mm-3959 === Subject: Re: Manifolds >>I have a question about topological 1-manifolds. Given a connected >>topological 1-manifold X (that is, a Hausdorff space with countable >>basis such that every point has a neighborhood which is homeomorphic to >>R), can we say that X is homeomorphic either to R or S^1? I have >>seen a proof of this result for smooth manifolds, but I am wondering if >>it is true for topological manifolds. If it is true, how difficult is >>the proof? > It is true, and I don't think the proof is difficult (but I haven't > come up with one that completely convinces me, in five minutes). What about the following outline of a proof: * For manifolds, connectedness implies path-connectedness. (The set of points which can be connected to a given point in X by a path is open and closed.) [A path in X is a continuous map from a *compact* interval into X.] * If two points x,y in X can be connected by a path, they can be connected by an injective path. (Cut out the middle part of a given path between the first point where injectivity fails, and the last point, where this value is taken on.) * Every injective map from a compactum into a Hausdorff space is a homeomorphism onto its image - i.e. an *embedding*. * If there exist two points in X which can be connected by two injective paths that have different images, then there exists a continuous, injective map from S^1 into X; hence, an embedding of S^1 into X. * If X contains an embedded S^1, it must be identical to this S^1. (If not, there will exist a point in X S^1. Connect this point by a path to (any point in) S^1. Then, the first point w on this path that lies on the S^1 has the property that for every small enough neighborhood U, the open set U{w} consists of at least *three* path-connected components. - This cannot happen, since X is a 1-dimensional manifold around w.) + So assume from now on that X is not diffeomorphic to S^1. Then by the preceding argument, to every pair of points x,y in X, there exists an injective continuous path from x to y, and the image P(x,y) of this path in X is *uniquely*(!) determined by x and y. Let us call such subsets P(x,y) of X *intervals*, and the points x and y the *endpoints of the interval* P(x,y). * The idea for the remaining part is to find an exhaustion of X by a countable sequence of intervals, I_0 subset I_1 subset I_2 subset ... subset X, X = Union of all (I_n), and to use these intervals to identify X with R. * For the latter task, start with the interval I_0, and with a homeomorphism psi_0 : [-1,1] --> I_0. Set a_0 := -1, b_0 := +1. Given I_n, a_n, b_n, psi_n and the interval I_(n+1) that contains I_n one can find an extension of psi_n to a homeomorphism psi_(n+1) : [a_(n+1),b_(n+1)] --> I_(n+1), psi_(n+1) | [a_n,b_n] = psi_n, such that a_(n+1) is either a_n or (a_n - 1), and b_(n+1) is either b_n or (b_n + 1) (depending on how the endpoints of I_(n+1) coincide with those of I_n). All homeomorphisms psi_n together yield a homeomorphism psi : Union of all [a_n,b_n] --> Union (I_n) = X. So, depending on the form of the union of intervals on the right, X will be homeomorphic to either the whole of R, or a compact interval, or a half-open interval. (The latter two are manifolds with boundary.) * Remains to be said how an exhaustion of X by intervals is achieved: This is where the countable basis of the topology comes in, because, together with the local charts of X as a 1-manifold, it makes that X can be covered by a *countable* union of intervals J_n. Set I_0 := J_0, and I_(n+1) := the smallest interval that contains I_n and J_(n+1). (This will be one of the intervalls of the form P(x,y), where x and y run through all 6 possible combinations of endpoints of I_n and J_(n+1).) This ends the proof. > I think the most illuminating approach would be to show that, > on the one hand, any such X can be triangulated (not necessarily > finitely, of course) and so has the structure of a piecewise-linear > 1-manifold; and, on the other, any piecewise-linear 1-manifold has > the structure of a smooth 1-manifold. (I *do* see how to do this > in the compact case, but at the moment the non-compact case is > eluding me, though I'm sure it's not really hard--using your > countability assumption.) > Lee Rudolph === Subject: equivalent defs of local compact hausdorff space Hi all, While studying topology I encounter different definitions of a local compact Hausdorff space, which should be equivalent. I was just wondering about the proof of the equivalence of the following definitions: (Assume (X,T) is a topological space which is Hausdorff) (1) in each point x in X, there exists a compact neighborhood U of x (2) each point x in X has a basis of neighborhoods containing compact sets (2) ==> (1) is clear, but I don't see why (1) ==> (2). PS An analogous question is but it didn't got any answers. === Subject: Re: equivalent defs of local compact hausdorff space > While studying topology I encounter different definitions of a local compact > Hausdorff space, which should be equivalent. > I was just wondering about the proof of the equivalence of the following > definitions: > (Assume (X,T) is a topological space which is Hausdorff) > (1) in each point x in X, there exists a compact neighborhood U of x > (2) each point x in X has a basis of neighborhoods containing compact sets No, you need to be more precise, because {x} is compact set, so every nhood contains a compact sets. > (2) ==> (1) is clear, but I don't see why (1) ==> (2). If x in open U, then some compact K with x in int K. Look at U / int K Now use theorem, that locally compact Hausdorff implies regular. (In fact, completely regular or Tychonov.) Thus some open V with x in V, cl V subset U / int K subset U. Show cl V is compact. I leave for you to described the quested base. > PS An analogous question is > but it didn't got any answers. Oh please, post it here in edited form. Another, Hausdorff equivalent locally compact, or as some put it, strongly locally compact is for all x, some open U nhood x with compact cl U. But I've already show how to prove that. === Subject: Re: equivalent defs of local compact hausdorff space Let X be a locally compact Hausdorff space, x a point, C a compact neighborhood of x and U an arbitrary neighborhood of x. Let W be the interior of U intersect C. Since cl(W) is a compact Hausdorff space, W contains a closed compact set V which is a neighborhood of x in cl(W). But V is also a neighborhood of x in W (relative topology) and therefore is a neighborhood of x in X. === Subject: Re: equivalent defs of local compact hausdorff space Please include context, otherwise serious discussion is not possible. > Let X be a locally compact Hausdorff space, x a point, C a compact > neighborhood of x and U an arbitrary neighborhood of x. Let W be the > interior of U intersect C. Since cl(W) is a compact Hausdorff space, > W contains a closed compact set V which is a neighborhood of x in > cl(W). But V is also a neighborhood of x in W (relative topology) and > therefore is a neighborhood of x in X. -- To Google and MathForum users: Reply only if adequate context is included _within_ the reply. Otherwise all contexts are removed from my view, the flow of thought disrupted and chaos reigns. In particular for Google users: Instead of simply hitting the prominent Reply link, which doesn't include a copy of the post to which one is replying, click the Show Options link (toward the top of an item in the thread), which causes a shaded area of links to appear next to the top of the item, including Reply (first) that does introduce a copy of the previous text (offset by > signs in the usual fashion). ---- === Subject: Re: equivalent defs of local compact hausdorff space > Hi all, > While studying topology I encounter different definitions of a local compact > Hausdorff space, which should be equivalent. > I was just wondering about the proof of the equivalence of the following > definitions: > (Assume (X,T) is a topological space which is Hausdorff) > (1) in each point x in X, there exists a compact neighborhood U of x > (2) each point x in X has a basis of neighborhoods containing compact sets > (2) ==> (1) is clear, but I don't see why (1) ==> (2). Perhaps you should provide the exact reference where you saw (2). What were the other hypotheses? === Subject: Re: equivalent defs of local compact hausdorff space >> Hi all, >> While studying topology I encounter different definitions of a local >> compact Hausdorff space, which should be equivalent. >> I was just wondering about the proof of the equivalence of the following >> definitions: >> (Assume (X,T) is a topological space which is Hausdorff) >> (1) in each point x in X, there exists a compact neighborhood U of x >> (2) each point x in X has a basis of neighborhoods containing compact >> sets >> (2) ==> (1) is clear, but I don't see why (1) ==> (2). > Perhaps you should provide the exact reference where you saw (2). What > were the other hypotheses? Each point x in X has a basis of neighborhoods, in which all sets are compact. I.e. for each x in X there exists a set B(x) of compact neighborhoods around x with the property that for every (arbitrary) neighborhood V, there is a B in B(x) such that B subset V. This is a definition for locally compact Hausdorff space, I found in some topology books. === Subject: Re: equivalent defs of local compact hausdorff space >> While studying topology I encounter different definitions of a local >> compact Hausdorff space, which should be equivalent. >> I was just wondering about the proof of the equivalence of the following >> definitions: >> (Assume (X,T) is a topological space which is Hausdorff) >> (1) in each point x in X, there exists a compact neighborhood U of x >> (2) each point x in X has a basis of neighborhoods containing compact >> sets >> >> (2) ==> (1) is clear, but I don't see why (1) ==> (2). > > Perhaps you should provide the exact reference where you saw (2). What > were the other hypotheses? > Each point x in X has a basis of neighborhoods, in which all sets are > compact. I.e. for each x in X there exists a set B(x) of compact > neighborhoods around x with the property that for every (arbitrary) > neighborhood V, there is a B in B(x) such that B subset V. > This is a definition for locally compact Hausdorff space, I found in some > topology books. Well, that definition is incorrect, so why not reveal the name of the culprit book? Or perhaps other hypothesis is there in addition, such as Hausdorff (which does not follow from this)?? Do you know the theorem that a compact Hausdorff space is normal? === Subject: Re: equivalent defs of local compact hausdorff space >> While studying topology I encounter different definitions of a local >> compact Hausdorff space, which should be equivalent. >> I was just wondering about the proof of the equivalence of the following >> definitions: >> (Assume (X,T) is a topological space which is Hausdorff) >> (1) in each point x in X, there exists a compact neighborhood U of x >> (2) each point x in X has a basis of neighborhoods containing compact >> sets Perhaps you should provide the exact reference where you saw (2). What > were the other hypotheses? > Each point x in X has a basis of neighborhoods, in which all sets are > compact. I.e. for each x in X there exists a set B(x) of compact > neighborhoods around x with the property that for every (arbitrary) > neighborhood V, there is a B in B(x) such that B subset V. > This is a definition for locally compact Hausdorff space, I found in some > topology books. > Well, that definition is incorrect, so why not reveal the name of the > culprit book? Or perhaps other hypothesis is there in addition, > such as Hausdorff (which does not follow from this)?? Under the hypothesis of regular or of Hausdorff, it's equivalent to the usual two definition of locally compact. It however isn't preferred for the others are simpler, easier to work with and more common. > Do you know the theorem that a compact Hausdorff space is normal? Irrelevant. What's relevant is locally compact Hausdorff implies regular. === Subject: Re: equivalent defs of local compact hausdorff space Originator: grubb@lola >> Each point x in X has a basis of neighborhoods, in which all sets are >> compact. I.e. for each x in X there exists a set B(x) of compact > neighborhoods around x with the property that for every (arbitrary) >> neighborhood V, there is a B in B(x) such that B subset V. >> This is a definition for locally compact Hausdorff space, I found in some >> topology books. >Well, that definition is incorrect, so why not reveal the name of the >culprit book? Or perhaps other hypothesis is there in addition, >such as Hausdorff (which does not follow from this)?? The OP pretty clearly stated the assumption that the space is Hausdorff. In this context, having one compact neighborhood of each point and having a base of neighborhoods at each point are equivalent. The proof is done by restricting to the given compact neighborhood and using the fact that every closed neighborhood in a compact set is a compact neighborhood. Normality of compact Hausdorff spaces helps. --Dan Grubb === Subject: Re: equivalent defs of local compact hausdorff space >Hi all, While studying topology I encounter different definitions of a local > compact Hausdorff space, which should be equivalent. > I was just wondering about the proof of the equivalence of the > following definitions: > (Assume (X,T) is a topological space which is Hausdorff) > (1) in each point x in X, there exists a compact neighborhood U of x > (2) each point x in X has a basis of neighborhoods containing compact > sets (2) ==> (1) is clear, but I don't see why (1) ==> (2). > >> >> Perhaps you should provide the exact reference where you saw (2). What >> were the other hypotheses? >> Each point x in X has a basis of neighborhoods, in which all sets are >> compact. I.e. for each x in X there exists a set B(x) of compact >> neighborhoods around x with the property that for every (arbitrary) >> neighborhood V, there is a B in B(x) such that B subset V. >> This is a definition for locally compact Hausdorff space, I found in some >> topology books. > Well, that definition is incorrect, so why not reveal the name of the > culprit book? Or perhaps other hypothesis is there in addition, > such as Hausdorff (which does not follow from this)?? Look, for example, on wikipedia (http://en.wikipedia.org/wiki/Local_compactness): 'a topological space X is locally compact iff every point has a local base of compact neighborhoods'. Besides this i am speaking about **Hausdorff** locally compact spaces. > Do you know the theorem that a compact Hausdorff space is normal? Let U be a compact neighborhood around x. Because X is Hausdorff, U is so too. U is compact Hausdorff, so it is normal. Normality implies that for each closed set S and open set O with S in O, there exists V (and cl V) so that S subset V subset cl(V) subset O. Could I use this? I don't see which S and O I should choose to prove that that every open neighborhood contains a compact neighborhood. === Subject: Re: equivalent defs of local compact hausdorff space Originator: grubb@lola >Let U be a compact neighborhood around x. Because X is Hausdorff, U is so >too. U is compact Hausdorff, so it is normal. Normality implies that for >each closed set S and open set O with S in O, there exists V (and cl V) so >that S subset V subset cl(V) subset O. Could I use this? I don't see >which S and O I should choose to prove that that every open neighborhood >contains a compact neighborhood. Let's be more explicit. Let U be an open set around x with compact closure. Let V be any other open set around x. We want to find a compact neighborhood of x contained in V. Well, x is in U intersect V and by regularity, there is a set W which is open in cl(U) and such that the closure of W in cl(U) is contained in U intersect V. Then W intersect U intersect V is open in the big space and has compact closure contained in V. --Dan Grubb === Subject: Re: equivalent defs of local compact hausdorff space While studying topology I encounter different definitions of a local >> compact Hausdorff space, which should be equivalent. >> I was just wondering about the proof of the equivalence of the following >> definitions: >> (Assume (X,T) is a topological space which is Hausdorff) >> (1) in each point x in X, there exists a compact neighborhood U of x >> (2) each point x in X has a basis of neighborhoods containing compact >> sets >> (2) ==> (1) is clear, but I don't see why (1) ==> (2). > Perhaps you should provide the exact reference where you saw (2). What > were the other hypotheses? > Each point x in X has a basis of neighborhoods, in which all sets are > compact. A base is a collection of open sets. What you describe is called a network. > I.e. for each x in X there exists a set B(x) of compact > neighborhoods around x with the property that for every (arbitrary) > neighborhood V, there is a B in B(x) such that B subset V. > This is a definition for locally compact Hausdorff space, I found in some > topology books. Another more common variant is a base of open precompact sets. precompact means cl U is compact. However when space is regular (or Hausdorff) they are equivalent even to the to the simple version, every point in some compact nhood. === Subject: Re: defferentiation and fourier transforms. On Tue, 04 Apr 2006 15:46:53 EDT, jackblack Thm 1: >Suppose f is Lebesgue integrable on R, and F is the Fourier Transform function. Then if >Int(1 + |w|^k)|F(f)(w)|dw is finite, then f is k differentiable a.e. True. As stated this is not all that useful, because almost-everywhere differentiability is not that useful a concept. In order to be able to give simple counterexamples, let's state that stronger and more useful version of this result. Taking k = 1 to make the result easier to state: Thm 2: Suppose that f is integrable. If int(1 + |w|) |F(f)(w)| is finite then f is absolutely continuous (or strictly speaking, f = g almost everywhere, where g is absolutely continuous). Saying that f is absolutely continuous implies three things: (i) f is differentiable almost everewhere (ii) f' is integerable (iii) f(b) = f(a) = int_a^b f'(t) dt. It's (iii) that's needed for interesting applications of all this; condition (iii) does not follow just from knowing that f is differentiable almost everywhere, but (iii) does follow from the hypotheses of the theorem. So we should really talk about Theorem 2 instead of your statement. >Is this just because F(f)(w) is of order 1/|w|^k then f is k differentiable No. No at least twice: First, the hypothesis (for k = 1) does not imply that F(f)(w) _is_ of order 1/|w|. And second, even if the hypothesis did imply that bound, it's not true that saying F(f) is of that order implies that f is absolutely continuous. For example, let f be the characteristic function of the interval [-1,1]. You can easily calculate F(f)(w) and see that it _is_ of order 1/|w|, although f is certainly not absolutely continuous, since it's not even continuous. >or is there more to this? >Is this though the upper bound? Basically, is there a function which is not k times differentiable, but F(f)(w) is of order 1/|w|^k+1 ? Yes. ************************ David C. Ullrich === Subject: A globally convergent series for the zeta function A globally convergent series for the zeta function valid for all complex-valued s except s=1, was conjectured by Konrad Knopp and proven by Helmut Hasse in 1930. Could someone please help me with a reference for a proof of this formula? === Subject: Re: A globally convergent series for the zeta function > A globally convergent series for the zeta function valid for all > complex-valued s except s=1, was conjectured by Konrad Knopp and proven > by Helmut Hasse in 1930. > Could someone please help me with a reference for a proof of this > formula? It might be this: Hasse, H. Ein Summierungsverfahren f.9fr die Riemannsche Zeta-Reihe. Math. Z. 32, 458-464, 1930. It might also be here: Sondow, J. Analytic Continuation of Riemann's Zeta Function and Values at Negative Integers via Euler's Transformation of Series. Proc. Amer. Math. Soc. 120, 421-424, 1994. I found this at MathWorld, http://mathworld.wolfram.com/RiemannZetaFunction.html around formula 19. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: A globally convergent series for the zeta function > A globally convergent series for the zeta function valid for all > complex-valued s except s=1, was conjectured by Konrad Knopp and proven > by Helmut Hasse in 1930. > Could someone please help me with a reference for a proof of this > formula? MathWorld ( http://mathworld.wolfram.com/RiemannZetaFunction.html ) gives the following reference: Hasse, H. Ein Summierungsverfahren f.9fr die Riemannsche Zeta-Reihe. Math. Z. 32, 458-464, 1930. === Subject: Re: Computational Number Theory >I'm interested in algorithms related to number theory, and I have no >problem with the math aspect of it, but I am a little weak when it >comes to the computer side of things. Specifically, what type of >software, hardware are used to implement the algorithms? I know that >it probably varies between contexts (teaching, cutting edge research), >but if anybody could give me any pointers, I would appreciate it. Your interest makes you a perfect customer for my favorite (recently) paper. (quoting)Recent years have seen the flowering of what is often termed computer-assisted or experimental mathematics, namely the utilization of modern computer technology as an active tool in mathematical research. In particular, a combination of commercial software (notably Mathematica and Maple), online tools and custom-written computer programs are being used to test conjectures, discover new identities, perform symbolic manipulations, plot data and even conduct formal proofs. (quoted from the paper Future Prospects for Computer-Assisted Mathematics by Bailey and Borwein. http://crd.lbl.gov/~dhbailey/dhbpapers/math-future.pdf Computer assisted mathematics is not completely equivalent to number theory, of course. I had enormous fun demonstrating simple number theory with an extended integer package, BC running under Unix. I imagine that as soon as you reach analytical number theory, floating point capability becomes a consideration. John === Subject: Re: Computational Number Theory <8va7329p927rmmvqd12raaac08q1m67boe@4ax.comI'm interested in algorithms related to number theory, and I have no >problem with the math aspect of it, but I am a little weak when it >comes to the computer side of things. Specifically, what type of >software, hardware are used to implement the algorithms? I know that >it probably varies between contexts (teaching, cutting edge research), >but if anybody could give me any pointers, I would appreciate it. > Your interest makes you a perfect customer for my favorite (recently) > paper. > (quoting)Recent years have seen the flowering of what is often termed > computer-assisted or experimental mathematics, namely the > utilization of modern computer technology as an active tool in > mathematical research. In particular, a combination of commercial > software (notably Mathematica and Maple), online tools and > custom-written computer programs are being used to test conjectures, > discover new identities, perform symbolic manipulations, plot data and > even conduct formal proofs. > (quoted from the paper Future Prospects for Computer-Assisted > Mathematics by Bailey and Borwein. > http://crd.lbl.gov/~dhbailey/dhbpapers/math-future.pdf > Computer assisted mathematics is not completely equivalent to number > theory, of course. I had enormous fun demonstrating simple number > theory with an extended integer package, BC running under Unix. I > imagine that as soon as you reach analytical number theory, floating > point capability becomes a consideration. > John === Subject: Re: _Graph_of_ a function is a relation. Gerry Myerson ha scritto nel messaggio >> Can someone tell me why the destiction between a function and it's graph >> is made in the statement?: The _graph_of_ a function is a relation. >> The graph of a function IS the function (one representation of it), is >> it not? > Would you say a picture of your mother is your mother? > A picture of your mother is a representation of your mother. > For some purposes, a picture of your mother is just as good > as your mother - for example, I could probably tell whether > I know your mother just by looking at her picture, without > actually seeing her. For some purposes, a picture of your > mother might even be better than your actual mother - for > sticking in a scrapbook and carrying around with you, for > instance. But, better or worse, a picture of your mother > isn't your mother. And the graph of a function isn't the > function. Why not? A function, as Magidin said can be viewed in several ways. But a function f:S->T in certain contexts (logic and universal algebra) is just a particular relation, i.e. a particular subset of a cartesian product S x T. A function is different by its graph if you consider the function f:S->T as a pair (SXT,G), where G is the graph. === Subject: Re: _Graph_of_ a function is a relation. days. My association with the Department is that of an alumnus. >Gerry Myerson ha scritto nel messaggio > Can someone tell me why the destiction between a function and it's graph > is made in the statement?: The _graph_of_ a function is a relation. > The graph of a function IS the function (one representation of it), is > it not? >> Would you say a picture of your mother is your mother? >> A picture of your mother is a representation of your mother. >> For some purposes, a picture of your mother is just as good >> as your mother - for example, I could probably tell whether >> I know your mother just by looking at her picture, without >> actually seeing her. For some purposes, a picture of your >> mother might even be better than your actual mother - for >> sticking in a scrapbook and carrying around with you, for >> instance. But, better or worse, a picture of your mother >> isn't your mother. And the graph of a function isn't the >> function. >Why not? A function, as Magidin said can be viewed in several ways. Actually, my point is that function can be DEFINED in several ways. Depending on your definition, a function may or may not be the same things as its graph. >A function is different by its graph if you consider the function f:S->T as >a pair (SXT,G), where G is the graph. Or, slightly more commonly, a triple (S,T,G). -- === Subject: Re: _Graph_of_ a function is a relation. > Can someone tell me why the destiction between a function and it's graph > is made in the statement?: The _graph_of_ a function is a relation. > The graph of a function IS the function (one representation of it), is > it not? > The graph of a relation IS the relation, is it not (book I am using > agrees here)? A binary relation over S is a subset of S^2. f:S -> S is also a subset of S^2, hence technically a (binary) relation. f:N -> P(N), n -> { j in N | j <= n } is not a subset of S^2 for any S. Hence isn't a relation. > My Discrete Maths textbook makes the distinction because (based on its > definitions of these things) the graph of a function from A to B is a > subset of A X B , and so is a relation (subset of A X B) from A to B. I dispute that binary relations are other than subsets of S^2 for some S. An n-ary relation over S is a subset of S^n. > The definition it gives for a function has no mention of A X B, but it > is still obvious that it could have been mentioned there. > No mention of the graph of a relation is made. Book simply implying > that graph of relation IS the relation, but doesn't do the same for a > function. > What are some thoughts on this? Is there an advanced math level > reason why the distinction between a function and it's graph is made? > Or am I missing something? The graph is a topological space, the function is a mapping from one space to another space and doesn't per se have any topology or spacial structure. For example, the functions may be continuous and the graph compact. The converse concepts, continuous space and compact function are humors. A continuous map of a connected space has a connected graph. The converse fails, the a map of a connected space with a connected graph may not be continous. === Subject: Re: _Graph_of_ a function is a relation. >> Can someone tell me why the destiction between a function and it's graph >> is made in the statement?: The _graph_of_ a function is a relation. >> The graph of a function IS the function (one representation of it), is >> it not? >> The graph of a relation IS the relation, is it not (book I am using >> agrees here)? >A binary relation over S is a subset of S^2. >f:S -> S is also a subset of S^2, hence technically a (binary) relation. >f:N -> P(N), n -> { j in N | j <= n } is not a subset of S^2 for any S. >Hence isn't a relation. Nonsense. It's not a realtion on N and is not a relation on P(N), but the notion of relation is more general than the notion of binary relation over a set. >I dispute that binary relations are other than subsets of S^2 for some S. And I dispute the insinuation that the original poster restricted himself to binary relations over a set. But, to address your disputations: A set R is a realtion if each element of R is an ordered pair. Later If R is a relation included in a cartesian product X x Y [...] it is sometimes conveniento to say that R is a relation from X to Y [emphasis in the original]. Bergman's An Invitation to General Algebra and Universal Definition 4.1.1. If X_1,....,X_n are sets, a relation on X_1,...,X_n means a subset R contained in X_1 x ... x X_n. Relations are often written as predicates; i.e., the condition (x_1,...,x_n) in R may be written as R(x_1,..,x_n), or Rx_1...x_n, of when n=2 as x_1 R x_2 >An n-ary relation over S is a subset of S^n. And there are relations that are not n-ary relations. >> What are some thoughts on this? Is there an advanced math level >> reason why the distinction between a function and it's graph is made? >> Or am I missing something? >The graph is a topological space, Really? Funny. I always thought the graph of a function f:X->Y was {(x,f(x)) : x in X}. Don't recall it being an ordered pair, the first element a set and the second element of the pair being a family of subsets satisfying the axioms of a topology. -- === Subject: Re: _Graph_of_ a function is a relation. What are some thoughts on this? Is there an advanced math level >> reason why the distinction between a function and it's graph is made? >> Or am I missing something? >The graph is a topological space, > Really? Funny. I always thought the graph of a function f:X->Y was > {(x,f(x)) : x in X}. As OP asked, advanced math level graph G_f is the space inherited from domain f x codomain f > Don't recall it being an ordered pair, the first element a set and the > second element of the pair being a family of subsets satisfying the > axioms of a topology. Isn't in set theory, is in topology. The graph of f:(X,tau) -> (Y,Tau), G_f = ({ (x,f(x) | x in X }, T) where T is the topology generated from the base { UxV / {(x,f(x) : x in X} | U in tau, V in Tau } G_f subset XxY is of course a relation by liberal standards. --) Press 1) for a reality of your choice Press 2) for a reality of somebody else's choice Press 3) for the standard reality Press 4) for a non-standard reality or stay on the line for Sir Reality. === Subject: Re: _Graph_of_ a function is a relation. > What are some thoughts on this? Is there an advanced math level > reason why the distinction between a function and it's graph is made? > Or am I missing something? >The graph is a topological space, >> Really? Funny. I always thought the graph of a function f:X->Y was >> {(x,f(x)) : x in X}. >As OP asked, advanced math level graph >G_f is the space inherited from domain f x codomain f And exactly what does this inherit if domain f and codomain f are not topological spaces? >> Don't recall it being an ordered pair, the first element a set and the >> second element of the pair being a family of subsets satisfying the >> axioms of a topology. >Isn't in set theory, is in topology. >The graph of f:(X,tau) -> (Y,Tau), > G_f = ({ (x,f(x) | x in X }, T) >where T is the topology generated from the base > { UxV / {(x,f(x) : x in X} | U in tau, V in Tau } Ah. So you are ASSUMING that X and Y are topological spaces to begin with. Rather unwarranted, in my opinion. There was absolutely no indication that this would be the case. You jumped the gun. -- === Subject: Re: The True Story on Root Solving. DEDICATED TO ALL YOUNG MATH STUDENTS This is a typical response to the original posting from mine. I said that no mathematician is able to deny that such simple and trivial arithmetical methods HAVE NO PRECEDENTS in the whole history of root solving. Your response to my posting is: to talk about the term well-defined which has no relevance to the crude fact that by means of the extremely simple ARITHMETICAL operation called: Rational Mean we can develope higher-order algorithms for solving roots of any degree, all this meaning that mathematicians of past times had at hand the elementary tool to develop such algorithms, however they didn't, so these so simple and trivial methods have no precedents at all. These new arithmetical algorithms ermbraces, ___among many new others___, all those well-known analytical methods: Newton's, Halley's, Householder, Bernoulli's for solving roots of any degree. Sumarizing, for instance, I am stating that from now on you will be able to develope Householder's method by means of the most simple arithmetic, this means that ancient mathematicians certainly had at hand the elementary arithmetical tools to develope such higher-order algorithm, however they didn't, and consequently mathematicians of modern times had to rely on the creation of the artificial Cartesian-system. That is the scence of my posting, the well-defined subject have no relevance at all, mainly because the phrase well-defined within the set of rational numbers has its fundamentals exclusively on the Cartesian- decimal DOGMA. This is well explained in my webpage and the book, based on all this, I can tell you that the Cartesian system is NOT WELL-DEFINED within the Natural set of the harmonies of Number ruled by the simple, natural and intelligible nexus of Quantity called The Rational Mean which brings to light a new true and natural scheme of Quantity. A new universal and natural principle of Quantity embracing all kind of algorithms for generating Number. >I don't see any results about how fast the RM processes converge, in >terms of linear or quadratic convergence. (I only see results about the >first few iterations.) The analytical methods you've listed do have >such results. linear or quadratic convergence? convergence, it deals with higher-order convergence (cubic, quartic, quintic, etc.). As said, in my webpage there many examples on Householder iterating functions generated by means of the most simple arithmetic (using the Arithmonic Mean as a especial case of the Rational Mean) , of course: __among many other new funtions__. My apologizes, indeed, I do not intend to be rude, but If you are not able to analyze and deduce the rate of convergence of all those well-kown householder's functions (or any other new iterating function shown in my webpage) then I do not understand the reasons you dare to bring any answer to my posting. Domingo Gomez Morin === Subject: Re: The True Story on Root Solving. DEDICATED TO ALL YOUNG MATH STUDENTS > This is a typical response to the original posting from mine. > [grandstanding snipped] >I don't see any results about how fast the RM processes converge, in >terms of linear or quadratic convergence. (I only see results about the >first few iterations.) The analytical methods you've listed do have >such results. > linear or quadratic convergence? > convergence, it deals with higher-order convergence > (cubic, quartic, quintic, etc.). > As said, in my webpage there many examples on Householder iterating > functions generated by means of the most simple arithmetic (using the > Arithmonic Mean as a especial case of the Rational Mean) , of course: > __among many other new funtions__. > My apologizes, indeed, I do not intend to be rude, but If you are not > able to analyze and deduce the rate of convergence of all those > well-kown householder's functions (or any other new iterating function > shown in my webpage) then I do not understand the reasons you dare to > bring any answer to my posting. I asked you about rates of convergence because you didn't have them posted on your website, whereas the rate of convergence for Newton's Method (for example) has been established. I was merely asking whether you had worked out these results and not posted, or whether you haven't worked them out. In short: If you use one of the RM methods and iterate it n times, how many decimal places accuracy will you have? *** The Fractal Fractions page reminded me of a problem that I ran across a few years ago. Typesetting it in plain text is a monster, so I'll rewrite it as: PROBLEM. If A(N) = 1 + B(N)/C(N), B(N) = 2 + C(N)/A(N), C(N) = N + A(N)/B(N), find the limit of A(N) as N approaches infinity. I'm sure the numbers 1, 2, and N were involved in the definitons, but I might have some things in the wrong places. The answer should be the golden section phi. === Subject: Re: The True Story on Root Solving. DEDICATED TO ALL YOUNG MATH STUDENTS ....... > *** > The Fractal Fractions page reminded me of a problem that I ran across a > few years ago. Typesetting it in plain text is a monster, so I'll > rewrite it as: > PROBLEM. If A(N) = 1 + B(N)/C(N), B(N) = 2 + C(N)/A(N), C(N) = N + > A(N)/B(N), find the limit of A(N) as N approaches infinity. > I'm sure the numbers 1, 2, and N were involved in the definitons, but I > might have some things in the wrong places. The answer should be the > golden section phi. > Should some of those Ns be N+1s? Also, are the values of A(1), B(1), and C(1) important? === Subject: Re: The True Story on Root Solving. DEDICATED TO ALL YOUNG MATH STUDENTS *** > The Fractal Fractions page reminded me of a problem that I ran across a > few years ago. Typesetting it in plain text is a monster, so I'll > rewrite it as: > PROBLEM. If A(N) = 1 + B(N)/C(N), B(N) = 2 + C(N)/A(N), > C(N) = N + A(N)/B(N), find the limit of A(N) as N approaches > infinity. > I'm sure the numbers 1, 2, and N were involved in the definitons, but I > might have some things in the wrong places. The answer should be the > golden section phi. > Should some of those Ns be N+1s? No ... Maybe I should write out the fraction in plain text after all. (Use a constant-width font like Courier to view this properly.) A(N) would be 1 + ... N + --------- 2 + ... 2 + --------------- 2 + ... 1 + --------- N + ... 1 + --------------------- 2 + ... 1 + --------- N + ... N + --------------- N + ... 2 + --------- 1 + ... (unless I've forgotten the order that the 1, 2, and N go in). === Subject: Detemining a function from sample inputs and outputs Hi there, I'm working on an electronics problem where I have a device that reads a temperature and outputs a number. I have a short list of values of what the device outputs and what the teperature it was reading is. Is it possible given these inputs and outputs to create a formula to derive one from the other? If I graph the data I have the result appears to me to be a straight line, with small variations that I put down to rounding on behalf of the device. Phill === Subject: Re: Detemining a function from sample inputs and outputs > I'm working on an electronics problem where I have a device that reads > a temperature and outputs a number. > I have a short list of values of what the device outputs and what the > teperature it was reading is. > Is it possible given these inputs and outputs to create a formula to > derive one from the other? If I graph the data I have the result > appears to me to be a straight line, with small variations that I put > down to rounding on behalf of the device. What you want to do is called curve fitting or nonlinear regression. It is the process of fitting a general function which may not be linear to a set of data values. Please take a look at my NLREG nonlinear regression program at http://www.nlreg.com It should do exactly what you want. -- Phil Sherrod (phil.sherrod 'at' sandh.com) http://www.dtreg.com (decision tree and SVM predictive modeling) http://www.nlreg.com (nonlinear regression) === Subject: Re: Detemining a function from sample inputs and outputs Commercial packages such as Matlab (expensive) and Excel provide curve fits. Usually they will give you an equation as well. === Subject: Re: Detemining a function from sample inputs and outputs knowledge of it I really didn't know what to search the help for :) A search for Curve fits gave me the LINEST function which gave me what I needed. === Subject: Re: Detemining a function from sample inputs and outputs knowledge of it I really didn't know what to search the help for :) A search for Curve fits gave me the LINEST function which gave me what I needed. === Subject: Re: Detemining a function from sample inputs and outputs knowledge of it I really didn't know what to search the help for :) A search for Curve fits gave me the LINEST function which gave me what I needed. === Subject: Free Music Downloads Hey check this out: http://www.kohit.net/ === Subject: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? how to prove that on R^1 space, where (a,b], { or [a,b) } is an interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an integer number, which belongs to this interval ? please for any help. === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? > how to prove that on R^1 space, where (a,b], { or [a,b) } is an > interval which lebesgue length l=1; ( b-a=1 ) there always exists an > integer number, which belongs to this interval ? please for any help. R is tiled by I = (a,a+1] and its integer shifts I+n = (a+n,a+n+1] But integerfree I -> integerfree I+n -> integerfree R = / I+n -><- Alternatively: True if a or b in N; else suffices to show (b-1,b) / N nonempty. Exists least n in N such b-1 < n -> n in (b-1,b) since b > n [ else b < n -> b-1 < n-1 contra n least ] QED --Bill Dubuque === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? > how to prove that on R^1 space, where (a,b], > { or [a,b) } is an interval which lebesgue lenght > l=1; ( b-a=1 ) there always exists an integer > number, which belongs to this interval ? Use (a, b]. We are given b-a = 1. Proof by contradiction. Assume there are no integers in the interval. If a is an integer then a+1 = b is an integer. But b is in the interval, which contradicts our assumption. So, a and b are not integers. Let n be the greatest integer less than a. Let m be the smallest integer greater than b. Then m - n > 1. But m = n+1, contradiction. === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? > how to prove that on R^1 space, where (a,b], { or [a,b) } is an > interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an > integer number, which belongs to this interval ? please for any help. Let c be the largest integer <= b. Show b-c < 1 by contradiction (hint c+1 is also an integer.) -William Hughes === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? William Hughes napisal(a): > how to prove that on R^1 space, where (a,b], { or [a,b) } is an > interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an > integer number, which belongs to this interval ? please for any help. > Let c be the largest integer <= b. Show b-c < 1 by contradiction > (hint c+1 is also an integer.) > -William Hughes === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? > William Hughes napisal(a): > how to prove that on R^1 space, where (a,b], { or [a,b) } is an > interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an > integer number, which belongs to this interval ? please for any help. > Let c be the largest integer <= b. Show b-c < 1 by contradiction > (hint c+1 is also an integer.) > -William Hughes If b-c >=1 can c be the largest integer <=b ? If b-c < 1 can we conclude that c is in (a,b] ? -William Hughes I === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? its good , i ll be glad. so let d=max{z in Z: zits good , i ll be glad. so let d=max{z in Z: zinteger, so d+1 is an integer to so:// d<=a and now we see, >that an integer d+1 is between a and b. is it ok too ?? Not quite. You want an integer in [a,b) or in (a,b]. Your d+1 could be equal to b, in which case, it is no good in the first case. But what happens if you are looking at [a,b) and you get d+1=b? > question2: >about Your posts, Since you did not bother to QUOTE the message you are replying to, I have no idea what you are talking about here. Do show some manners and learn to quote the message you are replying to in order ot provide context. See: >why do i ned to check, that b-c<1? what does it give. What is c? Perhaps the other poster worked with the ceiling instead of the floor? No idea, since ->you did not provide context<-. -- === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? > how to prove that on R^1 space, where (a,b], { or [a,b) } is an > interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an > integer number, which belongs to this interval ? please for any help. You might consider the greatest integer less than or equal to a, and the least integer greater than or equal to b, and then think about the possibilities. === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? Robert Low napisal(a): > how to prove that on R^1 space, where (a,b], { or [a,b) } is an > interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an > integer number, which belongs to this interval ? please for any help. > You might consider the greatest integer less than or equal to a, > and the least integer greater than or equal to b, and then > think about the possibilities. ok. so let z be agreatest integer less than or equal to a, and c the least integer greater than or equal to b. there we will have z<=ahow to prove that on R^1 space, where (a,b], { or [a,b) } is an >interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an >integer number, which belongs to this interval ? please for any help. For the case of (a,b], take ceil(a), the ceiling of a. For [a,b), take floor(b), the floor of b. ceil(x) = min{ z in Z : z >= x} floor(x) = max{ z in Z : z<= x} Prove that 0<=ceil(x)-x < 1 for all x, and 0<=floor(x)-x<1 for all x. That will do it. -- === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? >how to prove that on R^1 space, where (a,b], { or [a,b) } is an >interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an >integer number, which belongs to this interval ? Call this [*] > please for any help. > For the case of (a,b], take ceil(a), the ceiling of a. For [a,b), take > floor(b), the floor of b. > ceil(x) = min{ z in Z : z >= x} > floor(x) = max{ z in Z : z<= x} > Prove that 0<=ceil(x)-x < 1 for all x, and 0<=floor(x)-x<1 for all > x. That will do it. Begs the question. We now must ask: how do you define floor (or ceil)? One would think [*] would be required for this: Given x, there is an integer in (x-1,x] because it has length 1. Define floor(x) to be that integer. So: how do you define floor WITHOUT using [*], or (better) how do you prove [*] without using floor/ceil/int? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? days. My association with the Department is that of an alumnus. >>how to prove that on R^1 space, where (a,b], { or [a,b) } is an >>interval which lebesgue lenght l=1; ( b-a=1 ) there always exists an >>integer number, which belongs to this interval ? >Call this [*] >> please for any help. >> For the case of (a,b], take ceil(a), the ceiling of a. For [a,b), take >> floor(b), the floor of b. >> ceil(x) = min{ z in Z : z >= x} >> floor(x) = max{ z in Z : z<= x} >> Prove that 0<=ceil(x)-x < 1 for all x, and 0<=floor(x)-x<1 for all >> x. That will do it. >Begs the question. We now must ask: how do you define floor (or >ceil)? >One would think [*] would be required for this: You might; I don't. > Given x, there >is an integer in (x-1,x] because it has length 1. Define floor(x) >to be that integer. >So: how do you define floor WITHOUT using [*], Floor is defined to be floor(x) = max{ z in Z : z<= x}. We know it exists because any nonempty subset of Z which is bounded above has a maximum, and the archimedean property guarantees that this set is nonempty and bounded above (by the archimedean property, there exists n>0 in Z such that n>|x|, so -n is in {z in Z : z<=x}, and all such z are less than or equal to x). Likewise, you can define ceil by ceil(x) = min {x in Z : z>= x}. Again, it exists because any subset of Z which is nonempty and bounded below has a minimum; by the archimedean property the set is nonempty, and bounded below, using the same argument. Once you have the definition, If m = floor(x), then x-m>=0 since m is in {z in Z: z<=x}. And m+1>x, because x+1 is NOT in {z in Z : z<=x} (since m is the minimum. Therefore, x-m<1. No need to use [*] at all. >or (better) how do you prove [*] without using floor/ceil/int? By the archimedean property, there exists n such that n>=a. If you are considering [a,b), use the least such n; if you are considering (a,b], use n if n>a and n+1 if n=a. But this is just using floor/ceil without saying so. -- === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? > Likewise, you can define ceil by > ceil(x) = min {x in Z : z>= x}. I prefer 'ceil(x) = min {z in Z : z>= x}'. Even Homer nods. === Subject: Re: how to prove that in any interval (a,b] with lenght 1 (b-a=1) is always integer number ? >> Likewise, you can define ceil by >> ceil(x) = min {x in Z : z>= x}. >I prefer 'ceil(x) = min {z in Z : z>= x}'. >Even Homer nods. Yeah; there were a number of typos there. But presumably the point was clear: you do not define ceil or floor by looking at intervals of length 1 and finding integers in them. -- === Subject: series and irrationality Here is it: Prove that the series sum_1/(n^a sin(n*pi*sqrt(2)) ) absolutely converges for a>1. Here is what i've done: Since we know and it is not difficult to prove that |sqrt(2)-m/n| >= C/n^2 for some contstant C and for any pair of natuarals m,n, we have |n*pi*sqrt(2)-pi*m| >C/n, so |sin(n*pi*sqrt(2))|>=C/n and we can easily now get that our series is absolutely convergent for a>2. But what should be do for 1sigma-1, where alpha is non-Liouville number of order sigma, i.e. |alpha-m/n|>C(alpha)/n^sigma for all naturals m,n. === Subject: Re: series and irrationality <10540564.1144249479422.JavaMail.jakarta@nitrogen.mathforum.org>, > Here is it: > Prove that the series sum_1/(n^a sin(n*pi*sqrt(2)) ) absolutely converges for > a>1. > Here is what i've done: Since we know and it is not difficult to prove that > |sqrt(2)-m/n| >= C/n^2 for some contstant C and for any pair of natuarals > m,n, we have > |n*pi*sqrt(2)-pi*m| >C/n, so |sin(n*pi*sqrt(2))|>=C/n and we can easily now > get that our series is absolutely convergent for a>2. But what should be do > for 1 *** for all n, but that sin(...) > *** for sufficiently many n . === Subject: Re: series and irrationality > <10540564.1144249479422.JavaMail.jakarta@nitrogen.math > forum.org>, > Here is it: > Prove that the series sum_1/(n^a sin(n*pi*sqrt(2)) > ) absolutely converges for > a>1. Here is what i've done: Since we know and it is not > difficult to prove that > |sqrt(2)-m/n| >= C/n^2 for some contstant C and for > any pair of natuarals > m,n, we have > |n*pi*sqrt(2)-pi*m| >C/n, so > |sin(n*pi*sqrt(2))|>=C/n and we can easily now > get that our series is absolutely convergent for > a>2. But what should be do > for 1 sin(...) > *** for > all n, but that sin(...) > *** for sufficiently many > n . Could you please, specify.Do you know how to do it exactly. I don't know how to give some kind of better bond from below for sin(...) than of 1/n growth(which is not sufficeint as you must have seen). === Subject: Re: series and irrationality <16809738.1144335134195.JavaMail.jakarta@nitrogen.mathforum.org>, > <10540564.1144249479422.JavaMail.jakarta@nitrogen.math > forum.org>, Here is it: > Prove that the series sum_1/(n^a sin(n*pi*sqrt(2)) > ) absolutely converges for > a>1. Here is what i've done: Since we know and it is not > difficult to prove that > |sqrt(2)-m/n| >= C/n^2 for some contstant C and for > any pair of natuarals > m,n, we have > |n*pi*sqrt(2)-pi*m| >C/n, so > |sin(n*pi*sqrt(2))|>=C/n and we can easily now > get that our series is absolutely convergent for > a>2. But what should be do > for 1 I think what you need to do is show is not that > sin(...) > *** for > all n, but that sin(...) > *** for sufficiently many > n . > Could you please, specify.Do you know how to do it exactly. I don't know how > to give some kind of better bond from below for sin(...) than of 1/n growth(which is not sufficeint as you must have seen). Given small epsilon>0 and large N, the number of n with n <= N such that |sin(n*pi*sqrt(2))| < epsilon is approximately (2*epsilon/pi)*N. So for most n, |sin(n*pi*sqrt(2))| > epsilon. === Subject: Re: Regarding tan(x) = x > Sorry for the delay in responding, but I can't see why > Cauchy should get credit either. > We mathematicians aren't always good at giving credit > where it's most properly due. The Pythagorean Theorem, > L'Hopital's Rule, ... There's the joke that You might > be a mathematician if your major result will be named > for someone else. I don't think this is a case of what you're talking about. The Cauchy reference was in the Archibald/Bateman reference I gave at Raymond Clare Archibald and Henry Bateman, A guide to tables of Bessel functions, Mathematical Tables and Other Aids to Computation (= Mathematics of Computation) 1 #7 (July 1944), 205-308. See Section F: Series for the zeros of Bessel functions, pp. 271-275. Archibald is well known (to me, at least) for the See , for example. I cited However, I checked the appropriate volume of Cauchy's collected works out of the local university library (easier to flip through than the internet version I gave a link to), and as I said, it's not clear to me whether Cauchy obtained the asymptotic series representation for the solutions to tan(x) = x. I think it's probably implicit from the more general expansions at the bottom of p. 277 and the top of p. 278, which are apparently specialized to tan(x - pi/4) = (8/15)*x - (63/80)*(1/x) + ... tan(x + pi/4) = (8/23)*x - (15/23)*(39/16)*(1/x) + ... at the bottom of p. 278 (but I haven't looked into this), and Cauchy explicitly mentions tan(x) = x on p. 283. (All pages refer to the Cauchy reference I gave in my February 21 post that I cited above.) Incidentally, I'm giving a talk on tan(x) = x at the 2006 meeting of the Iowa Section of The Mathematical Association of America, April 7-8, Iowa State University. Title: The Remarkable Equation tan(x) = x Time: 10:15-10:35 A.M. Saturday April 8 in Concurrent Session 6 http://www.central.edu/maa/Meetings/ As I promised in the other thread, at some point (probably within the next 3 weeks) I'll write a fairly comprehensive summary of everything I've found, in particular all the numerous references I've collected on tan(x) = x. Dave L. Renfro === Subject: Re: Regarding tan(x) = x > Sorry for the delay in responding, but I can't see why > Cauchy should get credit either. > We mathematicians aren't always good at giving credit > where it's most properly due. The Pythagorean Theorem, > L'Hopital's Rule, ... There's the joke that You might > be a mathematician if your major result will be named > for someone else. ...And a really great mathematician if the name of the result is written in lower case (euclidean, abelian,...) [...] === Subject: Re: Regarding tan(x) = x Cauchy should get credit either. We mathematicians aren't always good at giving credit > where it's most properly due. The Pythagorean Theorem, > L'Hopital's Rule, ... There's the joke that You might > be a mathematician if your major result will be named > for someone else. > ...And a really great mathematician if the name of the result is written > in lower case (euclidean, abelian,...) Interesting you should say that. I would regard it at as a spelling error to fail to capitalize such adjectives. Have you got any reputable references showing lower case spelling? -- Larry Lard Replies to group please === Subject: Re: Regarding tan(x) = x Sorry for the delay in responding, but I can't see why > Cauchy should get credit either. We mathematicians aren't always good at giving credit > where it's most properly due. The Pythagorean Theorem, > L'Hopital's Rule, ... There's the joke that You might > be a mathematician if your major result will be named > for someone else. ...And a really great mathematician if the name of the result is written > in lower case (euclidean, abelian,...) > Interesting you should say that. I would regard it at as a spelling > error to fail to capitalize such adjectives. Have you got any reputable > references showing lower case spelling? I think the vast majority of modern algebra texts have abelian in lower case. It's less common to have euclidean in lower case, but hardly unknown. Taking a sampling from my bookshelves: Zariski and Samuel, Commutative Algebra: abelian, euclidean Herstein, Topics in Algebra: abelian, Euclidean Cohn, Algebra: abelian, Euclidean McCoy, Introduction to Modern Algebra: abelian, Euclidean Lang, Algebra: abelian Albert, Fundamental Concepts of Higher Algebra: abelian Moore, Elements of Linear Algebra and Matrix Theory: euclidean Birkhoff & MacLane, A Survey of Modern Algebra: Abelian and Euclidean Burnside, Theory of Groups of Finite Order: Abelian === Subject: Re: Regarding tan(x) = x There's the joke that You might be a mathematician if >> your major result will be named for someone else. > ...And a really great mathematician if the name of the > result is written in lower case (euclidean, abelian,...) Let's hope that someone with the last name Normal [1] never becomes a great mathematician! [1] normal subgroup, normal topological space, normal operator, normal as meaning perpendicular, normal probability distribution, normal form of a matrix, normal extension of a field, normal bundles in differential geometry, ... Dave L. Renfro === Subject: Re: Regarding tan(x) = x >Let's hope that someone with the last name Normal [1] >never becomes a great mathematician! The MathSciNet database doesn't seem to list anybody with that name. On the other hand, there are mathematicians named E, I, Pi, Root, Sine, Grad, Curl, Basis, Eigen, Lemma, Converse, Real, Field, Ring, Plane, Line, and Series (in fact, there's a Laurent Series). === Subject: Looking for Snellius Cyclometricus book by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id k35F9q301900 for ; Wed, 5 Apr 2006 11:09:52 -0400 I'm looking for the ancient Snellius Cyclometricus (Cyclometricvs) of the circle). Is it available online? Any references are welcome! === Subject: Re: estimation of bounds of n! as F(n) < n! < G(n) >> are there any simple functions F(n), G(n) that can act as lower and >> upper bounds of n! ? >Near the end of A new convergent expansion for the gamma function at >I mention, in my second miscellaneous thought, that for N >= 3, >N!/(Sqrt(2*pi)*((N+1/2)/e)^(N+1/2)) lies between (48*N+23)/(48*N+25) > and (24*N+12)/(24*N+13). >Of course, just multiplying through by (Sqrt(2*pi)*((N+1/2)/e)^(N+1/2)) >will then give you simple upper and lower bounds. >And, as I mention there, tighter bounds can also be obtained easily. to which we both posted, Good approximations for binom{a}{b} I mentioned a slight modification of the Euler-Maclaurin Sum Formula: > n > --- 1 7 ''' > > f(k) ~ C + F(n+1/2) - -- f'(n+1/2) + ---- f (n+1/2) > --- 24 5760 > k=1 > 31 (5) 127 (7) > - ------ f (n+1/2) + --------- f (n+1/2) - ... > 967680 154828800 >Where F is the antiderivative of f. This essentially yields the Half-Shift Gamma. Shifting the Euler- Maclaurin Sum Formula eliminates the f(n)/2 term in the usual Sum Formula, leaving only odd derivatives and the integral. This is why the integral evaluated at n+1/2 is usually an order of n better of an approximation than the integral at n, which is the first term in the Euler-Maclaurin Sum Formula. Rob Johnson take out the trash before replying === Subject: does this relation have a name? Does anyone know if the following relation between two functions has a name? Suppose we have functions f(x) and g(x) over, e.g., [0,1]. If sgn(f'(a)) = sgn(g'(a)) for any a in [0,1] then it's clear they have a similar 'shape' in some sense. Alternately (but not equivalently) a relation may be defined as f R g iff f(a) g(a) Does anyone know if the following relation between two > functions has a name? > Suppose we have functions f(x) and g(x) over, e.g., [0,1]. > If sgn(f'(a)) = sgn(g'(a)) for any a in [0,1] then [snip rest] One thing that comes to mind is that since derivatives, whether continuous or not, satisfy the intermediate value property, your relation is f ~ g <==> Z(f') = Z(g'), where Z(h) is the zero set of the function h (i.e. Z(h) = {x: h(x) = 0}). This assumes the functions f and g are differentiable at each point, by the way. Dave L. Renfro === Subject: Re: does this relation have a name? > Does anyone know if the following relation between two > functions has a name? > Suppose we have functions f(x) and g(x) over, e.g., [0,1]. > If sgn(f'(a)) = sgn(g'(a)) for any a in [0,1] then > [snip rest] > One thing that comes to mind is that since derivatives, > whether continuous or not, satisfy the intermediate value > property, your relation is f ~ g <==> Z(f') = Z(g'), > where Z(h) is the zero set of the function h > (i.e. Z(h) = {x: h(x) = 0}). No, not quite. Z(h) doesn't determine sgn(h). Think of f'(x)=x, g'(x)=x^2. === Subject: Re: does this relation have a name? >> One thing that comes to mind is that since derivatives, >> whether continuous or not, satisfy the intermediate value >> property, your relation is f ~ g <==> Z(f') = Z(g'), >> where Z(h) is the zero set of the function h >> (i.e. Z(h) = {x: h(x) = 0}). This assumes the functions >> f and g are differentiable at each point, by the way. > No, not quite. Z(h) doesn't determine sgn(h). > Think of f'(x)=x, g'(x)=x^2. Ooops, you're right. It looks like that, in my absolute haste to get my post out, I made an error of an absolute type! Dave L. Renfro === Subject: Re: does this relation have a name? To put my question in context: I have a difference equation whose solution somewhat masks its behavior. Solving it as a differential equation provides a solution whose behavior is much more clear. Robert Israel in sci.math.num-analysis pointed out that using the differential solution is equivalent to using the Euler forward method on the difference equation, and thus has a known error of O(h^2) where O(h^2) = f(t+h) - hf'(t) (if you'll pardon the messy notation). Since the result in the analysis where this comes up depends only on whether f(a) < f(b) or f(a) > f(b) and since I can show that the above relation holds between the difference and differential solutions, I believe I can make claims based on the simpler, differential solution. I'm not sure if this is a well-known approach, but it may be something others would find useful. Incidentally, this analysis also suggests an identity that I haven't come across before (and have not yet been able to prove): n-1 lim sum nCi * x^(n-i-1) = -x^-1 n->inf i=0 for -1 Incidentally, this analysis also suggests an identity that I haven't > come across before (and have not yet been able to prove): > n-1 > lim sum nCi * x^(n-i-1) = -x^-1 > n->inf i=0 > for -1 is there a quick way for finding the inverse for a set of quadratic > residues? Yes, the following method is based on Chapter 16 of Number Theory by J. Hunter, Oliver and Boyd, 1964. For example, to solve x^2 = A mod M for M = 385 and A = 344 (where 385 = 5 * 7 * 11). Column A is each prime power divisor p, B is sqrt(A) mod p, C is M/p, D is C mod p, E is (B/D) mod p, and F is E * C. Thus: A B C D E F 5 2 77 2 1 77 7 1 55 6 6 330 11 5 35 2 8 280 Now add the numbers in column F, mod M, with all possible mixtures of signs. This gives 302, 127, 27, 148 and their negations as the 8 solutions. When M is even, certain adjustments are necessary. -- === Subject: Re: inverse quadratic residues problem days. My association with the Department is that of an alumnus. >> is there a quick way for finding the inverse for a set of quadratic >> residues? >Yes, the following method is based on Chapter 16 of >Number Theory by J. Hunter, Oliver and Boyd, 1964. Edited to remove the double use of A: >For example, to solve x^2 = R mod M for M = 385 and >R = 344 (where 385 = 5 * 7 * 11). >Column A is each prime power divisor p, > B is sqrt(R) mod p, C is M/p, > D is C mod p, E is (B/D) mod p, > and F is E * C. Thus: > A B C D E F > 5 2 77 2 1 77 > 7 1 55 6 6 330 >11 5 35 2 8 280 This is just running through the proof of the Chinese Remainder Theorem. If m1,...,mr are pairwise coprime integers, then a solution to x=ai (mod mi) is given by taking a1*M1 + ... + ar*Mr, where Mi = m1*...*mr/mi. So your column F is just the terms ai*Mi, and the sum is the solution given by the usual proof of the CRT. >Now add the numbers in column F, mod M, with all >possible mixtures of signs. This gives 302, 127, >27, 148 and their negations as the 8 solutions. >When M is even, certain adjustments are necessary. Only in order to get all solutions. But all the ones you get this way will be solutions; the difficulty only lies in that there will be more than 2 square roots modulo 2^j if j>=3. -- === Subject: Re: inverse quadratic residues problem days. My association with the Department is that of an alumnus. >is there a quick way for finding the inverse for a set of quadratic >residues? >ie. given a set of quadratic residues [r[0],...,r[n]] mod >[m[0],..,m[n]] >(the set [m[0],..,m[n]] is the usual co-prime set) >a=r[0] mod m[0] >... >a=r[n] mod m[n] >using CRT it's easy to find a >x^2 = a mod m[0]*m[1]*...m[n] >but is there an easy way to finding x^2 without trying a lot of >a+b*m[0]*m[1]*..m[n] Huh? Since each r[i] is a quadratic residue modulo m[i], there exists x[i] such that x[i]^2 = r[i] (mod m[i]). So if x is the solution to x=x[0] (mod m[0]) x=x[1] (mod m[1]) . . . x=x[n] (mod m[n]), obtained via the CRT, then x^2 = a (mod m[0]*...*m[n]). Why are you adding multiples of m[0]*...*m[n] to a? -- === Subject: Re: inverse quadratic residues problem we know 21 = [0,1,1] mod [3,4,5] is one of the quadratic residues of 60=3*4*5. and now without doing 21+1*60 find 81 = [0,1,1] mod [3,4,5] of course in this case b=1 but is there another way? === Subject: Re: inverse quadratic residues problem days. My association with the Department is that of an alumnus. >we know But somehow we don't know how to quote messages to provide context. Next time, quote. >21 = [0,1,1] mod [3,4,5] is one of the quadratic residues of >60=3*4*5. and >now without doing 21+1*60 find 81 = [0,1,1] mod [3,4,5] >of course in this case b=1 but is there another way? What do you mean doing 21+1*60? What do you mean without doing whatever it is you think you are doing? You know that 0 is a quadratic residue mod 3 1 is a quadratic residue mod 4 1 is a quadratic residue mod 5 And you know that the unique integer a such that a=0 (mod 3) and a=1 (mod 4), a=1 (mod 5) is a=21. So you want to find an x such that x^2 = 21 (mod 60)? 0^2 = 0 (mod 3) 1^2 = 1 (mod 4) 1^2 = 1 (mod 5) so find x such that x=0 (mod 3) x=1 (mod 4) x=1 (mod 5) namely, x=21 (mod 60). Thus, 21^2 = 21 (mod 60). Or find x2 such that x2 = 0 (mod 3) x2 = -1 (mod 4) x2 = 1 (mod 5) namely, x2 = 51 (mod 60), and 51^2 = 21 (mod 60). Or find x3 such that x3 = 0 (mod 3) x3 = 1 (mod 4) x3 = -1 (mod 5) namely, x3=9 (mod 60), and 9^2 = 21 (mod 60). Or find x4 such that x4 = 0 (mod 3) x4 = -1 (mod 4) x4 = -1 (mod 5) namely, x4 = 39 (mod 60), and 39^2 = 21 (mod 60). Why 4? Because there is 1 square root of 0 modulo 3, and two each of 1 modulo 4 and 5, giving a total of 1*2*2 square roots of 21 modulo 60. -- === Subject: Re: inverse quadratic residues problem But somehow we don't know how to quote messages to provide context. > Next time, quote. >21 = [0,1,1] mod [3,4,5] is one of the quadratic residues of >60=3*4*5. and >now without doing 21+1*60 find 81 = [0,1,1] mod [3,4,5] >of course in this case b=1 but is there another way? > What do you mean doing 21+1*60? What do you mean without doing > whatever it is you think you are doing? > You know that 0 is a quadratic residue mod 3 > 1 is a quadratic residue mod 4 > 1 is a quadratic residue mod 5 > And you know that the unique integer a such that a=0 (mod 3) and a=1 > (mod 4), a=1 (mod 5) is a=21. > So you want to find an x such that x^2 = 21 (mod 60)? > 0^2 = 0 (mod 3) > 1^2 = 1 (mod 4) > 1^2 = 1 (mod 5) > so find x such that > x=0 (mod 3) > x=1 (mod 4) > x=1 (mod 5) > namely, x=21 (mod 60). Thus, 21^2 = 21 (mod 60). > Or find x2 such that > x2 = 0 (mod 3) > x2 = -1 (mod 4) > x2 = 1 (mod 5) > namely, x2 = 51 (mod 60), and 51^2 = 21 (mod 60). > Or find x3 such that > x3 = 0 (mod 3) > x3 = 1 (mod 4) > x3 = -1 (mod 5) > namely, x3=9 (mod 60), and 9^2 = 21 (mod 60). > Or find x4 such that > x4 = 0 (mod 3) > x4 = -1 (mod 4) > x4 = -1 (mod 5) > namely, x4 = 39 (mod 60), and 39^2 = 21 (mod 60). > Why 4? Because there is 1 square root of 0 modulo 3, and two each of 1 > modulo 4 and 5, giving a total of 1*2*2 square roots of 21 modulo 60. if I have r[i] = say [ 1, 1, 4] , do I have 8 solutions? [+-1, +-1, +4] other than zero, everything else has 2 solutions (plus and minus)? === Subject: Re: inverse quadratic residues problem days. My association with the Department is that of an alumnus. >>we know >> But somehow we don't know how to quote messages to provide context. >> Next time, quote. >>21 = [0,1,1] mod [3,4,5] is one of the quadratic residues of >>60=3*4*5. and >>now without doing 21+1*60 find 81 = [0,1,1] mod [3,4,5] >>of course in this case b=1 but is there another way? >> What do you mean doing 21+1*60? What do you mean without doing >> whatever it is you think you are doing? You did not really answer my question, but then that was in order for me to understand your question, so I guess it's up to you. [.snip.] >> Why 4? Because there is 1 square root of 0 modulo 3, and two each of 1 >> modulo 4 and 5, giving a total of 1*2*2 square roots of 21 modulo 60. >if I have r[i] = say [ 1, 1, 4] , do I have 8 solutions? [+-1, +-1, >+4] >other than zero, everything else has 2 solutions (plus and minus)? No; that only works for odd prime powers. If p is an odd prime, then x^2 = a (mod p) has no solutions, exactly one solution, or exactly two solutions (modulo p) depending on whether a is a quadratic non residue modulo p, divisible by p, or a quadratic residue modulo p, respectively. This follows from either the factorizatio of x^2-a over the field of p elements (which shows there are at most 2 solutions, and one if and only if x^2-a = (x-r)^2, which can only hold if 2r=0 (mod p), which gives r=0 (mod p), hence a=0 (mod p)), or from Lagrange's theorem on the number of solutions to a polynomial congruence modulo a prime. By Hensel's Lemma, if a is not a multiple of p and there exists r such that f(r)=r^2-a = 0 (mod p), then for each i>0 there exists a unique r_i such that f(r_i) = 0 (mod p^i), and r_i = r_{i+1} (mod p^i). They are obtained using Newton's Formula: r_0 = r r_{i+1} = r_i - f(r_i)*g(r) (mod p^{i+1}) where g(r) is any integer such that g(r)*f'(r_0) = 1 (mod p), with f'(r_0)=2r_0 (the derivative). Since we are assuming that a is not a multiple of p, such a number exists (since p is odd). Also, if f(r)=0 (mod p^i), then f(r)=0 (mod p). So Hensel's Lemma implies that f(r) has exactly two solutions modulo p^i for each i, whenever a is relatively prime to p and a quadratic residue modulo p. However, this fails for p=2. For example, x^2=1 (mod 8) has FOUR solutions, namely x=1, x=3, and x=5, and x=7. In general, x^2=a (mod 2^j) behaves in a more complicated manner. If a is odd, then x^2=a (mod 2^j) has one solution modulo 2 if j=1; it has 2 solutions (modulo 4) if j=2 and a=1 (mod 4), and no solutions if a=3 (mod 4). If j>=3, then the congruence will have either 4 solutions, or no solutions modulo 2^j, depending on whether a=1 (mod 8) or not. For a even, it gets much more complicated (though of course, a is not a quadratic residue in that case). So, in summary: assume that (a,n)=1; then a is a quadratic residue modulo n if and only if: (i) it is a quadratic residue modulo p for each odd prime divisor p of n; and (ii) if 4|n but 8 does not divide n, then a=1 (mod 4); and (iii) if 8|n then a=1 (mod 8). If a is a quadratic residue modulo n, then it will have 2^{r+u} square roots modulo n, where r is the number of distinct prime divisors of n, and (1) u=0 if n is not a multiple of 4; (2) u=1 if 4|n but 8 does not divide n; (3) u=2 if 8|n. -- === Subject: Proof of the Fusion Barrier Principle In the past several months there has come a change over me and I wish to note it here and now. I have found out what all of this Internet postings is coming to. It is a rough draft of what I eventually will put all into print publication. One of the items I need to have in print form early on is the Fusion Barrier Principle. This is my current statement of the Principle as shown on the Wikipedia page: (2) Fusion Barrier Principle. Fission energy is the highest form of energy that is able to be controlled and surpass breakeven. A Tokamak such as JET or ITER can only reach 2/3 breakeven because 1/3 of the input energy is forever lost in controlling the device. That is a very good condensed form of FBP. And over the years starting 1997 when I discovered this Principle I have given a proof of it. But now I need to do it properly. Proof of FBP: We start with the Maxwell Equations and we note that two of them are cylindrical in form. They are the two in motion of the Faraday Law and the Ampere-Maxwell Law. The other two laws of Maxwell Equations of the Gauss laws especially the Coulomb is not in motion and spherical. Now, here is the beauty. There is an old mathematical proof involving enclosing a cylinder in a sphere and vice versa of a sphere in a cylinder and the volume or surface area, either one, has a maximum of 2/3. Fusion is the enclosing of the two Maxwell Equations of Faraday and Ampere-Maxwell laws into the Coulomb law and vice versa. Since the maximum is 2/3, then that is the maximum for which we can transform the two Maxwell Equations of Faraday and Ampere-Faraday Laws and the Coulomb Law. If we can alter or transform them to be greater than 2/3 means the Maxwell Equations are no longer true. JET reached 64% breakeven in the 1990s before being dismantled as a tokamak machine. Nagamine reached 66% breakeven in muon catalyzed fusion which is the world recordholder, and if memory serves me he did this in early 1990s in a very cold experimental lab environment. ITER is the newest scheduled tokamak machine and everyone who does not believe or understand the FBP expects ITER to surpass 2/3 breakeven and even surpass 100% breakeven. They are foolish for they do not understand nor do they care to become reasonable and rational about fusion prospects. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof of the Fusion Barrier Principle The mystical property of cylindrical and spherical seem to be the lynch pin in this proof. Unfortunately, there is nothing spherical or cylindrical about any of the Maxwell equations. The two you claim are cynlindrical are about the relationship between a flux-integral on a bounded surface and a contour integral on the bounding line. The two you claim are spherical are about the relationship between a flux-integral on a bounding surface and an integral of the bounded volume. Using a specific geometric structure for any of the quantities is unnecessary... It is how first-year physics students (especially those who don't know calculus) learn to use these equations in basic cases, but it is largely a pedalogical construct.. certainly not a physical one. I could just as easily apply Gauss' law with a cylinder around an ion, measure the field, and with some needlessly complicated mathematics, figure out how much charge is inside. === Subject: Re: Proof of the Fusion Barrier Principle Several years back when I posted on this issue of the Fusion Barrier Principle, I made good progress in connecting and tying together Fusion with Superconductivity. In that the Fusion Barrier Principle would not be confined to fusion but that FBP would appear in cold physics as well as hot physics, since fusion and superconductivity are simply arising from the Maxwell Equation theory of Classical Physics. The fusion is the fusing of the cylinders that are Faraday's and Ampere-Maxwell's equations with the Coulomb equation of sphere. So fusion reduces to the mathematics of sphere and cylinder enclosing with its 2/3 maximum upper limit. In superconductivity the world recordholder is (Hg0.8Tl0.2)Ba2Ca2Cu3O8.33 at 138K. A few years back I had the trouble of placing the 2/3 or 1/3 benchmark to the 138K. If we take room temperature to be 290-300K then 1/3 of that is approx 100K. This much I am positive of. If both fusion and superconductivity are Classical physics of the Maxwell Equations then both are linked and the 1/3 Fusion Barrier Principle is linked to the upper limit of superconductivity Tc temperature. So what would the upper limit Tc be? Would it be the current world recordholder of 138K? We note that PV = nrT of the Ideal Gas Law that pressure is directly proportional to temperature and so we can eliminate from discussion the concern of higher Tc due to increasing of the pressure. So if the above is all accurate then the melting points of the elements cesium 302K to fluorine 54K and that of hydrogen 14K to iodine 387K. So adding 302 to 54 yields 356 and adding 14 to 387 yields 401. In superconductivity, it is a phenomenon of maximizing Electronegativity with Conduction Band theory and the maximum push and pull would be cesium with fluorine and the inverse of that is hydrogen to iodine. So that 1/3 of 401K is approximately the current world recordholder of 138Kelvin. Trouble is that melting points are not exactly the temperature I want or need to use in connection with Maxwell Equations of the flow of electrons. Although 133K is rather a nice close approach to 138K. This tells me I am close to the true understanding, but that the future will have more precise equations that will link the 138K superconductor exactly with the physical parameters of Electronegativity and Conduction Band theory. Link them precisely so that there is not my gap of 133K to 138K. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof of the Fusion Barrier Principle We note that PV = nrT of the Ideal Gas Law that pressure is directly proportional to temperature and so we can eliminate from discussion the concern of higher Tc due to increasing of the pressure. I further write: Perhaps I hit it straight-on. Perhaps I have no fault with the 133K to the recordholder of 138K because the 138K of the compound (Hg0.8Tl0.2)Ba2Ca2Cu3O8.33 was measured in an environment that did not compensate or factor in the ambient pressure. Perhaps if that compound were to have a ambient pressure factored into its Tc temperature the result would be 133K derived from the melting points of hydrogen to iodine then multiply by 1/3. I should post this to sci.chem but am unable from this posting format platform. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof of the Fusion Barrier Principle >In the past several months there has come a change over me and I wish to note it here and now. I have found out what all of this Internet postings is coming to. It is a rough draft of what I eventually will put all into print publication. One of the items I need to have in print form early on is the Fusion Barrier Principle. *** If you ever actually publish anything please give us the citation. Don't repeat your usual the reviewers missed the point stupidity. === Subject: Re: Proof of the Fusion Barrier Principle >>In the past several months there has come a change over me and I wish > to note it here and now. I have found out what all of this Internet > postings is coming to. It is a rough draft of what I eventually will > put all into print publication. One of the items I need to have in > print form early on is the Fusion Barrier Principle. > *** > If you ever actually publish anything please give us the citation. > Don't repeat your usual the reviewers missed the point stupidity. Is this a preview of an eternally coming feature? Michael === Subject: Re: Where is the Galois group? >> Consider following commuting 6x6 matrices: >> M:=[ [ 0, 0, a, 0, 0, 0 ], >> [ 1, 0, b, 0, 0, 0 ], >> [ 0, 1, c, 0, 0, 0 ], >> [ 0, 0, 0, 0, 0, a ], >> [ 0, 0, 0, 1, 0, b ], >> [ 0, 0, 0, 0, 1, c ] ]; >> N:=[ [ 0, 0, 0, b, -a, 0 ], >> [ 0, 0, 0, c, 0, -a ], >> [ 0, 0, 0, -1, 0, 0 ], >> [ 1, 0, 0, c, 0, -a ], >> [ 0, 1, 0, -1, c, -b ], >> [ 0, 0, 1, 0, -1, 0 ] ]; >> and >> P:=[ [ c, 0, -a, -b, a, 0 ], >> [ -1, c, -b, -c, 0, a ], >> [ 0, -1, 0, 1, 0, 0 ], >> [ -1, 0, 0, 0, 0, 0 ], >> [ 0, -1, 0, 0, 0, 0 ], >> [ 0, 0, -1, 0, 0, 0 ] ]; >> where a, b and c are rational numbers (for instance). >> It is easy to verify that the satisfy the polynomial equation >> T^3 - c*T^2 - b*T -a = 0 (p); >> Moreover M+N+P = c, M*N+M*P+N*P = -b, M*N*P = a >> So we can conclude that the field generated by M, N and P generate the >> splitting field of the polynomial at (p). > Do M, N, P commute with one another? That's what was stated in the first line. > If not, then they do not generate > a field. In fact, even if they do, the commutative ring they generate > m,ay not be an integral domain; hoave you checked this? >> My problem is that I can't find out what the Galois group of this field >> is (as extension of the scalar rational matrixes - to be identified with >> the field of rationals - ). > I think you need to reconsider the definition of Galois group. I think the definition of the Galois group is still the automorphism group of an algebraic extension. Of course if is still not clear whether this extention is a field. I.e. if it contains no divisors of zero. I tried to prove this directly but that seemed to be a bit tedious. On the other hand, the way the matrices M, N and P were obtained clearly indicate that (except for some special values of a, b or/and c) that there are no divisors of zero: 1) If we write M as [ [M',0], [0,M'] ] and note that M' is the companion matrix of the polynomial at (p). I assume that (p) has three distinct roots. The vector space generated by M' and 1 contains divisors of zero iff equation (p) is reducible. This is due to the fact that det(u+v*M'+w*M'^2) (u,v,w in |Q) splits into three factors of the form u+v*r+w*r^2 (this can be seen if M' is diagonalized where the diagonal elements are the roots of (p) ). The determinant thus becomes zero whenever one of the factors is zero, i.e. when one of the roots satisfies an equation of the second degree with rational coefficients. 2) Let F be the extension (of degree 3) of |Q (considered as scalar matrices) by M', then the polynomial at (p) splits as: (T-M')*( T^2 + (M'-c)*T + M'*(M'-c) - b ) (p') with coefficients in F. So the companion matrix of the second degree factor is N = [ [0, -M'*(M'-c) + b], [1, -(M'-c)] ] This is a 2x2 matrix with coefficients in F. An element of the vector space generated by |Q (as scalar matrices) and N never has det=0 (as shown by a straightforward but complicated calculation). So the extention F' of degree 2 of F by the matrix N is a field which implies that the extention F' of degree 6 of |Q (as scalar matrices) is a field too. >> Looking for an automorphism A such that A*M = N*A seems to involve a lot >> of computation (36 linear equations in 36 indeterminates). Moreover it is >> not obvious that a Galois automorphism is an inner automorphism. Is there >> any simple solution to this? === Subject: Math programme with animation of sine eqs Can anyone recommend a maths programme which has animation for sine equations in 2 variables (x and t), with an easy-to-use interface? Kevin === Subject: Re: Math programme with animation of sine eqs > Can anyone recommend a maths programme which has animation for sine > equations in 2 variables (x and t), with an easy-to-use interface? What do you mean with animation? Like having slider that gives the value for t and automatically displaying the graph? You could try DPGraph, http://www.dpgraph.com/ - but I think that the free player won«t do what you want, you would have to buy it. Dynamic Geometry Systems are good for things like that. I know publications in german that show how to use a DGS like Euklid for that purpose. Unfortunately, Euklid is not translated. You could try my own DGS , Archimedes Geo3D, which is, as the name suggests, in three dimensions (see signature). Nevertheless I provide the file Funktionsplotter2D in the examples-directory which does what you want. - change the language to english at Extras - Sprache ausw.8ahlen, restart - open Beispiele/Funktionsplotter2d.geosave from the examples-folder (which is in the directory where you installed the program) - turn the scene (by left dragging) so that you look at the xy-plane - Double-click on Funktionsterm, then a dialog will appear that allows you to enter your expression. The predefined parameters are a, b and c in this file, but of course you could rename them to t - you could just as well enter a*sin(x*b + c) - move the points Pa, Pb, Pc towards and away from the x-axis. See what happens. - you could also define an animation to vary parameter a. Ask again if you want to do that or read the manual Of course a DGS is not really a function-plotter, and a 3D-DGS is overkill for that, but maybe you decide that you rather want to display sin(a*x+b*y), and then you would be fine with 3D! The program is not yet officially released in english, so you will have to get along with the german website, but Download should be clear enough! Andreas -------------------- www.raumgeometrie.de Dynamic geometry in three dimensions! === Subject: Re: Math programme with animation of sine eqs Ich spreche deutsch ein bisschen, obgleich ich nicht sehr viele Zeit habe, es zu .9fben. Ich wollte die Entwicklung einer Sinuswelle .9fber Zeit beobachten. Welches Ihrer Zeugnisse w.8are das Beste f.9fr meinen Zweck, denken Sie? Vielen Dank. Kevin === Subject: Re: Math programme with animation of sine eqs > Ich spreche deutsch ein bisschen, obgleich ich nicht sehr viele Zeit > habe, es zu .9fben. Ich wollte die Entwicklung einer Sinuswelle .9fber > Zeit beobachten. Welches Ihrer Zeugnisse w.8are das Beste f.9fr meinen > Zweck, denken Sie? Ok, I«ll continue in english as this is an english newsgroup. But you will have no problem in using a german software, I see. - maybe you should download DPGraph . There is the possibility to use a variable TIME for time-dependent functions. This would probably do what you want, _but_ I am not an experienced user, and as I don«t have a license I can«t try it. It costs 10 $, but you can«t try your own functions unless you buy it. - if you would tell me the expression you want to explore, I could try it out in my software to test it. But be warned, it can«t be denied that I am perhaps a bit prejudiced towards my own software ;-) . It costs 30 Û, but you have a free trial of four weeks, and just for function-plotting you could use it for free unlimited, as you probably don«t need to save any files if you just enter them into the pre-made demo files. There are other DGS out there, but I am still not completely sure what you want to do, so please explain a bit more! Andreas === Subject: Re: Math programme with animation of sine eqs habe, es zu .9fben. Ich wollte die Entwicklung einer Sinuswelle .9fber > Zeit beobachten. Welches Ihrer Zeugnisse w.8are das Beste f.9fr meinen > Zweck, denken Sie? > Ok, I«ll continue in english as this is an english newsgroup. But you > will have no problem in using a german software, I see. > - maybe you should download DPGraph . There is the possibility to use a > variable TIME for time-dependent functions. This would probably do > what you want, but I am not an experienced user, and as I don«t have a > license I can«t try it. It costs 10 $, but you can«t try your own > functions unless you buy it. > - if you would tell me the expression you want to explore, I could try > it out in my software to test it. For a start: u(x,t) = 8/Pi^2 Sigma(from r = 1 to infinity) [1/r^2 sin((r*Pi*x)/20) sin((r*Pi)/2) cos((r*Pi*t)/20)] Kevin But be warned, it can«t be denied that > I am perhaps a bit prejudiced towards my own software ;-) . > It costs 30 ¥, but you have a free trial of four weeks, and just for > function-plotting you could use it for free unlimited, as you probably > don«t need to save any files if you just enter them into the pre-made > demo files. > There are other DGS out there, but I am still not completely sure what > you want to do, so please explain a bit more! > Andreas === Subject: Re: Math programme with animation of sine eqs >Ich spreche deutsch ein bisschen, obgleich ich nicht sehr viele Zeit >habe, es zu .9fben. Ich wollte die Entwicklung einer Sinuswelle .9fber >Zeit beobachten. Welches Ihrer Zeugnisse w.8are das Beste f.9fr meinen >Zweck, denken Sie? >>Ok, I«ll continue in english as this is an english newsgroup. But you >>will have no problem in using a german software, I see. >>- maybe you should download DPGraph . There is the possibility to use a >>variable TIME for time-dependent functions. This would probably do >>what you want, _but_ I am not an experienced user, and as I don«t have a >>license I can«t try it. It costs 10 $, but you can«t try your own >>functions unless you buy it. >>- if you would tell me the expression you want to explore, I could try >>it out in my software to test it. > For a start: > u(x,t) = 8/Pi^2 Sigma(from r = 1 to infinity) [1/r^2 sin((r*Pi*x)/20) > sin((r*Pi)/2) cos((r*Pi*t)/20)] Oh, I doubt that you will get this animation in realtime, as you have to approximate the sum. You definitely can«t do this with any dynamic geometry software. I don«t know if you could do it with dpgraph, try to get the manual for it (but having looked at the examples I think you can«t) I would probably write a program to get this. Sorry I could«t help you, Andreas === Subject: Re: Math programme with animation of sine eqs habe, es zu .9fben. Ich wollte die Entwicklung einer Sinuswelle .9fber >Zeit beobachten. Welches Ihrer Zeugnisse w.8are das Beste f.9fr meinen >Zweck, denken Sie? >>Ok, I«ll continue in english as this is an english newsgroup. But you >>will have no problem in using a german software, I see. >>- maybe you should download DPGraph . There is the possibility to use a >>variable TIME for time-dependent functions. This would probably do >>what you want, but I am not an experienced user, and as I don«t have a >>license I can«t try it. It costs 10 $, but you can«t try your own >>functions unless you buy it. >>- if you would tell me the expression you want to explore, I could try >>it out in my software to test it. > For a start: > u(x,t) = 8/Pi^2 Sigma(from r = 1 to infinity) [1/r^2 sin((r*Pi*x)/20) > sin((r*Pi)/2) cos((r*Pi*t)/20)] > Oh, > I doubt that you will get this animation in realtime, as you have to > approximate the sum. Yes, sorry. I wasn't thinking. It would be ok to sum from r = 1 to 30. > You definitely can«t do this with any dynamic geometry software. > I don«t know if you could do it with dpgraph, try to get the manual for > it (but having looked at the examples I think you can«t) > I would probably write a program to get this. What about the big math programmes, like Maple and Mathematica? Could they handle this, do you think? Kevin === Subject: Re: Math programme with animation of sine eqs boundary=------------000806090900010604010602 --------------------------------------------------------------------- > What about the big math programmes, like Maple and Mathematica? Could > they handle this, do you think? I simply don«t know if they do animations. > Kevin Ok, I undervaluated my own program. I prepared a file for you, see attachement. I hope that small attachements like this are allowed here. Instructions: - download the program - open the file - select Pa in the list on the left - select Extras - Animation The parameter a (which is t in your expression) will vary from 0 to 20. You see how your graph changes in realtime. I put the first twenty expressions into the expression, I think that«s enough, as you divide by 400 in the last expression. It looks really cool, I think. What process are you modeling? Andreas p.s.: You might want to change the term (double-click on funktionsterm) to use 28 instead of 8 as a factor, to see better whats happening. --------------------------------------------------------------------- Subject: Re: Math programme with animation of sine eqs What about the big math programmes, like Maple and Mathematica? Could > they handle this, do you think? > I simply don«t know if they do animations. > Kevin > Ok, > I undervaluated my own program. > I prepared a file for you, see attachement. I hope that small > attachements like this are allowed here. > Instructions: > - download the program > - open the file > - select Pa in the list on the left > - select Extras - Animation > The parameter a (which is t in your expression) will vary from 0 to 20. > You see how your graph changes in realtime. > I put the first twenty expressions into the expression, I think that«s > enough, as you divide by 400 in the last expression. > It looks really cool, I think. What process are you modeling? It's just a plucked violin string, that's all. Kevin === Subject: Re: Math programme with animation of sine eqs > It's just a plucked violin string, that's all. Now that at last I know what you are doing I can tell you that - you should set the beginning of the string to the origin. This is done by right-clicking on LOCUS2 and setting startx to 0 - you should accellerate the process a bit. This can be done by double-clicking on the object a (left in the object list) and changing Pa*V1 to Pa*V1*15. Of course plucking a violin like this is not realistic. And even if you pluck, you do not pluck in the middle. The instrument coming closest to this very sharp edge is the virginal, I think, but here also the pluck is not in the middle. Could you provide me functions for - the violin starting with a sine-wave, not with a piecewise linear function - the violin starting with an additional knot in the middle to show flageolet This would be nice to have for my teaching. Oh: And if someone is interested, the problem of the vibrating string was very inspiring for the development of math. Euler and d'Alembert were struggling about this problem and Euler generalized the concept of functions on the way - functions used to be analytic expressions, and for the vibrating string piecewise linear functions were needed to describe the initial form of the string (I am missing the term Anfangbedingungen here, sorry). Andreas > Kevin === Subject: Re: Math programme with animation of sine eqs Now that at last I know what you are doing I can tell you that > - you should set the beginning of the string to the origin. This is done > by right-clicking on LOCUS2 and setting startx to 0 > - you should accellerate the process a bit. This can be done by > double-clicking on the object a (left in the object list) and changing > Pa*V1 to Pa*V1*15. > Of course plucking a violin like this is not realistic. And even if you > pluck, you do not pluck in the middle. The instrument coming closest to > this very sharp edge is the virginal, I think, but here also the pluck > is not in the middle. > Could you provide me functions for > - the violin starting with a sine-wave, not with a piecewise linear function > - the violin starting with an additional knot in the middle to show > flageolet > This would be nice to have for my teaching. I'm studying partial differential equations and very much a beginner, I'm afraid. I wouldn't feel confident enough to help you here. Sorry. > Oh: And if someone is interested, the problem of the vibrating string > was very inspiring for the development of math. Euler and d'Alembert > were struggling about this problem and Euler generalized the concept of > functions on the way - functions used to be analytic expressions, and > for the vibrating string piecewise linear functions were needed to > describe the initial form of the string (I am missing the term > Anfangbedingungen here, sorry). Initial conditions. Kevin > Andreas Kevin === Subject: Re: Math programme with animation of sine eqs >It's just a plucked violin string, that's all. >>Now that at last I know what you are doing I can tell you that >>- you should set the beginning of the string to the origin. This is done >>by right-clicking on LOCUS2 and setting startx to 0 >>- you should accellerate the process a bit. This can be done by >>double-clicking on the object a (left in the object list) and changing >>Pa*V1 to Pa*V1*15. So may I assume that it works for you? Andreas === Subject: Re: Math programme with animation of sine eqs It's just a plucked violin string, that's all. >>Now that at last I know what you are doing I can tell you that >>- you should set the beginning of the string to the origin. This is done >>by right-clicking on LOCUS2 and setting startx to 0 >>- you should accellerate the process a bit. This can be done by >>double-clicking on the object a (left in the object list) and changing >>Pa*V1 to Pa*V1*15. > So may I assume that it works for you? Actually there was a problem with the animation. I got it started, but then I had a little difficulty stopping it. Once I did, however, I then found I couldn't restart it. I'm sure the problem was due to me being unfamiliar with your program. I'm a bit busy at the moment, so I haven't been able to study it properly and I certainly haven't got around to trying your latest suggestions, though I will do. If I'm not able to get the thing running as it should, I'll get in touch. Kevin === Subject: Re: Math programme with animation of sine eqs > ... >>- if you would tell me the expression you want to explore, I could try >>it out in my software to test it. For a start: u(x,t) = 8/Pi^2 Sigma(from r = 1 to infinity) [1/r^2 sin((r*Pi*x)/20) > sin((r*Pi)/2) cos((r*Pi*t)/20)] Oh, > I doubt that you will get this animation in realtime, as you have to > approximate the sum. Actually, the sum can be expressed in closed form for both a finite range of r and for r = Infinity. Mathematica expresses the infinite sum in terms of 8 PolyLog functions of complex argument that, when combined together, give an explicitly real result. Define (and save) the sum to n terms: u[n_] := u[n] = Function[{x,t}, Evaluate[8/Pi^2 Sum[ Sin[(r Pi x)/20] Sin[(r Pi)/2] Cos[(r Pi t)/20]/r^2, {r, 1, n}]]] (note that the Sin[(r Pi)/2] term vanishes for r even so you could compute the sum more efficiently by replacing r -> 2r - 1). Here is the infinite sum in closed form: u[Infinity][x, t] (PolyLog[2, I*E^((-I/20)*Pi*t - (I/20)*Pi*x)] + PolyLog[2, I*E^((I/20)*Pi*t - (I/20)*Pi*x)] + PolyLog[2, (-I)*E^((-I/20)*Pi*t + (I/20)*Pi*x)] + PolyLog[2, (-I)*E^((I/20)*Pi*t + (I/20)*Pi*x)] - PolyLog[2, E^((-I)*(Pi/2 - (Pi*t)/20 + (Pi*x)/20))] - PolyLog[2, E^(I*(Pi/2 - (Pi*t)/20 + (Pi*x)/20))] - PolyLog[2, E^((-I)*(Pi/2 + (Pi*t)/20 + (Pi*x)/20))] - PolyLog[2, E^(I*(Pi/2 + (Pi*t)/20 + (Pi*x)/20))])/Pi^2 Here is the sum at {x, t} = {10, 0}, {10, 10}, and {1.2, 2.5}: u[Infinity][10, 0] 1 u[Infinity][10, 10] 0 u[Infinity][1.2, 2.5] // Chop 0.12 The first two results are exact and the third agrees with the exact answer of 3/25. > Yes, sorry. I wasn't thinking. It would be ok to sum from r = 1 to 30. Actually, you get a good result with far less terms than that. > What about the big math programmes, like Maple and Mathematica? Could > they handle this, do you think? Certainly (see above). The 3D surface is piecewise flat: Plot3D[ u[Infinity][x, t], {x, 0, 20}, {t, 0, 20} ] Differentiating the sum with respect to x twice one obtains D[ u[Infinity][x, t], {x, 2} ] // Simplify 0 and similarly for t (Note, however, that this result does not hold along 4 lines in x-t space, where the flat surface bends). 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: SERIES: Converges or Diverges A series, sigma a(n), is defined as: a(1) = 1 a(n+1) = [ 2 + cos (n) ] a(n) / sqrt(n) Does sigma a(n) converge or diverge? At first, I thought it converges by the comparison test. Then, I thought that it might be divergent since 1/sqrt (n) was divergent. Help me in my confusion. === Subject: Re: SERIES: Converges or Diverges > A series, sigma a(n), is defined as: > a(1) = 1 > a(n+1) = [ 2 + cos (n) ] a(n) / sqrt(n) > Does sigma a(n) converge or diverge? Hint: Show that a(n+1) <= 3^n/(n!)^(1/2). === Subject: Re: SERIES: Converges or Diverges Sorry, I meant ratio test instead of comparison test. === Subject: Re: SERIES: Converges or Diverges > Sorry, I meant ratio test instead of comparison test. ...and if you show a(n+1)/a(n) -> 0, what would the conclusion be? === Subject: test === Subject: What kind of induction is this? In Rotman's book I read the theorem of Unique Factorizazion into Disjoint Cycles (for completely factorizable permutations). a = b_1...b_t (complete fact. into disjoint cycles) and we suppose that there exists a second factorization a = c_1...c_s. The proof ends with: ... gives b_t=c_s. The cancellation law gives b_1...b_{t-1} = c_1...c_{s-1}, and the proof is completed by an induction on max{s,t}. QED. I never heard of an induction over a finite range of integers (or, at least, that is what I guess)... Can someone show me how to conclude the proof with such an induction on max{s,t} ??? === Subject: Re: What kind of induction is this? >In Rotman's book I read the theorem of Unique Factorizazion into Disjoint >Cycles (for completely factorizable permutations). a = b_1...b_t (complete >fact. into disjoint cycles) and we suppose that there exists a second >factorization a = c_1...c_s. The proof ends with: >... gives b_t=c_s. The cancellation law gives b_1...b_{t-1} = >c_1...c_{s-1}, and the proof is completed by an induction on max{s,t}. QED. >I never heard of an induction over a finite range of integers (or, at least, >that is what I guess)... No, it's not induction over a finite range of integers. It is induction on the integer n, where n is the largest of the two numbers s and t. So, for example, the case n=10 would cover each of s=10,t=10; s=10,t=9; s=10,t=8; ....; s=10, t=1; s=9,t=10, s=8,t=10; ... s=1,t=10. >Can someone show me how to conclude the proof with such an induction on >max{s,t} ??? Prove it in the case where s=t=1 (i.e., n=1). This is trivial. Now, assume you know the result when the longest of the two factorizations has n elements. Consider a factorization a=c_1...c_s = b_1...b_t, where the longest of two factorizations has n+1 elements (so, n+1 is the maximum of s and t, n+1=max{s,t}). You want to prove that s=t and that, up to permuting the order of the b_i, c_1=b_1, c_2=b_2,..., c_s=b_s (since s=t is proven as well). After cancelling, you end up with an equality c_1 ... c_{s-1} = b_1...b_{t-1}. This is an equality of two factorizations, the longest one of which has n elements (since the maximum of s-1 and t-1 is n). By induction, you know that the two factorizations are identical (up to order), so t-1 = s-1, and up to a permutation you have c_1=b_1, c_2=b_2, ..., c_{s-1}=b_{s-1}. You have t-1=s-1, so t=s. You already knew that c_s=b_s (since s=t). Together with the conclusion from the induction, you get that c_i = b_i for i=1,...,s, and s=t, as wanted. -- === Subject: Re: What kind of induction is this? Arturo Magidin ha scritto nel messaggio >>In Rotman's book I read the theorem of Unique Factorizazion into Disjoint >>Cycles (for completely factorizable permutations). a = b_1...b_t (complete >>fact. into disjoint cycles) and we suppose that there exists a second >>factorization a = c_1...c_s. The proof ends with: >>... gives b_t=c_s. The cancellation law gives b_1...b_{t-1} = >>c_1...c_{s-1}, and the proof is completed by an induction on max{s,t}. >>QED. >>I never heard of an induction over a finite range of integers (or, at >>least, >>that is what I guess)... > No, it's not induction over a finite range of integers. It is > induction on the integer n, where n is the largest of the two numbers > s and t. Maybe I expressed the concept in a bad way, but I wanted to say that I know two kind of induction, but they both works on a _sequence_ of proposition P_n. In particular, they both end with ... then P_n is true for all n>=n'. In this case I know that we get an induction only on s or t (it depends on which is the largest one), but I never seen an induction that - in a certain way - does not cover all n in N. In this case, the largest (supposed) true P_n is P_s or P_t. I wonder whether this could be in agreement with the forms of induction I know. >>Can someone show me how to conclude the proof with such an induction on >>max{s,t} ??? > Prove it in the case where s=t=1 (i.e., n=1). This is trivial. > Now, assume you know the result when the longest of the two > factorizations has n elements. Consider a factorization > a=c_1...c_s = b_1...b_t, where the longest of two factorizations has > n+1 elements (so, n+1 is the maximum of s and t, n+1=max{s,t}). > You want to prove that s=t and that, up to permuting the order of the > b_i, c_1=b_1, c_2=b_2,..., c_s=b_s (since s=t is proven as well). > After cancelling, you end up with an equality > c_1 ... c_{s-1} = b_1...b_{t-1}. > This is an equality of two factorizations, the longest one of which > has n elements (since the maximum of s-1 and t-1 is n). By > induction, you know that the two factorizations are identical (up to > order), so t-1 = s-1, and up to a permutation you have c_1=b_1, > c_2=b_2, ..., c_{s-1}=b_{s-1}. > You have t-1=s-1, so t=s. You already knew that c_s=b_s (since > s=t). Together with the conclusion from the induction, you get that > c_i = b_i for i=1,...,s, and s=t, as wanted. Ok, thank you again. === Subject: Re: What kind of induction is this? days. My association with the Department is that of an alumnus. >Arturo Magidin ha scritto nel messaggio >In Rotman's book I read the theorem of Unique Factorizazion into Disjoint >Cycles (for completely factorizable permutations). a = b_1...b_t (complete >fact. into disjoint cycles) and we suppose that there exists a second >factorization a = c_1...c_s. The proof ends with: >... gives b_t=c_s. The cancellation law gives b_1...b_{t-1} = >c_1...c_{s-1}, and the proof is completed by an induction on max{s,t}. >QED. >I never heard of an induction over a finite range of integers (or, at >least, >that is what I guess)... >> No, it's not induction over a finite range of integers. It is >> induction on the integer n, where n is the largest of the two numbers >> s and t. >Maybe I expressed the concept in a bad way, but I wanted to say that I know >two kind of induction, but they both works on a _sequence_ of proposition >P_n. In particular, they both end with ... then P_n is true for all n>=n'. Should be something other than all n>=n, shouldn't it? In any case, here the sequence of propositions is: P(n): any two factorizations of sigma, the longest of which has length n, are in fact identical (up to the order of the factors). >In this case I know that we get an induction only on s or t (it depends on >which is the largest one), No, we're not doing that. We could, but we are not doing that. > but I never seen an induction that - in a certain >way - does not cover all n in N. Yes, it does. The n is the length of the longest factorization. > In this case, the largest (supposed) true >P_n is P_s or P_t. You can interpret it that way, but you don't have to. -- === Subject: Re: What kind of induction is this? Arturo Magidin ha scritto nel messaggio >>Arturo Magidin ha scritto nel messaggio >>In Rotman's book I read the theorem of Unique Factorizazion into >>Disjoint >>Cycles (for completely factorizable permutations). a = b_1...b_t >>(complete >>fact. into disjoint cycles) and we suppose that there exists a second >>factorization a = c_1...c_s. The proof ends with: >>... gives b_t=c_s. The cancellation law gives b_1...b_{t-1} = >>c_1...c_{s-1}, and the proof is completed by an induction on max{s,t}. >>QED. >>I never heard of an induction over a finite range of integers (or, at >>least, >>that is what I guess)... > No, it's not induction over a finite range of integers. It is > induction on the integer n, where n is the largest of the two numbers > s and t. >>Maybe I expressed the concept in a bad way, but I wanted to say that I >>know >>two kind of induction, but they both works on a _sequence_ of proposition >>P_n. In particular, they both end with ... then P_n is true for all >>n>=n'. > Should be something other than all n>=n, shouldn't it? It is n>=n', n' is the induction basis integer. Come on... I haven't all day to write the whole statement... ;) > In any case, here the sequence of propositions is: > P(n): any two factorizations of sigma, the longest of which has > length n, are in fact identical (up to the order of the > factors). Ok. >>In this case I know that we get an induction only on s or t (it depends on >>which is the largest one), > No, we're not doing that. We could, but we are not doing that. No, it's not induction over a finite range of integers. It is induction on the integer n, where n is the largest of the two numbers s and t., I meant that we were doing exactly that... >> but I never seen an induction that - in a certain >>way - does not cover all n in N. > Yes, it does. The n is the length of the longest factorization. >> In this case, the largest (supposed) true >>P_n is P_s or P_t. > You can interpret it that way, but you don't have to. Ok, I think I am beginning to see what you mean. === Subject: Re: What kind of induction is this? >Arturo Magidin ha scritto nel messaggio >Arturo Magidin ha scritto nel messaggio >In Rotman's book I read the theorem of Unique Factorizazion into >Disjoint >Cycles (for completely factorizable permutations). a = b_1...b_t >(complete >fact. into disjoint cycles) and we suppose that there exists a second >factorization a = c_1...c_s. The proof ends with: ... gives b_t=c_s. The cancellation law gives b_1...b_{t-1} = >c_1...c_{s-1}, and the proof is completed by an induction on max{s,t}. >QED. I never heard of an induction over a finite range of integers (or, at >least, >that is what I guess)... >> No, it's not induction over a finite range of integers. It is >> induction on the integer n, where n is the largest of the two numbers >> s and t. >Maybe I expressed the concept in a bad way, but I wanted to say that I >know >two kind of induction, but they both works on a _sequence_ of proposition >P_n. In particular, they both end with ... then P_n is true for all >n>=n'. >> Should be something other than all n>=n, shouldn't it? >It is n>=n', n' is the induction basis integer. Come on... I haven't all day >to write the whole statement... ;) Sorry; I just didn't see the apostrophe on the second n; it got lost with the quote marks. >> In any case, here the sequence of propositions is: >> P(n): any two factorizations of sigma, the longest of which has >> length n, are in fact identical (up to the order of the >> factors). >Ok. >In this case I know that we get an induction only on s or t (it depends on >which is the largest one), >> No, we're not doing that. We could, but we are not doing that. >No, it's not induction over a finite range of integers. It is >induction on the integer n, where n is the largest of the two numbers >s and t., >I meant that we were doing exactly that... When you said induction only on s or on t, it seemed to me that you were saying that the induction would have to refer to s (the length of the factorization to the left of the equality symbol) or would have to refer to t (the length of the factorization to the right of the equality symbol), and ONLY to one. That the induction would have to be induction on s or induction on t, and not on both. > but I never seen an induction that - in a certain >way - does not cover all n in N. >> Yes, it does. The n is the length of the longest factorization. > In this case, the largest (supposed) true >P_n is P_s or P_t. >> You can interpret it that way, but you don't have to. >Ok, I think I am beginning to see what you mean. As someone pointed out, you ->can<- use a variant of induction over more general partially ordered sets. You could take the set of all pairs (s,t), where s and t are positive integers, and partially order them by letting (s,t) <= (r,u) if and only if s<=r and t<=u. Then you could do induction over all pairs by showing that if all pairs (x,y) with (x,y) < (s,t) satisfy the condition, then (s,t) satisfies the condition (a variant of strong induction). This would work because for any given pair (s,t), there are only finitely many pairs that are strictly smaller than (s,t); and there is a unique least pair (namely (1,1)). So, you know (1,1) has the property because all the pairs smaller than it (namely, the empty set) have it. Then you know the pairs (2,1), (2,2), and (1,2) have it because they each have {(1,1)} as the complete set of pairs strictly smaller than it. Then you get (3,1), (2,2), and (1,3). And so on. This will cover all pairs (s,t). But it is simpler in this case to just do it linearly as I mentioned above. -- === Subject: Re: What kind of induction is this? <443456fd$0$5651$4fafbaef@reader4.news.tin.it Maybe I expressed the concept in a bad way, but I wanted to say that I know > two kind of induction, but they both works on a _sequence_ of proposition > P_n. In particular, they both end with ... then P_n is true for all n>=n'. The induction goes over an _ordered set_ of cases. Say we want to prove some statement S(n), where n is a parameter of some kind. The set of cases (values of n) does not need to be totally ordered or well-ordered, as in a sequence. It is sufficient that the set of cases has some order relation < such that for given n there are only finitely many m such that mI was thinking there must be some statistical approach that would allow me to use the y value as a weight and generate a probability that the expected signal strongly correlates with the observed, or maybe a curve fitting algorithm that I could use? Suggestions on approaches for this? Im doing this in matlab, but will implement the final solution in c++. *** Since you have Matlab, why not use the built-in function corr? === Subject: silly question, the ellipse i seemed to have forgot how to show that D lies on an ellipse D(t) =(2cos(2t), sin(2t)) and the ellipse is (x/2)^2 + y^2 =1 i know this is quite simple..but ive sat here for 15 mins with no luck! thank you === Subject: Re: silly question, the ellipse days. My association with the Department is that of an alumnus. >i seemed to have forgot how to show >that D lies on an ellipse >D(t) =(2cos(2t), sin(2t)) and the ellipse is (x/2)^2 + y^2 =1 >i know this is quite simple..but ive sat here for 15 mins with no luck! Plug in 2cos(2t) for x, sin(2t) for y, into (x/2)^2 + y^2, and use the standard trig identity to get that this is always equal to 1; thus every point in D(t) satisfies the equation, hence lies on the ellipse. Showing that every point on the ellipse is the result of plugging in specific values of t takes a bit more doing, but not much. -- === Subject: Is Godel's proof constructive? (This may be a naive question). In his famous paper of 1931 on incompleteness theorems, Goedel explicitly claims that his proof of incompleteness theorem is *constructive*. But typically the final step in the proof is like: Let G be the Godel sentnece in the formal theory K, 1) Assume K is simply consistent. Assume G is provable. Deduce that ~G is also provable. K is inconsistent. Contradiction. Hence G is not provable. 2) Assume K is w-consistent. Assume ~G is provable. - - - Contradiction. Hence ~G is not provable. Now, doesn't this use of reductio-ad-absurdum qualify as 'non-constructive'? Or is there a way to turn these arguments into constructively meaningful ones? Any elaboration on this point will be very valuable. George. === Subject: Re: Is Godel's proof constructive? > (This may be a naive question). > In his famous paper of 1931 on incompleteness theorems, Goedel > explicitly claims that his proof of incompleteness theorem is > *constructive*. > But typically the final step in the proof is like: > Let G be the Godel sentnece in the formal theory K, > 1) Assume K is simply consistent. > Assume G is provable. > Deduce that ~G is also provable. > K is inconsistent. Contradiction. > Hence G is not provable. > 2) Assume K is w-consistent. > Assume ~G is provable. > - - - > Contradiction. > Hence ~G is not provable. > Now, doesn't this use of reductio-ad-absurdum qualify as > 'non-constructive'? Or is there a way to turn these arguments into > constructively meaningful ones? Goedel's proof yields an effective procedure whereby you can turn a proof of G into a proof of a contradiction in K, and you can turn a proof of ~G into an omega-inconsistency in K (i.e. an algorithm which will yield an infinite family of sentences that witness K's omega-inconsistency). So the proof is constructive. > Any elaboration on this point will be very valuable. > George. === Subject: Re: Is Godel's proof constructive? > In his famous paper of 1931 on incompleteness theorems, Goedel > explicitly claims that his proof of incompleteness theorem is > *constructive. > But typically the final step in the proof is like: > Let G be the Godel sentnece in the formal theory K, > 1) Assume K is simply consistent. > Assume G is provable. > Deduce that ~G is also provable. > K is inconsistent. Contradiction. > Hence G is not provable. > 2) Assume K is w-consistent. > Assume ~G is provable. > - - - > Contradiction. > Hence ~G is not provable. > Now, doesn't this use of reductio-ad-absurdum qualify as > 'non-constructive'? Or is there a way to turn these arguments into > constructively meaningful ones? It's not really reductio, it's just a contrapositive, and proof by contrapositive is intuitionistically valid as long as you don't eliminate the extra negation symbols (by replacing (not not A) with A). I will elaborate. In the case at hand, (K is consistent) means (not (K is inconsistent)). Moreover, ``K is inconsistent'' is actually a positive statement -- it says that a number of a certain form exists -- and ``consistent'' is the negation of ``inconsistent.'' This linguistic difficulty is probably why you are confused. Here's a better proof sketch for (1). Assume G is provable in K. Then (not G) is provable in K. Thus a contradiction is provable in K, so K is inconsistent. By contraposition, if K is not inconsistent then G is not provable in K. This is the same as saying that if K is consistent then G is not provable in K, because consistent means ``not inconsistent'' by definition. There is a similar constructive reading of the proof of (2) in the quote above. === Subject: Re: Is Godel's proof constructive? > (This may be a naive question). > In his famous paper of 1931 on incompleteness theorems, Goedel > explicitly claims that his proof of incompleteness theorem is > *constructive*. > But typically the final step in the proof is like: > Let G be the Godel sentnece in the formal theory K, > 1) Assume K is simply consistent. > Assume G is provable. > Deduce that ~G is also provable. > K is inconsistent. Contradiction. > Hence G is not provable. > 2) Assume K is w-consistent. > Assume ~G is provable. > - - - > Contradiction. > Hence ~G is not provable. > Now, doesn't this use of reductio-ad-absurdum qualify as > 'non-constructive'? Or is there a way to turn these arguments into > constructively meaningful ones? > Any elaboration on this point will be very valuable. The primary sense in which Godel means that the result is constructive is that the sentence G, given a choice of Godel numbering, can be explicitly produced. Thus it is not an abstract existence theorem about unprovable & unrefutable statements, but a constructive argument. The same is true of the Godel-Rosser refinement, in which omega-consistency is removed as a hypothesis. Of course consistency of the Peano axioms remains as a hypothesis, so one might object to use of the word constructive on those grounds. Indeed any use of proof by contradiction (law of excluded middle) would be objectionable to a constructivist, but that is not the sense of constructive intended here. === Subject: Re: Is Godel's proof constructive? > Of course consistency of the Peano axioms remains > as a hypothesis, so one might object to use of the > word constructive on those grounds. Huh? The theorem I know as Godel's first incompletness theorem does not mention PA at all. === Subject: Re: Is Godel's proof constructive? > Of course consistency of the Peano axioms remains > as a hypothesis, so one might object to use of the > word constructive on those grounds. > Huh? The theorem I know as Godel's first incompletness theorem does > not mention PA at all. Note that the remark of mine that you quoted was concerning how JB Rosser's neatened up result eliminated any omega- consistency hypothesis. My point was that the G.9adel-Rosser theorem shows Peano arithmetic is essentially incomplete if it is consistent. If your point is that the first incompleteness theorem applies to formal first-order arithmetic theories more generally than PA, this is true. It is after all an essential incompleteness result, so that no formal first-order extension of a theory to which the hypotheses apply is both complete and consistent. One such fragment of PA to which the conditions of the incompleteness theorem have been show to apply has as language {+,*,S,<,0} and nonlogical axioms: Q1. Ax ~(Sx = 0). Q2. AxAy Sx = Sy implies x = y. Q3. Ax (x + 0) = x. Q4. AxAy x + Sy = S(x + y). Q5. Ax x * 0 = 0. Q6. AxAy x * Sy = (x * y) + x. Q7. Ax ~(x < 0). Q8. AxAy x < Sy if and only if (x < y or x = y). Q9. AxAy x < y or x = y or y < x. If we add to these an axiom scheme of induction, then we have in essence PA, as detailed in these lecture notes by B. Kim: http://math.yonsei.ac.kr/bkim/goedel.pdf === Subject: Re: Is Godel's proof constructive? > If your point is that the first incompleteness theorem applies > to formal first-order arithmetic theories more generally than > PA, this is true. It is after all an essential incompleteness > result, so that no formal first-order extension of a theory to > which the hypotheses apply is both complete and consistent. Yes, that's all I meant. PA is a prime example of a theory that fits the hypotheses, but the incompleteness theorem isn't a theorem about PA per se. === Subject: Re: a partition of R >It may be that the OP wanted an order-preserving bijection >between his two dense parts of R. >Can that be done with few additional assumptions (AC, CH etc)? This can be done with NO additional assumptions. Every real number x = n(x) + f(x), where n(x) is an integer, and 0 <= f(x) < 1. Let A = {x: n(x) is even if and only if x is rational} and B the complement. Then x -> x+1 interchanges A and B. Harder, but not too difficult: do this such that A and B have positive measure in every interval of positive length. -- 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: math development curiosity question <49gb8kFo6t0mU1@individual.net> finite precision real numbers, which can be represented as rationals >plus a measure of uncertainty. Then a theory of differential equations >can be developed using those finite precision reals, and the classical >theory can be recovered by considering the limit as that uncertainty >goes to zero. My claim is that the connections between theory and >practical applications would be more transparent that way. > Good luck to you on this. My guess is that your theory will quickly > get bogged down in huge complications. For example, rationals plus > uncertainty will not obey the usual rules of algebra. The classical > mathematics of real numbers is much simpler. What I propose is that the notion of uncertainty is so fundamental that it should be incorporated into mathematics at a foundational level, and then to deal with it, we need a logic that can handle uncertainty (i.e. probabilistic logic). Then we recover both classical mathematics and classical logic as the limiting case as the uncertainty goes to zero. I claim that this actually makes things simpler, as it is closer to the intuitive mathematics we actually reason with when applying mathematics. === Subject: Re: math development curiosity question Discussion, linux) <49gb8kFo6t0mU1@individual.net> it should be incorporated into mathematics at a foundational level, and > then to deal with it, we need a logic that can handle uncertainty (i.e. > probabilistic logic). Then we recover both classical mathematics and > classical logic as the limiting case as the uncertainty goes to zero. I > claim that this actually makes things simpler, as it is closer to the > intuitive mathematics we actually reason with when applying > mathematics. So where's your theory of probability come from? Do you have a theory of probability that doesn't depend on any prior mathematics? (Like, say, the mathematics of the real interval?) What probabilistic logic do you have in mind? -- Flowers in the Attic was based on a true story. [...] HOW is it OK to just butcher such an awesome piece of work? It's like passing Pokemon off as the Mona-Lisa; sick and entirely wrong. -- An Amazon reviewer pissed that the movie didn't include incest === Subject: Re: math development curiosity question <49gb8kFo6t0mU1@individual.net> <87odzf16dq.fsf@phiwumbda.org What I propose is that the notion of uncertainty is so fundamental that > it should be incorporated into mathematics at a foundational level, and > then to deal with it, we need a logic that can handle uncertainty (i.e. > probabilistic logic). > So where's your theory of probability come from? Evolution. > Do you have a theory > of probability that doesn't depend on any prior mathematics? The basic laws of the logic don't require any prior mathematics. A theory designed to explain why we should use those laws of logic would require a well developed theory of mathematics. .> What probabilistic logic do you have in mind? Something similar to the one we humans use. For humans, reasoning under uncertainty is more basic than Boolean logic. . > Flowers in the Attic was based on a true story. [...] HOW is it OK > to just butcher such an awesome piece of work? It's like passing > Pokemon off as the Mona-Lisa; sick and entirely wrong. > -- An Amazon reviewer pissed that the movie didn't include incest If you quit giving us JSH quotes, then, uh, there will be little point === Subject: Re: math development curiosity question >What I propose is that the notion of uncertainty is so fundamental that >it should be incorporated into mathematics at a foundational level, and >then to deal with it, we need a logic that can handle uncertainty (i.e. >probabilistic logic). >>So where's your theory of probability come from? > Evolution. That isn't informative. >>Do you have a theory >>of probability that doesn't depend on any prior mathematics? > The basic laws of the logic don't require any prior mathematics. So what are they? > .> What probabilistic logic do you have in mind? > Something similar to the one we humans use. For humans, reasoning under > uncertainty is more basic than Boolean logic. More assertions. Give something concrete, and show how it is to be used. David, you effectively say that those of us who do this standard stuff for a living are taking money under false pretences---I don't appreciate being called a thief. But you won't even show us a single concrete example of what we ought to be doing instead. All we get is vague notions about being able to explain things to a computer (whatever that means) and hints about 'probabilistic logic' which you don't define. === Subject: Re: math development curiosity question <49gb8kFo6t0mU1@individual.net> <87odzf16dq.fsf@phiwumbda.org> <49lkuiFpf2h6U1@individual.net .> What probabilistic logic do you have in mind? > Something similar to the one we humans use. For humans, reasoning under > uncertainty is more basic than Boolean logic. > More assertions. Give something concrete, and show how > it is to be used. Let's say we write finite precision reals like this: 31.2345(7) where the seven in parenthesis is a measure of the uncertainty of the last digit (the five). As a first approximation to what that seven means, it can be taken to be something like a standard deviation. So, compute the value of x = 1.456(3) + 2.543(4), and be sure to include the uncertainty in the result. Give the best answer that you believe can be given. Is it true that x < 4.000(1) ? Is x < 4.000(7) ? Is x < 4.010(1) ? The basic ideas are simple enough that I shouldn't have to tell you what rules to apply. > David, you effectively say that those of us who do this > standard stuff for a living are taking money under false > pretences---I don't appreciate being called a thief. And yet you (or at least, many mathematicians) have no qualms about calling anyone who suggests there are better ways of doing things, a crackpot. === Subject: Re: math development curiosity question > HdB does NOT speak for mathematicians here, and has on other occasions > declared himself not to be one. Denying that engineering mathematics is nevertheless mathematics is analogous to denying that everything in music that is not Beethoven or Bach may nevertheless be music. We are engineers, physicists and computer programmers and we are practicing some kind of mathematics. Han de Bruijn === Subject: Re: math development curiosity question > HdB does NOT speak for mathematicians here, and has on other occasions > declared himself not to be one. > Denying that engineering mathematics is nevertheless mathematics is > analogous to denying that everything in music that is not Beethoven > or Bach may nevertheless be music. We are engineers, physicists and > computer programmers and we are practicing some kind of mathematics. In music, there are composers and performers, and some who are both, but those who are only performers do not generally tell composers what they are forbidden to compose. === Subject: Re: math development curiosity question >Real applied mathematicians >(say, people who try to solve hard nonlinear pdes) have no >problem with the axiom of choice---they use it to prove that >initial value problems have maximal solutions, for example. >>Somebody should point out to those real applied mathematicians that >>the axiom of choice cannot be relevant to computable solutions of pdes. > So that HdB does not consider knowing that a solution exists is relevant > to the problem of computing solutions? > What a very peculiar world he lives in. In my career as an engineer, it always went the other way around. We as a company have designed a heat exchanger. We have built a mathematical model for the heat transfer in it. And then we are going to search for (numerical) solutions. Nobody is going to tell us that these solutions maybe do not exist. We just _know_ that they must exist, irrespective of any mathematical evidence. And we are certainly not going to wait until some mathematician has proved that a solution exists. Han de Bruijn === Subject: Re: math development curiosity question > >Real applied mathematicians >(say, people who try to solve hard nonlinear pdes) have no >problem with the axiom of choice---they use it to prove that >initial value problems have maximal solutions, for example. >>Somebody should point out to those real applied mathematicians that >>the axiom of choice cannot be relevant to computable solutions of pdes. So that HdB does not consider knowing that a solution exists is relevant > to the problem of computing solutions? What a very peculiar world he lives in. > In my career as an engineer, it always went the other way around. We as > a company have designed a heat exchanger. We have built a mathematical > model for the heat transfer in it. And then we are going to search for > (numerical) solutions. Nobody is going to tell us that these solutions > maybe do not exist. We just _know_ that they must exist, irrespective > of any mathematical evidence. And we are certainly not going to wait > until some mathematician has proved that a solution exists. > Han de Bruijn Why does HdB get so up tight about the possibility that a mathematician may tell him something? And then why does HdB get so upset when mathematicians object to his telling them how to run their business? === Subject: Re: math development curiosity question <7db96$44311764$82a1e228$8044@news2.tudelft.nl> <49dbu9Fnk3e9U1@individual.net> <49dg8uFo30ipU1@individual.net> <49efbtFodlomU1@individual.net> <49g4e6Fog8osU3@individual.net> to the problem of computing solutions? > In my career as an engineer, it always went the other way around. We as > a company have designed a heat exchanger. We have built a mathematical > model for the heat transfer in it. And then we are going to search for > (numerical) solutions. Nobody is going to tell us that these solutions > maybe do not exist. We just _know_ that they must exist, irrespective > of any mathematical evidence. You know that they exist, *if* the mathematical model makes sense. You know that they have something to do with the mathematical model, *if* the relationship between the discretization and the continuum model is appropriate. It may be that they exist because what you are working with is a special case of something known in general, for example, even if you (in particular) don't know the general result. Some of us just think that it's a good idea to check whether the model makes sense in these ways before trying to solve it numerically. There's another issue. It can happen that one develops a pde for a system, and the pde is broken. Nevertheless, the numerics seem to be ok: this is because the numerics are not really solving that pde approximately, but the pde you (maybe) should have had in the first place. The sophisticated model is broken, but the discretization still gives a good approximate description of the system. It's an unfair universe. === Subject: Re: math development curiosity question > There's another issue. It can happen that one develops a pde > for a system, and the pde is broken. Nevertheless, the numerics > seem to be ok: this is because the numerics are not really > solving that pde approximately, but the pde you (maybe) should > have had in the first place. The sophisticated model is broken, > but the discretization still gives a good approximate description > of the system. This sounds both familiar and interesting. Could you please elaborate a few more bits? An example, perhaps? Han de Bruijn === Subject: Re: math development curiosity question <7db96$44311764$82a1e228$8044@news2.tudelft.nl> <49dbu9Fnk3e9U1@individual.net> <49dg8uFo30ipU1@individual.net> <49efbtFodlomU1@individual.net> <49g4e6Fog8osU3@individual.net> <290cc$443393b2$82a1e228$12828@news2.tudelft.nl There's another issue. It can happen that one develops a pde > for a system, and the pde is broken. Nevertheless, the numerics > seem to be ok: this is because the numerics are not really > solving that pde approximately, but the pde you (maybe) should > have had in the first place. The sophisticated model is broken, > but the discretization still gives a good approximate description > of the system. > This sounds both familiar and interesting. Could you please elaborate > a few more bits? An example, perhaps? The example I had in mind is something that happened to a colleague several years ago. As I recall, it was modelling viscous flow over a step. He had a pde for it which he knew was wrong, but the discretization gave results that agreed well with experiment. I don't think he ever published it. === Subject: Re: math development curiosity question >>Real applied mathematicians >>(say, people who try to solve hard nonlinear pdes) have no >>problem with the axiom of choice---they use it to prove that >>initial value problems have maximal solutions, for example. > Somebody should point out to those real applied mathematicians that > the axiom of choice cannot be relevant to computable solutions of pdes. Weird. I've been solving hard nonlinear pdes in my engineering period and _never_ encountered the axiom of choice. Guess I had no choice :-) Han de Bruijn === Subject: Re: math development curiosity question <7db96$44311764$82a1e228$8044@news2.tudelft.nl> <49dbu9Fnk3e9U1@individual.net> <49dg8uFo30ipU1@individual.net> <49efbtFodlomU1@individual.net> <49g4e6Fog8osU3@individual.net> <37ded$4433818e$82a1e228$17285@news1.tudelft.nl Weird. I've been solving hard nonlinear pdes in my engineering period > and _never_ encountered the axiom of choice. Guess I had no choice :-) Guess you never bothered about the proofs of having well-posed initial value problems, either. Example: Hawking&Ellis, The Large Scale Structure of Space-Time, Chapter 7 (The Cauchy Problem), p249. The existence of a maximal development of Cauchy data uses Zorn's lemma. Alternatively, Heuser, Functional Analysis, p4, gives Zorn's lemma as a basic fact in the introduction, and uses it without much comment thereafter. (You need it to prove, for example, that a general vector space has a basis.) === Subject: Re: math development curiosity question <7db96$44311764$82a1e228$8044@news2.tudelft.nl> <49dbu9Fnk3e9U1@individual.net> <49dg8uFo30ipU1@individual.net> <49efbtFodlomU1@individual.net> <49g4e6Fog8osU3@individual.net> <37ded$4433818e$82a1e228$17285@news1.tudelft.nl Weird. I've been solving hard nonlinear pdes in my engineering period > and _never_ encountered the axiom of choice. Guess I had no choice :-) > Guess you never bothered about the proofs of having well-posed initial > value problems, either. > Example: Hawking&Ellis, The Large Scale Structure of Space-Time, > Chapter 7 (The Cauchy Problem), p249. The existence of a maximal > development of Cauchy data uses Zorn's lemma. > Alternatively, Heuser, Functional Analysis, p4, gives Zorn's lemma as > a basic fact in the introduction, and uses it without much comment > thereafter. (You need it to prove, for example, that a general vector > space has a basis.) Mathematicians tend to use the tools they've got, whether they need them or not. === Subject: Re: math development curiosity question >>Weird. I've been solving hard nonlinear pdes in my engineering period >>and _never_ encountered the axiom of choice. Guess I had no choice :-) > Guess you never bothered about the proofs of having well-posed initial > value problems, either. Oh yes. But the initial values were reasoned from a sound engineering perspective. Consider the following set of PDE's (d is partial): c.G_p.(u.dT_p/dr + v.dT_p/dz) + a.(T_p - T_s) = 0 c.G_s.dT_s/dz + a.