think about the problem a little bit before asking. (ii) While you're thinking about it, keep in mind exactly what it is you're trying to prove! Say you have a group, and a metric defined on that group. Exactly what does it mean to say that that metric makes the group into a topological group? ************************ David C. Ullrich === Subject: Re: How to negate defintion to prove that a function is not uniformly continuous? >Ken, I still have a few questions for your symbolic approach: >> Also remember ~(p -> q) is equivalent to p ^ ~q. >Why not > ~(p -> q) is equivalent to p -> ~q ??? Because it's _not_ equivalent to that. >I have also seen that and and imply used interchangeably in >books... No, you most certainly have not seen and and imply used interchangeably. You need to learn to read things _much_ more carefully. Really. A little while ago you were complaining about this book being unclear. If you're at the stage where you think that and and imply mean the same thing you have no chance at all here - the problem is not the book, the problem is that you're not even _close_ to being ready for material at this level. When you're confused about whether and and imply mean the same thing _any_ book is going to be unclear! You really _should_ take a lower-level itro to proof course or something first. >since you brought this up, are they the same? >> Your definition says >> (Ae>0)(Ed>0)(Ax)(Ay)((|x-y| < d) -> (|f(x)-f(y)| < e)). >> Negate all that step by step using the above rules to get >> (Ee>0)(Ad>0)(Ex)(Ey)((|x-y| < d) ^ ~(|f(x)-f(y)| < e)) >You changed the order of (Ae>0)(Ed>0) to (Ee>0)(Ad>0), but maintain the >order of other parts, why? >Also, where do you place such that? ************************ David C. Ullrich === Subject: Re: How to negate defintion to prove that a function is not uniformly continuous? > .... > Ken, I still have a few questions for your symbolic approach: > Also remember ~(p -> q) is equivalent to p ^ ~q. > Why not > ~(p -> q) is equivalent to p -> ~q ??? > I have also seen that and and imply used interchangeably in > books... since you brought this up, are they the same? No! Look at the truth tables for and and implies, and you'll see that they differ when p is false. Have you really seen them used interchangeably in books? > .... > Your definition says > (Ae>0)(Ed>0)(Ax)(Ay)((|x-y| < d) -> (|f(x)-f(y)| < e)). > Negate all that step by step using the above rules to get > (Ee>0)(Ad>0)(Ex)(Ey)((|x-y| < d) ^ ~(|f(x)-f(y)| < e)) > You changed the order of (Ae>0)(Ed>0) to (Ee>0)(Ad>0), but maintain the > order of other parts, why? I didn't change the order. In each case the first quantifier concerned e (epsilon) and the second concerned d (delta). You showed that you already knew the rules for negating quantifiers. > Also, where do you place such that? Just get the logical structure right first, and then add any such wording that makes the English sound sensible. Ken Pledger. === Subject: Re: geodesics in hyperbolic plane schrieb berthuffman@gmail.com : >> Its kind of hard to show my computations, but here is most of it: >> First, metric and its inverse on H >> g_11 = y^{-2} g_12 = g_21 = 0 g_22 = y^{-2} >> g^11 = y^2 g^12 = g^21 = 0 g^22 = y^2 >> chirstoffel symbols >> G_12^1 = G_22^2 = -y^{-1} >> G_11^2 = y^{-1} >> G_11^1 = G_12^2 = G_22^1 = 0 >> (not listing the symmetric symbols) >> if V=d(gamma)/dt=(v_1(t),v_2(t)) then the covariant derivative is >> DV/dt = sum_k (dv_k/dt + sum_{i,j} G_ij^k v_i v_j) X_k >> we see that (v_1(t),v_2(t)) = (0,1) >> so dv_k/dt = 0. also since v_1=0 we see that the G_ij^k v_i v_j will be >> zero unless i=j=2. so it seems we only get one contribution from the >> sum: >> DV/dt = G_22^2 v_2 v_2 X_2 = -y^{-1} X_2 >> where X_2 is the standard basis vector in the y direction i.e. @/@y. >> I know my notation is really sloppy but I think the basic idea is >> there. >> does this have anything to do with gamma not being parameterized >> correctly, or do I simply have a mistake in my computation? > Can anyone help with this computation? I know my work is really sloppy > so if someone could just point me the right direction (first of all, is > it true that the covariant derivative of the curve is zero?) Look at Robert Israel's answer, and ask yourself what the first point is that you do not understand. By the way: In this file http://www.math.utah.edu/~treiberg/Hilbert/Hilber.pdf there is on page 3 a different proof that vertical lines are geodesics. - The idea is to show directly that these lines are locally length-minimizing. === Subject: Re: geodesics in hyperbolic plane <438d9d5e$9$fuzhry+tra$mr2ice@news.patriot.net> 11/30/2005 >I know my notation is really sloppy but I think the basic idea is >there. Okay, now reparameterize by arc length and calculate the covariant derivative. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Integer Linear Programming does anyone know the author/developer of Integer Linear Programming? I need a paper/book as reference for my master's thesis. Chris === Subject: Re: Integer Linear Programming The first to develop an algorithm for integer linear programming was Ralph E. Gomory. His algorithm was called the cutting plane method. Look for his papers in the 1960's and 1970's. Unfortunately, his algorithm was most inefficient and today there are much better methods. Happy hunting! Ray Steiner === Subject: Re: Integer Linear Programming > does anyone know the author/developer of > Integer Linear Programming? Integer Linear Programming (or Integer Programming) is a type of optimization problem. I don't know if it can be said to have an author. Perhaps what you want is a classic algorithm paper. > I need a paper/book as reference for my master's thesis. Searching through the papers of George Dantzig in the 1950s should be a good start. If he didn't originate the classic algorithms, he probably helped develop them. You might also do a search on some of the classic IP problems such as the Traveling Salesman Problem. - Randy === Subject: Compare (1-1/(2^n))^p(n) with 1-1/p(n) I'm solving one probability problem where the probability of the event is (1-1/(2^n))^p(n), n - natural and p - any positive polynomial. I want to compare the behavior of this probability with 1-1/p(n) for sufficiently large n. Is it true that (1-1/(2^n))^p(n) > 1-1/p(n) if n are large enough? === Subject: Re: Compare (1-1/(2^n))^p(n) with 1-1/p(n) > I'm solving one probability problem where the probability of the event > is (1-1/(2^n))^p(n), n - natural and p - any positive polynomial. I > want to compare the behavior of this probability with 1-1/p(n) for > sufficiently large n. Is it true that (1-1/(2^n))^p(n) > 1-1/p(n) if n > are large enough? Sure. (1 - 1 / (2^n)) ^ p(n) > (1 - p(n) / (2^n)), so all we need show is that p(n) / 2^n < 1 / p(n) for sufficiently large n. This is equivalent to p(n) ^ 2 < 2^n for sufficiently large n, which is true, because an exponential function dominates any polynomial function as n goes to infinity. === Subject: Re: Compare (1-1/(2^n))^p(n) with 1-1/p(n) (1-1/x) ^ a > 1 - a/x - how could I forget about it! === Subject: Re: suitable interpolation polynomial? problem: I need ONE special coefficient of the interpolation polynomial, hoping not to be obliged to compute all data points necessary to build the whole interpolation polynomial. Unfortunately, it is not the constant coefficient. Also, I don't see how to simplify the occurring linear equation. > If the polynomial has integer coefficients and you know a bound for > them, you can do it with only one function evaluation. Thus suppose > your polynomial