mm-3839 === Subject: Proof of matrix result wanted The Wikipedia entry for Matrix Exponential quotes the result that in an algebraically closed field any n*n matrix M can be expressed uniquely as a sum of n*n matrices D + N where D is diagonalizable and N is nilpotent and D commutes with N Anyone know where a proof of this can be found, or feel up to finding a proof themselves? Jordan Normal forms of a matrix. Also, is the product of diagonalizable matrices diagonalizable? Ditto nilpotent? Finally, if the matrix is a differential operator, i.e. each element is a partial derivative operator, what can one say about the field containing the matrix? Or must one apply this matrix to a vector space before it makes sense to think of it as being associated with a field? John Ramsden === Subject: Re: Proof of matrix result wanted > The Wikipedia entry for Matrix Exponential quotes the result > that in an algebraically closed field any n*n matrix M can be > expressed uniquely as a sum of n*n matrices D + N where D is > diagonalizable and N is nilpotent and D commutes with N > Anyone know where a proof of this can be found, or feel up > to finding a proof themselves? > Jordan Normal forms of a matrix. > Also, is the product of diagonalizable matrices diagonalizable? > Ditto nilpotent? > Finally, if the matrix is a differential operator, i.e. each element > is a partial derivative operator, what can one say about the > field containing the matrix? Or must one apply this matrix > to a vector space before it makes sense to think of it as > being associated with a field? I'm not sure what you mean here. But a matrix of differential operators would act on a vector of functions. However you cannot say that the set of functions is itself a field. Rather, the whole matrix is some operator acting on an infinite dimensional vector space. With infinite dimensional vector spaces, questions like diagonalizability and such like become more tricky. However, if you think if differentiation acting on a space of functions (strictly speaking it will typically only be defined on a dense subspace), then you can usually diagonalize this. For example differentiation acting on L_2[0,1] diagonalizes with respect to the basis e_n(x)=exp(2pi inx). (In general you often take the eigenfunctions of the Laplacian.) Well, this is a huge subject, and I'm not quite sure where your knowledge is right now, but it would be a lot to chew. Stephen === Subject: Re: Proof of matrix result wanted >The Wikipedia entry for Matrix Exponential quotes the result >that in an algebraically closed field any n*n matrix M can be >expressed uniquely as a sum of n*n matrices D + N where D is >diagonalizable and N is nilpotent and D commutes with N >Anyone know where a proof of this can be found, or feel up >to finding a proof themselves? >Jordan Normal forms of a matrix. Well yes it does! Since the result you are trying to prove is not affected by replacing M by a similar matrix, you can assume M is in Jordan Normal Form. Then you can just write M as D + N, where D contains the diagonal entries of N and is 0 elsewhere, and N contains the off-diagonal of M and is zero elsewhere. e.g. [ 2,1 0 0 ] = [ 2 0 0 0 ] + [ 0 1 0 0 ] [ 0 2 0 0 ] [ 0 2 0 0 ] [ 0 0 0 0 ] [ 0 0 3 1 ] [ 0 0 3 0 ] [ 0 0 0 1 ] [ 0 0 0 3 ] [ 0 0 0 3 ] [ 0 0 0 0 ] It is easy to see that D and N commute, because all matrices are block sums of the Jordan Blocks, and D is scalar on each block. >Also, is the product of diagonalizable matrices diagonalizable? Not always: [ 1 2 ] X [ 1 0 ] = [ 1 1 ] [ 0 2 ] [ 0 1/2 ] [ 0 1 ] >Ditto nilpotent? Not always: [ 0 1 ] X [ 0 0 ] = [ 1 0 ] [ 0 0 ] [ 1 0 ] [ 0 0 ] >Finally, if the matrix is a differential operator, i.e. each element >is a partial derivative operator, what can one say about the >field containing the matrix? Or must one apply this matrix >to a vector space before it makes sense to think of it as >being associated with a field? Dunno! Derek Holt. === Subject: Re: Proof of matrix result wanted >>The Wikipedia entry for Matrix Exponential quotes the result >>that in an algebraically closed field any n*n matrix M can be >>expressed uniquely as a sum of n*n matrices D + N where D is >>diagonalizable and N is nilpotent and D commutes with N >>Anyone know where a proof of this can be found, or feel up >>to finding a proof themselves? >>Jordan Normal forms of a matrix. Sorry, I have not addressed the uniqueness part of the claim. I'll leave that to someone else! Derek Holt. >Well yes it does! Since the result you are trying to prove is not affected >by replacing M by a similar matrix, you can assume M is in Jordan Normal >Form. Then you can just write M as D + N, where D contains the diagonal >entries of N and is 0 elsewhere, and N contains the off-diagonal of M >and is zero elsewhere. e.g. >[ 2,1 0 0 ] = [ 2 0 0 0 ] + [ 0 1 0 0 ] >[ 0 2 0 0 ] [ 0 2 0 0 ] [ 0 0 0 0 ] >[ 0 0 3 1 ] [ 0 0 3 0 ] [ 0 0 0 1 ] >[ 0 0 0 3 ] [ 0 0 0 3 ] [ 0 0 0 0 ] >It is easy to see that D and N commute, because all matrices are block sums of >the Jordan Blocks, and D is scalar on each block. >>Also, is the product of diagonalizable matrices diagonalizable? >Not always: >[ 1 2 ] X [ 1 0 ] = [ 1 1 ] >[ 0 2 ] [ 0 1/2 ] [ 0 1 ] >>Ditto nilpotent? >Not always: >[ 0 1 ] X [ 0 0 ] = [ 1 0 ] >[ 0 0 ] [ 1 0 ] [ 0 0 ] >>Finally, if the matrix is a differential operator, i.e. each element >>is a partial derivative operator, what can one say about the >>field containing the matrix? Or must one apply this matrix >>to a vector space before it makes sense to think of it as >>being associated with a field? >Dunno! >Derek Holt. === Subject: Re: Proof of matrix result wanted >The Wikipedia entry for Matrix Exponential quotes the result >that in an algebraically closed field any n*n matrix M can be >expressed uniquely as a sum of n*n matrices D + N where D is >diagonalizable and N is nilpotent and D commutes with N >Anyone know where a proof of this can be found, or feel up >to finding a proof themselves? >Jordan Normal forms of a matrix. >Sorry, I have not addressed the uniqueness part of the claim. I'll leave >that to someone else! OK, I'll tackle that one. For each eigenvalue r of M, let V(r) = ker((M-rI)^n). Since D commutes with N (and therefore with M-rI), V(r) is invariant under D. Let D_r be the linear map v -> D v of V(r) to itself. I claim the only possible eigenvalue of D_r is r. For if d is an eigenvalue of D_r for eigenvector v, we have (M-dI)^n v = sum_{j=0}^n (n choose j) N^j (D-dI)^(n-j) v = 0 (the j=n term is 0 because N^n = 0, the others are 0 because (D-dI) v = 0), but we know ker((M-rI)^n) intersect ker((M-dI)^n) = {0} if r <> d. Now this implies V(r) is contained in ker((D-rI)^n). But since D is diagonalizable, ker((D-rI)^n) = ker(D-rI). Thus for v in V(r), D v = r v. This determines D (and thus also N) uniquely on each V(r), and thus also on the span of all the V(r), which is the whole space k^n (where k is the field). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Proof of matrix result wanted >The Wikipedia entry for Matrix Exponential quotes the result >that in an algebraically closed field any n*n matrix M can be >expressed uniquely as a sum of n*n matrices D + N where D is >diagonalizable and N is nilpotent and D commutes with N >Anyone know where a proof of this can be found, or feel up >to finding a proof themselves? >Jordan Normal forms of a matrix. >Sorry, I have not addressed the uniqueness part of the claim. I'll leave >that to someone else! Suppose M=D+N and M=D'+N' where D and D' are diagonal, N and N' are nilpotent, D commutes with N, and D' commutes with N'. D commutes with N implies D and N both commute with M. Similarly, D' and N' both commute with M. Also, since D and D' are diagonal, they commute with each other. Thus D commutes with M implies D commutes with D'+N', which implies D commute with N'. Similarly D' commutes with N. Then N commutes with M implies N commutes with D'+N', which implies N commutes with N'. Now D+N=D'+N' implies D-D'=N'-N. But D-D' is diagonal and N'-N is nilpotent (since N,N' commute). However if a diagonal matrix over a field is nilpotent then it must be the zero matrix. It follows that D=D' and N=N'. quasi === Subject: Re: Proof of matrix result wanted >>The Wikipedia entry for Matrix Exponential quotes the result >>that in an algebraically closed field any n*n matrix M can be >>expressed uniquely as a sum of n*n matrices D + N where D is >>diagonalizable and N is nilpotent and D commutes with N >>Anyone know where a proof of this can be found, or feel up >>to finding a proof themselves? >>Jordan Normal forms of a matrix. >>Sorry, I have not addressed the uniqueness part of the claim. I'll leave >>that to someone else! >Suppose M=D+N and M=D'+N' where D and D' are diagonal, N and N' are >nilpotent, D commutes with N, and D' commutes with N'. Unfortunately this is not enough, because we are only told that D and D' are diagonalizable, not that they are diagonal. We can assume that D is diagonal, but not a priori D'. Robert Israel has just posted a correct proof of uniqueness. Derek Holt. >D commutes with N implies D and N both commute with M. Similarly, D' >and N' both commute with M. >Also, since D and D' are diagonal, they commute with each other. >Thus D commutes with M implies D commutes with D'+N', which implies D >commute with N'. Similarly D' commutes with N. >Then N commutes with M implies N commutes with D'+N', which implies N >commutes with N'. >Now D+N=D'+N' implies D-D'=N'-N. But D-D' is diagonal and N'-N is >nilpotent (since N,N' commute). However if a diagonal matrix over a >field is nilpotent then it must be the zero matrix. >It follows that D=D' and N=N'. >quasi === Subject: Re: What's the defenition of Wedge product ? <445937a6@news1.ethz.ch> <4459d49f@news1.ethz.ch of the projection operator from the p-fold tensor product > to the p-fold exterior product of a vector space > contain a factor > 1/p! > in front of a certain sum of terms? Sorry I misunderstood the question. Obviously, if p divides n, then in characteristic p you can't divide by n! and hence you can't chop something up and spread it evenly over all n! permutations. You can't, likewise, average something over all n! permutations. === Subject: Re: What's the defenition of Wedge product ? schrieb Gene Ward Smith : [ ... ] > Sorry I misunderstood the question. Obviously, if p divides n, then in > characteristic p you can't divide by n! and hence you can't chop > something up and spread it evenly over all n! permutations. You can't, > likewise, average something over all n! permutations. But this only shows that the *usual definition* of the projection operator from the n-fold tensor product (x)^n V to the n-fold exterior product /^n V c (x)^n V for a vector space V over a field of characteristic p > 0 is not well-defined for p <= n, because there is a factor 1/n! in its definition. What I really wanted to know is: Under what condition on p, V, and n does there exist *a* canonical projection p: (x)^n V --> /^n V, projection meaning that p is the identity on the subset /^n V of (x)^n V ? === Subject: Re: f^ [n](x) = exp( n/ phi' (x) D ) o x > an other writing seems to be : > sum( k from 0 to infinity 1/ k! * { n / phi '(x) * d/dx } ^ k ) o > x . What is that x on the end. A function? A variable? If it is a variable, how do you compose with it? Should the ending be ... ^k)(x) and not ... ^k) o x ?? > I've tried the formula > exp( n/ phi' (x) D ) o x with a very simple > case f(x) = a*x +b , phi(x) =ln(x +b/(a-1)) / ln(a) > 1 / phi'(x) = (x + b/(a -1) )*ln(a) . > I believe you 're right saying : > a feeling it is a version > of the formula exp(aD)(g)(x) = g(x+a) which is a symbolic > way of writing the Taylor series. > Amiti.8es , Alain -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: f^ [n](x) = exp( n/ phi' (x) D ) o x >> an other writing seems to be : >> sum( k from 0 to infinity 1/ k! * { n / phi '(x) * d/dx } ^ k ) o >> x . > What is that x on the end. A function? A variable? > If it is a variable, how do you compose with it? > Should the ending be ... ^k)(x) and not ... ^k) o x ?? I think he meants that the operator is acting x. For example, one has the evolution operator psi(t) = e^(aHt)*psi(0) which tells how psi evolves in time or equivilently sum((aHt)^k/k!,k=0..infinity)*psi(0) (aHt)^k = a^k*t^k*H^k so sum((at)^k/k!*H^k*psi(0),k=0..infinity) and H acts on psi(0) if H^k = d^k/dx^k for k>0 and H^0 = 1 then the sum above is psi(0). So potentially he could be talking about some quantum mechanical system but I don't know since I don't know what phi and n represent.. the o is another mystery. === Subject: Re: f^ [n](x) = exp( n/ phi' (x) D ) o x <040520061021097177%edgar@math.ohio-state.edu.invalid> <040520061551393458%edgar@math.ohio-state.edu.invalid> How may it work ? Let us try a known simple phi(x) =1/x ;phi'(x) = -1/x^2 and exp(n/phi'(x) D ) = exp(-n*x^2 D) , D for d/dx Development : Id + (- n*x^2 D)/1! + (-n*x^2 D)^2 /2! + ... exp(-n*x^2D) o x = Id o x -(n*x^2 D)/1! o x + . , with a recurrence relation between terms: ti(x) = - n*x^2/ i *d/dx (t[i-1](x) ) , we obtain : x -n*x^2 +n^2*x^3 - ... = x*( 1 -nx +(nx)^2 - (nx)^3 ... ) or x / (1+n*x) = (x /(1 +x) )^ [n ] , [n ] iterated .88 suivre , Alain === Subject: Re: f^ [n](x) = exp( n/ phi' (x) D ) o x > How may it work ? > Let us try a known simple phi(x) =1/x ;phi'(x) = -1/x^2 > and exp(n/phi'(x) D ) = exp(-n*x^2 D) , D for d/dx > Development : > Id + (- n*x^2 D)/1! + (-n*x^2 D)^2 /2! + ... > exp(-n*x^2D) o x = Id o x -(n*x^2 D)/1! o x + . , > with a recurrence relation between terms: > ti(x) = - n*x^2/ i *d/dx (t[i-1](x) ) , we obtain : > x -n*x^2 +n^2*x^3 - ... = x*( 1 -nx +(nx)^2 - (nx)^3 ... ) > or x / (1+n*x) = (x /(1 +x) )^ [n ] , [n ] iterated > .88 suivre , > Alain So you are using x here with three different meanings. Probably an expert would actually do that, but if we want to understand this, maybe we shouldn't. One: x as a variable, for example: phi(x) = 1/x. Two: as a function, the identity function, for example when you write o x on the end. Three: as an operator, multiplication by x, for example in: exp(-n*x^2 D). So, for clarity, let's write these three different ways. Write x for a variable. Write id for the identity function of a real variable, id(x)=x. And write X for the operator multiply by x: if g is a function of a real variable, then Xg is the function defined by (Xg)(x) = x*g(x). Your formulas above also involve two other operators, D the derivative operator, so (Dg)(x) = g'(x); and the identity operator (not the same as the identity funtion id) call it Id: Id(g) = g. If psi is a function, write psi(X) for the operator multiplication by psi(x). Now: the formula will claim that exp(n/phi'(X) D ) id should be f^[n], where f^[n](x) = phi^{-1}(phi(x)+n). If we write the exponential symbolically as the series, say t(n,x) = sum(k=0 to infinity) t_k(x) n^k/k! , then we want t_k = (1/phi'(X) D)^k id , so that recursively t_k = (1/phi'(X) D) t_{k-1} and t_0 = id. So (taking n as a continuous variable) term-by-term differentiation of the series shows that t(n,x) satisfies (using @ for partial derivative sign): @t/@n = 1/phi'(x) * @t/@x , t(0,x) = x And the formula t(n,x) = phi^{-1}(phi(x)+n) does, indeed, formally satisfy this partial differential equation. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Number Sorting program? (not processed: recipient mail2news@dizum.com in exclude file) Mail-To-News-Contact: abuse@dizum.com What kind of free program is available to sort pages of numbers into manageable chunks? I need to sort many pages of this: 1111 1111 1111 1111 2222 2222 2222 2222 3333 3333 3333 3333 4444 4444 4444 4444 5555 5555 5555 5555 6666 6666 6666 6666 into pages of this: 1111 1111 1111 1111 2222 2222 2222 2222 3333 3333 3333 3333 4444 4444 4444 4444 5555 5555 5555 5555 6666 6666 6666 6666 or even this: 1111 1111 1111 1111 2222 2222 2222 2222 3333 3333 3333 3333 4444 4444 4444 4444 5555 5555 5555 5555 6666 6666 6666 6666 === Subject: Re: Number Sorting program? > What kind of free program is available to sort > pages of numbers into manageable chunks? > I need to sort many pages of this: > 1111 1111 1111 1111 > 2222 2222 2222 2222 > 3333 3333 3333 3333 > 4444 4444 4444 4444 > 5555 5555 5555 5555 > 6666 6666 6666 6666 > into pages of this: > 1111 1111 1111 1111 > 2222 2222 2222 2222 > 3333 3333 3333 3333 > 4444 4444 4444 4444 > 5555 5555 5555 5555 > 6666 6666 6666 6666 > or even this: > 1111 1111 1111 1111 > 2222 2222 2222 2222 > 3333 3333 3333 3333 > 4444 4444 4444 4444 > 5555 5555 5555 5555 > 6666 6666 6666 6666 That's not sorting a file, that's splitting a file into smaller parts. For that, try the program 'split'. If you really do need to sort, then try the program 'sort'. What times are these when one has to tell people that in order to sort things they should use 'sort', and in order to split things, they should use 'split'? Pointy-clicky days of hell, probably. Phil -- What is it: is man only a blunder of God, or God only a blunder of man? -- Friedrich Nietzsche (1844-1900), The Twilight of the Gods === Subject: Re: Number Sorting program? ........... > What times are these when one has to tell people that in order to > sort things they should use 'sort', and in order to split things, > they should use 'split'? Pointy-clicky days of hell, probably. > Phil ... and in order to search for a pattern they should use 'grep' and in order to list files they should use 'ls'? === Subject: Re: Number Sorting program? > What kind of free program is available to sort > pages of numbers into manageable chunks? > I need to sort many pages of this: > 1111 1111 1111 1111 > 2222 2222 2222 2222 > 3333 3333 3333 3333 > 4444 4444 4444 4444 > 5555 5555 5555 5555 > 6666 6666 6666 6666 > into pages of this: > 1111 1111 1111 1111 > 2222 2222 2222 2222 > 3333 3333 3333 3333 > 4444 4444 4444 4444 > 5555 5555 5555 5555 > 6666 6666 6666 6666 > or even this: > 1111 1111 1111 1111 > 2222 2222 2222 2222 > 3333 3333 3333 3333 > 4444 4444 4444 4444 > 5555 5555 5555 5555 > 6666 6666 6666 6666 How many pages at once? Is it in .txt format? This appears quite simple to do. If you have any version of Visual Basic available then go to comp.lang.basic.visual.misc and ask there. They will solve it, free of charge! HTH Mick. === Subject: Re: Set theory question - > The data is: > 8000 Trains, each with about 150 route locations in any order. > Locations are 6 character strings. I am trying to write a fast query > that takes me from A to B on any two trains that have some overlap in > their routes. There may be one or more shared locations in the train's > routes. > Dean That sounds like a mess. a good Optimization problem. Overlap is simply the stops at the same route locations. But should have a max wait time also. What else would overlap mean? === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >Here, we're talking about the accessibility of large integers and large >computations. It is absolutely essential that an AI know its own >limits, lest it get lost in some computation it can never complete. So >it must assess the feasibility of doing any computational experiment >before undertaking said experiment. >>Why can the AI not just stop? > It certainly can. But it must make intelligent choices about what > experiments are likely to succeed. Are you sure you want your computer mathematician thinking about which experiments will terminate? If it gets too interested in that question, it may ask itself Is there any algorithm I can use to find out whether an arbitrary algorithm with a given input ever terminates?, and rediscover one form of the Halting Problem. Patricia === Subject: Re: The Numbers Game > PS: I believe that the Acceptable Use Policy of your ISP (Internet > Service Portal) requires you to stay on topic on these boards. Mine doesn't. === Subject: Re: The Numbers Game On Thu, 4 May 2006 13:02:13 -0700, Richard Henry PS: I believe that the Acceptable Use Policy of your ISP (Internet >> Service Portal) requires you to stay on topic on these boards. >Mine doesn't. Then your Internet Service Portal must operate from a foreign country. Here in the US, ISPs are bound by a number of state and federal laws. === Subject: Re: The Numbers Game > On Thu, 4 May 2006 13:02:13 -0700, Richard Henry PS: I believe that the Acceptable Use Policy of your ISP (Internet >> Service Portal) requires you to stay on topic on these boards. >Mine doesn't. > Then your Internet Service Portal must operate from a foreign country. > Here in the US, ISPs are bound by a number of state and federal laws. It has been my experience that US state and federal laws are absolutley clueles about Usenet. === Subject: Re: The Numbers Game >> On Thu, 4 May 2006 13:02:13 -0700, Richard Henry Service Portal) requires you to stay on topic on these boards. >>Mine doesn't. >> Then your Internet Service Portal must operate from a foreign country. >> Here in the US, ISPs are bound by a number of state and federal laws. >It has been my experience that US state and federal laws are absolutley >clueles about Usenet. Then why are they reading posts through Yahoo and suing Google to read all its web searches? -- Chris McG. Harming humanity since 1951. What do you expect from a bunch of kiwi smoking sheep herders? -- oTTo === Subject: Re: The Numbers Game > Then your Internet Service Portal must operate from a foreign country. > Here in the US, ISPs are bound by a number of state and federal laws. Here in Yurp, we're bound by the Thurn und Taxis ordinances. === Subject: Re: The Numbers Game >> Then your Internet Service Portal must operate from a foreign country. >> Here in the US, ISPs are bound by a number of state and federal laws. > Here in Yurp, we're bound by the Thurn und Taxis ordinances. doesn't surprise me. my workplace is managed directly by Tristero. butting -- I am very new to programming drivers so if I sound un-knowledgeable then it's because I am. -- first4internet's Ceri Coburn on writing Sony's DRM rootkit === Subject: Re: The Numbers Game <2rqi52tsjbvo7hq5279evvkv8gn1lbu85l@4ax.com> So, you're a numerologist also? > What does stamp collecting have to do with anything? > ITYM coin collecting. If I had said numismatist. But I didn't say that, did I? > PS: I believe that the Acceptable Use Policy of your ISP (Internet > Service Portal) requires you to stay on topic on these boards. > Yeah, right. === Subject: Re: The Numbers Game >So, you're a numerologist also? > What does stamp collecting have to do with anything? >> ITYM coin collecting. >If I had said numismatist. Why would you say anything about collecting baseball cards in a thread about numbers? >But I didn't say that, did I? Why would you even *think* it? >> PS: I believe that the Acceptable Use Policy of your ISP (Internet >> Service Portal) requires you to stay on topic on these boards. >> Yeah, right. You two are off-topic on this. -- Chris McG. Harming humanity since 1951. What do you expect from a bunch of kiwi smoking sheep herders? -- oTTo === Subject: Re: The Numbers Game >>So, you're a numerologist also? > What does stamp collecting have to do with anything? > ITYM coin collecting. >>If I had said numismatist. > Why would you say anything about collecting baseball cards in a thread > about numbers? You obviously don't know much about baseball. --oTTo-- === Subject: Re: The Numbers Game >So, you're a numerologist also? > What does stamp collecting have to do with anything? >> ITYM coin collecting. >If I had said numismatist. >> Why would you say anything about collecting baseball cards in a thread >> about numbers? >You obviously don't know much about baseball. I said baseball cards, not baseball, AND YOU'RE OFF-TOPIC, TOO! What is it with you atheists? -- Chris McG. Harming humanity since 1951. What do you expect from a bunch of kiwi smoking sheep herders? -- oTTo === Subject: Re: The Numbers Game >>So, you're a numerologist also? > What does stamp collecting have to do with anything? ITYM coin collecting. >>If I had said numismatist. > Why would you say anything about collecting baseball cards in a thread > about numbers? >>You obviously don't know much about baseball. > I said baseball cards, not baseball, How do you think they keep track of positions in baseball? Huh? Well how do ya, punk? > AND YOU'RE OFF-TOPIC, TOO! What > is it with you atheists? My newsreader does it automatically. Read the FAQ. --oTTo-- === Subject: Re: The Numbers Game >You obviously don't know much about baseball. >> I said baseball cards, not baseball, > How do you think they keep track of positions in baseball? > Huh? Well how do ya, punk? With stamps? > My newsreader does it automatically. Read the FAQ. Mine too! === Subject: Re: The Numbers Game > I said baseball cards, not baseball, AND YOU'RE OFF-TOPIC, TOO! What > is it with you atheists? When the topic is the nonexistence of something, ANYTHING is off-topic. Also, trolling Wayne M. Delia. Dr. Delia, Dr. Fine, Dr. Delia! åR === Subject: Re: The Numbers Game >So, you're a numerologist also? > What does stamp collecting have to do with anything? >> ITYM coin collecting. >If I had said numismatist. >But I didn't say that, did I? I don't know; do you usually read out loud while typing? Dave lady philately's lovah DeLaney -- /DavidtDeLaneytpostingtfrom dbd@vic.com It's not the pot thattgrows the flower It's not the clock thattslows the hour tThe definition's plain for anyone to see Love istall it takes totmake a family - R&P. VISUALIZEtHAPPYNET VRbeable http://www.vic.com/~dbd/ - net.legends FAQ & Magic / I WUV you in all CAPS! --K. === Subject: Re: The Numbers Game <2rqi52tsjbvo7hq5279evvkv8gn1lbu85l@4ax.com> What does stamp collecting have to do with anything? >> ITYM coin collecting. >If I had said numismatist. >But I didn't say that, did I? > I don't know; do you usually read out loud while typing? If I'm using voice recognition software, yes. But then, I wouldn't be typing. > Dave lady philately's lovah DeLaney > -- > /David DeLaney posting from dbd@vic.com It's not the pot that grows the flower > It's not the clock that slows the hour The definition's plain for anyone to see > Love is all it takes to make a family - R&P. VISUALIZE HAPPYNET VRbeable <2rqi52tsjbvo7hq5279evvkv8gn1lbu85l@4ax.com> <11ek525g3vqigjho34rqkj49sult9np16f@4ax.com On 4 May 2006 09:51:25 -0700, mensanator@aol.compost >> On 3 May 2006 22:01:38 -0700, mensanator@aol.compost > > Oh, and an Applied Math Degree. >>So, you're a numerologist also? >> What does stamp collecting have to do with anything? >> PS: I believe that the Acceptable Use Policy of your ISP (Internet >> Service Portal) requires you to stay on topic on these boards. >Huh? In what way was I off topic? Did you read the original >message? If anything was off-topic it was your joke about >stamp collecting. > Joke? What joke? > I suggest you avail yourself of a dictionary, post haste. Webster's 9th: numerology: the study of the occult significance of numbers >Try to keep up. > You should take your own advice before you look even more foolish. Tis not I who is the fool. === Subject: Re: The Numbers Game > Webster's 9th: > numerology: the study of the occult significance of numbers Is that the one with gullible removed? >> You should take your own advice before you look even more foolish. > Tis not I who is the fool. ITYM 'Tis not I who am the fool. HTH ECT ECT === Subject: Re: The Numbers Game <2rqi52tsjbvo7hq5279evvkv8gn1lbu85l@4ax.com> <11ek525g3vqigjho34rqkj49sult9np16f@4ax.com> numerology: the study of the occult significance of numbers > Is that the one with gullible removed? One can be a practitioner without being a believer. I learned that in church. >> You should take your own advice before you look even more foolish. > Tis not I who is the fool. > ITYM > 'Tis not I who am the fool. > HTH ECT ECT === Subject: Re: The Numbers Game > PS: I believe that the Acceptable Use Policy of your ISP (Internet > Service Portal) requires you to stay on topic on these boards. >>Huh? In what way was I off topic? Did you read the original >>message? If anything was off-topic it was your joke about >>stamp collecting. >> Joke? What joke? >> I suggest you avail yourself of a dictionary, post haste. > Webster's 9th: > numerology: the study of the occult significance of numbers Cool. Now look up YHBT. Waves To Friends, --oTTo-- === Subject: Re: The Numbers Game On Thu, 4 May 2006 15:43:39 -0400, Otto Bahn >> PS: I believe that the Acceptable Use Policy of your ISP (Internet >> Service Portal) requires you to stay on topic on these boards. Huh? In what way was I off topic? Did you read the original >message? If anything was off-topic it was your joke about >stamp collecting. > Joke? What joke? > I suggest you avail yourself of a dictionary, post haste. >> Webster's 9th: >> numerology: the study of the occult significance of numbers >Cool. Now look up YHBT. D00d, that's not even a word. >Waves To Friends, WTF! === Subject: Re: Difficult question; Kuratowski embedding, wodyjslawski theorem > <16778706.1146722728638.JavaMail.jakarta@nitrogen.math > forum.org>, > <11060507.1146695559614.JavaMail.jakarta@nitrogen.math >> forum.org>, <6372235.1146671915176.JavaMail.jakarta@nitrogen.mathf > orum.org>, > >>(Wojdyslawski Thorem) >>3.The image h(X) of the Kratowski embedding >>h: X -> S is closed subset of >>convex hull Z in S defined by h(X). >> >> Write a member of Z as z = sum_{j=1}^m t_j >> h(b_j) >> where b_j in X, t_j > 0, sum_j t_j = 1. >> Hint: If this is the limit of a sequence >> h(a_n), >> show d(b_j, a_n) -> 0 as n -> infinity. >> What do you mean by if this is the limit of a >sequence h(a_n)? >Do you mean z is the limit of h(a_n) i.e. >h(a_n) -> sum_{j=1}^m t_j h(b_j) ? > > Yes. > >Could you explain more? > > You want to show h(X) is closed in Z, so you >> consider > a member z of Z that is the limit of a sequence > in > h(X), > and show that z is in h(X). >>I tried to prove h(a_n) -> sum_{j=1}^m t_j > h(b_j). >> >> That's not what you should be trying to prove, > it's >> what >> you assume: >> _if_ this is true, _then_ d(b_j, a_n) -> 0 as n -> infty. >> >I am sorry for my writing. But I mean that if I > assume >h(a_n) -> sum_{j=1}^m t_j h(b_j), >I can not come up to d(b_j, a_n) -> 0. > Further hint: > t_j d(b_j, a_n) <= sum_{j=1}^m t_j d(b_j, a_n) = z(a_n) I am confused. I am sorry but I can't come up with right answer even with this hint. Please show me the right answer. I would be appreciated. === Subject: Re: Difficult question; Kuratowski embedding, wodyjslawski theorem <30434375.1146773874872.JavaMail.jakarta@nitrogen.mathforum.org <16778706.1146722728638.JavaMail.jakarta@nitrogen.math > forum.org>, > <11060507.1146695559614.JavaMail.jakarta@nitrogen.math >> forum.org>, <6372235.1146671915176.JavaMail.jakarta@nitrogen.mathf > orum.org>, >(Wojdyslawski Thorem) >>3.The image h(X) of the Kratowski embedding >>h: X -> S is closed subset of >>convex hull Z in S defined by h(X). >> Write a member of Z as z = sum_{j=1}^m t_j >> h(b_j) >> where b_j in X, t_j > 0, sum_j t_j = 1. >> Hint: If this is the limit of a sequence >> h(a_n), >> show d(b_j, a_n) -> 0 as n -> infinity. > >What do you mean by if this is the limit of a >sequence h(a_n)? >Do you mean z is the limit of h(a_n) i.e. >h(a_n) -> sum_{j=1}^m t_j h(b_j) ? Yes. Could you explain more? You want to show h(X) is closed in Z, so you >> consider > a member z of Z that is the limit of a sequence > in > h(X), > and show that z is in h(X). >>I tried to prove h(a_n) -> sum_{j=1}^m t_j > h(b_j). >> That's not what you should be trying to prove, > it's >> what >> you assume: >> _if_ this is true, _then_ d(b_j, a_n) -> 0 as n -> infty. > >I am sorry for my writing. But I mean that if I > assume h(a_n) -> sum_{j=1}^m t_j h(b_j), I can not come up to d(b_j, a_n) -> 0. > Further hint: > t_j d(b_j, a_n) <= sum_{j=1}^m t_j d(b_j, a_n) = z(a_n) > I am confused. > I am sorry but I can't come up with right answer > even with this hint. > Please show me the right answer. > I would be appreciated. OK. To show h(X) is closed in Z: Consider a member z of Z that is the limit of a sequence h(a_n) in h(X). We want to show z is in h(X). We write z = sum_{j=1}^m t_j h(b_j) with 0 < t_j <= 1 and sum_{j=1}^m t_j = 1. For any epsilon > 0, there is N such that for n > N, ||h(a_n) - z|| < epsilon. This implies |h(a_n)(x) - z(x)| < epsilon for all x in X. In particular, take x = a_n, and note that h(a_n)(a_n) = d(a_n, a_n) = 0, while z(a_n) = sum_{j=1}^m t_j d(b_j, a_n), so for each j, t_j d(b_j, a_n) <= sum_{j=1}^m t_j d(b_j, a_n) < epsilon. Noting that t_j > 0, we see that d(b_j, a_n) -> 0 as n -> infinity, i.e. b_j is the limit of the sequence (a_n). By the uniqueness of limits of sequences, all b_j are the same, and so z = sum_{j=1}^m t_j h(b_1) = h(b_1) is in h(X). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Asymptotic Integral I spend some time with the following integral without much success: ,1 | ______ | | 2 | arccos(x) - x |1 - x 2 | ------------------------ j1(L x) dx | x ' 0 where j1 is the Bessel function of the first kind. Actually I am interested in this integral only for large L. I have some arguments (including numerical integration) that the leading term is pi/4. However I need the form of the next term. The big question is if it goes like 1/L or log(L)/L or even something else. Unfortunately, I have no idea how to attack this problem properly. It seems, most of the integral comes from small x but I cannot get it to work even using the asymptotic form of the Bessel function (for large or small arguments). Probably, you have to split the integral at some point (depending on L) and then use different evaluations for the two parts but I completely lack inspiration how to do this in practise. What I can offer is at least an acknowledgement in a physics paper on black hole entropy. Any ideas? Robert -- .oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo .oO Robert C. Helling School of Science and Engineering International University Bremen print Just another Phone: +49 421-200 3574 stupid .sign; http://www.aei-potsdam.mpg.de/~helling === Subject: Re: Asymptotic Integral If I deciphered your Integral correctly Mathematica gives the solution as int^1_0 ((ArcCos(x)-x * Sqrt(1-x^2))/x * BesselJ(1,L*x)^2) = pi/4 - pi/4 * HypergeometricPFQ[{1/2,1/2},{1,1,2},-L^2] -pi/64 * L^2 * HypergeometricPFQ[{3/2,3/2},{2,3,3},-L^2]. The Limit for L->Infinity seems to be just pi/4. === Subject: Re: Asymptotic Integral > If I deciphered your Integral correctly Mathematica gives the solution as > int^1_0 ((ArcCos(x)-x * Sqrt(1-x^2))/x * BesselJ(1,L*x)^2) = pi/4 - pi/4 * HypergeometricPFQ[{1/2,1/2},{1,1,2},-L^2] -pi/64 * L^2 * HypergeometricPFQ[{3/2,3/2},{2,3,3},-L^2]. The Limit for L->Infinity seems to be just pi/4. That's correct, but I need the next term. It either behaves like 1/L or like log(L)/L and currently I tend to believe the second. The behaviour is most important, although the coefficient would be nice as well. And yes, I had tried mathematica before but that seems to be unable to do the asymptotics of the generalized hypergeometric functions. Robert -- .oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo .oO Robert C. Helling School of Science and Engineering International University Bremen print Just another Phone: +49 421-200 3574 stupid .sign; http://www.aei-potsdam.mpg.de/~helling === Subject: Re: Asymptotic Integral > On Fri, 05 May 2006 05:44:48 EDT, Andreas Dieckmann as int^1_0 ((ArcCos(x)-x * Sqrt(1-x^2))/x * BesselJ(1,L*x)^2) = pi/4 - > pi/4 * HypergeometricPFQ[{1/2,1/2},{1,1,2},-L^2] -pi/64 * L^2 * > HypergeometricPFQ[{3/2,3/2},{2,3,3},-L^2]. The Limit for L->Infinity > That's correct, but I need the next term. It either behaves like 1/L > or like log(L)/L and currently I tend to believe the second. I tend to believe the second also. But I'm curious to know why you seem confident that It either behaves like 1/L or like log(L)/L. Couldn't it behave like something else? > The > behaviour is most important, although the coefficient would be nice as > well. And yes, I had tried mathematica before but that seems to be > unable to do the asymptotics of the generalized hypergeometric > functions. I wonder if Maple can handle the asymptotics. David === Subject: Re: Asymptotic Integral > On Fri, 05 May 2006 05:44:48 EDT, Andreas Dieckmann as int^1_0 ((ArcCos(x)-x * Sqrt(1-x^2))/x * BesselJ(1,L*x)^2) = pi/4 - > pi/4 * HypergeometricPFQ[{1/2,1/2},{1,1,2},-L^2] -pi/64 * L^2 * > HypergeometricPFQ[{3/2,3/2},{2,3,3},-L^2]. The Limit for L->Infinity > That's correct, but I need the next term. It either behaves like 1/L > or like log(L)/L and currently I tend to believe the second. > I tend to believe the second also. But I'm curious to know why you seem > confident that It either behaves like 1/L or like log(L)/L. Couldn't it > behave like something else? > The > behaviour is most important, although the coefficient would be nice as > well. And yes, I had tried mathematica before but that seems to be > unable to do the asymptotics of the generalized hypergeometric > functions. > I wonder if Maple can handle the asymptotics. Maple seems unable to do the definite integral, complaining that the function has a pole at 0. (which seems reasonable since lim_{x->0+}arccos(x)/x=oo). I wonder how Mathematica manages to come with a closed form answer for that. I also checked the indefinite integral, but no matter what I tried, Maple won't give an asymptotics series for the integrand on either x or L. Perhaps someone with more experience over in comp.soft-sys.math.maple can tackle it better. > David -- Ioannis === Subject: Re: Asymptotic Integral > On Fri, 05 May 2006 05:44:48 EDT, Andreas Dieckmann > If I deciphered your Integral correctly Mathematica gives the > solution as int^1_0 ((ArcCos(x)-x * Sqrt(1-x^2))/x * BesselJ(1, > L*x)^2) = pi/4 - pi/4 * HypergeometricPFQ[{1/2,1/2},{1,1,2},-L^2] > -pi/64 * L^2 * HypergeometricPFQ[{3/2,3/2},{2,3,3},-L^2]. The Limit That's correct, but I need the next term. It either behaves like 1/L > or like log(L)/L and currently I tend to believe the second. > I tend to believe the second also. But I'm curious to know why you seem > confident that It either behaves like 1/L or like log(L)/L. Couldn't > it behave like something else? > The > behaviour is most important, although the coefficient would be nice > as well. And yes, I had tried mathematica before but that seems to be > unable to do the asymptotics of the generalized hypergeometric > functions. > I wonder if Maple can handle the asymptotics. > Maple seems unable to do the definite integral, complaining that the > function has a pole at 0. (which seems reasonable since > lim_{x->0+}arccos(x)/x=oo). But we need to look at the integrand as a whole. Near x = 0, it is approximately L^2 Pi x/8, and so there is no pole there. David > I wonder how Mathematica manages to come with a closed form answer for > that. > I also checked the indefinite integral, but no matter what I tried, Maple > won't give an asymptotics series for the integrand on either x or L. > Perhaps someone with more experience over in comp.soft-sys.math.maple can > tackle it better. === Subject: Re: Asymptotic Integral I guess the expression pi/4-2*pi / L for the limit will not be far from the truth. Any limit I have seen on simpler hypergeometric functions went like inverse square root of the argument. === Subject: Re: Asymptotic Integral Ì Robert C. Helling ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > I spend some time with the following integral without much success: > ,1 > | ______ > | | 2 > | arccos(x) - x |1 - x 2 > | ------------------------ j1(L x) dx > | x > ' 0 > where j1 is the Bessel function of the first kind. Actually I am > interested in this integral only for large L. I have some arguments > (including numerical integration) that the leading term is > pi/4. However I need the form of the next term. The big question is if > it goes like 1/L or log(L)/L or even something else. > Unfortunately, I have no idea how to attack this problem properly. It > seems, most of the integral comes from small x but I cannot get it to > work even using the asymptotic form of the Bessel function (for large > or small arguments). Probably, you have to split the integral at some > point (depending on L) and then use different evaluations for the two > parts but I completely lack inspiration how to do this in practise. > What I can offer is at least an acknowledgement in a physics paper on > black hole entropy. Any ideas? Maple 9 says: j:=Int((arccos(x)-x*sqrt(1-x^2))/x*BesselJ(1,L*x)^2,x); / 2 1/2 2 | (arccos(x) - x (1 - x ) ) BesselJ(1, L x) j := | -------------------------------------------- dx | x / > series(%,x); 2 2 4 Pi L 2 L 3 Pi L 4 2 4 5 6 ----- x - ---- x - ----- x + (1/60 L + 1/40 L ) x + O(x ) 16 6 128 > Robert > -- .oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo .oO > Robert C. Helling School of Science and Engineering > International University Bremen > print Just another Phone: +49 421-200 3574 > stupid .sign; http://www.aei-potsdam.mpg.de/~helling -- Ioannis === Subject: Re: Is String Theory any better than Intelligent Design? Oh yeah, Mr. Dirk? And you are probably a very sane shrink who can give sanity certificates? Please spare us; for we are not entering here to read your nasty and completely argumentless comments. Have something to say? Argue your way in. You think I'm wrong? The cemetery is full of graves. The graves of worthless theories. Or, am I wrong? === Subject: a nice algebra problem A student (don't know his name) sent to me a few days ago the following elegant problem: Let R be a commutative ring with identity having n^2 elements (here n>1). Show that if R is not a field, then R must have at least n-1 zero divisors (here 0 is not considered a zero divisor). Equivalently, |U(R)|leq n^2-n, where U(R) is the group of units of R. I have a 8-lines proof, but it uses Sylow subgroups. Can you provide a *simpler*, more elementary proof? For I had to excuse myself for not being able to find one. The student said the problem is from some journal. === Subject: Re: a nice algebra problem === Subject: Re: a nice algebra problem > Let R be a commutative ring with identity having n^2 elements (here n>1). > Show that if R is not a field, then R must have at least n-1 zero divisors > I have a 8-lines proof, but it uses Sylow subgroups. Can you provide a > *simpler*, more elementary proof? R nonfield -> is non0 nonunit z -> f(r) = rz not onto=1-1 -> z zero-divisor -> f has image and kernel in Z = {zero-divisors} / 0 -> #R < #Z^2 QED --Bill Dubuque === Subject: Re: a nice algebra problem On 05 May 2006 01:46:07 -0400, Bill Dubuque Let R be a commutative ring with identity having n^2 elements (here n>1). >> Show that if R is not a field, then R must have at least n-1 zero divisors >> I have a 8-lines proof, but it uses Sylow subgroups. Can you provide a >> *simpler*, more elementary proof? >R nonfield -> is non0 nonunit z -> f(r) = rz not onto=1-1 -> z zero-divisor >-> f has image and kernel in Z = {zero-divisors} / 0 -> #R < #Z^2 QED >--Bill Dubuque You certainly don't waste any lines. Nice, quasi === Subject: Re: a nice algebra problem > A student (don't know his name) sent to me a few days ago the following elegant problem: > Let R be a commutative ring with identity having n^2 elements (here n>1). Show that if R is not a field, then R must have at least n-1 zero divisors (here 0 is not considered a zero divisor). Equivalently, |U(R)|leq n^2-n, where U(R) is the group of units of R. > I have a 8-lines proof, but it uses Sylow subgroups. Can you provide a *simpler*, more elementary proof? For I had to excuse myself for not being able to find one. The student said the problem is from some journal. It is a problem from a student contest a few years ago. No need for R to be commutative or with unit. An equivalent formuation is: If the ring R contains m>0 zero divisors, then R has at most (m+1)^2 elements. Proof. Let u,v be two non-zero elements such that uv=0. We generate more zero divisors in the following way. For an arbitrary x, the element xu is either 0 or also a zero divisor, since (xu)v=x(uv)=0. If xu=yu for some different elements x,y, then (x-y)u=0, and x-y is a zero divisor. This implies that 0 or an arbitrary zero divisor can be obtained at most m+1 times in the form xu. Thus, each of 0 and the m zero divisors is obtained at most m times, and the number of elements of R cannot exceed (m+1)^2. Mate === Subject: Re: geometry system bump === Subject: t test question I have a simple question, Let's say I have two sets of data corresponding to a control experiment and a treatment experiment. The control has two experiments that give the values 23 and 15 respectively while the two treatment experiments give the values 1006 and 1967. While there seems to be a real difference to me doing a t test on these values and converting to p gives a nonsignificant result .09 to be specific and a t of 3.04. Also someone else has also run tests and supposedly came up with an extremely low p value although I am not sure what statistic test they ran. Should I expect a low p value from this type of data? Did I make a mistake on the t test? === Subject: Re: t test question >I have a simple question, >Let's say I have two sets of data corresponding to a control experiment >and a treatment experiment. The control has two experiments that give >the values 23 and 15 respectively while the two treatment experiments >give the values 1006 and 1967. >While there seems to be a real difference to me doing a t test on these >values and converting to p gives a nonsignificant result .09 to be >specific and a t of 3.04. Also someone else has also run tests and >supposedly came up with an extremely low p value although I am not sure >what statistic test they ran. Should I expect a low p value from this >type of data? Did I make a mistake on the t test? You did not make a mistake on the t-test. This test tests for a translation of the mean, and the distribution of the mean in the treatment experiments is so diffuse, that any reasonable confidence range for the second is so broad as to include 0. However, assuming normality of each, the ratio of the spreads should have a standard Cauchy distribution under the hypothesis that the variances are equal, and the ratio of more than 120 is significant at the 1% level. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: t test question ETAtAhRAs3hNKBIX4yAb/o7BO4/i0sgmywIVAI1VzIyxXZBdZ8ymdfTNi1Cd4Pbf Let 1 and 2 be the treatments. If both treatments have the same variance, then: t = (m1-m2)/e; e = sqrt(s1^2/n1+s2^2/n2); m = mean, n = count, s = standard deviation for that treatment. Is that what you used? --OL === Subject: Re: JSH: How can they care? > I'm sure you people want to keep the faith, but come on. How can any > of these people you think are actually competent and excellent > mathematicians keep so quiet if they really were? > I like to try and contact some of those big names every once in a > while, as I try to find a way to break the impasse, which is why Mazur > and Granville got early drafts of a key paper of mine. > Recently, I sent them some stuff about n^2 - r while I was on my way to > figuring out my latest result, but got no reply THIS time. Gee, whenever I send e-mail to one of the big names, I get some feedback. For instance, in Richard Guy's book, he mentions that if the Goldbach Conjecture is true, then 5 is the only odd untouchable number. I pointed out that this is only true if the stronger statement: CONJECTURE. Every even number >= 8 can be written as the sum of two DISTINCT primes. is true. He mentioned that it was a good point. I've also sent e-mails to Neil Sloane, about the Online Encyclopedia of Integer Sequences. For instance, I gave a proof that the two sequences A100774 and A020639 are the same. (The old A100774 was: A100774: the sequence defined by a(1) = 1, a(2) = 2, and a(n + 1) = a(n) * Lpf(n+1) / Gpf (a(n)), where Gpf(n) is the greatest prime factor of n. It has since been replaced. A100774 is the Lpf(n) function, the least prime factor of n.) I sent an e-mail with the proof _on April 1_, and he took me seriously. And that's only two out of dozens of such incidents. > They've learned. Yes; they've learned not to wrestle with a pig: It gets you dirty, and the pig likes it. More succinctly, they know you're a crank, and they have better things to do. Like publish papers that WON'T cause a journal to go out of business. > If they reply, I talk about them on Usenet, when they sit quiet > afterwards. > I fear that they do not give a damn about mathematics, and why should > they? > It's screwed them over. > These people grew up being told mathematical ideas that I can shoot > down with simple quadratic equations were gold. > They built their careers on research that my research shows is invalid. > What do they have left? > They just have the faith of the world, which keeps them in their > positions, and keeps them getting paychecks. > What does Wiles have if the full story comes out? > Not only does he lose credit for proving FLT, but it's likely that ALL > of his research over his entire career goes out, as not being valid > mathematics. > These people get shot back to zero. > More and more I can understand why they would choose to sit quiet as in > their position, would I do any different? > Maybe luckily for me I've been disillusioned so many times in life that > it's hard for me to believe in anything any more, except what I can > personally and simply prove down to basic axioms so that there is > absolutely no room for error. > Then what happened to Wiles, Ribet, Taylor, Granville, Mazur and so > many of you cannot happen. > If you all had gone through your lessons, proving everything back to > basic axioms, you might possibly have found a flaw in ideal theory, and > saved yourselves a lot of grief. I don't do research in ideal theory. Are there any flaws in graph theory? --- Christopher Heckman === Subject: Re: JSH: How can they care? > I'm sure you people want to keep the faith, but come on. How can any > of these people you think are actually competent and excellent > mathematicians keep so quiet if they really were? > I like to try and contact some of those big names every once in a > while, as I try to find a way to break the impasse, which is why Mazur > and Granville got early drafts of a key paper of mine. > Recently, I sent them some stuff about n^2 - r while I was on my way to > figuring out my latest result, but got no reply THIS time. > They've learned. Yes, they've learned. They've learned that when you pet or feed a stray dog because you feel sorry for him, it just encourages him to hang around and keep bothering you. So it's better just to ignore him, especially when he's the type that snarls and bites and leaves messes on the carpet. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: a book to give the geography of the land of mathematics schrieb Rufi_Dukes : > can anyone suggest a good book for laying out the map of mathematics > for a highly motivated, intelligent but mathematically virginal adult, > wanting to get the lie of the land before launching myself on a > course of mathematical study? [ ... ] Concepts of Modern Mathematics by Ian Stewart Men of Mathematics by E. T. Bell === Subject: Re: a book to give the geography of the land of mathematics > Men of Mathematics by E. T. Bell I'm told that this book has many historical inaccuracies, but it sure is a fun book to read. === Subject: Re: a book to give the geography of the land of mathematics <445a80f2@news1.ethz.ch schrieb Rufi_Dukes : > can anyone suggest a good book for laying out the map of mathematics > for a highly motivated, intelligent but mathematically virginal adult, > wanting to get the lie of the land before launching myself on a > course of mathematical study? > [ ... ] > Concepts of Modern Mathematics by Ian Stewart > Men of Mathematics by E. T. Bell all of which have helped launch my little barque; now i'm off to the library to test its interloan capacities! cheers to yall, rufi === Subject: Re: Corners in metric spaces Let M = (X, d) be a metric space. >>Definition of corner point >>Let S be subset of X. p is a corner point of S in X iff >>there exists a sequence B_1, B_2, B_3,... of open balls such that: >>1. the elements in the sequence are pairwise disjoint >>3. for each B_i: >> 3.1 there exists (at least) two distinct points such that >> both are boundary points of the closure of S and >> both are boundary points of the closure of B_i >> 3.2 the intersection of B_i and boundary of S is empty >> 3.3 the intersection of the closure of B_i and the closure of >> B_i+1 is a singleton and not in the closure of S >>2. each sequence p_1, p_2, p_3, ... where each p_i is in B_i, has >> p as a limit point >>This definition works in many cases, for instance in R^2, with the >>usual metric, the solution set of y = |X| has a corner point at >>the origin while y = x^2 does not have corner points. In R^3 the >>set of corner points of the intersection of any two ellipsoids >>(degenerate or not) is cornerless. >>Definition of function C: C(S) is the set of corner points of S. >>Definition of cornerless set: S is cornerless <-> C(S) is empty. >>Definition of n-cornerless set: >>S is 0-cornerless <-> S is cornerless, >>S is n+1-cornerless <-> C(S) is n-cornerless >>Since the empty set is cornerless, >>S is n-cornerless -> S is n+1-cornerless >>I would like to have the following sentence true: >>Sentence 1: >>Let S_1 and S_2 be any two n-cornerless sets. >>Then their union is n+1-cornerless >>and their intersection is n+1-cornerless. >>I think that the following sentence should be true at least in >>some spaces, for instance R^2 with usual metric: >>Sentence 2: For any set S in M, >>for the functions CSeq_0(S) = S, CSeq_n+1(S) = C(CSeq_n(S)), >>there exists a n such that CSeq_n+1(S) = CSeq_n(S) >>(this is trivially true for n-cornerless (with finite n) sets) >> I have sets in R^2 that provide counterexamples (I think) to your >> sentences S1 and S2. We need the following two lemmas. >> Lemma A. Let x_0 < x_1 < ... be real numbers, with (x_n) -> p. Then >> (in R^2) the set S ={ (x_n, 0) : n=0,1,...