mm-3089 === === Subject: Re: A foggy day...speed limit lower in rain >ever see a blue circle with a red cross ? > No, that's a red circle with a red cross, on a blue background. > It means no stopping. > (Yeah, I know what you meant, I should have put a smiley in my previous post.) > If there was only one diagonal bar instead of two crossed bars it would mean no parking. Which raises the question, why do we need so many different signs for no stopping / parking? 1a. On a bezant azure, fimbriated gules a saltire of the second (no stopping); 1b. The same with the a bend rather than a saltire (no parking); 2. The letter P in black beneath a red diagonal line in a red circle (no parking); 3. Broken yellow lines along the edge of the road (no stopping, in towns); 4. Diagonal white lines on the far side of a continuous white line (no stopping, on the open road); 5. The letters NP in white on a blue ground. (no parking). Has anyone seen any others? When does stopping become parking? Michael === Subject: Re: A foggy day...speed limit lower in rain On Mon, 10 Jul 2006 10:24:00 +1200, snip--- >Which raises the question, why do we need so many different signs for no >stopping / parking? There's a difference between stopping and parking. -- Brian Dooley Wellington New Zealand === Subject: Re: A foggy day...speed limit lower in rain >Which raises the question, why do we need so many different signs for no >stopping / parking? > There's a difference between stopping and parking. That's all you can think of to post here??? Get off usenet you attention seeking wit.. === Subject: Re: Rudimentary Math Question? > Then what number squared results in -1 ? >> The number i. Do a search on imaginary numbers if you're interested. > You're forgetting about -i, as (-i)^2 = -1 as well. > Also, z^2 = i if z = (1 + i)/sqrt(2). > --- Christopher Heckman === Subject: Re: Rudimentary Math Question? You guys are all correct, but i is a definition, not a derivation, right? WDA end > Then what number squared results in -1 ? >> The number i. Do a search on imaginary numbers if you're interested. > You're forgetting about -i, as (-i)^2 = -1 as well. > Also, z^2 = i if z = (1 + i)/sqrt(2). > --- Christopher Heckman === Subject: Re: Rudimentary Math Question? You guys are all correct, but i is a definition, not a derivation, right? The number i is a result of the attempt to make every polynomial have a root (zero). x^2 + 1 is not zero for any real number z, so a non-real number was defined so that it made x^2 + 1 = 0. This is like working with the rational numbers and realizaing that you don't have a solution to x^2 - 2 = 0; that's where the square root of 2 (and its negative) comes from. --- Christopher Heckman > Then what number squared results in -1 ? >> The number i. Do a search on imaginary numbers if you're interested. > You're forgetting about -i, as (-i)^2 = -1 as well. > Also, z^2 = i if z = (1 + i)/sqrt(2). === Subject: Loose connectivity, factoring and residues The factoring problem can be easily approached using simple algebra. Start with x^2 - y^2 = S - 2*x*k where all are integers, as notice then you trivially have x^2 + 2*x*k + k^2 = y^2 + S + k^2 so x+k = sqrt(y^2 + S + k^2) and finding y is just a matter of factoring (S+k^2)/4. Now with just the explicit equation you end up with nothing but trivialities, but turning to congruences, you can now simply let x^2 - y^2 = 0 mod T which--this is important--now forces S - 2*x_res*k = 0 mod T where I put in x_res to emphasize that now it's congruences, so there is loose connectivity and an explicit value of x is not needed--just a residue. But now I can just solve for k, assuming 2, S and x are coprime to T: k = S*(2*x_res)^{-1} mod T where (2*x_res)^{-1} is the modular inverse of (2*x_res) mod T. That the modular inverse makes an appearance is critical, but more importantly I now have a way to find all the variables!!! That can be done by simply picking a residue for x_res and then picking S, like x_res = 1, and S =1, to get k. For instance if T=35, and I use x_res=S=1, then k = 18 mod 35, and k=18 will suffice. Then y is found by factoring (1+18^2)/4 and then you have x as well. Of course there will exist and x and y such that x^2 - y^2 = 0 mod T for any x_res you choose, which is trivial to prove, as that is equivalent to x^2 - y^2 = kT where k can be any integer. So an equation that is useless explicitly becomes quite powerful with modular algebra--introducing loose connectivity--leading to a general method for factoring. === Subject: Re: Loose connectivity, factoring and residues > The factoring problem can be easily approached using simple algebra. ... > For instance if T=35, and I use x_res=S=1, then k = 18 mod 35, and k=18 > will suffice. > Then y is found by factoring (1+18^2)/4 and then you have x as well. And how does one factor ( 1 + 18^2 ) / 4 = 81.25 ? Mike === Subject: Anyone Knows The Answer To This? What are the five odd numbers that equal 20? TIA. === Subject: Re: Anyone Knows The Answer To This? Distribution: world Timmy TIA. -5 + 15 + 5 + 3 + 1 + 1 Obviously no 5 positive odd integers can total 20 --- there sum is odd! === Subject: Re: Anyone Knows The Answer To This? || What are the five odd numbers that equal 20? | -5 + 15 + 5 + 3 + 1 + 1 | Obviously no 5 positive odd integers can total 20 --- there sum is | odd! There are six odd numbers (above), not five. ______________Gerard S. === Subject: Re: Anyone Knows The Answer To This? > What are the five odd numbers that equal 20? > TIA. I'd be happy to find just *one* odd number that equals 20, let alone five of them. === Subject: Re: Anyone Knows The Answer To This? > What are the five odd numbers that equal 20? 