is P(x) = sum_{j=0}^n a_j x^j, you want a_m, and > you know all |a_j| <= B for j <= m. Then find a prime p > 2B+1 and > evaluate P(p) - 2^(-1) mod p^{m+1}. > Write this in base p as sum_{j=0}^m c_j p^j for integers 0 <= c_j < p, > and the coefficients are a_j = c_j - (p-1)/2 for j <= m. Actually my polynomial is defined over a finite field. I said p-adic, because I was worried I wouldn't get any reply if the computation is not in characteristic 0. Additionally, it's easy to transfer any result to the corresponding finite field. Because I was not yet able to figure out why your method works, Mr. Israel, could you please tell me if it is getting easier/harder/impossible in characteristic >0? Best, Oswald Kluge === Subject: Re: suitable interpolation polynomial? > problem: I need ONE special coefficient of the interpolation > polynomial, hoping not to be obliged to compute all data points > necessary to build the whole interpolation polynomial. > Unfortunately, it is not the constant coefficient. > Also, I don't see how to simplify the occurring linear equation. > If the polynomial has integer coefficients and you know a bound for > them, you can do it with only one function evaluation. Thus suppose > your polynomial is P(x) = sum_{j=0}^n a_j x^j, you want a_m, and > you know all |a_j| <= B for j <= m. Then find a prime p > 2B+1 and > evaluate P(p) - 2^(-1) mod p^{m+1}. > Write this in base p as sum_{j=0}^m c_j p^j for integers 0 <= c_j < p, > and the coefficients are a_j = c_j - (p-1)/2 for j <= m. > Actually my polynomial is defined over a finite field. I said p-adic, > because I was worried I wouldn't get any reply if the computation is > not in characteristic 0. Additionally, it's easy to transfer any result > to the corresponding finite field. I don't understand why you'd say p-adic if you meant a finite field. There are people here who are quite comfortable working in finite fields. I'm not one of them. > Because I was not yet able to figure out why your method works, Mr. > Israel, could you please tell me if it is getting > easier/harder/impossible in characteristic >0? Working in a finite field changes matters completely, I think, and makes the method impossible. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: suitable interpolation polynomial? > Hi! I want to compute one special coefficient of a univariate > polynomial, the problem is I don't know the polynomial itself, only > it's degree and I can compute any value of the polynomial, but it's > very costly to do so. > For example, if you look at the Lagrange interpolation polynomial, you > can read off every coefficient immediately, but you would have to > compute all values defining the interpolation polynomial, because every > value shows up in every coefficient. > Are there interpolation polynomial forms that don't build up their > coefficients from all the values at once? My common sense tells me this > is impossible, but maybe you know better? Additionally, the polynomial > is defined over the p-adic numbers, so I'm not sure I can work with > trigonometric functions. > Any help would be appreciated! Everyone has refrained from making the trivial point that if the special coefficient you're after is the constant term then one value, that at zero, will suffice. If you're after, say, the coefficient of x^73, you could try to compute (1 / 2 pi) integral-around-unit-circle (f(z) / z^74) dz. What I really mean is compute a good numerical approximation to that integral. I don't know how hard that would be, and I don't know what the p-adic analogue would be. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: ? any simple relations between algebraic operation result of two self-adjoint square matrix > No, but the trace (which is the sum of the eigenvalues, counting > multiplicities) of C is the sum of the traces of A and B. > If A and B are hermitian, each eigenvalue of C is in the interval > [a_{min} + b_{min}, a_{max} + b_{max}], where a_{min} and a_{max} > are the minimum and maximum eigenvalues of A, b_{min} and b_{max} > the minimum and maximum eigenvalues of B. Found in Horn's Intro matrix analysis the proof for part of the above: abs( e-value(C)_Max ) <= abs( e-value(A)_max )+abs( e-value(B)_max ) for slef adjoint A and B matrices. But do not see the min bound. The above bound just remind me about if the below exist: max( e-value(A)_Max, e-value(B)_Max ) <= e-value( f(A,B) )_Max and min( e-value(A)_min, e-value(B)_min ) >= e-value( f(A,B) )_min Are there such functions that makes the above relations hold (either one holds or both hold)? Is it does, what is that function? by Cheng Cosine Dec/02/2k5 NC === Subject: Re: ? any simple relations between algebraic operation result of two self-adjoint square matrix >>Hi: >> Given two square matrices A and B of the same size. >>Are there any relations between the resulting matrix C come >>from the algebraic operations on A and B? >>For example, >>C = A+B >>Are there any simple relations between the eigenpair of C and the >>original matrices A and B? Say, the e-values of C is the sum of >>the corresponding e-values of A and B? > No, but the trace (which is the sum of the eigenvalues, counting > multiplicities) of C is the sum of the traces of A and B. But how to show this or which book gives the proof? Any physical interpretation for this? This seems to be an invariant under the addition operation. Any other invariant? > If A and B are hermitian, each eigenvalue of C is in the interval > [a_{min} + b_{min}, a_{max} + b_{max}], where a_{min} and a_{max} > are the minimum and maximum eigenvalues of A, b_{min} and b_{max} > the minimum and maximum eigenvalues of B. Agian, how to show this? Lastly, does this imply that C = weighted average of A1, A2, ..., An, where those N-by-N matrices An's are all self adjoint. Then the e-values of C fall in the interval [min(evalues of all A's), max(evalues of all A's)]? by Cheng Cosine Dec/01/2k5 NC === Subject: How gracefully do BCH codes degrade? Given random inputs J1, J2 such that the Hamming distance between J1 and J2 equals d. As a function of d, what is the probability that the BCH parity block for J1 will equal that of J2? For a (1023,963) code, the probability of matching parity blocks should be 0 for 0 Given random inputs J1, J2 such that the Hamming distance between J1 > and J2 equals d. As a function of d, what is the probability that the > BCH parity block for J1 will equal that of J2? you are missing the frame length, and the parity block length. d usally means min distance of the code. however, you seem to be asking the distribution of distances between codewords. Specifically, the number of a particular distance. That is a possion distribution, with the peak about 1/2 of the block length, and the part near 0 to min distance truncated, (it look like a normal distribution). > For a (1023,963) code, the probability of matching parity blocks should > be 0 for 0but I'm curious whether there > exist values of d for which the probability of collision is > significantly higher than the background level 2^-60. (I don't care > about insignificant differences like 2^-60 * 1.0000000001, nor about > tendencies that only apply to shorter parity blocks.) collision process is externally caused. are you talking about just the code detection capabilites? you may be refering to the code's undetected error rate which i think may be calculated for BCH. you can expand the binomial theroem and use weights on each term to determine undetected error for a code, and that will show you for a particular d, what its contribution is. Problem is finding the weights, but