mm-4829 === Subject: Re: Trust metrics and economic system > I really don't know how to explain this well...but I'll try... I am looking for a non-inflationary and non-deflationary system that > will account for, reward and punish the behaviors of a set of players. I consider the following actions: - Create a team > - Recruit a player > - Lose a player > - Disband a team - Join a team > - Leave a team Each player would start with a number of points p and the total number > of points in the system would be p * the number of players at all > times. If a player leaves the system, the total number of points in > the system would decline by p points. If a new player joins the > system, the total number of points in the system does increase by p > points. The effect of each action should promote long-lasting teams (the goal) > and the system should establish a trust metrics where the player with > the most points at a particular time is the player who has contributed > to the goal the most. A bad example (because I haven't figured out the kind of math I need > to grab to make this coherent and logical): Starting balance for each player: 8 points - Create a team = +8 points > - Recruit a player = +4 points > - Lose a player = -4 points > - Disband a team = -8 points * number of players on the team - 8 > points - Join a team = +4 points > - Leave a team = -4 points How would approach this problem? This sounds like it cannot really work: I, as a player, could just create teams as fast as possible to gain points, not even need to recruit/join as creating teams is the most rewarding action. The fastest wins. I'd guess you rather need to define what a team is supposed to accomplish (e.g., killing other team's players, as in any respectable game), and reward that... (Apologise if I have misinterpreted.) -LV === Subject: Re: Trust metrics and economic system I really don't know how to explain this well...but I'll try... I am looking for a non-inflationary and non-deflationary system that > will account for, reward and punish the behaviors of a set of players. I consider the following actions: - Create a team > - Recruit a player > - Lose a player > - Disband a team - Join a team > - Leave a team Each player would start with a number of points p and the total number > of points in the system would be p * the number of players at all > times. If a player leaves the system, the total number of points in > the system would decline by p points. If a new player joins the > system, the total number of points in the system does increase by p > points. The effect of each action should promote long-lasting teams (the goal) > and the system should establish a trust metrics where the player with > the most points at a particular time is the player who has contributed > to the goal the most. A bad example (because I haven't figured out the kind of math I need > to grab to make this coherent and logical): Starting balance for each player: 8 points - Create a team = +8 points > - Recruit a player = +4 points > - Lose a player = -4 points > - Disband a team = -8 points * number of players on the team - 8 > points - Join a team = +4 points > - Leave a team = -4 points How would approach this problem? This sounds like it cannot really work: I, as a player, could just > create teams as fast as possible to gain points, not even need to > recruit/join as creating teams is the most rewarding action. The > fastest wins. I'd guess you rather need to define what a team is supposed to > accomplish (e.g., killing other team's players, as in any > respectable game), and reward that... (Apologise if I have misinterpreted.) -LV The point credits/debits I assigned to each action were just to illustrate. I know they do not work. That's the point of my query. I am looking for a way to assign these point credits/debits in a coherent, mathematically sound manner. I can see many possibilities, but I was wondering if this kind of problem is well known in Mathematica. Another example to address the issue you pointed at: - Create a team = -8 points - Recruit a player = +1 points - Lose a player = -1 points - Disband a team = -2 points * number of players on the team points - Join a team = +1 points - Leave a team = -1 points This assignment of point credits/debits works better. I didn't verify if it's inflationary/deflationary, though. And again, I am more interested in learning the fundamental issues with the underlying a mathematical model. PS: When teams compete, there'll be another scoring mechanism, such as the one you describe. === Subject: Re: Trust metrics and economic system posting-account=5nuXVgoAAABKTZU1rPm9p_kaNNZydOIK Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) Reformulating in an attempt to be more clear... - Consider a set of players. - Consider the following actions: * Create a team * Recruit a player * Lose a player * Disband a team * Join a team * Leave a team - Consider a player started with set number of points. Example: 8 - Consider each action implies a credit or debit of points. Example: * Create a team = -8 points * Recruit a player = +1 points * Lose a player = -1 points * Disband a team = -2 points x number of players on the team * Join a team = +1 points * Leave a team = -1 points PROBLEM How to determine the initial number of points each player receives, How to assign point credits/debits to each action, so that this system entices durable, strong teams and reliable players? Additionally, the total number of points a player acquires should tell something about their trustworthiness and reliability to serve as team players and/or team leaders. Examples of behaviors this system should strongly discourage: - Creating or disbanding teams too often or in too great a number. - Team hopping. - Taking actions for the sole purpose of accumulating points. - Inflation of the number of points in the system. - Deflation of the number of points in the system. Well, I can come up with point credits/debits assignment that intuitively appear good to me, but I wonder if there is an optimal solution to this problem. I value intuition, but if there was a mathematical model behind my solution, I would be more confident I am not totally off course. In the example I listed above: Player's initial point balance: 8 points * Create a team = -8 points * Recruit a player = +1 points * Lose a player = -1 points * Disband a team = -2 points x number of players on the team * Join a team = +1 points * Leave a team = -1 points I can tell a player will initially be able to create a single team. Because creating a team costs 8 points and a player starts with 8 points. I can tell that inviting a player to their team, a team leader grants credit to a player. This is expressed by the fact that Join a team is worth +1 point. I can also tell that when a player agrees to join a team, it grants credit to the team leader. This is expressed by the fact that Recruit a player is worth +1 point. However, this point credits/debits assignment does nothing to prevent players from switching teams constantly. Also, this point credits/debits assignment is deflationary, since disbanding a team removes 8 points from the total number of points in the system. I was told to consider zero-sum games, Nash equilibrium, Prisoner's dilemma, and a few other pieces of game theory, but I fail to derive a practical way to use them. === Subject: Re: Trust metrics and economic system > Additionally, the total number of points a player acquires should > tell something about their trustworthiness and reliability to serve > as team players and/or team leaders. If you want the pool of points to remain constant, then any possible > loop in game states must have zero change in total points. For example, a player leaving a team, joining another, leaving that > one and returning to their original team is a loop: the state > afterward is exactly the same as the state before. æIf you penalize > the team hopping player there by deducting points, you must reward > other players or the system will be deflationary. Which players? [ Aim to discourage ... ] - Taking actions for the sole purpose of accumulating points. The system cannot discourage such actions. It can have no way of > divining purpose. All you can do is to construct the rules in such a > way that normal play is good enough that maximizing play is not > worthwhile. - Tim Very interesting point about cycles. Maybe my requirement on the number of points in the system remaining constant must be eliminated. === Subject: Re: Trust metrics and economic system posting-account=5nuXVgoAAABKTZU1rPm9p_kaNNZydOIK rv:1.9.0.6) Gecko/2009011912 Firefox/3.0.6,gzip(gfe),gzip(gfe) > Additionally, the total number of points a player acquires should > tell something about their trustworthiness and reliability to serve > as team players and/or team leaders. If you want the pool of points to remain constant, then any possible > loop in game states must have zero change in total points. For example, a player leaving a team, joining another, leaving that > one and returning to their original team is a loop: the state > afterward is exactly the same as the state before. æIf you penalize > the team hopping player there by deducting points, you must reward > other players or the system will be deflationary. Which players? [ Aim to discourage ... ] - Taking actions for the sole purpose of accumulating points. The system cannot discourage such actions. It can have no way of > divining purpose. All you can do is to construct the rules in such a > way that normal play is good enough that maximizing play is not > worthwhile. - Tim > Very interesting point about cycles. Maybe my requirement on the number of points in the system remaining > constant must be eliminated. As long as the system puts a limit on the length on any cycle, abuse should be averted. === Subject: Re: Trust metrics and economic system posting-account=5nuXVgoAAABKTZU1rPm9p_kaNNZydOIK rv:1.9.0.6) Gecko/2009011912 Firefox/3.0.6,gzip(gfe),gzip(gfe) I added one action Dismiss a player. The current point credits/debits assignment is geared towards enticing long lasting, stable teams. Starting balance: +8 - Create a team -8 - Recruit a player +1 - Dismiss a player -2 - Lose a player -1 - Disband a team -2 * number of players on the team - Join a team +1 - Leave a team -2 === Subject: Re: A functional-differential equation >Ok, I fixed it (I think). For completeness, I'll give the full construction, incorporating the >modifications ... Let f_0 : R -> R be an infinitely differentiable function such that >f_0 is not identically 0 on (0,1) but f_0 (and all its derivatives) >are identically 0 outside of (0,1). This can be accomplished using >bump functions. Let f_1, f_2, f_3, ... be the successive derivatives of f_0, and let >f_(-1), f_(-2), f_(-3), ... be the successive antiderivatives of f_0, >uniquely determined by the initial condition f_(-n)(0) = f_(-n+1)(1), >for all n in N. Define f as a piecewise function, as follows ... For n in Z, and x in [n,n+1), define f(x) = f_n(x-n). I claim f satisfies f'(x) = f(x+1) for all x in R. > It satisfies f'(x) = f(x+1) for all non-integer x, but I don't think your function is differentiable at non-positive integers. It's continuous, certainly, but to get a C^oo function by putting together little pieces you've got to make sure not just that the values line up at the endpoints, but that all the derivatives also line up at the endpoints. For example, you know that f_{-1}(1) = f_0(0), but how do you know that f_{-1}'(1) = f_0'(0)? RS === Subject: Re: A functional-differential equation >Ok, I fixed it (I think). >For completeness, I'll give the full construction, incorporating the >modifications ... >Let f_0 : R -> R be an infinitely differentiable function such that >f_0 is not identically 0 on (0,1) but f_0 (and all its derivatives) >are identically 0 outside of (0,1). This can be accomplished using >bump functions. >Let f_1, f_2, f_3, ... be the successive derivatives of f_0, and let >f_(-1), f_(-2), f_(-3), ... be the successive antiderivatives of f_0, >uniquely determined by the initial condition f_(-n)(0) = f_(-n+1)(1), >for all n in N. >Define f as a piecewise function, as follows ... >For n in Z, and x in [n,n+1), define f(x) = f_n(x-n). >I claim f satisfies f'(x) = f(x+1) for all x in R. > It satisfies f'(x) = f(x+1) for all non-integer x, but I don't think >your function is differentiable at non-positive integers. It's >continuous, certainly, but to get a C^oo function by putting together >little pieces you've got to make sure not just that the values line up >at the endpoints, but that all the derivatives also line up at the >endpoints. For example, you know that f_{-1}(1) = f_0(0), but >how do you know that f_{-1}'(1) = f_0'(0)? I already posted a quick reply, but it hasn't appeared yet (server glitch). In any case, in that reply, I think I misinterpreted your objection, so forget that reply. I definitely don't have time to show all the proof details, but I'm fairly confident (not 100%, but still confident) that the function I defined satisfies the requirements of the problem. As far as infinite differentiability goes, it's immediate that f_n is infinitely differentiable for all n in Z. f_0 is infinitely differentiable since that's how it was chosen. f_n is infinitely differentiable for all n in N, since the functions f_1, f_2, f_3, ... are just the successive derivatives of f_0. f_(-n) is infinitely differentiable for all n in N, since the functions f_(-1), f_(-2), f_(-3) ... are successive antiderivatives of f_0. The initial values for the antiderivatives were specified so as to insure the correct glueing for f, but infinite differentiability is not dependent on that. Thus, the functions f_n are defined on all of R, and each f_n is the derivative of f_(n-1). Now f is a piecewise function, so we _do_ have to worry about whether f is differentiable. However we don't have to worry about infinite differentiability since once we show that f is differentiable and that f'(x) = f(x+1) for all x in R, infinite differentiability is automatic. However infinite differentiability of f would then be irrelevant since, once we've verified the identity f'(x) = f(x+1), we are done. I'm very tired right now, so I still may be missing the point of your objection, and if so, I apologize -- please bear with me. Also, if you reply to this message, I probably won't have time to reply (or even look at it) until tomorrow night at the earliest. But please reply anyway. But let me ask you this ... If you think the piecewise function f as I defined it is not differentiable, can you specify a particular value of x where you think differentiability fails? On the other hand, if you agree that f is everywhere differentiable, can you specify a particular value of x where you think f'(x) is not equal to f(x+1)? quasi === Subject: Re: A functional-differential equation >Ok, I fixed it (I think). >For completeness, I'll give the full construction, incorporating the >modifications ... >Let f_0 : R -> R be an infinitely differentiable function such that >f_0 is not identically 0 on (0,1) but f_0 (and all its derivatives) >are identically 0 outside of (0,1). This can be accomplished using >bump functions. >Let f_1, f_2, f_3, ... be the successive derivatives of f_0, and let >f_(-1), f_(-2), f_(-3), ... be the successive antiderivatives of f_0, >uniquely determined by the initial condition f_(-n)(0) = f_(-n+1)(1), >for all n in N. >Define f as a piecewise function, as follows ... >For n in Z, and x in [n,n+1), define f(x) = f_n(x-n). >I claim f satisfies f'(x) = f(x+1) for all x in R. >It satisfies f'(x) = f(x+1) for all non-integer x, but I don't think >your function is differentiable at non-positive integers. It's >continuous, certainly, but to get a C^oo function by putting together >little pieces you've got to make sure not just that the values line up >at the endpoints, but that all the derivatives also line up at the >endpoints. For example, you know that f_{-1}(1) = f_0(0), but >how do you know that f_{-1}'(1) = f_0'(0)? I already posted a quick reply, but it hasn't appeared yet (server >glitch). In any case, in that reply, I think I misinterpreted your >objection, so forget that reply. I definitely don't have time to show all the proof details, but I'm >fairly confident (not 100%, but still confident) that the function I >defined satisfies the requirements of the problem. As far as infinite differentiability goes, it's immediate that f_n is >infinitely differentiable for all n in Z. f_0 is infinitely differentiable since that's how it was chosen. f_n is infinitely differentiable for all n in N, since the functions >f_1, f_2, f_3, ... are just the successive derivatives of f_0. f_(-n) is infinitely differentiable for all n in N, since the >functions f_(-1), f_(-2), f_(-3) ... are successive antiderivatives of >f_0. The initial values for the antiderivatives were specified so as >to insure the correct glueing for f, but infinite differentiability is >not dependent on that. Thus, the functions f_n are defined on all of R, and each f_n is the >derivative of f_(n-1). Now f is a piecewise function, so we _do_ have to worry about whether >f is differentiable. However we don't have to worry about infinite >differentiability since once we show that f is differentiable and that >f'(x) = f(x+1) for all x in R, infinite differentiability is >automatic. However infinite differentiability of f would then be >irrelevant since, once we've verified the identity f'(x) = f(x+1), we >are done. I'm very tired right now, so I still may be missing the point of your >objection, and if so, I apologize -- please bear with me. Also, if you reply to this message, I probably won't have time to >reply (or even look at it) until tomorrow night at the earliest. But >please reply anyway. But let me ask you this ... If you think the piecewise function f as I defined it is not >differentiable, can you specify a particular value of x where you >think differentiability fails? I thought it would fail to be differentiable at negative integers. But after thinking about it a bit more I see that I was wrong. Let DL(f,x) be the derivative from the left of f at x, and DR(f,x) the derivative from the right. There is no question that DL(f,x) and DR(f,x) exist at all x. What bothered me was that I didn't see why we should have DL(f,n) = DR(f,n) when n is a negative integer. In other words, you design the pieces so that the values of the function match at the endpoints, but I didn't see why that should force the values of the derivatives to also match. But of course if you unpack the definitions and check it it does work out. RS === Subject: Re: A functional-differential equation >Ok, I fixed it (I think). >For completeness, I'll give the full construction, incorporating the >modifications ... >Let f_0 : R -> R be an infinitely differentiable function such that >f_0 is not identically 0 on (0,1) but f_0 (and all its derivatives) >are identically 0 outside of (0,1). This can be accomplished using >bump functions. >Let f_1, f_2, f_3, ... be the successive derivatives of f_0, and let >f_(-1), f_(-2), f_(-3), ... be the successive antiderivatives of f_0, >uniquely determined by the initial condition f_(-n)(0) = f_(-n+1)(1), >for all n in N. >Define f as a piecewise function, as follows ... >For n in Z, and x in [n,n+1), define f(x) = f_n(x-n). >I claim f satisfies f'(x) = f(x+1) for all x in R. > It satisfies f'(x) = f(x+1) for all non-integer x, but I don't think >your function is differentiable at non-positive integers. It's >continuous, certainly, but to get a C^oo function by putting together >little pieces you've got to make sure not just that the values line up >at the endpoints, but that all the derivatives also line up at the >endpoints. For example, you know that f_{-1}(1) = f_0(0), but >how do you know that f_{-1}'(1) = f_0'(0)? No -- that's not what lining up means. Lining up means f_(-1)'(1) = f_0(1), but that's automatic since f_(-1) is a full antiderivative of f_0 on _all_ of R, not just on a piece. quasi === Subject: Re: A functional-differential equation >Ok, I fixed it (I think). >For completeness, I'll give the full construction, incorporating the >modifications ... >Let f_0 : R -> R be an infinitely differentiable function such that >f_0 is not identically 0 on (0,1) but f_0 (and all its derivatives) >are identically 0 outside of (0,1). This can be accomplished using >bump functions. >Let f_1, f_2, f_3, ... be the successive derivatives of f_0, and let >f_(-1), f_(-2), f_(-3), ... be the successive antiderivatives of f_0, >uniquely determined by the initial condition f_(-n)(0) = f_(-n+1)(1), >for all n in N. >Define f as a piecewise function, as follows ... >For n in Z, and x in [n,n+1), define f(x) = f_n(x-n). >I claim f satisfies f'(x) = f(x+1) for all x in R. >It satisfies f'(x) = f(x+1) for all non-integer x, but I don't think >your function is differentiable at non-positive integers. It's >continuous, certainly, but to get a C^oo function by putting together >little pieces you've got to make sure not just that the values line up >at the endpoints, but that all the derivatives also line up at the >endpoints. For example, you know that f_{-1}(1) = f_0(0), but >how do you know that f_{-1}'(1) = f_0'(0)? No -- that's not what lining up means. Lining up means f_(-1)'(1) = f_0(1), but that's automatic since f_(-1) >is a full antiderivative of f_0 on _all_ of R, not just on a piece. Ignore this reply -- it was done quickly, and I think it may be missing the point of your objection. Instead, consider my later reply. I'll have more time to look at this later tonight or tomorrow. quasi === Subject: WM lying as usual In case you have not yet understood my proof, here it is again: This IS NOT a PROOF, DUMBASS. It doesn't have any axioms or rules of inference. It is chock full of bull. An ACTUAL proof has STEPS that FOLLOW from previous steps AND AXIOMS, follow via the application of rules of inference. > I consider FISON {1, 2, 3, ..., n} and all smaller FISONs {1, 2, > 3, ..., k} for k < n. The union of these FISONs is FISON {1, 2, > 3, ..., n}. Exactly. So since you were ALREADY CONSIDERING this, WHY ARE YOU BOTHERING considering the smaller ones and unioning them? THIS ADDS NOTHING TO THE ORIGINAL FISON ending in n! This is so for every union of finitely many FISONs. It is so, for every FISON, that IT IS EQUAL TO ITSELF. That the smaller ones are all subsets of the one ending in n means that your CLAIMING TO BE TAKING THE UNION simply CANNOT MATTER AT ALL. *THAT* is provable. Given that it doesn't matter, WHY have you invoked this union? Given that they don't add anything, WHY have you invoked these smaller FISONs?? > And in fact there are not infinitely many FISONs for any n. FISONs IN GENERAL *ARE NEVER* *for* ANY particular n! If you mean the FISON *ending in* n, THEN SAY SO!! But of course, you won't. You will continue to use your own bull terminology because Virgil is stupid enough to tolerate it. What FISON is for 42? What FISON is for 69? What FISON is for a googolplex? YOU CAN' SAY, because the notion of ANY fison being for ANY number IS JUST BULL. === Subject: Re: -- strict local mimima and level curves > >(3) Suppose f : R^2 -> R is continuous, with f(0,0)=0 and >f(p)>0 for all p in R^2 {(0,0)}. Also assume that f has no local >minima other than (0.0). Must all level sets of f be bounded? >(4) Assuming the answer to (3) is no, and with the same hypothesis, >must there be at least one bounded level set, with level greater than >0? >Ok, I got it. >The answer to (4) (and hence also to (3)) is no. >Here's an example ... >Define a family of ellipses as follows ... >For 0 < t < 1, let C_t be the ellipse whose equation is > (x/a)^2 + (y/b)^2 = 1 > >where a=t/(1-t) and b=t. >Now define f as follows: > f(0,0) = 0 > > f(x,y) = t, if (x,y) is on C_t > > f(x,y) = 1/y, if y>=1 > >Then f satisfies the hypothesis of (3) (and (4)), but all level sets >are unbounded, except for level 0. More precisely, all nonempty level sets are unbounded, except for >level 0. > > Oops -- I forgot to define f for y<=-1. > > Let me fix it ... > > Define f as follows: > > f(0,0) = 0 > > f(x,y) = t, if (x,y) is on C_t > > f(x,y) = 1/y, if y>=1 > > f(x,y) = -y, if y <= -1. > > Now all level sets with levels greater than 0 are unbounded. > > quasi A more explicit example is f(x,y) = (x^2 + y^2)*e^x. If c > 0, {f = c} is given by the equation y^2 = c*e^(-x) - x^2, whose graph contains arbitrarily large negative x values. So all level sets with levels greater than 0 are unbounded here. As for local extrema, the critical points are (0, 0) and (-2, 0). The first one gives the strict absolute minimum, and the second is a saddle point. So other than (0,0), there are no other local extrema. I got the idea for this by thinking of Picard's theorem. Suppose you have a nonpolynommial entire function F that vanishes at 0, with F nonzero elsewhere. Then Picard tells us F takes on every complex value infinitely many times, with one possible exceptional value. The exceptional value here is obviously 0, so for each w in C {0}, there is a sequence z_n -> oo with F(z_n) = w. Thus |F| has all level sets with levels greater than 0 unbounded. Also, |F| has no local extrema except for the origin by the maximum principle. So this will give you a slew of examples, and you can always consider |F|^2 if you want something super smooth. (The example above comes from looking at |F|^2 with F(z) = ze^z, giving (x^2 + y^2)*e^(2x). I changed the 2x to x for simplicity.) === Subject: Re: Riemann integrable => Lebesgue integrable > How does one prove Riemann integrable => Lebesgue integrable ? A Riemann integrable function f is continuous a.e. If you know this result, the above follows immediately. === Subject: Re: NOTE SUBMISSION How unusual! An imbecile posting from mathforum.org! I have found out a integral bigger than the equals sine! More > impotent blithering has never been.... === Subject: Re: #217 Chapter 7, are AP-adics well-ordered or not? ; new book 2nd edition: New True Mathematics > > (some snipping) > > infinity. That's ellipsis. It has several meanings. -- hz === Subject: Re: Function ranging over all real values on every open interval posting-account=HR1dqAkAAAA9E7mXiqvduHAelsrIxH3e Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > How can we construct function from R to R which, on every open interval, takes every real value? Hope I'm clear :-) This was the Ponder This problem some years back. I will do one from (0,1) to (0,1). You can use your favorite bijection from (0,1) to R before and after. Let x be in (0,1). Express x in ternary. If there are in infinite number of 2's in the expansion, set f(x)=x and ignore it. Otherwise, let the last 2 be in the n place. Multiply x by 3^n and take the fractional part, which consists of only 0's and 1's. Now read it as binary. To prove this works, consider that you have a target value y that you want to find an x such that f(x)=y. Note that every interval contains a subinterval of the form ((3m+2)/3^n,(3m+3)/ 3^n). All the points in this subinterval will have a 2 in the n place. Append the binary expansion of y to the ternary expansion of (3m+2)/ 3^n. Read it in ternary and that is your x. === Subject: Re: Function ranging over all real values on every open interval > > How can we construct function from R to R which, on every > open interval, takes every real value? > > We can't, but if you want to know how to *prove* that such a function > exists, see example 1.2 from: > > s#PPA5,M1 > > > Jose Carlos Santos That example doesn't prove there's no constructive approach. === Subject: Re: Function ranging over all real values on every open interval > How can we construct function from R to R which, on every > open interval, takes every real value? > We can't, but if you want to know how to *prove* that such a function > exists, see example 1.2 from: > s#PPA5,M1 > > That example doesn't prove there's no constructive approach. I never said it did. Jose Carlos Santos === Subject: Re: Function ranging over all real values on every open interval > > How can we construct function from R to R which, on every > open interval, takes every real value? > We can't, but if you want to know how to *prove* that such a function > exists, see example 1.2 from: > ion > s#PPA5,M1 > > That example doesn't prove there's no constructive approach. > > I never said it did. > > > Jose Carlos Santos Your answer to how can we construct was We can't, but if you want to know how to *prove* that such a function exists ... What exactly did you mean by that then? === Subject: Re: Function ranging over all real values on every open interval > How can we construct function from R to R which, on every > open interval, takes every real value? > We can't, but if you want to know how to *prove* that such a function > exists, see example 1.2 from: > ion > s#PPA5,M1 > That example doesn't prove there's no constructive approach. > I never said it did. > > Your answer to how can we construct was We can't, but if you want > to know how to *prove* that such a function exists ... What exactly > did you mean by that then? I meant that such a function cannot be constructed but, in spite of that, its existence could be proved. Then I told the OP where he could see such a proof. If the OP had asked how one could construct a non-continuous function from the reals into the reals which preserves the sum I would have given him the same reply (with a different link, of course :-) ). The idea that the existence of such a proof would prove that such a function cannot be constructed never crossed my mind. I don't understand how you could have interpreted my reply that way. Jose Carlos Santos === Subject: Re: Function ranging over all real values on every open interval > > How can we construct function from R to R which, on every > open interval, takes every real value? > We can't, but if you want to know how to *prove* that such a function > exists, see example 1.2 from: > ct > ion > s#PPA5,M1 > That example doesn't prove there's no constructive approach. > I never said it did. > > Your answer to how can we construct was We can't, but if you want > to know how to *prove* that such a function exists ... What exactly > did you mean by that then? > > I meant that such a function cannot be constructed but, in spite of > that, its existence could be proved. That is false, and that is my point. See the example I gave. > Then I told the OP where he could > see such a proof. > > If the OP had asked how one could construct a non-continuous function > from the reals into the reals which preserves the sum I would have > given him the same reply (with a different link, of course :-) ). > > The idea that the existence of such a proof would prove that such a > function cannot be constructed never crossed my mind. I don't understand > how you could have interpreted my reply that way. It crossed my mind that you believed the proof you cited, which appeals to the existence of Hamel bases, surely indicates that no constructive example can be given. But you're right, I didn't voice my objection correctly. Let me try again: What led you to believe we can't construct such an example? > > Jose Carlos Santos === Subject: Re: Function ranging over all real values on every open interval > How can we construct function from R to R which, on every > open interval, takes every real value? > We can't, but if you want to know how to *prove* that such a function > exists, see example 1.2 from: > ct > ion > s#PPA5,M1 > That example doesn't prove there's no constructive approach. > I never said it did. > Your answer to how can we construct was We can't, but if you want > to know how to *prove* that such a function exists ... What exactly > did you mean by that then? > I meant that such a function cannot be constructed but, in spite of > that, its existence could be proved. > > That is false, and that is my point. See the example I gave. You are right and I was wrong. > Then I told the OP where he could > see such a proof. > If the OP had asked how one could construct a non-continuous function > from the reals into the reals which preserves the sum I would have > given him the same reply (with a different link, of course :-) ). > The idea that the existence of such a proof would prove that such a > function cannot be constructed never crossed my mind. I don't understand > how you could have interpreted my reply that way. > > It crossed my mind that you believed the proof you cited, which > appeals to the existence of Hamel bases, surely indicates that no > constructive example can be given. But you're right, I didn't voice my > objection correctly. Let me try again: What led you to believe we > can't construct such an example? What led to my belief was the fact that I was convinced that such a function could not be Lebesgue measurable. Yes, you constructed such a function, as well as two other posters. I wasn't thinking straight. :-( Jose Carlos Santos === Subject: Re: Free product solvable? > stronger: The free product of two non-identity > groups is solvable if and only if both groups are > cyclic of order 2. (that is, no finiteness > assumed). > > One can extend that result to free products with > amalgamation. Solvable H*_G K (free product of H and > K amalgamating a subgroup G of both) iff G is > solvable and of index 2 in H and K. > > im Sorry if this is a trivial question, but why do you need G to be solvable? Steve === Subject: Re: Free product solvable? <31793229.1233960754875.