mm-1149 === Subject: Imaginary Number help requested for eigenvector i Folks: been years since I dealt with imaginary numbers. I know that sqrt(-1) = (+i,-i) but thats about all I remember. I tried a practice problem to find the eigenvalues/eigenvector for (2 -1) (5 -2) After doing det(2-Lambda -1 ) (5 -2-Lambda) My 2 eigenvalues were +i, -i I can setup both equations using 1 where eigenvalue = i) (2-i -1 ) (x) = (0) (5 -2-i ) (y) (0) 2 where eigenvalue = -i) (2+i -1 ) (x) = (0) (5 -2+i ) (y) (0) Im stumped though on the steps required to get to the final answer. The books answer is: EigenVector1 Span(2 + i), EigenVector2 Span(2 -i) ( 5 ) ( 5 ) I can do basic eigenvectors but this imaginary number throws me off. Can anyone give me a hint on the in-between steps!!?!! === Subject: Re: Imaginary Number help requested for eigenvector > i Folks: Some folks dont follow by references in thread but by subject, so changing subject may not be a continuation of your thoughts. > I tried a practice problem to find the eigenvalues/eigenvector for > (2 -1) > (5 -2) > After doing det(2-Lambda -1 ) > (5 -2-Lambda) > My 2 eigenvalues were +i, -i > I can setup both equations using > 1 where eigenvalue = i) > (2-i -1 ) (x) = (0) > (5 -2-i ) (y) (0) 2x - ix - y = 0 *i 2ix + x - iy = 0 5x - 2y - iy = 0 - 2ix + x - iy = 0 = 4x - 2y - 2ix = 0 y = 2x - ix x = y/(2 - i) = y(2 + i)/5 Thus eigen directions ( 1 ) (2+i) (2-i) ( 5 ) normalize to unit length to obtain eigenvector proper by dividing by the length for the first sqr(1^2 + (2-i)(2+i)) = sqr 6 or for the second sqr((2+i)(2-i) + 5^2) = sqr 30 > 2 where eigenvalue = -i) > (2+i -1 ) (x) = (0) > (5 -2+i ) (y) (0) > Im stumped though on the steps required to get to the final answer. > The books answer is: > EigenVector1 Span(2 + i), EigenVector2 Span(2 -i) > ( 5 ) ( 5 ) > I can do basic eigenvectors but this imaginary number throws me off. Take it easy, dont attempt geometrical visualizations! > Can anyone give me a hint on the in-between steps!!?!! but always i is there, theres no getting around that. ;-) === Subject: Re: Imaginary Number help requested for eigenvector > (2-i -1 ) (x) = (0) > (5 -2-i ) (y) (0) You could do row reduction to wipe out the bottom equation and leave you with: (2-i)x - y = 0 y = (2-i)x and then pick any non-zero x you want. The book obviously picked x=(2+i) which gives y=5. I often just try to do these simply by inspection by setting x=1 and seeing what that gives me for y, and then scaling if I wanted to try to put it in a nicer form. So in this case, just looking at the original matrix you have above, setting x=1 immediately by inspection gives me y=2-i, and so the eigenvector [1 2-i]. Scale that by 2+i and you get the books answer: [2+i 5] Since a multiple of any eigenvector is also an eigenvector, any of these are the right answer (unless the book asked for a unit eigenvector or some such). > I can do basic eigenvectors but this imaginary number throws me off. Dont treat it like anything special. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Imaginary Number help requested for eigenvector > (2-i -1 ) (x) = (0) > (5 -2-i ) (y) (0) > You could do row reduction to wipe out the bottom equation and leave > you with: > (2-i)x - y = 0 > y = (2-i)x > and then pick any non-zero x you want. The book obviously picked > x=(2+i) which gives y=5. > I often just try to do these simply by inspection by setting x=1 and > seeing what that gives me for y, and then scaling if I wanted to > try to put it in a nicer form. So in this case, just looking > at the original matrix you have above, setting x=1 immediately > by inspection gives me y=2-i, and so the eigenvector [1 2-i]. Hi Rich: mentioned in class but his wording lost me (Maybe I was too tired to understand his wording). You cleared it up a little. Question: In the set x to 1 and solve for y case above. Do you only need to do that with just one of the two equations? I guess what Im saying is that theres 2 equations above. So with the set x to 1 and solve for y approach, can we just randomly pick out one equation of the two and do that? So for the first equation (as you mentioned), you get (1, 2-i) For the 2nd equation 5x + (-2-i)y=0, (setting x=1 & solve for y), you get (1, -5/(-2-i)). > Scale that by 2+i and you get the books answer: [2+i 5] Our book tends to make things much harder than it should be. At least for beginning linear algebra students like myself :-) This is why I get lost alot in this class. Invariably, the teacher always comes out with a different answer than the books answer. But the teachers answer is always correct as well (just written in a different form from the books answer). Because Im not math savvy, its difficult for me to tell if I came up with the right answer (as the book always tends to write their answer in a more complex form than needed). Its funny, I understand the process to find eigenvalues/eigenvectors but get thrown off by other variants (which I need to practice on). === Subject: Re: Imaginary Number help requested for eigenvector > Question: In the set x to 1 and solve for y case above. Do you only > need to do that with just one of the two equations? For a 2-component eigenvector, yes. Because the A-lambda*I matrix is singular by construction, one