mm-3629 === Subject: : A Dirichlet series Let S be the set of squarefree integers >=1. Evaluate sum_{x in S} x^(-2) Hint: It's not frightfully deep. === Subject: : Re: A Dirichlet series > Let S be the set of squarefree integers >=1. Evaluate > sum_{x in S} x^(-2) > Hint: It's not frightfully deep. On second thought, I think this one is too easy. Write T for the sum, and just verify zeta(4) T = zeta(2) straight from the definitions, making the customary obeisances to absolute convergence. Since zeta(2) = (pi^2) / 6 zeta(4) = (pi^4) / 90 we get the cosmic-looking conclusion T = 15 / (pi^2). === Subject: : Re: A Dirichlet series Am 22.05.2006 11:09 schrieb Larry Hammick: > Let S be the set of squarefree integers >=1. Evaluate > sum_{x in S} x^(-2) > Hint: It's not frightfully deep. Denominator of Euler-zeta-Product? Gottfried Helms === Subject: : New mathematics/physical sciences positions at http://jobs.phds.org, May 22, 2006 New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Quantitative Research Specialist: Jhirad Consulting, NY,NY. Major financial firm seeks an equities quant to spearhead asset allocation and risk research on a multibillion portfolio. Candidate will develop quantitative methods and tools to augement return and risk profile among equities in the US, Asia and Europe. Candidate will be... http://jobs.phds.org/jobs/JhiradConsult/quantitative * Product Specialist - Derivatives, Fixed Income: Jhirad Consulting, NY,NY. Prestigious financial firm is seeking fixed income and derivatives product managers. 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Role - Specify, fit and implement numerous flavours of... http://jobs.phds.org/jobs/huxley/mortgage-prepayment === Subject: : algebraic number on a circle, that is no root of 1 Hi Folks. Do exist an algebraic number (within the complex numbers) on the unit circle, that is no root of 1 ? In other words: I'm looking for a complex number x, that is algebraic over the rational numbers and that satisfy: |x| = 1 and x^n <> 1 for all natural numbers n > 0 . Holger. === Subject: : Re: algebraic number on a circle, that is no root of 1 > Hi Folks. > Do exist an algebraic number (within the complex numbers) > on the unit circle, that is no root of 1 ? Better than that: there are algebraic integers with this property. If you look at the roots of the irreducible polynomial x^10+x^9-x^7-x^6-x^5-x^4-x^3+x+1 you will find eight of its ten roots are algebraic integers on the unit circle which are not roots of unity. === Subject: : Re: algebraic number on a circle, that is no root of 1 <588021.1148290748366.JavaMail.jakarta@nitrogen.mathforum.org>, Holger > Hi Folks. > Do exist an algebraic number (within the complex numbers) > on the unit circle, that is no root of 1 ? > In other words: I'm looking for a complex number x, that > is algebraic over the rational numbers and that satisfy: > |x| = 1 and x^n <> 1 for all natural numbers n > 0 . > Holger. Or course. (3/5)+i(4/5) is not a root of unity. === Subject: : Re: algebraic number on a circle, that is no root of 1 <220520060755201440%anniel@nym.alias.net.invalid Do exist an algebraic number (within the complex numbers) > on the unit circle, that is no root of 1 ? > Or course. (3/5)+i(4/5) is not a root of unity. It's not hard to show that the powers of this number are dense in the unit circle. LH === Subject: : Re: algebraic number on a circle, that is no root of 1 Larry Hammick a .8ecrit : > Do exist an algebraic number (within the complex numbers) > on the unit circle, that is no root of 1 ? >> Or course. (3/5)+i(4/5) is not a root of unity. > It's not hard to show that the powers of this number are dense in the > unit circle. As fot *any* number on the unit circle which is not a root of unity > LH === Subject: : Is this true? Proof? Hi all! IĒd like to see a proof of this, if itĒs true: C,a,b integer. C product of two primes (I donĒt know if this is a necessary condition). If a^2-C divides b^2-C then b-c is a triangular number (a number of the form n*(n+1)/2; n integer) Thx :) === Subject: : Re: Is this true? Proof? > IĒd like to see a proof of this, if itĒs true: > C,a,b integer. > C product of two primes (I donĒt know if this is a necessary condition). > If a^2-C divides b^2-C then b-c is a triangular number (a number of the > form n*(n+1)/2; n integer) It's not true. Take C = 6 (= 2*3), a = 3, and b = 15. Then a^2 - C (= 3) | b^2 - C (= 219), but b - C = 9, which is not a triangular number. Jose Carlos Santos === Subject: : Re: Is this true? Proof? What if C=p*q , p,q primes <> 2? Jos.8e Carlos Santos escribi.97: >> IĒd like to see a proof of this, if itĒs true: >> C,a,b integer. >> C product of two primes (I donĒt know if this is a necessary condition). >> If a^2-C divides b^2-C then b-c is a triangular number (a number of >> the form n*(n+1)/2; n integer) > It's not true. Take C = 6 (= 2*3), a = 3, and b = 15. Then > a^2 - C (= 3) | b^2 - C (= 219), > but b - C = 9, which is not a triangular number. > Jose Carlos Santos === Subject: : Re: Is this true? Proof? > What if C=p*q , p,q primes <> 2? Please don't top-post. If you want to know what's that and why you shouldn't do it, read http://www.caliburn.nl/topposting.html or http://www.html-faq.com/etiquette/?toppost Now, take C = 15 (= 3*5), a = 4, and b = 17. Jose Carlos Santos === Subject: : Re: Constrained Optimisation (Lagrange) I agree with you that it would be pointless to ask more than checking a handful of cases in an exam. My lecturer has said that the problem will have more constraints than you would be able to reasonably check within the time - and it will be a two decision variable quadratic objective function. Maybe the trick he is referring to is simply discard some constraints at the beginning by checking that some dont intersect as you say. I'm presuming it will be mainly common sense otherwise he would have taught the technique! Chris. === Subject: : Re: Argument principle > Im doing some work on the argument principle, and I am not wholy > understanding it. > from what I understand > (1/(2*Pi*i) * integral (a closed path) f ' (z) / f(z) = z - p (where z > is the zeros in side the closed path, and p the poles -both with > multiplicities) > it also leads to this > (1/(2*Pi*i) * integral (a closed path) f ' (z) / f(z) = i * > (Triangle)_path g *arg f(z) > the (triangle is) integer multiple of 2*Pi > Im slightly confused to the term i * (Triangle)_path g *arg f(z) The integral of f'/f along the path _g_ is equal to 2*pi*i times the index of f o g with respect to 0; this follows from the definitions of integral of a function along a path and of index. But the index of f o g with respect to 0 is precisely how the argument of f(z) changes while _z_ follows the path _g_. I hope that this helps. Jose Carlos Santos === Subject: : Number of generators of an ideal generated by XY-WZ, X^2-WY, Y^2-XZ cannot be generated by 2 elements. -kira === Subject: : Re: Number of generators of an ideal Ah... Now that I've asked the question, my memory is coming back to me that I've once asked this question on this forum again. If my memory recalls correctly, there is trick where I can take the ideal J=(w,x,y,z)^3 and form I/J where I=(XY-WZ, X^2-WY, Y^2-XZ). Then, since J consists of all terms of degree 3, the quotient I/J will have only terms of degree 2; hence a vector space. The dimension of I/J is 3, one for each of the generators that are clearly linearly independent. So, if the ideal I were generated by 2 elements, then each of them can span at most 1 dimension since each will have the form a(XY-WZ)+b(X^2-WY)+c(Y^2-XZ) for some a,b,c in k. Then I/J would be at most of 2 dimensions. Contradiction. Sorry for posting again. I need to learn to try harder before asking for help. -kira > generated by > XY-WZ, X^2-WY, Y^2-XZ > cannot be generated by 2 elements. > -kira === Subject: : mean value like and differentiable function Let f be twice continuously differentiable fucntion on [a,b] sich that f(a) = f(b) = 0 and x_0 any fixed real from (a,b). How can i prove that there exists c from (a,b) which satisfies f(x_0) = (x_0-a)(x_0-b)*f''(c)/2. === Subject: : Re: mean value like and differentiable function >Let f be twice continuously differentiable fucntion on [a,b] sich that f(a) = f(b) = 0 and x_0 any fixed real from (a,b). >How can i prove that there exists c from (a,b) which satisfies >f(x_0) = (x_0-a)(x_0-b)*f''(c)/2. A hint is that [(f(x_0)/(x_0-a) - f(x_0)/(x_0-b)]/(b-a) can be written as the integral of f'' times a certain weight. (Start with f(x_0) = f(x_0) - f(a) = int_a_{x_0} f', and make a change of variables so the integral goes from 0 to 1. Do the same with f(b) - f(x_0). Subtract, then write f'(here) - f'(there) as the integral of f''.) Now if w(c) is positive and I = int_0^1 w(c) f''(c) dc then there must exist c such that f''(c) int_0^1 w = I, because if m and M are the minimum and maximum of f'' then m int_0^1 w <= I ,+ M int_0^1 w. ************************ David C. Ullrich === Subject: : Re: mean value like and differentiable function > On Mon, 22 May 2006 06:05:42 EDT, eugene >Let f be twice continuously differentiable fucntion > on [a,b] sich that f(a) = f(b) = 0 and x_0 any fixed > real from (a,b). >How can i prove that there exists c from (a,b) which > satisfies >f(x_0) = (x_0-a)(x_0-b)*f''(c)/2. > A hint is that > [(f(x_0)/(x_0-a) - f(x_0)/(x_0-b)]/(b-a) > can be written as the integral of f'' times a certain > weight. > (Start with f(x_0) = f(x_0) - f(a) = int_a_{x_0} f', > and > make a change of variables so the integral goes from > 0 to 1. > Do the same with f(b) - f(x_0). Subtract, then write > f'(here) - f'(there) as the integral of f''.) > Now if w(c) is positive and > I = int_0^1 w(c) f''(c) dc > then there must exist c such that f''(c) int_0^1 w = > I, > because if m and M are the minimum and maximum of f'' > then > m int_0^1 w <= I ,+ M int_0^1 w. > David C. Ullrich I was trying to use Taylor formula but failed. === Subject: : Dynamics of Preditor-Prey Differential Equations I would very much appreciate your help on how to understand the dynamics of this system of differential equations. You can find the system on my webpage, at this page... http://www.informatics.sussex.ac.uk/users/ctf20/dphil_2005/Coexistence.htm under the heading 2. Scavenger Cycles. Basically it is the standard preditor-prey equations, but I have decided to embed them within a re-cycling ecosystem model. I find 4 fixed points. But the system is 3D and so difficult to visualize. I would like to communicate with someone who is interested in analysing this system with me further, and perhaps publishing something in collaboration with me, if we find anything interesting. Chrisantha Fernando === Subject: : When is f(1,y) ^ [n] a p degree poynomial? n > 2 , p a positive integer, function f(x,y) :R^2 -> I [n] means nth iteration on first argument : Ex. [3] for f(f(f(1,y),y),y) ; x =1 I intervals on R to be defined . May we find two functions f, g such as f(1,y) ^ [n] + g(1,y) ^ [n+1] = 0 f(1,y) ^ [n] and g(1,y) ^ [n+1] two polynomials degree p in y , Alain === Subject: : Re: When is f(1,y) ^ [n] a p degree poynomial? alainverghote@yahoo.fr nous a r.8ecemment amicalement signifi.8e : > n > 2 , p a positive integer, function f(x,y) :R^2 -> I > [n] means nth iteration on first argument : > Ex. [3] for f(f(f(1,y),y),y) ; x =1 > I intervals on R to be defined . a lot of possibilities : f(x,y) = x + P(y) P(y) polynomial with degree p f(x,y) = (x-1)(x-P(y))u(x,y) + P(y) P(y) polynomial with degree p u(x,y) any function f(x,y) = x^a y^(p(a-1)/a^n-1) a positive not equal to 1 x and y positive ... > May we find two functions f, g such as > f(1,y) ^ [n] + g(1,y) ^ [n+1] = 0 > f(1,y) ^ [n] and g(1,y) ^ [n+1] two polynomials > degree p in y , f(x,y) = (x-1)(x-P(y))cos(x+y) + P(y) g(x,y) = (x-1)(x+P(y))sin(x+y) - P(y) > Alain -- Patrick === Subject: : primality tests - question about speed comparison > Given a prime-exponent Mersenne number M(n), you can: > a) run a base-2 probable-prime test, > i.e. test whether 2^(M(n) - 1) mod (M(n) = 1 > (which composites such as M(11) and M(23) will pass) > (1) Why did you choose base 2? > (2) Doing this is pointless. I'm using the speed of the 2prp test as a benchmark. > or > b) run the fastest-known primality test (for such numbers), > which no composites will pass > Roughly how much slower, as a function of n, is b) compared with a)? > It will not be slower at all. It will run at the same speed. such big M-primes are being discovered. How much slower than the prp-2 test is the fastest test for a 'general' number, i.e. to show primality for a number that isn't 'special' like a Mersenne number, Fermat number, etc.? E.g. if the prp-2 test takes n seconds, for a probable prime in the region of e^x, roughly how long will the fastest perfect primality test take? Chris === Subject: : Re: primality tests - question about speed comparison number, i.e. to show primality for a number that isn't 'special' like a > Mersenne number, Fermat number, etc.? E.g. if the prp-2 test takes n > seconds, for a probable prime in the region of e^x, roughly how long > will the fastest perfect primality test take? This question can not be answered. It is implementation and platform specific and the answer will vary according to both. It will also vary according to the size of the number. Currently, for numbers in the several hundreds to several thousands of digits, carefully tuned implementations of Cyclotomy Methods (the Cohen-Lenstra-Bosma-et.al methods) will be fastest. But this algorithm is NOT polynomial time. ECPP is polynomial time (if certain unproved assumptions are correct) but it does not become faster than Cohen-Lenstra until at least several thousand digits and the exact crossover point is implementation dependent. Also, if GRH is correct, then Miller-Rabin is also in P, and will be competitive with ECPP when convolution techniques are used for the arithmetic. But again, it is not known exactly whether this method or ECPP will be faster for very large (e.g. 10's of thousands of digits) numbers. === Subject: : 10 questions know how to answer them please do so step by step please. i need them today. Q:simplify the expression cscx cos(-x) Q: Find the amplitude and period of the graph of the following function: y=3Sin 2pix Q: Find the exact value of cos75 Q:Sketch one cycle of the graph of the function: .87 fx=4cos (pix/3) Q: Simplify the expression: .87 sin ((pi/2)-x)secx Q: Verify the identity: sinx/1+cosx = 1-cosx/sinx Q: simplify the expression, do no evaluate it sin30 cos40+cos30 sin 40 === Subject: : Re: 10 questions > know how to answer them please do so step by step please. i need them > today. You have probably been given some identities. csc(x) = 1/sin(x) sec(x) = 1/cos(x) cos(-x) = cos(x) sin(-x) = -sin(x) > Q:simplify the expression > cscx cos(-x) See if the above gives you a starting place. > Q: Find the amplitude and period of the graph of the following > function: > y=3Sin 2pix > Q:Sketch one cycle of the graph of the function: > .87 fx=4cos (pix/3) y = a*sin(bx) and y = a*cos(bx) have amplitude a and period b/(2*pi) > Q: Find the exact value of cos75 > Q: simplify the expression, do no evaluate it > sin30 cos40+cos30 sin 40 You probably have the identities for sin(a+b), sin(a-b), cos(a+b), cos(a-b). You will find they are useful for these two questions. Hint for the first one: Can you write cos(75) as cos(a+b) where a and b are things you know the sine and cosine of? - Randy === Subject: : Re: 10 questions > know how to answer them please do so step by step please. i need them > today. > You have probably been given some identities. > csc(x) = 1/sin(x) > sec(x) = 1/cos(x) > cos(-x) = cos(x) > sin(-x) = -sin(x) > Q:simplify the expression > cscx cos(-x) > See if the above gives you a starting place. > Q: Find the amplitude and period of the graph of the following > function: > y=3Sin 2pix > Q:Sketch one cycle of the graph of the function: > .87tfx=4cos (pix/3) > y = a*sin(bx) and y = a*cos(bx) have amplitude a and period b/(2*pi) If I'm not mistaken (consider b = 1), Randy meant to write period (2*pi)/b. The higher the coefficient b, the shorter the period. === Subject: : Re: 10 questions Randy is right , start reading your book of trigonometry ( Why not !), make plots of all given functions (got a hand screen calculator?) and think by yourself ... yes , you can Alain === Subject: : Re: Branch cuts, branch points >My guess is that they're talking about one of the >following possibilities: ... >2) several branch cuts meet at a point. >>But this I did think of, then dismiss: it would be a case >>where it should be noted that branch points are not >>necessarily the endpoints of branch cuts, not the other >>way around, wouldn't it? >I was thinking of e.g. an antiderivative of 1/(z^3-1), where >one choice of branch cuts would have three branch cuts coming >from the three roots of unity and meeting at 0. I guess you >could consider this as two branch cuts that happen to partly >coincide, but if you don't, 0 is an endpoint of a branch cut >and I don't think you can call it a branch point. You might be surprised at what I can call a branch point, if pressed. I would prefer, in fact, to concentrate not on one single cut at a time, but on the system of branch cuts as a whole. Following 1933), with minor modernizations of terminology, I call a graph (1-complex) C embedded (piecewise-smoothly, say) in a compact Riemann surface S a complete system of branch cuts for S just in case every component of the complement S-C is simply connected (Bliss would probably want S and S-C both to be connected, but I don't care). A complete system of branch cuts for a (possibly multiply) punctured Riemann surface S-{z_1,...,z_k} is C-{z_1,...,z_k} where C is a complete system of branch cuts for S. A complete system of branch cuts for a multivalued algebraic function f (poles allowed) on S is a complete system of branch cuts for S-{z_1,...,z_k}, where {z_1,...,z_k} is any finite subset of S containing all poles and all branchpoints of f. Then your example is a complete system of branch cuts (and is equivalent, in the obvious way, to the standard system of branch cuts {1,w,w^2}x[0,infty) for z^3-1, where 1+w+w^2=0). Logarithmic singularities (and non-algebraic multivalued meromorphic functions) complicate this only a little. ... Anyway, the point is that the decomposition into arcs of a general 1-complex, which typically has branch points of its own (in another sense: points of valence > 2), is typically not unique, so for complex analysts to focus on branch cuts that are arcs is a (historically understandable) habit that may no longer have any good justification. Lee Rudolph === Subject: : Re: Branch cuts, branch points simply because they are zeros of the logarithm > [or singularties of b(z)^(-c) if you like]. > Is this nomenclature wrong, and if so, can you suggest an alternative? A branch-point is a singularity of the argument so it is the argument you need to think about. === Subject: : Re: Sets and graphs Please, kindly accept my sincere apologies for my unacceptable and irresponsible behaviour in the earlier exchange. Shame on me and I shall always blame myself for the obvious ignorance and lack of manners I showed. I am truly very sorry. > A set is a graph with no nodes connected. > What about an edge that goes from a node back to the same node? By 'no nodes connected' I mean no cycles whatsoever. > Is it true? > No. A graph is a set of nodes and a set of edges. Yes, right. > Thus a graph with no edges, is a set of nodes and the empty set. Thus /a set/ of individuals and sets, maybe (?) > ---- :-) Tom :-) === Subject: : Re: Sets and graphs A set is a graph with no nodes connected. > What about an edge that goes from a node back to the same node? > By 'no nodes connected' I mean no cycles whatsoever. > Is it true? > No. A graph is a set of nodes and a set of edges. > Yes, right. > Thus a graph with no edges, is a set of nodes and the empty set. > Thus /a set/ of individuals and sets, maybe (?) A graph is a couplet (V,E) where V is a set whose members are commonly called nodes and E is a collection of edges. A edge is denoted as ({v1, v2}, n) a unordered pair of vertices v1, v2 and a positive integer n, the multiplicity of the edge. A set V of points can be mapped to the graph (V,nulset), which is a natural bijection between sets and graphs without edges. A simple graph, is a graph (V,E) with no multiple edges nor loops, ie a graph with for all e in E, some distinct v1,v2 in V with e = ({v1, v2}, 1) > :-) Tom :-) === Subject: : Re: Sets and graphs A set is a graph with no nodes connected. > What about an edge that goes from a node back to the same node? > By 'no nodes connected' I mean no cycles whatsoever. > Is it true? > No. A graph is a set of nodes and a set of edges. > Yes, right. > Thus a graph with no edges, is a set of nodes and the empty set. > Thus /a set/ of individuals and sets, maybe (?) > A graph is a couplet (V,E) where V is a set whose members are commonly > called nodes and E is a collection of edges. A edge is denoted as > ({v1, v2}, n) > a unordered pair of vertices v1, v2 and a positive integer n, the > multiplicity of the edge. A set V of points can be mapped to the > graph (V,nulset), which is a natural bijection between sets and > graphs without edges. > A simple graph, is a graph (V,E) with no multiple edges nor loops, ie a > graph with for all e in E, > some distinct v1,v2 in V with e = ({v1, v2}, 1) Yes, of course, Mr Elliot. :-) I really intended to say that graph theory, set theory, and e.g. formal languge theory represent different formulations of a single idea. And, in particular, that a graph with no edges is a set. This however entails further problems, as you have kindly presented them, whereas I still have not earned my reference formalism on which I could base my considerations. I seem to continue to bounce off of the infinite regress involved in the definitions of comparison (and substitution) or truth. I even fail to execute a single TM instruction. > :-) Tom :-) is! :-) Obviously, I (still) have no clue. But that only provides more stimulation! Tom === Subject: : Re: *BAN* Homosexual Marriage! using electron, positron, neutrino and anti-neutrino as well as the weak force interaction among them; it has theoretically estimated the radii of neutron and proton as well as the range and strength of so-called nuclear force, also has explained the reason for the feature of saturation and short-range of nuclear force and various nuclear reactions http://www.physicswd.com/eng/web/docroot/download/04011202E.zip === Subject: : Laplace transform of reciprocal? I need the Laplace transform for 1/t. It would appear at first glance that this would be Ei(s), where Ei denotes the exponential integral. Is that right? === Subject: : Re: Laplace transform of reciprocal? >I need the Laplace transform for 1/t. It would appear at first glance >that this would be Ei(s), where Ei denotes the exponential integral. >Is that right? No, the transform of 1/t doesn't exist. One source for more information is the online version of Abramowitz and Stegun at The Laplace transform of 1/(t+a), for a>0, is given at Info on the exponential integral and related functions is given at HTH === Subject: : Re: Laplace transform of reciprocal? > I need the Laplace transform for 1/t. It would appear at first glance > that this would be Ei(s), where Ei denotes the exponential integral. > Is that right? No. You can see that 1/t has no LT by looking at the LT of 1/(t+a) and then letting a --> 0. Maple 9.5 gives: > L1:=laplace(1/(t+a),t,s); L1 := (1/2) exp(a s) (2 Ei(1, a s) + 2 ln(a s) - 2 ln(s) - ln(a) + ln(1/a)) > limit(L1,a=0); infinity R.G. Vickson === Subject: : Re: Laplace transform of reciprocal? > I need the Laplace transform for 1/t. It would appear at first glance > that this would be Ei(s), where Ei denotes the exponential integral. > Is that right? 