} has exactly one corner, >> namely the point (p,0). >I though also this when made sets that are not n-cornerless for >some larger n than 1. I thought about how to get rid of that >sort of corners but did not come up with anything nice. I >then thought that maybe they are corners for a good reason but >I still have to find one. > If you're going to iterate the process of taking corners then you > almost certainly have to consider corners of non-connected point sets. > I'm not sure what one's intuition says about this situation. Subsets > of a line in R^2 possibly should not have corners, but similar sets, > like {(1/n,0) : n = 1,2,... } union {(0,1/n) : n = 1,2,..} possibly > should. According to my intuition, in a way, yes. My intuitive picture is that in your latter case, there is a point and a lot of stuff where one part of it is a wall and the other part is another wall. The walls meet at the point. It would be possible to change the 3.1 in the definition to: 3.1 there exists (at least) two distinct points such that both are boundary points of the closure of S and both are boundary points of the closure of B_i and new--> the poinds are distinct to any other such points for B_j, j<>i I intuitively do not find that very attractive. The original definition works in quite many cases, I sort of feel that things will get too messy with the new 3.1. However, I thougt now about a criteria that in my option may decide which one is better. If the sentences 1. and 2. are true for some n>1 in R^n, preferably for all n>0, with the alternative definition and not with the original then the alternative is the correct one. If that does not make difference then the one that works with Sentence 1. is better. Still I think that it is possible that a lot of nice things are true with the original definition but not with the alternative. >> I'm not going to attempt a rigorous proof. The idea is that you can >> find an increasing sequence (n_k) and circles C_k such that >> 1. n_{k+1} > n_k + 1 >> 2. the center of C_k is above the x-axis. >> 3. C_k meets the x-axis in the two points x_{n_k} and x_{n_k + 1} >> 4. C_k and C_{k+1} are tangent. >> 5. the radius of the C_k's tends to zero. >I may be misunderstanding and/or unfamiliar with the terminology >but if 1. implies that the difference between consecutive elements >in the sequence is at least one and 3. requires that two points p1, p2 >belong to the circles such that d(p1, p2) > 1, does this not >imply that their radius is at least 1/2 and contradicts the >requirement in 5 ? > n_k, k=1,2 .. is the sequence of _indices_. We want it to increase by > at least two at each step. For example, suppose {n_0, n_1 ... } = > {1, 5, 22, 24...}. Then we want our circles to sit on the pairs of > points (x_1,x_2), (x_5,x_6), (x_22,x_23), (x_24,x_25), ... >> Then if B_k is the ball with boundary C_k the B_k's and (p,0) satisfy >> the definition. >> OK, thank you, now I can see that much. I am sorry but this was not sufficient for me to understand further your idea. If I had to describe the words you used one by one I may say most of the time something relatively coherent but do not understand the idea. I suppose that understanding that would require a lot of googling and studying. >> Lemma B. Let alpha be a countable ordinal, s:alpha -> R a bounded >> order-preserving injection, S = { (s(eta),0) : eta < alpha }. Then the >> corners of S are exactly the points s(beta+omega), for every beta >> such that beta+omega <= alpha. >I have to first understand the circle sequence idea before proceeding. >I think that it may be nice sometimes to make a difference between >angled corners and not angled corners. Definition of angled corner >point: p is an angled corner point if it is a corner point and >if neccessarily for the sequence of open balls B_n, >the limit n->oo, radius_of(B_n)/radius_of(B_{n+1}) <> 1 >where radius_of(b) is the radius of an open ball b. >Being an angled corner point is (as are being a corner point >and corner rank) invariant under metrical equivalence. >The corner point at the union of two circles with finite radius >when their intersection is a singleton is not an angled corner. >The corners of a rectangle (non-degenerate) are angled corners. > [..] > You say these notions are invariant under metrical equivalence. I'm not > sure that's true. It depends, I suppose, on what exactly you mean by > metrical equivalence. If you mean that two metrics on a space are > equivalent if they induce the same topology, then it certainly isn't > true. Take for example the sup metric on R^2, d_sup((v,w),(x,y)) = > sup {|v-x|, |w-y|}. d_sup is equivalent to the euclidean metric, but > since balls are now squares with sides parallel to the coordinate > axes the corners of sets turn out to be completely different. Even if > you modify section 3.3 of your definition of corner you still don't get > the same as the euclidean corner. I studied the book Fractals Everywhere, Barsnley et. al, it had only one definition for equivalent metrics. Wikipedia says that there are various degrees of metric equivalence: http://en.wikipedia.org/wiki/Metric_space Indeed under topologically isomorphic (or homeomorphic) metrics the notions are not invariant (Fractals Everywhere said just homeomorphic). What I mean is (I repeat the idea from Fractals Everywhere): Two metrics d_1, d_2 on space X are equivalent iff there exists constants 0 < c_1 < c_2 < oo such that c_1*d_1(x, y) <= d_2(x, y) <= c_2*d_1(x, y) Two metric spaces (X_1, d_1) and (X_2, d_2) are equivalent iff there exists a bijective funtion h: X_1 -> X_2 such that d_2(h(x), h(y)) is a metric equivalent to d_1. The book says that deformation is bounded, there is nowhere arbitrarily large compression or stretching nor overlapping, folding or ripping. By invariant I mean: Let Space_1 = (M_1, d_1) and Space_2 = (M_2, d_s) be two equivalent metric spaces (as defined). Let S_1 be a subset of M_1. Let h be the bijective function. Let S_2 be a subset of M_2 such that p_1 in S_1 -> h(p_1) in S_2 and p_1 not in S_1 -> h(p_1) not in S_2 Then p_1 is a corner point in S_1 -> h(p_1) is a corner point in S_2, the same for angled corner point. I appreciate very much your help and contribution. I am quite busy now but on next week I should have time to think this again. what I would like to do next is to be able to prove existence or construct a non-empty set S in R^2 such that: 1. The boundary of S is equal to its corner points 2. All corners of S are non-angled (the conjecture was that Mandelbrot-set has those properties) in contrast to the Sierpinksi-triangle T which has the properties 3. (the boundary pf) T is equal to its corner points 4. All corners of T are angled and examine the properties and common things between sets that have 1. and 2. and between sets that have 3. and 4. and then perhaps between those two collection of sets. We Pretty === Subject: Re: Corners in metric spaces [snip > I am quite busy now but on next week I should have time to think this > again. > what I would like to do next is to be able to prove existence or > construct > a non-empty set S in R^2 such that: > 1. The boundary of S is equal to its corner points > 2. All corners of S are non-angled > (the conjecture was that Mandelbrot-set has those properties) > in contrast to the Sierpinksi-triangle T which has the properties > 3. (the boundary pf) T is equal to its corner points > 4. All corners of T are angled A bit more details before going to bed: In fact what happens when you take the corners of a set when the set is in a way regular (it may be the same as having finite corner rank), for instance balls and their unions is that the Hausdorff-dimension drops by one. I think this very nice feature. If you have a regular set and keep taking corners eventually you will reach empty set and zero dimensionality. It would be very nice for the definition to fullfill sentence 2. In fact I now think that if the alternative definition works then it is also almost surely correct. If the set has not finite corner rank then this decreasing of dimensionalty may break up. It may be even the other way aound. I am quite sure that if you keep only angled corners this happes and also because of the connection between dimensionality and corner rank the the dimensionality has to be non-integer (because is does not fit for the behaviour of regular sets). Now I think that this is very good reason that the dimensionality of the Sierpinski-triangle is not integer. The case if you keep non-angled corners may and I strongly think it is different. I you think side by side a lot of non-angled corners they sort of fill up the space and I think this is the reason why the boundary of Mandelbrot set has integer dimensionaly (this is not exact presentation but I think it is the reason). We Pretty === Subject: Re: Corners in metric spaces again. > what I would like to do next is to be able to prove existence or > construct > a non-empty set S in R^2 such that: > 1. The boundary of S is equal to its corner points > 2. All corners of S are non-angled > (the conjecture was that Mandelbrot-set has those properties) > in contrast to the Sierpinksi-triangle T which has the properties > 3. (the boundary pf) T is equal to its corner points > 4. All corners of T are angled > and examine the properties and common things between sets that > have 1. and 2. and between sets that have 3. and 4. and then perhaps > between those two collection of sets. It is hard to stop thinking this now but one more thing, it is of course trivial to say that some things are related in non-trivial way, I am not claiming this but I just mention it. If a set S_1 has properties 1. and 2. and there is an iteration algorithm for the set and S_2 has properties 3. and 4. with iteration algorithm, then I think that maybe there is a third iteration combines the dynamics of the iteration (of course there are trivial combinations). What I think that even I may be able to do is to search something that is iterated and looks like Sierpinski-triangle when looked from one point of view and like something (maybe Mandelbrot-set) with 3. and 4. In some way they are both very basic sets that defined by iteration. Wikipedia described the idea of homotopy, it would be nice to know the homotopy-group of Sierpinski-triangle. Also mention a fact (I do not suggest connection) that some time ago I noticed (other have also noticed) that if you color the numbers in Pascal - triangle such that black if it is divisible by number n, white otherwise, it looks like sort of Sierpinski-triangle. I had that time a feeling that it should have good explanation. We Pretty === Subject: Re: Corners in metric spaces Quick correction: Being a corner point is not invariant in general. The sentence below does not hold in general. It may hold for non-angled corners. >Then p_1 is a corner point in S_1 -> h(p_1) is a corner point in S_2, >the same for angled corner point. === Subject: how does cheb_eval.c work? Hi All, Does anyone happen to know how errors are computed in cheb_eval_c.c? is this in anyway related to the minimax theorem? === Subject: Re: Tiny FLT proof possible? I note that you conveniently left out any example solutions. If it is so trivial, surely you have found some. > My quadratic residue result may offer the route to a tiny proof of > Fermat's Last Theorem. > The result is that given naturals n_1, p a prime factor of n_1, n_2, C > = n_1 + n_2, and k a difference of factors of 2*C, it must be true that > (8*n_2 + k^2) is a square modulo p > so with x^n + y^n = z^n, you can let C = z^n, and k is a difference of > factors and there is a set of all such possibles. > The thing is, n_1 and n_2 can be ANY naturals that add to give z^n. > It's trivial to show then that > (8*z^n + k^2) would have to be a square modulo p > for all p, primes less than z^n. > A simple check of combinations using f_1 f_2 = z, with n=3, gives an > indication why that's impossible, as I think it actually gets blocked > by 5, or if not 5, 7, which has only three quadratic residues. > (f_1^3 - f_2^3), (f_1^2 f_2 - f_1 f_2^2), (f_1^3 f_2^3 - 1) > are just some combinations I can think of immediately with n=3, and if > FLT were false, for like p=7, it'd have to be true that all of those > squared would add to 8*z^2 mod 7 to give a quadratic residue of 7, i.e. > give 0, 1, 2 or 4. > This quadratic residue result of mine may be one of the most powerful > in number theory, offering routes to proving Goldbach's conjecture, and > probably all kinds of Diophantine equations. > A tiny proof of FLT that is just about as trivially easy as you can > get. Kind of strange. > James Harris === Subject: not positive-definite covariance matrix Hi. I am trying to apply the Cholesky decomposition to a covariance matrix. Matlab is giving me an error saying that it is not positive definite: >> chol(c) ??? Error using ==> chol Matrix must be positive definite. Since that is a covariance matrix I expect it to be positive definite, though. I am populating it using a Gaussian function, C(d) = C_0 * exp (-k * d^2), where d is distance between samples and the coefficients C_0, k are estimated from the observations. A positive definite function should yield a positive definite matrix. I am not sure if the Gauss function is positive definite itself, but since it is commonly employed as covariance function, I expect it to be so. Some more details about that covariance matrix: - its dimensions are 10-by-10 - when I try [R,p] = chol(c) I get p == 9 - it has one negative eigenvalue: -7.2e-009 - its conditioning number is very large: cond(c) == 3.1e+10 - its determinant is practically zero: det(c) == -1.8e-038 - it has full rank: rank(c) == size(c, 1) == size(c, 2) - I am able to invert it, inv(c), without warnings. - I am able to apply the LU decomposition, lu(c), to it. Would you have any thoughts on why that covariance matrix is _not_ positive definite, please? I am sending the data values below. Felipe. % The coefficients of the covariance function have the following values: K = 2.2901e-006; C_0 = 25.9934; % The covariance matrix is: % (I am sending the matrices in vector form so that breaklines in my post won't prevent you from copying and pasting the code in Matlab): c = zeros(10,10); c(:) = [... 25.9934 25.9666 25.9480 25.9495 25.8859 25.7550 25.8295 25.6329 25.5712 25.4802 25.9666 25.9934 25.9756 25.9216 25.9185 25.7926 25.8106 25.6493 25.6156 25.5376 25.9480 25.9756 25.9934 25.9588 25.9737 25.8939 25.8997 25.7868 25.7604 25.6971 25.9495 25.9216 25.9588 25.9934 25.9550 25.8964 25.9546 25.8324 25.7807 25.7123 25.8859 25.9185 25.9737 25.9550 25.9934 25.9627 25.9512 25.8923 25.8759 25.8294 25.7550 25.7926 25.8939 25.8964 25.9627 25.9934 25.9614 25.9697 25.9652 25.9396 25.8295 25.8106 25.8997 25.9546 25.9512 25.9614 25.9934 25.9482 25.9114 25.8669 25.6329 25.6493 25.7868 25.8324 25.8923 25.9697 25.9482 25.9934 25.9852 25.9702 25.5712 25.6156 25.7604 25.7807 25.8759 25.9652 25.9114 25.9852 25.9934 25.9886 25.4802 25.5376 25.6971 25.7123 25.8294 25.9396 25.8669 25.9702 25.9886 25.9934 ]; % The corresponding distance matrix, with which the covariance function is evaluated, is: d = zeros(10,10); d = [... 0 21.2208 27.6401 27.1707 42.5542 63.4404 52.5649 78.0986 84.5669 93.3130 21.2208 0 17.3072 34.7515 35.4951 58.2042 55.5116 76.2823 79.9613 87.8991 27.6401 17.3072 0 24.1130 18.2262 40.9362 39.7132 59.0327 62.7106 70.7560 27.1707 34.7515 24.1130 0 25.4183 40.4057 25.5564 52.0983 59.9067 68.9113 42.5542 35.4951 18.2262 25.4183 0 22.7126 26.6571 41.2569 44.4845 52.5753 63.4404 58.2042 40.9362 40.4057 22.7126 0 23.1892 19.9906 21.7818 30.0809 52.5649 55.5116 39.7132 25.5564 26.6571 23.1892 0 27.5860 37.1571 46.1521 78.0986 76.2823 59.0327 52.0983 41.2569 19.9906 27.5860 0 11.7663 19.7435 84.5669 79.9613 62.7106 59.9067 44.4845 21.7818 37.1571 11.7663 0 9.0577 93.3130 87.8991 70.7560 68.9113 52.5753 30.0809 46.1521 19.7435 9.0577 0 ]; === Subject: Re: not positive-definite covariance matrix > ... problem with matrix not being positive definite .... ----------------------------- I am very curious about your matrix d. You describe it as, d is distance between samples. What does that mean? To be speaking of a 10 x 10 covariance matrix, you would be referring to ten different variables obtained in each sample. What is the nature of these variables and how many samples of them are there? Exactly how do the samples relate to the matrix d - that is, how was d calculated? You have obtained matrix c directly from matrix d with the calculation: c = C_0 * exp (-K * d.^2). What makes you think that this c should constitute a valid covariance matrix and therefore be positive definite? (Remove xyzzy and .invalid to send me email.) Roger Stafford === Subject: Re: not positive-definite covariance matrix , > I am very curious about your matrix d. You describe it as, d is > ... ----- Sorry about the duplications. They were not intentional. (Remove xyzzy and .invalid to send me email.) Roger Stafford === Subject: Re: not positive-definite covariance matrix > C(d) = C_0 * exp (-k * d^2), > K = 2.2901e-006; Look at the correlation matrix you get from this (by the way, you want .^, not ^; probably just a transcription error though): C = C_0 * exp (-K * d.^2); s = sqrt(diag(C)); R = C ./ (s*s') R = Columns 1 through 6 1 0.99897 0.99825 0.99831 0.99586 0.99083 0.99897 1 0.99931 0.99724 0.99712 0.99227 0.99825 0.99931 1 0.99867 0.99924 0.99617 0.99831 0.99724 0.99867 1 0.99852 0.99627 0.99586 0.99712 0.99924 0.99852 1 0.99882 0.99083 0.99227 0.99617 0.99627 0.99882 1 0.99369 0.99297 0.99639 0.99851 0.99837 0.99877 0.98613 0.98676 0.99205 0.9938 0.99611 0.99909 0.98376 0.98546 0.99103 0.99181 0.99548 0.99891 0.98026 0.98246 0.9886 0.98918 0.99369 0.99793 Columns 7 through 10 0.99369 0.98613 0.98376 0.98026 0.99297 0.98676 0.98546 0.98246 0.99639 0.99205 0.99103 0.9886 0.99851 0.9938 0.99181 0.98918 0.99837 0.99611 0.99548 0.99369 0.99877 0.99909 0.99891 0.99793 1 0.99826 0.99684 0.99513 0.99826 1 0.99968 0.99911 0.99684 0.99968 1 0.99981 0.99513 0.99911 0.99981 1 You got ten variables, all of which are almost perfectly positively correlated. It's not surprising that the cov matrix is degenerate. Use a larger value for K (1e-3, say) and you'll get variables that are less strongly correlated. Hope this helps. - Peter Perkins The MathWorks, Inc. === Subject: Re: not positive-definite covariance matrix >> C(d) = C_0 * exp (-k * d^2), >> K = 2.2901e-006; >Look at the correlation matrix you get from this (by the way, you want .^, not >^; probably just a transcription error though): >C = C_0 * exp (-K * d.^2); >s = sqrt(diag(C)); >R = C ./ (s*s') >R = > Columns 1 through 6 > 1 0.99897 0.99825 0.99831 0.99586 0.99083 > 0.99897 1 0.99931 0.99724 0.99712 0.99227 > 0.99825 0.99931 1 0.99867 0.99924 0.99617 > 0.99831 0.99724 0.99867 1 0.99852 0.99627 > 0.99586 0.99712 0.99924 0.99852 1 0.99882 > 0.99083 0.99227 0.99617 0.99627 0.99882 1 > 0.99369 0.99297 0.99639 0.99851 0.99837 0.99877 > 0.98613 0.98676 0.99205 0.9938 0.99611 0.99909 > 0.98376 0.98546 0.99103 0.99181 0.99548 0.99891 > 0.98026 0.98246 0.9886 0.98918 0.99369 0.99793 > Columns 7 through 10 > 0.99369 0.98613 0.98376 0.98026 > 0.99297 0.98676 0.98546 0.98246 > 0.99639 0.99205 0.99103 0.9886 > 0.99851 0.9938 0.99181 0.98918 > 0.99837 0.99611 0.99548 0.99369 > 0.99877 0.99909 0.99891 0.99793 > 1 0.99826 0.99684 0.99513 > 0.99826 1 0.99968 0.99911 > 0.99684 0.99968 1 0.99981 > 0.99513 0.99911 0.99981 1 >You got ten variables, all of which are almost perfectly positively correlated. > It's not surprising that the cov matrix is degenerate. Use a larger value for >K (1e-3, say) and you'll get variables that are less strongly correlated. Something which may work better is to compute the elements of R to oodles of accuracy, and use the appropriate accuracy on the matrix. I believe in that case if the distances satisfy the triangle inequality the matrix will be positive definite in all cases. There are times to use far more accuracy in the computations than are in the data, and this is such a situation. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: not positive-definite covariance matrix > Hi. > I am trying to apply the Cholesky decomposition to a covariance matrix. > Matlab is giving me an error saying that it is not positive definite: > >> chol(c) > ??? Error using ==> chol > Matrix must be positive definite. > Since that is a covariance matrix I expect it to be positive definite, > though. I am populating it using a Gaussian function, > C(d) = C_0 * exp (-k * d^2), > where d is distance between samples and the coefficients C_0, k are > estimated from the observations. A positive definite function should > yield a positive definite matrix. I am not sure if the Gauss function > is positive definite itself, but since it is commonly employed as > covariance function, I expect it to be so. Looking at the numbers its not surprising it's ill-conditioned: the covariance just doesn't change much. So a Cholesky decomposition will give poor results. There is a second variant of Cholesky in MatLab which terminates early if the matrix is ill-conditioned, but using a reduced rank decomposition, as others have said, is likely to give a better approximation to the data. john2 === Subject: Re: not positive-definite covariance matrix > Some more details about that covariance matrix: > - its dimensions are 10-by-10 > - when I try [R,p] = chol(c) I get p == 9 > - it has one negative eigenvalue: -7.2e-009 > - its conditioning number is very large: cond(c) == 3.1e+10 > - its determinant is practically zero: det(c) == -1.8e-038 > - it has full rank: rank(c) == size(c, 1) == size(c, 2) > - I am able to invert it, inv(c), without warnings. > - I am able to apply the LU decomposition, lu(c), to it. > Would you have any thoughts on why that covariance matrix is _not_ > positive definite, please? The large condition number and tiny (negative) determinant suggest that your matrix is ill conditioned. Algorithms that compute the same quantity mathematically (inverse, determinant, eigenvalues) may give wildly different answers when applied to such a matrix. Another way to look at this: Mathematically, with infinite precision, your matrix may be positive definite. However, you are working with a finite precision representation of the matrix. There are no guarantees that the finite precision representation is also positive definite. === Subject: Re: not positive-definite covariance matrix ETAtAhRwJKcPe7BR+OdTU23RkicQxGU0HwIVAIK1nch0+Y4esTdkimKpN2vEvt4l Your negative eigenvalue is probably a zero eigenvalue with roundoff error. 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So I thought to myself, hey, why not use this result to try and factor? So I arbitrarily picked k=1, C = mT, so I'd have sqrt(1 + 8*mT) where T is the target to be factored, and I solved for m as a congruence relation and found that m = (r-1)(8*T)^{-1} mod n_1 where is a square modulo n_1, or as I like to say, is a quadratic residue of n_1. Notice here you can pick n_1 to be ANY natural number you choose. I tossed out floor(T/2) as a possibility, as, well, it was just an idea of the moment. I also have tended to just pick r=4, because just about any n_1 will have 4 as a quadratic residue, but you can pick something else. What the mathematics tells you is that ALL solutions for m that will give you that quadratic residue, must fit into the congruence relationship given. But how likely are you to find a solution? I don't know. It is possible, amazingly enough, and possibly unfortunately, that as T increases and size, and as you pick an ever large n_1, close to T, that the number of possible m's increases rapidly, as in, faster than linear. Or, they may drop rapidly. Or, they may stay roughly the same, while the quadratic residues increase linearly. Or the number of m's may increase at roughly the same rate as the number of quadratic residues. So the theory is not such a big deal as it's all simple. But the practicality is in question, where maybe, if you pick n_1 correctly, and go to bigger and bigger composites, um, you may find it easier to factor than with small numbers. So little test examples may give the wrong idea, but I'm still in guess mode. The problem, with T = p_1 p_2, where the p's are primes, can be considered to be finding the solution set xp_1 - yp_2 = 1 where m = xy, and it looks to me like it might go as the square, in turns of possible m's so that while you increase T linearly, the number of m's that will work, increase as the square, which if true means we're screwed, and someone could figure out a way to factor with this thing, where it would actually work better with RSA sized numbers than with smaller numbers. But I'm still just guessing and I'm not seeing posts that address the actual theoretical issues, so I think it's an open question. Then again, I think I could be reaching, hoping a nice idea that can't be practical can be, and that instead the odds of landing on an m decrease so rapidly that they are vanishingly small with a large T, which I think is the opinion that most would think should apply. After all, if that weren't true, then I couldn't just be chatting out a solution to the factoring problem on Usenet for DAYS, so it must not be true, right? James Harris === Subject: Re: My factoriing idea, theoretical issues Try a Google search on excludent factorization method It involves quadratic residues and multipliers. === Subject: Re: My factoriing idea, theoretical issues >I noted that you can use the result that with naturals n_1, p a prime > factor of n_1, n_2, > C = n_1 + n_2, and k a difference of factors of 2*C that > (8*C + k^2) mod n_1 must be a square modulo p > and, of course, that means that it is a square modulo n_1. could you explain that please ? === Subject: Re: My factoriing idea, theoretical issues <445ab716$0$16435$892e7fe2@authen.yellow.readfreenews.netI noted that you can use the result that with naturals n_1, p a prime > factor of n_1, n_2, > C = n_1 + n_2, and k a difference of factors of 2*C that > (8*C + k^2) mod n_1 must be a square modulo p > and, of course, that means that it is a square modulo n_1. > could you explain that please ? Sure. I hope you do not get upset with too much detail as I will assume you know nothing about quadratic residues and I admit I do not know a lot myself. Quadratic residues are just the square modulo some number, for example 4^2 =2 + 7*2 so 2 is a square modulo 7 or you can say that 2 is a quadratic residue of 7, though my understanding is that mathematicians like to say 2 is a quadratic residue modulo 7. 4^2 = 2 + 7*2 you can see both what a residue is and what a quadratic residue is, as that is written with congruence relations as 4^2 = 2 mod 7 which is just like clock time, as if you have a 24 hour clock, you are using time modulo 12. So 1300 hours is modulo 2400 hours. Or 11 am is modulo 12. What I found is that these residues are connections between sums and factorizations, and using that information I have my very own factoring method. I just discovered it last weekend. That is what I do. I discover things. Consider 5 + 7 = 12, so n_1 = 5, n_2 = 7, and C = 12. Multiply C times 2, and you have 24, and a difference of factors gives 2, where I took 6 and 4, and subtracted. So k = 2 is a possibility, and plugging into my formula above I have 8*12 + 4 = 100, and 100 mod 7 = 2 as required, as 2 is a square modulo 7. And, of course, because 100 is divisible by 5 it is a square modulo 5, as 100 = 0 + 5*5*4. I came up with the idea of factoring a composite T, by looking to find a natural number m such that 8*mT had 1 as a difference of factors, and found that with that idea I could find m as a congruence relation, showing where you'd have to search for an m that would work with a particular quadratic residue that I call r. The equation is m = (r-1)(8*mT)^{-1} mod n_1 where (8*mT)^{-1} is what's called a modular inverse. And that is just some number that multiplies times 8*mT to give a number with a residue of 1. For instance with the modular inverse of 5 mod 7 is 3, because 3*5 = 1 mod 7 as 3*5 = 1 + 7*2. That idea is perfect as the mathematics is without question correct. The only question is, does the idea work well enough to be effective in factoring large numbers? Unfortunately, it may be. And I say unfortunately as if it is too good of a solution, it will change the world, removing the Internet as it now stands, forcing an entirely new Internet, and in the process it will change the entire world. The problem is that mathematicians may have known almost a decade ago that such an event was possible, but hid that information from the world by not acknowledging it or its significance as a Ukrainian mathematician named Anatoly Plotknikov may have proven that such a thing was possible in a paper peer reviewed and published in 1996. They may have done so stupidly believing that even if some solution were possible it would be something very complicated or that it was so difficult to find that they were safe in having the world depend on a system that was inherently insecure--and proven to be back in 1996. If so, then the world as you and I know it may end because of a relatively small group of people who bet the world--and lost. James Harris === Subject: Re: My factoriing idea, theoretical issues > I noted that you can use the result that with naturals n_1, p a prime > factor of n_1, n_2, I love this show. Can always count on someone stupider than me proving their status daily. Hey Robert, look I'm not the dumbest person on usenet :-) Tom === Subject: Re: My factoriing idea, theoretical issues OpenPGP: id=A0E28D18 >> I noted that you can use the result that with naturals n_1, p a prime >> factor of n_1, n_2, > I love this show. Can always count on someone stupider than me proving > their status daily. Which is quite hard for someone who doesn't killfile JH and even replies to his postings. Could you and other repliers please stop doing that? Makes this ng more readable. === Subject: Re: Is platonism in math correct? symbols and very precise rules with which you can build certain symbolic > structures. These structures that can be used model some, if not all, > aspects of physical reality. Like term formalism, game formalism either solves or sidesteps difficult metaphysical and epistemological problems with mathematics. What is mathematics about? Nothing. What /are/ numbers, sets, and so on? They do not exist, or they might as well not exist. How is mathematics known? What is mathematical knowledge? It is knowledge of the rules of the game, or knowledge that certain moves that accord with these rules have been made. The equation '2^10 = 1024' and the theorem that for every natural number x there is a prime number y > x (in symbols, AxEy(y > x & y is prime)) each indicate the outcome of a certain play in accordance with the rules of arithmetic. [...] The game formalist, however, is left with a daunting problem. Why are the mathematical games so useful in the sciences? After all, no one even looks for useful applications of chess. Why think that the meaningless game of mathematics should have any applications? It clearly does, and we have to explain those applications. A similar problem arises for applications of mathematics within mathematics. Why is the game of complex analysis useful in the game of real analysis or arithmetic? This issue is all the more troubling for someone who is a game formalist about, say, complex analysis, but not about real analysis or arithmetic. [Shapiro. Stewart | 2000 | Thinking about mathematics: The philosophy of mathematics | Oxford / Oxford UP. (145+)] #PH === Subject: Re: Is platonism in math correct? >> I find it helps to think of mathematics as a kind of game played with >> symbols and very precise rules with which you can build certain symbolic >> structures. These structures that can be used model some, if not all, >> aspects of physical reality. > Like term formalism, game formalism Aw... so there is name for it then! Interesting. either solves or sidesteps > difficult metaphysical and epistemological problems with mathematics. > What is mathematics about? Nothing. What /are/ numbers, sets, and so > on? They do not exist, or they might as well not exist. How is > mathematics known? What is mathematical knowledge? It is knowledge of > the rules of the game, or knowledge that certain moves that accord with > these rules have been made. The equation '2^10 = 1024' and the theorem > that for every natural number x there is a prime number y > x (in > symbols, AxEy(y > x & y is prime)) each indicate the outcome of a > certain play in accordance with the rules of arithmetic. > [...] The game formalist, however, is left with a daunting problem. Why > are the mathematical games so useful in the sciences? After all, no one > even looks for useful applications of chess. Why think that the > meaningless game of mathematics should have any applications? It > clearly does, and we have to explain those applications. Engineers and scientists often construct wooden models to obtain useful insights into the real world. Nothing mysterious about that. And I don't think there is any need to attribute a different plane of existence for wooden models in general just because they have proven so useful in applications. Why then should the purely symbolic models of mathematicians be different in this regard? Dan Download my DC Proof software at http://www.dcproof.com === Subject: Re: Is platonism in math correct? <9yA6g.7532$I_1.5056@newsfe23.lga Like term formalism, game formalism either solves or sidesteps > difficult metaphysical and epistemological problems with mathematics. > What is mathematics about? Nothing. What /are/ numbers, sets, and so > on? They do not exist, or they might as well not exist. How is > mathematics known? What is mathematical knowledge? It is knowledge of > the rules of the game, or knowledge that certain moves that accord with > these rules have been made. The equation '2^10 = 1024' and the theorem > that for every natural number x there is a prime number y > x (in > symbols, AxEy(y > x & y is prime)) each indicate the outcome of a > certain play in accordance with the rules of arithmetic. > [...] The game formalist, however, is left with a daunting problem. Why > are the mathematical games so useful in the sciences? After all, no one > even looks for useful applications of chess. Why think that the > meaningless game of mathematics should have any applications? It > clearly does, and we have to explain those applications. [...] Please don't snip off references: [Shapiro S| 2000 | Thinking about mathematics: The philosophy of mathematics | Oxford / Oxford UP. (145+)] > Engineers and scientists often construct wooden models to obtain useful > insights into the real world. Nothing mysterious about that. And I don't > think there is any need to attribute a different plane of existence for > wooden models in general just because they have proven so useful in > applications. Why then should the purely symbolic models of mathematicians > be different in this regard? Technical models such as wooden 3D ones bear a representational relation to the real objects they are models of, whereas mere formal games, which lack such semantic 'aboutness', don't. Frege's main criticism of game formalism goes along these lines [in his Grundgesetze der Arithmetik 2 (1903)]: An arithmetic without thought as its content will also be without possibility of application. Why can no application be made of a configuration of chess pieces? Obviously, because it expresses no thought. If it did so and every chess move conforming to the rules corresponded to a transition from one thought to another, applications of chess would also be conceivable. Why can arithmetical equations be applied? Only because they express thoughts. How could we possibly apply an equation which expressed nothing and was nothing more than a group of figures, to be transformed into another group of figures in accordance with certain rules? [I]t is applicability alone which elevates arithmetic from a game to the rank of science. [Shapiro S | 2000 | Thinking about mathematics: The philosophy of mathematics | Oxford / Oxford UP. (147)] #PH === Subject: Re: Is platonism in math correct? > What is math's relation to the real world? Is math some higher ideal > that exists on a different plane from reality, such as in platonism, or > is there some fundamental connection between mathematics and reality? > http://en.wikipedia.org/wiki/Platonic_idealism > Just a site I wanted you to see. :P Here's a very good text on platonism: Encyclopedia of Philosophy | Zalta. Edward N (ed) | #PH === Subject: Re: HCN-prime connection <4458BBD7.38F5D378@pat7.com to fella who provided data list, I very much appreciate it. It was > unclear what you mean't by stopping when the list of factors breaks > down ? I stopped at 384 because In A. Flammenkamp's Table of 1200 Highly Composite Numbers starting at no. 385, we see a number in parentheses (22). What does that mean? Possibly it means the 22nd prime? no number divisors 2 3 5 71113171923293137414347535961677173 ------------------------------------------------------------------- 384 1.76e+037 199065600 8 4 4 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 385 2.04e+037 201326592 7 5 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) 386 2.11e+037 212336640 9 5 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 387 3.17e+037 222953472 8 6 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 388 3.21e+037 226492416 7 5 3 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) 389 4.22e+037 233570304 10 5 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 390 4.28e+037 235929600 9 4 3 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) . . . 1197 3.62e+088 4.28e015 10 6 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) 1198 4.83e+088 4.34e015 12 5 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) 1199 5.44e+088 4.45e015 9 7 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) But what does the (43) mean? Every factor from the 22nd to the 43rd prime? It would have helped if this were explained. I've also got the C program that generated the list, so maybe later I can fix it to produce a proper table. > anyway, I count that you've doubled the # entries I have, HOWEVER > what I need is not the next prime but rather the nearest prime, which > could be higher or lower. As example, for HCN=24, that prime is 23, > not 29. Any chance you can rerun it? Yeah, I'll have to work around the fact that my library has a next_prime but no prev_prime function. I'll post the results later. === Subject: Re: HCN-prime connection > to fella who provided data list, I very much appreciate it. It was > unclear what you mean't by stopping when the list of factors breaks > down ? > I stopped at 384 because > In A. Flammenkamp's Table of 1200 Highly Composite Numbers > starting at no. 385, we see a number in parentheses (22). It looks like Flammenkamp has links to much much longer tables at http://wwwhomes.uni-bielefeld.de/achim/highly.html > What does that mean? Possibly it means the 22nd prime? > no number divisors 2 3 5 71113171923293137414347535961677173 > ------------------------------------------------------------------- > 384 1.76e+037 199065600 8 4 4 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > 385 2.04e+037 201326592 7 5 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) > 386 2.11e+037 212336640 9 5 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > 387 3.17e+037 222953472 8 6 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > 388 3.21e+037 226492416 7 5 3 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) > 389 4.22e+037 233570304 10 5 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > 390 4.28e+037 235929600 9 4 3 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) > 1197 3.62e+088 4.28e015 10 6 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) > 1198 4.83e+088 4.34e015 12 5 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) > 1199 5.44e+088 4.45e015 9 7 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) > But what does the (43) mean? Every factor from the 22nd to the > 43rd prime? I think that's exactly what it means. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: HCN-prime connection > to fella who provided data list, I very much appreciate it. It was > unclear what you mean't by stopping when the list of factors breaks > down ? > > I stopped at 384 because > > In A. Flammenkamp's Table of 1200 Highly Composite Numbers > starting at no. 385, we see a number in parentheses (22). > It looks like Flammenkamp has links to much much longer tables > at http://wwwhomes.uni-bielefeld.de/achim/highly.html Sorry to be lazy - can someone confirm or deny that all (maximally) highly composite numbers are the product of a previously (maximally) highly composite number with either the next larger prime, or some previously prime already in its factorisation. I'm wondering if there's possibly some Dijkstraisation of the algorithm from usenet later today...) If Dijkstraisation is possible, then finding lists of thousands should take only milliseconds. However, I'm not 100% sure that such a relation exists. Maybe the bus will give me some inspiration... Later, Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: HCN-prime connection <4458BBD7.38F5D378@pat7.com> unclear what you mean't by stopping when the list of factors breaks > down ? > I stopped at 384 because > In A. Flammenkamp's Table of 1200 Highly Composite Numbers > starting at no. 385, we see a number in parentheses (22). > It looks like Flammenkamp has links to much much longer tables > at http://wwwhomes.uni-bielefeld.de/achim/highly.html I missed those the first time I saw that web page. > What does that mean? Possibly it means the 22nd prime? > no number divisors 2 3 5 71113171923293137414347535961677173 > ------------------------------------------------------------------- > 384 1.76e+037 199065600 8 4 4 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > 385 2.04e+037 201326592 7 5 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) > 386 2.11e+037 212336640 9 5 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > 387 3.17e+037 222953472 8 6 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > 388 3.21e+037 226492416 7 5 3 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) > 389 4.22e+037 233570304 10 5 3 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > 390 4.28e+037 235929600 9 4 3 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 (22) > . > . > . > 1197 3.62e+088 4.28e015 10 6 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) > 1198 4.83e+088 4.34e015 12 5 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) > 1199 5.44e+088 4.45e015 9 7 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1 (43) > But what does the (43) mean? Every factor from the 22nd to the > 43rd prime? > I think that's exactly what it means. > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Question > You say that 0.9999........ equals one. > Let me pose a simple argument against that. In the process of dividing > one by one > now say it is zero the remainder will be one, now put zero to the right > of that remainder > and of coarse a dot after the zero, the number after the dot will be 9 > and the remainder will be 1, the same thing will be repeated, and it is > always with a remainder of 1. WRONG! The first remainder is 0.1 since it is 0.9 + 0.1 = 1.0. The second is 0.01, since 0.99 + 0.01 = 1.00 The next is 0.001, since 0.999 + 0.001 = 1.000 So the remainders a la Zuhair, converge to zero, and the finite sequences of nines converge to 1. > Now do you think if you continue that infinitelly the remainder will be > zero. It converges to zero. > I don't think so The more fool you. === Subject: Re: Question <4bjv3oF120sraU1@individual.net> <4bkb11F11te9gU1@individual.net> <4bkmsbF11l70rU1@individual.net> <4blqbbF11n2mpU1@individual.net> Let me pose a simple argument against that. In the process of dividing > one by one > now say it is zero the remainder will be one, now put zero to the right > of that remainder > and of coarse a dot after the zero, the number after the dot will be 9 > and the remainder will be 1, the same thing will be repeated, and it is > always with a remainder of 1. > WRONG! > The first remainder is 0.1 since it is 0.9 + 0.1 = 1.0. > The second is 0.01, since 0.99 + 0.01 = 1.00 > The next is 0.001, since 0.999 + 0.001 = 1.000 > So the remainders a la Zuhair, converge to zero, and the finite > sequences of nines converge to 1. > Now do you think if you continue that infinitelly the remainder will be > zero. > It converges to zero. > I don't think so > The more fool you. I agree that it converges to zero, but the question is would it reach zero. I don't think it will REACH zero. there will be always a difference of 1/10^Aleph-0 which is bigger than zero. do you understand what I meant. Zuhair === Subject: Re: Question > You say that 0.9999........ equals one. Let me pose a simple argument against that. In the process of dividing > one by one > now say it is zero the remainder will be one, now put zero to the right > of that remainder > and of coarse a dot after the zero, the number after the dot will be 9 > and the remainder will be 1, the same thing will be repeated, and it is > always with a remainder of 1. > WRONG! > The first remainder is 0.1 since it is 0.9 + 0.1 = 1.0. > The second is 0.01, since 0.99 + 0.01 = 1.00 > The next is 0.001, since 0.999 + 0.001 = 1.000 > So the remainders a la Zuhair, converge to zero, and the finite > sequences of nines converge to 1. Now do you think if you continue that infinitelly the remainder will be > zero. > It converges to zero. I don't think so > The more fool you. > I agree that it converges to zero, but the question is would it reach > zero. > I don't think it will REACH zero. The mathematical value of a sequence, insofar as any sequence has a value, is its *limit*, not necessarily any of the values that it ever reaches. Whether any term in that sequence actually equals the limit is immaterial. According to ZuhairÍs analysis, decimals exclude the majority of reals. === Subject: Re: Representing Omega > For your information I've invented the Pseudobinary system which I > called unary ( wrongly) , and this system in the best numeral system at > all. It represents all rational numbers finitelly. I invented that > system before I knew anything about this star notation of Kauffman's, > and in reality my numeral system is much advanced than that one of > Kauffman's. If it is so good, why doesn't it sell? === Subject: Re: Representing Omega > It seems that you are not very well oriented to this kind of notation > ( the star notation) 0***...* is not a finite number It is not a number at all, but merely a supposed numeral, but it fails > even as that. A sequence that does not end, does not end, and no > prestidigitation by Zuhair will make it end. to me as you think > It is Zuhair who fails for failure to note the difference between a > number and a numeral, which was the point at issue, not who came up with > that particular numeral. > also not my manufacture, Louis H. Kauffman used ****.......* to > represent Omega+1 . The use of numerals to represent numbers precedes the now defunct ROman empire, but there is still today a distinction between the representation and the object represented. > All my question was a very simple one that is if we adopt a system with > zero in it what would be the best representation of omega, a question > which you Virgil as well as many who replied in this thread didn't > answer it clearly. For example you said that it is not 0***.... nor > 0***.....* it is the union of them I said not such thing. What I did say was that if one uses the von Neumann model of the naturals, then the union of all of them is the set of all of them. In this sense, one can say omega is represented by Union(0, 0*, 0**, ...) === Subject: Re: Representing Omega ( the star notation) 0***...* is not a finite number It is not a number at all, but merely a supposed numeral, but it fails > even as that. A sequence that does not end, does not end, and no > prestidigitation by Zuhair will make it end. to me as you think It is Zuhair who fails for failure to note the difference between a > number and a numeral, which was the point at issue, not who came up with > that particular numeral. > also not my manufacture, Louis H. Kauffman used ****.......* to > represent Omega+1 . > The use of numerals to represent numbers precedes the now defunct ROman > empire, but there is still today a distinction between the > representation and the object represented. > All my question was a very simple one that is if we adopt a system with > zero in it what would be the best representation of omega, a question > which you Virgil as well as many who replied in this thread didn't > answer it clearly. For example you said that it is not 0***.... nor > 0***.....* it is the union of them > I said not such thing. What I did say was that if one uses the von > Neumann model of the naturals, then the union of all of them is the set > of all of them. In this sense, one can say omega is represented by > Union(0, 0*, 0**, ...) yer and what the symbole of that is. i can say the same thing on the system of Louis H.Kauffman. Omega= Union ( , *,**,***,****,........) According to Louis H. Kauffman the symbole he gave for Omega is ****....... So Union ( ,*,**,***,****,...........) = *****...... In a similar manner , can I deduce that a system which contain zero in it would have the following result? Union ( 0, 0*, 0**, 0***, ..........) = 0****......... Zuhair === Subject: Re: Representing Omega > It seems that you are not very well oriented to this kind of > notation > ( the star notation) 0***...* is not a finite number It is not a number at all, but merely a supposed numeral, but it > fails > even as that. A sequence that does not end, does not end, and no > prestidigitation by Zuhair will make it end. attributed > to me as you think It is Zuhair who fails for failure to note the difference between a > number and a numeral, which was the point at issue, not who came up > with > that particular numeral. also not my manufacture, Louis H. Kauffman used ****.......* to > represent Omega+1 . > The use of numerals to represent numbers precedes the now defunct ROman > empire, but there is still today a distinction between the > representation and the object represented. All my question was a very simple one that is if we adopt a system with > zero in it what would be the best representation of omega, a question > which you Virgil as well as many who replied in this thread didn't > answer it clearly. For example you said that it is not 0***.... nor > 0***.....