1,1,1,1, and 16 Sixteen is condisered even by mathematicians, but in some contexts it is a very odd number. I think William may be faminiar with that riddle too. For example: don't you think 16 would be an odd number of sugar cubes to have in a hot drink. Bye. Jasen What's blue and white ans climbs trees. === Subject: Re: Anyone Knows The Answer To This? > What are the five odd numbers that equal 20? > 1,1,1,1, and 16 > Sixteen is condisered even by mathematicians, but > in some contexts it is a very odd number. > I think William may be familiar with that riddle too. All primes are odd, especially 2 because it's the only even prime. > For example: don't you think 16 would be an odd number of sugar > cubes to have in a hot drink. No. Some like it sweet. > What's blue and white ans climbs trees. A cold white man wanting to get out of the Bush. === Subject: Re: Anyone Knows The Answer To This? > What are the five odd numbers that equal 20? An odd number is of the form: 2k+1 The sum of five odd numbers looks like: = (2a+1 + 2b+1 + 2c+1 + 2d+1 + 2e+1) = 2(a + b + c + d + e + 2) + 1 That '+1' on the end there means that the sum of five odd numbers is always odd. There is no five odd numbers that sum to 20. Nicholas Sherlock -- http://www.sherlocksoftware.org === Subject: Re: Anyone Knows The Answer To This? > What are the five odd numbers that equal 20? > TIA. LOL === Subject: Logistic time-step porblem First of all I'm not sure if this belongs here, if not, could you suggest a more appropriate group. I'm having some problems researching the possibility of explicit solutions to military logistics problems. Although, I think the actual problem is more general. I have a time-step simulation tool that simulates strategic-lift (aircraft, ships, etc...) moving stuff from a start port to an end port. Obviously aircraft make many flights back and forth. While a time-step simulation suits this problem, it is quite expensive in time. The question I need answering is whether there exists an explicit solution to such a problem. I'm pretty sure it's not linear, and perhaps chaotic behaviour prevents such a solution. By the way, when I say solution I mean things like finding: a) How long it takes to move everything with various amounts of start-lift, port infrastructure (#berths, #AC handlers, etc...) and so on; b) The optimum number of the above to produce the quickest time; etc... I would be very grateful if you could point me in a direction to find if such a thing is possible and if so, whether the complexity of such a solution would make it impractical. confusing!) Chris Riddle === Subject: Roulettes Content-Length: 475 Originator: rusin@vesuvius What are the intrinsic and parametric equations [x(s),y(s)] traced by a point in 2D of a rigid curve having natural equation rho1=f(s) rolling on another fixed curve of natural equation rho2=g(s) ? and vice-versa? ( rho is radius of curvature, s arc length) In particular, What transformation/substitution of equations transforms a cycloid (rigid circle rolling on a fixed line) to an involute (rigid line rolling on a fixed circle)? And vice-versa? Narasimham === Subject: Some Very Trivial Ideas on Morse Theory Content-Length: 844 Originator: rusin@vesuvius Although I am a undergraduate student in Physics, I am always very interested in self-learning Mathematics. I learned of Morse theory some time ago [I thought it was very difficult, as I knew it is related with Floer Homology], and I am really fond of its elegance and power. I think it will be nice if we can do further generalizations of it to the Lie group-valued case [when I learned of Novikov's theories I have a bit more confidence on this], and maybe this will have some connection with principal G-bundle, characteristic class and gauge theory invariants such as Seiberg-Witten as well. So I spend some time to think about it and write some of my ideas here: http://hkusua.hku.hk/~h0399932/draft.png It's very trivial and very prime. I still have too much to learn. But I === Subject: Thoughts on Bessel functions Content-Length: 1146 Originator: rusin@vesuvius Consider the homogenous differential equation x y'' + (n+1) y' = y where n is a fixed complex number. If n is not a negative integer, we can develop one solution in series as B_n(x) = sum_i=0^infinity 1/(n+1)^(i) x^i/i! = sum_i=0^infinity x^i/(C(n+i,n) i!^2) Here a^(b) = Gamma(a+b)/Gamma(a) is the Pochhammer symbol, and C(a,b) the extended binomial coefficient function. The Bessel functions and modified Bessel functions of the first kind can both be expressed in terms of B_n. We have B_n(x) = n! I_n(2 sqrt(x))/x^(n/2) so that I_n(z) = 1/n! (z/2)^n B_n(z^2/4) Similarly, we have J_n(z) = 1/n! (z/2)^n B_n(-z^2/4) Outside of n equaling a negative integer, this function B is better behaved than the Bessel functions, since it is entire. It has a simpler differential equation. It expresses both the J and I functions in terms of a single function without requiring imaginary arguments. Arguably, it seems to me, it is a better way to start out the theory of Bessel functions. What I'm wondering is if anyone has actually done this, and if B_n has a name. Also, can anyone see why this approach isn't used, if in fact it is not? === Subject: Re: Thoughts on Bessel functions Content-Length: 2401 Originator: rusin@vesuvius >Consider the homogenous differential equation >x y'' + (n+1) y' = y >where n is a fixed complex number. If n is not a negative integer, we >can develop one solution in series as >B_n(x) = sum_i=0^infinity 1/(n+1)^(i) x^i/i! = sum_i=0^infinity >x^i/(C(n+i,n) i!^2) >Here a^(b) = Gamma(a+b)/Gamma(a) is the Pochhammer symbol, and C(a,b) >the extended binomial coefficient function. >The Bessel functions and modified Bessel functions of the first kind >can both be expressed in terms of B_n. We have >B_n(x) = n! I_n(2 sqrt(x))/x^(n/2) >so that > I_n(z) = 1/n! (z/2)^n B_n(z^2/4) >Similarly, we have >J_n(z) = 1/n! (z/2)^n B_n(-z^2/4) >Outside of n equaling a negative integer, this function B is better >behaved than the Bessel functions, since it is entire. It has a simpler >differential equation. It expresses both the J and I functions in terms >of a single function without requiring imaginary arguments. Arguably, >it seems to me, it is a better way to start out the theory of Bessel >functions. What I'm wondering is if anyone has actually done this, and >if B_n has a name. Also, can anyone see why this approach isn't used, >if in fact it is not? The B_n(z^2/4) is usually written (see 9.6.47 in the Handbook of Mathematical Functions, edited by Abramowitz and Stegun) as 0F1(n+1,z^2/4), so it is too close to a special case of hypergeometric and confluent hypergeometric functions to spend a new name on it. The use of Bessel functions is first of all tied to the fact that they appear as solutions to many differential equations in engineering and physics; so everybody would rather use the old Bessel function notation instead of any new B-Function notation if the solution to such a real-world problem is needed as a paper-saving argument. Also: simplicity of the differential equation is not a unique argument: if we use diffraction theory of circular