1/t has no Laplace transform. The LT is defined by an integral which blows up in this case. LH === Subject: : Dini's theorem I know the following version of Dini's theorem: Let T be a compact metric space and let f, f_1, f_2, ...: T --> IR be continuous functions with f_1 <= f_2 <= ... and f = sup f_n. Then (f_n) converges uniformly to f. My question is: If T=IR and I know that all f_n are bounded on IR cup {infty} (i.e. they have a limit lim_{t-->infty} f_n(t)=a_ninfty}f(t)=ametric space and let f, f_1, f_2, ...: T --> IR be continuous functions >with f_1 <= f_2 <= ... and f = sup f_n. Then (f_n) converges uniformly >to f. >My question is: If T=IR and I know that all f_n are bounded on IR cup >{infty} (i.e. they have a limit lim_{t-->infty} f_n(t)=a_nresp. lim_{t-->infty}f(t)=a{infty} (one point compactification) and I can continue the f_n and f >onto IR cup{infty} continuously by defining f_n(infty)=a_n resp >f(infty)=a. Then f_n and f are continous on IR cup {infty}. Does >this proceeding allow the application of Dini's theorem provieded that >the monotonicity condition in the theorem is fullfilled??? No. One of the hypotheses is not necessarily satisfied in this case. (Take an example: f_n(t) = 1/(1 + |t|/n)). These tend to 1 monotonically on R, and it should be clear that the convergence is not uniform on R, so your scheme can't work. Look at the example and see what goes wrong.) >pkg ************************ David C. Ullrich === Subject: : Re: Dini's theorem I think this is not a counterexample. The limit function is not continuous on the compactified space (=0 at infty, =1 on IR) as required by Dini. BR pkg === Subject: : Re: Fastest way through a traffic signal > You are looking for an optimum solution. You therefore need to decide >> precisely what quantity, or combination of quantities, you are trying >> to optimise. Fastest way through a traffic signal is too vague, and >> imprecise statements about tradeoffs between one quantity and another >> won't do either. ... > Anyway, after all that, I make the expected time between your points X > and Y - the expression to minimise - equal to (deep breath) > > v_max/a * Int{t_s -> infinity} P(t) dt > + Int{t_s -> infinity} t*P(t) dt > + (s_s/v_max + v_max/a) * Int{0 -> t_s} P(t) dt > + Int{0 -> t_s} (t + 1/(2*a*v_max)*v^2 - v/a -s/v_max)*P(t) dt > > where v is speed as a function of time (direct from the v-curve), and s > is distance as a function of time (which can obviously be got from the > v-curve). > I will have to take a little time to go through > your formula and see how it works, it's certainly > similar to my thoughts but subtly different I > If the signal changes at t = T then the time to get > from point X to point Y is given by the time slowing > plus the time to get back to v_max plus the time at > v_max to reach Y: > t_xy = T + (v_max - v(T))/a + (L - s(T) -(v_max^2 - v(T)^2)/(2a) ) / v_max > where > s(T) = integral of v(t) from 0 to T > (same as yours above). That assumes X is the point > where the optimum speed equals v_max. Your formula > should then be integral( t_xy(T) * P(T) ) from 0 to > infinity. Once I decode the ASCII, hopefully they'll > be the same. > Right. Your integral( t_xy(T) * P(T) ) looks to be the same as my > line 3 + line 4, except that I have integrated over 0 to t_s, rather > remainder of your integral, from t_s -> infinity, is the same as my > first two lines I think. Specifically, when T > t_s, v(T) = 0 and s = > s_s, so your t_xy becomes just > t_xy = T + v_max/a > which is the same as what I'm integrating in line 1 + line 2. > So, summing it all up, it looks like we might have arrived at the same > answer. Going back a little (I left the relevant quote at the top), I think we have now agreed the question, we have the mathematical expression for the quantity to be optimised. > I just split mine into rather messy fragments in the mistaken > belief that it would make it easier to see how to go about the > optimisation (largely because, for some unknown reason, I was thinking > at the time that my t_s was a known constant...!) I am familiar with how to optimise a single value by finding a turning point in the expression for the quantity as a function of the parameter and with numerical methods (like your program) but I'm not sure how you would go about finding an analytical solution to the optimum equation for v(t) from an equation for P(t) which is why I thought the question appropriate for sci.maths. Is there a way to solve this if we know for example that P(t) is Poisson or perhaps a normal distribution or is your numerical approach the only way to get a result? To be honest I am somewhat surprised that nobody has said this has been done before and pointed to some published analysis. George === Subject: : Re: Fastest way through a traffic signal > You are looking for an optimum solution. You therefore need to decide >> precisely what quantity, or combination of quantities, you are trying >> to optimise. Fastest way through a traffic signal is too vague, and >> imprecise statements about tradeoffs between one quantity and another >> won't do either. ... > Anyway, after all that, I make the expected time between your points X > and Y - the expression to minimise - equal to (deep breath) > > v_max/a * Int{t_s -> infinity} P(t) dt > + Int{t_s -> infinity} t*P(t) dt > + (s_s/v_max + v_max/a) * Int{0 -> t_s} P(t) dt > + Int{0 -> t_s} (t + 1/(2*a*v_max)*v^2 - v/a -s/v_max)*P(t) dt > > where v is speed as a function of time (direct from the v-curve), and s > is distance as a function of time (which can obviously be got from the > v-curve). > > I will have to take a little time to go through > your formula and see how it works, it's certainly > similar to my thoughts but subtly different I > > If the signal changes at t = T then the time to get > from point X to point Y is given by the time slowing > plus the time to get back to v_max plus the time at > v_max to reach Y: > > t_xy = T + (v_max - v(T))/a + (L - s(T) -(v_max^2 - v(T)^2)/(2a) ) / v_max > > where > > s(T) = integral of v(t) from 0 to T > > (same as yours above). That assumes X is the point > where the optimum speed equals v_max. Your formula > should then be integral( t_xy(T) * P(T) ) from 0 to > infinity. Once I decode the ASCII, hopefully they'll > be the same. > Right. Your integral( t_xy(T) * P(T) ) looks to be the same as my > line 3 + line 4, except that I have integrated over 0 to t_s, rather > remainder of your integral, from t_s -> infinity, is the same as my > first two lines I think. Specifically, when T > t_s, v(T) = 0 and s = > s_s, so your t_xy becomes just > t_xy = T + v_max/a > which is the same as what I'm integrating in line 1 + line 2. > So, summing it all up, it looks like we might have arrived at the same > answer. > Going back a little (I left the relevant quote at the top), I think > we have now agreed the question, we have the mathematical > expression for the quantity to be optimised. > I just split mine into rather messy fragments in the mistaken > belief that it would make it easier to see how to go about the > optimisation (largely because, for some unknown reason, I was thinking > at the time that my t_s was a known constant...!) > I am familiar with how to optimise a single value by finding a > turning point in the expression for the quantity as a function of > the parameter and with numerical methods (like your program) > but I'm not sure how you would go about finding an analytical > solution to the optimum equation for v(t) from an equation for > P(t) which is why I thought the question appropriate for > sci.maths. Is there a way to solve this if we know for example > that P(t) is Poisson or perhaps a normal distribution or is your > numerical approach the only way to get a result? There is a branch of calculus called calculus of variations which deals with exactly this. Whether it is powerful enough to cope with this problem I don't know. Since possibly not too many people are following this bit in detail, it might be worth starting again, just stating the expression to be minimised, and the conditions, and see if anyone has a clue about how to solve it analytically. You're probably way ahead of me here and have figured out all this already, but having split my integral more sensibly I think it can be written something like Int{0 -> infinity} (t + s_s/v_max + v_max/a) * P(t) * dt + Int{0 -> infinity} (v(t)^2/(2*a*v_max) - v(t)/a - s(t)/v_max) * P(t) * dt so that now the first integral is independent of v(t), and we really need to minimise just Int{0 -> infinity} (v(t)^2/(2*a*v_max) - v(t)/a - s(t)/v_max) * P(t) * dt where a, v_max and s_s are known constants P(t) is a known probability density function s(x) = Int{0 -> x} v(t) dt and we need to minimise this expression by choosing v(t) under the constraints that v <= v_max everywhere Int{0 -> infinity} v(t) dt = s_s possibly some limits on dv/dt (max acceleration/deceleration) any other constraints I forgot! I may have got some details of that wrong (and I'm still using my s_s where you preferred L), but I reckon if you strip down the problem and lay it out something along these lines, then it could be more inviting to a greater number of sci.math punters. > To be honest I am somewhat surprised that nobody has said > this has been done before and pointed to some published > analysis. > George === Subject: : Re: Fastest way through a traffic signal ... > Here's another complication. How far away can you see the signal? If > you can see it for some significant distance (i.e. rather more that the > point at which you need to start slowing down) then this could affect > the probability. > Does not matter in general. As a train driver you will *never* see a signal > switch from green to yellow or red. In typical block-system operations > you have a sequence of signals, each train triggers aspect changing when > it passes signals. In normal operation, the aspect of the signal a train > passes changes from green to red, the signal before changes from red to > yellow and the signal before that changes from yellow to green. Here > green means: pass at full speed; yellow means: pass at a speed that allows > you to stop at the next signal because it is red; and red means stop. > (There are systems with more aspects, but this is the basic one.) I can > imagine only a few ways that you may see an aspect change from red to green. > One of them is when the previous train branches off, the switches are > changed and the new way is fully clear. > And if three aspect signals are not used you have main signals and > advance signals, where the main signal only is able to show red and > green, but it is preceded by an advance signal that either shows > green or yellow. (Disregarding the possibility of branches, when > quote a few more aspects can be shown.) are only two aspect, red and green, but the advanced warning which is passed by an electrical signal through the rails has three values so that the speed can be dropped exactly as you described. The speed drops to the v_max we have been discussing on that advance notification. There are essentially no branches but going back from approaching the signal on red to running under yellow happens when the train in front clears the track section ahead that controls the signal. > In all those cases, when you see yellow, at any point you must be able to > stop in time. The best way is to coast along and brake if your speed > would become larger than the speed that is allowed at that point, and > apply brakes at that point. (And I may note that coasting along with > a train in general does not reduce speed very much. It is common > practise in the Netherlands that trains coast along from some 20 km > before their destination and have to apply brakes in order to stop.) Exactly right, and light brakes may be needed on downhill gradients. > In block operation it is best to come to a full stop before the red light, > wait until it turns yellow and speed up as the aspects allow you. You > know that in that case you are approximately two blocks behind the previous > train, and if you stay that way you will only see green. Or red as you approach each signal changing to yellow just before you reach it each time. George === Subject: : Re: Integral Operator confusion >>I'm pretty confused about integral operators. This is a proposition >>in the book and I don't know how to prove it. Let S = Integral here, u >>is the measure for X and v the measure for Y. >>Let k(x,y) be a measurable function on (X x Y) of absolute operator tpe >>(p,q) >> Presumably a typo for absolute operator type (p,q). What's >> the definition of that? >Yes, that was a typo for absolute operator type (p,q). A measurable >function k(x,y) on (X x Y) is an integral operator of absolute type >(p,q) if |k| > is of operator type (p,q). Great. Now what does it mean to say that |k| is of operator type (p,q)? (My best guess is that it means that the operator with |k| for its kernel defines a bounded mapping from L^p to L^q. If that's the definition of |k| has operator type (p,q) then the question you asked is trivial, from the fact that in general we have |int f| <= int |f|. Is that the definition?) >Most of this is just sailing over my head. >> and g be an element of L^p (v). Then for almost all x in X the >>integral f(x) = S_Y k(x,y) * g(y) dv exists and the function f belongs >>to L^p (u) with ||f||_p <= || |k| ||_(p,q) * ||g||_q. >>TMH >> ************************ >> David C. Ullrich >TMH ************************ David C. Ullrich === Subject: : Dini's theorem I know the following version of Dini's theorem: Let T be a compact metric space and let f, f_1, f_2, ...: T --> IR be continuous functions with f_1 <= f_2 <= ... and f = sup f_n. Then (f_n) converges uniformly to f. My question is: If T=IR and I know that all f_n are bounded on IR cup {infty} (i.e. they have a limit lim_{t-->infty} f_n(t)=a_ninfty}f(t)=a> Hi there >> I am having problems computing an integral >> Integral (0 to infinity) x^a / (1+x^2) dx where -1> for analytic fucntion z^a/(1+z^2) >> In the upper half plane, I can pick a branch f(z) = exp (a (log|z| + i >> arg (z)) >> so using contour integration I should be able to find a solution for my >> question >> B'______A'__0__A______B >> (being the real line in the complex plane, let G be a semi circle >> joining B to B' and g be a semi circle joing A' to A, both centred at >> 0, so now I have a contour B'A' u g u AB u G) >> with a theorem (jordan lemma i believe) the integrals of f(z) wrt to >> the paths G and g tend to 0 . >I don't know what the Jordan lemma is, It's for example in Churchill, I think under that name. I forget the exact hypotheses, it's a special case of the dominated convergence theorem (a special case with a trivial proof). > but those integrals do >tend to 0, assuming A -> 0+ and B -> oo (which you never stated). >> so I go about calculating the residue which occurs at z=i (z=-i not in >> my contour so dont have to worry about that) >> residue of f(z) at i := 1/2i *exp(a*Pi*i/2) >> so by the cauchy residue formula I have >> [integral (path AB) f(z) dz] + [integral (path G) f(z) dz] + [integral >> (path B'A') f(z) dz] + [integral (path g) f(z) dz ]=Pi*exp(a*Pi*i/2) >> so from the jordan lemma >> [integral (path AB) f(z) dz] +[integral (path B'A') f(z) dz] = >> Pi*exp(a*Pi*i/2) >No, the limit of the left hand side equals the right hand side. >> but now I'm a little stuck, im not sure what I have to do to complete >> the question >The limit of the AB integral is what you want. The B'A' integral >is a constant times the AB integral; figure out that constant and >you'll have it. >> (I know the solution is Pi/(2*cos(a*Pi/2)) >> any help in the next step would be much appreciated >> TIA ************************ David C. Ullrich === Subject: : Re: JSH: Crowd mentality, consensus, and fraud >If you look at a few things that I've done, it's impossible not to >wonder what's going on here that people are getting away with calling >me a crackpot, and acting as if I've accomplished nothing. If one looks at _all_ the posts you've made, including of course the ones you eventually cancelled, it's impossible not to notice that you've been threatening legal action off and on for years. We've heard it all before - we want to know when the trial starts! I haven't even got my subpoena yet. >You people live in your lie as long as you think you can, and when it >is revealed and the consequences come down, there will be no mercy >because people will read post after post, where people here from the >math community, showed their true colors. >Remember with Enron? Remember the taped phone calls? >Well, your posts here are the equivalent. >Your public statements are not only freely available for use by >reporters around the world, but they are also admissable in courts of >law around the world. >My role here has been, investigator for the prosecution. >My role was to use whatever means were necessary that were ethical and >within the bounds of the law to fully reveal the full extent of the >corruption within the math field. >The most important thing in any prosecution is an air-tight case. >I needed as much on the record as possible, as the best witnesses are >mathematicians themselves. >You are my voice. Your posts here are my work product. >James Harris ************************ David C. Ullrich === Subject: : Re: Riemann Mapping question >I'm having trouble getting my teeth into this problem. >Let f be the riemann mapping from the square with vertices 1+i, 1-i, >-1+i, -1-i >to the unit disk, determined by f(0)=0 and f ' (0)>0. >I would like to prove that this is a meromorphic function, and would >also like to find the poles and zeros of this map. >1) I can see that using the Riemann mapping theorem I can get a unique >map from the square to the circle, and that by applying schwarz >reflections (and showing they agree on the overlap) that I can get the >squares to cover the complex plane, which I think obviously (well >visually I can see this!) shows that this set of circles also cover the >complex plane. I have no idea what you're talking about, sets of circles covering the plane. I _do_ know what set of squares covering the plane you're talking about, I think... I suspect you've got something screwed up somewhere. Say Q_0 is the original square, and Q_1 is the square just to the right. Your original map takes Q_0 to the unit disk. Now if you use reflection to extend it to Q_0 union Q_1, can you describe in a few words how the extended map behaves on the union of those two squares? >The problem is, I'm uncertain whether this is showing that f DOES >actually extend to a meromorphic function >2) With the aid of a diagram I can see that the zeros of the map are >images of 0. >im pretty sure that not all the maps of 0 (2a+2b*i where a,b element of >the integers) are zeros, but im confused to how to progress with this >question. >any help would be much appreciated ************************ David C. Ullrich === Subject: : Re: Riemann Mapping question Let f be the riemann mapping from the square with vertices 1+i, 1-i, >-1+i, -1-i >to the unit disk, determined by f(0)=0 and f ' (0)>0. >I would like to prove that this is a meromorphic function, and would >also like to find the poles and zeros of this map. >1) I can see that using the Riemann mapping theorem I can get a unique >map from the square to the circle, and that by applying schwarz >reflections (and showing they agree on the overlap) that I can get the >squares to cover the complex plane, which I think obviously (well >visually I can see this!) shows that this set of circles also cover the >complex plane. > I have no idea what you're talking about, sets of circles covering > the plane. I _do_ know what set of squares covering the plane > you're talking about, I think... I really shouldnt do complex analysis at 4 in the morning! On reviewing this question this morning, I realised what I said about the circles makes NO sense at all! (they wouldnt be circles!) > I suspect you've got something screwed up somewhere. Say Q_0 is the > original square, and Q_1 is the square just to the right. Your > original map takes Q_0 to the unit disk. Now if you use reflection > to extend it to Q_0 union Q_1, can you describe in a few words how > the extended map behaves on the union of those two squares? the map doesnt act conformally on the union of the squares (2-1 map) so f ' (0) = 0 on the boundaries. I think that just tells me the zeros of the derivative of f. Intutively (though I cant explain this -would really appreciate if you could tell me why) I feel that the reflection of the centre of the square, is a pole. >The problem is, I'm uncertain whether this is showing that f DOES >actually extend to a meromorphic function >2) With the aid of a diagram I can see that the zeros of the map are >images of 0. >im pretty sure that not all the maps of 0 (2a+2b*i where a,b element of >the integers) are zeros, but im confused to how to progress with this >question. >any help would be much appreciated > ************************ > David C. Ullrich === Subject: : Re: Riemann Mapping question >>I'm having trouble getting my teeth into this problem. >>Let f be the riemann mapping from the square with vertices 1+i, 1-i, >>-1+i, -1-i >>to the unit disk, determined by f(0)=0 and f ' (0)>0. >>I would like to prove that this is