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > stronger: æThe free product of two non-identity > groups is solvable if and only if both groups are > cyclic of order 2. (that is, no finiteness > assumed). One can extend that result to free products with > amalgamation. Solvable H* G K (free product of H and > K amalgamating a subgroup G of both) iff G is > solvable and of index 2 in H and K. im Sorry if this is a trivial question, but why do you need G to be solvable? G is a subgroupp of the amalgamated product. If a group has a non- solvable subgroup, then it is non-solvable. That is also why H and K need to be solvable in the free product case. -- Arturo Magidin, sans .sig === Subject: Re: Free product solvable? > One can extend that result to free products with > amalgamation. Solvable H*_G K (free product of H > and K amalgamating a subgroup G of both) iff G is > solvable and of index 2 in H and K. > Sorry if this is a trivial question, but why do you > need G to be solvable? The clear answer is: > G is a subgroup of the amalgamated product. If a > group has a non-solvable subgroup, then it is non- > solvable. That is also why H and K need to be solvable > in the free product case. You might have been confused by two things not specifically addressed: In H *_G K you do quotient out by many copies of G, but you leave half of them, so that this group does contain G as a subgroup. Also we don't explicitly mention that both H and K are solvable, since they have an index 2 solvable subgroup (which must be normal) and so are solvable. === Subject: Re: Free product solvable? > One can extend that result to free products with > amalgamation. Solvable H*_G K (free product of H > and K amalgamating a subgroup G of both) iff G > is > solvable and of index 2 in H and K. > > Sorry if this is a trivial question, but why do > you > need G to be solvable? > > The clear answer is: > > G is a subgroup of the amalgamated product. If a > group has a non-solvable subgroup, then it is non- > solvable. That is also why H and K need to be > solvable > in the free product case. > > You might have been confused by two things not > specifically addressed: In H *_G K you do quotient > out by many copies of G, but you leave half of > them, > so that this group does contain G as a subgroup. > Also > we don't explicitly mention that both H and K are > solvable, since they have an index 2 solvable > subgroup (which must be normal) and so are solvable. Steve === Subject: Modular forms spanned by Eisenstein series over the some field Hi! I've heard that the space of modular forms of some weight k equals to the span of some eisenstein series over some field. Can anyone give me a detailed explanation of this? (eg. spanned by which eisenstein series and spanned over which field?) Galois === Subject: Re: order homomorphism theorem > To put the problem into perspective, let P be a partition of S. How > does one define an order for S/P, the partition parts, that's induced > from S? That is different from you original problem, because the partition > may not come from the kernel (relation) of a map to some other poset. > There may be no order relation such that the canonical map from S to S/P > is order preserving (take e.g. S = {0 < 1 < 2} and P = { {0,2} , {1} }. Under well known premises, a group can be partition into a group that > reflects the original group. What premises are needed for an ordered set \ > to be partition into an ordered set that reflects the original order?[...] One more remark: there may be more than one order on S/P such that the canonical map from S to S/P is order preserving. Then one may take the smallest such order. So it boils down to existence: Let ~ be the equivalence relation that corresponds to the partition. Define a mixed chain in S as a tripel (n , a , b) where n is a natural number and a,b: n --> S such that a(i) ~ b(i) and b(i) <= a(i+1) for all i in n. So you have a situation like this a(0) ~ b(0) <= a(1) ~ b(1) <= ... <= a(n-1) ~ b(n-1) A mixed chain (n, a, b) is said to go from x to y if x <= a(0) and b(n-1) <= y. A circuit is a mixed chain (n, a, b) going from a(0) to a(0), i.e. that satisfies b(n-1) <= a(0). By extending chains like this x' ~ x <= a(0) ~ b(0) <= ... <= a(n-1) ~ b(n-1) <= y ~ y' you can see that the relation there goes a chain from x to y is compatible with the equivalence relation. Therefore one can define a relation on S/~ via [x] <# [y] if there goes a chain from x to y Then one can check that x <= y ==> [x] <# [y] holds for all x,y in S and that the relation <# is reflexive and transitive. So the only obstacle to <# being an order on S/~ is the possible lack of antisymmetry. Now suppose [x] <# [y] and [y] <# [x], let's say via mixed chains (n, a, b) and (m, c, d) going from x to y and from y to x. Pasting the chains together gives x ~ x <= a(0) ~ ... ~ b(n-1) <= y ~ y <= c(0) ~ ... ~ d(m-1) <= x ~ x which is a circuit. So the necessary and sufficient condition for <# to be antisymmetric is: all elements in a circuit are equivalent. Observe that the previous convexity condition a ~ b and a <= c <= b force a ~ c ~ b is the special case of the circuit b ~ a <= c ~ c So we have the equivalent conditions (i) S/~ has an order such that the canonical S --> S/~ is order preserving (ii) all elements in a circuit are equivalent Moreover, if this holds then order <# is the smallest order on S/~ such that S --> S/~ is order preserving because every mixed chain from x to y x <= a(0) ~ b(0) <= ... <= a(n-1) ~ b(n-1) <= y will then give a (real) chain [x] <= [a(0)] = [b(0)] <= ... <= [a(n-1)] ~ [b(n-1)] <= y in any such order on S/~ so that [x] <= [y]. -- Marc === Subject: Re: #198 Chapter 7, The two large flaws in the Peano Axioms and how \ to fix them; new book 2nd edition: New True Mathematics sha1:mVm78lPd+T4Q0OlkZT0jHeuRs84= > Maybe, but if I recall correctly, he sometimes > æ 987...000...789, > and that can't be described thus. I've seen AP consistently writing the same digit before and after the > ellipsis, so that no ambiguity is ever possible. In fact, if I am not > mistaken, he has more than once restated the concept in this thread > too. You are just playing dumb: no big news, of course. Again, you're simply wrong. Even with the same digit before and after the ellipsis, notation like 0000....00100....00000 is very ambiguous. What does it mean? In what position is the 1? Furthermore, he discusses all possible digit sequences, so *obviously* the notation is ambiguous! Anything can go in the ellipsis. Finally, I don't believe that he is so consistent as you say (that the last digit before and after the ellipsis are the same), but I confess I haven't found a counterexample. Seems they came up various times that he's tried to define multiplication, but I didn't find such an example. -- When you go to class today, if your professor talks about algebraic number theory, or misuses Galois Theory[,] I want you to carefully notice how you feel. Hold on to that feeling so that you never forget it. --James S. Harris, on channeling rage via Galois theory. === Subject: Re: #198 Chapter 7, The two large flaws in the Peano Axioms and how \ to fix them; new book 2nd edition: New True Mathematics <87ocxfdrzo.fsf@phiwumbda.org> posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > Maybe, but if I recall correctly, he sometimes > æ 987...000...789, > and that can't be described thus. I've seen AP consistently writing the same digit before and after the > ellipsis, so that no ambiguity is ever possible. In fact, if I am not > mistaken, he has more than once restated the concept in this thread > too. You are just playing dumb: no big news, of course. Again, you're simply wrong. æEven with the same digit before and after > the ellipsis, notation like 0000....00100....00000 is very ambiguous. > What does it mean? In what position is the 1? Nobody has been using that notation: no more than one ellipsis is allowed, otherwise of course it is ambiguous. You keep injecting confusion. > Furthermore, he discusses all possible digit sequences, so *obviously* > the notation is ambiguous! æAnything can go in the ellipsis. Nope. Talking about all possible digit arrangements and the ambiguous notation you inject have nothing to do together. Again, think P-adics: at least that's something you should have respect for, I suppose. > Finally, I don't believe that he is so consistent as you say (that the > last digit before and after the ellipsis are the same), but I confess > I haven't found a counterexample. I appreciate you honesty. -LV === Subject: Re: #198 Chapter 7, The two large flaws in the Peano Axioms and how \ to fix them; new book 2nd edition: New True Mathematics sha1:QVULTEBLvdz4zpXY62zcIA7L4Wk= > Maybe, but if I recall correctly, he sometimes > æ 987...000...789, > and that can't be described thus. > I've seen AP consistently writing the same digit before and after the > ellipsis, so that no ambiguity is ever possible. In fact, if I am not > mistaken, he has more than once restated the concept in this thread > too. > You are just playing dumb: no big news, of course. > Again, you're simply wrong. æEven with the same digit before and after > the ellipsis, notation like 0000....00100....00000 is very ambiguous. > What does it mean? In what position is the 1? Nobody has been using that notation: no more than one ellipsis is > allowed, otherwise of course it is ambiguous. You keep injecting > confusion. Why don't you look up the following post from AP: Then let us know if I'm still the guy injecting confusion here. -- Jesse F. Hughes I will admit I can get giddy over these forays into ideas at the extreme edge. Being wrong can just add to the fun, oddly enough. --James S. Harris just wants to have fun. === Subject: Re: #198 Chapter 7, The two large flaws in the Peano Axioms and how \ to fix them; new book 2nd edition: New True Mathematics sha1:rI4xfyhmctOxgw37bJO5InX7Opw= > In Chaitin's Omega, it is perfectly clear what is intended to follow > the three dots: an omega sequence of digits. æWhich digits may be > unknown, but we know what type fills in the dots. æIn AP's stuff, it > is perfectly unclear. æConsider > æ 999...999 > Is there an omega sequence of 9's followed by three 9's? æ(Probably > not.) æIs there an omega sequence of 9's, followed by an omega^op > sequence of 9's? As far as I have seen, that's simply meant as an omega sequence (of > 9's, in this case), from start to end. Can't be! Omega sequences have no end. It makes no sense to write 999...999 to stand for an omega sequence of 9s. -- Jesse F. Hughes Do not click any hyperlinks that you do not trust. Type them in the Address bar yourself. -- Microsoft gives security advice. === Subject: Re: #198 Chapter 7, The two large flaws in the Peano Axioms and how \ to fix them; new book 2nd edition: New True Mathematics <87skmrdsfh.fsf@phiwumbda.org> posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > In Chaitin's Omega, it is perfectly clear what is intended to follow > the three dots: an omega sequence of digits. æWhich digits may be > unknown, but we know what type fills in the dots. æIn AP's stuff, it > is perfectly unclear. æConsider > æ 999...999 > Is there an omega sequence of 9's followed by three 9's? æ(Probably > not.) æIs there an omega sequence of 9's, followed by an omega^op > sequence of 9's? As far as I have seen, that's simply meant as an omega sequence (of > 9's, in this case), from start to end. Can't be! Omega sequences have no end. æIt makes no sense to write 999...999 to > stand for an omega sequence of 9s. æ You (I don't mean you only) seem to have problems with notation: I suppose you would have no objection in reading (9), but you cannot accept 9...9. -LV === Subject: Re: #198 Chapter 7, The two large flaws in the Peano Axioms and how \ to fix them; new book 2nd edition: New True Mathematics sha1:wGq8afRXbd4QfB+iTUsnbx7+RL4= > In Chaitin's Omega, it is perfectly clear what is intended to follow > the three dots: an omega sequence of digits. æWhich digits may be > unknown, but we know what type fills in the dots. æIn AP's stuff, it > is perfectly unclear. æConsider > æ 999...999 > Is there an omega sequence of 9's followed by three 9's? æ(Probably > not.) æIs there an omega sequence of 9's, followed by an omega^op > sequence of 9's? > As far as I have seen, that's simply meant as an omega sequence (of > 9's, in this case), from start to end. > Can't be! > Omega sequences have no end. æIt makes no sense to write 999...999 to > stand for an omega sequence of 9s. æ You (I don't mean you only) seem to have problems with notation: I > suppose you would have no objection in reading (9), but you cannot > accept 9...9. with a first digit, a last digit and an infinite number of digits in between. -- Jesse F. Hughes If anything is true in general about Usenet, it's that people can go on and on about just about anything. -- James Harris speaks the truth. === Subject: Re: #198 Chapter 7, The two large flaws in the Peano Axioms and how \ to fix them; new book 2nd edition: New True Mathematics posting-account=NBUN0woAAAAnKUwZeEs32wbbmpbmo_jZ Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > He refuses to define exactly what the '...' is supposed to mean beyond > the vacuous explanation of an infinite number of digits, offering no > mathematically sound basis or explanation of it. So once again, Tribble rejects the AP-reals just because > some of the require an ellipsis, the three dots '...', in > order to specify their digits. In that case, one can reject the standard reals for the > exact same reason! There are some standard reals whose > decimal digits can't be given without an ellpsis! One > well-known example is the uncomputable Chaitin's Omega. How deceitful. In Chaitin's Omega, it is perfectly clear what is intended to follow > the three dots: an omega sequence of digits. æWhich digits may be > unknown, but we know what type fills in the dots. æIn AP's stuff, it > is perfectly unclear. æConsider æ 999...999 Is there an omega sequence of 9's followed by three 9's? æ(Probably > not.) æIs there an omega sequence of 9's, followed by an omega^op > sequence of 9's? As far as I have seen, that's simply meant as an omega sequence (of > 9's, in this case), from start to end. And what are the start and end of the omega sequence 3.14159....., the digits of the decimal representation of pi? In fact, if you start 9999....9999999 from the left, when do you meet the sequence coming from the right? Given any 987234.....09234, and given that this indicates you have a calculable sequence of digits from both left and right, how can you not have the two sequences fly by each other - and note there's no particular nth digit from the left or mth digit from the right such that it represents a meeting place of the two sequences, since they would then both be finite. So they both grope towards each other never meeting. So to define any abc......xyz, you really can only claim this represents two distinct sequences that you choose to represent as one. What advantage do you gain by this? I'm not saying there isn't any, I just can't see it at the moment. >(Although, in analogy to the P- > adics, I am not 100% sure calling it an omega sequence is completely > correct: my own lack of proper knowledge, in any case. Just think P- > adics, and you won't get it too wrong.) Maybe, but if I recall correctly, he sometimes æ 987...000...789, and that can't be described thus. I've seen AP consistently writing the same digit before and after the > ellipsis, so that no ambiguity is ever possible. In fact, if I am not > mistaken, he has more than once restated the concept in this thread > too. You are just playing dumb: no big news, of course. -LV === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=UM3jRwkAAADTHFmJ20qgwageu031CeWA FunWebProducts; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; PeoplePal 3.0; eMusic DLM/4; CNPVersion2 - Congoo NetPass; FDM; .NET CLR 3.5.30729; .NET CLR 3.0.30618; RRHSO_BLD1),gzip(gfe),gzip(gfe) Fonzi physics? You mean you intend to discuss quantum field theory with Ralph and Potsie? === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=wigfZgkAAACDgITarXffzxJygX81YRSs Gecko/2008121622 Fedora/3.0.5-1.fc9 Firefox/3.0.5,gzip(gfe),gzip(gfe) > BACKGROUND Any analytical individual will pick up the mathematical inconsistency > of Special Relativity. æThe algebraic equations of the Lorentz > transformations are symmetric. æMeaning applied one way, t is > transformed to t' wherein t'>t. æHowever, in the obverse process, the > same equation transforms t' to t resulting in t>t'. æIn summary the > LTE implies t http://farside.ph.utexas.edu/teaching/em/lectures/node109.html http://en.wikipedia.org/wiki/Energy-momentum tensor http://en.wikipedia.org/wiki/Einstein field equations This mathematical fallacy shows up in the twin paradox, where only one > twin ages slowly, while the other is exempt from time dilation. æWhile > orthodox physics explains the slowly aging twin as due to effects of > acceleration (an explanation inadequate in itself), it cannot explain > why the ænormally aging twin is EXEMPT from time dilation. << Einstein's relativity principle states that: All inertial frames are totally equivalent for the performance of all physical experiments. In other words, it is impossible to perform a physical experiment which differentiates in any fundamental sense between different inertial frames. By definition, Newton's laws of motion take the same form in all inertial frames. Einstein generalized this result in his special theory of relativity by asserting that all laws of physics take the same form in all inertial frames. > http://farside.ph.utexas.edu/teaching/em/lectures/node108.html Where is there anything about slowing clocks or ageing twins. Both of Einstein's SR postulates seem to be true. Both your paragraphs seem to be false. Sue... Classical Electromagnetism: An intermediate level course http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html Visualizing Electricity and Magnetism at MIT http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm http://einstein.stanford.edu/highlights/status1.html === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=-Yr6sAoAAAArJwgdHLA4MBxm4oIdzdtZ InfoPath.1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8),gzip(gfe),gzip(gfe) NetApp/6.0.2P1) Back to the subject. In another thread, Eric has been pathetically trying to prop up a GPB positive result. Here is Eric being schooled on the fine art of data analysis. Eric palpitates: Everitt is the principle investigator of the project. The cited up already - you have written several hundred posts about this, and you need to be quiet. Strich calmly replies: You mean silenced? I must be touching a nerve. proper citation should be made): Author: C W F Everitt et al Title: Gravity Probe B data analysis status and potential for improved accuracy of scientific results Journal: Class. Quantum Grav. 2008;25(114002) Now Eric, try to focus. Look at title. Let me quote the good part. It states potential for improved accuracy of scientific results. What does POTENTIAL mean in this context? It means that whatever they are planning to accomplish, is NOT YET accomplished. This means what we are reading are PRELIMINARY RESULTS. Preliminary results of what? Let us read on. It states improved accuracy of scientific results. Improved from where? And this is the CLINCHER. It means 'improved' from the 'INITIAL ANALYSIS' which had shown a NEGATIVE RESULT. What we are seeing now is a SECONDARY ANALYSIS. Anything but the primary analysis intended in the ORIGINAL experimental design becomes a POST-HOC analysis [Eric, please review POST-HOC* so you can understand what I am telling your small coconut: http://en.wikipedia.org/wiki/Post-hoc Obviously the GPB result is NEGATIVE based on the ORIGINAL experimental design. This publication by Everitt is a preliminary paper on what is a POST-HOC analysis. --------------------------------------------------------------------- As it is with many research, the data may be analyzed in one way or in many ways. The data is INTENDED to be analyzed only in one way. That is the mark of a good experimental design. However, none is prevented from re-examining the data and analyzing it in other ways. This is called a POST-HOC* analysis, and in statistics, the results from this kind of analysis is not as valid compared to the intended analysis. What you have cited is a preliminary result of a post-hoc analysis. You may wait for the final result of this post-hoc analysis, but these results, whether positive or negative, is not statistically valid, and needs another experiment in its own right. *[A post-hoc analysis means that after examining the raw data, the examiner comes up with a new way of analyzing the data that is different from the original analysis in the original experimental design. Since the examiner has already seen the raw data, he can then tweak the new analysis to force a positive result. Example: suppose that you perform a quarters-tossing experiment and arrive at 51% heads and 49% tails; a researcher then examines the data and decides to throw away the last 50% of measurements, his reason being the coin was deformed in the later part and the result is not random; note that this was not part of the original experimental design; however, as he tosses out the last 50% of measurements, the remaining data shows that there are 30% heads and 70% tails, thereby showing that quarters are biased towards tails; the latter is a post-hoc analysis and is not valid compared to the pre-hoc analysis.] Strich, IQ200 === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Firefox/3.0.5,gzip(gfe),gzip(gfe) > Preliminary results of what? æLet us read on. æIt states improved > accuracy of scientific results. æImproved from where? æAnd this is > the CLINCHER. æIt means 'improved' from the 'INITIAL ANALYSIS' which > had shown a NEGATIVE RESULT. Except the analysis did not show a negative result. === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Firefox/3.0.5,gzip(gfe),gzip(gfe) Preliminary results of what? æLet us read on. æIt states improved > accuracy of scientific results. æImproved from where? æAnd this is > the CLINCHER. æIt means 'improved' from the 'INITIAL ANALYSIS' which > had shown a NEGATIVE RESULT. Except the analysis did not show a negative result. > Except in this case, whereas pretty much any but positive is negative. æPonzi physics uncovered. æNow what? (do it again and again until them > cows come home?) æ~ BG ...and, once again for the reading impaired, THE ANALYSIS DID NOT SHOW A NEGATIVE RESULT. === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Firefox/3.0.5,gzip(gfe),gzip(gfe) > Preliminary results of what? æLet us read on. æIt states improved > accuracy of scientific results. æImproved from where? æAnd this is > the CLINCHER. æIt means 'improved' from the 'INITIAL ANALYSIS' which > had shown a NEGATIVE RESULT. Except the analysis did not show a negative result. > Except in this case, whereas pretty much any but positive is negative. æPonzi physics uncovered. æNow what? (do it again and again until them > cows come home?) æ~ BG ...and, once again for the reading impaired, THE ANALYSIS DID NOT SHOW > A NEGATIVE RESULT. It also didn't show a sufficient positive result to qualify as based > upon their own empirical standards. æ~ BG What the hell is it with you people and this inability to READ? GP-B didn't meet design sensitivity but it did - unquestionably - observe the effects in question. === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=PTS84AoAAACr67p51zvy0Hlr3LkoIUcc 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; InfoPath.2),gzip(gfe),gzip(gfe) > What the hell is it with you people and this inability to READ? GP-B > didn't meet design sensitivity but it did - unquestionably - observe > the effects in question. The college drop-out needs to know that he cannot throw a fit just because he cannot get what he wants. Why is your whining not applied to accept the null results of the MMX, but GP-B does? There is a lot of rooms for double standards when dealing with politics but none in science. Again, the college drop-out just has to learn that. === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Firefox/3.0.5,gzip(gfe),gzip(gfe) [snip stupidity] The woo-woo brigade is in full force. Crackpotter makes the 4th moron who is unable to click a link, read the abstract, and admit being wrong. === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Firefox/3.0.5,gzip(gfe),gzip(gfe) What the hell is it with you people and this inability to READ? GP-B > didn't meet design sensitivity but it did - unquestionably - observe > the effects in question. The college drop-out needs to know that he cannot throw a fit just > because he cannot get what he wants. æWhy is your whining not applied > to accept the null results of the MMX, but GP-B does? æThere is a lot > of rooms for double standards when dealing with politics but none in > science. æAgain, the college drop-out just has to learn that. æ http://www.iop.org/EJ/abstract/0264-9381/25/11/114002 Moron. === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=PTS84AoAAACr67p51zvy0Hlr3LkoIUcc SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; Media Center PC 5.0; \ InfoPath.2),gzip(gfe),gzip(gfe) > The college drop-out needs to know that he cannot throw a fit just > because he cannot get what he wants. Why is your whining not applied > to accept the null results of the MMX, but GP-B does? There is a lot > of rooms for double standards when dealing with politics but none in > science. Again, the college drop-out just has to learn that. http://www.iop.org/EJ/abstract/0264-9381/25/11/114002 These guys had already wasted almost one billion dollars. You got to be out of your mind for me to spend an extra $30 to tell me which I have already known that the whole project is indeed a fiasco. MMX showed null results, and physicists lived with them. In doing so, that revolutionized the development of physics. GP-B also showed null results, and it is time for the scientists to accept just that and move on. It is also time to devolutionize the development of physics. The blatant and shameless usurp of Einstein Dingleberries in the development of physics in the last one hundred years needs to find a closure. === Subject: Re: PONZI physics... and the Gravity Probe B experiment > The college drop-out needs to know that he cannot throw a fit just > because he cannot get what he wants. Why is your whining not applied > to accept the null results of the MMX, but GP-B does? There is a lot > of rooms for double standards when dealing with politics but none in > science. Again, the college drop-out just has to learn that. \ http://www.iop.org/EJ/abstract/0264-9381/25/11/114002 > These guys had already wasted almost one billion dollars. You got to > be out of your mind for me to spend an extra $30 to tell me which I > have already known that the whole project is indeed a fiasco. > MMX showed null results, and physicists lived with them. In doing so, > that revolutionized the development of physics. GP-B also showed null > results, and it is time for the scientists to accept just that and > move on. It is also time to devolutionize the development of > physics. The blatant and shameless usurp of Einstein Dingleberries in > the development of physics in the last one hundred years needs to find > a closure. Moron. > Eric Gisse is simply pro Ponzi, no matters what the consequences, and > as always without any speck of remorse for their Ponzi past, present > or future. As long as it's our hard earned loot that's getting spent > and there's not even a speck of concern for the environmental impact > of such missions, because they also never apply the true all-inclusive > (birth to grave) cost. > > ~ BG The Lense-Thirring effect already showed up in ground-based satellite laser ranging tracking data to Lageos I and II. Gravity probe B did not improve the situation, they way I understand it Gravity probe B is unfortunately a failure. Q -- No signature === Subject: Re: PONZI physics... and the Gravity Probe B experiment > The college drop-out needs to know that he cannot throw a fit just > because he cannot get what he wants. Why is your whining not applied > to accept the null results of the MMX, but GP-B does? There is a lot > of rooms for double standards when dealing with politics but none in > science. Again, the college drop-out just has to learn that. http://www.iop.org/EJ/abstract/0264-9381/25/11/114002 > These guys had already wasted almost one billion dollars. You got to > be out of your mind for me to spend an extra $30 to tell me which I > have already known that the whole project is indeed a fiasco. > MMX showed null results, and physicists lived with them. In doing so, > that revolutionized the development of physics. GP-B also showed null > results, and it is time for the scientists to accept just that and > move on. It is also time to devolutionize the development of > physics. The blatant and shameless usurp of Einstein Dingleberries in > the development of physics in the last one hundred years needs to find > a closure. Moron. > Eric Gisse is simply pro Ponzi, no matters what the consequences, \ and > as always without any speck of remorse for their Ponzi past, present > or future. As long as it's our hard earned loot that's getting spent > and there's not even a speck of concern for the environmental impact > of such missions, because they also never apply the true all-inclusive > (birth to grave) cost. > ~ BG > The Lense-Thirring effect already showed up in ground-based satellite > laser ranging tracking data to Lageos I and II. Gravity probe B did not > improve the situation, they way I understand it Gravity probe B is > unfortunately a failure. > Q > -- > No signature > > As are many public funded research projects that either get spun into > positive hype and spendy infomercials that makes their conditional > laws of physics no matters how subjective or evidence excluding, seem > perfectly objective and pretty much exactly of what they had expected. > > How about an honest side by side listing of the good, the bad and the > ugly kinds of public funded research, so that we can see exactly where > our hard earned public loot has gone. I'm talking about the all- > inclusive and thus birth to grave cost, and not merely of the original > funding budget. > > ~ BG Sometimes money is spend on a project that fails, at best you can hope that it does not happen again. -- No signature === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Firefox/3.0.5,gzip(gfe),gzip(gfe) The college drop-out needs to know that he cannot throw a fit just > because he cannot get what he wants. æWhy is your whining not applied > to accept the null results of the MMX, but GP-B does? æThere is a lot > of rooms for double standards when dealing with politics but none in > science. æAgain, the college drop-out just has to learn that. æhttp://www.iop.org/EJ/abstract/0264-9381/25/11/114002 These guys had already wasted almost one billion dollars. Over nearly 40 years, and developed some useful technology in the process. Given that you are universally unhappy with how science works, I don't find that your complaint has any more merit than the others. >æYou got to > be out of your mind for me to spend an extra $30 to tell me which I > have already known that the whole project is indeed a fiasco. Yes I can see how this would appear to be an insurmountable problem. Going to a library is clearly not an option, as that's too much effort . Emailing the PI and asking nicely is not an option as you are simply unable to be civil. Asking me nicely for a copy is not an option for the same reason, even though my flexible moral compass would allow me to give you a copy if you simply asked nicely without being a ing douchebag about it. But I have to wonder, if you have no access to the basic literature, where do you get off making claims about results you clearly do not have access to? A true mystery of the ages - I will consult my giant pine cone in lieu of a magic 8 ball. MMX showed null results, and physicists lived with them. æIn doing so, > that revolutionized the development of physics. Sure, but you reject the results so I'm less than clear what your point is. You don't accept special or general relativity, and your position on Einstein is clear so that means you also think quantum mechanics is wrong. > æGP-B also showed null > results, and it is time for the scientists to accept just that and > move on. The result isn't null - read the literature. Oh wait you haven't...hmm! But if you haven't read the literature, how do you know? Are you sole- sourcing your information based off what a loon says even though the details are clearly available for those who can lift their hands to click around on the Stanford site? > æIt is also time to devolutionize the development of > physics. æThe blatant and shameless usurp of Einstein Dingleberries in > the development of physics in the last one hundred years needs to find > a closure. æ Moron. How's the fight going, by the way? Do anything other than post your complaints to USENET? === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=PTS84AoAAACr67p51zvy0Hlr3LkoIUcc 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; InfoPath.2),gzip(gfe),gzip(gfe) > [snip crap] What is your argument? === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008122219 Gentoo Firefox/3.0.5,gzip(gfe),gzip(gfe) [snip crap] What is your argument? Crackers don't matter. === Subject: Re: PONZI physics... and the Gravity Probe B experiment posting-account=PTS84AoAAACr67p51zvy0Hlr3LkoIUcc 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; InfoPath.2),gzip(gfe),gzip(gfe) > Please go back writing children's books. [...] you made such an awful goof about relative simultaneity. > [...] As I said, you need to go back writing children's books before embarrassing yourself any further. === Subject: Accounting solution manuals / test banks posting-account=8cWRQwoAAAAhaenOECdoYi5uBwUSHq5w Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) Hello I have available the following test bank / solutions available - $25 each West Federal Taxation 2008: Corporations, Partnerships, Estates, and Trusts - Solution Manual West Federal Taxation 2008: Corporations, Partnerships, Estates, and Trusts - Test Bank West Federal Taxation 2007 : Comprehensive - Test Bank West Federal Taxation 2008 : Comprehensive Appendix E - Solution Also have test bank Advanced Financial Accounting (7e) - Jeffrey, Baker and King for $50 Please email me at masterp7184 [at] yahoo [dot] com === Subject: Inroads on the ( Generalized ) Collatz Problem > ... CGT stems from certain inferences about base 2 > which must ultimately occur toward the inception of some > endgame. So there is a kind of Collatz Problem aspect to Go, > which is still unresolved (also known as the 3x + 1 problem). This paper is invaluable: ( version September 2008 ) http://www.numbertheory.org/pdfs/survey.pdf Here is some UBASIC to get started -- may need 2+ Ghz. -------------- 10 S=2^32:randomize 20 for T=2 to 100:M=prm(T) 30 F=1:for P=1 to T-1:F*=prm(P):next:D=F^50 40 J=0:for I=0 to 99999:K=0:C=int(rnd*S):' --- collatz.ub 50 while and{num(C)<>1,K<2000}:inc K:C=(1+num(C)*M)//D:wend 60 J+=num(C)<>1:next:print M,J,J/I 70 next -------------- Some improvement though (obscure) loop-detection is possible. Are the zero entries actual or merely failures to observe any loops? A few odd composites are included to illustrate their behavior -- more prone to (pathological?) looping probability: 3 0 5 0 7 0 (9) 0.675... 11 0.206... 13 0.167... (15) 0.98.. 17 0.098... 19 0.00023.. (21) 0.95... 23 0.0147.. 29 0 31 0.131... 37 0.004... 41 0 43 0.00005... 47 0.00322... 53 0.00014... 59 0.1568... 61 0.07... 67 0.023... 71 0 73 0.0052... 79 0 83 0.0198... 89 0 97 0.0075... 101 0.00377... 103 0.00033.. 107 0 109 0 113 0.0071... 