pupils in optics, we need the Fourier transforms of Chebyshev polynomials and immediately face the integral representations of the Bessel functions, so simplicity of integral representations is also a nice feature. Also: what would be the linearly independent counterpart of B_n? This must be of equivalent simplicity to convince us to switch so we can patch solutions to the DE across boundaries :) === Subject: Re: Thoughts on Bessel functions Content-Length: 2052 Originator: rusin@vesuvius > The B_n(z^2/4) is usually written (see 9.6.47 in the Handbook of > Mathematical Functions, edited by Abramowitz and Stegun) as 0F1(n+1,z^2/4), > so it is too close to a special case of hypergeometric and confluent > hypergeometric functions to spend a new name on it. I found by consulting Watson that A. George Greenhill called it the Bessel-Clifford function, however. > The use of Bessel functions is first of all tied to the fact that they > appear as solutions to many differential equations in engineering and physics; > so everybody would rather use the old Bessel function notation instead > of any new B-Function notation if the solution to such a real-world problem > is needed as a paper-saving argument. Greenhill also pointed out that sometimes the new Bessel functions (which are actually older than Bessel, as Lagrange used them) sometimes work better in some engineering situations. Also: simplicity of the differential > equation is not a unique argument: if we use diffraction theory of circular > pupils in optics, we need the Fourier transforms of Chebyshev polynomials and > immediately face the integral representations of the Bessel functions, so > simplicity of integral representations is also a nice feature. Once again, sometimes these become simpler, and sometimes they are not. > Also: what would be the linearly independent counterpart of B_n? This must > be of equivalent simplicity to convince us to switch so we can patch > solutions to the DE across boundaries :) Here's an integral representation for it: x^(-n/2) K_n(2 sqrt(x)) = (1/2) int_0^infinity exp(-t-x/t) dt/t^(n+1) defined for Re(x) > 0. This not only defines the other solution, x^(-n/2)K_n(2 sqrt(x)), it provides an example of when the integral representation is simpler in the new form. If we don't normalize the hypergeometric way, but instead use x^(-n/2)I_n(2 sqrt(x)), we get another very nice feature, by the way: a function with which is holomorphic on C^2; in other words, in both variables n and x. === Subject: Call for Papers: Graphs, Mappings and Combinatorics at the Minnesota State Fair Content-Length: 1651 Originator: rusin@vesuvius Leonardo's Basement (www.leonardosbasement.org) a Minneapolis based school for Art, Science, Mathematics and Technology has been most honorably charged with the assignment of do some math stuff for us by the Minnesota State Fair Organizing Committee. To this end we are issuing a call for papers on the topic of Graphs, Mappings and Combinatorics at the Minnesota State Fair. Papers submitted need not pertain specifically to the Minnesota State Fair but should generally embrace the topic of a state fair for example: 1 - animal sizes 2 - crowd movement 3 - corndog consumption 4 - are people having fun? 5 - hypotheses on why some amusement rides do not exist 6 - puzzle challenges 7 - puzzle solutions 8 - anything having to do the SOMA Cube Puzzle, because we are making a huge one (it is kind of how we got the gig) Dissertations that are tongue in cheek, completely unprovable, incomprehensible or just plain funny to read out loud, are of course most welcome. Please submit by August 1, 2006. Selected papers will be presented on top of the world's largest SOMA Cube Puzzle (assembled from 4ftx4ftx4ft cubes) on September 4, 2006. Presentation preference will be shown to participants who: 1 - submit 2 - can show up 3 - have an appetite for the absurd At this time we are not able to reimburse travel or lodging expenses. However if you have a truly gonzo mathematics presentation I will be most encouraged to figure something out. Please send your submission to my special no spam email: epvnospam-nospam1@yahoo.com If you need to snail mail please contact me at the above email address and I will provide a mailing address. === Subject: Computing the Gerstenhaber Bracket Content-Length: 538 Originator: rusin@vesuvius I've been working on doing some computations in Hochschild cohomology, and all the definitions I've seen of the Gerstenhaber bracket define it either hopelessly abstractly or in terms of the bar resolution. I have a different resolution I've been using to compute HH^*, and I was hoping to be able to compute the Gerstenhaber bracket (or HH^2 to a specific formal deformation of the algebra) without having to compute an explicit quasi-isomorphism to the bar resolution. Any ideas or references would be appreciated. Thanx, Aaron === Subject: Re: Computing the Gerstenhaber Bracket Content-Length: 776 Originator: rusin@vesuvius > I've been working on doing some computations in Hochschild cohomology, > and all the definitions I've seen of the Gerstenhaber bracket define it > either hopelessly abstractly or in terms of the bar resolution. I have a > different resolution I've been using to compute HH^*, and I was hoping > to be able to compute the Gerstenhaber bracket (or HH^2 to a specific > formal deformation of the algebra) without having to compute an explicit > quasi-isomorphism to the bar resolution. Any ideas or references would > be appreciated. I do not think anyone has figured out how to compute the bracket from an arbitrary projective resolution. Indeed, most computations that I am aware of explicitly construct those dreaded quasi-isomorphisms. -- m === Subject: Re: Computing the Gerstenhaber Bracket Content-Length: 1230 Originator: rusin@vesuvius Aaron Bergman ha escrit: > I've been working on doing some computations in Hochschild cohomology, > and all the definitions I've seen of the Gerstenhaber bracket define it > either hopelessly abstractly or in terms of the bar resolution. I have a > different resolution I've been using to compute HH^*, and I was hoping > to be able to compute the Gerstenhaber bracket (or HH^2 to a specific > formal deformation of the algebra) without having to compute an explicit > quasi-isomorphism to the bar resolution. Any ideas or references would > be appreciated. I don't know if I understand the problem: do you mean you have a resolution that allows you the computation of the Gerstenhaber bracket and you want an explicit quasi-isomorphism with the bar resolution because your are not sure that your resolution actually computes Hochschild cohomology? If this is the case, you don't need this explicit quasi-isomorphism: since Hochschild cohomology is a derived functor (an Ext; Weibel, An introduction to homological algebra, lemma 9.1.3), all you have to check is if your resolution is projective (in an apropriate sense). Then, the usual theorem of comparison of projective resolutions does the work for you. Agust.92 Roig === Subject: A conjecture on class numbers Content-Length: 661 Originator: rusin@vesuvius Hello all, Just two curious conjectures here: Let m,n be any element of {0,1,2,....