for a code with a min distance of 13 the first 13 terms are zero, and the 14th is not. > Also, are there any nonzero inputs with glaringly obvious patterns that > tend to hash to 0? A glaringly obvious pattern would be something > like 10101010.... or 111...000.... That is external to the code, and done by another layer. > The reason I ask these question relates to a mailing list discussion of > the suitability of BCH codes for hashing board positions in go (a > strategy game). It seems none of us on the list have both the > background and interest to evaluate BCH adequately. If BCH codes can > guarantee uniqueness for hashes of the disproportionate number of > similar positions (i.e. low Hamming distance inputs) being analyzed, > while having a negligible effect on the collision rate for less similar > ones, then that's about as good as one can hope for. The collision > rate tends to be acceptably low using purely random codes anyhow, but > we've discussed the matter enough that a definitive solution would be > nice. The code can be simpler, any prime CRC, 64 or 128 bit long, or longer. Parity block length k is the key factor (2^k or 2^k-1) For more robustness, you can add another code on top of it all. === Subject: arc length formed by a circle in a circle I've done a fair amount of searching without finding an answer, so I hope some of you may be able to help me out. Start out with two circles, one large (A) and one small (B). Circle B sits inside circle A such that they are tangent. Both diameters are known. Now from the center point of A, we draw two lines outward so that these are tangent to each side of circle B. I'm trying to determine the arc length between these two lines of circle A. Equivalently, I would like to know the angle created by these two lines. I feel that this should be trivial, but it's been many years since I've used geometry and trigonometry. Hopefully one of you may see what I don't. -Jason === Subject: Re: arc length formed by a circle in a circle > Start out with two circles, one large (A) and one small (B). Circle B > sits inside circle A such that they are tangent. Both diameters are > known. Now from the center point of A, we draw two lines outward so > that these are tangent to each side of circle B. I'm trying to > determine the arc length between these two lines of circle A. This is not possible if the center of A is inside circle B. So you need to include the premise radius B < 1/2 radius A. > Equivalently, I would like to know the angle created by these two > lines. -- To Google and MathForum users: Reply only if adequate context is included _within_ the reply. Use the quote feature. Otherwise, if you insist upon using quick reply, don't reply, as all the contexts are removed from view, the flow of thought messed up and chaos reigns. === Subject: Re: arc length formed by a circle in a circle > I've done a fair amount of searching without finding an answer, so I > hope some of you may be able to help me out. > Start out with two circles, one large (A) and one small (B). Circle B > sits inside circle A such that they are tangent. Both diameters are > known. Now from the center point of A, we draw two lines outward so > that these are tangent to each side of circle B. I'm trying to > determine the arc length between these two lines of circle A. > Equivalently, I would like to know the angle created by these two > lines. > I feel that this should be trivial, but it's been many years since I've > used geometry and trigonometry. Hopefully one of you may see what I > don't. > -Jason circle is less than half of the diameter of the larger circle. Assuming that: Call C the centre of A, D the centre of B. Also, call r the radius of the smaller circle and R the radius of the larger one. Let T be the point of contact of one of the tangent lines with the smaller circle. Then the angle DTC is right, and from symmetry, the half-angle between the tangent lines has sine equal to r/(R-r). Hope it helps, ZVK(Slavek) === Subject: Re: arc length formed by a circle in a circle Let the centre of the big circle be O, that of the wee circle P, the radius of the big circle R and that of the wee circle r. The way the lines are described, we must have 2*r < R (or O would be inside the wee circle). If the lines touch the wee circle at A and B then the triangle OAP has a right angle at A, and the length of AP is r. Since the circles share a tangent, the length of OP is R-r. So the sine of the angle at O is r/(R-r), and the arc length you want is R*2*arsin(r/(R-r)) Duncan === Subject: Re: arc length formed by a circle in a circle if R radius of A, and r is radius of B with r < R/2, and z is the angle formed by the two lines, then sin(z/2) = r/(R - r). (remember tangent line to a circle at point P, are orthogonal to radius passing through P). -- Quotes from The Weather Man: Robert Spritz: Do you know that the harder thing to do, and the right thing to do, are usually the same thing? Easy doesn't enter into grown-up life... to get anything of value, you have to sacrifice. === Subject: Re:Another theme of concatenated integers. >>** I just spotted an error, it should be 277! ** >> at end of post >>And now not only a triangle number connection to >>1234567891011..n, but a factorial connection as well! >>1234567891011 == r (mod 11!) >>1234567891011121314151617181920 >>212223242526272829 == r (mod 29!) >>As the terms in these integers gets larger watch the trailing digits of (r) >>See how they start to match up with the trailing digits of 12345..n >>and increase in count as you add more terms to the integer. >>Below a much larger example -- >>12345678910111213141516171819202122232425262728293 >>03132333435363738394041424344454647484950515253545 >>55657585960616263646566676869707172737475767778798 >>08182838485868788899091929394959697989910010110210 >>31041051061071081091101111121131141151161171181191 >>20121122123124125126127128129130131132133134135136 >>13713813914014114214314414514614714814915015115215 >>31541551561571581591601611621631641651661671681691 >>70171172173174175176177178179180181182183184185186 >>18718818919019119219319419519619719819920020120223 >>20420520620720820921021121221321421521621721821922 >>02212222232242252262272282292302312322332342352362 >>37238239240241242243244245246247248249250251252253 >>25425525625725825926026126226326426526626726826927 >>0271272273274275276277 == r (mod 277!) >>(r) here has a 68 trailing digit match. >>I bet no one can explain this one! >>Dan >The factorials have more and more factors of 10, so >when you mod out >by a high factorial, you are also modding out by a high >power of 10, >so the digit match is automatic. >In the prime factorization of 277!, there are 273 >factors of 2, and 68 >factors of 5. It follows that 277! has 68 factors of >10, and therefore >68 trailing zeros. >Hence, the fact that you got a 68 digit match is not at >all >surprising. >quasi Nice explanation about a factorial divisor! I would have lost that bet.:-( Still no one has acknowledged the triangle number connection to these integers 1234567891011...? Is it that I am not explaining it correctly or is there an obvious explanation that I am missing? Dan >277!