127 0 131 0 137 0 . . . at `C' = 3,5,7,29,41... loop probability ( `J/I' ) appears to be zero. \ at least from almost 3,000,000 trials on `C' = 29, randoms < 2^128. Blowup will not occur on this generalized Collatz mapping so positive integers reach their halting condition, terminating at 1. Care must be taken with such inferences, noting the many primes on the list that demonstrate obvious non-zero loop overlaps. Well, there are certain clean places to play and others not so clean in terms of contributions to strategy and overall concept. What follows is John Conway's Fractran processor. Setting loop limits at 1-14 in line #30, and entering `2' at input will yield prime number generator, in order. When iterating on fraction list for loop limits 15-24 there is an illustration of the 3x+1 proof -- the sequence converted to orbits in the powers of 2 (see paper). Perhaps there is a generalizing method of illustrating a Fractran code for which Collatz Primes produce a halting condition. -------------- 10 dim A(24):for I=1 to 24:read A(I):next:' conway's fractran for prime# 20 input W:P=W ' `P' should be 2 for the prime# generator 40 next I:stop 50 if W<>2 then print P;:if P=2 then stop else goto 30 60 if bitcount(P)=1 then print len(P)-1; 70 goto 30 75 ' ---------- 1 to 14 prime generator 80 data 17//91,78//85,19//51,23//38,29//33,77//29,95//23,77//19 90 data 1//17,11//13,13//11,15//14,15//2,55//1 95 ' ---------- 15 to 24 3x+1 proof 97 ' -- http://www.emis.de/journals/DMTCS/pdfpapers/dm050103.pdf 100 data 1//11,136//15,5//17,4//5,26//21,7//13,1//7,33//4,5//2,7//1 -------------- - jb ------------------------------------------------------------------- Transparent Frogs Discovered In Colombia http://ecoworldly.com/2009/02/05/transparent-frogs-discovered-in-colombia/ ------------------------------------------------------------------- === Subject: Re: Inroads on the ( Generalized ) Collatz Problem > May also need to have UBASIC explained as not everyone > is familiar with it. While true, our discussion could get easily sidetracked by such details which are amply available as documentation in the UBASIC help file. > What does prm(T) function do? Prime? T-th prime, within reasonable computation limits. > What does num(C) do? Returns the numerator from (if rational type) `C' . > What does M//D do? Integer division? Produces type rational (simplified by their gcd). > Why does print M,J,J/I only print two numbers? And which two? Good call. I omitted column `J' from the output. > And what are these numbers supposed to mean? Though an interesting question I suppose it qualifies as a philosophy investigation rather than an engineering idea. Correction for `M' = 71: 21 of 3.98 million trials were nonzero, indicating possible looping (or sequences longer than 2,000) a result placing into jeopardy other zero assumptions on the list. And, thank you for your interest. I examined your post made to rec.games.go on June 9, 2007, but obtained different results: -------------------- === Subject: Re: X beats Y, Y beats Z, Z beats X [ ... ] > Now when playing a game where high number wins > A vs C or B vs C is evenly matched (each side wins > equally often although in some pairings not every > match is won due to tying and the B vs C never ties). I think you mean to talk about an -expectation- because in practice I do not find that each side wins equally often... > But if all three play simultaneously, C wins about > 52% of the time compared to A and B which win about > 24% each. Instead I obtain A winning about 32% of the trials, B wins 32.9% of the trials, and C wins 35% of the trials. This occurs rather consistently, using 1,000,000 trials. > This is due to A and B tying which automatically > makes C the winner. `C' could still have rolled less than either `A' or `B'. > And since C can never tie B, C has an advantage. `C' has an advantage over 'A', but so does `B' it seems. ----------------------------------------- 10 dim A%(6,6):randomize: ' 3-WayDice --- 3Way.Ub 20 for I=1 to 6:for J=1 to 6:read A%(I,J):next:next 30 R2=13 : for K=0 to 999999 40 R1=A%(1,int(1+rnd*6))+A%(2,int(1+rnd*6)) 50 ' R2=A%(3,int(1+rnd*6))+A%(4,int(1+rnd*6)) 60 R3=A%(5,int(1+rnd*6))+A%(6,int(1+rnd*6)) 70 R1r2+=R1>R2:R2r1+=R2>R1 72 R1r3+=R1>R3:R3r1+=R3>R1 74 R2r3+=R2>R3:R3r2+=R3>R2 80 next : Z=R1r2+R2r1+R1r3+R3r1+R2r3+R3r2 100 print r1 v r2 ;R1r2-R2r1,r1 v r3 ;R1r3-R3r1,r2 v r3 ;R2r3-R3r2 105 print using(3,4),r1= ;(R1r2+R1r3)/Z, 106 print using(3,4),r2= ;(R2r1+R2r3)/Z,r3= ;(R3r1+R3r2)/Z 110 stop 200 data 1,2,3,4,5,6,7,8,9,10,11,12 205 data 6,6,6,6,6,6,7,7,7,7,7,7 210 data 1,2,3,10,11,12,4,5,6,7,8,9 ----------------------------------------- - jb ----------------------------------------------------------------- French fighter planes grounded by computer virus http://www.telegraph.co.uk/news/worldnews/europe/france/4547649/French-fight\ er-planes-grounded-by-computer-virus.html ----------------------------------------------------------------- === Subject: Draft paper submission deadline extended: TMFCS-09 posting-account=NjTN6QoAAACcj_0TitH5yQen5d07uP1E .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) Draft paper submission deadline extended: TMFCS-09 The deadline for draft paper submission at the 2009 International Conference on Theoretical and Mathematical Foundations of Computer Science (TMFCS-09) (website: http://www.PromoteResearch.org ) is extended due to numerous requests from the authors. The conference will be held during July 13-16 2009 in Orlando, FL, USA. The conference will take place at the same time and venue where several other international conferences are taking place. The other conferences include: ´ International Conference on Artificial Intelligence and Pattern Recognition (AIPR-09) ´ International Conference on Automation, Robotics and Control Systems (ARCS-09) ´ International Conference on Bioinformatics, Computational Biology, Genomics and Chemoinformatics (BCBGC-09) ´ International Conference on Enterprise Information Systems and Web Technologies (EISWT-09) ´ International Conference on High Performance Computing, Networking and Communication Systems (HPCNCS-09) ´ International Conference on Information Security and Privacy (ISP-09) ´ International Conference on Recent Advances in Information Technology and Applications (RAITA-09) ´ International Conference on Software Engineering Theory and Practice (SETP-09) ´ International Conference on Theory and Applications of Computational Science (TACS-09) The website http://www.PromoteResearch.org contains more details. DevipSr Publicity committee CC: Stephen J. Herschkorn === Subject: Re: real roots of P(x) = x^n + a x^2 + b >Let n is odd natural number. I need to prove that the polynomial P(x) >= x^n + a x^2 + b has at most 3 real roots. > Is it O.K. to invoke Descartes' Rule of Signs? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey === Subject: Re: lattice points in a circle posting-account=lxTw-goAAABZ0peiD1HYaXeP9eKmZSOS AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.46 Safari/525.19,gzip(gfe),gzip(gfe) A point (m,n) in the plane R^2 is a lattice point if both >coordinates m and n are integers. æProve that the number >lattice points inside any circle centered at the origin >is a number of the form 4k+1 for some integer k. Let k represent the number of lattice points (m,n), where >m,n in Z, that lie on the x-axis. Because any point along >the circumference of a circle to the center is equidistant, >A rotation of pi/2 radians about the origin, yields the same >number of lattice points along the positive y-axis due to symmetry. >If we rotate twice more, each by pi/2 radians about the origin, >then we will cover the negative x and y axis with an equal >number of lattice points. æTherefore, 4 such rotations >yields 4k lattice points. æHowever, (0,0) is also contained >in the circle, thus there are 4k + 1 lattice points >inside any circle. How does the above stand as a proof? æPathetic, adequate, >etc? æThe more proofs I do the better I'll become, I hope, and >so any feedback on how to improve the above is much >appreciated. Where in your proof did you mention the lattice points strictly inside > the quadrants? Of course, the argument is essentially the same. By rotational symmetry (or alternatively, by algebraic symmetry of the > polynomial x^2 + y^2), the number of lattice points strictly inside > the quadrants is the same for each quadrant, hence the total is a > multiple of 4. Then add the 4 equal counts for pieces on the axes, > excluding the origin, and finally, add the origin. My only point is that you seem to have focused only on the lattice > points on the axes. quasi Ah yes, that is much better. -- conrad === Subject: PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA PET PRODUCTS FACTORY PET BED Dog bed supplier in china cat tree factory china pet product manufacturer car dog bed pen posting-account=JfxyvAoAAABu_ghKkbkNWDuekhvXB1Xc Mister X; GTB5; TencentTraveler 4.0; Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ),gzip(gfe),gzip(gfe) CHINA PET BEDS WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG BED PET BED dog bed manufacturer pet products cat tree factory iron dog bed china pet factory pet CHINA PET BEDS WWW.PETBED-CATTREE.COM PET BED CHINA Pet Factory Boat Car dog bed factory cat tree manufacturer pet products supplier in china pet factory dog CHINA PET BEDS WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG BED PET BED dog bed cat tree factory pet products manufacturer wrought iron dog bed pen in china CHINA PET BEDS WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG BED PET BED dog bed manufacturer cat tree pet products factory cat furniture supplier in china pet CHINA PET BEDS WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG BED PET BED cat tree factory dog bed manufacturer pet products supplier in china dog bed pen cat PET BEDS CHINA WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG BED PET BED dog bed manufacturer pet products factory cat tree furniture supplier iron dog bed pen PET BEDS CHINA WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG BED PET BED dog bed cat tree manufacturer pet products factory iron dog bed car dog bed pen china PET BEDS CHINA WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG BED PET BED dog bed factory china pet bed manufacturer cat tree supplier in china pet factory cat bed PET BEDS CHINA WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG BED PET BED dog bed cat trees manufacturer pet products factory wrought iron dog bed pen in china PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA PET PRODUCTS FACTORY PET BED Dog bed factory cat tree manufacturer pet products factory iron dog bed supplier in china === Subject: CHINA PET BEDS WWW.PETBED-CATTREE.COM PET BED CHINA PET FACTORY DOG \ BED PET BED dog bed manufacturer cat tree factory iron dog bed car dog bed pen supplier in china posting-account=JfxyvAoAAABu_ghKkbkNWDuekhvXB1Xc Mister X; 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6th Edition <23902277.1233725568559.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=WL_GnQoAAADBwXe7_iAFpBywDZDz-bSs 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > On Feb 4, 5:32æam, whoisstep...@gmail.com i have the complete instructors manual for Hibbeler 6th edition structural analysis in electronic format. æemail me at whoisstep...@gmail.com If you still have this solution manual please let me know > Mark Hi Mark, yes i still have this manual. email me at whoisstephen@gmail.com and i will be able to provide you with any details you like. === Subject: Re: Moon Rock Found In Arctic - More Evidence Supporting Apollo Hoax Theory posting-account=u3pAnQkAAAAzxnLVBonSyCKugd0mZCzf Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) >http://www.astronomy.com/asy/default.aspx?c=a&id=4526 > What sort of physicist or mathematician can't understand that non- > parallel sun-shadows implies a local light-source implying movie-set > staged pictures? > æ æ What sort of photographer doesn't realize that shadows on curved > or rough surfaces look different from shadows on flat surfaces? æWhat > sort of hiker doesn't realize that natural surfaces, especially > uneroded volcanic surfaces, are often curved or rough? What sort of > person who has entered a dark room doesn't know that diffuse > reflectance can illuminate an entire room, even if it is in shadow? > We already know what a liar thinks, so you don't have to keep > reminding us via your spew. > Our Selene/moon is upon average dark as coal (not as light grayish and > xenon arc lamp spectrum illuminated as a certain terrestrial guano > island). > æ æ æGood. Coal has diffuse reflection. There is no such thing as a > soil or rock so dark it won't reflect any light. Paint the walls of a > room black, turn out the lights, and open the door from outside. The > light ricocheting around the walls will still allow one to see the > walls. > æ æ æIf you shine a light on a piece of slate, almost as dark as coal, > you will see the surface. If the surface of the moon was smooth, one > might even see a specular reflection. However, you don't see much in > the way of specular reflection not because the surface is so dark. The > reason is that the soil reflects diffusely. > æ æ What are you calling dark as coal? I don't think there is any > surface which reflects less than 2% or the light incident on it. For > most materials, the specular reflectance (i.e., the imaging type of > reflection) is negligible. However, most sufaces reflect upwards of > 10% of the light incident on them. Including surfaces most people > think of as dark. > æ æ æVarious optical companies sell devices for absorbing light. There > is a type of horn, shaped so that light entering it must ricochet many > times before leaving. The inside is painted black. This device > reflects less than 1% of the light entering it. æHowever, please note > that it is the multiple reflection that wittles down the remitted > light. So far as I know, there is no moon crater that has this > particular shape. > æ æLight bouncing off one side of the crater will spread and make the > shaded part visible. If you don't believe it, buy a cup made of black > plastic. Shine a light into it. Not only will you see the surface by > the diffuse reflected light, you may even see a little specular > reflection due to the flat surfaces. Look at a black car in the sun. > Black cars are very shiny. If you roughen the surface, you get rid of > the specular reflection. However, the specular reflectivity that is > lost goes into the diffuse reflectivity. > æ æ There is no such thing as a completely black surface. In fact, the > index of refraction of a material increases with absorptivity. This is > an unavoidable consequence of electrodynamics (see the > æ æ To make a totally black absorber, you have to make a small hole > in a container which the inside surface is completely black. > æ æ æYou need some courses, or at least do experiments, in optics. As > a first project, I suggest you search for a material surface that > absorbs more than 95% of the light incident on it. Lamp black is > probably the darkest surface you will find. There may even be some > measurements available on the Web. > æ æ Also note that the eye can adjust to up to five orders of > magnitude in intensity. So the eye can see very small amounts of > reflected light. > æ æ You sound so much like my antiSemitic employers. I painted a > container black to absorb the light, and then was accused of > witchcraft because witches use black. Where was ZOG when I really > needed it? > æ æ Furthermore, how can the moon be absolutely black, as you imply, > and be covered with salt?-) > Damn! You actually answered his point. Wow. You will now be accused of > being in league with GW Bush and/or Adolf Hitler and there will be > references to Nazi/Zionists as well. > Your totally out-of-context and otherwise incorrectly founded rants > are noted. How is it out of context to point out that people who disagree with > you are accused of being in league with GW Bush AND Adolf Hitler (at > the same time no less). I'm impressed you stopped bringing the Borg > and incest-clones into it. Furthermore, how is it an incorrectly > founded rant when anyone doing a Google groups search under Brad > Guth and Apollo Moon Hoax will find that all you ever do is insult > people and make claims that you never seem to back up with facts and > figures? As soon as you have something/anything objective and thus based > entirely upon the regular laws of physics, we'll talk. æUntil then you > have nothing but more of the same old mainstream subjective science, > that which oddly can't be peer replicated outside of your DARPA and > NASA cabal. No. We can talk when you provide some verifiable evidence for your > technobabble rants. A simple verifiable cite to anything you've > claimed on this thread alone would be sufficient. What it all boils down to is little Brad Guth is angry because NASA > won't take his opinions on potential life on Venus seriously. If you don't like the regular laws of physics and can't be the least > bit bothered with considering the best available science, that's not > my responsibility. No, your responsibility is to provide verifiable evidence for the claims you've continued to post. Something you claim to have done, but quite frankly I've never seen anything verifiable posted by you... just a lot of hot air. > In the mean time, I'm just returning the warm and fuzzy topic/author > stalking and bashing favor with all the love and affection that I can > muster. I guess you have to stick with what you're good at. God knows it's not actually holding a real conversation. > The matter of FACT that our NASA has screwed up multiple times isn't > my fault. Nobody said it was. > If I were BHO, I'd fire the entire lot and restart our NASA from > scratch, and the very next time a public servant tells an honest hard > working American to get lost is when that person plus any of their > collaborating staff gets fired without one cent of retirement or any > benefits whatsoever, much less given any kind of employment transfer > privileges. If you were in charge, you'd be sending missions to Venus to find extraterrestrials that aren't there and firing everyone who had the good sense to argue with you about it. > You seem to be employed by either NASA or DARPA, as otherwise you > wouldn't have been so intent upon stalking and systematically trashing > each and every word I had to offer. æIn fact, there's little if any > difference between yourself and the likes of Art Deco or Saul Levy, > which means you must also be Jewish if not a devout Zionist. Wrong and wrong again. This is something you're really good at, being wrong that is. > Just because your PC or MAC has been made firmware unable to find > anything on its own, especially if such evidence puts those Apollo > missions at risk, as such is also not my fault. æRemember that I'm not > the one claiming as having those marvelous fly-by-rocket landers and > having spent a day on our naked moon without ever so much as a scratch > or any traumatized cell of DNA. It's not my job to find your evidence. It's yours to provide it. > For some reason James Alfred Van Allen was fully aware of the risks, > and yet he wasn't permitted to publicly speak about or much less > publish his expertise on such matters of space travels outside of our > protective magnetosphere, or much less near and/or upon our naked > Selene/moon that's as bad if not worse off than any GSO environment. > It seems Van Allen was actually highly in favor of using rad-hard > robotics, therefore if you trash me you are also trashing Van Allen, > because thus far I 100% concur with his research and findings. Bull. Robots wouldn't have gotten one/tenth the funding NASA got for the entire space program (Mercury, Gemini and Apollo) And again, you've made a claim (paste) > protective magnetosphere, or much less near and/or upon our naked > Selene/moon that's as bad if not worse off than any GSO environment. (/paste) without posting anything that verifies it. But that's simply par for you, isn't it? === Subject: Re: Moon Rock Found In Arctic - More Evidence Supporting Apollo Hoax Theory posting-account=u3pAnQkAAAAzxnLVBonSyCKugd0mZCzf Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > You clearly are a devout brown-nosed minion or worse of the mainstream > status quo. æNext time you post a reply, please flush. > æ~ BG > Well, I'm sorry Darwin123, you didn't get a GW Bush vis a vis Adolf > Hitler reference or a Zionist/Nazi slam. You will have to settle for > the consolation prize of a devout brown-nosed minion reference. > Some of us still need silly objective evidence of those fly-by-rocket > controlled soft landings, rad-hard EVAs and their safe return without > so much as a hitch, any least bit fogged or even UV secondary reaction > kind of Kodak moment. æGot any of that? > Oops! æ My mistake, as I forgot that your Zionist Nazi laws of physics > work differently while anywhere near or on the moon. > æ~ BG > Hey Dawin123! I won the zionist nazi reference! yay me! > Seriously Brad, such as what? Short of taking you to the moon and > letting you touch the LM yourself, what would satisfy you? > Independently obtained objective science that can be peer replicated, > of which JAXA and ISRO should each have more than accomplished by now, > that is if there was anything remotely similar to our fairly extensive > and highly reflective Apollo mission remains, and/or of their > extensively modified lunar surface terrain to image upon (other than > depicting a few artificial impact sites). OK, less technobabble, you're losing Disneygeek! What specifically > would be the nature of the independently obtained objective science? > Who would have to be the group that would independently obtain > objective science? How do you peer replicate a moon program without > having another moon program to do it? And, let's say China or India > actually send something up and go over to one of our landing sites and > send back pictures, what would be to stop you from claiming THOSE were > faked? > As I've also said before; æShow me your R&D prototype fly-by-rocket > lander demo of that era, and I'll show you mine. æOf course mine would > have incorporated at least two fairly powerful momentum reaction > wheels, in addition to having a much lower inert CG, plus at least > half again as much fuel. You of course had the benefit of hindsight. If in fact I believed your > R&D was in any way shape or form practical... which, since I don't > know your qualifications in astro-engineering, I don't! The point I'm trying to make, is that we still have no such viable fly- > by-rocket lander that's suitable for that kind of complex mission. > However, as of today it should be obtainable, though just not > demonstrated because ???????? æ~ BG A hundred years ago, you could buy a Stanley Steamer, today you can't. > Does that mean the Stanley Steamer was a conspiracy and never really > existed? We had a clutzy system put in place 40 years ago that worked > as well as it was expected to. Then the focus changed from manned > landings to reusable spacecraft and the Apollo stuff was put in the > back because it wasn't expected to be needed again. The good old Stanley Steamer is 100% objective proof positive as you > can get, unlike our Apollo ruse/sting of the century whereas most of > everything mission critical has either been systematically disposed > of, lost or being kept as taboo/nondisclosure rated. æ~ BG You can show me a Stanley Steamer, but you can't buy one now. Is that proof that the Steamer never really existed as a car and all you have are models? No of course not. Truth is, there is nothing that would convince you that we'd gone to the moon. If the Chinese or the Indians send their little robots to shoot our 40 year old landing sites, you'd claim they were gotten to and the footage was faked. === Subject: Graph theory posting-account=lxTw-goAAABZ0peiD1HYaXeP9eKmZSOS AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.46 Safari/525.19,gzip(gfe),gzip(gfe) I'm working over a proof of V - E = 1 where V is the number of vertices and E is the number of edges and the whole thing is in terms of trees. Now, my definition of a tree is a connected graph with no cycles. But I've also come across a definition that said: a connected graph with no nontrivial cycles. Which made me think, what is a trivial cycle and how can a tree have one? Beyond that, the tack to proving the above is starting with a tree T and removing one of the edges, reducing the problem to two subtrees that have been proved. But before this, one must specify how you end up with two trees if you remove an edge from a tree. i.e., the questions that must be answered are how do you know the components yielded will be a tree? How do you know that if you remove a single edge it produces only two trees? Why not three or it may shatter in many pieces., etc. Here is my current reasoning that the removal of an edge from a tree T will yield two subtreets, T_1 and T_2: if a tree is a connected graph with no cycles, then there is no loop that exists in the tree. furthermore, there is only a one directional path from any vertex u to any vertex v in the tree where u!=v. As a result, an edge only connects two vertices by a single path. For if there were more than one path, then there would be more than a single directional path from some vertex u to some vertex v which contradicts the definition of a tree. Therefore, the removal of an edge from a tree will yield two components which by definition are trees that were contained in a larger tree. If they were not trees, then the original would not be a tree which contradicts the whole thing. Any obvious weak points or pitfalls that were not addressed? -- conrad === Subject: Re: Graph theory posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc10 Firefox/3.0.5,gzip(gfe),gzip(gfe) > I'm working over a proof of V - E = 1 > where V is the number of > vertices and E is the number of > edges and the whole thing is > in terms of trees. æNow, my > definition of a tree is a connected > graph with no cycles. æBut I've > also come across a definition that > said: a connected graph with > no nontrivial cycles. æWhich made > me think, what is a trivial cycle > and how can a tree have one? Beyond that, the tack to proving > the above is starting with a tree > T and removing one of the edges, > reducing the problem to two > subtrees that have been proved. But before this, one must specify > how you end up with two trees > if you remove an edge from a tree. i.e., the questions that must be > answered are how do you know > the components yielded will > be a tree? æHow do you know > that if you remove a single edge > it produces only two trees? > Why not three or it may > shatter in many pieces., etc. Here is my current reasoning > that the removal of an edge from > a tree T will yield two subtreets, > T 1 and T 2: if a tree is a connected graph with > no cycles, then there is no loop > that exists in the tree. furthermore, > there is only a one directional > path from any vertex u to any > vertex v in the tree where u!=v. > As a result, an edge only > connects two vertices by a single > path. æFor if there were more than > one path, then there would be > more than a single directional > path from some vertex u to some > vertex v which contradicts the > definition of a tree. Therefore, > the removal of an edge from > a tree will yield two components > which by definition are trees that > were contained in a larger tree. If > they were not trees, then the original > would not be a tree which contradicts > the whole thing. Any obvious weak points or pitfalls > that were not addressed? -- > conrad Instead of removing any egde, observe that since the graph T is finite, there is least one vertex of degree 1, so removing it and the edge which touches it gives you another graph T', which is also a tree (because any cycles in T' are also cycles in T) and connected, and use induction. -- m === Subject: Re: Graph theory > I'm working over a proof of V - E = 1 where V is the number of vertices > and E is the number of edges and the whole thing is in terms of trees. Huh? You're wanting to show that for a tree graph that V = E + 1 ? > Now, my definition of a tree is a connected graph with no cycles. But > I've also come across a definition that said: a connected graph with no > nontrivial cycles. Which made me think, what is a trivial cycle and > how can a tree have one? > A trivial cycle can be an edge from a vertex back to itself. It could also be two edges between the same two points. Perhaps that author requires all graphs to be simple, ie no loops and no double edges. Your tree has no cycles by definition. > Beyond that, the tack to proving the above is starting with a tree T and > removing one of the edges, reducing the problem to two subtrees that > have been proved. But before this, one must specify how you end up with two trees if you > remove an edge from a tree. > Correct. > i.e., the questions that must be answered are how do you know the > components yielded will be a tree? How do you know that if you remove a > single edge it produces only two trees? Why not three or it may shatter > in many pieces., etc. > Use the definition and use theorem, that if an edge e is removed from a connected graph G and Ge is still connected, then G contains a cycle. > Here is my current reasoning that the removal of an edge from a tree T > will yield two subtreets, T_1 and T_2: if a tree is a connected graph with no cycles, then there is no loop > that exists in the tree. furthermore, there is only a one directional > path from any vertex u to any vertex v in the tree where u!=v. What? Now the graph is directed? No. > As a result, an edge only connects two vertices by a single path. For > if there were more than one path, then there would be more than a single > directional path from some vertex u to some vertex v which contradicts > the definition of a tree. Therefore, the removal of an edge from a tree > will yield two components which by definition are trees that were > contained in a larger tree. If they were not trees, then the original > would not be a tree which contradicts the whole thing. Any obvious weak points or pitfalls > that were not addressed? > Much too informal including changing G from a graph to a directed graph.. If e is an edge between v and w of a connected graph G and if Ge is connected, then there's a (undirected) path between v and w. Thus that path and e are a cycle. That e is a loop from v to v, in the event that G isn't simple, is a special case. === Subject: Re: 60 seconds ~ 2pi radians posting-account=_m9LngoAAAAtlhPgHQ2dCa-Pag64iiXh Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) If it is believed that you can't deny something exists without first > accepting that thing exists to deny, i.e. without contradiction then > it is an attempt to create a foundationalist philosophy based on a > single, undeniable truth which seemingly implies that it is fixed and > assured; but this supposed first principle that we cannot deny that > something exists without first accepting that thing exists to deny, > i.e. without contradiction depends upon a logical structure which is > really a second postulate whch unjustifiably makes claim that the > capacity to judge correctly, to distinguish the true from the false > and to determine what is contradictory or non-contradictory cannot be > mistaken or has no chance for error, THEN this first principle remains > theoretical, but theoretically the best theory. No nothing like that, theory is entirely man made and is dependent > upon man's mind. > Like your theory about how reality just is? > However reality isn't, reality is what it is, reality is absolute and > because you cant deny existence without contradiction then that only > leaves one thing to do, to identify it, i.e. what does the existent > exist as. > I disagree. Your trying to create a foundationalist philosophy based on a single, undeniable truth which you seem to imply is fixed and assured. Your first principle that we cannot deny that something exists without first accepting that thing exists to deny, i.e. without contradiction depends upon a logical structure which is really a second postulate. This second postulate is theoretical. > Matter, matter's nature or man's ideas. And if an existent is > identified as one of man's ideas, e.g god, then the next stage is to > determine what is the trigger of that idea, is it matter and matter's > nature or just another of man's ideas. ...Causes and effects are discovered, not by reason but through experience, when we find that particular objects are constantly conjoined with one another. We tend to overlook this because most ordinary causal judgments are so familiar; we've made them so many times that our judgment seems immediate. But when we consider the matter, we realize that ñan (absolutely) unexperienced reasoner could be no reasoner at allî (EHU, 45n). Even in applied mathematics, where we use abstract reasoning and geometrical methods to apply principles we regard as laws to particular cases in order to derive further principles as consequences of these laws, the discovery of the original law itself was due to experience and observation, not to a priori reasoning. Even after we have experience of causal connections, our conclusions from those experiences aren't based on any reasoning or on any other process of the understanding. They are based on our past experiences of similar cases, without which we could draw no conclusions at all. But this leaves us without any link between the past and the future. How can we justify extending our conclusions from past observation and experience to the future? The connection between a proposition that summarizes past experience and one that predicts what will occur at some future time is surely not an intuitive connection; it needs to be established by reasoning or argument. The reasoning involved must either be demonstrative, concerning relations of ideas, or probable, concerning matters of fact and existence... ...all of us - ordinary people, infants, even animals - ñimprove by experience,î forming causal expectations and refining them in the light of experience... ...When we examine experience to see how expectations are actually produced, we discover that they arise after we have experienced ñthe constant conjunction of two objects;î only then do we ñexpect the one from the appearance of the other.î But when ñrepetition of any particular act or operation produces a propensity to renew the same act or operation\.83we always say, that this propensity is the effect of Customî. So the process that produces our causal expectations is itself causal. Custom or habit ñdetermines the mind\.83to suppose the future conformable to the past.î But if this background of experienced constant conjunctions was all that was involved, then our ñreasoningsî would be merely hypothetical. Expecting that fire will warm, however, isn't just conceiving of its warming, it is believing that it will warm. Belief requires that there also be some fact present to the senses or memory, which gives ñstrength and solidity to the related idea.