February 2013 - mathfun@mailman.xmission.com

[math-fun] Convolution
by Hans Havermann 25 Feb '13

25 Feb '13

Bill Gosper's sharing of the 4x4 "nice little sliding block head exercise" got me thinking about Stewart Coffin's 4x4x4 Convolution puzzle. John Rausch shares a partially assembled Convolution on his Puzzle World website: http://www.johnrausch.com/puzzleworld/puz/img/lg/convolution_6.jpg Three pieces (#1 at 9 o'clock, #2 at 7 o'clock, #3 at 5 o'clock) need to be added to the assemblage. The black 'cubies' end up in the corners of the finished cube. Also, piece #1 and piece #3 are seriously foreshortened (what look like one-cubie extensions going to the back are actually two-cubie extensions). The number of cubies in each of the three unassembled pieces is 9 (to fit into the 27 empty spaces of the unfinished cube). Can you complete the construction in your head? Bonus: What's wrong with one of the cubies?

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[math-fun] Siebeck/Marden, inellipses, & complex cubics
by Henry Baker 24 Feb '13

24 Feb '13

Oh, what affine day to solve a cubic. An equilateral triangle centered at the origin in the complex plane, but at an arbitrary orientation angle theta, has 3 vertices: r*(cos(theta)+i*sin(theta)), r*(cos(theta+2pi/3)+i*sin(theta+2pi/3)), r*(cos(theta-2pi/3)+i*sin(theta-2pi/3)). This equilateral triangle comes with its circumcircle |z|=r, and its incircle |z|=r/2. We can draw these parametrically: r*(cos(t)+i*sin(t)), r/2*(cos(t)+i*sin(t)). If we now "squish" the complex plane such that the x axis is _expanded_ by a real factor a>=1, and the y axis is _contracted_ by a real factor b, our triangle becomes distorted: r*(a*cos(theta)+i*b*sin(theta)), r*(a*cos(theta+2pi/3)+i*b*sin(theta+2pi/3)), r*(a*cos(theta-2pi/3)+i*b*sin(theta-2pi/3)). This distortion squishes the circumcircle & incircle into ellipses, the second of which is the Steiner inellipse: the subject of the Siebeck/Marden theorem. r*(a*cos(t)+i*b*sin(t)) r/2*(a*cos(t)+i*b*sin(t)). If we form the complex polynomial with the roots of our distorted triangle, we get the equation: (z-r*(a*cos(theta)+i*b*sin(theta)))* (z-r*(a*cos(theta+2pi/3)+i*b*sin(theta+2pi/3)))* (z-r*(a*cos(theta-2pi/3)+i*b*sin(theta-2pi/3))) = z^3 + (3/4)r^2(b^2-a^2)z - (r^3/4)*(cos(3*theta)*a*(a^2+3b^2)+i*sin(3*theta)*b*(b^2+3a^2)) = 0. Although we could independently scale a,b,r separately, this parameterization of the space is redundant; for convenience, we choose a=s*a', b=s*b', such that a'^2-b'^2=1, and r^2*(a^2-b^2) = r^2*((s*a')^2-(s*b')^2) = r^2*s^2*(a'^2-b'^2) = (r*s)^2*(a'^2-b'^2) = (r*s)^2. Thus, without loss of generality, we could have chosen a,b,r such that a^2-b^2=1 to begin with. This choice makes the coefficient of the linear term z to be -(3/4)r^2, which is a _negative real number_. As an example, choose a=1.18034, b=0.5, so our cubic is z^3 - (3/4)r^2 - r^3(cos(3*theta)*0.559 + i*sin(3*theta)*0.5) We would now like to invert this process, and go from a monic cubic equation to a squished triangle of roots. We ask the question: given a monic cubic equation with complex coefficients, how do we determine the parameters theta, r, a, b, which describe the roots of this equation? We are given the equation P(z)=z^3+p*z+q=0. Because the coefficient of the z^2 term is zero, the centroid of the roots is at the origin. We need to _rotate_ this equation so that the principle axis is located on the real axis. Thanks to Siebeck/Marden, this principle axis goes through the roots of P'(z)/3=z^2+p/3=0. Without loss of generality, we can _rotate_ and _scale_ our given equation P(z)=0, such that P(z)=z^3-(3/4)z+q'=0, which conveniently places the roots of P'(z)/3=z^2-1/4 at +-(1/2). This rotation and scaling transforms the arbitrary complex number q into another complex number q'. We know that q' = - (1/4)*(cos(3*theta)*a*(a^2+3b^2)+i*sin(3*theta)*b*(b^2+3a^2)) but how can we ever disentangle theta, a, b from this mess? The constraint a^2-b^2=1 provides a broad hint: choose parameter phi such that a=cosh(phi), b=sinh(phi), which guarantees that a^2-b^2 = cosh(phi)^2 - sinh(phi)^2 = 1. With this substitution, our rotated & scaled P(z) now looks like: P(z) = z^3 -(3/4)z - (1/4)(cos(3*theta)*cosh(3*phi)+i*sin(3*theta)*sinh(3*phi)) = z^3 -(3/4)z - (1/4)cos(3*theta - 3*i*phi) = z^3 -(3/4)z - (1/4)cos(3*(theta-i*phi)) = z^3 -(3/4)z + q' so, q' = -(1/4) cos(3*(theta-i*phi)) and theta-i*phi = acos(-4q')/3, where acos is the _complex_ arccosine function. Thus, a single call to a complex arccosine function produces a real angle theta, and a real parameter phi, and a,b are recovered from a=cosh(phi), b=sinh(phi). Using this parameterization, we can easily recover the _eccentricity_ of our ellipses (= circles distorted by a,b). The foci f for the circumellipse are +-1 (the Steiner inellipse is 1/2 the size), so f=a*e, or e=f/a=1/a. >From our numerical example above with a=1.18034, b=0.5, we now see that phi=asinh(0.5), a=cosh(asinh(0.5))=1.18034, b=sinh(asinh(0.5))=0.5, and e = f/a = 1/cosh(phi) = 0.8944. The _string length_ for our circumellipse is 2*(a+f)=2a(1+e)=2*a+2=4.36068. (Our string is looped around the foci at +-1.) We see that our ellipse flattens to a pancake (e=1) when phi=0, in which case we get the degenerate form of the cubic _with 3 real roots_. Our ellipse converges (very quickly) to a circle (e=0) when phi->oo; however, this is a very large circle because the string length is >> distance between the foci.