=``(2)^273)(5)^68*``(7)^44*``(11)^27*``(13)^22* >(17)^16*``(19)^14*``(23)^12*``(29)^9*``(31)^8*``(37)^7*(41)^6*``(43)^6*``(4 7)^5*``(53)^5*``(59)^4*``(61)^4*``(67)^4*``(71)^3*``(73)^3*``(79)^3*``(83)^3 * ``(89)^3*``(97)^2*``(101)^2*``(103)^2*``(107)^2*``(109)^2*``(113)^2*``(127)^ 2 *``(131)^2*``(137)^2*``(139)*``(149)*``(151)>*``(157)*``(163)*``(167)*``(173 ) *``(179)*``(181)*``(191)>*``(193)*``(197)*``(199)*``(211)*``(223)*``(227)*`` ( 229)>*``(233)*``(239)*``(241)*``(251)*``(257)*``(263)*``(269)>*``(271)*``(27 7 ) >quasi === Subject: Re: Some intuitive explanations of Special, Unitary and Orthogonal? schrieb Kane Peng: > This may sounds weird.... Being a student of Physics, I know the > definitions, and I have a bit of knowledge on Lie Group, Manifold, > Differential Geometry, Homotopy, Homology, Cohomology, Fiber Bundle, > Characteristic Class, Category Theory... I just wonder what is an > intuitive geometric / topological explanation of Special, Unitary > and Orthogonal, for I always feel I do not really understand them. > Consider some basic stuff, all of us know SO(3) is the group of > rotations in R^3, SO(3) is diffeomorphic to RP^3, SU(2) is > diffeomorphic to S^3, SU(2) is a double-cover of SO(3), SL(2,C) has > some subtle structure ... We have nice explanation for each of them, > but is there a way being able to come up with all these kind of facts You might want to search the web for quaternions and Clifford algebra. === Subject: Question about Maple (a color problem) First excuse me for speaking so bad english but nevermind. Question : --------------- How change the color of the characters with Maple ? Example : --------------- I d like to see the 1 number like a brown characters and the 4 number like a green characters Anthony === Subject: Re: frobenius norm >> If you use ||E||_F = sqrt( Trace( E'*E)) then its >> straightforward to >> compute that ||u*v'||_F = ||u||_F*||v||_F if one of >> u,v is a vector >> (' denotes conjugate transpose) > Can you illustrate a proof of this straightforward > idea? if v is a vector, since E' = v*u', sqr( ||E||_F) = Trace( u*v'*v*u') = v'*v*Trace(u*u') (v'*v is a scalar) = v'*v*Trace(u'*u) (Trace(A*B)=Trace(B*A)) = sqr( ||v||_F)*sqr(||u||_F). If u is a vector, then ||E||_F = sqrt( Trace(E'*E)) = sqrt( Trace( E*E')) = ||E'||_F, and we can apply the above to E'. === Subject: Re: frobenius norm >> If you use ||E||_F = sqrt( Trace( E'*E)) then its >> straightforward to >> compute that ||u*v'||_F = ||u||_F*||v||_F if one > of >> u,v is a vector >> (' denotes conjugate transpose) >> > > > Can you illustrate a proof of this > straightforward > idea? > if v is a vector, since E' = v*u', > sqr( ||E||_F) = Trace( u*v'*v*u') = v'*v*Trace(u*u') > (v'*v is a > scalar) = v'*v*Trace(u'*u) (Trace(A*B)=Trace(B*A)) > = sqr( ||v||_F)*sqr(||u||_F). > If u is a vector, then ||E||_F = sqrt( Trace(E'*E)) = > sqrt( Trace( E*E')) > = ||E'||_F, and we can apply the above to E'. I'm sorry, but I don't understand/follow your proof, Duncan. why are you using sqrt(||E||_F)? if you are trying to get rid of the sqrt on the trace term, wouldn't you square (||E||_F) and not sqrt it? Also, how do you go from v'*v*Trace(u'*u)= sqr( ||v||_F)*sqr(||u||_F)? === Subject: The proof of Dirac delta function with equivalent function... Hi there, I'd thought two days.... How to porve that d^4 -------|x|^3 = 12.83å(x) where .83å(x) is delta function and |x| is absolute function dx^4 First, the inner product of test function .83Í(x) is .81.87 <.83Í(x), d^4|x|^3/dx^4> = .81.8d .83Í(x)*d^4|x|^3/dx^4 dx -.81.87 integration by parts, let u=.83Í(x), du=.83Í'(x)dx and dv=d^4|x|^3/dx^4 dx will v=d^3|x|^3/dx^3 ??????? I am confused.... ok, if it follows, then the inner product becomes uv - .81.8dvdu= .81.87 .81.87 .83Í(x)*d^3|x|^3/dx^3 | - .81.8d(d^3|x|^3/dx^3)*.83Í'(x)dx -.81.87 -.81.87 now we separate the intevals into (-.81.87,0) and (0,.81.87) then we have d^3 0 d^3 .81.87 .81.87 .83Í(x)*--------- (-x^3) | + .83Í(x)*--------- (x^3) | - .81.8d(d^3|x|^3/dx^3)*.83Í'(x)dx dx^3 -.81.87 dx^3 0 -.81.87 this equals to 0 .81.87 .81.87 .83Í(x)*(-6) | + .83Í(x)*6 | - .81.8d(d^3|x|^3/dx^3)*.83Í'(x)dx -.81.87 0 -.81.87 ^^^^^^^^^^(a)^^^^^^^^^^^^^^^^^^ ^^^^^^^(b)^^^^^^^^^^^^^^^^ if we substitute both (a) and (b) and compare with 12.83å(x), One of them must be zero but I don't know why and how? Maybe I am wrong everywhere I proved..... === Subject: The proof of Dirac delta function with equivalent function.(I'd modified prob) Hi there, I'd thought two days.... How to porve that d^4 -------|x|^3 = 12.83å(x) where .83å(x) is delta function dx^4 and |x| is absolute function First, the inner product of test function .83Í(x) is .81.87 <.83Í(x), d^4|x|^3/dx^4> = .81.8d .83Í(x)*d^4|x|^3/dx^4 dx -.81.87 integration by parts, let u=.83Í(x), du=.83Í'(x)dx and dv=d^4|x|^3/dx^4 dx will v=d^3|x|^3/dx^3 ??????? I am confused.... ok, if it follows, then the inner product becomes uv - .81.8dvdu= .81.87 .81.87 .83Í(x)*d^3|x|^3/dx^3 | - .81.8d(d^3|x|^3/dx^3)*.83Í'(x)dx -.81.87 -.81.87 ^^^^^^(A)^^^^^^^^ now we separate the intevals into (-.81.87,0) and (0,.81.87) then we have d^3 0 d^3 .81.87 .81.87 .83Í(x)*--------- (-x^3) | + .83Í(x)*--------- (x^3) | - .81.8d(A)dx dx^3 -.81.87 dx^3 0 -.81.87 this equals to 0 .81.87 .81.87 .83Í(x)*(-6) | + .83Í(x)*6 | - .81.8d(d^3|x|^3/dx^3)*.83Í'(x)dx -.81.87 0 -.81.87 ^^^^^^^^^^(a)^^^^^^^^^^^^^^^^^^ ^^^^^^^(b)^^^^^^^^^^^^^^^^ if we substitute both (a) and (b) and compare with 12.83å(x), One of them must be zero but I don't know why and how? Maybe I am wrong everywhere I proved..... .89.96.95¢ === Subject: Re: The proof of Dirac delta function with equivalent function.(I'd modified prob) sorry, I repost the problem,, Hi there, I'd thought two days.... How to porve that d^4 -------|x|^3 = 12delta(x) dx^4 where |x| is absolute function First, the inner product of test function g(x) is = S g(x)*d^4|x|^3/dx^4 dx from -INF to INF where S is integral symbol and INF means infinite integration by parts, let u=g(x), du=g'(x)dx and dv=d^4|x|^3/dx^4 dx will v=d^3|x|^3/dx^3 ??????? I am confused.... ok, if it follows, then the inner product becomes uv - .bcvdu= INF INF g(x)*d^3|x|^3/dx^3 | - S (d^3|x|^3/dx^3)*g'(x)dx -INF -INF ^^^^^^(A)^^^^^^^^ now we separate the intevals into (-INF,0) and (0,INF) then we have d^3 0 d^3 INF INF g(x)*--------- (-x^3) | + g(x)*--------- (x^3) | - S(A)dx dx^3 -INF dx^3 0 -INF this equals to 0 INF INF g(x)*(-6) | + g(x)*6 | - S (d^3|x|^3/dx^3)*g'(x)dx -INF 0 -INF ^^^^^^^^^^(a)^^^^^^^^^^^^^^^^^^ ^^^^^^^(b)^^^^^^^^^^^^^^^^ if we substitute both (a) and (b) and compare with 12delta(x), One of them must be zero but I don't know why and how? Maybe I am wrong everywhere I proved..... === Subject: The proof of Dirac delta function with equivalent function.(I'd modified prob) Hi there, I'd thought two days.... How to porve that d^4 -------|x|^3 = 12.83å(x) where .83å(x) is delta function dx^4 and |x| is absolute function First, the inner product of test function .83Í(x) is .81.87 <.83Í(x), d^4|x|^3/dx^4> = .81.8d .83Í(x)*d^4|x|^3/dx^4 dx -.81.87 integration by parts, let u=.83Í(x), du=.83Í'(x)dx and dv=d^4|x|^3/dx^4 dx will v=d^3|x|^3/dx^3 ??????? I am confused.... ok, if it follows, then the inner product becomes uv - .81.8dvdu= .81.87 .81.87 .83Í(x)*d^3|x|^3/dx^3 | - .81.8d(d^3|x|^3/dx^3)*.83Í'(x)dx -.81.87 -.81.87 ^^^^^^(A)^^^^^^^^ now we separate the intevals into (-.81.87,0) and (0,.81.87) then we have d^3 0 d^3 .81.87 .81.87 .83Í(x)*--------- (-x^3) | + .83Í(x)*--------- (x^3) | - .81.8d(A)dx dx^3 -.81.87 dx^3 0 -.81.87 this equals to 0 .81.87 .81.87 .83Í(x)*(-6) | + .83Í(x)*6 | - .81.8d(d^3|x|^3/dx^3)*.83Í'(x)dx -.81.87 0 -.81.87 ^^^^^^^^^^(a)^^^^^^^^^^^^^^^^^^ ^^^^^^^(b)^^^^^^^^^^^^^^^^ if we substitute both (a) and (b) and compare with 12.83å(x), One of them must be zero but I don't know why and how? Maybe I am wrong everywhere I proved..... .89.96.95¢ === Subject: Re: The proof of Dirac delta function with equivalent function.(I'd modified prob) === Subject: Re: The proof of Dirac delta function with equivalent function.