î In these circumstances, belief is as unavoidable as is the feeling of a passion; it is ña species of natural instinct,î ñthe necessary result of placing the mindî in this situation. http://plato.stanford.edu/entries/hume/ > As it happens man had an idea > that the universe must have been created, but it would need a creator > and then along come god, slow walking low tall talking god along come > lonely lanky god. Things can exist that aren't real, but rather they are used by man to > help him identify things like the real from the illusioned. This is > where chazzzz and his Kantian ilk get all tripped up, they say god > doesn't exist, which is a contradiction in terms, god does exist but > only as an idea. I suppose I could accept that theory if you don't claim what you say > is certain. If you say it is certain because it is necessary to > conceive in the first place that begs the question since your > repeating what you propose as a conclusion with no other support. The basic problem can be summarised as follows. Less words Mortal, the basic problems of human survival can be summed > up as the failure to identify and eliminate contradictions in man's > identification of that which exists without contradiction, reality. > What reason do we have for supposing that the inductive principle is true?, the majority of people who have not reflected carefully on the issue will say something like, Because it has been proved to be true in the past. The reasoning of such a reply may be spelled out in more detail as follows. Whenever we have observed instances in the past, and have drawn conclusions about (at that time) future unobserved instances, our conclusions invariably turn out later to be confirmed via direct observation (i.e. once we get round to observing those formerly unobserved instances). Since it has always been the case that unobserved instances have been found to resemble observed instances, we can confidently conclude that (at least probably) all unobserved instances will resemble observed instances. The trouble with this answer, as Hume was at pains to point out, is that it begs the question. The reply itself takes the form of an inductive argument - reasoning about the future on the basis of the past - and thus must presuppose the very thing it aims to establish: the inductive principle. It is therefore guilty of fallacious circular reasoning. > Not having to prove a negative is formally called an appeal to > ignorance don't get aroused, the ignorance part is not directed at > you, its just the name logicians have agreed to call it. Oh rubbish, can you prove you have never murdered anyone? > What do you mean by prove? In many legal cases if a person can prove that they were at another location when the murder happened, they prove a negative. > MG === Subject: Finding an example posting-account=z0CEbwoAAAC4Yfd6z5bsJTekd_ucAbBB Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Notation used: l^p space: Each element in the l^p space is a sequence x=(x1,x2,.81c..) of numbers such that |x1|^p + |x2|^p +.81c.81c. < .81.87 ( Or .87.94|xn|^p < .81.87 ) Give an element x belonging to lp space for p>1 but x does not belong to l^1 space. === Subject: Re: Finding an example posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze Filter 1.2.0.72; GTB5; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Notation used: > l^p space: Each element in the l^p space is a sequence x=(x1,x2,.....) > of numbers such that |x1|^p + |x2|^p +....... < .81.87 ( Or .87.94|xn|^p < .81.87 ) Give an element x belonging to lp space for p>1 but x does not belong > to l^1 space. (1, 1/2, 1/3, ...). Dave === Subject: Re: lattice of topology types > I'm trying to get a big picture view of properties of topologies. I'm > looking for a diagram that show the property inclusions like 'all T1 > topologies are T0 topologies' or 'all Hilbert spaces are Banach > spaces'. But in (annotated) picture form. Searching, I can't seem to > find anything online. Is there a good reference online (or off) for > this? The set of topologies for a give set S, is a complete, complemented, non-distributive lattice. The infinum of a collection of topologies for S is the set intersection of the topologies. That is easy. The very hard part is showing it's a complemented lattice. Complements are not unique because the lattice is not distributive. === Subject: Re: WARNING: Dangerous Meme Alert!! Don't Drink The Obama Kool-aide posting-account=_m9LngoAAAAtlhPgHQ2dCa-Pag64iiXh Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) I do not think memes have any specific physical characterstic by which > they can be reckognized one from the other. They could be like methods that have material shells or ways of doing things. Maybe the way of treating horses spread or military styles of behavior. The outer shell of a meme might be the repeatable way it is told by people where its material shell would be in memory or on paper, a story. Maybe the meme spreads as summarized rules for doing something which would be hard to learn in detail. You raise some good points though. A rule of thumb is a principle with broad application that is not intended to be strictly accurate or reliable for every situation. It is an easily learned and easily applied procedure for approximately calculating or recalling some value, or for making some determination. http://en.wikipedia.org/wiki/Rule_of_thumb ...Generally grandmasters survey the chess board and forecast the pieces only one move ahead. Then they select the most plausible play or two and investigate its consequences deeper. At every move ahead the number of choices to consider explodes exponentially, yet great human players will concentrate only on a few of the most probable countermoves at each rehearsed turn. Occasionally they search far ahead when they spot familiar situations they know from experience to be valuable or dangerous. But in general, grandmasters (and now Deep Thought) work from rules of thumb. For instance: Favor moves that increase options; shy from moves that end well but require cutting off choices; work from strong positions that have many adjoining strong positions. Balance looking ahead to really paying attention to what's happening now on the whole board. Every day we confront similar tradeoffs. We must anticipate what lies around the corner in business, politics, technology, or life. However, we never have sufficient information to make a fully informed decision. We operate in the dark. To compensate we use rules of thumb or rough guidelines. Chess rules of thumb are actually pretty good rules to live by. (Notes to my daughters: Favor moves that increase options; shy away from moves that end well but require cutting off choices; work from strong positions that have many adjoining strong positions. Balance looking ahead to really paying attention to what's happening now on the whole board.) Common sense embodies a positive myopia. Rather then spend years developing a company employee manual that anticipates every situation that might arise-yet be out of date the moment it is printed-how much better to adopt positive myopia and not look so far ahead. Devise some general guidelines for the events that seem sure to arise on the next move and treat extreme cases if and when they come up. To navigate through rush-hour traffic in an unfamiliar city we can either plan detailed routes through the town on a map-thinking far ahead-or adopt a heuristic such as Go west until we hit the river road, then turn left. Usually, we do a bit of both. We refrain from looking too far ahead, but we do look immediately in front. We meander west, or uphill, or downtown, while using the map to evaluate the next immediate turn ahead, wherever we are. We employ limited look-ahead guided by rules of thumb. Out of Control - The Rise of Neo-Biological Civilization Kevin Kelly - 1996 http://www.kk.org/ http://www.amazon.com/Out-Control-Rise-Neo-Biological-Civilization/dp/020157\ 7933/ > The meme meme for instance has a specific sound in English but what > that sound triggers inside an English-speaking mind, might be > different in any individual.It is in fact a spiritual entity, which > is a set of connections inside a human brain that joins neurons which > represent something in the outside world. A meme would be identical in > two brains, if the connections inside those brains would connect to > other intems that all represented the same things in the outside > world. However, this is unlikely. To - for instance - the word meme is > also connected to the name Robert Aunger, while this name did > previously not exists in my memory at all. The OP seems to think memes can be like drugs, and theoretically they > can, > however some drugs are nescessary for survival (dopamine is produced > by the human body for a good reason) > and so human beings would swiftly become exinct if it wasn't for many > indispensible memes. By the way, without the Obama meme, Obama's mission will probably > fail, > so if you do not wish the USA well, you shoulf reject it, > If however you wish it well, it might be better to embrace it. No we can't does by itself not bring you anywhere. Love, (another indispensible meme) Peter van Velzen > February 2009 > Trang > Thailand === Subject: Continuous iteration (not tetration) posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) Hi. I was playing around with a method to iterate the function f(x) = x^2 - 2 in a continuous and natural manner. Namely, I tried this: f^n(x) = sum_{k=0...inf} (-1)^(2^(n-1) - k) U(2^(n-1), k) x^(2k) U(n, k) = (2n C(n + k, n - k))/(n + k) where C is the binomial coefficient (extends to real/complex values via Gamma function). However, when one substitutes fractional and real values of n in there, the resulting map does not behave like a flow: f^1.5(f^1.5(x)) != f^3(x), for example. What's up here, anyway? === Subject: Re: Continuous iteration (not tetration) Am 07.02.2009 06:19 schrieb mike3: > Hi. I was playing around with a method to iterate the function f(x) = > x^2 - 2 in a continuous and natural manner. Namely, I tried this: f^n(x) = sum_{k=0...inf} (-1)^(2^(n-1) - k) U(2^(n-1), k) x^(2k) > U(n, k) = (2n C(n + k, n - k))/(n + k) where C is the binomial coefficient (extends to real/complex values > via Gamma function). However, when one substitutes fractional and real > values of n in there, the resulting map does not behave like a flow: > f^1.5(f^1.5(x)) != f^3(x), for example. What's up here, anyway? > Hmm, I've no idea how you did arrive at the formula, so I don't know how to approach that problem as you stated it. However, I tried my method using matrices and fixpoint-shift as I described in the U-tetration / Lucas-lehmer-thread in january. For f(x) = x^2 - 2 I replace g(x) = x^2 + 4x such that f(x) = g(x-2) + 2 f(f(x)) = g(g(x-2)) + 2 ... valid for integer iterations. I implement this by matrices F for f(x) and G for g(x) respectively. Then I interpolate the h'th fractional iterates by h'th fractional power of G, which leads to fÁh(x) = gÁh(x-2) +2 With this interpolation I get, using x = 1 fÁ3 (1) = -1 // which is then a fixpoint fÁ1.5(1) = -1.96780518962 fÁ1.5(fÁ1.5(1)) -> -1 (???) where the partial sums (using Euler-acceleration) for the last series at indexes k=59 to 64 are ... [-0.95633853043] [-0.95860386158] [-0.95924500120] [-0.96093526176] [-0.96181062959] [-0.96317678166] ... still very slowly converging. However, for other values of x, for instance x=3 I get, using 64 terms for series (Euler-summation) fÁ3 ( 3 ) = 2207 -------------------------------- fÁ1.5( 3 ) = 15.2786549738 = x1 fÁ1.5( x1 ) = 2206.99999999 and for x=4 in the next example I used 96 terms due to bad approximation (in spite of Euler-summation) fÁ3 ( 4 ) = 37634 -------------------------------- fÁ1.5( 4 ) = 41.4920069092 fÁ1.5( x1 ) = 37633.9128037 // approximation improving when more terms ----------------------------------------------------- The powerseries for gÁ1.5(x) which I got begins with g(x) = 8 x + 14/3 x^2 +28/45 x^3 -1/90 x^4 + 2/2025 x^5 -17/133650 x^6 + 17/868725 x^7 -697/208494000 x^8 + 287/469111500 x^9 + O(x^10) Gottfried === Subject: Re: Continuous iteration (not tetration) <6v54caFid9p6U2@mid.dfncis.de> posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) > Am 07.02.2009 06:19 schrieb mike3:> Hi. I was playing around with a method to iterate the function f(x) = > x^2 - 2 in a continuous and natural manner. Namely, I tried this: f^n(x) = sum {k=0...inf} (-1)^(2^(n-1) - k) U(2^(n-1), k) x^(2k) > U(n, k) = (2n C(n + k, n - k))/(n + k) where C is the binomial coefficient (extends to real/complex values > via Gamma function). However, when one substitutes fractional and real > values of n in there, the resulting map does not behave like a flow: > f^1.5(f^1.5(x)) != f^3(x), for example. What's up here, anyway? Hmm, > æI've no idea how you did arrive at the formula, so I don't know > æhow to approach that problem as you stated it. æHowever, I tried my method using matrices and fixpoint-shift > æas I described in the U-tetration / Lucas-lehmer-thread in january. æFor æ æf(x) = x^2 - 2 æI replace æ æg(x) = x^2 + 4x æsuch that æ æf(x) = g(x-2) + 2 > æ æf(f(x)) = g(g(x-2)) + 2 > æ æ... > æ valid for integer iterations. æI implement this by matrices F for f(x) and G for g(x) > ærespectively. > æThen I interpolate the h'th fractional iterates by h'th fractional > æpower of G, which leads to æ æ fÁh(x) = gÁh(x-2) +2 æWith this interpolation I get, using x = 1 æ æ fÁ3 æ(1) æ æ æ æ= -1 æ æ// which is then a fixpoint æ æ fÁ1.5(1) æ æ æ æ= æ-1.96780518962 > æ æ fÁ1.5(fÁ1.5(1)) -> -1 æ æ (???) æwhere the partial sums (using Euler-acceleration) > æfor the last series at indexes k=59 to 64 are > æ æ æ... > æ æ [-0.95633853043] > æ æ [-0.95860386158] > æ æ [-0.95924500120] > æ æ [-0.96093526176] > æ æ [-0.96181062959] > æ æ [-0.96317678166] > æ æ æ... > æstill very slowly converging. However, for other values of x, for instance x=3 I get, using > 64 terms for series (Euler-summation) æ æ fÁ3 æ( 3 æ) æ = æ2207 > æ æ -------------------------------- > æ æ fÁ1.5( 3 æ) æ = æ15.2786549738 = x1 > æ æ fÁ1.5( x1 ) æ = æ2206.99999999 and for x=4 in the next example I used 96 terms > due to bad approximation (in spite of Euler-summation) æ æ fÁ3 æ( 4 æ) = æ37634 > æ æ -------------------------------- > æ æ fÁ1.5( 4 æ) = æ41.4920069092 > æ æ fÁ1.5( x1 ) = æ37633.9128037 > æ æ æ æ æ æ // approximation improving when more terms ----------------------------------------------------- The powerseries for gÁ1.5(x) which I got begins with æ g(x) = æ8 x + 14/3 x^2 +28/45 x^3 -1/90 x^4 + 2/2025 x^5 > æ æ æ æ æ-17/133650 x^6 + 17/868725 x^7 -697/208494000 x^8 > æ æ æ æ æ+ 287/469111500 x^9 + O(x^10) Gottfried I am sorry you do not understand the parameteric method. Firstly you verify x = (t+1/t) =f(t) x^2 =(t+1/t)^2 = t^2 + 2 + 1/t^2 So t^2 + 1/t^2 = x^2 - 2 = h(x) Thence x-> h(x) corresponds to t->t^2 (square) Or h(f(t)) = f(t^2) this last equation must be NOTICE... h^2(f(t)) = h(h(f(t)))= h(f(t^2)) = f(t^4) and iteratively h^[n](f(t)) = f(t^(2^n)) .88 suivre, Alain === Subject: Re: Continuous iteration (not tetration) Two small corrections: Am 07.02.2009 10:59 schrieb Gottfried Helms: = x1 inserted in the following: > > fÁ3 ( 4 ) = 37634 > -------------------------------- > fÁ1.5( 4 ) = 41.4920069092 = x1 > fÁ1.5( x1 ) = 37633.9128037 > // approximation improving when more terms > > The powerseries for gÁ1.5(x) which I got begins with > > g(x) = 8 x + 14/3 x^2 +28/45 x^3 -1/90 x^4 + 2/2025 x^5 > -17/133650 x^6 + 17/868725 x^7 -697/208494000 x^8 > + 287/469111500 x^9 + O(x^10) > means also f(x) = g(x-2) + 2 f(x) = 2+ 8*(x-2) + 14/3*(x-2)^2 +28/45*(x-2)^3 -1/90*(x-2)^4 + 2/2025*(x-2)^5 ... (and so on) Gottfried === Cancel-Key: sha1:tkdRYPkv3sqQBE6+7jS7VQwXTgA= === Subject: Re: Continuous iteration (not tetration) posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.6) Gecko/2009011912 Firefox/3.0.6,gzip(gfe),gzip(gfe) > Hi. I was playing around with a method to iterate the function f(x) = > x^2 - 2 in a continuous and natural manner. Namely, I tried this: f^n(x) = sum {k=0...inf} (-1)^(2^(n-1) - k) U(2^(n-1), k) x^(2k) > U(n, k) = (2n C(n + k, n - k))/(n + k) where C is the binomial coefficient (extends to real/complex values > via Gamma function). However, when one substitutes fractional and real > values of n in there, the resulting map does not behave like a flow: > f^1.5(f^1.5(x)) != f^3(x), for example. What's up here, anyway? when i have seen infinite sum continuous iterations they tend to try to mimic the property of finite sums this is common in ramanujan for instance what i have seen when you have a finite sum n-1 --- f(n) = / a(j) --- j=0 the natural continuous extensions tend to look like oo oo --- --- f(x) = / a(j) - / a(k) --- --- j=0 k=x this is due to generalisations of the difference operator that help maintain the desired algebraic structure and in many cases this can be formalised maybe this trick works here? otherwise the polynomial has an immediate representation as a matrix operator so the exponentiation trick can be applied -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Continuous iteration (not tetration) posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) Hi. I was playing around with a method to iterate the function f(x) = > x^2 - 2 in a continuous and natural manner. Namely, I tried this: f^n(x) = sum {k=0...inf} (-1)^(2^(n-1) - k) U(2^(n-1), k) x^(2k) > U(n, k) = (2n C(n + k, n - k))/(n + k) where C is the binomial coefficient (extends to real/complex values > via Gamma function). However, when one substitutes fractional and real > values of n in there, the resulting map does not behave like a flow: > f^1.5(f^1.5(x)) != f^3(x), for example. What's up here, anyway? when i have seen infinite sum continuous iterations > æ they tend to try to mimic the property of finite sums this is common in ramanujan > æ for instance what i have seen when you have a finite sum æ æ æ æ æn-1 > æ æ æ æ æ--- > æ æ æ æ æ > æ f(n) = / æ æa(j) > æ æ æ æ æ--- > æ æ æ æ æj=0 the natural continuous extensions tend to look like æ æ æ æ æ oo æ æ æ æ æ æ oo > æ æ æ æ æ--- æ æ æ æ æ æ--- > æ æ æ æ æ æ æ æ æ æ æ æ > æ f(x) = / æ æa(j) æ - æ/ æ a(k) > æ æ æ æ æ--- æ æ æ æ æ æ--- > æ æ æ æ æj=0 æ æ æ æ æ æk=x this is due to generalisations of the difference operator > that help maintain the desired algebraic structure > æ and in many cases this can be formalised maybe this trick works here? otherwise > æ the polynomial has an immediate representation as a matrix operator > so the exponentiation trick can be applied -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar Bonjour Everybody, Known methods to obtain direct iterates of trinoms are : conjugates of x^2 , Chebyschev polynomials and conjugates, Combinations of exponentials ai*exp(+/-2^x), Trigonometric equivalent expressions, And parametrization: Example x =f(t) = (t+1/t) h(x) =f(t^2)= (t^2+ 1/t^2) Here, you easily find h(x) = x^2 -2 thence h^[n](x) = , n any positive integer What about h^[r](x) , any real? Amicalement, Alain === Subject: Re: Kouznetsov-like limit tetration formula for _different bases_. posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > Hi. You may remember my (brief) thread Kouznetsov'sSimpleTetration > Formula?. Well, I recently managed to put together a similarKouznetsov-like formula (it turns out it was hinted at on the page), > but for a wider range of bases . For example, instead of just eta = e^ > (1/e), it appears to work for the entire interval (e^-e, e^(1/e)) > although not at the endpoints and not at base 1. And yes, that's e^-e > you saw there, which is less than 1. Whereas the e^(1/e) formula involved correcting the function f(z) = > e - (2e)/z, this involves f(z) = F - exp(-beta*z), where F is the > fixpoint of the iteration (i.e. ^oo B, where B is the base, or -W(-ln > (B))/ln(B) in terms of Lambert's W-function), and beta is a constant > that I figured is equal to log(log(F))/log(B).Kouznetsovmentions > something much similar on his Ksexp page, where he discusses > exponential asymptotic, saying how thetetrationis asymptotic to an > exp. Then using the very sameKouznetsoviterated-log technique, one > obtains the following formula: ^z b = lim {n->oo} log b^n(F - exp(-beta*(z + z0(n) + n))) where z0(n) = log(F - ^n b)/beta - n (i.e. that value of z such that log b^n(F - exp(-beta + z + n)) equals > 1, so that log b^n(F - exp(-beta*(z + z0(n) + n))) will equal 1 when z > = 0) which seems to work for all complex towers except the singularities. It also seems the formula works for convergent imaginary/complex bases > as well. That is, for the entire red region shown in theTetration > Fractal:http://www.tetration.org/Fractals/Atlas/index.html > (except perhaps at the region's boundary, which is where e^-e and e^(1/ > e) lie.) It does not seem to work outside of it. Just inside. However, might it > be possible to use analytic continuation? I noticed that taking a > Taylor expansion for the half-height tower w.r.t. the base, near e^(1/ > e), then substituting, one could get convergence in a somewhat larger > radius -- and higher-than-e^(1/e) bases could be tetrated. Perhaps > this could be done to the whole z-plane? Although it would be crap for > computation. What was really interesting, though, was what was revealed upon making > a graph.Kouznetsovwas right -- it is indeed periodic in the interval > (1, e^(1/e)) with imaginary period. Check out this graph for base b = > (5/4)^(4/5), and complex height z: http://img218.imageshack.us/img218/8569/firsttetgraphzo8.jpg Note the multiple rows of singularities, and the branchcuts extending > toward the left, connecting them. Then also check out this interesting graph, for base b = 1/4, which > is less than 1, also for complex tower z: http://img218.imageshack.us/img218/8678/secondttetgrapheg6.jpg As you can see, the singularities are no longer confined to the left > part of the plane below Re(z) <= -2. Interestingly the singularities > in the right part don't form a straight line along the real axis but > instead are on a slight diagonal. You can also see that there are > branchcuts coming off the singularities (look at the graph of the > imaginary part), only they stripe the plane instead of connecting > the singularities. (This is all done using the principal branch for > log.) Note that there may also be singularities along the negative > real axis due to thetetrationfunctional equation, although I'm not > sure -- I can't seem to resolve them with a graph, although the > functional equation suggests it is undefined at those points, hence a > singularity, but it doesn't seem to explode near them, or if it does > it must be in an incredibly tight region. How come this interesting behavior occurs, anyway? (It sure seems that tetration is a far richer operation than its predecessor operations, by the way.) And this suggests also that one of Kouznetsov's uniqueness conditions, that is, no discontinuities/ singularities/branchpoints/ etc. for Re(z) > -2, does not hold for all bases -- not even positive real ones -- as for e^-e < b < 1 the singularities (branchpoints) and branchcuts for ^z b appear in the halfplane Re(z) > -2 as well. The point where b = 1 appears to be some sort of singularity in the base- parameter: attempts to expand a Taylor series in b at some point b 0 for some noninteger tower z appear to yield a radius of convergence of |1 - b 0|. === Subject: Complex plane integral? posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) Hi. Is there any use in integrating a complex function over a 2- dimensional region of the complex plane, as opposed to just along a line? For example, integrating f(z) = exp(z) over the unit disk. === Subject: Basic number theory question posting-account=Y44pzAoAAAAjjL9rsY86pnFg1GgTdijA Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) Suppose I have fixed positive integers k and j, and additionally an integer a >= j+1, such that k | (a^2 - j - j). Supposing I have a value of a that satisfies the divisibility, how can I calculate in terms of a, k, and j, the minimum possible n>0 such that k | (a+n)^2 - j - j. It doesn't matter if a2 = a+n actually _works_ or not (although that would certainly be ideal), it only matters that it's a starting point allowing me to ignore many intermediary values. I'm doing this in a computational setting and the values of a, k, and j are huge so it is really not feasible for me to iterate through every single value of a looking for the next one that works. === Subject: Re: Basic number theory question posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) > Suppose I have fixed positive integers k and j, and additionally an > integer a >= j+1, such that k | (a^2 - j - j). Supposing I have a value of a that satisfies the divisibility, how can > I calculate in terms of a, k, and j, the minimum possible n>0 such > that k | (a+n)^2 - j - j. It doesn't matter if a2 = a+n actually works or not (although that > would certainly be ideal), it only matters that it's a starting point > allowing me to ignore many intermediary values. æI'm doing this in a > computational setting and the values of a, k, and j are huge so it is > really not feasible for me to iterate through every single value of a > looking for the next one that works. > If k divides a^2-j-j and (a+n)^2-j-j then it must divide the difference 2an+n^2 = (2a+n)*n If k is even, it follows that n must be even. If k is divisible by a prime p, then n must me = 0 mod p or = -2a mod p. If you have the factorization of k, this eliminates a lot. In fact, if p^r || k (i.e. p^r|k and not p^{r+1}|k), there must be an s, 0<= s <=r such that n = 0 mod p^s and n = -2a mod p^{r-s} Now if p^t || 2a, this implies s and r-s cannot both be >t. This implies that n=0 mod p^min{t,[r/2]}. At least you can easily enumerate all solutions for n mod p^r. These can be combined with the chineses remainder theorem to the set of all solutions mod k ad in fact they will all work. === Subject: Re: Basic number theory question posting-account=Y44pzAoAAAAjjL9rsY86pnFg1GgTdijA Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) Suppose I have fixed positive integers k and j, and additionally an > integer a >= j+1, such that k | (a^2 - j - j). Supposing I have a value of a that satisfies the divisibility, how can > I calculate in terms of a, k, and j, the minimum possible n>0 such > that k | (a+n)^2 - j - j. It doesn't matter if a2 = a+n actually works or not (although that > would certainly be ideal), it only matters that it's a starting point > allowing me to ignore many intermediary values. æI'm doing this in a > computational setting and the values of a, k, and j are huge so it is > really not feasible for me to iterate through every single value of a > looking for the next one that works. > If k divides æa^2-j-j æand æ(a+n)^2-j-j then it must divide the > difference > 2an+n^2 = (2a+n)*n > If k is even, it follows that n must be even. > If k is divisible by a prime p, then n must me = 0 mod p or = -2a mod > p. > If you have the factorization of k, this eliminates a lot. Let's say I have the factorization of k (I don't, but I suppose it wouldn't be too bad to compute, my k's are only about a maximum of 4 million or so), and furthermore I know that every prime factor of k is congruent to 1 mod 4 (in fact, that is the definition of k in my actual problem, it's the sequence of all numbers prime and not prime that do not have any prime factors congruent to 0, 2, or 3 mod 4). Can I use this to give me an even better result? === Subject: Re: Basic number theory question > Suppose I have fixed positive integers k and j, and additionally an > integer a >= j+1, such that k | (a^2 - j - j). > Is a^2 - j - j a typo for a^2 - j - 1? > Supposing I have a value of a that satisfies the divisibility, how can > I calculate in terms of a, k, and j, the minimum possible n>0 such > that k | (a+n)^2 - j - j. > k | (a^2 - j - 1 + 2an + n^2) k | 2an + n^2; the result is independent of j. k | n(2a + n) n = k is one solution. Hence in finite time, the smallest required n could be calculated. If 2a <= k, then n = k - 2a is a smaller solution. I conjecture that prime k will give the larger n's. > It doesn't matter if a2 = a+n actually _works_ or not (although that > would certainly be ideal), it only matters that it's a starting point > allowing me to ignore many intermediary values. What do you mean by works? > I'm doing this in a computational setting and the values of a, k, and j > are huge so it is really not feasible for me to iterate through every > single value of a looking for the next one that works. === Subject: Re: Basic number theory question <20090206220018.E64895@agora.rdrop.com> posting-account=Y44pzAoAAAAjjL9rsY86pnFg1GgTdijA Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > Suppose I have fixed positive integers k and j, and additionally an > integer a >= j+1, such that k | (a^2 - j - j). Is a^2 - j - j a typo for a^2 - j - 1? I retract my statement, it actually -was- a typo, but it was a typo for a^2 - j^2 - j === Subject: Re: Basic number theory question <20090206220018.E64895@agora.rdrop.com> posting-account=Y44pzAoAAAAjjL9rsY86pnFg1GgTdijA Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > Suppose I have fixed positive integers k and j, and additionally an > integer a >= j+1, such that k | (a^2 - j - j). Is a^2 - j - j a typo for a^2 - j - 1? No, a^2 - j^2 - j was correct. Supposing I have a value of a that satisfies the divisibility, how can > I calculate in terms of a, k, and j, the minimum possible n>0 such > that k | (a+n)^2 - j - j. k | (a^2 - j - 1 + 2an + n^2) k | 2an + n^2; æthe result is independent of j. k | n(2a + n) n = k is one solution. æHence in finite time, > the smallest required n could be calculated. If 2a <= k, then n = k - 2a is a smaller solution. I conjecture that prime k will give the larger n's. It doesn't matter if a2 = a+n actually works or not (although that > would certainly be ideal), it only matters that it's a starting point > allowing me to ignore many intermediary values. What do you mean by works? I just mean that when this algorithm or method tells me what to use for n, it doesn't matter if k | (a+n)^2 - j^2 - j (although my goal is obviously to eventually find such an n), I just want an n such that for 0 < s < n, k does not divide (a+s)^2 - j^2 - j. In my current algorithm, after finding an a that works, I try (a+1)^2 - j^2 - j, then (a+2)^2 - j^2 - j, etc. Being able to start this iteartive testing at any number other than 1 would be an improvement. === Subject: Re: A new definition for Life in >On Feb 5, 7:00æpm, Matt Silberstein >I'm sorry, I seem to have missed the part of the Origin of Species >where Darwin says this. > It is likely that you missed lots of parts of Origins. And likely you > missed lots of the development of the science of evolutionary biology > in the 150 years since Darwin published. Regardless of how you came to > your state of knowledge, though, John J is right: evolution is the > change in inherited characteristics in a population of reproducing > organisms over time. Natural selection does change the composition of > the surviving populations, but evolution is change. That change can be > improvement by some standard or corruption, it can be simpler or more > complex, but if it is change in inherited characteristics it is > evolution. As I've noted before, Matt, you appear to have a deep, indeed, a >profound lack of interest in reality. You sound like a physicist. Yes, I know that's ad hominem, but it's relevant. All you care about >are your own fantasies. The evidence means nothing to you. As was pointed out to me there is little point it engaging with you. You don't add anything, you aren't interesting, and it does not look like anyone is swayed by your posts. -- Matt Silberstein Do something today about the Darfur Genocide http://www.beawitness.org http://www.darfurgenocide.org http://www.savedarfur.org Darfur: A Genocide We can Stop === Subject: Re: A new definition for Life <7f8qo4db1g5d4369roi4delruefcniuqu2@4ax.com> posting-account=5ayZ-goAAABGZmmwx8zZEwz6gU2OuVSd CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) On Feb 6, 11:54æpm, Matt Silberstein > in > As was pointed out to me there is little point it engaging with you. > You don't add anything, you aren't interesting, and it does not look > like anyone is swayed by your posts. I'm not entirely sure if you're just playing Devil's advocate or you are innately perverse, Matt. === Subject: Re: A new definition for Life > On Feb 5, 7:00 pm, Matt Silberstein > > > > >I'm sorry, I seem to have missed the part of the Origin of Species >where Darwin says this. >It is likely that you missed lots of parts of Origins. And likely you >missed lots of the development of the science of evolutionary biology >in the 150 years since Darwin published. Regardless of how you came to >your state of knowledge, though, John J is right: evolution is the >change in inherited characteristics in a population of reproducing >organisms over time. Natural selection does change the composition of >the surviving populations, but evolution is change. That change can be >improvement by some standard or corruption, it can be simpler or more >complex, but if it is change in inherited characteristics it is >evolution. > > > As I've noted before, Matt, you appear to have a deep, indeed, a > profound lack of interest in reality. You sound like a physicist. Statements like this show that you have no interest in truth. This is either jealousy, ignorance or intentional stupidity. > > Yes, I know that's ad hominem, but it's relevant. All you care about > are your own fantasies. Now you are contradicting yourself. Physicists are only interested in the truth. You cranks do not care about truth at all. The evidence means nothing to you. > You are the one ignoring the evidence. You are either a troll or incurably stupid. > > > === Subject: Deriving weighted least squares posting-account=_KcbFAkAAACPcKINjRxCKEAs3AA9RdEJ Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) Hi Guys, It's been a long time since I've done any of these here joindey-up sums and I am looking for some pointers. I want to do a least-squares on the following... -- {Aijk - (C0 + Cw.G.Wijk - Ni + Nj)}^2 -- ----------------------------------- + K Ni^2 / 1 + B1ijk + B2ijk / -- -- over N, Cw and C0 (C0 is a constant) - yeah, I got that exprewssion from a book. If memory serves, I need to take the derivative and solve it where this equals zero. I haven't done derivatives for a VERY long time. Can anyone point me at a good source for getting back up to speed on that? TIA Steve === Subject: real roots of Mandelbrot polynomials If M_1(x)= x and M_(n+1)(x)=M_n(x)^2+x, the 2^(n-1) complex roots of M_n follow the Mandelbrot set (each root is the center of one of the connex components of the interior of M) Is there any way to find the number of real roots of M_n, or at least to get a good asymptotic estimation ? === Subject: Re: real roots of Mandelbrot polynomials posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) On 7 Feb., 07:39, Denis Feldmann æ follow the Mandelbrot set (each root is the center of one of the > connex components of the interior of M) Is there any way to find the > number of real roots of M n, or at least to get a good asymptotic > estimation ? First it is clear that