* it is the union of them > I said not such thing. What I did say was that if one uses the von > Neumann model of the naturals, then the union of all of them is the set > of all of them. In this sense, one can say omega is represented by > Union(0, 0*, 0**, ...) > yer and what the symbole of that is. > i can say the same thing on the system of Louis H.Kauffman. > Omega= Union ( , *,**,***,****,........) > According to Louis H. Kauffman the symbole he gave for Omega is > ****....... > So Union ( ,*,**,***,****,...........) = *****...... > In a similar manner , can I deduce that a system which contain zero in > it would have the following result? > Union ( 0, 0*, 0**, 0***, ..........) = 0****......... > Zuhair Apparently Zuhair can deduceanything he wants to deduce, without let or hindrance from logic. === Subject: zero-divisors aecd086c689/e365d3da1264633d?lnk=st&q=4D+field&rnum=12#e365d3da1264633d> If you can get the URL above to work you can see that Robin Chapman æ and John Rickard found divisors of zero for B_5 numbers in about a æ day. I don't know how they found them.æ How would you find them with æ Mathematica? Here are some URLs for background. 27fcb3e2ea0/788d899ebbf4cbeb?lnk=st&q=B+numbers&rnum=8#788d899ebbf4cbeb> That post above doesn't make it past the comp.soft-sys.mathematica æ moderator. Someone wants to have the sci.math.symbolic group moderated too. Here is some more that would probably be blocked by moderators of many æ groups I'd like to post to: 1) Six dimensions to three dimensions with Synergetics Coordinates is æ shown in the MathForum geometry.research URL above if you can get it æ to work. A six-dimensional point is shown as thirty tetrahedrons. What is the mathematical terminology for that kind of representation æ in three dimensions of a regular simplex in higher dimensions? 2) The attention given to quaternions and not to four dimensional fields æ over the rational numbers, or over the integers congruence modulo a æ prime number, by so many people, suggests that there is some kind of æ dead end for rational numbers or integers congruence modulo a prime; æ that is, something you need irrational numbers for as coordinates. æ What can be done with exact rational numbers or integers congruence æ modulo a prime as coordinates and what can't be done? 3) The graphic at: which reminds me of: That's either just a coincidence or a clue from Bucky Fuller. The five vertexes of the edge length five pentatope are the five fifth roots of unity (B[1,1,1,1,-4]) as BuckyNumbers. Do you lose anything important when you map the fifth roots of unity to the vertexes ofæa pentagon on the unit circle? The Discrete Fourier Transform: using BuckyNumber roots of unity instead of Complex numbers, surprises me. Like Bucky said, I have not forgotten that I have mentioned these things before. ææ Cliff Nelson Dry your tears, there's more fun for your ears, Forward Into The Past 2 PM to 5 PM, Sundays, California time, at: Don't be a square or a blockhead; see: === Subject: Re: zero-divisors >The attention given to quaternions and not to four dimensional fields æ >over the rational numbers, or over the integers congruence modulo a æ >prime number, by so many people, suggests that there is some kind of æ >dead end for rational numbers or integers congruence modulo a prime; æ It is in fact true that a dead end of sorts occurs in dimension 4 over the reals (that is, there are no associative division algebras in higher dimensions) but there is no such dead end over Q nor over F_p . In other words, CN has it exactly backwards. This is especially humorous since the algebras described earlier in his post contain or are related to the field Q[X]/(X^4+X^3+X^2+X+1) of the fifth roots of unity, which is indeed a dimension-4 extension of the rational numbers which has attracted a fair bit of attention. dave === Subject: Re: zero-divisors >The attention given to quaternions and not to four dimensional fields æ >over the rational numbers, or over the integers congruence modulo a æ >prime number, by so many people, suggests that there is some kind of æ >dead end for rational numbers or integers congruence modulo a prime; æ > It is in fact true that a dead end of sorts occurs in dimension 4 > over the reals (that is, there are no associative division algebras > in higher dimensions) but there is no such dead end over Q nor > over F_p . In other words, CN has it exactly backwards. > This is especially humorous since the algebras described earlier in > his post contain or are related to the field Q[X]/(X^4+X^3+X^2+X+1) > of the fifth roots of unity, which is indeed a dimension-4 extension of > the rational numbers which has attracted a fair bit of attention. > dave The attention given to quaternions and not to four dimensional fields æ over the rational numbers, or over the integers congruence modulo a æ prime number, by so many people, suggests that there is some kind of æ dead end for rational numbers or integers congruence modulo a prime; æ that is, something you need irrational numbers for as coordinates. æ What can be done with exact rational numbers or integers congruence æ modulo a prime as coordinates and what can't be done? I was referring to a dead end in coordinate geometry. Cliff Nelson Dry your tears, there's more fun for your ears, Forward Into The Past 2 PM to 5 PM, Sundays, California time, at: http://www.kspc.org/ Don't be a square or a blockhead; see: http://bfi.org/node/574 === Subject: Re: JSH: Factoring problem, trivial? > Good point. But you're overlooking the fact that if I'm the supreme > ruler of the universe then pigs fly. There is a god named Ull, which sort of resembles the surname Ullrich: ULL: Nordic God of Frost Glitter and Skiing which he invented. see for example: http://www.godchecker.com/pantheon/norse-mythology.php?deity=ULL or David Bernier === Subject: Re: JSH: Factoring problem, trivial? [...] Right. For a real hoot, try factoring 71 * 73 = 5183 this way -- it > takes way over a million(!) tries to hit N=13045683. > [jstevh@msn.com] >> It may have atypical behavior, factoring better with LARGE numbers that >> have large primes than with small ones. > [David C. Ullrich] > Could be. > This is really sort of pathetic. Maybe if we only try it on a number > that's too large to fit in the combined memory and disk space of all > the computers on the planet it would work perfectly! Too bad we'll > never know. > There's no point trying to explain it to James, but, ya, we will know ;-) Hey I did the derivation. I know exactly how it works and why. Reading your post, you don't, so I'll explain, clearing the air for people you might confuse. The method finds a rational solution to sqrt(1+ 8*mT) where T is the target composite, and m is found using my research on quadratic residues as given ANY natural n_1, where n_1 is coprime to 8*T, it can be shown that m = (r-1)(8*T)^{-1} mod n_1 where (8*T)^{-1} is the modular inverse of 8*T modulo n_1, and r is a quadratic residue of n_1. What makes this method so damn scary is that n_1 can be ANY composite that you choose. I arbitrarily said floor(T/2) but I just tossed that out there without thinking much about it. And those are the basics without extra. The poster I'm replying to claims that the method skips solutions, and yes, that's true as it goes looking for m's such that the quadratic residue will be what you give it. It does so perfectly. If an m will work that won't give the quadratic residue you asked for, then of course, the mathematics will skip it!!! Now with the mathematics covered, consider what the poster said, and understand my frustation in dealing with this pathetic person. > Stripped of fluff, the method is searching for an integer i such that > 1 + i*T > is a square, where T is the composite to be factored. Then from Remember specifically the search is for a rational sqrt(1 + 8*mT) where T is the composite to be factored. What he says is not incorrect, but is vague to the point of silliness. > r^2 = 1 + i*T > it follows at once that > (r+1)*(r-1) = i*T Here his notation can be confusing as I use r in my equations for a quadratic residue. I think that was possibly deliberate in an effort to be confusing. Also note that so far he hasn't said crap of mathematical interest. > So _maybe_ (not always, as James claimed in his orginal writeup) > gcd(r+1, T) > and > gcd(r-1, T) > will reveal non-trivial factors of T. > Nothing wrong with that, so far as it goes. You didn't say anything!!! Nothing about quadratic residues. Nothing about how solutions are searched for. Nothing of any interest from my original posting. > Now James doesn't try all i, he tries i only of a special form, and the i's > so chosen follow an arithmetic progression. This skips some i's that can't > possibly lead to a square, but also skips i's that _do_. Empirical evidence > suggests it skips way too many such i's (grossly increasing the amount of > search needed over what simpler approaches achieve), and in any case it's > just a linear search. How that behaves is obvious: sometimes you get lucky > early, most times you don't, and your luck, on average predictably gets > worse the larger the composite. That's enough. Nothing about quadratic residues or how the mathematics actually chooses solutions. At best the poster manages a pathetic criticism claiming it's just an arithmetic progression. Maybe he doesn't know what a quadratic residue is. James Harris === Subject: Re: JSH: Factoring problem, trivial? [spared sci.crypt] [Tim Peters] >> ... >> There's no point trying to explain it to James, but, ya, we will know ;-) [jstevh@msn.com] > Hey I did the derivation. I know exactly how it works and why. As I said, there's no point trying to explain it to you. I'd thank you for the long confirmation, except that wasn't in doubt :-) > ... [nothing of interest] ... >> Stripped of fluff, the method is searching for an integer i such that >> 1 + i*T >> is a square, where T is the composite to be factored. Then from > Remember specifically the search is for a rational > sqrt(1 + 8*mT) > where T is the composite to be factored. _Your_ search is, yes. It's possible (but seemingly not for you) to gain insight by considering a more general family of methods of which yours is just one. In that respect, it's definitely true that my message didn't consider James Harris to be the center of the universe. By doing so, my hope was to make this of _some_ interest to people other than you. > What he says is not incorrect, but is vague to the point of silliness. You used a wrong word there. While it is darned simple, it was also correct, clear, and free of irrelevant bull: your method _is_ a particular special case of the more-general approach I described there. Of course I know that your method restricts i to the form 8*m where m is congruent to (r-1)*(8*T)^-1 modulo n_2 -- so what? Something else I know, but you don't (because you refuse to try your method, are unable to analyze its performance, and have no valid intuitions about how these kinds of methods actually work), is that trying _all_ i in 1 + i*T works much better than the specialization you arrived at (& by factors of 100s for even tiny composites -- and, no, that doesn't get better as composites get larger). Someone actually interested in factoring more than in feeding James Harris's fantasy life would in fact be far better off using 1+i*T directly than using the specialization you keep going on about from a base of willfully total ignorance about how it performs. As a _practical_ method for factoring, neither is actually worth consideration, so the most interesting thing about your method in truth is that it manages to do so much worse than the dead-simplest method of the same general form (that being trying all naturals for i). That may be of general interest to others, not really the specifics of _how_ you managed to create a method of this form most notable solely for how poorly it performs. Who cares that you dug yourself this hole by using a weak result about modular squares specifically? I sure don't. If you had decided to use only prime i, or only i that were expressible as the sum of two cubes, or ... you would probably have ended up with a similarly impractical method. There are an endless number of ways to screw up, and picking over a poor method's bones isn't usually interesting unless there was an _a priori_ reason to expect it to have done better. There was no such reason in this case. If you think it's interesting to pick over your method's bones, that's fine. I don't. > ... [nothing of interest] ... === Subject: Re: JSH: Factoring problem, trivial? If it doesn't GREAT!!! > I'm only looking at factoring as a way to force things, and every time > I think I might have something, I hate it, as it seems so stupid that > it's needed. > If I do succeed, who gets hurt? > Lots of people who don't deserve it, didn't ask for it, who will suffer > because they trusted a lot of supposedly smart people who lied to them > a lot. > Wrong is good with me. I'm thinking about taking a nice long break > from this crap. > And now I draw a line in the sands of time ... > _______________________________________ > It's Wed May 3 06:32:00 EDT 2006 Yeah, funny, I have said it before. Thing is, I have often enjoyed what I do, as I brainstorm ideas and get to argue with people from all over the world. Trouble is, I'm wondering if I'm getting diminishing returns, as I am less motivated to make new discoveries, and more concerned about the impact of my current ones. I think solving the factoring problem would be proof of incomprehensible stupidity within the ranks of mathematicians and cryptanalysts, if Plotnikov's paper is right, and it would be a horrible blow to a world that trusted. But being wrong about the factoring problem is just fine. Then things are just like before! I practice extreme mathematics. That involves brainstorming--coming up with LOTS of ideas, many of them useless--and talking out those ideas in public to get heated criticism so that the crappy ideas are gotten rid of faster. Some people hate the process with a passion, and fight to control my postings, even though it's Usenet, I often explain what I do--like now--and they can ignore me. Like I ignore most postings on these newsgroups. So I know it can be done. I concluded that they are control freaks, giving in to territorial needs from their reptilian brains telling them they can control what cannot be controlled as they see it as their territory. Or they're just assholes. James Harris === Subject: Re: JSH: Factoring problem, trivial? what I do, as I brainstorm ideas and get to argue with people from all > over the world. > Trouble is, I'm wondering if I'm getting diminishing returns, as I am > less motivated to make new discoveries, and more concerned about the > impact of my current ones. I wouldn't worry about it. I don't think there is any danger of your returns diminishing below where they are now. - Randy === Subject: Re: JSH: Factoring problem, trivial? what I do, as I brainstorm ideas and get to argue with people from all > over the world. > Trouble is, I'm wondering if I'm getting diminishing returns, as I am > less motivated to make new discoveries, and more concerned about the > impact of my current ones. > I wouldn't worry about it. I don't think there is any danger of > your returns diminishing below where they are now. > - Randy Oh yeah, like how I just figured out yet another factoring method, after a few days of brainstorming? Or how I related quadratic residues, sums and factoring? No, by diminishing returns I mean the fear to thrill ratio. The fear is rising as I contemplate a world that may just be stupid on a scale I never anticipated, and the thrill of discovery does not quite compare with worrying about breaking nations because a small group of people too arrogant to understand that reality does not care about their opinions of themselves, bet the world, and lost. James Harris === Subject: Series divergence Consider g(x) = Sum_n(2,oo) [(cos (2*pi*n*x))/(n ln n)] How to show as x goes to zero, A*( ln (ln (1/|x|)) < g(x), with a positive constant. === Subject: Re: Series divergence <20650639.1146795503660.JavaMail.jakarta@nitrogen.mathforum.org>, > Consider > g(x) = Sum_n(2,oo) [(cos (2*pi*n*x))/(n ln n)] > How to show as x goes to zero, A*( ln (ln (1/|x|)) < g(x), with a positive > constant. Here's an idea that might work: Let's ignore the 2Pi. Suppose k is a large integer and 1/(k+1) <= x <= 1/k. Write the sum as Sum(n=2,k) cos(nx)/[nln(n)] + Sum(n=k+1,oo) cos(nx)/[nln(n)] = I + II. Just for simplicity, take x = 1/k. Then I >= cos(1)*Sum(n=2,k) 1/[nln(n)], which is like int_[2,k] dx/[xln(x)] = ln(ln(k)) - ln(ln(2)) = ln(ln(1/x)) - ln(ln(2)). Note II = Re Sum(n=k+1,oo) exp(inx)/[nln(n)]. Sum by parts and then take absolute values to get (I think) something on the order of 1/|exp(i/k) - 1|*Sum(n=k+1,oo) 1/[n^2ln(n)]. The first factor is like 1/(1/k) = k, and the sum is like int_[k,oo) dx/[x^2ln(x)] <= int_[k,oo) dx/x^2 = 1/k. In other words, II is bounded independent of k. If all this is right, it will work for the range 1/(k+1) <= x <= 1/k, giving the desired ln(ln(1/|x|)) lower bound. === Subject: Re: JSH: Tiny FLT proof possible? > [emphases mine] >> My quadratic residue result ****may**** offer the route to a tiny proof of >> Fermat's Last Theorem. >> This quadratic residue result of mine ****may**** be one of the >> most powerful in number theory, offering routes to proving >> Goldbach's conjecture, and probably all kinds of Diophantine >> equations [and] a tiny proof of FLT that is just about as trivially >> easy as you can get. Kind of strange. > And tomorrow's EuroMillions lottery numbers ****may**** be 3, 16, 23, > 35, 47, with lucky stars 1 and 9. I know which event _I_ think more > probable. Sorry, I'm new here, but I'd like to say 'lay off'. I've seen a couple of jstevh's posts (the factoring, and now this) and he seems enthusiastic, why not support his work rather than dissing his use of may (which i know is not the most mathematical of terms, but he does speek of mathematical experimentation, which is generally incorrect, but can lead to some interesting results even when it contains mistakes). -- burton samograd kruhft .at. gmail kruhft.blogspot.com www.myspace.com/kruhft metashell.blogspot.com === Subject: Re: JSH: Tiny FLT proof possible? <87mzdx2oqh.fsf@gmail.com [emphases mine] >> My quadratic residue result ****may**** offer the route to a tiny proof of >> Fermat's Last Theorem. >> This quadratic residue result of mine ****may**** be one of the >> most powerful in number theory, offering routes to proving >> Goldbach's conjecture, and probably all kinds of Diophantine >> equations [and] a tiny proof of FLT that is just about as trivially >> easy as you can get. Kind of strange. > And tomorrow's EuroMillions lottery numbers ****may**** be 3, 16, 23, > 35, 47, with lucky stars 1 and 9. I know which event _I_ think more > probable. > Sorry, I'm new here, but I'd like to say 'lay off'. I've seen > a couple of jstevh's posts (the factoring, and now this) and he seems > enthusiastic, why not support his work rather than dissing his use of > may (which i know is not the most mathematical of terms, but he does > speek of mathematical experimentation, which is generally incorrect, > but can lead to some interesting results even when it contains > mistakes). I think ten years or so of him constantly telling people they're liars, dishonest, going to lose their jobs (and in a couple of instances actually contacting employers or state law enforcement officials) in reaction to people pointing out logic errors starts to wear thin after awhile. As you say, you're new here. Watch the progression of events with anyone who actually tries to correct a proof. What you're seeing now is the residue after people have tried and failed multiple times to get him to accept counterexamples to his latest major result. - Randy === Subject: Re: JSH: Tiny FLT proof possible? <87mzdx2oqh.fsf@gmail.com [emphases mine] >> My quadratic residue result ****may**** offer the route to a tiny proof of >> Fermat's Last Theorem. >> This quadratic residue result of mine ****may**** be one of the >> most powerful in number theory, offering routes to proving >> Goldbach's conjecture, and probably all kinds of Diophantine >> equations [and] a tiny proof of FLT that is just about as trivially >> easy as you can get. Kind of strange. > And tomorrow's EuroMillions lottery numbers ****may**** be 3, 16, 23, > 35, 47, with lucky stars 1 and 9. I know which event _I_ think more > probable. > Sorry, I'm new here, but I'd like to say 'lay off'. Ahh, to be a newbie again, what sweet innocence that would be :) > I've seen > a couple of jstevh's posts (the factoring, and now this) and he seems > enthusiastic, Being new, you've missed James' best stuff. Most of it has been deleted from the Google archive, but there are others: eg . > why not support his work rather than dissing his use of > may (which i know is not the most mathematical of terms, but he does > speek of mathematical experimentation, which is generally incorrect, > but can lead to some interesting results even when it contains > mistakes). Sure it can. Unfortunately he doesn't do any. He does five lines of symbolic manipulation, then posts and says hey this may solve famous_problem_354, and then you would all be DOOMED!. -- Larry Lard Replies to group please === Subject: Re: JSH: Tiny FLT proof possible? <87mzdx2oqh.fsf@gmail.com [emphases mine] >> My quadratic residue result ****may**** offer the route to a tiny proof of >> Fermat's Last Theorem. >> This quadratic residue result of mine ****may**** be one of the >> most powerful in number theory, offering routes to proving >> Goldbach's conjecture, and probably all kinds of Diophantine >> equations [and] a tiny proof of FLT that is just about as trivially >> easy as you can get. Kind of strange. > And tomorrow's EuroMillions lottery numbers ****may**** be 3, 16, 23, > 35, 47, with lucky stars 1 and 9. I know which event _I_ think more > probable. > Sorry, I'm new here, but I'd like to say 'lay off'. I've seen > a couple of jstevh's posts (the factoring, and now this) and he seems > enthusiastic, why not support his work rather than dissing his use of > may I originally was supportive of JSH's work. However, most of the time he's simply wrong. (He claimed to have posted a factoring algorithm, saying that it might lead to the downfall of the world economies and anything else depending on security. This was in February.) One exception is that he found equations for the prime-counting function, but it has been discovered already. He claims that ideal theory (part of abstract algebra) is wrong, but when I asked him to post a proof, he never did. He prefers being a martyr. --- Christopher Heckman === Subject: Re: JSH: Tiny FLT proof possible? [Proginoskes] > I originally was supportive of JSH's work. However, most of the time > he's simply wrong. (He claimed to have posted a factoring algorithm, > saying that it might lead to the downfall of the world economies and > anything else depending on security. This was in February.) Heh. I bet you're thinking about February of _this_ year. Take a peek at a year earlier; this is worth quoting at some length because it shows there's nothing new under the JSH sun: === Subject: JSH: Nearly done ... Yup, that's why I call it a super sieve, as in picking b_1 and b_2 for you, the algebra checks against all rationals, the entire set, and that's an infinite set! Now we are beyond brilliant into the arena of almost impossible to imagine, with a technique for factoring, which loops through the entire field of rationals in searching for a solution. So must x reveal a non-trivial factor of M? The answer, amazingly enough, depends on quadratic residues!!! ... I'm a person not just making claims--I have the full demonstration of everything I say here, and I've worked it out, and put it out for people to consider. So, how can that be possible? How can what I say here be true, and the mathematical establishment not pay attention? There's something wrong with them. I have other big results and they've tried to ignore those, and I've talked about some of them here where people have usually lied about my work. It's weird. But it's the reality. With pure math people can lie. That's just a fact. If some group of mathematicians write gibberish, and declare it to be a proof of something, then basically that goes over, if they are considered to be mathematicians of note. Sure, mathematicians say that's not the way it is, but that's the way it is. And if an amateur mathematician makes major discoveries, it goes the other way. Mainstram mathematicians just band together and ignore their work, like they will try to do with my latest discovery, which they also try to do because they're very stupid. The factoring problem is quite important in the real world. It's not pure math, so some of those people you respect and admire may soon be in jail, hated worldwide, and villified by people who will not be able to comprehend their behavior. I find it hard to understand myself. You do not believe that will happen. It does not matter. It will happen. There are probably only a few days before many of your heroes are cast down, humiliated, ripped from their positions, and put up on public display as enemies of humanity itself, as people too blind to act in the best interests of society, and too dumb to realize they wouldn't get away with it. You do not believe but it will happen, and the foundations of your society will not just be shaken--they will be shattered--as I promised years ago. The end was never in doubt. You cannot betry the truth, and you cannot block mathematics. You betrayed mathematics itself, so you will be destroyed by it. It is the just solution, the most logical, the most rational one, and one of absolute perfection. Those who betrayed the field of mathematics, who sullied that field with their lies, and their belief that social rules could win that they could make up truth and get away with it are about to learn just how powerful of a field it truly is as a lesson that humanity will never forget. No one will be able to forget. You will learn this time. You will learn, never to make these mistakes again, and the lesson will be complete. James Harris His best rants are behind him ;-) Unfortunately, that factoring method didn't actually work well despite the apocalyptic rhetoric, and he seemed to stop writing programs to _test_ his ideas then. There followed a seemingly endless succession of surrogate factoring variations, all untested by James, most carefully tested by me at the time, with none of any promise. This went on (& on), until James finally couldn't resist writing a program of his own again, and rediscovered what everyone else had been telling him for lo those many months about _all_ his variations: === Subject: SFT: Experimentation starts ... In any event, I did verify though that as you use *integer* factors to get your surrogate, the factoring percentage drops as the size of the number increases. ... And he had a reason for that (the theorem works over all rationals, and focusing on integers is a human choice ...), but gave up quickly: the next day he announced his retirement: sci.math,alt.math.recreational,alt.math.undergrad,sci.skeptic === Subject: JSH: Letting it drop ... Math was fun for a while, occupied my attention for years, but now, it's been there, done that time. ... He didn't post again for ... don't really remember, but it was months. One major difference with _this_ round of purported factoring miracles is that James appears utterly determined not to test any part of his ideas against reality this time, no way no how. Another is that he dismisses the, umm, unencouraging testing results of others entirely this time. Sounds like he finally found a way he can't lose :-) > One exception is that he found equations for the prime-counting > function, but it has been discovered already. He claims that ideal > theory (part of abstract algebra) is wrong, but when I asked him to > post a proof, he never did. He probably doesn't remember. The notorious liar Nora Baron eventually convinced him that some of what he was saying directly contradicted a theorem of Dedekind's. I'd be truly surprised if he remembered _which_ theorem, but the conclusion was obvious: if Dedekind's purported theorem contradicted what James said, Dedekind must have been wrong. It follows that ideal theory is overthrown. QED. And why _should_ he remember more than that? It's a very simple proof ;-) > He prefers being a martyr. I don't believe he believes that -- but yes :-) === Subject: Re: JSH: Tiny FLT proof possible? >> I originally was supportive of JSH's work. However, most of the time >> he's simply wrong. (He claimed to have posted a factoring algorithm, >> saying that it might lead to the downfall of the world economies and >> anything else depending on security. This was in February.) > Heh. I bet you're thinking about February of _this_ year. Take a peek at a > year earlier; this is worth quoting at some length because it shows > there's nothing new under the JSH sun: === > Subject: JSH: Nearly done Funny, really funny. But there's somethinge strange about this. Why didn't James delete this embarassing post from Google? It's still here: Jose Carlos Santos === Subject: Re: JSH: Tiny FLT proof possible? <87mzdx2oqh.fsf@gmail.com [emphases mine] >> My quadratic residue result ****may**** offer the route to a tiny proof of >> Fermat's Last Theorem. >> This quadratic residue result of mine ****may**** be one of the >> most powerful in number theory, offering routes to proving >> Goldbach's conjecture, and probably all kinds of Diophantine >> equations [and] a tiny proof of FLT that is just about as trivially >> easy as you can get. Kind of strange. > And tomorrow's EuroMillions lottery numbers ****may**** be 3, 16, 23, > 35, 47, with lucky stars 1 and 9. I know which event _I_ think more > probable. > Sorry, I'm new here, but I'd like to say 'lay off'. I've seen > a couple of jstevh's posts (the factoring, and now this) and he seems > enthusiastic, why not support his work rather than dissing his use of > may (which i know is not the most mathematical of terms, but he does > speek of mathematical experimentation, which is generally incorrect, > but can lead to some interesting results even when it contains > mistakes). Didn't you see the post where he called you an asshole? > -- > burton samograd kruhft .at. gmail > kruhft.blogspot.com www.myspace.com/kruhft metashell.blogspot.com === Subject: Re: JSH: Tiny FLT proof possible? <87mzdx2oqh.fsf@gmail.com Sorry, I'm new here, but I'd like to say 'lay off'. > -- Yes, you're clearly new here. === Subject: Re: How do I calculate odds? >> Unless you're a multi-millionaire in the >> first place, the situation is very definitely asymmetrical. >> Play the lottery with the money you have left AFTER you buy insurance! >Now, the other guy thinks its more important to buy health or fire >insurance... but as you have demonstrated that its not much different than >playing the lotto. Tell me, what sound does an idea make when it goes over your head? === Subject: Re: How do I calculate odds? > For example, I do not believe in god yet hundreds of millions do.. do I > have any right to call them all idiots? (even if I believe > mathematically its ridiculuous) Might I ask a question on this mathematical probability of God thing? If I recall, in high school, they told us that when a cell divides, it forms two of the original type of cell. If a skin cell divides, you have two skin cells, etc. It cannot divide and form another type of cell. That would be a mutation which is known to be almost (if not completely) universally fatal in nature. When a human egg is fertilized, it begins a division process, and for several divisions, the cells follow the scenario above. They are all the same type of cell. Then, one day, the cells divide and, lo and behold, they are different. There are somehow skin cells, nerve cells, bone cells, muscle cells, fat cells, etc. The question: What causes that non-fatal cell mutation to happen at the same time almost every time with relatively few failures? I know medical science has a name for the process. I don't even care what the name is, but you can tell me if you like. Heck, I could have named it something. But, HOW do a cells of type X divide into multiple types of cells? Just something to think about. We can address the failures (surely God wouldn't have a system that fails) later. But, for now, one thing at a time. What would be the mathematical probability of a cell of type X dividing into two cells of types X and Y, or Y and Z, on schedule almost every time? I suggest VERY near 0. === Subject: Re: How do I calculate odds? On Thu, 4 May 2006 16:41:10 -0500, nativetexan_1 >Just something to think about. We can address the failures (surely God >wouldn't have a system that fails) later. But, for now, one thing at a >time. What would be the mathematical probability of a cell of type X >dividing into two cells of types X and Y, or Y and Z, on schedule almost >every time? I suggest VERY near 0. I haven't been following this thread so I may be missing context... Disclaimer: I'm not a biologist. Doesn't cell division follow a script controlled by its DNA much like a computer program follows a sequence of instructions? So I would suggest instead of VERY near 0 instead that your probability would be VERY near 1. That's how a properly functioning program works; same input to the same program gets you the same output. Of course, if a transistor goes bad, the program crashes (bad mutation). --Lynn === Subject: Re: Dave Rusin, Fermat, and Fermat's Last Theorem <27625461.1146745157960.JavaMail.jakarta@nitrogen.mathforum.org>, > > I will offer $100 for a counterexample -- any photo > showing me and Fermat > in the same room at the same time. You may draw > your own conclusions. > > I tried, but they were too large to fit in this > message. > not lose The only reason I'm replying is so that DR can be amused at how you've missed the point. > You still can provide their prime factorization, which I think will be > suitable to be fit in only in few lines! Whoosh! -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Dave Rusin, Fermat, and Fermat's Last Theorem <18553986.1146751005681.JavaMail.jakarta@nitrogen.mathforum.org Frankly speaking, and when I posted my threads about FLT and finally my short proof that was based on a conjecture that no mankind ever can guess, as quasi explained,I thought many mathematician will be able to prove the missing part only the second day and share me the proof, but since I'm not a mathematician and my job as a civil engineer and time doesn't allow me to go further > And I hoop you can imagine now the importance of guessing instead of denying me all my efforts to settle up this issue > But,it seems I have to do all the thing alone > I wonder now how you will tolorate the COMING! > B.Karzeddin One of the main rules of publishing is that you have to read everything that has been published on the subject before you publish. I know you did no such thing. You are not Fermat, he was a mathematical genius. You are a civil engineer that maybe does not even know congruent arithmetic. I was not closing all door. I was poiting you to the reality of the difficulty of the problem, and hoping you would see what are the chances you found a proof. === Subject: Re: Beliefs Create Reality, Even In Mathematics. I agree that mathematics is an approximation of truth. I also think that mathematics can give best approximation of truth. As time goes by, the approximation becomes better and better. If maths can tell us the exact truth without any error, maths will lose its vitality. === Subject: fourier problem help, Using the expansion of problem 1b, show that: 1 / sin(z) = 1 / z + infSum( (-1)^n * [1 / (z - n*pi) + 1 / (z + n*pi)] ). where z is any number which is not a multiple of pi. The exapansion for 1b was: cos(a*x) = 2 sin(a*pi) / pi [ 1/(2a) + infSum( [(-1)^n * a * cos(nx)] / [a^2 - n^2] ) === Subject: Re: fourier problem help, > Using the expansion of problem 1b, show that: > 1 / sin(z) = 1 / z + infSum( (-1)^n * [1 / (z - n*pi) + 1 / (z + n*pi)] ). > where z is any number which is not a multiple of pi. > The exapansion for 1b was: > cos(a*x) = 2 sin(a*pi) / pi [ 1/(2a) + infSum( [(-1)^n * a * cos(nx)] / > [a^2 - n^2] ) pi*cos(ax)/2/sin(a*pi)/a - 1/2/a^2 = sum((-1)^n*cos(nx)/(a^2-n^2)) 1/sin(z)/2/z - 1/2/z^2 = sum((-1)^n/(z^2 - (n*pi)^2) Try picking two values for a and x in the first equation so it will get you the second equation.... Look specifically at the summand and try and see how you can get the other summand. Hints are: Its very easy for x and for a it forms a very simple ratio. This problem is only a simple algebra problem but looks tough cause of all the extraneous stuff. [ === Subject: Re: fourier problem help, > pi*cos(ax)/2/sin(a*pi)/a - 1/2/a^2 = sum((-1)^n*cos(nx)/(a^2-n^2)) > 1/sin(z)/2/z - 1/2/z^2 = sum((-1)^n/(z^2 - (n*pi)^2) How did you get [1/sin(z)/2/z - 1/2/z^2] from [1 / sin(z) = 1 / z ]? > Try picking two values for a and x in the first equation so it will get > you the second equation.... Look specifically at the summand and try and > see how you can get the other summand. Hints are: Its very easy for x and > for a it forms a very simple ratio. T Do you mean I should set z = some ratio of x and a? his problem is only a simple algebra problem but > looks tough cause of all the extraneous stuff. === Subject: Re: fourier problem help, >> pi*cos(ax)/2/sin(a*pi)/a - 1/2/a^2 = sum((-1)^n*cos(nx)/(a^2-n^2)) >> 1/sin(z)/2/z - 1/2/z^2 = sum((-1)^n/(z^2 - (n*pi)^2) > How did you get [1/sin(z)/2/z - 1/2/z^2] from [1 / sin(z) = 1 / z ]? simple algebra? Do you know algebra? Do you know that the sum is linear? I re >> Try picking two values for a and x in the first equation so it will get >> you the second equation.... Look specifically at the summand and try and >> see how you can get the other summand. Hints are: Its very easy for x and >> for a it forms a very simple ratio. T > Do you mean I should set z = some ratio of x and a? well, a = b/c where b and c are two things(elements, variables, constants, etc..). look, this is not even really a math problem. You are given two graphics sum((-1)^n*cos(nx)/(a^2-n^2)) sum((-1)^n/(z^2 - (n*pi)^2) and you must compare them and see how they are different. its obvious... let me remove what is the same to see what is different sum((-1)^n*cos(nx)/(a^2-n^2)) = ---------cos(nx)/(a----------) sum((-1)^n/(z^2 - (n*pi)^2) = ------------(z-----pi--) this should give a huge clue as to what to do, remember, you can choose a, z, x to be whatever you want... obviously since we are trying to get z then a and x need to be a function of z. (although that doesn't imply that the function has to have z since it can be constant w.r.t. z). Notice that the sums are the same and hence do not change the problem... the (-1)^n are in both... so we are really looking at cos(nx)/(a^2-n^2) and 1/(z^2 - (n*pi)^2 need to figure out how to set x and a to get the escond one. It should be blatently obvious what to do with x... and pretty obvious what one could do with a... try what you think and see what happens. I will not do the problem for you since its way to easy and it won't help you any.... but if you spend 5 mins thinking about it I'm sure you will get it and realize how easy it is. === Subject: Surjective Morphism Let V=Z(xz-y^2, yz-x^3, z^2-x^2y) in A^3. Prove that map b:A^1 -> V defined by b(t)=(t^3, t^4, t^5) is a surjective morphism. I stared by assuming that if (x,y,z) != (0,0,0) then t=x/y. What happens next? === Subject: Re: Surjective Morphism > Let V=Z(xz-y^2, yz-x^3, z^2-x^2y) in A^3. > Prove that map b:A^1 -> V defined by b(t)=(t^3, t^4, t^5) is a > surjective morphism. > I stared by assuming that if (x,y,z) != (0,0,0) then t=x/y. > What happens next? Wrong way, you probably meant t := y/x. Actually, the wording of the original problem is a bit misleading, what you want is that the image of the morphism b: A^1 --> A^3 defined by b(t)=(t^3, t^4, t^5) coincides with V. Obviously it is contained in V. First you should check, that for each point (0,0,0) =/= (x,y,y) in V, you have xyz =/=0 i.e., every coordinate is nonzero. Then show y/x = z/y and (y/x)^3 = x . Marc === Subject: Re: Surjective Morphism >Let V=Z(xz-y^2, yz-x^3, z^2-x^2y) in A^3. >Prove that map b:A^1 -> V defined by b(t)=(t^3, t^4, t^5) is a >surjective morphism. >I stared by assuming that if (x,y,z) != (0,0,0) then t=x/y. >What happens next? I wouldn't start that way. Try this instead. Show directly that (x,y,z) in V implies x^4=y^3 and x^5=z^3. Thus, y=x^(4/3) and z=x^(5/3) (assuming every element of the base field has at least one cube root in the base field). Now let t=x^(1/3) (any cube root if more than one) and express x,y,z in terms of t. The above outline will clinch the proof provided your field has at least one cube root in the field for every field element. I think the proof will go through if the field has at least one 4th root for every element. Similarly for 5th roots. However the result also holds over some fields which are not closed under roots. For example, it's clear that it holds over the rational numbers. In fact. let K be any field whose multiplicative group K^* is isomorphic to the direct product of the form G x H where every element of G is a cube in G and H is free abelian. Then I think the result holds over the field K. A analogous statement can be made replacing cube with 4th power or 5th power. Now maybe the result holds over all fields in some obvious way, in which case, I'm making a mess of things. quasi === Subject: Re: Surjective Morphism >Let V=Z(xz-y^2, yz-x^3, z^2-x^2y) in A^3. >Prove that map b:A^1 -> V defined by b(t)=(t^3, t^4, t^5) is a >surjective morphism. >I stared by assuming that if (x,y,z) != (0,0,0) then t=x/y. >What happens next? > I wouldn't start that way. But it works well (except one should use y/x instead of x/y) > Now maybe the result holds over all fields in some obvious way, in > which case, I'm making a mess of things. If you look at the defining equations you will see, that for every point of V, if _some_ coordinate is zero then _all_ are zero. So you can assume that (x,y,z) with xyy =/= 0 and the above start with t := y/x = z/x works well (just compute t^3). Marc === Subject: Re: Surjective Morphism <38nl52dhovmv279ltvadchjkmhu3emseul@4ax.comLet V=Z(xz-y^2, yz-x^3, z^2-x^2y) in A^3. >Prove that map b:A^1 -> V defined by b(t)=(t^3, t^4, t^5) is a >surjective morphism. >I stared by assuming that if (x,y,z) != (0,0,0) then t=x/y. >What happens next? > I wouldn't start that way. > Try this instead. > Show directly that (x,y,z) in V implies x^4=y^3 and x^5=z^3. > Thus, y=x^(4/3) and z=x^(5/3) (assuming every element of the base > field has at least one cube root in the base field). > Now let t=x^(1/3) (any cube root if more than one) and express x,y,z > in terms of t. > The above outline will clinch the proof provided your field has at > least one cube root in the field for every field element. > I think the proof will go through if the field has at least one 4th > root for every element. Similarly for 5th roots. > However the result also holds over some fields which are not closed > under roots. For example, it's clear that it holds over the rational > numbers. > In fact. let K be any field whose multiplicative group K^* is > isomorphic to the direct product of the form G x H where every element > of G is a cube in G and H is free abelian. Then I think the result > holds over the field K. > A analogous statement can be made replacing cube with 4th power or > 5th power. > Now maybe the result holds over all fields in some obvious way, in > which case, I'm making a mess of things. > quasi Why does (x,y,z) in V implies x^4=y^3 and x^5=z^3? === Subject: Re: Surjective Morphism >>Let V=Z(xz-y^2, yz-x^3, z^2-x^2y) in A^3. >>Prove that map b:A^1 -> V defined by b(t)=(t^3, t^4, t^5) is a >>surjective morphism. >>I stared by assuming that if (x,y,z) != (0,0,0) then t=x/y. >>What happens next? >> I wouldn't start that way. But even if you did start that way, it would be t=y/x, not t=x/y. quasi === Subject: Re: Surjective Morphism >>Let V=Z(xz-y^2, yz-x^3, z^2-x^2y) in A^3. >>Prove that map b:A^1 -> V defined by b(t)=(t^3, t^4, t^5) is a >>surjective morphism. >>I stared by assuming that if (x,y,z) != (0,0,0) then t=x/y. >>What happens next? >> I wouldn't start that way. >> Try this instead. >> Show directly that (x,y,z) in V implies x^4=y^3 and x^5=z^3. >> Thus, y=x^(4/3) and z=x^(5/3) (assuming every element of the base >> field has at least one cube root in the base field). >> Now let t=x^(1/3) (any cube root if more than one) and express x,y,z >> in terms of t. >> The above outline will clinch the proof provided your field has at >> least one cube root in the field for every field element. >> I think the proof will go through if the field has at least one 4th >> root for every element. Similarly for 5th roots. >> However the result also holds over some fields which are not closed >> under roots. For example, it's clear that it holds over the rational >> numbers. >> In fact. let K be any field whose multiplicative group K^* is >> isomorphic to the direct product of the form G x H where every element >> of G is a cube in G and H is free abelian. Then I think the result >> holds over the field K. >> A analogous statement can be made replacing cube with 4th power or >> 5th power. >> Now maybe the result holds over all fields in some obvious way, in >> which case, I'm making a mess of things. >> quasi >Why does (x,y,z) in V implies x^4=y^3 and x^5=z^3? (x,y,z) in V implies (1) xz = y^2 (2) yz = x^3 (3) z^2 = x^2*y Consider 2 cases, case 1: z=0 Show that z=0 implies x=y=0, hence x^4=y^3 and x^5=z^3. case 2: z is nonzero Show that z nonzero implies x,y are both nonzero. To get x^4=y^3, multiply equations (1) and (2) together and simplify. To get x^5=z^3, multiply equations (2) and (3) together and simplify. quasi === Subject: Re: Surjective Morphism >Let V=Z(xz-y^2, yz-x^3, z^2-x^2y) in A^3. >Prove that map b:A^1 -> V defined by b(t)=(t^3, t^4, t^5) is a >surjective morphism. >I stared by assuming that if (x,y,z) != (0,0,0) then t=x/y. >What happens next? >>I wouldn't start that way. >>Try this instead. >>Show directly that (x,y,z) in V implies x^4=y^3 and x^5=z^3. ... further discussion deleted ... > Why does (x,y,z) in V implies x^4=y^3 and x^5=z^3? Note that (x,y,z) in V implies these identities hold: y^2 = xz, x^3 = yz, and z^2 = x^2 y Then you get this x^4 = x x^3 = xyz, and from that, x^4 = y xz = y y^2 = y^3. So: x^4 = y^3. Next, take x^5 = x^2 x^3 = x^2 (y z) = (x^2 y) z = z^2 z = z^3 It's just a matter of manipulating formulas to make it possible to take advantage of the identities that define V. Dale === Subject: Re: geometry algorithm - clarrifications >This is a good description of what I need: >Input: a collection of polygons, and a bigger rectangle. >Output: a placement of polygons such that: >a) If not all polygons can enter in the rectangle (because their total >area is bigger than the area of the given rectangle) I'm interested in >selecting some polygons such that the total uncovered area is minimal. Note that IF the sum of the areas of the polygons exceeds the area of the rectangle THEN the polygons can't all fit in the rectangle. But the converse is not true, so your because... is a little out of place. >b) If all polygons can fit into that rectangle I need a placement such >that from the uncovered area I can extract a rectangle with a maximal >area. The idea behind this formulation is that I don't want too many >small pieces of waste (uncovered area), but I want few and big parts >(which I can later reuse). >The polygons are to be placed into the rectangle at any orientation >including mirror images (ie flipped over) and without overlapping. >This problem is a generalization of the case where all polygons are >rectangles. In that case is very easy to combine them, ... but in the >case of polygons with any shape I have no idea on how to efficiently >solve this problem. When there were only two polygons this was a doable computation to solve exactly -- there are three degrees of freedom for each polygon (2 translation and one rotation) and I see now you are also allowing us to reverse the orientation of the polygons relative to each other. So that makes two 6-d.o.f. problems to solve exactly, and I'm sure fast code could be concocted to carry this out. But if you will have many polygons this is going to become quickly infeasible -- among other things you will have 2^(n-1) possible combinations of relative orientation to consider, so already that's exponential growth for a naive algorithm. On the other hand, you can speed this up quite a bit if you don't mind some sloppiness: simply find for each polygon the smallest bounding circle, and then look for the optimal packing of the circles within rectangles. The advantage of this is that rotation and reflection don't affect the circles at all. The disadvantage is that the packing will never be any better than the _worst_ possible orientation of the polygons would allow within those disks -- in particular for non-convex polygons you lose all ability to fit outies into innies. If you apply this approach to the 2-polygon case you find that the optimal rectangle will be an R x (R+r) rectangle, where r < R are the two radii of the bounding circles for the polygons. That's pretty close to optimal when the polygons are nearly regular and have lots of sides, but fails misearbly for a couple of long, skinny rectangles. I don't know what kind of algorithms you can use for the simplified problem of placing n disks so that they don't overlap. Getting the absolute minimum smallest rectangle is hard even when the disks all have the same radius; but at least in that case a regular packing will permit the determination of a rectangle with very little wasted space. When the n disks have wildly different radii I don't quite know what to advise. It seems like a greedy algorithm would allow you to put all the disks into a rectangle whose total area is within a fixed multiple of the sum of the areas of the disks. Is that good enough? So if you really have to do this with a large number of polygons, you should hope that the polygons are pretty similar to disks ! A mathematician once told me how he was hired by the military to fill a large sphere with small spheres (buckshot) to make the big sphere as heavy as possible. After a lot of geometric analysis and computation it occurred to him to ask the guys what they had done before. Answer: load the big sphere into a pickup truck, fill it with as much shot as possible, then drive around on a bumpy road and shovel in more shot as the small spheres settled. He never could improve upon that algorithm. Nowadays computers can simulate a bumpy road very quickly. Start with a moderately good placement as indicated above, then jiggle the polygons and make the enclosing rectangle shrink as you go. If you want to try harder for a lower amount of waste, you can take a jump away from a local minimum, i.e. enlarge the rectangle a bit and drive over a really big pothole, then try the old trick some more. An organized method for doing this is called simulated annealing. Sometimes it works well. dave === Subject: Re: geometry algorithm - clarrifications <8nrb521hu0ect1fd333sekceg4dec79spm@4ax.com> Input: a collection of polygons, and a bigger rectangle. > Output: a placement of polygons such that: Define placement or at least clarify, any orientation and mirror images. > a) If not all polygons can enter in the rectangle (because their total > area is bigger than the area of the given rectangle) I'm interested in > selecting some polygons such that the total uncovered area is minimal. This reminds of the nap sack problem, a difficult, NP-hard problem that has been given much consideration. As I recall, the greedy algorithm was disadvised. > b) If all polygons can fit into that rectangle I need a placement such > that from the uncovered area I can extract a rectangle with a maximal > area. The idea behind this formulation is that I don't want too many > small pieces of waste (uncovered area), but I want few and big parts > (which I can later reuse). Interesting. Some sort of reverse image nap sack problem. > The polygons are to be placed into the rectangle at any orientation > including mirror images (ie flipped over) and without overlapping. This is better placed first where I indicated. > This problem is a generalization of the case where all polygons are > rectangles. In that case is very easy to combine them, ... but in the > case of polygons with any shape I have no idea on how to efficiently > solve this problem. Oh, that problem is easy? I've doubts it's easy for more than two or more than a few rectangles. Have you tried to solve your problem with the limited input of rectangles of different sizes? Even simpler is to solve it with squares of different sizes. What happens as the number