a meromorphic function, and would >>also like to find the poles and zeros of this map. >>1) I can see that using the Riemann mapping theorem I can get a unique >>map from the square to the circle, and that by applying schwarz >>reflections (and showing they agree on the overlap) that I can get the >>squares to cover the complex plane, which I think obviously (well >>visually I can see this!) shows that this set of circles also cover the >>complex plane. >> I have no idea what you're talking about, sets of circles covering >> the plane. I _do_ know what set of squares covering the plane >> you're talking about, I think... >I really shouldnt do complex analysis at 4 in the morning! On reviewing >this question this morning, I realised what I said about the circles >makes NO sense at all! (they wouldnt be circles!) >> I suspect you've got something screwed up somewhere. Say Q_0 is the >> original square, and Q_1 is the square just to the right. Your >> original map takes Q_0 to the unit disk. Now if you use reflection >> to extend it to Q_0 union Q_1, can you describe in a few words how >> the extended map behaves on the union of those two squares? >the map doesnt act conformally on the union of the squares (2-1 map) No, it's 1-1 on the union of those two squares! Or maybe it's 2-1 at a few boundary points or something, but I get the idea that you think that f(Q_1) = f(Q_0). That's definitely not so. What _is_ f(Q_1)? >so f ' (0) = 0 on the boundaries. I have no idea what f'(0) = 0 on the boundaries means. But never mind... > I think that just tells me the >zeros of the derivative of f. >Intutively (though I cant explain this -would really appreciate if you >could tell me why) >I feel that the reflection of the centre of the square, is a pole. That's true. That's _immediate_ from the Schwarz reflection principle! Can you state exactly the version of the reflection principle that you're using here? >>The problem is, I'm uncertain whether this is showing that f DOES >>actually extend to a meromorphic function >>2) With the aid of a diagram I can see that the zeros of the map are >>images of 0. >>im pretty sure that not all the maps of 0 (2a+2b*i where a,b element of >>the integers) are zeros, but im confused to how to progress with this >>question. >>any help would be much appreciated >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: : Re: Riemann Mapping question I'm having trouble getting my teeth into this problem. >>Let f be the riemann mapping from the square with vertices 1+i, 1-i, >>-1+i, -1-i >>to the unit disk, determined by f(0)=0 and f ' (0)>0. >>I would like to prove that this is a meromorphic function, and would >>also like to find the poles and zeros of this map. >>1) I can see that using the Riemann mapping theorem I can get a unique >>map from the square to the circle, and that by applying schwarz >>reflections (and showing they agree on the overlap) that I can get the >>squares to cover the complex plane, which I think obviously (well >>visually I can see this!) shows that this set of circles also cover the >>complex plane. >> I have no idea what you're talking about, sets of circles covering >> the plane. I _do_ know what set of squares covering the plane >> you're talking about, I think... >I really shouldnt do complex analysis at 4 in the morning! On reviewing >this question this morning, I realised what I said about the circles >makes NO sense at all! (they wouldnt be circles!) >> I suspect you've got something screwed up somewhere. Say Q_0 is the >> original square, and Q_1 is the square just to the right. Your >> original map takes Q_0 to the unit disk. Now if you use reflection >> to extend it to Q_0 union Q_1, can you describe in a few words how >> the extended map behaves on the union of those two squares? >the map doesnt act conformally on the union of the squares (2-1 map) > No, it's 1-1 on the union of those two squares! Or maybe it's > 2-1 at a few boundary points or something, but I get the idea > that you think that f(Q_1) = f(Q_0). That's definitely not so. > What _is_ f(Q_1)? f(Q_0) = Q_1, where Q_1 is Q_0 reflected wrt the right hand side of Q_0 so at the intersection of Q_1 and Q_0 it is just the shared boundary the right hand side of Q_1 is the f(applied to the Q_0 left hand side boundary) its basically a piece of paper folded opened out. (I hope you can see what im trying to say) >so f ' (0) = 0 on the boundaries. > I have no idea what f'(0) = 0 on the boundaries means. > But never mind... at the boundaries the map is no longer conformal (from above they overlap 2 -1 map at the boundaries) > I think that just tells me the >zeros of the derivative of f. >Intutively (though I cant explain this -would really appreciate if you >could tell me why) >I feel that the reflection of the centre of the square, is a pole. > That's true. That's _immediate_ from the Schwarz reflection principle! > Can you state exactly the version of the reflection principle that > you're using here? my version of the schwarz reflection principle Q_0, Q_1 are domains s.t Q_0 n Q_1 =g, where g is an arc of a circle or a straight line S_g(Q_0) =Q_1 (will define S_q below) suppose f:Q_0 u g ---> Q_1 u g cts ly so that it is holomophic in Q_1 definition of the schwartz reflection suppose g is a straight line or arc of a circle S_g(z) = h^-1 conjugate (h(z)) where h is a fractional linear transformation which takes g to a segment of the real line, h(z) c Real line ah, I see what you mean by it is immediate from the schwarz reflection principle that f, will have a pole (as f(0)=0) => meromophic. I think i've over complicated the problem. Because it all seems rather simple now. (i hope), With this function f, i get a checkboard pattern. so above, below, to the left, to the right, of every square with a zero, they have poles, and vice versa (hopefully from the diagram below, you can see what im thinking) Q_30 ; Q_31 ; Q_32 : Q_33 Q_20 ; Q_21 ; Q_22 : Q_23 Q_10 ; Q_11 ; Q_12 : Q_13 Q_00 ; Q_01 ; Q_02 : Q_03 say Q_00 has a zero, then does Q_02, Q_20, etc to the right and above Q_00 (Q_01, Q_10) they have poles. I really appreciate all the time you have put in to helping me understand this >>The problem is, I'm uncertain whether this is showing that f DOES >>actually extend to a meromorphic function >>2) With the aid of a diagram I can see that the zeros of the map are >>images of 0. >>im pretty sure that not all the maps of 0 (2a+2b*i where a,b element of >>the integers) are zeros, but im confused to how to progress with this >>question. >>any help would be much appreciated >> ************************ >> David C. Ullrich > ************************ > David C. Ullrich === Subject: : Re: (nondirected) Colimit in the category of rings > The coproduct of a pair of rings is not the tensor product. The only > tensor product which makes sense is that of the Z-algebras, and this is > not the same (it is the coproduct of the Z-algebras though). This is another mistake; Z-algebras = rings. The tensor product is the coproduct in the category of commutative R-algebras over a commutative ring. If the algebras are allowed to be noncommutative, the tensor product is universal in the cones Ri->S, where f1(x1)f2(x2)=f2(x2)f1(x1) ; this is in Jacobson. If the cones are allowed to be arbitrary, to get a universal cone one must take arbitrary words rather than just pairs as in the tensor product. === Subject: : Re: (nondirected) Colimit in the category of rings >> The coproduct of a pair of rings is not the tensor product. The only >> tensor product which makes sense is that of the Z-algebras, and this is >> not the same (it is the coproduct of the Z-algebras though). >This is another mistake; Z-algebras = rings. The tensor product is the >coproduct in the category of commutative R-algebras over a commutative >ring. If the algebras are allowed to be noncommutative, the tensor >product is universal in the cones Ri->S, where >f1(x1)f2(x2)=f2(x2)f1(x1) ; this is in Jacobson. If the cones are >allowed to be arbitrary, to get a universal cone one must take >arbitrary words rather than just pairs as in the tensor product. What's a ne? Lee Rudolph === Subject: : Re: Lying about the distributive property > ... > Like > > 7*(h(x) + b)*(g(x) + d) = (f(x) + 7*b)*(g(x) + d) > > which is actually easy in a way as you can just divide g(x) + d from > both sides, so let's make it slightly harder and have > > 7*P(x) = (f(x) + 7*b)(g(x) + d) > > 7*(x - 7) = (sqrt(7x) + 7)(sqrt(7x) - 7). > Yeah, and you picked d not coprime to 7, so that both have sqrt(7) as a > factor if you're in the ring of algebraic integers, but so? Ok, let me try something different (although you did *not* have that restriction in your original formulation). Take P(x) = (x + 1)(x + 8) and further: f(x) = 0 if x = 0 else it is x + 1 g(x) = 0 if x = 0 else if is 7x - 1 we now have: 7*P(x) = (f(x) + 7)(g(x) + 8). Is f(x) divisible by 7? All of f(x), g(x) and P(x) satisfy the requirements you have set. But only for x = 0 the left factor is divisible by 7, for other x it is the right factor that is divisible by 7. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: Torkel =?ISO-8859-1?Q?Franz=E9n_is_dead?= > Prevent confusions about Godel's Theorem? Just open his last text and > he talks about Godel's Theorem not requiring self-reference, then cites > a proof that refers to a certain set not being r.e. who's proof is > where the self-reference occurs. Poor Torkel makes one attempt at > being original and he flubs it. It seems you read Torkel's book the same way you read the various books on recursion theory and incompleteness you listed some time back. Here's the relevant passage from Torkel's book: G.9adel's proof of the first incompleteness theorem and Rosser's strenghtened version have given many the impression that the theorem can only be proved by constructing self-referential statements in the language of S, or even that only strange self-referential statements are known to be undecidable in elementary arithmetic. To counteract such impressions, we need only introduce a different kind of proof of the first incompleteness theorem. One argument is as follows. By the Matiyasevich-Robinson-Davis-Putnam theorem, which has been referred to a couple of times already, there is no algorithm that given any Diophantine quation D(x_1, ..., x_n) = 0 will decide whether or not the equation has a solution. But then there can be no theory that correctly decides every statement of the form the Diophantine equation D(x_1, ..., x_n) = 0 has at least one solution. ... A reader who is disinclined to digest the reasoning in these proofs of the first incompleteness theorem should just keep in mind the essential point that incompleteness is not restricted to arithemtical translations of strange self-referential statements. Indeed, we know that systems S to which G.9adel's theorem applies have undecidable statements of the form the Diophantine equation D(x_1, ..., x_n) = 0 has no solutions. A few paragraphs later Torkel also notes that there is a proof due to Kripke that relies on neither self-refential statements or recursion theory, but rather on non-standard models of arithmetic. Presumably you think that all this is balderdash and the MRDP theorem relies on self-reference in some way. This is not so, and the existence of a r.e. but non-recursive set is proved by diagonalization, involving no self-reference of any sort. Why you think Kripke's proof relies on self-reference I don't understand, as it doesn't even involve diagonalization in any obvious sense. Probably you simply don't have the first idea about what you're talking about. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: : Re: Plot of f(kt) against f(t) > I have two non-linear functions; f(t) and g(kt). I am predicating that > f(t) approx g(kt) and would like to estimate the constant k, given the > numerical values of f(t) and g(kt) for all t. One idea I had was to > obtain a scatter plot of g(kt) against f(t). Does anyone know what the > properties of such a plot would be? Could I fit a smooth profile to > the points and subsequently use them to estimate k? > Peter. An idea (untested!) that occurred to me was to form the linear correlation coefficient r(f(t),g(kt)) and to maximize r(k) (assuming that r is positive and close to +1; otherwise, minimize r(k)). N. Shamsundar University of Houston === Subject: : Re: Plot of f(kt) against f(t) > I have two non-linear functions; f(t) and g(kt). I am predicating that > f(t) approx g(kt) and would like to estimate the constant k, given the > numerical values of f(t) and g(kt) for all t. One idea I had was to > obtain a scatter plot of g(kt) against f(t). Does anyone know what the > properties of such a plot would be? Could I fit a smooth profile to > the points and subsequently use them to estimate k? > Peter. An idea (untested!) that came to me was that you could consider the linear correlation coefficient r(f(t),g(kt)) to be a nonlinear function of k, and maximize r w.r.t. k. N. Shamsundar University of Houston === Subject: : alternative definition of group? Let * be a binary operation on G, which satisfies (1) (x*y)*z = x*(y*z) for all x, y, z in G (2) for all x, y in G, there exists z such that x*z=y (3) for all x, y in G, there exists z such that z*x=y Show that G is a group (here the definition of group is the classical one: it has identity element and inverse) === Subject: : Re: alternative definition of group? > Let * be a binary operation on G, which satisfies > (1) (x*y)*z = x*(y*z) for all x, y, z in G > (2) for all x, y in G, there exists z such that x*z=y > (3) for all x, y in G, there exists z such that z*x=y > Show that G is a group (here the definition of group is the classical > one: it has identity element and inverse) Let e_x be an element of G such that e_x*x = x. If _y_ as any element of G, take _z_ in G such that y = x*z. Then e_x*y = e_x*(x*z) = (e_x*x)*z = x*z = y. So, if e = e_x, this proves that, for all _y_ in G, e*y = y. The same argument proves that there's some e' in G such that, for all _y_ in G, y*e' = y. But then e = e*e' = e'. So, there's an identity in G. To get an inverse of an element _x_ in G, take x' such that x*x' = e. You know that there's some element x'' in G such that x''*x = e. Then x'' = x''*e = x''*(x*x') = (x''*x)*x' = e*x' = x'. So, x' is an inverse for _x_. Jose Carlos Santos === Subject: : Re: alternative definition of group? > Let * be a binary operation on G, which satisfies > (1) (x*y)*z = x*(y*z) for all x, y, z in G > (2) for all x, y in G, there exists z such that x*z=y > (3) for all x, y in G, there exists z such that z*x=y > Show that G is a group (here the definition of group is the classical > one: it has identity element and inverse) I believe you need to add the assumption that G contains at least one element. Otherwise the statements (1)-(3) could be vacuously true. === Subject: : Re: Question About Proof that Pi is Irrational > Is there a proof of pi being irrational WITHOUT resorting to a proof > that it is transcendental? > Yes, there is a very simple proof of the irrationality of pi > that uses nothing more advanced than integration by parts. > It's due to Niven, and can be found in many introductory > Number Theory textbooks. It can also be found in calculus textbooks. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: : Re: Question About Proof that Pi is Irrational >> Is there a proof of pi being irrational WITHOUT resorting to a proof >> that it is transcendental? >> Yes, there is a very simple proof of the irrationality of pi >> that uses nothing more advanced than integration by parts. >> It's due to Niven, and can be found in many introductory >> Number Theory textbooks. > It can also be found in calculus textbooks. And also at , where it is phrased in convenient singable form. Kevin Wald wald@math.uchicago.edu | No, *super*natural logarithms. http://www.math.uchicago.edu/~wald | -- Willow, By Napier's Bones === Subject: : Re: Question About Proof that Pi is Irrational > Is there a proof of pi being irrational WITHOUT resorting to a proof > that it is transcendental? The standard proof is to find a series that converges faster than allowed (Liouville style). -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: : Re: Question About Proof that Pi is Irrational Michael Ejercito (in part): >> Is there a proof of pi being irrational WITHOUT resorting >> to a proof that it is transcendental? > The standard proof is to find a series that converges > faster than allowed (Liouville style). First, he asked for a proof WITHOUT resorting to a proof that it is transcendental. Second, pi is not a Liouville number -- it's Liouville-Roth constant is not infinity, but is instead at most 8.0161: http://mathworld.wolfram.com/IrrationalityMeasure.html http://bbs.sachina.pku.edu.cn/Stat/Math_World/math/l/l336.htm http://users.cs.dal.ca/~jborwein/pi-culture.pdf [see p. 12] (First error is excusable since your native language is German.) Dave L. Renfro === Subject: : Re: parametrization of 3x3 covariance matrix > Consider a 3x3 covariance matrix C with matrix elements > c(i,j) = c(j,i) (i,j = 1,2,3). > Can I write, without loss of generality, that: > c(i,i) >= 0 > and > c(i,j) = p(i) * sqrt( c(i,i) * c(j,j) ) > with > -1 <= p(i) <= +1 . > Is Det(C) >= 0 automatic for this parametrization? > Onyee There's an interesting example in Bierman's book on Kalman Filtering: if P = ( 1 0.5-d 0.5-d) ( 0.5-d 1 -0.5+d) ( 0.5-d -0.5+d 1 ) Then the correlations (your p(i,j) lie strictly between -1 and 1 iff -0.5 < d < 1.5. But det( P) = 2d*(1.5-d)*(1.5-d), so that for P to be positive definite (and so a proper covariance matrix) we must also have d>0. Duncan === Subject: : Re: parametrization of 3x3 covariance matrix Am 22.05.2006 15:33 schrieb Duncan Muirhead: >> Consider a 3x3 covariance matrix C with matrix elements >> c(i,j) = c(j,i) (i,j = 1,2,3). >> Can I write, without loss of generality, that: >> c(i,i) >= 0 >> and >> c(i,j) = p(i) * sqrt( c(i,i) * c(j,j) ) >> with >> -1 <= p(i) <= +1 . >> Is Det(C) >= 0 automatic for this parametrization? >> Onyee > There's an interesting example in Bierman's book on Kalman > Filtering: if > P = ( 1 0.5-d 0.5-d) > ( 0.5-d 1 -0.5+d) > ( 0.5-d -0.5+d 1 ) > Then the correlations (your p(i,j) lie strictly between > -1 and 1 iff -0.5 < d < 1.5. But det( P) = 2d*(1.5-d)*(1.5-d), so > that for P to be positive definite (and so a proper covariance > matrix) we must also have d>0. > Duncan also the arccos(r_ij) of the correlations r_ij require analoguous conditions. Since the arccos()-functions returns the length of distances between the head of two vectors (the vectors, of which the correlation is in question and is expressed by their cosine) on the sphere-surface, the sum of 2 lengthes cannot be smaller than the maximal length, since they must form a triangle on the surface. So transforming the correlation elementwise from r_ij -> d_ij getting 0 r_12 r_13 r_21 0 r_23 r_31 r_32 0 ---> by d_ij = arccos(r_ij)---> 0 d_12 d_13 d_21 0 d_23 d_31 d_32 0 you can check contradictions by simply summing d_12 + d_13 which must be > d_23, for instance. For greater matrices I have used that sometimes to detect the most likely position, if by transmitting correlation-matrices via pen&paper or by rounding, an error was introduced. Gottfried Helms === Subject: : Re: JSH: Fraud question is pertinent <87fyj4xy21.fsf@phiwumbda.org> <87r72mwrew.fsf@phiwumbda.org> Your paper was never sent out for review. After a few >> months it was accidentally put in a to-be-published file >> instead of a reject file, as intended. There was no peer >> review. Argyros' attempt after the fact to send you Dale >> Hall's note as if it were a review was ludicrous and utterly >> dishonest. >> This seems to be pure speculation. Have you any evidence that >> something like this happened? > Yes. > Uh huh. Sure. And what is that evidence? A copy of a note from Argyros. Marcus. > Just to be clear: It's the best story I could come up with isn't > evidence. > -- > Jesse F. Hughes > It's easy folks. Just talk about my approach to your favorite > mathematician. If they can't be interested in it, they've > demonstrated a lack of mathematical skill. -- James Harris === Subject: : Re: JSH: Fraud question is pertinent > Your paper was never sent out for review. After a few >months it was accidentally put in a to-be-published file >instead of a reject file, as intended. There was no peer >review. Argyros' attempt after the fact to send you Dale >Hall's note as if it were a review was ludicrous and utterly >dishonest. >>This seems to be pure speculation. Have you any evidence that >>something like this happened? > Yes. >>Uh huh. Sure. And what is that evidence? > A copy of a note from Argyros. > Marcus. >>Just to be clear: It's the best story I could come up with isn't >>evidence. >>-- >>Jesse F. Hughes >>It's easy folks. Just talk about my approach to your favorite >>mathematician. If they can't be interested in it, they've >>demonstrated a lack of mathematical skill. -- James Harris I'm curious about this. As the unsolicited editor in the story, I had wondered what had actually transpired between receipt and acceptance of this abomination. My assumption had been that it was passed around, with no one able to make heads or tails of what the paper actually said, but no one having the guts to label the paper as the complete and total gibberish it is. I have never been satisfied with this. Your version does seem to be an interpretation of events that fits the facts reasonably, but I've never seen any corroborating evidence. Perhaps you could attempt to be a little less cryptic, without betraying any confidential information? For instance, are you at all acquainted (personally or professionally) with Argyros? Could you describe this note of his (i.e., was it a note to you, the note he Dale === Subject: : Re: JSH: Fraud question is pertinent Discussion, linux) <87fyj4xy21.fsf@phiwumbda.org> <87r72mwrew.fsf@phiwumbda.org> Your paper was never sent out for review. After a few > months it was accidentally put in a to-be-published file > instead of a reject file, as intended. There was no peer > review. Argyros' attempt after the fact to send you Dale > Hall's note as if it were a review was ludicrous and utterly > dishonest. >> This seems to be pure speculation. Have you any evidence that > something like this happened? > Yes. >> Uh huh. Sure. And what is that evidence? > A copy of a note from Argyros. That would count as evidence, I suppose. Though, one would like to know more: a note to whom? What did it say? And so on. Argyros doesn't seem like a disinterested observer of the whole affair. -- [I want to] stand at the pinnacle of human achievement with no one else in all of history even close, no human being having faced what I have--and survived. Because when all is said and done, make no mistake, the simple truth is, I am better. --James S. Harris === Subject: : Re: Semiperimeter center or hopeless center of a triangle? joshipura@gmail.com escribi.97: > I am not a mathematician but I love the way you all here respond to my > curiosity. > I never heard of a center based on an idea, which I'd like to call it > a hopeless point or semiperimeter point. (This is not > isoperimetric point as mentioned at > http://faculty.evansville.edu/ck6/tcenters/recent/isoper.html ) > This point if exists for a triangle ABC may be at an intersection of > lines AXa, BXb and CXc, where point Xa is on BC such that |XaB|+|BA| = >> XaC|+|CA| = perimeter/2. > In other words, it is hopeless to find shorter route Xa-B-A or Xa-C-A > because distance from Xa to A via B or via C is the same. > What could be properties of such semiperimeter center? > If it exists, I'd like to call it the Bangalore point. Bangalore > downtown has triangular blocks and one has to always search for the > shortest distance around to reach the other end of the triangle. It is the Nagel Point (http://faculty.evansville.edu/ck6/tcenters/class/nagel.html), The points Xa, Xb and Xc are the tangency points of the ex-circles (tangents to one side and the prolongation of t other two). -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: : Re: Indra's Net of Pearls Kewlest kewl of all the Heavens hahaha :-)) === Subject: : Re: Minimum Spanning Tree on a Grid of Points > I want to find the minimum spanning tree of a grid of n*n points. > And an n*m grid. >> While finding minimum spanning trees in the general case is NP > complete, >> There are fast algorithms to determine minimum spanning trees in >> polynomial time. See, eg., Kruskal's Algoirthm, or Prim's Algorithm. >> Maybe your meaning of the term minimum spanning tree is different from >> the standard? >> Yes, sorry. I'll try again: >> While finding the Steiner tree in the general case in NP complete... > This might be worth a look: > MR1369336 (96i:90070) > Harris, Frederick C., Jr.