(I'd modified prob) >On 2 Dec 2005 09:18:48 -0800, Harry901.tw >Probably stuff that appeared just fine on his >screen but which becomes the incomprehensible >mess below at this end. >Try reposting in plain _text_ Yeah, anyone who's unable to use plain text to post an incomprehensible mess probably isn't ready for sci.math. Lee Rudolph === Subject: Re: The proof of Dirac delta function with equivalent function.(I'd modified prob) On 2 Dec 2005 09:18:48 -0800, Harry901.tw Probably stuff that appeared just fine on his screen but which becomes the incomprehensible mess below at this end. Try reposting in plain _text_ >Hi there, I'd thought two days.... >How to porve that > d^4 >-------|x|^3 =3D 12=CE=B4(x) where =CE=B4(x) is delta function >dx^4 and |x| is absolute function >=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D >First, the inner product of test function =CF=88(x) is > =E2=88=9E ><=CF=88(x), d^4|x|^3/dx^4> =3D =E2=88=AB =CF=88(x)*d^4|x|^3/dx^4 dx > -=E2=88=9E >integration by parts, let u=3D=CF=88(x), du=3D=CF=88'(x)dx >and dv=3Dd^4|x|^3/dx^4 dx will v=3Dd^3|x|^3/dx^3 ??????? >I am confused.... >ok, if it follows, then the inner product becomes >uv - =E2=88=ABvdu=3D > =E2=88=9E =E2=88=9E >=CF=88(x)*d^3|x|^3/dx^3 | - =E2=88=AB(d^3|x|^3/dx^3)*=CF=88'(x)dx > -=E2=88=9E -=E2=88=9E > ^^^^^^(A)^^^^^^^^ >now we separate the intevals into (-=E2=88=9E,0) and (0,=E2=88=9E) >then we have > d^3 0 d^3 =E2=88=9E =E2=88= >=9E >=CF=88(x)*--------- (-x^3) | + =CF=88(x)*--------- (x^3) | - =E2=88= >=AB(A)dx > dx^3 -=E2=88=9E dx^3 0 -=E2= >=88=9E >this equals to > 0 =E2=88=9E =E2=88=9E >=CF=88(x)*(-6) | + =CF=88(x)*6 | - =E2=88=AB(d^3|x|^3/dx^3)*=CF= >=88'(x)dx > -=E2=88=9E 0 -=E2=88=9E >^^^^^^^^^^(a)^^^^^^^^^^^^^^^^^^ ^^^^^^^(b)^^^^^^^^^^^^^^^^ >if we substitute both (a) and (b) and compare with 12=CE=B4(x), >One of them must be zero but I don't know why and how? >Maybe I am wrong everywhere I proved..... >=E5=9B=9E=E8=A6=86 David C. Ullrich === Subject: super exponential growth? I know in exponential growth the growth rate is some constant times the population. But what is it called when this constant increases proportionally with the population? Is there any special term for this? Any examples of nature/life that demonstrate this? === Subject: Re: super exponential growth? > I know in exponential growth the growth rate is some constant times the > population. But what is it called when this constant increases > proportionally with the population? Is there any special term for this? > Any examples of nature/life that demonstrate this? Ray Kurzweil, in _The Singularity is Near_, calls this the law of accelerating returns. I'm not convinced by his derivation, but he derives a formula for W, the world knowledge as it pertains to designing and building computational devices, as W = p * e^(q * e^(r * t)), where t is time and p, q, and r are constants. The derivation is unassailable, but the assumptions are not, IMO. Rick === Subject: Re: super exponential growth? > I know in exponential growth the growth rate is some constant times the > population. But what is it called when this constant increases > proportionally with the population? Eh? Then it is not a constant. Is there any special term for this? > Any examples of nature/life that demonstrate this? If you meant relative rate, you will end up with a differential equation dy/dt = k * y^2 and the solution family is well-known; it is y(t) = y(0) / (1 - k * y(0) * t) It has a guaranteed singularity: in finite time 1/(k*y(0)), it diverges to infinity. This is not supposed to happen in nature, so this model does not deserve a name (AFAIK). The scenario changes in limited living space; for simplicity, let the disincentive to growth be a constant multiple of (1 - y(t)/N), where N is the carrying capacity of the living space. Then the ODE dy/dt = k*y^2*(1-y/N) has an implicit solution family log(y/(N-y)) - N/y = k*N*t + C so that if y(0) is between 0 and N, y(t) stays there, too. But I do not know if this model has (or deserves) a name. === Subject: Re: super exponential growth? > I know in exponential growth the growth rate is some constant times the > population. But what is it called when this constant increases > proportionally with the population? Is there any special term for this? > Any examples of nature/life that demonstrate this? So, if p(t) is population as a function of time, then we want p ' (t) = (k * p) * p. That is, k*p is the constant of growth, which is proportional to the population. If we are given that p(0) = p0, then the solution is p(t) = p0 / (1 - p0 * k * t). The interesting thing about this solution is that at the finite time t = 1 / (k * p0), the population tends to infinity. This is very different than in regular exponential population growth, where the population never tends to infinity at a finite time. That being said, there are some population models that are governed by this growth pattern, at least on a local scale. Suppose there is some animal which wanders around randomly in a fixed region. Whenever two such animals meet, they mate. So, the growth rate of the population is proportional to the number of chance encounters between animals, which is proportional to the square of the number of animals. === Subject: Re: super exponential growth? > I know in exponential growth the growth rate is some constant times the > population. But what is it called when this constant increases > proportionally with the population? Is there any special term for this? > Any examples of nature/life that demonstrate this? Not from the nature or life science, but very similar concept exists in economics where one uses discrete compound interest rate i normally called as compound interest rate or more technically effective compound interest rate. Now if you consider that the interest rate itself was continuous that would grow from a value of say r over the installment to be effectively equal to the discrete value i, then such r is called as nominal interest rate where i=(e^r) - 1. In your example of exponential growth in population given by eqn say y=ae^(bt), b is analogous to i the effective interest rate. (note that b is a constant and so is i) In short, there are two terms to differentiate two different concepts in growth - effective and nominal. -- Respectfully, Mohan Pawar MIO Instruments LLC === Subject: Re: Tim's golden Coordinates (polysigned news) ... Inside a cube is a tetra: connect four of the eight corners by diagonals on the sides. This connects to the thread determinants of nxnxn matrices Hero === Subject: Re: Bound on standard deviation ? >> > I have a random variable which is taking values between O and 1 > according to an unknown probability distribution. The question is : > knowing the expectation value, what is the maximum possible standard > deviation ? > To make it clear, I want to maximize int(p(x)(x-m)^2 dx) under the > constraints : > - p(x)=0 for x<0 and x>1; > - p(x) >= 0; > - int(p(x)dx) = 1; > - int(p(x)x dx) = m; > Intuitively, it seems that the maximum is m(1-m) which corresponds to > a probability distribution with 2 dirac peaks in 0 and 1, with > amplitudes (1-m) and m respectively. But how to proove it ? > >> Note that int_0^1 p(x) x^2 dx <= int_0^1 p(x) x dx = m so int_0^1 p(x) >> (x-m)^2 dx = int_0^1 p(x) x^2 dx - m^2 <= m - m^2. > A very nice proof. Note that we do not need a density p. Expressed > otherwise, 0 <= X <= 1 implies X^2 <= X, which implies E[X^2] <= EX. > More generally, if X has Bernoulli(p) distribution, 0 <= Y <= 1, > EY = p, and h: [0,1] --> R