of the rows must be a multiple of the other, and since there are only two rows, you need only pick one row, and it does not matter which. That said, Id still check the answer in both rows, to guard against making an arithmetic error. Once you get to eigenvectors that will have three or more components, its not this easy. I will again often try to eyeball something quickly, but if that fails, row-reduce the A-lambda*I matrix to the point where you can see which variables are completly free (i.e. they completely fall out of the row-reduced matrix). Then pick simple, non-zero values (like 1 :-) for those free variables and work out what the other variables have to me. To use your 2x2 example, when you row-reduce, you get [2-i -1][x] [0] [ 0 0][y] = [0] In that case, you can view either variable as completely free, pick some non-zero value for it, and solve for the other one. But imagine some 3x3 case where the row-reduced matrix is: [a b c][x] [0] [0 d e][y] = [0] [0 0 0][z] [0] In this case Id view z as completely free, set it to 1 (or some other convenient value), use the 2nd row to get a value for y, then use the 1st row to get a value for x, then check my tentative eigenvector against the un-reduced matrix. > I guess what Im > saying is that theres 2 equations above. So with the set x to 1 and > solve for y approach, can we just randomly pick out one equation of the > two and do that? In the 2x2 case, yes. Not in general. See above. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Imaginary Number help requested for eigenvector > For a 2-component eigenvector, yes. Because the A-lambda*I matrix is > singular by construction, one of the rows must be a multiple of the > other, That should be ...is singular by construction, in the 2x2 case one of the rows must be a multiple of the other... In 3x3 and higher, singularity of the matrix does not imply that any one row need be a multiple of any another row. The linear dependancy can be more complex than that. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Sorry about the bad alignment in the post Hi Folks: Just realized the alignment looked bad so I apologize if it looked confusing: The books answer was: Span of (2 + i) ( 5 ) and Span of (2 - i) ( 5 ) Im not sure how they came up with that answer. thank you... === Subject: Re: proof Set . > Pls help... > The number of Subsets of a Set with n elements is 2^n? > pls help me, i like poof by induction if you can, or what other proof. Think of it like this. Say there are 4 elements in the set. Each subset can be represented like this: 0000, 0001, 0010, 1110, etc. In other words 0000 says that for each element of the set, none of them are in this particular subet. 1110 says that the first, second, and third elements are in the subset, and the fourth isnt. Thinking about it that way may make the problem clearer. === Subject: Re: Transfer Function -- Steve, I-Net+ Free PC Tech Support - http://www.webzila.com DLL Files - http://www.dllplanet.com Live Support - http://www.webzila.com/livesupport >> I need to determine a single tranfer function equivalent to the sytem of >> the >> following figure: >> http://users3.ev1.net/~srudenko/img/transferfunction.jpg >> Here is what I came up for the answer. Could someone verify it? >> G(s) = Y(s) / X(s) so then... >> G(s) = G1(s)G4(s) / [ 1+G1(s) [G2(s) + G3(s)] ] > Looks good to me. >> The above solution was my answer. Here is what my friend came up with: >> G(s) = G1(s) / [ 1+G1(s) [G2(s) + G3(s)] ] >> Whos answer is correct? >> I think mine is because my friends answer doesnt even seem to have G4(s) >> in >> it. > Even if yours were wrong, his is definitely wrong. Given the > diagram, G4 has to be in the final answer. > -- > Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: I need help proving: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i44BJxJ23820; hye i tried to crack it but could not if u find the proof then please let me know too ------------------------------ http://www.zonelinks.com ------------------------------ The Power Link Dimension >a1^3 + a2^3 + a3^3......+an^3 >(a1 + a2 + a3.....an)^2 >for any kind of sequence: a1,a2,a3,...,an. === Subject: Power Series by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i44BK0t23884; The Power Series(from n=0 to infinity) ∑ (x)^n represents the function 1/(1-x) in the open interval (-1,1). By differentiation from this or otherwise, obtain a power series expansion for 1/(1-x)^2. === Subject: Re: Power Series by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i44ER3i13560; >The Power Series(from n=0 to infinity) ∑ (x)^n represents the >function >1/(1-x) in the open interval (-1,1). By differentiation from this or >otherwise, obtain a power series expansion for 1/(1-x)^2. Special characters dont show up well on this newgroup. You mean: Sum(n=0 to infinity) x^n = 1/(1-x) since that is the sum of the geometric series (for -1< x< 1). By differentiation from this or otherwise Hey, you are given a hint, use it! 1/(1-x)= (1-x)^(-1) and the derivative of that is -(1-x)^(-2)(-1)= 1/(1-x)^2. Since, inside their radius of convergence, we can differentiate power series term by term, 1/(1-x)^2= Sum(n=0 to infinity) (x^n)= Sum(n=1 to infinity) (n x^(n-1))= Sum(i=0 to infinity) ((i+1)x^i). Notice that n x^(n-1)= 0 for n=0- thats why I started the sum at n=1 rather than n=0. Then I let i= n-1 so the sum would be indexed by the power of x. === Subject: Re: Power Series Adjunct Assistant Professor at the University of Montana. >The Power Series(from n=0 to infinity) ∑ (x)^n represents the >function >1/(1-x) in the open interval (-1,1). By differentiation from this or >otherwise, obtain a power series expansion for 1/(1-x)^2. The derivative of 1/(1-x) is -1/(1-x)^2. The derivative of a power series is equal to the power series of the derivatives on the interval of absolute convergence. So -1/(1-x)^2 = sum_{n=1}^{infinity} nx^{n-1} for x in (-1,1). and therefore, you can get the power series for 1/(1-x)^2 from this. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Power Series > The Power Series(from n=0 to infinity) #8721; (x)^n represents the > function Huh? > 1/(1-x) in the open interval (-1,1). By differentiation from this or > otherwise, obtain a power series expansion for 1/(1-x)^2. === Subject: Re: Power Series > The Power Series(from n=0 to infinity) #8721; (x)^n represents the > function > Huh? I believe the #8721; (at least thats what I saw) is a capital sigma. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Power set by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i44ER3f13556; >Please. provide me proof of. >The number of subsets of a set with n Elements is 2^n? A set with 0 elements (the empty set) has exactly 1= 2^0 subset: the empty set, so the statement is true for n=0. Suppose any set with N elements has 2^N subsets and let A be a set with N+1 elements. Call one of those elements x. How many subsets of A DO NOT contain x? Thats the same as asking how many subsets there are of A-{x}. Since A-{x} has N elements, there are 2^N subsets of A that do not contain x. How many subsets of A DO contain x? Every subset of A that contains x is simply a set that does not contain x UNION {x}. Since there are 2^N subsets of A that DO NOT contain x, there are also 2^N subsets of A that DO contain. The total number of subsets of A is 2^n+ 2^n= 2(2^N)= 2^(N+1). That is, the statement is true for n=0 and whenever it is true for n= N, it is also true for n= N+1. By induction, the statement is true for all non-negative integers n. === Subject: Re: Power set > Please. provide me proof of. > The number of subsets of a set with n Elements is 2^n? There are C(n,0) subsets with 0 elements, C(n,1) with 1 element, etc., so a total of C(n,0) + C(n,1) + ... + (C,n) = (1 + 1)^n by the Binomial Theorem. -- Jim Heckman === Subject: Re: Power set > Please. provide me proof of. > The number of subsets of a set with n Elements is 2^n? You can do this very simply by induction. === Subject: Re: trigonometry by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i44Ed0O15053; I believe you mean to prove if a = arctan(1/2), b = arctan(1/3), then a+b=pi/4. This is easly seen with the identity tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b)) With tana = 1/2 and tanb = 1/3, then tan(a+b)= (1/2+1/3)/(1-1/2*1/3) = 5/6 / (5/6) =1 Thus, a+b = arctan(1) = pi/4 So arctan(1/1) = arctan(1/2)+arctan(1/3) Note that arctan(1/3) = arctan(1/5)+arctan(1/8) arctan(1/8)=arctan(1/13)+arctan(1/21) .... So pi/4 = arctan(1/2)+arctan(1/5)+arctan(1/13)+... Note that 1,1/2,1/3,1/5,1/8,1/13, etc are fibonacci numbers, so we get a formula for pi based on the odd fibonacci numbers. >>suppose that a= tan-1(2) and B= tan-1(3). How would you show that >>tan(a+B)= pi/4. > You wouldnt. Its not true. tan(a+B)= (tan(a)+ tan(B))/(1+ >tan(a)tan(B). In this case, that would be (2+3)/(1+6)= 5/7= 0.71428 >while >pi/4= 0.7854. === Subject: I need help badly!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i44GLUp27637; If p is a prime. How to show that there are p^2 distinct monic quadratic polynomilas over Zp. Also, how to show that p(p+1)/2 of these polynomials are reducible, and p(p-1)/2 of them are irreducible over Zp. === Subject: Re: I need help badly!!! > If p is a prime. How to show that there are p^2 distinct monic > quadratic polynomilas over Zp. Also, how to show that p(p+1)/2 of > these polynomials are reducible, and p(p-1)/2 of them are irreducible > over Zp. You know, you are the third person to ask this, that strikes me as odd, whats the deal? anyway this was my response to show that there are p^2 distinct monic polynomials, just not that they are all in the form x^2+ax+b where a and b are in Zp. any reducible polynomial can be uniquely factored up to rearrangement into two terms of the form (x+c)(x+d) where c and d are in Zp. first count when they are distinct, p choose 2 = p!/(p-2)!2!=p(p-1)/2, then there are the p possibilities where c=d. so p(p+1)/2 to finish, subtract the number of reducible polynomials from p^2, the total number of polynomials === Subject: Algebra equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i44Hi8d04413; I am horrible at algebra and I need help forming an equation. I need to calculate a salary bonus for 2 individuals (different bonus structure of course) that is based on net income after the bonus and after taxes. I have come up with the bonus calculation igoring the tax effect, as follows: b=n-.15b-(.075(b-750,000)-30000), where b equals net income after bonus and n equals net income before bonus. Therefore b-n equals the bonus amount. How do I input a tax computation (@39%) in this equation that takes into account the deductibility of the bonus? Brian === Subject: Re: inverse of the ithprime function >The ithprime function in Maple assigns to each prime a natural number. No, it assigns a prime to each natural number. >What is a good Maple program for the inverse function? That is, a function q such that q(ithprime(j)) = j? Try numtheory[pi]. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: new versions of Fermat available by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i44BK4024154; A new version of the polynomial and matrix oriented CAS Fermat has been posted. It is version 3.3 for Mac and Windows. This version provides remarkable speedups in basic one variable polynomial arithmetic and g.c.d. Since those algorithms are used by other parts of Fermat, the speedup propagates throughout the system to some extent. Problem 3 on the Fermat Tests Page [see web site below] saves 30%. Some computations over Z/pZ save 60%. A second feature is a new implementation of LaGrange interpolation for determinant of sparse matrices. For Linux it is version 3.4. In addition to the above, ground ring Z/pZ is implemented for primes up to 2^31-1. Robert H. Lewis Department of Mathematics Fordham University New York City www.bway.net/~lewis === Subject: Re: Logistic map function Content-Length: 1563 Originator: rusin@vesuvius > Is it possible to exist continuous function f, such that: > f(x + 1, y) = y * (1 - f(x, y)) * f(x, y). > If exists f, how can be expressed? > Leonel Express it as a power series around a fixed point. To do that you need to find the various partial derivatives of f(x,y) at the fixed point. By repeatedly halving the iteration Ôcount, limits can be found which determine partials with respect to x. The process is rather tedious, but it is the only way I know of to find such a function. (It seems reasonable that this process should work, although I havent actually tried it.) The method is hinted at in this answer to a similar question: One way to find h(x) such that h(h(x))=g(x) is to substitute the portion of the half-iteration power series found so far into itself, then the discrepancy of the next term in the result is (a_1 + a_1^n) times the next term for h(x), where a_1 is the linear term of h(x). Since that earlier post, I noticed a post about Schroeders equation, which is an equation of a similar sort that uses multiplication to get to the next iteration instead of adding one. A Google search turned up this paper: http://www.mth.msu.edu/~shapiro/Pubvit/Downloads/RieszExpo/ RieszExpo.html To find my old post, I searched for 165677473, and this came up also: === Subject: Re: Logistic map function Content-Length: 2236 Originator: rusin@vesuvius >Is it possible to exist continuous function f, such that: >f(x + 1, y) = y * (1 - f(x, y)) * f(x, y). > The obvious one is f(x,y) = 0. Thats probably not the one youre > looking for. > Note that for y > 0, the function g(z) = y (1-z) z is a homeomorphism > of (-infinity,0] onto itself, while for y < 0 it is a homeomorphism > of [0,infinity) onto itself. > Take f(x,0) = 0 for all x. > Take f(x,y) for y > 0 to be an arbitrary continuous function with > values in (-infinity,0] and boundary values 0 at y = 0 and > f(1,y) = y (1 - f(0,y)) f(0,y); similarly for y < 0 but with values > in [0, infinity). Use the functional equation to define inductively > f(x,y) for n < x <= n+1 for positive integers n. For n <= x < n+1 > with n <= -1 we take f(x,y) to be the unique negative solution of the > functional equation when y > 0 and the unique positive solution of > the functional equation when y < 0. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 If f(x, y) = x * y * y * (1 - y) + y * (1 - x) => (1) f(x, 0) = x * 0 * 0 * (1 - 0) + 0 * (1 - x) = 0 + 0 = 0 (2) f(0, y) = 0 * y * y * (1 - y) + y * (1 - 0) = 0 + y = y (3) f(1, y) = 1 * y * y * (1 - y) + y * (1 - 1) = y * y * (1 - y) + y * 0 = y * y * (1 - y) + 0 = y * y * (1 - y) (4) y * (1 - f(0, y)) * f(0, y) = y * (1 - y) * y = y * y * (1 - y) = f(1, y) (5) f(2, y) = 2 * y * y * (1 - y) + y * (1 - 2) = 2 * y * y * (1 - y) - y (6) y * (1 - f(1, y)) * f(1, y) = y * (1 - y * y * (1 - y)) * y * y * (1 - y) which is not equal symbolically to f(2, y). I want to know an explicit formula (or algorithm) of analytic function which satisfy the requested functional equation. According to mathworld.com there are explicit solutions only for f(x, -2), f(x, 2) and f(x, 4). Is it possible to exist general solution(s) - f(x,y) defined for every real x & y? Leonel === Subject: Re: Logistic map function Content-Length: 1781 Originator: rusin@vesuvius >>Is it possible to exist continuous function f, such that: >>f(x + 1, y) = y * (1 - f(x, y)) * f(x, y). >If f(x, y) = x * y * y * (1 - y) + y * (1 - x) => This does not satisfy your functional equation, so whats the point in mentioning it? >I want to know an explicit formula (or algorithm) of analytic function >which satisfy the requested functional equation. According to >mathworld.com there are explicit solutions only for f(x, -2), f(x, 2) >and f(x, 4). Is it possible to exist general solution(s) - f(x,y) >defined for every real x & y? AFAIK Mathworld is correct in that the general solution for natural numbers x, with arbitrary starting value f(0,y), can not be written in closed form except for those values of y. That doesnt say there arent particular explicit solutions. As I told you, f(x,y) = 0 is an explicit solution. The other fixed point for y <> 0, f(x,y) = 1 - 1/y, is another explicit solution (but of course not continuous at y=0). I very much doubt that youll find another explicit solution that is continuous on the plane. Note that any continuous explicit solution that is not constant in x would give you an explicit solution to the recursion x_{n+1} = y x_n (1 - x_n) for x_0 in some interval. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Urgent: Help on the definite positive matrix? Content-Length: 1174 Originator: rusin@vesuvius My question is whether we can prove that (I-SCDS) is positive definite in space R(n-1)x(n-1) or not. I know, CD should have all eigenvalues with negative real parts, however, it may not be poistive semi definite. Do you mean that you can prove it? Please let me know. Fan > Matrix C(nxn) is symmetric, negative semi definite, and the diagonal > elements equal to the minus of sum of all off diagonal elements. > If matrix D(nxn) is positive definite, show that the matrix (I-SCDS) is > positive definite in space R(n-1)x(n-1) where I is the indentity > matrix in (n-1)x(n-1) and > S = [ I -e]; where e = [ 1 1 ... 1] is (n-1)x1 vector and S is a > nx(n-1) matrix. > I think that show that ... can be replaced with > show that the matrix SCDS is negative semidefinite in .... > The original statement must remain true even for a SCDS > with elements so large that I is overwhelmed. === Subject: Re: Urgent: Help on the definite positive matrix? Content-Length: 465 Originator: rusin@vesuvius > My question is whether we can prove that (I-SCDS) is > positive definite in space R(n-1)x(n-1) or not. I know, > CD should have all eigenvalues with negative real parts, however, > it may not be poistive semi definite. Do you mean that you can > prove it? Please let me know. No. I just meant that the problem could be restated in terms of a slightly simpler expression. === Subject: Re: Zero-divisors Content-Length: 622 Originator: rusin@vesuvius > Im looking for examples of rings R (finite or otherwise, commutative > or otherwise) that have the following property: > There exists a constant K so that for every zero-divisor x in R there > are exactly K zero-divisors ys in R such that xy=0. > Has anybody met such rings or has any thoughts how to construct them? > THanks, > Felix. The ring of 2x2 matrices over a field with q elements has this property with K = q^2-1. (I assume you dont allow x or y to be 0). --Edwin === Subject: Peano spaces and dimension Originator: grubb@lola Content-Length: 187 Originator: rusin@vesuvius Let X be a compact, connected, locally connected metric space with covering dimension at least 2. Does X have to contain a disk, i.e. a homeomorph of the disk in the plane? --Dan Grubb === Subject: Paper published by Algebraic and Geometric Topology Content-Length: 661 Originator: rusin@vesuvius The following paper has been published: Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-16.abs.html Title: Cubulating spaces with walls Author(s): Bogdan Nica Abstract: We describe a correspondence between spaces with walls and CAT(0) cube complexes. Secondary: 20E42 Keywords: Space with walls, Median graph, CAT(0) cube complex Author(s) address(es): Department of Mathematics, McGill University Montreal, Canada H3A 2K6 Email: bnica@math.mcgill.ca === Subject: Re: Urgent: Help on the definite positive matrix? Content-Length: 627 Originator: rusin@vesuvius > My question is whether we can prove that (I-SCDS) is > positive definite in space R(n-1)x(n-1) or not. I know, > CD should have all eigenvalues with negative real parts, however, > it may not be poistive semi definite. Do you mean that you can I assume you mean negative semi-definite. > prove it? Please let me know. No. I just meant that the problem could be restated in terms of a slightly simpler expression. Since CD is not necessarily positive semi-definite, the solution, if any, will depend on the special structure of S. === Subject: Re: Urgent: Help on the definite positive matrix? Content-Length: 1812 Originator: rusin@vesuvius > My question is whether we can prove that (I-SCDS) is >> positive definite in space R(n-1)x(n-1) or not. I know, >> CD should have all eigenvalues with negative real parts, however, >> it may not be poistive semi definite. Do you mean that you can >I assume you mean negative semi-definite. >> prove it? Please let me know. >No. I just meant that the problem could be >restated in terms of a slightly simpler expression. >Since CD is not necessarily positive semi-definite, >the solution, if any, will depend on the special structure of S. (after some unsuccessful tries of proof => never try to proof something false.) I restate your question (hopefully I got it right): given C the negative of a symmetric L-matrix with all row (hence also all column) sums equal to zero D symmetric positive definite C, D both n times n S=[ I_{n-1}, e], e = vector of all ones show that I_{n-1}-S*C*D*S is positive definite counterexample: C=[ -4 1 1 