(1-NV-C) > An introduction to Steiner minimal trees on grids. (English. English > summary) > Proceedings of the Twenty-sixth Southeastern International Conference on > Combinatorics, Graph Theory and Computing (Boca Raton, FL, 1995). > Congr. Numer. 111 (1995), 3--17. > 90C35 (05C05 05C35) It's excellent. Alun Harford === Subject: : regarding the correlation of two binary matrices I am a cs student who's also doing a project in this field. My project is: Input: satellite images from MSG European satellite. Those are colored images which come along with text files that physically represent every pixel in the image (i mean that every pixel has a record in the text file which says what is the IR, temp' reflectance & more in that specific pixel) - deriving the clouds from the complete image & building in c++ a very large 2D matrix which represent the image in binary: 1 for cloud existence & 0 otherwise. - correlating between two sequential images & deriving the clouds direction vector. That way i can speculate the clouds general motion (direction). My questions for you: 1.Can you suggest a better way to preform this actions? 2. I need to correlate between 2 binary matrices via 'cross correlation' method & i really don't know how to implement it in c++ since it's very complexed (honestly, i prefer finding a source file & invest my time surfing) 3. Based on your experience in this field: Does cross correlation is the best way to compute direction vector for those matrices (which reform every second..)? what about other block matching methods? Can you write down something that helps? I appreciate any help! Peace & sun shine forever, === Subject: : Re: Calculus XOR Probability Robert Low said: > Is the set of triples {x,y,z} such that y=2x+3z a set of pairs? Nope. > Is it a line? Yep. > Huh? How is this a line? I'd imagine a set of triples, if each element > is a real number, to be a subset of R^3, and unless I'm much mistaken > the set above forms a plane. Perhaps you can enlighten us, Tony, how > you get a line from this... > Look at it sideways on? You could stick your eye on the plane and it would appear as a line. I should have said x=2y=3z. Then you have a set of triples, not pairs, defining a line. -- Smiles, Tony === Subject: : Re: Calculus XOR Probability Robert Low said: > Perhaps you can enlighten us, Tony, how > you get a line from this... > Look at it sideways on? > Oh yeah? Where from? > known mathematical universe, so he must be able to look > at R^3 from outside, too. Higher dimensional geometry has been a passion of mine since my mid teens, but that's a whole other matter. :) -- Smiles, Tony === Subject: : Re: Calculus XOR Probability > Robert Low said: > Perhaps you can enlighten us, Tony, how you get a line from > this... > Look at it sideways on? > Oh yeah? Where from? known mathematical universe, so he must be able to look at R^3 from > outside, too. > Higher dimensional geometry has been a passion of mine since my mid > teens, but that's a whole other matter. :) TO's passion for geometry, of every dimension, seems to have been more honored in the breech than in th'observance. === Subject: : Re: Calculus XOR Probability stephen@nomail.com said: > Mike Kelly said: > Everything you just said implies each ELEMENT is finite and says > nothing about the size of the set. > No? In an actually infinite set, such as the reals in [0,1], there are elements >> (points) with an infinite number of elements between them. Unless there exist >> such pairs of elements, the set is not actually infinite in my book, so the set >> of finite naturals is not truly infinite to me. >> You're confusing dense with infinite. > I think that, for Tony, actually infinite implies dense. >> In Tony's book {1, 2, 3, ... Big'Un} is an actually infinite set >> because there are an infinite number of elements between 1 >> and Big'Un. However I do not think he thinks it is a dense >> set. >> By Tony's definition { 1, 2, 3, ..., w } (where w stands for omega) >> is an actually infinite set. {1, 2, 3, ... } is not >> actually infinite, but if you add a single element it >> becomes actually infinite. > I don't think you can fool Tony quite that easily. The set {1, 2, 3, > ... } just goes off into an indefinite fog to the right, but none of > the numbers in it become Tinfinite, so it's Tfinite. Adding a single > element won't change that. Similarly, putting two > Tfinite-but-Tunbounded sequences back to back doesn't work: > Oh I am quite sure he will change his definition of actually infinite > to exclude this example. Of course, his definition of actually infinite, > like all his definitions of infinite is somewhat circular, so it > is not really a definition at all. Hand waving. > He said: >> In an actually infinite set, such as the reals in [0,1], there are elements >> (points) with an infinite number of elements between them. > This is meaningless because it does not define what he means > by an infinite number of elements between two elements. > Using the standard definition of infinite, there are an infinite > number of elements between 1 and w. But Tony is using some > different definition of infinite which he is unable to > articulate and only he can comprehend. Is w a member of the set? No. So, that statement doesn't apply to the elements of the set, does it? > notion of the number line - something much bigger than the > mathematical set of integers, because it just goes on and on, even > after reaching Tinfinity and stuff like that. If you look at a > (normal) number line in front of you, you can't see the ends (because > they don't exist), and step back as far as you like, you still can't > see them (the (nonexistent) ends). But just magically step back, uh, > infinitely far, and then the only reason you can't see them might be > because your glasses aren't good enough. Get new glasses - zoom in to > the right - whatever - you'll then be able to start building more stuff > on the end of the original unending line. I'm channelling again... > pzzzt. > Brian Chandler > http://imaginatorium.org > Tony's whole idea is based on the number line. It is rather > silly for him to be arguing about set theory, because his > theory is really not about sets. He is trying to present > some sort of number line theory. In his theory, all numbers > are on the number line. Every point on the line is a number, > and every number is a point (or sub point, or sub-sub point or > whatever). No matter where you are on the line, there are points > to the left and right. This is why he thinks omega is a contradiction. > The idea of a number without a predecessor is impossible in > his number line. Of course no one has ever claimed that omega > is on the number line, or that it has to be, but according > to Tony it must be. If it's supposed to be a size of the set, a count of elements, then it's some kind of hyperreal value. It's a quantity xor it's not on the real line. > Tony's line never ends, but when you talk about it you > have to pick arbitrary end points for the purpose of discussion, > and these endpoints can be infinitely far from each other. > This is where his theory quickly degenerates into an > incoherent jumble. It's not a matter of that, but choosing a point infinitely far away as a unit, just like we choose a point an infinite number of points from the origin as a finite unit. > The way his theory handles sets is to consider all the > numbers in some segment of his number line that have > some property. This of course requires you to pick > some end points. If you want to talk about all the > prime numbers, you have to pick some upper bound, such > as Big'Un first. Of course there are apparently prime > numbers larger than Big'Un, but you have to ignore them, > despite the fact you wanted to talk about all the prime numbers. > You can always make Big'Un larger apparently, to get more > of them, because it is just a variable, or something. It's a variable bound, without which one cannot gauge the realtive sizes of boundless sets. > Sets of non-numbers really are not handled at all, other > than by vigorous hand waving. Symbolic sets are very well handled, thank you, and other inductively defined sets can be related to these two types. > Stephen -- Smiles, Tony === Subject: : Re: Calculus XOR Probability > stephen@nomail.com said: In Tony's book {1, 2, 3, ... Big'Un} is an actually infinite set > because there are an infinite number of elements between 1 > and Big'Un. However I do not think he thinks it is a dense > set. >> By Tony's definition { 1, 2, 3, ..., w } (where w stands for omega) > is an actually infinite set. {1, 2, 3, ... } is not > actually infinite, but if you add a single element it > becomes actually infinite. >> I don't think you can fool Tony quite that easily. The set {1, 2, 3, >> ... } just goes off into an indefinite fog to the right, but none of >> the numbers in it become Tinfinite, so it's Tfinite. Adding a single >> element won't change that. Similarly, putting two >> Tfinite-but-Tunbounded sequences back to back doesn't work: >> Oh I am quite sure he will change his definition of actually infinite >> to exclude this example. Of course, his definition of actually infinite, >> like all his definitions of infinite is somewhat circular, so it >> is not really a definition at all. > Hand waving. ??? Do you know what hand waving means? Your use of the word here suggest that you do not. Anyway, your definition of actually infinite contains the word infinite, which is why I said your definition was somewhat circular. Your definition of actually infinite is meaningless until you define what an infinite number of elements between two elements. >> He said: > In an actually infinite set, such as the reals in [0,1], there are elements > (points) with an infinite number of elements between them. >> This is meaningless because it does not define what he means >> by an infinite number of elements between two elements. >> Using the standard definition of infinite, there are an infinite >> number of elements between 1 and w. But Tony is using some >> different definition of infinite which he is unable to >> articulate and only he can comprehend. > Is w a member of the set? No. So, that statement doesn't apply to the elements > of the set, does it? w is a member of the set {1, 2, 3, ..., w }, which is the set I am talking about. You need to pay attention to what people actually write. Stephen === Subject: : Re: Calculus XOR Probability > stephen@nomail.com said: >> Mike Kelly said: >> Everything you just said implies each ELEMENT is finite and says >> nothing about the size of the set. > No? In an actually infinite set, such as the reals in [0,1], there are elements > (points) with an infinite number of elements between them. Unless there exist > such pairs of elements, the set is not actually infinite in my book, so the set > of finite naturals is not truly infinite to me. >> You're confusing dense with infinite. > I think that, for Tony, actually infinite implies dense. >> In Tony's book {1, 2, 3, ... Big'Un} is an actually infinite set >> because there are an infinite number of elements between 1 >> and Big'Un. However I do not think he thinks it is a dense >> set. >> By Tony's definition { 1, 2, 3, ..., w } (where w stands for omega) >> is an actually infinite set. {1, 2, 3, ... } is not >> actually infinite, but if you add a single element it >> becomes actually infinite. > I don't think you can fool Tony quite that easily. The set {1, 2, 3, > ... } just goes off into an indefinite fog to the right, but none of > the numbers in it become Tinfinite, so it's Tfinite. Adding a single > element won't change that. Similarly, putting two > Tfinite-but-Tunbounded sequences back to back doesn't work: >> Oh I am quite sure he will change his definition of actually infinite >> to exclude this example. Of course, his definition of actually infinite, >> like all his definitions of infinite is somewhat circular, so it >> is not really a definition at all. > Hand waving. >> He said: >> In an actually infinite set, such as the reals in [0,1], there are elements >> (points) with an infinite number of elements between them. >> This is meaningless because it does not define what he means >> by an infinite number of elements between two elements. >> Using the standard definition of infinite, there are an infinite >> number of elements between 1 and w. But Tony is using some >> different definition of infinite which he is unable to >> articulate and only he can comprehend. > Is w a member of the set? No. So, that statement doesn't apply to the elements > of the set, does it? >> > I think it's possible to put together a sort of picture of Tony's > notion of the number line - something much bigger than the > mathematical set of integers, because it just goes on and on, even > after reaching Tinfinity and stuff like that. If you look at a > (normal) number line in front of you, you can't see the ends (because > they don't exist), and step back as far as you like, you still can't > see them (the (nonexistent) ends). But just magically step back, uh, > infinitely far, and then the only reason you can't see them might be > because your glasses aren't good enough. Get new glasses - zoom in to > the right - whatever - you'll then be able to start building more stuff > on the end of the original unending line. I'm channelling again... > pzzzt. > Brian Chandler > http://imaginatorium.org >> Tony's whole idea is based on the number line. It is rather >> silly for him to be arguing about set theory, because his >> theory is really not about sets. He is trying to present >> some sort of number line theory. In his theory, all numbers >> are on the number line. Every point on the line is a number, >> and every number is a point (or sub point, or sub-sub point or >> whatever). No matter where you are on the line, there are points >> to the left and right. This is why he thinks omega is a contradiction. >> The idea of a number without a predecessor is impossible in >> his number line. Of course no one has ever claimed that omega >> is on the number line, or that it has to be, but according >> to Tony it must be. > If it's supposed to be a size of the set, a count of elements, then it's some > kind of hyperreal value. It's a quantity xor it's not on the real line. >> Tony's line never ends, but when you talk about it you >> have to pick arbitrary end points for the purpose of discussion, >> and these endpoints can be infinitely far from each other. >> This is where his theory quickly degenerates into an >> incoherent jumble. > It's not a matter of that, but choosing a point infinitely far away as a unit, > just like we choose a point an infinite number of points from the origin as a > finite unit. >> The way his theory handles sets is to consider all the >> numbers in some segment of his number line that have >> some property. This of course requires you to pick >> some end points. If you want to talk about all the >> prime numbers, you have to pick some upper bound, such >> as Big'Un first. Of course there are apparently prime >> numbers larger than Big'Un, but you have to ignore them, >> despite the fact you wanted to talk about all the prime numbers. >> You can always make Big'Un larger apparently, to get more >> of them, because it is just a variable, or something. > It's a variable bound, without which one cannot gauge the realtive sizes of > boundless sets. >> Sets of non-numbers really are not handled at all, other >> than by vigorous hand waving. > Symbolic sets are very well handled, thank you, and other inductively defined > sets can be related to these two types. Did you come up with a way to handle my set of curves on the Euclidean plane passing through a specified point? I don't recall you being able to classify that as either symbolic or inductively defined. Matt === Subject: : Re: Calculus XOR Probability Virgil said: > Is the set of triples {x,y,z} such that y=2x+3z a set of pairs? Nope. > Is it a line? Yep. > Huh? How is this a line? I'd imagine a set of triples, if each element > is a real number, to be a subset of R^3, and unless I'm much mistaken > the set above forms a plane. Perhaps you can enlighten us, Tony, how > you get a line from this... > > Look at it sideways on? Oh yeah? Where from? > Possibly from (3*Big'un, 6*Big'un, 2^Big'un), from which point he can, > one hopes, never return to any real space. Brian Chandler > http://imaginatorium.org Sorry, that should read y=2x=3z. A line in n-dimensional space is a set of n- tuples such that each is related to every other through a linear equation. So, my point stands, that the definition of a line as a set of pairs only applies in 2D space, and is therefore not general. -- Smiles, Tony === Subject: : Re: Calculus XOR Probability cbrown@cbrownsystems.com said: > Matt Gutting said: > Virgil said: > cbrown@cbrownsystems.com said: > For the last time, no. If the limit of the staircase is anything > different from the diagonal, which it is, then there is no > contradiction. >> There is no mathematically valid model in which the limit of the >> sequence of staircase functions is anything but the diagonal function. >> If TO wished to claim otherwise, then he must create and present to us >> the entire system in which he claims his allegations hold, as they do >> not hold in any current system. > > Okay. Here goes. > > Rather than a set of points, let us define both the staircase and the diagonal > as sequences of segments .... > In other words, rather than saying there is no natural number x such > that 2*x = 3, let us say there is no rational number x such that 2*x > = 3. In other words, let's defined a type of limit that lends itself to linear measure. You asked for a better definition of limit for this purpose. Well, here it is. Your discrepancy of sqrt(2) is explained yet again. > Does it surprise you that these two statements provide different > answers? Does it surprise you that your directionless point-wise definition of the limit gives an incorrect measure and mine does not? It shouldn't. > ... defined as a pair of reals which represent the x and y > coordinate differences between subsequent points. Let us compare the two thus > in a segment-wise manner, maintaining the same number of segments in each, and > see if the segments which describe the staircase approach those that describe > the diagonal. Where n=1, we have two segments to the staircase, {0,1} and > {1,0}, for a total change of {1,1}. Dividing the diagonal into two segments we > have {1/2,1/2} and {1/2,1/2}, also for a total change of {1,1}. Now, as n > increases we have {1,1}=sum(x=1->n: {1/n,0}+{0,1/n}) for the diagonal, and sum > {1,1}=(x=1->n: {1/2n,1/2n}+{1/2n,1/2n}) for the diagonal. While the locations > of the points in each segment become arbitrarily close, the vectors defining > the segments of which the lines are made never become close, but are always at > a 45 degree angle to their corresponding segments in the other line. > > When you look at the distance traveled, you sum all the x components of the > vectors in each line and sum all the y components, and you get {1,1} in both > cases, and the distance is sqrt(2). > > When you look at the lengths of each, you sum the length of each vector in the > line. For the staircase we have sum(x=1->n: 1/n+1/n)=2. For the diagonal we > have sum(x=1->n: 1/sqrt(2)+1/sqrt(2))=sqrt(2). Because of the difference in > vector direction, even at the infinitesimal scale, the staircase is longer than > the diagonal. > > Is that an entire enough system for you? :D > No, because all that I see you have done is to note (using > unnecessarily obfuscatory language) that the limit of the length of the > staircases is 2; which no one is disagreeing with; and that the length > of the diagonal is sqrt(2), which also no one is disagreeing with. And the legth of the limit of the staircases is 2, which you DO disagree with because you fail to see the obvious *linear* difference between that and the straight diagonal. > What is missing is a statement of /exactly what you mean/ by the > length of (the limit of the staircases) is {whatever you propose}. The limit of the staircases is the series Sum(n->oo: {1/n,0},{0,1/n}). That's n repetitions of a step with length 2/n, for a total length of 2. > In order for me to understand your answer, you must first state > /exactly what you mean/ by the limit of the staircases; which