of squares increases? Have you complete solutions, ie algorithms in math/English (please no computer code yet) for the n-square and n-rectangle problems for any values of n or for a meaningful range of values, say <= 50? Those would make good pilot problems. Oh what the heck, for minimal computer time, and hazard of greatest error, use the greedy algorith, or a modification of it to weed out the worse choices, ie the pig pork decisions politicians like to propose. === Subject: Re: geometry algorithm - clarrifications <8nrb521hu0ect1fd333sekceg4dec79spm@4ax.com> enclosing rectangle is unique, right? No, two 1x2 OO.OO = 4 00 = 4 00 === Subject: Re: geometry algorithm - clarrifications <8nrb521hu0ect1fd333sekceg4dec79spm@4ax.com> What are you talking about, William? Laura clearly specified in her first post that the enclosing rectangle was minimal. If there are two minimal rectangles that don't coincide, then one extends outside the other. Thus if these minimally enclose a polygon, the polygon must be entirely enclosed within one rectangle's interior, but also appear in the region of the other that is outside the first. A contradiction. > So when the polys are aligned so that we have a possible solution, the > enclosing rectangle is unique, right? > No, two 1x2 > OO.OO = 4 > 00 = 4 > 00 === Subject: Re: geometry algorithm - clarrifications <8nrb521hu0ect1fd333sekceg4dec79spm@4ax.com> > What are you talking about, William? Figure it out for yourself. Minimal rectangles don't have to be unique. However she changed the problem, to providing an enclosing rectangle and how to efficiently fill it. > Laura clearly specified in her first post that the enclosing rectangle > was minimal. If there are two minimal rectangles that don't coincide, > then one extends outside the other. Thus if these minimally enclose a > polygon, the polygon must be entirely enclosed within one rectangle's > interior, but also appear in the region of the other that is outside > the first. A contradiction. > So when the polys are aligned so that we have a possible solution, the > enclosing rectangle is unique, right? No, two 1x2 > OO.OO = 4 > 00 = 4 > 00 === Subject: Re: geometry algorithm - clarrifications <8nrb521hu0ect1fd333sekceg4dec79spm@4ax.com> [Should remove sci.logic from newsgroups list] > What are you talking about, William? > Figure it out for yourself. Minimal rectangles don't have to be unique. > However she changed the problem, to providing an enclosing rectangle and > how to efficiently fill it. Presumably you meant for Minimal rectangles don't have to be unique to apply to a problem where one seeks a configuration of polygons that can be contained in a smaller rectangle than is possible with any other configuration. Your example with 2 1x2 rectangles enclosed within either a 2x2 square or a 1x4 rectangle shows that there can be multiple arrange- ments of polygons that allow equal-area minimal bounding rectangles. waveletter is arguing that for a given configuration, there is one and only one minimal-area enclosing rectangle, which is true if you require its sides be parallel to x-y axes. But waveletter's attempted proof (in next paragraph quoted below) is incomplete: it fails to mention that each of the four sides of a minimal enclosing rectangle must touch a contained polygon, which must be noted to support the claim that part of a polygon must appear in the region of the other that is outside the first. -jiw > Laura clearly specified in her first post that the enclosing rectangle > was minimal. If there are two minimal rectangles that don't coincide, > then one extends outside the other. Thus if these minimally enclose a > polygon, the polygon must be entirely enclosed within one rectangle's > interior, but also appear in the region of the other that is outside > the first. A contradiction. > So when the polys are aligned so that we have a possible solution, the > enclosing rectangle is unique, right? No, two 1x2 OO.OO = 4 00 = 4 > 00 === Subject: Re: geometry algorithm - clarrifications <8nrb521hu0ect1fd333sekceg4dec79spm@4ax.com> <445AE5EE.978B7470@pat7.com> Hi James: What? Is that what William is arguing? But that's obvious, and it's banal, and it's obviously not what Laura posed in her original, intriguing post. sure, William: 2x2 = 4 and 2x(1x2) = 4, but none of these polygons is concave, as in the original problem. And, yes, James is correct that I left out a detail--and I thought it kind of obvious--that the enclosing rectangle has to touch the enclosed, mated concave polygons. If it doesn't touch, then there is an sub-rectangle that still encloses the mated pair (I don't define this, but I leave it as something that can be done independently by all of and has a smaller area. So, OK, with James's insight added to my original remarks, I still say that in the problem, clearly stated by Laura--and clearly motivated by practical problems and the research literature--there is a unique minimal bounding rectangle. So Laura's problem is valid and interesting and well-posed, if a bit difficult. Right or wrong? --Ron > [Should remove sci.logic from newsgroups list] > What are you talking about, William? Figure it out for yourself. Minimal rectangles don't have to be unique. > However she changed the problem, to providing an enclosing rectangle and > how to efficiently fill it. > Presumably you meant for Minimal rectangles don't have to be unique to > apply to a problem where one seeks a configuration of polygons that can > be contained in a smaller rectangle than is possible with any other > configuration. Your example with 2 1x2 rectangles enclosed within either > a 2x2 square or a 1x4 rectangle shows that there can be multiple arrange- > ments of polygons that allow equal-area minimal bounding rectangles. > waveletter is arguing that for a given configuration, there is one and > only one minimal-area enclosing rectangle, which is true if you require > its sides be parallel to x-y axes. But waveletter's attempted proof > (in next paragraph quoted below) is incomplete: it fails to mention > that each of the four sides of a minimal enclosing rectangle must touch > a contained polygon, which must be noted to support the claim that part > of a polygon must appear in the region of the other that is outside > the first. > -jiw > Laura clearly specified in her first post that the enclosing rectangle > was minimal. If there are two minimal rectangles that don't coincide, > then one extends outside the other. Thus if these minimally enclose a > polygon, the polygon must be entirely enclosed within one rectangle's > interior, but also appear in the region of the other that is outside > the first. A contradiction. So when the polys are aligned so that we have a possible solution, the > enclosing rectangle is unique, right? No, two 1x2 OO.OO = 4 00 = 4 > 00 === Subject: Re: reals and union <23150162.1146742585616.JavaMail.jakarta@nitrogen.mathforum.org> has a countable cover of pairwise disjoint closed sets, at most > one of those sets is nonvoid. Thus lemma: If C is a countable closed partition of [-n,n], ie a countable partition of closed sets, then C = { [-n,n] } If C is a countable closed partition of R, then C = { R }. C1 = { K / [-1,1] | K in C } - {nulset} is a countable closed partition of [-1,1]. C1 = { [-1,1] }. Some K1 in C with [-1,1] = K1 / [-1,1] [-1,1] subset K1 Cn = { K / [-n,n] | K in C } - {nulset} is a countable closed partition of [-n,n]. Cn = { [-n,n] }. Some Kn in C with [-n,n] = Kn / [-n,n] [-1,1] subset [-n,n] subset Kn = K1. Conclude for all n, [-n,n] subset K1. K1 = R; C = { R }. Thus the problem of showing R has no non-trivial countable closed partition, reduces to showing [0,1] has no non-trivial countable closed partition. This method of proof will suffice to show that if Hausdorff S = /_n K_n where for all n in N, K_n is a compact connected subset and for all n, K_n subset K_(n+1), ie the K_n's are upward nested, then there is no non-trivial countable closed partition of S. === Subject: Re: reals and union <3408221.1146719310288.JavaMail.jakarta@nitrogen.mathforum.org>, > Could you please help me with the following problem: > Prove that R cannot be written as a countable union of disjoint closed sets. Suppose E_1, E_2, ... are disjoint nonempty closed subsets of R whose union is R. Clearly R(E_1 U E_2) has a bounded component (a,b). Because the E_n's exhaust R, one of them intersects (a,b). Let E_n_1 be the first one that does; clearly n_1 > 2. Now E_n_1 is closed and doesn't contain a, so it contains a point b_1 in (a, b) closest to a. We then have E_k disjoint from (a, b_1) for k <= n_1. Now let E_n_2 be the first of the E's that intersects (a, b_1). E_n_2 contains a point a_1 in (a, b_1) closest to b_1. We now have E_k disjoint from (a_1, b_1) for k <= n_2. Continue ... We create a < a_1 < a_2 < ... < b_2 < b_1 < b and a sequence 1 < 2 < n_1 < n_2 < ... such that E_k is disjoint from (a_m, b_m) for k <= n_(m+1). The closed intervals [a_m, b_m] are nested and therefore have non-empty intersection. Any x in this intersection cannot belong to any E_k, contradiction. === Subject: Re: complex roots of polynomial degree 4 > I've got a polynomial f(s)=s^4+s^3*(2y)+s^2*(1+4*x^2)+s*(8x^2y)+x^2, > where 0.5<=x<=2, 0<=y<=0.5. How can I proof that this polynomial have > only complex roots with Re, Im not equal to zero ? In Mathematica you would input Reduce[s^4 + (2 y) s^3 + (4 x^2 + 1) s^2 + (8 x^2 y) s + x^2 == 0 && 1/2 <= x <= 2 && 0 <= y <= 1/2, Reals] False The output proves that there are no real solutions. > Are there any methods to find this roots in explicit form ?? You could use Solve[s^4 + (2 y) s^3 + (4 x^2 + 1) s^2 + (8 x^2 y) s + x^2 == 0] but the output in terms of explicit radicals is not really useful. A better representation of the solutions is in terms of Root objects. The advantages are: -Faster to obtain. -Numerically more stable to evaluate. -In general, radical formulations are prone to numeric problems. Root objects do not have this liability. -When the roots of an irreducible cubic are all real but not rational, the so-called casus irreducibilis shows that they still must be expressed in terms of I (mathworld.wolfram.com/CasusIrreducibilis.html). This means that numeric evaluation will give small imaginary parts unless, by happenstance, they exactly cancel. Small numeric error from round-off makes this unlikely. -For sufficiently complicated algebraics, it is often faster to evaluate the Root form numerically, at least at high precision. -Polynomial combinations of Root objects simplify using RootReduce. -Derivatives of Root objects with respect to a parameter are expressed in terms of Root objects. This is useful for eigenvalue sensitivity analysis. > I've tried to find they > with numerical methods, and after find relation between they Re, and Im > parts and variables x, and y, but I've had difficulties near x=y=0. But x = 0 is not in 0.5<=x<=2? Paul _______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul === Subject: Re: complex roots of polynomial degree 4 >>I've got a polynomial f(s)=s^4+s^3*(2y)+s^2*(1+4*x^2)+s*(8x^2y)+x^2, >>where 0.5<=x<=2, 0<=y<=0.5. How can I proof that this polynomial have >>only complex roots with Re, Im not equal to zero ? > In Mathematica you would input > Reduce[s^4 + (2 y) s^3 + (4 x^2 + 1) s^2 + (8 x^2 y) s + x^2 == 0 && > 1/2 <= x <= 2 && 0 <= y <= 1/2, Reals] > False > The output proves that there are no real solutions. Does Mathematica have a Why command? I find a terse False unconvincing. Martin Cohen === Subject: Re: TO's largest natural > Mike Kelly said: > But the smallest infinity of standard theories isn't an infintie > value like you've been talking about. It describes an equivalence > class. > Yes, I know, but it's called the size of the set, and yet, doesn't satisfy > very > many intuitions about what a set size is. You can add an infinite number of > additional elements to an infinite set without changing the cardinality. I > find > that objectionable. And TO suggests that adding one more item anywhere in an endless sequence makes it more endless? > And, it's referred to as the smallest infinite ordinal > number, as if it's some kind of quantity, but it's really not. Sure it is, for those for whom quantity is not totally intuitional. > So, with all > due respect to von Neumann, I wouldn't give his ordinals such a central place > in the theory. They are not exclusively his, they are a joint effort. And all finite cardinals and finite ordinals, uncluding TO's, work in precisely the same way. So by TO's own axiom of infinite inductivity, they must work the same way for infinites as for finites, by injection and bijection. > > The english-language definition of finite is something like bounded > or limited. The idea that the naturals are bounded but unbounded is > probably causing some of my confusion here. > Well, that's understandable. Infinite means literally without end, and the > standard set theoretical deifninition classifies the set of finite naturals > as > countably infinite. My issue with that is that, when no two elements in the > ordered set can ever have any infinite number of other elements between them, > then I don't see as there is an infinite number of elements in the set. But endless does not mean in the middle must be endless, only that one end does not exist. > There's > no infinite _number_ of elements. As TO has insisted that there are no infinite numbers at all, either he is saying that all sets are finite or he is contradicting himself. > > Are you saying that sets cannot have properties their elements do not? > So you cannot have an infinite set of finite values? I think that's > faulty logic. > > Question : The rationals in [0,1] are countably infinite. Yet there are > only internally infinite, if I understand you right? > Okay, you're new to this discussion. When I see this question I often get > rather irked, because it's usually from someone I have explained this to half > a > dozen times. So, I'll try to patiently explain my position without getting > annoyed. :) > We're talking about whether the set of finite naturals is infinite, right? > the set theoretical definition we would say yes, since it is possible to form > an injection from the set into a proper subset, like the evens. However, from > the perspective of the set having an actually infinite quantity of elements > within it, I would say it does not qualify. What constitutes having an infinite quantity of elements within it? > The standard alternative, the von Neumann ordinals, rely on a model starting > with 0 as the empty set, and defines each natural as the set of all its > predecessors. It then defines the first limit ordinal, omega, as the set of > all > finite naturals, and declares it infinite, since it's larger than all > finites. > The problem with this is that, in every finite case, the maximal element is > always one less than the set size, and there is nothing that changes that in > the infinite case. Sure there is! To is being deliberately obstreperous about this because vN naturals so easily and definitely show him wrong. The union of all vN naturals does not contain a maximal element at all, so that it is different in that respect from every finite natural. > Essentially, omega is the successor to the largest finite TO keeps claiming that this sort of impossibility must exist. Where he originally got this stupid idea he won't say, but he keeps returning to it again and again. No one but TO believes the union of all vN naturals has a largest member, so TO should give up on trying to imply otherwise. === Subject: Re: Consequence of MVT. > I am trying to prove a consequence of MVT. The proof I saw in a book > only states the intervals as open intervals, all I have done it change > this to ANY interval. Please tell me if this spoils the proof in any > way. If it does, what do I have to add to make it complete? > Theorem > If f'(x) = 0 for all x in an interval I, then f is constant on I. > Proof > Let x1, x2 be any two numbers in I with x1 < x2. > Since f is differentiable on I, it is also differentiable on [x1, x2]. The sqrt function is differentiable on (0, 1] but not at 0. === Subject: Re: Book on Godel Proof for Beginner > Wasn't there a book by Raymond Smullyan > which begins with easy logic puzzle questions about lying knights and such, > then step by step raises the complexity level of the puzzles, until it > finally ends up with a proof of a version of Goedel's incompleteness theorem!? > Unfortunately, I don't remember the title, but maybe someone else does... Forever Undecided: a puzzle Guid to Godel, 1987, Oxford University Press. ISBN 0-19-280141-4 (paperback). Out of print. === Subject: Equivalent formulations of Fermat's Last Theorem It's not hard to come up with some of these simply by transforming the Fermat curve; for instance, for odd n the rational points on x^n+y^n=1 correspond to the rational points on (x^2+y^2)^n = x^n+y^n by inversion in the unit circle (for even n the curve is reducible, but you basically get the same thing.) But I'm interested in other equivalent formulations, or consequences, of FLT. I recall seeing somewhere a hyperelliptic curve which was supposed to have rational points iff a corresponding Fermat curve did, for instance. === Subject: A self-contained approximation to the inverse log-factorial function Daniel Lichtblau and David W. Cantrell have published elegant approximations to the inverse factorial function, exploiting the fact that the Lambert W function can be used to invert certain approximations to the factorial: For a particular application I needed the inverse of the log-factorial: y = ln(x!) and I decided (as a challenge) to go for a self-contained approximation, that is, one that makes no use of the Lambert W function. The approximation presented below has a |rel. error| < 2.41E-9 for x>=1. The |rel. error| decreases nicely for larger x: x |rel. error| ----------------------- 1.5 2.40E-09 2 1.24E-09 5 1.88E-10 10 1.50E-11 100 2.46E-13 1000 2.59E-14 10000 2.18E-15 100000 1.45E-16 Algorithm --------- For y < C1 (ca. 5, for the values of constants C1 thru C9, see below) an initial approximation to the inverse log-factorial is used that is inspired by the inverse parabola-like behavior near the x-minimum (C4 being close to 0.5): x = C2 (y+C3)^C4 + C5 For larger values of y, the initial approximation is somewhat similar to the the Lichtblau/Cantrell approximations: x = y / (ln(y-C6) + C7/(1 + C8 ln(y))-C9) This combined initial approximation is then improved by means of 3 Newton iterations using a truncated Stirling series as an approximation to the log-gamma function. Fortunately, it turns out that the error introduced by using the truncated series can to a large extent be compensated by optimizing constants C1 thru C9. First x is converted into a suitable argument for the gamme function: x = x+1 Then the following Newton improvement is performed 3 times: y' = (x-0.5) ln(x)-x + 0.5 ln(2 Pi) + 1/x ( 1/12 + 1/x^2 (-1/360 + 1/x^2 1/1260)) d = ln(x)-0.5/x + 1/x^2(-1/12 + 1/x^2 ( 1/120 + 1/x^2 -1/252)) x = x-(y'-y)/d Finally, the improved x is returned as an argument for the factorial: x = x-1 The constant values are: C1 = 5.1683197 C2 = 2.2607111 C3 = 0.1091978 C4 = 0.5251431 C5 = 0.7032945 C6 = 0.5465472 C7 = 4.5033124 C8 = 0.2299674 C9 = 3.6406814 Note that these values were optimized for exactly 3 Newton iterations. Any other number of iterations requires a special optimization. Knud Thomsen Geologist, Denmark === Subject: Re: A self-contained approximation to the inverse log-factorial function Here is a Pascal implementation of the above approximation. function InvLnFacAppr(Y: Double): Double; {--------------------------------------------------------------} { Returns an approximation to the inverse of y = ln(x!) for } { x >= 1. } { |Rel. error| < 2.41E-9. } { } { The |rel. error| decreases for larger x: } { } { x |rel. error| } { -------------------- } { 1.5 2.40E-09 } { 2 1.24E-09 } { 5 1.88E-10 } { 10 1.50E-11 } { 100 2.46E-13 } { 1000 2.59E-14 } { 10000 2.18E-15 } { 100000 1.45E-16 } { } { Knud Thomsen 2006. } {--------------------------------------------------------------} const // Constants optimized for 3 Newton iterations on truncated Stirling series // (partly compensating for Stirling truncation errors) C1 = 5.1683197; C2 = 2.2607111; C3 = 0.1091978; C4 = 0.5251431; C5 = 0.7032945; C6 = 0.5465472; C7 = 4.5033124; C8 = 0.2299674; C9 = 3.6406814; var X,T,Y_,D: Double; I: Integer; begin if Y <877j52d995.fsf@nonospaz.fatphil.org> <445a466c.94288770@netnews.att.net That's the > problem with idiotic rambling. It's difficult to be sure exactly who > the idiotic rambler is. I don't think 'idiotic' is always the right word. But I think mathematics (though less so for philosophy of mathematics) is a subject - perhaps the quintessential one - in which we can more readily detect that the spiel is nonsensical, or error ridden, or stubborn and foolish, or utterly uniformed. MoeBlee === Subject: Re: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= On 5 May 2006 01:54:10 -0700, MoeBlee in >> That's the >> problem with idiotic rambling. It's difficult to be sure exactly who >> the idiotic rambler is. >I don't think 'idiotic' is always the right word. Pejorative terms are almost invariably self referential and usually self reverential. Next to scientists idiots always make the best idiot detectors. > But I think >mathematics (though less so for philosophy of mathematics) is a subject >- perhaps the quintessential one - in which we can more readily detect >that the spiel is nonsensical, or error ridden, or stubborn and >foolish, or utterly uniformed. Well this is indeed curious.Certainly mathematics is an unusually well structured edifice. That does not however mean that mathematicians are comparably well structured. They rant and rave, whine and snivel, and jump up and down like scalded monkeys just like everyone else when their own ox is gored. I take it you've beeen sequestered in a Tibetan monastery the last couple of years or you would have noticed my own revisionist contributions on such topics as universal truth, finite tautological regressions, derivatives of cross products in angular momentum, the absence of any single real number line in formal terms, irrational and transcendental numbers, the analytical origin of Planck's constant, Turing computability and intelligence, and the non computability of numbers etc. etc. etc. ad nauseum all of which were posted to sci.math and all of which received uniformly rave reviews, the emphasis here being on rave as a code word for reactionary among mathematkers of every stripe. Unfortunately Torkel had little to say regarding these topics and confined his remarks to anonymous and ambiguous references to fnoffling so I don't have much quarrel with anything he said since he didn't really say it. But plenty of others have and I've had plenty to say about what they chose to say and I've said plenty about it. The problem seems to be divorcing mathematics from the opinions of mathematikers and what has been demonstrated from what hasn't. I notice you leave yourself an escape hatch above in distinguishing math from the philosophy of math. But to me it's all science and those who prefer to indulge their own flights of fancy in mathematics are no better qualified to judge one from the other than idiotic ramblers. ~v~~ === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <87mzedmszc.fsf@nonospaz.fatphil.org> <877j52d995.fsf@nonospaz.fatphil.org> <445a4140.92965203@netnews.att.net I think we're bound by the dictum de mortuis nil nisi bonum. > Personally I see nothing wrong with discusssing the issue critically > in terms you lay out here. But not specifically in relation to what > Torkel may or may not have said and what he may or may not have > intended by what he said addressed to various individuals. My own > discussions with him were pretty much couched in terms of fnoffling > by which I understood him to disparage what I was saying. But the > manipulative skill he demonstrated was just clever enough to avoid > the petty mindedness exhibited by so many who disagree on these > groups. If you wish to pursue the topic further I suggest you show the > good manners to abstract the issue from Torkel himself. At least then > we can get down to brass tacks of demagoguery. Torkel could be funny. > I'll give him that even if we didn't necessarily agree. What demagoguery? If there are brass tacks to be gotten down to here, then those would not be in examination of the Form of Demagoguery but in particular examples of purported demagoguery in particular posts. If there is a suggestion that Franzen's postings recommend the subject of demagoguery, then I'd like to know who it is you think used demagoguery and what posts you adduce as examples. And I don't think we have to observe some special propriety given that the man is recently dead. I can't speak for other appreciators of Franzen, but for me, I would find it less appropriate to have a subject broached only to cloud as suggestion or insinuation than to have it spoken frankly. MoeBlee === Subject: Re: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= On 5 May 2006 01:42:15 -0700, MoeBlee in >> I think we're bound by the dictum de mortuis nil nisi bonum. >> Personally I see nothing wrong with discusssing the issue critically >> in terms you lay out here. But not specifically in relation to what >> Torkel may or may not have said and what he may or may not have >> intended by what he said addressed to various individuals. My own >> discussions with him were pretty much couched in terms of fnoffling >> by which I understood him to disparage what I was saying. But the >> manipulative skill he demonstrated was just clever enough to avoid >> the petty mindedness exhibited by so many who disagree on these >> groups. If you wish to pursue the topic further I suggest you show the >> good manners to abstract the issue from Torkel himself. At least then >> we can get down to brass tacks of demagoguery. Torkel could be funny. >> I'll give him that even if we didn't necessarily agree. >What demagoguery? You've never seen or experienced demagoguery on the usenet? You've got to be kidding. I see almost nothing but appeals to popular prejudice. Maybe that's because I deal in original thoughts and you don't. > If there are brass tacks to be gotten down to here, >then those would not be in examination of the Form of Demagoguery but >in particular examples of purported demagoguery in particular posts. If >there is a suggestion that Franzen's postings recommend the subject of >demagoguery, then I'd like to know who it is you think used demagoguery >and what posts you adduce as examples. As noted above I have no information on the subject vis-a-vis Torkel and certainly wouldn't bring it up here if I thought I had.My exhanges with Torkel were limited to fnoffling. > And I don't think we have to >observe some special propriety given that the man is recently dead. And I think we do since he is no longer among us. Especially in the context of commemorative comments. > I >can't speak for other appreciators of Franzen, but for me, I would find >it less appropriate to have a subject broached only to cloud as >suggestion or insinuation than to have it spoken frankly. There is no insinuation on my part. I said exactly what I had to say in commemorative terms and strongly urged others to remove their critical discussions to other venues. ~v~~ === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <87mzedmszc.fsf@nonospaz.fatphil.org> <877j52d995.fsf@nonospaz.fatphil.org> <445a4140.92965203@netnews.att.net And I don't think we have to > observe some special propriety given that the man is recently dead. I > can't speak for other appreciators of Franzen, but for me, I would find > it less appropriate to have a subject broached only to cloud as > suggestion or insinuation than to have it spoken frankly. Seeing as he spent most of his career arguing on usenet groups, bringing it almost to an art form, having an argument about him seems a pretty good way of showing respect. In my view. === Subject: Re: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= On 5 May 2006 03:07:28 -0700, William of Ockham >> And I don't think we have to >> observe some special propriety given that the man is recently dead. I >> can't speak for other appreciators of Franzen, but for me, I would find >> it less appropriate to have a subject broached only to cloud as >> suggestion or insinuation than to have it spoken frankly. >Seeing as he spent most of his career arguing on usenet groups, >bringing it almost to an art form, having an argument about him seems a >pretty good way of showing respect. In my view. Well that's certainly a reasonable objective. My problem is that I only have the one series of comments regarding fnoffling to draw on so there really isn't much I can contribute except to note the fact of wittiness apparent even in those ambiguous observations. On the more general subject of demagoguery and character assassination however there is a wealth of material available though not necessarily related to Torkel. ~v~~ === Subject: Re: Be bad to be successful. <0OL5g.948$VV2.72688@news20.bellglobal.com> To be successful one needs not to be good as defined > by > scriptures. >>What do you think ? >By scriptures do you mean the bible? The bible is not > relevant > to >reality at all. >That's bull > How is the bible relavent to any part of reality? Because people who believe in the bible are part of > reality, > and their beliefs affect their actions. Note that this applies to belief in ANYTHING, not just the > bible. The End of Belief is the Beginning of Intelligence > Unity encourages seekers not to believe anything they > have > not > verified > with experience. Check what you know by empirical > verification, > and > what > you > just believe. > For example: > Is Jesus [or ...] the only way to God? > Is unwavering belief in Jesus essential? > Is the truth what your scriptures say? > Is your effort going to help you to be free? > Do you have free will? > Are you creating your reality with your thoughts? > Is your suffering somehow your fault? > Is the universe a dream or illusion? > Are you already enlightened but have just forgotten it? > Do you value belief above enquiry? > Drop the Lot > The spiritual information that you acquire from teachers > and > old > scriptures is best taken as a hypothesis for you to verify. > Unverified > spiritual beliefs hypnotize you and lure you into believing > all > kinds of > irrational ideas. It takes courage to admit you do not know > if > your > beliefs > are accurate - that you are just a blind believer. > These messages are not intended to replace your > unverified > beliefs, > but > to provoke a thirst for truth; they are an invitation to > apply > the > words. > Application is much easier said than done. > Each of these messages is a challenge to help you to grow > in > truth > and > consciousness, not more information to add to your > prodigious > collection > of > borrowed spiritual knowledge. > Drop the lot and become a seeker of truth with a single > hypothesis > to > verify in the laboratory of your inner world: God is all > there > is > and I > am > part of Him. Start today. > Intelligence > The relinquishing of unverified belief heralds the > awakening > of > spiritual intelligence. Your intelligence is sharpened by > enquiry. > Enquiry > requires an open mind; self-enquiry is the essence of > religion. > Life is innately intelligent and thus the source of life > is > a > superior > intelligence. This insight allows you to trust in > life-as-God. > Intelligence > understands the beauty of trust. > Allow life to live you without the unnecessary baggage of > unverified > beliefs. The end of belief is the awakening of trust in > life-as-God. > Trust > is intelligence. -- from the book UNITY -The Dawn of Conscious Civilization > by > Maitreya > Ishwara > -- LOL I believe this Maitreya Ishwara is a sherpa, > and you are his donkey... hee haw > -- ...tethered to a pole. Appearances > -- Appearances or not, there is no doubt > you are peddleing Maitreya Ishwara > wares. In your mind. > -- > > Do not believe it! > Otherwise you will have to verify it > And that will require the verification of your own mind first > A task that could consume your whole life > If you suppose verification must be absolute then yes. > Good thing though I posted it I don't believe any of it. > -- > > Most peddlers do not believe in the products they peddle Even so - if verification is not 100% -- it simply is not verification As Seng-tsan sang in the Song of Trusting the Heart A hairsbreadth's miss is as the distance between sky and earth. === Subject: Re: Be bad to be successful. <0OL5g.948$VV2.72688@news20.bellglobal.com> To be successful one needs not to be good as defined > by > scriptures. >>What do you think ? >By scriptures do you mean the bible? The bible is not > relevant > to >reality at all. >That's bull > How is the bible relavent to any part of reality? Because people who believe in the bible are part of > reality, > and their beliefs affect their actions. Note that this applies to belief in ANYTHING, not just the > bible. The End of Belief is the Beginning of Intelligence > Unity encourages seekers not to believe anything they > have > not > verified > with experience. Check what you know by empirical > verification, > and > what > you > just believe. > For example: > Is Jesus [or ...] the only way to God? > Is unwavering belief in Jesus essential? > Is the truth what your scriptures say? > Is your effort going to help you to be free? > Do you have free will? > Are you creating your reality with your thoughts? > Is your suffering somehow your fault? > Is the universe a dream or illusion? > Are you already enlightened but have just forgotten it? > Do you value belief above enquiry? > Drop the Lot > The spiritual information that you acquire from teachers > and > old > scriptures is best taken as a hypothesis for you to verify. > Unverified > spiritual beliefs hypnotize you and lure you into believing > all > kinds of > irrational ideas. It takes courage to admit you do not know > if > your > beliefs > are accurate - that you are just a blind believer. > These messages are not intended to replace your > unverified > beliefs, > but > to provoke a thirst for truth; they are an invitation to > apply > the > words. > Application is much easier said than done. > Each of these messages is a challenge to help you to grow > in > truth > and > consciousness, not more information to add to your > prodigious > collection > of > borrowed spiritual knowledge. > Drop the lot and become a seeker of truth with a single > hypothesis > to > verify in the laboratory of your inner world: God is all > there > is > and I > am > part of Him. Start today. > Intelligence > The relinquishing of unverified belief heralds the > awakening > of > spiritual intelligence. Your intelligence is sharpened by > enquiry. > Enquiry > requires an open mind; self-enquiry is the essence of > religion. > Life is innately intelligent and thus the source of life > is > a > superior > intelligence. This insight allows you to trust in > life-as-God. > Intelligence > understands the beauty of trust. > Allow life to live you without the unnecessary baggage of > unverified > beliefs. The end of belief is the awakening of trust in > life-as-God. > Trust > is intelligence. -- from the book UNITY -The Dawn of Conscious Civilization > by > Maitreya > Ishwara > -- LOL I believe this Maitreya Ishwara is a sherpa, > and you are his donkey... hee haw > -- ...tethered to a pole. Appearances > -- Appearances or not, there is no doubt > you are peddleing Maitreya Ishwara > wares. In your mind. > -- > > Do not believe it! > Otherwise you will have to verify it > And that will require the verification of your own mind first > A task that could consume your whole life > If you suppose verification must be absolute then yes. > Good thing though I posted it I don't believe any of it. > -- > > Most peddlers do not believe in the products they peddle Even so - if verification is not 100% -- it simply is not verification As Seng-tsan sang in the Song of Trusting the Heart A hairsbreadth's miss is as the distance between sky and earth. === Subject: Re: Surfaces of hyperspheres On 3 May 2006 09:24:31 -0700, Michael Hennebry : > Draw uniformly at random a point of norm 1 in Euclidean n-space. : : Since otherwise the answer is obvious, : let's assume that alpha is in the open interval (0, 1). : We want twice the probability that the x-coordinate > alpha. : This is proportional to the n-volume of the : appropriate section of the unit n-sphere. Why volume? Doesn't that imply that you consider random points of norm _at most_ 1; rather than norm exactly 1, as I am interested in? : It is the convex hull of the union of the origin : and the lens-shaped portion : { p in R^n: |p|<=1 & p[1]> alpha} . : The n-volume is : alpha*S(n-1)*(1-alpha^2)^((n-1)/2)/n + : 1 : INT S(n-1)*(1-x^2)^((n-1)/2) dx , : alpha : : where S(n-1) is the n-1-volume : of a unit n-1-sphere. : The first term is n-volume of a cone : defined by the origin and an n-1-sphere. : The other is the n-volume of the lens-sphaped portion. : -- :-- Hans Georg http://www.ii.uib.no/~georg/ `This Universe never did make sense; I suspect that it was built on government contract.' (Heinlein) === Subject: Re: Filters and Nets <040520060939468225%edgar@math.ohio-state.edu.invalid> === Subject: Re: Filters and Nets > Let X be a set and F a filter on X. > Denote by s_F the associated net, i.e. the directed set for s_F > is D = { (A,x) : A in F, x in A } > where (A,x) <= (B,y) means B subset A (<= is the preorder > on D), and s_F : (A,x) |--> x. > Huh? For all A in F, chose a x_A in A. > as it does not require the Axiom of Choice. Ok, s_F:D -> X, (A,x) -> x, is a net because F a filter and since for all U,V in F, U / V in F, F is updirected by reverse subset. > Mate I know that there are also alternative definitions for subnets, > but I think that it would be preferable to keep the Kelley's one. Whoops, subnets are mushy. ---- === Subject: Just a pain in the ass Differential Equation (for me at least) y'[x] = e^(y[x]^2 - x^2) and y[0]=0 or what I tried in Mathematica: DSolve[{y'[x] == [ExponentialE]^(y[x]^2 - x^2), y[0] == 0}, y[x], x] Can anyone tell me how it is solved? === Subject: Re: Just a pain in the ass Differential Equation (for me at least) gesareli nous a r.8ecemment amicalement signifi.8e : > y'[x] = e^(y[x]^2 - x^2) and y[0]=0 > or what I tried in Mathematica: > DSolve[{y'[x] == [ExponentialE]^(y[x]^2 - x^2), y[0] == 0}, y[x], x] > Can anyone tell me how it is solved? y' = e^(y^2-x^2) ==> y' e^(-y^2) = e^(-x^2) ==> erf(y) = erf(x) + c y(0)=0 ==> c=0 ==> erf(y) = erf(x) ==> y=x -- patrick === Subject: Re: Just a pain in the ass Differential Equation (for me at least) >y'[x] = e^(y[x]^2 - x^2) and y[0]=0 >or what I tried in Mathematica: >DSolve[{y'[x] == [ExponentialE]^(y[x]^2 - x^2), y[0] == >0}, y[x], x] >Can anyone tell me how it is solved? y(x)=x I don't know of a systematic way how to arrive at the solution ; got it just by chance. Best wishes Torsten. === Subject: Re: Sheaves and Stalks and Dense spaces <4459B092.6000507@web.de I'am racking my brain to follow your arguments. I see that the locus U > where f and g coincide is open and dense in X. But what is your > reasoning to conclude that U the whole space? > I could think of some irreducible space X in which every non-void open > set is dense, but not closed. > Am I missing something here? > J. Ok, right.. that part of the proof was wrong.. so it probably won't :) Jose Capco === Subject: Re: Sheaves and Stalks and Dense spaces >> I'am racking my brain to follow your arguments. I see that the locus U >> where f and g coincide is open and dense in X. But what is your >> reasoning to conclude that U the whole space? >> I could think of some irreducible space X in which every non-void open >> set is dense, but not closed. >> Am I missing something here? >> J. > Ok, right.. that part of the proof was wrong.. so it probably won't > :) > Jose Capco The claim does not hold in general - as we have found out together with Ryan. But if we think of the phenomenon of analytic continuation (like in complex calculus, Riemannian surfaces etc.) the assertion *is* true in this case. In sheaf theoretic terms this means that the locus of coincidence of two sections is both open (in general) *and* closed (only in particular cases). I hope this can be helpful. Best wishes, J. === Subject: Re: Sheaves and Stalks and Dense spaces > I suppose a correct reformulation > would be, two sections whose germs agree at every closed point agree > everywhere, since the support of their difference is closed and hence > empty. Don't you need a lemma on a topological space X saying that any open subset U of X containing every closed point of X is X itself? I am able to prove this for Noetherian or quasi-compact T1-spaces. How does this lemma hold in general topological spaces? J. === Subject: Re: Sheaves and Stalks and Dense spaces <4459B092.6000507@web.de> <445A64A7.4020708@web.de I suppose a correct reformulation > would be, two sections whose germs agree at every closed point agree > everywhere, since the support of their difference is closed and hence > empty. > Don't you need a lemma on a topological space X saying that any open > subset U of X containing every closed point of X is X itself? > I am able to prove this for Noetherian or quasi-compact T1-spaces. How > does this lemma hold in general topological spaces? Well, you don't even need quasi-compact for T1, since all points are closed there, so an open set containing all of them must be the whole space. I suppose in general it's not true, and here's an example: Let X be the natural numbers, whose open sets are all initial segments in the usual order. Every open set excludes infinitely many points, so no point is closed. Therefore the condition is trivially satisfied for any open set, yet the topology is not the trivial topology. I guess I was thinking of schemes, which do satisfy my lemma, since any ideal of any ring is contained in a maximal ideal, and schemes are locally the spectra of rings. This implies that every closed set contains a closed point, hence every open set misses one. I've spent quite a while thinking about it but I can't figure out a simple criterion on X for the lemma to hold. You could ask for every nonempty closed set to contain a closed point, for example, which I suppose is about as simple as it gets, but isn't one of the standard axioms. Well, the theorem is true if and only if this holds. For if it fails, we can push forward a sheaf on some closed set with no closed points, and get a counterexample on X. In order to save face, I will now mutter something about no reasonable person wanting to study a space without closed points. -- Ryan Reich ryan.reich@gmail.com === Subject: Re: Sheaves and Stalks and Dense spaces > I suppose a correct reformulation > would be, two sections whose germs agree at every closed point agree > everywhere, since the support of their difference is closed and hence > empty. >> Don't you need a lemma on a topological space X saying that any open >> subset U of X containing every closed point of X is X itself? >> I am able to prove this for Noetherian or quasi-compact T1-spaces. How >> does this lemma hold in general topological spaces? > Well, you don't even need quasi-compact for T1, since all points are > closed there, so an open set containing all of them must be the whole > space. spaces. I should have better written Kolmogorov space. > I suppose in general it's not true, and here's an example: Let > X be the natural numbers, whose open sets are all initial segments in > the usual order. Every open set excludes infinitely many points, so > no point is closed. Therefore the condition is trivially satisfied > for any open set, yet the topology is not the trivial topology. I > guess I was thinking of schemes, which do satisfy my lemma, ... unfortunately not this strictly. Please see my last comment below. > since any > ideal of any ring is contained in a maximal ideal, and schemes are > locally the spectra of rings. This implies that every closed set > contains a closed point, hence every open set misses one. > I've spent quite a while thinking about it but I can't figure out a > simple criterion on X for the lemma to hold. You could ask for every > nonempty closed set to contain a closed point, for example, which I > suppose is about as simple as it gets, but isn't one of the standard > axioms. I agree that it look pretty simple, but it not that easy to distinguish those topological spaces satisfying this closed point property from those which do not. I will try below ... > Well, the theorem is true if and only if this holds. For if > it fails, we can push forward a sheaf on some closed set with no > closed points, and get a counterexample on X. > In order to save face, I will now mutter something about no reasonable > person wanting to study a space without closed points. That seems to be the case, but aren't we looking for a set of closed points which is rich enough to carry as much information as the whole space in topological terms? This reminds me of Jacobson spaces which are in line with Jacobson rings in commutative algebra. Here is what I can provide for the claim on the existence of closed points: If X is (i) Noetherian and Kolmogorov or (ii) quasi-compact and Kolmogorov, then any non-void closed subset of X contains a closed point. This implies the lemma I mentioned in my previous posting. Assumption (i) is certainly satisfied by any affine, hence quasi-compact, scheme, but not by a scheme or an arbitrary topological space, in general. This is why some general theorems on schemes need some condition on the underlying topological space (such as Noetherian, locally Noetherian, quasi-compact, ...). Here is a sketch of the proof of the claim above: If V is a non-void closed subset of X then the system VV of non-void closed subsets of V has minimal elements (w.r.t. inclusion); for (i) this is because X is Noetherian, for (ii) by Zorn's lemma and because X, hence V is quasi-compact. By Kolmogorov's axiom, any minimal non-void closed subset of X is a singleton. Best wishes, J. === Subject: Re: Sheaves and Stalks and Dense spaces <4459B092.6000507@web.de> <445A64A7.4020708@web.de> <445AECBB.5040502@web.de I suppose a correct reformulation > would be, two sections whose germs agree at every closed point agree > everywhere, since the support of their difference is closed and hence > empty. >> Don't you need a lemma on a topological space X saying that any open >> subset U of X containing every closed point of X is X itself? >> I am able to prove this for Noetherian or quasi-compact T1-spaces. How >> does this lemma hold in general topological spaces? > Well, you don't even need quasi-compact for T1, since all points are > closed there, so an open set containing all of them must be the whole > space. > spaces. I should have better written Kolmogorov space. > I suppose in general it's not true, and here's an example: Let > X be the natural numbers, whose open sets are all initial segments in > the usual order. Every open set excludes infinitely many points, so > no point is closed. Therefore the condition is trivially satisfied > for any open set, yet the topology is not the trivial topology. I > guess I was thinking of schemes, which do satisfy my lemma, > ... unfortunately not this strictly. Please see my last comment below. Unfortunately yes. They may not satisfy your conditions, true, but every nonempty closed set in any scheme contains a closed point, because any such set intersects some affine scheme nontrivially, and this is true of affine schemes because of the existence of maximal ideals. > since any > ideal of any ring is contained in a maximal ideal, and schemes are > locally the spectra of rings. This implies that every closed set > contains a closed point, hence every open set misses one. > I've spent quite a while thinking about it but I can't figure out a > simple criterion on X for the lemma to hold. You could ask for every > nonempty closed set to contain a closed point, for example, which I > suppose is about as simple as it gets, but isn't one of the standard > axioms. > I agree that it look pretty simple, but it not that easy to distinguish > those topological spaces satisfying this closed point property from > those which do not. I will try below ... I was being facetious; I don't think it's at all a simple condition, which is why I was disappointed not to have uncovered a better one. > Well, the theorem is true if and only if this holds. For if > it fails, we can push forward a sheaf on some closed set with no > closed points, and get a counterexample on X. > In order to save face, I will now mutter something about no reasonable > person wanting to study a space without closed points. > That seems to be the case, but aren't we looking for a set of closed > points which is rich enough to carry as much information as the whole > space in topological terms? This reminds me of Jacobson spaces which are > in line with Jacobson rings in commutative algebra. You know, thinking more about the condition I stated above, that every closed set contain a closed point, although it may not be easy to check it's not obviously less so than checking that the closed points are dense, which is a reasonable condition I tried hard to make work for a while. > Here is what I can provide for the claim on the existence of closed > points: If X is (i) Noetherian and Kolmogorov or (ii) quasi-compact and > Kolmogorov, then any non-void closed subset of X contains a closed point. > This implies the lemma I mentioned in my previous posting. Assumption > (i) is certainly satisfied by any affine, hence quasi-compact, scheme, > but not by a scheme or an arbitrary topological space, in general. This > is why some general theorems on schemes need some condition on the > underlying topological space (such as Noetherian, locally Noetherian, > quasi-compact, ...). > Here is a sketch of the proof of the claim above: If V is a non-void > closed subset of X then the system VV of non-void closed subsets of V > has minimal elements (w.r.t. inclusion); for (i) this is because X is > Noetherian, for (ii) by Zorn's lemma and because X, hence V is > quasi-compact. By Kolmogorov's axiom, any minimal non-void closed subset > of X is a singleton. Yes, this is the obvious method of proof (note that it is absolutely equivalent to the proof of the existence of maximal ideals for rings), which I suppose suggests that your axioms are the natural ones. Here's a thought that I had yesterday also. Consider a partial ordering on the points of X, given by x < y if x is in the closure of y (the specialization ordering, perhaps?). Equivalently, if y is in every neighborhood of x. T0 is a necessity here because otherwise this isn't actually a partial ordering; it says precisely that two points in the same neighborhoods as each other are equal. So let's require that X is T0. Any minimal element of this partial order is obviously a closed point, for otherwise its closure would contain another point, which would be smaller. Now, asking that every closed set contain a closed point is equivalent to asking that the closure of any point contain one (since we could produce such a point in a general nonempty closed set by looking at the closure of any point in it), and that is just asking that every point of this partial order lie above a minimal element. By Zorn's Lemma and the Hausdorff maximal principle (one for each direction, both of course equivalent), this is the same as asking that every descending chain be bounded below. Now, asking that any descending sequence of closures of points have a nonempty intersection is almost like asking for local quasi-compactness. Certainly it's true if this holds since then any point is contained in a quasi-compact closed set, hence its closure is. It very nearly implies that the closure of any point is itself quasi-compact; to verify this, all we need to do is verify that any descending chain of closed subsets of such a set has nonempty intersection. By taking irreducible components of the elements of the chain, we can reduce this to a question about descending irreducible closed subsets. This practically forces the next axiom: every closed irreducible subset of X is quasi-compact. This implies that every point lies above a closed point, and conversely, if we also assume that X is a Zariski space (every closed irreducible set has a generic point) then the latter implies that every irreducible set is quasi-compact. So, the minimal reasonable set of hypotheses on X in order that every closed set contain a closed point is: X is T0 and every closed, irreducible subset is quasi-compact. If X is Zariski then the two statements are equivalent. -- Ryan Reich ryan.reich@gmail.com === Subject: simple optimization Hi there, I have a function of 1 parameters that generate three curves. My goal is to optimize the parameter so that the 3 curves overlap as best as possible. I tried to minimize the residuals, but not to avail as the magnitude of the curve change with the parameters, hence it always converge to the biggest parameter. Does somebody have any ideas on how to do that? Dietmar === Subject: Re: simple optimization > Hi there, > I have a function of 1 parameters that generate three curves. My goal > is to optimize the parameter so that the 3 curves overlap as best as > possible. Just to clarify: do you have three functions f1(x,a), f2(x,a) and f3(x,a) that contain a common parameter, 'a', and produce threee graphs when you plot the f(x,a) against x for any fixed a? I will assume this is what you mean, but if not, please say so. OK, so now what do you mean when you say the curves overlap? Do the curves y = 0 and y = sin(x) overlap? Certainly, they cross one another frequently and never get too far apart, but is that what you mean? What is your definition of overlap? If you want to maximize the amount of overlap, don't you need to measure it in some way? Also, even if we can minimize the overlap between two curves, how do we extend that to three curves? > I tried to minimize the residuals, but not to avail as the > magnitude of the curve change with the parameters, hence it always > converge to the biggest parameter. I have no idea what this means. Do you mean that the curves grow farther apart as you increase the parameter? > Does somebody have any ideas on how to do that? You do need to explain the nature of your problem in more detail. R.G. Vickson > Dietmar === Subject: Quadratic that computers have a hard time with Hello all, related publication) which discussed a particular quadratic that computers were having trouble with. The basic idea was that the discriminant was very close to zero, and whether it showed as positive, negative, or zero depended on round-off errors in the computer. I'm not Matt === Subject: Re: Quadratic that computers have a hard time with > Hello all, > related publication) which discussed a particular quadratic that > computers were having trouble with. The basic idea was that the > discriminant was very close to zero, and whether it showed as positive, > negative, or zero depended on round-off errors in the computer. I'm not For reference suppose we have ax^2 + bx + c = 0 to solve. trick for obtaining best accuracy in real roots of a real quadratic. Compute one root using the sign in front of the square root that agrees with -b, so that in the numerator of the quadratic formula we combine two terms of the same sign (without cancellation). Then compute the other root by dividing c/a by the first one. Your description of the issues suggests however a concern with a discriminant that is so close to zero as to create an uncertainty about the actual sign of discriminant b^2 - 4ac. In this case we would know that the roots are nearly (if not exactly) equal, and we would know their approximate value accurately. What we would have difficulty with is knowing the difference between the two roots with accuracy. === Subject: Re: Quadratic that computers have a hard time with Have a look at Solving a Quadratic Equation on a Computer by George E. Forsythe, published in the book _The Mathematical Sciences_, MIT Press, 1969, pp138 to 152. === Subject: integer polynomials approximation I have the following question: We have a closed interval [a,b]. Can we approximate a continious on [a,b] function with polynomials from Z[X] ? Some cases: 1). if a,b both are integers, then obviously the neccessary condition is f(a), f(b) from Z. I know the proof using Bernstein polynomials of the fact that these condition are also sufficient. 