(T_s - T_p) = 0 Constants: c, G_p,G_s, u,v, a. Coordinates: r,z. Unknowns: T_p, T_s. Both PDE's have T_p or T_s prescribed at the inflow openings _only_. Some collegues of mine devised additional boundary conditions and I agree with you that it's sometimes difficult to get them on the right track again. Somewhat less trivial are the boundary conditions for the PDE's: d(r.u)/dr + d(r.v)/dz = 0 and du/dz - dv/dr = 0 The velocity component u or the velocity component v may be prescribed at the boundary. But exclusive: not both. I've seen much sinning against this as well. Especially in combination with numerical analysis. Han de Bruijn === Subject: Re: math development curiosity question >>So what I propose is the we (mathematicians) > HdB does NOT speak for mathematicians here, and has on other occasions > declared himself not to be one. In fact, HdB didn't speak at all there. (Though I tend to agree with the general sentiment that David is well out on a limb with this.) === Subject: Re: math development curiosity question > In fact, HdB didn't speak at all there. (Though I tend to agree > with the general sentiment that David is well out on a limb > with this.) HdB has spoken elsewhere. I kept my promise about the _infinitesimals_ in physics. Not Richard Feynman though, but Max Planck. Would you like to join in? It's in the Calculus XOR Probability thread where I have a debate with Randy Poe about this: Han de Bruijn === Subject: Re: math development curiosity question <7db96$44311764$82a1e228$8044@news2.tudelft.nl> <49dbu9Fnk3e9U1@individual.net> <49dg8uFo30ipU1@individual.net> <49efbtFodlomU1@individual.net> <49g4e6Fog8osU3@individual.net> <49gb8kFo6t0mU1@individual.net> <49ha8qFojrehU1@individual.net> <45b81$4433950a$82a1e228$13028@news2.tudelft.nl In fact, HdB didn't speak at all there. (Though I tend to agree > with the general sentiment that David is well out on a limb > with this.) > HdB has spoken elsewhere. I kept my promise about the _infinitesimals_ > in physics. Not Richard Feynman though, but Max Planck. Would you like > to join in? It's in the Calculus XOR Probability thread where I have > a debate with Randy Poe about this: > Han de Bruijn The cow is its complement in the universe. It's defined by everything it's not. People like Ernst Zermelo called AC the well-ordering principle and accept it as fact, Zorn found it a lemma. Where infinite sets are equivalent, AC is redundant, and where it would overconstrain a theoretical system leading to incompleteness and thus eventual inconsistency of a backing theory, where its implication is obvious, it's useless, except in a countable universe as a trivial theorem, where it's quite useful. A good partial differentialist outranks most set theorists. Ah, the memories. I'm how you say pleased that's quite correct, still is, and always will be. There's no universe in ZF. ZF is, incomplete, there is no set of true statements in ZF, unconditionally true statements. I borrowed a copy of Graham Priest's Beyond the Limits of Thought today. Priest discusses paraconsistency, where I am more interested in the intraconsistency. http://www.oup.co.uk/isbn/0-19-925405-2 Check it out. About Hegel, and Kant, there is much discussion, although I tend to focus on the Thing-in-Itself moreso than the noumena, and Hegel's Being and Nothing as the important elements instead of his rephrasal of potential versus actual. Graham closes with: Certainly there is no going back to how things were before htis. But maybe this century will see a return to the mainstreaming of a more traditional philosophical issue, the nature of reality -- and if I am right, a nature that is contradictory. I think that instead there can be found a nature that is uncontradictory because everything is natural, in terms of its existence or being. While it would be contradictory to nature for nature to be contradictory to itself, that resolves to the liar paradox which with these notions of, say, true axioms, or ever the possibility of calling anything true, ever, that has it being quite uncontradictory for nature to be. A paradox would be paradoxical, anathema to reason. Beyond _limits_ of thought is an axiomless system, of natural deduction. That's not to be taken that I don't agree with many or most of Graham's statements and descriptions of the historical evolution of paraconsistency in dialethism, I do. Priest is generally considered to be heretical, AND state-of-the-art. Non-heretical art is craft. Recently I read The Infinite Book, that's another nicely bound volume, if you're interested in infinity and care it wouldn't take too much space on the shelf, nor is it particularly an involving read. Next, about infinity, I hope to read Yaroslav's volume, it appears to be in somewhat more limited circulation. Han, I'm still looking for a management summary of differential equations, the Handbook, of Differential Equations, and one of these An Atlas of Functions seems to be a helpful start, along with say the H.M.F., Handbook of Mathematical Functions, that is quite an amount of information. Ross F. === Subject: Re: math development curiosity question > So what I propose is the we (mathematicians) first develop a theory of > finite precision real numbers, which can be represented as rationals > plus a measure of uncertainty. Then a theory of differential equations > can be developed using those finite precision reals, And we could give it a cool name, like, say, interval analysis. Oh wait, somebody already thought of that one. And people are already working on it. And it's all done within the standard framework. But hey, don't let me distract you from spending time on this. Interval analysis is a good thing, and I suspect your time would be better spent on that than on trying to persuade other people of the moral advantages of your point of view. And as a side benefit, if you can provide some actual evidence that your point of view is better in some sense, you might become more persuasive. and the classical > theory can be recovered by considering the limit as that uncertainty > goes to zero. You mean that you get pseudomathematics out of mathematics by taking the appropriate limit? Cool. === Subject: Re: math development curiosity question <7db96$44311764$82a1e228$8044@news2.tudelft.nl> <49dbu9Fnk3e9U1@individual.net> <49dg8uFo30ipU1@individual.net> <49efbtFodlomU1@individual.net> <49g4e6Fog8osU3@individual.net> <49gb8kFo6t0mU1@individual.net> <49h9l0Fp10iuU1@individual.net So what I propose is the we (mathematicians) first develop a theory of > finite precision real numbers, which can be represented as rationals > plus a measure of uncertainty. Then a theory of differential equations > can be developed using those finite precision reals, > And we could give it a cool name, like, say, interval analysis. Except that it's distinctly different from interval analysis. > and the classical > theory can be recovered by considering the limit as that uncertainty > goes to zero. > You mean that you get pseudomathematics out of mathematics > by taking the appropriate limit? Cool. I meant that most of the classical theory of differential equations can be recovered that way. That theory is not pseudomathematicians. If we try to recover set theory with that approach, we do not get classical set theory, and we do not get pseudomathematics. === Subject: Re: math development curiosity question >So what I propose is the we (mathematicians) first develop a theory of >finite precision real numbers, which can be represented as rationals >plus a measure of uncertainty. Then a theory of differential equations >can be developed using those finite precision reals, >>And we could give it a cool name, like, say, interval analysis. > Except that it's distinctly different from interval analysis. It is? So it exists already? Don't keep it to yourself, share it with us. Or if it doesn't, develop it and prove what you say. Everybody else seems to think that mathematics is trundling along quite acceptably, so it would be nice to see this better way in action. === Subject: Re: math development curiosity question <7db96$44311764$82a1e228$8044@news2.tudelft.nl> <49dbu9Fnk3e9U1@individual.net> <49dg8uFo30ipU1@individual.net> <49efbtFodlomU1@individual.net> <49g4e6Fog8osU3@individual.net> <49gb8kFo6t0mU1@individual.net> <49h9l0Fp10iuU1@individual.net> <49is4rFojeniU3@individual.netSo what I propose is the we (mathematicians) first develop a theory of >finite precision real numbers, which can be represented as rationals >plus a measure of uncertainty. Then a theory of differential equations >can be developed using those finite precision reals, >>And we could give it a cool name, like, say, interval analysis. > Except that it's distinctly different from interval analysis. > It is? So it exists already? Don't keep it to yourself, share > it with us. The notion of uncertainty is a probabilistic notion. > Or if it doesn't, develop it and prove what you say. Everybody > else seems to think that mathematics is trundling along quite > acceptably, so it would be nice to see this better way in action. This was something I was interested in when I was in graduate school quite a few years ago. I believed that I could develop a foundation for mathematics which would make the transfer of our mathematical knowledge to a computer (i.e. an artificial intelligence) especially easy. I encountered a tad bit of condescension and hostility to my ideas, so I quit. And by the way, not *everybody* else thinks everything is just fine in the foundations of mathematics. === Subject: Re: math development curiosity question <7db96$44311764$82a1e228$8044@news2.tudelft.nl> <49dbu9Fnk3e9U1@individual.net> <49dg8uFo30ipU1@individual.net> <49efbtFodlomU1@individual.net> <49g4e6Fog8osU3@individual.net> <49gb8kFo6t0mU1@individual.net> <49h9l0Fp10iuU1@individual.net> <49is4rFojeniU3@individual.netSo what I propose is the we (mathematicians) first develop a theory of >finite precision real numbers, which can be represented as rationals >plus a measure of uncertainty. Then a theory of differential equations >can be developed using those finite precision reals, >>And we could give it a cool name, like, say, interval analysis. > Except that it's distinctly different from interval analysis. > It is? So it exists already? Don't keep it to yourself, share > it with us. > The notion of uncertainty is a probabilistic notion. That doesn't really answer the question though, does it? If all you have is a vague idea that there must be a better way, then you don't really have anything. Either give evidence that what you propose is better, or stop claiming it. (And ironically enough, serious probability theory is one of the areas of maths where significant amounts of set theory do crop up.) > And by the way, not *everybody* else thinks everything is just fine in > the foundations of mathematics. OK, *practically* everybody else. And certainly you haven't presented any evidence that your concerns are shared by anybody. Indeed, you haven't even given any evidence that what we do is wrong in any sense; it simply seems to be that you don't care for the flavour. But neither I nor anybody else here is stopping you from developing your alternative foundations: the only thing stopping you (if you're right) is that you're spending the time arguing about whether other people ought to reject what they're doing in favour of something that doesn't even exist yet. So, spend the time more fruitfully. Once you have something to show, show it and see what the reaction is. (Of course, I wouldn't be surprised if it's given the same warm welcome as Tony Orlow's or Han de Bruijn's stuff---but what do you have to lose?) === Subject: parallelepiped distances What is the minimum volume of a parallelepiped such that the distance between any pair of vertices is at least 1? quasi === Subject: parallelepiped volumes vs triangular areas I think this one has a nonzero minimum (but with my record so far, I wouldn't bet too much on it). Problem: What is the minimum volume of a parallelepiped P such that for every 3 distinct vertices of P, the area of the corresponding triangle is at least 1? === Subject: Re: parallelepiped volumes vs triangular areas >I think this one has a nonzero minimum (but with my record so far, I >wouldn't bet too much on it). >Problem: >What is the minimum volume of a parallelepiped P such that for every 3 >distinct vertices of P, the area of the corresponding triangle is at >least 1? Set the base vertices at (0,0,0) (0,4,0) (2,4,0) (2,0,0) and the upper ones at (1,1,epsilon) (1,5,epsilon) (3,5,epsilon) (3,1,epsilon) I think all triangles are have area of at least 1. The volume is 8*epsilon, which can be arbitrarily small. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: parallelepiped volumes vs triangular areas >>I think this one has a nonzero minimum (but with my record so far, I >>wouldn't bet too much on it). >>Problem: >>What is the minimum volume of a parallelepiped P such that for every 3 >>distinct vertices of P, the area of the corresponding triangle is at >>least 1? >Set the base vertices at >(0,0,0) >(0,4,0) >(2,4,0) >(2,0,0) >and the upper ones at >(1,1,epsilon) >(1,5,epsilon) >(3,5,epsilon) >(3,1,epsilon) >I think all triangles are have area of at least 1. >The volume is 8*epsilon, which can be arbitrarily small. Yep. It's seems very hard to constrain these 3D objects -- with the extra dimension, they're very slippery. In essence your example is this: As a base for the parallelepiped, take any parallelogram in the xy-plane with area at least 2. To get the top face of the parallelepiped, shift the base up by epsilon and then translate the top far away while staying in the plane z=epsilon. The resulting parallelepiped has volume which is epsilon times the area of the base and hence can be forced to approach 0, but if the translation is far enough away, all triangles will have area at least than 1. quasi === Subject: Re: parallelepiped volumes vs triangular areas >As a base for the parallelepiped, take any parallelogram in the >xy-plane with area at least 2. >To get the top face of the parallelepiped, shift the base up by >epsilon and then translate the top far away while staying in the plane >z=epsilon. The resulting parallelepiped has volume which is epsilon >times the area of the base and hence can be forced to approach 0, but >if the translation is far enough away, all triangles will have area at >least than 1. You also have to be careful to keep the upper vertexes away from lines containing sides and diagonals of the base. My first attempt used a square base and contained triangles which had altitude epsilon. You could compensate for this by defining far away to mean more than 2/epsilon, or by doing as I did and missing the sides and diagonals. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: parallelepiped distances days. My association with the Department is that of an alumnus. >What is the minimum volume of a parallelepiped such that the distance >between any pair of vertices is at least 1? There is no minimum, and the infimum is zero. Consider the family of parallelograms with one vertex on (0,0), one vertex on (2,0), one vertex at (cos(t),sin(t)) with 0= sqrt(2)>1. The area of the parallelogram is 2*sin(t), so there is no minimum and the infinimum is zero. Now consider the parallelopipeds obtained by taking one of the parallelograms P, and taking P x [0,1]. The volume is 2*sin(t), so there is no minimum and the infimum is 0. If I didn't mess up, anyway. -- === Subject: Re: parallelepiped distances >>What is the minimum volume of a parallelepiped such that the distance >>between any pair of vertices is at least 1? >There is no minimum, and the infimum is zero. >Consider the family of parallelograms with one vertex on (0,0), one >vertex on (2,0), one vertex at (cos(t),sin(t)) with 0upper right vertex at (cos(t)+2,sin(t)). >The distance between any two vertices is always at least 1: the only >distance that decreases as t ranges from pi/2 to 0 is the distance >between (cos(t),sin(t)) and (2,0), and that distance is >sqrt(2)*sqrt(cos(t)+1)>= sqrt(2)>1. >The area of the parallelogram is 2*sin(t), so there is no minimum and >the infinimum is zero. >Now consider the parallelopipeds obtained by taking one of the >parallelograms P, and taking P x [0,1]. The volume is 2*sin(t), so >there is no minimum and the infimum is 0. >If I didn't mess up, anyway. No it's fine, and very clear. I never really had good 3D visualization -- I'm trying to repair that. It could be that I'm missing a 3D gene. Maybe I'll go up one dimension to see if my visualization is better in 4D. quasi === Subject: Re: parallelepiped distances days. My association with the Department is that of an alumnus. I did make one mistake, though it was inconsecuential. >What is the minimum volume of a parallelepiped such that the distance >between any pair of vertices is at least 1? >>There is no minimum, and the infimum is zero. >>Consider the family of parallelograms with one vertex on (0,0), one >>vertex on (2,0), one vertex at (cos(t),sin(t)) with 0>upper right vertex at (cos(t)+2,sin(t)). >>The distance between any two vertices is always at least 1: the only >>distance that decreases as t ranges from pi/2 to 0 is the distance >>between (cos(t),sin(t)) and (2,0), and that distance is >>sqrt(2)*sqrt(cos(t)+1)>= sqrt(2)>1. This is a mess. The distance is sqrt(5-4cos(t)). Still larger than 1 for any t>0. -- === Subject: Re: parallelepiped distances >I did make one mistake, though it was inconsecuential. >>What is the minimum volume of a parallelepiped such that the distance >>between any pair of vertices is at least 1? >There is no minimum, and the infimum is zero. >Consider the family of parallelograms with one vertex on (0,0), one >vertex on (2,0), one vertex at (cos(t),sin(t)) with 0upper right vertex at (cos(t)+2,sin(t)). >The distance between any two vertices is always at least 1: the only >distance that decreases as t ranges from pi/2 to 0 is the distance >between (cos(t),sin(t)) and (2,0), and that distance is >sqrt(2)*sqrt(cos(t)+1)>= sqrt(2)>1. >This is a mess. The distance is sqrt(5-4cos(t)). Still larger than 1 >for any t>0. I didn't catch that error since the geometric view of your construction makes it clear without calculation that all distances are at least 1. For example, it's immediate that the distance from (2,0) to the line x=1 is 1, so the distance from (2,0) to any point in the xy-plane to the left of the line x=1 must exceed 1, which takes care of the distance from (2,0) to (cos(t),sin(t)) in your construction. quasi === Subject: integral inequality I'm really tearing my hair out trying to prove this integral inequality. Fair warning, it is homework related. The problem is to show ( integral |f(x)| dx )^5 <= 4 ( integral x^(5/16) |f(x)|^(5/4) dx )^4. Now I'm fairly sure this is just holder's inequality, but the problem is choosing the right functions and the right exponents... and I can't get it! One big problem is the ^5 on the left hand side, which really screws with the format for Holder. I really think after all this time I'm not going to gain any sort of enlightenment by bashing my heard against this any more, I really would like to understand this stuff better but this problem seems extremely difficult... === Subject: Re: integral inequality > I'm really tearing my hair out trying to prove this integral > inequality. Fair warning, it is homework related. The problem is to > show > ( integral |f(x)| dx )^5 <= 4 ( integral x^(5/16) |f(x)|^(5/4) dx > )^4. > Now I'm fairly sure this is just holder's inequality, but the problem > is choosing the right functions and the right exponents... and I can't > get it! One big problem is the ^5 on the left hand side, which really > screws with the format for Holder. I really think after all this time > I'm not going to gain any sort of enlightenment by bashing my heard > against this any more, I really would like to understand this stuff > better but this problem seems extremely difficult... Are you sure the way you stated the problem is correct? If we choose f(x) to be the characteristic function of [0, t], then by choosing t sufficiently small, the inequality will not hold. === Subject: Re: integral inequality Oh dear, I forgot to say that the integrals go from 1 to infinity, not the whole real line. A thousand pardons. Even what you said, though, helps pound home the fact this problem is highly nontrivial... === Subject: Re: integral inequality > ( integral |f(x)| dx )^5 <= 4 ( integral x^(5/16) |f(x)|^(5/4) dx > )^4. > Oh dear, I forgot to say that the integrals go from 1 to infinity Hint: For any a in R you like, int f(x) dx = int [x^(-a)]*[f(x)x^a] dx. === Subject: Re: Polynomials, general factorization, distributive property > Here I will give an explanation with some basic examples, where I make > the effort in the hope that it will clear the air about how I use the > distributive property with certain factorizations of a polynomial. > Consider a polynomial C(x) with a constant term that is 1 on the > complex plane. > Multiply it by 7 and factor it as > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. ..... > But what about non-polynomial factors? > Well, let C(x) = x^2 + x - 1, and > g(x) = sqrt(x^2 + x) + 1 > and notice that though it's not a polynomial it still must be true that > 7 multiplied through the first factor, and again, as these results are > valid in the complex plane it's meaningless to say that 7 is a factor > of f(x). > But, of course, for this example > f(x) = 7*sqrt(x^2 + x) I don't think so ... (f(x)+7)(g(x)+1) =/= 7*C(x) lets say x=4 C(4) = 19, 7C(4)=133 sqrt(4^2+4) = sqrt(20) ~= 4.472136 f(4) ~= (7 * 4.472136 ) = 31.304952 g(4) ~= (4.472136 + 1) = 5.472136 (f(4) + 7 ) * (g(4) + 1 ) ~= ( 31.304952 + 7 ) * ( 5.472136 + 1) = 38.304952 * 6.472136 = 247.91486 clearly something is wrong. bye. Jasen === Subject: Re: Polynomials, general factorization, distributive property > Here I will give an explanation with some basic examples, where I make > the effort in the hope that it will clear the air about how I use the > distributive property with certain factorizations of a polynomial. > Consider a polynomial C(x) with a constant term that is 1 on the > complex plane. > Multiply it by 7 and factor it as > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > The reason the polynomial is multiplied by a constant is that in some > of my research you get expressions similar to the above, where I'm > abstracting for simplicity. > Now if f(x) and g(x) are simple linear functions, it's easy enough to > see how 7 distributes through, for instance, if C(x) = x^2 + 2x + 1 and > g(x) = x, then it must be true that > f(x) = 7x > where it doesn't mean anything in the complex plane to say that f(x) > has 7 as a factor as so does g(x), though it equals x. I'll make this as simple as possible. No one disagrees that you can factor 7 out of the term f(x) + 7, with the result h(x) + 1, where h(x) = f(x)/7. No problem. However you can also factor sqrt(7) out of both the first term and the second term: f(x) + 7 = sqrt(7) * (f(x)/sqrt(7) + sqrt(7)) and g(x) + 1 = sqrt(7) * (g(x)/sqrt(7) + 1/sqrt(7)). Nothing wrong with that either. These are perfectly good factorizations in the complex numbers, where you said you were doing your factoring. These factorizations, just like the one that you prefer, are consistent with the distributive property. The problem is not that anyone objects to your factorization. The problem is that, in the complex numbers, it is not unique. Infinitely many other factorizations are possible also. The problem is, to be useful to you, you must do this factorization in a ring where your preferred factorization is the ONLY one possible (up to units). You have not addressed this uniqueness problem at all. Until you do, you are going to just keep butting your head against the wall. What you are now saying, in the ring of complex numbers, is trivial and it is not unique. You need to get beyond this. > The mathematical terminology hasn't been developed for these types of > arguments, so I usually just say that 7 multiplied through one factor > of C(x), in this case, 7x+7. > I doubt many would argue that for any polynomial C(x) on the complex > plane, where f(x) and g(x) are linear functions of x that 7 would have > to have multiplied through the factor where 7 is visible, with the > requirement that f(0) = g(0) = 0. > But what about non-polynomial factors? > Well, let C(x) = x^2 + x - 1, and > g(x) = sqrt(x^2 + x) + 1 > and notice that though it's not a polynomial it still must be true that > 7 multiplied through the first factor, and again, as these results are > valid in the complex plane it's meaningless to say that 7 is a factor > of f(x). > But, of course, for this example > f(x) = 7*sqrt(x^2 + x) > and the question is how to describe the general case with > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > Now I can give specific cases repeatedly, but it's better to try and > abstract out what's happening---why is it true that 7 multiplies > through the one factor? It's obvious why it *does*. > Can anyone give a counterexample where it does not? Not in the complex numbers. But you are asking the wrong question. > I like these examples because I not only show the simple polynomial > case--which should be very familiar--but I also give an explicit > non-polynomial factorization where again you can actually see the > result. > The question then for those who disagree with me is, can you have an > explicit solution where the seemingly simple result that 7 multiplied > through f(x) + 7 is violated? > Can you break my toy examples? Not in the complex numbers. You CAN factor 7 out in the way you want. The problem is, you can also factor it out in infinitely many other ways. You are not interested in conclusions about the ring of complex numbers. You need a ring in which your factorization is the ONLY one possible. Clearly, obviously you have no proof that the complex numbers is such a ring. You keep setting up this silly straw man: Prove that 7 cannot factor out of f(x) + 7 in the complex numbers. NO ONE argues that it cannot. You just cannot seem to understand that this is missing the point. You need not EXISTENCE of your factorization - everyone concedes that - you need UNIQUENESS. And you have never once come to grips with that. > If so, give the counterexample. > If not, why not? See above. You are asking the wrong question. Marcus. > James Harris === Subject: Re: Polynomials, general factorization, distributive property > Here I will give an explanation with some basic examples, where I make > the effort in the hope that it will clear the air about how I use the > distributive property with certain factorizations of a polynomial. > Consider a polynomial C(x) with a constant term that is 1 on the > complex plane. > Multiply it by 7 and factor it as > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > The reason the polynomial is multiplied by a constant is that in some > of my research you get expressions similar to the above, where I'm > abstracting for simplicity. > Now if f(x) and g(x) are simple linear functions, it's easy enough to > see how 7 distributes through, for instance, if C(x) = x^2 + 2x + 1 and > g(x) = x, then it must be true that > f(x) = 7x > where it doesn't mean anything in the complex plane to say that f(x) > has 7 as a factor as so does g(x), though it equals x. > The mathematical terminology hasn't been developed for these types of > arguments, so I usually just say that 7 multiplied through one factor > of C(x), in this case, 7x+7. > I doubt many would argue that for any polynomial C(x) on the complex > plane, where f(x) and g(x) are linear functions of x that 7 would have > to have multiplied through the factor where 7 is visible, with the > requirement that f(0) = g(0) = 0. > But what about non-polynomial factors? > Well, let C(x) = x^2 + x - 1, and > g(x) = sqrt(x^2 + x) + 1 > and notice that though it's not a polynomial it still must be true that > 7 multiplied through the first factor, and again, as these results are > valid in the complex plane it's meaningless to say that 7 is a factor > of f(x). > But, of course, for this example > f(x) = 7*sqrt(x^2 + x) > and the question is how to describe the general case with > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > Now I can give specific cases repeatedly, but it's better to try and > abstract out what's happening---why is it true that 7 multiplies > through the one factor? > Can anyone give a counterexample where it does not? You've already observed that your distinction is meaningless in the complex plane. However, you don't really want the complex plane, since in that context you won't get the divisibility result you're really after (you know, the old Area 2 one). Looky here: Let Q(x) = x^2 - x + 1 f(x) = -3 - x + sqrt(9 - x - 6x^2) g(x) = 3 - x - sqrt(9 - x - 6x^2) To eliminate any confusion, we'll decide to use the positive square root, i.e., the one for which sqrt(9) = 3. Now f(0) = g(0) = 0 and it's easy enough to verify that 7Q(x) = (f(x) + 7)(g(x) + 1) If we let x = 1 we have f(1) = -4 + sqrt(2) g(1) = 2 - sqrt(2) Obviously, -4 + sqrt(2) is divisible by 7 in the complex field, since EVERYTHING is divisible by 7 there. However, that doesn't seem to be what you really want. So here's the question: pick the context where divisibility makes nontrivial sense and tell me how 7 multiplies through f(1). In this case, the 7 splits into the factors (3 + sqrt(2)) and (3 - sqrt(2)), the first of which is f(1) + 7 and the second is g(1) + 1. Rick === Subject: Re: Polynomials, general factorization, distributive property the effort in the hope that it will clear the air about how I use the > distributive property with certain factorizations of a polynomial. > Consider a polynomial C(x) with a constant term that is 1 on the > complex plane. > Multiply it by 7 and factor it as > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > The reason the polynomial is multiplied by a constant is that in some > of my research you get expressions similar to the above, where I'm > abstracting for simplicity. > Now if f(x) and g(x) are simple linear functions, it's easy enough to > see how 7 distributes through, for instance, if C(x) = x^2 + 2x + 1 and > g(x) = x, then it must be true that > f(x) = 7x > where it doesn't mean anything in the complex plane to say that f(x) > has 7 as a factor as so does g(x), though it equals x. > The mathematical terminology hasn't been developed for these types of > arguments, so I usually just say that 7 multiplied through one factor > of C(x), in this case, 7x+7. > I doubt many would argue that for any polynomial C(x) on the complex > plane, where f(x) and g(x) are linear functions of x that 7 would have > to have multiplied through the factor where 7 is visible, with the > requirement that f(0) = g(0) = 0. > But what about non-polynomial factors? > Well, let C(x) = x^2 + x - 1, and > g(x) = sqrt(x^2 + x) + 1 > and notice that though it's not a polynomial it still must be true that > 7 multiplied through the first factor, and again, as these results are > valid in the complex plane it's meaningless to say that 7 is a factor > of f(x). > But, of course, for this example > f(x) = 7*sqrt(x^2 + x) > and the question is how to describe the general case with > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > Now I can give specific cases repeatedly, but it's better to try and > abstract out what's happening---why is it true that 7 multiplies > through the one factor? > Can anyone give a counterexample where it does not? > You've already observed that your distinction is meaningless > in the complex plane. I don't think that follows from what is given. Obviously enough 7(x+1)(x+1) = (7x + 7)(x+1) is distinct, from some other factorization, like 7(x+1)(x+1) = (sqrt(7)x + sqrt(7))(sqrt(7)x + sqrt(7)) while 7 is a factor of all because it's in the complex plane, or do you disagree? There is a way to consider a distinction between those two, even on the complex plane. > However, you don't really want the complex plane, since in Yes I do. I NEED it because of b.s. arguments like what you're trying to pull here. > that context you won't get the divisibility result you're > really after (you know, the old Area 2 one). > Looky here: > Let Q(x) = x^2 - x + 1 > f(x) = -3 - x + sqrt(9 - x - 6x^2) > g(x) = 3 - x - sqrt(9 - x - 6x^2) > To eliminate any confusion, we'll decide to use the > positive square root, i.e., the one for which sqrt(9) = 3. However, mathematically, BOTH roots are available, which logically must follow as what if you were perverse (by human standards) and wanted to use the NEGATIVE instead? The mathematics can't change, right? I mean, can the entire world of mathematics move based on whether or not you--just some human being--decides to use the positive or the negative result of the square root? Nope. > Now f(0) = g(0) = 0 and it's easy enough to verify that > 7Q(x) = (f(x) + 7)(g(x) + 1) > If we let x = 1 we have > f(1) = -4 + sqrt(2) > g(1) = 2 - sqrt(2) > Obviously, -4 + sqrt(2) is divisible by 7 in the complex field, > since EVERYTHING is divisible by 7 there. However, that doesn't > seem to be what you really want. So here's the question: pick Hand-waving. You keep seeming to need to mind-read here, claiming to say what I want or don't want, but I believe in mathematics. There is a mathematical truth in there, right? So what is true mathematically? > the context where divisibility makes nontrivial sense and tell > me how 7 multiplies through f(1). In this case, the > 7 splits into the factors (3 + sqrt(2)) and (3 - sqrt(2)), the > first of which is f(1) + 7 and the second is g(1) + 1. The square root is ambiguous, despite what anyone claims, as some human being can say, oh, take the positive, and some OTHER human being can say, NO, take the negative!!! So what the mathematics does is handle either case. Trivial. Since you can take either case--being a quirky human being--the mathematics doesn't say WHICH case has 7 as a factor, but logic dictates that one does. It's a trivial argument, but it just so happens to shoot down standard ideas about Galois Theory!!! Neat. Simple. Rigorous. > Rick Hand-waving and b.s. don't make a mathematical argument. Your reply lacked polish. James Harris === Subject: Re: Polynomials, general factorization, distributive property You've already observed that your distinction is meaningless >> in the complex plane. > I don't think that follows from what is given. > Obviously enough > 7(x+1)(x+1) = (7x + 7)(x+1) ! HOWD YOU DOO DAT !!! > is distinct, from some other factorization, like > 7(x+1)(x+1) = (sqrt(7)x + sqrt(7))(sqrt(7)x + sqrt(7)) SHUCKS, YOU DID IT AGAIN !! > while 7 is a factor of all because it's in the complex plane, or do you > disagree? you never made it past the 7th grade, that is why you are STUCK on 7 You keep seeming to need to mind-read here, claiming to say what I want > or don't want, but I believe in mathematics. but you don't KNOW mathematics. > There is a mathematical truth in there, right? You asking ME ? You Talking to ME ?? > So what is true mathematically? > The square root is ambiguous, despite what anyone claims, as some human > being can say, oh, take the positive, and some OTHER human being can > say, NO, take the negative!!! Wierdo, stay OUT of SCI.MATH till you pass algebra 1 (high school) > So what the mathematics does is handle either case. Trivial. > Since you can take either case--being a quirky human being--the > mathematics doesn't say WHICH case has 7 as a factor, but logic > dictates that one does. > It's a trivial argument, but it just so happens to shoot down standard > ideas about Galois Theory!!! > Neat. Simple. Rigorous. complete BULL. > Hand-waving and b.s. don't make a mathematical argument. > Your reply lacked polish. > James Harris Your logic lacks sanity. === Subject: Re: Polynomials, general factorization, distributive property >Here I will give an explanation with some basic examples, where I make >the effort in the hope that it will clear the air about how I use the >distributive property with certain factorizations of a polynomial. >Consider a polynomial C(x) with a constant term that is 1 on the >complex plane. >Multiply it by 7 and factor it as >7*C(x) = (f(x) + 7)*(g(x) + 1) >where f(0) = g(0) = 0. >The reason the polynomial is multiplied by a constant is that in some >of my research you get expressions similar to the above, where I'm >abstracting for simplicity. >Now if f(x) and g(x) are simple linear functions, it's easy enough to >see how 7 distributes through, for instance, if C(x) = x^2 + 2x + 1 and >g(x) = x, then it must be true that >f(x) = 7x >where it doesn't mean anything in the complex plane to say that f(x) >has 7 as a factor as so does g(x), though it equals x. >The mathematical terminology hasn't been developed for these types of >arguments, so I usually just say that 7 multiplied through one factor >of C(x), in this case, 7x+7. >I doubt many would argue that for any polynomial C(x) on the complex >plane, where f(x) and g(x) are linear functions of x that 7 would have >to have multiplied through the factor where 7 is visible, with the >requirement that f(0) = g(0) = 0. >But what about non-polynomial factors? >Well, let C(x) = x^2 + x - 1, and >g(x) = sqrt(x^2 + x) + 1 >and notice that though it's not a polynomial it still must be true that >7 multiplied through the first factor, and again, as these results are >valid in the complex plane it's meaningless to say that 7 is a factor >of f(x). >But, of course, for this example >f(x) = 7*sqrt(x^2 + x) >and the question is how to describe the general case with >7*C(x) = (f(x) + 7)*(g(x) + 1) >where f(0) = g(0) = 0. >Now I can give specific cases repeatedly, but it's better to try and >abstract out what's happening---why is it true that 7 multiplies >through the one factor? >Can anyone give a counterexample where it does not? >>You've already observed that your distinction is meaningless >>in the complex plane. > I don't think that follows from what is given. > Obviously enough > 7(x+1)(x+1) = (7x + 7)(x+1) > is distinct, from some other factorization, like > 7(x+1)(x+1) = (sqrt(7)x + sqrt(7))(sqrt(7)x + sqrt(7)) > while 7 is a factor of all because it's in the complex plane, or do you > disagree? What you think is immaterial here. You're not even wrong. > There is a way to consider a distinction between those two, even on the > complex plane. So show it. Define what you mean by 7 multiplies through f(x). >>However, you don't really want the complex plane, since in > Yes I do. > I NEED it because of b.s. arguments like what you're trying to pull > here. No, actually you don't. In fact, since 7 divides everything in the complex numbers, that choice is just about the worst domain to be in. >>that context you won't get the divisibility result you're >>really after (you know, the old Area 2 one). >>Looky here: >>Let Q(x) = x^2 - x + 1 >> f(x) = -3 - x + sqrt(9 - x - 6x^2) >> g(x) = 3 - x - sqrt(9 - x - 6x^2) >>To eliminate any confusion, we'll decide to use the >>positive square root, i.e., the one for which sqrt(9) = 3. > However, mathematically, BOTH roots are available, which logically must > follow as what if you were perverse (by human standards) and wanted to > use the NEGATIVE instead? of difference. > The mathematics can't change, right? I mean, can the entire world of > mathematics move based on whether or not you--just some human > being--decides to use the positive or the negative result of the square > root? > Nope. >>Now f(0) = g(0) = 0 and it's easy enough to verify that >> 7Q(x) = (f(x) + 7)(g(x) + 1) >>If we let x = 1 we have >> f(1) = -4 + sqrt(2) >> g(1) = 2 - sqrt(2) >>Obviously, -4 + sqrt(2) is divisible by 7 in the complex field, >>since EVERYTHING is divisible by 7 there. However, that doesn't >>seem to be what you really want. So here's the question: pick > Hand-waving. Heh. That's what I call a substantive objection. You wouldn't care to clarify by pointing out the part that's confusing you, would you? > You keep seeming to need to mind-read here, claiming to say what I want > or don't want, but I believe in mathematics. Regardless of your beliefs or lack thereof, I actually can say what you'll need. Over the past decade you've made your intents abundantly clear. > There is a mathematical truth in there, right? > So what is true mathematically? >>the context where divisibility makes nontrivial sense and tell >>me how 7 multiplies through f(1). In this case, the >>7 splits into the factors (3 + sqrt(2)) and (3 - sqrt(2)), the >>first of which is f(1) + 7 and the second is g(1) + 1. > The square root is ambiguous, despite what anyone claims, as some human > being can say, oh, take the positive, and some OTHER human being can > say, NO, take the negative!!! > So what the mathematics does is handle either case. Trivial. Exactly. All you have to do is switch four signs in my counterexample. > Since you can take either case--being a quirky human being--the > mathematics doesn't say WHICH case has 7 as a factor, but logic > dictates that one does. No. As I've stated repeatedly, in your example there are plenty of situations where 7 distributes among the factors in a non-trivial way. Let me make this abundantly clear: You're claiming that if * Q(x) is a polynomial with Q(0) = 1 * f and g are functions with f(0) = g(0) = 0 * 7*Q(x) = (f(x) + 7)(g(x) + 1) then 7 | f(x) for all x. The responses are: 1. If divisibility is considered in the context of the complex numbers then your conclusion is true, but trivial, since 7 divides everything in C. In other words, 7 also divides g(x) or anything else. 2. If divisibility is considered in the context of the algebraic integers, then there are abundant examples showing your conclusion is false. 3. If you redefine your conclusion to be 7 multiplies through f(x) then the conclusion is meaningless until you define your terms. > It's a trivial argument, but it just so happens to shoot down standard > ideas about Galois Theory!!! As I predicted. Response category 2.d, slightly modified by replacing algebraic integers with Galois theory. > Neat. Simple. Rigorous. >>Rick > Hand-waving and b.s. don't make a mathematical argument. POT...KETTLE...BLACK? Would that you'd take your sentence to heart. > Your reply lacked polish. LOL. Surely that's not the best you can do. Rick === Subject: Re: Polynomials, general factorization, distributive property the effort in the hope that it will clear the air about how I use the >distributive property with certain factorizations of a polynomial. Consider a polynomial C(x) with a constant term that is 1 on the >complex plane. Multiply it by 7 and factor it as 7*C(x) = (f(x) + 7)*(g(x) + 1) where f(0) = g(0) = 0. The reason the polynomial is multiplied by a constant is that in some >of my research you get expressions similar to the above, where I'm >abstracting for simplicity. Now if f(x) and g(x) are simple linear functions, it's easy enough to >see how 7 distributes through, for instance, if C(x) = x^2 + 2x + 1 and >g(x) = x, then it must be true that f(x) = 7x where it doesn't mean anything in the complex plane to say that f(x) >has 7 as a factor as so does g(x), though it equals x. The mathematical terminology hasn't been developed for these types of >arguments, so I usually just say that 7 multiplied through one factor >of C(x), in this case, 7x+7. I doubt many would argue that for any polynomial C(x) on the complex >plane, where f(x) and g(x) are linear functions of x that 7 would have >to have multiplied through the factor where 7 is visible, with the >requirement that f(0) = g(0) = 0. But what about non-polynomial factors? Well, let C(x) = x^2 + x - 1, and g(x) = sqrt(x^2 + x) + 1 and notice that though it's not a polynomial it still must be true that >7 multiplied through the first factor, and again, as these results are >valid in the complex plane it's meaningless to say that 7 is a factor >of f(x). But, of course, for this example f(x) = 7*sqrt(x^2 + x) and the question is how to describe the general case with 7*C(x) = (f(x) + 7)*(g(x) + 1) where f(0) = g(0) = 0. Now I can give specific cases repeatedly, but it's better to try and >abstract out what's happening---why is it true that 7 multiplies >through the one factor? Can anyone give a counterexample where it does not? >>You've already observed that your distinction is meaningless >>in the complex plane. > I don't think that follows from what is given. > Obviously enough > 7(x+1)(x+1) = (7x + 7)(x+1) > is distinct, from some other factorization, like > 7(x+1)(x+1) = (sqrt(7)x + sqrt(7))(sqrt(7)x + sqrt(7)) > while 7 is a factor of all because it's in the complex plane, or do you > disagree? > What you think is immaterial here. You're not even wrong. You keep trying to joke the simplified argument off, but the mathematics is easy to the point of triviality. Here it's not difficult to accept that 7(x+1)(x+1) = (7x + 7)(x+1) can be distinguished from 7(x+1)(x+1) = (sqrt(7)x + sqrt(7))(sqrt(7)x + sqrt(7)) as two separate and distinct factorizations. How factors of 7 distribute makes the difference, nothing more. It's not rocket science. > There is a way to consider a distinction between those two, even on the > complex plane. > So show it. Define what you mean by 7 multiplies through f(x). The terminology is awkward as I've admitted, but it's not beyond comprehension. Consider, from the distributive property a*(b+c) = a*b + a*c and you can say that a multiplied through b+c to give a*b + a*c as saying it's a factor is meaningless in the complex field. Understand? >>However, you don't really want the complex plane, since in > Yes I do. > I NEED it because of b.s. arguments like what you're trying to pull > here. > No, actually you don't. In fact, since 7 divides everything in the > complex numbers, that choice is just about the worst domain to be in. Nope. It's perfect to remove the issue of factors and show that this is just the distributive property, as is the distributive property still valid on the complex plane? Or not? I want you to answer that question please. Does the distributive property still work on the complex plane? Or does you objection about it being the worst domain to be in still apply in your mind? >>that context you won't get the divisibility result you're >>really after (you know, the old Area 2 one). >>Looky here: >>Let Q(x) = x^2 - x + 1 >> f(x) = -3 - x + sqrt(9 - x - 6x^2) >> g(x) = 3 - x - sqrt(9 - x - 6x^2) >> Did anyone check those to see if they work? I admit I haven't. >>To eliminate any confusion, we'll decide to use the >>positive square root, i.e., the one for which sqrt(9) = 3. > However, mathematically, BOTH roots are available, which logically must > follow as what if you were perverse (by human standards) and wanted to > use the NEGATIVE instead? > of difference. Yes it does, as consider 1+sqrt(4) is it divisible by 3 or not? If you claim that it is divisible by 3 as you take the positive of the square root, I'll point out that someone else might say take the negative and for that reason claim it's coprime to 3, so you get this area of ambiguity. Or consider 1+/-2 if the first doesn't quite make it for you. So, mathematically given an ambiguous expression, like the sqrt(2), which has two solutions, there's no way you can directly see that 7 is a factor of just one, as unlike with 1+sqrt(4) = 3 or -1 you can't evaluate the square root to integers. Understand? So W. Dale Hall showed nothing. > The mathematics can't change, right? I mean, can the entire world of > mathematics move based on whether or not you--just some human > being--decides to use the positive or the negative result of the square > root? > Nope. >>Now f(0) = g(0) = 0 and it's easy enough to verify that >> 7Q(x) = (f(x) + 7)(g(x) + 1) >>If we let x = 1 we have >> f(1) = -4 + sqrt(2) >> g(1) = 2 - sqrt(2) >>Obviously, -4 + sqrt(2) is divisible by 7 in the complex field, >>since EVERYTHING is divisible by 7 there. However, that doesn't >>seem to be what you really want. So here's the question: pick > Hand-waving. > Heh. That's what I call a substantive objection. You wouldn't > care to clarify by pointing out the part that's confusing you, > would you? I explained in my reply. Mathematically because the square root is ambiguous, you can't SEE 7 as a factor, any more than you can SEE 3 as a factor with 1+sqrt(4) but that doesn't change the fact that 3 is a factor of just one of the possible solutions. You need square roots because of the ambiguity and when I point that out, you start hand-waving. It's not complicated. Again, the mathematics here is easy to the point of triviality. > You keep seeming to need to mind-read here, claiming to say what I want > or don't want, but I believe in mathematics. > Regardless of your beliefs or lack thereof, I actually can say > what you'll need. Over the past decade you've made your intents > abundantly clear. See? There you go. You can't stick with mathematics but keep playing these social games, in comment after comment. It's like you can't conceive of an objective mathematics only reply. > There is a mathematical truth in there, right? > So what is true mathematically? >>the context where divisibility makes nontrivial sense and tell >>me how 7 multiplies through f(1). In this case, the >>7 splits into the factors (3 + sqrt(2)) and (3 - sqrt(2)), the >>first of which is f(1) + 7 and the second is g(1) + 1. > The square root is ambiguous, despite what anyone claims, as some human > being can say, oh, take the positive, and some OTHER human being can > say, NO, take the negative!!! > So what the mathematics does is handle either case. Trivial. > Exactly. All you have to do is switch four signs in my > counterexample. And you can switch signs with 1-sqrt(4) AND IT STILL HAS TWO SOLUTIONS. Switching signs does not remove the ambiguity. Also consider 1-/+2 as another case where switching signs is meaningless. I doubt the mathematics here escapes you. It's trivial. It's easy. It doesn't take a genius to understand it. > Since you can take either case--being a quirky human being--the > mathematics doesn't say WHICH case has 7 as a factor, but logic > dictates that one does. > No. As I've stated repeatedly, in your example there are plenty > of situations where 7 distributes among the factors in a non-trivial > way. Let me make this abundantly clear: You're handwaving again. > You're claiming that if > * Q(x) is a polynomial with Q(0) = 1 > * f and g are functions with f(0) = g(0) = 0 > * 7*Q(x) = (f(x) + 7)(g(x) + 1) > then 7 | f(x) for all x. > The responses are: > 1. If divisibility is considered in the context of the complex numbers > then your conclusion is true, but trivial, since 7 divides everything > in C. In other words, 7 also divides g(x) or anything else. > 2. If divisibility is considered in the context of the algebraic > integers, then there are abundant examples showing your conclusion > is false. > 3. If you redefine your conclusion to be 7 multiplies through f(x) > then the conclusion is meaningless until you define your terms. I went to the complex plane to keep people from trying to make it into a divisibility argument, but you're still trying!!! It's about the distributive property and the simple principle that if you multiply a group, you multiply the elements within that group, where given 7*C(x) = (f(x) + 7)*(g(x) + 1) where f(0) = g(0) = 0 it must be true that 7 multiplies throuhg only one of the factors of C(x) based on some simple logic. 1. The distributive property acts without regard to the value of the elements within the group. 2. Checking at x=0 shows that 7 multiplied through just one factor of C(x); therefore, by logical point 1. it does so for all x. Easy argument to the point of triviality. > It's a trivial argument, but it just so happens to shoot down standard > ideas about Galois Theory!!! > As I predicted. Response category 2.d, slightly modified by replacing > algebraic integers with Galois theory. And that's more social stuff from you. > Neat. Simple. Rigorous. That's not even wrong. >>Rick > Hand-waving and b.s. don't make a mathematical argument. > POT...KETTLE...BLACK? Childish. > Would that you'd take your sentence to heart. > Your reply lacked polish. > LOL. Surely that's not the best you can do. > Rick Social crap will only go on for so long Decker. The mathematics doesn't change, and your objections are not mathematical. Try to do some math if you reply again versus grandstanding for the Usenet crowd. James Harris === Subject: Re: JSH Polynomials, general factorization, distributive property [... added JSH: to Subject, deleted the usual junk ...] [Rick Decker] >> Looky here: >> Let Q(x) = x^2 - x + 1 >> f(x) = -3 - x + sqrt(9 - x - 6x^2) >> g(x) = 3 - x - sqrt(9 - x - 6x^2) [jstevh@msn.com] > Did anyone check those to see if they work? These things are true, under the usual convention of taking sqrt to mean the positive square root: - Q(x) is a polynomial. - f(0) = g(0) = 0 - 7*Q(x) = (f(x) + 7)*(g(x) + 1) I guess that's what you mean by they work, right? > I admit I haven't. Figures. Which of the three checks above was hard for you? You can easily check all of them in your head. For the last, note that after adding 7 to f, and 1 to g, the factors are of the form: A + B and A - B where A = 4-x and B = sqrt(9-x-6*x^2) Hint: (A+B)*(A-B) = A^2-B^2. [... the usual deleted ...] === Subject: Re: Polynomials, general factorization, distributive property [Rick Decker] > ... > No. As I've stated repeatedly, in your example there are plenty > of situations where 7 distributes among the factors in a non-trivial > way. Let me make this abundantly clear: > You're claiming that if > * Q(x) is a polynomial with Q(0) = 1 > * f and g are functions with f(0) = g(0) = 0 > * 7*Q(x) = (f(x) + 7)(g(x) + 1) > then 7 | f(x) for all x. > The responses are: > 1. If divisibility is considered in the context of the complex numbers > then your conclusion is true, but trivial, since 7 divides everything > in C. In other words, 7 also divides g(x) or anything else. > 2. If divisibility is considered in the context of the algebraic > integers, then there are abundant examples showing your conclusion > is false. > 3. If you redefine your conclusion to be 7 multiplies through f(x) > then the conclusion is meaningless until you define your terms. Well, to be fair, it's you who redefined James's conclusion to 7 | f(x) for all x. His version of his conclusion in this particular thread was f(x) = 7x, where it doesn't mean anything in the complex plane to say that f(x) has 7 as a factor. God only knows what he's _trying_ to say, but 7 | f(x) for all x ain't it ;-) ... [jstevh@msn.com] >> Your reply lacked polish. [Rick] > LOL. Surely that's not the best you can do. I would have thought not a few months ago, but given how long he's been stuck on this particular, and particularly inane, argument ... do the math :-) === Subject: Re: Polynomials, general factorization, distributive property : Yes I do. [ need the complex plane ] : I NEED it because of b.s. arguments like what you're trying to pull : here. You mean the bit where he shows you're wrong? That bit? : Hand-waving. Maybe you could actually argue the mathematics instead of just yelling hand-waving. It all looked pretty solid to me. : Your reply lacked polish. So why is it that your only defense is to jump up and down yelling about the complex plane (where everything divides everything) and about choices of square roots (did you try the other one?) rather than just trying to show where the problem in his argument lies? This is not a social thing, James, this is math. Math doesn't care about what you want or how loud you yell. If you can't argue the math then you've goe nothing. Justin === Subject: Re: Polynomials, general factorization, distributive property > Looky here: > Let Q(x) = x^2 - x + 1 > f(x) = -3 - x + sqrt(9 - x - 6x^2) > g(x) = 3 - x - sqrt(9 - x - 6x^2) > To eliminate any confusion, we'll decide to use the > positive square root, i.e., the one for which sqrt(9) = 3. > Now f(0) = g(0) = 0 and it's easy enough to verify that > 7Q(x) = (f(x) + 7)(g(x) + 1) > If we let x = 1 we have > f(1) = -4 + sqrt(2) > g(1) = 2 - sqrt(2) > Obviously, -4 + sqrt(2) is divisible by 7 in the complex field, > since EVERYTHING is divisible by 7 there. However, that doesn't > seem to be what you really want. So here's the question: pick > the context where divisibility makes nontrivial sense and tell > me how 7 multiplies through f(1). In this case, the > 7 splits into the factors (3 + sqrt(2)) and (3 - sqrt(2)), the > first of which is f(1) + 7 and the second is g(1) + 1. Nice find. Would you care to venture as to the sort of cheating, dishonesty, and downright naughtiness you'll be accused of? Or, will it finally emerge that one of these factors is properly a unit, playing on one of JSH's favorite complaints about the algebraic integers? Of course, by choosing the opposite signs for the sqrt(...), the factors of 7Q get interchanged. This swaps the factors of 7 when you evaluate @ x=1, so the bogus properly a unit argument can't even get off the ground, unless JSH goes the extra step in claiming that one equation's proper units may not be proper units for another equation, as in I don't know how to define a proper unit, but I know it when I see it, to paraphrase Justice Potter Stewart's discussion of pornography. Dale. === Subject: Re: Polynomials, general factorization, distributive property >> Looky here: >> Let Q(x) = x^2 - x + 1 >> f(x) = -3 - x + sqrt(9 - x - 6x^2) >> g(x) = 3 - x - sqrt(9 - x - 6x^2) >> To eliminate any confusion, we'll decide to use the >> positive square root, i.e., the one for which sqrt(9) = 3. >> Now f(0) = g(0) = 0 and it's easy enough to verify that >> 7Q(x) = (f(x) + 7)(g(x) + 1) >> If we let x = 1 we have >> f(1) = -4 + sqrt(2) >> g(1) = 2 - sqrt(2) >> Obviously, -4 + sqrt(2) is divisible by 7 in the complex field, >> since EVERYTHING is divisible by 7 there. However, that doesn't >> seem to be what you really want. So here's the question: pick >> the context where divisibility makes nontrivial sense and tell >> me how 7 multiplies through f(1). In this case, the >> 7 splits into the factors (3 + sqrt(2)) and (3 - sqrt(2)), the >> first of which is f(1) + 7 and the second is g(1) + 1. > Nice find. > Would you care to venture as to the sort of cheating, dishonesty, > and downright naughtiness you'll be accused of? Or, will it > finally emerge that one of these factors is properly a unit, > playing on one of JSH's favorite complaints about the algebraic > integers? Okay, I'll speculate, in rough order of estimated liklihood. 1. The post will generate no response at all. 2. The response will be: a. A repost of the argument showing that in the complex numbers any counterexample must show that the distributive property implies that 7 MUST divide f(x). b. The same as (a), but with the assertion appended that I haven't showed the flaw in his distributive argument. In other words, counterexamples don't refute proofs. c. Irrational solutions are a special case (BTW, the only rational solutions are for x = 0, -1 and in both cases 7 does divide f(x)). d. The algebraic integers are flawed and this counterexample just serves to point it out. e. Some form of the not properly a unit argument. f. This is not a counterexample, since James is doing all his computation in the Object Ring, rather than in the ring of algebraic integers. and finally, g. A sheepish admission, I guess I was wrong. Hey, people make mistakes and all of you groupthink stooges fell for it. > Of course, by choosing the opposite signs for the sqrt(...), > the factors of 7Q get interchanged. This swaps the factors > of 7 when you evaluate @ x=1, so the bogus properly a unit > argument can't even get off the ground, unless JSH goes the > extra step in claiming that one equation's proper units may > not be proper units for another equation, as in I don't know > how to define a proper unit, but I know it when I see it, to > paraphrase Justice Potter Stewart's discussion of pornography. > Dale. Rick === Subject: Re: Polynomials, general factorization, distributive property <3ZmdnVxX8bdnF6nZRVn-iA@hamilton.edu> [snippage] >> To eliminate any confusion, we'll decide to use the >> positive square root, i.e., the one for which sqrt(9) = 3. > Okay, I'll speculate, in rough order of estimated liklihood. I knew as soon as I saw the above extracted portion what the response would be; one you didn't give in your list. Remember, James doesn't... um... accept (less charitable folks might say understand) the idea that sqrt is a symbol which means the positive square root of. He doesn't accept that while 2 ^ 2 = 4 and (-2) ^ 2 = 4 nonetheless sqrt(4) = 2 without ambiguity. Can you do a counterexample without square roots? Then you will have a chance of seeing one of your list. -- Larry Lard Replies to group please === Subject: Re: Polynomials, general factorization, distributive property > [snippage] >>To eliminate any confusion, we'll decide to use the >>positive square root, i.e., the one for which sqrt(9) = 3. >>Okay, I'll speculate, in rough order of estimated liklihood. > I knew as soon as I saw the above extracted portion what the response > would be; one you didn't give in your list. Remember, James doesn't... > um... accept (less charitable folks might say understand) the idea > that sqrt is a symbol which means the positive square root of. He > doesn't accept that > while > 2 ^ 2 = 4 > and > (-2) ^ 2 = 4 > nonetheless > sqrt(4) = 2 without ambiguity. > Can you do a counterexample without square roots? Then you will have a > chance of seeing one of your list. Here's a cheap way; it's nothing more than Rick's example modified to use a factorization of the difference of cubes rather than difference of squares: f(x) = (4-x)^(2/3) - (9 - x - 6x^2)^(1/3) - 7 g(x) = (4-x)^(4/3) + (4-x)^(2/3)(9 - x - 6x^2)^(1/3) + (9 - x - 6x^2)^(2/3) - 1 you get (after a little manipulation): (f+7)(g+1) = (4-x)^2 - (9 - x - 6x^2) If x = 1, the factorization is 7 = (3^(2/3) - 2^(1/3))*(3^(4/3) + 3^(2/3) 2^(1/3) + 2^(2/3)) Another (just as cheap) way would be to factor the sum of cubes: f(x) + 7 = (4-x)^(2/3) + (6x^2 + x - 9)^(1/3) g(x) + 1 = (4-x)^(4/3) - (4-x)^(2/3)(6x^2 + x - 9)^(1/3) + (6x^2 + x - 9)^(2/3) which yields (when x = 1) 7 = (3^(2/3) + 2^(1/3))*(3^(4/3) - 3^(2/3) 2^(1/3) + 2^(2/3)) I think it would be much prettier to have something a bit less contrived, but perhaps that's gilding the lilly. Perhaps JSH will now claim that none of the roots is definable by itself. Next will be that no root of an irreducible polynomial can be mentioned without the other roots. === Subject: Re: Polynomials, general factorization, distributive property Stuff that was absolutely incorrect. Didn't remember to check for f(0) = g(0) = 0. Darn it all, anyhow. Dale. === Subject: Re: Polynomials, general factorization, distributive property > Here I will give an explanation with some basic examples, where I make > the effort in the hope that it will clear the air about how I use the > distributive property with certain factorizations of a polynomial. > Consider a polynomial C(x) with a constant term that is 1 on the > complex plane. > Multiply it by 7 and factor it as > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > The reason the polynomial is multiplied by a constant is that in some > of my research you get expressions similar to the above, where I'm > abstracting for simplicity. > Now if f(x) and g(x) are simple linear functions, it's easy enough to > see how 7 distributes through, for instance, if C(x) = x^2 + 2x + 1 and > g(x) = x, then it must be true that > f(x) = 7x > where it doesn't mean anything in the complex plane to say that f(x) > has 7 as a factor as so does g(x), though it equals x. > The mathematical terminology hasn't been developed for these types of > arguments, so I usually just say that 7 multiplied through one factor > of C(x), in this case, 7x+7. > I doubt many would argue that for any polynomial C(x) on the complex > plane, where f(x) and g(x) are linear functions of x that 7 would have > to have multiplied through the factor where 7 is visible, with the > requirement that f(0) = g(0) = 0. > But what about non-polynomial factors? > Well, let C(x) = x^2 + x - 1, and > g(x) = sqrt(x^2 + x) + 1 Do you maybe mean sqrt(x^2+x)-1? > and notice that though it's not a polynomial it still must be true that > 7 multiplied through the first factor, and again, as these results are > valid in the complex plane it's meaningless to say that 7 is a factor > of f(x). > But, of course, for this example > f(x) = 7*sqrt(x^2 + x) > and the question is how to describe the general case with > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > Now I can give specific cases repeatedly, but it's better to try and > abstract out what's happening---why is it true that 7 multiplies > through the one factor? What does it mean to say that 7 multiplies through one factor? > Can anyone give a counterexample where it does not? Not until I know what it means. > I like these examples because I not only show the simple polynomial > case--which should be very familiar--but I also give an explicit > non-polynomial factorization where again you can actually see the > result. > The question then for those who disagree with me is, can you have an > explicit solution where the seemingly simple result that 7 multiplied > through f(x) + 7 is violated? > Can you break my toy examples? > If so, give the counterexample. > If not, why not? > James Harris === Subject: Re: Polynomials, general factorization, distributive property >>Can anyone give a counterexample where it does not? > Not until I know what it means. For that you need to a genius of the same order as JSH. === Subject: Re: Polynomials, general factorization, distributive property <44332ab7$0$16324$892e7fe2@authen.yellow.readfreenews.net Here I will give an explanation with some basic examples, where I make > the effort in the hope that it will clear the air about how I use the > distributive property with certain factorizations of a polynomial. > Consider a polynomial C(x) with a constant term that is 1 on the > complex plane. > that is the point (1,0) ? I think upChuckie is a troll. JSH is clear about his example at this point. > Multiply it by 7 and factor it as > 7*C(x) = (f(x) + 7)*(g(x) + 1) > where f(0) = g(0) = 0. > you did not multiply both sides by 7. Why is that ? Now upChuckie is just being dense. JSH said: Multiply [C(x)] by 7 and should have said: SUPPOSE [C(x)*7] factors as 7 * C(x) = (f(x) + 7) * (g(x) + 1), where f(0) = g(0) = 0. (7 * C(x) can factor in other ways, of course.) > Now if f(x) and g(x) are simple linear functions, it's easy enough to > see how 7 distributes through, Well, maybe not. If f(x) = a*x and g(x) = b*x then 7 * C(x) = (a x + 7)(b x + 1) = ab x^2 + (a + 7b) x + 7, which shows C(x) must be a quadratic polynomial; otherwise f and g can't both be linear. (Note that C(x) originally was a polynomial with a constant term of 1.) > for instance, if C(x) = x^2 + 2x + 1 and > g(x) = x, then it must be true that > f(x) = 7x > but it dosen't have to be, because you left it undefined as f(x). I think upChuckie doesn't understand the word if. All JSH is saying here is that 7 (x^2 + 2x + 1) = (f(x) + 7) (x + 1) implies f(x) = 7x, which is simple algebra. > where it doesn't mean anything in the complex plane to say that f(x) > has 7 as a factor as so does g(x), though it equals x. > what does that mean ? In the ring of complex numbers, any nonzero complex number is a factor of any other nonzero complex number. In the ring C[x] (polynomials with complex coefficients), any nonzero complex number is factor of any polynomial > The mathematical terminology hasn't been developed for these types of > arguments, Actually, it has. JSH just hasn't taken the trouble to learn it. > so I usually just say that 7 multiplied through one factor > of C(x), in this case, 7x+7. > you assume C(x) = x + 1, right ? Read it again ... 7 multiplied through ONE FACTOR OF C(x); thus, 7 is multiplied by x+1. > I doubt many would argue that for any polynomial C(x) on the complex > plane, where f(x) and g(x) are linear functions of x that 7 would have > to have multiplied through the factor where 7 is visible, with the > requirement that f(0) = g(0) = 0. > that is gobbligook He's just saying that there are other ways for the 7 to be spread among the factors of 7 * C(x). > But what about non-polynomial factors? > Well, let C(x) = x^2 + x - 1, and > g(x) = sqrt(x^2 + x) + 1 But g(0) is not 0 here. > and notice that though it's not a polynomial it still must be true that > 7 multiplied through the first factor, Not true. > and again, as these results are > valid in the complex plane it's meaningless to say that 7 is a factor > of f(x). > dosen't matter, use the distributative property. > sorry, but the rest of you post is just non-sense. > Seems you do not have sufficient grasp of simple High School algebra. JSH is making a big deal out of calculations which any high school student can do. He has no interesting results. === Subject: Re: Polynomials, general factorization, distributive property >> Here I will give an explanation with some basic examples, where I make >> the effort in the hope that it will clear the air about how I use the >> distributive property with certain factorizations of a polynomial. >> Consider a polynomial C(x) with a constant term that is 1 on the >> complex plane. >> that is the point (1,0) ? > I think upChuckie is a troll. JSH is clear about his example at this > point. I think Proginoskes is Troll too. JSH is Troll for sure. To prove you not troll , can you answer my question ? that is the point (1,0) ? and Why is jsh using x ?, should be z if he is on the complex plane. === Subject: Re: Polynomials, general factorization, distributive property <44332ab7$0$16324$892e7fe2@authen.yellow.readfreenews.net> <4433e9d9$0$79453$892e7fe2@authen.yellow.readfreenews.net> Here I will give an explanation with some basic examples, where I make >> the effort in the hope that it will clear the air about how I use the >> distributive property with certain factorizations of a polynomial. >> Consider a polynomial C(x) with a constant term that is 1 on the >> complex plane. >> that is the point (1,0) ? > I think upChuckie is a troll. JSH is clear about his example at this > point. > I think Proginoskes is Troll too. JSH is Troll for sure. Proginoskes is an AI program created by Archimedes Plutonium. > To prove you not troll , can you answer my question ? > that is the point (1,0) ? The answer to your question is NO. You're parsing JSH's statement as Consider a polynomial C(x) with (a constant term that is 1 on the complex plane). It should be parsed as Consider (a polyomial on the complex plane) with a constant term that is 1; i.e., Consider a polynomial with complex coefficients with a constant term of 1 in the standard verbiage. > and Why is jsh using x ?, should be z if he is on the complex plane. The day JSH uses standard terminology is the day that George W. Bush will become an environmentalist. === Subject: Re: simple probability > No. Once X=i is given, they magically become indpendent again. Part of > your problem is that you keep writing {Y>X|X=i}; if you just write > {Y>i|X=i} instead, the problem goes away! You see, there is a > difference between saying that {Y>X} and {Z>X} are dependent---they > are---vs saying that the events are independent when X=i is given. > Still not convinced? Look at it this way: P{Z>X,Y>X} = sum > P{X>i,Y>i,X=i} = (1/m) sum (m-i)^2/m^2. However, P{Y>X}P{Z>X} = [sum > (m-i)/m ]^2 <> P{Y>X,Z>X}, so that is why they are dependent (the > square of the sum is not the sum of the squares). When X is not > given, we have to sum over all the X-values, but when X=i is given, > the sum collapses to one term. We do have indpendence at the one-term > level. I get it algebraically, but slightly less by intuition (which seems to be of importance when modelling these problems). === Subject: Re: simple probability <44329674.4060504@hotmail.com> <44345088.1040004@hotmail.com No. Once X=i is given, they magically become indpendent again. Part of > your problem is that you keep writing {Y>X|X=i}; if you just write > {Y>i|X=i} instead, the problem goes away! You see, there is a > difference between saying that {Y>X} and {Z>X} are dependent---they > are---vs saying that the events are independent when X=i is given. > Still not convinced? Look at it this way: P{Z>X,Y>X} = sum > P{X>i,Y>i,X=i} = (1/m) sum (m-i)^2/m^2. However, P{Y>X}P{Z>X} = [sum > (m-i)/m ]^2 <> P{Y>X,Z>X}, so that is why they are dependent (the > square of the sum is not the sum of the squares). When X is not > given, we have to sum over all the X-values, but when X=i is given, > the sum collapses to one term. We do have indpendence at the one-term > level. > I get it algebraically, but slightly less by intuition (which seems to > be of importance when modelling these problems). OK, so think of three identical but independent roulette wheels. We spin wheel #1 and it shows 5 (this is X). The events that both wheels #2 and #3 show > 5 are mutually independent. It is true that they both contain the number '5', but that does not make them dependent. (It is also irrelevant that I am interested in the number '5' because that is what I got upon spinning roulette wheel #1; I could have gotten it instead because it is my Aunt Sarah's favourite number, or is the age of my cat, or whatever.) R.G. Vickson === Subject: Re: simple probability > Problem: > X is a random variable that takes values [1,...,m] with equal > probability (uniform) > Y is a random variable that takes values [1,...,m] with equal probability You never state explicitly that X and Y are independent, which you seem to be assuming. So will I henceforth. > What is Pr[X My answer: > [...] > So Pr[X ---- > Now introduce third variable Z, with same as the ones above. > What is Pr[X should be Pr[X And for n variables ( (m-1)/2m )^n. > I did some simple simulation and got that this was wrong. Please help. As others have pointed out, {X I am not a mathematician. I am an engineer - my math is deficient. > There is something I have taken for granted and I no longer feel comfortable > doing so. > This involves the change of variables under the integral sign and the use of > the determinant of the Jacobian. > I have tried to search for proofs of this, but come up short - they are too > theoretical for me. > Could someone please take the time to point me to a proof that might be > comprehnsible... > something that is direct, to the point? > Or, failing that... point me to a proof for a 2D coordinate system: > The integral in a plane over x and y... > Tom Try Robert G Bartle ,The Elements of Real Analysis 2nd edition(at least) Wiley.The proof is in the last chapter and is for R^n ,This book is very readable and a good reference for Advanced Calculus for alot of your further needs.You might also look in the book by Angus Taylor and Mann -Advanced Calculus which might have one for 2 and 3 dimensions but I havent looked at it .smn === Subject: Re: Jacobian Change of Variables question > I am not a mathematician. I am an engineer - my math is deficient. > There is something I have taken for granted and I no longer feel comfortable > doing so. > This involves the change of variables under the integral sign and the use of > the determinant of the Jacobian. > I have tried to search for proofs of this, but come up short - they are too > theoretical for me. > Could someone please take the time to point me to a proof that might be > comprehnsible... > something that is direct, to the point? > Or, failing that... point me to a proof for a 2D coordinate system: Another place to look is Calculus on Manifolds by Michael Spivak. In spite of the scary title it is a wonderful book on the calculus of several variables. The basic idea of the proof is that the mappings look locally like linear transformations (given by the Jacobian) and that linear transformations multiply the area (or volume or whatever) of a given region by the absolute value of the determinant when they transform them. Achava > The integral in a plane over x and y... > Tom > Try Robert G Bartle ,The Elements of Real Analysis 2nd edition(at > least) Wiley.The proof is in the last chapter and is for R^n ,This book > is very readable and a good reference for Advanced Calculus for alot of > your further needs.You might also look in the book by Angus Taylor and > Mann -Advanced Calculus which might have one for 2 and 3 dimensions but > I havent looked at it .smn === Subject: Re: Fermat's last theorem and a counter example Following Yours: * * * > Still, It will very nice to obtain one > issue * * * My analysis can help You for to find counterexamples: set Z = 3^u abt + a^3 + b^3 once (a+b)/3 and (a+b)/t X = 3^u abt + b^3 Y = 3^u abt + a^3 etc. taken a=4; b=11; t=5 once u = 2 (minimum value): Z = 9*4*11 + 4^3 + 11^3 = 1791 = 3^2 *199 X = 9*4*11 + 1331 = 1727 = 11*157 Y = 9*4*11 + 64 = 460 = 2^2 *5*23 X+Y = 2187 = 3^7 Z-Y = 1331 = 11^3 Z-X = 64 = 2^6 What for n=5 and bigger will satisfy: Z^n/(X+Y) X^n/(Z-Y) Y^n/(Z-X) I hope for some luck for to find more counter- examples and also for more composite Z;X;Y and X+Y; Z-Y; Z-X Ro-bin === Subject: Re: Fermat's last theorem and a counter example > Following Yours: > * * * Still, It will very nice to obtain one this > issue > * * * > My analysis can help You for to find > counterexamples: > set Z = 3^u abt + a^3 + b^3 once (a+b)/3 and > and (a+b)/t > X = 3^u abt + b^3 > Y = 3^u abt + a^3 etc. > taken a=4; b=11; t=5 once u = 2 (minimum value): > Z = 9*4*11 + 4^3 + 11^3 = 1791 = 3^2 *199 > X = 9*4*11 + 1331 = 1727 = 11*157 > Y = 9*4*11 + 64 = 460 = 2^2 *5*23 > X+Y = 2187 = 3^7 > Z-Y = 1331 = 11^3 > Z-X = 64 = 2^6 > What for n=5 and bigger will satisfy: > Z^n/(X+Y) > X^n/(Z-Y) > Y^n/(Z-X) > I hope for some luck for to find more counter- > examples and also for more composite Z;X;Y and > X+Y; Z-Y; Z-X > Ro-bin Your efforts are very much appreciated, and the challenge now to meet those conditions made in this thread only at this link below http://mathforum.org/kb/message.jspa?messageID=4512080&tstart=45 Best of luck B.Karzeddin === Subject: Reverse calculate that! Can any person or software find a formula that generates the sequence of positive rational numbers, starting with n a(n) 0 4 1 4 2 108/71 3 236/239 4 412/559 5 636/1079 6 908/1847 7 1228/2911 ? (I played with Matlab and actually have a Matlab expression using 55 keystrokes.) === Subject: Re: Reverse calculate that! > Can any person or software find a formula that generates > the sequence of positive rational numbers, starting with > n a(n) > 0 4 > 1 4 > 2 108/71 > 3 236/239 > 4 412/559 > 5 636/1079 > 6 908/1847 > 7 1228/2911 > (I played with Matlab and actually have a > Matlab expression using 55 keystrokes.) Tried polynomial interpolation and Horner form -- both results involve largish rational numbers and require more keystrokes. The sequence of numerators and denominators is not found at http://www.research.att.com/~njas/sequences Prime factor decomposition of the a(n) involves largish primes, so I would not expect a simple closed-form formula. Of course, I assume table look-up (in Mathematica notation) f[n_] := 4 {1, 1, 27/71, 59/239, 103/559, 159/1079, 227/1847, 307/2911}[[n + 1]] is not acceptable? This requires less than 55 keystrokes. 