is convex, then Eh(X) >= Eh(Y). We say > that the Bernoulli is maximal in the sense of stochastic variablility > ordering on the unit interval. See Ross, Stochastic Processes. Benjamin === Subject: Re: sequence of continuous functions <43907c17$0$24390$892e7fe2@authen.yellow.readfreenews.net> I shouldn't have said norm - I am just interested in convergence in the topology of pointwise convergence, i.e. the product topology. === Subject: Re: sequence of continuous functions >> On 30 Nov 2005 20:50:22 -0800, boyandshark >>If you have a sequence of real valued continuous functions over a >>compact metric space X, is it always possible to find a convergent >>subsequence in some norm. For instance, is it always possible to find a >>convergent subsequence that converges pointwise to a possibly >>noncontinuous function over X? >> Clearly not. For example consider f_n(x) = n. >> ************************ >> David C. Ullrich >Hmm... What if we added the additional restriction that for fixed x, >the sequence {f_n(x)} is bounded? Does there exist a sequence of >functiontions satisfying this which has no convergent subsequence? You still haven't specified what norm you're talking about. If we're talking about the uniform norm, which seems like the appropriate one, then yes, there are such counterexamples. The condition you really want is equicontinuity. A function is continuous if for every epsilon > 0 there exists delta > 0 such that etc. A _family_ of functions is equicontinuous if the same delta works for all the functions in the family. The Ascoli-Arzela Theorem (or one version of it) says that if you have a uniformly bounded and equicontinuous sequence on a compact metric space then there is a convergent subsequence. ************************ David C. Ullrich === Subject: Re: sequence of continuous functions <011220051133277549%bruck@math.usc.edu> hypothesis by adding the condition that the sequence of continuous functions (I guess they don't even have to be continuous) is uniformly bounded, then it is possible to find a subsequence that converges pointwise to a possibly noncontinuous limit? === Subject: Re: sequence of continuous functions > hypothesis by adding the condition that the sequence of continuous > functions (I guess they don't even have to be continuous) is uniformly > bounded, then it is possible to find a subsequence that converges > pointwise to a possibly noncontinuous limit? No, no, no---a subNET. Read up on Moore-Smith convergence, nets, subnets. --Ron Bruck === Subject: Re: ineqality > One can w.l.o.g. assume that > a <= b <= c, > and set > b = a + x1 > c = a + x1 + x2 > with new non-negative variables x1 and x2. > When this substitution is applied to the difference > (a^3 + b^3 + c^3 + 6abc)^2 - 3 (ab + bc + ca)^3, > one obtains a polynomial in a, x1, and x2 > with only non-negative coefficients, > 6*x1*x2^5+15*x1^2*x2^4+x2^6+42*a*x1*x2^4+12*x1^4*x2^2 > +19*x1^3*x2^3+27*a^2*x2^4+6*a*x2^5+90*a^2*x1^3*x2 > +171*a^2*x1^2*x2^2+126*a^2*x1*x2^3+30*a*x1^4*x2 > +84*a*x1^3*x2^2+96*a*x1^2*x2^3+27*a^4*x2^2+48*a^3*x2^3 > +90*a^3*x1^2*x2+126*a^3*x1*x2^2+27*a^4*x1^2+60*a^3*x1^3 > +45*a^2*x1^4+12*a*x1^5+3*x1^5*x2+x1^6+27*a^4*x1*x2, > which *obviously* cannot take on negative values for a,x1,x2>=0. > Hence: > (a^3 + b^3 + c^3 + 6abc)^2 - 3 (ab + bc + ca)^3 >= 0 > for all a,b,c >= 0. > Q.E.D. Nice! Is it possible to do without setting b = a+x_1 and c=x+x_2? I think it should be... -- Peper === Subject: Re: ineqality >> One can w.l.o.g. assume that >> a <= b <= c, >> and set >> b = a + x1 >> c = a + x1 + x2 >> with new non-negative variables x1 and x2. >> When this substitution is applied to the difference >> (a^3 + b^3 + c^3 + 6abc)^2 - 3 (ab + bc + ca)^3, >> one obtains a polynomial in a, x1, and x2 >> with only non-negative coefficients, >> 6*x1*x2^5+15*x1^2*x2^4+x2^6+42*a*x1*x2^4+12*x1^4*x2^2 >> +19*x1^3*x2^3+27*a^2*x2^4+6*a*x2^5+90*a^2*x1^3*x2 >> +171*a^2*x1^2*x2^2+126*a^2*x1*x2^3+30*a*x1^4*x2 >> +84*a*x1^3*x2^2+96*a*x1^2*x2^3+27*a^4*x2^2+48*a^3*x2^3 >> +90*a^3*x1^2*x2+126*a^3*x1*x2^2+27*a^4*x1^2+60*a^3*x1^3 >> +45*a^2*x1^4+12*a*x1^5+3*x1^5*x2+x1^6+27*a^4*x1*x2, >> which *obviously* cannot take on negative values for a,x1,x2>=0. >> Hence: >> (a^3 + b^3 + c^3 + 6abc)^2 - 3 (ab + bc + ca)^3 >= 0 >> for all a,b,c >= 0. >> Q.E.D. > Nice! > Is it possible to do without setting b = a+x_1 and c=x+x_2? I think it > should be... I think so too, but please GO AWAY and DON'T BOTHER US ANY FURTHER with your problems from the current polish math olympiad http://www.om.edu.pl/zadania.php http://www.om.edu.pl/zadania/om/om57_1.pdf (No.10) until December, 5th. Someone should inform the people in charge about your recent activities === Subject: Re: On limitations of the Mautsch Principle (was Re: ineqality) How shocking! ;-) After two internet-free nights, I return to this newsgroup which becomes increasingly useless with every new message in this threat: :-( [ ... ] >However, to defend the principle as it is, how about this ... [ ... ] > Conjecture: > Let f(x_1, x_2, ..., x_n) be a [ homogeneous ] > symmetric polynomial with real coefficients such that > for all x_1, x_2, ..., x_n>=0, > f(x_1, x_2,..., x_n) >= 0 with equality iff x_1 = x_2 = ... = x_n. > Then in the expansion of the polynomial > f(y_1, y1+y_2, ..., y_1+y_2+...+y_n) > there are no negative coefficients. >>Counter examples: >>Dave Rusin's >> ( x^2 + y^2 )^2 - 3 x y (2 x^2 - 3 x y + 2 y^2) >> x y ( x^2 + y^2 )^2 - 3 x^2 y^2 (2 x^2 - 3 x y + 2 y^2) >>and Thomas Mautsch's >> (x+y-z)^2 (y+z-x)^2 (z+x-y)^2 > The above examples are _not_ counterexamples to my conjecture. > A key requirement in my conjecture is that equality hold iff all > variables are equal. Neither of the above examples satisfies that > requirement. Once you have a counterexample like Dave Rusin's from above, you can construct very many counterexamples with special properties: Take for example Dave's example P(x,y) = ( x^2 + 3 - 3 x y)^2, and add to it a constant epsilon times an arbitrary symmetric polynomial Q(x,y) of the same homogeneity as P(x,y) that satisfies Q(x,y) > 0 for all x,y>=0, except (0,0). Take e.g. Q(x,y) = x^6 + y^6. Then for all small enough epsilon>0, the resulting polynomial P(x,y) + epsilon Q(x,y) will also be a counter example to my original conjecture, but with the additional property that it is only zero at (x,y)=(0,0), i.e. for x=y. Then this polynomial, and also the products (x-y)^(2k) * ( P(x,y) + epsilon Q(x,y) ) will be counterexamples to your conjecture. In addition, the polynomials P(x,y) + epsilon Q(x,y) from the example above are irreducible over the reals; and I also have the strong feeling that you can achieve similar counterexamples for symmetric polynomials in more than two variables. Remains the question, how I was tricked to believe that this symmetry-breaking trick for symmetric inequalities could be sort of universal. - Well, I obviously did not check all non-negative symmetric polynomials systematically, and concentrated on problems from math contests instead (e.g. many inequalities for triangle sides a,b,c - with the substitution a=(x+y), b=(y+z), c=(z+x) and x,y,z>=0). >>I think you need the modified requirement >> *only* for all x_1, x_2, ..., x_n >= 0: >> f(x_1, x_2,..., x_n) >= 0 with equality iff x_1 = x_2 = ... = x_n. > No, that's far too restrictive. > For example, that would make the method not applicable to any sum of > squares. > Also, the method applies perfectly to the A-M GM inequalities for all > n, Do you have a short proof for this fact, or just experimental evidence? - That would be nice! If the symmetry-breaking principle