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[math-fun] Venn diagram
by James Propp 23 Feb '13

23 Feb '13

In case you ever wondered whether it's really possible to engage in nano-engineering in an environmentally responsible way, here's a poster in which some nano-engineers inadvertently confess that there's no way to do it: http://jamespropp.org/venn.jpg Jim Propp

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Re: [math-fun] nice little sliding block head exercise
by Bill Gosper 22 Feb '13

22 Feb '13

Scott H.> If moving 1 piece pushes/moves N-1 other pieces, does that count as 1 move or N moves? N. As Neil explained it to me, that's one "action". A "move" is one piece along any path. A "step" is one piece moving one square. --rwg Your sneaky moderator used rotation and claims six moves vs nine. This is so unthinkably illegal that I doubt there is even a term for such a maneuver. A "deviation"? A "tort"?

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[math-fun] nice little sliding block head exercise
by Bill Gosper 22 Feb '13

22 Feb '13

Neil is tabulating the "hardest" 4x4s according to polyomino size histogram. Some of the short ones are tricky. E.g., a four-piecer: http://gosper.org/4x4-3,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0.gif . Five moves to get the tetromino to the lower right, where a move is of one piece along any path. --rwg

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Re: [math-fun] Fermat's Last Theorem question
by Adam P. Goucher 22 Feb '13

22 Feb '13

> The controversy arises from Wiles' use of Grothendieck's development > of algebraic geometry, and in particular of so-called "cohomological > number theory". As a matter of technical convenience, Grothendieck > assumes the existence of a "universe", which is, essentially, a set > that can model ZFC. The existence of such a set implies the > consistency of ZFC, so (by Godel Incompleteness) it is an assumption > strictly stronger than ZFC. By the Law of Excluded Middle, either ZFC is consistent (in which case it admits a model, and Wiles' proof works) or inconsistent (in which case, by the Principle of Explosion, we can derive any statement, including FLT). So, ZFC --> FLT. Sincerely, Adam P. Goucher

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[math-fun] Fermat's Last Theorem question
by Henry Baker 22 Feb '13

22 Feb '13

Q about the proof of FLT: I realize that Fermat himself wouldn't recognize Wiles's proof, but would Fermat recognize the assumptions? I.e., does the Wiles proof require any "exotic" assumptions -- e.g., axiom of choice, generalized Riemann hypothesis, etc. ?