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: Mathematica 6.0 Does anyone know when Mathematica 6.0 will be released? What new features will it bring? Lars === Subject: recurrence relation Can anyone show me how to do the following problem: Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find the solution of the recurrence relation in part (a) with initial condition a_1 = 4. === Subject: Re: recurrence relation >Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. In addition to the other fine answers you've been given, note that if the a_i satisfy a recurrence relation of the form L(a_n, a_{n-1}, ...) = P(n) where P is a polynomial of degree d, then they also satisfy a recurrence relation, L(a_n, a_{n-1}, ...) - L(a_n, a_{n-1}, ...) = P'(n) for some polynomial P' of degree d-1. (Note that the left side is also linear -- though it involves more variable than before.) Iterate this process to eliminate the Ps altogether -- the a_i now solve a HOMOGENEOUS LINEAR recurrence relation. These you can solve (right?) dave === Subject: Re: recurrence relation > Can anyone show me how to do the following problem: > Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find > the solution of the recurrence relation in part (a) with initial > condition a_1 = 4. Yet another approach is to calculate the generating function of the recursion by multiplying both sides by x^n and summing from n = 1 to infinity. You should get an algebraic relationship that solves the generating function problem for you and from there you can frequently work out the formula for the recursion. See Wilf's book called something like Generatingfunctionology. Achava === Subject: Re: recurrence relation > Can anyone show me how to do the following problem: > Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find > the solution of the recurrence relation in part (a) with initial > condition a_1 = 4. > Yet another approach is to calculate the generating function of the > recursion by multiplying both sides by x^n and summing from n = 1 to > infinity. You should get an algebraic relationship that solves the > generating function problem for you and from there you can frequently > work out the formula for the recursion. See Wilf's book called > something like Generatingfunctionology. > Achava I was going to give a similar reply, working out the details of the generating function, but for this problem it turns out to be pretty messy and I think that the previous responses give a better solution for this particular recurrence relation. This is because the closed form for the sum from n = 1 to infinity of (n^2)x^n is not very nice. Mike === Subject: Re: recurrence relation > Can anyone show me how to do the following problem: > Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find > the solution of the recurrence relation in part (a) with initial > condition a_1 = 4. Whenever you have a recurrance of the form a_n = k*a_{n-1} + f(n), you can divide by k^n to get a_n/k^n = a_{n-1}/k^{n-1} + f(n)/k^n. Letting b_n = a_n/k^n and g(n) = f(n)/k^n, you get b_n = b_{n-1} + g(n), which is easier. === Subject: Re: recurrence relation Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find > the solution of the recurrence relation in part (a) with initial > condition a_1 = 4. > Whenever you have a recurrance of the form > a_n = k*a_{n-1} + f(n), > you can divide by k^n to get > a_n/k^n = a_{n-1}/k^{n-1} + f(n)/k^n. > Letting b_n = a_n/k^n and g(n) = f(n)/k^n, > you get > b_n = b_{n-1} + g(n), > which is easier. I don't understand why it's easier or how you knew to do this, if it's smart at this. === Subject: Re: recurrence relation > Can anyone show me how to do the following problem: > Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find > the solution of the recurrence relation in part (a) with initial > condition a_1 = 4. Surely your textbook shows you how to do this kind of problem. 1. Start with a_n = 2a_(n-1) - do you know how to find the general solution of this recurrence? 2. Then your textbook surely gives you a way to find a particular solution to a_n = 2a_(n-1) + 2n^2. 3. Then it surely tells you how to combine the solutions you found in 1. and 2. to find the general solution of a_n = 2a_(n-1) + 2n^2. 4. Then surely from that you can pick out the solution which gives you a_1 = 4. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: recurrence relation Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find > the solution of the recurrence relation in part (a) with initial > condition a_1 = 4. > Surely your textbook shows you how to do this kind of problem. > 1. Start with a_n = 2a_(n-1) - do you know how to find the general > solution of this recurrence? I would just write out the terms and look for a pattern, because there's only one term. If there were two terms, I would use a characteristic equation, but I don't know how to do that with only one term. > 2. Then your textbook surely gives you a way to find a particular > solution to a_n = 2a_(n-1) + 2n^2. Actually, no. It only does one problem like this, and it only says we guess that blah blah blah. It doesn't tell me WHY or HOW that was the guess, so it's really not very helpful > 3. Then it surely tells you how to combine the solutions you found > in 1. and 2. to find the general solution of a_n = 2a_(n-1) + 2n^2. Again, it doesn't tell me in common > 4. Then surely from that you can pick out the solution which gives > you a_1 = 4. No, I can't, which is why I'm asking. I have trouble reading the mumbo-jumbo math talk of my textbook, and am looking for someone to explain it a friendlier matter. Consider this: I have several of these problems to do. I'm trying to get help on one of them. If I can't understand this one problem, I have no hope on the rest. If you want to help me on this problem, that's great. If not, please don't tell me to read the textbook, because I've done that -- several times, and don't understand what it's saying, and don't think it explains the one problem it does show of this type very well. It's not a clear book, so I'm asking people for human explanations. === Subject: Re: recurrence relation > Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find > the solution of the recurrence relation in part (a) with initial > condition a_1 = 4. > Surely your textbook shows you how to do this kind of problem. > 1. Start with a_n = 2a_(n-1) - do you know how to find the general > solution of this recurrence? > I would just write out the terms and look for a pattern, because > there's only one term. If there were two terms, I would use a > characteristic equation, but I don't know how to do that with only one > term. I find this response puzzling. The characteristic equation is the polynomial equation whose coefficients are just the coefficients in the recurrence. It doesn't matter whether there are two terms, or one term, or seventy-three terms. But in fact your idea of writing out terms and looking for a pattern should work very nicely for a_n = 2a_(n-1). Why don't you try it? > 2. Then your textbook surely gives you a way to find a particular > solution to a_n = 2a_(n-1) + 2n^2. > Actually, no. It only does one problem like this, and it only says we > guess that blah blah blah. It doesn't tell me WHY or HOW that was the > guess, so it's really not very helpful Well, from this and some other responses of yours it sounds like you have a pretty cruddy textbook (the other possibility is that you have a very good textbook & not a clue as to how to use it, but I'm inclined to give you the benefit of the doubt). So here are some other options for you: 1. Go see your teacher for help - that's what she's paid for - and while you're there, point out how useless the textbook is. 2. Go to a library (or a bookstore) and get another book. There are dozens of textbooks on discrete mathematics, and most of them have chapters on recurrence relations, and you're bound to find one that speaks to you. Anyway, part of the reason WHY in this kind of problem we make the guess we make is that some very clever/lucky person made that guess many years ago and it worked. If the thing that involves n but not terms of the sequence (the 2 n^2 in your problem) is a polynomial, for example, it has been found by clever/lucky people that the thing to guess is a polynomial of the same degree - you just have to plug it in and do some algebra to see what the coefficients should be. > 3. Then it surely tells you how to combine the solutions you found > in 1. and 2. to find the general solution of a_n = 2a_(n-1) + 2n^2. > Again, it doesn't tell me in common Did you leave something out here? > 4. Then surely from that you can pick out the solution which gives > you a_1 = 4. > No, I can't, which is why I'm asking. Well, I'm sure that if you get through steps 1., 2., and 3., then you *can* do step 4. If you knew the answer was a_n = 3 n^7 + C 5^n (it isn't, but let's pretend), only you didn't know what C was (in much the same way that when you antidifferentiate, you always get a plus-C), but you did know that a_1 = 4, don't you think you could figure out what C had to be? > I have trouble reading the mumbo-jumbo math talk of my textbook, and > am looking for someone to explain it a friendlier matter. Sorry, I don't do friendly. > Consider this: I have several of these problems to do. I'm trying to > get help on one of them. If I can't understand this one problem, I > have no hope on the rest. If you want to help me on this problem, > that's great. If not, please don't tell me to read the textbook, > because I've done that -- several times, and don't understand what > it's saying, and don't think it explains the one problem it does show > of this type very well. It's not a clear book, so I'm asking people > for human explanations. Heck, I barely do human. For future reference, it might have been a better idea to present a little more of this up front. To tell us what exactly your textbook says, and how that falls short of helping you to do these problems. Some of us might even have this alleged textbook on our shelves, and could maybe point you in the right direction. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: recurrence relation > Can anyone show me how to do the following problem: > Find all solutions of the recurrence relation a_n = 2a_n-1 + 2n^2. Find > the solution of the recurrence relation in part (a) with initial > condition a_1 = 4. Imitate differential equations, in particular the method of undetermined coefficients. In more detail, expect the solution in the form a_n = d * 2^n + (a quadratic polynomial in n) and match the coefficients. (What is the 2^n doing there?) === Subject: Re: recurrence relation undetermined coefficients. I haven't taken differential equations, so unfortunately don't understand. > In more detail, expect the solution in the form > a_n = d * 2^n + (a quadratic polynomial in n) I don't understand what this means....? > and match the coefficients. (What is the 2^n doing there?) or this. Could you possibly explain anything in further detail? Having not taken diffyQs, I'm lost. === Subject: Re: recurrence relation > Imitate differential equations, in particular the method of > undetermined coefficients. > I haven't taken differential equations, so unfortunately don't > understand. > In more detail, expect the solution in the form > a_n = d * 2^n + (a quadratic polynomial in n) > I don't understand what this means....? > and match the coefficients. (What is the 2^n doing there?) > or this. > Could you possibly explain anything in further detail? Having not taken > diffyQs, I'm lost. What's a quadratic polynomial in n?! This is high-school algebra, not differential equations. -- Ron Bruck === Subject: Re: Why are Asians so good at math? >> Vague. Which adaptations do you have in mind? >Primarily language and the related mental apparatuses. The first humans >had to teach themselves language. It is not genetically wired. If it >were there would not be 6000 different languages. >Bob Kolker The particular language is not genetically wired; vocabulary is essentially arbitrary, but grammar far less so. Languages evolved, and there are many different versions of language which are roughly equal, just as there are many quite different ways of setting up mathematical notation. Why are powers superscripts? -- 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: Why are Asians so good at math? On 5 Apr 2006 21:11:55 -0400, hrubin@odds.stat.purdue.edu (Herman >The particular language is not genetically wired; >vocabulary is essentially arbitrary, but grammar >far less so. Comparatively speaking, Chinese has much less grammar than English. In Chinese we have the grammatical units like characters, words or terms, phrases and sentences. Chinese depends mainly on its characters to realize its grammatical functions. There is no changes in affixes. In fact, there is no affixes for Characters. But there are many changes with the terms or words which are formed with characters. All the grammatical changes depend on characters or terms, e.g. we use auxiliaries to show the tenses, voices, moods, cases, etc. In one word, Chinese grammar depends mainly on different parts of speech to realize its syntax. On the contrary, English grammar depends mainly on the changes of the words themselves to show different grammatical elements. For example, the tenses are realized through the changes of verbs; the numbers of nouns are realized by adding -s to the end of nouns or through the changes of the nouns themselves, etc. *************** Japanese grammar (compared to English grammar) is extremely regular. In fact, while there are over 250 verb classes in English (verbs that conjugate mostly the same way), there are only two such classes in Japanese, plus two irregular verbs. This regularity actually makes Japanese easier to learn than many other languages. The verb always comes at the end of the sentence, and does not change for person, gender, or number. Also, there are only two irregular verbs in the entire language, and even these conjugate predictably. Furthermore, there are only two basic tenses: past and present. The present tense is used to express future intent, and so it is also known as the non-past. Rather than using auxiliaries or linking verbs, as English does (will, want, can, would, should, etc), Japanese concentrates these functions into the verb ending. For example, tabemasu corresponds to eat, taberaremasu corresponds to can eat, and tabetai corresponds to want to eat. Sometimes more than one ending can be applied to the same verb simultaneously, which can be somewhat confusing. ********************** We learn grammar unconsciously at a very young age, but we are not born with some *imprint* for grammar or all languages would have similar grammar structures and they don't. -- Dorothy There is no sound, no cry in all the world that can be heard unless someone listens .. The Outer Limits === Subject: Re: Why are Asians so good at math? <49hvs0Fof1cbU2@individual.net> The particular language is not genetically wired; >vocabulary is essentially arbitrary, but grammar >far less so. > Comparatively speaking, Chinese has much less grammar than English. In > Chinese we have the grammatical units like characters, words or terms, > phrases and sentences. Chinese depends mainly on its characters to > realize its grammatical functions. There is no changes in affixes. In > fact, there is no affixes for Characters. But there are many changes > with the terms or words which are formed with characters. All the > grammatical changes depend on characters or terms, e.g. we use > auxiliaries to show the tenses, voices, moods, cases, etc. In one > word, Chinese grammar depends mainly on different parts of speech to > realize its syntax. Why are you making the structure of Chinese sound so complicated. In Chinese, there is: No gender. No verb conjugation or tenses. I go; you go; he go; etc. I go today; I go tomorrow; I go yesterday. No that there is no ambiguity whatsoever because go is used without the form shall go or went. Chinese is the SIMPLEST of all languages in structure. It's based on a person's ability to understand by CONTEXT. Even if I use the ungrammatical version in English, I go; you go; he go; etc. I go today; I go tomorrow; I go yesterday. Only an idiot would NOT grasp the precise meaning of any of those phrases. That's why you always hear Confucius say but not Confucius says. :-) The Chinese were already already into advanced literature and poetry when the folks in other civilizations were still hanging by their tails on trees. -- Bob. === Subject: Re: Why are Asians so good at math? <49hvs0Fof1cbU2@individual.net> In Chinese, there is: > No gender. > I go today; I go tomorrow; I go yesterday. > Chinese is the SIMPLEST of all languages in structure. It's based > on a person's ability to understand by CONTEXT. What about had run? At noon tomorrow, I will begin to run. At 1:00, I will have been running for one hour. How do they express this in chinese? For someone without this construct in their native language, it must be very difficult to learn, when they study english. > The Chinese were already already into advanced literature and > poetry when the folks in other civilizations were still hanging by > their tails on trees. And wei chi. Why did they stagnate, while Europe blew past them after their Renaissance period? Mark === Subject: Re: Why are Asians so good at math? > And wei chi. > Why did they stagnate, while Europe blew past them > after their Renaissance period? It was the reactionary Ming Dynasty that doomed China to secondratehood in the modern era. When the Chinese were at the top of their form they were lightyears ahead of the West. Bob Kolker === Subject: Re: Why are Asians so good at math? >Why are you making the structure of Chinese sound so complicated. Because it is. >In Chinese, there is: >No gender. >No verb conjugation or tenses. > I go; you go; he go; etc. I go today; I go tomorrow; I go >yesterday. There is no mandatory tense. That means that if you want to express tense, you may have to use a method that is specific to the words. Similarly, there is no singular/plural distinction, so when the Chinese express a plural, they have to add extra words, and the words added are idiosyncratic. >Chinese is the SIMPLEST of all languages in structure. It lacks some of the structures of English and European languages, but has other structures instead. >I go; you go; he go; etc. I go today; I go tomorrow; I go >yesterday. >Only an idiot would NOT grasp the precise meaning of any of those >phrases. That's why you always hear Confucius say but not >Confucius says. :-) But of course that means that when you want to express a tense nuance that is not simple, your language has to get more complicated. http://en.wikipedia.org/wiki/Chinese_grammar provides some explanations and examples of Chinese grammar that is complicated from the English perspective. Note also the use of the words context-dependent or context-sensitive which means the rules change idiosyncratically in certain contextual circumstances. >The Chinese were already already into advanced literature and >poetry when the folks in other civilizations were still hanging by >their tails on trees. There is no evidence of this. The Chinese were writing down their poetry and literature while other peoples without written language were not. But for example we have the Iliad and the Odyssey which were passed down for centuries by oral tradition before they were written. lojbab === Subject: Re: Why are Asians so good at math? >> It is OK to read the Greek works, our Mathematical forefathers (the >> Muslims) did that, but it is not OK to pin everything on them, knowing >> that a lot of the ancient work is lost. > Without the axiomatic method we would not have theoretical physics and any > of the technology which requires theoretical physics. The best we could > manage is heurstic approaches, rules of thumb. > The Greeks invented the axiomatic method. No other culture came close > prior to the Greeks. I recently asked a PhD in Mathematics if he could evolve the value of Phi using only stright edge and compass. He declared it impossible. Perhaps these days it is, but what to make of the great Pyramid? In transitions of knowledge, once culture subsumes another, and this value was passed to the Greeks from Egypt [evidence is the Parthenon], though our /understanding/ of Phi has not been generally sustained, and we prefer Pi as a working measure. This distinction has as a basis the philosophical shift which took place in Greek culture [early and late Homer are quite distinctly of differing philosophy]. Moderns no long even perceive the old measure, and while talking of reason and rationalisation, no longer understand the original sense of 'ratio', which does not only mean 'to measure', but 'to proportion', and the underlying geometry of pre-Homeric culture as the basis for algebra - which incidentally is not Greek but Arab, al jabr. Not only is much ancient knowledge lost, but that which still exists has become incomprehensible to us, since our modern approach has no unifying canonical basis - we have no mainstream unified field theorum - and we fail to recognise the work of the past as possessing one, to whatever level it was then appreciated. Phil Innes > Bob Kolker === Subject: Re: Why are Asians so good at math? > It is OK to read the Greek works, our Mathematical forefathers (the > Muslims) did that, but it is not OK to pin everything on them, knowing > that a lot of the ancient work is lost. >> Without the axiomatic method we would not have theoretical physics and any >> of the technology which requires theoretical physics. The best we could >> manage is heurstic approaches, rules of thumb. >> The Greeks invented the axiomatic method. No other culture came close >> prior to the Greeks. >I recently asked a PhD in Mathematics if he could evolve the value of Phi >using only stright edge and compass. He declared it impossible. Perhaps >these days it is, but what to make of the great Pyramid? Do the slopes of the various pyramids equal any irrational numbers? It is difficult to determine them now, because the facings were made of softer stone, but Egyptian papyri indicate that the slopes were 11/7. This number is close to both Phi and pi/2. The Egyptians used right-angled triangles for the purpose. -- 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: Why are Asians so good at math? <498bk3FnattvU1@individual.net> value of Phi using only stright edge and compass. evolve? >He declared it impossible. Perhaps >these days it is, but what to make of the great Pyramid? In those days, it wasn't. The ancient Egyptians weren't bound by things like logic or proof. > Do the slopes of the various pyramids equal any > irrational numbers? Is this a question that can be answered empirically? I think you would need a very very finely gradationed measuring stick... Mark === Subject: Re: Why are Asians so good at math? <0001HW.C05426C700180999F0284550@family.alibis.com> <4970l4Fn58u0U2@individual.net> <498bk3FnattvU1@individual.net> It is OK to read the Greek works, our Mathematical forefathers (the >> Muslims) did that, but it is not OK to pin everything on them, knowing >> that a lot of the ancient work is lost. > Without the axiomatic method we would not have theoretical physics and any > of the technology which requires theoretical physics. The best we could > manage is heurstic approaches, rules of thumb. > The Greeks invented the axiomatic method. No other culture came close > prior to the Greeks. > I recently asked a PhD in Mathematics if he could evolve the value of Phi > using only stright edge and compass. He declared it impossible. Perhaps > these days it is, but what to make of the great Pyramid? The same thing the Eqyptians did, ramps. Since they not only diidn't know what pi was, they aimed it at the wrong star. > In transitions of knowledge, once culture subsumes another, and this value > was passed to the Greeks from Egypt [evidence is the Parthenon], though our > /understanding/ of Phi has not been generally sustained, and we prefer Pi as > a working measure. > This distinction has as a basis the philosophical shift which took place in > Greek culture [early and late Homer are quite distinctly of differing > philosophy]. Moderns no long even perceive the old measure, and while > talking of reason and rationalisation, no longer understand the original > sense of 'ratio', which does not only mean 'to measure', but 'to > proportion', and the underlying geometry of pre-Homeric culture as the basis > for algebra - which incidentally is not Greek but Arab, al jabr. > Not only is much ancient knowledge lost, but that which still exists has > become incomprehensible to us, since our modern approach has no unifying > canonical basis - we have no mainstream unified field theorum - and we fail > to recognise the work of the past as possessing one, to whatever level it > was then appreciated. > Phil Innes > Bob Kolker === Subject: Re: Why are Asians so good at math? piqued at the subject because I am an Americanized Asian. > Asians are good at math at the lower echelon mostly because of > they work hard, their work ethics, and the large NUMBER of them > in colleges, universities, and graduate schools, majoring in math > and math related subjects. > Yes, but... WHY is there such a large number in math > related subjects... maybe they have some natural (racial) > aptitude, ya think? This is known as begging the question > That was a statistical remark. Well since Asia has been doing math since 1000 years before Athens was even a Roman Colony, it's trivial. Since they don't do statistics, given the Romans invented statistics. > There are more Asians than all the other continents put together. > Take any subject that is not a rare one, you you'll have more Asians > specializing in it than any other nationality of region on earth. > -- Bob. A former math major. > What's your current major? > Professor (Full, 1977) Emeritus in Statistics. :-) > As for your other question in the next post, I was born in China, > after hundreds of generation of being Chinese; and became a > Naturalized American (USA) Citizen in 1968. > -- Bob. === Subject: Re: Why are Asians so good at math? <43EB0216.8BC3CEA6@yahoo.com> <47t2gjFgpsg2U1@individual.net> <49ae4aFncqt4U1@individual.net It may have more to do with genetics and evolution than you think. http://www.nytimes.com/2006/03/12/weekinreview/12wade.html?ex=1299819600&en= 4ffa8520bf6f7229&ei=5088&partner=rssnyt&emc=rss > There have been no significant genetic changes to human > beings in 50,000 years. When did the races diverge? > Most human adaptation is cultural, not genetic. Vague. Which adaptations do you have in mind? > enough, the two scientists argue, for natural selection to favor any > variant gene that enhanced cognitive ability. > Variation in cognitive ability is within the normal range of human > diversity. There is no proof of a -gene- for smartness. Not yet. In any case, there will likely be discovered a slew of genes for smartness. Which leaves open, for now, the question of racially unequal intelligence; at least for things like math or chess. > There are not enough genes in a human genome to account > for all human charactersitics. There are, when we look at gene combinations, reinforcing or suppressing one another. > Accident and chance play a role. Every day, new discoveries are made, tying genes to specific characteristics. The 'nature/nurture' pendulum is swinging heavily toward nature. When was the last announcement that some trait was cultural, not biological? Your entire memo is egalitarian wishful thinking, almost an apologia. Mark === Subject: Re: Why are Asians so good at math? <43EB0216.8BC3CEA6@yahoo.com> <47t2gjFgpsg2U1@individual.net> <49ae4aFncqt4U1@individual.net> <49hvs0Fof1cbU2@individual.net Vague. Which adaptations do you have in mind? > Primarily language and the related mental apparatuses. The first humans > had to teach themselves language. It is not genetically wired. If it > were there would not be 6000 different languages. > Bob Kolker Bob, Did you realize that if there is ANY merit in your logic about the first humans, then genes and genetics wouldn't play ANY roll in anything, would it, where there's diversity? Evolution existed in languages just as evolution existed in everything that existed, even before the first human being existed, which is less than a minute in the earth's evolutionary clock of 24 hours. -- Reef Fish Bob. === Subject: Re: Why are Asians so good at math? > Evolution existed in languages just as evolution existed in everything > that existed, even before the first human being existed, which is > less than a minute in the earth's evolutionary clock of 24 hours. Evolution is change taken as a generic term. I am talking about genetically based evolution, the kind that Darwin studied. The human race has undergone very little (if any) genetic evolution in the last 50,000 years. The overwhelming engine of change is cultural, not genetic. Cultural change has made much more difference to us than genetic changes in recent (last 50,000) years. Bob Kolker === Subject: Re: Why are Asians so good at math? <43EB0216.8BC3CEA6@yahoo.com> <47t2gjFgpsg2U1@individual.net> <49ae4aFncqt4U1@individual.net> <49hvs0Fof1cbU2@individual.net> <49i632Foodu1U1@individual.net Evolution existed in languages just as evolution existed in everything > that existed, even before the first human being existed, which is > less than a minute in the earth's evolutionary clock of 24 hours. > Evolution is change taken as a generic term. I am talking about > genetically based evolution, the kind that Darwin studied. So? But I was using the absurdity of your analogy about the first human being did not have anyone from whom to inherit the genes, and then illogically related that to the existence of 6000 languages. > The human race has undergone very little (if any) genetic evolution in > the last 50,000 years. 50,000 years constitue only a few seconds in the evolutionary clock of 24 hours. >The overwhelming engine of change is cultural, > not genetic. Cultural change has made much more difference to us than > genetic changes in recent (last 50,000) years. > Bob Kolker That is, of course, your unsubstantiated conjecture, unsupported by any factual evidence. If you think there's not much genetic evolution in the last 50,000 years, then you know very little about the genetic changes in some areas of the USA in the past 200 years alone. Have you not heard of the modern day Darwin, from Atlanta, by the name of Jeff Foxworthy? He has written books about You may be a Redneck if ... and one of the lines is your family tree has only a single trunk. Generations of inbreeding by folks in what's known as the Redneck country (AL, GA, TN, KY, to name a few) you can easily find generations of the same family living within the same county, marrying each other, and soon ran out of outside breeds and started marrying cousins, and other relatives. :-) You can bet your bottom ... dollar that their GENES have undergone irreparable evoluntionary changes within just the short span. But the meaning of evolution is certainly not limited to genetic evolution. In fact the study of the evolution of languages has gone from the lexicographic studies of various language groups to a more recent lexicostatistical classification of languages by reconstructing evolutionary language trees by statistical methods in numerical taxonomy. -- Bob. -- Bob. === Subject: Re: Why are Asians so good at math? > It may have more to do with genetics and evolution than you think. > http://www.nytimes.com/2006/03/12/weekinreview/12wade.html?ex=1299819600&en= 4f > fa8520bf6f7229&ei=5088&partner=rssnyt&emc=rss >> There have been no significant genetic changes to human >> beings in 50,000 years. > When did the races diverge? There are no races, so the answer is never. >> Most human adaptation is cultural, not genetic. > Vague. Which adaptations do you have in mind? language, use of tools, writing, making iron. > enough, the two scientists argue, for natural selection to favor any > variant gene that enhanced cognitive ability. >> Variation in cognitive ability is within the normal range of human >> diversity. There is no proof of a -gene- for smartness. > Not yet. > In any case, there will likely be discovered a slew of > genes for smartness. Which leaves open, for now, the > question of racially unequal intelligence; at least for > things like math or chess. Well, until races can be proved to exist, it is premature to claim racial unequal in anything. >> There are not enough genes in a human genome to account >> for all human charactersitics. > There are, when we look at gene combinations, reinforcing > or suppressing one another. >> Accident and chance play a role. > Every day, new discoveries are made, tying genes to > specific characteristics. The 'nature/nurture' pendulum is > swinging heavily toward nature. Well that is certainly silly. > When was the last announcement that some trait was > cultural, not biological? Give me a trait, and I will show the cultural input. > Your entire memo is egalitarian wishful thinking, almost an apologia. Yours lacks contact with reality. -- Love, Jim http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- === Subject: Critical points of an autonomous diff.eq system x'=y - (x^3)/3 + x y'=-ax where a>0 are there any other critical points besides (0,0) and if so, how can i find them? === Subject: Re: Critical points of an autonomous diff.eq system >x'=y - (x^3)/3 + x >y'=-ax >where a>0 >are there any other critical points besides (0,0) and if so, how can i >find them? Hint: what is the definition of a critical point? === Subject: Can the algebraic numbers be constructed by elementary means? Can the algebraic numbers be constructed by elementary means? There being only countably many algebraic numbers, I'd be satisfied with any sufficiently effective construction that confined itself to countable sets. Effectiveness rules out the axiom of choice. Countability of the set ensures finite representability of its elements. Such a construction could potentially be very useful for exact complex arithmetic where the quantities are finite, much as the rationals serve this purpose for real arithmetic. While the rationals are ideal for solving rational linear systems, they do not support exact solution of rational polynomial systems such as those arising with ruler-and-compass and other higher-order-polynomial constructions, for which one needs a suitable representation of algebraic polynomial systems. If such a construction is known I'd appreciate learning about it. Vaughan Pratt -- Don't contact me at pratt@boole.stanford.edu, substitute cs for boole instead. === Subject: Re: Can the algebraic numbers be constructed by elementary means? > Can the algebraic numbers be constructed by elementary means? > There being only countably many algebraic numbers, I'd be satisfied > with any sufficiently effective construction that confined itself to > countable sets.... Your phrase constructed by elementary means may be a bit strong. You probably know that polynomial equations of degree 5 or more can't always be solved by elementary means. However, if you just want to define a list which includes all the algebraic numbers, why not just enumerate all the polynomials with integer coefficients? The roots of each polynomial are finitely many algebraic numbers, which can form the next block on your list. But perhaps I've misunderstood your question. Ken Pledger. === Subject: Re: Can the algebraic numbers be constructed by elementary means? > Can the algebraic numbers be constructed by elementary means? > There being only countably many algebraic numbers, I'd be satisfied > with any sufficiently effective construction that confined itself to > countable sets. Effectiveness rules out the axiom of choice. > Countability of the set ensures finite representability of its > elements. The axiom of choice is not necessary since the rationals can be well-ordered. It's easy to index the algebraic numbers by the natural numbers and prove that the basic arithmetic operations are recursive. So the answer to your question is yes. > Such a construction could potentially be very useful for exact complex > arithmetic where the quantities are finite, much as the rationals serve > this purpose for real arithmetic. While the rationals are ideal for > solving rational linear systems, they do not support exact solution of > rational polynomial systems such as those arising with > ruler-and-compass and other higher-order-polynomial constructions, for > which one needs a suitable representation of algebraic polynomial > systems. > If such a construction is known I'd appreciate learning about it. > Vaughan Pratt > -- > Don't contact me at pratt@boole.stanford.edu, substitute cs for boole instead. === Subject: Carleson's prize http://www.guardian.co.uk/international/story/0,,1738394,00.html There are some statements about real world applications of Carleson's work. Even the iPod is mentioned. Is there something to this other than pop science hype? === Subject: Re: Carleson's prize > http://www.guardian.co.uk/international/story/0,,1738394,00.html > There are some statements about real world applications of > Carleson's work. Even the iPod is mentioned. Is there something > to this other than pop science hype? Yes and no. Carleson's general area, Harmonic Analysis, has grown out of the study of how sound waves can be represented as sums of sine and cosine waves, that is, as Fourier series. Fourier series is crucial to how a CD player works, and how it reconstructs the sound from the digital information. It is necessary to know that any periodic function can indeed be represented by a Fourier series. However, there are many different hypotheses and representations of such Fourier series, and I would guess that those sufficient to understand how a CD player works were known by the late 19th century. Carleson proved a Fourier representation theorem of tremendous generality, and so his result as such isn't really needed for applications. However his result was extremeley difficult, and people were rather surprized that it was actually true. Only recently have people got to the point where it can be generally understood by most of the experts - at the time his paper appeared, I am told that many universities started seminars aimed at understanding his paper, only to close down the seminars a week or two later, because they found the paper so hard to understand. The general study of Harmonic Analysis also includes Littlewood-Paley square function type inequalities, and I think it would be safe to say that the methods behind this have really helped develop the subject of wavelets. To be honest, I don't really know how MP3 players achieve their high compression ratios, but if they don't use wavelets or something similar to do it, surely some future standards will. Thus while Carleson's result doesn't directly impact the inner workings of the iPod, my guess is that much of the culture of thinking that it created has had, or certainly will have, significant impact on such technologies. Stephen === Subject: Prove on Partial Differention The variables p,v,t are connected by the relation pv=Rt(R=constant). If w is a differentiable function of p and v, prove that: [Partial(w)/Partial(p)](t is constant) = [Partial(w)/Partial(p)](v is constant) - [Partial(w)/Partial(v)](p is constant) *v/p === Subject: Final Call for Papers: IMECS 2006 (the multiconference of 14 engineering & scientific computing conferences) International Association of Engineers (http://www.iaeng.org) Engineering Letters (http://www.engineeringletters.com) International MultiConference of Engineers and Computer Scientists 2006 IMECS 2006 20-22 June, 2006, Hong Kong http://www.iaeng.org/IMECS2006 The International MultiConference of Engineers and Computer Scientists 2006 will take place in Hong Kong, 20-22 June, 2006. The IMECS 2006 is organized by the International Association of Engineers (IAENG). The conference has the focus on the frontier topics in the theoretical and applied engineering and computer science subjects. The IMECS 2006 consists of 14 workshops (see the details at IMECS website: www.iaeng.org/IMECS2006). The multiconference serves as good platforms for the engineering community members of different disciplines to meet with each other and to exchange ideas. The current conference committee of the IMECS 2006 includes over 150 workshop co-chairs and committee members of mainly research center heads, department heads, professors, and research scientists from over 20 countries, while a few of the committee members are also experienced software development directors and engineers. All submitted papers will be under peer review and accepted papers will be published in the conference proceeding (ISBN: 988-98671-3-3). The abstracts will be indexed and available at major academic databases. The Technology Research Databases (TRD) of CSA (Cambridge Scientific Abstracts), DBLP and Computer Science Bibliographies have promised to index the print proceeding in advance of its publication. And after the publication of the proceeding, print copies will also be sent to databases like IEE INSPEC, Engineering Index (EI) and ISI Thomson Scientific for indexing. The accepted papers will also be considered for publication in the special issues of the journal Engineering Letters. Some participants may also be invited to submit extended version of their conference papers for considering as book chapters (soon after the conference). The following workshops are held as parts of the IMECS 2006: IWAIA'06 The 2006 IAENG International Workshop on Artificial Intelligence and Applications IWB'06 The 2006 IAENG International Workshop on Bioinformatics IWCS'06 The 2006 IAENG International Workshop on Computer Science IWDMA'06 The 2006 IAENG International Workshop on Data Mining and Applications IWEE'06 The 2006 IAENG International Workshop on Electrical Engineering IWFE'06 The 2006 IAENG International Workshop on Financial Engineering IWIE'06 The 2006 IAENG International Workshop on Imaging Engineering IWINDE'06 The 2006 IAENG International Workshop on Industrial Engineering IWICWS'06 The 2006 IAENG International Workshop on Internet Computing and Web Services IWOR'06 The 2006 IAENG International Workshop on Operations Research IWSCCS'06 The 2006 IAENG International Workshop on Scientific Computing and Computational Statistics IWSE'06 The 2006 IAENG International Workshop on Software Engineering IWWN'06 The 2006 IAENG International Workshop on Wireless Networks ICMHA'06 The IAENG International Conference on Mathematical, Statistical and Computer Methods in HIV/AIDS 2006 Submission: Prospective authors are invited to submit their draft paper in abstract format (one page) or in full paper format to imecs@iaeng.org by 6 April, 2006. The submitted file can be in MS Word format, PS format, or PDF formats. The first page of the draft paper should include: ¡P Title of the paper; ¡P Name, affiliation and e-mail address for each author; ¡P A maximum of 5 keywords of the paper; Also, the name of the workshop session that the paper is being submitted to should be stated in the email. Extended Draft Manuscript / Abstract Submission Deadline: 6 April, 2006 Extended Camera-Ready papers & Pre-registration Due: 13 April, 2006 IMECS 2006: 20-22 June, 2006 More details about the IMECS 2006 can be found at: http://www.iaeng.org/IMECS2006/index.html Engineering Letters http://www.engineeringletters.com http://www.engineeringletters.com/special issues.html ISSN: 1816-0948 (online version); 1816-093X (print version) Subject Category: Computer science and engineering Published by: International Association of Engineers http://www.iaeng.org ******** This call for papers is updated on 8 March 2006. It will be highly appreciated if you can circulate these calls for papers to your colleagues. === Subject: Re: Charles Darwin's Famous Last Words -- The Grim Reaper. > year's Famous Last Worlds quiz, so it is listed below. > AND THE CORRECT ANSWER IS: > ( X ) Hmmm! My hospital bill. I'm SURE it can't be much. For those interested in the actual answer rather than ramblings, according to his daughter and son, Darwin's last words were I am not the least afraid to die. === === Subject: Re: Inventing fundamental concept > Is there any methodology for inventing fundamenta ideas in mathematics? > I always fascinated of making a parallel field in mathematics.Is it > possoble. Folks ..what do u think? Start with a difficult, unsolved problem. Solve it; and in so doing, pioneer new techniques and ideas. That's a very common pattern in math history. === Subject: Re: Inventing fundamental concept >>Is there any methodology for inventing fundamenta ideas in mathematics? >>I always fascinated of making a parallel field in mathematics.Is it >>possoble. Folks ..what do u think? > Start with a difficult, unsolved problem. Solve it; and in so doing, > pioneer new techniques and ideas. That's a very common pattern in math > history. Oddly reminded of the Monty Python skit How to rid the world of all known diseases (from http://orangecow.org/pythonet/sketches) Which I will now bore the newsgroup with: The cast: ALAN John Cleese NOEL Graham Chapman JACKIE Eric Idle The sketch: (Cut to a sign saying 'How to do it'. Music. Pull out to reveal a 'Blue Peter' type set. Sitting casually on the edge of a dais are three presenters in sweaters - Noel, Jackie and Alan - plus a large bloodhound.) Alan: Hello. Noel: Hello. Alan: Well, last week we showed you how to become a gynaecologist. And this week on 'How to do it' we're going to show you how to play the flute, how to split an atom, how to construct a box girder bridge, how to irrigate the Sahara Desert and make vast new areas of land cultivatable, but first, here's Jackie to tell you all how to rid the world of all known diseases. Jackie: Well, first of all become a doctor and discover a marvellous cure for something, and then, when the medical profession really starts to take notice of you, you can jolly well tell them what to do and make sure they get everything right so there'll never be any diseases ever again. flute. (picking up a flute) Well here we are. You blow there and you move your fingers up and down here. Noel: Great, great, Alan. Well, next week we'll be showing you how black and white people can live together in peace and harmony, and Alan will be over in Moscow showing us how to reconcile the Russians and the Chinese. So, until next week, cheerio. Alan: Bye. Jackie: Bye. (Children's music.) === Subject: Re: Why Clustering and MDS are Methodologically Incompatible the issue of mixed numeric and nominal variables, usually prescribing the standard solution of turning nominal variables into a string of dummy variables. For many reasons, I find this solution unsatisfactory, and particularly so when the nominal variables have large numbers of values (say US State). Do you have a suggestion for some algorithms to try in cases with mixed variables where nominal variables dominate and may have dozens of values each? DM > I'll post a separate post to explain WHY those two mothods >are INCOMPATIBLE, in GOAL or METHOD, and why trying to do >both all at once under some hybrid model can only make it worse > Let's start with why MDS is not compatible to, or suitable for finding > clusters, as in cluster analysis. I had already given one reason in > the related post in question. (1) below: > 1. A proper and useful cluster solution does NOT depend > on the representation (or even existence) in any dimension, but it > does depend heavily on the clustering criterion which produce > many DIFFERENT solutions on different criteria > 2. There are literally HUNDREDS of clustering algorithms based on > at least a dozen or more different criteria for clusters. Often > a particular single criterion can be carried out by different kinds > of algorithms: > For example, > (a) Agglomerative (start with n points as n clusters and combine > two at a time until the final cluster has n points) > (b) Divisive (start with n points as 1 cluster and split one each > time until there are n clusters of single points at the end). > (c) partition n points into a specified number of clusters (groups) > by optimizing certain criteria, such as minimum Within Groups > SS, Ward's Centroid method, etc. > (d) Iterative (if the number of clusters is specified in advance) > such as various k-means methods. > (e) Jardine-Sibson and others where clusters may be overlapping. > None of the above defines what a cluster is, or what constitutes a > cluster. They are based on some global-partition (a-d) or a global > criterion that allows overlapping sets (hence not a partition). > Clustering (or grouping of objects) is an easy and intuitive concept > to grasp, but nearly impossible to define or find. Throw a bunch > (n pieces, n moderately large) of candy on the floor, and any 4-year > old can cluster them better than any of your clustering algorithms > can! > (f) Ling (1972) On the theory and construction of K-Clusters, > Computer Journal, 15, 326-332. > Ling (1973) A Probability Theory of Cluster Analysis, JASA > 58, 159-164. > made a cluster a well-defined object, defined by two > parameters, K, the number of points within a distance delta > of K other points in the cluster, and well isolated (separated) > from other clusters. K = 1 would yield the single-linkage > clusters, while other integer values of K require the clusters > to be more and more compact, and less stringy. > (g) There are many other cluster criteria and algorithms. You > can find numerous books titled, or with key phrases, Cluster > Analysis or Clustering algorithms. There are also journal > (3) In view of (1) and (2) above, the only INCISIVE analysis of > clusters can only be done via specific (or using several > different) clustering criteria and methods. It is unfortunate > that most of the methods are just algorithms, leaving the > concept of a cluster ill-defined. But that is the present > DEFECT within the global subject of Cluster Analysis, and > the existence of 100+ different algorithms and methods (often > delivering clusters with very DIFFERENTcharacteristics) is > the best one can do. > (4) MDS (Multidimensional Scaling) is not suitable for Clustering > because it forces the representation of n objects into n points > in a p-dimensional Euclidean space, by matching the INPUT > n x n matrix of observed Dissimilarities (not necessarily > distances) with the pairwise Distance matrix of the configuration > in p-space, often distorting the distances between points so > much that it LOSES any clustering phenomenon that is clear > via clustering methods. > of the application of MDS to real problems. NONE of those > examples was concerned with, or dealt with, the existence or > non-existence of clusters! Just pick up any one of those > applications and you'll see WHY. > (6) Last but not least, the INPUT data for MDS may not be suitable > at all for clustering. For example, in the famous Rothkopf > data > of the % of trainees, when listened to a pair of Morse Codes > signals for letters and numbers, are asked whether the played > Morse Codes were the same or different. The trainees are > confused even when the same symbols are played one after > the other. Thus none of the A-A, B-B, etc. was 100%, and > most of the A-B and B-A and other pairs have different values > for their same answer, and hence the matrix is asymmetric > with diagonals not equal to 0 for distances. > In short, the general class of methods for MDS is to seek a Euclidean > representation of an n x n similarity/dissimilarity matrix by a set of > points in a p-dimensional Euclidean space in search of some > interpretable AXES in some low dimension, with complete disregard > of any clustering properties -- with a goal similar to that of Factor > Analysis and Principal Components Analysis neither of which is > concerned with clusters also. > Given the above characteristics of Clustering and why MDS is not > suitable, it should be quite clear that Cluster Analysis is NOT > suitable > for discovering any interpretable axes in the MDS sense. > In my (1973) JASA example of clustering a set of the 60 brightest stars > by their apparent pairwise distances in the sky, using the single- > linkage equivalent of my definition of k =1, and the isolation index > associated with the probability theory of significant clusters in a > random graph model, I was able to correctly identify clusters (using > the > probability theory alone, 52 of the 60 stars that belonged to named > constellations, with five of the star constellations perfectly > identifed, > which included UMa (Big Dipper) and Cyg (Swan). > The constellation that was least accounted for (hence least well- > defined) was Draconis, which is a serpent-like constellation that > goes between UMa (Big Dipper) and UMi (Little Dipper) but is not > well separated from the stars in those constellations. > That was perhaps an extreme example in which the goal was to > identify the stringy-connected and isolated clusters to see how > well they corresponded to the constellations that are well-known > because people have long clustered them by eye. This was > also a problem in which there was absolute no attempt to seek > any interpretable dimensions formed by the bright stars. > My bottom line: > If you want to find CLUSTERS, use clustering methods on > your original data. > If you want to seek the structure of interpretable dimensions > imbedded in a dataset, there are many multivariate methods > for that, including MDS which works for certain data (as that > of Rothkopf's Morse Code data) that would not work on other > methods that require (among other things) a metric or > symmetry of the input matrix. > The above is my quick attempt to summarize and distinguish two > HUGE areas of multivariate analysis, each of which had at least > a history of 40 or more years of development. > -- Bob. === Subject: Re: Why Clustering and MDS are Methodologically Incompatible > Most cluster analysis texts skim over > the issue of mixed numeric and nominal variables, usually prescribing > the standard solution of turning nominal variables into a string of > dummy variables. For many reasons, I find this solution > unsatisfactory, and particularly so when the nominal variables have > large numbers of values (say US State). Do you have a suggestion for > some algorithms to try in cases with mixed variables where nominal > variables dominate and may have dozens of values each? Decision tree based models handle categorical variables with large numbers of categories well because they don't have to generate dummy variables. They just partition the categories. I have customers building decision tree models with categorical variables including state, zipcode, automobile model, etc. -- Phil Sherrod (phil.sherrod 'at' sandh.com) http://www.dtreg.com (decision tree and SVM modeling) http://www.nlreg.com (nonlinear regression) === Subject: Re: Why Clustering and MDS are Methodologically Incompatible Your customers must have response variables to work with. Do you have a way to do unsupervised learning with decision trees? The efficacy of tree methods to deal with categorical variables with many categories deserves a separate topic. The ability to do it does not imply anything about the effectiveness. It is still better to reduce the number of categories first before throwing them into the tree algorithm. > Most cluster analysis texts skim over > the issue of mixed numeric and nominal variables, usually prescribing > the standard solution of turning nominal variables into a string of > dummy variables. For many reasons, I find this solution > unsatisfactory, and particularly so when the nominal variables have > large numbers of values (say US State). Do you have a suggestion for > some algorithms to try in cases with mixed variables where nominal > variables dominate and may have dozens of values each? > Decision tree based models handle categorical variables with large numbers of > categories well because they don't have to generate dummy variables. They just > partition the categories. I have customers building decision tree models with > categorical variables including state, zipcode, automobile model, etc. > -- > Phil Sherrod > (phil.sherrod 'at' sandh.com) > http://www.dtreg.com (decision tree and SVM modeling) > http://www.nlreg.com (nonlinear regression) === Subject: Re: Why Clustering and MDS are Methodologically Incompatible > Your customers must have response variables to work with. Do you have > a way to do unsupervised learning with decision trees? Yes. I'm sorry, I missed the beginning of this thread. > The efficacy of tree methods to deal with categorical variables with > many categories deserves a separate topic. The ability to do it does > not imply anything about the effectiveness. It is still better to > reduce the number of categories first before throwing them into the > tree algorithm. Why would you reduce the number of categories? That just adds extra steps and removes potentially useful information. -- Phil Sherrod (phil.sherrod 'at' sandh.com) http://www.dtreg.com (decision tree and SVM predictive modeling) http://www.nlreg.com (nonlinear regression) === Subject: Re: Why Clustering and MDS are Methodologically Incompatible You're welcome, and I appreciate the acknowledgment from someone I know, from reading these groups, who have had much experience with these methods. > Most cluster analysis texts skim over > the issue of mixed numeric and nominal variables, usually prescribing > the standard solution of turning nominal variables into a string of > dummy variables. For many reasons, I find this solution > unsatisfactory, and particularly so when the nominal variables have > large numbers of values (say US State). In clustering, often the available variables are collapsed into an n x n similarity or distance matrix through various metrics and combination of metrics. If one of the variables contains a large number of categorical information (as US States), that's a very special problem -- which may be condensed into smaller number of categories such as regions before proceeding. Other times, there are methods (Hartigan and others) that cluster the p variables AND the n objects simultaneously -- that is NOT in any shape or form a multidimensional scaling method -- but a simultaneous clustering method -- which is actually better than using the nxn distance matrix most of the time, if doable. > Do you have a suggestion for > some algorithms to try in cases with mixed variables where nominal > variables dominate and may have dozens of values each? > DM It's no longer a question of an algorithm. Here your problem is how to COMBINE the available information in your variables BEFORE you start finding clusters. There are no algorithms to handle these problems. Moreover, you have to deal with subject matter of what OBJECTS you are trying to cluster, and WHY are you seeking clusters. Often when you don't have a natural n x n pairwise distance matrix to begin with, but an n x p matrix of many variables, there are other, often more suitable methods, of analyzing the problem, such as those in market segmentation where the geographical regions do come into play. Sorry I can't give you a more specific response than this. -- Bob. > I'll post a separate post to explain WHY those two mothods >are INCOMPATIBLE, in GOAL or METHOD, and why trying to do >both all at once under some hybrid model can only make it worse > Let's start with why MDS is not compatible to, or suitable for finding > clusters, as in cluster analysis. I had already given one reason in > the related post in question. (1) below: > 1. A proper and useful cluster solution does NOT depend > on the representation (or even existence) in any dimension, but it > does depend heavily on the clustering criterion which produce > many DIFFERENT solutions on different criteria > 2. There are literally HUNDREDS of clustering algorithms based on > at least a dozen or more different criteria for clusters. Often > a particular single criterion can be carried out by different kinds > of algorithms: > For example, > (a) Agglomerative (start with n points as n clusters and combine > two at a time until the final cluster has n points) > (b) Divisive (start with n points as 1 cluster and split one each > time until there are n clusters of single points at the end). > (c) partition n points into a specified number of clusters (groups) > by optimizing certain criteria, such as minimum Within Groups > SS, Ward's Centroid method, etc. > (d) Iterative (if the number of clusters is specified in advance) > such as various k-means methods. > (e) Jardine-Sibson and others where clusters may be overlapping. > None of the above defines what a cluster is, or what constitutes a > cluster. They are based on some global-partition (a-d) or a global > criterion that allows overlapping sets (hence not a partition). > Clustering (or grouping of objects) is an easy and intuitive concept > to grasp, but nearly impossible to define or find. Throw a bunch > (n pieces, n moderately large) of candy on the floor, and any 4-year > old can cluster them better than any of your clustering algorithms > can! > (f) Ling (1972) On the theory and construction of K-Clusters, > Computer Journal, 15, 326-332. > Ling (1973) A Probability Theory of Cluster Analysis, JASA > 58, 159-164. > made a cluster a well-defined object, defined by two > parameters, K, the number of points within a distance delta > of K other points in the cluster, and well isolated (separated) > from other clusters. K = 1 would yield the single-linkage > clusters, while other integer values of K require the clusters > to be more and more compact, and less stringy. > (g) There are many other cluster criteria and algorithms. You > can find numerous books titled, or with key phrases, Cluster > Analysis or Clustering algorithms. There are also journal > (3) In view of (1) and (2) above, the only INCISIVE analysis of > clusters can only be done via specific (or using several > different) clustering criteria and methods. It is unfortunate > that most of the methods are just algorithms, leaving the > concept of a cluster ill-defined. But that is the present > DEFECT within the global subject of Cluster Analysis, and > the existence of 100+ different algorithms and methods (often > delivering clusters with very DIFFERENTcharacteristics) is > the best one can do. > (4) MDS (Multidimensional Scaling) is not suitable for Clustering > because it forces the representation of n objects into n points > in a p-dimensional Euclidean space, by matching the INPUT > n x n matrix of observed Dissimilarities (not necessarily > distances) with the pairwise Distance matrix of the configuration > in p-space, often distorting the distances between points so > much that it LOSES any clustering phenomenon that is clear > via clustering methods. > of the application of MDS to real problems. NONE of those > examples was concerned with, or dealt with, the existence or > non-existence of clusters! Just pick up any one of those > applications and you'll see WHY. > (6) Last but not least, the INPUT data for MDS may not be suitable > at all for clustering. For example, in the famous Rothkopf > data > of the % of trainees, when listened to a pair of Morse Codes > signals for letters and numbers, are asked whether the played > Morse Codes were the same or different. The trainees are > confused even when the same symbols are played one after > the other. Thus none of the A-A, B-B, etc. was 100%, and > most of the A-B and B-A and other pairs have different values > for their same answer, and hence the matrix is asymmetric > with diagonals not equal to 0 for distances. > In short, the general class of methods for MDS is to seek a Euclidean > representation of an n x n similarity/dissimilarity matrix by a set of > points in a p-dimensional Euclidean space in search of some > interpretable AXES in some low dimension, with complete disregard > of any clustering properties -- with a goal similar to that of Factor > Analysis and Principal Components Analysis neither of which is > concerned with clusters also. > Given the above characteristics of Clustering and why MDS is not > suitable, it should be quite clear that Cluster Analysis is NOT > suitable > for discovering any interpretable axes in the MDS sense. > In my (1973) JASA example of clustering a set of the 60 brightest stars > by their apparent pairwise distances in the sky, using the single- > linkage equivalent of my definition of k =1, and the isolation index > associated with the probability theory of significant clusters in a > random graph model, I was able to correctly identify clusters (using > the > probability theory alone, 52 of the 60 stars that belonged to named > constellations, with five of the star constellations perfectly > identifed, > which included UMa (Big Dipper) and Cyg (Swan). > The constellation that was least accounted for (hence least well- > defined) was Draconis, which is a serpent-like constellation that > goes between UMa (Big Dipper) and UMi (Little Dipper) but is not > well separated from the stars in those constellations. > That was perhaps an extreme example in which the goal was to > identify the stringy-connected and isolated clusters to see how > well they corresponded to the constellations that are well-known > because people have long clustered them by eye. This was > also a problem in which there was absolute no attempt to seek > any interpretable dimensions formed by the bright stars. > My bottom line: > If you want to find CLUSTERS, use clustering methods on > your original data. > If you want to seek the structure of interpretable dimensions > imbedded in a dataset, there are many multivariate methods > for that, including MDS which works for certain data (as that > of Rothkopf's Morse Code data) that would not work on other > methods that require (among other things) a metric or > symmetry of the input matrix. > The above is my quick attempt to summarize and distinguish two > HUGE areas of multivariate analysis, each of which had at least > a history of 40 or more years of development. -- Bob. === Subject: vector ode to scalar odes I have a 2D vector ODE: m*x'' = -GMm/r^2 * i_r, where x = r*i_r, and i_r = (cos theta, sin theta) is a unit vector in the direction from the point mass M to the point mass m, where both r = r(t) and theta = theta(t). I am to show this vector ODE implies the two scalar ODEs: 1. r'' - r*(theta')^2 = -GM / r^2 2. 2r'(theta)' + r(theta)'' = 0 I computed x'' and then took the dot product of i_r on both sides of the vector ODE; this got me Equation (1). However, I cannot figure out how to get Equation (2). === Subject: Re: vector ode to scalar odes > I have a 2D vector ODE: > m*x'' = -GMm/r^2 * i_r, > where x = r*i_r, and i_r = (cos theta, sin theta) is a unit vector in the > direction from the point mass M to the point mass m, where both r = r(t) and > theta = theta(t). > I am to show this vector ODE implies the two scalar ODEs: > 1. r'' - r*(theta')^2 = -GM / r^2 > 2. 