you > have not done in the above paragraphs. Is the limit of the staircases > a function? Is it a real number? A set of line segments? A set of pairs > of pairs in R^2 x R^2? A white elephant? I stated already it's a sequence of line segments. See above, defined as a pair of reals which represent the x and y coordinate differences between subsequent points. Each of those pairs represents a line segment. > The closest you get is this cryptic comment: Because of the difference > in vector direction, even at the infinitesimal scale, the staircase is > longer than the diagonal. But this doesn't tell me what the limit of > the staircases is; it simply mentions several (undefined) properties > you propose it to have. It's a staircase with oo stairs, each 2/oo long, given riser and tread. What is your question? > For example, presumably there is some point p = (a,b) in R^2 that is in > the limit of the staircases. Does that point satisfy b = 1 - a, or does > it not? The tread of one step meets the riser of the next at a point on the diagonal. Where the riser meets its tread, that corner is NOT on the diagonal, even if it may be only an infinitesimal difference away, and consider coincident with the line according to stard finitist limits. > Given that point p, what is the vector direction, at the infinitesimal > scale associated with it? Can we deduce it from the values of a and b? > For example, how do I determine the vector direction, at the > infinitesimal scale at the point (1/2,1/2) (which I presume is in the > limit of the staircases)? The point (1/2,1/2) is in every staircase for n>1, for sure. The direction of the tread before it is horizontal, and the diretcion of the riser after that point is vertical. Remember, directions are not defined for points, but for segments. That point has not direction of its own, hence the need to look at the limit, not of the points, but of the segments. > Given two points p and q in R^2 which are in the limit, how do I > determine whether p and q have the same or different vector > directions, at the infinitesimal scale? Points do not have directions, ultimately. The segment {1/2,0} is horizontal, and {0,1/2} is vertical. > Once you have addressed these questions, we can suppose that your > definition of the limit of the staircases is a mathematical object > called L. /Then/ I can evaluate a statement you might make of the > form the length of L is {whatever you propose}. Are you sure you won't ask the alreayd answered questions, again? > Until then, you haven't defined what you mean by the length of (the > limit of the staircases); all you have defined is the limit of (the > length of the staircases); and at least in its result, we are all in > agreement: the limit of the length of the staircases is 2, and the > length of the diagonal is sqrt(2). But you disagree that the limit of the staircases is anything other than the diagonal, whereas I have demonstrated a form of limit which shows clearly that there's a difference, and which accounts precisely for the error. > The remainder of your definition leaves me as desirous of a > definition as before: you have simply introduced new, undefined terms > to define a previously undefined term. This renders your definition no > more meaningful than it was before, mathematically speaking. Your failure to understand what I've defined as a limit isn't my problem. > First of all, it's not that the points become the same set in the limit. It's > that the limits of the two sets of points are identical (the same set of > points). Nothing becomes anything in the limit. > Your objection is semantic? Take it to alt.picky.english. > No, his objection is that you imply that for each point in the limit, > there is some /unique/ continuous /curve/ of points which can be > identified as becoming the point in the limit. > That is not a feature of the definition of limit I gave (nor is it, > in general, a feature of various other definitionsof limit I have > seen). Your definition of limit is inappropriate for this measure, and mine, whether you can understand the concept of defining a curve as the limit of a sequence of segments or not, is the correct notion of limit for this purpose, and does NOT give an erroneous answer. > It does not require a /unique/ sequence to be identified with each > point in the limit; simply that /at least one/ such sequence exists for > a point to be considered a limit point of the sequence of sets of pairs > in R^2. > Nor does it generate a continuous /curve/ of points which is associated > with a particular point in the limit; it provides a discrete /sequence/ > of points which converges to a point in the limit. Yes, and your directionless points are not adequate for measuring the curve. > > Second, the statement Virgil is making is not a leap; it's a consequence of > the definition of limit. Unless you have a different definition. > I just offered one that explains the discrepancy. Did you read any of it? Is > this supposed to explain why my limit definition doesn't make sense, as you > claimed in your next post to have shown? Nice hand waving. > Aside from the fact that you have not even provided a definition of > what kind of mathematical object the limit of the staircases is (a > set? a real number? an equation?), I don't see how your discussion of > limit above applies to anything that is not a collection of segments; > which is to say your definition is (at best) simply providing an > example, not providing a proper definition. I said first off it was a sequence of segments. If you can't follow the very first statement of the argument, then get some ritalin or something. Claiming I haven't said what I started out saying is disingenuous. > For example, let C_n = {(a,b): b = sin(n*a)/n^2}. According to my > definition of limit, lim n->oo {C_n} = {(a,b): b = 0}, as you should be > able to see for yourself by applying my definition. Yeah, and that'll get you another ty measure, which you will continue to blame on infinitididdit. This creationist math is deplorable. > Could you walk us through how your definition of limit applies to this > sequence? What do you claim you mean by lim n->oo {C_n} in this case? > Is it a set? Is it a function? Is it a real number? It would be a sequence of cycles, as wavelength and amplitude approach 0. It will also have infinitesimal nonlinearities to it, and the limit will appear to have a length with an error of pi. Big surprise. -- Smiles, Tony === Subject: : Re: Calculus XOR Probability > cbrown@cbrownsystems.com said: >> What is missing is a statement of /exactly what you mean/ by the >> length of (the limit of the staircases) is {whatever you propose}. > The limit of the staircases is the series Sum(n->oo: {1/n,0},{0,1/n}). That's n > repetitions of a step with length 2/n, for a total length of 2. You're assuming again that you can interchange the sum and the limit process. The length of the limit of staircases need not equal the limit of the length of staircases with the standard definition. Do you mean to say that the limit of the staircases is a series? That's how your sentence is phrased, but it doesn't seem to make sense. You're apparently making a sequence of geometric figures (staircases), then stating that the limit is an infinite series, presumably evaluated in the same way that infinite series typically are - although you need to be clearer about the meaning of n->oo (1/n,0),(0,1/n). How does the limit of a sequence of geometric figures get to be a sequence of real numbers? Or is that what you meant? >> In order for me to understand your answer, you must first state >> /exactly what you mean/ by the limit of the staircases; which you >> have not done in the above paragraphs. Is the limit of the staircases >> a function? Is it a real number? A set of line segments? A set of pairs >> of pairs in R^2 x R^2? A white elephant? > I stated already it's a sequence of line segments. See above, defined as a > pair of reals which represent the x and y coordinate differences between > subsequent points. Each of those pairs represents a line segment. So, each staircase is a sequence of line segments. How do you decide that the limit is also a sequence of line segments? >> The closest you get is this cryptic comment: Because of the difference >> in vector direction, even at the infinitesimal scale, the staircase is >> longer than the diagonal. But this doesn't tell me what the limit of >> the staircases is; it simply mentions several (undefined) properties >> you propose it to have. > It's a staircase with oo stairs, each 2/oo long, given riser and tread. What is > your question? *My* question is, since you haven't actually defined oo, how can you tell whether oo or 2/oo exist? >> For example, presumably there is some point p = (a,b) in R^2 that is in >> the limit of the staircases. Does that point satisfy b = 1 - a, or does >> it not? > The tread of one step meets the riser of the next at a point on the diagonal. > Where the riser meets its tread, that corner is NOT on the diagonal, even if it > may be only an infinitesimal difference away, and consider coincident with the > line according to stard finitist limits. >> Given that point p, what is the vector direction, at the infinitesimal >> scale associated with it? Can we deduce it from the values of a and b? >> For example, how do I determine the vector direction, at the >> infinitesimal scale at the point (1/2,1/2) (which I presume is in the >> limit of the staircases)? > The point (1/2,1/2) is in every staircase for n>1, for sure. The direction of > the tread before it is horizontal, and the diretcion of the riser after that > point is vertical. Remember, directions are not defined for points, but for > segments. That point has not direction of its own, hence the need to look at > the limit, not of the points, but of the segments. How do you know that the limit of the segments exists, and that it is a segment? >> Given two points p and q in R^2 which are in the limit, how do I >> determine whether p and q have the same or different vector >> directions, at the infinitesimal scale? > Points do not have directions, ultimately. The segment {1/2,0} is horizontal, > and {0,1/2} is vertical. Okay, so how about the infinitesimal scale? >> Once you have addressed these questions, we can suppose that your >> definition of the limit of the staircases is a mathematical object >> called L. /Then/ I can evaluate a statement you might make of the >> form the length of L is {whatever you propose}. > Are you sure you won't ask the alreayd answered questions, again? I still have questions about your answers to the questions. >> Until then, you haven't defined what you mean by the length of (the >> limit of the staircases); all you have defined is the limit of (the >> length of the staircases); and at least in its result, we are all in >> agreement: the limit of the length of the staircases is 2, and the >> length of the diagonal is sqrt(2). > But you disagree that the limit of the staircases is anything other than the > diagonal, whereas I have demonstrated a form of limit which shows clearly that > there's a difference, and which accounts precisely for the error. I don't see a clear definition of limit. Can you fill in the blanks here: DEFINITION: The limit of a _____ (insert name of mathematical object) is a ___ (insert name of a mathematical object) satisfying the following criteria: ______. Both blanks have to be filled with terms which either are agreed upon generally, or are defined in turn according to the template provided. Once you can fill in those blanks, then we have something we can talk about. Until then, your definition is not sufficiently well-formed to be able to discuss anything related to it. === Subject: : Re: Calculus XOR Probability Matt Gutting said: > David R Tribble said: >> MoeBlee said: >> On the other hand, this is a theorem in set theory (assuming that 'oo' >> is some already defined constant symbol and that '<' is some already >> defined 2-place relation symbol): >> Am(Ax(x>m -> f(x) > g(x)) -> (oo>m -> f(oo)>g(oo))) >> It's uninteresting, but at least correct. If there is anything >> (whatever oo is), then it would depend on whether the set of real >> numbers is a subset of the domains of the functions f and g and whether >> oo is a member of the domains of the functions f and g. > Yes, and if the functions f anf g are defined over all the reals, then they can > be considered to be defined in the limit as well, and formulaic expressions of > oo can therefore be compared and ordered in this way. That's what's > interesting. If 2*x2, and oo>2, then 2*oo> That's a neat trick, considering that oo is not a member of the reals. >> Apparently, defining a simple relation between the members of a >> set (such as defining 2x2 in R) can call into existence >> new values in the set that previously did not exist in the set before >> the relation was defined. If we are talking about the hyperreals, then that's not happening. You can > insist there are no infinite values on the real line, but that's just refusal > to consider a new idea. Values aren't called into existence. If those values > exist, then the formulaic expression applies to them. > Yes, but first, if you're not calling the values into existence, you have to > prove that they exist. > Matt No, I can make a statement that depends on whether they exist or not. (E oo A neN oo>n) ^ (E meN A n>m f(n)>g(n)) -> f(oo)>g(oo). So, IF an infinite value exists, and f(n)>g(n) for all n greater than some m, then we can conclude that for this infinite value oo, f(oo)>g(oo). Tada! Infinite induction. -- Smiles, Tony === Subject: : Re: Calculus XOR Probability <446c80c3$0$24200$88260bb3@free.teranews.com> Yes, and if the functions f anf g are defined over all the reals, then they can >> be considered to be defined in the limit as well, and formulaic expressions of >> oo can therefore be compared and ordered in this way. That's what's >> interesting. If 2*x2, and oo>2, then 2*oo> That's a neat trick, considering that oo is not a member of the reals. >> Apparently, defining a simple relation between the members of a >> set (such as defining 2x2 in R) can call into existence >> new values in the set that previously did not exist in the set before >> the relation was defined. >> If we are talking about the hyperreals, then that's not happening. You can >> insist there are no infinite values on the real line, but that's just refusal >> to consider a new idea. Values aren't called into existence. If those values >> exist, then the formulaic expression applies to them. Matt Gutting said: >> Yes, but first, if you're not calling the values into existence, you have to >> prove that they exist. > No, I can make a statement that depends on whether they exist or not. (E oo A > neN oo>n) ^ (E meN A n>m f(n)>g(n)) -> f(oo)>g(oo). So instead of proving that oo exists in the system of standard arithmetic, you instead have added an axiom that defines oo. The result, of course, is no longer standard arithmetic, but a system based on the set R U {oo}, a.k.a. the real projective line. (See http://en.wikipedia.org/wiki/Real_projective_line) > So, IF an infinite value > exists, and f(n)>g(n) for all n greater than some m, then we can conclude that > for this infinite value oo, f(oo)>g(oo). Tada! Infinite induction. And here is your axiom that defines what you call infinite induction. Can we assume that these are the first two axioms of the T-arithmetic system? Of course, you're still missing axioms (or theorems) that define your standard unit infinity Big'Un (is it the same as oo?), Little'Un, and the basic arithmetic operations that apply to them. === Subject: : Re: Calculus XOR Probability > Matt Gutting said: > If we are talking about the hyperreals, then that's not > happening. WE are talking about the standard reals. TO doesn't even understand them yet. TO knows too little about hyperreals to do anything but make an ass of himself. > You can insist there are no infinite values on the > real line, but that's just refusal to consider a new idea. The real line already exists in current systems. If TO want something else, he will have to provide the system from which it can spring. TO has not done so and does not seem able to do so in the foreseeable future. > Values aren't called into existence. If those values exist, > then the formulaic expression applies to them. Yes, but first, if you're not calling the values into existence, > you have to prove that they exist. No, I can make a statement that depends on whether they exist or not. Any statement about any thing depends on whether that thing exists or not. But claims requiring the existence of things which don't, are false. > (E oo A neN oo>n) ^ (E meN A n>m f(n)>g(n)) -> f(oo)>g(oo). So, IF an > infinite value exists, and f(n)>g(n) for all n greater than some m, > then we can conclude that for this infinite value oo, f(oo)>g(oo). > Tada! Infinite induction. And if 2 = 1, Bertrand Russell is the Pope. === Subject: : Re: Calculus XOR Probability > Matt Gutting said: > David R Tribble said: >> MoeBlee said: >> On the other hand, this is a theorem in set theory (assuming that 'oo' >> is some already defined constant symbol and that '<' is some already >> defined 2-place relation symbol): >> Am(Ax(x>m -> f(x) > g(x)) - >> (oo>m -> f(oo)>g(oo))) >> It's uninteresting, but at least correct. If there is anything >> (whatever oo is), then it would depend on whether the set of real >> numbers is a subset of the domains of the functions f and g and >> whether >> oo is a member of the domains of the functions f and g. > Yes, and if the functions f anf g are defined over all the reals, then > they can > be considered to be defined in the limit as well, and formulaic > expressions of > oo can therefore be compared and ordered in this way. That's what's > interesting. If 2*x2, and oo>2, then 2*oo> That's a neat trick, considering that oo is not a member of the reals. >> Apparently, defining a simple relation between the members of a >> set (such as defining 2x2 in R) can call into existence >> new values in the set that previously did not exist in the set before >> the relation was defined. > > If we are talking about the hyperreals, then that's not happening. You > can > insist there are no infinite values on the real line, but that's just > refusal > to consider a new idea. Values aren't called into existence. If those > values > exist, then the formulaic expression applies to them. Yes, but first, if you're not calling the values into existence, you have > to > prove that they exist. Matt No, I can make a statement that depends on whether they exist or not. (E oo A > neN oo>n) ^ (E meN A n>m f(n)>g(n)) -> f(oo)>g(oo). So, IF an infinite value > exists, and f(n)>g(n) for all n greater than some m, then we can conclude > that > for this infinite value oo, f(oo)>g(oo). Tada! Infinite induction. But if they don't exist, the whole thing is a pile of do do. Until TO can come up with a system in which they can exist (and standard set theories do NOT allow TO_infinities) it IS a pile of do do.. === Subject: : Re: Calculus XOR Probability Matt Gutting said: > Virgil said: > MoeBlee said: > There is nothing wrong with saying that, for functions f and g > and > a real number m, if f(x)>g(x) for all x>m, that f(oo)>g(oo). >> You may have no trouble saying it but you might have some trouble >> proving it. And what cannot be proved is mere speculation. > oo>m, hence f(x)>g(x) for all x>m -> f(oo)>g(oo). QED >> QED as end of a proof from WHAT axioms and definitions? > of all x>m. >> Not until oo is defined, it isn't, and it is not yet defined as a >> natural or a real or any other object for which > is defined. Sure, it's a number greater than any finite. A n e N oo>n! Duh. > That's a description of oo, but not a definition of it. For example, > it doesn't allow one to distinguish between oo, 2 * oo, and 0.5 * oo. > Since you declare these three to be distinct entities, your definition > of oo must allow you to make the distinctions. Put E oo before it, and it's a definition of a number greater than any finite. One can then go on to define arithmetic operations o this unit the same way as arithmetic operations on finite units, although calculations are restriced to rational portions of these units. > I have specifically stated that infinite values are to be compared > formulaically, and that there is no single point of infinity, but a whole > spectrum of such formulaically related infinities. > I'm not clear here. Do you mean that there are many infinite numbers, or that > the one number, oo, can take on multiple values? If the former, then as > I said your definition of oo must allow you to distinguish it from these others, > and I don't see how your definition of oo as a number greater than any finite > does that. If the latter, on the other hand, then oo is not well defined, and > it's not possible to assign a meaning to it. The number oo is generally a vague notion of any infinite value. In Bigulosity, I've declared Big'un as a unit infinity, with properties similar to 1, but on a relatively infinite scale. Big'un is a specific infinite value, as opposed to absolute oo. > Matt -- Smiles, Tony === Subject: : Re: Calculus XOR Probability > Matt Gutting said: > That's a description of oo, but not a definition of it. For > example, it doesn't allow one to distinguish between oo, 2 * oo, > and 0.5 * oo. Since you declare these three to be distinct > entities, your definition of oo must allow you to make the > distinctions. > Put E oo before it, and it's a definition of a number greater than > any finite. Claiming that something exists does not define anything until one shows that what one claims is possible. TO is essentially trying to say that he has a number larger than any number. And claiming that something exists, without proofs, as TO keeps doing, does not establish that anything does exist. > One can then go on to define arithmetic operations o this > unit the same way as arithmetic operations on finite units But no one has done so. Note that for real numbers, the operations of arithmetic are not only defined, but their usual properties are all proved to be valid. So that even defining similar operations for TO's nonexistent numbers without proofs that they have the properties he expects of them, is mere handwaving. > Yes, that was pretty simplistic, Virgil, and obviously nothing I > have proposed. I have specifically stated that infinite values > are to be compared formulaically, and that there is no single > point of infinity, but a whole spectrum of such formulaically > related