mm-3949 === 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