} 1. Let F(n) be a square-free number of form 8n+3. If F(n) is prime, then -8F(n) has class number h(d) of form 4m+2. If composite, then of form 4m. Ex. F(n) = 11 with h(-8*11) = 2; F(n) = 35 with h(-8*35) = 4. 2. Let F(n) be a square-free number of form 8n+5. If F(n) is prime, then both -4F(n) and -8F(n) have class numbers of form 4m+2. If composite, then of form 4m. Ex. F(n) = 13 with h(-4*13) = 2, h(-8*13) = 6; F(n) = 85 with h(-4*85) = 4, h(-8*85) = 12. True or not? P.S. Numbers 8n+1 and 8n+7 are not that well-behaved. --Titus === Subject: how to efficiently solve a relatively large scale linear equation system? Content-Length: 766 Originator: rusin@vesuvius I am doing a project relating to solve relatively large scalce linear equation system, here is my equations: (A + T' *B * T) * X = (c + T' *d) Unknowns: X (N*1 vector, N is between 100 and 1000) Knowns: A-----N*N symmetric narrow band matrix with band width= 5 (actually it comes from assembling stiffness matrix of scale 6*6) B-----N*N symmetric narrow band matrix with band width= 2 T------N*N lower triangular matrix with first 3 elements in diagnol are zeros. (this means this is a singular matrix) T'------transpose of T c-------N*1 vector d-------N*1 vector I need to solve this equation system in C++ efficiently and the speed for solving the equation is CRITICAL for me. Anybody has any idea? Helen === Subject: Re: A curious identity Content-Length: 692 Originator: rusin@vesuvius Not if L = 1. If I'm interpreting this right, all terms of S_1 are |i-i|, hence zero. It follows that you should require n >= 2 (requiring 2n+1 to be prime does not of itself eliminate n = 1), and place L in G{1}. The primality of 2n+1 enforces that |i-j| with i and j in the indicated order occurs in *one* S_L sum for any (i,j) in GxG, and that all zero elements |i-i| occur specifically in the L = 1 sum. The sum of all the absolute values is twice a tetrahedral number, specifically (n-1)*n*(n+1)/3. If this partitions equally among the (n-1) values of L that have the nonzero terms, we will get n*(n+1)/3 for each. FWIW, my *guess* is that the modified proposition is true. --OL === Subject: Re: characterization of W^{1,p} norms Content-Length: 1479 Originator: rusin@vesuvius > I have a question concerning W^{1,p} norms: > is it true that the W^{1,p} norm of a function $v$ > can be > characterized as > int u v + nabla u nabla v > with $u$ ranging in W^{1,q}? > (q is the dual exponent of p) > A possible proof is the followinig: a standard > representation theorem > for the dual of W^{1,p} in N dimensions states that > any dual F can be > represented through a N+1 ple of L^q functions f_i, > so that > =int f_0 u + (f_1,...f_N) nabla u > So, if I want to compute the norm of u, I must test > its gradient > against all vector fields. On the other hand the > vector field > (f_1,...,f_N) can be decomposed into a sum of a > solenoidal field and a > gradient. When testing with such decomposition, the > solenoidal field is > calceled out, and we remain with the scalar product > of two gradients. > Any opinions? > joe > Joe I was flipping through a book and saw something along the lines of sup( int_Omega grad u grad v ) >= c || u || where the norm is the W_0^{1,p}(Omega) one and we are supping over v in W_0^{1,q} with norm of v <=1. Here q is conjugate of p. They said result followed from some Zygmund - Calderon theorem but I never figure out which one. If you have this figure out I would really like to know some of the details or where to look. I am in some open , bounded subset Omega of R^n . craig version=3.1.0 Received: from relay6.cso.uiuc.edu (relay6.cso.uiuc.edu [128.174.5.12]) by mail.math.niu.edu (8.13.6/8.13.6) with ESMTP id k687SqCa016283 for ; Sat, 8 Jul 2006 02:28:53 -0500 (CDT) Received: from chx400.switch.ch (chx400.switch.ch [130.59.10.2]) by relay6.cso.uiuc.edu (8.13.6/8.13.6) with ESMTP id k687SrJp019556 for ; Sat, 8 Jul 2006 02:28:54 -0500 (CDT) Received: from geraldo.cc.utexas.edu ([146.6.70.83]) by chx400.switch.ch with esmtp (Exim 3.20 #1) id 1Fz7FA-0007Rn-00 for sci-math-research@moderators.isc.org; Sat, 08 Jul 2006 09:28:48 +0200 Received: (from news@localhost) by geraldo.cc.utexas.edu (8.12.11/8.12.11/cc-solaris.mc-1.7) id k687SkSu028854 for sci-math-research@moderators.isc.org; Sat, 8 Jul 2006 02:28:46 -0500 (CDT) === Subject: Re: Computing the Gerstenhaber Bracket Status: RO Content-Length: 1431 > Aaron Bergman ha escrit: > I've been working on doing some computations in Hochschild cohomology, > and all the definitions I've seen of the Gerstenhaber bracket define it > either hopelessly abstractly or in terms of the bar resolution. I have a > different resolution I've been using to compute HH^*, and I was hoping > to be able to compute the Gerstenhaber bracket (or HH^2 to a specific > formal deformation of the algebra) without having to compute an explicit > quasi-isomorphism to the bar resolution. Any ideas or references would > be appreciated. > I don't know if I understand the problem: do you mean you have a > resolution that allows you the computation of the Gerstenhaber bracket > and you want an explicit quasi-isomorphism with the bar resolution > because your are not sure that your resolution actually computes > Hochschild cohomology? No. I have an explicit projective resolution that isn't the bar resolution. I was hoping to be able to compute the Gerstenhaber bracket from it, but judging from the other response, I guess that's not so likely. The parenthetical comment is somewhat unclear, so, to elaborate, I was also hoping to describe the explicit deformation for a given element of HH^2 using my resolution, but