a root of M n is also a root of M {k*n}, thus for composite n some root are known from lower degrees and leave us with much less primitive roots (i.e. roots not already belonging to a lower degree).. Next note that once a root of M n is non-real, i.e. you have found a complex bubble of the Mandelbrot set, this will produce an accordingly big number of complex sub-bubble solutions for M {kn}, in fact phi (k) if I'm not mistaken. Thus we have, I think, the following recursions general roots = 2^(n-1) primitive roots = 1/2 Sum {d|n} mu(d) * 2^{n-d} nonreal primitive roots = Sum {d|n} phi(d) * primitve roots(n/d) - Sum {d|n, phi(d) odd} real primitve roots(n/d) real primitive roots = primitive roots - nonreal primitive roots I think this still ignores complex satellites and tehrefore overestimates the number of primitive real solution. Again, if I'm not mistaken the recursions lead to n | primitive real roots ---+---------------------- 1 | 1 2 | 1 3 | 1 4 | 4 5 | 11 6 | 21 but that seems to be too high. In fact M 1 = x M 2 = x^2 +x = x*(x+1) M 3 = x^4 +2x^3 +x^2 +x = x*(x^3+2x^2+x+1) M 4 = x^8 +4x^7 +6x^6 + 6x^5 +5x^4 +2x^3 + x^2 +x = x*(x+1)* (x^6+3x^5+3x^4+3x^3+2x^2+1) and x^6+3x^5+3x^4+3x^3+2x^2+1 has only two real roots, not four. I don't know right now if the satellites (i.e. what corresponds two primitve roots that do not belong to sub-bubbles) are always distributed in such a manner that at most one is real? Or that at most a share of 1/phi(n) is real? Or whatever ... === Subject: Re: real roots of Mandelbrot polynomials hagman a .8ecrit : > On 7 Feb., 07:39, Denis Feldmann If \ M_1(x)= x and M_(n+1)(x)=M_n(x)^2+x, the 2^(n-1) complex roots of M_n > follow the Mandelbrot set (each root is the center of one of the > connex components of the interior of M) Is there any way to find the > number of real roots of M_n, or at least to get a good asymptotic > estimation ? > > First it is clear that a root of M_n is also a root of M_{k*n}, thus > for > composite n some root are known from lower degrees and leave us > with much less primitive roots (i.e. roots not already belonging > to a lower degree).. > Much less ? On M_(p*q), with p and q prime, we get exactly 2^(p-1) + 2^(q -1)-1 old roots, for 2^(pq -1) in all... Not such a gain :-) > Next note that once a root of M_n is non-real, i.e. you have found > a complex bubble of the Mandelbrot set, this will produce an > accordingly > big number of complex sub-bubble solutions for M_{kn}, in fact phi > (k) > if I'm not mistaken. > True (but quite hard to prove, btw), but again, not much useful for M_n if n is prime. > Thus we have, I think, the following recursions > > general roots = 2^(n-1) > primitive roots = 1/2 Sum_{d|n} mu(d) * 2^{n-d} > nonreal primitive roots = Sum_{d|n} phi(d) * primitve roots(n/d) > - Sum_{d|n, phi(d) odd} real primitve roots(n/d) > real primitive roots = primitive roots - nonreal primitive roots > > I think this still ignores complex satellites and tehrefore > overestimates > the number of primitive real solution. > Again, if I'm not mistaken the recursions lead to > > n | primitive real roots > ---+---------------------- > 1 | 1 > 2 | 1 > 3 | 1 > 4 | 4 > 5 | 11 > 6 | 21 > > but that seems to be too high. > > In fact > M_1 = x > M_2 = x^2 +x = x*(x+1) > M_3 = x^4 +2x^3 +x^2 +x = x*(x^3+2x^2+x+1) > M_4 = x^8 +4x^7 +6x^6 + 6x^5 +5x^4 +2x^3 + x^2 +x = x*(x+1)* > (x^6+3x^5+3x^4+3x^3+2x^2+1) > and x^6+3x^5+3x^4+3x^3+2x^2+1 has only two real roots, not four. > > I don't know right now if the satellites (i.e. what corresponds two > primitve roots > that do not belong to sub-bubbles) are always distributed in such a > manner > that at most one is real? No, this looks wrong (or I dont understand your meaning) . Actually, the real roots of M_n between -2 and -1.999, say, are quite numerous, and at least the leftmost of them (say the 3 leftmost are not satellites of previous bubbles (if the position of the leftmost root of M_(n-1) was -2+e, the exact position of the three leftmost roots of M_n are -2+e/4, -2+9e/4 and -2+25e/4) Or that at most a share of 1/phi(n) is real? > Or whatever ... > === Subject: CHINA PET BED WWW.PETBED-CATTREE.COM CAT TREE dog bed China Pet Products PET BED Boat Plane Car dog bed CHINA DOG BED China pet bed \ supplier WWW.PETBED-CATTREE.COM China pet products factory China Pet Dog bed \ cat tree Wrought iron car china dog pet bed pen factory Manufacturers pet supplies furniture pet supplier pet product posting-account=OMXU5QoAAAD8TJWbpISY4Zhh_xMLlIIr Mister X; GTB5; TencentTraveler 4.0; Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ),gzip(gfe),gzip(gfe) CHINA PET BED WWW.PETBED-CATTREE.COM CAT TREE dog bed China Pet Products PET BED Boat Plane Car dog bed CHINA DOG BED China pet bed supplier WWW.PETBED-CATTREE.COM China pet products factory China Pet Dog bed cat tree Wrought iron car china dog pet bed pen factory Manufacturers pet supplies furniture pet supplier pet product PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA PET BED FACTORY PET BED Pet products dog bed factory cat tree manufacturer in china pet factory pet bed supplier PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA PET BED FACTORY PET BED Pet products cat tree manufacturer Dog bed factory in china iron dog bed pen factory pet PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA CAT TREE FACTORY PET BED DOG BED Car dog bed factory pet product manufacturer pet bed supplier in china pet factory dog bed PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA CAT TREE FACTORY PET BED DOG BED Pet bed factory pet product factory iron dog bed car dog bed manufacturer in china pet bed PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA CAT TREE FACTORY PET BED DOG BED Pet bed dog bed factory china pet products manufacturer in china car dog bed pen supplier PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA CAT TREE FACTORY PET BED DOG BED Pet bed Pet products manufacturer wrought iron dog bed car pet bed factory in china pet PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA CAT TREE FACTORY PET BED DOG BED Pet bed supplier dog bed pet product in china pet factory iron dog bed car dog bed factory PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA CAT TREE FACTORY PET BED DOG BED Pet bed manufacturer Dog bed pet products manufacturer Cat furniture factory in china pet PET BEDS CHINA WWW.PETBED-CATTREE.COM CHINA CAT TREE FACTORY PET BED DOG BED Pet bed dog bed pet product factory wrought iron dog bed in china pet supplier pet dog bed === Subject: Re: Tesla magnetic wall: is this real? > http://www.youtube.com/watch?v=m8N6kIiRRHQ No it is not real. The hand sync, for example, is off. I've done quite a lot of computer graphics and I am pretty sure I can see where they have rubbed out the hand poking through the loop as it goes up and down. === Subject: Re: I am an Aether scientist. > This nonsense: > http://en.wikibooks.org/wiki/File:Rel2.gif Why is it nonsense? > But the above silliness does NOT apply to this universe > THERFORE both Lorentz and instein's math has false > assumptions in it. 'Silliness' is not a valid concept in physics. That is the only word that can do justice to lack of simultaneity. Talking about words...look up ''physics''...it's not math nor Einstein's illusive delusions. Spirit of Truth === Subject: Re: The modern mathematical concept of infinity is ... posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > On 6 Feb., 12:18, David Formosa (aka ? the Platypus) > Please show that the concept of actual infinity leads to (A & ~A) otherwise > your claim that it is not consistent is unsupported. > The concept of actual infinity says that the set of all positive even > numbers has cardinality aleph 0 > 2n for every n in N. > On the other hand a, non-empty set consisting of only positive even > natural numbers cannot have a cardinal number that is larger than > every even number. This can easly be obtained from the sequence of > initial segments > 2 > 2, 4 > 2, 4, 6 > ... So WM is back to posting this week. We know that WM is an ultrafinitist who is strongly opposed to the Axiom of Infinity. Whenever I try to defend finitistic beliefs, a supporter of the Axiom of Infinity argues that the axiom is necessary for the mathematics of the sciences. This tends to lead to a dead-end discussion. But WM (unlike Albrecht or HdB) has argued that Infinity is not necessary for the sciences, at least not physics, if there is a lack of reconciliation between WM, who believes (related to) an upper bound on the largest number needed for the sciences, and the mathematical majority, who believes that Infinity is necessary for the sciences. (As an aside, note that AP has recently named an upper bound on the largest number needed for physics as well -- 10^500, related to the factorial of the number of nucleons in an atom of the element for which AP is named.) But here in this thread, WM is arguing for finitism, not necessarily ultrafinitism. Of course, he's hoping to prove that ZF is inconsistent, by trying to derive a contradiction in the theory. He has attempted several times to establish an induction schema from which he'd derive that contradiction. In other threads, I've mentioned schemata similar to WM's, from which I'd like to derive the existence of nonstandard natural numbers -- but have so far failed to do so without deriving a contradiction. But here, WM wants there to be a contradiction, so that he could prove ZF inconsistent. Notice that WM actually has given a correct proof of something (but not what WM wants to prove, of course). He's proved that: ZF-Infinity |- (WM's Induction Schema -> ~Infinity) but of course, as the Infinity defenders already know, since WM never proved his Induction Schema in ZF, he hasn't proved ZF to be inconsistent at all. But wouldn't it be easier for a finitist not to keep trying to come up with new induction schemata from which to derive ~Infinity, but just assume ~Infinity as an axiom? Of course, but WM seeks an intuitive reason why one should reject the Axiom of Infinity. It's similar to what opponents of the Continuum Hypothesis have done. Rather than accept ~CH as an axiom, such set theorists seek more intuitive axioms that imply ~CH. Thus in this thread, WM is attempting to do the same with Infinity that many set theorists have done with CH. And so far, he's failed. No one is accepting WM's Induction Schema (or Quantifier Dyslexia, or other problems) as intuitive axioms that imply ~Infinity at all. === Subject: Re: The modern mathematical concept of infinity is ... posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > We know that WM is an ultrafinitist who is strongly opposed > to the Axiom of Infinity. Charitable to say the least. WM's ideas are more than a little confused. He claims that there are a finite number of integers, but no largest integer. He will accept potential infinity, but notactual infinity (he cannot really define either term). WM is strongly opposed to axiomatics. His claim is that the assumption of actual infinite sets leads to contradictions. In fact what WM thinks of as contradictions are results he does not like. The only contradictions he achieves are by word games. something (but not what WM wants to prove, of course). He's > proved that: ZF-Infinity |- (WM's Induction Schema -> ~Infinity) Well, it depends on what you think WM's Induction Schema is. Every induction argument he has give works fine for finite sets. WM then makes the great leap and says the results must hold for infinite sets (after all infinite sets are made from finite sets). So the proof is more like ZF-Infinity |- (WM's continuity at infinity assumptions -> ~Infinity) or in words if infinite sets have to have the same properties as finite sets then infinite sets are inconsistent. - William Hughes === Subject: Re: The modern mathematical concept of infinity is ... posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > So WM is back to posting this week. We know that WM is an ultrafinitist who is strongly opposed > to the Axiom of Infinity. Give up with sci.math: get into alt.journalism.pseudo-histor-socio- psycho-gossip. Write about the super-standard math of Amy Winehouse and the intra-finitistic logic of Rob Lowe. Better: haven't you got a life of your own? Go live it. -LV === Subject: lwal grabs wrong end of stick again posting-account=IBUqVwoAAADepmzxVr9iEYD5Z0A483SY rv:1.9.0.1) Gecko/2008070206 Firefox/3.0.1,gzip(gfe),gzip(gfe) > We know that WM is an ultrafinitist who is strongly opposed > to the Axiom of Infinity. No, what we know is that enheim is an idiot and a fraud. enheim's ideas are far too incoherent to be dignified with a tag like ultrafinitist. === Subject: Re: The modern mathematical concept of infinity is ... Supersedes: <1rc0jhqwzd1ra.ksmr4a037wnd.dlg@40tude.net > We know that WM is an ultrafinitist who [bla bla]. > No, WE don't know that, asshole. Actually, we CAN'T know it, since it is simply false that he's an /ultrafinitist/ (of any sort); WM is just a VERY CONFUSED _mathematical crank_ (and that's ***NOT*** the same, idiot). Herb === Subject: Re: The modern mathematical concept of infinity is indefensible [...] > Now, please, show me a physical infinity. The density of a singularity. === Subject: Re: The modern mathematical concept of infinity is indefensible sha1:jhj9v0auDOEQFQcuGEQUa+0W5p4= > [...] > Now, please, show me a physical infinity. The density of a singularity. Han: Singularities don't exist. Respondent: Why not? Han: Because there's no such thing as a physical infinity. Respondent: How do you know? Han: I've never seen one. Show me a physical infinity. Respondent: The density of a singularity. (Repeat) -- Jesse F. Hughes Liberals have hijacked science for long enough. Now it's our turn. === Subject: Re: The modern mathematical concept of infinity is indefensible [...] > No, but _this_ is implementable as a first approximation! There are many > n for which your fundamental fact is actually _true_ even in computers. > But an infinite set can _never_ be stored, due to lack of space. PA is implementable in computers. The axioms can be quite naturally modeled within lambda calculus and LC is mechanically implementable. === Subject: Re: The modern mathematical concept of infinity is ... > My interest is the number of natural numbers. > The number of natural numbers from 1 to n > > the set F_n > > cannot be more than the > cardinal number of the FISON {1, 2, 3, ..., n}. This is finite for > every n. > > A statement about the F_n > > The set of natural numbers from 1 to n is the FISON {1, 2,3 ..,n}. > > A statement about the F_n > > This also is finite for every n. > > A statment about the F_n > > These results do not depend on the question whether there is a last n > in N. > My sequence (F_n) with > F_n = U[k =< n] {1, 2, 3, ..., k} > has no last n. > > A statement about the F_n > > Not at all. It does not stop at any point. > But each FISON of my infinite sequence (F_n) has a largest natural > number, > > A statment about the F_n. > > because every n that is available is a largest number of the > sequence preceding it. That cannot be avoided in reasonable > reasoning. > > Up to this point everything is correct. However, you have said > nothing about the set of all natural numbers. You have only > said things about the F_n. The set of all natural numbers is > the union of the F_n. > > Therefore I have shown that the set of all natural numbers is not > actually infinite > > No, you have shown nothing about the set of all natural numbers. > You have only shown things about the *elements* of the set > of all natural numbers. > > - William Hughes Or certain proper subsets of N. === Subject: Re: The modern mathematical concept of infinity is ... > On 6 Feb., 12:18, David Formosa (aka ? the Platypus) [...] Actual infinity is not sound and not consistent and math lose it's > ability of prediction with the use of the concept of actual infinity. Please show that the concept of actual infinity leads to (A & ~A) otherwise > your claim that it is not consistent is unsupported. > > The concept of actual infinity says that the set of all positive even > numbers has cardinality aleph_0 > 2n for every n in N. > On the other hand a, non-empty set consisting of only positive even > natural numbers cannot have a cardinal number that is larger than > every even number. WM again assumes without proof that what holds for finite sets necessarily holds for sets which are not finite in order to deduce that all set are finite. > > A: The set of all positive even numbers has cardinal number larger > than every member. > ~A: No set of positive even numbers can have a cardinal numer larger > than every member. WM's ~A only holds for finite sets (which, however ordered, must always have a first and a last member), but does not hold for non-finite sets (which can be ordered so as not to have both a a first and a last member, for example, non-empty open intervals of rationals or reals). The set of positive even integers has neither a first nor a last under the standard ordering, and the set of even naturals has no last, so that WM's assertions are not binding on them. > > This is one of several contradictions of set theory. It is really only a conflict between WM's set theory and standard set theories. But as WM's is incoherent and full of internal contradictions, he can only impose it on those who are in thrall to him, his poor students. === Subject: Re: The modern mathematical concept of infinity is ... > As my proof shows, every term of the infinite sequence (F_n) of FISONs > F_n is finite and covers all the preceding terms. Therefore, up to > every finite number n, all is covered by a single FISON. We do not get > out of this finite domain because N contains only finite numbers. > Whether the sequence is finite or infinite is not relevant. On the contrary, it is the one relevant issue! Which is, no doubt, why WM wishes to slide past it. The union of a finite sequence of FISONs necessarily equals its maximal member, so is always another FISON. The union of an infinite sequence of FISONs necessarily is not a FISON, as it provably can have no maximal member, which every FISON must by definition have. === Subject: Re: The modern mathematical concept of infinity is ... > Ultimately, it's because WM doesn't understand the difference > between > 1. All FISONs have property P > and > 2. The set of all FISONs has property P > > If property P describes the cardinal number of FISONs then there is no > difference. The cardinal number of {1, 2} is the same as the cardinal > number of {{1}, {1, 2}}. An example is not a proof. And in any case, here is a counterexample: All dice have 6 faces, but the set of all dice, not being a die, does not have any faces. > > As I have shown, Not to the satisfaction of anyone but WM. > The statement about the cardinal number of a FISON is automatically a > statement about the cardinal number of a subset of N. A statement > about the cardinal number of all FISONs is automatically a statement > about the cardinal number of all natural numbers. Your ambiguity is showing again. What do you mean by the cardinal number of all FISONs? Do you man the cardinal number of each FISON separately of the cardinal number of the set of all FISONs? Similarly, do you mean the cardinal of each natural separately or the cardinal of the set of all naturals? Until you learn to state what you mean with a good deal more precision, your meanings get lost in your ambiguity, and mean nothing. > > My proof is flawed. As usual. === Subject: Re: The modern mathematical concept of infinity is ... > It is about every n in N and hence about every FISON.If it is possible > to state something for every natural number n in N, then my procedure > states that for every natural number n in N the union of all FISONs > that are smaller or equal to FISON n is the same as FISON n. Everything you said so far is correct. If N is nothing but its numbers, then that holds for N. This, of course, is nonsense. In fact, it is utterly meaningless! > What holds for N? > > It holds for every number including all preceding numbers, that is, it > holds for N, that a single FISON covers all FISONs. For any single FISON f, there is a largest member n of that FISON, and then n+1 is in N but not in that FISON, so for every FISON there is a larger FISON, so no FISON can contain all others. * That the union of all FISONs smaller or equal to FISON N is the same > as FISON N? You must first show that the phrase the FISON N is meaningful. > > It is the result of my proof. It is the result of WM's delusion. It is easily shown that WM's delusion leads to a contradiction for any standard definition of N and FISONs, since in such an N there cannot be a largest standardly defined FISON. > I > know what you mean when you write the FISON n. It is your own > illucid way of describing the Finite Initial Segment {1,...,n}. > Which Finite Initial Segment of naturals do you mean when you write > FISON N? > > I am not able to determine a fixed number. Because there cannot be any such n in N for which there is no n+1 in N. At least for any standardly defined N. > N is potentially infinite. There is no such thing as a potentially infinite SET. Nothing in sethood prohibits a set from being either finite or actually infinite, but the requirements of sethood prohibit potential infiniteness. > What I have been able to do is to show that N is not actually > infinite, i.e. has a larger cardinality than every natural number can > measure. Not without assuming things bar N from being a set at all. > > Yes. I call N the assembly that contains all natural numbers and > nothing else. Therefore the sequence (F_n, with n in N) covers all > natural numbers. And every element of the sequence covers all > preceding elements. Therefore one element of the sequence covers N. That last claim does not follow, at least not without additional assumptions which have not been made explicit, and which do not hold for any standard definition of N. > This element cannot be determined because N is potentially infinite. Then it is not a set a all. > But it has been shown that it is not actually infinite., i.e., larger > than any FISON. No it has not. WM overlooked making explicit the assumptions that would allow him to bridge his non sequitur above. Until he does that, and gets everyone to agree to those assumptins, his alleged proof has a fatal gap. > === Subject: Re: The modern mathematical concept of infinity is ... > > A union of FISONs that covers every n in N covers N. I absolutely agree. Which union of FISONs do you claim covers every n > in N? The rest of us have said that the union U { F_k | k in N } is a union of FISONs that cover N, but it is an infinite union. There > is no n such that U { F_k | k <= n } covers N. You seem to think otherwise, so for *which* n does the > finite union of n FISONs cover N? > > My proof uses an infinite sequence (F_n) of FISONs > F_n = U { F_k | k <= n }. That does not answer the question. WM has claimed,in effect, that there is an n for which the FISON ending in n contains N, but does not name that n. My reaction is to demand of WM that he put up or shut up. > I do not pretend that I can give a fixed n. if ther is no fixed n, then in any sort of standard mathematics there is no n at all, since those naturals are all fixed. Apparently in WM's world, some naturals are allowed to play hide and seek. > Therefore I use an > infinite sequence. But it is clear that there is no part of N outside > of every F_n, because the sequence covers every n and all its > preceding numbers from 1 on. > > So the opinion of most of you has been falsified that N is larger than > every FISON. N is certainly larger than any fison in our world, so it is quite a puzzlement how N is necessarily greater than any FISON but somehow not greater than every FISON. Could it be WM's Quantifier Dyslexia rearing its ugly head again? > I did not say that > N is missing an element. > > And I did not blame you to have said so, but you should say what n is > not in my sequence of FISONs. > You > > Here, let me give it to you more slowly. For each n, we have the following property: (E k in F_n)(A m in F_n)( m <= k ). However, ~(E k in N)(A m in N)( m <= k ), since whenever k is in N, so is k + 1 and k < k + 1. Thus, we've found a property which distinguishes each F_n from N, > > but not the complete infinite sequence. Even that. No member of the infinite sequence of FISONs is a member of N and no member of N is a member of that infinite sequence of FISONs. > > namely, the property P defined by P(S) <-> (E k in S)(A m in S)( m <= k ). > > Your reasoning concerns only single elements F_n of my infinite > sequence (F_n). It concerns each member of that sequence. > But my proof is not based upon any fixed n. That it is not about naturals at all and is irrelevant. > n assumes every natural > number. Therefore we have for the sequence (F_n) > > ~(E k in N)(A m in (F_n))( m <= k ). That is logically equivalent to claiming an F_n such that (A k in N) (E m in (F_n)) ( m > k) But k = n+1 is an immediate and total counterexample of your claim, and thus disproves it. > > If you claim that N is not covered by the *infinite sequence* (F_n), > then prove it. What we claim is that our N is not covered by any finite subsequence. > P(F_n) is true (for every n), but P(N) is false, so it cannot be the > case that F_n = N. > > Either N c (F_n) or for some m we have m is in N but m is not in > (F_n). > The first is true. Not in our N. And in our n, we show for some m we have m is in N but m is not in (F_n) merely by taking m =n+1. > If m is in N, then there is an n > m and m is in F_m. Quire true, but a bit pointless. > > Probably you have missed the argument. > > 1) N c (F_n) (unless you find any m in N but not in (F_n)) Given any n, take m = n+1 giving m in N but not in F_n, ergo NOT N c (F_n). So no, we have not missed it, but it doesn't support WM's foulishness. === Subject: Re: The modern mathematical concept of infinity is ... > > Which FISON do I have ignored? All but finitely many of them. > > Give a number. The largest natural plus one. Since you insist that there is a largest natural, you can hardly object to my using it. And for everyone but WM every natural has a successor natural. > This is so for every union of finitely many FISONs. But that leaves out the vast majority of them. Every union leaves out the majority? it is not every union since you omit most unions. > > Say which FISON I omit. The FISON ending in the largest natural plus one. > > There are infinitely many FISONS containing any n. That is not interesting for my proof. On the contrary, it shows your proof is not a proof at all. > > In what way? That you do not see it proves you willful blindness. > But of course I > consider every n in N, and there is always the same result: The union > of all these FISONs U[k =< n] {1, 2, 3, ..., k} is one FISON, namley > FISON {1, 2, 3, ..., n}. Which still omits most FISONS and most naturals For all n? Since you only consider unions of FISONs which are subsets of a given > FISON, and omit all those which are not, > > Which one have you in mind, for instance? Since you must choose your n first in order to determine which union you are forming, for that n you omit infinitely many FISONs from your union. Thus for any choice of n you omit infinitely many FISONs, and for no choice of n can you include all FISONs. > > But if you start with N, which is not a FISON itself, the union of the > set of all FISONs in N is N, which is not a FISON. > > It is true, infinity cannot be treated in a sound manner. At least not by such dunderheads as WM. But there are those who can. > The result > depends on how you start. But my starting point is as justified as > yours. You are allowed to choose any axioms you want, but are not allowed to impose your choices on others, unless the poor bastards are your students. In fact, given any non-empty set of FISONs with no maximal member by > inclusion, its union is the non-FISON N. > > As I have shown this premise is wrong. It is only wrong if one insists on making contrary assumptions, as WM does. It is not wrong in the absence of such assumptions. > The existence of your given > set can be contradicted when starting as I did. You start by assuming as axioms, and without proof, things that I, and most others here, do not accept. Since you cannot force those axioms on any but your enslaved students, they are the only ones that you can browbeat into submission. > Or try to explain why my proof does not cover all n (and all > smaller natural numbers) in N. It is not that they do not cover each n individually, or even for each > FISON individually but that they do not hold for N, nor for the set of > all FISONs. > > My proof covers all n in a FISON, and as the magnitude of the FISONs > considered is not limited, my proof covers all n from 1 on with > unlimited magnitude. Perhaps unlimited in the sense that one cannot name the limit, but always limited in the sense of having a largest natural or FISON. None of what you claim is valid unless there is a largest natural and corresponding largest FISON, even if unnamed and unnameable. >If your N is different, then it must contain > something else than finite numbers. It is what my N does NOT contain that makes the difference. My N does not contain a last/largest natural. > The problem is that, for some n, your non-proof ignores everything from > n+1 onwards But I do not stop at some n! Each of your proofs does exactly that. > Each assumes a largest n for the extent of the proof and denies the > relevance of any larger ones. Thus is invalid to generalize for any set > of naturals which does not have a largest member. > > Each natural number is the largest of a FISON. There are no other > natural numbers. Each natural is the smallest of an infinite sequence of naturals. We need no other natural numbers. > If your N is different, then it must contain > something else. My N, along with the N's for nearly everyone, differs from WM's only in what it does NOT contain. it does NOT contain a largest/last natural. And since WM is quite unable to specify an alleged last/largest for his own N, I suspect that his does not contain one either, even though he stubbornly refuses to admit it. All your proofs presume a largest natural, > > Not at all. At all! Each FISON of my infinite sequence (F_n) with > F_n = U[k =< n] {1, 2, 3, ..., k} > assumes that it has a largest natural. That cannot be avoided in > reasonable reasoning. But my sequence does not stop at any point. So > my prof covers all natural numbers from 1 on. Your proof does not prove what it claims to prove, since it never looks at all of N, but only at the contents of one FISON. > æI claim that my proof covers all n. It only covers sets of naturals which have a largest member, therefore > need not hold for N. > > All natural numbers have this property: each one is finite. But that does not in any way require that the set of objects all of which are finite itself be finite, any more that a set of objects all of which are red must be a red set. Following WM's form of argument, the odd sized set {2,3,6} would have to be an even set and the even sized set {1,3,5,7} would have to be an odd set. === Subject: Re: The modern mathematical concept of infinity is ... Ultimately, it's because WM doesn't understand the difference > between 1. All FISONs have property P and 2. The set of all FISONs has property P I am satisfied with 1. > If 2 is different, then your set of all FISONs contains more than > all FISONs. But then it is not the set of all FISONs - and therefore > not of interest for me. Nope. æThe set of all FISONs has elements that are FISONs. æSo the > set of all FISONS does not contain more than all FISONs. æHowever, > proving > that P is true for every element of a set does not prove that P > is true for the set. 1 and 2 are not the same. > > My interest is the number of natural numbers. > The number of natural numbers from 1 to n cannot be more than the > cardinal number of the FISON {1, 2, 3, ..., n}. This is finite for > every n. > The set of natural numbers from 1 to n is the FISON {1, 2,3 ..,n}. > This also is finite for every n. > These results do not depend on the question whether there is a last n > in N. > My sequence (F_n) with > F_n = U[k =< n] {1, 2, 3, ..., k} > has no last n. Not at all. It does not stop at any point. > But each FISON of my infinite sequence (F_n) has a largest natural > number, because every n that is available is a largest number of the > sequence preceding it. That cannot be avoided in reasonable > reasoning. > > Therefore I have shown that the set of all natural numbers is not > actually infinite (meaning that it has more elements than every > natural number). Actually what you have show is the exact opposite of what you have claimed to show, namely you have shown that the 'number' of naturals and, equally, the number of FISONs, are both infinite (having more elements than any natural number). === Subject: Re: Is there a universal entropy formula that allows comparing any \ kind of distribution? posting-account=McZ3aQkAAADz6LV-boDe1LcriRhf3lj3 Gecko/2009011913 Firefox/2.0.0.9;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) > JG vd Galienhttp://www.satoconor.com February 6, 2009 sci.math > === > Subject: Relative Pareto Entropy? > for the Pareto distribution is: Entropy=H(f)=ln(k/m)-(1/k)-1 (1.) Pareto power law Pr(X>x)=(m/x)^k x >= m (2.) DESC ----------------------------------------------------------------- > 14631269990896.7 æ æ æ æ1.51169289996617 æ æ æ æ-30.2394419661841 > 190216746.650989 æ æ æ æ1.52160949886678 æ æ æ æ-18.9867072547286 > 129105441.026909 æ æ æ æ1.50485197727519 æ æ æ æ-18.6029282762211 > 124975271.288802 æ æ æ æ2.04976925028009 æ æ æ æ-18.438039428074 > 101723785.718849 æ æ æ æ1.34057316935514 æ æ æ æ-18.3987248685568 > 94681989.1438299 æ æ æ æ1.77588786356754 æ æ æ æ-18.2286352317757 > 74691758.9517272 æ æ æ æ2.30751568745425 æ æ æ æ-17.859342335023 > 58658272.2889092 æ æ æ æ2.12707565168039 æ æ æ æ-17.6623620412572 > 58281608.6310413 æ æ æ æ1.46257550899264 æ æ æ æ-17.8168728288427 > 54922716.7695079 æ æ æ æ1.68399109291601 æ æ æ æ-17.7064436099137 still want to be able to compare the distributions with each other. m has an enourmous impact on the absolute entropy of (1.) I believe that m is irrelevant in the formula if you want relative > entropy which you can compare with other Pareto distributions. Relative Pareto Entropy=ln(k)-(1/k)-1 (3.) (1.) can be used to compare distributions from one source. But I want > to > compare distributions from many sources (up to 4000 sources) with each > other. Questions: > 1) Can one calculate a (relative) Pareto entropy value that can be > compared to any Pareto > distribution? > 2) Is there some kind of standarized mathematical entropy for > comparing any kind > of distribution? Like thermodynamical entropy can be compared to any > kind of > physical and chemical system because it has a unit (joule/Kelvin). > 3) Is then (3.) the correct formula for relative Pareto entropy? > 4) (1.) entropy correlates very well with the standard deviation (sd) > of the normal distribution. > Like when (1.) gets smaller (more negative) sd becomes greater. So can > one say that when > (1.) gets smaller there comes more randomness in the distribution? > 5) mean/sd=SNR=signal-to-noise-ratio. SNR also becomes greater when > (1.) gets smaller > in the samples. Does this contradict with the getting more randomness > in the sample hypothesis? Johan Gerard van der Galien. When Boltzmann connected physical entropy with probability, he has given a simple example of the most probable partition of velocity quantas between molecules, giving to the system most degrees of freedom. This is measured by the polynomial coefficient, and the logarithm of its inverse is the measure of entropy H(m). For molecules, due to their numbers, it is possible to exploit Stirling aproximation, and simplify the final formula. Entropy would be maximal, if each molecule in the system had its own velocity (0, 1, 2, 3, ...). Due to their number, it is not possible, it would lead to too high temperatures. Therefore, Maxwell-Boltzmann distribution of velocities of molecules is observed. Due to the relation between velocities of molecules and their kinetic energy, we observe simultaneously normal distribution of their energies. kunzmilan === Subject: New version of my factorization applet posting-account=O-WGjQ8AAAAU7UvFJ5CqF6gThuf-tX2q .