2 ; 1 -5 1 3 ; 1 1 -6 4 ; 2 3 4 -9]; >> eig(C) ans = -12.4188 -6.3868 -5.1944 -0.0000 v=rand(4,4); v=v+v; [V,L]=eig(v); D=V*diag([1,2,4,16])*V; D D = 7.3152 3.3135 4.3326 4.3309 3.3135 6.3995 2.6780 2.2294 4.3326 2.6780 5.3139 2.9348 4.3309 2.2294 2.9348 3.9714 eig(D) ans = 1.0000 2.0000 4.0000 16.0000 S S = 1 0 0 1 0 1 0 1 0 0 1 1 A=S*C*D*S; eig(A) ans = 11.0948 -27.7833 -15.0847 hence I-A has an eigenvalue -10.0948 ..... hth peter === Subject: Time integral of Wiener processes Epigone-thread: braycoxstrim Content-Length: 110 Originator: rusin@vesuvius Can anyone tell me the integral wrt time from zero to one of the product of two independent Wiener processes? === Subject: Re: Time integral of Wiener processes Content-Length: 551 Originator: rusin@vesuvius >Can anyone tell me the integral wrt time from zero to one of the >product of two independent Wiener processes? int_0^1 X(t)Y(t) dt is Gaussian with mean zero and variance E (int_0^1 X(t)Y(t) dt)^2 = E int_0^1 int_0^1 X(t)X(s)Y(t)Y(s) dt ds = int_0^1 int_0^1 E[X(t)X(s)Y(t)Y(s)] ds dt = int_0^1 int_0^1 E[X(t)X(s)]E[Y(t)Y(s)] ds dt = int_0^1 int_0^1 min(s,t)min(s,t) ds dt = 1/6 (if I calculated the integral correctly) tommi -- Maniaan liittyy matikkamenestys -- UL 100 13.5.03 === Subject: Re: Peano spaces and dimension Content-Length: 702 Originator: rusin@vesuvius >Let X be a compact, connected, locally connected metric space >with covering dimension at least 2. Does X have to contain a >disk, i.e. a homeomorph of the disk in the plane? >--Dan Grubb Surely youve thought of the following, so it must be just stupid: What happens if you start with a closed disk and remove a dense sequence of open disks? Surely if the radii are small enough whats left still has covering dimension 2? (Maybe you start with a square and remove open squares that form some regular pattern to make it easier to show that the dimension of whats left is 2...?) ************************ David C. Ullrich === Subject: Re: Peano spaces and dimension Content-Length: 1110 Originator: rusin@vesuvius >Let X be a compact, connected, locally connected metric space >with covering dimension at least 2. Does X have to contain a >disk, i.e. a homeomorph of the disk in the plane? >--Dan Grubb > Surely youve thought of the following, so it must be just stupid: > What happens if you start with a closed disk and remove a > dense sequence of open disks? Surely if the radii are small > enough whats left still has covering dimension 2? (Maybe > you start with a square and remove open squares that > form some regular pattern to make it easier to show that > the dimension of whats left is 2...?) > David C. Ullrich Looks to me like the remainder has a base of sets with totally disconnected boundaries, which would mean the space is one dimensional. Covering dimension is a topological invariant, unlike Hausdorff dimension, so the size of the removed sets is not relevant (I think). ---Dan Grubb === Subject: Re: Bernsteins Inequality for fractional derivatives? Content-Length: 3073 Originator: rusin@vesuvius >I needed to look up the result in Stein to make sure >it said what I thought. See below: >>Its been a while since I dealt with this type of problem, so please >>forgive me if I am wrong. > The PDE guy had a question. I came up with an > answer - not quite trivial, but not all that > hard. Seems to me the following must be well > known. But I dont know the literature at all - > hes more familiar with the literature and he > thinks it may not be well-known. Opinions? > > Say F denotes the Fourier transform in R^n. > Define a fractional partial derivative operator > D_1^a by > > F(D_1^a f)(x) = |x_1|^a F(f)(x). > > Suppose a > 0. The question was whether > > ||D_1^a f||_p <= c ||f||_p > > for f with F(f) supported in the unit ball. > (The answer is yes.) >>I do not know if the result is well known stated as it is, but is an easy >>consequence of one of the classical multiplier theorems. The problem is >>essentially one-dimensional, so I will assume that f is a function of one >>real variable. Let k(x) be a cut-off function, of class C^oo, with support >>in [-2,2] and equal to 1 on [-1,1], and let m(x) = k(x)*|x|^a. If F(f) kas >>support in [-1,1], then >> D_1^a f = F^(-1)( m*F(f)) >Yes. >>The multiplier m satisfies the conditions Theorem 3 in chapter III of >>Steins book Singular integrals and differentiability properties of >>functions, so that it is bounded on L^p, 1Right. But thats only for 1 < p < oo. Sorry I wasnt specific in >my original post; in fact the inequality holds for 1 <= p <= oo >(just as the standard Bernsteins inequality does). >Which is of course because F(m) is in L^1 - I guess what I >really should have asked is how obvious is it that F(m) is >integrable. (Its not hard, certainly less deep than the >standard multiplier theorems for 1 < p < oo, but I dont >recall ever actually seeing an inequality in print from >which it follows directly...) >>Chapter V in the same book deals with Riesz potentials, which are >>essentially of the same type but with a<0. >>Julian Aguirre >>UPV/EHU At least for anyone who has been a Stein student, it is obvious that F(m) is in L^1. Stein would frequently use the fact that the fourier transform of |x|^a is formally C |x|^{-n-a}, where C depends