perhaps that is also hard. Thanx, Aaron === Subject: function semi-algebraic over real closed sub-field? Content-Length: 579 Originator: rusin@vesuvius Hi. Getting right to the point: Let F be a real-closed subfield of the reals R. Let f : [0,1]^n -> R denote a continuous real function, semi-algebraic (over R), satisfying that f(x) lies in F for every rational x (i.e. on Q). Does this imply f to be semi-algebraic over F? (It does under the additional hypothesis that f be a rational function: solve the interpolation problem for (x,f(x)), x in Q ) If it does not, how about if f is differentiable or analytic? Martin === Subject: Re: characterization of W^{1,p} norms Content-Length: 1479 Originator: rusin@vesuvius > I have a question concerning W^{1,p} norms: > is it true that the W^{1,p} norm of a function $v$ > can be > characterized as > int u v + nabla u nabla v > with $u$ ranging in W^{1,q}? > (q is the dual exponent of p) > A possible proof is the followinig: a standard > representation theorem > for the dual of W^{1,p} in N dimensions states that > any dual F can be > represented through a N+1 ple of L^q functions f_i, > so that > =int f_0 u + (f_1,...f_N) nabla u > So, if I want to compute the norm of u, I must test > its gradient > against all vector fields. On the other hand the > vector field > (f_1,...,f_N) can be decomposed into a sum of a > solenoidal field and a > gradient. When testing with such decomposition, the > solenoidal field is > calceled out, and we remain with the scalar product > of two gradients. > Any opinions? > joe > Joe I was flipping through a book and saw something along the lines of sup( int_Omega grad u grad v ) >= c || u || where the norm is the W_0^{1,p}(Omega) one and we are supping over v in W_0^{1,q} with norm of v <=1. Here q is conjugate of p. They said result followed from some Zygmund - Calderon theorem but I never figure out which one. If you have this figure out I would really like to know some of the details or where to look. I am in some open , bounded subset Omega of R^n . craig === Subject: Large deviations principles of Non-Freidlin-Wentzell type. Originator: israel@math.ubc.ca (Robert Israel) We consider potential type dynamical systems in finite dimensions with two meta-stable states. They are subject to two sources of perturbation: a slow external periodic perturbation of period T and a small Gaussian random perturbation of intensity s, and therefore mathematically described as weakly time inhomogeneous diffusion processes. A system is in stochastic resonance provided the small noisy perturbation is tuned in such a way that its random trajectories follow the exterior periodic motion in an optimal fashion, i.e. for some optimal intensity s(T). The physicists.89 favorite measures of quality of periodic tuning .9a and thus stochastic resonance .9a such as spectral power amplification or signal-to-noise ratio have proven to be defective. They are notrobust w.r.t. effective model reduction, i.e. for the passage to a simplified finite state Markov chain model reducing the dynamics to a pure jumping between the meta-stable states of the original system. An entirely probabilistic notion of stochastic resonance based on the transition dynamics between the domains of attraction of the meta-stable states .9a and thus failing to su er from this robustness defect .9a was proposed before in the context of one-dimensional diffusions. It is investigated for higher dimensional systems here, by using extensions and refinements of the Freidlin-Wentzell theory of large deviations for time homogeneous di usions. Large deviation principles developed forweakly time inhomogeneous di usions prove to be key tools for a treatment of the problem of diffusion exit from a domain and thus for the approach of stochastic resonance via transition probabilities between meta-stable sets. http://www.geocities.com/jaykovf1/Jakov3.pdf === Subject: How to solve this PDE (laplace equation)? Originator: israel@math.ubc.ca (Robert Israel) I am looking for suggestions on how to solve this differential equation in Q dimensions: nabla^2_a F[a,b,c] = R[a,b]^{2-Q} R[a,c]^{2-Q} with boundary condition F -> 0 as any of the |a| or |b| or |c| going to infinity. Here F[a,b,c] is a function of 3 Q-dimensional vectors a,b,c, nabla^2_a is the laplacian with respect to the vector a, and R[x,y] is the Euclidean distance between x and y R[x,y] = (sum_{i=1}^Q (x^i - y^i)^2 )^{1/2} Up to constant factors I believe the solution can be expressed as an integral F[a,b,c] propto int R[x,a]^{2-Q} R[x,b]^{2-Q} R[x,c]^{2-Q} d^{Q}x we can also re-write this integral in fourier space F[a,b,c] propto int int exp[i vec{k}.(a-c) + i vec{q}.(a-b)] / [ vec{k}.vec{k} vec{q}.vec{q} (vec{q}+vec{k}).(vec{q}+vec{k}) ] d^{Q}k d^{Q}q but is it possible to perform these integrals explicitly? I'm also interested if I could solve a generalized version of this equation -- with arbitrary number of variables nabla^2_{a_1} F[a_1,a_2,...,a_n] = R[a_1,a_2]^{2-Q} R[a_1,a_3]^{2-Q} ... R[a_1,a_n]^{2-Q} === Subject: Fixed points of a weakly continuous map f:H->H satisfying f(f(x))=x Originator: israel@math.ubc.ca (Robert Israel) Let H be an infinite-dimensional Hilbert space. Does there exist a mapping f:H--->H so that: i) f is weakly continuous, ii) f(f(x))=x for all x in H, iii) f has no fixed points (for no x f(x)=x) or every mapping f:H--->H satisfying i) and ii) must have a fixed point? === Subject: a limit problem Originator: israel@math.ubc.ca (Robert Israel) Define that (n)_r=n(n-1)...(n-r+1), what's the result of the limit: lim_{n->infty, r->infty, r/n->p} (n-k)_r / (n)_r. is it e^{-kp} ? how to prove it? === Subject: Re: a limit problem Originator: israel@math.ubc.ca (Robert Israel) > Define that (n)_r=n(n-1)...(n-r+1), > what's the result of the limit: > lim_{n->infty, r->infty, r/n->p} (n-k)_r / (n)_r. > is it e^{-kp} ? > how to prove it? First, using (n)_r = n!