2). If [a,b] doesn't contain any integers, i also know the proof. I'm interested what happens in other cases: for example if [a,b] does contain at least 1 integer and one of the a,b is not an integer. === Subject: Re: integer polynomials approximation > I have the following question: > We have a closed interval [a,b]. Can we approximate a continious on [a,b] function with polynomials from Z[X] ? > Some cases: > 1). if a,b both are integers, then obviously the neccessary condition is f(a), f(b) from Z. I know the proof using Bernstein polynomials of the fact that these condition are also sufficient. > 2). If [a,b] doesn't contain any integers, i also know the proof. > I'm interested what happens in other cases: > for example if [a,b] does contain at least 1 integer and one of the a,b is not an integer. Am I missing something? What is the range of these functions? How good does the approximation need to be? Suppose that (for any a, b), the function is constantly 1.5. What would be a good Z[X] approximation? -- Se.87n O'Leathl.97bhair === Subject: Re: integer polynomials approximation > I have the following question: > We have a closed interval [a,b]. Can we approximate > a continious on [a,b] function with polynomials from > Z[X] ? > Some cases: > 1). if a,b both are integers, then obviously the > neccessary condition is f(a), f(b) from Z. I know the > proof using Bernstein polynomials of the fact that > these condition are also sufficient. > 2). If [a,b] doesn't contain any integers, i also > know the proof. > I'm interested what happens in other cases: > for example if [a,b] does contain at least 1 > integer and one of the a,b is not an integer. > Am I missing something? What is the range of these > functions? How > good does the approximation need to be? The goodness of approximation is the usual one: for any epsilon >0 we can find a polynomial from Z[X] of degree n such that for all x from [a,b] : |f(x)-p(x)| <= epsilon. > Suppose that (for any a, b), the function is > constantly 1.5. What > would be a good Z[X] approximation? Ah, that's the point. Not every function can be approximated for all [a,b]. For example it is obvious that if [a,b] contains integers the NECCESSARY condition for f is that f( Z in [a,b] ) in Z. The question was, whether this condition is also SUFFICIENT. By the way, there are cases where we mustn't impose any restrictions on f: when [a,b] doesn't contain integers, every continious function can be approximated in above stated sense. === Subject: Re: integer polynomials approximation <31143260.1146837948729.JavaMail.jakarta@nitrogen.mathforum.org I have the following question: We have a closed interval [a,b]. Can we approximate > a continious on [a,b] function with polynomials from > Z[X] ? Some cases: > 1). if a,b both are integers, then obviously the > neccessary condition is f(a), f(b) from Z. I know the > proof using Bernstein polynomials of the fact that > these condition are also sufficient. 2). If [a,b] doesn't contain any integers, i also > know the proof. I'm interested what happens in other cases: > for example if [a,b] does contain at least 1 > integer and one of the a,b is not an integer. Am I missing something? What is the range of these > functions? How > good does the approximation need to be? > The goodness of approximation is the usual one: > for any epsilon >0 we can find a polynomial from Z[X] of degree n such that for all x from [a,b] : |f(x)-p(x)| <= epsilon. That is the obvious meaning of good but it just seemed that only a small class of functions could be so approximated by Z[X] polynomials. (See below) > Suppose that (for any a, b), the function is > constantly 1.5. What > would be a good Z[X] approximation? > Ah, that's the point. Not every function can be approximated for all [a,b]. For example it is obvious that if [a,b] contains integers the NECCESSARY condition for f is that f( Z in [a,b] ) in Z. The question was, whether this condition is also SUFFICIENT. > By the way, there are cases where we mustn't impose any restrictions on f: when [a,b] doesn't contain integers, every continious function can be approximated in above stated sense. I was about to ask: So if a = 0.25 and b = 0.75, what are some good approximations to my function. Then I was going what if a = 1.25 and b = 1.75. Then I clicked and realised how it could be done. I should have thought a bit harder before my first post. It would have helped if I had looked up Bernstein polynomial. Sorry. An interesting question. I'll think about it but I doubt that I will manage to help you before anyone else does. -- Se.87n O'Leathl.97bhair === Subject: Re: integer polynomials approximation one can prove that if |b-a|<4 and that f(Z) is included in Z then it also works. I will try to find an exact reference for this result .. Bye. === Subject: commuative algebra ? Hi all, I studied yesterday the Gelfand transformation on unital commutative algebra. (that is the links between maximal ideal, characters, spectrum, etc..). Nevertheless I cannot see where one uses the assumption that the Algebra is commutative ? fedor === Subject: Re: commuative algebra ? > Hi all, > I studied yesterday the Gelfand transformation on unital commutative > algebra. (that is the links between maximal ideal, characters, > spectrum, etc..). Nevertheless I cannot see where one uses the > assumption that the Algebra is commutative ? [...] By algebra, do you mean a C*-algebra? David Bernier === Subject: Re: commuative algebra ? no, I only meant commutative Banach algebra. Why ? is it absurd ? === Subject: Re: Prime Buddies I'd love to show y'all how smug and arrogant you are. However, since > your all both judge and jury, I can't find an appellate court. Submit your proof to a journal. They qualify as an appellate court. Your personal attacks [smug and arrogant] make you look like a crank. Your work has been reviewed by professional mathematicians and been found wanting. Whether you like this, that is how mathematics gets accepted: peer review. === Subject: Re: Prime Buddies >I don't think the twin prime problem is any older than (what we used >to call) Fermat's Last Theorem. That also resisted solution for >centuries. Yet, eventually we solved it, and without changing our >thinking - we just had to push our thinking farther and deeper than >we had previously done. >If history was hinting, it's a good thing Wiles didn't take the hint. > I actually don't like Wiles' solution either. Which may be of no > suprise to you. I think there is a more simple and more eloquent > solution available which has yet to be discovered. Do you also believe in the Tooth Fairy? > However, I grant that the Fermat theorem could be solved by standard > techniques, brilliantly applied. I strongly suspect that the prime > twin problem can't. The justification I use is in the way that Pat(n) > develops over time... recursively combining complexity upon complexity. > I think it's a new field of math: Randomics that is needed. Maybe you should first read up on what people already know about randomics. Have you read Knuth yet? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Prime Buddies <0NK5g.12543$yU5.8540@news.get2net.dk> <445783C5.E85BC85A@pat7.com> I actually don't like Wiles' solution either. Which may be of no >> suprise to you. I think there is a more simple and more eloquent >> solution available which has yet to be discovered. >Do you also believe in the Tooth Fairy? There are two camps on this matter. (this matter = What did Fermat mean by 'remarkeable proof' written in the margin next to this problem?) 1) Fermat thought he knew the answer but actually didn't. 2) Fermat knew the answer but didn't have time to clarify it and formulate it before his death. Of people in group 2) most don't believe that Fermat had in mind what Wiles had in mind. We'll never know. === Subject: Re: Prime Buddies >> I actually don't like Wiles' solution either. Which may be of no >> suprise to you. I think there is a more simple and more eloquent >> solution available which has yet to be discovered. >Do you also believe in the Tooth Fairy? > There are two camps on this matter. > (this matter = What did Fermat mean by 'remarkeable proof' written in > the margin next to this problem?) > 1) Fermat thought he knew the answer but actually didn't. > 2) Fermat knew the answer but didn't have time to clarify it and > formulate it before his death. > Of people in group 2) most don't believe that Fermat had in mind what > Wiles had in mind. > We'll never know. The people in group 2 believe in the Tooth Fairy. I guess we'll never know about her, either. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Prime Buddies <0NK5g.12543$yU5.8540@news.get2net.dk> <445783C5.E85BC85A@pat7.com> Wiles had in mind. > We'll never know. We don't know what Fermat had in mind but we do know it wasn't what Wiles had in mind. === Subject: Ramanujan and Fifth Power Identities Hello all, Ramanujan gave a set of identities, a^4 + b^4 + c^4 = 2d^(2k) which he claimed could be found for any positive integer k. It turns out we can generalize the above using analogous equations for _third_ and _fifth_ powers which can also be valid for any k. Details at: http://www.geocities.com/titus_piezas/ramanujan_page14.html --Titus === Subject: Re: Ramanujan and Fifth Power Identities > Ramanujan gave a set of identities, > a^4 + b^4 + c^4 = 2d^(2k) That is not an identity. > which he claimed could be found for any positive integer k. It turns > out we can generalize the above using analogous equations for _third_ > and _fifth_ powers which can also be valid for any k. Details at: > http://www.geocities.com/titus_piezas/ramanujan_page14.html Post them here. === Subject: Re: Another way of calculating the Lorentz factor > I have discovered that the Lorentz factor can also be calculated as > follows: > gamma = sqrt(beta * (1/beta + sum(beta^n))) > where > gamma is the Lorentz factor > beta is v / c > v is the velocity > c is the speed of light > n is an odd integer ranging from 1 to infinity I can't see how to reduce this to 1/sqrt(1-beta^2), and I don't know PC _____________________________________________ PeCh@MailAPS.org / Skype: PeterJChristensen Physical arguments & plain geometry -> LT : www.PeterChristensen.eu/phys/LT.htm (Under construction, only constructive comments, please) === Subject: Re: Another way of calculating the Lorentz factor > I have discovered that the Lorentz factor can also be calculated as > follows: > gamma = sqrt(beta * (1/beta + sum(beta^n))) > where > gamma is the Lorentz factor > beta is v / c > v is the velocity > c is the speed of light > n is an odd integer ranging from 1 to infinity > I can't see how to reduce this to 1/sqrt(1-beta^2), I gave the reduction (in excessive detail) at David === Subject: Re: Another way of calculating the Lorentz factor > I have discovered that the Lorentz factor can also be calculated as > follows: > gamma = sqrt(beta * (1/beta + sum(beta^n))) > where > gamma is the Lorentz factor > beta is v / c > v is the velocity > c is the speed of light > n is an odd integer ranging from 1 to infinity > I can't see how to reduce this to 1/sqrt(1-beta^2), and I don't know > PC > _____________________________________________ > PeCh@MailAPS.org / Skype: PeterJChristensen > Physical arguments & plain geometry -> LT : > www.PeterChristensen.eu/phys/LT.htm > (Under construction, only constructive comments, please) I can't remember exactly how I derived this equation. I know it was from a mistake in a calculation I was doing. I was using a spreadsheet. I was comparing the ratio of 2 wavelengths (Compton and de Broglie) undergoing length contraction, and the result was unexpected. I started to manipulate the result by multiplying (and the other operations) it by certain values, and I eventually got the expected result. I looked at the terms from my manipulation and realised that they were all beta to the power of an odd integer. Well it was something like that anyway, but unfortunately I deleted the results of the manipualtions, and corrected the original mistake. I just tried to repeat the process but failed, as I can't remember the original mistake. It should'nt be that hard to work out though, as it only involved the Compton wavelength, beta, and gamma. === Subject: Non-standard topology on R^2 Can anyone introduce a topology on R^2 so that R^2 is a linear topological space and is homeomorphic to R^2{(0,0)}? === Subject: Re: Non-standard topology on R^2 > Can anyone introduce a topology on R^2 so that R^2 is a linear topological space and is homeomorphic > to R^2{(0,0)}? As Lee pointed out, you can't have both of these. Since the first condition by itself is no biggie, let's go for the second. 1. Let f:R^2 ---> R^2 {(0,0)} (the punctured plane) be a bijection. 2. Define U in R^2 to be open iff f(U) is open in the punctured plane. Then f is a homeomorphism. Probably not exactly what you wanted, right? Dale === Subject: Re: Non-standard topology on R^2 <1NK6g.73238$H71.35454@newssvr13.news.prodigy.com Can anyone introduce a topology on R^2 so that R^2 is a linear topological space and is homeomorphic > to R^2{(0,0)}? > As Lee pointed out, you can't have both of these. You can, if the (linear) topology is not Hausdorff. Mate === Subject: Re: Non-standard topology on R^2 >Can anyone introduce a topology on R^2 so that R^2 is a linear topological space and is homeomorphic >to R^2{(0,0)}? I am certain that you mean the R^2 in so that R^2 is a linear topological space to be the retopologized R^2. I am not sure, however, whether you mean the R^2 in is homeomorphic to R^2{(0,0)} to be the standard R^2 or the retopologized R^2. In the first case (assuming that by a linear topological space you mean a topological vectorspace over R or C), then the answer is unequivocally no: a topological vectorspace is contractible, but the standard R^2{(0,0)} is not. In the second case (on the same assumptions), the answer is no if you also assume finite dimensionality, but sometime yes if you don't. The proof in the finite-dimensional case is the same. However, little-l-2 minus one point is homeomorphic to little-l-2, and both are equinumerous with R^2. I'm sure I'm assuming Choice wildly all over the place, and am agnostic, ignorant, and apathetic about the situation where Choice is not assumed or is denied. Lee Rudolph === Subject: Re: Non-standard topology on R^2 > Can anyone introduce a topology on R^2 so that R^2 is a linear topological space and is homeomorphic > to R^2{(0,0)}? The topology cannot be Hausdorff, because it would be the usual (Euclidean) one. So, take the indiscrete topology. Any bijection from R^2 to R^2 {0} is a homeomorphism. Mate === Subject: math problem I have difficulty in finding solution to this problem Start Elements: derivative[f(x)] = e^[(f(x)^2) - (x^2) ] f(0)=0 ToFind: the f(x) Please let me know your thoughts about possible solutions? === Subject: Re: math problem v1riish@hotmail.com nous a r.8ecemment amicalement signifi.8e : > I have difficulty in finding solution to this problem > Start Elements: > derivative[f(x)] = e^[(f(x)^2) - (x^2) ] > f(0)=0 > ToFind: > the f(x) > Please let me know your thoughts about possible solutions? You already had solution in previous post : f' = e^[f^2 - x^2] ==> f' e^(-f^2) = e^(-x^2) ==> erf(f) = erf(x) + c f(0) = 0 ==> c=0 ==> erf(f(x)) = erf(x) And, since erf(x) is an injection : f(x) = x -- patrick === Subject: Re: What is the meaning of life ? The meaning is up to the individual with no one correct answer. Our lives are an improbability. The cosmic sludge that is the core of all life, lucked out being on a planet, a specific distance that happened to orbit a specific sun which made life possible to exist. Even more spectacular is that life did come forth. You are a cosmic lottery ticket somehow meant to live despite the odds. Some chose meaning in abdicating authority over themselves to others that decide meaning. Some find meaning in living life like there is no tomorrow with endless laundry lists of goals, while others see endless tomorrows never accomplishing those things which were one day meant to be done. Again, meaning is up to the lottery ticket that hit. Ramona > Post your views. === Subject: Re: What is the meaning of life ? Since there can be no 'meaning' without consciousness (what is the value of anything without direct experience?), the meaning of life must revolve around consciousness. The question then, can be rephrased as: What is the telos (or directed function) of consciousness? Taking the very reasonable philosophical position of functionalism, consciousness appears to be a mathematical process - or computation of some kind. Based on the conjecture of neuroscientists Koch and Crick that consciousness is an 'executive summary' i.e qualia are symbols summarizing the mass of associations making up meaningful 'concepts', we can make a reasonable guess as to what kind of mathematical computation consciousness could be: Consciousness simply appears to be mathematics modelling itself i.e Consciousness is mathematics reflecting on mathematics. If mathematical entities are 'patterns', then consciousness consists of 'meta-patterns', or 'patterns within the patterns'. The up-shot of these reasonable guesses is that consciousness appears to be intimately connected with three things: (1) The creation of knowledge (2) The integration of knowledge (3) The self-reflectivity of knowledge Where 'knowledge' is defined as 'pattern' or assocation of universal applicability. The function of consciousness then is simply these three things: the creation of knowledge, the integration of knowledge and the self-reflectivity of knowledge. Since I argued that resolving the meaning of life is synonymous to resolving the function of consciousness, the meaning of life then, is precisely equivalent to the 3 aforementioned conditions. === Subject: Re: What is the meaning of life ? <2OL5g.949$VV2.72514@news20.bellglobal.com> Asking the question is missing it. >> Not possible. Ignoring maybe. >> What is the meaning of life? >> LMAO! > Achoo >> Seeking life's meaning >> In books peppered with concepts >> He sneezed out his soul. >> -- > LOL > If only that were possible... > Even so, it appears my love is not as arcane as yours. >> Appearances... >> -- >> > With what? >> -- >> ~Stumper > LOL Ask the one who asked the question to verify it for you. > Just look. >> -- >> ~Stumper > How does it see? > Someone trying to verify something with something else. > -- > ~Stumper Well, if it is so difficult for you, perhaps you shoud give it up. === Subject: Re: What is the meaning of life ? > > Asking the question is missing it. >> Not possible. Ignoring maybe. >> What is the meaning of life? >> LMAO! > Achoo >> Seeking life's meaning >> In books peppered with concepts >> He sneezed out his soul. >> -- > LOL > If only that were possible... > Even so, it appears my love is not as arcane as yours. >> Appearances... >> -- >> > With what? >> -- >> ~Stumper > LOL Ask the one who asked the question to verify it for you. > Just look. >> -- >> ~Stumper > How does it see? >> Someone trying to verify something with something else. >> -- >> ~Stumper > Well, if it is so difficult for you, > perhaps you shoud give it up. Who is there to give up what? Better just be aware. -- ~Stumper === Subject: Re: What is the meaning of life ? <2OL5g.949$VV2.72514@news20.bellglobal.com> Asking the question is missing it. >> Not possible. Ignoring maybe. >> What is the meaning of life? >> LMAO! > Achoo >> Seeking life's meaning >> In books peppered with concepts >> He sneezed out his soul. >> -- > LOL > If only that were possible... > Even so, it appears my love is not as arcane as yours. >> Appearances... >> -- >> > With what? >> -- >> ~Stumper > LOL Ask the one who asked the question to verify it for you. > Just look. >> -- >> ~Stumper > How does it see? > Someone trying to verify something with something else. >> -- >> ~Stumper > Well, if it is so difficult for you, > perhaps you shoud give it up. > Who is there to give up what? The who is The one that asked the question The what is The appendage that the who is grasping to > Better just be aware. If you knew how to see You would know it could not be otherwise > -- > ~Stumper === Subject: Re: What is the meaning of life ? <2OL5g.949$VV2.72514@news20.bellglobal.com> Asking the question is missing it. Not possible. Ignoring maybe. What is the meaning of life? LMAO! Achoo Seeking life's meaning > In books peppered with concepts > He sneezed out his soul. > -- LOL > If only that were possible... > Even so, it appears my love is not as arcane as yours. Appearances... > -- > > Do not believe ...verify! Love is stronger than Fear. > -- > > ...talking about love is missing it. Scary. > -- > You think. > What are we talking about? > -- > > LOL Exactly! === Subject: Discrete Hyperbolic Surface Model The following simple 3D model can be constructed using thick paper or plastic sheet to demonstrate continuous co-planarity around a saddle point. Four edges OP,OR,OQ and OS of four isosceles trapeziums with are placed and attached side by side using a cello tape making the total angle at saddle point O more than 360 degrees. There is symmetry about the plane ROS. Their circumcenters are at C1, C2, C3, and C4 which are co-planar by virtue of symmetry. The four trapeziums have equal value of circumradiii O-C1, O-C2, O-C3, O-C4 = a. However, they are not in general co-planar. An isosceles trapezium needs 3 dimensions (two parallel sides and parallel distance h between them). Instead (a, alpha, h) are taken here. I shall be much obliged for indication of a simple way to do the vector dihedral calculations to find relation between angles QOP and ROS in terms of (alpha, beta, h12/a, h34/a ratios), so that saddle point O also lies in the plane of (O-C1, O-C2, O-C3, O-C4). A surface so triangulated has an interesting local property: Any saddle point has its four neighbor points co-planar, but not so when a fifth non-neighbor is included. Hope the hyperbolic geometry exercise would provide some fun too. A sketch is at: http://img255.imageshack.us/img255/5543/saddlpt1en.jpg Narasimham G.L. === Subject: Tables of fractional order Bessel functions Does anyone know of a readily available published reference that provides tables of Bessel functions of the first kind of fractional (1/2, 3/2, 5/2, etc.) order? Excel (naturally) does not provide a canned function that yields values, so I need to use the closed form-type formulas for the first few, and after that, rely on a recursion formula. It would be helpful to have tabulated values to === Subject: Re: Tables of fractional order Bessel functions > Does anyone know of a readily available published reference that > provides tables of Bessel functions of the first kind of fractional > (1/2, 3/2, 5/2, etc.) order? Excel (naturally) does not provide a > canned function that yields values, so I need to use the closed > form-type formulas for the first few, and after that, rely on a > recursion formula. It would be helpful to have tabulated values to Handbook of Mathematical Functions by Abramowitz and Stegun, chapter 10 (Bessel Functions of Fractional Order). === Subject: Re: Looking for a Journal > I am looking for an appropriate journal, most likely > algebra, .................. > Any suggestions for a journal ......... > ............are welcome and greatly > appreciated. > Jonathan Groves The Web site of contains the following: 1. List of more than 500 printed journals of mathematics, at the address http://www.ams.org/mathweb/mi-journals5.html with internet sites containing tables of contents, abstracts, information about submissions and subscriptions, and in some cases, electronic versions of papers. 2. List of electronic mathematics research journals at the address http://www.ams.org/mathweb/mi-journals2.html 3. List of new journals of mathematics at the address http://www.ams.org/mathweb/mi-newjs.html Jean-Claude Evard Western Kentucky University Department of Mathematics E-mail: Jean-Claude.Evard@wku.edu === Subject: Re: Looking for a Journal According to Dr. Enochs here at UK, journals that publish original research papers that can be understood by good undergraduate math majors do exist since there are fairly many open problems and research questions that can be understood by good undergraduates. I will take Gerry's advice and Dr. Enochs' advice and try to search our library here at UK for suitable journals. However, you are certainly welcome to mention any such journals you know about. Jonathan Groves === Subject: Re: Looking for a Journal <32480547.1146843112123.JavaMail.jakarta@nitrogen.mathforum.org According to Dr. Enochs here at UK, journals that publish > original research papers that can be understood by good > undergraduate math majors do exist since there are fairly > many open problems and research questions that can be > understood by good undergraduates. > I will take Gerry's advice and Dr. Enochs' advice and > try to search our library here at UK for suitable journals. > However, you are certainly welcome to mention any such > journals you know about. On September 24, 2005 I posted a rather large collection of elementary and expository journals in sci.math that you might want to look through. I included internet links to information whenever I was able to find one. http://mathforum.org/kb/message.jspa?messageID=3974282 Another list is by Stanley Rabinowitz. I don't know if his list includes everything on my list (I did notice that many of those on my list that weren't on his list, when I made my post, later showed up on his list a couple of days after my post appeared), but I'm pretty sure his list includes some items that mine doesn't. However, his web page will take a long time to load if you're using a phone dial-up internet connection. http://www.mathpropress.com/elementaryJournals.html Dave L. Renfro === Subject: Question about orthogonal tensor maps Hi folks, I am interrested where could I find references about the Riccati-like IVP: (d/dt)R+omega*R=0_3, R(0)=I_3 where omega is a skew-symmetric tensor in so_3 and, of course, R is an orthogonal proper tensor in SO_3. Somebody told me this is called the Darboux equation or Darboux problem. I could not find any references by searching the www and I would be very grateful if someone could help me by offering some references to this IVP (that is fundamental in attitude kinematics). At least its real name... so I can find papers dealing with it. I am interested in what types of solutions may be found (I know for example that this equation has a time-explicit solution when vector omega* is constant, and also when it has fixed direction. Here vector omega* represents the 3-dimmensional vector associated to the skew-symmetric tensor omega). Vlad === Subject: Followup: degree of solutions of a *system* of polynomial equations? Thx for your reply! Also, to follow up on why Mathematica was taking forever to find the minimal polynomials for algebraic solutions returned by Solve, it was because I was just using RootReduce for the task. The functions in package NumberTheory`AlgebraicNumberFields` will do the same thing much faster. === Subject: numeri algebrici qualcuno mi sa dare una dimostrazione del seguente fatto : il campo dei numeri algebrici .8f algebricamente chiuso. vorrei sapere come dimostrarlo. grazie. === Subject: Re: numeri algebrici days. My association with the Department is that of an alumnus. >qualcuno mi sa dare una dimostrazione del seguente fatto : >il campo dei numeri algebrici =E8 algebricamente chiuso. >vorrei sapere come dimostrarlo. Don't know if you know enough English, but: If a is algebraic over the field of algebraic numbers, then there exists a polynomial with algebraic number coefficients such that a is a root. Let b_1,...,b_n be the coefficients. Then K=Q[b_1,...,b_n] is finite over Q, and by assumption K(a) is finite over K, so K(a) is finite over Q, hence a is algebraic over Q, hence a is in the field of algebraic numbers. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: numeri algebrici il campo dei numeri algebrici =E8 algebricamente chiuso. >vorrei sapere come dimostrarlo. > Don't know if you know enough English, but: > If a is algebraic over the field of algebraic numbers, then there > exists a polynomial with algebraic number coefficients such that a > is a root. Let b_1,...,b_n be the coefficients. Then K=Q[b_1,...,b_n] > is finite over Q, and by assumption K(a) is finite over K, so K(a) > is finite over Q, hence a is algebraic over Q, hence a is in the > field of algebraic numbers. I don't know enough about the notation to follow this. OP was asking for a proof that the algebraic numbers are algebraically closed, which I assume means closed under addition and multiplication (question 1)? What does Q[b_1,...,b_n] mean and how does this proof establish closure (question 2)? And it seems to me that I've seen a proof that a+b and a*b are algebraic by explicit construction of polynomials that have a+b and a*b as roots. Could you give a hint about that construction (question 3) and is the above actually equivalent to that (question 4)? - Randy === Subject: Re: numeri algebrici >>qualcuno mi sa dare una dimostrazione del seguente fatto : >>il campo dei numeri algebrici =E8 algebricamente chiuso. >>vorrei sapere come dimostrarlo. >> Don't know if you know enough English, but: >> If a is algebraic over the field of algebraic numbers, then there >> exists a polynomial with algebraic number coefficients such that a >> is a root. Let b_1,...,b_n be the coefficients. Then K=Q[b_1,...,b_n] >> is finite over Q, and by assumption K(a) is finite over K, so K(a) >> is finite over Q, hence a is algebraic over Q, hence a is in the >> field of algebraic numbers. >I don't know enough about the notation to follow this. >OP was asking for a proof that the algebraic numbers >are algebraically closed, which I assume means closed >under addition and multiplication (question 1)? closed means that every nonconstant polynomial has a root; or equivalently, that every root of a nonconstant polynomial with coefficients in the field is already in the field. If he meant is closed under the algebraic operations, which is what you are asking here, then of course I did not address that. That would require you to show that a+b is algebraic when a and b are algebraic, and that ab is algebraic when a and b are. > What >does Q[b_1,...,b_n] mean and how does this proof establish >closure (question 2)? Q[b_1,...,b_n] is the smallest subring of some ring containing Q which contains b1, ..., bn. The ring of all polynomial expressions in b_1,...,b_n with coefficients in Q. It is equal to the smallest subfield of C that contains Q and contains b_1,..,b_n. under multiplication: it addresses algebraic closure. That any number which is algebraic over the algebraic numbers is itself an algebraic number. > And it seems to me that I've seen >a proof that a+b and a*b are algebraic by explicit construction >of polynomials that have a+b and a*b as roots. Could >you give a hint about that construction (question 3) and >is the above actually equivalent to that (question 4)? No, the above is not equivalent to that; it addresses a completely different question. As to the constructions, it's been outlined before. Here is an explicit construction from Niven, Zuckerman, Montgomery, Introduction to the Theory of Numbers fifth Edition, Theorem 9.12 Suppose a satisfies a^m + r_{m-1} a^{m-1} + ... + r_1 a + r_0 = 0 and b satisfies b^n + s_{n-1} b^{n-1} + ... + s_1 b + s_0 = 0. Let k = n*m. Define the complex number c1,...,ck to be the numbers a^i*b^j, 0<=i qualcuno mi sa dare una dimostrazione del seguente fatto : >>il campo dei numeri algebrici =E8 algebricamente chiuso. >>vorrei sapere come dimostrarlo. >> Don't know if you know enough English, but: >> If a is algebraic over the field of algebraic numbers, then there >> exists a polynomial with algebraic number coefficients such that a >> is a root. Let b_1,...,b_n be the coefficients. Then K=Q[b_1,...,b_n] >> is finite over Q, and by assumption K(a) is finite over K, so K(a) >> is finite over Q, hence a is algebraic over Q, hence a is in the >> field of algebraic numbers. >I don't know enough about the notation to follow this. >OP was asking for a proof that the algebraic numbers >are algebraically closed, which I assume means closed >under addition and multiplication (question 1)? > closed means that every nonconstant polynomial has a root; He definitely said algebraically closed. So I'm mistranslating the ENGLISH. And after I posted I realized he couldn't have been asking what I thought, since he said the field of algebraic numbers. > equivalently, that every root of a nonconstant polynomial with > coefficients in the field is already in the field. OK. That's another result I've heard quoted but would be interested in the proof, so I'm still interested in the translation of your proof. > If he meant is closed under the algebraic operations, which is what > you are asking here, then of course I did not address that. That would > require you to show that a+b is algebraic when a and b are algebraic, > and that ab is algebraic when a and b are. > What >does Q[b_1,...,b_n] mean and how does this proof establish >closure (question 2)? > Q[b_1,...,b_n] is the smallest subring of some ring containing Q which > contains b1, ..., bn. The ring of all polynomial expressions in > b_1,...,b_n with coefficients in Q. It is equal to the smallest > subfield of C that contains Q and contains b_1,..,b_n. > under multiplication: it addresses algebraic closure. That any number > which is algebraic over the algebraic numbers is itself an algebraic > number. > And it seems to me that I've seen >a proof that a+b and a*b are algebraic by explicit construction >of polynomials that have a+b and a*b as roots. Could >you give a hint about that construction (question 3) and >is the above actually equivalent to that (question 4)? > No, the above is not equivalent to that; it addresses a completely > different question. OK. Sorry. > As to the constructions, it's been outlined before. Here is an > explicit construction from Niven, Zuckerman, Montgomery, Introduction > to the Theory of Numbers fifth Edition, Theorem 9.12 > Suppose a satisfies > a^m + r_{m-1} a^{m-1} + ... + r_1 a + r_0 = 0 > and b satisfies > b^n + s_{n-1} b^{n-1} + ... + s_1 b + s_0 = 0. > Let k = n*m. Define the complex number c1,...,ck to be the numbers > a^i*b^j, 0<=i ac = a^{u+1}b^v = some c_w if u+1 ac = a^{u+1}b^v = -(r_{m-1}a^{m-1}+...+r_1 a + r_0)b^t if u=m-1. > In either case, we have rational constants h_{j,1},...,h_{j,k} such > that > ac_j = h_{j,1}c_1 + ... + h_{j,k}c_k. > There are also rational constants w_{j,1},..., w_{j,k} such that > bc_j = w_{j,1}c_1 + ... + w_{j,k}c_k. > So > (a+b)c_j = (h_{j,1}+w_{j,1})c_1 + ... + (h_{j,k}+w_{j,k})c_k > These latter equations can be thought of as a system of homogeneous > linear equations in c_1,...,c_k with coefficients in Q; the c_i are > not all zero, and the characteristic polynomial will have a+b as an > eigenvalue, hence is a root of the characteristic polynomial. Ah. This is one reason why I asked. I saw someone give a linear algebra-based proof and I was trying to remember the more elementary proof, unsuccessfully. I didn't realize they're the same proof. > We have > ab*c_j = a(w_{j,1}c_1 + ... + w_{j,k}c_k) > = w_{j,1}(ac_1) + ... + w_{j,k}(ac_k) > so substituting into each ac_i we get again equations for ab*c_j, > leading to a matrix whose characteristic polynomial has ab as a root. Long ago I attempted to learn number theory by signing up for a course taught from the book by Takashi Ono, but I found I didn't have the vocabulary to understand the class discussions and eventually dropped the course. Perhaps it's time for another look at Ono to see if I can learn this on my own. - Randy === Subject: Re: numeri algebrici A field F is algebraically closed if every polynomial with coefficients in F has all its roots in F (question 1). === Subject: Godel's Theorem in ZF(C) Is there an equivalent of Godel's proof for his second incompleteness theorem using ZF(C) axioms and notation? I tried reading the original proof, but it uses the PM notations and system which all looks like cuneiform to me. On the other hand, I'm familiar with ZF(C) so I'd rather not make the effort to learn PM because of the redundancy. Any ideas? === Subject: Need Help.. I am Disable right from the beginning of my childhood from Polio which affected my both legs. I need your onlyyyyyy 1 CLICK on my URL for MY http://www.ozwebresources.com/idevaffiliate/idevaffiliate.php?id=269 === Subject: semilinear transformations I'm trying to understand something about the group $Gamma$L(n,q) - so this is the semidirect product of GL(n,q) with the cyclic group C_k where q=p^k. I'm trying to figure out whether if x is in $Gamma$L(n,q) has prime order r, and x does not lie in GL(n,q) - can we conjugate by something in GL(n,q) to get something in C_k? === Subject: Re: A theorem of plane geometry schrieb Gene : > Given a chord of circle, not a diameter, the pole of the chord is the > intersection of the tangents to the endpoints of the chord. The theorem > I have in mind is this: given two nonintersecting (including on the > boundry) chords, the line drawn between the poles of the chords > intersects the circle (defining another chord.) Even more is true, and well-known in plane geometry: The line connecting the poles of two lines in the plane is the polar of the point of intersection of these lines. - > Does this theorem have a name, or is it any kind of recognized theorem > or lemma of plane geometry? In high-brow terms you'd say that this is the fact, that the map induced by the polarity relation with respect to a circle is an *involutive correlation*. === Subject: mistake in book probability book? I've been working a little through Probability by A. N. Shiryaev 2nd ed. On page 30 he talks about the probably of a what I call a basis A_k. A_k = {w: a_k = 1} where w = (a_1,...,a_n), a_k in {0,1} and represents the outcome of the n trials with each having 2 possibilities. (a bernoulli trial) Obviously A_k^c = {w: w_k = 0} and if the probability of a success(a_k = 1) is p, then P(A_k) = p and P(A_k^c) = q. (i.e., the probability of the kth trial being a success or failure) Now he states and a similar caculation shows that P(A_k^c) = q and that, for k != 1, P(A_k A_l) = p^2, P(A_k A_l^c) = pq, P(A_k^c A_l) = q^2 and its easy to deduce from this that AA_k and AA_l are independent[algebras] for k != l. The problem is that I see in no way how he got those probabilities P(A_k A_l) = p^2, P(A_k A_l^c) = pq, P(A_k^c A_l) = q^2 It would seem by symmetry alone would would have (assuming first two are correct) P(A_k A_l) = p^2, P(A_k A_l^c) = pq, P(A_k^c A_l) = pq and P(A_k^c A_l^c) = 1 - p^2 i.e., since P(A_k^c A_l^c) = P((A_k U A_l)^c) = 1 - P(A_k U A_l) = 1 - P(A_k) - P(A_l) + P(A_k A_l) = 1 - p - p + p^2 = p^2 - 2p + 1 = q^2 and P(A_k^c A_l) = P((A_k U A_l^c)^c) = 1 - P(A_k U A_l^c) = 1 - (P(A_k) + P(A_l^c) - P(A_k A_l^c) = 1 - P(A_k) - P(A_l^c) + P(A_k A_l^c) = 1 - p - q + pq = pq Is this right? I can't find any flaws with it but seems it would be a blatent error in this book. I'm suspicious of other potential errors too(I chaulked them up to my ignorance but maybe I was wrong). I'm wondering if this book is worth working through(as I'm just up to page 30 and probably found about 5 potential errors). It seems like a pretty thorough book and is relatively complicated for me(I have to work through it slow since I forgot alot of the basic stuff). Just that sometimes when I work through something it just makes absolutely no sense how he got the result. Most of the time I just feel he used some trick that I forgot or can't see but in this case I have to bet he's wrong because I can't see how it could violate the symmetry of the problem like that. Most likely he ment P(A_k^c A_l^c) = q^2 but forgot the ^c on one of them. Seems thats how I can explain most of errors I've seen(some are just typos too). Anyways, Jon === Subject: number theoretical question This one is for those who are much more informed on number theoretical questions: consider the Diophantine equation p^q-r^s=1, where p,q,r,s are primes. How far is this from the Catalan equation (solved recently) a^n-b^m=1 ? I mean: can the Catalan equation be reduced to the indicated one? If yes, I have a follow up. If no, I'd like to go to bed. And since I was diagnosed with insomnia , you guess what I would like to read as an answer... === Subject: Re: Calculus XOR Probability <81550$44585bbb$82a1e228$25054@news1.tudelft.nl> instead were concentrating of reverse-engineering the results to correct them, > or something. Did I dream that? Maybe you should read more about Bottom Up versus Top Down in: http://hdebruijn.soo.dto.tudelft.nl/QED/klassiek.htm Han de Bruijn === Subject: Re: Calculus XOR Probability <81550$44585bbb$82a1e228$25054@news1.tudelft.nl> instead were concentrating of reverse-engineering the results to correct them, > or something. Did I dream that? > Maybe you should read more about Bottom Up versus Top Down in: > http://hdebruijn.soo.dto.tudelft.nl/QED/klassiek.htm That's the wrong web page. Must be: http://hdebruijn.soo.dto.tudelft.nl/QED/index.htm Han de Bruijn === Subject: Re: Calculus XOR Probability <8764ko7o4w.fsf@phiwumbda.org> <6d8e4$44585cd6$82a1e228$25107@news1.tudelft.nl> about Tony's infinite numbers? Can you describe what Big'un is? Has he learned anything about > your 'aftermath'? Can he define 'aftermath'? I have learned that it's impossible to idealize the probability of selecting an arbitrary natural from {1,2,3, ... ,n} to all naturals. I'm sure that Tony has learned a lot in those heated debates as well. I cannot describe what Big'un is. Tony has learned some about my 'aftermath': > Distant objects may appear to be stationary, even when moving at great speeds. > When you are far, far behind, you are in no place to judge the progress of those ahead > of you. Yeah, sure. How much ahead of me you all are. Especially in 'sci.math'. Han de Bruijn === Subject: Re: Calculus XOR Probability >> What are you learning Han? What do you think Tony has learned? Have you learned anything >> about Tony's infinite numbers? Can you describe what Big'un is? Has he learned anything about >> your 'aftermath'? Can he define 'aftermath'? > I have learned that it's impossible to idealize the probability of > selecting an arbitrary natural from {1,2,3, ... ,n} to all naturals. Which is something that nearly everybody else already knew. Claiming some sort of superiority because you have learned something that everybody else already knew comes off as a bit odd, if not downright kooky. Distant objects may appear to be stationary, even when moving at great speeds. >> When you are far, far behind, you are in no place to judge the progress of those ahead >> of you. > Yeah, sure. How much ahead of me you all are. Especially in 'sci.math'. > Han de Bruijn Well you have now learned something that everybody else already knew, so I suppose you are catching up. Of course that is assuming that we all have not learned some new stuff in the meanwhile, and I doubt that is a safe assumption. Stephen === Subject: Re: Calculus XOR Probability <8764ko7o4w.fsf@phiwumbda.org> <6d8e4$44585cd6$82a1e228$25107@news1.tudelft.nl> What are you learning Han? What do you think Tony has learned? Have you learned anything >> about Tony's infinite numbers? Can you describe what Big'un is? Has he learned anything about >> your 'aftermath'? Can he define 'aftermath'? > I have learned that it's impossible to idealize the probability of > selecting an arbitrary natural from {1,2,3, ... ,n} to all naturals. > Which is something that nearly everybody else already knew. > Claiming some sort of superiority because you have learned > something that everybody else already knew comes off as > a bit odd, if not downright kooky. Since everybody else seems to know everything ... How about infinitesimals? Do they belong to mathematics or not? Han de Bruijn === Subject: Re: Calculus XOR Probability cbrown@cbrownsystems.com said: >> So, I feel justified, until I see a good counterexample, that >> equality between quantitative expresseions proven inductively hold in the >> infinite case. > So, your idea of a mathematical argument is: n*(n+1)/2 is the sum of > the naturals from 1 to n, because you haven't seen a good > counter-example? > I think you are misreading Tony here. Tony is claiming that > because 1 + 2 + .... n = n*(n+1)/2 when n is finite, it is also > true when n is infinite, and he will continue to believe in > general that any equality that holds for the finite case > also holds in the infinite case until you provide a counter example > where it is clear that the equality does not hold in the infinite case. > Well, one might then ask: why does he believe it in the finite case to > /start/ with? Because he read it in a book? Or because it can be proven > by a mathematical argument? I thought about it in two ways. First: 1. for n=1, sum(x=1->n: x)=1 = f(1) 2. n(n+1)/2 + (n+1) = n(n+1)/2 + 2(n+1)/2 = (n+2)(n+1)/2 = f(n+1) 3. Therefore, for all n in n, sum(x=1->n: n)=n(n+1)/2 Second, list the naturals, in unary. 