infinities. I'm not clear here. Do you mean that there are many infinite > numbers, or that the one number, oo, can take on multiple values? > If the former, then as I said your definition of oo must allow you > to distinguish it from these others, and I don't see how your > definition of oo as a number greater than any finite does that. > If the latter, on the other hand, then oo is not well defined, and > it's not possible to assign a meaning to it. > The number oo is generally a vague notion of any infinite value. In > Bigulosity, I've declared Big'un as a unit infinity Declarations don't cut it. Definitions (if mathematically adequate, which TO's are not) and proofs of existence ( of which TO has none) and proofs that expected properties are valid (of which TO has none), are basic requirements which TO still fails to meet. === Subject: : Re: Calculus XOR Probability > Matt Gutting said: > Virgil said: > MoeBlee said: > There is nothing wrong with saying that, for functions f and g > and > a real number m, if f(x)>g(x) for all x>m, that f(oo)>g(oo). >> You may have no trouble saying it but you might have some trouble >> proving it. And what cannot be proved is mere speculation. > oo>m, hence f(x)>g(x) for all x>m -> f(oo)>g(oo). QED >> QED as end of a proof from WHAT axioms and definitions? > of all x>m. >> Not until oo is defined, it isn't, and it is not yet defined as a >> natural or a real or any other object for which > is defined. > Sure, it's a number greater than any finite. A n e N oo>n! Duh. >> That's a description of oo, but not a definition of it. For example, >> it doesn't allow one to distinguish between oo, 2 * oo, and 0.5 * oo. >> Since you declare these three to be distinct entities, your definition >> of oo must allow you to make the distinctions. > Put E oo before it, and it's a definition of a number greater than any finite. No, it's an assertion that the object described exists. Matt === Subject: : Re: Calculus XOR Probability > Matt Gutting said: > Sure, it's a number greater than any finite. A n e N oo>n! Duh. That's a description of oo, but not a definition of it. For > example, it doesn't allow one to distinguish between oo, 2 * oo, > and 0.5 * oo. Since you declare these three to be distinct > entities, your definition of oo must allow you to make the > distinctions. > Put E oo before it, and it's a definition of a number greater than > any finite. One can then go on to define arithmetic operations o this > unit the same way as arithmetic operations on finite units, although > calculations are restriced to rational portions of these units. The 'finite' real numbers can all be proven to exist merely from axims and basic definitions. TO's 'infinite numbers' cannot, or at least have not, been proved to exist. In the real numbers, sums and products, differences and quotients whose divisor is not zero can all be proved to exist and have the usual properties. None of TO's stuff has been, or can be, proven to work as he says it will without some system of axioms and definitions which does not yet exist. And TO's handwaving will not bring such a system into existence. Yes, that was pretty simplistic, Virgil, and obviously nothing I > have proposed. I have specifically stated that infinite values > are to be compared formulaically, and that there is no single > point of infinity, but a whole spectrum of such formulaically > related infinities. I'm not clear here. Do you mean that there are many infinite > numbers, or that the one number, oo, can take on multiple values? > If the former, then as I said your definition of oo must allow you > to distinguish it from these others, and I don't see how your > definition of oo as a number greater than any finite does that. > If the latter, on the other hand, then oo is not well defined, and > it's not possible to assign a meaning to it. > The number oo is generally a vague notion of any infinite value. In > Bigulosity, I've declared Big'un as a unit infinity, with properties > similar to 1, but on a relatively infinite scale. Big'un is a > specific infinite value, as opposed to absolute oo. TO needs a system, not merely more handwaving, but provides us only with more unsystematic handwaving. === Subject: : Re: Calculus XOR Probability >> Matt Gutting said: >> Do you have an infinite set that doesn't have an order of generation, >> besides >> the sets of all reals in an interval or altogether? > Sure. How about the set of all curves on the Euclidean plane passing > through > a given point at least once? I don't see a particular way of generating > this > set. >> Matt >> How exactly are you defining curve? > A set S of points with the following property: For all x in S and all e >> 0, > there exists a y in S such that dist(x,y) <= e, and there exists a z not > in > S such that dist(x,z) <= e (where dist() is the standard Euclidean metric > on the plane). > Hmmm.... wouldn't the set {(a,b): a, b rational} then qualify as a >> curve by this definition? > Why not use the standard definition: a continuous image of a real > interval. >> Because I'm trying to stay away from quantitative objects, and using the >> fewest possible references to numbers. >> Matt > That TO always screws up quantitative objects doesn't mean that others > can't use them properly. Point taken. How about I keep the definition I gave and rename the set? Matt === Subject: : extention of homomorphism to place - help I need to understand the following: Let R be an integral domain, P a prime ideal of R, and K the field of fraction of R. Let R_p be the locallisation of R relatively to P. Then R_p/(P) is a field, and we have an homomorphism f:x--> x+P of R_p to this field. As is well known, f can be extended to a place of K. My question is: is such an extention unique? same question if we suppose also R integrally closed. Thx === Subject: : Algebraic Fractal Geometry Is there a theory of algebraic fractal geometry ? I need it in proof of Riemann Hypothesis. We Pretty === Subject: : Re: Algebraic Fractal Geometry | | _____________________ | /| /| | | | ||__|| | Do not feed the | | / | --Mgt. | | / |_____________________| | / _ || | / |____ || | / | | | |____/ || | / |_|_|/ | _|| | / / |____| || | / | | | --| | | | | |____ --| | * _ | |_|_|_| | -/ |*-- _-- _ | || | / _ | / ` |* / _ /- | | | | * ___ c_c_c_C/ C_c_c_c____________ | === Subject: : Re: [OT] Lucid Dreams > I had a dream almost 35 years ago that I still remember pretty clearly. > At the time I was in high school and my mother woke me up to get ready > for school. I wanted to stay in bed for just a few more minutes and > of course, fell asleep again immediately. But I dreamed that I got up, > got dressed, ate breakfast, and went to school. I had gone through > my first two classes and was taking a test in the third class when my > mother woke me up again. Not only did I not want to get up, but since > I had a clear memory of the dream, it felt like I was having to do the > first three classes all over again. :-) Yes, I think we've all been there many times... > http://dlazechk.dl.funpic.org/mergedbedtimedreams.html ah... how sweet :-) Dirk Vdm === Subject: : God Prefers Women! http://www.msnbc.msn.com/id/12893635/site/newsweek/page/2/ And so began modern division betwen men and women...poor babies... === Subject: : Re: God Prefers Women! > http://www.msnbc.msn.com/id/12893635/site/newsweek/page/2/ > And so began modern division betwen men and women...poor babies... How easy the infantile are convinced by stitching things together to make a story.I know more about the geometric facets of Christian writing than anyone alive yet I enjoyed the breathless fiction woven into scattered facts and real people from art,history,symbols ect. The other side of Christianity is far more satisfying and is closer to the real secrets behind an existence and the power and majesty which encompasses ours,To huddle together and try to describe the natural furniture was always the choice of those who feared the journey of going out to meet nature at its most raw. The real texts of Christianity are not hidden ,what they say offends those who cherish the easy life with easy answers.For the genuine astronomer who promotes celestial phenomena for the majesty of scale in all its interweaving parts and cycles it can take its toll but it is worth it.For those who reduced astronomy to the page, the theory or the equation,the cosmos can only be a means to an end of self promotion and self congratulation. C H A P T E R ~ XX How, seeing that the Life of Christ is most bitter to Nature and Self, Nature will have none of it, and chooseth a false careless Life, as is most convenient to her. Now, since the life of Christ is every way most bitter to nature and the Self and the Me (for in the true life of Christ, the Self and the Me and nature must be forsaken and lost, and die altogether), therefore, in each of us, nature hath a horror of it, and thinketh it evil and unjust and a folly, and graspeth after such a life as shall be most comfortable and pleasant to herself, and saith, and believeth also in her blindness, that such a life is the best possible. Now, nothing is so comfortable and pleasant to nature, as a free, careless way of life, therefore she clingeth to that, and taketh enjoyment in herself and her own powers, and looketh only to her own peace and comfort and the like. And this happeneth most of all, where there are high natural gifts of reason, for that soareth upwards in its own light and by its own power, till at last it cometh to think itself the true Eternal Light, and giveth itself out as such, and is thus deceived in itself, and deceiveth other people along with it, who know no better, and also are thereunto inclined. === Subject: : Re: God Prefers Women! > How easy the infantile are convinced by stitching things together to make a story. How easy is it? I would have said that given umpteen monkeys, infinity and a lot of typwriters; the most likely result would be, not a comprehensive and complete version of the Encyclopedia Britannica but dead monkeys and rusted, broken typewriters. But there is nothing like a daft idea to turn facts on their heads. === Subject: : Horn of 1/X dilemma When you spin the function 1/X around on the X-axis, and take the volume, you notice that the volume is finite. However, the surface area is infiinite. Another words, this shape can hold a finite amount of paint, but it requires infinity gallons to paint it. Please explain? === Subject: : Re: Horn of 1/X dilemma > When you spin the function 1/X around on the X-axis, > and take the volume, you notice that the volume > is finite. However, the surface area is infiinite. > Another words, this shape can hold a finite amount > of paint, but it requires infinity gallons to paint it. Volumes are computed by multiplying three dimensions together, areas by multiplying two dimensions. When you let all but one of the dimensions get smaller and smaller, it's easier to get convergence when you have TWO small quantities among your factors (the volume computation) than when you have just ONE small quantity among your factors (the area computation). Interestingly, very few people seem surprised at the same phenomena one dimension lower, where a curve of finite length can bound a finite area. For example, the area in the first quadrant bounded by x = 0, x = 1, y = 0, and y = x + x*sin(1/x). Dave L. Renfro === Subject: : Re: Horn of 1/X dilemma Now that is an interesting dilemma. This is in contrast to the axiom of determinacy and transfinite numbers. The axiom of determinacy treats what is internal to the set as infinite, yet externally finite to the game. Two sets with equivalent infinite strategy and infinite moves will have an end game of one being the winner, provided that there is perfect information update. This denotes inequality when two sets are completely and inexhaustibly equal. This is in contrast to the axiom of choice where two perfectly infinite and equivalent sets will run equal to each other. The way out of this dilemma is to rigidly assort what is internal which is inexhaustible and what is external which is finite. Transfinite numbers are the same. They are running internally infinite, but externally finite within the boundary of the set. This is not a contradiction as long as you rigidly assort them to their respective categories. The externality is based on a generalization. It is not a logical sum. It is one of the key mistakes I am finding in a lot of math papers. They confuse logical sum with generalizations. Infinity no matter how bounded within a function rule cannot add up to 1. This is where you Hilbert's grand hotel paradox. They think infinity will add up to a logical sum enough to double. Infinity cannot do this. But if you could bind it externally then it becomes finite, then you can do this. In any case, I think the way out of this is to invert my solution. Internal volume is finite. External surface area is infinite. But it is infinite insofar as the math can be a finite generalization within a boundary. Then you can paint the surface with a finite number of paint. I hope this is adequate. Sorry for being overly drawn out. B.T. === Subject: : Re: Horn of 1/X dilemma > This is in contrast to the axiom of determinacy > and transfinite numbers. > The axiom of determinacy treats what is internal > to the set as infinite, yet externally finite > to the game. While Gabriel's Horn may be somewhat semantically analogous to certain ideas in set theory and logic (e.g. existence of countable models), I don't think the analogy can be exploited to obtain any useful mathematical insight. Incidentally, some of the history of Gabriel's Horn is discussed in the following Historia-Matematica thread, which began on May 9, 2006: [HM] Gabriel's Horn http://mathforum.org/kb/thread.jspa?threadID=1379857 Dave L. Renfro === Subject: : Re: Horn of 1/X dilemma > When you spin the function 1/X around on the X-axis, and take the > volume, you notice that the volume is finite. However, the surface > area is infiinite. > Another words, this shape can hold a finite amount of paint, but it > requires infinity gallons to paint it. > Please explain? Does the coating of paint have any thickness? === Subject: : Re: Horn of 1/X dilemma >> When you spin the function 1/X around on the X-axis, and take the >> volume, you notice that the volume is finite. However, the surface >> area is infiinite. >> Another words, this shape can hold a finite amount of paint, but it >> requires infinity gallons to paint it. >> Please explain? > Does the coating of paint have any thickness? you beat me to it. I have a can of paint that has zero thickness when applied. One gallon covers all. Only takes an infinite amount of time to put it on, and very tiny paint brushes for the point. === Subject: : Discursion rather than discussion Am I right in thinking that the author meant this: The likelihood of severe weather increases as the strength of these fields increase, especially in the way they increase in relation to each other. ...the likelihood of severe weather increases as the strength/intensity of these fields/parameters increase -not only increase, but increase in the context of preferred juxtaposition of the fields relative to one another. I'm bad enough at maths without having to shuffle through that sort of verbage. === Subject: : Re: Hyper-sphere partitioning question beeworks@hotmail.com says... >If I understand you correctly, you want to find a partitioning >consisting of congruent cells (equal area and equal shape)AND such >that one can choose the cells to have an arbitarily small bounding >radius. Is that right? >Any congruent cell partitioning (CCP) would have to exhibit the >symmetry of a regular polytope. As suggested elsewhere in this >thread, the regular polytope of symmetry need not be a regular >polytope of Sn. Using the orange slice method, the symmetry could >be a regular polytope on Sm where m <= n. >That is not to say that the CCP must be a regular or orange slice >polytope. For example, one could start with a regular polytope and >then place a new vertex in the center of each cell (on Sn of course). >One could then place a new vertex at the center of each face of the >cell, and then at the center of each side of each face and so on. >Doing so would subdivide a p-cell regular polytope into a (2pq)-cell >CCP, where q is the number of vertices of a cell. However, further >subdivision along these lines would result in cells that are not all >congruent. >Having played the above subdivision out, I would conjecture that >the greatest number of cells that a CCP could have would in fact >be 2pq, where p is the number of cells in the polytope and q is >the number of vertices of each cell. Of course p and q depend of m, >the dimension of the polytope. (I.e. the vertices of a regular >polytope of dimension m lies on Sm.) > [..] there are _many_ ways of tiling S^2 with congruent polygons, including many that are irregular (do not have polytope symmetry). The first and third reference below in particular have some amazing pictures of irregular tilings by congruent triangles. -RS. Robert J. MacG. Dawson and Blair Doyle Tilings of the Sphere with Right Triangles I: The Asymptotically Right Families Robert J. MacG. Dawson and Blair Doyle Tilings of the Sphere with Right Triangles II: The (1,3,2), (0,2,n) Subfamily Yukako Ueno, Yoshio Agaoka Classification of Tilings of the 2-Dimensional Sphere By Congruent Triangles === Subject: : Re: Lucid Dreams I don't know if any of you play around with these types of things or > not, but have any of you had any lucid dreams lately? Do you know what > lucid dreams are? If not, let me know and I'll explain it to you. > But, my main purpose in writing this is I just want/need to know if any > of you are having lucid dreams. > There is no such thing as a lucid (or conscious) dream. > You just *dream* that you know that you are dreaming. > There's nothing mysterious about it. You can dream > anything, and that obviously includes knowing that you > are dreaming. > I've had several series of dreams where I changed the outcome. If > there's no such thing as a conscious dream, then my unconscious must be > able to make decisions. > BTW, the math I dream about is always gibberish. I've tried peeking at > slides (of a talk which I was giving) or writing down what's on a > blackboard, and I've never returned with anything, even when it was > supposedly my work! > --- Christopher Zzz Heckman. I think all math holds true in dreams also. I can count, add, divide, subtract. Also equations of geometry holds true in dream. Problem is with physics. I can fly on earth. Physics collapse in dream.Math does not collapse. === Subject: : Re: Lucid Dreams > > I don't know if any of you play around with these types of things or > not, but have any of you had any lucid dreams lately? Do you know what > lucid dreams are? If not, let me know and I'll explain it to you. > But, my main purpose in writing this is I just want/need to know if any > of you are having lucid dreams. > > There is no such thing as a lucid (or conscious) dream. > You just *dream* that you know that you are dreaming. > There's nothing mysterious about it. You can dream > anything, and that obviously includes knowing that you > are dreaming. > I've had several series of dreams where I changed the outcome. If > there's no such thing as a conscious dream, then my unconscious must be > able to make decisions. > BTW, the math I dream about is always gibberish. I've tried peeking at > slides (of a talk which I was giving) or writing down what's on a > blackboard, and I've never returned with anything, even when it was > supposedly my work! > --- Christopher Zzz Heckman. > I think all math holds true in dreams also. I can count, add, divide, > subtract. Also equations of geometry holds true in dream. > Problem is with physics. I can fly on earth. Physics collapse in > dream.Math does not collapse. Not sure about that. Look at for instance James Harris, who is in a permantent state of dreaming. In his dreams you and I and everyone else is just playing a role. Now, we can't say that his math does not collapse, can we? Dirk Vdm === Subject: : Re: is it easy? hi > >given a set of integers X={d1,d2,d3,...,dn}, and another integer N, >is it easy to know there is an (integer) linear combination of the set >X,such that >c1*d1+c2*d2+...