/(n-r)!, you see that (n-k)_r / (n)_r = (n-r)_k / (n)_k. Write this out, and you get (n-r)/n * (n-r-1)/(n-1) * ... * (n-r-k+1)/(n-k+1). The limit as n,r increases with r/n -> p is then (1-p) * (1-p) * ... * (1-p) = (1-p)^k. Einar === Subject: Re: Maximal entropy from Bayes' law? Originator: israel@math.ubc.ca (Robert Israel) It has been a while for me since posting to sci.math.research.. worked out the power (beta) of an HIV test example, which differs in a striking manner from the picture you get looking at the specificity and selectivity of the antibody test. (The forward probability, I think that is called.) I collected some links here and have tried to fit them into the OP with a few quotes from the linked documents. > I was walking on the beach with Chris Lee and James Dolan last > weekend, and we came across a problem about statistical inference. I don't know about you, but I love it when I come across things on the beach. :) > Suppose we repeatedly observe a system in a well-understood but > stochastic way and get some frequency distribution of outcomes. > What's our best guess for the probability that the system is in a > given state? I'm hoping that under some conditions the Bayesian > approach gives results that match Jaynes' maximum entropy method. In general, http://mathworld.wolfram.com/InverseProblem.html but specifically, http://citeseer.ifi.unizh.ch/12595.html A Comparison Of Two Approaches: Maximum Entropy On The Mean (MEM) And Bayesian Estimation (BAYES) For Inverse Problems may be of use. So now the question is What are the conditions? > Let's make the problem very finitistic to keep it simple. > Suppose X is a finite set of states and Y is a finite set > of observation outcomes. Suppose we have a function > f: Y x X -> [0,1] > giving the conditional probability f(y|x) for the outcome of > an observation to equal y given that the system's state is x. That seems more like a, or even *the* general observation problem. What we see is not always what is there. > We can think of f as a stochastic function from states to > observation outcomes, sending each point of X to a probability > measure on Y. By linearity, f gives a map from probability measures > on X to probability measures on Y. I'll also call this f. > Given a probability measure q on Y, what's our best guess > for a probability measure p on X with f(p) = q? Assume such > a measure exists. How is what we see colored by our expectation of what is there? How do we take the color of our expectations away from what we see? Zen provides answers to the philosophical problem. Math provides answers to actual problems. http://en.wikipedia.org/wiki/Prior_probability provides: Some attempts have been made at finding probability distributions in some sense logically required by the nature of one's state of uncertainty; these are a subject of philosophical controversy. > One approach is Jaynesian: maximize entropy! In other words, > look at the probability measures p with f(p) = q and choose > the p with maximum entropy. > Of course to define entropy we need a prior measure on X; > let's use counting measure. So, we are trying to maximize > sum_{x in X} p(x) ln(p(x)) > over p satisfying the linear constraint f(p) = q. This is something > people know how to do. http://en.wikipedia.org/wiki/Prior_probability provides: Another idea, championed by Edwin T. Jaynes, is to use the principle of maximum entropy. The motivation is that the Shannon entropy of a probability distribution measures the amount of information contained the distribution. The larger the entropy, the less information is provided by the distribution. Thus, by maximizing the entropy over a suitable set of probability distributions on X, one finds that distribution that is least informative in the sense that it contains the least amount of information consistent with the constraints that define the set. > Another approach is Bayesian: use Bayes' rule! The conditional > probability for any measurement outcome y given the state x is f(y|x). > Use counting measure on X as our prior, and use Bayes' law to > update this prior when we do an observation and get the outcome y. http://mathworld.wolfram.com/BayesTheorem.html > In fact, let's imagine that we keep doing observations, getting > outcomes distributed according to the probability measure q, and > use these to keep updating our probability measure on X. In the > limit of infinitely many observations, this measure should converge > (almost surely) to some fixed measure on X. > Do we get the same answer using both approaches? Once again, from http://en.wikipedia.org/wiki/Prior_probability : ----- ...Edwin T. Jaynes has published an argument (Jaynes 1968) based on Lie groups that suggests that the prior for the proportion p of voters voting for a candidate, given no other information, should be p ^(- 1) * (1 - p) ^(- 1) If one is so uncertain about the value of the aforementioned proportion p that one knows only that at least one voter will vote for Smith and at least one will not, then the conditional probability distribution of p given this information alone is the uniform distribution on the interval [0, 1], which is obtained by applying Bayes' Theorem to the data set consisting of one vote for Smith and one vote against, using the above prior. --------- Now that is something similar to what I did to work out the HIV testing example. I think I could do or at least follow that, but would have some difficulty as the population parameter p (not probability but proportion) wasn't covered in MTH 241 Statistics. But Bayes's rule was. I don't know my right foot from a Lie group yet. > If so, we might say we had a derivation of the maximal entropy > principle from Bayes' law... or vice versa. > Feel free to fix my question; my description of the Bayesian > approach is pretty sketchy, and I may have something screwed up. Oh, I can't do that, but I hope those who read these inserted links will have something to add. I'm just a machinist studying math at the community college! http://mathworld.wolfram.com/BayesianAnalysis.html This, I think, is more helpful, clear, rigorous, and general than the cents... Finally, http://citeseer.ifi.unizh.ch/mohammad-djafari96full.html A Full Bayesian Approach for Inverse Problems Doug Goncz Replikon Research Seven Corners, VA 22044-0394 === Subject: Re: Maximal entropy from Bayes' law? Content-Length: 1412 Originator: rusin@vesuvius Sorry that this reply is so broken up into separate posts. I keep having new thoughts about it. > I was walking on the beach with Chris Lee and James Dolan last > weekend, and we came across a problem about statistical inference. > Suppose we repeatedly observe a system in a well-understood but > stochastic way and get some frequency distribution of outcomes. > What's our best guess for the probability that the system is in a > given state? I'm hoping that under some conditions the Bayesian > approach gives results that