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; FDM; InfoPath.1),gzip(gfe),gzip(gfe) Hello folks, Yesterday I uploaded to the Web server the new version of my factorization applet which can be seen at http://www.alpertron.com.ar/ECM.HTM The new code is very fast. For example it can factor 10^59+213 (60 digits) in 24 segundos and 10^71-1 (71 digits) in just 4 minutes 22 seconds in a Core 2 Duo 1.86 GHz based PC. It also has a new batch factorization mode which can be used to factor or determine primality of several related numbers by just typing a line. For instance, typing x=10^36;x=x+1;i-100;x the applet will factor all numbers from 10^36 to 10^36+99. Dar.92o Alpern Buenos Aires - Argentina === Subject: Re: New version of my factorization applet It's impressive! I gave it 10 minutes to factor 100!+1 but that just isn't long enough-- im === Subject: Re: New version of my factorization applet <14689844.1234015718055.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=O-WGjQ8AAAAU7UvFJ5CqF6gThuf-tX2q .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; FDM; InfoPath.1),gzip(gfe),gzip(gfe) > It's impressive! I gave it 10 minutes to factor 100!+1 > but that just isn't long enough-- im Your number equals 101 x 14303 x 149239 x 350433007170616328107072379 x P121, where P121 is a 121-digit prime number. The applet can find the 27-digit number using the ECM method but this requires more time. Dar.92o Alpern === Subject: Question on subcategories posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/2008111318 Ubuntu/8.04 (hardy) Firefox/3.0.4,gzip(gfe),gzip(gfe) suppose F:C-->D, G:D-->C are two functors such that GF=Id_C. Is it true that F is fully faithful and injective on objects, i.e. a full embedding? S. === Subject: Re: Question on subcategories posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) suppose F:C-->D, G:D-->C are two functors such that GF=Id_C. Is it > true that F is fully faithful and injective on objects, i.e. a full > embedding? S. You really mean GF = Id_C and not just up to natural transformation? In that case, F is clearly injective on objects, since F(A)=F(B) implies A=GF(A)=GF(B)=B. Similarly, it is also clear that F is faithful. But F need not be full: Let C have one object A and one morphism (which is id_A). Let D be the category of whatever, pick an object B in Obj(D) and define: F(A) = B, F(id_A) = id_B G(X) = A for all X in Obj(D), G(f:X->Y)=id_A for all f in Mor(D). Clearly, G and F are functors with GF = Id_C. As soon as B has non-trivial endomorphisms, F is not full. hagman === Subject: Re: Question on subcategories posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/2008111318 Ubuntu/8.04 (hardy) Firefox/3.0.4,gzip(gfe),gzip(gfe) > suppose F:C-->D, G:D-->C are two functors such that GF=Id C. Is it > true that F is fully faithful and injective on objects, i.e. a full > embedding? S. You really mean æGF = Id C and not just up to natural transformation? > In that case, F is clearly injective on objects, since F(A)=F(B) > implies > A=GF(A)=GF(B)=B. > Similarly, it is also clear that F is faithful. > But F need not be full: > Let C have one object A and one morphism (which is id A). > Let D be the category of whatever, pick an object B in Obj(D) > and define: > F(A) = B, F(id A) = id B > G(X) = A for all X in Obj(D), G(f:X->Y)=id A for all f in Mor(D). > Clearly, G and F are functors with GF = Id C. > As soon as B has non-trivial endomorphisms, F is not full. hagman the functor category Fun(A,C) for a small category A. F is the discrete functor and G evaluates at an fixed a in A. Then GF=Id C. But F is not full even in this special situation, right? S. === Subject: Re: Question on subcategories posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) suppose F:C-->D, G:D-->C are two functors such that GF=Id C. Is it > true that F is fully faithful and injective on objects, i.e. a full > embedding? S. You really mean æGF = Id C and not just up to natural transformation? > In that case, F is clearly injective on objects, since F(A)=F(B) > implies > A=GF(A)=GF(B)=B. > Similarly, it is also clear that F is faithful. > But F need not be full: > Let C have one object A and one morphism (which is id A). > Let D be the category of whatever, pick an object B in Obj(D) > and define: > F(A) = B, F(id A) = id B > G(X) = A for all X in Obj(D), G(f:X->Y)=id A for all f in Mor(D). > Clearly, G and F are functors with GF = Id C. > As soon as B has non-trivial endomorphisms, F is not full. hagman the functor category Fun(A,C) for a small category A. F is the > discrete functor and G evaluates at an fixed a in A. Then GF=Id C. > But F is not full even in this special situation, right? S. Sorry, I dont't seem to recall what the discrete functor is. Is it F: C->Fun(A,C) as given on objects by F(c)(x)=c, F(c)(f:x->y) = id c? (For otherwise I don't see how to guarantee F(c)(a)=c etc.) === Subject: Re: Question on subcategories posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/2008111318 Ubuntu/8.04 (hardy) Firefox/3.0.4,gzip(gfe),gzip(gfe) > suppose F:C-->D, G:D-->C are two functors such that GF=Id C. Is it > true that F is fully faithful and injective on objects, i.e. a full > embedding? S. You really mean æGF = Id C and not just up to natural transformation? > In that case, F is clearly injective on objects, since F(A)=F(B) > implies > A=GF(A)=GF(B)=B. > Similarly, it is also clear that F is faithful. > But F need not be full: > Let C have one object A and one morphism (which is id A). > Let D be the category of whatever, pick an object B in Obj(D) > and define: > F(A) = B, F(id A) = id B > G(X) = A for all X in Obj(D), G(f:X->Y)=id A for all f in Mor(D). > Clearly, G and F are functors with GF = Id C. > As soon as B has non-trivial endomorphisms, F is not full. hagman the functor category Fun(A,C) for a small category A. F is the > discrete functor and G evaluates at an fixed a in A. Then GF=Id C. > But F is not full even in this special situation, right? S. Sorry, I dont't seem to recall what the discrete functor is. > Is it F: C->Fun(A,C) as given on objects by > F(c)(x)=c, F(c)(f:x->y) = id c? > (For otherwise I don't see how to guarantee F(c)(a)=c etc.) Yes, exactly. I am sorry for being not precise. === Subject: Parabola confusion on TI-85 calc posting-account=hlsuLgoAAABdRR1Wo-FQwG-MHXfO4_GV .NET CLR 2.0.50727; .NET CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) Hello all, I'm learning about parabolas and my book's asked me to sketch out the following function: (2x - 1)(x - 3). The vertex of the parabola is ((1/2) + 3) / 2 = 7/4. And plugging that into the function (multiplying the factors): 2x^2 - 7x + 4 to get -17/8 or -2.125. I wanted to verify my answer on my graphing calculator (Texas Instruments TI-85) so entered y1 = (2x - 1)(x - 3), had a look at the graph (looked fine at first glance), and then did the FMIN function to find the least value of the function. It gave me -3.125 which is not what I got! I put the expanded version of the function in to y1: 2x^2 - 7x + 4, and FMIN gives me -2.125. I've had a look at the manual by can't seem to see if I'm doing anything wrong - am I making a silly mistake here? Pete === Subject: Re: Parabola confusion on TI-85 calc > Hello all, > I'm learning about parabolas and my book's asked me to sketch out the > following function: (2x - 1)(x - 3). > The vertex of the parabola is ((1/2) + 3) / 2 = 7/4. > And plugging that into the function (multiplying the factors): > 2x^2 - 7x + 4 to get -17/8 or -2.125. Check your multiplication. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: Parabola confusion on TI-85 calc > I'm learning about parabolas and my book's asked me to sketch out the > following function: (2x - 1)(x - 3). > > The vertex of the parabola is ((1/2) + 3) / 2 = 7/4. > > And plugging that into the function (multiplying the factors): > > 2x^2 - 7x + 4 to get -17/8 or -2.125. It happens that (2x - 1)(x - 3) = 2x^2 -7x +3. Jose Carlos Santos === Subject: Re: Parabola confusion on TI-85 calc <6v5e5pFi5c72U1@mid.individual.net> posting-account=hlsuLgoAAABdRR1Wo-FQwG-MHXfO4_GV .NET CLR 2.0.50727; .NET CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) I'm learning about parabolas and my book's asked me to sketch out the > following function: (2x - 1)(x - 3). The vertex of the parabola is ((1/2) + 3) / 2 = 7/4. And plugging that into the function (multiplying the factors): 2x^2 - 7x + 4 to get -17/8 or -2.125. It happens that (2x - 1)(x - 3) = 2x^2 -7x +3. > Jose Carlos Santos DOH! Of course it does! Think I need a break from this for a couple of hours! Pete === Subject: Re: How many sum of two number is not divisible by their difference === > Subject: Re: How many sum of two number is not divisible by their > difference > <24600515.1233911741882.JavaMail.jakarta@nitrogen.mathforum.org \ <4189142.1233912385522.JavaMail.jakarta@nitrogen.mathforum.org \ posting-account=cvz5-QoAAABVNzogw177Plx_25TguPUZ > X-HTTP-UserAgent: Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; > .NET > CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.2; > MS-RTC LM > 8),gzip(gfe),gzip(gfe) > Please give me some idea or solution. :) > How many different positive integers not exceeding > 2009 can be chosen at > most such that the sum of any two of them is not > divisible by their > difference. = set P > Some numbers whose sum is divisible by their > difference are :D >1) (n , n + 1) æwhere 1 < = n < = 2008. 2) (n, n+ 2) æwhere 1 < = n < = 2007. > Set P does not contain these numbers.- Hide quoted text - - Show quoted text - Generalizing ... (k, mk + (m+1)), where 1 <= k <= 2009 and 1 <= mk + (m+1) <= 2009 --------------------- A more general approach ... Let (p, q), with p < q, be a pair satisfying the conditions of the > problem. æThen (p + q) = r * (q - p) for some Natural r. æSolve for q in terms of p > and r: q = p * (r + 1)/(r - 1) q must be a Natural. æSo r-1 must divide p * (r + 1). ær-1 divides r+1 > only if r = 2 or 3. æSo two sets of solutions are q = p * 3, or (p, 3p), 1 <= p <= 2009/3 > q = p * 2, or (p, 2p), 1 <= p <= 2009/2 Other solutions are found as follows: æFor each p, 1 <= p <= 2009, > find r such that r+1 divides p and 1 <= q <= 2009. - MO- Hide quoted text - - Show quoted text - Correction: Generalizing: (k, mk + (m-1)), where 1 <= k <= 2009 and 1 <= mk + (m+1) <= 2009 - MO- Hide quoted text - - Show quoted text - Correction to the Correction: Never mind about my attempts at generalization of D&D's solution. The > more general approach above says it all. - MO We know that r-1 divide p(r+1) But why you know that r-1 must divide (r+1) .? Can you explain me more. ? TIA === Subject: Re: How many sum of two number is not divisible by their difference posting-account=cvz5-QoAAABVNzogw177Plx_25TguPUZ CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.2; MS-RTC LM 8),gzip(gfe),gzip(gfe) > === > Subject: Re: How many sum of two number is not divisible by their > difference > <24600515.1233911741882.JavaMail.jaka...@nitrogen.mathforum.org æ \ æ <4189142.1233912385522.JavaMail.jaka...@nitrogen.mathforum.org æ æ \ posting-account=cvz5-QoAAABVNzogw177Plx 25TguPUZ > X-HTTP-UserAgent: Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; > .NET > æ æ CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.2; > æ æ MS-RTC LM > æ æ 8),gzip(gfe),gzip(gfe) > Please give me some idea or solution. :) > How many different positive integers not exceeding > 2009 can be chosen at > most such that the sum of any two of them is not > divisible by their > difference. = set P > Some numbers whose sum is divisible by their > difference are :D >1) (n , n + 1) æwhere 1 < = n < = 2008. > 2) (n, n+ 2) æwhere 1 < = n < = 2007. > Set P does not contain these numbers.- Hide quoted text - > - Show quoted text - Generalizing ... (k, mk + (m+1)), where 1 <= k <= 2009 and 1 <= mk + (m+1) <= 2009 --------------------- A more general approach ... Let (p, q), with p < q, be a pair satisfying the conditions of the > problem. æThen (p + q) = r * (q - p) for some Natural r. æSolve for q in terms of p > and r: q = p * (r + 1)/(r - 1) q must be a Natural. æSo r-1 must divide p * (r + 1). ær-1 divides r+1 > only if r = 2 or 3. æSo two sets of solutions are q = p * 3, or (p, 3p), 1 <= p <= 2009/3 > q = p * 2, or (p, 2p), 1 <= p <= 2009/2 Other solutions are found as follows: æFor each p, 1 <= p <= 2009, > find r such that r+1 divides p and 1 <= q <= 2009. - MO- Hide quoted text - - Show quoted text - Correction: Generalizing: (k, mk + (m-1)), where 1 <= k <= 2009 and 1 <= mk + (m+1) <= 2009 - MO- Hide quoted text - - Show quoted text - Correction to the Correction: Never mind about my attempts at generalization of D&D's solution. æThe > more general approach above says it all. - MO We know that r-1 divide p(r+1) But why you know that r-1 must divide (r+1) .? Can you explain me more. ? TIA- Hide quoted text - - Show quoted text - r-1 does not necessarily divide r+1, but if it does, the (p,q) with q = p * (r+1)/(r-1) is a solution to the original problem. Again, the 3 we find the solutions (p, 3p) and (p, 2p). Next, for each value of p between 1 and 2009 inclusive, search for values of r such that r-1 divides p. For example, with p = 30, r-1 divides p if r = 2, 3, 4, 6, 7, 11, and 16. We already counted the solutions r = 2, 3. The other solutions would be (30, 50), for r = 4 (30, 42), for r = 6 (30, 40), for r = 7 (30, 36), for r = 11, and (30, 34), for r = 16 - MO === Subject: Re: Q on Riemann domains >Could you remind us what a Riemann domain is? All the properties are contained in this: >Suppose M is a Riemann domain with projection map P -> C^n. P is a local >homeomorphism. I found an easy counterexample to the first question - a connected riemann domain that's mapped onto a ball in C, where P is not injective. And a proof for the second question: in a riemann domain, there is a distance to the boundary defined by d(z)=sup(r: P maps a neighborhood of z homeomorphically to a ball radius R). So the question is whether this is really a distance to the boundary or whether P stops being injective instead. I'm somewhat proud of my proof showing P can't stop being injective before it runs into the edge of the domain. Laura === Subject: Re: Q on Riemann domains > >Could you remind us what a Riemann domain is? > > All the properties are contained in this: > >Suppose M is a Riemann domain with projection map P -> C^n. >P is a local homeomorphism. Do you mean you consider this sentence to be a definition or even description of a Riemann domain? === Subject: Re: How to find a if there is a zero or root between two points quickly Hi. I was curious on this technique, and tried to following along from the link below. I put something together with Mathematica. Since I wasn't sure how long the chain was, I fixed-pointed it until it ended with Indeterminate (0/0). (Quiet was used so as to not display error messages) Not tested very well, but I do see that it counts double root only once. http://mathworld.wolfram.com/SturmFunction.html SturmChain[equ_, var_] := Module[{}, v = {equ, D[equ, var]}; v = FixedPointList[FullSimplify[ {#[[2]], #[[2]]* PolynomialQuotient[#[[1]], #[[2]], var] - #[[1]]}] &, v] // Quiet; v = DeleteCases[v, {_, Indeterminate}]; v[[All, 1]] ] CountRoot[equ_, {x_, a_, b_}] := Module[{g}, g = SturmChain[equ, x]; g = {g /. x -> a, g /. x -> b}; g = (Count[Abs[Differences[Sign[#1]]], 2] &) /@ g; First[Abs[Differences[g]]] ] I counted a difference of -1 to +1 =2 as a change in sign. There may be a better way. poly1 = -1 - 3 x + x^5; This is from the example on the above link: SturmChain[poly1, x] { -1 - 3 x + x^5, -3 + 5 x^4, 1 + (12 x)/5, 59083/20736 } CountRoot[poly1, {x, -2, 0}] 2 CountRoot[poly1, {x, 0, 2}] 1 Here's an example of a double root. For some reason, this algorithm counts 3 once. The Mathematica built-in function counts 3 twice. poly2 = Expand[(x - 5)*(x - 3)*(x - 3)*(x - 2)*(x - 1)] -90 + 213 x - 184 x^2 + 74 x^3 - 14 x^4 + x^5 This is ok: CountRoot[poly2, {x, 0, 2.5}] 2 This counts 3 once! CountRoot[poly2, {x, 2.5, 3.5}] 1 The Mathematica built-in function counts 3 twice: CountRoots[poly2, {x, 2.5, 3.5}] 2 I don't know how to improve on this. > On Feb 4, 9:19 am, spellu...@fb04373.mathematik.tu-darmstadt.de (Peter > =A0> I need to determine if there is a zero or root between two points > =A0> I have wxMaxima and Mathcad and I can find the symbolic roots but th= > e > =A0> answers are very messy. > Hi. =A0Just throwing this out @ Mathematica. > This says there is one real root between 0 and 1.5 inclusive. > Don't know the underlying algorithm thou. > poly =3D Expand[(x - 5)*(x - 4)*(x - 3)*(x - 2)*(x - 1)] > x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 > CountRoots[poly, {x, 0, 1.5}] > This is for a real time embedded application. Right now I am > wondering why the results for the Sturm chain are different > symbolically and numerically. > Peter Nachtwey > because the Sturm chain algorithm (Euclid) is numerically unstable. > symbolically done, everything is o.k., but the roundoff effects > are very strong having the effect that, interpreted in backward analysis, > you deal with a considerably different polynomial. > hth > peter > However the example he presents involves integer > coefficients and integer roots, well separated, > without multiplicity > 1. > I've not plowed through his numeric code, but > because it is significantly more complicated > than the one symbolic call to CountRoots, > it has a much greater likelihood of a bug. > I am pretty sure you are right. > > Symbolically my Sturm chain is: > (%o14) v0-v1 > (%o15) a0-a1 > (%o16) -(c5*(20*v0-20*v1)+(a1-a0)*c4)/(20*c5) > (%o17) a1-a0 > > Numerically my Sturm chain is: > (%o50) 0 > (%o51) 1000 > (%o52) -8625/226 > (%o53) 1245825000/2247001 > for the numbers I provided above one can set that %o51 and %o53 should > have opposite signs but they don't numerically. > > Peter Nachtwey > > === Subject: Re: How to find a if there is a zero or root between two points quickly I forgot to mention that the only reason I mention this function CountRoots from Mathematica is that maybe the op can find a similar function in the math programs he is using: > I have wxMaxima and Mathcad ... Dana === Subject: Re: How to find a if there is a zero or root between two points quickly posting-account=SIMyZAoAAAAhUsAAG9RTx1O9_n6inDDv Gecko/2008120122 Firefox/3.0.5 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > I forgot to mention that the only reason I mention this function > CountRoots from Mathematica is that maybe the op can find a similar > function in the math programs he is using: æ> I have wxMaxima and Mathcad ... Dana This is for an embedded application. Much of the code is first written on a CAS before being implemented in C. The first thing I did after being enlightened about the Sturm chain was to do a search an I found the same example on the Wolfram site (%i1) f0: x^5-3*x-1; f1: diff(f0,x,1); f2: -remainder(f0,f1); f3: -remainder(f1,f2); (%o1) x^5-3*x-1 (%o2) 5*x^4-3 (%o3) (12*x+5)/5 (%o4) 59083/20736 That was easy so I raised the high bar. Now the goal is to find the maximum interval that has no zeros between the beginning and time. wxMaxima appears to have a bug so that the numerical solution did not work so I copied the equations to Mathcad and tried to solve for the interval. The solution was very messy so I doubt this method can be used in an embedded application. I have another solution that is easy to implement and executes quickly but isn't optimal. Peter Nachtwey Now determine if there is a zero between t0 and t1 in the velocity. This would indicate that the motion profile will overshoot the set point. First start with the position polynomial. The position polynomial is only used to generate the velocity and acceleration polynomial. Only two divides are required to get a constant quotient. (%i15) eq0: x0+v0*t+(a0/2)*t^2+c3*t^3+c4*t^4+c5*t^5-x1$ /* Position */ eq1: diff(eq0,t,1)-v1; /* Velocity, find the zeros between 0 and t1 */ eq2: diff(eq1,t,1)-a1; /* Acceleration */ eq3: -remainder(eq1,eq2); eq4: -remainder(eq2,eq3); (%o16) -v1+v0+5*c5*t^4+4*c4*t^3+3*c3*t^2+a0*t (%o17) 20*c5*t^3+12*c4*t^2+6*c3*t-a1+a0 (%o18) -(c5*(20*v0-20*v1)+(30*c3*c5-12*c4^2)*t^2+((5*a1+15*a0) *c5-6*c3*c4)*t+(a1-a0)*c4)/(20*c5) (%o19) -20*c5*t^3-12*c4*t^2-6*c3*t+a1-a0 === Subject: Re: How to find a if there is a zero or root between two points quickly posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) > CountRoots from Mathematica is that maybe the op can find a similar > function in the math programs he is using: æ> I have wxMaxima and Mathcad ... Dana This is for an embedded application. æMuch of the code is first > written on a CAS before being implemented in C. æThe first thing I did > after being enlightened about the Sturm chain was to do a search an I > found the same example on the Wolfram site (%i1) f0: x^5-3*x-1; > f1: diff(f0,x,1); > f2: -remainder(f0,f1); > f3: -remainder(f1,f2); > (%o1) x^5-3*x-1 > (%o2) 5*x^4-3 > (%o3) (12*x+5)/5 > (%o4) 59083/20736 That was easy so I raised the high bar. Now the goal is to find the > maximum interval that has no zeros between the beginning and time. Erm, doesn't this simply mean you need to find the first root? === Subject: Re: How to find a if there is a zero or root between two points quickly > t to mention that the only reason I mention this function > CountRoots from Mathematica is that maybe the op can find a similar > function in the math programs he is using: > > æ> I have wxMaxima and Mathcad ... > > Dana > > This is for an embedded application. æMuch of the code is first > written on a CAS before being implemented in C. æThe first thing I did > after being enlightened about the Sturm chain was to do a search an I > found the same example on the Wolfram site > > (%i1) f0: x^5-3*x-1; > f1: diff(f0,x,1); > f2: -remainder(f0,f1); > f3: -remainder(f1,f2); > (%o1) x^5-3*x-1 > (%o2) 5*x^4-3 > (%o3) (12*x+5)/5 > (%o4) 59083/20736 > > That was easy so I raised the high bar. Now the goal is to find the > maximum interval that has no zeros between the beginning and time. > > Erm, doesn't this simply mean you need to find the first root? The book The Numerical Treatment of a Single Nonlinear Equation by Alston Householder in 1970 is really about solving polynomial roots. It gives the usual Sturm's theorem as well as other facts. Like alternate constructions of a Sturm's sequence as the GCD of the polynomial and its derivative is the common but not the only possibility. It also gives the Budan and Fourier theorem which is easier to compute but less precise than the Sturm's theorem. Lots of stuff that never makes it into a book with a single chapter on polynomial roots. Google throws up quite a bit of interesting stuff for Budan Fourier that does not seem to show for Sturm theorem or Strum sequence. === Subject: Re: How to find a if there is a zero or root between two points quickly posting-account=wy-WGAoAAABUfYlyvtPTFsKujAiLOBd1 Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > I forgot to mention that the only reason I mention this function > CountRoots from Mathematica is that maybe the op can find a similar > function in the math programs he is using: æ> I have wxMaxima and Mathcad ... Dana in wxmaxima (a free open-source program that runs on windows, linux, ...): (%i3) ? nroots; -- Function: nroots (

, , ) Returns the number of real roots of the real univariate polynomial

in the half-open interval `(, ]'. The endpoints of the interval may be `minf' or `inf'. infinity and plus infinity. `nroots' uses the method of Sturm sequences. (%i1) p: x^10 - 2*x^4 + 1/2$ (%i2) nroots (p, -6, 9.1); (%o2) 4 .... This program has been in Macsyma/Maxima probably for decades. If you know in advance the degree of the polynomial and most of the coefficients, and some other information and you are operating under some real-time constraint, you might not want to use this program. It presumably will tell you the answer, though the example uses a floating-point number which is a questionable strategy. 9.1 is slightly less than 91/100. Closer to 9.09999847. But the polynomial p given in the example has roots nowhere near 9.1 RJF === Subject: Re: How to find a if there is a zero or root between two points quickly boundary=----=_NextPart_000_0011_01C98931.BC00C660 --------------------------------------------------------------------- > I need to determine if there is a zero or root between two points > I have wxMaxima and Mathcad and I can find the symbolic roots but the > answers are very messy. > > Hi. Just throwing this out @ Mathematica. > This says there is one real root between 0 and 1.5 inclusive. > Don't know the underlying algorithm thou. > > poly = Expand[(x - 5)*(x - 4)*(x - 3)*(x - 2)*(x - 1)] > > x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 > > CountRoots[poly, {x, 0, 1.5}] > > 1 > Are you sure? y = x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 has 5 roots: x1=1 x2=2 x3=3 x4=4 x5=5 * See also * http://home.arcor.de/janch/janch/ news/20090207-polynomial/ -- Microsoft-kompatibel/optimiert f.9fr IE7+OE7 === Subject: Re: How to find a if there is a zero or root between two points quickly > > Dana DeLouis schrieb im Newsbeitrag > I need to determine if there is a zero or root between two points > I have wxMaxima and Mathcad and I can find the symbolic roots but the > answers are very messy. > Hi. Just throwing this out @ Mathematica. > This says there is one real root between 0 and 1.5 inclusive. > Don't know the underlying algorithm thou. > poly = Expand[(x - 5)*(x - 4)*(x - 3)*(x - 2)*(x - 1)] > x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 > CountRoots[poly, {x, 0, 1.5}] > 1 > > > Are you sure? > > y = x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 > > has 5 roots: > > x1=1 > x2=2 > x3=3 > x4=4 > x5=5 And how many of those are between 0 and 1.5? Phil -- I tried the Vista speech recognition by running the tutorial. I was amazed, it was awesome, recognised every word I said. Then I said the wrong word ... and it typed the right one. It was actually just detecting a sound and printing the expected word! -- pbhj on /. === Subject: Re: How to find a if there is a zero or root between two points quickly boundary=----=_NextPart_000_007A_01C9895C.29C1E7E0 --------------------------------------------------------------------- > > Dana DeLouis schrieb im Newsbeitrag > I need to determine if there is a zero or root between two points > I have wxMaxima and Mathcad and I can find the symbolic roots but the > answers are very messy. > Hi. Just throwing this out @ Mathematica. > This says there is one real root between 0 and 1.5 inclusive. > Don't know the underlying algorithm thou. > poly = Expand[(x - 5)*(x - 4)*(x - 3)*(x - 2)*(x - 1)] > x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 > CountRoots[poly, {x, 0, 1.5}] > 1 > > > Are you sure? > > y = x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 > > has 5 roots: > > x1=1 > x2=2 > x3=3 > x4=4 > x5=5 > > And how many of those are between 0 and 1.5? > Just one solution: x1 = 1 c 9 * x^9 +c 8 * x^8 + c 7 * x^7 + c 6 * x^6 + c 5 * x^5 + c 4 * x^4 + c 3 * \ x^3 + c 2 * x^2 + c 1 * x + c 0 = 0 c 0 = -120 c 1 = 274 c 2 = -225 c 3 = 85 c 4 = -15 c 5 = 1 c 6 = 0 c 7 = 0 c 8 = 0 c 9 = 0 If solving as non-linear equation and using start value = 0 then you will find the 1st solution for sure for this example. * See * http://home.arcor.de/janch/janch/ news/20090207-polynomial/ -- Microsoft-kompatibel/optimiert f.9fr IE7+OE7 === Subject: Re: How to find a if there is a zero or root between two points quickly <1Gvhl.1029$vb.1010@bignews4.bellsouth.net> <498d8efe$0$31330$9b4e6d93@newsspool4.arcor-online.net> posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) I need to determine if there is a zero or root between two points > I have wxMaxima and Mathcad and I can find the symbolic roots but the > answers are very messy. Hi. æJust throwing this out @ Mathematica. > This says there is one real root between 0 and 1.5 inclusive. > Don't know the underlying algorithm thou. poly = Expand[(x - 5)*(x - 4)*(x - 3)*(x - 2)*(x - 1)] x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 CountRoots[poly, {x, 0, 1.5}] 1 Are you sure? y = x^5 - 15*x^4 + 85*x^3 - 225*x^2 + 274*x - 120 has 5 roots: x1=1 > x2=2 > x3=3 > x4=4 > x5=5 Erm, yes, wasn't this just his point? He explicitly constructed poly to have these five roots and indeed exactly one of these is between 0 and 1.5 Back to the topic I guess that Mathematica uses Sturm chains as well for CountRoots - what could be better? It involves only derivatives of polynomials and polynomial division * See also > *http://home.arcor.de/janch/janch/ news/20090207-polynomial/ -- > æ æ æ æ æ æ æ æMicrosoft-kompatibel/optimiert f.9fr IE7+OE7 Before optimizing for a single user agent, I suggest getting rid of the more than 100 validation errors: http://validator.w3.org/check?verbose=1&uri=http%3A%2F%2Fhome.arcor.de%2Fjan\ ch%2Fjanch%2F news%2F20090207-polynomial%2F and http://schneegans.de/sv/?url=http%3A%2F%2Fhome.arcor.de%2Fjanch%2Fjanch%2F news%2F20090207-polynomial%2F ;) === Subject: Re: ? understanding the world by math posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080614i-0etch1),gzip(gfe),gzip(gfe) On Feb 5, 4:27 pm, Mariano Su.87rez-Alvarez > On Feb 5, 3:08 pm, victor meldrew ...@yahoo.co.uk Hi Victor. I really appreciate you taking a look at those weird > probablility patterns. > They are from a standard rand() gnu function so nothing too fancy. Just relying on a C random generator. What distribution > is that meant to simulate? It is intended to be a geometrical survey. By taking random values and > scaling them to unit magnitude an (n-1)-sphere in n-signed is covered > as a square or whatever the graph is spec'd as; the resulting graphs > are strictly in terms of magnitude. We could likewise study some > angular measures as raw data without getting too carried away. I guess > finally somebody sees that these probability graphs are well behaved > and have content. Somehow this is the product speaking. It does seem > to whimper out after awhile and become more redundant. Still these are > all well behaved systems algebraicially. There is nothing actually > broken about the higher sign numbers. They just don't do what alot of > people think they should in terms of distance behaviors. Whether they > play a part in physics or not the support for spacetime remains in the > systems whose distance behaviors under product are straight and > linear, or whatever terminology ought to go here. I do see the work that you point out. Still Victor you insist upon an > RxC representation whilst you instantiate in RxRxRxR for P4. I've never instantiated in R x R x R x R in my life :-( > Yeah, whatever. You know what I mean. Does he? I for one have *absolutely* no idea > what you mean by that! I have considered and computed on the simplistic RxC product of two > values (r1,z1), (r2,z2) where of course the r is the real component > and the z is the complex component, their product being: > (r1,z1)(r1,z2) = ( r1 r2, z1 z2 ) Very simple, n'est-ce pas? however this product is not quite the dead ringer for P4 that it seems > on first investigation. Well there is an isomorphism, that's all. Well, then, so what? > Why are people thinking this a gag? > When you want isometric correspondence then we just don't know do we? > Yet already the polysign built the complex numbers, the real > numbers, ... > Really, Mariano's defense has been poor, but good enough for you to > come in and save the day. FWIW... I did not try to defend anything. I presented you some facts > (which are, in fact, essentially the very same Victor gave you, > apart from the fact that I made explicit the isomorphisms and > a few other details), and that was it. Whatever it was that > was poor, it was not a defense. -- m Hah !!! Let the onlooker decide what you've communicated. Information station MSA defeated. 2009020110:31:01est unposted For future bait with MarianoSaNan Mariano so des ka? Mariano sa. Mariano so des; Marianosa. Mariano eche ban. Marianosa eche san? Marianosa nan ban. Marianosanan... Mariano S. A. (NAN) Flagging it in Flogging it out Slogging it in Slagging it out Structured types are such big frights un till the dis crete work is done. To say there is an error language Until there is an error language There is an error language The NAN is done INF and NAN are just a who and when or just a when and who but when the two are two perhaps then real work can be done. So Tommy makes a plea and Marianosanan nanny ban not an eche ban sends along a very very strong construction endorsed by g. To see the two as free their lack of simplicity Their cryptic ways their cryptic stays they're stuck on every thorn waiting to be reborn meanwhile bleeding dry. Preserved and older forms. Like in a book of one. The one the eche ban. The eche ban can never be undone. So you say you've got a phaser that can outdue my tazer and I tell you mine is only set to stun. The big drop is coming. - T === Subject: Re: ? understanding the world by math posting-account=IBUqVwoAAADepmzxVr9iEYD5Z0A483SY rv:1.9.0.1) Gecko/2008070206 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Mariano so des ka? > Mariano sa. > ...... This is even worse than galathea's blank verse :-( === Subject: Re: ? understanding the world by math posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080614i-0etch1),gzip(gfe),gzip(gfe) On Feb 7, 11:35 am, victor_meldrew_...