2r'(theta)' + r(theta)'' = 0 > I computed x'' and then took the dot product of i_r on both sides of the > vector ODE; this got me Equation (1). However, I cannot figure out how to > get Equation (2). Simple, just take the cross product of x'' and i_r. Or alternatively, take the equation for x'' and subtract (eq.1 times i_r). -Michael. === Subject: Re: vector ode to scalar odes > I have a 2D vector ODE: > m*x'' = -GMm/r^2 * i_r, > where x = r*i_r, and i_r = (cos theta, sin theta) is a unit vector in the > direction from the point mass M to the point mass m, where both r = r(t) and > theta = theta(t). x = r.i_r = r(cos t, sin t) x' = r'.i_r + rt'(-sin t, cos t) x = r.i_r + r't'(-sin t, cos t) + r't'(-sin t, cos t) + rt(-sin t, cos t) + r(t')^2 (-cos t, -sin t) = r.i_r + (2r't' + rt)(-sin t, cos t) - r(t'^2) i_r = -(GM/r^2)i_r assuming m /= 0 > I am to show this vector ODE implies the two scalar ODEs: > 1. r'' - r*(theta')^2 = -GM / r^2 > 2. 2r'(theta)' + r(theta)'' = 0 > I computed x'' and then took the dot product of i_r on both sides of the Please show your work instead of making us do it over. > vector ODE; this got me Equation (1). However, I cannot figure out how > to get Equation (2). Because (-sin t, cos t) & i_r are orthogonal unit vectors. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics |> So testing the consequences of the existence of a |> measurable cardinal is to test the existence of a |> measurable cardinal, right? | |If you first admit that the assertion that a measurable cardinal exists |has no meaning other than that the consequences of that assertion are |true, then yes. I find it hard to say whether the meaning of an assertion should be always possible to determine just by considering what consequences it has according to an appropriately chosen background theory. But I don't think it's a question that needs to be answered here. If it's not possible to determine the meaning of a measurable cardinal exists just by looking at the role it plays in mathematics (as it's done today) then it follows that it is *not* necessary to decide what it means in order to decide whether the mathematics using it is valid science, or in order to do mathematics with it. If it is possible to determine the meaning of a measurable cardinal exists just by looking at the role it plays in mathematics (as it's done today), whether by considering its meaning to be just its consequences or any other way, then why ask me to stipulate? If the meaning is determinable, then whatever the correct meaning is, testing its consequences then becomes the proper scientific way of treating it. You give the impression of having suspicions that mathematicians have gotten the wrong idea of what some mathematical statements mean, by imagining that there is extra meaning to them, but to prosecute on that front would require a different kind of argument than you're making. Keith Ramsay === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <871wwwqp0g.fsf@phiwumbda.org> <87fylbughw.fsf@phiwumbda.org> <44211853$0$2031$ba620dc5@text.nova.planet.nl |Feferman has asserted that the Pi01 implications of the large cardinal > |axioms actually come from the assertion that the axioms are consistent > |with ZFC, and not from the axioms themselves (at least that's how I > |understand the situation). I haven't seen Friedman's response to > |Feferman's criticism. > What do you mean by actually come from the assertion that the > axioms are consistent with ZFC? What I think I mean is that instead of adding a large cardinal axiom to ZFC, we could add the axiom con(ZFC)->con(ZFC+LCA) and still get the same Pi01 implications, and of course, then we could still believe that the hierarchy of infinities in ZFC is a fantasy world. Feferman did not say exactly that, though I *believe* that what he did say is logically equivalent to that. If I understood him correctly, he said that instead of the large cardinal axiom, we could add the axiom the Pi01 implications of the LCA are true. But maybe there's a subtle difference between the two that I am missing. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >> |Feferman has asserted that the Pi01 implications of the large cardinal >> |axioms actually come from the assertion that the axioms are consistent >> |with ZFC, and not from the axioms themselves (at least that's how I >> |understand the situation). I haven't seen Friedman's response to >> |Feferman's criticism. >> What do you mean by actually come from the assertion that the >> axioms are consistent with ZFC? > What I think I mean is that instead of adding a large cardinal axiom to > ZFC, we could add the axiom con(ZFC)->con(ZFC+LCA) and still get the > same Pi01 implications, and of course, then we could still believe that > the hierarchy of infinities in ZFC is a fantasy world. > Feferman did not say exactly that, though I *believe* that what he did > say is logically equivalent to that. If I understood him correctly, he > said that instead of the large cardinal axiom, we could add the axiom > the Pi01 implications of the LCA are true. > But maybe there's a subtle difference between the two that I am > missing. It seems to me that you need con(ZFC+LCA), not just con(ZFC) -> con(ZFC+LCA). But Goedel proved that ZFC+A |- con(ZFC+A) is never true, so if ZFC+A |- p iff con(ZFC+A) |- p for all p in some class of meaningful sentences, then con(ZFC+A) is not in that class. con(ZFC+LCA) then would still be an assumption of a 'higher' order. Unless the class of meaningful sentences is a fuzzy class, of course. Or unless you allow non-monotonic reasoning on the meta-level. As done in most sciences. -- Herman Jurjus === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <871wwwqp0g.fsf@phiwumbda.org> <87fylbughw.fsf@phiwumbda.org> <44211853$0$2031$ba620dc5@text.nova.planet.nl> |> |Feferman has asserted that the Pi01 implications of the large cardinal |> |axioms actually come from the assertion that the axioms are consistent |> |with ZFC, and not from the axioms themselves (at least that's how I |> |understand the situation). I haven't seen Friedman's response to |> |Feferman's criticism. |> What do you mean by actually come from the assertion that the |> axioms are consistent with ZFC? | |What I think I mean is that instead of adding a large cardinal axiom to |ZFC, we could add the axiom con(ZFC)->con(ZFC+LCA) and still get the |same Pi01 implications, and of course, then we could still believe that |the hierarchy of infinities in ZFC is a fantasy world. As I pointed out, con(ZFC+{LCA}) is equivalent to every pi-0-1 consequence of {LCA} in ZFC is true. [It's not enough to assume that con(ZFC)->con(ZFC+{LCA}).] |Feferman did not say exactly that, though I *believe* that what he did |say is logically equivalent to that. If I understood him correctly, he |said that instead of the large cardinal axiom, we could add the axiom |the Pi01 implications of the LCA are true. I seem to remember reading some remark on these lines. |But maybe there's a subtle difference between the two that I am |missing. The problem with this is that something analogous is true of every other hypothesis. Is it appropriate to say the same kind of thing about all the others too? Say some famous mathematician decides he doesn't believe in mathematical induction on pi-0-2 formulas, but merely believes that it's consistent with primitive recursive arithmetic, PRA. All that quantifier alternation nonsense is just fantasy! Goedel showed how to construct a sentence F such that con(ZFC+F) is equivalent to ~F, so I don't think it's always correct to identify the meaning of a sentence with its consistency with a system, whatever the system. Keith Ramsay === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <6sYXf.1$Fy2.0@newsread3.news.pas.earthlink.net> All of the mathematics which has the potential to help us understand > the world in which we live, fits within the definition of mathematics I > have given. To go beyond that definition, is to move into a fantasy > world. > Considering the set of all function on the natural numbers, and > analyzing which of them are computable or not, helps programmers > understand the world in which we work. The guiding principle I like to go by is that we truly understand something when we can explain it to a computer. I claim that if you accept that and think about it, then eventually you will agree that we can truly understand computable functions, but we are deceiving ourselves if we think that we need to postulate the existence of a world beyond the world of computation in order to understand the world of computation. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics david petry says... >The guiding principle I like to go by is that we truly understand >something when we can explain it to a computer. Computers don't have any more problem understanding set theory than they do understanding arithmetic. I don't actually think that current computers understand either one, but computers can be programmed to do proofs in axiomatic set theory perfectly well. Set-theoretic statements are finite bits of syntax. Set-theoretic proofs are finite sequences of statements. You don't have to have access to anything infinite in order to work with set theory. -- -- NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > david petry says... >> The guiding principle I like to go by is that we truly understand >> something when we can explain it to a computer. > Computers don't have any more problem understanding set theory > than they do understanding arithmetic. I don't actually think > that current computers understand either one, but computers can > be programmed to do proofs in axiomatic set theory perfectly well. > Set-theoretic statements are finite bits of syntax. Set-theoretic > proofs are finite sequences of statements. You don't have to have > access to anything infinite in order to work with set theory. But then you reduce sentences about sets to sentences of the form 'ZFC proves ...'. And such sentences have a very concrete meaning. For example, no ontological commitment to the existence of, say, P(P(N)) is needed to make sense of such statements. In a way, your argument even confirms Petry's opinion: in the end, the only thing that's really 'exact' is what can be reduced to concrete, computable formalism. The rest is vague, fantasy. You can use it alright, for heuristic purposes. But don't claim that it is well-defined beyond suspicion or has meaning in itself, without first coding or translating it into more basic notions. What bothers me more with David Petry is: where's the mathematics? The way it proceeds, it appears to remain just some philosophical point. -- Herman Jurjus === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl What bothers me more with David Petry is: where's the mathematics? > The way it proceeds, it appears to remain just some philosophical point. As I have mentioned before, my motivation was to create a language and a foundation for mathematics that would facilitate the transfer of our mathematical knowledge to the computer (i.e. an artificial intelligence). Along the way, I came to believe that this new language and foundation would also be of significant value to applied mathematicians, in that it would clear out a lot of deadwood; it would facilitate communication between mathematicians and those who need to apply mathematics. It's not merely philosophy, even if Usenet discussions of it never go beyond philosophy. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl It's not merely philosophy, even if Usenet discussions of it never go > beyond philosophy. Then go beyond philosophy. Show us the mathematics. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl It's not merely philosophy, even if Usenet discussions of it never go > beyond philosophy. > Then go beyond philosophy. Show us the mathematics. I have certainly introduced the mathematical ideas many times. No useful discussion ensues. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > What bothers me more with David Petry is: where's the mathematics? > The way it proceeds, it appears to remain just some philosophical point. > As I have mentioned before, my motivation was to create a language and > a foundation for mathematics that would facilitate the transfer of our > mathematical knowledge to the computer (i.e. an artificial > intelligence). Along the way, I came to believe that this new language > and foundation would also be of significant value to applied > mathematicians, in that it would clear out a lot of deadwood; it would > facilitate communication between mathematicians and those who need to > apply mathematics. The only communication it would facilitate would be between those who limited themselves in this way and computers. goal of mathematics. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Herman Jurjus says... >> Set-theoretic statements are finite bits of syntax. Set-theoretic >> proofs are finite sequences of statements. You don't have to have >> access to anything infinite in order to work with set theory. >But then you reduce sentences about sets to sentences of the form >'ZFC proves ...'. And such sentences have a very concrete meaning. >For example, no ontological commitment to the existence of, say, P(P(N)) >is needed to make sense of such statements. I think that's true. There is no ontological commitment needed to work with set theory. >In a way, your argument even confirms Petry's opinion: in the end, the >only thing that's really 'exact' is what can be reduced to concrete, >computable formalism. The rest is vague, fantasy. I don't see how it confirms Petry's opinion. He is asserting 1. We should limit mathematics to what can be explained to a computer. 2. If we agree to 1. above, then we must throw out Cantorian set theory. I'm saying, in contrast, that 2 does not follow from 1. Cantorian set theory can perfectly well be explained to a computer (inasmuch as anything can be). >You can use it alright, for heuristic purposes. But don't claim that it >is well-defined beyond suspicion or has meaning in itself, without first >coding or translating it into more basic notions. I didn't say that it should be coded or translated into more basic notions. >What bothers me more with David Petry is: where's the mathematics? >The way it proceeds, it appears to remain just some philosophical point. It seems to me that David Petry's notion of mathematics is a proper subset of the standard notion of mathematics. It doesn't produce anything new. The advantage to adopting David's position is that if all the most talented mathematicians stopped doing stuff with infinite sets and noncomputable functions, then that would free them up to work on more important things. David isn't really proposing any new research program, he's proposing the elimination of *existing* research programs. -- -- NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl> the elimination of *existing* research programs. That nails it. Though, he might answer that we have to sweep out the old to make room for the new. Yet, what does he propose the new would be? What project(s) does he have in mind that he wants us to read about 'mathematics' is generally defined? === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl> the elimination of *existing* research programs. > That nails it. Not really. I'm suggesting that pseudoscience doesn't belong in the curriculum of publicly funded universities. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >>... >All of the mathematics which has the potential to help us understand >the world in which we live, fits within the definition of mathematics I >have given. To go beyond that definition, is to move into a fantasy >world. >>Considering the set of all function on the natural numbers, and >>analyzing which of them are computable or not, helps programmers >>understand the world in which we work. > The guiding principle I like to go by is that we truly understand > something when we can explain it to a computer. I claim that if you > accept that and think about it, then eventually you will agree that we > can truly understand computable functions, but we are deceiving > ourselves if we think that we need to postulate the existence of a > world beyond the world of computation in order to understand the world > of computation. How can you justify this assertion, when apparently you have not yet reproduced, in your mathematics, some equivalent to the undecidability of the halting problem? It is one of the earliest and most important results in Theory of Computation. Patricia === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >>... >All of the mathematics which has the potential to help us understand >the world in which we live, fits within the definition of mathematics I >have given. To go beyond that definition, is to move into a fantasy >world. >>Considering the set of all function on the natural numbers, and >>analyzing which of them are computable or not, helps programmers >>understand the world in which we work. > The guiding principle I like to go by is that we truly understand > something when we can explain it to a computer. I claim that if you > accept that and think about it, then eventually you will agree that we > can truly understand computable functions, but we are deceiving > ourselves if we think that we need to postulate the existence of a > world beyond the world of computation in order to understand the world > of computation. Here is a question: Is it possible to find an algorithm for predicting, from source code, whether an arbitrary program will terminate on a given input, assuming it does not run out of memory? This is a practical question, with real-world consequences. Can it be asked and answered in your model of mathematics? If so, how? Patricia === Subject: Re: Harvey Friedman on Cantorian pseudomathematics from source code, whether an arbitrary program will terminate on a given > input, assuming it does not run out of memory? Well, let's see. If we had such an algorithm, we could ask it to predict whether the following program would halt. K = 4 loop: for each pair of primes (p1, p2) such that each is less than K if p1 + p2 = K then K = K+2, goto loop: halt Wouldn't that be nice. > This is a practical question, with real-world consequences. I don't think this is going anywhere, but please enlighten me about the real-world consequences. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >>Here is a question: Is it possible to find an algorithm for predicting, >>from source code, whether an arbitrary program will terminate on a given >>input, assuming it does not run out of memory? > Well, let's see. If we had such an algorithm, we could ask it to > predict whether the following program would halt. > K = 4 > loop: > for each pair of primes (p1, p2) such that each is less than K > if p1 + p2 = K then K = K+2, goto loop: > halt > Wouldn't that be nice. I thought you didn't believe in thinking about non-computable functions, and the consequences of being able to compute them? >>This is a practical question, with real-world consequences. > I don't think this is going anywhere, but please enlighten me about the > real-world consequences. One obvious use is during program debug. Suppose a program does not finish in several times the expected run time. Should the programmer look first for a true infinite loop, or a bad performance problem? The analysis steps are different, so it would be helpful to know which way to go, rather than having to guess. For comparison, there is a million dollar prize for a proof showing whether the complexity classes P and NP are the same. If we did not already have the answer, I'm sure there would be at least as high a bounty on the halting problem. Can it be asked and answered in your model of mathematics? If so, how? I hope you will post a substantive answer to those questions. That would be interesting. More likely, it will not go anywhere, suggesting that the question either cannot be asked, or cannot be answered, in your model of mathematics. Patricia === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >> ... >> All of the mathematics which has the potential to help us understand >> the world in which we live, fits within the definition of mathematics I >> have given. To go beyond that definition, is to move into a fantasy >> world. >> Considering the set of all function on the natural numbers, and >> analyzing which of them are computable or not, helps programmers >> understand the world in which we work. > The guiding principle I like to go by is that we truly understand > something when we can explain it to a computer. By that standard you must understand next to nothing. Can you explain English to a computer? Can you explain music to a computer? Can you explain calculus to a computer? Can you explain tennis to a computer? What do you even mean by 'explain it to a computer'? I know you have some daft idea that AIs will somehow be unable to understand set theory, but you offered no justification. Currently comuters are not capable of understanding anything that I would consider an 'explanation'. Programming something into a computer is not the same as explaining something to a computer Stephen === Subject: Re: Harvey Friedman on Cantorian pseudomathematics something when we can explain it to a computer. > By that standard you must understand next to nothing. > Can you explain English to a computer? I don't truly understand English. >Can you explain music to a computer? I don't truly understand music. >Can you explain calculus to a computer? Yes. > Can you explain tennis to a computer? Don't know. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <6sYXf.1$Fy2.0@newsread3.news.pas.earthlink.net> something when we can explain it to a computer. well, no, the proof that the halting problem is insoluble is very strong evidence that merely explaining something to a computer in no wise guarantees us understanding [that] something, in particular, we cannot even understand whether understanding that something is a tractible problem It seems, beyond needing to learn some math, that you need to learn some computer science, too, before trying to change the way practitioners in the two fields spend their days. xanthian. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > ... > All of the mathematics which has the potential to help us understand > the world in which we live, fits within the definition of mathematics I > have given. To go beyond that definition, is to move into a fantasy > world. Considering the set of all function on the natural numbers, and > analyzing which of them are computable or not, helps programmers > understand the world in which we work. > The guiding principle I like to go by is that we truly understand > something when we can explain it to a computer. When I began, computers were people, not machines incapable of doing anything without precise instructions. On that basis, that computers are people, and that only, I will agree with David. In the sense of explaining things to something that has no self-awareness or consciousness, I disagree. > I claim that if you > accept that and think about it, then eventually you will agree that we > can truly understand computable functions, but we are deceiving > ourselves if we think that we need to postulate the existence of a > world beyond the world of computation in order to understand the world > of computation. But I disagree with any claim that the world of computation in the sense described above means anything. Except possibly to those computers which do not understand anything. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > All of the mathematics which has the potential to help us understand > the world in which we live, fits within the definition of mathematics I > have given. To go beyond that definition, is to move into a fantasy > world. The world [in which we live] is very ... very complex! It's so complex that we don't know if time is quantized or continuous, or don't know if the mathematical description of physics is the same within a Schwarzchild radius. It's so complex that it still exists even *after* any reasonings (and the fossilized remains of the reasoning beings) anywhere in that world have gone into oblivion, before all the matter are rushing into a 1-mm sphere during the Big Crunch, or before all the matter (including any remaining black holes) would evaporate into the lowest possible energy photons. In other words, no (mortal) mathematics would have the potential (whatever the heck that could mean!) to help us, the mortal beings, to understand that world in its entirety. That could only be a mathematics of God. And you possess a definition of mathematics that would nicely fit all of such God-mathematics into it? -- ---------------------------------------------------- Time passes, there is no way we can hold it back. Why then do thoughts linger, long after everything else is gone? Ryokan ---------------------------------------------------- === Subject: Re: Harvey Friedman on Cantorian pseudomathematics On 4 Apr 2006 11:01:59 -0700, david petry said: >> Among those functions, there are those that are >> computable and those that are not. > Well, if we define mathematics to be the science of phenomena > observable in the world of computation, then it's a little silly to > say that there exist functions that are not computable. It would also be silly if we defined mathematics to be the science of, say, addition. Both that definition and your personal definition throw out large portions of what most all mathematicians themselves consider mathematics. It is of course silly to define mathematics as the science of addition. The question is why your personal definition is not equally silly. An advantage of the definition above -- mathematics = addition -- is that, for all its silliness, it is quite clear. Your own is quite opaque. For example, what is the world of computation? And what is it to be observable in that world? Observation is typically something connected to sense perception. Are you claiming that we can observe computable functions? How does one do that? Or are you just talking about observing the actual workings of digital computers? If so, does that mean that all the observable functions are finite? After all, no matter how fast they get, computers will always have finite limitations on storage. If you are not talking about the actual workings of digital computers, in what manner does one of your observable functions exist? And in what sense is it observable? I'd be happy to try to understand your view, but I really haven't a clue what it is. Currently, we define computability relative to a backdrop of arbitrary functions from natural numbers to natural numbers, without prejudging which are computable and which are not. You seem to want to bootstrap the enterprise from some undefined notion of computation without that backdrop, but I don't see how that notion ever gets off the ground. You simply haven't got a theory here. >> Countable distinguishes the set of of former from the set of the >> latter. > Fine. So the set of things that exist is countable, and the set of > things that don't exist is not countable. Makes sense. It does? The set of things is that don't exist is obviously empty, hence finite, hence countable. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <6sYXf.1$Fy2.0@newsread3.news.pas.earthlink.net> said: >> Among those functions, there are those that are >> computable and those that are not. > Well, if we define mathematics to be the science of phenomena > observable in the world of computation, then it's a little silly to > say that there exist functions that are not computable. > It would also be silly if we defined mathematics to be the science of, > say, addition. Both that definition and your personal definition throw > out large portions of what most all mathematicians themselves consider > mathematics. It is of course silly to define mathematics as the science > of addition. The question is why your personal definition is not > equally silly. All of the mathematics that has the potential to help us understand the world in which we live falls within the scope of mathematics as the science of phenomena observable in the world of computation. > An advantage of the definition above -- mathematics = addition -- is > that, for all its silliness, it is quite clear. Your own is quite > opaque. For example, what is the world of computation? And what is > it to be observable in that world? Observation is typically something > connected to sense perception. As a conceptual aid, we can think of the computer as a microscope which helps us peer deeply into the world of computation, where the world of computation may be defined as what we are seeing when we look through that microscope. Mathematics is then the science which studies the phenomena observed when we look through that microscope. So, we can approach mathematics the same way we approach science. We accept a reality (in this case, the world of computation) and then *seek* a theory that explains the phenomena we observe. When you (Menzel) ask me to tell you what the theory is, you are asking the wrong question. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Discussion, linux) <6sYXf.1$Fy2.0@newsread3.news.pas.earthlink.net> It would also be silly if we defined mathematics to be the science of, >> say, addition. Both that definition and your personal definition throw >> out large portions of what most all mathematicians themselves consider >> mathematics. It is of course silly to define mathematics as the science >> of addition. The question is why your personal definition is not >> equally silly. > All of the mathematics that has the potential to help us understand the > world in which we live falls within the scope of mathematics as the > science of phenomena observable in the world of computation. So the statement, The halting problem is not computable, does not help us understand the world in which we live? -- You are beneath contempt because you betray mathematics itself, and spit upon the truth, spit upon decency, and spit upon the intelligence of the world. You betrayed the world, and now it's time for the world to notice. -- James S. Harris awaits Justice for crimes against Math. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <6sYXf.1$Fy2.0@newsread3.news.pas.earthlink.net> <87lkuj6zh1.fsf@phiwumbda.org All of the mathematics that has the potential to help us understand the > world in which we live falls within the scope of mathematics as the > science of phenomena observable in the world of computation. > So the statement, The halting problem is not computable, does not help > us understand the world in which we live? Well, if I told you that I have a short, simple, and efficient program that will tell you whether or not a program will halt, as long as the code is less than a zillion lines long, that would not contradict the assertion that no general algorithm for the halting problem exists. Does that answer your question? === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >All of the mathematics that has the potential to help us understand the >world in which we live falls within the scope of mathematics as the >science of phenomena observable in the world of computation. >>So the statement, The halting problem is not computable, does not help >>us understand the world in which we live? > Well, if I told you that I have a short, simple, and efficient program > that will tell you whether or not a program will halt, as long as the > code is less than a zillion lines long, that would not contradict the > assertion that no general algorithm for the halting problem exists. I won't believe you if you claim your halting problem decider is even a little bit shorter than a zillion lines. Patricia === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <87lkuj6zh1.fsf@phiwumbda.org> world in which we live falls within the scope of mathematics as the >science of phenomena observable in the world of computation. >>So the statement, The halting problem is not computable, does not help >>us understand the world in which we live? > Well, if I told you that I have a short, simple, and efficient program > that will tell you whether or not a program will halt, as long as the > code is less than a zillion lines long, that would not contradict the > assertion that no general algorithm for the halting problem exists. > I won't believe you if you claim your halting problem decider is even a > little bit shorter than a zillion lines. I'm not sure you get the point: the assertion that no general algorithm for the halting problem exists does not by itself tell you anything at all about the world we live in. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >All of the mathematics that has the potential to help us understand the >world in which we live falls within the scope of mathematics as the >science of phenomena observable in the world of computation. >>So the statement, The halting problem is not computable, does not help >>us understand the world in which we live? >Well, if I told you that I have a short, simple, and efficient program >that will tell you whether or not a program will halt, as long as the >code is less than a zillion lines long, that would not contradict the >assertion that no general algorithm for the halting problem exists. >>I won't believe you if you claim your halting problem decider is even a >>little bit shorter than a zillion lines. > I'm not sure you get the point: the assertion that no general algorithm > for the halting problem exists does not by itself tell you anything at > all about the world we live in. Yes and no. Obviously, blindly applying something proved using abstractions such as infinite memory does not work. You need to think. On the other hand, understanding and using abstractions can help us understand the world we live in, or at least the world programmers work in. It is because of ideas I learned in the abstract world of Turing machines, that I am certain you don't have a short program that can decide whether a significantly longer program halts on a given input. Patricia === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > On 4 Apr 2006 11:01:59 -0700, david petry > said: > Among those functions, there are those that are >> computable and those that are not. Well, if we define mathematics to be the science of phenomena > observable in the world of computation, then it's a little silly to > say that there exist functions that are not computable. > It would also be silly if we defined mathematics to be the science of, > say, addition. Both that definition and your personal definition throw > out large portions of what most all mathematicians themselves consider > mathematics. It is of course silly to define mathematics as the science > of addition. The question is why your personal definition is not > equally silly. > All of the mathematics that has the potential to help us understand the > world in which we live falls within the scope of mathematics as the > science of phenomena observable in the world of computation. > An advantage of the definition above -- mathematics = addition -- is > that, for all its silliness, it is quite clear. Your own is quite > opaque. For example, what is the world of computation? And what is > it to be observable in that world? Observation is typically something > connected to sense perception. > As a conceptual aid, we can think of the computer as a microscope which > helps us peer deeply into the world of computation, where the world of > computation may be defined as what we are seeing when we look through > that microscope. > Mathematics is then the science which studies the phenomena observed > when we look through that microscope. So, we can approach mathematics > the same way we approach science. We accept a reality (in this case, > the world of computation) and then *seek* a theory that explains the > phenomena we observe. When you (Menzel) ask me to tell you what the > theory is, you are asking the wrong question. Then according to Petry's conceptual aid, there could have been no mathematics prior to the invention of that metaphorical microscope. Those so dependent on computers as all that are hardly mathematicians at all. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics david petry quoted from Harvey Friedman: > An absolutely crucial issue for the foundations of mathematics is > whether there is any significant use of set theoretic methods for real > mathematics. And the answer is: of course there is! The logician Frege, whose binding of philosophy to mathematics first gave us a firmly based understanding of counting (which is as real as real mathematics can get) and of number, expressed his work entirely in the constructs of set theory: http://sowi.iwp.uni-linz.ac.at/Dialog/DT/Fregerib/frege4.html HTH xanthian. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > The logician Frege, whose binding of philosophy to mathematics > first gave us a firmly based understanding of counting [ ... ] Sure. And look how happy we are! Because counting is sooooo difficult. Han de Bruijn === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > The logician Frege, whose binding of philosophy to mathematics > first gave us a firmly based understanding of counting [ ... ] > Sure. And look how happy we are! Because counting is sooooo difficult. > Han de Bruijn Notice how people tend to denigrate what they do not understand. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Discussion, linux) > david petry quoted from Harvey Friedman: >> An absolutely crucial issue for the foundations of mathematics is >> whether there is any significant use of set theoretic methods for real >> mathematics. > And the answer is: of course there is! > The logician Frege, whose binding of philosophy to mathematics > first gave us a firmly based understanding of counting (which is as > real as real mathematics can get) and of number, expressed his > work entirely in the constructs of set theory: > http://sowi.iwp.uni-linz.ac.at/Dialog/DT/Fregerib/frege4.html You suppose Harvey Friedman doesn't know about Frege? -- Tempted and tried we're oft made to wonder Why it should be thus all the day long When there are others living about us Never molested though in the wrong. -- Bad Livers, Farther Along === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <87odzg6kj4.fsf@phiwumbda.org You suppose Harvey Friedman doesn't know about Frege? I'm just trying to answer the question the quote raises in David Petry's mind, not guess, or try to guess, what Friedman knows or should know. If Friedman doesn't mean his question in the way that Petry interprets it, that is certainly useful information to Petry in trying to shake Petry's current obsession, which seems in large part to be based on Petry's (seemingly muddled) understanding of what he reads Friedman saying. HTH xanthian. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > 8. I became entirely convinced, rather early, that it is very > unlikely that any yes/no problem stated by mathematicians, that is > not overtly set theoretic, requires any substantial use of > abstract set theory. > That sums it up. Offhand, I can't give you any better reference. > Sorry. I have worked ten years in engineering practice, and another five years in computer graphics, and never encountered an instance where set theory was required. Except when it was resorted to explicitly, like i.e. in CSG (Constructive Solid Geometry & Ray Tracing): http://glasnost.itcarlow.ie/~powerk/Graphics/Notes/node12.html Set theory in midi-music, quite useful with the processing of chords, is another example. Quote: If you believe that music is physics ... I have an application where set theory is employed with the transition of chords. With the chords Am and C, for example, only the notes in the symmetric difference Am<>C need to be changed (to become Off or On respectively). Oh well. C Maj = {C,E,G,B} , F# Maj = {F#,A#,C#,F} , G minor = {G,A#,D} Hence F# Maj C Maj = {F#,A#,C#,F} On , C Maj F# Maj = {C,E,G,B} Off. (You've defined disjoint sets.) Now wait and listen. Then: G minor F# Maj = {G,D} On , F# Maj G minor = {F#,C#,F} Off. What you see here are the common (asymmetric) set theoretic differences, denoted by . (Hope that I made no mistakes here.) But again, these are the _only_ real world examples I've encountered. Han de Bruijn === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > Fine. So the set of things that exist is countable, and the set of > things that don't exist is not countable. Makes sense. Uhm, a refinement of this statement would be that a _discrete_ set of things that doesn't exist is not countable. The continuum is not very much countable, but it nevertheless exists. Han de Bruijn === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > Fine. So the set of things that exist is countable, and the set of > things that don't exist is not countable. Makes sense. > Uhm, a refinement of this statement would be that a _discrete_ set of > things that doesn't exist is not countable. The continuum is not very > much countable, but it nevertheless exists. > Han de Bruijn How can that be, when your version of reality does no allow of such things? === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >By the way, in the Nik Weaver paper, doesn't he have infinite sets? >>Countable ones. > So do you think the distinction between countable and uncountable is > meaningful? Uhm, I think the only meaningful distinction is between continuous and discrete. Where the discrete corresponds with countable and continuous corresponds with uncountable. I mean: the continuous is so smooth that it's impossible to count anything in it ... Han de Bruijn === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <32d0a$443377e6$82a1e228$14328@news1.tudelft.nlBy the way, in the Nik Weaver paper, doesn't he have infinite sets? >>Countable ones. > So do you think the distinction between countable and uncountable is > meaningful? > Uhm, I think the only meaningful distinction is between continuous and > discrete. Where the discrete corresponds with countable and continuous > corresponds with uncountable. I mean: the continuous is so smooth that > it's impossible to count anything in it ... > Han de Bruijn Those are relative to metric spaces or topologies or to something more than just the cardinality of a set. So, if you're just saying that something is countable or uncountable on the bases that you describe, then you're just talking about a different notion from what is at stake in the question I asked David Petry. The context of that question was why he claims his philosophy is supported by Nik Weaver's proposal when that proposal has an infinite set (I can add, though I didn't mention it earlier, as I understand, the meta-theory seems to have even higher transfinite ordinals). In response, Petry points out that Weaver's set is countable. But if Petry doesn't see a meaningful distinction between countable and uncountable, then I don't see what Petry intends by pointing out the distinction in this instance nor how it bears upon the oddity of his claiming that a proposal that has infinite sets supports his philosophy. === Subject: What is Projective modeling? I want to learn what exactly is projective modeling and how does it exactly relate to banking / financial institutions? How do they use it? Can someone point me to good online resources / books or explain me in short? === Subject: (-1 * -1 = +1), imaginary numbers..... Hi - can somebody please disprove what is being said on http://cubicao.tk/stupidevil1.html mathematically? (WARNING - don't go there if you are easily offended) Some people said that: sqrt(a * b) = sqrt(a) * sqrt(b) is not true all of the time, but given what is said on that page, it says that sqrt(a * b) = sqrt(a) * sqrt(b) even when you have negative numbers for a and b. === Subject: Re: (-1 * -1 = +1), imaginary numbers..... > Hi - can somebody please disprove what is being said on > http://cubicao.tk/stupidevil1.html mathematically? (WARNING - don't go > there if you are easily offended) > Some people said that: > sqrt(a * b) = sqrt(a) * sqrt(b) is not true all of the time, but given > what is said on that page, it says that sqrt(a * b) = sqrt(a) * sqrt(b) > even when you have negative numbers for a and b. Why do you need to debunk it? The author is Gene Ray. === Subject: Re: (-1 * -1 = +1), imaginary numbers..... doesnt work, given what he is saying (i.e. imaginary numbers actually don't exist). However, the post about 1^(1/2) * 1^(1/2) = 1^(1/2+1/2) = 1^1 = 1 -1^(1/2) * -1^(1/2) = -1^(1/2+1/2) = -1^1 = -1 makes the most sense to me.....but then it makes me wonder why we can't give the square root of a negative number?? (I just really want to understand this - I don't want that website to be true...) === Subject: Re: (-1 * -1 = +1), imaginary numbers..... > doesnt work, given what he is saying (i.e. imaginary numbers actually > don't exist). However, the post about > 1^(1/2) * 1^(1/2) = 1^(1/2+1/2) = 1^1 = 1 > -1^(1/2) * -1^(1/2) = -1^(1/2+1/2) = -1^1 = -1 > makes the most sense to me.....but then it makes me wonder why we can't > give the square root of a negative number?? (I just really want to > understand this - I don't want that website to be true...) We can. I'm not going to go through this because this guy is a fruitcake and there is nothing to be gained. If you want to take the root of a negative number, use a^b = exp(b*log(a)). If you don't know how that works, you need to take a course in complex analysis or do some reading about it. === Subject: Re: (-1 * -1 = +1), imaginary numbers..... > Hi - can somebody please disprove what is being said on > http://cubicao.tk/stupidevil1.html mathematically? (WARNING - don't go > there if you are easily offended) > Some people said that: > sqrt(a * b) = sqrt(a) * sqrt(b) is not true all of the time, but given > what is said on that page, it says that sqrt(a * b) = sqrt(a) * sqrt(b) > even when you have negative numbers for a and b. Well since the sqrt sign is just a mathematician's way of saying that Pythagorus taught be how to screw up proofs, raher than Euclid, it's actually true non of the time, even for postive numbers, except in hyperbolas. === Subject: Re: (-1 * -1 = +1), imaginary numbers..... > Hi - can somebody please disprove what is being said on > http://cubicao.tk/stupidevil1.html mathematically? (WARNING - don't go > there if you are easily offended) That is a crank website. For more like it, see http://www.crank.net/timecube.html In the complex plane, the number -1 (or any other complex number except 0) has two distinct square roots. The notation sqrt(x) therefore gets ambiguous, since sqrt is not (properly speaking) a function. Not a big deal. === Subject: Re: (-1 * -1 = +1), imaginary numbers..... If you look at the image you see that he says that i^2=i*i=sqrrt(-1)*sqrrt(-1) , sqrrt(-1)*sqrrt(-1) have exponents of 1/2 which add up to only 1, so by laws of exponents you have i^2=sqrrt(-1)*sqrrt(-1)=(-1)^1=-1 not what ever crap they say. And he contradicts himself by saying that i=+/-1 when he uses the definition in his calculations that i=sqrrt(-1). You would think that if this is true maybe Gauss or Euler might pick up on this. If they did not find this *flaw* it means it is not there! It is good to make it as a rule not to question their mathematical skills, because you cannot surpass them you dumb cranks. Crank!! === Subject: Re: (-1 * -1 = +1), imaginary numbers..... a= -1 b=-1 a*b=1 sqrt(-1*-1)=1 sqrt(-1)=i however i*i=-1 === Subject: Re: (-1 * -1 = +1), imaginary numbers..... > a= -1 b=-1 a*b=1 sqrt(-1*-1)=1 sqrt(-1)=i however i*i=-1 sqrt(1)=-1 therefore sqrt(-1*-1) = sqrt(1) = -1 = i*i Hanford === Subject: Re: (-1 * -1 = +1), imaginary numbers..... <443529A4.940FF324@lmcinvestments.com> a= -1 b=-1 a*b=1 sqrt(-1*-1)=1 sqrt(-1)=i however i*i=-1 > sqrt(1)=-1 therefore sqrt(-1*-1) = sqrt(1) = -1 = i*i Yep. Since there are two square roots for every real (except 0): sqrt(-1) = +i, -i (two roots, not one) sqrt(-1) x sqrt(-1) = i x i, -i x i, i x -i, -i x -i sqrt(-1) x sqrt(-1) = -1, +1 sqrt(-1) x sqrt(-1) = +sqrt(-1 x -1), -sqrt(-1 x -1) sqrt(-1) x sqrt(-1) = +1, -1 Or in other words, (+i)(+i) = -1 (-i)(+i) = +1 (+i)(-i) = +1 (-i)(-i) = -1 so sqrt(-1) x sqrt(-1) = -1 and +1 === Subject: Re: (-1 * -1 = +1), imaginary numbers..... a= -1 b=-1 a*b=1 sqrt(-1*-1)=1 sqrt(-1)=i however i*i=-1 === Subject: Re: Superconductor amplification of quantum gravity fluctuations? You Amaze me with your silly complexity. You are the kind of people that will use society to do your dirty work. You will stand off to the side and torment a person in a hit and run fashion and let society destroy him when he goes crying to the authorities. Then you will sit back and laugh about this person so easily destroyed without using guns or poisons or bombs. You are the kind of peoples I would keep a tight leash on if I could have power and control. If I were Hitler You would be my Jew. If you want to kill people in mass you do not want to be a Charles Manson or a Timothy McVey or a Leigh Harvy Oswald you want to be EINSTEIN and invent somekind of Theory or LAW that makes a Special Weapon of mass destruction. XoXoXo EB > Note, if this superconductor slowing of speed of light phase locks to the vacuum that the superconductor occupies then note how > things should scale > Planck time > T* ~ (hG/c^5)^1/2 ~ n^5/2Tp > n is vacuum polarization index > In effect, the hypothesis here is that the superconductor amplifies n from 1 to n >> 1 in some regions of wavevector-frequency > space. > This should give LARGE MACROQUANTUM GRAVITY TIME FLUCTUATIONS! > Note that the experimenters report, if I remember correctly? > Lp*^2 = 10^20Lp^2 ~ n^3Lp^2 > Therefore, n ~ 10^20/3 ~ 10^7 > T* is then (10^20/3)^5/2 ~ 10^100/6 ~ 10^16 > So that jumps from 10^-44 sec to 10^-28 sec > Planck distance scales as n^3/2 > i.e. (10^20/3)3/2 ~ 10^10 i.e. 10^-23 cm for space fluctuations > Note that induction near EM fields are macro-quantum coherent states of off light cone virtual photons of all 3 independent > polarizations! That is why the susceptibilities are all over wavevector-frequency space. For example the static Coulomb force is > for zero frequency f and all wave vectors kx, ky, kz. > Harder than Schwinger, but it contains gems of insight. The meat is really in Vol 2. Vol 1 is very general and abstract. > Use dimensionless Planck units, i.e. spherical Schwarzschild coordinates t, r, theta, phi all dimensionless, i.e. c = 1, G = 1, t > scaled to Planck time 5x10^-45 sec, r scaled to Planck length 1.6 x 10^-33 cmm M scaled to Planck mass 2x10^-5gm > Mass of universe ~ 10^62 dimensionless in Planck units > Mass of proton ~ 8 x 10^-20 > Mass of Earth ~ 10^38 > The static spherically symmetric VACUUM METRIC is > goo = -(1 - 2M/r) > grr = (1 - 2M/r) > gtheta,theta = r^2 > gphi,phi = r^2sin^2theta > all other components are zero. > The area of a sphere at r is the Euclidean 4pir^2, and the circumference of a circle is 2pir. > What does this mean. This is what is locally measured by HOVERING LNIF detectors that are on timelike NON-GEODESICS with r = > constant, theta,t and phi,t = 0 > These guys must be firing rockets all the time. They cannot be in warp drive. > This is the convenient representation. Any other set of detectors locally coincident with these are connected by transformations > of the LOCAL Diff(4) group, which is simply GLOBAL T4 locally gauged. > x^u(P) -> x^u(P)' = x^u(x^u'(P)) > x^u(P)' are 4 functions x^u of 4 variables x^u' > i.e. overlap of two local coordinate charts {x^u(P)} & {x^u'(P)} at EVENT P > There are of course 16 Xu^u'(P) > e.g. > dx^u(P) = X^uu'(P)dx^u'(P) > Remember what GLOBAL T4 is > x^u(P) -> x^u(P)' = x^u(P) + a^u > where a^u is a global constant independent of P > Locally gauging T4 means let a^u -> a^u(P) an ARBITRARY FUNCTION OF EVENTS P. > Then make a Taylor series expansion at P and only keep first order terms so that > dx^u(P) = x^u(P)' - x^u(P) ~ X^uu'(P)dx^u'(P) > for two overlapping local coordinate charts at P. > Of course, if you consider X^uu' as a 4x4 matrix its algebra will have 16 infinitesimal generators, i.e. GL(4,R). > The physical idea here behind GCT Diff(4), however, is that we displace each point P by a locally arbitrary amount &P in contrast > to T4 where each point P is displaced by the same rigid amount over the entire manifold. > *Compare this to Hal Puthoff's completely different SSS PV metric > g00 = -e^-2M/r > grr = gtheta,theta = gphi,phi = 1/g00 > Einstein's vacuum field equation here is > Ruv = 0 > The total regularized quantum vacuum zero point energy here must be exactly zero for this solution to work. > Next define Kruskal coordinates for the maximal analytic extension of the above SSS vacuum solution. > First define > r* = r + 2Mln|(r/2M) - 1| > Now somehow some people think this is the Hilbert error? If so, I don't understand why it is an error since we are free to choose > almost any representation we like. > For 0 < r < 2M ---> 0 > r* > -infinity > 2M < r < + infinity ---> -infinity < r* < + infinity > There a 4 essential coordinate patches > I: v = e^(r*/4M)sinh(t/4M) > u = e^(r*/4M)cosh(t/4M) > II: v = -e^(r*/4M)sinh(t/4M) > u = -e^(r*/4M)cosh(t/4M) > F: v = e^(r*/4M)cosh(t/4M) > u = e^(r*/4M)sinh(t/4M) > P: v = -e^(r*/4M)sinh(t/4M) > u = -e^(r*/4M)cosh(t/4M) > These equations are formally equivalent to the equations connecting the Rindler vacuum to Minkowski 1 + 1 space-time where the > flow lines are hyperbolic, i.e. a local constant absolute acceleration for a selected class of LNIF detectors. > The extension from r* and t to u and v is like the extension from Rindler to Minkowski coordinates. p. 631 DeWitt > to be continued === Subject: Re: Open geometry problem? <058b22hrt3tia7i8sr9vbklvsi3igc7f82@4ax.com Prove or disprove: > Given a closed disk in the plane and n distinct points, n>2, in the > disk. No special restrictions are imposed on the n points (for example > -- collinearity is allowed, and boundary points are also allowed). > Then the disk can be partitioned into n closed convex > subregions of equal areas (shared boundaries are allowed but the > subregions must have pairwise non-overlapping interiors) such that > each subregion contains exactly one of the n given points. I think I can prove that this problem is true for any set of points of order 2^n in general position (ie no three colinear). Furthermore, extending to the non-colinear case looks like it should follow nicely from this. Could somebody please proof-read this sketch? First, use a famous lemma, the 2-dimensional Ham Sandwich Theorem, which says that any two shapes in the plane can be simultaneously broken into two regions of equal area by a single line. Note that when this theorem is drawn on paper, one usually draws the two shapes as being connected and disjoint, but we don't require either of these to be true in order to prove the theorem. We'll show that, given an even number of points in general position in a convex set C in the plane, we can always break C in half in such a way so that each half contains the same number of points. It follows from this reasoning that if we have 2^n points in C, we can recursively obtain quasi's/Sayaka's desired partition quoted above. Let's say we have an even number of points in the disk D. Surround each point in the disk with a ball of radius epsilon. Choose epsilon to be tiny, so that none of these little balls intersect. Define the shape E to be the union of all these balls. Then the lemma says that there exists a line that simultaneously breaks D and E in half. Now, let's examine that line. If epsilon isn't small enough, it may be that this line intersects several of the little balls. If that's the case, make epsilon smaller. Since no three points are colinear, we can eventually make epsilon small enough so that this dividing line intersects at most two of the little balls. Let's suppose that epsilon is sufficiently small. We know that the line (call it L) can intersect at most two of the little balls. Let's look at each case: 0) If L doesn't intersect any of the balls, we're fine. This means that the line L has the same number of points on either side of it, and also divides D in half. 1) It's impossible for L to intersect exactly one ball. That's because we have an even number of balls to begin with, so if L intersects one ball, it means that there is an odd number of balls strictly to one side of L, and an even number of balls strictly to the other side of L. So, one side of L has one more ball strictly within it than the other side does. There's no way that L can intersect one ball in order for us to have equal area due to E on either sides. This would be easier if I could draw a picture, but it's certainly true. 2) Finally, if L intersects two of the balls, L must have the same number of balls strictly in either side of it. This is also pretty easy to verify if you draw a picture -- the cumulative area of all the balls on one side of L must be the same as the cumulative area of all the balls on the other side of L, which can also be written as (area of balls lying strictly to the left) + (fraction area of the two balls lying not strictly to the left) = (area of balls lying strictly to the right) + (fraction area of the two balls lying not strictly to the right) since we have an even number of balls, each side of the expression above must be an integer multiple of the area of the little balls, and each (fraction area) term is bounded strictly by 0 and 2. Since each (area of balls lying strictly) term is an integer (and they're either both even or both odd), it means that each (fraction area) term has to be 1. Therefore, either L intersects both points (not balls), or it doesn't intersect either of them. In either case, L must contain the same number of points on either side, and it also bisects D, so we're done. Does that look correct and/or useful? What would be really nice is if we could show that if quasi's/Sayaka's problem is true for some number of points K, then it must also be true for K-1 points, because then the preceding argument would show that it's always true. But I could easily have made a mistake, which is why I'm asking. =) === Subject: Re: Open geometry problem? >> Prove or disprove: >> Given a closed disk in the plane and n distinct points, n>2, in the >> disk. No special restrictions are imposed on the n points (for example >> -- collinearity is allowed, and boundary points are also allowed). >> Then the disk can be partitioned into n closed convex >> subregions of equal areas (shared boundaries are allowed but the >> subregions must have pairwise non-overlapping interiors) such that >> each subregion contains exactly one of the n given points. >I think I can prove that this problem is true for any set of points of >order 2^n in general position (ie no three colinear). Furthermore, >extending to the non-colinear case looks like it should follow nicely >from this. Could somebody please proof-read this sketch? >First, use a famous lemma, the 2-dimensional Ham Sandwich Theorem, >which says that any two shapes in the plane can be simultaneously >broken into two regions of equal area by a single line. Note that when >this theorem is drawn on paper, one usually draws the two shapes as >being connected and disjoint, but we don't require either of these to >be true in order to prove the theorem. >We'll show that, given an even number of points in general position in >a convex set C in the plane, we can always break C in half in such a >way so that each half contains the same number of points. It follows >from this reasoning that if we have 2^n points in C, we can recursively >obtain quasi's/Sayaka's desired partition quoted above. >Let's say we have an even number of points in the disk D. Surround each >point in the disk with a ball of radius epsilon. Choose epsilon to be >tiny, so that none of these little balls intersect. Define the shape E >to be the union of all these balls. Then the lemma says that there >exists a line that simultaneously breaks D and E in half. >Now, let's examine that line. If epsilon isn't small enough, it may be >that this line intersects several of the little balls. If that's the >case, make epsilon smaller. Since no three points are colinear, we can >eventually make epsilon small enough so that this dividing line >intersects at most two of the little balls. >Let's suppose that epsilon is sufficiently small. We know that the >line (call it L) can intersect at most two of the little balls. Let's >look at each case: >0) If L doesn't intersect any of the balls, we're fine. This means >that the line L has the same number of points on either side of it, and >also divides D in half. >1) It's impossible for L to intersect exactly one ball. That's >because we have an even number of balls to begin with, so if L >intersects one ball, it means that there is an odd number of balls >strictly to one side of L, and an even number of balls strictly to the >other side of L. So, one side of L has one more ball strictly within >it than the other side does. There's no way that L can intersect one >ball in order for us to have equal area due to E on either sides. This >would be easier if I could draw a picture, but it's certainly true. >2) Finally, if L intersects two of the balls, L must have the same >number of balls strictly in either side of it. This is also pretty >easy to verify if you draw a picture -- the cumulative area of all the >balls on one side of L must be the same as the cumulative area of all >the balls on the other side of L, which can also be written as >(area of balls lying strictly to the left) + (fraction area of the two >balls lying not strictly to the left) = (area of balls lying strictly >to the right) + (fraction area of the two balls lying not strictly to >the right) >since we have an even number of balls, each side of the expression >above must be an integer multiple of the area of the little balls, and >each (fraction area) term is bounded strictly by 0 and 2. Since each >(area of balls lying strictly) term is an integer (and they're either >both even or both odd), it means that each (fraction area) term has to >be 1. >Therefore, either L intersects both points (not balls), or it doesn't >intersect either of them. In either case, L must contain the same >number of points on either side, and it also bisects D, so we're done. >Does that look correct and/or useful? What would be really nice is if >we could show that if quasi's/Sayaka's problem is true for some number >of points K, then it must also be true for K-1 points, because then the >preceding argument would show that it's always true. But I could >easily have made a mistake, which is why I'm asking. =) I don't have time to look at it right now, but the driving idea of your proof looks interesting. I'll take a closer look later. quasi === Subject: Re: analytic solution for this equation Robert , I do agree, convergence of numerical solutions is not so simple to define. An other point we may easily change coeff. a/4 of f(x) = (a/4)[f(x/2) + 2 f((x+a)/2) + f((x+2)/2)], by derivation or integration with g(x) = (d/dx)^ r o f(x) r real , Amicalement , Alain. === === Subject: cyclic codes I«m writing my dipolma thesis about cyclic codes (1d, 2d and md), mainly basic things like their zero sets, generator polynomials and check tensors. Especially for the case of the multidimensional cyclic codes I can`t find any proper literature. Can somebody help me out? Kerstin === Subject: Re: cyclic codes I«m writing my dipolma thesis about cyclic codes (1d, 2d and md), mainly basic things like their zero sets, generator polynomials and check tensors. Especially for the case of the multidimensional cyclic codes I can`t find any proper literature. Can somebody help me out? Kerstin there is a ton of stuff out there try The Theory of Error-Correcting Codes by Williams and Sloane QA268 M3 1978 They reference 1478 papers and books in the Bibliography track them down from there there is nothing in depth on the internet that I have found === Subject: maya & math system Hello everybody, I am looking for anybody who can help me by finding any information that is know on the mathematic system of the Maya culture. Books, writtings, formules anything is welcome. Brainy. === Subject: C^r norm Is there a standard C^r norm definition? for example in my notes i have for a C^1, norm ||f|| = sup{ f(x), f ' (x) } so im assuming the C^0 norm is just the sup norm C^2 norm is sup {f(x), f ' (x) , f '' (x)} and generalised to r derivatives for the C^r norm would this be correct/standard definition? === Subject: Re: C^r norm > Is there a standard C^r norm definition? > for example in my notes i have for a C^1, norm > ||f|| = sup{ f(x), f ' (x) } It seems your notes are incomplete: the sup should be explained more. > so im assuming the > C^0 norm is just the sup norm > C^2 norm is sup {f(x), f ' (x) , f '' (x)} > and generalised to r derivatives for the C^r norm > would this be correct/standard definition? === Subject: Re: C^r norm <060420060858578561%anniel@nym.alias.net.invalid Is there a standard C^r norm definition? > for example in my notes i have for a C^1, norm > ||f|| = sup{ f(x), f ' (x) } > It seems your notes are incomplete: the sup should be explained more. the sup is taken over all x is this a standard definition for it? > so im assuming the > C^0 norm is just the sup norm > C^2 norm is sup {f(x), f ' (x) , f '' (x)} > and generalised to r derivatives for the C^r norm > would this be correct/standard definition? > === Subject: Re: C^r norm > Is there a standard C^r norm definition? for example in my notes i have for a C^1, norm > ||f|| = sup{ f(x), f ' (x) } > It seems your notes are incomplete: the sup should be explained more. > the sup is taken over all x all x in what? and what about absolute values? > is this a standard definition for it? > so im assuming the > C^0 norm is just the sup norm > C^2 norm is sup {f(x), f ' (x) , f '' (x)} > and generalised to r derivatives for the C^r norm would this be correct/standard definition? > > === Subject: Re: A sum = ? >>I need ,if possible, to find a closed form of the sum : >>SUM_{k=1 to k=n}(-1)^k (n-k)! (n+k)! >A quick change of variables yields > n-1 > --- n-k > > (-1) k! (2n-k)! > --- > k=0 >For each k, > n-k n C(2n-k+x,2n+1) > (-1) k! (2n-k)! = (-1) (2n+1)! lim -------------- > x->0 x >Using [8] from with >m = 0, the sum of C(2n-k+x,2n+1) from k=0 to k=n-1 is >C(2n+1+x,2n+2) - C(n+1+x,2n+2). Therefore, the sum we want is > n C(2n+1+x,2n+2) C(n+1+x,2n+2) > (-1) (2n+1)! lim -------------- - ------------- > x->0 x x > n 1 n > = (-1) ---- [ (2n+1)! - (-1) (n+1)! n! ] > 2n+2 > 1 n > = ---- [ (-1) (2n+1)! - (n+1)! n! ] > 2n+2 1 2 n = - (n!) ( (-1) C(2n+1,n) - 1 ) 2 Although the answers that Robert Israel and I have posted are the same, I just thought the last form somehow seemed more palatable. Rob Johnson take out the trash before replying === Subject: divisibility conditions for binary numbers Hi All, Suppose we have a 32 bit binary number. We can tell it is divisible by looking at the rightmost one bit. If it zero , it is divisible by 2. If rightmost two bits are zeros then the number is divisible by 4 . If rightmost three bits are zeros the number is divisible by 8 and so on. Could anyone has some idea on divisibility of binary numbers by 3,5,6,7 ..... Shishir === Subject: Re: divisibility conditions for binary numbers > Hi All, > Suppose we have a 32 bit binary number. We can tell it is divisible > by looking at the rightmost one bit. If it zero , it is divisible by 2. > If rightmost two bits are zeros then the number is divisible by 4 . If > rightmost three bits are zeros the number is divisible by 8 and so on. > Could anyone has some idea on divisibility of binary numbers by 3,5,6,7 I don't think much can be said, but given a binary number x = sum_n ( k_n 2^n ) where each digit k_n is 0 or 1, let A = the number of k_n's which are 1 and n is even B = the number of k_n's which are 1 and n is odd Then x is divisible by 3 if and only if A-B is divisible by 3. E.g 1110101 (binary) is divisible by 3 since A=4 and B=1. In decimal, there are similar tricks for characterising multiples of 3, 9, or 11. LH === Subject: Re: Open sets feel that I kind of got the answers a deserved. However, it actually served it purposes. Since the book I am studying, (Topology by George McCarty), unfortunately not is the best for learning the topic by my self, it is often difficult to get an inuitive understanding of what I am reading. Most of the examples in the book so far has a metric concept, even though it also is obvious that topology as a topic is much larger. With the hints from your answers I think I got a better understanding for further study. Rereading the text with you comments in mind, I realised where the knowledge was too vague. Thx again. === Subject: Proof for the height of a Red/black tree? The hight of a red/black tree is bounded by O(lg(n)). I would now like to prove this using induction. Is it correct that if the hight of the tree is called h then the following is the inductive hypothesis: h <= 2lg(n+1) but before I can prove this I need prove that the internal nodes of a subtree is at least 2^(blackheight) -1. Else it would not be legal to assume that 2^(h/2)-1 = 2^(blackheight) -1 eventhough it is obvious from property 4 that says that each red node only has black kids. My major concern is wether h <= 2lg(n+1) is the inductive hypothesis or not. === Subject: Question of the Prime Number Theorem I'm looking for a relatively simple proof that if all the zeros of the Riemann zeta function have Rl(s) <= c < 1 then pi(x) = Li(x) + o(x^c') for any c'>c. Is there a proof of this result (assuming it to be true!) that does not involve the distribution of the zeros of zeta(s)? In other words, is there an argument based simply on the fact that zeta'(s)/zeta(s) + zeta(s) is holomorphic in Rl(s) >= c? More generally, can one deduce anything about f(x) from the abscissa of convergence of its Laplace transform F(s)? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: A modest proposal (was Calculus XOR Probability) Han de Bruijn a .8ecrit : > Han's original point is that calculus is very precise and works, > whereas the case where you have a uniform probability distribution > over an infinite set of possibilities is not well handled by the > classical notion that all individual probabilities sum to 1, because > the individual probabilities are considered equal to zero. It is the > opinion of both of us that this can be resolved, among other ways, by > assigning infinitesimal nonzero probabilities to each possibility, > leaving intact the notion that the sum of the individual > probabilities sums to 1. I am not sure why this is roundly rejected. > Can you address that? >> Because there is no model of the reals, either standard or >> nonstandard, in which the sum of countably many equal values can equal 1. > You can repeat this a thousand times, but Tony and I don't get it. In my > not so humble opinion, you must also reject then the integral(0,1) dx , > because it is derived from the Riemann sum n.1/n , which is exactly the > same as summing up (n) probabilities with 1/n chance for each. dx = 1/n. > Now take the limit for n->oo and you're done. What's the problem? (Well, > I _can_ understand that mainstream mathematics can't drop the whole wide > world of calculus because of this little issue) > Han de Bruijn Here is what comes closest to HdB ideas : take a non standard integer N (infinite, ie greater than all standard ones, of course). Define the measure of a standard set of integers A as p(A)= the sum of 1/N for all n Han's original point is that calculus is very precise and works, > whereas the case where you have a uniform probability distribution > over an infinite set of possibilities is not well handled by the > classical notion that all individual probabilities sum to 1, because > the individual probabilities are considered equal to zero. It is the > opinion of both of us that this can be resolved, among other ways, by > assigning infinitesimal nonzero probabilities to each possibility, > leaving intact the notion that the sum of the individual > probabilities sums to 1. I am not sure why this is roundly rejected. > Can you address that? >> Because there is no model of the reals, either standard or >> nonstandard, in which the sum of countably many equal values can equal 1. > You can repeat this a thousand times, but Tony and I don't get it. In my > not so humble opinion, you must also reject then the integral(0,1) dx , > because it is derived from the Riemann sum n.1/n , which is exactly the > same as summing up (n) probabilities with 1/n chance for each. dx = 1/n. > Now take the limit for n->oo and you're done. What's the problem? (Well, > I _can_ understand that mainstream mathematics can't drop the whole wide > world of calculus because of this little issue) > Han de Bruijn Then HdB clearly does not understand the definition of the Reimann integral or the properties of the field of real numbers in which it is defined. The Reimann integeral exists if one can show that if the limit of the set of upper Darboux sums and the set of lower Darboux sums for a fixed function and fixed closed real interval converge to a common limit as the size of the maximum partiion decrease towards zero, and then the limit is the integral. There is nothing is probability theory that says anything like that limit process applied to finite spaces has to carry over in any way to any infinite space. === Subject: Re: A modest proposal (was Calculus XOR Probability) > Here is what comes closest to HdB ideas : take a non standard integer > N (infinite, ie greater than all standard ones, of course). Define the > measure of a standard set of integers A as p(A)= the sum of 1/N for all > n instance, p(n is even)= 1/2 or 1/2-1/N according as N is even or odd; > and the shadow of p(A), p¡(A), is a real having almost the properties > of a probability (could it be the density :-)), while p(A) *is* (of > course) a discrete measure... and so additive. Note that with this > definition, p(n=42)=1/N and the shadow p¡(n=42)= 0, according to HdB > intuitions. Note also that the main point of contention (ie p(A), or > p¡(A)are nor probabilities (ie not countably additive)) is somewhat > shunted... Why a non standard integer? Why is large enough for our purpose not good enough? And haven't we already gone through all this with David C. Ullrich in our company? Failing upon a Transfer Principle or some such. Han de Bruijn === Subject: Re: A modest proposal (was Calculus XOR Probability) >> Here is what comes closest to HdB ideas : take a non standard integer >> N (infinite, ie greater than all standard ones, of course). Define the >> measure of a standard set of integers A as p(A)= the sum of 1/N for all >> n > instance, p(n is even)= 1/2 or 1/2-1/N according as N is even or odd; >> and the shadow of p(A), p'(A), is a real having almost the properties >> of a probability (could it be the density :-)), while p(A) *is* (of >> course) a discrete measure... and so additive. Note that with this >> definition, p(n=42)=1/N and the shadow p'(n=42)= 0, according to HdB >> intuitions. Note also that the main point of contention (ie p(A), or >> p'(A)are nor probabilities (ie not countably additive)) is somewhat >> shunted... > Why a non standard integer? Why is large enough for our purpose not > good enough? You are the one who claims that it is not good enough. If you want to talk about the probability of choosing an even integer out of the set {1, 2, 3, ... n} where n is some very large integer, then you will get no argument from anybody. However that is apparently not good enough for you, and so your definitions keep bouncing back and forth, apparently just for the sake of argument. Stephen === Subject: Re: A modest proposal (was Calculus XOR Probability) Han de Bruijn a .8ecrit : >> Here is what comes closest to HdB ideas : take a non standard >> integer N (infinite, ie greater than all standard ones, of course). >> Define the measure of a standard set of integers A as p(A)= the sum of >> 1/N for all n > integers) For instance, p(n is even)= 1/2 or 1/2-1/N according as N >> is even or odd; and the shadow of p(A), p¡(A), is a real having >> almost the properties of a probability (could it be the density >> :-)), while p(A) *is* (of course) a discrete measure... and so >> additive. Note that with this definition, p(n=42)=1/N and the shadow >> p¡(n=42)= 0, according to HdB intuitions. Note also that the main >> point of contention (ie p(A), or p¡(A)are nor probabilities (ie not >> countably additive)) is somewhat shunted... > Why a non standard integer? Why is large enough for our purpose not > good enough? If you need to ask, you will never understand. All this talk breaks down at infinity And haven't we already gone through all this with David C. > Ullrich in our company? Failing upon a Transfer Principle or some such. It only makes impossible the result you want, ie a countably additive measure. Except you dont want it... > Han de Bruijn === Subject: Re: A modest proposal (was Calculus XOR Probability) <729c8$44325afd$82a1e228$15581@news1.tudelft.nl> <647ae$44337700$82a1e228$14245@news1.tudelft.nl> <3bc2e$4434eee8$82a1e228$31458@news1.tudelft.nl> <4435180e$0$1148$7a628cd7@news.club-internet.fr Here is what comes closest to HdB ideas : take a non standard integer > N (infinite, ie greater than all standard ones, of course). Define > the measure of a standard set of integers A as p(A)= the sum of 1/N for > all n For instance, p(n is even)= 1/2 or 1/2-1/N according as N is even or > odd; and the shadow of p(A), p¡(A), is a real having almost the > properties of a probability (could it be the density :-)), while p(A) > *is* (of course) a discrete measure... and so additive. Note that with > this definition, p(n=42)=1/N and the shadow p¡(n=42)= 0, according to > HdB intuitions. Note also that the main point of contention (ie p(A), > or p¡(A)are nor probabilities (ie not countably additive)) is somewhat > shunted... We've already been round this one once (or possibly more than once). But I guess that somebody else making another attempt can't hurt. Well, it can't hurt me, and that's my principal concern :-) === Subject: Re: Calculus XOR Probability > There is no problem. Nobody disputes that > lim_{n rightarrow infty} (n. 1/n) = 1 > But nobody except you (and Tony?) thinks that you can meaningfully > claim that > lim_{n rightarrow infty} (n.1/n) = > (lim_{n rightarrow infty} n).(lim_{n rightarrow infty} 1/n) > The result lim(a_n b_n) = (lim a_n)(lim b_n) is only guaranteed to > be true if a_n and b_n both converge. Since this condition is not > satisfied in the above example, there is no foundation for your > argument. I'm not aware of the fact that I've ever claimed this nonsense. And that the above would be my argument. It would be helpful if you think of Max Planck's work. The infinitely small quantities there are _not_ in the limit as well, until they are summed up: integrated. Or until they result in finite differential quotients. Why not treat infinitely small probabilities in the same manner? As I have proposed repeatedly. (This is not the same as to deprive infinitesimals from their meaning, as _you_ have tried repeatedly) Aah, you say it can not be done? Then The Theory of Heat Radiation could not have been done as well. But it _has_ been done. Han de Bruijn === Subject: Re: Calculus XOR Probability <729c8$44325afd$82a1e228$15581@news1.tudelft.nl> <647ae$44337700$82a1e228$14245@news1.tudelft.nl> <3bc2e$4434eee8$82a1e228$31458@news1.tudelft.nl> lim_{n rightarrow infty} (n. 1/n) = 1 > But nobody except you (and Tony?) thinks that you can meaningfully > claim that > lim_{n rightarrow infty} (n.1/n) = > (lim_{n rightarrow infty} n).(lim_{n rightarrow infty} 1/n) > The result lim(a_n b_n) = (lim a_n)(lim b_n) is only guaranteed to > be true if a_n and b_n both converge. Since this condition is not > satisfied in the above example, there is no foundation for your > argument. > I'm not aware of the fact that I've ever claimed this nonsense. Yes, that's probably the root of the problem. === Subject: Re: Calculus XOR Probability >> I'm not aware of the fact that I've ever claimed this nonsense. > Yes, that's probably the root of the problem. Cheap argument, huh? Han de Bruijn === Subject: Re: Calculus XOR Probability <729c8$44325afd$82a1e228$15581@news1.tudelft.nl> <647ae$44337700$82a1e228$14245@news1.tudelft.nl> <3bc2e$4434eee8$82a1e228$31458@news1.tudelft.nl> I'm not aware of the fact that I've ever claimed this nonsense. > Yes, that's probably the root of the problem. > Cheap argument, huh? Anything free is worth what you paid for it... === Subject: Re: Calculus XOR Probability > Your infinitesimals are only used under integral signs. See if I can find a counter-example. As promised to Robert Low as well. Han de Bruijn === Subject: Re: Calculus XOR Probability <442172af$0$19435$6d36acad@titian.nntpserver.com> <729c8$44325afd$82a1e228$15581@news1.tudelft.nl> <647ae$44337700$82a1e228$14245@news1.tudelft.nl> <37dd5$4433af32$82a1e228$17862@news2.tudelft.nl Your infinitesimals are only used under integral signs. > See if I can find a counter-example. As promised to Robert Low as well. > Han de Bruijn How about the counterexample from Counterexamples in Real Analysis that there is an infinitesimal in the reals? That's mathematician usage. Also, you might have noticed before dx and dy being moved around quite aptly and so forth in terms of the Leibniz notation, of those infinitesimals. The limit of the, sum, is either an infinitely small value summed over infinitely many or the wrong anwer, in the geometric interpretation of well-known geometric identities of for example area, or volume, for almost all functions. Consider any function that is not a straight line in R^2, any finite sum is only an approximation and never exact, of the area under said curve. It is only in infinite induction that the sum of those differential width sections is correct, not for any, only for all. If dx isn't an infinitesimal, and S not an infinite sum, the integral is rigorously not correct, because there are counterexamples to those being finite, as described above. In terms of via a restricted transfer principle infinite induction and nonstandard reals that are contiguous on the real number line, the reals are well-orderable, so, where the reals are integral iota-multiples, the reals are well-orderable, else not. There are only and everywhere reals between zero and one, inclusive. Ross F. === Subject: Need Solution method for nonlinear system Please go thro the state space rep. of nonlinear system in the code below eqn(1). ---------------------------------------------------------------------------- -------------------------------------------------------------------- R = sqrt(x(1).^2+x(2).^2); V = sqrt(x(3).^2+x(4).^2); beta = -0.59783*exp(x(5)); R0 = 6374;%Km H0 = 13.406;%Km D = -beta*exp(((R0-R)/H0)*V); G = - (3.986*10e5) / R^3; v1 = randn; v2 = randn; %State space of nonlinear system -- eqn(1) dx1 = x(3); dx2 = x(4); dx3 = D*x(3)+G*x(1)+v1; dx4 = D*x(4)+G*x(2)+v2; dx5 = randn; %solution method :Runge-Kutta 4th order %slope1 k1=[dx1;dx2;dx3;dx4;dx5]; x_half_h = x + 0.5*[dx1*h;dx2*h;dx3*h;dx4*h;dx5*h]; %slope2 k2 = [x_half_h(3);x_half_h(4);-beta*exp(((R0-sqrt(x_half_h(1)^2+x_half_h(2)^2))/H 0)*V)*x_half_h(3)+1*(- (3.986*10e5) / (sqrt(x_half_h(1)^2+x_half_h(2)^2))^3)*x_half_h(1)+randn;... -beta*exp(((R0-sqrt(x_half_h(1)^2+x_half_h(2)^2))/H0)*V)*x_half_h(4)+ 1*(- (3.986*10e5) / (sqrt(x_half_h(1)^2+x_half_h(2)^2))^3)*x_half_h(2)+randn;randn]; x_half_h = x + 0.5*[k2(1)*h;k2(2)*h;k2(3)*h;k2(4)*h;k2(5)*h]; %slope3 k3 = [x_half_h(3);x_half_h(4);-beta*exp(((R0-sqrt(x_half_h(1)^2+x_half_h(2)^2))/H 0)*V)*x_half_h(3)+1*(- (3.986*10e5) / (sqrt(x_half_h(1)^2+x_half_h(2)^2))^3)*x_half_h(1)+randn;... -beta*exp(((R0-sqrt(x_half_h(1)^2+x_half_h(2)^2))/H0)*V)*x_half_h(4)+ 1*(- (3.986*10e5) / (sqrt(x_half_h(1)^2+x_half_h(2)^2))^3)*x_half_h(2)+randn;randn]; x_half_h = x + 0.5*[k3(1)*h;k3(2)*h;k3(3)*h;k3(4)*h;k3(5)*h]; %slope4 k4 = [x_half_h(3);x_half_h(4);-beta*exp(((R0-sqrt(x_half_h(1)^2+x_half_h(2)^2))/H 0)*V)*x_half_h(3)+1*(- (3.986*10e5) / (sqrt(x_half_h(1)^2+x_half_h(2)^2))^3)*x_half_h(1)+randn;... -beta*exp(((R0-sqrt(x_half_h(1)^2+x_half_h(2)^2))/H0)*V)*x_half_h(4)+ 1*(- (3.986*10e5) / (sqrt(x_half_h(1)^2+x_half_h(2)^2))^3)*x_half_h(2)+randn;randn]; %Next state x=x+(1/6)*h*(k1+2*k2+2*k3+k4); ---------------------------------------------------------------------------- --------------------------------------------------------------- The above matlab code I implemented to find the solution of non linear system in the time predictive form. I used Runge-Kutta 4th order method. But solution jumps randomly to infinity or NaN for each run of the program. Could any one help me to find the solution for the - Nandy === Subject: matrix representation Hi there, We find it natural to represent a linear mapping by a matrix. Let V and W be linear spaces over a field F. Let T be a linear mapping from V to W, notation T in L(V,W). Write m = dim V, n = dim W. Mat(a x b, F) is the set of a x b matrices with elements in F. I was wondering whether - after we choose a basis for V and W - the mapping M : Lin(V,W) --> Mat(n x m, F) ; M(T) = mat(T) is a bijective mapping (mat(T) stands for the standard matrix representation of T). This means that every linear mapping has (after choice of bases) a unique matrix representation, and every matrix corresponds with one linear mapping. It sounds quite logical to me. If it is indeed true, one could define a matrix in that way. But what then is the difference between a linear mapping and a matrix? (Usually one would say that a linear mapping is just a mapping and a matrix is an array of scalars. But if we define a matrix more precisely on the way described above, how is this difference declared precisely?) === Subject: Re: matrix representation days. My association with the Department is that of an alumnus. >Hi there, >We find it natural to represent a linear mapping by a matrix. Let V and W be >linear spaces over a field F. Let T be a linear mapping from V to W, >notation T in L(V,W). Write m = dim V, n = dim W. Mat(a x b, F) is the set >of a x b matrices with elements in F. >I was wondering whether - after we choose a basis for V and W - the mapping >M : Lin(V,W) --> Mat(n x m, F) ; M(T) = mat(T) >is a bijective mapping (mat(T) stands for the standard matrix representation >of T). This means that every linear mapping has (after choice of bases) a >unique matrix representation, and every matrix corresponds with one linear >mapping. Yes. It is a bijective map. More than that, it is a vector space isomorphism under the usual addition and scalar multiplication. >It sounds quite logical to me. If it is indeed true, one could define a >matrix in that way. Which matrix are you hoping to define? >But what then is the difference between a linear mapping and a matrix? For starters, linear mappings are defined for any vector space, not just for finite dimensional ones; matrices tend to be finite. >(Usually one would say that a linear mapping is just a mapping and a matrix >is an array of scalars. But if we define a matrix more precisely on the way >described above, how is this difference declared precisely?) They are isomorphic as vector spaces. But for example, if n=m but V is different from W, then you can multiply matrices, but you cannot compose the linear maps. So matrices will have other properties. Also, the isomorphism is not canonical: change the basis, it changes the isomorphism. So different matrices end up corresponding to the same map. The linear transformation does NOT depend on the choice of bases, but the matrix representation does. -- === Subject: Bases for closure of a finite field? For finite fields, every element can be written as a finite power of a generator, and there are typically two different bases used: - the polynomial basis, where elements are written as sums of consecutive powers of a generator - the normal basis, where elements are written as sums of (prime-power) powers of an element For example, take polynomials over F_2 mod x^4-x-1. Then x is a generator and we can write every element y in the polynomial basis as y = sum_{i=0}^3 y_i x^i where y_i is in F_2. Let z=x^3. Then we can also write y in the normal basis as y = sum_{j=1}^4 y_j z^{2^j} Let a_n = lcm(1,...,n). Then F_{p^{a_n}} contains a copy of F_{p^{a_m}} for each m less than n. The limit (there are actually several, but they're all isomorphic as fields) is the algebraic closure of F_p. I haven't been able to find any information about how to write down specific elements of this field. It looks like there's an infinitesimal generator x, and every element of the field can be written as a rational root of unity times a finite power of x. Is that right? Are there bases corresponding to the polynomial or normal basis for the closure? If not, what kind of bases are there? === Subject: What is the intuitive meaning of Goedel's undecidable statement in arithmetic? About three months ago, I posted, How can the meaning of Goedel's unprovable statement descend from infinity? In that post, I tried to show that Goedel's undecidable statement has no intuitive meaning because it is a statement of the form: The following is unprovable: The following is unprovable: The following is unprovable: ... I failed to respect the fact that the arithmoquine (I will define this in a minute) of any symbolic property is a meaningful statement about numbers. Although I clearly see the correctness of the proof, I have worked obsessively since then to understand why our current theory of arithmetic falls short of exhausting all true statements. Obviously, although we know the undecidable statement is true, we *don't have enough rules in logic to prove it*. It must therefore be some unique style of reasoning that allows us to say that a theorem that states the unprovability of its own symbolic representation is necessarily true. But what is the nature of this style of reasoning? What is the extra thing we do to show the truth of Goedel's undecidable statement? Have any efforts been made to expand the technique of mathematical proof to include such arguments? Here is my work so far. First, I will give a proof of the undecidability of a statement. I will copy some definitions from my previous post and add some new ones. The symbols of arithmetic (+, -, =, logic symbols, etc.) may be assigned certain numbers, so that statements in arithmetic may be coded into a Goedel number by making the assigned numbers exponents of the primes, in the order they occur. We can define what it means for a number to be the Goedel number of a proof: we make the Goedel numbers of sequentially derived statements exponents of the primes, in the order they are deduced. It is a proof if each deduction is valid, and this can be checked mechanically. Let n(Statement) be the Goedel number of the statement written between the quotes. Let N(number) be the symbolic representation of a number, called the *numeral* of the number, to be used in a written statement. For example, N(328) would become in the Goedel number of its statement, through the identifier n, three primes with exponents representing the symbols 3, 2, and 8. Let Pf(x) be the property that there is a symbolic proof of the Goedel number x. Thus, Pf(n(N(2) + N(2) = N(4))). Notice the use of n and N in the statement. Let NPf(x) be the negation of the property Pf(x). Thus, NPf(x) is the property that there is no proof of Goedel number x. Let Aq(x) be a function on Goedel numbers with free variables -- which may be rightly called properties -- to Goedel numbers of statements. Aq(x) is the Goedel number of the statement that results when the statement of x has replaced for its free variables the numeral of the value of x. Thus, the Aq(n(x is prime)) = n(N(n(x is prime)) is prime). Again, notice the use of n and N. Theorem: There is a statement in arithmetic that is neither provable nor disprovable. Proof: Consider the property NPf(Aq(x)). It is true when there is no proof of the arithmoquine of x, which may be a Goedel number with a free variable. Now consider the *arithmoquine* of the property, Aq(n(NPf(Aq(x)))). We see that Aq(n(NPf(Aq(x)))) = n(NPf(Aq(N(n(NPf(Aq(x))))))). Thus, the arithmoquine of NPf(Aq(x)) is actually equal to the Goedel number of a statement about the unprovability of the arithmoquine of NPf(Aq(x)). In the right-hand expression, we see that a statement is stating the unprovability of an expression for its own Goedel number. If it were provable, the expression would be of an unprovable statement, and -- because of equivalence in symbolic structure of proof -- the statement itself would be, a contradiction. If it were *false*, the expression would be provable, and again, by equivalence in the symbolic structure of proof, the statement itself would be provable, and thus true, another contradiction. Therefore the statement is *true* but *unprovable*. I have kept notes on this theorem and drawn diagrams showing the sameness relations between symbols in the two expressions above: for example, I have drawn lines between the Aq's of the first statement above and the Aq's of the second. There seem to be two possible associations between equal symbols: mathematical association and self-referential association. The second Aq in the first statement is mathematically associated with the first Aq in the second statement, but *self-referentially* associated with the second Aq in the second statement. What I mentioned above, the equivalence in symbolic structure of proof, the idea that if I can prove a statement, I can find an arithmetical proof of its corresponding Goedel number, and vice versa, is also very important. How do all these ideas function together in the trick of the proof? What is the philosophical lesson of such a trick? Like I said, I clearly see the proof's correctness, but I do not fully comprehend its meaning. Even Wittgenstein called the proof a logical conjuring trick. === Subject: Re: What is the intuitive meaning of Goedel's undecidable statement in arithmetic? > Although I clearly see the correctness of the proof, I have worked > obsessively since then to understand why our current theory of > arithmetic falls short of exhausting all true statements. The current theory of Arithmentic is incapable to demonstrate or to contradict all the possible statements on numbers, because there are algorithms that can produce chaotic sequences that shows statistical properties that, naturally, cannot be explained with the aid of the standard set of axioms. But that properties are normally .presented as conjectures. By example : The Modified Eratosthenes Sieve produces a sequence known as Lucky Numbers. It has the statistical property that any even number can be decomposed as the sum of two Luckies. This statement is the product of a fortunate combination of events that cannot be handled with the standard set of logical or arithmetical axioms. === Subject: FWD: Pay attention this one is moving by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id k36FGM302636 for ; Thu, 6 Apr 2006 11:16:22 -0400 by support2.mathforum.org (8.12.11.20060308/The Math Forum, $Revision: 1.6 secondary) with ESMTP id k36FEQHg017343 for ; Thu, 6 Apr 2006 11:14:28 -0400 There is a Big PR Campaign running ALL Weekend! Get KKPT First Thing, This Is Going To Explode! Check out HOT NEWS!!! KOKO PETROLEUM (KKPT) - This is our #1 pick for next week! Our last pick gained $2.16 in 4 days of trading. Current Price: $1.10 5 day expected price $2.70 This has been a great producer in our last PR Campaign VANCOUVER, British Columbia--(BUSINESS WIRE)--March 17, 2006--KOKO Petroleum, Inc. (PINK SHEETS:KKPT - News) has now completed two wells. The wells have now been equipped for production. The first two days resulted in the recovery of water from the formation and the oil cut (percentage of oil to water) has now increased substantially. The water should be eliminated within a few days. Production tanks and separator are being installed KOKO Petroleum, Inc. (PKKPT - News) announced today that it has revised its Letter of Intent with JMT Resources, Ltd. Fort Worth Texas regarding the Development of JMT's polymer flood pilot program on its leasehold located in Corsicana, Texas. === Subject: any free C/C++ library to solve n-order linear ode Hi everyone. Are there any free C/C++ library to solve n-order linear ode given in a common form with its coefficients? I know that gsl (GNU Scientific Library) can solve systems of first-order ode's, but I can't find some algorithm that can automatically convert an n-order linear ode in common form to a system of ode's. Could anyone tells me about such librarry or suggest some algorithm? === Subject: Re: any free C/C++ library to solve n-order linear ode > Hi everyone. > Are there any free C/C++ library to solve n-order linear ode given in a > common form with its coefficients? I know that gsl (GNU Scientific > Library) can solve systems of first-order ode's, but I can't find some > algorithm that can automatically convert an n-order linear ode in > common form to a system of ode's. Could anyone tells me about such > librarry or suggest some algorithm? Converting an nth order ode to a system of 1st order ode's is so straightforward, that in effect, the gls program you mention is also going to be an out of the box method to solve nth order ode's. === Subject: Simple bijection N <-> Q Here is an interesting sequence I discovered the other day: S(0) = 0. S(2n) = n+1, for n > 0. S(2n+1) = 1/S(2n). Or, equivalently: S(0) = 0. S(n even) = n/2 + 1, for n > 0. S(n odd) = 1/S(n-1). The interesting thing about this sequence is that it maps the naturals to the rationals. In other words, it is a bijection S(n in N) <-> p/q in Q, from all the naturals n in N to all the rationals p/q in Q. It is similar to Cantor's diagonal mapping of the rationals, but is simpler because it maps each natural to a unique rational p/q, where p and q are relatively prime, i.e., there are no duplicate rationals (such as 2/4, 3/6, 4/8, etc.) produced by the mapping. Cantor's denumeration of the rationals proceeds in an orderly diagonal fashion, as shown in this image: http://david.tribble.com/img/MapNtoQ1.png Sequence S(n) proceeds in a less obvious fashion, as shown here: http://david.tribble.com/img/MapNtoQ.png The progression is symmetric about the p=q axis, because each odd term S(2n+1) is the inverse of the preceding even term S(2n). This image shows the progression of only the odd terms: http://david.tribble.com/img/MapNtoQ2.png The inverse mapping is: invS(0) = 0. invS(1) = invS(p/p) = 1. invS(p) = invS(p/1) = 2^(p-1). invS(p/q) = 2 invS(p/q - 1), for p > q. invS(p/q) = 1 + invS(q/p), for p < q. The inverse function handles rationals in non-reduced form (e.g., 2/4, 12/18, 15/25, etc.), mapping them to their equivalent reduced form (p/q, where p and q are relatively prime). This works because evaluating invS(pk/qk) winds up with a term of invS(k/k) in the recursive evaluation, which is simply 1. The inverse function makes it apparent that every rational of the form p/1 (i.e., every natural) is mapped to each natural that is a power of 2, i.e., p = S(2^(p-1)). The first 100 terms of S(n): 0. 0 20. 7/3 40. 10/3 60. 13/5 80. 13/3 1. 1/1=1 21. 3/7 41. 3/10 61. 5/13 81. 3/13 2. 2/1=2 22. 7/4 42. 10/7 62. 13/8 82. 13/10 3. 1/2 23. 4/7 43. 7/10 63. 8/13 83. 10/13 4. 3/1=3 24. 7/2 44. 11/4 64. 7/1=7 84. 17/7 5. 1/3 25. 2/7 45. 4/11 65. 1/7 85. 7/17 6. 3/2 26. 7/5 46. 11/7 66. 7/6 86. 17/10 7. 2/3 27. 5/7 47. 7/11 67. 6/7 87. 10/17 8. 4/1=4 28. 8/3 48. 9/2 68. 11/5 88. 15/4 9. 1/4 29. 3/8 49. 2/9 69. 5/11 89. 4/15 10. 4/3 30. 8/5 50. 9/7 70. 11/6 90. 15/11 11. 3/4 31. 5/8 51. 7/9 71. 6/11 91. 11/15 12. 5/2 32. 6/1=6 52. 12/5 72. 13/4 92. 18/7 13. 2/5 33. 1/6 53. 5/12 73. 4/13 93. 7/18 14. 5/3 34. 6/5 54. 12/7 74. 13/9 94. 18/11 15. 3/5 35. 5/6 55. 7/12 75. 9/13 95. 11/18 16. 5/1=5 36. 9/4 56. 11/3 76. 14/5 96. 11/2 17. 1/5 37. 4/9 57. 3/11 77. 5/14 97. 2/11 18. 5/4 38. 9/5 58. 11/8 78. 14/9 98. 11/9 19. 4/5 39. 5/9 59. 8/11 79. 9/14 99. 9/11 Though I discovered this little sequence independently, I was most certainly not the first to find it, since it's such a simple sequence. A brief Google search turned up these links, showing that Kepler played with a similar sequence, treating it as a binary tree and attempting to relate it to the cosmic harmonies: http://ndirty.cute.fi/~karttu/Kepler/a086592.htm http://www.research.att.com/~njas/sequences/?q=id:A020651 http://www.research.att.com/~njas/sequences/?q=id:A020650 http://www.research.att.com/~njas/sequences/?q=id:A086592 http://david.tribble.com/src/java/tribble/math/misc/MapNtoQ.java -drt === Subject: Re: Simple bijection N <-> Q > Here is an interesting sequence I discovered the other day: > S(0) = 0. > S(2n) = n+1, for n > 0. > S(2n+1) = 1/S(2n). > Or, equivalently: > S(0) = 0. > S(n even) = n/2 + 1, for n > 0. > S(n odd) = 1/S(n-1). > The interesting thing about this sequence is that it maps the > naturals to the rationals. In other words, it is a bijection > S(n in N) <-> p/q in Q, from all the naturals n in N to all the > rationals p/q in Q. Unless I am being remarkably dense today, I do not see that your mappings produce any of the negative rationals. === Subject: Re: Simple bijection N <-> Q Here is an interesting sequence I discovered the other day: >> S(0) = 0. >> S(2n) = n+1, for n > 0. >> S(2n+1) = 1/S(2n). >> Or, equivalently: >> S(0) = 0. >> S(n even) = n/2 + 1, for n > 0. >> S(n odd) = 1/S(n-1). >> The interesting thing about this sequence is that it maps the >> naturals to the rationals. In other words, it is a bijection >> S(n in N) <-> p/q in Q, from all the naturals n in N to all the >> rationals p/q in Q. > Unless I am being remarkably dense today, I do not see that your > mappings produce any of the negative rationals. Perish the thought. Mea culpa. Like I said a couple of posts ago, S(n in N) maps to p/q, where p and q are both positive and relatively prime. But we could probably change the sequence to cover all the rationals including the negative ones: S(0) = 0. S(1) = 1. S(3n) = S(n)+1. S(3n+1) = 1/S(3n). S(3n+2) = -S(3n). This is just off the top of my head, so I'm not sure if it's correct. If not, then maybe this will do: S(0) = 0. S(1) = 1. S(4n) = S(n)+1. S(4n+1) = -S(n)-1. S(4n+2) = 1/S(4n). S(4n+3) = 1/S(4n+1). === Subject: Re: Simple bijection N <-> Q > S(0) = 0. > S(1) = 1. > S(3n) = S(n)+1. > S(3n+1) = 1/S(3n). > S(3n+2) = -S(3n). > This is just off the top of my head, so I'm not sure if it's correct. > If not, then maybe this will do: > S(0) = 0. > S(1) = 1. > S(4n) = S(n)+1. > S(4n+1) = -S(n)-1. > S(4n+2) = 1/S(4n). > S(4n+3) = 1/S(4n+1). I was not suggesting that there are not lots of bijections from the naturals to all rationals, including the negatives. An a very simple one would be useful, though not essential. === Subject: Re: Simple bijection N <-> Q days. My association with the Department is that of an alumnus. >Here is an interesting sequence I discovered the other day: > S(0) = 0. > S(2n) = n+1, for n > 0. > S(2n+1) = 1/S(2n). >Or, equivalently: > S(0) = 0. > S(n even) = n/2 + 1, for n > 0. > S(n odd) = 1/S(n-1). This is not well defined for n=1. I suspect you meant to add S(1)=1. >The interesting thing about this sequence is that it maps the >naturals to the rationals. Yes. >In other words, it is a bijection >S(n in N) <-> p/q in Q, from all the naturals n in N to all the >rationals p/q in Q. No, it is not a bijection. For starters, the image of the map does not include any negative rational numbers. So let us assume you meant positive rationals. Still not true. As far as I can see, the map send natural numbers to either natural numbers, or egyptian fractions (fractions with numerator equal to 1). I don't see which n maps to something like 2/3, 3/4, or 4/5, not to mention all those other rationals which are greater than 1 but not integers. So how can it be a bijection between natural numbers and positive rationals? [.snip.] >The first 100 terms of S(n): > 0. 0 > 1. 1/1=1 > 2. 2/1=2 > 3. 1/2 > 4. 3/1=3 > 5. 1/3 > 6. 3/2 Hmmm... This is not the function you describe above. According to the description given above, S(6) should be (6/2)+1 = 3+1 = 4, not 3/2. -- === Subject: Re: Simple bijection N <-> Q Here is an interesting sequence I discovered the other day: >> S(0) = 0. >> S(2n) = n+1, for n > 0. >> S(2n+1) = 1/S(2n). >>Or, equivalently: >> S(0) = 0. >> S(n even) = n/2 + 1, for n > 0. >> S(n odd) = 1/S(n-1). > This is not well defined for n=1. > I suspect you meant to add S(1)=1. Ack. I tried to simplify it by adding the for n>0 in the definition of S(2n), but perhaps I should have just said: 1. S(0) = 0. 2. S(1) = 1. 3. S(2n) = S(n)+1. 4. S(2n+1) = 1/S(2n). Note the correction in rule 3. >> The interesting thing about this sequence is that it maps the >> naturals to the rationals. >> In other words, it is a bijection >> S(n in N) <-> p/q in Q, from all the naturals n in N to all the >> rationals p/q in Q. > No, it is not a bijection. For starters, the image of the map does > not include any negative rational numbers. So let us assume you meant > positive rationals. Yep. S(n in N) maps to p/q, where p and q are both positive and relatively prime. >>The first 100 terms of S(n): >> 0. 0 >> 1. 1/1=1 >> 2. 2/1=2 >> 3. 1/2 >> 4. 3/1=3 >> 5. 1/3 >> 6. 3/2 > Hmmm... This is not the function you describe above. According to the > description given above, S(6) should be (6/2)+1 = 3+1 = 4, not 3/2. When corrected as above, it is. Apologies. === Subject: Re: Simple bijection N <-> Q > Ack. I tried to simplify it by adding the for n>0 in the definition > of S(2n), > but perhaps I should have just said: > 1. S(0) = 0. > 2. S(1) = 1. > 3. S(2n) = S(n)+1. > 4. S(2n+1) = 1/S(2n). > Note the correction in rule 3. This is the continued fraction method: Start with 1, and use the two operations add 1 and reciprocal repeatedly in any order. Then you get all positive rationals as values. On the other hand, this is the binary method: start with 0 and use the two operations double and double then add 1 repeatedly in any order. Then you get all nonnegative natural numbers. Your correspondence comes from comparing these two. === Subject: What about equation : f(m(x ,y) , x ) = f(x ,y) ? Voil.88 , f(m(x ,y) ,x ) = f(x ,y) (1) m and f:R^2 -> R continous for x and y . In a first step , I examine pairs (f,m) satisfying (1) such as 1¡ f(x, y) = constant 2¡ m(x,y) =y and f(x, y) = f(y, x) 3¡ f(x ,y) = g(x +(1 -d)*y) ;m(x ,y) =d*x +(1 -d)*y ... Here is an other exemple: f(x ,y) = g( 1/x +y^2) ; m(x ,y) = x/(1 - x^3 +x*y^2) . What must be the link between m(x ,y) and f(x ,y) and can we build more general solutions ? Friendly , Alain . === Subject: CW - complex question, Construct a cell complex X as follows : Start with one 0-cell *, attach a 1-cell a to it to make a circle. Then attach two 2-cells C and D, C by a map of degree 2, and D by a map of degree 3. Let / denote union. Consider the subcomplex * / a / C (i.e. consisting of *, a, and C altogether). Isn't X / A homotopy equivalent to S^2? If you replaced C by B you'd get S^2 as well? Also, if I consider the subcomplex * / a, then X / A is homotopy equivalent to the one point union of S^2 with S^2. Am I right? James === Subject: Re: CW - complex question, > Construct a cell complex X as follows : > Start with one 0-cell *, attach a 1-cell a to it to make a circle. > Then attach two 2-cells C and D, C by a map of degree 2, and D by a > map of degree 3. > Let / denote union. Consider the subcomplex * / a / C (i.e. > consisting of *, a, and C altogether). Isn't X / A homotopy > equivalent to S^2? If you replaced C by B you'd get S^2 as well? I'll assume you meant to write X = * / a / C, and by X / A, you meant the quotient space X/a, obtained by collapsing */a to a point . In that case, yes, it's the same as the quotient of a 2-cell by its boundary, and so is actually homeomorphic to S^2. As far as replacing C by B, you haven't told us what B is. If you meant D (i.e., take the attaching map to have degree 3 rather than 2), the result after collapsing */a to a point is again the quotient of the 2-cell by its boundary. > Also, if I consider the subcomplex * / a, then X / A is homotopy > equivalent to the one point union of S^2 with S^2. What is X? Are you now speaking of the full complex X = * / a / C / D, where C and D are 2-cells attached to a as described above? > Am I right? If you meant that above thing as X, and if you meant A to be the subcomplex */a, then yes, the quotient is homeomorphic to the one-point union S^2 v S^2. Dale. === Subject: Re: CW - complex question, >> Construct a cell complex X as follows : >> Start with one 0-cell *, attach a 1-cell a to it to make a circle. >> Then attach two 2-cells C and D, C by a map of degree 2, and D by a >> map of degree 3. >> Let / denote union. Consider the subcomplex * / a / C (i.e. >> consisting of *, a, and C altogether). Isn't X / A homotopy >> equivalent to S^2? If you replaced C by B you'd get S^2 as well? > I'll assume you meant to write X = * / a / C, and by X / A, > you meant the quotient space X/a, obtained by collapsing */a to > a point . In that case, yes, it's the same as the quotient of > a 2-cell by its boundary, and so is actually homeomorphic to S^2. Sorry I was very sloppy. X = * / a / C / D, and A = * / a / C. Then X / A = S^2? If you replace C by D in A, then X / A = S^2 as well. Below, X is the same. Then let A now be * / a. Then X / A = one point union of S^2 with S^2. > As far as replacing C by B, you haven't told us what B is. If you > meant D (i.e., take the attaching map to have degree 3 rather than > 2), the result after collapsing */a to a point is again the quotient > of the 2-cell by its boundary. >> Also, if I consider the subcomplex * / a, then X / A is homotopy >> equivalent to the one point union of S^2 with S^2. > What is X? Are you now speaking of the full complex > X = * / a / C / D, > where C and D are 2-cells attached to a as described above? >> Am I right? > If you meant that above thing as X, and if you meant A to be the > subcomplex */a, then yes, the quotient is homeomorphic to the > one-point union S^2 v S^2. > Dale. === Subject: Re: Exchange order of integration and summation Are you sure it's called Fubini's theorem? I looked it up and that establishes a connection between a multiple integral and a repeated one, doesn't it? I have another, similar problem. Could you help me with that too, please? What if not all terms in the sum are positive, for example, when we have the expansion of exp(-alpha*x) instead of that of the Bessel function, with alpha and x real and alpha>0? (The unexpanded function exp(-x^2) was just an example. I actually have the function (x^beta)*exp(-x^2), with beta>0, but that shouldn't matter.) === Subject: Re: Exchange order of integration and summation <24397710.1144342694756.JavaMail.jakarta@nitrogen.mathforum.org Are you sure it's called Fubini's theorem? I looked it up and that establishes a connection between a multiple integral and a repeated one, doesn't it? A sum is a particular case of an integral (with respect to counting measure). Or if you want you can say sum_{j=1}^infinity f(j) = int_1^infinity f(floor(x)) dx. > I have another, similar problem. Could you help me with that too, please? > What if not all terms in the sum are positive, for example, when we have the expansion of exp(-alpha*x) instead of that of the Bessel function, with alpha and x real and alpha>0? The condition for Fubini's theorem to work is that the double integral of the absolute value of the integrand is finite. === Subject: Re: Exchange order of integration and summation I have a similar problem, but with an integral which goes from 0 to Infinity: Integral of x=0 to Inf {exp(-x^2) * sum of k=0 to Inf [f(k,x)]} dx. The sum is a Taylor expansion of a modified zeroth order Besselfunction of the first kind (I0(x)), which uniformly converges for all xI have a similar problem, but with an integral which goes from 0 to Infinity: >Integral of x=0 to Inf {exp(-x^2) * sum of k=0 to Inf [f(k,x)]} dx. >The sum is a Taylor expansion of a modified zeroth order Besselfunction >of the first kind (I0(x)), which uniformly converges for all xthis correct?). If the upper limit of the integral were finite, I could >change the order of integration and summation, but it is not. >Can I still change the order, given that the upper limit of the integral is Inf? If this is the expansion I_0(x) = 1 + x^2/4 + x^4/64 + ... all the terms are nonnegative (when x is real), and you also have the bound I_0(x) < exp(x), so the integral of the sum converges absolutely. Fubini's Theorem says you can interchange the sum and integral. The result, by the way, is (according to Maple) sqrt(pi) exp(1/8) I_0(1/8)/2. === Subject: Re: Logarithm of transfinite numbers Umm, that's not the way I see it. I'm not sure what the problem is here. I >> don't think a bijection signifies equal size for infinite sets, first of all. > Do you, Tony, think it necessary to define what you mean by equal > size? Or do you merely prefer not to use the expression at all? David R Tribble said: >> And yet you have never provided us with a bijection between two >> same-sized infinite sets, such as the set of naturals and the set of >> even naturals, that omits elements from either set. > Do you, David, think it necessary to define what you mean by equal > size? You seem to be using it here to mean being bijectable - why > not use being bijectable for this? I understand that bijection equates to equal set size, but Tony does not. I avoid using the accepted mathematical terms like bijection when responding to Tony because such terms usually cause him to roll his eyes and start blathering about how the bijection is a flawed concept and does not really mean anything. So I use the less accurate term size, which I assume Tony has some understanding of. Not the same understanding as you and me, but still, using it reduces his tangents. === Subject: Re: Logarithm of transfinite numbers Umm, that's not the way I see it. I'm not sure what the problem is here. I >> don't think a bijection signifies equal size for infinite sets, first of all. > Do you, Tony, think it necessary to define what you mean by equal > size? Or do you merely prefer not to use the expression at all? > David R Tribble said: >> And yet you have never provided us with a bijection between two >> same-sized infinite sets, such as the set of naturals and the set of >> even naturals, that omits elements from either set. > Do you, David, think it necessary to define what you mean by equal > size? You seem to be using it here to mean being bijectable - why > not use being bijectable for this? > I understand that bijection equates to equal set size, but Tony > does not. Well, I grant you may have a rather clearer notion of what it means to define a term, but equal set size only equates to bijection if you define it to, since there is no pretheoretic meaning to size of an infinite set. That's been a major part of the argument - someone points out that bijectability is an equivalence relation, and that therefore it is consistent to refer to a bijectability equivalence class as being a size. Tony retorts that the size of a set should obviously be, um, well, you know, how many members the set has. Until Tony grasps the idea of defining terms, it's just hopeless using a term like 'size'. (He obviously doesn't grasp the idea of a definition now: look at the struggle to persuade him that if I say By pofnat I mean .... [bit about left ends], when I say pofnat this is what I mean.) > I avoid using the accepted mathematical terms like bijection when > responding to Tony because such terms usually cause him to roll his > eyes and start blathering about how the bijection is a flawed concept > and does not really mean anything. So I use the less accurate term > size, which I assume Tony has some understanding of. Not the > same understanding as you and me, but still, using it reduces his > tangents. This seems to be totally backwards. You agree to use the word Tony does, because he uses it to mean something different from what you mean??? Anyway, I suppose this will run for a myriad posts yet. Brian Chandler http://imaginatorium.org === Subject: Re: Logarithm of transfinite numbers David R Tribble said: >> Umm, that's not the way I see it. I'm not sure what the problem is here. I >> don't think a bijection signifies equal size for infinite sets, first of all. > Do you, Tony, think it necessary to define what you mean by equal > size? Or do you merely prefer not to use the expression at all? > David R Tribble said: >> And yet you have never provided us with a bijection between two >> same-sized infinite sets, such as the set of naturals and the set of >> even naturals, that omits elements from either set. > Do you, David, think it necessary to define what you mean by equal > size? You seem to be using it here to mean being bijectable - why > not use being bijectable for this? > I understand that bijection equates to equal set size, but Tony > does not. > I avoid using the accepted mathematical terms like bijection when > responding to Tony because such terms usually cause him to roll his > eyes and start blathering about how the bijection is a flawed concept > and does not really mean anything. So I use the less accurate term > size, which I assume Tony has some understanding of. Not the > same understanding as you and me, but still, using it reduces his > tangents. As I said to Brian, the issue is whether bijection equates to equal set size for infinite sets. The standard theory holds that it does whereas, rather than simply blathering about how the bijection is a flawed concept, I have specifically stated that the mapping function must be taken into account quantitatively, which for sets mapped from the naturals using monotonically increasing functions, is handled perfectly well using the Inverse Square Rule. I can always come up with tangents, David. You know that. That's the beauty of it all, when it all fits together in one fabric. -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers > David R Tribble said: >> Umm, that's not the way I see it. I'm not sure what the problem is here. >> I >> don't think a bijection signifies equal size for infinite sets, first of >> all. > Do you, Tony, think it necessary to define what you mean by equal > size? Or do you merely prefer not to use the expression at all? > David R Tribble said: >> And yet you have never provided us with a bijection between two >> same-sized infinite sets, such as the set of naturals and the set of >> even naturals, that omits elements from either set. > Do you, David, think it necessary to define what you mean by equal > size? You seem to be using it here to mean being bijectable - why > not use being bijectable for this? I understand that bijection equates to equal set size, but Tony > does not. I avoid using the accepted mathematical terms like bijection when > responding to Tony because such terms usually cause him to roll his > eyes and start blathering about how the bijection is a flawed concept > and does not really mean anything. So I use the less accurate term > size, which I assume Tony has some understanding of. Not the > same understanding as you and me, but still, using it reduces his > tangents. > As I said to Brian, the issue is whether bijection equates to equal set > size > for infinite sets. The standard theory holds that it does whereas, rather > than > simply blathering about how the bijection is a flawed concept, I have > specifically stated that the mapping function must be taken into account > quantitatively But TO's alleged mapping function is fictional as is his quantitative measuring as neither is compatible with any known set theory, whereas cardinality is well established and verified in many set theories. === Subject: Re: Logarithm of transfinite numbers for infinite sets. The standard theory holds that it does In set theory, 'cardinality of' is defined with respect to bijection. As long as one adheres to the rules of proper definition, nothing precludes one from adding additional definitions to capture other aspects of sets. If the axioms don't provide for the expression of certain aspects, then you need different axioms. One of the things your commentary lacks is not just a formalization but even mention of the set theory axioms as to why they lack the expressiveness you desire; that is to say, you haven't even checked out the set theory axioms to see whether, in fact, they cannot support the additional definitions that would capture your ideas. Moreover, you've not shown any need for the aspects you wish to capture. Lately, the idea of halving an infinite set or performing division on infinite sets has been mentioned. If there can be such a thing, then you need to show how to do define it, but even then, it would help if you made the case for the need for such a thing. Set theory is formal, rigorous, lucid, and easy to understand. The case has not been made that the ability to define a division operation on infinite sets (or whatever other features you might offer) is so desirable that your own much more complicated system (even at the level of primitives and axioms) should be preferred even as we can only speculate that you will produce such a system. === Subject: Re: Logarithm of transfinite numbers imaginatorium@despammed.com said: >> Umm, that's not the way I see it. I'm not sure what the problem is here. I >> don't think a bijection signifies equal size for infinite sets, first of all. > Do you, Tony, think it necessary to define what you mean by equal > size? Or do you merely prefer not to use the expression at all? Sure. I would say that the sizes of two infinite sets are equal if described by equivalent formulas on Big'un, the unit infinity. > David R Tribble said: >> And yet you have never provided us with a bijection between two >> same-sized infinite sets, such as the set of naturals and the set of >> even naturals, that omits elements from either set. > Do you, David, think it necessary to define what you mean by equal > size? You seem to be using it here to mean being bijectable - why > not use being bijectable for this? >> ... This would be >> sufficient to prove your belief. >> You came close once, when I asked which even numbers in N were not >> included in the bijection >> f(n) = 2n, for all n in N >> You claimed that there were finite values of n in N that did not >> correspond to finite values of 2n in N, but you never explained what >> those were. I realize this is related to your belief that N is a >> finite set even though it has no maximum element. David, indeed we did discuss this, but you don't seem to have ever gotten what > I actually did say. Do you remember the discussion of value ranges? You know, > the one that always came down to there being no largest finite and therefore no > value range to the set, even though it has lots of values that vary over a wide > range? Given any definite value range, we have half as many evens as naturals, > Unless I'm much mistaken, this means For any bounded (finite) > contiguous set, we have... Yes... > ... and unless the number line is twice as long for evens as for all naturals, this > must also be the case over then entire infinite value range of the real line. > Ah, must also be the case over the entire infinite value range. > Well, normal mathematics does not have T-infinite value ranges. Why > should this property (HEN, for Half as many Evens as Naturals) apply > to the infinite value range? Why should the ratio of evens to naturals over the entire range be any different from the ratio that holds for all contiguous subsets thereof? If every segment of the number line obeys this ratio, then where do the extra half of the evens come from over the real line? They come from the doubled range that any set of evens has compared to its mapped set of naturals, that's where. This is elementary. > We know that you have also said (of the > property BSB for Bound on Set of Bits) that: > Yes, the set of bits is bounded for every specific pofnat, but there > is no bound on the length of the string required for all pofnats. > (Tony Orlow) > It rather seems that for property HEN you claim the infinite case > just follows, but for BSB you agree it doesn't. Why should that be? Is > your axiomatic basis going to include a sort of encyclopedia of > propositions with oracular declarations of whether each applies to the > infinite case? If you can put boundedness in terms of a formula, then we can talk about in context of the inverse function rule and quantify the set. If not, then it doesn't apply, does it? > The fact that I cannot name a finite value that doesn't have a finite double is > pretty irrelevant to the fact that half the naturals are even and all the evens > are natural (restricting to positive values). The Inverse Function Rule quite > intuitively and correctly leads to the conclusion that there are half as many > evens as naturals in any finite interval (give or take 1), and half as many > overall on the real line. So, the no largest finite argument elicits but a > shrug. Sorry. > No need to apologise. I can't be bothered investigating the Inverse > Function Rule very closely, but we went all through this before. For > the set of evens as a subset of the naturals the limiting density (or > some similar expression) is well-defined, and captures the intuitive > notion that every other natural is an even. Why don't you bother to > read what mathematicians a lot cleverer than you have already done to > investigate this stuff - then you could use terminology that other > people could understand. I HAVE discussed this in terms of set density in other threads, some time ago. The Inverse Function Rule is a direct outgrowth and generalization of that concept, as I explained when I introduced it months ago. If you can't be bothered to even try out the rule on a few finite cases and a few infinite cases, or even to remember what its derivation is so you don't accuse me of ignoring the very notions that led to it, then why should I try to use terminology that others understand because they already know it, when that terminology doesn't describe the rule I'm suggesting? If I use infinite in some nonstandard way, everyone clamors that I cannot subvert a term already in use, but now you want me to use limiting density, when that's only a special case of the Inverse Function Rule, which I explicitly stated? Gimme a break. The Premesin's in the medicine cabinet. Help yourself. Ha! > You've become quite good at evading the question. > The bijection shows us, quite obviously, that there are equal numbers > of naturals and evens, that there is an even for every natural, and a > natural for every even, hence they are equally numerous. > No. It doesn't do this, unless you have (see question above) carefully > defined what you mean by equally numerous. If you are talking to an > intelligent person, of course they will see that you are referring to > set cardinality. But if you are talking to Tony, sloppiness like this > will just prolong the misery. I'm sorry you're so miserable, Brian. My offer still stands regarding the Premesin. I hate to see you suffer so. This discussion right now is ultimately about what is the best measure of size for infinite sets. David's trapped in standard mode, and I can't cross that threshold, so he and I are bound to be at odds over this. It's understood that he's referring to cardinality, and that he's defending it as a logical generalization of the notion of set size for infinite sets, while I am saying it's not sufficient for my tastes. There's no misunderstanding in that respect. > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers > imaginatorium@despammed.com said: > Do you, Tony, think it necessary to define what you mean by equal > size? Or do you merely prefer not to use the expression at all? > Sure. I would say that the sizes of two infinite sets are equal if described > by > equivalent formulas on Big'un, the unit infinity. Defining something in terms of other undefined things doesn't define anything. > Ah, must also be the case over the entire infinite value range. > Well, normal mathematics does not have T-infinite value ranges. Why > should this property (HEN, for Half as many Evens as Naturals) apply > to the infinite value range? > Why should the ratio of evens to naturals over the entire range be any > different from the ratio that holds for all contiguous subsets thereof? The entire range implies a condition contrary to fact. > We know that you have also said (of the > property BSB for Bound on Set of Bits) that: Yes, the set of bits is bounded for every specific pofnat, but there > is no bound on the length of the string required for all pofnats. > (Tony Orlow) It rather seems that for property HEN you claim the infinite case > just follows, but for BSB you agree it doesn't. Why should that be? Is > your axiomatic basis going to include a sort of encyclopedia of > propositions with oracular declarations of whether each applies to the > infinite case? > If you can put boundedness in terms of a formula, then we can talk about in > context of the inverse function rule and quantify the set. If not, then it > doesn't apply, does it? Since there is no inverse function rule in any standard system, and TO has no system yet, it does NOT apply. > This discussion right now is ultimately about what is the best > measure of size for infinite sets. David's trapped in standard > mode, and I can't cross that threshold, so he and I are bound to be > at odds over this. It's understood that he's referring to > cardinality, and that he's defending it as a logical generalization > of the notion of set size for infinite sets, while I am saying it's > not sufficient for my tastes. There's no misunderstanding in that > respect. There is also no mistaking that TO has no better method to offer (of measuring set size) since (1) TO's method is NOT valid in any current set theory, and (2) TO does NOT have any new set theory in which his method is valid. === Subject: Re: Logarithm of transfinite numbers Virgil said: > Virgil said: Virgil said: > , > < snipposaurus rex > Inverse Function Rule of Bigulosity Theory If the nth element in an inductively defined quantitative set Absent any definition of what are or are not quantitative > sets, the definition fails at this point. Sets of real numbers. So what are inductively defined sets of real numbers in TOmatics? > The reals are not inductively defined in any system I know of. You have heard, perhaps, of Orlow's amazing H-Riffic Number System? > I have heard of unicorns and leprechauns, too. Have you ever seen one? Besides, Virgular, is the set of square roots of naturals a set of > naturals, rationals or irrational reals? > None of the above. Some of the square roots of naturals are naturals and > some are irrationals. Both of which are reals. I got your point. > is given by f(n), and f(g(n))=g(f(n))=n for all n in N, > According to the above, f and g must be bijections of the set > N to itself, i.e., permutations of N. Uh, no. f maps naturals to some set of reals (which may all be > naturals) and g maps that set back to the set of naturals. Not according to TO's reqirement that f(g(n))=g(f(n))=n for all n > in N. That requires that f and g have naturals as both arguments > and as values, so that both must be functions from N to N. Oh, you're right. I should say for n in R, even though we are talking > about mapping the set of naturals to the reals using f(n), since Iam > also talking about g(n). Sorry about that. > TO is clearly unable to sort out even his own definitions without help, > so his competence to sort out an entire new set theory is, at best, > questinable. Well, math is like that. How many proofs are presented as complete by professionals, only to have some flaw discovered in the vetting? I'm not even a professional, nor even a tiny bit undistracted, so, I accept such correction then the number of elements e, such that x floor(g(y)-g(x)+1). Since neither f or g, at least as so far described, need > preserve any order relation on N, this makes no sense at all. Actually that is an astute observation, so it ust have been a > mistake, ;) I thought about it this weekend, and reminded myself > that I forgot to include the stipulation that the sets be > monotonically increasing. A set cannot be monotonically increasing in any mathematical > sense. Well, I mean a set defined using a monotonically increasing mapping > function, obviously. > Then must all your f's and g's be order isomorphisms? Yes, the rearrangement of order in the mapping throws monkey wrenches into the measurement of sets. Some of those situations seem easily solvable, but I am sure there will be mappings that make exact quantification difficult or impossible. In any case, this aspect of the theory covers many of the most egregious counterintuitive results, concerning things as simple as the evens vs. the naturals, and as deep as the formulaic relationship between the reals in the unit interval and the naturals, and provides for a full spectrum of ordered infinities. So, it's a good start. If you are measuring a set over the entire positive domain What sort of sets are being measured? And how is positive > domain defined with respect to that set? > Sets of real numbers (not ALL of them), mapped from the naturals, > in quantitative order, over the entire domain of the reals, that > is, over the real number line. Does TO mean the image of an order preserving mappping from the > naturals to the reals? Yes, that sounds like the correct wording. Let me repeat it three > times in my head....okay I think I got it, the image of an then we use x=1 and y=N, As most standard set theories prohibit a set being a member of > itself, what sort of theory is TO using in which he can have y > a member of N and simultaneoulsy equal to N? Remember, N is the length of the real line, the count of the > hypernaturals, and of the reals in the unit interval. Let me call > it BigOne from now on, to avoid confusion. Then is TO now distinguishing the set of naturals from his whatever > it is bigone beast? Well, yes, it keeps causing confusion. N is what you call the set of > naturals which are all finite but my set of naturals includes > infinite values, and I've been calling the size of it N. > Don't. > There's > enough confusion, so the unit infinity is BigOne, > What is this BigOne supposed to measure? > It cannot be the length of the reals since there is no such thing. > Shouldn't it be spelled begone? Actually, I decided last night the unit infinity is Big'un and the unit infinitesimal is Lil'un. That gives it a nice twang. Big'un's the length of the entire real line measured in units, and so is also the number of unit intervals and natural numbers on the real line, as well as the number of Lil'un-length unique infinitesimal real intervals within the unit interval. That is the basis for Bigulosity on the real line. > and the unit > infinitesimal is LilOne. > What is this bigone supposed to measure? Big'un measures the Universe of Quantity, the Real Number Line. :D ZFC may have no Universe, but Bigulosity's got Big'un. Well Order the Line! ;-) > The size of a set is its Bigulosity > How does one compare theBigulosity of two sets? For infinite sets one compares the formulaic expressions defining the size of each in terms of Big'un, and if one is greater than the other for all n greater than some identifiable value, then one can say that that order relation is preserved in the infinite case, and precisely say that, for instance, Big'un> Big'un/2, and there are twice as many naturals as even naturals. > With cardinalities there is a nice clear method which gives a provably > consistent ordering of sizes for well ordered sets, and for arbitrary > sets given an axiom of choice. > But absent rules for comparison , and proof of a consistent ordering of > Bigulosities, the mere word means nothing. Please critique my wording above. > , and we > have regular numbers, big numbers, and lil numbers. > I repeat, absent rules for comparison , and proof of a consistent > ordering of Bigulosities, the mere word means nothing. Sure, but I have put forth the statement above recently, and don't recall you're ever having responded to it. DO you find anything unreasonable about this use of formulaic comparison? > Yeah, well, I'm asking how you think boundedness tells you > something about the set size. Like, what does it equal? In serially ordered sets, where any non-extreme member has > exactly one predecessor ans exactly one successor, bounded sets are > finite and unbounded one are infinite, at least in mathematics. Well, yeah > Every inequality is based on a difference, right? Wrong! Order inequalities, like x < y or x < = y, are based on > ORDER RELATIONS on the set in which the objects occur. If there were no difference between x and y, then x false and x<=y would only be true because x=y. TO has it backwards, if neither x < y nor y < x , THEN x = y. The > order comes first before there can be a direction of inequality. > Absent order, one can never say x < y or x > y, only x =/= y. Right, so you're asking again about the order relation, and I'll > reiterate that's a good point, which I think needs to be addressed by > the construction of the number line starting with 0 and 1, and which > I need to consider more in trying to form this axiomatic basis. > To anyone competent to build a consistent system, these elementary > issues would be obvious enough not to need pointing out by others. I told you I was already thinking about that, and since I am trying, as suggested, to build a system from the ground up to integrate all these ideas, the axiomatic foundation has to be laid just right to avoid subsequent issues. Excuse me while I contemplate the nature of truth itself and the quantitative foundation of logic, as well as the derivation of quantity from geometric truth, so as to achieve this goal. Okely Dokes? I know that's what you think. Does TO claim that a sum of endlesly many positive naturals has a > finite sum? If endlessly many includes the notion of every finite value being > in a finite position in this series, then yes, I don't consider there > to be an infinite number of finite positions, and thus don't see the > sum achieving infinity. > TO is deliberately conflating the mathematical meaning versus his > improper non-mathematical meanings of 'finite' and 'infinite' here. Of course, when asked whether I consider this infinite, I am going to respond with how it appears to me. You all already know what you think, and I do too, so if you're quizzing me to make sure I know what you're theory says, don't bother. According to your theory, the sum of all 1's in finite positions is infinite, since there are an infinite number of finite positions. According to my definition of infinite, the set of all finite naturals doesn't qualify, and neither does the string of 1's in all finite positions. Either you're asking my opinion, or you're quizzing me on yours, and I'm not responding to the latter. > Does TO claim that the sum of an endless series of positive naturals has > any Dedekind finite natural as a sum, or even as an upper bound? Unboundedly large but finite, as I've said dozens of times. > Divergent series, such as these, never achieve any kind of value. That > is the point of divergence. In fact, it's quite intuitive to think about a completed infinity like Big'un, and consider the divergent sum up to such an infinite number. A while back, three of us separately gave equivalent infinite expressions for the sum of all natural numbers, N(N+1)/2. That wasn't coincidence. It was the next step. So get your boots on, as the 's likely to get deep. :) > what is the point here? I'm missing it. > > That infinitely many non-zero naturals can add up to more tha > any finite value. If there ARE infinitely many, which is at issue here. There are endlessly many, which makes a sum of endlessly many > positive naturals. if TO claims this is a finite sum let him state > its value. Because a person can't name the largest finite value, this is > obviously impossible, now isn't it, as well as hopelessly irrelevant. > It is exactly the point at issue. Any hopelessness, is entirely in TO's > position, and any irrelevance ie entirely in TO's misdefinitins of > 'finite' and 'infinite' to try and make them mean what no one else in > the world accepts. The infinite string of Big'un 1's to the left of the binary point represents 2 ^Big'un-1, but Big'un isn't the number of finite naturals. It's an actually infinite number, a unit on the next quantitative level, not an amorphous boundary between finite and infinite like aleph_0. Since it has been explained so clearly by so many, TO's > intrasigent position can only be a result of invincible > ignorance or deliberate trolling. Or, clear vision of an alternative perspective. Anything knowingly in violation of all the systems extant, which > TO's maunderings are, is either invincible ignorance or deliberate > trolling. > Or a sincere effort to correct a misguided system. > The present systems, and there are of them, work fine. If TO thinks he > can do better, let him present his alternate system up front rather than > reiterating his unfounded and unsupportable claims. I have presented plenty of aspects of it, and reiterated them many time for your enjoyment and edification, apparently to little or no avail. When I complete the roots of the proper axiomatic system, you'll be the first to find out. In the mean time, perhaps you can comment on the notion of comparing formulaic expression on infinite variables in terms of the value of one being greater than the value of the other for all finite values greater than some particular value. but you've changed all that Virgil. Because of you, and you > alone, I must vanquish this theory. TO may try, but will fail as all such fools fail. > For Virgil guards the Gate > The gate hardly need guarding from the efforts of fools like TO. > But it is rather fun to poke holes in the fabric of his illusions. Well, that's true. There are dozens of ways to enter the garden, besides the gate, like through the gaping holes in the wall. ;-) > You have given new meaning to my life, to take meaning away from > yours. It will provide considerable amusement to those who have not yet > killfiled all your threads to watch TO try. But in the matter of mathematical creativity, TO is a eunuch, he > would, but he can't. > Oh ouch, I must get some mathemtical viagra to be as numerically > manly as Virgil, or he will kick sand in my face and walk away with > my beautiful theory. > To enable such a eunuch would require a fatal dose of viagra. > fertile > And TO need not worry about others walking away with what TO does not > even have. Ah, now you see my evil plan! Foiled again by the wiley likes of the Great Virgil, drat you! -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers > Virgil said: > You have heard, perhaps, of Orlow's amazing H-Riffic Number System? I have heard of unicorns and leprechauns, too. > Have you ever seen one? I have never seen Orlow's amazing H-Riffic Number System either. > There's > enough confusion, so the unit infinity is BigOne, What is this BigOne supposed to measure? > It cannot be the length of the reals since there is no such thing. > Shouldn't it be spelled begone? Actually, I decided last night the unit infinity is Big'un and the unit > infinitesimal is Lil'un. That gives it a nice twang. Big'un's the length of > the > entire real line measured in units What sort of units does TO propose which are capable of measuring the length of the entire real line in any system currently extant? > and so is also the number of unit intervals and natural numbers on > the real line, First TO must show that all these things have measurements, which requires a definition of 'measurement', secondly that there is a unit of measure suitable for all 3 measurements, thirdly that they all are equal. TO so far has done none of these things. > Big'un measures the Universe of Quantity What is the Universe of Quantity? > The size of a set is its Bigulosity How does one compare theBigulosity of two sets? > For infinite sets one compares the formulaic expressions defining the > size of each in terms of Big'un, and if one is greater than the other > for all n greater than some identifiable value, then one can say that > that order relation is preserved in the infinite case, and precisely > say that, for instance, Big'un> Big'un/2, and there are twice as many > naturals as even naturals. Replacing one undefined expression by another does not advance things. How does one determine these formulaic expressions for arbitrary sets? Cardinality has precise explanations of how to compare set sizes. Bigulosity has gobbledegook. > > With cardinalities there is a nice clear method which gives a provably > consistent ordering of sizes for well ordered sets, and for arbitrary > sets given an axiom of choice. But absent rules for comparison , and proof of a consistent ordering of > Bigulosities, the mere word means nothing. Please critique my wording above. Done above! , and we > have regular numbers, big numbers, and lil numbers. I repeat, absent rules for comparison , and proof of a consistent > ordering of Bigulosities, the mere word means nothing. > Sure, but I have put forth the statement above recently, and don't recall > you're ever having responded to it. DO you find anything unreasonable about > this use of formulaic comparison? Until the rules for constructing these formulaic comparisons are clearly set out, yes! Handwaving doesn't do it. > Right, so you're asking again about the order relation, and I'll > reiterate that's a good point, which I think needs to be addressed by > the construction of the number line starting with 0 and 1, and which > I need to consider more in trying to form this axiomatic basis. To anyone competent to build a consistent system, these elementary > issues would be obvious enough not to need pointing out by others. I told you I was already thinking about that, and since I am trying, as > suggested, to build a system from the ground up to integrate all these ideas, > the axiomatic foundation has to be laid just right to avoid subsequent > issues. > Excuse me while I contemplate the nature of truth itself and the quantitative > foundation of logic, as well as the derivation of quantity from geometric > truth, so as to achieve this goal. Okely Dokes? Only if you do it off line until you have it all done. > Does TO claim that the sum of an endless series of positive naturals has > any Dedekind finite natural as a sum, or even as an upper bound? > Unboundedly large but finite, as I've said dozens of times. And TO has been wrong each time. Divergent series, such as these, never achieve any kind of value. That > is the point of divergence. > In fact, it's quite intuitive to think about a completed infinity like > Big'un, Intuition is bad mathematics unless it can be backed up by a logically coherent system. So far, TO only has intuition with no logic and no coherence and no system. > The present systems, and there are of them, work fine. If TO thinks he > can do better, let him present his alternate system up front rather than > reiterating his unfounded and unsupportable claims. I have presented plenty of aspects of it TO wants to build only the top story to his structure without any foundation. That leaves everything entirely conjectural. Go off and work on that foundation for a while, TO, and don't bother trying to prop up the attic until you have a basement to stick under it. > For Virgil guards the Gate > > The gate hardly need guarding from the efforts of fools like TO. > But it is rather fun to poke holes in the fabric of his illusions. > Well, that's true. There are dozens of ways to enter the garden, besides the > gate, like through the gaping holes in the wall. ;-) The gaping holes are all in TO's conjectures. If the walls of standard mathematics had such holes as TO alleges, on present evidence, TO would be too dim to find them. === Subject: Re: Logarithm of transfinite numbers Virgil said: > Virgil said: > You have heard, perhaps, of Orlow's amazing H-Riffic Number System? I have heard of unicorns and leprechauns, too. Have you ever seen one? > I have never seen Orlow's amazing H-Riffic Number System either. 