match Jaynes' maximum entropy method. If we observe a number n of y's coming from a single x (denoting the set of outcomes by Y - the set is sufficient since the y samples are independent given x), then P(x | Y) propto P(Y | x) P(x) . If we take P(x) to be uniform, then we have P(x | Y) propto prod_{y in Y) f(y, x)^{n_{y}} , where n_{y} is the number of occurrences of y in the n observations. As n gets large, we will have n_{y} ~ n. f(y, x) so ln P(x | Y) ~ sum_{y in Y} n f(y, x) ln f(y, x) so assuming we evenutally observe every possible value of y, we will have P(x | Y) ~ e^{- n H[f](x)} where H[f](x) is the entropy of f(y, x) as a distribution on y. As n gets very large, P(x | Y) will concentrate on those x's which minimize H[f](x), which is not the maximum entropy result, at least not in the way you defined it in your post. Ian. === Subject: Re: Maximal entropy from Bayes' law? Content-Length: 3273 Originator: rusin@vesuvius > I was walking on the beach with Chris Lee and James Dolan last > weekend, and we came across a problem about statistical inference. > Suppose we repeatedly observe a system in a well-understood but > stochastic way and get some frequency distribution of outcomes. > What's our best guess for the probability that the system is in a > given state? I'm hoping that under some conditions the Bayesian > approach gives results that match Jaynes' maximum entropy method. Further to my previous post: as you have probably noticed, the situation you are discussing is more complex in one way, and more simple in another, than the one Jaynes is discussing in the reference I gave. Jaynes is discussing the case where there is a deterministic relation between your x and y (rather than my x and y - sorry my notation was confusing). In other words, f(y, x) = delta(y, F(x)) for some function F. Then, given , the mean of F under an unknown measure on X, maximum entropy furnishes a measure on X. Alternatively, given the average of y over m samples of y (equivalently, the average of F(x) over m samples of x), Bayes' theorem furnishes a posterior measure on X^{m}. Marginalizing gives a measure on X. As m -> infty, the two measures agree, essentially because the second method counts the number of ways that (m - 1) x's can produce a given average value of F given that the m^{th} value is fixed. Your situation involves a stochastic function f, which, what is more, is not measured directly. There are two ways to think about constructing a measure on X, as you say. One is to take the maximum entropy measure on X under the constraint q = f(p). (It should be noted, though, that you do not have q. You have samples from q.) The other is to look at n samples of y coming from a single unknown value x, and to construct the posterior measure on x. It is hard to see how the latter will produce a maximum entropy distribution, though, as it does not 'count the number of ways' of producing something, or at least, I cannot see how it does. The combination of the two cases would involve m different points in Y^{n} coming from m points in X (so for each x sample, there are n samples from f(y, x)). Bayes theorem would then construct a measure on X^{m}, which could be marginalized to given a measure on X. Jaynes' argument suggests that it is in the case m -> infty (and probably, because you have a stochastic function and not a deterministic one, n -> infty as well) that the equivalence will exist. To put it another, I think equivalent, way: the constraint q = f(p) is equivalent to a large number of constraints of Jaynes' type, because q(y) = int dx f(y, x) p(x) = , where f_{y}(x) = f(y, x). To fit in with his argument then, requires being given, for each y, the average value of f_{y} over m samples of x, and then to let m -> infty. In practice, the values of f_{y} are not available directly, but something similar would be the average of the frequencies of occurrence of y across the different samples of x. For these to be 'precise' measures of the average values of the f_{y} though, indeed to guarantee having an average value for every y, means taking n -> infty. This seems equivalent to the previous paragraph. Ian. === Subject: Re: Maximal entropy from Bayes' law? Originator: israel@math.ubc.ca (Robert Israel) entropy is a shockingly strong assumption. It should only be used if A) Your probability measure really is being generated by an astronomically large number of microstates interacting. I'm told this sometimes happens in physics. 1)>Boltzmann, trying to explain his formula for thermodynamical entropy, published (1877) the following table: (7,0,0,0,0,0,0) (6,1,0,0,0,0,0) (5,2,0,0,0,0,0) (4,3,0,0,0,0,0) (5,1,1,0,0,0,0) (4,2,1,0,0,0,0) (3,3,1,0,0,0,0) (3,2,2,0,0,0,0) (4,1,1,1,0,0,0) (3,2,1,1,0,0,0) (2,2,2,1,0,0,0) (3,1,1,1,1,0,0) (2,2,1,1,1,0,0) (2,1,1,1,1,1,0) (1,1,1,1,1,1,1) He calculated entropy of each partition using the polynomial coefficient. He suggested, that the logarithm of it is the equal to entropy. It is thus possible to use it even for small sets. Shannon realization of previous table in information theory: (a,a,a,a,a,a,a) (a,a,a,a,a,a,b) (a,a,a,a,a,b,b) (a,a,a,a,b,b,b) (a,a,a,a,a,b,c) (a,a,a,a,b,b,c) (a,a,a,b,b,c,c) (a,a,a,b,b,b,c) (a,a,a,b,b,c,c) (a,a,a,a,b,c,d) (a,a,a,b,b,c,d) (a,a,b,b,c,c,d) (a,a,a,b,c,d,e) (a,a,b,c,d,e,f) (a,b,c,d,e,f,g) Here, another polynomial coefficient is applicable, connected with another measure of entropy. Both entropies are additive, nevertheless they can not solve the mixing problem, how to distinguish strings: 000000111111 010101010101 111010110000. kunzmilan === Subject: Re: Maximal entropy from Bayes' law? Content-Length: 1420 Originator: rusin@vesuvius > I was walking on the beach with Chris Lee and James Dolan last > weekend, and we came across a problem about statistical inference. > Suppose we repeatedly observe a system in a well-understood but > stochastic way and get some frequency distribution of outcomes. > What's our best guess for the probability that the system is in a > given state? I'm hoping that under some conditions the Bayesian > approach gives results that match Jaynes' maximum entropy method. Pages 38 - 41 discuss a relationship between maximum entropy and Bayes' theorem. As I understand it, it is the following. Let X be the 'state space', Y = X^{n}, and F be a function from X to somewhere suitable. 