+cn*dn=N ? > I don't know if by easy you mean easy in theory or easy > because there's an efficient algorithm. If the coefficients ci > are not restricted to be non-negative integers, then it's very easy > indeed, in the first sense: N can be written in that form if and > only if it is divisible by the greatest common divisor of X. > how do you see N can be written in that form <=> it is divisible by > the gcd of X? > how do you then go and actually find a non-trivial solution? > (or please point me to find info on it.) See http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm http://en.wikipedia.org/wiki/Euclidean_algorithm http://www.geocities.com/scroussette/gcd.txt === Subject: : proper weighting for a nonlinear regression fit Hello all, I'm trying to fit a curve to correlation data, but the problem is that I don't know what weighting I should apply to get a good chi-squared fit. The data fed to the correlator is sampled logarithmically, so that the sampling time for the data at the low end of the correlation curve is very short compared to the sampling times for the data at the high end. This of course means that the data at the low end has been averaged over more intervals than the data at the high end. Is there a good way to weight the residuals for the best possible fit? === Subject: : Re: after beginner's algebra, where to? (which I doubt), an analytic geometry course. Then you would be ready > for calculus. There are introductory material in analytic geometry in some pre-calculus books. I've learned te analytic geometry as a part of calculus (the textbook name was 'Calculus and Analytic Geometry', forgot the authors. === Subject: : Re: after beginner's algebra, where to? >> after completing a beginner's course in algebra i was >> hoping someone here could tell me two things: >[snip rest] >Since beginning algebra means different things >to different people I'm sure it must come after A course in Arithmetic, so he's pretty far along... dave === Subject: : Re: after beginner's algebra, where to? pretty far along... I guess you first take A Course in Arithmetic. Then you study Basic Number Theory, which gives you the background for understanding Basic Algebra. After that you can take Advanced Algebra at a JC and learn about logs, exponents and trig functions. === Subject: : Re: after beginner's algebra, where to? Since beginning algebra means different things >> to different people [...] > I'm sure it must come after A course in Arithmetic, > so he's pretty far along... Especially if it's this book: A Course in Arithmetic by Jean-Pierre Serre Dave L. Renfro === Subject: : Re: after beginner's algebra, where to? If your goal is to do physics, then you must know that the math physicists know is tought a litlle different. Learn the basic 3 calculus course like math majors, which are far from where you are, then get a book on mathematical physics. Becuase you do not need all the things a mathematician might learn for physics. And you will need the basic 3 courses in physics as well. The best way to do this is for you to go to a community college. === Subject: : Re: after beginner's algebra, where to? the math physicists know is tought a litlle different. > Learn the basic 3 calculus course like math majors, > which are far from where you are, then get a book on > mathematical physics. Becuase you do not need all > the things a mathematician might learn for physics. > And you will need the basic 3 courses in physics as > well. The best way to do this is for you to go to > a community college. Some colleges take 4 semesters to cover elementary calculus (Univ. of NC at Chapel Hill used to, for example) and some take 2 semesters to cover this material (MIT and CalTech, for example). What are the basic 3 courses in physics? If you're talking about the standard 2 semester upper level sequences in classical mechanics, quantum mechanics, and electromagnetic theory, I don't think there are *any* community colleges that offer this. On the other hand, if you're talking about a 2 semester introductory course in calculus-based physics (such as Halliday and Resnick's Fundamentals of Physics text is designed for), followed by a one semester sophomore level course in modern physics, then I suppose you can find this at community colleges. My point is that for a medium such sci.math you should be a little more explicit or else your point could be misunderstood. I'm sure this was Dave Rusin's point in this thread, by the way. Dave L. Renfro === Subject: : Re: Power law regime > Supposed I have a distribution and I find that it fits a power-law > regime (y = ax^k). What does it tell me actually? What is the > significance of this observation in general? If k is not a small integer or very simple rational, x had better be dimensionless, otherwise the fit would be a delusion. Indeed, the concoction of dimensionless groups of physical properties is a necessary part of the derivation of empirical formulae in various technical areas. -- === Subject: : Re: Set Covering > Consider a barrel with 14 numbered balls and 8 are > drawn without replacement. We choose 3 numbers > and hope to match all 3 with the drawn 8. > (14,8,3,3) > Say we take the following 5 lines as the beginning > of a comprehensive cover not necessarily optimal. > (a) Threes(1, 1) = 1: Threes(1, 2) = 2: Threes(1, 3) = 12 > (b) Threes(2, 1) = 3: Threes(2, 2) = 6: Threes(2, 3) = 9 > (c) Threes(3, 1) = 7: Threes(3, 2) = 8: Threes(3, 3) = 9 > (d) Threes(4, 1) = 10: Threes(4, 2) = 11: Threes(4, 3) = 12 > (e) Threes(5, 1) = 1: Threes(5, 2) = 13: Threes(5, 3) = 14 > Is there a generalized formula for computing the amount of > fresh cover that each subsequent line of 3 gives? > eg . (a) gives fresh cover 14-3C 8-3 = 11C5 = 462 > (b) gives fresh cover 462 - (14-6C8-6) = 434 > Now from simulation I can get the fresh cover for (c), 351 > (d),325 and (e), 304. but was wondering if there is a > general formula that could be applied to get the same. > TIA Mick You can use the inclusion-exclusion principle: The number of 8-subsets covered by line (a) alone is (14-3)C(8-3) = 11C5 = 462, as you computed. The number of 8-subsets covered by lines (a) and (b) is 2*((14-3)C(8-3)) - (14-6)C(8-6) = 2*(11C5) - 8C2 = 2*462 - 28 = 896. So the number of new 8-subsets covered is 896 - 462 = 434, as you computed. The number of 8-subsets covered by lines (a), (b), and (c) is 3*((14-3)C(8-3)) - (2*((14-6)C(8-6)) + 1*((14-5)C(8-5))) +1*((14-8)C(8-8))= 3*(11C5) - (2*(8C2) + 9C3) + 6C0 = 3*462 - (56 + 84) + 6 = 1247. So the number of new 8-subsets covered is 1247 - 896 = 351, as you computed. Similarly, the number of 8-subsets covered by lines (a), (b), (c), and (d) turns out to be 1572, yielding 1572 - 1247 = 325 new 8-subsets covered, as you computed. Similarly, the number of 8-subsets covered by lines (a), (b), (c), (d), and (e) turns out to be 1876, yielding 1876 - 1572 = 304 new 8-subsets covered, as you computed. A general formula for n balls (14 in your example), k drawn (8 in your example), t chosen (3 in your example) on each of r lines (5 in your example) is sum [j = 1 to r] sum [I subset {1,2,...,r}: |I| = j] (-1)^(j+1) * ((n - |union[i in I] A_i|) C (k - |union[i in I] A_i|)), where A_i is the set of balls in line i. In your example, A_1 = {1,2,12}, A_2 = {3,6,9}, etc. This formula gives the total number of k-subsets covered. If you want just the new k-subsets covered by line s, you should take r = s and restrict the inner sum to j-subsets I containing s. By the way, a lower bound for the number of lines in a cover is ceiling(14C8 / 11C5) = ceiling(3003 / 462) = ceiling(6.5) = 7. It looks like you're working on a greedy algorithm for a set covering problem, where at each stage you choose a 3-subset that covers the largest number of new 8-subsets. Rob Pratt === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper <080520061306438590%anniel@nym.alias.net.invalid> See http://arxiv.org/abs/math.GM/0312309 >> Craig >> Tell us again after a reputable journal has refereed it. He should consider submitting it to http://www.crank.net . They have a wonderfully amusing section on cranky pseudo-mathematics. :-) === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper <2083b$44604a1b$82a1e228$29148@news2.tudelft.nl> <4460c19a$0$9256$ed2619ec@ptn-nntp-reader01.plus.net> <4468a646$0$11868$3b214f66@aconews.univie.ac.at> Mathematicians have developed a proof method that they call proof by induction. This method allows them to prove an infinite number of statements, without writing down an infinite number of proofs. It is (very) easy to prove the following statement by induction (without talking about what the algorithm does at each iteration): If the Collatz algorithm starts with a power of two, then it will eventually halt at one. This shows that your Theorem 2 is blatantly wrong. --Gerhard The fact that you have completely ignored my challenge to you to pick any number and prove that the Collatz algorithm halts at one for that number *without talking about what the algorithm does at each iteration* indicates to me that you know that Theorem 2 is correct, but you just don't want to admit it. For instance, take the number 32. I bet that you will not be able to give a rigorous proof that the Collatz algorithm halts with 32 as input without specifying directly or indirectly that the reason the algorithm halts at one is because it divides 32 by 2 five times. Craig === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper # # The fact that you have completely ignored my challenge to you to pick # any number and prove that the Collatz algorithm halts at one for that # number *without talking about what the algorithm does at each # iteration* indicates to me that you know that Theorem 2 is correct, but # you just don't want to admit it. There are many reasons why Theorem 2 must be wrong. One reason is that it claims something about proofs, without specifying the underlying axiom system: Neither its statement nor the argument do depend on a particular axiom system. --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper <4460c19a$0$9256$ed2619ec@ptn-nntp-reader01.plus.net> <4468a646$0$11868$3b214f66@aconews.univie.ac.at> <4471f8b6$0$28520$3b214f66@aconews.univie.ac.at> There are many reasons why Theorem 2 must be wrong. One reason is that it claims something about proofs, without specifying the underlying axiom system: Neither its statement nor the argument do depend on a particular axiom system. --Gerhard Again, that is just avoiding my challenge. You are just hiding behind the notion of axiom systems to save face. When a mathematician says I proved Theorem X, why should he or she be allowed to claim this without referring to a specific axiom system, yet when I say that Collatz is unprovable, you claim that I must refer to the specific axiom system? That is a double standard. Craig === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper # # Again, that is just avoiding my challenge. You are just hiding behind # the notion of axiom systems to save face. Don't be ridiculous. We know since Goedel that provability is not an absolute concept, but heavily depends on the underlying axiom system. And you claim something on the provability of a mathematical statement in absolute terms... --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper # # Again, that is just avoiding my challenge. You are just hiding behind # the notion of axiom systems to save face. When a mathematician says # I proved Theorem X, why should he or she be allowed to claim this # without referring to a specific axiom system, yet when I say that # Collatz is unprovable, you claim that I must refer to the specific # axiom system? That is a double standard. You should at least learn about the Peano system. If you keep on reading, this will eventually lead you to so-called inductive proofs. That's a fascinating concept that is going to revolt you view of mathematical proofs, and in particular your view of Theorem 2. --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper <2083b$44604a1b$82a1e228$29148@news2.tudelft.nl> <4460c19a$0$9256$ed2619ec@ptn-nntp-reader01.plus.net> <4468a646$0$11868$3b214f66@aconews.univie.ac.at If the Collatz algorithm starts with a power of two, > then it will eventually halt at one. > For instance, take the number 32. I bet that you will not be able to > give a rigorous proof that the Collatz algorithm halts with 32 as input > without specifying directly or indirectly that the reason the algorithm > halts at one is because it divides 32 by 2 five times. (1) Given a number n divisible only by 2, the Collatz algorithm's first step will find a number m such that m <2083b$44604a1b$82a1e228$29148@news2.tudelft.nl> <4460c19a$0$9256$ed2619ec@ptn-nntp-reader01.plus.net> <4468a646$0$11868$3b214f66@aconews.univie.ac.at> (1) Given a number n divisible only by 2, the Collatz algorithm's first step will find a number m such that m <2083b$44604a1b$82a1e228$29148@news2.tudelft.nl> <4460c19a$0$9256$ed2619ec@ptn-nntp-reader01.plus.net> <4468a646$0$11868$3b214f66@aconews.univie.ac.at (1) Given a number n divisible only by 2, the Collatz algorithm's > first > step will find a number m such that m (2) Eventually, any such process must end at 1. > You don't need to know specifically what Collatz does. > This is not a rigorous mathematical proof that 32 causes Collatz to > halt at one. It doesn't even talk about 32. I think you have fallen into an eye-of-the-beholder fallacy. Is there any such thing as showing directly or indirectly that 32 is not divisible by any prime other than 2? If I demonstrate this, will you read into my proof that 32 = 2^5 and therefore directly or indirectly I showed what the Collatz sequence for 32 looks like at each step? Here are just two of the ways I think prove that 32 is some power of 2, without establishing that 32 = 2^5. First: Square 2 repeatedly until the result exceeds 32; we will get 4, then 16, then 256. Divide 256 by 32 and you get zero remainder; because 256 is a power of 2, so is its divisor 32. Second: Make a list of all primes less than 32, e.g. by say Eratosthenes Sieve: 2,3,5,7,11,13,17,19,23,29,31 Of these, only 2 exactly divides 32. Representation of 32, by the Fund. Thm. of Arithmetic, as a unique product of prime powers consists simply of some power of 2. (No other prime can appear with positive exponent, since no prime factor of 32 can exceed 32.) === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper <2083b$44604a1b$82a1e228$29148@news2.tudelft.nl> <4460c19a$0$9256$ed2619ec@ptn-nntp-reader01.plus.net> <4468a646$0$11868$3b214f66@aconews.univie.ac.at> I think you have fallen into an eye-of-the-beholder fallacy. Is there any such thing as showing directly or indirectly that 32 is not divisible by any prime other than 2? If I demonstrate this, will you read into my proof that 32 = 2^5 and therefore directly or indirectly I showed what the Collatz sequence for 32 looks like at each step? Here are just two of the ways I think prove that 32 is some power of 2, without establishing that 32 = 2^5. First: Square 2 repeatedly until the result exceeds 32; we will get 4, then 16, then 256. Divide 256 by 32 and you get zero remainder; because 256 is a power of 2, so is its divisor 32. Second: Make a list of all primes less than 32, e.g. by say Eratosthenes Sieve: 2,3,5,7,11,13,17,19,23,29,31 Of these, only 2 exactly divides 32. Representation of 32, by the Fund. Thm. of Arithmetic, as a unique product of prime powers consists simply of some power of 2. (No other prime can appear with positive exponent, since no prime factor of 32 can exceed 32.) Very clever. But 32=2^5 is hidden in such proofs: 2^(2*2*2)/32=8=2^3, so 32=2^(2*2*2-3). (Note that 2*2*2-3=5.) And the Sieve procedure establishes that 16=2^4 and 32/2=16 (without actually saying it explicitly). What you are trying to do can't be done for even the simplest type of number 32, which is a power of 2. So what gives you or anyone else the idea that you can do this for more complicated types of numbers which are not powers of 2? And if you can't do it for more complicated types of numbers, then how are you going to do it for all numbers? You can't, because it's an impossible task. You can mention all types of techniques like induction, proof by contradiction, existential proofs all you like, but they are not going to fix this problem like they do for other easier mathematics problems. Collatz is a math problem which logic cannot address. The only way you can know for certain that Collatz is true is through prophesy. Craig === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper <2083b$44604a1b$82a1e228$29148@news2.tudelft.nl> <4460c19a$0$9256$ed2619ec@ptn-nntp-reader01.plus.net> <4468a646$0$11868$3b214f66@aconews.univie.ac.at Is there any such thing as showing directly or indirectly that > 32 is not divisible by any prime other than 2? > If I demonstrate this, will you read into my proof that 32 = 2^5 > and therefore directly or indirectly I showed what the Collatz > sequence for 32 looks like at each step? > Here are just two of the ways I think prove that 32 is some > power of 2, without establishing that 32 = 2^5. A new math game! How about this: (1) For every other integer, 1, 3, 5, ... up to 31, I show that n is invertible mod 32 by giving the inverse. For instance, for 3 I give 11, subtract 32 from 33 and get 1. For 5 I give 13, subtract 32 twice from 65 and get 1. And so forth. (2) I show 2 is not invertible mod 32 by checking all the possibilities. If I'm too lazy to do that I show it is a zero divisor since 2*16=32. represented by the odd integers. (4) Using (3) I show that 32 cannot be divided by any prime > 2 (since those represent units.) === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper # (1) Given a number n divisible only by 2, the Collatz algorithm's # first # step will find a number m such that m <2083b$44604a1b$82a1e228$29148@news2.tudelft.nl> <4460c19a$0$9256$ed2619ec@ptn-nntp-reader01.plus.net> <4468a646$0$11868$3b214f66@aconews.univie.ac.at (1) Given a number n divisible only by 2, the Collatz algorithm's > first > step will find a number m such that m (2) Eventually, any such process must end at 1. > You don't need to know specifically what Collatz does. > This is not a rigorous mathematical proof that 32 causes Collatz to > halt at one. It doesn't even talk about 32. Details could be added, of course. It doesn't need to talk about 32 beyond noting no odd prime divides it. === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper # #> In the Feinstein model of proof, the only feasible way of proving #> statements on a number of objects is TO EXPLICITLY WRITE DOWN ONE #> SEPARATE STATEMENT for every considered object. # # So the fundamental theorem of arithmetic is also unprovable? I guess you are right. I don't think that there are a lot of things that one can actually prove in the Feinstein-model-of-proof. --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: : Re: Collatz 3n+1 conjecture is unprovable paper <44699eb1$0$12126$3b214f66@aconews.univie.ac.at> <4471dd6e$0$28520$3b214f66@aconews.univie.ac.at # #> In the Feinstein model of proof, the only feasible way of proving > #> statements on a number of objects is TO EXPLICITLY WRITE DOWN ONE > #> SEPARATE STATEMENT for every considered object. > # So the fundamental theorem of arithmetic is also unprovable? > I guess you are right. > I don't think that there are a lot of things that one > can actually prove in the Feinstein-model-of-proof. You could prove the Monster is a finite simple group not belonging to any of a certain list of families of such groups. === Subject: : antiderivative of a function on an interval I'm using Stewart's Calculus book to self-learn Calculus. When it covers derivative of a function, it does not mention intervals in the definition. When it covers antiderivative of a function, the definition has the notion of interval interwoven with it... defn: A function F is called an antiderivative of f on an interval I if F'(x) = f(x) for all x in I The rest of the sections keeps bringing back this interval issue. For example... d ln|x| = 1/x for all x <> 0 dx .. the general antiderivative of f(x) = 1/x is ln|x| + C on any interval that doesn't contain 0. In particular, this is true on each of (-oo,0) and (0, oo) I don't know why is going through the trouble of stating this, when it is quite evident from looking at 1/x or ln|x| that 0 can not be used. Why mention all this as antiderivative on ____ interval ? Is there a reason behind this? The book then goes onto cover other topics all the way through FTC. Then we arrive at a indefinite integral section that ties back to the previous antiderivative section. Once again this interval things pops up. The section shows a table of indefinite integrals then below it... Recall from ... that the most general antiderivative ON A GIVEN INTERVAL (italicized for somer reason) is obtained by adding a constant to a particular antiderivative. (Now all bold) WE ADOPT THE CONVENSION THAT WHEN A FORMULA FOR A GENERAL INDEFINITE INTEGRAL IS GIVEN, IT IS VALID ONLY ON AN INTERVAL. What's going on? I know why intervals with continuity are important for definite integrals, but what's the big deal with antiderivatives? Also, what is all this stuff from the book saying? Anyone have experience with Stewart's book must have run across all this. For differentiation, we don't go through all this trouble of mentioning intervals and such, as it's implied... y = |x| ==> dy/dx = |x|/x It's obvious that dy/dx is only valid on (-oo, 0) and (0, oo). Stewart doesn't go through the trouble of saying this for derivatives. dy/dx is simply the DERIVATIVE of y (nobody says dy/dx is the DERIVATIVE on interval _____ of y). What am I missing about this interval business with antiderivatives? If anyone knows, please shed some light on this. === Subject: : Curtis' Miracle Octad Generator Simple question on this webpage: http://log24.com/theory/geometry.html In the box How the MOG Works they indicate that each brick considered by itself forms an octad. Does this mean a solid brick of eight black cells, which can fall in three different positions? If so, this seems clumsy to me, since it is not a brick from any of the bricks in the main diagram. I take it 759 are the blocks of a Steiner (5,8,24) system === Subject: : Re: Non-commutative polynomials >Let P=P(X_1,...,X_n) be a polynomial in n non-commutative variables >(say, on the complex field). Suppose that for any 2x2 complex matrices >A_1,...,A_n, we have P(A_1,...,A_n)=0. How to prove then that P is >(abstractly) the zero polynomial ? A ring is called a PI ring (for Polynomial Identity) if there is a nonzero polynomial P of the type you describe. Many familiar classes of rings are of this type, including commutative rings (P = X Y - Y X) and matrix rings over a field. Also of interest are polynomials which do not vanish identically on a ring but whose image is contained in the center of the ring (Central polynomials). Here is a reference to get you started: (80j:16013) Jacobson, N. Some recent developments in the theory of algebras with polynomial identities. Topics in algebra (Proc. 18th Summer Res. Inst., Austral. Math. Soc., Austral. Nat. Univ., Canberra, 1978), pp. 8--46, Lecture Notes in Math., 697, Springer, Berlin, 1978. dave === Subject: : Re: When a mathematician makes a mistake... >> How embarrassing is it if a mathematician submits a paper to a >> preprint archive and then it is found (either by the author or >> someone else) that there is a mistake? Not embarassing at all. Making mistakes in mathematics is an everyday occurrence, and these things happen. We learn through our mistakes, and it makes us better mathematicians for it. Refusing to admit you have erred, or steadfastly resisting peer reviews on your work, is the time to feel embarassed. The hallmark of crankhood is precisely this, for thinking you have accomplished something, while ignoring other mathematicians who say you have not. Any regualr sci.math member can point to many examples of this, whether it be proving FLT, disproving Collatz, contradicting Cantor's theorems, or what have you. Jonathan Hoyle === Subject: : Re: When a mathematician makes a mistake... I evaluate visiting mathematics and science lecturers after they submit to preprint archives. There are always mistakes. They have to know and fix it. It is the reason why math people should have a good sense of humor. :-) I am a mathematician with a math disability. I am not dyslexic. Computers now handle things so that I can do my concept work. I am highly forgiving of people who honestly acknowledge that I pointed out something important. Some people get ruffled and strike back, but at least they know. They are not real or even honest mathematicians if they strike at good and honest critique. It is highly unethical not to commit to rectifying scholarly mistakes. If