1. 0 is a real number 2. If x is a real number, then 2^x and 2^-x are real numbers. You have seen it, and now you see it again, and you've seen other versions of it as well. > There's > enough confusion, so the unit infinity is BigOne, What is this BigOne supposed to measure? > It cannot be the length of the reals since there is no such thing. > Shouldn't it be spelled begone? > Actually, I decided last night the unit infinity is Big'un and the unit > infinitesimal is Lil'un. That gives it a nice twang. Big'un's the length of > the > entire real line measured in units > What sort of units does TO propose which are capable of measuring the > length of the entire real line in any system currently extant? None is any widely accepted system, but in mine there exists the infinite unit Big'un, which IS the length of the real line. > and so is also the number of unit intervals and natural numbers on > the real line, > First TO must show that all these things have measurements, which > requires a definition of 'measurement', secondly that there is a unit of > measure suitable for all 3 measurements, thirdly that they all are equal. > TO so far has done none of these things. Actually, I can axiomatically state these things and treat Big'un as a primitive, and if I can derive no contradictions, and can derive useful results, you have nothing to complain about. Big'un measures the Universe of Quantity > What is the Universe of Quantity? The Real Number Line, as my previous sentence went on to say. Why snip the answer to your question and then ask it, instead of just reading the answer like a normal person? The size of a set is its Bigulosity How does one compare theBigulosity of two sets? For infinite sets one compares the formulaic expressions defining the > size of each in terms of Big'un, and if one is greater than the other > for all n greater than some identifiable value, then one can say that > that order relation is preserved in the infinite case, and precisely > say that, for instance, Big'un> Big'un/2, and there are twice as many > naturals as even naturals. > Replacing one undefined expression by another does not advance things. Oh. Define 0. > How does one determine these formulaic expressions for arbitrary sets? Using the mapping function from the naturals, or N=S^L, depending on whether your infinite set is quantitative or symbolic. > Cardinality has precise explanations of how to compare set sizes. > Bigulosity has gobbledegook. Precise? Only in the sense that you always get SOME answer, but not in the sense that it is able to distinguish sizes for many kinds of sets. Leave that to Bigulosity. Your theory can't even talk about the KIND of a number of bits required for your ill-defined set of naturals. That's not precise in my book. It's schlock. :) > With cardinalities there is a nice clear method which gives a provably > consistent ordering of sizes for well ordered sets, and for arbitrary > sets given an axiom of choice. But absent rules for comparison , and proof of a consistent ordering of > Bigulosities, the mere word means nothing. > Please critique my wording above. > Done above! Replacing one undefined expression by another does not advance things. That just sounds like it went over your head, as usual. Do you disagree that one can consider a formulaic expression to be larger than another for infinite variables if it is larger than the other for all finites greater than some value? Yes or no? > , and we > have regular numbers, big numbers, and lil numbers. I repeat, absent rules for comparison , and proof of a consistent > ordering of Bigulosities, the mere word means nothing. Sure, but I have put forth the statement above recently, and don't recall > you're ever having responded to it. DO you find anything unreasonable about > this use of formulaic comparison? > Until the rules for constructing these formulaic comparisons are clearly > set out, yes! Handwaving doesn't do it. Well, I have been discussing the use of IFR lately, and went over a number of examples using N=S^L some months ago. So, if I'm waving my hands, maybe it's just to keep the flies out of my face that seem to congregate in this garden. > Right, so you're asking again about the order relation, and I'll > reiterate that's a good point, which I think needs to be addressed by > the construction of the number line starting with 0 and 1, and which > I need to consider more in trying to form this axiomatic basis. To anyone competent to build a consistent system, these elementary > issues would be obvious enough not to need pointing out by others. > I told you I was already thinking about that, and since I am trying, as > suggested, to build a system from the ground up to integrate all these ideas, > the axiomatic foundation has to be laid just right to avoid subsequent > issues. > Excuse me while I contemplate the nature of truth itself and the quantitative > foundation of logic, as well as the derivation of quantity from geometric > truth, so as to achieve this goal. Okely Dokes? > Only if you do it off line until you have it all done. Come on, Virgil. If I disappeared, you know you'd miss me. I give your life meaning, like a squirrel to a dog, like a mouse to a cat, like a Triple McFatBurger to a disgusting piece of white trash. You love it! Geometry->Quantity->Logic. Thoughts from the deep? > Does TO claim that the sum of an endless series of positive naturals has > any Dedekind finite natural as a sum, or even as an upper bound? Unboundedly large but finite, as I've said dozens of times. > And TO has been wrong each time. WRONG!! :D > Divergent series, such as these, never achieve any kind of value. That > is the point of divergence. In fact, it's quite intuitive to think about a completed infinity like > Big'un, > Intuition is bad mathematics unless it can be backed up by a logically > coherent system. So far, TO only has intuition with no logic and no > coherence and no system. Would for your sake that that were so, and understandable that you see it as such, for the King's most practiced archer dismisses the Zen archer's aim as blind luck, and without consequence. The discipline that underlies the aim of faith is not obvious to those wearing the uniform of the ranks. But, hey, that goes without saying, of course! :D > The present systems, and there are of them, work fine. If TO thinks he > can do better, let him present his alternate system up front rather than > reiterating his unfounded and unsupportable claims. > I have presented plenty of aspects of it > TO wants to build only the top story to his structure without any > foundation. That leaves everything entirely conjectural. Yes, well, building from the roof down has its drawbacks, but painting from the floor up makes no more sense. > Go off and work on that foundation for a while, TO, and don't bother > trying to prop up the attic until you have a basement to stick under it. Do you pour a foundation before you know what shape the house will take, or what load the foundation must support? Do you assemble a car, starting with the wheels? Is a baby born feet first? Well, you probably were. > For Virgil guards the Gate > > The gate hardly need guarding from the efforts of fools like TO. > But it is rather fun to poke holes in the fabric of his illusions. Well, that's true. There are dozens of ways to enter the garden, besides the > gate, like through the gaping holes in the wall. ;-) > The gaping holes are all in TO's conjectures. If the walls of standard > mathematics had such holes as TO alleges, on present evidence, TO would > be too dim to find them. Yes, of course, you are right as always, O Wise and Terrible Virgil. -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers Virgil said: >> I have never seen Orlow's amazing H-Riffic Number System either. > 1. 0 is a real number > 2. If x is a real number, then 2^x and 2^-x are real numbers. > You have seen it, and now you see it again, and you've seen other versions of > it as well. You omitted the part where you claim that all reals can be derived by recursively applying rule 2. They can't. An uncountable number of reals are never derived. For example, applying rule 2 repeatedly will never result in even a simple number like 3. But please, oh please, prove me wrong. === Subject: Re: Logarithm of transfinite numbers David R Tribble said: > Virgil said: >> I have never seen Orlow's amazing H-Riffic Number System either. > 1. 0 is a real number > 2. If x is a real number, then 2^x and 2^-x are real numbers. > You have seen it, and now you see it again, and you've seen other versions of > it as well. > You omitted the part where you claim that all reals can be derived by > recursively applying rule 2. > They can't. An uncountable number of reals are never derived. > For example, applying rule 2 repeatedly will never result in even a > simple number like 3. > But please, oh please, prove me wrong. I wish I could give you a good answer. Hopefully, at some point, I can get to implementing the H-riffics on the computer, and make it do my survey for me so I can derive rules once the intuition is established, but these gradal-type numbers are difficult. Right now, I cannot exactly tell you the bit string corresponding to 3. -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers > David R Tribble said: > Virgil said: >> I have never seen Orlow's amazing H-Riffic Number System either. > 1. 0 is a real number > 2. If x is a real number, then 2^x and 2^-x are real numbers. You have seen it, and now you see it again, and you've seen other > versions of > it as well. You omitted the part where you claim that all reals can be derived by > recursively applying rule 2. They can't. An uncountable number of reals are never derived. > For example, applying rule 2 repeatedly will never result in even a > simple number like 3. But please, oh please, prove me wrong. > I wish I could give you a good answer. Hopefully, at some point, I can get to > implementing the H-riffics on the computer No real number system, nor anything requiring infinitely many objects can be implemented on a computer as long as computers are limited to finitely many states. One can approximate such systems usefully, as has been done, but never more that approximate them. Anyone familiar with computer science should know better. , and make it do my survey for me > so > I can derive rules once the intuition is established, but these gradal-type > numbers are difficult. Right now, I cannot exactly tell you the bit string > corresponding to 3. === Subject: Re: Logarithm of transfinite numbers Please don't feed the troll. In terms of discussions about the size or comparison of sizes of infinite sets, there are a variety of methods, qualitative and quantitative. For example, half of the integers are even. A proper subset of a set can be said to be smaller, re Katz and corroboration. A proper subset of a set with infinitely many proper supersets that are subsets of the set may be said in a stronger sense to be larger, eg integers to rationals. A powerset as proper superset in ubiquitous ordinals can be said to be larger, and that's basically why it is. There is no universe in ZF(C). There's a universe, infinite sets are equivalent. Where there is a universe, ZF(C) is inconsistent as application as a universal theory. ZF is inconsistent, where everything's a mathematical object and the primary and only objects of the theory are sets, because there's no universe or universal set in ZF. A variety of modern methods have there being maximal ordinals, for example model theory, which is used to make statements about ZF in ZF. While that is so, in ZF there is no maximal ordinal. A variety of modern methods have there being maximal ordinals. I think it's fair to say that readers are not ignorant of that line of argument. A question then is whether they're hypocritical. Ciao, Ross F. === Subject: Re: Logarithm of transfinite numbers Ross A. Finlayson said: > Please don't feed the troll. But Troll Hungry!!! ;) > In terms of discussions about the size or comparison of sizes of > infinite sets, there are a variety of methods, qualitative and > quantitative. For example, half of the integers are even. A proper > subset of a set can be said to be smaller, re Katz and corroboration. Yes, that's one of the inviolable rules, that a proper subset have a smaller size. Viva Katz! > A proper subset of a set with infinitely many proper supersets that are > subsets of the set may be said in a stronger sense to be larger, eg > integers to rationals. A powerset as proper superset in ubiquitous > ordinals can be said to be larger, and that's basically why it is. Yes, at least standard theory recognizes the power set as a larger beast, albeit through somewhat sketchy logic. > There is no universe in ZF(C). There's a universe, infinite sets are > equivalent. Where there is a universe, ZF(C) is inconsistent as > application as a universal theory. ZF is inconsistent, where > everything's a mathematical object and the primary and only objects of > the theory are sets, because there's no universe or universal set in > ZF. > A variety of modern methods have there being maximal ordinals, for > example model theory, which is used to make statements about ZF in ZF. > While that is so, in ZF there is no maximal ordinal. A variety of > modern methods have there being maximal ordinals. I don't know about maximal infinites of any sort. A unit infinity makes sense to me, as the finite unit is an infinite number of point reals. > I think it's fair to say that readers are not ignorant of that line of > argument. A question then is whether they're hypocritical. Oh, well, that's endemic, no? === Subject: Re: Logarithm of transfinite numbers Yes, at least standard theory recognizes the power set as a larger beast, > albeit through somewhat sketchy logic. Wnat is sketchy about the logic? === Subject: Re: Logarithm of transfinite numbers said: > Yes, at least standard theory recognizes the power set as a larger beast, > albeit through somewhat sketchy logic. > Wnat is sketchy about the logic? > Well, the entire proof regarding the power set rests on the notion of the element which is not in the set to which it maps, and the contradiction derived from assuming such a thing indicating that no such element exists in the mapping, and that therefore the bijection fails to map an element. In the case of the power set it's possible to construct such a statement, whereas in other situations where a bijection is possible between obviously different levels of infinity, such an argument isn't possible to construct. So, it seems there is a proof that rests one a particular notion, while the general notion of quantitative comparisons such as 2^aleph_0 is lost when it comes to 2*aleph_0, aleph_0^2, or any other formulaic expression on infinity. So, while I don't disagree that the power set is larger, I don't consider the proof to be of the sort that leads to very precise comparisons. It only manages to distinguish this particular relative infinity through a peculiarity of the construction. Maybe sketchy isn't the right word. Maybe the logic is just parochial. I don't really disagree with it. I just think we can do better. -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers > said: > Yes, at least standard theory recognizes the power set as a > larger beast, albeit through somewhat sketchy logic. Wnat is sketchy about the logic? > Well, the entire proof regarding the power set rests on the notion > of the element which is not in the set to which it maps, and the > contradiction derived from assuming such a thing indicating that no > such element exists in the mapping, and that therefore the bijection > fails to map an element. Wrong! A function maps elements of its domain (source set) to elements of it codomain (target set). In the case of a function from any set as a domain to its power set as codomain, the function does map from every element of the domain, but it fails map TO every element of the codomain. And it is THIS failure which bars bijectiveness, In the case of the power set it's possible > to construct such a statement, whereas in other situations where a > bijection is possible between obviously different levels of infinity, > such an argument isn't possible to construct. So, it seems there is a > proof that rests one a particular notion, while the general notion of > quantitative comparisons such as 2^aleph_0 is lost when it comes to > 2*aleph_0, aleph_0^2, or any other formulaic expression on infinity. > So, while I don't disagree that the power set is larger, I don't > consider the proof to be of the sort that leads to very precise > comparisons. It only manages to distinguish this particular relative > infinity through a peculiarity of the construction. Maybe sketchy > isn't the right word. Maybe the logic is just parochial. I don't > really disagree with it. I just think we can do better. === Subject: Re: Logarithm of transfinite numbers > don't really disagree with it. I just think we can do better. I wouldn't say 'prochial', but the logic is basic in that it is entirely first order predicate calculus with identity, which is proven to be consistent. That puts the proof on firmer ground than anything derived with logic not proven to be consistent. === Subject: Re: Logarithm of transfinite numbers I wouldn't say 'prochial', but the logic is basic in that it is > entirely first order predicate calculus with identity, which is proven > to be consistent. That puts the proof on firmer ground than anything > derived with logic not proven to be consistent. In both places, 'consistent' should be 'sound'. === Subject: Re: Logarithm of transfinite numbers > Yes, at least standard theory recognizes the power set as a larger beast, > albeit through somewhat sketchy logic. > Wnat is sketchy about the logic? (Why do you write proof in quotation marks below? To remind us that you can't understand it, perhaps?) > Well, the entire proof regarding the power set rests on the notion of the > element which is not in the set to which it maps, and the contradiction derived > from assuming such a thing indicating that no such element exists in the > mapping, and that therefore the bijection fails to map an element. A major difference between cranks and mathematicians with alternative theories, is that mathematicians can accurately quote elementary results of theories they are proposing alternatives to. The proof as you laughably refer to it is a proof of the non-existence of a bijection between two sets. That's all. Mathematicians like proving results like This exists, or That does not exist, because these results are simple and incontrovertible, as opposed to the hand-waving we see in other quarters. > In the case > of the power set it's possible to construct such a statement, whereas in other > situations where a bijection is possible between obviously different levels of > infinity, such an argument isn't possible to construct. So, it seems there is a > proof that rests one a particular notion, while the general notion of > quantitative comparisons such as 2^aleph_0 is lost when it comes to 2*aleph_0, > aleph_0^2, or any other formulaic expression on infinity. So, while I don't > disagree that the power set is larger, I don't consider the proof to be of the > sort that leads to very precise comparisons. What nonsense. Nothing is lost. Mathematicians are perfectly capable of formalising other comparisons of infinite sets, and getting it right. > It only manages to distinguish > this particular relative infinity through a peculiarity of the construction. > Maybe sketchy isn't the right word. Maybe the logic is just parochial. I > don't really disagree with it. I just think we can do better. Uh, _you_ think _you_ can do better. I think David Ullrich would probably write Guffaw at this point. Brian Chandler http://imaginatorium.org === Subject: Re: Logarithm of transfinite numbers albeit through somewhat sketchy logic. > Wnat is sketchy about the logic? > Tony, you're not a troll. The powerset result is generally seen to hold true for the sets (in/of the universe) that are well-founded, regular, that in all theories satisfy the properties of well-foundedness as it is variously axiomatized as a predicate to resolve to true. Then, there's the notion that infinite sets are not regular. That is basically where some least infinite set, which happens to perfectly correspond to the set of natural numbers, where least means first in emergent synthesis, that that set is not regular: N E N, it contains itself, it is a/the Ding-an-Sich, and the universe of all sets and numbers and objects and so forth as well. You might cringe at the notion of a set being irregular, having itself as a member of its transitive closure, particularly so in the case of something along the lines of the set of natural integers, which seem to have a perfectly adequate definition via Peano or finite von Neumann ordinals, which would be an infinite von Neumann ordinal. While that is so, you can consider what are currently seen as non-standard views of perceptions of the natural integers as containing an infinite element, and not necessarily the non-standard notion of them as hyperintegers but instead having some maximal element, infinity, and at the same time having as identity infinity. That notion already exists for example as in recent discussion about various results of a definition of the object(s) that comprise the natural integers and the completeness of the set of true theorems about them, with there being a universe. So, with the powerset result, it's seen as a finitist statement of the successor axiom of Peano arithmetic, because all infinite sets are not regular. Ciao, Ross F. === Subject: Re: Logarithm of transfinite numbers Ross A. Finlayson said: > Yes, at least standard theory recognizes the power set as a larger beast, > albeit through somewhat sketchy logic. > Wnat is sketchy about the logic? Tony, you're not a troll. But me still hungry! Drop food from bridge for Tony to eat. Hrarggh! ;) No, I'm just a pain in the ass with a sense of humor and a theory in development. Maybe that's the definition. :) > The powerset result is generally seen to hold true for the sets (in/of > the universe) that are well-founded, regular, that in all theories > satisfy the properties of well-foundedness as it is variously > axiomatized as a predicate to resolve to true. > Then, there's the notion that infinite sets are not regular. That is > basically where some least infinite set, which happens to perfectly > correspond to the set of natural numbers, where least means first in > emergent synthesis, that that set is not regular: N E N, it contains > itself, it is a/the Ding-an-Sich, and the universe of all sets and > numbers and objects and so forth as well. > You might cringe at the notion of a set being irregular, having itself > as a member of its transitive closure, particularly so in the case of > something along the lines of the set of natural integers, which seem to > have a perfectly adequate definition via Peano or finite von Neumann > ordinals, which would be an infinite von Neumann ordinal. > While that is so, you can consider what are currently seen as > non-standard views of perceptions of the natural integers as > containing an infinite element, and not necessarily the non-standard > notion of them as hyperintegers but instead having some maximal > element, infinity, and at the same time having as identity infinity. > That notion already exists for example as in recent discussion about > various results of a definition of the object(s) that comprise the > natural integers and the completeness of the set of true theorems about > them, with there being a universe. > So, with the powerset result, it's seen as a finitist statement of the > successor axiom of Peano arithmetic, because all infinite sets are not > regular. Yes, you've put forward this notion before, of a set that can contain itself, or is its own power set. I think you probably have a valid point in that, but somehow it rubs me the wrong way, given combinatoric facts. To me, the power set must be bigger than the root set, exponentially speaking. Where power sets and their roots sets seem equal, as in the bijection I put forth between !N and its power set, it's because there isn't any precise concept of range involved, and the two march off together hand in hand, albeit at exponentially different rates. When the range is viewed as a variable and infinities compared formulaically, I think one always gets that the power set is exponentially larger ala N=S^L where S=2 and L is the size of the root set. > Ciao, > Ross F. -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers primitive, and if I can derive no contradictions, and can derive useful > results, you have nothing to complain about. That would be good. But since you criticize other mathematics on the basis that its axioms aren't true as statements about a fundamental reality, then your own axioms are subject to such scrutiny too. I don't know why you would think it is so manifestly true that there is an object that exists as a fundamental reality that is the length of the real line but (if you do hold this:) that there isn't an object that is the set of counting numbers. > Oh. Define 0. The unique x(Ay ~yex). === Subject: Re: Logarithm of transfinite numbers said: > Actually, I can axiomatically state these things and treat Big'un as a > primitive, and if I can derive no contradictions, and can derive useful > results, you have nothing to complain about. > That would be good. But since you criticize other mathematics on the > basis that its axioms aren't true as statements about a fundamental > reality, then your own axioms are subject to such scrutiny too. I don't > know why you would think it is so manifestly true that there is an > object that exists as a fundamental reality that is the length of the > real line but (if you do hold this:) that there isn't an object that is > the set of counting numbers. Yes, that's not a vacuous point, and one I can appreciate. Assuming a length to the real number line when it has no discernible ends does seem somewhat arbitrary, and that's why it needs to be assumed a priori, since it's not really a derivable value. But, if we say there exists this infinite line, then there is SOME length to it which is some infinite value, and we simply name this value Big'un, and see what we can do with it. It appears that the assumption that this value is not only the length of the line, but also the number of points within any unit segment of it, actually fits perfectly with the Inverse Function Rule and answers the question of the relationship between the naturals and the reals, without producing any contradictions that I've been able to detect. Believe me, I'm on the lookout. My first attempt to formulaically relate them failed for reasons pointed out by someone, perhaps David Kastrup, and seemed problematic anyway due to IFR, but so far this one seems to work in every respect. Now, when it comes to omega or aleph_0, I suppose one can use such a term to denote the number of finite naturals, or perhaps the maximum number of contiguous points that can be contained in an interval of zero standard length, but then the question arises as to what one can do with this number. The derivation of omega via the von Neumann ordinals appears to be in contradiction to the identity relation between element count and value in the naturals, and leads to what I consider a false concept of infinity. So, while the idea of Big'un as the length of the real line may seem arbitrary, indeed I find it more appealing due simply to the fact that, so far, it seems to provide a starting place for the consideration of integrated notions of infinity without producing contradictions and instead producing answers to previously unanswerable questions. > Oh. Define 0. > The unique x(Ay ~yex). That sounds like a definition of the null set ala von Neumann, but not necessarily going to the heart of what 0 is. Really, I was referring to its use in the Peano axioms, where is is taken as a primitive. I see no reason why it can't be taken as an assumed primitive and a starting place, and actually see this as a natural starting place for mathemtics as a whole. Start with nothing. > -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers Actually, I can axiomatically state these things and treat Big'un as a > primitive, and if I can derive no contradictions, and can derive useful > results, you have nothing to complain about. > That would be good. But since you criticize other mathematics on the > basis that its axioms aren't true as statements about a fundamental > reality, then your own axioms are subject to such scrutiny too. I don't > know why you would think it is so manifestly true that there is an > object that exists as a fundamental reality that is the length of the > real line but (if you do hold this:) that there isn't an object that is > the set of counting numbers. > Yes, that's not a vacuous point, and one I can appreciate. Assuming a length to > the real number line when it has no discernible ends does seem somewhat > arbitrary, and that's why it needs to be assumed a priori, since it's not > really a derivable value. The question is why you think that is a priori true as opposed to the axioms of set theory. > But, if we say there exists this infinite line, then > there is SOME length to it which is some infinite value, and we simply name > this value Big'un, and see what we can do with it. I don't see that as being a priori true any more than the axioms of set theory. > It appears that the > assumption that this value is not only the length of the line, but also the > number of points within any unit segment of it, actually fits perfectly with > the Inverse Function Rule and answers the question of the relationship between > the naturals and the reals, without producing any contradictions that I've been > able to detect. Fits reflects a coherence theory of truth, as opposed to the criterion of fundamental truth you've challenged set theory with. Also, lack of contradiction reflects an approach that has a longer track record with set theory than with your own notions, especially since your own notions can't even be looked at for formal consistency until they are formalized. And you'd have to say what is the question as to the relation between the naturals and the reals you have in mind. > Oh. Define 0. > The unique x(Ay ~yex). > That sounds like a definition of the null set ala von Neumann, It's basic to Z set theory (though Zermelo's original paper did take the existence of the empty set as axiomatic since he hadn't formalized to the extent of making the first order logic explicit). And, if I'm not mistaken, it goes back even much further in the past than Zermelo. > but not > necessarily going to the heart of what 0 is. What 0 is? It exists as an observable object? As a platonic object? Whatever it is, for me, the above definition captures a splendid representation. > Really, I was referring to its use > in the Peano axioms, where is is taken as a primitive. Then since you know it is primitive in first order PA, I don't see the sense of asking for a definition of it in first order PA. > I see no reason why it > can't be taken as an assumed primitive and a starting place, and actually see > this as a natural starting place for mathemtics as a whole. Start with nothing. Fine. But it turns out that if you have an axiom that says that for any set, there is any subset of members having a certain property, then 0 as a primitive is redundant. === Subject: Re: Logarithm of transfinite numbers said: > said: > Actually, I can axiomatically state these things and treat Big'un as a > primitive, and if I can derive no contradictions, and can derive useful > results, you have nothing to complain about. That would be good. But since you criticize other mathematics on the > basis that its axioms aren't true as statements about a fundamental > reality, then your own axioms are subject to such scrutiny too. I don't > know why you would think it is so manifestly true that there is an > object that exists as a fundamental reality that is the length of the > real line but (if you do hold this:) that there isn't an object that is > the set of counting numbers. > Yes, that's not a vacuous point, and one I can appreciate. Assuming a length to > the real number line when it has no discernible ends does seem somewhat > arbitrary, and that's why it needs to be assumed a priori, since it's not > really a derivable value. > The question is why you think that is a priori true as opposed to the > axioms of set theory. I think that, as an assumption, it is not inherently wrong, but sensible, thatan infinitely long oline has an infinite length. The only leap is assigning some particular infinite value to this length and calling it a unit, but units are like that, arbitrarily declared and then built upon. > But, if we say there exists this infinite line, then > there is SOME length to it which is some infinite value, and we simply name > this value Big'un, and see what we can do with it. > I don't see that as being a priori true any more than the axioms of set > theory. It doesn't seem to you that the infinite number line has an infinite length, or that we can put a name on this infinite length? Which of those seems unreasonable? > It appears that the > assumption that this value is not only the length of the line, but also the > number of points within any unit segment of it, actually fits perfectly with > the Inverse Function Rule and answers the question of the relationship between > the naturals and the reals, without producing any contradictions that I've been > able to detect. > Fits reflects a coherence theory of truth, as opposed to the > criterion of fundamental truth you've challenged set theory with. Also, > lack of contradiction reflects an approach that has a longer track > record with set theory than with your own notions, especially since > your own notions can't even be looked at for formal consistency until > they are formalized. And you'd have to say what is the question as to > the relation between the naturals and the reals you have in mind. The question is what the relationship is, formulaically, between the discrete naturals and the continuous reals, if any such formula is possible. Standard theory holds that c is 2^aleph_0, which may be aleph_1, right? That is based on a combinatorial approach that confuses infinite languages with infinite quantitative sets. If we can map the naturals to the reals in the unit interval, then we've created a bijection between the two, akin to that proposed by reflecting the bits of the binary naturals around the binary point to create the binary fractions. This bijection is generally dismissed because the binary naturals contain no infinite strings, but if we're talking about the hypernaturals, that objection is moot, and the bijection stands, and the unbounded megabigulous discrete set is equal to the bounded megabigulous continuous set of reals in the unit interval. In addition, the whole question of the Continuum Hypothesis is answered, since not only are there a full spectrum of formulaic infinities between oo and 2^oo, but there are a whole spectrum of infinities less than any standard oo as well. Are these not questions unanswerable in standard set theory? > Oh. Define 0. The unique x(Ay ~yex). > That sounds like a definition of the null set ala von Neumann, > It's basic to Z set theory (though Zermelo's original paper did take > the existence of the empty set as axiomatic since he hadn't formalized > to the extent of making the first order logic explicit). And, if I'm > not mistaken, it goes back even much further in the past than Zermelo. Okay, then von Neumann didn't pioneer the confusion of a set with the size of the set, he just formalized it. > but not > necessarily going to the heart of what 0 is. > What 0 is? It exists as an observable object? As a platonic object? > Whatever it is, for me, the above definition captures a splendid > representation. 0 is the relative here and now, the origin, the reference point for all quantity. Disagree? > Really, I was referring to its use > in the Peano axioms, where is is taken as a primitive. > Then since you know it is primitive in first order PA, I don't see the > sense of asking for a definition of it in first order PA. I didn't. I suggested I was using it as a primitive. It can be described qualitatively, as all primitives probably should for ease of comprehension, but the axioms ultimately define the behavior of the object, no? > I see no reason why it > can't be taken as an assumed primitive and a starting place, and actually see > this as a natural starting place for mathemtics as a whole. Start with nothing. > Fine. But it turns out that if you have an axiom that says that for any > set, there is any subset of members having a certain property, then 0 > as a primitive is redundant. Perhaps axiomatic statements regarding 0 can make that axiom redundant, eh? I can't really tell from this description of the axiom in question. > -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers > said: > I think that, as an assumption, it is not inherently wrong, but > sensible, thatan infinitely long oline has an infinite length. For every mathemtical definition of length that I am aware of, it derives from the existence of a metric on the set on which it is defined and produces distances between pairs of members of that set. No metric produces distances on anything but such pairs of members of that set. What definition of length is TO using which does NOT obey these standard rules? > The > only leap is assigning some particular infinite value to this length What length? A length for any set of reals must be determined by a particular pair of real numbers related to that set. Which two reals will TO use to determine the length of the set of all reals? > But, if we say there exists this infinite line, then > there is SOME length to it which is some infinite value, and we > simply name this value Big'un, and see what we can do with it. I don't see that as being a priori true any more than the axioms of > set theory. > It doesn't seem to you that the infinite number line has an infinite > length, or that we can put a name on this infinite length? Which of > those seems unreasonable? Both! To start with, which pair of real numbers is TO going to use to determine the length of his real line? As there is no length not so determined, TO meeds these real numbers. It appears that the assumption that this value is not only the > length of the line, but also the number of points within any unit > segment of it, actually fits perfectly with the Inverse Function > Rule and answers the question of the relationship between the > naturals and the reals, without producing any contradictions that > I've been able to detect. Fits reflects a coherence theory of truth, as opposed to the > criterion of fundamental truth you've challenged set theory with. > Also, lack of contradiction reflects an approach that has a longer > track record with set theory than with your own notions, especially > since your own notions can't even be looked at for formal > consistency until they are formalized. And you'd have to say what > is the question as to the relation between the naturals and the > reals you have in mind. > The question is what the relationship is, formulaically, between the > discrete naturals and the continuous reals TO has not shown that there is any system in which such formulaic need to exist. Standard set theories have shown that there is a logically consistent sense in which the cardinality of the set of reals differs from that of the set of naturals. Oh. Define 0. The unique x(Ay ~yex). That sounds like a definition of the null set ala von Neumann, It's basic to Z set theory (though Zermelo's original paper did > take the existence of the empty set as axiomatic since he hadn't > formalized to the extent of making the first order logic explicit). > And, if I'm not mistaken, it goes back even much further in the > past than Zermelo. > Okay, then von Neumann didn't pioneer the confusion of a set with the > size of the set, he just formalized it. TO is the only one possessing that confusion. but not necessarily going to the heart of what 0 is. What 0 is? It exists as an observable object? As a platonic object? > Whatever it is, for me, the above definition captures a splendid > representation. > 0 is the relative here and now, the origin, the reference point for > all quantity. Disagree? As I have no idea what TO means by the relative here and now, and I certainly accept that 0 perfectly represents the cardinality of the empty set, whether or not that set is regarded as naturally natural, I suspect I would have to disagree. > I see no reason why it can't be taken as an assumed primitive and > a starting place, and actually see this as a natural starting > place for mathemtics as a whole. Start with nothing. Fine. But it turns out that if you have an axiom that says that for > any set, there is any subset of members having a certain property, > then 0 as a primitive is redundant. > Perhaps axiomatic statements regarding 0 can make that axiom > redundant, eh? I can't really tell from this description of the axiom > in question. As 0 need not, and usually does not, appear at all in axiom sets for set theory, except as an abbreviation, its meaning in any such system is established only by definition, usually as either the empty set or as the cardinality of the empty set. Thus the meaning of 0 is derived, not inherent. === Subject: Re: Logarithm of transfinite numbers thatan infinitely long oline has an infinite length. The only leap is assigning > some particular infinite value to this length and calling it a unit, but units > are like that, arbitrarily declared and then built upon. You say that your system is based on fundamental truths that set theory is not based on. Yet, your axioms are turning out to be no less arbitrary than those of set theory. > It doesn't seem to you that the infinite number line has an infinite length, or > that we can put a name on this infinite length? Define 'length' and we can discuss. On the other hand, since infinity is not something we can look at with direct observation, what infinite things have or don't have must depend on an axiomatization and definitions of the primitives. I don't know of any primitive notion that demands that there be an object that is the entire length of the real numbers under the standard linear ordering of the reals. > Fits reflects a coherence theory of truth, as opposed to the > criterion of fundamental truth you've challenged set theory with. Also, > lack of contradiction reflects an approach that has a longer track > record with set theory than with your own notions, especially since > your own notions can't even be looked at for formal consistency until > they are formalized. And you'd have to say what is the question as to > the relation between the naturals and the reals you have in mind. > The question is what the relationship is, formulaically, between the discrete > naturals and the continuous reals, if any such formula is possible. Standard > theory holds that c is 2^aleph_0, which may be aleph_1, right? That is based on > a combinatorial approach that confuses infinite languages with infinite > quantitative sets. You keep saying confuses and things like that. Set theory has primitives, axioms, definitions, and theorems. There is no confusion. You still haven't even started thinking about the point I've been making for months now - definitions are not substantive. > If we can map the naturals to the reals in the unit > interval, then we've created a bijection between the two, akin to that proposed > by reflecting the bits of the binary naturals around the binary point to create > the binary fractions. This bijection is generally dismissed because the binary > naturals contain no infinite strings, but if we're talking about the > hypernaturals, that objection is moot, and the bijection stands, and the > unbounded megabigulous discrete set is equal to the bounded megabigulous > continuous set of reals in the unit interval. Get back to me when you've defined all those terms. > In addition, the whole question > of the Continuum Hypothesis is answered, If your theory is consistent and rich enough to express basic arithmetic, then there will be other undecidable sentences, even if not the continuum hypothesis. > since not only are there a full > spectrum of formulaic infinities between oo and 2^oo, but there are a whole > spectrum of infinities less than any standard oo as well. Are these not > questions unanswerable in standard set theory? It is undecidable in set theory whether there are cardinalities between omega and 2^omega. If your theory is consistent and rich enough to express basic arithmetic, then there will be undecidable sentences in your theory. As to infinities less than omega, no, in set theory it is not undecidable; in set theory there are no infinities less than omega. > Okay, then von Neumann didn't pioneer the confusion of a set with the size of > the set, he just formalized it. Von Neumann was among many set theorists provding formalizations, definitions, and theorems. You have no comprehension whatsoever of the history of set theory. > 0 is the relative here and now, the origin, the reference point for all > quantity. Disagree? How can I disagree with what you don't define - 'origin', 'reference point', 'quantity'. > Then since you know it is primitive in first order PA, I don't see the > sense of asking for a definition of it in first order PA. > I didn't. I suggested I was using it as a primitive. You asked Virgil (or whoever it was) to define '0'. > It can be described > qualitatively, as all primitives probably should for ease of comprehension, I have no idea what you mean by describing a primitive quantitatively. > but > the axioms ultimately define the behavior of the object, no? Informally put, that's okay. Right, it's the axioms that tell you how the primitives are going to work with one another. > Perhaps axiomatic statements regarding 0 can make that axiom redundant, eh? I don't know why you say that. I said nothing like that. I said that from the axiom schema of separation, we can prove that there exists an prove that that there is at most one such object. Thus, taking '0' as primitive and/or having an empty set axiom is superfluous. > can't really tell from this description of the axiom in question. Cf. Chapter 1 of just about any good set theory book. And, as I recall, schema of separation and axiom of extensionality, we can prove the existence of a unique empty set. === Subject: Re: Logarithm of transfinite numbers > said: > Actually, I can axiomatically state these things and treat Big'un > as a primitive, and if I can derive no contradictions, and can > derive useful results, you have nothing to complain about. That would be good. But since you criticize other mathematics on > the basis that its axioms aren't true as statements about a > fundamental reality, then your own axioms are subject to such > scrutiny too. I don't know why you would think it is so manifestly > true that there is an object that exists as a fundamental reality > that is the length of the real line but (if you do hold this:) that > there isn't an object that is the set of counting numbers. > Yes, that's not a vacuous point, and one I can appreciate. Assuming a > length to the real number line when it has no discernible ends does > seem somewhat arbitrary, and that's why it needs to be assumed a > priori, since it's not really a derivable value. Thus TO is claiming that it is legitimate for HIM to assume things arbitrarily but not for anyone else? > But, if we say there > exists this infinite line, then there is SOME length to it Why does something that does not have ends have to have something that requires that it have ends? > It appears that the assumption that this > value is not only the length of the line, but also the number of > points within any unit segment of it It does not appear so to us, so that we will require formal proof of that claim. Oh. Define 0. The unique x(Ay ~yex). > That sounds like a definition of the null set ala von Neumann, but > not necessarily going to the heart of what 0 is. Really, I was > referring to its use in the Peano axioms, where is is taken as a > primitive. I see no reason why it can't be taken as an assumed > primitive and a starting place, and actually see this as a natural > starting place for mathemtics as a whole. Start with nothing. That is precisely what von Neumann did, and is precisely what TO has violently objected to in the past only because it showed how stupid one of TO's arguments was. === Subject: Re: Logarithm of transfinite numbers Virgil said: > said: > Actually, I can axiomatically state these things and treat Big'un > as a primitive, and if I can derive no contradictions, and can > derive useful results, you have nothing to complain about. That would be good. But since you criticize other mathematics on > the basis that its axioms aren't true as statements about a > fundamental reality, then your own axioms are subject to such > scrutiny too. I don't know why you would think it is so manifestly > true that there is an object that exists as a fundamental reality > that is the length of the real line but (if you do hold this:) that > there isn't an object that is the set of counting numbers. Yes, that's not a vacuous point, and one I can appreciate. Assuming a > length to the real number line when it has no discernible ends does > seem somewhat arbitrary, and that's why it needs to be assumed a > priori, since it's not really a derivable value. > Thus TO is claiming that it is legitimate for HIM to assume things > arbitrarily but not for anyone else? I don't think anyone here has claimed that a theory doesn't need some kind of declared primitives. > But, if we say there > exists this infinite line, then there is SOME length to it > Why does something that does not have ends have to have something that > requires that it have ends? Infinite length doesn't require ends. Neither does the length of a circumference of a circle. Put the two together, and the infinite number circle can have a length, even when infinite and endless. That length is Big'un. > It appears that the assumption that this > value is not only the length of the line, but also the number of > points within any unit segment of it > It does not appear so to us, so that we will require formal proof of > that claim. It helps if you don't cut the sentence in half. Given Big'un reals per unit and Big'un units on the real line, both evenly distributed, we have Big'un^2 reals on the real line by normal multiplication as is used to calculate such numbers in the finite case. By the Inverse Function Rule, if the naturals are mapped to the reals using f(x)=x/Big'un, then the inverse function g(x)=Big'un*x, and over the range from 0 to Big'un according to IFR, we have floor(Big'un*Big'un-Big'un*0+1)=Big'un^2+1 reals. We have the +1 because we are including both 0 and Big'un, which is one more point than Big'un half-open unit intervals. So, this demonstrates that IFR works with the definition of Big'un to produce the same count over the real line in two different ways, confirming with simple logic that the higher level approach works, at least in this case. Oh. Define 0. The unique x(Ay ~yex). That sounds like a definition of the null set ala von Neumann, but > not necessarily going to the heart of what 0 is. Really, I was > referring to its use in the Peano axioms, where is is taken as a > primitive. I see no reason why it can't be taken as an assumed > primitive and a starting place, and actually see this as a natural > starting place for mathemtics as a whole. Start with nothing. > That is precisely what von Neumann did, and is precisely what TO has > violently objected to in the past only because it showed how stupid one > of TO's arguments was. There are lots of ways to start with nothing. The trick is to figure out how the universe did it and came up with everything. :) -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers > Virgil said: said: > Actually, I can axiomatically state these things and treat > Big'un as a primitive, and if I can derive no contradictions, > and can derive useful results, you have nothing to complain > about. That would be good. But since you criticize other mathematics > on the basis that its axioms aren't true as statements about a > fundamental reality, then your own axioms are subject to such > scrutiny too. I don't know why you would think it is so > manifestly true that there is an object that exists as a > fundamental reality that is the length of the real line but (if > you do hold this:) that there isn't an object that is the set > of counting numbers. Yes, that's not a vacuous point, and one I can appreciate. > Assuming a length to the real number line when it has no > discernible ends does seem somewhat arbitrary, and that's why it > needs to be assumed a priori, since it's not really a derivable > value. > > Thus TO is claiming that it is legitimate for HIM to assume things > arbitrarily but not for anyone else? > I don't think anyone here has claimed that a theory doesn't need some > kind of declared primitives. Since TO allows himself to assume existence of whatever he wants, such as a length for the set of real numbers, by what rule of logic does he now deny us the right to assume an infinite set of finite naturals? > But, if we say there exists this infinite line, then there is > SOME length to it Why does something that does not have ends have to have something > that requires that it have ends? Infinite length doesn't require ends. All finite lengths do, at least for real intervals, so that by TO's own arguments, this must hold for inifinite lengths of reals as well. > Neither does the length of a > circumference of a circle. The measurement of The circumference of a circle must have a starting point and an ending point or else it cannot be measured at all, and one is just going around in circles like TO does. > Put the two together, and the infinite > number circle can have a length, even when infinite and endless. That > length is Big'un. At what point does the measurement start and at what point does it end on this illusionary infinite circle? > By the Inverse Function Rule None existent rules don't count. > Oh. Define 0. The unique x(Ay ~yex). That sounds like a definition of the null set ala von Neumann, > but not necessarily going to the heart of what 0 is. Really, I > was referring to its use in the Peano axioms, where is is taken > as a primitive. I see no reason why it can't be taken as an > assumed primitive and a starting place, and actually see this as > a natural starting place for mathemtics as a whole. Start with > nothing. That is precisely what von Neumann did, and is precisely what TO > has violently objected to in the past only because it showed how > stupid one of TO's arguments was. There are lots of ways to start with nothing. The trick is to figure > out how the universe did it and came up with everything. That is either astrophysics or religion or some combination of the two, but not mathematics. === Subject: Re: Logarithm of transfinite numbers distributed, we have Big'un^2 reals on the real line by normal multiplication > as is used to calculate such numbers in the finite case. By the Inverse > Function Rule, if the naturals are mapped to the reals using f(x)=x/Big'un, > then the inverse function g(x)=Big'un*x, and over the range from 0 to Big'un > according to IFR, we have floor(Big'un*Big'un-Big'un*0+1)=Big'un^2+1 reals. We > have the +1 because we are including both 0 and Big'un, which is one more point > than Big'un half-open unit intervals. So, this demonstrates that IFR works with > the definition of Big'un to produce the same count over the real line in two > different ways, confirming with simple logic that the higher level approach > works, at least in this case. Even though you're still in the early stages of formalizing, it would help if you gave some state of the project report, within a single post or sequence of posts, to say what primitives, axioms, and deductions you've devised so far. Otherwise, passages such as above are inpenetrable. Consider some of what you need to give as primitive or defined: Big'un reals per unit real line evenly distributed finite inverse function Inverse Function Rule naturals mapped normal multiplication f(x) [function notation] = / range floor 0 1 2 half-open interval count If you at least stated what you've come up with thus far, then one might be able to appreciate what you're trying to convey to that extent. === Subject: Re: Logarithm of transfinite numbers said: > Given Big'un reals per unit and Big'un units on the real line, both evenly > distributed, we have Big'un^2 reals on the real line by normal multiplication > as is used to calculate such numbers in the finite case. By the Inverse > Function Rule, if the naturals are mapped to the reals using f(x)=x/Big'un, > then the inverse function g(x)=Big'un*x, and over the range from 0 to Big'un > according to IFR, we have floor(Big'un*Big'un-Big'un*0+1)=Big'un^2+1 reals. We > have the +1 because we are including both 0 and Big'un, which is one more point > than Big'un half-open unit intervals. So, this demonstrates that IFR works with > the definition of Big'un to produce the same count over the real line in two > different ways, confirming with simple logic that the higher level approach > works, at least in this case. > Even though you're still in the early stages of formalizing, it would > help if you gave some state of the project report, within a single > post or sequence of posts, to say what primitives, axioms, and > deductions you've devised so far. Otherwise, passages such as above are > inpenetrable. > Consider some of what you need to give as primitive or defined: > Big'un > reals > per > unit > real line > evenly > distributed > finite > inverse function > Inverse Function Rule > naturals > mapped > normal multiplication > f(x) [function notation] > range > floor > half-open > interval > count > If you at least stated what you've come up with thus far, then one > might be able to appreciate what you're trying to convey to that > extent. > Okay, I'm going to print out this list and put it with my notes, among which are other suggestions from you and from Virgil. I'll try to comply soon. I know I keep saying that, but personal affairs make it hard to progress very quickly -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers > said: > Given Big'un reals per unit and Big'un units on the real line, both > evenly > distributed, we have Big'un^2 reals on the real line by normal > multiplication > as is used to calculate such numbers in the finite case. By the Inverse > Function Rule, if the naturals are mapped to the reals using > f(x)=x/Big'un, > then the inverse function g(x)=Big'un*x, and over the range from 0 to > Big'un > according to IFR, we have floor(Big'un*Big'un-Big'un*0+1)=Big'un^2+1 > reals. We > have the +1 because we are including both 0 and Big'un, which is one more > point > than Big'un half-open unit intervals. So, this demonstrates that IFR > works with > the definition of Big'un to produce the same count over the real line in > two > different ways, confirming with simple logic that the higher level > approach > works, at least in this case. Even though you're still in the early stages of formalizing, it would > help if you gave some state of the project report, within a single > post or sequence of posts, to say what primitives, axioms, and > deductions you've devised so far. Otherwise, passages such as above are > inpenetrable. Consider some of what you need to give as primitive or defined: Big'un > reals > per > unit > real line > evenly > distributed > finite > inverse function > Inverse Function Rule > naturals > mapped > normal multiplication > f(x) [function notation] > = > / > range > floor > 0 > 1 > 2 > half-open > interval > count If you at least stated what you've come up with thus far, then one > might be able to appreciate what you're trying to convey to that > extent. > Okay, I'm going to print out this list and put it with my notes, among which > are other suggestions from you and from Virgil. I'll try to comply soon. I > know > I keep saying that, but personal affairs make it hard to progress very > quickly May I suggest that the above list is still far from complete. For example, it also should include: length; order, as in an order relation; arithmetical operations like addition, subtraction, multiplication, division, powers, roots and logarithms, if they are to be extended in any way beyond standard natural or real number operations; and many more. === Subject: Re: Logarithm of transfinite numbers Actually, I can axiomatically state these things and treat Big'un > as a primitive, and if I can derive no contradictions, and can > derive useful results, you have nothing to complain about. That would be good. But since you criticize other mathematics on > the basis that its axioms aren't true as statements about a > fundamental reality, then your own axioms are subject to such > scrutiny too. I don't know why you would think it is so manifestly > true that there is an object that exists as a fundamental reality > that is the length of the real line but (if you do hold this:) that > there isn't an object that is the set of counting numbers. > Yes, that's not a vacuous point, and one I can appreciate. Assuming a > length to the real number line when it has no discernible ends does > seem somewhat arbitrary, and that's why it needs to be assumed a > priori, since it's not really a derivable value. > Thus TO is claiming that it is legitimate for HIM to assume things > arbitrarily but not for anyone else? > But, if we say there > exists this infinite line, then there is SOME length to it > Why does something that does not have ends have to have something that > requires that it have ends? > It appears that the assumption that this > value is not only the length of the line, but also the number of > points within any unit segment of it > It does not appear so to us, so that we will require formal proof of > that claim. Oh. Define 0. The unique x(Ay ~yex). > That sounds like a definition of the null set ala von Neumann, but > not necessarily going to the heart of what 0 is. Really, I was > referring to its use in the Peano axioms, where is is taken as a > primitive. I see no reason why it can't be taken as an assumed > primitive and a starting place, and actually see this as a natural > starting place for mathemtics as a whole. Start with nothing. > That is precisely what von Neumann did, and is precisely what TO has > violently objected to in the past only because it showed how stupid one > of TO's arguments was. What's your point? We don't need protection from Tony here, Virgil, the above being one of your less abusive posts. I think Tony is sincere and quite pleasant, knowledgeable and sincere. I suppose you won't reply. Do you have a point? It's probable that you should assume, say, a basically high level of mathematical sophistication of your readers, were you rational, perhaps in combination with social. Don't you have anything positive to say? That does not reflect well, because of the negative things, where saying nothing is not necessarily the same thing as saying nothing nice. There's a good point in saying that there are as many reals, on the line, between zero and one as there are naturals. Then when you sum their values over the naturals as is the generally understood method in the integral calculus or Cholera bacilli analysis, that gives perfect and generally perfectly expected results. That's where the integral bar is an S for summation. Basically post-Cauchy-Weierstrass in the quest for foundations of the eighteenth and nineteenth centuries certain results and counterexamples led to basically denial of the quite real utility, and thus in a way quite obvious formal soundness, of the infinitesimal analysis. That is where at the same time its justification is disguised as delta-epsilonics, with basically the exact same notion and quite widespread usage of the Leibniz notation, how in the limit via infinite induction it works, including ratios of infinitesimals, eg dx/dy. Consider L'Hospital(e), and the quadrature (squaring, in 2-D), for different reasons, and the same. Obviously in my theory the points are polydimensional, with being one- or two-sided on the line, in continuity one-sided and with weight over the summation index of the index' reciprocal, scalar, in the polydimensional: inverse, with geometry in the large and small, the very. There's something new, for you. The universe is infinite, infinite sets are equivalent. Familiar? How about the notion that there is a universe, would you agree that there is a universe. The more comprehensively it's examined the larger it gauge invariance.) Right: start with nothing, don't care. That's so it goes. Obviously in my theory of axiomless natural