did not work, then this might shatter the idea of Bar-Natan http://www.math.toronto.edu/~drorbn/projects/ArithGeom/ to find a geometric proof of the AGM inequality for n numbers by filling n-dimensional hypercubes of side length (a1 + a2 + ... + an) with n^n disjoint boxes of dimension a1 x a2 x ... x an, each... > but your restriction would not allow those examples. > As far as I can see, my conjecture stands. > In fact, I have a sketch of a proof, but it's late, so I'll defer it. > But I'm pretty confident about it (I would even bet on it). > Counterexamples are welcome, but please check to make sure the > hypothesis of the conjecture are satisfied. > Also, note the latest, revised version which requires that the > polynomial be homogeneous as well as symmetric. === Subject: Re: On limitations of the Mautsch Principle (was Re: ineqality) On 2 Dec 2005 02:31:58 +0100, Thomas Mautsch >How shocking! ;-) >After two internet-free nights, I return to this newsgroup >which becomes increasingly useless >with every new message in this threat: > :-( >[ ... ] >>However, to defend the principle as it is, how about this ... >[ ... ] >> Conjecture: >> >> Let f(x_1, x_2, ..., x_n) be a >[ homogeneous ] >> symmetric polynomial with real coefficients such that >> for all x_1, x_2, ..., x_n>=0, >> >> f(x_1, x_2,..., x_n) >= 0 with equality iff x_1 = x_2 = ... = x_n. >> >> Then in the expansion of the polynomial >> >> f(y_1, y1+y_2, ..., y_1+y_2+...+y_n) >> >> there are no negative coefficients. >Counter examples: >Dave Rusin's > ( x^2 + y^2 )^2 - 3 x y (2 x^2 - 3 x y + 2 y^2) > x y ( x^2 + y^2 )^2 - 3 x^2 y^2 (2 x^2 - 3 x y + 2 y^2) >and Thomas Mautsch's > (x+y-z)^2 (y+z-x)^2 (z+x-y)^2 >> The above examples are _not_ counterexamples to my conjecture. >> A key requirement in my conjecture is that equality hold iff all >> variables are equal. Neither of the above examples satisfies that >> requirement. >Once you have a counterexample like Dave Rusin's from above, >you can construct very many counterexamples with special properties: >Take for example Dave's example > P(x,y) = ( x^2 + 3 - 3 x y)^2, >and add to it a constant epsilon >times an arbitrary symmetric polynomial Q(x,y) >of the same homogeneity as P(x,y) >that satisfies > Q(x,y) > 0 for all x,y>=0, except (0,0). >Take e.g. > Q(x,y) = x^6 + y^6. >Then for all small enough epsilon>0, >the resulting polynomial > P(x,y) + epsilon Q(x,y) >will also be a counter example to my original conjecture, >but with the additional property that it is only zero >at (x,y)=(0,0), i.e. for x=y. >Then this polynomial, and also the products > (x-y)^(2k) * ( P(x,y) + epsilon Q(x,y) ) >will be counterexamples to your conjecture. I don't believe it -- let's see it explicitly. To break my conjecture you have to find a polynomial f(x_1,x_2,..,x_n), where n>=2, with real coefficients, such that: (1) f is symmetric and homogeneous (2) for all x_1,x_2...,x_n>=0, f(x_1,x_2,..,x_n)>=0 with equality iff x_1=x_2...=x_n (3) in the expansion of f(y_1,y_1+y_2,...,y_1+y_2+...+y_n) there is at least one term with a negative coefficient. I have a proof of my conjecture in sketch form, so I won't believe it's broken until I see an actual counterexample. quasi === Subject: Re: On limitations of the Mautsch Principle (was Re: ineqality) >>[ ... ] >However, to defend the principle as it is, how about this ... >>[ ... ] > Conjecture: > > Let f(x_1, x_2, ..., x_n) be a >>[ homogeneous ] > symmetric polynomial with real coefficients such that > for all x_1, x_2, ..., x_n>=0, > > f(x_1, x_2,..., x_n) >= 0 with equality iff x_1 = x_2 = ... = x_n. > > Then in the expansion of the polynomial > > f(y_1, y1+y_2, ..., y_1+y_2+...+y_n) > > there are no negative coefficients. >> >>Counter examples: >>Dave Rusin's >> ( x^2 + y^2 )^2 - 3 x y (2 x^2 - 3 x y + 2 y^2) >> x y ( x^2 + y^2 )^2 - 3 x^2 y^2 (2 x^2 - 3 x y + 2 y^2) >>and Thomas Mautsch's >> (x+y-z)^2 (y+z-x)^2 (z+x-y)^2 > > The above examples are _not_ counterexamples to my conjecture. > > A key requirement in my conjecture is that equality hold iff all > variables are equal. Neither of the above examples satisfies that > requirement. >>Once you have a counterexample like Dave Rusin's from above, >>you can construct very many counterexamples with special properties: >>Take for example Dave's example >> P(x,y) = ( x^2 + 3 - 3 x y)^2, ^^^^^^^^^^^^^^^^^^ I messed this one up. - Dave's example is of course P(x,y) = ( x^2 + y^2 - 3 x y)^2. >>and add to it a constant epsilon >>times an arbitrary symmetric polynomial Q(x,y) >>of the same homogeneity as P(x,y) >>that satisfies >> Q(x,y) > 0 for all x,y>=0, except (0,0). >>Take e.g. >> Q(x,y) = x^6 + y^6. >>Then for all small enough epsilon>0, >>the resulting polynomial >> P(x,y) + epsilon Q(x,y) >>will also be a counter example to my original conjecture, >>but with the additional property that it is only zero >>at (x,y)=(0,0), i.e. for x=y. >>Then this polynomial, and also the products >> (x-y)^(2k) * ( P(x,y) + epsilon Q(x,y) ) >>will be counterexamples to your conjecture. > I don't believe it -- let's see it explicitly. > To break my conjecture you have to find a polynomial > f(x_1,x_2,..,x_n), where n>=2, with real coefficients, such that: > (1) f is symmetric and homogeneous > (2) for all x_1,x_2...,x_n>=0, f(x_1,x_2,..,x_n)>=0 > with equality iff x_1=x_2...=x_n > (3) in the expansion of f(y_1,y_1+y_2,...,y_1+y_2+...+y_n) there is at > least one term with a negative coefficient. > I have a proof of my conjecture in sketch form, so I won't believe > it's broken until I see an actual counterexample. I gave the example for n=2 above; here is my Maple code for it: f:= (x,y) -> (100 * (x^2 - 3*x*y + y^2)^2 + x^4 + y^4)*(x-y)^2 : simplify( f(x1,x2)-f(x2,x1) ) ; 0 expand( f(y1,y1+y2) ) ; 102*y2^2*y1^4+204*y2^3*y1^3-94*y2^4*y1^2-196*y2^5*y1+101*y2^6 ^^^!! ^^^!! And here is a counterexample for n=3: f:= (x,y,z) -> (100*(x+y-z)^2*(y+z-x)^2*(z+x-y)^2 + x^6+y^6+z^6) * ((x-y)^2+(x-z)^2+(y-z)^2) : simplify( f(x1,x2,x3)-f(x2,x1,x3) ) ; 0 simplify( f(x1,x2,x3)-f(x3,x2,x1) ) ; 0 expand( f(y1,y1+y2,y1+y2+y3) ) ; 1720*y1^4*y2^3*y3+4*y2^8+206*y1^6*y2^2+60*y2^6*y1^2 +24*y2^7*y1+860*y2^4*y1^4+80*y2^5*y1^3+824*y2^3*y1^5 +202*y3^8+206*y1^6*y3^2+16*y2^7*y3+46*y2^6*y3^2+412*y1^5*y3^3 -170*y1^4*y3^4-760*y1^3*y3^5-170*y1^2*y3^6 +900*y3^4*y2^4+1682*y3^5*y2^3+1844*y3^6*y2^2+1014*y3^7*y2 +412*y3^7*y1+82*y3^3*y2^5+1550*y1^4*y2^2*y3^2+690*y1^4*y2*y3^3 +206*y1^6*y2*y3+180*y2^5*y1^2*y3-1240*y2^4*y1^2*y3^2 +84*y2^6*y1*y3+204*y2^5*y1*y3^2+200*y2^4*y1^3*y3-1280*y2^3*y1^3*y3^2 +1236*y1^5*y2^2*y3+1236*y1^5*y2*y3^2-2120*y1^3*y2^2*y3^3 -2240*y1^3*y2*y3^4-2780*y2^3*y1^2*y3^3-3070*y2^2*y1^2*y3^4 -1650*y1^2*y2*y3^5+300*y1*y3^3*y2^4+1100*y1*y3^4*y2^3 +1392*y1*y3^5*y2^2+1272*y1*y3^6*y2 LOTS of negative coefficients, I am afraid! Example of machine proof of an inequality [LONG], I especially liked that the symmetry-breaking method also allowed to prove inequalities for which equality was obtained at points where not all variables are equal, for example the inequality a^4 + b^4 + c^4 <= 32 R^4 between three triangle sides a = (x+y), b = (y+z), c = (z+x) and the circumcenter radius R = abc/(4sqrt((xyz(x+y+z))) (as a polynomial inequality in the variables x,y,z >= 0). === Subject: Re: On limitations of the Mautsch Principle (was Re: ineqality) On 2 Dec 2005 13:55:35 +0100, Thomas Mautsch >[ ... ] >>However, to defend the principle as it is, how about this ... >[ ... ] >> Conjecture: >> >> Let f(x_1, x_2, ..., x_n) be a >[ homogeneous ] >> symmetric polynomial with real coefficients such that >> for all x_1, x_2, ..., x_n>=0, >> >> f(x_1, x_2,..., x_n) >= 0 with equality iff x_1 = x_2 = ... = x_n. >> >> Then in the expansion of the polynomial >> >> f(y_1, y1+y_2, ..., y_1+y_2+...