1) Given the constraint = f, one can maximize entropy over measures on X to get a measure on X. 2) Given that (1/n) sum_{i = 1}^{n} F(y_{i}) = f for an unknown sample y in Y, one can use Bayes' theorem to compute the posterior measure for y, and then marginalize to get the posterior measure for x in X. As n -> infty, these two procedures agree. The Darwin-Fowler method in statistical physics does more or less the same thing to get from the microcanonical to canonical ensemble I think. I am not sure if this really addresses your question, but perhaps it will be helpful. Ian. === Subject: Re: Maximal entropy from Bayes' law? Content-Length: 2869 Originator: rusin@vesuvius In my understanding, the orthodox Bayesian solution to your set-up would involve a prior on p, the probability measure on X. If you wish, you may call this P(p) a hyper-prior. Next, you need to formalize what you mean by a best guess. Here are two possiblities: 1) The p maximizing P(p) subject to the constraint f(p) = q. This may not be unique. 2) The p' minimizing the expected loss function EL(p') = sum_{p | f(p) = q} L(p', p) P(p) where L is some loss function specifying how embarassed you would feel about guessing p' when the real answer is p. Note that the expected loss minimizing p' isn't guaranteed to satisfy f(p') = q (but it might anyhow). If you are willing to bend the rules a bit, you can fit maximum entropy into this framework. Just think of maximum entropy as an unnormalized hyper-prior P(p) that says P(p) is infinitely greater than P(q) whenever p has higher entropy than q. (You could make this precise with probability measures valued in non-Archimedean fields, but where's the fun in that?) entropy is a shockingly strong assumption. It should only be used if A) Your probability measure really is being generated by an astronomically large number of microstates interacting. I'm told this sometimes happens in physics. B) You don't care. Or perhaps more precisely, when you'd rather compute a partition function than come up with a loss function and do the integral in above possibility #2. > Another approach is Bayesian: use Bayes' rule! The conditional > probability for any measurement outcome y given the state x is f(y|x). > Use counting measure on X as our prior, and use Bayes' law to > update this prior when we do an observation and get the outcome y. > In fact, let's imagine that we keep doing observations, getting > outcomes distributed according to the probability measure q, and > use these to keep updating our probability measure on X. In the > limit of infinitely many observations, this measure should converge > (almost surely) to some fixed measure on X. I'm skeptical about this converging; I don't see any reason why it wouldn't just hop around all of the various p where p = f(q). But let's see. We'll do n observations, y_1 .. y_n, and see what happens to E( P( x | y_1 .. y_n ) ) = sum_{y_1} .. sum_{y_n} P(y_1 .. y_n ) P( x | y_1 .. y_n ) By Bayes' law, E( P( x | y_1 .. y_n ) ) = sum_{y_1} .. sum_{y_n} P(x) P( y_1 .. y_n | x ) Let's assume the y_i are independent, E( P( x | y_1 .. y_n ) ) = P(x) sum_{y_1} .. sum_{y_n} P( y_1 | x ) .. P( y_n | x ) Re-arranging, E( P( x | y_1 .. y_n ) ) = P(x) ( sum_y f(y|x) )^n But wait! sum_y f(y|x) = 1, so E( P( x | y_1 .. y_n ) ) = P(x) So I guess it does converge, just to the boring P(x) and not to the interesting maximum entropy solution. -Thomas C === Subject: Re: Maximal entropy from Bayes' law? Content-Length: 3191 Originator: rusin@vesuvius > In my understanding, the orthodox Bayesian solution to your > set-up would involve a prior on p, the probability measure on X. > If you wish, you may call this P(p) a hyper-prior. > Next, you need to formalize what you mean by a best guess. > Here are two possiblities: > 1) The p maximizing P(p) subject to the constraint f(p) = q. > This may not be unique. > 2) The p' minimizing the expected loss function > EL(p') = sum_{p | f(p) = q} L(p', p) P(p) > where L is some loss function specifying how embarassed you > would feel about guessing p' when the real answer is p. Note > that the expected loss minimizing p' isn't guaranteed to satisfy > f(p') = q (but it might anyhow). So if the loss function is delta(p', p), we are doing MAP estimation, and we are back to (1). > If you are willing to bend the rules a bit, you can fit maximum > entropy into this framework. Just think of maximum entropy > as an unnormalized hyper-prior P(p) that says P(p) is infinitely > greater than P(q) whenever p has higher entropy than q. The 'infinite difference' seems unnecessary to me. If the prior is simply e^H(p), then the MAP estimate will be to maximize H(p) subject to the constraint f(p) = q, which is exactly the maximum entropy solution suggested by John Baez. Or am I missing something? > entropy is a shockingly strong assumption. It should only be > used if > A) Your probability measure really is being generated by an > astronomically large number of microstates interacting. I'm > told this sometimes happens in physics. The relation between Bayes' theorem and maximum entropy that I outline in my posts does suggest that maximum entropy should only be applied when the 'measurement' you have is a very good estimate at least of the mean of p. It is an interesting question when this assumption is justified (deviations from it would mean deviations from the canonical ensemble in statistical physics, for example, because the Bayes' theorem argument should be applied with finite m). In physical systems, perhaps it is because of the time-averaging that would be involved in any real measurement? > Another approach is Bayesian: use Bayes' rule! The conditional > probability for any measurement outcome y given the state x is f(y|x). > Use counting measure on X as our prior, and use Bayes' law to > update this prior when we do an observation and get the outcome y. > In fact, let's imagine that we keep doing observations, getting > outcomes distributed according to the probability measure q, and > use these to keep updating our probability measure on X. In the > limit of infinitely many observations, this measure should converge > (almost surely) to some fixed measure on X. > I'm skeptical about this converging; I don't see any reason why it > wouldn't just hop around all of the various p where p = f(q). I do not think the rest of your post (snipped) is analysing what John Baez was suggesting. The result certainly converges, according to my interpretation of what he was saying. Ian. Ian.