1 11 111 1111 11111 111111 ...... See how it forms a diagonal half of a square? The side of that square is n, and the area is n^2, so the 1's cover half of that, or n^2/2. But, the 1's also cover the diagonal, half of which belongs to the other side, containing n 1's. So, we subtract half of those n 1's from n^2/2 and get n^2/2-n/2 = (n^2-n)/2 = n(n+1)/2. > Similarly, I ask: why should I believe him when he says that it holds > when n is infinite (whatever that may mean)? Because he read it in a > book? Or because it can be proven by a mathematical argument? I can assert that inductive proof of equalities holds in the infinite case, but you don't believe that's true. So, I offer this idea, as a visualization for you. COnsider that each of those 1's is an infinitesimal square, and that you have an infinite number of them covering half a square. If there are n rows of such squares and n columns, there are n^2 overall, and the set covers half the square. It's harder to see the additional n/2 elements in this picture, because n/2 is infinitesimal compared to n^2/2, but you can see it in the example above. The diagonal line, when included in the area of the triangle, increases its are infinitesimally. -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Matt Gutting said: > Virgil said: > > > If f(2) and g(2) are both defined, how can this be unless 2 is in the >> domain of both? > > Unless the set generated by applying f to the naturals contains 2, 2 will > not > be a value used as a parameter to g. It need not be in S, which is the > domain > of g. However, this means g(2) is not a natural. It's still a real > number. :) > > If 2 is not in S, and S is the domain of g, then g(2) is not defined. A > function > is only defined over its domain. > > Matt > > > The domains and codomains of the functions f and g are R. That means f(x) for > x > in R returns a y in R as a result. If N is a proper subset of R then S will > also be a proper subset of R, because for every unique r in R but not in N > there is a unique f(r) in R but not in S. g(2) will always be defined, but > may > not be in N. If that's the case, then 2 isn't in S. That's very basic. > Doesn't > anyone get this? I know I'm not crazy. Am I dreaming? Tony knows so many things that are no so, perhaps he is dreaming. === Subject: Re: Calculus XOR Probability > Virgil said: > > Distances between points depends only on their locations, so that > is all one has to pay any attention to. > But, when two points have locations that cannot be distinguished, how > do you derive direction from the set of two points? You don't, because you don't have two points unless you can distinguish between them. > > The length of a broken or polygonal line depends only on the > location of of the joints and not at all on the direction of the > segments between joints. > What is the slope of the staircase at x=0. It's infinite. Why does TO seem to think that changing the direction between points has any effect on the distance between them? He must never have come across circles. Does that > ever change as the staircase approaches the limit? No. That point at > x=0 always has slope infinite, as opposed to the slope of the > diagonal. The direction between 0,0 and 0,1/n never changes. But that direction is irrelevant to the total length. One can straighten out the staircase without changing its. > > So that something that only takes into account the locations of > those joints includes everything needed. > As long as those locations are ON the curve, yes. But they need not be at any particular points of the curve. In particular to approximate the length of the staircse, one need not take a point at any of the corner points of the staircase, and the LUB of such approximations will still be the correct length. > > No, the locations of the staircase in the limit are > indistinguishable from the locations of the points on the > diagonal. > > Then, based on those LOCATIONS, they have the same lengths. > No, that depends on direction. Not in mathematics, it doesn't. In TOmatics, anything may go, but mathematics is much stricter, things must make sense. > I don't think you can ever get this. > When you reduce the difference in locations down to an infinitesimal > level, they appear to be the same object, but the difference in > direction does not also have a limit of 0, but is unchanged. In the limit, the horizontal and vertical directions disappear, or get averaged out to 45 degree directions (TO has long argued that 0*oo = 1, but now when that appears to be the case, a zero slope combined with an infinite slope gives a slope of 1), TO now rejects his own claim. > > > * Do you mean that it fails to have some property which is > mathematically required before one is allowed to use the word > limit? If so, what property is that? > > No, it fails to have a property that is required to use that > limit to measure the diagonal. > > WRONG! Only the locations of the joints if any, are relevant, and > in the limit there are no longer any joints except the endpoints of > the entire diagonal. > And the locations on the joints are supposed to be point ON THE CURVE > to ensure parallel orientation and proper measure. Only if directions were being measured, which they are not. When only distances are being measured, only distances count. > > In other words, each element must have a point parallel to the > curve in order for the measure to work. > > That may be a TO requirement, but it is not a mathematical one. The > only mathematical requirement for a curve to have a length is the > the lengths of its polygonal approximations have a finite upper > bound, and then that upper bound is then defined to be the length. > > If the endpoints are on the curve. Actually all the points of the polygonal approximation are required to be on the curve, and taking the LUB of lengths assures that the endpoints of the curve are included in some of them. > > So, where your type of limit does NOT measure direction, it > cannot be used as a measure of a curve. Surely you can understand > that. > > Why should we understand what is not true? There is only one way, > mathematically, to measure the length of a curve, as has been > described above. In many, but not all, instances it reduces to > evaluating a definite integral > Yeah, and that works fine, when the points are on the curve. One takes a finite sequence of points along the curve, finds the metric distance between consecutive points of that sequence, and adds up those distances. That is polygonal approximation to the length of the curve. If the set of all such polygonal approximations is bounded, then its least upper bound is, by definition, the length of that curve. Any measure of the length of a curve in a metric space that does not agree with this result is wrong. > > Uh, if length is not a type of measure, then I don't know what > is. > > Arc length is the upper limit, provided that limit exists, on the > lengths of polygonal approximations to the curve. > And that would be some real value associated with the curve, a > measure of it. See above! > > I mean, take the formula for the one, and perform algebraic > operations on it to transform it into the formula for the other. > Derive it formulaically. > > Been there, done that. > No you haven't. Stop lying. Let the one step staircase go from (0,0) up to (0,1) over to (1,1). The n-step stair then goes from (0,0) to (0,1/n) to (1/n,1/n) to (1/n,2/n) and so on to ((n-1)/n,(n-1)/n) to ((n-1)/n,1) to (1,1). Each is of length exactly 2. The n step stair lies between lines y = x and y = x + 1/n, though with points on both lines. As n -> oo, line y = x + 1/n has line y = x as its limit, and any points caught in between get squeezed onto line y = x. So that the limit of the steps is the diagonal. === Subject: Re: Calculus XOR Probability Virgil said: > Virgil said: > > Distances between points depends only on their locations, so that > is all one has to pay any attention to. > > But, when two points have locations that cannot be distinguished, how > do you derive direction from the set of two points? > You don't, because you don't have two points unless you can distinguish > between them. Very good. So, if the neighboring points cannot be distinguished, and change in direction at that level is lost. > > > The length of a broken or polygonal line depends only on the > location of of the joints and not at all on the direction of the > segments between joints. > > What is the slope of the staircase at x=0. It's infinite. > Why does TO seem to think that changing the direction between points has > any effect on the distance between them? He must never have come across > circles. I ahve come across them and around them. Is the diameter equal to half the circumference? It's shorter? COuld that have anythign to do with NOT changing direction? > Does that > ever change as the staircase approaches the limit? No. That point at > x=0 always has slope infinite, as opposed to the slope of the > diagonal. The direction between 0,0 and 0,1/n never changes. > But that direction is irrelevant to the total length. > One can straighten out the staircase without changing its. I assume you meant length. That's true, even in the limit. > > > So that something that only takes into account the locations of > those joints includes everything needed. > > As long as those locations are ON the curve, yes. > But they need not be at any particular points of the curve. No, but they need to be somewhere on the curve. > In particular to approximate the length of the staircse, one need not > take a point at any of the corner points of the staircase, and the LUB > of such approximations will still be the correct length. > > No, the locations of the staircase in the limit are > indistinguishable from the locations of the points on the > diagonal. > > Then, based on those LOCATIONS, they have the same lengths. > > No, that depends on direction. > Not in mathematics, it doesn't. In TOmatics, anything may go, but > mathematics is much stricter, things must make sense. > I don't think you can ever get this. > When you reduce the difference in locations down to an infinitesimal > level, they appear to be the same object, but the difference in > direction does not also have a limit of 0, but is unchanged. > In the limit, the horizontal and vertical directions disappear, or get > averaged out to 45 degree directions (TO has long argued that 0*oo = 1, > but now when that appears to be the case, a zero slope combined with an > infinite slope gives a slope of 1), TO now rejects his own claim. It appears that way, but they never get any more parallel to the diagonal. I think yours is a false assumption. > > > > * Do you mean that it fails to have some property which is > mathematically required before one is allowed to use the word > limit? If so, what property is that? > > No, it fails to have a property that is required to use that > limit to measure the diagonal. > > WRONG! Only the locations of the joints if any, are relevant, and > in the limit there are no longer any joints except the endpoints of > the entire diagonal. > > And the locations on the joints are supposed to be point ON THE CURVE > to ensure parallel orientation and proper measure. > Only if directions were being measured, which they are not. When only > distances are being measured, only distances count. So, you measure with your ruler at a 45 degree angle to what you're measuring? > > > In other words, each element must have a point parallel to the > curve in order for the measure to work. > > That may be a TO requirement, but it is not a mathematical one. The > only mathematical requirement for a curve to have a length is the > the lengths of its polygonal approximations have a finite upper > bound, and then that upper bound is then defined to be the length. > > > If the endpoints are on the curve. > Actually all the points of the polygonal approximation are required to > be on the curve, and taking the LUB of lengths assures that the > endpoints of the curve are included in some of them. > > > So, where your type of limit does NOT measure direction, it > cannot be used as a measure of a curve. Surely you can understand > that. > > Why should we understand what is not true? There is only one way, > mathematically, to measure the length of a curve, as has been > described above. In many, but not all, instances it reduces to > evaluating a definite integral > > Yeah, and that works fine, when the points are on the curve. > One takes a finite sequence of points along the curve, finds the metric > distance between consecutive points of that sequence, and adds up those > distances. That is polygonal approximation to the length of the curve. > If the set of all such polygonal approximations is bounded, then its > least upper bound is, by definition, the length of that curve. > Any measure of the length of a curve in a metric space that does not > agree with this result is wrong. > > > Uh, if length is not a type of measure, then I don't know what > is. > > Arc length is the upper limit, provided that limit exists, on the > lengths of polygonal approximations to the curve. > > And that would be some real value associated with the curve, a > measure of it. > See above! > > > I mean, take the formula for the one, and perform algebraic > operations on it to transform it into the formula for the other. > Derive it formulaically. > > Been there, done that. > > No you haven't. Stop lying. > Let the one step staircase go from (0,0) up to (0,1) over to (1,1). > The n-step stair then goes from (0,0) to (0,1/n) to (1/n,1/n) to > (1/n,2/n) and so on to ((n-1)/n,(n-1)/n) to ((n-1)/n,1) to (1,1). > Each is of length exactly 2. > The n step stair lies between lines y = x and y = x + 1/n, though with > points on both lines. > As n -> oo, line y = x + 1/n has line y = x as its limit, > and any points caught in between get squeezed onto line y = x. > So that the limit of the steps is the diagonal. Yes, I have seen that argument, but you did not symbolically state the formula for the staircase, and then symbolically derive the formula for the diagonal from it. That I would consider proof, as long as it's valid. -- Smiles, Tony === Subject: Re: Calculus XOR Probability cbrown@cbrownsystems.com said: > cbrown@cbrownsystems.com said: > using it to measure distance. Distance and length are real numbers. > Is lim n->oo {C_n} a real number, or is it a set of points? > It's a sequential set of points, that is a line of some sort, with a real > measure called length. No, it's not. {C_n} is a /sequence/ of /sets of points/. A sequence of elements from some set X is essentially a function f : N -> X. We write f_n instead of f(n), so we don't get confused and think that f(3/2) might have some meaning. We then write {f_n} to indicate the whole function, rather than its value at some particular n. (lim n->oo {C_n}) is defined to be a particular /set of points/. There is no requirement that this limit be a line of some sort, with a real measure called length. For example, if C_n = {(x,y): x^2 + y^2 <= 1/n + 1}, the limit of this sequence the unit disk: {(x,y): x^2 + y^2 <= 1}. That, to the best I can determine, is not a line of some sort, with a real measure called length: it is simply a set of points. These facts can be deduced from the /definition/ of limit of a sequence of sets of points that I gave previously. I don''t know what you mean by a line of some sort, with a real measure of length, so I can't otherwise comment on whether or not your statement is correct or incorrect; it is meaningless to me, mathematically speaking; although as regular English usage it seems contradicted by the example I just gave. Can you provide a definition of what it means for a set C to be a line of some sort, with a real measure of length? Can you /then/ show how /my definition/ of lim n->oo {C_n} satisifies /your definition/ of a line of some sort, with a real measure of length? If you can, then I would accept that the lim n->oo {C_n} is a line of some sort with a real measure of length. If you can't then I'm afraid your statement is meaningless to me, mathematically speaking. I meant that lim n->oo {C_n} is a set of points; that's why I defined it that way, and not some other way. > To measure distance or length of a set of points in R^2 with the usual > metric, I use the usual approach, just like everyone else. > That's good, because this approach doesn't work. When you prove the length in > the limit is 2 instead of sqrt(2), that's an indication that the object in the > limit is NOT exactly the same as the diagonal line... Well, since this point still seems unclear to you, I will amplify on all the definitions used in my argument: Forget the diagonal line for the moment. Forget figures in the x/y plane. These phrases are unneccesary to my argument, and are confusing you by their usual connotations. By R I mean the real numbers - not the real numbers + infinitesimals + infinite numbers. By R^2, I mean the set of all ordered pairs (a,b) where a and b are real numbers. I will call these pairs in R^2 or simply pairs. By a set of pairs in R^2 I mean, well, a set of pairs in R^2. If A and B are two sets of pairs in R^2, then A = B if, and only if, for every pair p, p in A if and only if p in B. If A = B, and f is any function that takes a set of pairs in R^2 to a real number, then f(A) = f(B). Thus if A = B, A is identical to B in every way except by name. By a metric on R^2, I mean that I define a function dist(p,q) for all pairs p = (p_a, p_b) and q = (q_a, q_b) in R^2 by: dist(p,q) = sqrt((p_a-q_a)^2 + (p_b-q_b)^2). Thus dist is a function from two pairs in R^2, that yields a real number. I observe the following, which you are free to prove me wrong about: * dist(p,q) = 0 if, and only if, p = q. * dist(p,q) = dist(q,p) for all pairs p, q. * For all pairs p,q, and r, dist(p,q)+dist(q,r) >= dist(p,r). Thus, the function dist is a (toplogical) metric by the usual definition: it satisfies the three conditions above. That makes R^2, with reference to the function dist, a metric space as defined by the metric dist. (One could define other functions which would also be a metric on R^2. By the usual metric in R^n for any n, I mean that the dist function we want to use is the same as euclidean distance in R^n between the two points p and q). ----------------------- What do I mean by C is a curve? Suppose C is a set of pairs in R^2. Call such a set a curve if there is a function f: R -> R^2 such that the point (1) the pair f(x) is in C for all x in R; (2) for all pairs p in C, there exists a unique x in R such that p = f(x); which I will write as f^-1(p). (2) if e is a real number greater than 0, and p is a point in C, then there exists a real number d such that if q is a pair in C, and dist(p, q) < e, then |f^-1(p) - f^-1(q)| < d. (This is what is I mean by there exists a continuous function f from the real line to C.) ------------------------ Let C be the set of pairs defined by {p = (a,b) : b = 1 - a}. Is C a curve? Well, let f be the function defined as f(x) = (x, 1-x). (1) Is f(x) in C for all x in R? (2) If p is a point in C, is there a unique x in R with f(x) = p? (3) if e is a real number greater than 0, and p is a point in C, is there a real number d such that for all q in C with dist(p,q) < e, it is always the case that |f^-1(p) - f^-1(q)| < d? The answer to each of the above is yes (feel free to prove me wrong); so C is a curve, because the function f exists and satisfies the three rules I gave in the definition of C is a curve. -------------------------- Now suppose we define a segment C_[u,v] of a curve C as the subset of C defined by {p : p in C and u <= f^-1(p) <= v}. Let C be the set of pairs defined by {p = (a,b) : b = 1 - a}; and let D = C_[0,1]; then D = {p = (a,b) : b = 1-a, a,b>=0} (feel free to prove me wrong). -------------------- If D = C_[u,v] is a segment of a curve C, then define the length of D as follows: (1) Call a finite sequence of n reals {a_0, a_1, ..., a_(n-1)} a joint set if a_0 = u, a_(n-1) = v, and a_i < a_j for all i, j. So a joint set is basically a finite set of points in [0,1] that includes 0 and 1; and where the points in the sequence are in order. For example if [u,v] = [0,1], (0, 1/2, 3/4, 7/8, 1) is a joint set, but (0, 3/4, 1/2, 7/8, 1) (2) Recall that because weare saying that C is a curve, there is a continuous function f : [0,1] -> D. Define the joint length of a joint set J = {a_0, a_1,..., a_(n-1)} as the sum dist(f(a_0), f(a_1)) + dist(f(a_1), f(a_2)) + ... + dist(f(a_(n-2), f(a_(n-1)). (3) Define the length of D as the /smallest/ real number which is /greater than/ or equal to the joint length of any joint set. ---------------------- Do you now understand what I mean by the length of the set D of pairs in R^2? It's simply a real number, that I can define for a segment D of any set of pairs C that also satisfies the definition of curve. Can you see that by this definition, the length of the segment D = {(a,b) : b = 1 - a, a,b >= 0}, which is simply a set of pairs, is the real number sqrt(2)? If not, the proof is not difficult. Can you see that by this definition the length of the nth staircase is always 2, regardless of n? Again, the proof is not difficult; but it is tedious. That is what I mean by the usual method of measuring the length of a curve. It is a well-defined method of turning any set of pairs in R^2 which meets certain requirements, into a real number; which we then call the length of that curve. to be parallel to the diagonal in order for the usual metric to work? No, not really; but that is irrelevant to my argument. I'm not talking about needing to be parallel in order for the usual metric to work. I'm talking about two different well-defined functions: (1) A limit function: whose domain is /any/ sequence {C_n} of sets of pairs in R^2, which allows us to unambiguously calculate a unique set of pairs C, which we write as C = lim n->oo {C_n}. (2) A length function: as described above, whose domain is sets of pairs C which obey the definition segment of a curve given above, and which yields a unique real number which we call the length of the set of pairs C. My claim is that, under these definitions, it is not always the case that length(lim n->oo {C_n}) = lim n->oo {length(C_n)}. In particular, I can explicitly provide a sequence {C_n} where length(C_n) = 2 for all n, thus lim n->oo {length(C_n)} = 2; but length(lim n->oo {C_n}) = sqrt(2). That would break your infinite induction principle, would it not? the concept of an inductive proof of equality holding in the infinite case is > invalid, because the error in your staircase example is easily explainable in > the way I've been describing. Your description of the error appears to be based on a misconception: you think that the length of a curve can only be unambigously defined when the definition of curve is something other than a set of points in R^2 with some additional restrictions. But I /have/ given above a definition of length which is unambiguously defined for sets of pairs in R^2 meeting certain restrictions. Can you make the mental leap betwen sets of pairs in R^2 and sets of points in R^2? Just substitute the word point for pair, and euclidean distance for dist. That's what R^2 with the usual metric means. You may complain that then my example isn't really figures in the x/y plane anymore. Fine; since it is irrelevant to providing you with a counter-example to infinite induction, continue to call them sets of pairs in R^2. (1) Do you think I can't define a sequence of staircases {C_n} as a sequence of sets of pairs in R^2? I have already done this for you at least once: see, for example, the end of: (2) Do you think that the limit of that sequence is not the set of pairs D = {(a,b) : b = 1-a, a,b <= 0}? Again this limit is proven at the same link as above. (3) Do you think that the length of the set D is other than sqrt(2)? (Sketch of proof: select n points in the interval [0,1] including 0 and 1. Order them as them as the sequence {a_0, a_1, ..., a_(n-1)}. Let f(x) = (x, 1-x). Calculate the sum i = 1 to n-1 dist(f(a_(i-1), f(i)). Conclude that this sum is constant independent of n, and equals sqrt(2). Therefore, since sqrt(2) is the smallest real number >= sqrt(2), the length of D is sqrt(2).) (4) Do you think that it is not the case that for any of the staircases in {C_n}, the length of that set is other than 2? Do you require another tedious proof of this rather obvious fact as well? (Sketch of proof: show that the length of a segment of a curve is the sum of the lengths of any finite disjoint union of smaller segments of the segment of the curve. Show that any tread or riser of the nth staircase has length 1/n, via argument similar to the one given above. The staircase is the disjoint union of the n treads and n risers. Thus the length is n*(1/n + 1/n) = 2). (5) Do you think that the length of lim n->oo {C_n} = lim n->oo (the length of the set C_n); in other words, do you think that sqrt(2) = 2? (6) Doesn't that contradict the principle of infinite induction? I'm concerned, you proved it had length 2. That is because you are not using the same definition of limit, length, and curve that I am using (and you are also not following the proof). To that end I gave a full definittion of length above; it applies to sets of pairs in R^2 which are also curves, not sets of points only if they have directions or a sequence in the limit, unless by those things you mean exactly and only a set of pairs in R^2 which is a segment of a curve. If the limit of the staircases is the set D of pairs in R^2, there is no difference between measuring (the length of) the staircase in the limit, and measuring (the length of) of the set of pairs D; they are /defined/ to be the same thing. The length of a set D of pairs in R^2 is always the same, no matter how you get that set of pairs, whether by formula or some other means such as the limit of a sequence. That definition is what I meant by the usual method. > Finally, I use the principle of infinite induction to /deduce/ that > length(limit)=limit(length); i.e.: > Yes, that's very nice, and I have no problem with that. You're measuring the > length of an infinite number of infinitesimal steps, not a straight line. I'm measuring the length of a set of pairs in R^2. That set of pairs D is always the same thing; it isn't sometimes a line, and at other times a fractal line: it is simply a set of pairs in R^2. Two sets of pairs are equal if and only if each contains the same set of pairs. Period. > ... the principle of infinite induction claims that I /can always/ > use the limit of the lengths of the curves to measure the length of > the limit of the curves. > Yeah, and that worked out pretty good, didn't it? If you consider 2 = sqrt(2) working out pretty good, I suppose. I consider it a contradiction, that implies that your principle of induction doesn't hold in every case. > It is this /third/ assertion (premise B, infinite induction) which > causes a problem, not the first two. > No, you're just confused because you think that location is all there is to a > set of points. You're confused because you think it is sensible, when someone is making an argument about sets of points in R^2, to insert comments about some other thing: points with directions. This is what I mean by 2*x = 3 is satisfied by the natural number 3/2. Sets of points with directions, whatever they might be, are /different things/ than sets of points, just as rational numbers are /different things/ than natural numbers. My argument is regarding the latter, not the former; and I /explicitly stated/ that. > When you're talking about a curve, you're talking about an > uncountable sequence of points, and there is a connectedness between successive > points that has direction. No, that's not what I'm talking about, and that's why your objections are meaningless to me, mathematically speaking. When I'm talking about a curve, I'm talking about a /set/ of /points/ in /R^2/; such that there exists a continuous mapping to the real line, under the metric topology induced by the euclidean distance between points. And that's all I mean; nothing else. I don't know /what/ you're talking about when you're talking about a curve; but it really doesn't matter. Because my argument is not about what /you mean/ by a curve; it's about what /I have defined a curve to mean for the purposes of this argument/. It is one thing to fail to understand what /I/ mean by a curve, and consider my argument meaningless. You can always ask for clarifcation of any terms I use; and I gave ample clarifications above. If there are still terms you find unclear, feel free to ask for clarification. But it is entirely another thing to confuse my argument with someone else's /different/ argument, because you fail to apply the definitions I give in my argument, and/or substitute your own definitions instead. Until you learn this distinction, you will never be able to make a mathematcial argument yourself. > One can just as easily view a curve as a set of > infinitesimal segments between points, as a set of points separated by > infinitesimal segments. One can just as easily consider it anything at all one likes: a road that isn't straight, the shape of a woman's hip, a kind of baseball pitch, or a particularly obscure turn of phrase. And that would be then be a /different argument/. This is /this argument/; so such flights of fancy are no more relevant than stating we can view 3/2 is a natural number. > The third assertion is simply false: limit(length of curve) is not > equal to length(limit of curves), where limit, length, and curve > refer to objects in the domain of R^2 with the usual metric. Therefore, > the principle of infinite induction /does not apply/ in the case > where we are working in R^2 with the usual metric; and I think you'll > find that it doesn't apply in most situations. > What do you mean by the usual metric. Maybe this is the problem here. Is the > metric usually wrong? I gave you a definition of metric above; hopefully this answers your question. The usual metric in R^n is the function yielding the euclidean distance between two points. I have no idea what you mean by a metric being wrong. It's simply a function, obeying some properties. > There is no reason to tack on additional ill-defined (or even > well-defined) concepts like therefore there are points with no, some, > and all directions, at infinitesimally different locations that are > not part of my argument, and are certainly not a part of R^2 with the > usual metric, which is, after all, the usual domain of curves, > including all the familar the results of calculus. > Yeah, well, as I understand it, the usual metric is generally used parallel to > whatever one is measuring. Well, as you understand it is not what I mean why I say the usual metric. So I gave you a definition of the usual metric. Do you now understand what I mean by the usual metric? It's function, that is also a (toplogical) metric: the usual metric is the function which gives the euclidean distance between two points. You could use other metrics if you like; some of the more bizzare ones are quite interesting. That would be of course a /different/ argument than the one I am giving. > To add these other features of points in is an amusing diversion; but > it is a diversion: it is like saying the natural number satisfying > '2*x = 3' is x = 3/2. Fine, but 3/2 is not a natural number. > No, it's nothing like that. Yes, it is /exactly/ like that. When you say the usual metric is generally used parallel to whatever one is measuring, you are saying 3/2 is a natural. A metric is not something that you use parallel to something that you are measuring. It is simply a function d over a set X of the form d: X^2 -> R that yields a real number greater than or equal to 0, which obeys some additional rules for all p,q, and r in X: * d(p,q) = 0 if, and only if, p = q. * d(p,q) = d(q,p) * For all p,q, and r, d(p,q)+d(q,r) >= d(p,r). Thus 2*x = 3 for some natural is to x = 3/2 as the length of a set of points can be determined via a metric is to but a metric is generally used parallel to whatever one is measuring Similarly, when you claim that the length of D = {(a,b): b = 1- a, a,b>= 0} is anything other than 2, you are not refering to my argument's definition of length which is defined on /sets/ of /points/; you are refering to /some other/ definition of length that is used on /some other/ kind of figure in the x/y plane (which apparently consists of sets of points with directions and other notions). Thus 2*x = 3 for some natural x is to x = 3/2 as the length of the set D of points in R^2 is sqrt(2) is to no; it must be 2, because D is actually a set of points with directions. > As far as I'm concerned, the error of sqrt(2) is the inverse of the > cosine of the 45 degree angle each riser and tread makes with the diagonal. The > accurate measure of curves depends directly on parallel metrics. How do statements like this address any of the logical steps in my argument? How does inverse cosine apply to a /set/ of /pairs/ in R^2? > This is not true only in R^2 with the usual metric; in fact, you have > to work very carefully for your infinite induction to hold in a > paricular domain. > I don't think you (or Han) are being that careful, in general, when you > invoke it; and that is the point I have been trying to make. > Well, I think you're wrong, and I don't think you're example illuminated > anything other than basic calculus ideas of approximation. It's increasingly hard to imagine how you expect any mathematical argument to convince you of something about which you have already decided it is simply true. Suppose I have a set X. Suppose I define a limit on sequences of X such that lim n->oo {x_n} is well-defined for all sequences of elements in X. Suppose I define a function f : X -> R (the real numbers). Now, suppose as a /result/ of these /definitions/, I can explicitly provide a sequence {x_n}, such that I can provide a proof that f(x_n) = 1 for all n, but I can also prove that f(lim n->oo {x_n}) = 0. I'm not claiming at the moment I have neccesarily done this; I'm just asking you to /suppose/ that I had, for the sake of argument. Which, if any, of the following would you then claim? (1) Therefore, infinite induction does not hold in this case. (2) Therefore, lim n->oo {x_n} is actually not a member of X, it is something else. (3) Therefore, f(x_n) is actually not 1 for some natural n. (4) Therefore, lim n->oo {x_n} is not the infinite case. (5) {Insert other possible objection here}. You are currently claiming (2), in our non-hypothetical example of the staircases. Suppose I can convince you that (2) is false (i.e., by restricting our discussion solely to sets of pairs in R^2). If (4) is what you would then claim, then what /is/ the infinite case in your principle of infinite induction? > It has nothing to do > with probability over an infinite set, and didn't show why infinite induction > of equalities ever fails. It's just your assumption of usual metric that > fails. That fails /how/? * Fails to be a function of the form (R^2) x (R^2) -> R? * Fails to be a (topological) metric? * Fails to provide an unambiguous definition of length for any set of pairs in R^2 which is also a curve by my definition? * Fails to unambiguously define whether or not any pair p is a limit point of some sequence {C_n} of sets of pairs in R^2? * Fails to unambiguously define a set of pairs in R^2 which is the limit of any sequence {C_n} of sets of pairs in R^2? * Fails in /any way/ to obey the /premises/ of your assertion? Or, is it that it simply fails to support your assertion that infinite induction is always applicable, while still satisfying your premises? In that case, why would you continue to assert your principle? It would be false. different lines, which are at right angles to each other? > Use your mind's eye. Each side is parallel to one of the axes. This isn't Poetry. It's Mathematics. You don't use your minds eye to define terms; you use precise, unambiguous mathematical definitions. That's part of what makes mathematics a different thing than poetry. > According to the definitions I gave, each of the above formulas > identically yields the single point (0,0) in R^2 with the usual metric. > In other words, the figure in the x/y plane described by {(x,y) : > f(x,y)=c} for all the various functions f is identical: they are not > merely indistinguishable, they are equal. > Single points don't have any metric anyway. Since it is clear that at this point, you didn't know what a metric is, your statement is meaningless. Hopefully now you can see that if X = {a} is a singleton set; the only possible metric on X would be f(a,a) = 0. > You claim that, essentially, they are different: but you are using > logic equivalent to but there /is/ a natural number such that 2*x =3; > it is x = 3/2. > No, that is about as related to what I am saying as your example is to the > original topic. Like, not. In your post of April 26, you said: Now, I would pose this suggestion again. Inductive proof is considered generally invalid in the limit, that is, in the infinite case. The method of inductive proof is only considered to prove a property for all FINITE n. However, in this sense of a limit, my suggestion is that an equality between expressions proven inductively holds for all n, finite or infinite. I don't think that causes any problems such as what you've been suggesting, do you? My example of the staircases is related to thetopic, in that I am attempting to answer the question you posed by providing you with a believable counte-rexample to the principle of infinite induction espoused above that causes the same sorts of problems as I suggested. This is hampered by your inexperience at understanding what is meant /mathematically/ by a believable counter-example; i.e., a /proof/ that a counter-example exists which satisfies your premises, and yet contradicts your assertion. This in turn exposes your inexperience at understanding what is meant by a proof in general; and in particular, what is meant by a mathematical definition. Thus these long and doubtless frustrating explanations. Ah well; perhaps you'll pick it up as you go along. circle have? (groan) /I/ don't claim the circle has a direction at all; that phrase is meaningless to me. You seem to claim that it does or at least may a direction; and that's why I asked for a /definition/ of the term. I'm not a mind reader; so as you have given no definition that I can apply to, for example, the direction of an ellipse, aside from phrases like just use your mind's eye, the only way for me to determine it currently is to say ask Tony; that phrase is meaningless to me, mathematically speaking. distinguishability of those points. So what? So what!?! So therefore, my argument shows that infinite induction doesn't hold in every sense; in particular, it doesn't hold in the standard sense of R^2 with the usual metric (without points with directions, however you may choose to define them). That's what you /asked me/ to /provide/. My argument is based on: R^2, the usual metric, the standard definitions of limit, length, and curve; and the /logical deductions/ which follow from these /definitions/. One logical deduction is length(limit of curves) is not equal to limit of(length of nth curve). And thus, your assertion of infinite induction fails to hold in the standard case. The fact that my argument /doesn't apply/ in some /other case/ is irrelevant to the argument as a disproof of the assertion: the principle of infinite induction holds in every case. It disproves it by providing a counter-example under particular assumptions: R, R^2, euclidean distance, sets of points, limits. Therefore, the conclusion is: There exist conditions (assumptions and definitions) under which your principle doesn't apply. Therefore, you cannot simply assert that your principle is true, without also /proving/ that it is not false under the /particular/ assumptions and definitions /you/ are considering: e.g., R + infinitesimals + other ad hoc additions, infinite probability distributions, etc.: /regardless/ of how you choose to define those terms, and whose definitions I'm not particularly interested in. Because my example shows that it is /at least possible/ that under your definitions, your principle will /also/ not hold. > You're usual metric didn't work > for precisely the reasons I've explained, and you can't blame it on infinite > induction. Your critique of my proof has so far hinged on your assertion that there are two different objects under consideration: the real diagonal line, and the thing which is the limit of the staircases, which may /look like/ the diagonal line,and indeed be indistinguishable from the diagonal line, but /isn't/ the diagonal line (as well as numerous other misunderstandings of metric and so on) Hopefully, by figures in the x/y plane, you can now see that I am talking solely about pairs in R^2, using the function dist() as a metric. And that by limit, curve and length, I mean very specific and limited statements about sets of pairs in R^2 and sequences of same. Thus there is exactly one set I mean by the set of pairs D = {(a,b) : b = 1- a, a,b >= 0}. It has length sqrt(2). It is the same set (by set the limit of the staircases (redefined as sets of pairsin R^2). And yet the nth staircase, which is a set of pairs in R^2, always has length 2, regardless of n. So under those assumptions, infinite induction fails: sqrt(2) = length(D) = length(limit {C_n})) <> limit {length(C_n)} = 2. Therefore, infinite induction doesn't always hold. And that is what you /asked me/ to prove. Therefore you must /prove/ that your principle /does/ hold when you use it in any other argument. If, that is, what you want to do is make a mathematical argument. Otherwise, you can say whatever you like; but it would be off-topic in sci.math, except for entertainment's sake. Expect to then be mocked and maligned by some posters, at the minimum: this is, after all, usenet. have any direction. A point on a curve has direction as long as it doesn't have > a derivative discontinuity, that is, a corner. A curve is a sequence of points, > or of segments between points with direction. How is what you just described relevant to sets of pairs in R^2? If it isn't relevant to that, then what you are talking about is not what my argument is about. My argument is about sets of pairs in R^2; and well-defined functions on those sets such as length and limit. Derivative continuity and sequence of points are not relevant to the discussion, unless you can express those terms as unambiguous statements regarding sets of pairs in R^2, and the definitions of length, curve and limit that I use in the argument. is there a natural number such that 2*x =3? > is to > yes, 3/2 > as > is there a difference in R^2 with the usual metric between the > figures x^2 + y^2 = 0 and lim n->oo {y=x,-1/n< x< 1/n}? > is to > yes; they have different directions because they > are different elements of X^2 x P(X union {none}){}. > Straw man. That's not my argument. It is indeed your argument: you argue by use of undefined terms which are not relevant to the argument, or redefine terms that are defined in the argument to mean something else (or both simultaneously, as is usually the case). Unless you can rephrase your statement A curve is a sequence of points, or of segments between points with direction as a statement limited exclusively to sets of pairs in R^2, the definition of such a set being a curve, the defined length function on those sets which are curves, the defined limit function on sequences of those sets, and the usual mathematical operations on R and rules of inference, then you are /not talking about/ sets of pairs in R^2; you are talking about /some other thing/ than what my argument is talking about. 3/2 is a not a natural number. It's not my job to guess what you mean by a statement like Directions as tangent lines are elements of curves that exist conceptually at each point on the curve. (Although I admit it is entertaining as a sidebar to try to formalize such vague statements into actual meaningful mathematical statements.) Such a statement doesn't stand as a refutation of my argument, until you can translate it into a statement about pairs in R^2, the defined limit function, the defined length function, the definition of curve, and the usual mathematical operations on R. /Then/ you can say therefore since {whatever you end up with}, we see that there is a contradiction between {some statement in my arguiment about the set of pairs in R^2 defined by D} and {some other statement of my argument}. Until then, you are talking about something else; such as 3/2 in the context of the natural numbers. And as such, your statements are meaningless, mathematically speaking. example is obvious. I'm am not convinced. The /problem/ is that you don't know what is meant by a mathematical argument, i.e., a proof. If you did, you would have seen weeks ago that it is obvious that your principle of infinite induction is contradicted by my argument: which is in fact a proof that there exist a counter-example to your assertion that satisfies its premises, as you requested. > Then you're being clueless. You should be able to parse that statement. Or, did > you read probabilities as possibilities again? Remedial English is down the > hall to the left. This is Remedial Mathematics. The current topic is what is a valid > mathematical argument?. Stay in your seat until the bell rings. If that's your attitude, the discussion's over. Have a nice day. Ah well. Sorry you feel that way. See ya round then! > I'm not going to tolerate that kind of condescending nonsense. But I should tolerate being told to that I am clueless, and that I should go to Remedial English? You don't consider that a condescending or otherwise insulting remark, which is not to the point of the argument? It seems to me if you are going to dish out jovial insults of this sort, you should be willing to accept them as well. > I believe I have > a valid point, and that your refutation by counterexample was invalid. We've > strayed far from the topic, and now you want to make this a remedial class? It is not being condescending for me to observe that your knowledge of what makes a valid mathematical argument is sorely lacking; anymore than to observe that if you ignored red lights and stop signs and then parked your car on my lawn, I would consider your ability to drive properly to be sorely lacking. Nor is it condescending for me to observe that I know a lot more about mathematical arguments than you do. Which is not to say that aren't other people who know a lot more about mathematical arguments than I do. You may find that my judgement stings or is unfair, and I'm sorry if you feel that way; particularly since it seems to impede your ability to learn. But it's my honest judgement, and I feel very confident in these observations; observations that appear to be shared by most other observers. If you're interested in mathematics, then you need to /learn/ some mathematics; because what you are doing now is not mathematics, it is some other thing. Since you are posting in sci.math, I assume you are interested in learning mathematics. If you aren't, why /are/ you posting here? === Subject: Re: Calculus XOR Probability cbrown@cbrownsystems.com said: > cbrown@cbrownsystems.com said: > cbrown@cbrownsystems.com said: > cbrown@cbrownsystems.com said: When you say there's a problem with my limit, what do you mean? I mean that it doesn't take into account anything but location, and you're > using it to measure distance. > Distance and length are real numbers. Is lim n->oo {C_n} a real number, or is it a set of points? > It's a sequential set of points, that is a line of some sort, with a real > measure called length. > No, it's not. > {C_n} is a /sequence/ of /sets of points/. C_n for n in N is a sequence of sequential sets of elements, which can either be the corners of the steps, or the line segments which connect them. > A sequence of elements from some set X is essentially a function f : N > -> X. We write f_n instead of f(n), so we don't get confused and > think that f(3/2) might have some meaning. We then write {f_n} to > indicate the whole function, rather than its value at some particular > n. > (lim n->oo {C_n}) is defined to be a particular /set of points/. There > is no requirement that this limit be a line of some sort, with a real > measure called length. If that is the type of limit you're using, you have no business claiming it gives a valid measure, then blaming the fact that it doesn't on infinite induction. > For example, if C_n = {(x,y): x^2 + y^2 <= 1/n + 1}, the limit of this > sequence the unit disk: {(x,y): x^2 + y^2 <= 1}. That, to the best I > can determine, is not a line of some sort, with a real measure called > length: it is simply a set of points. Uh, yeah, that's not a line, but a disk. If you replace the <= with a =, then it'll be the circumference, which is a line of some sort. What you have there is a 2D space of some sort. You're going to need two directions on your points to measure that. ;) > These facts can be deduced from the /definition/ of limit of a > sequence of sets of points that I gave previously. > I don''t know what you mean by a line of some sort, with a real > measure of length, so I can't otherwise comment on whether or not your > statement is correct or incorrect; it is meaningless to me, > mathematically speaking; although as regular English usage it seems > contradicted by the example I just gave. > Can you provide a definition of what it means for a set C to be a > line of some sort, with a real measure of length? E A (endpoints) E B E x<>A -> E y: x=successor(y) E x<>B -> E y: y=successor(x) x=successor(y) <-> y=predecessor(x) Length=Sum(x=successor(A)->B: x-predecessor(x)) where subtraction of points indicates distance, which depends on the dimension of space. > Can you /then/ show how /my definition/ of lim n->oo {C_n} satisifies > /your definition/ of a line of some sort, with a real measure of > length? Takin the corners of the stairs as your sequence of points, with n stairs, each is 1/n in height and width, and when you sum all those up, you get 2. yes, even if n is infinite. It's the same as the probability problem in that respect, except you've thrown in the conflation of location proximity with identity. > If you can, then I would accept that the lim n->oo {C_n} is a line of > some sort with a real measure of length. > If you can't then I'm afraid your statement is meaningless to me, > mathematically speaking. I meant that lim n->oo {C_n} is a set of > points; that's why I defined it that way, and not some other way. Each pair of contiguous points has a distance between them that contributes to the overall length. When the direction at that point is not towards the endpoint, then you are going to get a less than straight line, with length above that given by Pythagorean theorem and the two endpoints. > To measure distance or length of a set of points in R^2 with the usual > metric, I use the usual approach, just like everyone else. > That's good, because this approach doesn't work. When you prove the length in > the limit is 2 instead of sqrt(2), that's an indication that the object in the > limit is NOT exactly the same as the diagonal line... > Well, since this point still seems unclear to you, I will amplify on > all the definitions used in my argument: > Forget the diagonal line for the moment. Forget figures in the x/y > plane. These phrases are unneccesary to my argument, and are confusing > you by their usual connotations. > By R I mean the real numbers - not the real numbers + infinitesimals + > infinite numbers. > By R^2, I mean the set of all ordered pairs (a,b) where a and b are > real numbers. I will call these pairs in R^2 or simply pairs. > By a set of pairs in R^2 I mean, well, a set of pairs in R^2. > If A and B are two sets of pairs in R^2, then A = B if, and only if, > for every pair p, p in A if and only if p in B. > If A = B, and f is any function that takes a set of pairs in R^2 to a > real number, then f(A) = f(B). Thus if A = B, A is identical to B in > every way except by name. > By a metric on R^2, I mean that I define a function dist(p,q) for all > pairs p = (p_a, p_b) and q = (q_a, q_b) in R^2 by: > dist(p,q) = sqrt((p_a-q_a)^2 + (p_b-q_b)^2). > Thus dist is a function from two pairs in R^2, that yields a real > number. > I observe the following, which you are free to prove me wrong about: > * dist(p,q) = 0 if, and only if, p = q. > * dist(p,q) = dist(q,p) for all pairs p, q. > * For all pairs p,q, and r, dist(p,q)+dist(q,r) >= dist(p,r). > Thus, the function dist is a (toplogical) metric by the usual > definition: it satisfies the three conditions above. That makes R^2, > with reference to the function dist, a metric space as defined by the > metric dist. > (One could define other functions which would also be a metric on R^2. > By the usual metric in R^n for any n, I mean that the dist function > we want to use is the same as euclidean distance in R^n between the two > points p and q). Okay, but what happens in the limit, in your estimation? When the distances between successive points become less than any real number, you seem to think that the distance is zero or something. But then, how do you account for the length of the line? How do those zero lengths sum to the overall length of the line? If you simply want to forget about infinitesimal line segments, then you really can't handle this object. This works well for measuring the diagonal, but not for summing infinitesimal segments, such as the staircase in the limit, without some notion of infinitesmals. > ----------------------- > What do I mean by C is a curve? > Suppose C is a set of pairs in R^2. Call such a set a curve if there > is a function f: R -> R^2 such that the point > (1) the pair f(x) is in C for all x in R; > (2) for all pairs p in C, there exists a unique x in R such that p = > f(x); which I will write as f^-1(p). > (2) if e is a real number greater than 0, and p is a point in C, then > there exists a real number d such that if q is a pair in C, and dist(p, > q) < e, then |f^-1(p) - f^-1(q)| < d. > (This is what is I mean by there exists a continuous function f from > the real line to C.) Okay. You're defining a linear continuum as a curve. Basically, you're mapping the real line to any curve. That's fine. Except, how do you determine which r in R any given point corresponds to, besides the first? > ------------------------ > Let C be the set of pairs defined by {p = (a,b) : b = 1 - a}. Is C a > curve? Well, let f be the function defined as f(x) = (x, 1-x). > (1) Is f(x) in C for all x in R? If one uses the x coordinate as the real x in R, does that mean your curve must have infinite domain? No, that's jus one way to do it, but okay... > (2) If p is a point in C, is there a unique x in R with f(x) = p? > (3) if e is a real number greater than 0, and p is a point in C, is > there a real number d such that for all q in C with dist(p,q) < e, it > is always the case that |f^-1(p) - f^-1(q)| < d? > The answer to each of the above is yes (feel free to prove me wrong); > so C is a curve, because the function f exists and satisfies the > three rules I gave in the definition of C is a curve. Okay. > -------------------------- > Now suppose we define a segment C_[u,v] of a curve C as the subset > of C defined by {p : p in C and u <= f^-1(p) <= v}. Alright, as long as we can determine where f(u) and f(v) lie. > Let C be the set of pairs defined by {p = (a,b) : b = 1 - a}; and let D > = C_[0,1]; then D = {p = (a,b) : b = 1-a, a,b>=0} (feel free to prove > me wrong). No, I can live with that. > -------------------- > If D = C_[u,v] is a segment of a curve C, then define the length of > D as follows: > (1) Call a finite