@yahoo.co.uk Mariano so des ka? > Mariano sa. > ...... This is even worse than galathea's blank verse :-( You can see that I am really not so clever. Still, anything to keep up the humor is wise I think. Think of all the students bored in math class... They know they've been eating a pile of crap... They can feel it. Now they can know it. - Tim === Subject: Re: A mathematical assistant that you'd like? > On 31 Jan, 14:50, Arnold Neumaier How would \ you propose to cope with the problem that the same > mathematical result may exist in many variants differing slightly in > notation, phrasing, exact asssumptions? > I also have an idea about how such a language might me written. If you > have an online Arabic-English dictionary you look up words using a > hashing algorithm. Hashing algorithms, as we all know look up in a > constant time independently of the amount of information in them. > Could theorems be placed in a hashing algorithm? > How would you find the theorem that ''every number not leaving 7 as > remainder when divided by 8 is the sum of three primes'' in such a > dictionary? > I would write it > X>5;X%8!=7;X=P(1)+P(2)+P(3); The problem is: would everyone else contributing to the dictionary have written it that way so that a search would locate it if you pose this as the question? > The fact that a lot of theorems appear in different forms and in > differing degrees of generality need not deter us too much. What you > would have to do is to include ALL manifestations of any theorem. The > most general manifestation would be the named theorem and all other > manifestations would simply have pointers to that theorem X=P(1)+P(2)+P > (3) would point to Goldbach. > > http://en.wikipedia.org/wiki/Goldbach's_conjecture > > http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=eu\ c lid.facm/1229618748 > > and perhaps Goldbach would point to the Extended Riemann hypothesis. > The proof given in the reference above being made available. BTW - You > work at an academic institution, you will probable have access to the > body of the paper which I don't. Is it right? I can't say at this > stage. No, our library doesn't subscribe to this journal. But judging from the abstract, the contents does not touch essentials of the discussion anyway. > http://en.wikipedia.org/wiki/Generalized_Riemann_hypothesis > > The search routines employed by Google show that such a search > technique is possible. QED will need to use Sawzall. > > http://en.wikipedia.org/wiki/Sawzall_(programming_language) > > Each theorem, proof etc will be given a 64 bit address. I think we are > reasonably safe with 64 bits even with the whole of mathematics. Yes. Even if all variants of the same theorem will get different id's. > The > assignment will be made by QED. The user just enters a theorem and (if > known) a proof. Just as you simply construct your website, you don't > worry about spiders and retrieval. In QED (ideally) you would have a > website, you would then enter something into your website. The spider > in addition to the task it performs at present of looking at words > would look for words that had meaning in QED. In QED a user would have > first to define what the quantities meant. For example :- > > REAL x; > ROOT -> ax^2 +bx +c > > would not have a solution if b^2 - 4ac was negative. And would you rely on that names always have the same meaning for every author? What if someone labels the statement 2+2=4 the ''Rienamm hypothesis'', and then gives a proof for it? > COMPLEX x; would however ALWAYS have roots. It might be possible to > use context to infer things about the quantities. Rigorously every > variable used has to have its properties defined strictly. A > statistical interpreter (or translator) could make this for you. In this way, no two theories created by different people will have much in common. How do you recognize that their theorems are equivalent? > We can see that the whole theory of QED is inextricably linked with > concepts of search and machine translation. But conceptual search, not search for words. > (Trivial for things like Calculus > and Matrix theory) > Not even elementary number theory is trivial. See, e.g., > C. Zinn, > Understanding Mathematical Discourse > http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.44.8407 > I think this is because the right keywords are not there (Yet) Searching for keywords will not b good enough if you look for a lemma that might help you in your attempted proof of something but have no idea how someone else ould have called such a lemma. > I have given you a lot to think about. I will probably publish this > all in a blog some time. If you do, please let me know the location of the blog. Arnold Neumaier === Subject: Re: Complex plane integral Sure. For example, the extended Cauchy integral formula involves integrating the partial derivative of the function wrt zbar over the plane region enclosed by a curve. It's basically the same as integrating over a 2D real plane. If the function is analytic then the derivative wrt zbar is 0. Laura === Subject: Re: Complex plane integral Sure. For example, the extended Cauchy integral formula involves integrating the partial derivative of the function wrt zbar over the plane region enclosed by a curve. It's basically the same as integrating over a 2D real plane. If the function is analytic then the derivative wrt zbar is 0. Laura === Subject: Re: Polysign numbers posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080614i-0etch1),gzip(gfe),gzip(gfe) On Feb 6, 10:59 am, victor_meldrew_...@yahoo.co.uk Yes, that's correct. To the best of my knowledge my website is a > complete rendition and all that I write here is diarrhea. I was wondering about the nasty smell. It has taken > down into areas that need more work. How about you doing some of that (mathematical) work? In further defense of magnitude I would present to you the gorilla > conjecture: > If a gorilla can sort sticks by length with out any mathematical > training at all then can we accept magnitude as fundamental? Guy the gorilla sez Golden rocks! I'm in it for the long haul. Caught a ride on > the freight train... Wasn't going my way, but what the hey, I'm going > somewhat that way. Having gone too far I'll just hop off, trace the > tracks back to the last good crossing where a road sign reads the way > with a nice neat arrow pointing my way. I read it on the way but was a > bit on the edge as to whether I read it right. If I read it wrong then > the next train would be a real pain at 30 knots on the hour. Zzzzzzzzzzzzzzzzzzzz.... Nice to meet you on the street Victor. Sod off, Tim the dim! OK. So you like magnitude as primitive. When we get to tacking on sign it starts to get tacky, and this tacky feeling is what I think most people dislike in the polysign. The recovery of a serious effort is in the observation that sign is dimension. Finally then the seriousness of the sign stands freely. Sign is dimension, just off by one. Hence an argument that the human race is off by one. Thusly time goes overlooked as a primitive arithmetic structure and occupies a paradoxical and often talked about quality that has remained mysterious. Then again the whole geometry can be taken as unidirectional in that it no longer requires the direct bidirectional feature that the cartesian space constructed. Instead we see a balancing form of space. - Tim === Subject: Re: Polysign numbers posting-account=IBUqVwoAAADepmzxVr9iEYD5Z0A483SY rv:1.9.0.1) Gecko/2008070206 Firefox/3.0.1,gzip(gfe),gzip(gfe) > OK. So you like magnitude as primitive. Do I? When I figure out what that means, I'll let you know. > The recovery of a serious effort is in > the observation that sign is dimension. And what about the observation that your polysigns form a cyclic group. > Hence an argument that the human race is off by one. How blessed is the human race! that it has Tim Golden to point out the errors of its ways! === Subject: Re: Polysign numbers posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080614i-0etch1),gzip(gfe),gzip(gfe) On Feb 7, 11:37 am, victor_meldrew_...@yahoo.co.uk OK. So you like magnitude as primitive. Do I? When I figure out what that means, I'll let you know. The recovery of a serious effort is in > the observation that sign is dimension. And what about the observation that your polysigns form a cyclic > group. Sure, this is fine. But the form was built back in time as the real number. Under my ordinary education the real number is presented as real analysis without any attention to group theory. These old rules are the ones that I wish to challenge. In that most modern mathematical theory treats the reals as fundamental and that I have inserted a clause beneath this level then I must challenge the accumulation of thinking since that time. This goes so far as to challenge the physicists concept of isotropic space, which is arguably a mathematical construction. So I cast my net far and wide and so what I have caught is an assortment of fish. This assortment is rather familiar yet in its new context the fishiness of the polysign is clearly there for the well educated man. I must then cast my doubt upon the well educated man. He has overlooked a fundamental principle for far too long. What a shame and what a sham. Now he can make good. And then we can make better. The puzzle will remain open unconditionally no matter how tantalizing and pristine the current system is preached to be. Again drewel and foam coming forth in a frothy spray of misperception and through the haze I see it all... - Tim Hence an argument that the human race is off by one. How blessed is the human race! that it has Tim Golden to point out > the errors of its ways! === Subject: Re: Polysign numbers posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080614i-0etch1),gzip(gfe),gzip(gfe) On Feb 6, 3:16 am, victor_meldrew_...@yahoo.co.uk On Feb 5, 3:08 pm, victor_meldrew_...@yahoo.co.uk Just relying on a C random generator. What distribution > is that meant to simulate? It is intended to be a geometrical survey. Do you mean it's simulating a geometric distribution? I do see the work that you point out. Still Victor you insist upon an > RxC representation whilst you instantiate in RxRxRxR for P4. I've never instantiated in R x R x R x R in my life :-( Yeah, whatever. You know what I mean. Whatever what? I really don't know what you mean; you introduced > R x R x R x R into this thread, not me, but you never > explained its purpose. Alright then. You are inline with the simplex reasoning, you just haven't bothered to admit it yet. Do you understand that when we declare (x,x,x,x,x...) = 0 as in sum over s ( s x ) = 0 or - x + x * x ... @ x = 0 (All of the above meaning the same thing) that this inherently implies the simplex balance? Then my criticism or your usage of barycentric coordinates is false. Still, it should be pointed out that this 4D version lays there. Especially since Mariano's e0,e1, ... notation is pretty much there. The point in polysign is to get to the crux without the old crap. The truth of polysign is that sign is a general phenomenon having to do directly with dimension. This nuance is what gets avoided yet it seems you are part way down the rabbit hole. Now go find the cocoa puffs. And look out for the false mimics; those round little rabbit turds that Mariano chokes on of his own volition. As an aside, it really would help if you could post > in a much closer approximation to the English language. When you want isometric correspondence then we just don't know do we? We do know; at least we who are competent to have worked it out > do. It's easy to write down the distance function on R x C > corresponding to yours on P_4, viz. the square of your > distance corresponds to the map (x,z) |--> (1/3)(x^2 + 2|z|^2). > It's easy to do this for all P_n. Hmmm. So we're in P4, so lets say I've dropped a point off for your procedure: y(P4) = y = - 1 + 2 # 3 . Now I've got to get y into your x,z form. Have you inherently provided me a transform for this? I think not since I need a spec of a plane in P4 for your z space and a line spec in P4 for your x. Since this line passes through the origin then one point might suffice to state the coordinate system. Since it is not specified I should only assume that it is # 1 + 1. Is this coherent with what you are doing? The Mario representation would be e0 + e2 for your x unit vector direction, scaled to unity, and not to be confused with orthogonal unit vectors. This should set the x component positive reference frame direction. This is how I understand the equivalence to be layed out. If I am off I trust you'll correct me and at least reflect this back to your own context for a parallel communication. Thusly we might actually come to some agreement on some peculiar thing even if not applicable for the whole system. We'll still need to spec the plane's reference frame. But upon carrying all of this out what will your isometric product look like? I myself do not withhold my own misunderstanding. You claim to understand this and I would welcome a complete rendition here. It need not be extraordinary. What has been done in the past is not so clean and simple and sweet as you have portrayed it. Still you stick like a glue to your RxC space, yet all your mumblings have not really provided much. Because the four-signed numbers are three dimensional then transforms exist from them to any other three dimensional system that covers the volume of its space. Thusly you may as well state your distance function in terms of RxRxR. Where are the benefits of the form RxC? They are in the product aren't they? To express such products we need to spec a transform from your RxC (x,z) to P4 (y) and vice versa. Then upon studying the product relationship we can observe any equivalence or difference. Your one-liner map (x,z) |--> (1/3)(x^2 + 2|z|^2). doesn't really mean much to me without the background information that I am asking of you. It seems to me that you are in some midstep position with this equation which is supposed to relate to distance in P4. As I've seen the word distance used differently here by Mariano and also Tommy recently I am a bit miffed that there is no universal agreement on distance. I see that we can define strange distance functions and play with them, but the ordinary sense of distance does not need this level of confusion. I am operating on the assumption that you are discussing ordinary distance and especially since P4 does carry a physical graphing capability that we could then plot such positions out and literally measure these distances with a measuring tape. Do you balk at this? Then perhaps we ought to step down to P3, where such distances can be plotted in the plane of the paper. Unfortunately without a specification of the transform you are using statements like your one above are sitting floating in air without much beneath them. When you say that something is obvious and so how could it go misunderstood such as the instantiation of the P4 product on (R,C), well, the holes in the thinking collapse it under instantiation. I've already worked on this at http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html I must assume that you are withholding information, and I do not mean to say that this is with any ill intent. I recommend you get right into it here as you see it and I'll try to follow along. If you have an isometric isomorphism from P4 to (R,C) then others will take interest in that. I'm surprised lwal hasn't spoken up yet here, but it wouldn't surprise me if lwal does check in here and if you can do what uit sounds like you claim then I'd point him to your post directly. It should be a fairly complete rendition though I think. Please note: optional long winded text at end of post was ommitted since this one is a bit breezy up front, though without spacing out the thoughts sufficiently wide and reinforcing them with such redundancy I doubt that an onlooker would see much wrong with your statement. That is not even to say that there is anything wrong with it, but that it must carry with it some additional information, and if we are going to start dabbling in twisting and morphing the metric of the space or folding up its vertices then that is your bad, not mine. Whose space will be twisted? P4, or RxC? The conveyance of your attachment to RxC and the challenge this poses to its purity can be squashed by the likes of you simply by putting down your information here. - Tim === Subject: Re: Polysign numbers posting-account=IBUqVwoAAADepmzxVr9iEYD5Z0A483SY rv:1.9.0.1) Gecko/2008070206 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Alright then. You are inline with the simplex reasoning Am I? That's nice! > Then my criticism or your usage of barycentric coordinates is false. As I hadn't used barycentric coorinates anywhere in this thread, this means that your criticism must be false. > This nuance is what gets avoided yet it > seems you are part way down the rabbit hole. Now go find the cocoa > puffs. You keep your cocoa puffs down your rabbit's hole? That explains much! > I think not since I need a spec of a plane in > P4 for your z space and a line spec in P4 for your x. Is that a spec as in Grothendieck's EGA? > Your one-liner > æ æmap (x,z) |--> (1/3)(x^2 + 2|z|^2). > doesn't really mean much to me Couldn't you work it out? I already indicated the isomorphism between your P n and R^a x C^b. It's now just a question of transporting your distance function from one to t'other. > carry a physical graphing capability This is sci.math, not sci.physics. > æ æhttp://bandtechnology.com/PolySigned/Deformation/P4T3Comparison .html As an algebra over R, P 4 and R x C are isomorphic (is closely resembles a Goldenballs for is isomorphic?). > I must assume that you are withholding information Oh yes, I have taught you everything you know, but I have not taught you everything I know! > that is your bad My what? === Subject: Which Ramanujan Expansions are Besicovitch Almost Periodic? Given a ramanujan expansions of the form: a(x) = sum_{q=1}^{infty} a_q c_q(x) where a_q = lim_{T -> infty} 1/{2T}int_{x=-T}{T} a(x) c_q(x) dx c_q(x) = sum_{k=1}{q} e^{2 pi i k/q x} and the summation over k is limited to k which are relatively prime to q I am trying to determine which conditions place a(x) in which space of almost periodic functions. Here are a few conditions I know of. If the convergence is uniform, then a(x) is in the set uniform almpst period \ (u.a.p.) functions (actually the subset of u.a.p.; limit periodic) If sum{k=1}{infty} phi(k) (a_q)^2, where phi(q) is the Euler totient function, then a(x) is in B^2 a.p. I.e. the constants are square summable as per the Besicovitch version of the Riesc-Fischer theorem for B^2 a.p. functions. In their 1940 paper (http://www.zblmath.fiz-karlsruhe.de/MATH/general/erdos/cit/06301273.htm), Erdos, KAC, KAmpen, Wintner, proved if a(n) [a(x) when evaluated for positive integers] is multiplicative and subject to a summability condition, \ a(x) is B1 a.p. or B^2 a.p. depending on the summability condition. As near as a I can tell there is no necessary condition for inclusion in B^2 \ a.p. only sufficient conditions. Does anyone know of other sufficient conditions on a(x) or a_q which place a(x) in B^2 a.p.? For example, if a(x) is of bounded variation for all x, is a(x) in B^2 (or a \ crisper subset)? If it helps, here is a general map of almost periodic function spaces I made \ to aid in my own understanding: http://www.washburnresearch.org/NumberTheory/MapOfAlmostPeriodicSpaces.pdf Again the question of the post is what other conditions on a Ramanujan-Fourier expansion or its constants are sufficient to prove that \ the resulting function is in B^ a.p. (or some other space of a.p. functions)? === Subject: Re: Q on Riemann domains A riemann domain is similar to a Riemann surface except that it can be higher-dimensional. Also because of the global projection map P (which makes it a complex manifold), a Riemann domain can't be compact. I don't know why they want to exclude compact surfaces like CP^n. It would seem as relevant as Riemann surfaces that are compact. Laura === Subject: Numerical analysis question Hi! I need a hint to prove this: If f is a continuous function having fourth derivative on the interval [a, b] and T(m) denotes the composite trapezium rule approximation to int_a^b f(x) dx using m subintervals then [T(m) [CapitalEth] T(2m)]/[T(2m) [CapitalEth] T(4m)] -> 4 as m -> infinity RSimmons === Subject: Re: Numerical analysis question I've tried your suggestion but to obtain the result I must have [f''(eta_n) \ - (1/4)f''(mu_n)]/[(1/4)f''(mu_n) - (1/16)f''(xi_n)] -> 4 as m -> infinity, \ where f''(eta_n), f''(mu_n) and f''(xi_n) are, respectively, the error term \ of T(m), the error term of T(2m) and the error term of T(4m). RSimmons === Subject: Re: Numerical analysis question posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze Filter 1.2.0.72; GTB5; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Hi! I need a hint to prove this: If f is a continuous function having fourth derivative > on the interval [a, b] and T(m) denotes the composite > trapezium rule approximation to int a^b f(x) dx using m subintervals then [T(m) [CapitalEth] T(2m)]/[T(2m) [CapitalEth] T(4m)] -> 4 as m -> infinity Do you have a formula for the error of approximation of the trapezium rule? If not, try to derive one using Taylor's Theorem. Then see what happens when you insert your error term into your ratio. Dave === Subject: Numerical analysis question Hello! I need a hint to prove this: If f is a continuous function having fourth derivative on the interval [a, b] and T(m) denotes the composite trapezium rule approximation to int_a^b f(x) dx using m subintervals then [T(m) - T(2m)]/[T(2m) - T(4m)] -> 4 as m -> infinity RSimmons === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem posting-account=HfPrdwoAAACwxUqNrgJwfq54YvdU5mcg Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) I have proof of Ferma theorem, which results from the proof of Bill > Conjecture, (You may verify yourself). If you are interesting in the > proof of Bill Conjecture, I can post it here in several days. Can I > post in HTML? Else I need some days to put it anywhere in the web. I'm > not scientist and have not perfect English. So I couldn't place > to the post office. E A proof is of course only correct if it's TeX'ed, not HTML'ed. æ:) I think RTF may be good to. === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) > I have proof of Ferma theorem, which results from the proof of Bill > Conjecture, (You may verify yourself). If you are interesting in the > proof of Bill Conjecture, I can post it here in several days. Can I > post in HTML? Else I need some days to put it anywhere in the web. I'm > not scientist and have not perfect English. So I couldn't place > to the post office. E A proof is of course only correct if it's TeX'ed, not HTML'ed. æ:) I think RTF may be good to. No, RTF proofs usually have more counter-examples than TeX'ed ones. ;) === Subject: Re: Elegant 17th-Century Proof of Fermat's Last Theorem > On Feb 6, 9:00æpm, hagman On 4 Feb., 02:36, Zurikela I have proof of Ferma theorem, which results > from the proof of Bill > Conjecture, (You may verify yourself). If you > are interesting in the > proof of Bill Conjecture, I can post it here in > several days. Can I > post in HTML? Else I need some days to put it > anywhere in the web. I'm > not scientist and have not perfect English. So > I couldn't place > illl now and couldn't get > to the post office. E A proof is of course only correct if it's TeX'ed, > not HTML'ed. æ:) I think RTF may be good to. > > No, RTF proofs usually have more counter-examples > than TeX'ed ones. ;) Coming back to any source: fax, scanner, digital or any photo of some such manuscript: FLT and Bill's conjectures at once so realy it is some gold fever. Would we see some correct mathematics at all ??? I am theremore skeptical once offering here from 21 Januar some quite elementary proof but only for FLT and from n=5. Case for n=3 remains far from some elementary grip and where it is to dream about Bill's simplification ? Ro-Bin === Subject: a^(p-1) = 1 mod.(p^b) Let p = prime number, a, b, integer >0, if a=1 mod.(p^b) then a^(p-1) = 1 mod.(p^b) is always verified. Vincenzo Librandi vincenzo.librandi@tin.it === Subject: Re: a^(p-1) = 1 mod.(p^b) posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) On 7 Feb., 18:16, Vincenzo Librandi 0, if a=1 mod.(p^b) then a^(p-1) = 1 mod.(p^b) is always verified. More generally, if a = 1 mod b then a^c = 1 mod b === Subject: Re: a^(p-1) = 1 mod.(p^b) Author: hagman === Subject: Re: a^(p-1) = 1 mod.(p^b) > Let p = prime number, a, b, integer >0, > if a=1 mod.(p^b)> then > a^(p-1) = 1 mod.(p^b) > is always verified. >More generally, >if a = 1 mod b then a^c = 1 mod b therefore, a^(p-1) = 1 mod.(p^b) is always verified ? (is not !) Vincenzo Librandi === Subject: f(x,y) question I am studying some functions of two variables. For the function: f(x,y) = sinx - cosy Is this function continuous for all x and y? I think since sinx is continuous for all x, and cosy is continuous for all y, then the above f(x,y) function is automatically continuous, since f(x,y) is not creating any division by zero situations. continuous function +/- continuous function = continuous function that's what I'm thinking, but I'm not sure. for example, at pi/2, and 3pi/2, the cosine polarity changes. and at 0pi and pi, the sine polarity changes, but I don't think that matters in this scenario. === Subject: Re: f(x,y) question > I am studying some functions of two variables. > > For the function: f(x,y) = sinx - cosy It would have been more clear if you had written sin(x) - cos(y). > Is this function continuous for all x and y? Yes. > I think since sinx is continuous for all x, and cosy is continuous for > all y, then the above f(x,y) function is automatically continuous, Right. > since f(x,y) is not creating any division by zero situations. .. but not for this reason. There are other possible causes of lack of continuity besides that one. Actually, division by zero generates points at which the function is not defined, not points of discontinuity. > continuous function +/- continuous function = continuous function Right. > that's what I'm thinking, but I'm not sure. > > for example, at pi/2, and 3pi/2, the cosine polarity changes. and at > 0pi and pi, the sine polarity changes, but I don't think that matters > in this scenario. Not a little bit. Jose Carlos Santos === Subject: Bijections on S_n Are there any algorithmically natural bijections into S_n? Given a left(or right) rotation r_n of S_n then forms a cyclic subgroup. The projection of p in S_n by removing an element(say the last) forms a homomorphism between S_n/ and S_(n-1) hence we can form a chain ultimtely allowing us to parameterize S_n. The question is are there any simple ways? Normally this is done recurseively but is there a non recursive method? Since is cyclic and just the group of rotations one would expect that \ we could start out with some random permutation p and form the coset p \ then modify p in a predicable way and repeat the process. But the real question is if there is a bijection between S_n and N or R or some other simple set so that a permutation can be computed in a short time. \ No tables or recursions allowed. f(x) = x^2 allows us to compute f quite easily... is there a similar function that would give us a permutation. i.e, can we find a function f(k,n) that turns the kth permutation of size n? \ There are obviously many function that can do this(simply all the ways to permutate the set of permutatations or n!!) Is there at least one that is algorithmically simple and doesn't require recursion? One might say it is impossible and if it is it is probably related to the fact that Z_n! is not isomorphic to S_n but maybe one can use the structure \ of S_n in some way? Maybe this is related to galios theory too? In that there is no simple algorithms(which really needs to be well defined) which might have to do with A_n being simple for n >= 5. === Subject: Re: Bijections on S_n posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008121621 Ubuntu/8.04 (hardy) Firefox/3.0.5,gzip(gfe),gzip(gfe) > Are there any algorithmically natural bijections into S_n? Given a left(or right) rotation r_n of S_n then forms a cyclic > subgroup. The projection of p in S_n by removing an element(say the last) forms a > homomorphism between S_n/ and S_(n-1) Aha? A group homomorhism? Do you think is normal? hence we can form a chain ultimtely allowing us to parameterize S_n. The question is are there any simple ways? Normally this is done > recurseively but is there a non recursive method? Since is cyclic and just the group of rotations one would expect that > we could start out with some random permutation p and form the coset p some other simple set so that a permutation can be computed in a short time. > No tables or recursions allowed. f(x) = x^2 allows us to compute f quite > easily... is there a similar function that would give us a permutation. i.e, can we find a function f(k,n) that turns the kth permutation of size n? > There are obviously many function that can do this(simply all the ways to > permutate the set of permutatations or n!!) Is there at least one that is algorithmically simple and doesn't require > recursion? One might say it is impossible and if it is it is probably related to the > fact that Z_n! is not isomorphic to S_n but maybe one can use the structure > of S_n in some way? Maybe this is related to galios theory too? In that > there is no simple algorithms(which really needs to be well defined) which > might have to do with A_n being simple for n >= 5. Is this not using recursion / sufficienly iterative for your purpose? void permute(int data[], int k) { int n=2; while (k>0) { int r=k % n; int t=data[r]; data[r]=data[0]; data[0]=t; k = k/n; ++n; } } === Subject: Intersection Form. Clueless. Hi: Please be patient here with me, I am pretty clueless on this one: I am working with the intersection form in 4-d, and I am trying to understand how we get a matrix out of this. I understand that we take cocycles in H^2 and cup them, then we evaluate the resulting cocycle on the fundamental class. Still, how do we get a matrix out of this?. And how is this matrix 0 on torsion elements?. (and what is a torsion cocycle?). In a purely algebraic way, this is confusing too, as Z has no torsion. I have tried using Poincare duality, but I still don't see it. === Subject: Re: Intersection Form. Clueless. posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc10 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Hi: Please be patient here with me, I am pretty clueless > on this one: I am working with the intersection form in 4-d, and I am trying to understand how we get a matrix out of this. I understand that we take cocycles in H^2 and cup them, then we evaluate the resulting cocycle on the fundamental class. Still, how do we get a matrix out of this?. If you are working over a field: Every time you have a F-vector space V and a bilinear form q : V x V --> F, you can get a matrix A = (a {i,j}) by picking a basis {v 1, ..., v n} for V and setting a {i,j} = q(v i, v j). That's just linear algebra. In your case, V = H^2(X) and q is given by q(a, b) = < a cup b, fundamental class> for all a, b in H^2(X). Over a ring which is not a field, you will not be able to do this because H^2(X) is not going to have a basis in general. > And how is this matrix 0 on torsion elements?. (and what is a torsion cocycle?). A torsion cocycle is a cocycle which is torsion (unless whatever it is youare reading is making a different convention, of course) This is not the same thing, in general, as a cocycle whose cohomology class is torsion. though. > In a purely algebraic way, this is confusing too, as Z has no torsion. You have not given context to see what this has to do with anything. My mind reading capalities, though, suggest that what you are reading means cocycle whose class in cohomology is torsion. > I have tried using Poincare duality, but I still don't see it. You have tried using duality to do what, exactly? -- m === Subject: Re: Intersection Form. Clueless. > Hi: Please be patient here with me, I am pretty > clueless > on this one: I am working with the intersection form in 4-d, and I am trying to understand how we get a matrix out > of this. I understand that we take cocycles in H^2 and cup > them, then we evaluate the resulting cocycle on the > fundamental class. Still, how do we get a matrix out of this?. > > If you are working over a field: Sorry. I am working over Z. This is what makes the form special. > > Every time you have a F-vector space V and a bilinear > form > q : V x V --> F, you can get a matrix A = (a_{i,j}) > by picking > a basis {v_1, ..., v_n} for V and setting a_{i,j} = > q(v_i, v_j). > That's just linear algebra. > > In your case, V = H^2(X) and q is given by > > q(a, b) = < a cup b, fundamental class > for all a, b in H^2(X). > > Over a ring which is not a field, you will not be > able > to do this because H^2(X) is not going to have a > basis > in general. Yes. We have H^2(X;Z). This is part of why I am confused. I am not clear on why this would be a V.Space. > > And how is this matrix 0 on torsion elements?. (and > what is a torsion cocycle?). > > A torsion cocycle is a cocycle which is torsion > (unless whatever it is youare reading is making > a different convention, of course) This is not > the same thing, in general, as a cocycle whose > cohomology class is torsion. though. > > In a purely algebraic way, this is confusing too, as Z has no torsion. > > You have not given context to see what this has > to do with anything. My mind reading capalities, > though, suggest that what you are reading means > cocycle whose class in cohomology is torsion. > Yes, sorry, the coefficients are in Z-integers, and the form is supposed to be Z-valued. > I have tried using Poincare duality, but I still > don't see it. > > You have tried using duality to do what, exactly? > Yes, sorry again. I am trying to understand the cup product in cohomology from the perspective in homology. All I understand --which is little at this point -- is that when we translate from cocycles to cycles ( using Poincare duality), cuping the two cocycles and evaluating at [M] as above, is equivalent to finding the intersection number of the (Poincare-) equivalent cycles. > -- m === Subject: Re: Intersection Form. Clueless. <15242436.1234041390762.