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Re: [math-fun] Fermat's Last Theorem question
by Adam P. Goucher 22 Feb '13

22 Feb '13

Warren, Goodstein's theorem can be stated in Peano arithmetic using two quantifiers (for all n there exists a solution t to a Diophantine equation modelling a Turing machine that checks whether the nth Goodstein process halts within t steps), but cannot be proved in Peano arithmetic. The same applies to Kruskal's tree theorem (which is strictly harder) and the Robertson-Seymour theorem (which is even harder still). Fermat's Last Theorem can be stated using one quantifier (there does not exist a solution t to a Diophantine equation modelling a Turing machine that systematically checks all quadruples {a,b,c,n} of naturals and halts whenever a^(n+2) + b^(n+2) = c^(n+2)), so I suppose it's potentially 'easier' than the aforementioned three theorems. Anyway, you can't define the reals in Peano arithmetic, since it's a first-order theory and therefore incapable of quantifying over sets. Second-order logic with the Peano axioms *can* define and prove many facts about the reals (maybe that's what you meant), but this is not the same as Peano arithmetic. Sincerely, Adam P. Goucher http://cp4space.wordpress.com > ----- Original Message ----- > From: Warren Smith > Sent: 02/22/13 04:37 AM > To: math-fun > Subject: [math-fun] Fermat's Last Theorem question > > Peano's axioms suffice to define the integers, and hence (?) > real & complex numbers, eh? > And you are claiming you'd be "surprised" if Fermat's Last Thm > was proven using those alone?!?! I'm freaking surprised if more is needed. > In fact, I'm shaken. > > -- > Warren D. Smith > http://RangeVoting.org <-- add your endorsement (by clicking > "endorse" as 1st step) > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Fermat's Last Theorem question
by Warren Smith 22 Feb '13

22 Feb '13

Peano's axioms suffice to define the integers, and hence (?) real & complex numbers, eh? And you are claiming you'd be "surprised" if Fermat's Last Thm was proven using those alone?!?! I'm freaking surprised if more is needed. In fact, I'm shaken. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Marden/Siebeck theorem
by Henry Baker 21 Feb '13

21 Feb '13

The Siebeck/Marden theorem doesn't construct the ellipse itself, only the foci. Therefore, characteristics such as the _eccentricity_ of the ellipse aren't evident, since having only the foci don't tell us how long the string is. However, the triangle with its Steiner inellipse is an affine-transformed equilateral triangle with its incircle. Since we know that the centroid of the triangle is also the center of the Steiner inellipse, as well as the midpoint of the foci, we can translate the triangle to put the centroid at the origin, and then rotate the triangle to put the foci on the real axis. (If the foci coincide with one another, we are already done, as the triangle is already equilateral.) We can further standardize the triangle by stretching isotropically to place the foci at +-1. We can now easily compute the affine transformation, because the lengths of the vectors from the centroid to the 3 vertices should be equal for an equilateral triangle. We can compute a factor which stretches only in the Y dimension which will make all three of these lengths equal. Since the foci are at +-1, this stretch factor is just the eccentricity of the ellipse. (Note: since we have 2 equations in one unknown, we could actually use a least-squares method to get a numerically accurate solution for the stretch factor; this also handles the case where one of the equations may be degenerate.) We have thus also answered the question about the locus of points of all the triangles _having the same Steiner inellipse_: since these triangles are all affine-transformed from equilateral triangles _using the same affine transform_, this locus of points is an ellipse with the same eccentricity as the given Steiner inellipse. This affine transformation also shows that inellipse is tangent to the 3 triangle sides at their _midpoints_. Note that if the 3 triangle vertices are complex numbers that are the solution of a cubic polynomial, all of the steps up until the last one can be performed _without factoring the cubic polynomial_. I think it may be possible to compute the asymmetric stretch (=eccentricity) without factoring the cubic, but I don't see it right off hand. Note that the line of the foci is the _principle axis_ not just for the ellipse, but also for the triangle itself, as the line of the foci is the _least-squared error line through the triangle vertices_. Does this mean that if you throw up a triangle in a vacuum so that it spins, it will spin on the axis of its line of foci ?

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