Re: [math-fun] analytic function for new Airbnb logo?
HGB>I wanted an approximation that was entire over the complex plane, so the straight lines and discontinuities would have to be fudged. If I take the image, convert it to black-and-white, and erode the boundaries to a single pixel width, I will get a "continuous" curve that goes around the origin twice. If I consider this curve in (r,theta) coordinates, I can now consider the Fourier transform. I'd like to reset the origin in such a way that the "fundamental" is zero -- i.e., the lowest frequency (which takes 2 revolutions) has a coefficient of zero. (There are other possible locations for the origin: the balancing point, the center of gyration, etc.) That having been done, I could alternatively use (log(r),theta) coordinates, which have some nicer properties, because it is the actual complex log function. Fourier analysis can approximate the discontinuous changes in curvature; hopefully modifying these coefficients a little will remove these discontinuities with a minimum of ringing. <HGB Ringing only happens at true discontinuities (e.g. http://www.tweedledum.com/rwg/gibbs.htm ) A nice example of going around twice, mixing circular arcs and straight segments, is to draw, with a single set of amplitudes and phases (for which I have closed forms) alternately an equilateral triangle and its inscribed circle: gosper.org/Incirc_1.mp4 (The file isn't corrupt--download it.) --rwg The sides of your rounded edge squares are impressively flat. But not as flat as those tan(sin())-sin(tan()) games we played. At 03:48 PM 7/30/2014, Bill Gosper wrote: HGB> My approach would be to take the Fourier transform of the polar plot, rwg>Defined as piecewise linear|circular? and then simplify to a small number of coefficients. Of course, it will take 2 circuits around the origin to complete the figure. The sides are perhaps a little too flat, which will blow up the number of coefficients required. I guess if you transform [0,4pi) to [0,1) you could use some sort of Cheby polynomial approximation. It would be nice to be able to transform it back to p(x,y)=0, where p is a polynomial in x,y. At 02:56 PM 7/29/2014, Whitfield Diffie wrote: HGB>Anyone game for an analytic function to approximate Airbnb's new logo? They should manufacture a paper clip and give it out at trade shows. Whithttp://www.underconsideration.com/brandnew/archives/airbnb_logo_detail.png rwg> You can get some impressively straight segments with theta fcns: gosper.org/halph.pdf (10MB?!) or see http://gosper.org/thetpak.html (scrolled 69%). (.3281... is √(one ninth constant), which I claim should be named Halphen's. It's the largest q for which the curve is convex.)) But a problem with this logo is that it switches curvatures discontinuously. <rwg Another possibility: modulate the arcspeed (slow down for curves, speed up for straightaways) before taking the Fourier.
It's well known and easy to prove that if NxN matrices A and B commute, then (*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A) . But AB = BA is not a necessary condition for (*), as googling will readily reveal. Does anyone know necessary and sufficient conditions on A and B for (*) to hold? If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ? I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real. --Dan
Well, if you believe Mathematica, there are no such 2x2 real matrices: A = {{a11, a12}, {a21, a22}}; B = {{b11, b12}, {b21, b22}}; Solve[MatrixExp[A].MatrixExp[B](*==MatrixExp[A+B]*)== MatrixExp[B].MatrixExp[A] && A.B != B.A, {a11, a12, a21, a22, b11, b12, b21, b22}, Reals] // Timing {17.191310, {}} Charles Greathouse Analyst/Programmer Case Western Reserve University On Mon, Aug 4, 2014 at 9:17 PM, Dan Asimov <dasimov@earthlink.net> wrote:
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I probably should have tried that yesterday. Instead. I left out the condition A.B != B.A and just tried to find the general solution for exp(A)exp(B) = exp(A+B) = exp(B)exp(A), and it kept reporting that that was beyond the ability of Solve. So finally I set the lower left elements of A and B to 0, and precomputed the exp() matrices, calling them AA, BB, ABAB, to get: ----- In[26]:= Solve[{AB==ABAB,BA==ABAB},{a,b,d,e,f,h},Reals] Solve::nsmet: This system cannot be solved with the methods available to Solve. h a d a e h a + e b E (E - E ) E (E - E ) f d + h Out[26]= Solve[{{{E , -------------- + --------------}, {0, E }} == a - d e - h a + e d + h a + e (E - E ) (b + f) d + h
{{E , -(-------------------------)}, {0, E }}, True},
-a + d - e + h
{a, b, d, e, f, h}, Reals]
But it's reasonable that there may be no real solutions, since I haven't seen any. Now on to the complex (and quaternion and octonion?) ones. --Dan On Aug 5, 2014, at 7:48 AM, Charles Greathouse <charles.greathouse@case.edu> wrote:
Well, if you believe Mathematica, there are no such 2x2 real matrices:
A = {{a11, a12}, {a21, a22}}; B = {{b11, b12}, {b21, b22}}; Solve[MatrixExp[A].MatrixExp[B](*==MatrixExp[A+B]*)== MatrixExp[B].MatrixExp[A] && A.B != B.A, {a11, a12, a21, a22, b11, b12, b21, b22}, Reals] // Timing
{17.191310, {}}
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Mon, Aug 4, 2014 at 9:17 PM, Dan Asimov <dasimov@earthlink.net> wrote:
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Charles, Just wondering, from your Mathematica comment (*==MatrixExp[A+B]*): Is there some reason that the truth of exp(A)exp(B) = exp(B)exp(A) implies that these terms are also equal to exp(A+B) ??? --Dan On Aug 5, 2014, at 7:48 AM, Charles Greathouse <charles.greathouse@case.edu> wrote:
Well, if you believe Mathematica, there are no such 2x2 real matrices:
A = {{a11, a12}, {a21, a22}}; B = {{b11, b12}, {b21, b22}}; Solve[MatrixExp[A].MatrixExp[B](*==MatrixExp[A+B]*)== MatrixExp[B].MatrixExp[A] && A.B != B.A, {a11, a12, a21, a22, b11, b12, b21, b22}, Reals] // Timing
{17.191310, {}}
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Mon, Aug 4, 2014 at 9:17 PM, Dan Asimov <dasimov@earthlink.net> wrote:
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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No. But if the system can't be solved without that constraint then clearly it won't become solvable when adding it. It's not just fancy that caused me to comment it out -- with it in Mathematica claims ignorance. Charles Greathouse Analyst/Programmer Case Western Reserve University On Tue, Aug 5, 2014 at 4:54 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Charles,
Just wondering, from your Mathematica comment (*==MatrixExp[A+B]*):
Is there some reason that the truth of
exp(A)exp(B) = exp(B)exp(A)
implies that these terms are also equal to exp(A+B) ???
--Dan
On Aug 5, 2014, at 7:48 AM, Charles Greathouse < charles.greathouse@case.edu> wrote:
Well, if you believe Mathematica, there are no such 2x2 real matrices:
A = {{a11, a12}, {a21, a22}}; B = {{b11, b12}, {b21, b22}}; Solve[MatrixExp[A].MatrixExp[B](*==MatrixExp[A+B]*)== MatrixExp[B].MatrixExp[A] && A.B != B.A, {a11, a12, a21, a22, b11, b12, b21, b22}, Reals] // Timing
{17.191310, {}}
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Mon, Aug 4, 2014 at 9:17 PM, Dan Asimov <dasimov@earthlink.net> wrote:
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Taking A to be diagonal [[a,0],[0,b] and B to be arbitrary [[x,y],[z,w]], Maple finds four families of matrices A,B: [[4*I*Pi+b, 0], [0, b]], [[x, y], [0, x+2*I*Pi]] [[-4*I*Pi+b, 0], [0, b]], [[x, y], [0, x-2*I*Pi]] [[4*I*Pi+b, 0], [0, b]], [[x, 0], [z, x+2*I*Pi]] [[-4*I*Pi+b, 0], [0, b]], [[x, 0], [z, x-2*I*Pi]] satisfying exp(A)exp(B) = exp(A+B) = exp(B)exp(A) with AB != BA. When I try the Jordan matrix A:=[[a,1],[0,b]] with arbitrary B = [[x,y],[z,w]] I get an error message when I try to solve the equations. On Tue, Aug 5, 2014 at 10:48 AM, Charles Greathouse < charles.greathouse@case.edu> wrote:
Well, if you believe Mathematica, there are no such 2x2 real matrices:
A = {{a11, a12}, {a21, a22}}; B = {{b11, b12}, {b21, b22}}; Solve[MatrixExp[A].MatrixExp[B](*==MatrixExp[A+B]*)== MatrixExp[B].MatrixExp[A] && A.B != B.A, {a11, a12, a21, a22, b11, b12, b21, b22}, Reals] // Timing
{17.191310, {}}
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Mon, Aug 4, 2014 at 9:17 PM, Dan Asimov <dasimov@earthlink.net> wrote:
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Hmmm. Just happened on this abstract from the 1998 ICM (in Berlin): ----- Michaelis Walter J University of New Orleans New Orleans LA U S A The exp onential matrices of noncommuting matrices may com mute We show how to construct entire families of noncommuting by real matrices A and B for which eA eB eAB and yet eA eB eB eA Though it is known that esA etB etB esA s t R A B B A and that eAt eBt eABt t R A B B A our examples provide noncommuting by real matrices A and B for which eAt eB eB eAt t R We also give examples of noncommuting by real matrices and equivalently of by complex matrices A and B such that eA eB eAB I the identity matrix and thus eA eB eAB BA ee ----- Apologies, but I don't have time to fix the screwed-up characters right now. But it claims to find real matrices A,B with AB!= BA, both 2x2 and 4x4, and exp(A)exp(B) = exp(B)exp(A) (no claim about exp(A+B)). If anyone can e-mail me a copy of his paper that would be greatly appreciated. --Dan On Aug 4, 2014, at 6:17 PM, Dan Asimov <dasimov@earthlink.net> wrote:
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Hi, Dan. You might find this page relevant... http://www.ams.org/journals/proc/2013-141-03/S0002-9939-2012-11396-6/
And does this help? http://ac.els-cdn.com/0024379589905545/1-s2.0-0024379589905545-main.pdf?_tid... On 2014-08-05 18:03, Mike Speciner wrote:
Hi, Dan. You might find this page relevant... http://www.ams.org/journals/proc/2013-141-03/S0002-9939-2012-11396-6/
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Here is just one of many possible examples of [exp(A), exp(B)] = 0 for [A, B] != 0, using commutator notation. The Pauli spin matrices are σ1 = [[0,1],[1,0]], σ2 = [[0,-i],[i,0]], σ3 = [[1,0],[0,-1]]. They satisfy σ1 σ2 = -σ2 σ1 = i σ3, and the rest by cyclic permutation. Let u be a real unit vector, u1^2 + u2^2 + u3^2 = 1. Let U = i(u1 σ1 + u2 σ2 + u3 σ3). It's easy to see that the eigenvalues of U are +i and -i. Then it follows from the series expansion, and substituting U^2 = -1, that exp(tU) = cos t + U sin t. Thus if t is a nonzero multiple of π, and A = tU, exp(A) is plus or minus the identity, and so commutes with every matrix, while A itself does not. The second part of the problem asks for exp(A) exp(B) = exp(A+B) = exp(B) exp(A), where by the symmetry of the middle expression, one equality implies the other. Let v be another real unit vector, orthogonal to u, and similarly construct V. Let A = 3πU, B = 4πV, so that A+B has "length" 5π. Then exp(A) exp(B) = (-1)^3 (-1)^4 = (-1)^5 = exp(A+B). -- Gene
________________________________ From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Monday, August 4, 2014 6:17 PM Subject: [math-fun] exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan
participants (6)
-
Bill Gosper -
Charles Greathouse -
Dan Asimov -
Eugene Salamin -
Mike Speciner -
W. Edwin Clark