sequence of n reals {a_0, a_1, ..., a_(n-1)} a joint > set if a_0 = u, a_(n-1) = v, and a_i < a_j for all i, j. So a joint > set is basically a finite set of points in [0,1] that includes 0 and > 1; and where the points in the sequence are in order. For example if > [u,v] = [0,1], (0, 1/2, 3/4, 7/8, 1) is a joint set, but (0, 3/4, 1/2, > 7/8, 1) > (2) Recall that because weare saying that C is a curve, there is a > continuous function f : [0,1] -> D. Define the joint length of a > joint set J = {a_0, a_1,..., a_(n-1)} as the sum > dist(f(a_0), f(a_1)) > + dist(f(a_1), f(a_2)) + ... > + dist(f(a_(n-2), f(a_(n-1)). > (3) Define the length of D as the /smallest/ real number which is > /greater than/ or equal to the joint length of any joint set. Okay, but that's only for a finite sequence so far..... > ---------------------- > Do you now understand what I mean by the length of the set D of pairs > in R^2? It's simply a real number, that I can define for a segment D of > any set of pairs C that also satisfies the definition of curve. > Can you see that by this definition, the length of the segment D = > {(a,b) : b = 1 - a, a,b >= 0}, which is simply a set of pairs, is the > real number sqrt(2)? If not, the proof is not difficult. > Can you see that by this definition the length of the nth staircase > is always 2, regardless of n? Again, the proof is not difficult; but it > is tedious. > That is what I mean by the usual method of measuring the length of a > curve. It is a well-defined method of turning any set of pairs in R^2 > which meets certain requirements, into a real number; which we then > call the length of that curve. Well, thank you for that lengthy explanation. It makes sense. It's just that I find myself wondering, if this rigorous definition of the length of the curve gives the result that the length of the staircase in the limit is 2, then why do you question that result, and not the exact equivalence between that curve and the diagonal? I think that result is ultimately correct. > to be parallel to the diagonal in order for the usual metric to work? > No, not really; but that is irrelevant to my argument. > I'm not talking about needing to be parallel in order for the usual > metric to work. I'm talking about two different well-defined > functions: > (1) A limit function: whose domain is /any/ sequence {C_n} of sets of > pairs in R^2, which allows us to unambiguously calculate a unique set > of pairs C, which we write as C = lim n->oo {C_n}. > (2) A length function: as described above, whose domain is sets of > pairs C which obey the definition segment of a curve given above, and > which yields a unique real number which we call the length of the set > of pairs C. > My claim is that, under these definitions, it is not always the case > that length(lim n->oo {C_n}) = lim n->oo {length(C_n)}. That depends, as you yourself stated, on how you define your limit. No? > In particular, I can explicitly provide a sequence {C_n} where > length(C_n) = 2 for all n, thus lim n->oo {length(C_n)} = 2; but > length(lim n->oo {C_n}) = sqrt(2). That's only because you equate the two objects which have obviously different measures. > That would break your infinite induction principle, would it not? No, for the same reason I have reiterated. They are two different objects. The error isn't in infinite induction, but the failure to make this distinction. > the concept of an inductive proof of equality holding in the infinite case is > invalid, because the error in your staircase example is easily explainable in > the way I've been describing. > Your description of the error appears to be based on a misconception: > you think that the length of a curve can only be unambigously defined > when the definition of curve is something other than a set of points > in R^2 with some additional restrictions. > But I /have/ given above a definition of length which is > unambiguously defined for sets of pairs in R^2 meeting certain > restrictions. And guess what? The length is 2. Huh! > Can you make the mental leap betwen sets of pairs in R^2 and sets of > points in R^2? Just substitute the word point for pair, and > euclidean distance for dist. That's what R^2 with the usual > metric means. Yes, I know, a pair of Cartesian coordinates. > You may complain that then my example isn't really figures in the x/y > plane anymore. No, I understand this is on the Cartesian plane, and therefore each point is a pair of real coordinates. > Fine; since it is irrelevant to providing you with a counter-example to > infinite induction, continue to call them sets of pairs in R^2. > (1) Do you think I can't define a sequence of staircases {C_n} as a > sequence of sets of pairs in R^2? I have already done this for you at > least once: see, for example, the end of: That seems to point to this message. :( > (2) Do you think that the limit of that sequence is not the set of > pairs D = {(a,b) : b = 1-a, a,b <= 0}? Again this limit is proven at > the same link as above. > (3) Do you think that the length of the set D is other than sqrt(2)? > (Sketch of proof: select n points in the interval [0,1] including 0 and > 1. Order them as them as the sequence {a_0, a_1, ..., a_(n-1)}. Let > f(x) = (x, 1-x). Calculate the sum i = 1 to n-1 dist(f(a_(i-1), f(i)). > Conclude that this sum is constant independent of n, and equals > sqrt(2). Therefore, since sqrt(2) is the smallest real number >= > sqrt(2), the length of D is sqrt(2).) When you define the points using y=1-x, then this works. > (4) Do you think that it is not the case that for any of the staircases > in {C_n}, the length of that set is other than 2? Do you require > another tedious proof of this rather obvious fact as well? > (Sketch of proof: show that the length of a segment of a curve is the > sum of the lengths of any finite disjoint union of smaller segments of > the segment of the curve. Show that any tread or riser of the nth > staircase has length 1/n, via argument similar to the one given > above. The staircase is the disjoint union of the n treads and n > risers. Thus the length is n*(1/n + 1/n) = 2). Yes, the length reamins at 2, even in the limit. > (5) Do you think that the length of lim n->oo {C_n} = lim n->oo (the > length of the set C_n); in other words, do you think that sqrt(2) = > 2? No, I think lim(staircase)<>diagonal, using any limit suitable for measuring length. > (6) Doesn't that contradict the principle of infinite induction? No, it contradicts the notion that the diagonal is an infinitesimal staircase. > I'm concerned, you proved it had length 2. > That is because you are not using the same definition of limit, > length, and curve that I am using (and you are also not following > the proof). According to your definitions, the length is 2, as long as you don't claim it's the diagonal. > To that end I gave a full definittion of length above; it applies to > sets of pairs in R^2 which are also curves, not sets of points only if > they have directions or a sequence in the limit, unless by those > things you mean exactly and only a set of pairs in R^2 which is a > segment of a curve. Didn't you prove that way that the length is also 2? > If the limit of the staircases is the set D of pairs in R^2, there is > no difference between measuring (the length of) the staircase in the > limit, and measuring (the length of) of the set of pairs D; they are > /defined/ to be the same thing. No, one is defined to be a pointwise limit of the other. > The length of a set D of pairs in R^2 is always the same, no matter > how you get that set of pairs, whether by formula or some other means > such as the limit of a sequence. > That definition is what I meant by the usual method. And it seems to support your original proof, does it not? Finally, I use the principle of infinite induction to /deduce/ that > length(limit)=limit(length); i.e.: > Yes, that's very nice, and I have no problem with that. You're measuring the > length of an infinite number of infinitesimal steps, not a straight line. > I'm measuring the length of a set of pairs in R^2. That set of pairs > D is always the same thing; it isn't sometimes a line, and at other > times a fractal line: it is simply a set of pairs in R^2. Two sets of > pairs are equal if and only if each contains the same set of pairs. > Period. If and only if equal. Equality is a funny thing. It depends on distinguishability, and that depends on the method of measure. > ... the principle of infinite induction claims that I /can always/ > use the limit of the lengths of the curves to measure the length of > the limit of the curves. > Yeah, and that worked out pretty good, didn't it? > If you consider 2 = sqrt(2) working out pretty good, I suppose. I > consider it a contradiction, that implies that your principle of > induction doesn't hold in every case. No, it implies that the diagonal is different from the fractal diagonal. It is this /third/ assertion (premise B, infinite induction) which > causes a problem, not the first two. > No, you're just confused because you think that location is all there is to a > set of points. > You're confused because you think it is sensible, when someone is > making an argument about sets of points in R^2, to insert comments > about some other thing: points with directions. This is what I mean > by 2*x = 3 is satisfied by the natural number 3/2. If I want to talk about splitting 2 pies between 3 people, and want to restrict the conversation to whole numbers, that's a pretty hopeless coversation, isn't it? If you want to bring arc length into the conversation, but don't want to talk about the methods that make approximating arc length work, then what's the point? > Sets of points with directions, whatever they might be, are /different > things/ than sets of points, just as rational numbers are /different > things/ than natural numbers. My argument is regarding the latter, not > the former; and I /explicitly stated/ that. Yes, and then you went on to make conclusions about the length of the staircase on the assumption that your set of points was sufficient to measure that. If you want to split your pies like that, then you better get a knife. In other words, don't talk about limits of curves that don't provide a good measure, and then blame that on something else. You wanna talk about measure? Points don't cut it. You need the lines between points to get measure. > When you're talking about a curve, you're talking about an > uncountable sequence of points, and there is a connectedness between successive > points that has direction. > No, that's not what I'm talking about, and that's why your objections > are meaningless to me, mathematically speaking. > When I'm talking about a curve, I'm talking about a /set/ of /points/ > in /R^2/; such that there exists a continuous mapping to the real line, > under the metric topology induced by the euclidean distance between > points. And that's all I mean; nothing else. Not much difference ultimately. > I don't know /what/ you're talking about when you're talking about a > curve; but it really doesn't matter. Because my argument is not about > what /you mean/ by a curve; it's about what /I have defined a curve to > mean for the purposes of this argument/. > It is one thing to fail to understand what /I/ mean by a curve, and > consider my argument meaningless. You can always ask for clarifcation > of any terms I use; and I gave ample clarifications above. If there are > still terms you find unclear, feel free to ask for clarification. > But it is entirely another thing to confuse my argument with someone > else's /different/ argument, because you fail to apply the definitions > I give in my argument, and/or substitute your own definitions instead. > Until you learn this distinction, you will never be able to make a > mathematcial argument yourself. Okay, so your limit is exactly the diagonal, even though your rigorous definition of arc length gives a different value than for the diagonal. I can see we'll never agree on this. These kinds of counterexamples are all over the place, disproving all sorts of things on the basis of something else. > One can just as easily view a curve as a set of > infinitesimal segments between points, as a set of points separated by > infinitesimal segments. > One can just as easily consider it anything at all one likes: a road > that isn't straight, the shape of a woman's hip, a kind of baseball > pitch, or a particularly obscure turn of phrase. > And that would be then be a /different argument/. This is /this > argument/; so such flights of fancy are no more relevant than stating > we can view 3/2 is a natural number. > The third assertion is simply false: limit(length of curve) is not > equal to length(limit of curves), where limit, length, and curve > refer to objects in the domain of R^2 with the usual metric. Therefore, > the principle of infinite induction /does not apply/ in the case > where we are working in R^2 with the usual metric; and I think you'll > find that it doesn't apply in most situations. > What do you mean by the usual metric. Maybe this is the problem here. Is the > metric usually wrong? > I gave you a definition of metric above; hopefully this answers your > question. The usual metric in R^n is the function yielding the > euclidean distance between two points. > I have no idea what you mean by a metric being wrong. It's simply a > function, obeying some properties. No, the usual metric isn't wrong, but you're not really using it on the staircase, when you ignore the infinitesimal right angles in it. The distances between elements approach 0, and at that point you think the corners disappear, but they don't. There is no reason to tack on additional ill-defined (or even > well-defined) concepts like therefore there are points with no, some, > and all directions, at infinitesimally different locations that are > not part of my argument, and are certainly not a part of R^2 with the > usual metric, which is, after all, the usual domain of curves, > including all the familar the results of calculus. > Yeah, well, as I understand it, the usual metric is generally used parallel to > whatever one is measuring. > Well, as you understand it is not what I mean why I say the usual > metric. So I gave you a definition of the usual metric. Do you now > understand what I mean by the usual metric? Sure, and as you stated, d(x,z)<=d(x,y)+d(y,z). In this case, d(x,z)=(d(x,y)+d (y,z))/sqrt(2). > It's function, that is also a (toplogical) metric: the usual metric is > the function which gives the euclidean distance between two points. You > could use other metrics if you like; some of the more bizzare ones are > quite interesting. That would be of course a /different/ argument than > the one I am giving. > To add these other features of points in is an amusing diversion; but > it is a diversion: it is like saying the natural number satisfying > '2*x = 3' is x = 3/2. Fine, but 3/2 is not a natural number. > No, it's nothing like that. > Yes, it is /exactly/ like that. > When you say the usual metric is generally used parallel to whatever > one is measuring, you are saying 3/2 is a natural. A metric is not > something that you use parallel to something that you are > measuring. I am saying that approximating a curve is done parallel to the curve, or done wrong. > It is simply a function d over a set X of the form d: X^2 -> R that > yields a real number greater than or equal to 0, which obeys some > additional rules for all p,q, and r in X: > * d(p,q) = 0 if, and only if, p = q. > * d(p,q) = d(q,p) > * For all p,q, and r, d(p,q)+d(q,r) >= d(p,r). > Thus > 2*x = 3 for some natural > is to > x = 3/2 > as > the length of a set of points can be determined via a metric > is to > but a metric is generally used parallel to whatever one is > measuring > Similarly, when you claim that the length of D = {(a,b): b = 1- a, > a,b>= 0} is anything other than 2, you are not refering to my > argument's definition of length which is defined on /sets/ of /points/; > you are refering to /some other/ definition of length that is used on > /some other/ kind of figure in the x/y plane (which apparently > consists of sets of points with directions and other notions). > Thus > 2*x = 3 for some natural x > is to > x = 3/2 > as > the length of the set D of points in R^2 is sqrt(2) > is to > no; it must be 2, because D is actually a set of points with > directions. > As far as I'm concerned, the error of sqrt(2) is the inverse of the > cosine of the 45 degree angle each riser and tread makes with the diagonal. The > accurate measure of curves depends directly on parallel metrics. > How do statements like this address any of the logical steps in my > argument? How does inverse cosine apply to a /set/ of /pairs/ in R^2? Given p1, p2 and p3 in C, the distance between p1 and p3 will be the sum of d (p1,p2) and d(p2,p3) times the cosine of the angle between the segments p1p2 and p2p3. So, if the individual segments are summed, you will get an answer that is the distance between the endpoints, divided by this cosine. This is not true only in R^2 with the usual metric; in fact, you have > to work very carefully for your infinite induction to hold in a > paricular domain. I don't think you (or Han) are being that careful, in general, when you > invoke it; and that is the point I have been trying to make. > Well, I think you're wrong, and I don't think you're example illuminated > anything other than basic calculus ideas of approximation. > It's increasingly hard to imagine how you expect any mathematical > argument to convince you of something about which you have already > decided it is simply true. > Suppose I have a set X. > Suppose I define a limit on sequences of X such that lim n->oo {x_n} is > well-defined for all sequences of elements in X. > Suppose I define a function f : X -> R (the real numbers). > Now, suppose as a /result/ of these /definitions/, I can explicitly > provide a sequence {x_n}, such that I can provide a proof that f(x_n) = > 1 for all n, but I can also prove that f(lim n->oo {x_n}) = 0. > I'm not claiming at the moment I have neccesarily done this; I'm just > asking you to /suppose/ that I had, for the sake of argument. I would then conclude that whatever notion of limit you are using is inadequate for measuring whatever f(x_n) is supposed to be measuring. > Which, if any, of the following would you then claim? > (1) Therefore, infinite induction does not hold in this case. Not when there is another obvious explanation for the error. > (2) Therefore, lim n->oo {x_n} is actually not a member of X, it is > something else. No, I wouldn't say that. > (3) Therefore, f(x_n) is actually not 1 for some natural n. I would say, given that it's inductively provable that it is, then it is. > (4) Therefore, lim n->oo {x_n} is not the infinite case. I would say there is an issue between your limit and f(x). > (5) {Insert other possible objection here}. > You are currently claiming (2), in our non-hypothetical example of > the staircases. Maybe I read that wrong. I would say that there is a discrepancy between your notion of limit and the measure you are evaluating. > Suppose I can convince you that (2) is false (i.e., by restricting our > discussion solely to sets of pairs in R^2). > If (4) is what you would then claim, then what /is/ the infinite case > in your principle of infinite induction? When you are proving an equality between real expressions, then it holds in all cases, and nothing magical happens at oo. > It has nothing to do > with probability over an infinite set, and didn't show why infinite induction > of equalities ever fails. It's just your assumption of usual metric that > fails. > That fails /how/? > * Fails to be a function of the form (R^2) x (R^2) -> R? > * Fails to be a (topological) metric? > * Fails to provide an unambiguous definition of length for any set of > pairs in R^2 which is also a curve by my definition? > * Fails to unambiguously define whether or not any pair p is a limit > point of some sequence {C_n} of sets of pairs in R^2? > * Fails to unambiguously define a set of pairs in R^2 which is the > limit of any sequence {C_n} of sets of pairs in R^2? > * Fails in /any way/ to obey the /premises/ of your assertion? > Or, is it that it simply fails to support your assertion that infinite > induction is always applicable, while still satisfying your premises? > In that case, why would you continue to assert your principle? It would > be false. It's not the usual metric that fails. I misspoke. It's the false equivalence between the diagonal and the staircase in the limit. > different lines, which are at right angles to each other? > Use your mind's eye. Each side is parallel to one of the axes. > This isn't Poetry. It's Mathematics. You don't use your minds eye to > define terms; you use precise, unambiguous mathematical definitions. > That's part of what makes mathematics a different thing than poetry. So, you also didn;'t understand what a square parallel to the axes might mean? What else could it possibly mean? > According to the definitions I gave, each of the above formulas > identically yields the single point (0,0) in R^2 with the usual metric. > In other words, the figure in the x/y plane described by {(x,y) : > f(x,y)=c} for all the various functions f is identical: they are not > merely indistinguishable, they are equal. > Single points don't have any metric anyway. > Since it is clear that at this point, you didn't know what a metric > is, your statement is meaningless. Hopefully now you can see that if X > = {a} is a singleton set; the only possible metric on X would be f(a,a) > = 0. Uh, right, a measure of zero. You claim that, essentially, they are different: but you are using > logic equivalent to but there /is/ a natural number such that 2*x =3; > it is x = 3/2. > No, that is about as related to what I am saying as your example is to the > original topic. Like, not. > In your post of April 26, > you said: > Now, I would pose this suggestion again. Inductive proof is considered > generally invalid in the limit, that is, in the infinite case. The > method of inductive proof is only considered to prove a property for > all FINITE n. However, in this sense of a limit, my suggestion is that > an equality between expressions proven inductively holds for all n, > finite or infinite. I don't think that causes any problems such as what > you've been suggesting, do you? > My example of the staircases is related to thetopic, in that I am > attempting to answer the question you posed by providing you with a > believable counte-rexample to the principle of infinite induction > espoused above that causes the same sorts of problems as I suggested. Perhaps you could present an example that isn't confounded by metrics and other issues. I doubt it. > This is hampered by your inexperience at understanding what is meant > /mathematically/ by a believable counter-example; i.e., a /proof/ > that a counter-example exists which satisfies your premises, and yet > contradicts your assertion. It was never my premise that the staircase in the limit is exactly the diagonal in every sense. > This in turn exposes your inexperience at understanding what is meant > by a proof in general; and in particular, what is meant by a > mathematical definition. > Thus these long and doubtless frustrating explanations. Ah well; > perhaps you'll pick it up as you go along. > circle have? (groan) > /I/ don't claim the circle has a direction at all; that phrase is > meaningless to me. You seem to claim that it does or at least may a > direction; and that's why I asked for a /definition/ of the term. I'm > not a mind reader; so as you have given no definition that I can apply > to, for example, the direction of an ellipse, aside from phrases like > just use your mind's eye, the only way for me to determine it > currently is to say ask Tony; that phrase is meaningless to me, > mathematically speaking. > distinguishability of those points. So what? > So what!?! > So therefore, my argument shows that infinite induction doesn't hold > in every sense; in particular, it doesn't hold in the standard sense > of R^2 with the usual metric (without points with directions, however > you may choose to define them). That's what you /asked me/ to > /provide/. > My argument is based on: R^2, the usual metric, the standard > definitions of limit, length, and curve; and the /logical deductions/ > which follow from these /definitions/. > One logical deduction is length(limit of curves) is not equal to limit > of(length of nth curve). And thus, your assertion of infinite > induction fails to hold in the standard case. > The fact that my argument /doesn't apply/ in some /other case/ is > irrelevant to the argument as a disproof of the assertion: the > principle of infinite induction holds in every case. > It disproves it by providing a counter-example under particular > assumptions: R, R^2, euclidean distance, sets of points, limits. > Therefore, the conclusion is: There exist conditions (assumptions and > definitions) under which your principle doesn't apply. Or, there exist definitions of limit which do not preserve measure. > Therefore, you cannot simply assert that your principle is true, > without also /proving/ that it is not false under the /particular/ > assumptions and definitions /you/ are considering: e.g., R + > infinitesimals + other ad hoc additions, infinite probability > distributions, etc.: /regardless/ of how you choose to define those > terms, and whose definitions I'm not particularly interested in. > Because my example shows that it is /at least possible/ that under your > definitions, your principle will /also/ not hold. It's either infinite induction, or pointwise limits with proper measure. One of those two concepts is at work here. > You're usual metric didn't work > for precisely the reasons I've explained, and you can't blame it on infinite > induction. > Your critique of my proof has so far hinged on your assertion that > there are two different objects under consideration: the real > diagonal line, and the thing which is the limit of the staircases, > which may /look like/ the diagonal line,and indeed be > indistinguishable from the diagonal line, but /isn't/ the diagonal > line (as well as numerous other misunderstandings of metric and so > on) > Hopefully, by figures in the x/y plane, you can now see that I am > talking solely about pairs in R^2, using the function dist() as a > metric. And that by limit, curve and length, I mean very > specific and limited statements about sets of pairs in R^2 and > sequences of same. Yes, points are pairs, and disatnces are pairs of pairs.You have your pairs in your limit, but not your pairs of pairs which are required by the dist() metric. It's those pairs of pairs which are distinguishable from those in the diagonal by always having a corresponding element equal from one pair to the other. > Thus there is exactly one set I mean by the set of pairs D = {(a,b) : b > = 1- a, a,b >= 0}. It has length sqrt(2). It is the same set (by set > the limit of the staircases (redefined as sets of pairsin R^2). And yet > the nth staircase, which is a set of pairs in R^2, always has length > 2, regardless of n. > So under those assumptions, infinite induction fails: sqrt(2) = > length(D) = length(limit {C_n})) <> limit {length(C_n)} = 2. > Therefore, infinite induction doesn't always hold. And that is what > you /asked me/ to prove. > Therefore you must /prove/ that your principle /does/ hold when you use > it in any other argument. No, you must disprove it with a pure example. Yours had another explanation. I maintain that it holds. > If, that is, what you want to do is make a mathematical argument. > Otherwise, you can say whatever you like; but it would be off-topic in > sci.math, except for entertainment's sake. Expect to then be mocked and > maligned by some posters, at the minimum: this is, after all, usenet. Yeah, like that hasn't happened once or twice. > have any direction. A point on a curve has direction as long as it doesn't have > a derivative discontinuity, that is, a corner. A curve is a sequence of points, > or of segments between points with direction. > How is what you just described relevant to sets of pairs in R^2? If > it isn't relevant to that, then what you are talking about is not what > my argument is about. If it's about a metric, the it's about distance between those pairs, not the pairs in isolation. > My argument is about sets of pairs in R^2; and well-defined functions > on those sets such as length and limit. Derivative continuity and > sequence of points are not relevant to the discussion, unless you can > express those terms as unambiguous statements regarding sets of pairs > in R^2, and the definitions of length, curve and limit that I use > in the argument. > figures x^2 + y^2 = 0 and lim n->oo {y=x,-1/n< x< 1/n}? is to yes; they have different directions because they > are different elements of X^2 x P(X union {none}){}. > Straw man. That's not my argument. > It is indeed your argument: you argue by use of undefined terms which > are not relevant to the argument, or redefine terms that are defined in > the argument to mean something else (or both simultaneously, as is > usually the case). > Unless you can rephrase your statement A curve is a sequence of > points, or of segments between points with direction as a statement > limited exclusively to sets of pairs in R^2, the definition of such a > set being a curve, the defined length function on those sets which > are curves, the defined limit function on sequences of those sets, > and the usual mathematical operations on R and rules of inference, then > you are /not talking about/ sets of pairs in R^2; you are talking about > /some other thing/ than what my argument is talking about. 3/2 is a not > a natural number. > It's not my job to guess what you mean by a statement like > Directions as tangent lines are elements of curves that exist > conceptually at each point on the curve. (Although I admit it is > entertaining as a sidebar to try to formalize such vague statements > into actual meaningful mathematical statements.) > Such a statement doesn't stand as a refutation of my argument, until > you can translate it into a statement about pairs in R^2, the defined > limit function, the defined length function, the definition of curve, > and the usual mathematical operations on R. > /Then/ you can say therefore since {whatever you end up with}, we see > that there is a contradiction between {some statement in my arguiment > about the set of pairs in R^2 defined by D} and {some other statement > of my argument}. > Until then, you are talking about something else; such as 3/2 in the > context of the natural numbers. And as such, your statements are > meaningless, mathematically speaking. Fine, I don't have time right now to formulate this for you. I'll think about it and respond later, but my immediate thought is that the pairs that define the points in the curve can be replaced by a sequence of pairs that denote the x and y changes of a series of segments, which in the limit approach the curve. In this sense, the pairs that define the staircase, one of the elements of which is always 0, do not approach the pairs that define the diagonal, which always have x=-y. > example is obvious. I'm am not convinced. > The /problem/ is that you don't know what is meant by a mathematical > argument, i.e., a proof. > If you did, you would have seen weeks ago that it is obvious that your > principle of infinite induction is contradicted by my argument: which > is in fact a proof that there exist a counter-example to your assertion > that satisfies its premises, as you requested. Then you're being clueless. You should be able to parse that statement. Or, did > you read probabilities as possibilities again? Remedial English is down the > hall to the left. This is Remedial Mathematics. The current topic is what is a valid > mathematical argument?. Stay in your seat until the bell rings. If that's your attitude, the discussion's over. Have a nice day. > Ah well. Sorry you feel that way. See ya round then! > I'm not going to tolerate that kind of condescending nonsense. > But I should tolerate being told to that I am clueless, and that I > should go to Remedial English? You don't consider that a > condescending or otherwise insulting remark, which is not to the point > of the argument? > It seems to me if you are going to dish out jovial insults of this > sort, you should be willing to accept them as well. > I believe I have > a valid point, and that your refutation by counterexample was invalid. We've > strayed far from the topic, and now you want to make this a remedial class? > It is not being condescending for me to observe that your knowledge of > what makes a valid mathematical argument is sorely lacking; anymore > than to observe that if you ignored red lights and stop signs and then > parked your car on my lawn, I would consider your ability to drive > properly to be sorely lacking. > Nor is it condescending for me to observe that I know a lot more about > mathematical arguments than you do. Which is not to say that aren't > other people who know a lot more about mathematical arguments than I > do. > You may find that my judgement stings or is unfair, and I'm sorry > if you feel that way; particularly since it seems to impede your > ability to learn. But it's my honest judgement, and I feel very > confident in these observations; observations that appear to be shared > by most other observers. > If you're interested in mathematics, then you need to /learn/ some > mathematics; because what you are doing now is not mathematics, it is > some other thing. > Since you are posting in sci.math, I assume you are interested in > learning mathematics. If you aren't, why /are/ you posting here? -- Smiles, Tony === Subject: Re: Calculus XOR Probability >> cbrown@cbrownsystems.com said: Another fine post. Your patience in trying to explain things to Tony is remarkable. >> Then you're being clueless. You should be able to parse that statement. Or, did >> you read probabilities as possibilities again? Remedial English is down the >> hall to the left. >> This is Remedial Mathematics. The current topic is what is a valid >> mathematical argument?. Stay in your seat until the bell rings. >> If that's your attitude, the discussion's over. Have a nice day. >> Ah well. Sorry you feel that way. See ya round then! >> I'm not going to tolerate that kind of condescending nonsense. > But I should tolerate being told to that I am clueless, and that I > should go to Remedial English? You don't consider that a > condescending or otherwise insulting remark, which is not to the point > of the argument? > It seems to me if you are going to dish out jovial insults of this > sort, you should be willing to accept them as well. This seems to be indicative of the general lack of consistency in Tony's thought process. He is apparently free to make insulting remarks about remedial education, but if anyone else does it is condescending nonsense which he will not tolerate. Is it any wonder he cannot put together a consistent mathematical argument? Stephen === Subject: Re: Calculus XOR Probability > Mike Kelly said: > > Well, only you seem to want to treat it like a number. Everyone > else knows that it isn't a regular number, it's a cardinal number. > It describes an equivalence class. You seem to have the idea of > what Aleph_0 actually is almost completely right in this post. So > why all the confusion elsewhere? > > mike. > > > It's not confusion. It's basic disgreement with the structure of the > system and the conclusions it produces. But there are no mathematical reasons, only TO's vaunted intuition, as basis for his disagreement, while there are specific proofs that set theory works the way it is advertised to work. > I get how the system works. I > disagree that it's the best we can do, or that cardinality is > anything like a good analog of size for infinite sets, and I'm trying > to offer improvements So far all TO's suggestions have been more in the nature of disasters than improvements. === Subject: Re: Calculus XOR Probability > Virgil said: > > Virgil said: > , > > What, specifically, prevents the same axioms appied to > infinite or infinitesimal units wouldn't work the same way as > with finite units? > > > No one has to show that TOmatics doesn't work, TO has to show > it does. > > The position of standard mathematics is that nothing other than > an axiom system has to be accepted until it is proved, and > axiom systems are only accepted for purposes of argument. > > Every mathematical theorem is really of the form: > Given such and such a set of axioms then.... > > So until TO can give his set of axioms, and the proofs, he has > nothing. > > > If by nothing you mean a set of complementary concepts, then > you're right. > > Where is this alleged set of allegedly complimentary concepts that > somehow does not depend on standard set theory, with all its > Cantorian consequences, as its unacknowledged basis? > Sipping tea and exchanging compliments. A mad tea party with TO presiding as Mad Hatter, and the March Zuhair in attendance. > > Without a specific set as domain and another as codomain > for a given function, there is no way to tell whether it is > invertible or not. > > > The invertible functions of IFR map reals to reals. > > Does TO mean that they are both bijections of R to itself? > > No, TO means they are a bijection between the subset of the reals > called the naturals and another subset of the reals. > > Then they are not really functions from the reals to the reals but > functions between the naturals and some other Peano set. > No, they are functions from the reals to the reals. Just because the > one set contains only naturals, that doesn't change where the > function is defined. > Look, you could have a function that only works on naturals and > always produces a non-natural, and an inverse function of that, and > it might work, but what I am suggesting is MORE restrictive than > that, for starters. > > > When measuring a > discrete set of reals, we compare to the set of naturals, > which is the unit, or standard, discrete set. > > The set of integers is also a discrete set > > That's true, and that could also be used as a standard infinite > set. I've thought about which would be better, without much > conclusion either way. > > > > So, if we map the naturals in > N to reals in S through f (n), > > Then the domain at issue is the set of naturals, not the set of > reals. > > > Right, the domain of f in the bijection is the naturals, but the > inverse relationship with g covers all the reals, and the domain > of g is the set of all reals mapped from naturals by f. > > That makes it some countable Peano subset of the reals. > Yeah, like the squares, evens, cube roots, etc. And the square roots, the negative integers, the natural sines of the naturals, and so forth. > > then applying g(n) to the values in S > > And the set S, not the reals is the codomain of f and the > domain of g and there is no reason to suppose that there is any > x for which f(x) and g(x) are both defined, and certainly no x > for which f(g(x)) = g(f(x)) = x, as TO requires. > > > Huh? That's gobbledygook. > > Then it is TO's gobbledegook, since he is the one who demands that > f(g(x)) = g(f(x) = x for all x. > I command it be so. Make it so, Number One. > > There is no reason to assume that f(x) and g(x) are > both naturals or are equal, but if the domain of g is f(n) for > all n in N, then g(f(n)) is defined as an element n of N. Are you > saying there is no such thing as an inverse function, and > therefore no such thing as a quantitative formulaic bijection? > > I am saying that TO's requirement that f(g(x)) = g(f(x) = x for > all x is nonsense, and until he clears it up and makes some > alternate, but more sensible, statement of what he means, anything > derived from his requirement is nonsense. > > No, put the word real before x, and it's perfectly fine. > > I can't even tell what convoluted excuse you have this time. > > it is the nonsense of TO's own statements that are the convolution > causing the problem. > > > will result in a set of reals which all happen to be > naturals. Is that so hard to picture? > > Quite impossible in the way TO describes it. > > It's impossible, not because of the way I describe it, but > because of whom I'm describing it to. > > > If g(x) and f(x) are both defined for any x, then that x must > be in the domain of both f and g. > > No, that is not correct. > > If f(2) and g(2) are both defined, how can this be unless 2 is in > the domain of both? > Unless the set generated by applying f to the naturals contains 2, 2 > will not be a value used as a parameter to g. It need not be in S, > which is the domain of g. However, this means g(2) is not a natural. > It's still a real number. :) Then why does TO keep insisting on f(g(x)) = g(f(x)) = x for all real x? > No, the requirement that f(g(x))=g(f(x))=x for all real x may be > slightly more restrictive than it needs to be, but as long as that > requirement is satisfied, when all inputs to f are natural Then all inputs to g must also be natural as they have the same domain and same codomain. === Subject: Re: Calculus XOR Probability Virgil said: nothing. > > > If by nothing you mean a set of complementary concepts, then > you're right. > > Where is this alleged set of allegedly complimentary concepts that > somehow does not depend on standard set theory, with all its > Cantorian consequences, as its unacknowledged basis? > > Sipping tea and exchanging compliments. > A mad tea party with TO presiding as Mad Hatter, and the March Zuhair in > attendance. Hahaha! Oh, that was a good one, Virgil. One lump or two? :) BTW, I doubt Zuhair gets it. > > > Without a specific set as domain and another as codomain > for a given function, there is no way to tell whether it is > invertible or not. > > > The invertible functions of IFR map reals to reals. > > Does TO mean that they are both bijections of R to itself? > > No, TO means they are a bijection between the subset of the reals > called the naturals and another subset of the reals. > > Then they are not really functions from the reals to the reals but > functions between the naturals and some other Peano set. > > No, they are functions from the reals to the reals. Just because the > one set contains only naturals, that doesn't change where the > function is defined. > > Look, you could have a function that only works on naturals and > always produces a non-natural, and an inverse function of that, and > it might work, but what I am suggesting is MORE restrictive than > that, for starters. > > > > When measuring a > discrete set of reals, we compare to the set of naturals, > which is the unit, or standard, discrete set. > > The set of integers is also a discrete set > > That's true, and that could also be used as a standard infinite > set. I've thought about which would be better, without much > conclusion either way. > > > > So, if we map the naturals in > N to reals in S through f (n), > > Then the domain at issue is the set of naturals, not the set of > reals. > > > Right, the domain of f in the bijection is the naturals, but the > inverse relationship with g covers all the reals, and the domain > of g is the set of all reals mapped from naturals by f. > > That makes it some countable Peano subset of the reals. > > Yeah, like the squares, evens, cube roots, etc. > And the square roots, the negative integers, the natural sines of the > naturals, and so forth. Right. Well, you should use the positive square roots, so it's a function. And sines of the naturals, I suppose, would be a function, but it would be rather difficult, I think, to invert in practice. Any set of reals (I guess that's what I mean by quantitative set) which is mapped from N progresses through its range at some rate. The slower it progresses, the denser it can said to be at that point, which is to say it has more elements per unit of distance on the line. That translates, intuitively, into the set being bigger. So, the faster rising function produces the smaller set. > > > then applying g(n) to the values in S > > And the set S, not the reals is the codomain of f and the > domain of g and there is no reason to suppose that there is any > x for which f(x) and g(x) are both defined, and certainly no x > for which f(g(x)) = g(f(x)) = x, as TO requires. > > > Huh? That's gobbledygook. > > Then it is TO's gobbledegook, since he is the one who demands that > f(g(x)) = g(f(x) = x for all x. > > I command it be so. Make it so, Number One. > > > There is no reason to assume that f(x) and g(x) are > both naturals or are equal, but if the domain of g is f(n) for > all n in N, then g(f(n)) is defined as an element n of N. Are you > saying there is no such thing as an inverse function, and > therefore no such thing as a quantitative formulaic bijection? > > I am saying that TO's requirement that f(g(x)) = g(f(x) = x for > all x is nonsense, and until he clears it up and makes some > alternate, but more sensible, statement of what he means, anything > derived from his requirement is nonsense. > > > No, put the word real before x, and it's perfectly fine. > > > I can't even tell what convoluted excuse you have this time. > > it is the nonsense of TO's own statements that are the convolution > causing the problem. > > > will result in a set of reals which all happen to be > naturals. Is that so hard to picture? > > Quite impossible in the way TO describes it. > > It's impossible, not because of the way I describe it, but > because of whom I'm describing it to. > > > If g(x) and f(x) are both defined for any x, then that x must > be in the domain of both f and g. > > No, that is not correct. > > If f(2) and g(2) are both defined, how can this be unless 2 is in > the domain of both? > > Unless the set generated by applying f to the naturals contains 2, 2 > will not be a value used as a parameter to g. It need not be in S, > which is the domain of g. However, this means g(2) is not a natural. > It's still a real number. :) > > Then why does TO keep insisting on f(g(x)) = g(f(x)) = x for all real x? Because that covers all possibilities of a quantitative set, if f and g are valid over all reals, and allows for compound functions and transitive bijections, without having to worry whether the mapping function covers all of the codomain of another. If they all accept all reals, and only return reals, then there is no problem with domains and codomains being valid. See? > No, the requirement that f(g(x))=g(f(x))=x for all real x may be > slightly more restrictive than it needs to be, but as long as that > requirement is satisfied, when all inputs to f are natural > Then all inputs to g must also be natural as they have the same domain > and same codomain. Why on earth do you keep saying that? As functions in general, which could be used to map any set of reals to some other, they are valid for all reals, and always return real values. When applied to a mapping from the naturals, which we use as a standard discrete set, that original mapping has a domain in N because that's the set we started with. But, the mapping function itself can take non-natural reals as well, so the domain of the *function* is the reals, while the domain of the *mapping* is the naturals. Say we map set N to S using f(x) and map S to T using g(x). Then N is mapped to T using g(f(x)). Say fi(x) and gi(x) are the inverse functions of f(x) anf g (x), then the inverse of g(f(x)) is fi(gi(x)). We don't want to have to worry about whether a function is defined at a value that another returns, so we are going to require, for now, that functions be defined over R, both in and out. -- Smiles, Tony === Subject: Re: Calculus XOR Probability > cbrown@cbrownsystems.com said: > Is lim n->oo {C_n} a real number, or is it a set of points? > It's a sequential set of points, that is a line of some sort, with a real > measure called length. What does TO mean by a sequential set of points. The usual meaning of a sequence is a set ordered in such a way that all but possibly one member has an immediate successor, all but possibly one has an immediate predecessor, and that any lessor member is linked to any greater through a finite set of successors. Since TO cannot impose this structure on the points of a line, he must mean something else. > > So why do you claim that I am using the limit to measure distance or > length? I am using the limit to take the limit of a sequence of curves, > nothing more. > > To measure distance or length of a set of points in R^2 with the usual > metric, I use the usual approach, just like everyone else. > That's good, because this approach doesn't work. It has worked perfectly for everyone for thousands of years. Why should it stop working now? > When you prove the length in the limit is 2 instead of sqrt(2), > that's an indication that the object in the limit is NOT exactly the > same as the diagonal line As he does not prove any such thing, TO is way off base. What he does prove is that TO's version of arc length behaves that way, but that is quite different from any actual version of arc length. > and the reason for the > discrepancy is easily seen to be due to the angle at which each of the > approximating segments intersects the diagonal it's supposedly measuring. Angles are irrelevant. For each finite sequence of points along a piece of an arc, one can find the point-to-point length without any attention being paid to any angle, but only to the distances between successive points. If the set of all such polygonal lengths has a least upper bound, that LUB is the length of the curve. I do not know what sort of measure of a curve TO is talking about which does not fit that definition. And neither does TO. > Do you understand what I'm saying about the elements of the staircase needing > to be parallel to the diagonal in order for the usual metric to work? We can easily understand what you are saying, but not why you are saying it, because it isn't true. > No, you are misunderstanding me. I reject your example as demonstrating that > the concept of an inductive proof of equality holding in the infinite case is > invalid, because the error in your staircase example is easily explainable in > the way I've been describing. But since that easy explanation is quite false, its ease does not excuse its irrelevance. > > I use the limit to find the limit of a sequence of curves. I don't > measure anything with it; nor have I claimed that you /should/ be > able to measure anything with it. > Fine so far. > > I use the /usual method/ to measure the lengh of any curve, both > before and after taking the limit, based on sets of points in R^2 using > the usual metric. > What is the usual method for measuring the staircase in the limit? As far as > I'm concerned, you proved it had length 2. As TO is totally unfamiliar with distinguishing between valid and invalid proofs, his judgement of what constitutes a valid one is just another of his irrelevancies. > > Finally, I use the principle of infinite induction to /deduce/ that > length(limit)=limit(length); i.e.: > Yes, that's very nice, and I have no problem with that. You're measuring the > length of an infinite number of infinitesimal steps, not a straight line. > > THE PRINCIPLE OF INFINITE INDUCTION is what I (erroneously) use to > measure the length of the limit of the curves; > Don't be so hard on yourself. You got the right answer. You did very well. > > because the principle of infinite induction claims that I /can always/ > use the limit of the lengths of the curves to measure the length of > the limit of the curves. > Yeah, and that worked out pretty good, didn't it? > > It is this /third/ assertion (premise B, infinite induction) which > causes a problem, not the first two. > No Yes! TO's alleged principle of infinite induction has been shown to be invalid but impeccable logic several times, but TO's unreasoning faith in it persists. > What do you mean by the usual metric. Maybe this is the problem here. Is > the > metric usually wrong? The usual metric for the Cartesian plane is that for points whose (x,y) coordinates are (a1,b1) and (a2, b2), the distance between them is sqrt( (a1-a2)^2 + (b1-b2)^2 ). Similar formulae hold in Cartesian spaces of other dimensions. > Yeah, well, as I understand it, the usual metric is generally used parallel > to > whatever one is measuring. Parallelism is irrelevant, and need not even be defined in the space in question, so long as it is a metric space. TO seems in every mathematical question