+y_n) >> >> there are no negative coefficients. > >Counter examples: >Dave Rusin's > ( x^2 + y^2 )^2 - 3 x y (2 x^2 - 3 x y + 2 y^2) > x y ( x^2 + y^2 )^2 - 3 x^2 y^2 (2 x^2 - 3 x y + 2 y^2) >and Thomas Mautsch's > (x+y-z)^2 (y+z-x)^2 (z+x-y)^2 >> >> The above examples are _not_ counterexamples to my conjecture. >> >> A key requirement in my conjecture is that equality hold iff all >> variables are equal. Neither of the above examples satisfies that >> requirement. >Once you have a counterexample like Dave Rusin's from above, >you can construct very many counterexamples with special properties: >Take for example Dave's example > P(x,y) = ( x^2 + 3 - 3 x y)^2, > ^^^^^^^^^^^^^^^^^^ >I messed this one up. - Dave's example is of course > P(x,y) = ( x^2 + y^2 - 3 x y)^2. >and add to it a constant epsilon >times an arbitrary symmetric polynomial Q(x,y) >of the same homogeneity as P(x,y) >that satisfies > Q(x,y) > 0 for all x,y>=0, except (0,0). >Take e.g. > Q(x,y) = x^6 + y^6. >Then for all small enough epsilon>0, >the resulting polynomial > P(x,y) + epsilon Q(x,y) >will also be a counter example to my original conjecture, >but with the additional property that it is only zero >at (x,y)=(0,0), i.e. for x=y. >Then this polynomial, and also the products > (x-y)^(2k) * ( P(x,y) + epsilon Q(x,y) ) >will be counterexamples to your conjecture. >> I don't believe it -- let's see it explicitly. >> To break my conjecture you have to find a polynomial >> f(x_1,x_2,..,x_n), where n>=2, with real coefficients, such that: >> (1) f is symmetric and homogeneous >> (2) for all x_1,x_2...,x_n>=0, f(x_1,x_2,..,x_n)>=0 >> with equality iff x_1=x_2...=x_n >> (3) in the expansion of f(y_1,y_1+y_2,...,y_1+y_2+...+y_n) there is at >> least one term with a negative coefficient. >> I have a proof of my conjecture in sketch form, so I won't believe >> it's broken until I see an actual counterexample. >I gave the example for n=2 above; here is my Maple code for it: > f:= (x,y) -> (100 * (x^2 - 3*x*y + y^2)^2 + x^4 + y^4)*(x-y)^2 : > simplify( f(x1,x2)-f(x2,x1) ) ; > 0 > expand( f(y1,y1+y2) ) ; > 102*y2^2*y1^4+204*y2^3*y1^3-94*y2^4*y1^2-196*y2^5*y1+101*y2^6 > ^^^!! ^^^!! Good thing I didn't bet on my claim. I was very confident about it too, but as your example clearly shows, I was dead wrong. I'll have to look over my proof sketch. I may have proved something, but exactly what, I'm not sure. quasi === Subject: Re: On limitations of the Mautsch Principle (was Re: ineqality) > On 2 Dec 2005 13:55:35 +0100, Thomas Mautsch > ^^^!! ^^^!! > Good thing I didn't bet on my claim. > I was very confident about it too, but as your example clearly shows, > I was dead wrong. > I'll have to look over my proof sketch. I may have proved something, > but exactly what, I'm not sure. It *is* confusing, isn't it :-) Dirk Vdm === Subject: Re: ineqality On 28 Nov 2005 02:12:45 +0100, Thomas Mautsch (mautsch@math.ethz.ch) >The solution suggested here seems to be incomplete: >> there is: >> ab + bc + ca = 3 >> prove that: >> a^3 + b^3 + c^3 + 6abc >= 9 >[ ... ] >> >> The key is to invoke compactness. >> >> A trick that works here is to reverse the roles of the constraint and >> the objective function. >This trick is unnecessary. - There is no gain >from interchanging objective function and constraint: >> The gain is compactness. In switched form, the feasible region is >> compact for a,b,c>=0. >> In the original form, the feasible region is not compact for a,b,c>=0. >> Thus, although the 2 forms are equivalent as far as proving the >> inequality, the switched form has the advantage of a compact feasible >> region, thus guaranteeing both a global max and a global min. >You are so right. - I simply overlooked this fact. >Sorry for the bother! >Yet, isn't it also true that if we can show >that the differentiable function > a^3 + b^3 + c^3 + 6abc >has on the connected smooth(!) region > {(a,b,c): a>0, b>0, c>0, ab + bc + ca = 3} >a unique critical point which is a local minimum, >then this critical point has to be a global minimum?! >> Let >> >> f(a,b,c) = a^3 + b^3 + c^3 + 6*a*b*c >> >> g(a,b,c) = a*b + b*c + c*a >> It's easy to see that S={(a,b,c) | f(a,b,c)=9 and a,b,c>=0} is >> compact, hence g must attain a maximum on S. >(Wait! - You have to show that f has no critical points on the level set > {(a,b,c): f(a,b,c)=9}.) >> Applying the method of Lagrange multipliers and restricting to >> a,b,c>0, we get a unique local maximum for g at a=b=c=1 with value > ^^^^^^ >> g(1,1,1)=3. >To obtain *this* result is the *real trick* in your proposed solution: >To do so, you have to solve a system of three polynomial equations, >none of which is linear. - Nowadays you can use computers for this, >but I don't see an easy way how you might have solved this system >by hand. >> Nowadays, I don't do anything by hand if I don't have to, unless I can >> do it in my head. To solve the system of equations, I used Maple's >> Grobner basis package >[ ... ] >!!! >Using Maple, one can also check that the point (a,b,c) = (1,1,1) >is indeed *the only critical point* of the function f >on the regular two-dimensional region > {(a,b,c): a>0, b>0, c>0, g(a,b,c)=3} >[... ] >The remaining part of what you write is then again >more or less unnecessary: >> By analyzing the boundary, we show that all values of g on the >> boundary of S are less than 3, hence g achieves a global max value of >> 3 at (1,1,1). >> This avoids the need for the Hessian. >Depends on personal taste, I suppose. - >Under certain circumstances, I prefer to use the Hessian; >e.g. if a smooth function on an interval has only one critical point, >it is sometimes quicker to see that this point is a local extremum >(by looking at the second derivatives, > or the monotonicity of the first derivative), >than to calculate the values the function takes on at the boundary. >[ ... ] >> Thus, we can conclude that the local optimum value at (1,1,1) is also >> the global optimum value, thus establishing the inequality. >Once you have shown (using the augmented Hessian), > ^^^^^^^^^ >that there is only one extremum of the function g on the surface {f=9}, >and that this extremum is a local maximum, >this maximum has to be a global one. Local max does not imply global max, even if it's the only extremum. The point of my switching the feasible region and objective function and then extending the feasible region to include it's boundary was to get a compact domain. With that simple trick, the existence a global max is guaranteed. I still had to worry about the boundary, and I did, but since the global max is then known to exist, it has to be either be at a critical point in the interior, or on the boundary. Since there is a unique critical point in the interior, once I show that the values on the boundary are all less than the value at the critical point, the global max must be at the critical point. No need to look at Hessians. If the dimension of the feasible region is greater than 1 then, even with a unique critical point, and even if you know it's a local max, you still don't know it's a global max unless you analyze the global behavior. quasi === Subject: Re: ineqality > If the dimension of the feasible region is greater than 1 then, even > with a unique critical point, and even if you know it's a local max, > you still don't know it's a global max unless you analyze the global > behavior. I was afraid you might say that. - Do you know a simple counter example (with simply-connected region)? Thomas