this is found out, then you will lose your math reputation. That is a sensitive issue. When I point out mistakes, I usually do it via private e-mail. It is up to them to fix it, or answer to it, or rebut it. They then send it back to the organization that is sponsoring the preprint archive or the lecture. Those who will not send back will have private ethical issues. I find mistakes in every single math paper, and not just trivial ones. There is always a private e-mail on my end. To not get a reaction from me means that there is something wrong with the paper. I don't make my fellow colleagues feel stupid. It is bad for business. Plus, it will cause politics. To share with colleagues and to ask for reflection is part of the business. A lot of these reflections will mean you have to amend. Math is really an involving enterprise. It is not Plato. It is not something that is eternal. It is changing. That is my mathematical enterprise. I track the conceptual changes in it. B.T. > How embarrassing is it if a mathematician submits a paper to a preprint archive and then it is found (either by the author or someone else) that there is a mistake? > (I mean one that is difficult to see even by experts, not something completely ridiculous). > Do people think they were lazy or stupid not to have realised? > That they can't be good mathematicians and are letting everyone down? Or is it quickly forgotten as something fairly respectable and routine? === Subject: : Re: When a mathematician makes a mistake... I agree with Phil === Subject: : Re: When a mathematician makes a mistake... <170520060851590687%edgar@math.ohio-state.edu.invalid> The great mathematcian Jean Dieudonne has said that when he gets to the part of a proof where it says It is easy to show or This is immediate... he realizes that is where the mistake is. He admits he own errors have been exactly at that point. === Subject: : Re: When a mathematician makes a mistake... <4d11kpF18d8j2U1@individual.net> <446cdbe5$11$fuzhry+tra$mr2ice@news.patriot.net> <446e2e83$14$fuzhry+tra$mr2ice@news.patriot.net 05/19/2006 >I had never say that some one atacks Ramanujan, you say that. > atack colleagues with so much agresivity, rejecting all the content in > the mathem.87tical work of this colleagues.? >Evidence here, are not axioms, are theorems. > How are theorems evidence for axioms? Theorems are not evidence for axioms, Theorems are evidence by their own, in fact Theorems includes axioms in its assumption part. > Shmuel (Seymour J.) Metz, SysProg and JOAT right to publicly post or ridicule any abusive E-mail. Reply to > domain Patriot dot net user shmuel+news to contact me. Do not > reply to spamtrap@library.lspace.org === Subject: : power iterated functions DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=s1024; d=yahoo.de; === b=1rt9fRuQCFb1yN/HFvy0QagAqY42UwtfzGLTeyCS7718xk7clYy0z3yLvvkz7MiywI8iNLx/iA z Wla00XuszT3evf8cxHJ3Se0/PxO1ta9IGeBVmgA9huwn00VTlkZ0XhEQ9RUJBko6WK2HuxR/QeXC y 4wE9JEyEb/i6jJ6miCg= ; Originator: bergv@math.uiuc.edu (Maarten Bergvelt) let R1 be the real numbers greater than 1. I am currently investigating the following set PF of functions :R1->R1. Let PF be the smallest set such that 1. the identity function id(x)=x is in PF and for each two functions f,g in PF the following functions are in PF 2. the power f^g 3. the function concatenation fog 4. the inverse function f~ The inverse function can be taken because by induction all functions remain strictly increasing and continuous (and so bijective). For example: x^(x^x)=(id^(id^id))(x) is in PF, its inverse h is in PF (i.e. the unique function with h^(h^h)=id) and x^{h(x)} is in PF. Now my question is whether f(x) equals g(x) only at finitely many arguments x. Because f~ o g also in PF we can reduce the question to whether each function f of PF has finitely many fixed points (f~ o g (x) = x iff g(x) = f(x)). Which I would heavily conjecture. A small subset of the functions of PF are the functions x^{x^{p(x)}} where p(x) is a polynomial with positive integer coefficients. For example x^{x^{x+3x^2}}=(((x^{x^x})^{(x^x)^x})^{(x^x)^x})^{(x^x)^x} two functions of that kind are equal if their polynoms are equal x^{x^{p(x)}} = x^{x^{q(x)}} <=> p(x)=q(x) And the polynoms are equal at only at finitely many points. So for this (very) small subset at least the conjecture is satisfied. === Subject: : Re: power iterated functions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > let R1 be the real numbers greater than 1. > I am currently investigating the following set PF of functions :R1->R1. > Let PF be the smallest set such that > 1. the identity function id(x)=x is in PF > and for each two functions f,g in PF the following functions are in PF > 2. the power f^g > 3. the function concatenation fog > 4. the inverse function f~ > The inverse function can be taken because by induction all functions > remain strictly increasing and continuous (and so bijective). > For example: x^(x^x)=(id^(id^id))(x) is in PF, > its inverse h is in PF (i.e. the unique function with h^(h^h)=id) > and x^{h(x)} is in PF. > Now my question is whether f(x) equals g(x) only at finitely many arguments > x. > Because f~ o g also in PF we can reduce the > question to whether each function f of PF has finitely many fixed > points (f~ o g (x) = x iff g(x) = f(x)). Which I would heavily conjecture. > A small subset of the functions of PF are the functions x^{x^{p(x)}} > where p(x) is a polynomial with positive integer coefficients. For > example > x^{x^{x+3x^2}}=(((x^{x^x})^{(x^x)^x})^{(x^x)^x})^{(x^x)^x} > two functions of that kind are equal if their polynoms are equal > x^{x^{p(x)}} = x^{x^{q(x)}} <=> p(x)=q(x) > And the polynoms are equal at only at finitely many points. So for this > (very) small subset at least the conjecture is satisfied. How do you generate polynomials (x^{x^{x+3x^2}})? I don't understand this. === Subject: : Re: power iterated functions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) << let R1 be the real numbers greater than 1. I am currently investigating the following set PF of functions :R1->R1. Let PF be the smallest set such that 1. the identity function id(x)=x is in PF and for each two functions f,g in PF the following functions are in PF 2. the power f^g 3. the function concatenation fog 4. the inverse function f~ The inverse function can be taken because by induction all functions remain strictly increasing and continuous (and so bijective). For example: x^(x^x)=(id^(id^id))(x) is in PF, its inverse h is in PF (i.e. the unique function with h^(h^h)=id) and x^{h(x)} is in PF. . . . . . . ---------------------------------------------------------------------------- ------------------------------------------------ I don't know what induction you're thinking of, but it cannot work as stated. Since 1) says f(x) := x is in your set, 2) implies that g(x) := x^x is also in your set. But g has a local minimum (at 1/e) and hence cannot be one-to-one on (0, oo), and so does not have an inverse function. Thus 1), 2), and 4) taken together lead to a contradiction. --Daniel Asimov === Subject: : Re: power iterated functions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Since 1) says f(x) := x is in your set, 2) implies that g(x) := x^x > is also in your set. But g has a local minimum (at 1/e) and hence > cannot be one-to-one on (0, oo), and so does not have an inverse that's why I only considering the functions on R1=(1,oo). === Subject: : Lie Algebra Classification Question Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Is there an easy classification, with a reasonably easy proof, of all finite-dimensional complex Lie Algebras L with one-dimensional derived subalgebra [L,L]? Fulton's book classifies all three-dimensional Lie Algebras according to the size of their derived subalgebra, but I was wondering if this aspect of the classification could be extended to n-dimensional algebras. === Subject: : curvatures Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Is it true that if gamma and delta are two curves of lengths L and M, respectively, whose curvatures satisfy k_gamma(tL) = k_delta(tM) forall tin [0,1] then gamma(tL) = delta(tM) as well? (probably modulo translations and rotations). === Subject: : Re: Chinese mathematicians added the final brick to Poincare's Conjecture? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Recently it is massively hyped on Chinese news media that two Chinese > mathematicians added the final brick to Poincare's Conjecture. I can't > find any international news media confirming it, except China's > government news agency Xinhua Agency: > http://english.people.com.cn/200606/04/eng20060604_270860.html > The English Wikipedia and MathWorld don't mention it either. > Do these Chinese mathematicians (Cao and Zhu) really make a critical > contribution to the Conjecture? the complete proof of the Poincar.8e Conjecture, which had puzzled mathematicians around the world, said Professor Shing-Tung Yau, a mathematician at Harvard University -- then it is most likely true. Shing-Tung Yau is an eminent differential geometer who won the Fields Medal in 1982. He also seems to be the first internationally renowned mathematician who has been willing to assert that the Poincar.8e Conjecture has been proved. --Daniel Asimov === Subject: : projective geometry Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Suppose I have a smooth curve in R^2 so that any line hits it at at most two points and so that it obeys the following hexagon rule: if six points P(1), P(2),... P(6) on the curve define a hexagon so that two of the three pairs of opposite sides are parallel (i.e., the line through P(1) and P(2) is parallel to the line through P(4) and P(5), etc.), then the third pair of sides are parallel. Is such a curve necessarily a conic? These properties indeed hold for conics, but I believe only for conics. I figure this must be known or easy (but not === Subject: : Re: power iterated functions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Addendum: The induction is as follows: f~ and fog is obviously again strictly increasing and continuous. f^g is continuous. It is also strictly increasing because generally a^x is strictly increasing for a>1 and x^a is strictly increasing for a>0. For x How do you generate polynomials (x^{x^{x+3x^2}})? > I don't understand this. Those polynomials are amongst others the outcome when not using the inverse functions at all in the generation of PF. We have the two laws: 1. x^{ab} = (x^a)^b 2. x^{a+b} = (x^a)(x^b) together follows x^{x^{a+b}) = x^{(x^a)(x^b)}=(x^{x^a})^{x^b}. So one could say + becomes ^ if it goes 2 steps down. Multiplication * becomes ^ when only going one step down. Now x+3x^2 = x + xx + xx + xx is a combination of + and * and so at most two steps down they become ^ and so x^{x^{x+3x^2}} is part of PF. Or explicitly: x^{x^{x+xx+xx+xx}}=x^{ (x^x)(x^{xx})(x^{xx})(x^{xx}) } = (x^{x^x})^{(x^x)^x}^{(x^x)^x}^{(x^x)^x} === Subject: : Re: curvatures Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Is it true that if gamma and delta are two curves of lengths L and M, > respectively, whose curvatures satisfy > k_gamma(tL) = k_delta(tM) forall tin [0,1] > then > gamma(tL) = delta(tM) > as well? (probably modulo translations and rotations). Short answer: no. If gamma and delta are line segments parametrized arbitrarily, the curvature hypothesis vacuously holds. Longer answer: sort of. Curvature, a geometric quantity, doesn't depend on parametrization, so you must equate the curvatures at points that correspond for some geometric reason (say, by being the same distance from an end of the curve). Perhaps you're thinking of the fundamental theorem of plane curves: If gamma and delta are (sufficiently) smooth *plane* curves parametrized by arc length (or, more generally, having the same speed functions) and if k_gamma(t) = k_delta(t) for all t, then (up to an isometry of the plane) gamma(t) = delta(t). The higher-dimensional version is similar, except in R^n there are (n-1) curvature functions that must match. The version in R^3 usually gets taught in first-semester differential geometry, and should be covered in any undergrad text. If memory serves, one of Spivak's volumes (probably Volume 2) contains the R^n version. Andrew D. Hwang Dept of Mathematics and CS College of the Holy Cross Worcester, MA, 01610-2395, USA === Subject: : Re: curvatures Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >> Is it true that if gamma and delta are two curves of lengths L and M, >> respectively, whose curvatures satisfy >> k_gamma(tL) = k_delta(tM) forall tin [0,1] >> then >> gamma(tL) = delta(tM) >> as well? (probably modulo translations and rotations). > Short answer: no. If gamma and delta are line segments parametrized > arbitrarily, the curvature hypothesis vacuously holds. > Longer answer: sort of. Curvature, a geometric quantity, doesn't depend > on parametrization, so you must equate the curvatures at points that > correspond for some geometric reason (say, by being the same distance > from an end of the curve). Perhaps you're thinking of the fundamental > theorem of plane curves: > If gamma and delta are (sufficiently) smooth *plane* curves > parametrized by arc length (or, more generally, having the same speed > functions) and if k_gamma(t) = k_delta(t) for all t, then (up to an > isometry of the plane) gamma(t) = delta(t). > The higher-dimensional version is similar, except in R^n there are > (n-1) curvature functions that must match. The version in R^3 usually > gets taught in first-semester differential geometry, and should be > covered in any undergrad text. If memory serves, one of Spivak's > volumes (probably Volume 2) contains the R^n version. > Andrew D. Hwang > Dept of Mathematics and CS > College of the Holy Cross > Worcester, MA, 01610-2395, USA curves in R^2. No lines allowed, but even with the lines, they are equivalent modulo rotation and scale. I also didn't mention many other things. Please see my other reply in this thread. === Subject: : Re: curvatures Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Is it true that if gamma and delta are two curves of lengths L and M, >respectively, whose curvatures satisfy >k_gamma(tL) = k_delta(tM) forall tin [0,1] >then >gamma(tL) = delta(tM) >as well? (probably modulo translations and rotations). I'm assuming that these are curves in the plane (otherwise you certainly need torsion as well as curvature to have any hope of characterizing the curve), regarded as functions from [0,L] and [0,M] respectively to R^2. Now gamma(tL) = delta(tM) would say that the images of the curves are the same, which would mean that the lengths must be the same. So either you didn't really mean L and M to be the lengths, or you forgot about scaling, and the conclusion was meant to be gamma(tL)/L = delta(tM)/M (modulo an isometry: not just translations and rotations are allowed, but reflections as well). But the curvatures should be scaled too, and in the opposite direction: gamma(tL)/L = delta(tM)/M would mean L k_gamma(tL) = M k_delta(tM). If theta is the counterclockwise angle from the positive x axis to the tangent vector to the curve, and we parametrize by arc length, then the curvature kappa = |theta'|. For a smooth curve whose curvature is never 0, kappa determines theta' up to sign, and thus theta up to sign and an additive constant, and thus the curve itself up to an isometry. But if the curvature is 0 at some point, it can't distinguish between a case where theta' is positive on both sides of that point and one where theta' changes sign. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: : Re: curvatures Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >>Is it true that if gamma and delta are two curves of lengths L and M, >>respectively, whose curvatures satisfy >>k_gamma(tL) = k_delta(tM) forall tin [0,1] >>then >>gamma(tL) = delta(tM) >>as well? (probably modulo translations and rotations). > I'm assuming that these are curves in the plane (otherwise you certainly > need torsion as well as curvature to have any hope of characterizing the > curve), regarded as functions from [0,L] and [0,M] respectively to R^2. > Now gamma(tL) = delta(tM) would say that the images of the curves are > the same, which would mean that the lengths must be the same. So > either you didn't really mean L and M to be the lengths, or you > forgot about scaling, and the conclusion was meant to be > gamma(tL)/L = delta(tM)/M (modulo an isometry: not just translations > and rotations are allowed, but reflections as well). But the curvatures > should be scaled too, and in the opposite direction: > gamma(tL)/L = delta(tM)/M would mean L k_gamma(tL) = M k_delta(tM). All these assumptions are correct. I was referring to plane curves, and yes, I mistakenly neglected scale. That's what I get for working at 4am. It should be gamma:[0,L]-> R^2 delta:[0,M]-> R^2 and I want to define a distance invariant under similarity transforms (scale, rotation and translation), more or less as the L^2 norm between the curvatures k_g and k_d: d(gamma,delta) = int_0^1 (L*k_g(tL) - M*k_d(tM))^2 dt Of course, since d is scaling invariant, one can't expect gamma(tL)=delta(tM), but the length 1 versions should be equal. gamma(tL)/L=delta(tM)/M. Is this metric known to geometers? Can we define the appropriate manifold of curves, compute geodesics, etc.? > If theta is the counterclockwise angle from the positive x axis > to the tangent vector to the curve, and we parametrize by arc length, > then the curvature kappa = |theta'|. For a smooth curve whose > curvature is never 0, kappa determines theta' up to sign, and thus > theta up to sign and an additive constant, and thus the curve itself > up to an isometry. But if the curvature is 0 at some point, it can't > distinguish between a case where theta' is positive on both sides of > that point and one where theta' changes sign. I thought of using k=dottheta, but I was being thrown off by the different arclengths. Now, with the appropriate scaling, it's trivial to show gamma(tL)/L = delta(tM)/M. The k=0 case is not a problem if k is the signed curvature (as it should): k= for then, if theta is defined by dotgamma = (costheta,sintheta), it follows k=dottheta (no | |). === Subject: : Re: Lie Algebra Classification Question Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Is there an easy classification, with a reasonably easy proof, of all > finite-dimensional complex Lie Algebras L with one-dimensional derived > subalgebra [L,L]? Yes. This actually works over any field K of characteristic different from 2. Every such algebra is the direct product of an abelian Lie algebra and either: - the (2n+1)-dimensional Heisenberg Lie algebra (n>0), or - the 2-dimensional non-abelian Lie algebra. Proof. Set [L,L]=Kz First case: [L,L] is not central in L. Pick x such that [x,z] is nonzero. Then is a 2-dimensional, non-abelian ideal in L. As the 2-dimensional Lie algebra is complete (every automorphism is inner) with trivial center, is a direct factor. As L has 1-dimensional derived subalgebra, the direct factor must be abelian. Second case: [L,L] is central in L. For all u,vin L, set [u,v]=B(u,v)x. Then B is a bilinear alternating form. We can decompose V as a direct sum W+X, so that the restriction of B on W is symplectic, and X is the kernel of B (i.e. the center of L). If X' is a complement subspace of [L,L] in X, then X' is abelian, and L is isomorphic to the direct product of H=(W+[L,L]) and X'. Taking a standard basis for the 2n-dimensional symplectic space W, one readily obtains that H is isomorphic to the (2n+1)-dimensional Heisenberg Lie algebra, defined as having a basis (x1,...,xn,y1,...,yn,z), where all brackets are zero, except those fo the form [xi,yi]=z. -- Yves === Subject: : Re: Lie Algebra Classification Question Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >> Is there an easy classification, with a reasonably easy proof, of all >> finite-dimensional complex Lie Algebras L with one-dimensional derived >> subalgebra [L,L]? > Yes. > This actually works over any field K of characteristic different from > 2. > Every such algebra is the direct product of an abelian Lie algebra and > either: > - the (2n+1)-dimensional Heisenberg Lie algebra (n>0), or > - the 2-dimensional non-abelian Lie algebra. > Proof. Set [L,L]=Kz > First case: [L,L] is not central in L. Here is an elementary way to deal with the first case. Let L' = . We may find y so that [x,y] = x. Let z be any element of L; suppose that [x,z] = ax, [y,z] = bx. Then z + bx - ay commutes with x and y. So L decomposes as a direct sum + C_L(). Since the derived algebra of C_L() is contained in , C_L() is abelian. Advertisement: this question appears as one of the harder problems in the book `Introduction to Lie Algebras' by Karin Erdmann and me, recently published in Springer's Undergraduate Mathematics Series. > Second case: [L,L] is central in L. > For all u,vin L, set [u,v]=B(u,v)x. Then B is a bilinear alternating > form. We can > decompose V as a direct sum W+X, so that the restriction of B on W is > symplectic, > and X is the kernel of B (i.e. the center of L). > If X' is a complement subspace of [L,L] in X, then X' is abelian, and L > is > isomorphic to the direct product of H=(W+[L,L]) and X'. Taking a > standard basis > for the 2n-dimensional symplectic space W, one readily obtains that H > is > isomorphic to the (2n+1)-dimensional Heisenberg Lie algebra, defined as > having > a basis (x1,...,xn,y1,...,yn,z), where all brackets are zero, except > those fo the form > [xi,yi]=z. > -- > Yves Mark Wildon === Subject: : Obituary: Ed Hook Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I have just read on a topology discussion list I subscribe to (http://www.lehigh.edu/~dmd1/algtop.html) that Ed Hook has died. === === Subject: : Ed Hook obit Word has been received of the February 21 death of Edward C. Hook, a 1970 Ph.D. student of Bob Stong.89s at Virginia. He was 60. Hook.89s dissertation .8bStable Equivariant Bordism.8a was published in Mathematische Zeitschrift (1972) in a joint paper with Theodor Br.9acker. After a postdoctoral appointment at M.I.T., Dr. Hook was on the Fordham faculty. He then joined Control Data. For the past 15 years, he was a senior software engineer with the NASA Advanced Supercomputing Division at the Ames Research Center, California. Ed Hook is survived by three University of Virginia alumnae: his wife Marda and daughters Karen and Elizabeth. I didn't really know Ed, but had interacted with him on occasion in and had a good deal of respect for his mathematics. I was saddened by the news of his passing. Dale.