to have an unerring ability to single out one irrelevant property as being the essential one in his warped version of mathematics. The distance between two points does not depend at all on the direction between them. If it did, then rotating the space in which the points occur would change the distance between them. === Subject: Re: Calculus XOR Probability Virgil said: > cbrown@cbrownsystems.com said: > Is lim n->oo {C_n} a real number, or is it a set of points? > > It's a sequential set of points, that is a line of some sort, with a real > measure called length. > What does TO mean by a sequential set of points. > The usual meaning of a sequence is a set ordered in such a way that > all but possibly one member has an immediate successor, all but possibly > one has an immediate predecessor, and that any lessor member is linked > to any greater through a finite set of successors. > Since TO cannot impose this structure on the points of a line, he must > mean something else. I mean everything above except for the limitation of finiteness on the number of successors between any two points. > > So why do you claim that I am using the limit to measure distance or > length? I am using the limit to take the limit of a sequence of curves, > nothing more. > > To measure distance or length of a set of points in R^2 with the usual > metric, I use the usual approach, just like everyone else. > > That's good, because this approach doesn't work. > It has worked perfectly for everyone for thousands of years. Why should > it stop working now? > When you prove the length in the limit is 2 instead of sqrt(2), > that's an indication that the object in the limit is NOT exactly the > same as the diagonal line > As he does not prove any such thing, TO is way off base. That was the concluson Chas logically drew, erroneously or not. > What he does prove is that TO's version of arc length behaves that way, > but that is quite different from any actual version of arc length. It wasn't MY version of arc length. That was supposed to be Chas' example of how lim(x->oo: f(x))=f(lim(x->: x)) (or something similar) doesn't necessarily hold. He says that's the problem with the proof, and I say it's because Chas is using a version of limit that doesn't lend itself to accurate measure of arc length. So, what are you talking about? I didn't create this example. > and the reason for the > discrepancy is easily seen to be due to the angle at which each of the > approximating segments intersects the diagonal it's supposedly measuring. > Angles are irrelevant. For each finite sequence of points along a piece > of an arc, one can find the point-to-point length without any attention > being paid to any angle, but only to the distances between successive > points. If the set of all such polygonal lengths has a least upper > bound, that LUB is the length of the curve. Yeah, if the segments are parallel at some point to the curve. > I do not know what sort of measure of a curve TO is talking about which > does not fit that definition. And neither does TO. There's a lot you don't know, despite your opinion of yourself. > > Do you understand what I'm saying about the elements of the staircase needing > to be parallel to the diagonal in order for the usual metric to work? > We can easily understand what you are saying, but not why you are saying > it, because it isn't true. If you say that, then you obviously don't understand what I'm saying. > > No, you are misunderstanding me. I reject your example as demonstrating that > the concept of an inductive proof of equality holding in the infinite case is > invalid, because the error in your staircase example is easily explainable in > the way I've been describing. > But since that easy explanation is quite false, its ease does not excuse > its irrelevance. Your declaration of falseness is false. That handwaving nonsense is pathetic, Virgil. Really, you can do better than that. Why don't you go measure a sine wave using vertical elements, and tell me how that works out? Show me one example where the segments are not parallel to the curve, which actually works, or admit you're full of crap. > > > I use the limit to find the limit of a sequence of curves. I don't > measure anything with it; nor have I claimed that you /should/ be > able to measure anything with it. > > Fine so far. > > > I use the /usual method/ to measure the lengh of any curve, both > before and after taking the limit, based on sets of points in R^2 using > the usual metric. > > What is the usual method for measuring the staircase in the limit? As far as > I'm concerned, you proved it had length 2. > As TO is totally unfamiliar with distinguishing between valid and > invalid proofs, his judgement of what constitutes a valid one is just > another of his irrelevancies. Irrelevancies? Whatever. The proof is valid. The limit is not of a sort which can be used to measure arc length of the diagonal, however. They are slightly different objects, one with length 2, and one with length sqrt(2). > > > Finally, I use the principle of infinite induction to /deduce/ that > length(limit)=limit(length); i.e.: > > Yes, that's very nice, and I have no problem with that. You're measuring the > length of an infinite number of infinitesimal steps, not a straight line. > > > THE PRINCIPLE OF INFINITE INDUCTION is what I (erroneously) use to > measure the length of the limit of the curves; > > Don't be so hard on yourself. You got the right answer. You did very well. > > > because the principle of infinite induction claims that I /can always/ > use the limit of the lengths of the curves to measure the length of > the limit of the curves. > > Yeah, and that worked out pretty good, didn't it? > > > It is this /third/ assertion (premise B, infinite induction) which > causes a problem, not the first two. > > No > Yes! TO's alleged principle of infinite induction has been shown to be > invalid but impeccable logic several times, but TO's unreasoning faith > in it persists. If you consider Chas' argument impeccable despite my far more exact argument which explains the precise error, then you really have no clue about the nature of analytical thought. That's too bad. Good luck with that. No wonder Transylfinity manages to persist. > > What do you mean by the usual metric. Maybe this is the problem here. Is > the > metric usually wrong? > The usual metric for the Cartesian plane is that for points whose (x,y) > coordinates are (a1,b1) and (a2, b2), the distance between them is > sqrt( (a1-a2)^2 + (b1-b2)^2 ). > Similar formulae hold in Cartesian spaces of other dimensions. Oh, well, that's just the Pythagorean distance formula in Cartesian coordinates. According to that, the staircase has length 2. Since it never travels in the diagonal direction, even in the limit, it makes sense that the length does not become sqrt(2). The slope is always oo at 0. The diagonal, of course, is sqrt(2), with slope everywhere equal to 1. > > Yeah, well, as I understand it, the usual metric is generally used parallel > to > whatever one is measuring. > Parallelism is irrelevant, and need not even be defined in the space in > question, so long as it is a metric space. > TO seems in every mathematical question to have an unerring ability to > single out one irrelevant property as being the essential one in his > warped version of mathematics. > The distance between two points does not depend at all on the direction > between them. If it did, then rotating the space in which the points > occur would change the distance between them. The sky is blue, Virgil, and love is good. (sigh) And puppies are cute. -- Smiles, Tony === Subject: Re: Calculus XOR Probability cbrown@cbrownsystems.com said: > Is lim n->oo {C_n} a real number, or is it a set of points? It's a sequential set of points, that is a line of some sort, with a real > measure called length. > What does TO mean by a sequential set of points. > The usual meaning of a sequence is a set ordered in such a way that > all but possibly one member has an immediate successor, all but possibly > one has an immediate predecessor, and that any lessor member is linked > to any greater through a finite set of successors. > Since TO cannot impose this structure on the points of a line, he must > mean something else. > I mean everything above except for the limitation of finiteness on the number > of successors between any two points. > So why do you claim that I am using the limit to measure distance or > length? I am using the limit to take the limit of a sequence of curves, > nothing more. To measure distance or length of a set of points in R^2 with the usual > metric, I use the usual approach, just like everyone else. That's good, because this approach doesn't work. > It has worked perfectly for everyone for thousands of years. Why should > it stop working now? Tony, why don't you just ignore Virgil? This could be a huge step forward for you and for all of us. While almost everything Virgil says is (of course) correct, almost none of it is designed to do anything except goad you into spouting more nonsense. The above is a classic example: your response to Chas's point (...this approach doesn't work) completely misses the point - so following this up with wittering about whether something has worked for thousands [huh??] of years is totally irrelevant, and an immensely feeble argument anyway. (Dowsing has also worked for thousands of years...) > When you prove the length in the limit is 2 instead of sqrt(2), > that's an indication that the object in the limit is NOT exactly the > same as the diagonal line > As he does not prove any such thing, TO is way off base. > That was the concluson Chas logically drew, erroneously or not. Not. (I mean not erroneously - you need have no doubt about that.) > It wasn't MY version of arc length. That was supposed to be Chas' example of > how lim(x->oo: f(x))=f(lim(x->: x)) (or something similar) doesn't necessarily > hold. Yes, that's exactly what it is. Chas has spelled out, in painstaking detail, exactly why a property (here length) of the limit of a sequence of objects is not the same as the limit of the sequence of the same property of the same objects, using normal mathematical meanings for terms such as limit and sequence. It would be nice to think you might read his post above rather carefully, because you might just learn something. If as you keep claiming you actually understand how the system works (you don't, or you wouldn't use those words), you would be able to tell us some new idea you have - for a definition of a Tlimit, or Tsequence, or something. But you can't. > He says that's the problem with the proof, and I say it's because Chas is > using a version of limit that doesn't lend itself to accurate measure of arc > length. A version of limit? He's using the mathematical meaning of limit. Do keep the hilarity level up by giving us your version of limit. >[Tony Orlow] There's a lot you don't know, despite your opinion of yourself. Oh, bugger! I've already used the word hilarity. > Do you understand what I'm saying about the elements of the staircase needing > to be parallel to the diagonal in order for the usual metric to work? Define metric. (Hint: it's in Chas's post, so you'll have to invent your own, to make it different.) > No, you are misunderstanding me. I reject your example as demonstrating that > the concept of an inductive proof of equality holding in the infinite case is > invalid, because the error in your staircase example is easily explainable in > the way I've been describing. I think this does suggest something interesting. You obviously think that maths is basically about calculation and formulae, and I suspect you think that proofs are something like what mathematicians call heuristics. To get full marks on the paper you need to write some stuff down, so you pull out a likely looking heuristic, and try it - if it gives the correct answer, great, otherwise move on and try the next heuristic. That's about the only way I can make sense of your claim above. (Reminiscent of JSH, in a way - he tries to prove theorems by producing strong examples, you know like All odd numbers are prime. Proof: well, 3, 5, 7, 11, and 57 are pretty convincing examples; how could it be otherwise?) > I use the limit to find the limit of a sequence of curves. I don't > measure anything with it; nor have I claimed that you /should/ be > able to measure anything with it. Fine so far. > I use the /usual method/ to measure the lengh of any curve, both > before and after taking the limit, based on sets of points in R^2 using > the usual metric. What is the usual method for measuring the staircase in the limit? As far as > I'm concerned, you proved it had length 2. Define measuring the staircase in the limit. There is a sequence of staircases, there is the limit of the sequence of staircases (all these are sets of points in R^2); then there is the length of various things. > If you consider Chas' argument impeccable despite my far more exact argument > which explains the precise error, then you really have no clue about the nature > of analytical thought. Hmm. Another keeper? Brian Chandler http://imaginatorium.org === Subject: Re: Calculus XOR Probability Matt Gutting said: > Matt Gutting said: > Virgil said: >> If f(2) and g(2) are both defined, how can this be unless 2 is in the >> domain of both? > Unless the set generated by applying f to the naturals contains 2, 2 will not > be a value used as a parameter to g. It need not be in S, which is the domain > of g. However, this means g(2) is not a natural. It's still a real number. :) >> If 2 is not in S, and S is the domain of g, then g(2) is not defined. A function >> is only defined over its domain. >> Matt > > The domains and codomains of the functions f and g are R. That means f(x) for x > in R returns a y in R as a result. If N is a proper subset of R then S will > also be a proper subset of R, because for every unique r in R but not in N > there is a unique f(r) in R but not in S. g(2) will always be defined, but may > not be in N. If that's the case, then 2 isn't in S. That's very basic. Doesn't > anyone get this? I know I'm not crazy. Am I dreaming? > Perhaps I'm confused, but I appear to see a line in the text above which seems > to have been written by you and which says S ... is the domain of g. Yes, where the domain of f is N. In general, the domains of f anf g are both R. Is the domain of x^2 R? Yes. What's the codomain of n^2 for n in N? Is it still R, or some countable subset thereof? > Matt > *** For really sucky service, try http://www.teranews.com *** -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Matt Gutting said: > Perhaps I'm confused, but I appear to see a line in the text above which > seems > to have been written by you and which says S ... is the domain of g. > Yes, where the domain of f is N. In general, the domains of f anf g are both > R. > Is the domain of x^2 R? Yes. But that function, f(x) = x^2 with domain R is not invertible. TO's ignorance trips him up again. === Subject: Re: Calculus XOR Probability > Matt Gutting said: > Matt Gutting said: > Virgil said: >> If f(2) and g(2) are both defined, how can this be unless 2 is in the >> domain of both? > Unless the set generated by applying f to the naturals contains 2, 2 will not > be a value used as a parameter to g. It need not be in S, which is the domain > of g. However, this means g(2) is not a natural. It's still a real number. :) >> If 2 is not in S, and S is the domain of g, then g(2) is not defined. A function >> is only defined over its domain. >> Matt > The domains and codomains of the functions f and g are R. That means f(x) for x > in R returns a y in R as a result. If N is a proper subset of R then S will > also be a proper subset of R, because for every unique r in R but not in N > there is a unique f(r) in R but not in S. g(2) will always be defined, but may > not be in N. If that's the case, then 2 isn't in S. That's very basic. Doesn't > anyone get this? I know I'm not crazy. Am I dreaming? >> Perhaps I'm confused, but I appear to see a line in the text above which seems >> to have been written by you and which says S ... is the domain of g. > Yes, where the domain of f is N. In general, the domains of f anf g are both R. > Is the domain of x^2 R? Yes. What's the codomain of n^2 for n in N? Is it still > R, or some countable subset thereof? The codomain of n^2, for n in N, is a subset of N (and therefore a countable subset of R). It is not R. Matt === Subject: Re: Calculus XOR Probability Matt Gutting said: > Matt Gutting said: > Matt Gutting said: > Virgil said: >> If f(2) and g(2) are both defined, how can this be unless 2 is in the >> domain of both? > Unless the set generated by applying f to the naturals contains 2, 2 will not > be a value used as a parameter to g. It need not be in S, which is the domain > of g. However, this means g(2) is not a natural. It's still a real number. :) >> If 2 is not in S, and S is the domain of g, then g(2) is not defined. A function >> is only defined over its domain. >> Matt > The domains and codomains of the functions f and g are R. That means f(x) for x > in R returns a y in R as a result. If N is a proper subset of R then S will > also be a proper subset of R, because for every unique r in R but not in N > there is a unique f(r) in R but not in S. g(2) will always be defined, but may > not be in N. If that's the case, then 2 isn't in S. That's very basic. Doesn't > anyone get this? I know I'm not crazy. Am I dreaming? >> Perhaps I'm confused, but I appear to see a line in the text above which seems >> to have been written by you and which says S ... is the domain of g. > > Yes, where the domain of f is N. In general, the domains of f anf g are both R. > Is the domain of x^2 R? Yes. What's the codomain of n^2 for n in N? Is it still > R, or some countable subset thereof? > The codomain of n^2, for n in N, is a subset of N (and therefore a countable > subset of R). It is not R. Right, it's a countable subset of R. > Matt -- Smiles, Tony === Subject: Re: Calculus XOR Probability >> Matt Gutting said: >> Matt Gutting said: >> Virgil said: > If f(2) and g(2) are both defined, how can this be unless 2 is in the > domain of both? >> Unless the set generated by applying f to the naturals contains 2, 2 will not >> be a value used as a parameter to g. It need not be in S, which is the domain >> of g. However, this means g(2) is not a natural. It's still a real number. :) > If 2 is not in S, and S is the domain of g, then g(2) is not defined. A function > is only defined over its domain. Matt >> The domains and codomains of the functions f and g are R. That means f(x) for x >> in R returns a y in R as a result. If N is a proper subset of R then S will >> also be a proper subset of R, because for every unique r in R but not in N >> there is a unique f(r) in R but not in S. g(2) will always be defined, but may >> not be in N. If that's the case, then 2 isn't in S. That's very basic. Doesn't >> anyone get this? I know I'm not crazy. Am I dreaming? > Perhaps I'm confused, but I appear to see a line in the text above which seems > to have been written by you and which says S ... is the domain of g. >> Yes, where the domain of f is N. In general, the domains of f anf g are both R. >> Is the domain of x^2 R? Yes. What's the codomain of n^2 for n in N? Is it still >> R, or some countable subset thereof? > The codomain of n^2, for n in N, is a subset of N (and therefore a countable > subset of R). It is not R. > Matt You may be confusing codomain and range. The range is a subset of the codomain. The codomain is whatever you decide it is. The natural interpretation of f(n^2) for n in N is that f is a mapping from N to N, in which case the domain is N and the codomain is N. The range is a proper subset of N. However you could have the following functions f1: N -> N ; f1(x)=x^2 f2: N -> Z ; f2(x)=x^2 f3: N -> R ; f3(x)=x^2 f4: R -> R ; f4(x)=x^2 f5: {0, 1, 2, 3} -> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ; f5(x)=x^2 f6: {0, 1, 2, 3} -> {0, 1, 4, 9} ; f6(x)=x^2 These are all different functions, with different codomains (except f3 and f4 which have the same codomain). If you look at it from a programming angle, the code for the first four would be different: unsigned f1(unsigned x) { return x*x; } int f2(unsigned x) { return x*x; } float f3(unsigned x) { return x*x; } float f4(float x) { return x*x; } f5 and f6 look real similar, but f6 is a bijection, and f5 is not. Determining whether a function is a bijection or not requires that the codomain be specified. Just given that f(x)=x^2 is not enough to determine if the function is a bijection. Stephen === Subject: Re: Calculus XOR Probability > Matt Gutting said: > Matt Gutting said: > Virgil said: >> If f(2) and g(2) are both defined, how can this be unless 2 is in the >> domain of both? > Unless the set generated by applying f to the naturals contains 2, 2 will not > be a value used as a parameter to g. It need not be in S, which is the domain > of g. However, this means g(2) is not a natural. It's still a real number. :) >> If 2 is not in S, and S is the domain of g, then g(2) is not defined. A function >> is only defined over its domain. >> Matt > The domains and codomains of the functions f and g are R. That means f(x) for x > in R returns a y in R as a result. If N is a proper subset of R then S will > also be a proper subset of R, because for every unique r in R but not in N > there is a unique f(r) in R but not in S. g(2) will always be defined, but may > not be in N. If that's the case, then 2 isn't in S. That's very basic. Doesn't > anyone get this? I know I'm not crazy. Am I dreaming? >> Perhaps I'm confused, but I appear to see a line in the text above which seems >> to have been written by you and which says S ... is the domain of g. > Yes, where the domain of f is N. In general, the domains of f anf g are both R. > Is the domain of x^2 R? Yes. What's the codomain of n^2 for n in N? Is it still > R, or some countable subset thereof? >> The codomain of n^2, for n in N, is a subset of N (and therefore a countable >> subset of R). It is not R. >> Matt > You may be confusing codomain and range. The range is a subset of > the codomain. The codomain is whatever you decide it is. > The natural interpretation of f(n^2) for n in N is that f is a mapping > from N to N, in which case the domain is N and the codomain is N. > The range is a proper subset of N. > However you could have the following functions > f1: N -> N ; f1(x)=x^2 > f2: N -> Z ; f2(x)=x^2 > f3: N -> R ; f3(x)=x^2 > f4: R -> R ; f4(x)=x^2 > f5: {0, 1, 2, 3} -> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ; f5(x)=x^2 > f6: {0, 1, 2, 3} -> {0, 1, 4, 9} ; f6(x)=x^2 > > These are all different functions, with different codomains > (except f3 and f4 which have the same codomain). If you > look at it from a programming angle, the code for the first > four would be different: > unsigned f1(unsigned x) { return x*x; } > int f2(unsigned x) { return x*x; } > float f3(unsigned x) { return x*x; } > float f4(float x) { return x*x; } > f5 and f6 look real similar, but f6 is a bijection, and f5 > is not. Determining whether a function is a bijection or not > requires that the codomain be specified. Just given that > f(x)=x^2 is not enough to determine if the function is a > bijection. > Stephen Matt === Subject: Re: Calculus XOR Probability Matt Gutting said: > Virgil said: > The length of a broken or polygonal line depends only on the location >> of of the joints and not at all on the direction of the segments >> between joints. > > What is the slope of the staircase at x=0. It's infinite. Does that ever change > as the staircase approaches the limit? No. That point at x=0 always has slope > infinite, as opposed to the slope of the diagonal. The direction between 0,0 > and 0,1/n never changes. > But you're taking into account multiple points (2 points). The sequence of > staircases converges pointwise; you don't *have* a staircase as the limit. Uh, yeah, you do have an infinite sequence of infinitesimal steps, and the slope at x=0 remains infinite. > Virgil: >> That may be a TO requirement, but it is not a mathematical one. The only >> mathematical requirement for a curve to have a length is the the lengths >> of its polygonal approximations have a finite upper bound, and then >> that upper bound is then defined to be the length. > > Tony: > If the endpoints are on the curve. > Whether the endpoints are on the curve or not, as long as the series of > lengths is convergent. Gee, the series of your lengths on the staircase converges to the same value it has in all finite cases. Some point on the segments must be parallel to the curve at a point on the curve perpendicular to a point in that segment, or your measure is off. > Matt > *** Read all about James Harris on http://www.teranews.com *** -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Matt Gutting said: > But you're taking into account multiple points (2 points). The > sequence of staircases converges pointwise; you don't *have* a > staircase as the limit. > > Uh, yeah, you do have an infinite sequence of infinitesimal steps, > and the slope at x=0 remains infinite. On the contrary, WE do not. TO may have mathematical impossibilities for dinner as often as he can swallow them, but we do not have to have them at all. > > Virgil: >> That may be a TO requirement, but it is not a mathematical one. >> The only mathematical requirement for a curve to have a length >> is the the lengths of its polygonal approximations have a >> finite upper bound, and then that upper bound is then defined >> to be the length. > > > Tony: > If the endpoints are on the curve. > > Whether the endpoints are on the curve or not, as long as the > series of lengths is convergent. > Gee, the series of your lengths on the staircase converges to the > same value it has in all finite cases. Some point on the segments > must be parallel to the curve at a point on the curve perpendicular > to a point in that segment, or your measure is off. Each individual approximation may be off, but if the LUB of those approximations is on, that is all that counts. At least in math. What goes on outside of math, mathematicians, as mathematicians, can afford to ignore. === Subject: Re: Calculus XOR Probability > Matt Gutting said: > Virgil said: >> The length of a broken or polygonal line depends only on the location >> of of the joints and not at all on the direction of the segments >> between joints. > What is the slope of the staircase at x=0. It's infinite. Does that ever change > as the staircase approaches the limit? No. That point at x=0 always has slope > infinite, as opposed to the slope of the diagonal. The direction between 0,0 > and 0,1/n never changes. >> But you're taking into account multiple points (2 points). The sequence of >> staircases converges pointwise; you don't *have* a staircase as the limit. > Uh, yeah, you do have an infinite sequence of infinitesimal steps, and the > slope at x=0 remains infinite. I hate to repeat myself, but in the absence of a precise, non-circular definition of infinite, the truth value of the above statement can't be evaluated. >> Virgil: >> That may be a TO requirement, but it is not a mathematical one. The only >> mathematical requirement for a curve to have a length is the the lengths >> of its polygonal approximations have a finite upper bound, and then >> that upper bound is then defined to be the length. >> Tony: > If the endpoints are on the curve. >> Whether the endpoints are on the curve or not, as long as the series of >> lengths is convergent. > Gee, the series of your lengths on the staircase converges to the same value > it has in all finite cases. Some point on the segments must be parallel to the > curve at a point on the curve perpendicular to a point in that segment, or your > measure is off. How can a point be parallel, or perpendicular, to anything? Matt === Subject: Re: Calculus XOR Probability Matt Gutting said: > Matt Gutting said: > Virgil said: >> The length of a broken or polygonal line depends only on the location >> of of the joints and not at all on the direction of the segments >> between joints. > What is the slope of the staircase at x=0. It's infinite. Does that ever change > as the staircase approaches the limit? No. That point at x=0 always has slope > infinite, as opposed to the slope of the diagonal. The direction between 0,0 > and 0,1/n never changes. >> But you're taking into account multiple points (2 points). The sequence of >> staircases converges pointwise; you don't *have* a staircase as the limit. > > Uh, yeah, you do have an infinite sequence of infinitesimal steps, and the > slope at x=0 remains infinite. > I hate to repeat myself, but in the absence of a precise, non-circular > definition of infinite, the truth value of the above statement can't be > evaluated. Oh come on. Infinite slope isn't even quaestionable. The stair is always vertical at the point x=0. > >> Virgil: >> That may be a TO requirement, but it is not a mathematical one. The only >> mathematical requirement for a curve to have a length is the the lengths >> of its polygonal approximations have a finite upper bound, and then >> that upper bound is then defined to be the length. >> Tony: > If the endpoints are on the curve. >> Whether the endpoints are on the curve or not, as long as the series of >> lengths is convergent. > > Gee, the series of your lengths on the staircase converges to the same value > it has in all finite cases. Some point on the segments must be parallel to the > curve at a point on the curve perpendicular to a point in that segment, or your > measure is off. > > How can a point be parallel, or perpendicular, to anything? The segment is parallel to the curve at a point. > Matt > *** Matt orders all his beanie babies from http://www.teranews.com *** -- Smiles, Tony === Subject: Re: Calculus XOR Probability Matt Gutting said: > Mike Kelly said: > But the smallest infinity of standard theories isn't an infintie >> value like you've been talking about. It describes an equivalence >> class. > > Yes, I know, but it's called the size of the set, and yet, doesn't satisfy very > many intuitions about what a set size is. You can add an infinite number of > additional elements to an infinite set without changing the cardinality. I find > that objectionable. And, it's referred to as the smallest infinite ordinal > number, as if it's some kind of quantity, but it's really not. So, with all > due respect to von Neumann, I wouldn't give his ordinals such a central place > in the theory. > *Some* people call it the size of the set; it certainly corresponds to the > size of the set when the set is finite. In my mind, it's not exactly the same > thing. Well, you have a good mind then. :) > > positions within the set finite, there are no two elements with an infinite > number of elements between them. So, despite the boundlessness which makes > injection into a proper subset possible, I can't consider this set to be > actually infinite in size. >> What's the difference between not finite and actually infinite? > The way I see it, a dimensionless quantity can be finite, greater than finite > and infinite, or less than finite and infinitesimal. I see countably > infinite as unboundedly large but finite. >> The english-language definition of finite is something like bounded >> or limited. The idea that the naturals are bounded but unbounded is >> probably causing some of my confusion here. > > Well, that's understandable. Infinite means literally without end, and the > standard set theoretical deifninition classifies the set of finite naturals as > countably infinite. My issue with that is that, when no two elements in the > ordered set can ever have any infinite number of other elements between them, > then I don't see as there is an infinite number of elements in the set. There's > no infinite _number_ of elements. > Considering that you still don't have a definition of infinite (the two you > have provided - or is it three? - either are circular, or don't go into enough > detail to allow me to distinguish infinite numbers from finite, I'm not sure > what exactly you mean by an infinite _number_ of elements. I mean a number of elements such that any sequential ordering of them would have pairs of elements with more than any finite number of elements between them. No such pairs exist in the standard naturals. > >> Are you saying that sets cannot have properties their elements do not? >> So you cannot have an infinite set of finite values? I think that's >> faulty logic. >> Question : The rationals in [0,1] are countably infinite. Yet there are >> only internally infinite, if I understand you right? > > Okay, you're new to this discussion. When I see this question I often get > rather irked, because it's usually from someone I have explained this to half a > dozen times. So, I'll try to patiently explain my position without getting > annoyed. :) > > the set theoretical definition we would say yes, since it is possible to form > an injection from the set into a proper subset, like the evens. However, from > the perspective of the set having an actually infinite quantity of elements > within it, I would say it does not qualify. COnsider this inductive argument. > As a base case of a set of consecutive naturals with a given size, consider the > set {1}. It contains 1 element, 1, so it has size 1, which is also its maximal > element. As we add successive elements, with each one we increment the set size > by adding an element, and simultaneously increment the maximal element by > adding the incremented value of the previous maximal element. So, with each > additional element incrementing both the set size and the maximal value, both > properties of the set remain forever equal. This means that, if one's > definition of infinite is consistently applied to both of these values, one > cannot become infinite unless the other does. So, the set size cannot be > infinite without an infinite maximal element. There's really no way to avoid > this logic except to bury one's head in the sand. > All the sets you're considering are sets with maximal elements. The natural > numbers do not form such a set. I don't see that the argument applies. Well, what I am talking about are the elements of the set of naturals, and noting that none of them is infinitely beyond the beginning of the set. > > No. It does't allocate a block of memory and then return a pointer to >> that memory. All the pointers are already defined, before you look at >> them. > > Function Successor(x) > { Successor(x+1) > } > > Are you saying this recursive function will generate the naturals all > immediately, or immediately run out of memory? I've heard that before. All the > elements spring from the axioms all at once. That's not very analytical. > The successor function isn't a generating function for the natural numbers; > it's a relation between each natural number and another one. > I'm not sure what you mean by not very analytical. Analysis means literally cutting apart. When you just say the set springs from the axioms, that doesn't explain much about the nature of it. It doesn't look at how the set grows incrementally. > > So you cannot perform *any* calculations of *any* kind without numbers? > > Tony: > Can you? > > Define calculation. I can certainly do calculations on groups of permutations. > Or on sets of, say, letters. You can do calculations using various quantitative measures of such things. By calculation I mean specifically a process that determines a quantitative result given quantitative values as input. > > Why should there be a number, at all? > > Tony: > Because there are points, and you can count their number. It's just that you'll > never finish counting, so the number is infinite. > If I'm counting something, the number of things I count is the last word I > say when I stop counting. If there isn't such a last word, how can I say > there's a corresponding number? Yopu can compare sets formulaically over the infinite domain using an infinite unit of measure. > It's not my fault if we can't get past the notion of an infinite number of > elements in a set. > If you could define infinite number and give axioms for the behavior of > infinite numbers, we could get past that notion. Perhaps the rules regarding whether results of arithmetic operators are finite, infinite or infinitesimal given inputs of those types would suffice? > > What does length mean to you? > > Tony: > It means distance between two points, measured in some unit intverals end to > end. > Then the length of the real line must mean the distance between two points > on the real line, measured in some unit intverals [sic] end to end, which > in turn appears to mean that the real line has two ends. Yes, you can picture Big'un as a conceptual end to the line, just like 12:00 noon was the conceptual end to the ball adding phase of the vase problem. > Mike: >> What about >> 2, 4, 6, 8, 10, 12, ... >> 1, 2, 3, 4, 5, 6, ... > Tony: > Over the same value range, the first has half as many elements. > But there's no value range here, as there isn't any maximal element. Ho hum. Wake up! Within any GIVEN value range, the first has half as many elements. No Largest Finite!!! (GONG) Huyah huyah huyah Ommmmm.....ega! > The idea that the sequences >> 0, 1, 2, 3, ... >> 1, 2, 3, 4, ... >> 2, 3, 6, 8, ... >> have diferent lengths is neither intuitive not consistent. > > Even though the first is a proper superset of the others? Hmmm.... > They line up right nice... :-) Yeah, one within the other. > Matt -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Matt Gutting said: > Considering that you still don't have a definition of infinite (the two you > have provided - or is it three? - either are circular, or don't go into > enough > detail to allow me to distinguish infinite numbers from finite, I'm not > sure > what exactly you mean by an infinite _number_ of elements. > I mean a number of elements such that any sequential ordering of them would > have pairs of elements with more than any finite number of elements between > them. No such pairs exist in the standard naturals. Since every standard meaning of sequential ordering prohibits TO's requirement, he has managed to eliminate all infinite sets from his set theory. > All the sets you're considering are sets with maximal elements. The natural > numbers do not form such a set. I don't see that the argument applies. > Well, what I am talking about are the elements of the set of naturals, and > noting that none of them is infinitely beyond the beginning of the set. No real or rational is infinitely beyond the beginning at 0. > When you just say the set springs > from the axioms, that doesn't explain much about the nature of it. It doesn't > look at how the set grows incrementally. The set of naturals doesn't grow. Sets don't grow, they just are. The set of naturals springs full grown from the axioms of Peano. > You can do calculations using various quantitative measures of such things. > By > calculation I mean specifically a process that determines a quantitative > result given quantitative values as input. But that presumes the very numbers which you cannot have without the set theory which you reject. > Yopu can compare sets formulaically over the infinite domain using an > infinite > unit of measure. Is he some relation to Yehudi? And no one else can do it. === Subject: log question I have a exercise in a book: Simplify: log - log12 + log2 Notice that in the first term there's no number in the power position. My friend thinks that log = log 1 (by convention), but I can't find that rule anywhere, and she doesn't remember where she saw it. I know that by convention if the *base* position is left blank, it's 10, so it seems at least credible that there's a similar convention for the power position. So, is the question a typo error, or is there a rule that log = log 1 (or some other number). === Subject: Re: probability and contradiction > Virgil said: > > Virgil said: > > > The stuff involving whether the child > is first- or second-born doesn't make a difference to whether it's a > boy or girl. > > Actually, I believe records show the incidence of first children being > male is slightly higher that for subsequent children. > > > Okay, that may be true, but I swear, Virgil, if I said the sky was blue, > you'd > say it wasn't, and if I said 1<>2 you'd prove them equal. > > I only correct TO when he is wrong. > > Besides not mentioning some possibly confounded result about first borns > being > 50.001% male, was I wrong in anything else I said? At least I didn't confirm > that the answer was 2/3. :) I have never claimed that TO was NEVER right, particularly about things outside mathematics. It is just that his incidence of errors about things mathematical is humongous for one trying to present himself as at all knowledgeable about things mathematical. That, and his resistance to being corrected when he is wrong, scuttles any chances he might have had to progress very far in mathematics. Unless it is all mere trolling. === Subject: Re: probability and contradiction > Suppose you meet me on the street corner while I'm with my son and I tell > you that I have another child at home. So what is the probability that my > other child is a girl. I look at the reduced sample space which is bb, bg, > gb, where the x in xy is the sex of the 1st born and y is the sex of the > 2nd born. Now I conclude that prob(other child is a girl ) = 2/3. (I seem > to remember this result from my probability class years ago). The probability space is {bb, gb, gg}. The probably of getting both a boy and a girl is 1/2. This is exactly what you are asking. Note that the gg event is impossible because you told me it was since you have atleast 1 son. You are treating it has if he just asked you What ist he probability that I have atleast 1 girl. This is called conditional probability: P(of atleast 1 girl) = 2/3 P(of a girl given that atleast one is a boy) = 1/2 = (1/3)/(2/3) The key here that is since you knwo one is a boy your real sample space for the question is {bb, bg}. i.e., really you don't even care that the first child is a boy.... it in no way reflects what the second child is(atleast we are assuming that). > Now suppose you meet me on the street corner while I'm with my son and I > tell you that this child is my 1st born and I have another child at home. > What is the prob that my other child is a girl. Well the sample space is > bb, bg and the p(other child is a girl) = 1/2. > Similarly if you meet me on the street corner while I'm with my son and I > tell you that this child is my 2nd born and I have another child at home. > What is the prob that my other child is a girl. Well the sample space is > bb, gb and the p(other child is a girl) = 1/2. Now you have changed the sample space... you are taking into account order. one has as a new sample space { b1b2, b1g2, g1b2, g1g2} or just {bb, bg, gb, gg} You are asking what is the probability of the second born being a girl... we see there are 2 possibilities out of 4 so its 1/2. (which is the exact same you asked before in some sense.) > Note that the last 2 cases had the same result, namely p(other child is a > girl) = 1/2. > But there is where the contradiction lies since if you meet me on the > street corner while I'm with my son and I tell you that this child is my > ??? born and I have another child at home. (??? means that you could not > understand what I said ( the noisy train was passing)). So the p(other > child is a girl) = 1/2. But wait a minute if you couldn't hear whether the > son with me was my 1st born or 2nd born then this reduces to the 1st case > above and the p(other child is a girl) = 2/3. Well which is it, 1/2 or > 2/3??????????????? There is no contradition. Sometimes it might be better to assume you are wrong if it doesn't make sense then to assume theres some flaw with mathematics or nature. You can use the sample space {bb, bg, gb, gg} to solve the first problem too: You say What is the probability of the child at home being a girl given that atleast one is a boy... counting we see there it could be bg or gb... i.e., 2/4 = 1/2 The second problem states What is the probably of the second born being a girl given that the first born is a boy? i.e., bg out of the sample space {bb, bg} = 1/2. Note that it doesn't matter which question you ask becaues ultimately you know with probability 1 that atleast one of the children is a boy... hence that even is certain and cannot in any way influence how you treat figuring out what the second one is. I mean, hell, its almost like asking Whats the probability that my son here is a boy, or Whats the probability that the sky is blue given that my 49th child is a boy and has green hair with two heads.. You just have to realize that these are independent trials and then everything will fall into place and its simply asking Whats the chance of my child being a girl? = 1/2. (Doesn't matter what kinda hoopla you put in front of it cause it won't change anything(well, there are things you can say that will make it change but)). Jon === Subject: Re: probability and contradiction > [...] >The fact that you heard or not that this is my >*****born child >does not change it. [...] >So yes, the probability is 2/3, and not 1/2. >>Astonishing. > By which you presumably mean nonsense? No, I would actually go for astonishing :-) After you think about it for a while, it is quite astonishing the subtlety in the difference between the two problems. If you pick a family with two children at random, and the family is found to have one boy, then the probability that the family also has one girl is 2/3. If you pick *a child* at random among the families with two children, and the child is found to be a boy, then the probability that the other child in the family is a girl is 1/2. This is the case here, where you're running into the person with the one randomly selected child. It is astonishing the subtle distinction between running into a person and seeing that person with a boy, and the fact that you randomly selected a family and that family is found to have one boy... I mean, really, the distinction is, I'd say, unbelievably subtle -- and it is quite amazing that that subtle distinction changes so radically the result. I now feel a bit frustrated, because I heard this problem many years ago, from one of my friends that enjoy puzzles and tricky questions -- my initial reaction was that well duh, the probability has to be 1/2, but then figured that that was too easy, and that there had to be some tricky detail ... After a few hours of thought, I actually solved it (I mean, I fell for the misleading logic, and reached the correct result of 2/3, which was confirmed by my friend). I felt so happy thinking that it was one of those cases when the apparently obvious answer is incorrect, for some twisted -- yet absolutely irrefutable -- reason. You know, kind of like the puzzle of the three doors, one of them hiding a prize -- the apparently obvious answer is that well DUH! any choice that you do has 1/3 probability of being correct, so changing the selection can not change the probability -- and it turns out to be incorrect (the problem we have in this thread feels soooo similar to this one...) Oh well, I'll let my friend know that we've been wrong all these years! :-) A little bit in my defense -- notice that my main claim when responding to the OP was that the fact that you hear or not that the child is the ****born (*if* that made a difference) has nothing to do with the correct result; it simply has to do with one's ability to determine the correct result (the fact that you didn't hear all the conditions of the problem does not change the problem, and therefore doesn't change the result -- it only changes your ability to determine the solution) Carlos -- === Subject: Re: probability and contradiction > [...] >The fact that you heard or not that this is my >*****born child >does not change it. [...] So yes, the probability is 2/3, and not 1/2. >>Astonishing. > > > By which you presumably mean nonsense? > No, I would actually go for astonishing :-) > After you think about it for a while, it is quite > astonishing > the subtlety in the difference between the two > problems. > If you pick a family with two children at random, and > the family > is found to have one boy, then the probability that > the family > also has one girl is 2/3. If you pick *a child* at > random > among the families with two children, and the child > is found > to be a boy, then the probability that the other > child in the > family is a girl is 1/2. > This is the case here, where you're running into the > person > with the one randomly selected child. > It is astonishing the subtle distinction between > running > into a person and seeing that person with a boy, and > the > fact that you randomly selected a family and that > family is > found to have one boy... I mean, really, the > distinction > is, I'd say, unbelievably subtle -- and it is quite > amazing that that subtle distinction changes so > radically > the result. > I now feel a bit frustrated, because I heard this > problem > many years ago, from one of my friends that enjoy > puzzles > and tricky questions -- my initial reaction was that > well > duh, the probability has to be 1/2, but then figured > that > that was too easy, and that there had to be some > tricky > detail ... After a few hours of thought, I actually > solved it (I mean, I fell for the misleading logic, > and > reached the correct result of 2/3, which was > confirmed > by my friend). I felt so happy thinking that it was > one > of those cases when the apparently obvious answer is > incorrect, for some twisted -- yet absolutely > irrefutable -- > reason. > You know, kind of like the puzzle of the three doors, > one > of them hiding a prize -- the apparently obvious > answer is > that well DUH! any choice that you do has 1/3 > probability > of being correct, so changing the selection can not > change > the probability -- and it turns out to be incorrect > (the > problem we have in this thread feels soooo similar to > this one...) > Oh well, I'll let my friend know that we've been > wrong > all these years! :-) > A little bit in my defense -- notice that my main > claim > when responding to the OP was that the fact that you > hear > or not that the child is the ****born (*if* that made > difference) has nothing to do with the correct > result; > it simply has to do with one's ability to determine > the > correct result (the fact that you didn't hear all the > conditions of the problem does not change the > problem, > and therefore doesn't change the result -- it only > changes your ability to determine the solution) > Carlos > -- Can you show the result formally? Perhaps using conditioning. === Subject: Re: is dis-interest in mathematics evolutionarily preferred? > It is the contribution of *mathematicians in general* > Nobody is quibbling about whether math has made useful contributions. > Every other topic like basket weaving or > physical education has also made useful contributions too. The issue > is whether LOVE of creating more and more > abstract mathematics is evolutionarily preferred. The answer, I > suggest, is No. Even if you are right, so what? First of all, not all of us think sex is the meaning of life. I therefore question your harsh use of the term evolutionary suicide. Choosing a life of either celibacy or mathematical study isn't tantamount to killing oneself. It's just a way of life. Second, it is *because we help you* in making your technology and luxury possible that you should respect us by not calling us losers. We are your friends. Don't displace your aggression onto us. It is with our mathematics that scientists and engineers are creating technological paradise. Sit back and enjoy the fun. === Subject: Re: is dis-interest in mathematics evolutionarily preferred? On 3 May 2006 00:18:25 -0700, magic math tricks > No, fitness is decided by the all-mighty dollar. No, it's decided by the ability to survive. (Survival of the fittest. Remember?) Mathematics is thousands of years old and therefore is fitter than you are. -- http://hertzlinger.blogspot.com === Subject: Re: is dis-interest in mathematics evolutionarily preferred? false. evolution applies to a gene pool -- it does not apply to the individual phenotypes that manifest the gene pool. thus, the apt comparison is between mathematics and humans of a particular phenotype (e.g., MDs). Humans of my phenotype are much more fit than mathematicians.