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008121622 Fedora/3.0.5-1.fc10 Firefox/3.0.5,gzip(gfe),gzip(gfe) > Hi: Please be patient here with me, I am pretty > clueless > on this one: I am working with the intersection form in 4-d, and I am trying to understand how we get a matrix out > of this. I understand that we take cocycles in H^2 and cup > them, then we evaluate the resulting cocycle on the > fundamental class. Still, how do we get a matrix out of this?. If you are working over a field: æ Sorry. I am working over Z. This is what makes > æ the form special. Every time you have a F-vector space V and a bilinear > form > q : V x V --> F, you can get a matrix A = (a {i,j}) > by picking > a basis {v 1, ..., v n} for V and setting a {i,j} = > q(v i, v j). > That's just linear algebra. In your case, V = H^2(X) and q is given by æ q(a, b) = < a cup b, fundamental class > for all a, b in H^2(X). Over a ring which is not a field, you will not be > able > to do this because H^2(X) is not going to have a > basis > in general. æ Yes. We have H^2(X;Z). This is part of why I > æ am confused. I am not clear on why this would > æ be a V.Space. One way to make sense of this is the following: let M be a finitely generated abelian group, and let phi : M x M --> Z be a bilinear form. Then it is easy to see that the torsion subgroup t(M) is contained in the radical of phi (that is, phi(t, x) = phi(x, t) = 0 for all t in t(M) and all x in M), so that phi induces a bilinear form psi on the quotient M/t(M) which is a finitely generated free abelian group. You can get a matrix for psi... -- m === Subject: Wittgenstein and the philosophy of mathematics (was Re: Sieg Heidegger!) > You have nothing to say about the critique of Heidegger's > thinking as per being as opposed to substance for a > correct rendering of *ousia* æYou are a student of philosophy, > aren't you? æWhy would this not pique your interest? I *was* a student of philosophy. Metaphysics is a dead subject to me at the moment. I was interested in metaphysics many years ago, but I have lost interest. So I will have to decline your offer to debate the details of Heidegger's metaphysics. I do have some residual interest in philosophy, especially the philosophy of mathematics. For example, I recently picked up Wittgenstein, Finitism and the Foundations of Mathematics by Mathieu Marion because of, well, several reasons. I was more interested in Wittgenstein before (I have nearly all of his published works), and I have always wanted to one day get to the bottom of his philosophy of mathematics, which seems to me to be uniquely conventionalist. This comes out very clearly especially in his lectures, e.g. Wittgenstein's Lectures on the Foundations of Mathematics, Cambridge 1939 , ed. Cora Diamond. One of my old philosophy teachers, Georg Henrik von Wright, attended those very lectures. He is also mentioned several times in Marion's book. The context is the relationship between the views of Hermann Weyl, Brouwer and Wittgenstein on the philosophy of mathematics, so all those things together made me sufficiently interested to get the book. It's not at the top of my reading list at the moment, however. Crossposted to sci.math and sci.logic in case someone might be interested in Mathieu Marion's book. I recall that the late Torkel Franzen, a regular sci.math poster, also said that Wittgenstein's philosophy of mathematics is uniquely conventionalist, setting it apart from most other philosophies of mathematics. === Subject: Re: JSH factors 15 <145c6a97-0a90-4f8c-9622-f4d7932dcd9f@o40g2000prn.g Forget about minimizing anything. If you write r+c and r-c in terms of v and \ D you'll find that the only way your method will give nontrivial factors of D is when you've been lucky enough to pick v so that v or 2v+1 has a factor in common with D. In other words, your method is simply an obfuscated version of trial division. Sorry, Rick === Subject: Re: JSH factors 15 <498aa523$1@dnews.tpgi.com.au> <498b4937$1@dnews.tpgi.com.au> (snip for brevity) > Picking u and v allows you to traverse every possible solution in > rationals to x^2 - Dy^2 = 1. > > If x = r/c, where r and c are integers and D is a composite then the > primary requirement that r+c or r-c have a gcd with D is that abs(r-c) > or abs(r+c) be less than abs(D). > > If you expand out with y = s/c, you have > > (r-c)(r+c) = Ds^2 > > so if you traverse every solution in integers, then there must exist > cases where you have small enough r, c and s, such that you meet that > requirement above. > > That is a mathematical absolute. > > So any method of smoothly traversing values for u and v through a > minimum will yield a solution. > > That is also a mathematical absolute. > > That is, the factoring problem is turned into a minima problem with some > rather simple algebraic equations, so it must succeed as that is an > easier problem with lots of techniques available. > > That is an assessment of the current state of the art. I believe that > mathematicians have worked minima problems to a high degree of skill. > > Therefore, regardless of what you stated in your other post, this > approach cannot fail because quite simply, it transforms the factoring > problem into finding minima's. > > The only mathematical objection which could invalidate that assessment > is if you cannot traverse through every solution in rationals to x^2 - > Dy^2 = 1, using u and v in rationals. > > Is that your claim? > > > James Harris Let's see if I can explain myself sufficiently clearly. Remember the goal is not to reach a solution eventually but to reach a solution efficiently. So you have to select values of v that result in a factorisation in x^2-Dy^2-1. Note that selecting v determines u. What I claim is that to correctly select v you need to already know the factorisation of D. Otherwise you have an infinite range of rationals to choose from and no means of narrowing the choice. If you could traverse through every solution in rationals then yes, you can factor D. However each solution matches one to one to a factorisation and each factorisation matches one to one to a value of v. So only by knowing the factorisation of D can you actually generate the values of V you require to factor D. Take our friend 1111 for example. The factorisation we want is (11,101). To generate a solution that factors D for 11 I need to set v to 5. But I only know this because I know 1111's factorisation. How otherwise can I set 5? Picking up from this formula above: (r-c)(r+c) = Ds^2 You are absolutely correct. Since it would be good to minimize s let's set s to 1. Thus D = (r-c)(r+c) and x=r/c and y=1/c. Generating this set of x and y would factor D. However you need to know the value of c ahead of time. But if you know c then you know that D has a pair of factors with a difference of 2c. If you know this then you have enough information to factor D already. Back to 1111. If 1111 = (r-c)(r+c) then c = 45 and r = 56 and s=1 is a solution. Presumably we could pick v to generate these values but we would have to know to actually generate c=45. But if I know that p exists such that p(p+2*45) = 1111 it is a simply quadratic to determine that p=11. I don't see it as a minima problem at all, which presumably is a reflection of my knowledge not your algorithm. However nothing in this thread or in your blog post indicates a means to determine v apart from trial and error which will simply be too slow and in fact is functionally the same as trial division. Obviously you are aware of research in the area of minima that I don't know. Remember I'm a programmer not a mathematician so in order to implement the algorithm I will need the specs of how to narrow down the search for a correct v in an efficient manner. === Subject: Re: JSH factors 15 <498aa523$1@dnews.tpgi.com.au> <498b4937$1@dnews.tpgi.com.au> <498caa23$1@dnews.tpgi.com.au> (snip for brevity) > Picking u and v allows you to traverse every possible solution in > rationals to x^2 - Dy^2 = 1. > If x = r/c, where r and c are integers and D is a composite then the > primary requirement that r+c or r-c have a gcd with D is that > abs(r-c) or abs(r+c) be less than abs(D). > If you expand out with y = s/c, you have > (r-c)(r+c) = Ds^2 > so if you traverse every solution in integers, then there must exist > cases where you have small enough r, c and s, such that you meet that > requirement above. > That is a mathematical absolute. > So any method of smoothly traversing values for u and v through a > minimum will yield a solution. > That is also a mathematical absolute. > That is, the factoring problem is turned into a minima problem with > some rather simple algebraic equations, so it must succeed as that is > an easier problem with lots of techniques available. > That is an assessment of the current state of the art. æI believe > that mathematicians have worked minima problems to a high degree of > skill. > Therefore, regardless of what you stated in your other post, this > approach cannot fail because quite simply, it transforms the > factoring problem into finding minima's. > The only mathematical objection which could invalidate that > assessment is if you cannot traverse through every solution in > rationals to x^2 - Dy^2 = 1, using u and v in rationals. > Is that your claim? > James Harris > Let's see if I can explain myself sufficiently clearly. Remember the > goal is not to reach a solution eventually but to reach a solution > efficiently. So you have to select values of v that result in a > factorisation in x^2-Dy^2-1. Note that selecting v determines u. > What I claim is that to correctly select v you need to already know the > factorisation of D. Otherwise you have an infinite range of rationals > to choose from and no means of narrowing the choice. > > Which contradicts with my point that with x and y solved out as > functions of v, finding an absolute non-zero minima for x MUST factor D. > If you don't address that issue you've blown past the proof. > > Let x = r/c where r and c are non-zero integers. You have that r and c > are functions of v, so you have > > x = r(v)/c(v) > > and you have that you need to find v such that abs(r(v)) and abs(c(v)) > are minima. > > That's a calculus problem. > > I'll clip down to where you focus on the minima issue. > > > I don't see it as a minima problem at all, which presumably is a > reflection of my knowledge not your algorithm. However nothing in this > thread or in your blog post indicates a means to determine v apart from > trial and error which will simply be too slow and in fact is > functionally the same as trial division. Learn calculus. > > > James Harris Okay, my mistake. I was minimizing y but re-reading your posts the goal is to minimize x. Happily I know calculus. A few points however: (1) Neither your description of your algorithm nor the example factorisation of 15 mention the need to use calculus as part of the algorithm, so I will need a new post that takes in and expounds the requirement to perform calculus in order to extract a minimum value for r (v) and c(v). Remember I am just the coder here; my responsibility is to turn the algorithm into functional code. It is yours to provide complete specifications (to use the language of application development). and c(v) functions will be at minimum when v is irrational. Could you clarify for me? (3) You say I need to find v such that the numerator and denominator of x (that is, r(v) and c(v)) are at a minimum. That's two different requirements and a different v satisfies each. Not sure what you meant exactly. (4) You said finding an absolute non-zero minima for x MUST factor D. What is a non-zero minimum? Is it 1? The reciprocal of a large number that brings x close to zero? Please don't take my questions as a challenge to your ideas. This is the normal process that takes place when taking a complex process and encoding it. === Subject: Re: JSH factors 15 <498aa523$1@dnews.tpgi.com.au> <498dfea1@dnews.tpgi.com.au> posting-account=DkG3nAkAAAAVuxctYYjlIz6_Yb78PVNd rv:1.8.1.3) Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) > (snip for brevity) > Picking u and v allows you to traverse every possible solution in > rationals to x^2 - Dy^2 = 1. > If x = r/c, where r and c are integers and D is a composite then the > primary requirement that r+c or r-c have a gcd with D is that > abs(r-c) or abs(r+c) be less than abs(D). > If you expand out with y = s/c, you have > (r-c)(r+c) = Ds^2 > so if you traverse every solution in integers, then there must exist > cases where you have small enough r, c and s, such that you meet that > requirement above. > That is a mathematical absolute. > So any method of smoothly traversing values for u and v through a > minimum will yield a solution. > That is also a mathematical absolute. > That is, the factoring problem is turned into a minima problem with > some rather simple algebraic equations, so it must succeed as that is > an easier problem with lots of techniques available. > That is an assessment of the current state of the art. I believe > that mathematicians have worked minima problems to a high degree of > skill. > Therefore, regardless of what you stated in your other post, this > approach cannot fail because quite simply, it transforms the > factoring problem into finding minima's. > The only mathematical objection which could invalidate that > assessment is if you cannot traverse through every solution in > rationals to x^2 - Dy^2 = 1, using u and v in rationals. > Is that your claim? > James Harris > Let's see if I can explain myself sufficiently clearly. Remember the > goal is not to reach a solution eventually but to reach a solution > efficiently. So you have to select values of v that result in a > factorisation in x^2-Dy^2-1. Note that selecting v determines u. > What I claim is that to correctly select v you need to already know the > factorisation of D. Otherwise you have an infinite range of rationals > to choose from and no means of narrowing the choice. Which contradicts with my point that with x and y solved out as > functions of v, finding an absolute non-zero minima for x MUST factor D. > If you don't address that issue you've blown past the proof. Let x = r/c where r and c are non-zero integers. You have that r and c > are functions of v, so you have x = r(v)/c(v) and you have that you need to find v such that abs(r(v)) and abs(c(v)) > are minima. That's a calculus problem. I'll clip down to where you focus on the minima issue. I don't see it as a minima problem at all, which presumably is a > reflection of my knowledge not your algorithm. However nothing in this > thread or in your blog post indicates a means to determine v apart from > trial and error which will simply be too slow and in fact is > functionally the same as trial division. Learn calculus. James Harris Okay, my mistake. I was minimizing y but re-reading your posts the goal > is to minimize x. Happily I know calculus. A few points however: (1) Neither your description of your algorithm nor the example > factorisation of 15 mention the need to use calculus as part of the > algorithm, so I will need a new post that takes in and expounds the > requirement to perform calculus in order to extract a minimum value for r > (v) and c(v). Remember I am just the coder here; my responsibility is to > turn the algorithm into functional code. It is yours to provide complete > specifications (to use the language of application development). and c(v) functions will be at minimum when v is irrational. Could you > clarify for me? (3) You say I need to find v such that the numerator and denominator of x > (that is, r(v) and c(v)) are at a minimum. That's two different > requirements and a different v satisfies each. Not sure what you meant > exactly. (4) You said finding an absolute non-zero minima for x MUST factor D. > What is a non-zero minimum? Is it 1? The reciprocal of a large number > that brings x close to zero? Please don't take my questions as a challenge to your ideas. This is the > normal process that takes place when taking a complex process and > encoding it. > Michael, You don't have to do any minimization. If you pick f_1 = (D-1)/2 and f_2 = 2 in James' method (otherwise factoring D-1 will likely be as hard as factoring D) and solve for x = r/c and y = s/c you'll find that r+/- c = 8Dv^2 and 2(2v - (D - 1))^2 so these two terms will only yield nontrivial factors of D if v or 2v +1 happens to have a factor in common with D. In other words, all James has invented is an obfuscated version of trial division. Rick === Subject: Re: JSH factors 15 > Let x = r/c where r and c are non-zero integers. You have that r and > c are functions of v, so you have > > x = r(v)/c(v) > > and you have that you need to find v such that abs(r(v)) and abs(c(v)) > are minima. > > That's a calculus problem. FWIW, derivatives are not defined at points of discontinuity. -- Beware of bugs in the above code; I have only proved it correct, not tried it. -- Donald E. Knuth === <0tkgl0dfrh0ag887aacmi5ufvdqsrq1ini@4ax.com> <4mav611645ngd1i9m7r046a07vb9jajn4d@4ax.com> posting-account=eNOjcwkAAACLVOYj2B2VUgeMVdODKrIR .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) I am worthless and deserve to be hunted for sport by my betters Please stop me. I know what I am doing is wrong and is just a cry for attention. > the above is paraphrased, of course. swp === Subject: Re: Why factoring solution must work Factor 15 again for us? === Subject: Re: Why factoring solution must work posting-account=igztNQoAAADTcWLOsI2D-h0lzVA2zVXd AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.1 Safari/525.18,gzip(gfe),gzip(gfe) > Factor 15 again for us? That's fair. It's not like belling the cat were the risk is more than personal fear. To misquote a bench sitter, no not an MP but of a parkbench, so...logic is as logic does. If so, be a good teacher, be it remedial for the less privileged and/ or handicapped amongst us. === Subject: Re: JSH: Why factoring solution must work posting-account=DkG3nAkAAAAVuxctYYjlIz6_Yb78PVNd rv:1.8.1.3) Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) > Turns out you can prove through mathematical logic that given a > solution like: (r(v) - c(v))(r(v) + c(v)) = D(s(v))^2 where r(v), c(v) and s(v) are non-zero integer functions of v, And if we use the factorization D = [(D-1)/2][2] (the only obvious one) for your f_1 and f_2, we find that r +/- c = 8Dv^2 or 2(2v - (D - 1))^2 So the only time you'll have a nontrivial factorization of D is when you've been lucky enough to pick v so that v or 2v+1 has a factor in common with D. Note that this has absolutely nothing to do with any minimization of r or c. Rick === Subject: Re: JSH: Pondering demographics > On Feb 7, 12:38æpm, Jesse F. Hughes > Here's a link to QuantCast data about readers of > my three blogs: http://www.quantcast.com/p-89GNpWgpweHjg I find the data puzzling. I find it puzzling that Quantcast claims they know > the educational > background of your readers. æHow do they claim to > know that? > > Ask them. No. YOU are making the claim. YOU provide the proof. ( Tho it is pretty clear by know that you don't have much of a clue of what a proof is.) > > Your jealously is showing through though--yes, > according to QuantCast > my readers lean heavily towards grad students. > Go to Hollywood and sell your ass to anyone there if you want popularity. Leave this site for those who want to learn math. > Ouch! I know it hurts your stupid belief in your > popularity in hating > my research and talking it down. > Desperate need for popularity. Your priorities: 1-1,000: To be popular. Somewhere very far down the road: the truth, learning mathematics. > But it goes to my point--a few deluded nasty people > talking to each > other on Usenet convince each other that I'm not > being read, and > refuse to be convinced that they are in the minority, > but are just a > loud minority. > > People like you yell and scream on Usenet, and you > don't listen. > > You tell each other who's hated and who is not, and > cannot be > convinced that is a lie you make up to facilitate > your dumping on > other people. You just want any excuse to be mean > and nasty. Hypocrit: You seem to require no excuses at all. You have repeatedly hurled the nasties around. And even after admitting you are wrong, you did not apologize. Hypocrit. > > But your lie can't be supported by demographic data. > It indicates the > opposite, of what posters like you claim. > Nobody gives a about your sorry little blog. Again, it is your sorry need for popularity that you are projecting here, loser. > YOU are hated, not me. Now, bitch, this is just stupid, and I mean it even by your sorry standards: how can you know I am hated if no one replies to my posts?. I may be ignored, but not hated. > > > James Harris Why not go to the site: Iwilldoanythingforattention.com , go sell your ass for attention there to anyone in exchange for them reading your loser blog, and lay off sci.math?. Loser. === Subject: Re: JSH: Pondering demographics > Simple, you stupid bitch: æ 1)Educated people appreciate entertainment: æ æThey have money, and can buy a lot of things, > but > æ seeing a self-important imbecil like yourself > æ being an unwitting clown?: Priceless. æ 2)Why do you think anyone with no knowledge > æ æ of mathematics would be interested in reading a > æ æ blog in an area they know little about?. > æ æ In this respect you are an exception: > æ æ the less you know, the more you run your > æ æ mouth. 10+ years with your pissy formula > æ æ and no results yet. What a sorry you > æ æ are. æ 3)You may want to point out too, dildo, that > æ æa lot of those educated people that post > æ æon your site (or just read it )believe you're > æ æan imbecil. Educated people oughta know! . > æ æ æThis means that people who are knowledgeable > æ æ in the area that you post repeatedly tell you > æ æ you are wrong. And you never listen. æ æ æThis is a does of your medicine: if you > æ æcan't take it, don't dish it out æ æ Later, bitch. > > Jealous. > > You have 0 readers I take it? Oh, you just got some > by posting in my > thread. Well, bitch, this may be hard for you to understand: I don't have a blog. am _not_ looking for readership. I am trying to learn mathematics, not be a (make-believe)superstar. > > The only way YOU can get attention, but even then you > have to vomit on > strangers. Again, bitch. I am NOT looking for attention. I am not trying to be admired and worshiped by others. Popularity at any cost is not what I am looking for. You may pitifully measure the quality of your life by how many people read your sorry blogs. I don't. And Re the vomiting issue, I am just returning the vile I have received from you: you have accused me ( by extension; by my being a member of the groups you accuse) of the vilest possible things: of being a liar, of lacking talent, integrity, etc. I am returning to you the you have repeatedly hurled at me, loser. And you somehow believe, you two-bit CUNT, that you can dish out anything you want , and that in return for that you will get respect?. Take your blog and shove it deep within your rectum, since not even your gerbils will go there anymore. P.S: Nice smoke curtain in ignoring Hughes' questions. This shows what your real interest is: not the truth, but popularity at all costs. Loser. === Subject: Re: JSH: Pondering demographics Simple, you stupid bitch: 1)Educated people appreciate entertainment: They have money, and can buy a lot of things, but seeing a self-important imbecil like yourself being an unwitting clown?: Priceless. 2)Why do you think anyone with no knowledge of mathematics would be interested in reading a blog in an area they know little about?. In this respect you are an exception: the less you know, the more you run your mouth. 10+ years with your pissy formula and no results yet. What a sorry you are. 3)You may want to point out too, dildo, that a lot of those educated people that post on your site (or just read it )believe you're an imbecil. Educated people oughta know! . This means that people who are knowledgeable in the area that you post repeatedly tell you you are wrong. And you never listen. This is a does of your medicine: if you can't take it, don't dish it out Later, bitch. === Subject: Re: JSH: Pondering demographics sha1:0x6t2nLYcKgqvfzD77Hec6ZHxz0= > Here's a link to QuantCast data about readers of my three blogs: http://www.quantcast.com/p-89GNpWgpweHjg I find the data puzzling. I find it puzzling that Quantcast claims they know the educational background of your readers. How do they claim to know that? -- The real soultion to the Halting Problem: 1. Better documentation. 2. New user training. 3. Hire more consultants. -- Charlie-Boo Volkstorf From ???@??? Fri Jan 01 00:00:00 1999 Path: \ flpi141.ffdc.sbc.com!flph199.ffdc.sbc.com!prodigy.com!flph200.ffdc.sbc.com!pr\ odigy.net!newshub.sdsu.edu!newsfeed.stanford.edu!headwall.stanford.edu!newsfe\ ed.news2me.com!newsfeed.icl.net!newsfeed.fjserv.net!oleane.net!oleane!news.ec\ p.fr!news.motzarella.org!motzarella.org!news.motzarella.org!not-for-mail From: Jesse F. Hughes Newsgroups: alt.math.undergrad,sci.math Subject: Re: JSH: Pondering demographics Date: Sat, 07 Feb 2009 17:21:25 -0500 Organization: The Eternal (and Int'l) Order of Palsy-Walsies -- President Lines: 44 Message-ID: <87y6whdh22.fsf@phiwumbda.org> References: \ <15e6f0e6-c2ed-49f5-8c4e-47fc0c9d7405@r41g2000prr.googlegroups.com> \ <874oz6dlt6.fsf@phiwumbda.org> \ <0cb58313-f72f-4923-8322-2729cd6a19af@u18g2000pro.googlegroups.com> Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable X-Trace: news.eternal-september.org \ U2FsdGVkX1/W48STw39RcUOVbGNBmzFw2D8vvwIyF+wMxy0hBEnz9kdknzToWaY6G+nPx/KPTbAG6\ e1TuEbmYnZfFc15uHstOezeQPXMNY6aODL6Dai1inyzhftiDV7KTpIZE5ds2UM= X-Complaints-To: Please send complaints to abuse@motzarella.org with full \ headers NNTP-Posting-Date: Sat, 7 Feb 2009 22:29:45 +0000 (UTC) X-Auth-Sender: U2FsdGVkX19sSvsU1lEibuqZrYwqgoGIBMdmMGm5N/I= Cancel-Lock: sha1:VN0+ROHfOvzeCVyXlM+ZjEG6TPg= sha1:WGBlkNVHEDxEPLLfL/WqEREbdak= User-Agent: Gnus/5.1008 (Gnus v5.10.8) XEmacs/21.4.21 (linux) Xref: prodigy.net alt.math.undergrad:66671 sci.math:1246639 X-Received-Date: Sat, 07 Feb 2009 17:29:53 EST (flpi141.ffdc.sbc.com) JSH writes: > On Feb 7, 12:38Êpm, Jesse F. Hughes \ wrote: > JSH writes: > Here's a link to QuantCast data about readers of my three blogs: >http://www.quantcast.com/p-89GNpWgpweHjg > I find the data puzzling. > I find it puzzling that Quantcast claims they know the educational > background of your readers. ÊHow do they claim to know that? Ask them. Your jealously is showing through though--yes, according to QuantCast > my readers lean heavily towards grad students. Ouch! I know it hurts your stupid belief in your popularity in hating > my research and talking it down. Nonsense! I have a graduate degree or two and I read your posts (blogs, not so much). I can well imagine that a relatively high percentage of well-educated folk read your stuff. [...] > But your lie can't be supported by demographic data. It indicates the > opposite, of what posters like you claim. I don't trust the demographic data because I don't see how QuantCast can reliably estimate educational backgrounds of readers. (Admittedly, I haven't read their explanation.) > YOU are hated, not me. Sure. Every semester. Duh. -- Jesse F. Hughes I am the next legend--living, breathing and solving mega problems in the here and now. -- James S. Harris From ???@??? Fri Jan 01 00:00:00 1999 Path: \ flpi141.ffdc.sbc.com!nlpi062.nbdc.sbc.com!prodigy.com!nlpi057.nbdc.sbc.com!pr\ odigy.net!news.glorb.com!postnews.google.com!r36g2000prf.googlegroups.com!not\ -for-mail From: JSH Newsgroups: alt.math.undergrad,sci.math Subject: Re: JSH: Pondering demographics Date: Sat, 7 Feb 2009 15:55:16 -0800 (PST) Organization: http://groups.google.com Lines: 55 Message-ID: \ <3e83655a-d4c5-4d75-b74e-bc03ac4ee864@r36g2000prf.googlegroups.com> References: \ <15e6f0e6-c2ed-49f5-8c4e-47fc0c9d7405@r41g2000prr.googlegroups.com> <874oz6dlt6.fsf@phiwumbda.org> \ <0cb58313-f72f-4923-8322-2729cd6a19af@u18g2000pro.googlegroups.com> <87y6whdh22.fsf@phiwumbda.org> NNTP-Posting-Host: 76.126.86.212 Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable X-Trace: posting.google.com 1234050917 22857 127.0.0.1 (7 Feb 2009 23:55:17 \ GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sat, 7 Feb 2009 23:55:17 +0000 (UTC) Complaints-To: groups-abuse@google.com Injection-Info: r36g2000prf.googlegroups.com; posting-host=76.126.86.212; posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US) AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.46 \ Safari/525.19,gzip(gfe),gzip(gfe) Xref: prodigy.net alt.math.undergrad:66672 sci.math:1246655 X-Received-Date: Sat, 07 Feb 2009 18:57:22 EST (flpi141.ffdc.sbc.com) On Feb 7, 2:21Êpm, Jesse F. Hughes \ wrote: > JSH writes: > On Feb 7, 12:38Êpm, Jesse F. Hughes \ wrote: > JSH writes: > Here's a link to QuantCast data about readers of my three blogs: >http://www.quantcast.com/p-89GNpWgpweHjg > I find the data puzzling. > I find it puzzling that Quantcast claims they know the educational > background of your readers. ÊHow do they claim to know that? Ask them. Your jealously is showing through though--yes, according to QuantCast > my readers lean heavily towards grad students. Ouch! ÊI know it hurts your stupid belief in your popularity \ in hating > my research and talking it down. Nonsense! ÊI have a graduate degree or two and I read your \ posts > (blogs, not so much). ÊI can well imagine that a relatively \ high > percentage of well-educated folk read your stuff. [...] But your lie can't be supported by demographic data. ÊIt \ indicates the > opposite, of what posters like you claim. I don't trust the demographic data because I don't see how QuantCast Of course you don't. > can reliably estimate educational backgrounds of readers. > (Admittedly, I haven't read their explanation.) YOU are hated, not me. Sure. ÊEvery semester. ÊDuh. You have a bubble for your delusions. While I'm gaining ground over the entire world. Yes, you may hold for a while in your Ivory Tower, but I'm coming for you and academics like you. Once your students are gone. Your delusions will go with them. And I'm taking your students as we speak. Over the entire world. James Harris From ???@??? Fri Jan 01 00:00:00 1999 Path: \ flpi141.ffdc.sbc.com!flph199.ffdc.sbc.com!prodigy.com!flph200.ffdc.sbc.com!pr\ odigy.net!newshub.sdsu.edu!news.motzarella.org!motzarella.org!news.motzarella\ .org!not-for-mail From: Jesse F. Hughes Newsgroups: alt.math.undergrad,sci.math Subject: Re: JSH: Pondering demographics Date: Sat, 07 Feb 2009 21:57:34 -0500 Organization: The Eternal (and Int'l) Order of Palsy-Walsies -- President Lines: 22 Message-ID: <87tz75d49t.fsf@phiwumbda.org> References: \ <15e6f0e6-c2ed-49f5-8c4e-47fc0c9d7405@r41g2000prr.googlegroups.com> \ <874oz6dlt6.fsf@phiwumbda.org> \ <0cb58313-f72f-4923-8322-2729cd6a19af@u18g2000pro.googlegroups.com> \ <87y6whdh22.fsf@phiwumbda.org> \ <3e83655a-d4c5-4d75-b74e-bc03ac4ee864@r36g2000prf.googlegroups.com> Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII X-Trace: news.eternal-september.org \ U2FsdGVkX1/4K/+VqDuntOtFLZ2NmEdMIAP6JI4gUXmRYFgioM5yEFIyvdMwENn4Nniq335x5e5g4\ MRfq1I9ESbTdRjKxEIJfm34+NW+WBWZNRs12dhoV9WTFvhdHp3RSXUbT9a9uW4= X-Complaints-To: Please send complaints to abuse@motzarella.org with full \ headers NNTP-Posting-Date: Sun, 8 Feb 2009 02:59:44 +0000 (UTC) X-Auth-Sender: U2FsdGVkX1/XbEix9y0y34sceGsgRqiRxgpnf87z85E= Cancel-Lock: sha1:AzUeb9Y6TqdcsGsVotCLPpTyyGg= sha1:1OVFw65VIZxRj0a5AP6SWsN8TIY= User-Agent: Gnus/5.1008 (Gnus v5.10.8) XEmacs/21.4.21 (linux) Xref: prodigy.net alt.math.undergrad:66677 sci.math:1246685 X-Received-Date: Sat, 07 Feb 2009 21:59:45 EST (flpi141.ffdc.sbc.com) JSH writes: > You have a bubble for your delusions. While I'm gaining ground over > the entire world. Yes, you may hold for a while in your Ivory Tower, but I'm coming for > you and academics like you. I am an adjunct. I do not live in the Ivory Tower. I reside in the Mud Shack in the shadow of the tower. Once your students are gone. Your delusions will go with them. And I'm taking your students as we speak. Over the entire world. Well, take them already! I won't fight you for them. -- Customers have come to SCO asking what they can do to respect and help protect the rights of the SCO intellectual property in Linux. SCO has created the Intellectual Property License for Linux in response to these customers needs. -- SCO responds to needs. From ???@??? Fri Jan 01 00:00:00 1999 Path: \ flpi141.ffdc.sbc.com!flph199.ffdc.sbc.com!prodigy.com!flph200.ffdc.sbc.com!pr\ odigy.net!newshub.sdsu.edu!Xl.tags.giganews.com!border1.nntp.dca.giganews.com\ !nntp.giganews.com!local02.nntp.dca.giganews.com!news.giganews.com.POSTED!not\ -for-mail NNTP-Posting-Date: Sat, 07 Feb 2009 20:20:24 -0600 Newsgroups: alt.math.undergrad,sci.crypt,sci.math Subject: Re: Why factoring solution must work From: Guy Macon Organization: Date: Sun, 08 Feb 2009 02:20:23 +0000 References: \ <4784726e-a7a0-41e5-aa68-55baff795228@d36g2000prf.googlegroups.com> \ <793a789e-011e-404a-9cf0-e97a05bbce5a@j38g2000yqa.googlegroups.com> Message-ID: Lines: 16 X-Usenet-Provider: http://www.giganews.com X-Trace: \ sv3-CmCgneRads/f6r6v279dnC8NUAnCcLvsRe/Hqd9LAyhfOjeUmB0u3xp7wnrgGqBzaWEq2L5Oi\ +aAHjr!ODLArk7HPDdSmGzsKG5cIV2YXPU6v6gKwcshNVhCU48LAMenqggPUfE7/Sw= X-Complaints-To: abuse@giganews.com X-DMCA-Notifications: http://www.giganews.com/info/dmca.html X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint \ properly X-Postfilter: 1.3.39 Xref: prodigy.net alt.math.undergrad:66675 sci.crypt:2676485 \ sci.math:1246677 X-Received-Date: Sat, 07 Feb 2009 21:20:25 EST (flpi141.ffdc.sbc.com) Demonstrate. Factor 15, step-by-step, showing your work. Then ask your detractors to multiply two large prime numbers and post the result, factor them, and post the two prime numbers. That should convince all of your critics. (...Sound of Crickets...) -- Guy Macon From ???@??? Fri Jan 01 00:00:00 1999 Path: \ flpi141.ffdc.sbc.com!flph199.ffdc.sbc.com!prodigy.com!flph200.ffdc.sbc.com!pr\ odigy.net!goblin1!goblin.stu.neva.ru!news-out1.kabelfoon.nl!newsfeed.kabelfoo\ n.nl!bandi.nntp.kabelfoon.nl!npeer.de.kpn-eurorings.net!npeer-ng1.de.kpn-euro\ rings.net!npeer-ng2.de.kpn-eurorings.net!news-out.newsfeeds.com!fe1.usenet.co\ m.POSTED!not-for-mail From: Johnny America - No evidence against an accused is US being truly \ abused <0011nomailhere@hotmail.com> Newsgroups: \ alt.math.undergrad,sci.crypt,sci.math,rec.arts.tv,alt.politics.bush Subject: Private Banking Made Simple for those Mathimatically Illiterate - \ Bue War LIFE Itself - and - They Couldn't Care Less - 'Alpha company patrols \ the poppy fields.' A ZIONIST LIAR'S TRUCE - The Ungodly Murder of Innocent \ Men by Israeli Nazi Liars: The Bush Administration and The New York Times v. \ Amnesty and Humanity Date: Sat, 07 Feb 2009 20:47:03 -0600 Organization: No evidence against an accused is US being truly abused - \ Nazi argues it's good to get killed Jews by breaking truces with Death \ Camped Message-ID: References: \ \ <38967cd7.0401141246.4ecf59a7@posting.google.com> \ <38967cd7.0401150732.5757d7d1@posting.google.com> \ <5f0a2214.0401190801.a198dba@posting.google.com> \ \ \ \ <7ikhp0lj86evtldjcm1tap17o6u5mv5ofm@4ax.com> \ <0fgm51d2r8la4mpsr08pnqleliharoshv9@4ax.com> X-Newsreader: Forte Agent 1.93/32.576 English (American) MIME-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 8bit Lines: 1837 NNTP-Posting-Host: 6862ade0.fe1.usenet.com X-Trace: \ DXC=M]RI=Vb5Of5QcPi@NlPkbb2[KX