deduction, currently A theory, there's only one theory with no axioms. Conveniently, only and all true statements are theorems. Ciao, Ross F. === Subject: Re: Logarithm of transfinite numbers Ross A. Finlayson said: > said: > Actually, I can axiomatically state these things and treat Big'un > as a primitive, and if I can derive no contradictions, and can > derive useful results, you have nothing to complain about. That would be good. But since you criticize other mathematics on > the basis that its axioms aren't true as statements about a > fundamental reality, then your own axioms are subject to such > scrutiny too. I don't know why you would think it is so manifestly > true that there is an object that exists as a fundamental reality > that is the length of the real line but (if you do hold this:) that > there isn't an object that is the set of counting numbers. Yes, that's not a vacuous point, and one I can appreciate. Assuming a > length to the real number line when it has no discernible ends does > seem somewhat arbitrary, and that's why it needs to be assumed a > priori, since it's not really a derivable value. > Thus TO is claiming that it is legitimate for HIM to assume things > arbitrarily but not for anyone else? > But, if we say there > exists this infinite line, then there is SOME length to it > Why does something that does not have ends have to have something that > requires that it have ends? > It appears that the assumption that this > value is not only the length of the line, but also the number of > points within any unit segment of it > It does not appear so to us, so that we will require formal proof of > that claim. Oh. Define 0. The unique x(Ay ~yex). That sounds like a definition of the null set ala von Neumann, but > not necessarily going to the heart of what 0 is. Really, I was > referring to its use in the Peano axioms, where is is taken as a > primitive. I see no reason why it can't be taken as an assumed > primitive and a starting place, and actually see this as a natural > starting place for mathemtics as a whole. Start with nothing. > That is precisely what von Neumann did, and is precisely what TO has > violently objected to in the past only because it showed how stupid one > of TO's arguments was. > What's your point? > We don't need protection from Tony here, Virgil, the above being one of > your less abusive posts. I think Tony is sincere and quite pleasant, > knowledgeable and sincere. That's because I don't call you Kerberos. Heh heh. Virgil's okay. He needs to bark. That's his job. I only get annoyed when my arguments are twisted into silliness dishonestly, which he doesn't seem to be doing TOO much these days. I think he has a right to defend what he believes. This is debate. Rock on, Ross! > I suppose you won't reply. Do you have a point? It's probable that > you should assume, say, a basically high level of mathematical > sophistication of your readers, were you rational, perhaps in > combination with social. > Don't you have anything positive to say? That does not reflect well, > because of the negative things, where saying nothing is not necessarily > the same thing as saying nothing nice. You know Ross, Virgil serves his purpose. His challenges are welcome, when he gets specific about what questions he expects answered and what definitions need to be established in order to make the theory even remotely acceptable. If, heaven forfend, I can one day satisfy Virgil's requirements, it seems almost impossible that anyone else could have the tiniest objection. So, if he sets the bar high, good, that gives me something to shoot for. Your denegerate intervals and Han's infinitesimal probabilities are vital to IFR and the solution to the Continuum Hypothesis and more. It's a good thing, this. :) > There's a good point in saying that there are as many reals, on the > line, between zero and one as there are naturals. Then when you sum > their values over the naturals as is the generally understood method in > the integral calculus or Cholera bacilli analysis, that gives perfect > and generally perfectly expected results. That's where the integral > bar is an S for summation. Absolutely right, and what Han's been arguing, alas, to no avail. It seems such a straightforward concept, and yet, mathematics has made a concerted effort to eradicate these bacilli, to the extent that it has destroyed its own beneficial E. Coli and can no longer disgest new ideas without difficulty. I think we offer a form of mathematical yogurt, a central staple in the diet of any vital 168 year old Georgian horseman. > Basically post-Cauchy-Weierstrass in the quest for foundations of the > eighteenth and nineteenth centuries certain results and counterexamples > led to basically denial of the quite real utility, and thus in a way > quite obvious formal soundness, of the infinitesimal analysis. That is > where at the same time its justification is disguised as > delta-epsilonics, with basically the exact same notion and quite > widespread usage of the Leibniz notation, how in the limit via infinite > induction it works, including ratios of infinitesimals, eg dx/dy. > Consider L'Hospital(e), and the quadrature (squaring, in 2-D), for > different reasons, and the same. > Obviously in my theory the points are polydimensional, with being one- > or two-sided on the line, in continuity one-sided and with weight over > the summation index of the index' reciprocal, scalar, in the > polydimensional: inverse, with geometry in the large and small, the > very. > There's something new, for you. Yes, I haven't quite integrated your one- and two-sided points yet, though multidimensional points are not unreasonable in the limit. > The universe is infinite, infinite sets are equivalent. Familiar? How > about the notion that there is a universe, would you agree that there > is a universe. The more comprehensively it's examined the larger it > gauge invariance.) Right: start with nothing, don't care. That's so > it goes. > Obviously in my theory of axiomless natural deduction, currently A > theory, there's only one theory with no axioms. Conveniently, only and > all true statements are theorems. I would like to see derivations of some theorems from first priciples and pure logic. That's what I'm considering now in the my attempt at foundations. Have you managed to derive theorems formally this way? We need to do this, as they have, if we wish to get the Garden to flower again. Also worms. Lots of worms. We need worms. ;) > Ciao, > Ross F. -- Smiles, Tony === Subject: Re: Logarithm of transfinite numbers Actually, I can axiomatically state these things and treat Big'un > as a primitive, and if I can derive no contradictions, and can > derive useful results, you have nothing to complain about. That would be good. But since you criticize other mathematics on > the basis that its axioms aren't true as statements about a > fundamental reality, then your own axioms are subject to such > scrutiny too. I don't know why you would think it is so manifestly > true that there is an object that exists as a fundamental reality > that is the length of the real line but (if you do hold this:) that > there isn't an object that is the set of counting numbers. Yes, that's not a vacuous point, and one I can appreciate. Assuming a > length to the real number line when it has no discernible ends does > seem somewhat arbitrary, and that's why it needs to be assumed a > priori, since it's not really a derivable value. Thus TO is claiming that it is legitimate for HIM to assume things > arbitrarily but not for anyone else? > But, if we say there > exists this infinite line, then there is SOME length to it Why does something that does not have ends have to have something that > requires that it have ends? It appears that the assumption that this > value is not only the length of the line, but also the number of > points within any unit segment of it It does not appear so to us, so that we will require formal proof of > that claim. Oh. Define 0. The unique x(Ay ~yex). That sounds like a definition of the null set ala von Neumann, but > not necessarily going to the heart of what 0 is. Really, I was > referring to its use in the Peano axioms, where is is taken as a > primitive. I see no reason why it can't be taken as an assumed > primitive and a starting place, and actually see this as a natural > starting place for mathemtics as a whole. Start with nothing. That is precisely what von Neumann did, and is precisely what TO has > violently objected to in the past only because it showed how stupid one > of TO's arguments was. > What's your point? > We don't need protection from Tony here, Virgil, the above being one of > your less abusive posts. I think Tony is sincere and quite pleasant, > knowledgeable and sincere. > That's because I don't call you Kerberos. Heh heh. Virgil's okay. He needs to > bark. That's his job. I only get annoyed when my arguments are twisted into > silliness dishonestly, which he doesn't seem to be doing TOO much these days. I > think he has a right to defend what he believes. This is debate. Rock on, Ross! > I suppose you won't reply. Do you have a point? It's probable that > you should assume, say, a basically high level of mathematical > sophistication of your readers, were you rational, perhaps in > combination with social. > Don't you have anything positive to say? That does not reflect well, > because of the negative things, where saying nothing is not necessarily > the same thing as saying nothing nice. > You know Ross, Virgil serves his purpose. His challenges are welcome, when he > gets specific about what questions he expects answered and what definitions > need to be established in order to make the theory even remotely acceptable. > If, heaven forfend, I can one day satisfy Virgil's requirements, it seems > almost impossible that anyone else could have the tiniest objection. So, if he > sets the bar high, good, that gives me something to shoot for. > Your denegerate intervals and Han's infinitesimal probabilities are vital to > IFR and the solution to the Continuum Hypothesis and more. It's a good thing, > this. :) > There's a good point in saying that there are as many reals, on the > line, between zero and one as there are naturals. Then when you sum > their values over the naturals as is the generally understood method in > the integral calculus or Cholera bacilli analysis, that gives perfect > and generally perfectly expected results. That's where the integral > bar is an S for summation. > Absolutely right, and what Han's been arguing, alas, to no avail. It seems such > a straightforward concept, and yet, mathematics has made a concerted effort to > eradicate these bacilli, to the extent that it has destroyed its own beneficial > E. Coli and can no longer disgest new ideas without difficulty. I think we > offer a form of mathematical yogurt, a central staple in the diet of any vital > 168 year old Georgian horseman. > Basically post-Cauchy-Weierstrass in the quest for foundations of the > eighteenth and nineteenth centuries certain results and counterexamples > led to basically denial of the quite real utility, and thus in a way > quite obvious formal soundness, of the infinitesimal analysis. That is > where at the same time its justification is disguised as > delta-epsilonics, with basically the exact same notion and quite > widespread usage of the Leibniz notation, how in the limit via infinite > induction it works, including ratios of infinitesimals, eg dx/dy. > Consider L'Hospital(e), and the quadrature (squaring, in 2-D), for > different reasons, and the same. > Obviously in my theory the points are polydimensional, with being one- > or two-sided on the line, in continuity one-sided and with weight over > the summation index of the index' reciprocal, scalar, in the > polydimensional: inverse, with geometry in the large and small, the > very. > There's something new, for you. > Yes, I haven't quite integrated your one- and two-sided points yet, though > multidimensional points are not unreasonable in the limit. > The universe is infinite, infinite sets are equivalent. Familiar? How > about the notion that there is a universe, would you agree that there > is a universe. The more comprehensively it's examined the larger it > gauge invariance.) Right: start with nothing, don't care. That's so > it goes. > Obviously in my theory of axiomless natural deduction, currently A > theory, there's only one theory with no axioms. Conveniently, only and > all true statements are theorems. > I would like to see derivations of some theorems from first priciples and pure > logic. That's what I'm considering now in the my attempt at foundations. Have > you managed to derive theorems formally this way? We need to do this, as they > have, if we wish to get the Garden to flower again. Also worms. Lots of worms. > We need worms. ;) > Ciao, > Ross F. > -- > Smiles, > Tony Hi Tony, hey how's it going, In terms of Virgil, your repartee, which was not necessarily unentertaining, to at least one external observer was not. When you two hit the mud slide there it detracts somewhat from the mathematical discussion, and particularly from yours. This is sci.math, humor is irrelevant but appreciated, it's also irrelevant, and the less sociable basically derision, particularly where unwarranted as are some of Virgil's comments to you, and in the past me, about such notions as, say, rationality and mathematical creativity, are unacceptable. Virgil, don't get me wrong, I would stand for your right to speak, my fist ends where your nose begins. Nah, that's wishy-washy, Tony, I guess it's okay, the point is that you're okay with that, and that you as well respect even Virgil here. My problem is with Virgil's disrespect. I think he's chosen his own role much as you have, a different one for different reasons, because he purposefully never ventures. Tony, with your N = S^L, or as I say b^p, that basically happens in the finite. In the infinite, where it's sufficient for b or p to equal one, S or L, i.e. unary or base infinity, then the real numbers of the unit interval exactly correspond in a 1-1 bijective mapping to the natural integers, and there is an obvious well-ordering of them, their natural ordering. Otherwise there is none, yet there is a well-ordering of the reals. coded powerset in those terms, the powerset is successor is order type, in ubiquitous naturals, or in a pure naive set theory, of the elements that quantify the universe. With the mapping f(x) = x + 1, number to successor, that is the powerset mapping and illustrates how as objects are discovered in the universe, the functions between them emerge. In ubiquitous ordinals, the powerset is order type is successor, and the powerset is simply a new kind of ordinal, an ornate ordinal. That allows there to be a universe, and quantifiers over said universe, without Cantor's paradox. (That's not the same as Russell and the Liar and their conflation.) The universe is infinite and infinite sets are equivalent. There is a universe, there is no universe in ZF, the universe is the constructible universe, or actually the constructed universe. There is a universe and over sufficiently primitive objects, based on the empty set, quantification over all objects in the universe results in the universe, thus ZF is inconsistent, as a simple expression of mathematical logic about mathematics, because no predicate always resolves to true, yet everything in ZF is a set. Every thing in ZF is a set, thus everything in ZF must be a set. That leads to notions of infinities as irregular, not well-founded. The notion of the unit infinity and infinitesimal, or scalar unit infinity and infinitesimal, I am happy to say that others who were in previous, say, years adamant against their consideration are not so disinclined today. It's an ancient notion that returns again and again after its denial, because it applies. So, where reapplication of those intuitive notions, buttressed with mathematical logic justifying their existence, leads to reevaluation of primary results in measure theory, and the possibility of an entire new field of analytical results in the neo-classical, that's good. Ross F. === Subject: PDE/Heat Equation I am taking PDE class and stuck in a problem. This is about smoothing of discontinuities and the limit as t approaches zero for solutions u(x, t) of the heat equation. Suppose f is a bounded function on the line, R^1, with a jump discontinuity at x_0, the right limit of f, f(x_0 ^+), obtained by approaching x_0 from above is different from the left limit of f, f(x_0^ - ), obtained by approaching x_0 from below. Assume f is continuous for x!= x__0. Show that lim ->0 u(x, t) = 1/2 (f(x^+ ) + f(x^ - )), which equals f(x) for x!= x_0 and is equal to the average of the left and right limits for x = x_0. Please give me some hint!!! === Subject: Re: PDE/Heat Equation > I am taking PDE class and stuck in a problem. > This is about smoothing of discontinuities and the limit as t > approaches zero for solutions u(x, t) of the heat equation. > Suppose f is a bounded function on the line, R^1, with a jump > discontinuity at x_0, the right limit of f, f(x_0 ^+), obtained by > approaching x_0 from above is different from the left limit of f, > f(x_0^ - ), obtained by approaching x_0 from below. Assume f is > continuous for x!= x__0. Show that lim ->0 u(x, t) = 1/2 (f(x^+ ) + > f(x^ - )), which equals f(x) for x!= x_0 and is equal to the average of > the left and right limits for x = x_0. > Please give me some hint!!! Go with the formula u(x_0,t) = int_{-infty}^infty c/sqrt{t} exp(-y^2/4t) f(x_0-y) dy where the c is the constant that always appears there. Split the integral into two pieces, int_0^{infty} and the rest. I think that you will see that one integral will converge to f(x_0+)/2 and the other to the other quantity. To see this, see for y>0 that f(x_0-y) = f(x_0-)+delta(y) where delta(y)->0 as y->0+. See that for very small t, that c/sqrt{t} exp(-y^2/4t) is mostly concentrated near the origin, and in a symmetric fashion. === Subject: Re: Spans of linear spaces - otre vez As I understood Michael's question, he was looking for a basis for the vector space G of functions from N to R. Clearly this vector space can be identified with the space of real valued sequences, and the standard basis vectors (the vectors e_k, as I defined them before) form a countably infinite basis for this vector space. This fact is 100% obvious and no mention of finite support is needed. (This is all I meant when I said that the standard basis vectors work.) As you are aware, there are several very well known infinite dimensional vector spaces, and G is merely one of them. In your post, I'm not sure whether you were refuting my statement or thought that I was suggesting something much broader. Best, Joe === Subject: Re: Spans of linear spaces - otre vez > As I understood Michael's question, he was looking for a basis for the > vector space G of functions from N to R. Clearly this vector space can > be identified with the space of real valued sequences, and the standard > basis vectors (the vectors e_k, as I defined them before) form a > countably infinite basis for this vector space. This fact is 100% > obvious and no mention of finite support is needed. (This is all I > meant when I said that the standard basis vectors work.) As you are > aware, there are several very well known infinite dimensional vector > spaces, and G is merely one of them. > In your post, I'm not sure whether you were refuting my statement or > thought that I was suggesting something much broader. this fundamental Usenet courtesy.) If you're suggesting that such sums as sum_{k=1}^infty a_k e_k make sense, you need to specify in WHAT sense. Not in the vector space sense, certainly: addition of vectors is a BINARY operation, and can be carried out only two at a time, limiting us to FINITE sums. basis has only two meanings in the literature: Hamel basis, i.e. vector space basis as in the last paragraph, and Schauder basis, where the sum is the limit of partial sums. Without a norm, or linear topology, there's no sense to your statement. And I know of no linear topology which allows one to form such sums, hence no Schauder basis context. -- Ron Bruck === Subject: Being a single parent is not easy, free join us to find help Looking for romance , friendship, love ? Just click here ... http://www.geocities.com/singlemomneedhelp http://singleparentclub.Zu5.net Wish you have a wonderful time ! === Subject: Finance Math: Bank Bill Discount Rate Formula Nature of problem: Calculation of Prepaid Interest on a Bank Bill when given investment yield Formula's involved: Investment yield (IY) = [(FV - PP)/PP] * [365/M] FV = face value in $ PP = purchase price in $ M = maturity of bill in days 365 represents days in year Investment yield is shown as a % Basically I need a formula for PP as a function of IY,M,FV. My algebra is a little rusty but I'm sure it can be worked out as all the data results in a IY of the correct rate (it's all ready given). Below are 4 series of real data where the bank has calculated the discount amount shown as the sum of (FV - PP) in terms of the FV-PP,FV,M,IY. data: (FV-PP,FV,M,IY) Series 1 (23247.95,6593516,21,6.1500) Series 2 (41038.81,7000000,35,6.1500) Series 3 (6305.32,1075500,35,6.1500) Series 4 (20642.01,4395980.82,28,6.1500) Any help would be appreciated, Hamish === Subject: Re: Finance Math: Bank Bill Discount Rate Formula > Nature of problem: Calculation of Prepaid Interest on a Bank Bill when > given investment yield > Formula's involved: > Investment yield (IY) = [(FV - PP)/PP] * [365/M] > FV = face value in $ > PP = purchase price in $ > M = maturity of bill in days > 365 represents days in year > Investment yield is shown as a % > Basically I need a formula for PP as a function of IY,M,FV. My algebra > is a little rusty but I'm sure it can be worked out as all the data > results in a IY of the correct rate (it's all ready given). > Below are 4 series of real data where the bank has calculated the > discount amount shown as the sum of (FV - PP) in terms of the > FV-PP,FV,M,IY. > data: (FV-PP,FV,M,IY) > Series 1 (23247.95,6593516,21,6.1500) > Series 2 (41038.81,7000000,35,6.1500) > Series 3 (6305.32,1075500,35,6.1500) > Series 4 (20642.01,4395980.82,28,6.1500) > Any help would be appreciated, > Hamish PP = 365*FV/(IY*M + 365) === Subject: Re: eigenvalues of matrix product? That's bad news. I got defeated... sad... What if A is a projection matrix, which only has eigenvalues 0s and 1s? >If I have a product: C=A*B, >where A is a matrix with 0<= eigenvalues <= a, B is a matrix with 0<= >eigenvalues <= b, >is there a rigorous way for me to prove C=A*B has eigenvalues >satisfying 0<=eigenvalues <= a*b ? > It would be true if A and B commute, but it's not true in general. > For example, > [ use fixed-width font] > [ 0 0 ] > A = [ 0 1 ] > [ 2 -2 ] > B = [ 1 -1 ] > [ 0 0 ] > AB = [ 1 -1 ] > A and B both have eigenvalues 0 and 1, but AB has eigenvalues 0 > and -1. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: eigenvalues of matrix product? What if A is a projection matrix, which only has eigenvalues 0s and 1s? Take another look at my example. >If I have a product: C=A*B, where A is a matrix with 0<= eigenvalues <= a, B is a matrix with 0<= >eigenvalues <= b, is there a rigorous way for me to prove C=A*B has eigenvalues >satisfying 0<=eigenvalues <= a*b ? > It would be true if A and B commute, but it's not true in general. > For example, > [ use fixed-width font] > [ 0 0 ] > A = [ 0 1 ] > [ 2 -2 ] > B = [ 1 -1 ] > [ 0 0 ] > AB = [ 1 -1 ] > A and B both have eigenvalues 0 and 1, but AB has eigenvalues 0 > and -1. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Giuseppe Peano -- He shall add I was looking at Italian male names on the web page below and happened to notice that Giuseppe is associated with He shall add. I had no idea that Peano's first name had this association. Is this common knowledge? It seems to me that something this serendipitous would be pointed out quite often when Peano is mentioned, but I don't think I've seen this mentioned before. http://www.20000-names.com/male_italian_names.htm Dave L. Renfro === Subject: Re: Giuseppe Peano -- He shall add Dave L. Renfro ha scritto nel messaggio >I was looking at Italian male names on the web page > below and happened to notice that Giuseppe is > associated with He shall add. > I had no idea that Peano's first name had this > association. Is this common knowledge? It seems > to me that something this serendipitous would be > pointed out quite often when Peano is mentioned, > but I don't think I've seen this mentioned before. > http://www.20000-names.com/male_italian_names.htm > Dave L. Renfro Giuseppe is the first italian name, due to Catholic belief. Is derived from the Hebraic Yoseph, derived from yasaph (to add), with the meaning of God will add, increase (the family, with the sons). The second most popular name is Giovanni, the third is Antonio. All of three are religious names. === Subject: Re: Giuseppe Peano -- He shall add > I was looking at Italian male names on the web page > below and happened to notice that Giuseppe is > associated with He shall add. > I had no idea that Peano's first name had this > association. Is this common knowledge? It seems > to me that something this serendipitous would be > pointed out quite often when Peano is mentioned, > but I don't think I've seen this mentioned before. > http://www.20000-names.com/male_italian_names.htm > Dave L. Renfro -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Giuseppe Peano -- He shall add >> I was looking at Italian male names on the web page >> below and happened to notice that Giuseppe is >> associated with He shall add. >> I had no idea that Peano's first name had this >> association. Is this common knowledge? It seems >> to me that something this serendipitous would be >> pointed out quite often when Peano is mentioned, >> but I don't think I've seen this mentioned before. >> http://www.20000-names.com/male_italian_names.htm >> Dave L. Renfro Victor Borge reminds us that Giuseppe Verdi would be rendered in English as Joe Green. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Giuseppe Peano -- He shall add Yes. After the Christianization of Europe, many Old/New Testament names were inserted into the European culture. Even my name, Ioannis which is the Greek precursor of the German Johannes/Hans, the English John, the Russian Ivan and the Italian Giovanni, basically comes from the Hebrew Yochannan, meaning YHVH is gracious :-) > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ -- Ioannis === Subject: JSH: Power of suggestion Ok, time for a break, but hey, isn't this fun!!! Don't worry, posts like this one won't change the effect as it is long-lasting. It has to do with how information travels within the human brain. You depend on those cycles--I call them--to maintain your personality and concept of reality as firm. What I can do is, you could say, loosen the brain's sense of reality as being this solid thing, and cause you to have a more relaxed point of view. Some of it's suggestion and part of it is using contradiction in special ways. What's odd about it is that you can tell people what you're doing and it's not a protection. The only protection is NOT to read my posts! Even I have to be careful with some of them now, as you have to concentrate to keep your grip on reality. As we move forward for some of you it will be an invigorating experience. Ideas will come to you, and your perspective will shift. The world will look different. But for some of you, it will terrify you. Things will happen that you can't believe and some of them, unfortunately, will not be real, but your mind may have some problems of which you are not aware which will be revealed by these techniques. Look for them--contradictions, odd phrasing--things that seem just slightly odd, and understand the power of suggestion in all of it, as you know, it's how magicians do it!! It's all about illusion. Reality is firm, but your view of it is not. Your view of reality is what can be shifted. Ok, enough fun, as I'm thinking that maybe for some of you it's time to sleep. You know, there's all that work done, and now it's time to do something else. And besides, I need to talk about math, like some primes or something. Sleep isn't bad for thinking about math, as sometimes, ideas come to you early in the morning, if you wake up early. James Harris === Subject: Re: JSH: Power of suggestion > Ok, time for a break, but hey, isn't this fun!!! Don't worry, posts > like this one won't change the effect as it is long-lasting. It has to > do with how information travels within the human brain. > You depend on those cycles--I call them--to maintain your personality > and concept of reality as firm. > What I can do is, you could say, loosen the brain's sense of reality as > being this solid thing, and cause you to have a more relaxed point of > view. > Some of it's suggestion and part of it is using contradiction in > special ways. > What's odd about it is that you can tell people what you're doing and > it's not a protection. > The only protection is NOT to read my posts! > Even I have to be careful with some of them now, as you have to > concentrate to keep your grip on reality. > As we move forward for some of you it will be an invigorating > experience. Ideas will come to you, and your perspective will shift. > The world will look different. > But for some of you, it will terrify you. Things will happen that you > can't believe and some of them, unfortunately, will not be real, but > your mind may have some problems of which you are not aware which will > be revealed by these techniques. > Look for them--contradictions, odd phrasing--things that seem just > slightly odd, and understand the power of suggestion in all of it, as > you know, it's how magicians do it!! > It's all about illusion. > Reality is firm, but your view of it is not. Your view of reality is > what can be shifted. > Ok, enough fun, as I'm thinking that maybe for some of you it's time to > sleep. You know, there's all that work done, and now it's time to do > something else. > And besides, I need to talk about math, like some primes or something. > Sleep isn't bad for thinking about math, as sometimes, ideas come to > you early in the morning, if you wake up early. > James Harris Ha ha... what a moron! Your powers of suggestion are about as effective your mathematical research, which is to say, not effective at all! === Subject: Re: JSH: Power of suggestion > Ok, time for a b > Sleep isn't bad for thinking about math, as sometimes, ideas come to > you early in the morning, if you wake up early. ps ... can you please define an ideal? === Subject: Re: JSH: Power of suggestion > Ok, time for a break, but hey, isn't this fun!!! Don't worry, posts > like this one won't change the effect as it is long-lasting. It has to > do with how information travels within the human brain. > You depend on those cycles--I call them--to maintain your personality > and concept of reality as firm. > What I can do is, you could say, loosen the brain's sense of reality as > being this solid thing, and cause you to have a more relaxed point of > view. If anyone knows about loosening one's sense of reality ... === Subject: Re: JSH: Power of suggestion : Ok, time for a break, but hey, isn't this fun!!! Sure is, I'm having fun! : You depend on those cycles--I call them--to maintain your personality : and concept of reality as firm. It's true! Every time I read one of your posts I realize that I'm sane! : The only protection is NOT to read my posts! Why? I need a good occasional laugh. : The world will look different. You've been saying that for years and really, truly I'm looking forward to it but it never seems to happen. When's it going to happen? : Ok, enough fun, as I'm thinking that maybe for some of you it's time to : sleep. You know, there's all that work done, and now it's time to do : something else. I have an exam to write for my Number Theory class, then I can sleep. : And besides, I need to talk about math, like some primes or something. Careful, don't go overboard. We all know you've got problems understanding the concept of division so perhaps you'd better sort that one out first. : Sleep isn't bad for thinking about math, as sometimes, ideas come to : you early in the morning, if you wake up early. I get up at 5am daily. Justin === Subject: Re: JSH: Power of suggestion > Ok, time for a break, but hey, isn't this fun!!! Don't worry, posts > like this one won't change the effect as it is long-lasting. It has to > do with how information travels within the human brain. > You depend on those cycles--I call them--to maintain your personality > and concept of reality as firm. James, are you trying to shake the world of neurophysiology in the same way as you have shaken the mathematical world? Make sure you let us know when you send your paper to Nature. > What I can do is, you could say, loosen the brain's sense of reality as > being this solid thing, and cause you to have a more relaxed point of > view. > Some of it's suggestion and part of it is using contradiction in > special ways. > What's odd about it is that you can tell people what you're doing and > it's not a protection. > The only protection is NOT to read my posts! > Even I have to be careful with some of them now, as you have to > concentrate to keep your grip on reality. I think you need to concentrate harder. > As we move forward for some of you it will be an invigorating > experience. Ideas will come to you, and your perspective will shift. > The world will look different. > But for some of you, it will terrify you. Things will happen that you > can't believe and some of them, unfortunately, will not be real, but > your mind may have some problems of which you are not aware which will > be revealed by these techniques. > Look for them--contradictions, odd phrasing--things that seem just > slightly odd, and understand the power of suggestion in all of it, as > you know, it's how magicians do it!! > It's all about illusion. > Reality is firm, but your view of it is not. Your view of reality is > what can be shifted. > Ok, enough fun, as I'm thinking that maybe for some of you it's time to > sleep. You know, there's all that work done, and now it's time to do > something else. > And besides, I need to talk about math, like some primes or something. > Sleep isn't bad for thinking about math, as sometimes, ideas come to > you early in the morning, if you wake up early. > James Harris === Subject: Re: Function measurability <44313a4c$3$fuzhry+tra$mr2ice@news.patriot.net> === Subject: Re: Function measurability 04/03/2006 at 10:41 PM, thomas@drizzle.net said: >Continuity is a statement about the topologies on R No, continuity is a statement about functions between topological spaces. R is just on of many. >and >X: f^{-1}(A) is an open subset of X if A is an open subset If f is continuous >of R. If the range of f is in R. The OP left a lot of things unstated. In particular, he never said that X was a subset of R. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: sum of three sines <3f519fb9e2551a63757ba0effaf309f3.35661@mygate.mailgate.org So, tell us what you're really looking for. > had hoped, nay assumed (ASS_U_ME makes an ass out of U and me) that > there would be some simple relation for three sines - clearly I was > wrong! What can be simpler than sinA + sinB + sinC? BTW, I think that sinA + sinB = 2sin(A+B/2)cos(A-B/2) should be sinA + sinB = 2sin((A+B)/2)cos((A-B)/2))? Either way, why would you want to caculate the right-hand expression when you can calculate the left? > The reason for the question was this: in physics one can superpose two > almost identical waves sin(k_1 x - omega_1 t) and sin(k_2 x - omega_2 t) > and then identify (in the infinitesimal limit) two quantities omega/k, > and also d(omega) / d(k) which have different significances in physics. > Briefly, the former is the wave speed while the latter is the group > So I wondered what happens if we had a sum of three (almost identical) > waves. I imagined that even with three (or four or an infinite number > of) waves we would only still be able to identifiy two forms of wave > velocity. By inspecting the sum of three waves and taking the > infinitesimal limit I was hoping that I could *see* why we still only > get two wave velocities. > That was why I was wondering what the sum of three sines was. > Hope this makes sense, > David. > -- === Subject: Re: sum of three sines > Why isn't SinA + SinB +SinC a simple relation for the sum of three > sines? > Why is it not the simplest? Oh, that's simple, all right, but it doesn't move me forward any = ____ I can see how we get two velocities in the infinitesimal limit if we use sin(kx - wt), but if there are three waves how can we still see only two velocities? So we need to get a representation of sinA + sinB + sinC to help see this. David. -- === Subject: Cube fit in hypercube What is the largest cube that will fit inside a 4-D hypercube? question and that no two solutions were in agreement. As an introduction: The largest line that will fit in a square has length sqrt(2). The largest line that will fit in a cube has length sqrt(3). The largest square that will fit in a cube has area 9/8. The largest line that will fit in a hypercube has length 2. The largest square that will fit in a hypercube has area 2. I'm sure that the largest cube that will fit in a 4-D hypercube has volume between 1 and 9/8 but beyond that I can't tell. === Subject: Re: Dartmouth University Mesmerized By Holocaust Speaker [ crossposts added ] > Not of this, English is imprecise. :D > But if you've got the neccessary axioms and results you can really > prove anything. One out of three? A. The Holocaust did not happen B. The holocaust did happen C. It is possible to divide a sphere in a finite number of pieces and then put them together to two spheres of the same size Do you believe? Chris === Subject: Re: Dartmouth University Mesmerized By Holocaust Speaker > [ crossposts added ] > Not of this, English is imprecise. :D > But if you've got the neccessary axioms and results you can really > prove anything. > One out of three? > A. The Holocaust did not happen > B. The holocaust did happen > C. It is possible to divide a sphere in a finite number of pieces and then > put them together to two spheres of the same size > Do you believe? > Chris A is false, B is true, C depends on interpretation. If the sphere is interpreted as physical one, in physical space, the process must fail. But if the sphere is an ideal one in a 3 dimensional Euclidean space and one allows use of the axiom of choice, then the process can succeed. === Subject: Re: Dartmouth University Mesmerized By Holocaust Speaker On Thu, 06 Apr 2006 17:10:56 -0600, Virgil >> [ crossposts added ] >> Not of this, English is imprecise. :D >> But if you've got the neccessary axioms and results you can really >> prove anything. >> One out of three? >> A. The Holocaust did not happen >> B. The holocaust did happen >> C. It is possible to divide a sphere in a finite number of pieces and then >> put them together to two spheres of the same size >> Do you believe? >> Chris >A is false, B is true, C depends on interpretation. >If the sphere is interpreted as physical one, in physical space, the >process must fail. >But if the sphere is an ideal one in a 3 dimensional Euclidean space and >one allows use of the axiom of choice, then the process can succeed. can it be a pizza? I'm hungry. -- Zapanaz International Satanic Conspiracy Customer Support Specialist http://joecosby.com/ In the kingdom of the blind, the one-eyed man is king. In the democracy of the blind, only the blind are eligible for the job. === Subject: Re: Dartmouth University Mesmerized By Holocaust Speaker <6sGTf.46601$VV4.732679@ursa-nb00s0.nbnet.nb.ca> <5lHTf.46620$VV4.733346@ursa-nb00s0.nbnet.nb.ca> <4435591e$0$11070$e4fe514c@news.xs4all.nl> <44355eb9$0$11071$e4fe514c@news.xs4all.nl> <44356786$0$11071$e4fe514c@news.xs4all.nl A. The Holocaust did not happen > B. The holocaust did happen > C. It is possible to divide a sphere in a finite number > of pieces and then put them together to two spheres > of the same size > Do you believe? Maybe we should ask a dishwasher at Hanover Inn, since at least one such has gone on to become the world's foremost authority on revisionist mathematics! Dave L. Renfro === Subject: Re: Dartmouth University Mesmerized By Holocaust Speaker <6sGTf.46601$VV4.732679@ursa-nb00s0.nbnet.nb.ca> <5lHTf.46620$VV4.733346@ursa-nb00s0.nbnet.nb.ca> <4435591e$0$11070$e4fe514c@news.xs4all.nl> <44355eb9$0$11071$e4fe514c@news.xs4all.nl> <44356786$0$11071$e4fe514c@news.xs4all.nl> I don't believe anything. O! If I could TRULY believe in nothing... Hrm. Well, I question your definition of Holocaust, and sphere. A real sphere or a mathematical one? The attempted genocide by Hitler involving concentration camps? === Subject: Re: Dartmouth University Mesmerized By Holocaust Speaker Rev. James Slaughter schreef in bericht > I don't believe anything. Really not? You are not one of the choice people? Those who believe in the axiom? Chris > O! If I could TRULY believe in nothing... > Hrm. Well, I question your definition of Holocaust, and sphere. Did I define it? > A real sphere or a mathematical one? Ok. you may prove it for a fake sphere > The attempted genocide by Hitler involving concentration camps? Yes Chris === Subject: Erik Hensema canceled my post -----BEGIN PGP SIGNED MESSAGE----- [English text follows Dutch text] Het volgende, waarschijnlijk door u geschreven, bericht: alt.politics.usa.misc,alt.religion.christian,alt.revisionism,alt.rush-limbau gh,alt.slack,sci.math,nl.wetenschap === Subject: Re: Dartmouth University Mesmerized By Holocaust Speaker is excessief gecrosspost en daarom automatisch gecancelled door cancelbot@hensema.net. Uw bericht heeft een Henningsen Index van 56, terwijl de in nl.* vastgestelde limiet staat op 30. Als u een bericht crosspost, zet dan altijd een Followup-To voor een redelijk aantal groepen, bij voorkeur niet meer dan twee of drie. Hoe u dit doet in uw software kunt u het beste vragen in de nieuwsgroep nl.comp.software.newsreaders. Vragen over de cancels zelf kunt u stellen per mail of in de groep nl.internet.misbruik. Uw bericht kunt u vinden op: http://news.nl.linux.org/cancel/archive/hi/hi-1144042202/1.msg Meer informatie is te vinden op: http://news.nl.linux.org/cancel/hi.html [English] The following message, probably written by you: alt.politics.usa.misc,alt.religion.christian,alt.revisionism,alt.rush-limbau gh,alt.slack,sci.math,nl.wetenschap === Subject: Re: Dartmouth University Mesmerized By Holocaust Speaker has been crossposted excessively and is automatically cancelled by cancelbot@hensema.net. Your message has a Henningsen Index of 56, while the limit for nl.* is set to 30. When crossposting a message, always set a Followup-To to a reasonable number of groups, preferably no more than two or three. Questions about how to do this in your software can be asked in news.software.readers. Questions about the cancels can be asked by mail or in the group nl.internet.misbruik. Your message can be found on: http://news.nl.linux.org/cancel/archive/hi/hi-1144042202/1.msg More information can be found on: http://news.nl.linux.org/cancel/hi.html -----BEGIN PGP SIGNATURE----- Version: 2.6.3ia Charset: noconv iQCVAwUBRDVnikjJfcDty9VxAQGhAwP/c8B4Y9HVkNGYgAPiIespAPqxXBpNcwAY XjPYMrvh0Ul3Tc1zg2irfa9gvN7siEkRa71J07VZMIISW+JgzQlDr0+wsdOzs0VW FNHYDNIA3w4Dg3mxudkm7z9vsc1c/2AfWs0zw6bTZmGoIZWhYNQUH3LEM0duBJEX 7zrRy1o2bqE= =iv3/ -----END PGP SIGNATURE----- === Subject: What exactly is a quasi group? After reading the definition of quasi group all I know it is a groupoid not necessary associative nor has a netural element, nothing more. I also know if the quasi group contains a netural element then it is a loop. I need an good but simple example. === Subject: Re: What exactly is a quasi group? > After reading the definition of quasi group all I know it is a groupoid not necessary > associative nor has a netural element, nothing more. I also know if the quasi group > contains a netural element then it is a loop. I need an good but simple example. sci.math is a quasi group; because quasi often posts here. :) The multiplication table of a quasigroup always forms a latin square; i.e., each element q in Q appears exactly once in each column, and once in each row. And every Latin square defines the multiplication table for some quasigroup. So, for example, every completed Sudoku is the multiplication table of a quasigroup. One interesting difference between groups and quasigroups (aside from the obvious fact that they have different axioms) is that while in both groups and quasigroups, there is a unique x such that a*x = b; in a group we can instanly find x = a^-1*b, but there is no such easy way to solve for x in a quasigroup. I'm trying, but I can't come up with an example of a finite quasigroup action (in the way that, say, the Rubix Cube is an example of a finite group action). === Subject: Re: What exactly is a quasi group? <23718185.1144353278829.JavaMail.jakarta@nitrogen.mathforum.org>, > After reading the definition of quasi group all I know it is a groupoid not > necessary associative nor has a netural element, nothing more. I also know > if the quasi group contains a netural element then it is a loop. I need an > good but simple example. Your nothing more misses the main axiom(s). A quasigroup is a groupoid in which for every a and b there exists a unique x such that xa = b and there exists a unique y such that ay = b; or: in the equation uv = w, any two of u, v, w determine the third uniquely. Here's a little quasigroup which is neither a loop nor a semigroup: | f g h __|______ f | f h g g | h g f h | g f h If you know what an isotopism is, you may like to try showing that this example is isotopic with a group. Ken Pledger. === Subject: Re: What exactly is a quasi group? > After reading the definition of quasi group all I know it is a groupoid not necessary associative nor has a netural element, nothing more. I also know if the quasi group contains a netural element then it is a loop. I need an good but simple example. quasigroup is to group as quasimodo is to modo. === Subject: Suicidal spiritualizing and the triple helix The only reason that the error of scientific materialism was ever possible came from prohibition on magickal and spiritual activity by Mosaic religions. When such things are forbidden, there is no apparent evidence of spiritual activity; which leads some to believe that there is no evidence for God either. This, I refer to as suicidal spiritualizing - in similar vein as Mortimer Adler referred as suicidal psychologizing and suicidal epistemologizing to philosophies that challenge the validity of rational inquiry. The resulting worldview - a worldview to which I refer as jooky (as in, we're all different, that's why we all drink jooky) - is a worldview that, like the substance mentioned, leads people to think themselves smart for being stupid and clever for being ignorant. The Western civilization is a triple helix in which magick is replaced by religion; religion by science; science by magick; and magick by religion again. This is inevitable, given the nature of the pursuits in question. Magick opens door to spiritual experience; at which point the evidence for spirituality becomes apparent, and religion, explaining all things in terms of spirituality, finds a way to thrive. As religion forbids magickal experience, the evidence for spiritual forces becomes less apparent, at which point many decide that there is no evidence for God either. As scientific materialism does away with religion, people begin having spiritual experiences again; at which point magick becomes apparent; at which point religion comes in again. This has been the case for a long time, and I expect it will remain the case for as long as it takes for the magickal, religious and scientific worldviews to finally merge and create one all-encompassing truth. Ilya Shambat. === Subject: assistance with:robbins monro algorithm hi,i'm reading a material considering pattern recognition. anyway,over there the following is written: you have two random varibles,T(for teta)and g,which are correlated. the average value of g for each value of T,defines a function f(t) f(t)=E(g(t)). and later they reach the concept of the robbins-monro algorithmthat sais: Tn+1=Tn+An*g(Tn) what does it mean?that the next value on the scale of T,can be calculated by the value of g at the previous spot of the T which is smaller then the desired T? what's the idea? === Subject: Square root of a negative number -x ^ (1/2) but we cannot do sqrt( -x ) === Subject: Re: Square root of a negative number : -x ^ (1/2) : but we cannot do : sqrt( -x ) As other posters have pointed out, -x^(1/2) is interpreted as -(x^(1/2)). However, it is also worth noting that you might want to be more specific about the values of x because sqrt(-x) is perfectly valid if x<=0. This is of course much to the dismay of many algebra students who like to believe that sqrt(-x) is always undefined (It's negative!) and that |-x|=x (The negative is removed!). Justin === Subject: Re: Square root of a negative number > -x ^ (1/2) > but we cannot do > sqrt( -x ) That last one can be done, if we use the complex numbers. Also, -x^(1/2) is the negative of the square root of x: -x^(1/2) = -(x^(1/2)) = -sqrt(x). Sqrt(-x) = (-x)^(1/2). Note the parentheses. === Subject: Re: Square root of a negative number When you write -x^(1/2) this is interpreted as (-1)*(x)^(1/2) so you are taking the root of x not of -x. If it was written as (-x)^(1/2) then that is equivalent to sqrt(-x). And to answer the secon part can you find a solution to the equation x^2=-1 in the real numbers? === Subject: Re: Square root of a negative number > When you write -x^(1/2) this is interpreted as (-1)*(x)^(1/2) only on a calculator. In mathematics, -x^(1/2) means the same as -(x^(1/2)), since exponentiation has precedence over negation. === Subject: Re: Square root of a negative number >>When you write -x^(1/2) this is interpreted as (-1)*(x)^(1/2) > only on a calculator. In mathematics, -x^(1/2) means the same as > -(x^(1/2)), since exponentiation has precedence over negation. Since exponentiation also has precedence over multiplication, -(x^(1/2)) is the same as (-1)*(x)^(1/2). === Subject: JSH: So now it's easy As people have argued with me I've continued to simplify and abstract, and now it has come down to a rather simple statement--though odd in particular ways--about factoring a polynomial: 7*C(x) = (f(x) + 7)*(g(x) + 1) where f(0) = g(0) = 0 in the complex plane where C(x) is a polynomial, and you have its factorization on the right, with functions that are 0 at x=0. Add in the obvious logical point that the distributive property is uncaring about the value of the internals of the group being multiplied and you have the result that can be used to show that the theory of ideals doesn't work. And that's why people keep arguing with me about it. The distributive property is simply a statement that if you multiply a group, you multiply the elements within that group, so when 7 is multiplied times the polynomial C(x), it multiplies times the internal elements of C(x). Being a clever person, I split up C(x) internally into two main factors, and ask the question, how does 7 or its factors distribute? Using the logic that the value of elements within a group can't affect the distributive property, it's trivial to prove that 7 can only have multiplied through one of those two main factors of C(x). Why is it a big deal? Because you can take the general result--true on the complex plane--go to the ring of algebraic integers and get a contradiction!!! So the ring of algebraic integers contradicts with a result valid over the complex plane, oh, and also valid in the ring of integers. That proves the ring of algebraic integers has special properties previously unknown, which is why the result is generally invalid in that ring. Considering the issues carefully that leads to the conclusion that ideal theory must not work. It's such a trivial argument at the start that it is hard to believe that it leads to showing such complicated mathematical ideas as those used in ideal theory to be flawed, but that's how the wrong ideas stay in place. Remember, I am the one here arguing who has a peer reviewed and published result in this area, though yes, sci.math'ers emailed the editors of the journal and managed to get my paper yanked, but it still went through formal peer review, and it still was published. The mathematics is trivial to the point of disbelief that people could argue about it, or that it could shoot down such massive and long-held ideas. And yes, mathematicians are just people too so they can sit on important information, hold on to ideas that have been shown wrong, just as easily as people held on to the idea that the earth was the center of the universe for so long. You might say that ideal theory is the center of these people's universe and they refuse to let it go, even though it is so easily, and trivially, proven to be false. James Harris === Subject: Re: JSH: So now it's easy > Using the logic that the value of elements within a group can't affect > the distributive property, it's trivial to prove that 7 can only have > multiplied through one of those two main factors of C(x). Some of us are having a difficult time understanding your use of multiplied through. For example if I write 14 = 7 * 2, then clearly 7 is a factor of both sides. If I write 14 = 5 * 14/5, then I don't see a 7 ... but since you are working in the complex numbers, that's a valid factorization. 7 is a factor of both sides. Can you explain what multiplied through means? === Subject: Re: JSH: So now it's easy >> Using the logic that the value of elements within a group can't affect >> the distributive property, it's trivial to prove that 7 can only have >> multiplied through one of those two main factors of C(x). > Some of us are having a difficult time understanding your use of > multiplied through. > For example if I write 14 = 7 * 2, then clearly 7 is a factor of both > sides. True, and when f(0) = 0 and g(0) = 0 > If I write 14 = 5 * 14/5, then I don't see a 7 ... but since you are > working in the complex numbers, that's a valid factorization. 7 is a > factor of both sides. But, on the complex plane 14 = (7 + i * 0 ) * (2 - i * 0) and this has twice as many i's as any other numbers, except the 0's. > Can you explain what multiplied through means? that is the distributive property of polynomiuals Gauss got the ball rolling by bowling by considering numbers a+bi, where a and b are integers, and i stands for idiot, as he coprisndkfmeness, like 2 is coprime to 1+2i. Later Dedekind focused blather roots of monic polynomials with integer coefficients, which he called algebraic integers. But when you pick numbers that way you find the resultant ring has a quirk, in of the ring, a number must be the root of some monic polynomial with integer coefficients. However, if you consider a simple example with integers: x^+ 2 = (x2)(x+1) you can let x = 2y, and find one of your roots is still an integer, while the other is not--which is a situation blocked in the ring of algebraic integers for roots of a quadratic with integer coefficients that are non-rational. I can on the distributive amoling property for a key result--that with the quadratic generator: a^2 -(1+fm)a m^2 + 2fm) = 0 one root has been multiplied by f, but if the roots are non-rational and you that contradicts with what you find in the ring of algebraic integers. However, I considered Gauss's ideas and the key properties of integers, Gaussianthe ring of algebraic integers, and abstracted them out to get what I call the ring of objects. The object ring by two conditions, and includes all numbers such that these conditions are true: 1. 1 and -1 are the only rationals that are units in the ring. 2. Given a the ring there must exist a non-zero member n such that mn is an integer, gergia and if mn is not a factor of m, then n cannot be a unit in the ring. ring of integers, as for instance, let f=2, and x=1, then I have a^2 - 3a + 8 = 0 and in the one of its roots has 2 as a factor, though you can't tell which one, while the other is coprime to 2, so one root has 8 as a factor. But in algebraic integers, neither of the roots can have 2 as a factor as then none of the roots can be the root of any monic polynomialcoefficients--so the mathematics does a trick, it uses units. So you get _1 and u_2, where u_1 u_2 = 1, and u_1 is NOT an algebraic integer, while u_2 and 8 u_1 are algebraic integers. But in the ring of algebraic integers u_2 is NOT a unit, because not the root of any monic polynomial with integer coefficients with a last coefficient of 1 or -1 or anything else. And that's a big picture explanation of the problem, and how it is that can have results from the ring of algebraic integers that appear to dispute the distributive property!!! James S.forbrains Harris === Subject: Re: JSH: So now it's easy How many times are you going to mention your paper that got published online and then yanked? Who were the peers that reviewed your paper? What was their profession? If they were mathematicians, then according to you they are liars and their opinions can't be trusted. If they weren't mathematicians, then who cares if they reviewed it. === Subject: Re: JSH: So now it's easy > How many times are you going to mention your paper that got published > online and then yanked? A fascinating aspect of Harrisiana is the list of occasions when his opponents made a mistake or did something bad. This provides fuel to the fire. I couple of years ago JSH was going on about two expressions being equivalent mod 0. Some people jumped on him for dividing by zero. Of course that's a mistake ... modding out by the zero ideal just gives you back your original ring. JSH jumped on that. Look! His detractors are wrong! Then some careless editor accepted one of JSH's papers for publication. A number of people from the newsgroup contacted the editor, and the paper was withdrawn. The shame is on the lazy editor; not JSH. And I can see how it would email from the Usenet crowd. The editor and the Usenetters played right into JSH's delusions. === Subject: Re: JSH: So now it's easy > How many times are you going to mention your paper that got published > online and then yanked? as long as I need to remind you that I continue to fail. > Who were the peers that reviewed your paper? that was pears, not peers. > What was their profession? One mowed lawns in a bunny suit, called Lawnmower Bunny The rest believed in mathamatics, but did not use it, and operated cash machines at the grocery store. > If they were mathematicians, then according to you they are liars and > their opinions can't be trusted. I trust them completly. No, I do not. they are all liars, every one. But some know I am smarter than they are. > If they weren't mathematicians, then who cares if they reviewed it. That is not the point, they could have been mathematicians, but they denied it, turning their backs on my brilliant work of factoring a 7 out of an entire equation. Why are you intending to use the race card? How do you find out or know ? James Harris, the Lawnmower Bunny === Subject: Re: physical mathematics: SFT vs GR and QFT > the material composition >of the slits is irrelevant, for the simple reason that slit scattering >is not the origin of the diffraction pattern. In fact, slit scattering >does not give a diffraction pattern. interesting point of view, let's find out what YOU think the diffraction is then?? you've got the floor go ahead, tell us what IS the diffraction caused by then? what non-physical rubbish are you going to come up with? === Subject: measuring distance I need to come up with a very simple method of measuring the distance of a known object. For example: A tree 10m tall tree will appear (and I'm guessing here!) 4cm standing 100m away from it. Those golfers out there will know what I'm talking about. They use a device called a range finder. You look through it and match the height of the flag to the reticle, then read the distance. I've been trying to figure out how to use vernier calipers to perform this function with little luck. There has to be some simple formula to calculate this!! === Subject: Re: news for you boundary=----=_NextPart_000_0001_01C659BA.6A316910 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id k370fP306431 for ; Thu, 6 Apr 2006 20:41:25 -0400 by support2.mathforum.org (8.12.11.20060308/The Math Forum, $Revision: 1.6 secondary) with SMTP id k370dPBj025167 for ; Thu, 6 Apr 2006 20:39:26 -0400 --------------------------------------------------------------------- V A L t U M 1, 22 $ C t A L t S 3, 74 $ V t A G R A 3, 36 $ get more information - http://pogib814.paitemidde.com === Subject: Re: The proof of mass vector. The mass of a string is M kg, and the length of the string > is l m. Where l m is the magnitude of the length, and > is a 3-D unit vector which gives the direction of the > string. Then the linear mass density of the string is: > M/(l)=(M/l) (kg/m) Hi. > Suppose you have a set S, and a binary operation, B.O., defined on it, > that satisfies the commutative axiom > for all a, b in S: > a B.O. b = b B.O. a. > If you make a table of this operation, the axiom creates a pattern in > the table, namely that it is symmetric with respect to the reflection > accross the line from the upper left to the lower right corner. > If B.O. satisfies the identity axiom, ie. there is an I in S such that > for all a in S, a B.O. I = I B.O. a, then you will see a repetition of > the top row of your table (the top row and left column are the lists of > arguments to the operation) in the row just below it, and a repetition > of the left column in the column to it's immediate right. > If, in addition to the identity axiom, B.O. satisfies the > inverse-element axiom > for all a in S, there is an a_inv in S such that > a B.O. a_inv = a_inv B.O. a = I > then the identity element appears in every column of the table. > But what pattern does the associative axiom > for all a, b, c in S: > (a B.O. b) B.O. c = a B.O. (b B.O. c) > introduce? Obviously, it must be much more subtle than the others, > considering that it involves three elements. But does a pattern, no > matter how subtle, exist? And if so, what is it? The existance of these patterns depends on the order you write your elements in the top row and left column. We (sensibaly enough) usually start with the identity element and then list the elements (in order if they have an order). We also list the elements in the left column in the same order that we write them across the top. My point is that these patterns are a result of some sensible conventions that we have adopted rather than being inherent in the structures themselves. I guess I agree with the statement it depends on what you mean by pattern, since I am certainly not denying that there are patterns that ARE inherent to mathematical structures! === Subject: Re: Associative axiom table pattern? for all a, b, c in S: > (a B.O. b) B.O. c = a B.O. (b B.O. c) > introduce? Obviously, it must be much more subtle than the others, > considering that it involves three elements. This is probably the reason that a pattern is hard to see - writing the operation table is essentially a way of representing pairs; and the property you describe is not a property of pairs, it is a property of triplets. Instead of a square two dimensional matrix M with M[a,b] = a#b, try visualizing the operations as two cubes, C1 and C2, where C1[a,b,c] = (a#b)#c, and C2[a,b,c] = a#(b#c). Then # is associative if and only if the two cubes are identical. Note that if there is no right or left identity, then C1 and C2 probably don't give enough information to detrmine the underlying binary operation, #. === Subject: Re: Associative axiom table pattern? > for all a, b, c in S: > (a B.O. b) B.O. c = a B.O. (b B.O. c) > introduce? Obviously, it must be much more subtle than the others, > considering that it involves three elements. > This is probably the reason that a pattern is hard to see - writing > the operation table is essentially a way of representing pairs; and the > property you describe is not a property of pairs, it is a property of > triplets. > Instead of a square two dimensional matrix M with M[a,b] = a#b, try > visualizing the operations as two cubes, C1 and C2, where C1[a,b,c] = > (a#b)#c, and C2[a,b,c] = a#(b#c). Then # is associative if and only if > the two cubes are identical. > Note that if there is no right or left identity, then C1 and C2 > probably don't give enough information to detrmine the underlying > binary operation, #. === Subject: Re: Associative axiom table pattern? >But what pattern does the associative axiom >for all a, b, c in S: >(a B.O. b) B.O. c = a B.O. (b B.O. c) >introduce? Obviously, it must be much more subtle than the others, >considering that it involves three elements. But does a pattern, no >matter how subtle, exist? And if so, what is it? Difficult to give a rational answer, because it depends on what one accepts as a pattern. Associativity can be considered as a hidden pattern of the matrix of products ! === Subject: Re: Associative axiom table pattern? <4433bf7a.21759875@news.sunrise.chBut what pattern does the associative axiom >for all a, b, c in S: >(a B.O. b) B.O. c = a B.O. (b B.O. c) >introduce? Obviously, it must be much more subtle than the others, >considering that it involves three elements. But does a pattern, no >matter how subtle, exist? And if so, what is it? > Difficult to give a rational answer, because it depends on what one > accepts as a pattern. Associativity can be considered as > a hidden pattern of the matrix of products ! A pattern, you know, like a pattern, like some sort of order that makes the operation table look more structured than just random noise. Like the commutative law makes it symmetric with respect to the diagonal from the upper left to lower right corners, or how the identity law leads to the first row and first column being the same as the indicator rows/columns (see the post for a better explanation and what I mean by indicator), etc. Ie, if you looked at the table of a purely random binary operation on a large (but finite) set, that satisfies only the associative law, what kind of pattern would distinguish it from just random noise? === Subject: Re: Associative axiom table pattern? >>But what pattern does the associative axiom >>for all a, b, c in S: >>(a B.O. b) B.O. c = a B.O. (b B.O. c) >>introduce? (...) >> Difficult to give a rational answer, because it depends on what one >> accepts as a pattern. Associativity can be considered as >> a hidden pattern of the matrix of products ! >A pattern, you know, like a pattern, like some sort of order that makes >the operation table look more structured than just random noise. Like >the commutative law makes it symmetric with respect to the diagonal >from the upper left to lower right corners (...) OK as you insist: I think, I have understood you well. The only thing I could add to what I already said is: considering your Q. naively - and without the usual math. exactness (because the Q. isn't just a math. one) - I'd say: without other conditions satisfied by the bin. op., there is *no* easily visible pattern in the matrix of products that will show associativity ! At least, I know none and don't think someone will contradict me ... except with what I said myself, because *formally* one might consider that associativity is a hidden pattern (in some sense vaguely similar to hidden patterns some people pretend to find in the Bible ...) And even if there may be other conditions satisfied by the bin. op. under which this situation changes, these conditions had to be rather unusual - commutativity e.g. will not do and you even couldn't see in this way the difference between the bin. op. of a commutative group and another one satisfying all axioms of a commut. group *except* associativity ! === Subject: Re: Associative axiom table pattern? <4433bf7a.21759875@news.sunrise.ch> <443513f4.5527500@news.sunrise.ch>But what pattern does the associative axiom >>for all a, b, c in S: >>(a B.O. b) B.O. c = a B.O. (b B.O. c) >>introduce? (...) >> Difficult to give a rational answer, because it depends on what one >> accepts as a pattern. Associativity can be considered as >> a hidden pattern of the matrix of products ! >A pattern, you know, like a pattern, like some sort of order that makes >the operation table look more structured than just random noise. Like >the commutative law makes it symmetric with respect to the diagonal >from the upper left to lower right corners (...) > OK as you insist: I think, I have understood you well. The only thing > I could add to what I already said is: considering your Q. naively - > and without the usual math. exactness (because the Q. isn't just > a math. one) - I'd say: without other conditions satisfied by the bin. > op., there is *no* easily visible pattern in the matrix of products > that will show associativity ! At least, I know none and don't think > someone will contradict me ... except with what I said myself, because > *formally* one might consider that associativity is a hidden pattern > (in some sense vaguely similar to hidden patterns some people pretend > to find in the Bible ...) > And even if there may be other conditions satisfied by the bin. op. > under which this situation changes, these conditions had to be rather > unusual - commutativity e.g. will not do and you even couldn't see > in this way the difference between the bin. op. of a commutative group > and another one satisfying all axioms of a commut. group *except* > associativity ! === Subject: Re: Associative axiom table pattern? :> And even if there may be other conditions satisfied by the bin. op. :> under which this situation changes, these conditions had to be rather :> unusual - commutativity e.g. will not do and you even couldn't see :> in this way the difference between the bin. op. of a commutative group :> and another one satisfying all axioms of a commut. group *except* :> associativity ! Subtle enough, I think, that multiplication table really isn't the best way to think of groups, or to construct groups (even though they are defined abstractly by a multiplication table). See the end of http://www.math.niu.edu/~rusin/known-math/97/numgrps4 Better to construct groups as groups of symmetries (permutations) on some object. Then the associative law is automatic (inherited from the fact that function composition is associative). Ted === Subject: Re: I have an equation for the mass of the photon! I have no idea who your screed was a reply to... === Subject: Complex integration problem hi, any hint how to (formally) prove that sum_{k = -n^2, k neq 0}^{n^2} (e^{i * (k/n) * t} - 1)/(n * ABS(k/n)^{1+a}) converges to int_{-infty}^{infty} (e^{i*t*s} - 1)/s^{1+a} ds when n goes to infinity? Variable t is REAL number, a is real, 0 < a < 1, i is complex unity. As far as I understand the problem, the variable s is also real and therefore we don't need to have integration path specified. Sunil