on a and how you normalize the fourier transform. Using a C^oo cutoff function maintains the decay and smooths out the singularity at 0. Therefore, F(m)(x) dies away like |x_1|^{-1-a} in the x_1 direction, and faster than any power of |x_k| in the x_k direction, for k <> 1. This means that F(m) is in L^1 for any a > 0. Rob Johnson take out the trash before replying === Subject: Re: Bernsteins Inequality for fractional derivatives? Content-Length: 4980 Originator: rusin@vesuvius >>I needed to look up the result in Stein to make sure >>it said what I thought. See below: >Its been a while since I dealt with this type of problem, so please >forgive me if I am wrong. >> The PDE guy had a question. I came up with an >> answer - not quite trivial, but not all that >> hard. Seems to me the following must be well >> known. But I dont know the literature at all - >> hes more familiar with the literature and he >> thinks it may not be well-known. Opinions? Say F denotes the Fourier transform in R^n. >> Define a fractional partial derivative operator >> D_1^a by F(D_1^a f)(x) = |x_1|^a F(f)(x). >> >> Suppose a > 0. The question was whether ||D_1^a f||_p <= c ||f||_p >> >> for f with F(f) supported in the unit ball. >> (The answer is yes.) >I do not know if the result is well known stated as it is, but is an easy >consequence of one of the classical multiplier theorems. The problem is >essentially one-dimensional, so I will assume that f is a function of one >real variable. Let k(x) be a cut-off function, of class C^oo, with support >in [-2,2] and equal to 1 on [-1,1], and let m(x) = k(x)*|x|^a. If F(f) kas >support in [-1,1], then > D_1^a f = F^(-1)( m*F(f)) >>Yes. >The multiplier m satisfies the conditions Theorem 3 in chapter III of >Steins book Singular integrals and differentiability properties of >functions, so that it is bounded on L^p, 1>Right. But thats only for 1 < p < oo. Sorry I wasnt specific in >>my original post; in fact the inequality holds for 1 <= p <= oo >>(just as the standard Bernsteins inequality does). >>Which is of course because F(m) is in L^1 - I guess what I >>really should have asked is how obvious is it that F(m) is >>integrable. (Its not hard, certainly less deep than the >>standard multiplier theorems for 1 < p < oo, but I dont >>recall ever actually seeing an inequality in print from >>which it follows directly...) >Chapter V in the same book deals with Riesz potentials, which are >essentially of the same type but with a<0. >Julian Aguirre >UPV/EHU Lets simplify things by restricting to the case n = 1, as Julian suggested. >At least for anyone who has been a Stein student, it is obvious that >F(m) is in L^1. Stein would frequently use the fact that the fourier >transform of |x|^a is formally C |x|^{-n-a}, where C depends on a and >how you normalize the fourier transform. Hmm. If (n = 1 and) -1 < a < 0 then the Fourier transform of |x|^a _is_ C |x|^{-1-a}, in the sense of tempered distributions. In his book Stein only states this fact for -1 < a < 0 (the notation is different, I think). Just by coincidence one of the postdocs here asked me the other day why there was that restriction on a in Stein, which is why I have the following comment ready on the spur of the moment: But for example if a = 0 then the Fourier transform of |x|^a is a delta function, not a power of |x|, and while Im not sure offhand what the Fourier transform of |x|^a is for a > 0, it cant be C |x|^{-1-a}, because thats not a tempered distribution (since its _positive_ and not locally integrable I dont see how theres any way in the world it can be interpreted as a tempered distribution; the integral of |x|^{-1-a} times a positive test function is +infinity regardless of how you interpret the integral.) If we were talking about _odd_ functions with the same homogeneity that would be different. Does formally mean something other than in the sense of tempered distributions here? >Using a C^oo cutoff function >maintains the decay and smooths out the singularity at 0. Therefore, >F(m)(x) dies away like |x_1|^{-1-a} in the x_1 direction, and faster >than any power of |x_k| in the x_k direction, for k <> 1. This means >that F(m) is in L^1 for any a > 0. Or it could be that the Fourier transform of |x|^a is a distribution with compact support plus something that looks like |x|^{-1-a} near infinity, or something? (Hmm, no, thats not true for a = 2: |x|^2 = x^2, so the FT of |x|^2 is the second derivative of a delta function. Which of course means that if you multiply |x|^2 by a smooth cutoff function you do get something thats obviously L^1, but it doesnt make the argument above right, in fact _that_ FT dies faster than any power of x at infinity...) >Rob Johnson This is a somewhat philosophical question I guess. Ive soon read two > college courses in scientific computing without understanding why some > iterative methods converge to the wanted value. My books just say that they > DO converge, not why. > So, what is the essence of the phenomena of convergence for iterative > methods (in general)? What makes an iteration tend to converge to the > solution? Some iterative algorithms do not converge. I can give lots of examples if you like. Certain ones, such as the iterative solution of a Volterra integral equation, converge for any reasonable kernel or inhomogeneous term. OTOH, iteration of certain Fredholm equations is guaranteed divergent. The basic idea is that an equation x = y + M(x), where M is a (known) mapping of a space (containing the objects x) onto itself, can be solved iteratively if the mapping has a norm that is smaller than 1. In less highfalutin language it means that each time you iterate, the result lies in a more and more restricted part of the space. On the other hand, if the norm is greater than 1, the iteration generally diverges because the successive iterates get mapped all over the place. Here is an example of a (possibly) divergent iteration: apply Newtons method to finding the inverse of a number, i.e solve f(x) a - 1/x = 0 where a > 0. (This method was widely used to implement division on machines that had no divide instruction, back in the old days.) The Newtonian iteration is x_{n+1} = ( 2 - a*x_n ) * x_n . Note that it uses only multiplication and addition (subtraction). If a > 1 and you choose x_0 > 1 this will diverge. It also diverges if 0 < a < 1 and you choose x_0 < 1. So it is necessary to augment the algorithm with a (cheap) method for making good initial guesses. Your homework assignment: Try it on a hand calculator to see how it works. Take a = 3 and x_0 = 1 and x_0 = 0.25 . -- Julian V. Noble Professor Emeritus of Physics ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ For there was never yet philosopher that could endure the toothache patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1. === Subject: Re: Why iterative methods converge > So, what is the essence of the phenomena of convergence for iterative > methods (in general)? What makes an iteration tend to converge to the > solution? There are different ideas. Some iterative methods can abe written as fixed-point iterations, so if you can prove contraction of sorts, youll get convergence. My expertise is in methods for linear systems, and there there are two very different ideas. One is that of contraction (youve probably seen the Jacobi method): you prove that your operator reduces your error in every iteration, so ultimately youll find a solution. The theory concerns itself both with finding the conditions for this contraction, and increasing the speed at which it converges. The other notion (maybe youve heard of the Conjugate Gradients method?) is that of orthogonality. Every next error is orthogonal to the previous. Theoretically this means that since youre in a finite dimensional space, ultimately youll converge. More interesting and practical is that orthogonality implies projection, implies minimization and so again you have a process that steadily converges. (Im sweeping a million footnotes under the carpet here.) However, the minimization here is a corrolary of the orthogonalization process. So..... Im not sure that there is one unified notion of convergence of iteration processes. V. -- email: lastname at cs utk edu homepage: cs utk edu tilde lastname === Subject: Re: Why iterative methods converge john > This is a somewhat philosophical question I guess. Ive soon read two > college courses in scientific computing without understanding why some > iterative methods converge to the wanted value. My books just say that they > DO converge, not why. > So, what is the essence of the phenomena of convergence for iterative > methods (in general)? What makes an iteration tend to converge to the > solution? For the why iterative algorithms converge, this is a quite readable introduction: http://www.physics.arizona.edu/~restrepo/475A/Notes/sourcea/ node21.html -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: Why iterative methods converge > This is a somewhat philosophical question I guess. Ive soon read two > college courses in scientific computing without understanding why some > iterative methods converge to the wanted value. My books just say that they > DO converge, not why. > So, what is the essence of the phenomena of convergence for iterative > methods (in general)? What makes an iteration tend to converge to the > solution? Look at newtons method [which is a classical method for finding roots and thus solving equations]. Say you want to know what y = ln(x) is. Then e^y = x so a root for e^y - x would be a solution (e.g. e^y - x = 0) Newtons method [without going into too much detail] uses the fact that the tangent to the equation [in this case] e^y - x at y will cross the Try this experiment [with graph paper]. Graph y = (x - 1)^2 which we know has a root at x=1. Now pick the point x=5 and draw the tangent to the line at that point. Find out where the tangent hits the [should be between 1 < x < 5]. Now go straight up from where the tangent hits the x-axis to the curve. Repeat the process. Eventually you will draw a tangent to the point on the x-axis at x=1. The tangent to the curve at x=1 is the y=0 line which is when you end the procedure. Now as to *why* this works you will have to learn what divided differences and various other methods are [not my forte]. Any good introduction to numerical analysis text will cover these algorithms in decent depth. For instance a decent text is Applied Numerical Analysis [ISBN 0-201-87072-X]. Its what my community college classes used. Tom