November 2010 - mathfun@mailman.xmission.com

[math-fun] Stuff I didn't know about pendulums
by Bill Gosper 01 Dec '10

01 Dec '10

First of all, a spherical bob (on a massless string) centered L below the pivot is a compound pendulum. Its period, for a given amplitude, depends on its radius R! It acts like a point mass centered at L+(3/5)*R^2/L rather than L. Unless it's hollow, when it's just L+R^2/L. So you can tell if it's hollow by swinging it. For a unit n-cube hanging by a vertex, this "center of oscillation" is 2/3 down the main diagonal, for all n. Wikipedia claims the center of oscillation is also the center of percussion. So if you hang a bat by the knob, and match oscillations with a ball on an adjustably long string, the ball will hang down to the sweet spot. If you're the sort of person who swings bats by their knob. Mathematica 7 clams on the ODE of motion: {\[Theta]''[t] == -g*Sin[\[Theta][t]]/L, \[Theta][0] == 0} but Macsyma returns an integral for t[\Theta] which Mma knows is an EllipticF. Testing by taking the limit for small \[Theta]'[0] insists the period approaches 149/30 Sqrt[L/g] instead of 2 pi Sqrt[L/g]! The bug is in Limit. Mma then solves for Theta, \[Theta][t] -> 2*JacobiAmplitude[ (1/2)*t*Derivative[1][\[Theta]][0],(4*g)/(L*Derivative[1][\[Theta]][0]^2)] with period \[CapitalLambda] == 4*Sqrt[L/g]*EllipticK[Sin[Subscript[\[Theta], max]/2]^2] and amplitude Subscript[\[Theta], max] == 2*ArcSin[(1/2)*Sqrt[L/g]*Derivative[1][\[Theta]][0]] The differentiator and simplifiers eventually confess that these solve the ODE. But had I thought to search "pendulum" in the Mma doc, it would have led me to an "application" of JacobiAmplitude where it is claimed, without derivation, to describe an overswung pendulum. This would have been a clue. I'm surprised this all isn't more obvious in, e.g., Wikipedia or MathWorld. Or DSolve. --rwg

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Re: [math-fun] equilateral elliptical pentagons
by Henry Baker 30 Nov '10

30 Nov '10

I've been continuing to work on this problem, if only to sharpen my Macsyma skills. ellipse: (x/a)^2+(y/b)^2=1 The equilateral triangle inscribed in a general ellipse can be handled by Macsyma. If a=1, then the length of the side does depend upon b and upon the location of the first of the points. If a=1, b=1/sqrt(3), then the equilateral triangle goes through the major & minor axes, and is one half of the inscribed quadrilateral for this a,b. The inscribed pentagon has a regular solution, in which the sides don't cross one another, and a degenerate solution in which they do cross (when a/=b). Using ascii art, the two cases look something like these: /\ / \ | | |__| /\ /\ /__\__/__\ \/ For example, if a=1, b=2, the first case is approximately like (1,0), (0.5798,+-1.6296), (-.9072,+-.8414), with side length 1.683. the second case is approximately like (1,0), (-.9926,+-.2428), (0.865,+-1.004), with side length 2.007. This answers the question whether the crossing case has smaller perimeter in the negative. BTW, it is easier in Macsyma to solve for the y coordinates first, and then the x coordinates. I don't know how to set up a set of polynomial equations in which the degenerate solutions don't occur. Also, when solving the equations for general a,b, you effectively get the solutions for _all conics_, since I don't know how to eliminate the solutions in which a, b are complex numbers.

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[math-fun] PrimeLatz conjecture (a picture, a loop and a b-file)
by Eric Angelini 29 Nov '10

29 Nov '10

Latest update: http://www.cetteadressecomportecinquantesignes.com/PrimeLatz.htm Best, É.

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Re: [math-fun] The PrimeLatz conjecture
by Phil Carmody 29 Nov '10

29 Nov '10

From: Bill Thurston > This is true for all **positive** integers N, and false for > all integers less than some negative number -C. > For large enough n, there are at least 3 primes between n > and 2n (from this wikipedia article, > Erd?s proved that for any positive integer k, there is a > natural number N such that for all n > N, there are at > least k primes between n and 2n. An equivalent statement had > been proved earlier by Ramanujan (see Ramanujan prime).) > Therefore, if the kth member of a PrimeLatz sequence is > sufficiently large, either the k+1st element or the k+2nd > element is smaller than n. If the kth member of a PrimeLatz sequence, a_k, is sufficiently large, and prime, then the k+1st element will be less than 7*a_k. The k+2nd element will be less than 7/2*a_k, and may itself be prime, causing subsequent terms to be even bigger. Just from the premises you've quoted, you've not proved the conclusion you claim. Phil

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[math-fun] The PrimeLatz conjecture
by Eric Angelini 27 Nov '10

27 Nov '10

Hello math-Fun, ... any taker? > Start from any integer N. > If N is even, divide N by 2. > If N is odd, add to N the next 3 prime numbers > N. > Iterate. > Conjecture: all N's will enter in a loop, sooner or later. http://www.cetteadressecomportecinquantesignes.com/PrimeLatz.htm Best, É.

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[math-fun] Buffon bijection?
by Veit Elser 26 Nov '10

26 Nov '10

When Buffon's needle is dropped at random on parallel lines spaced by the needle's length, it crosses a line with probability 2/pi. Add another set of parallel lines, perpendicular to the first, and with the same spacing. The probability that the needle crosses two lines, one from each set, is not equal to (2/pi)^2, as it would if the two types of crossing were independent, but 1/pi. So there is the curious fact that the probability of crossing a horizontal line and a vertical line is equal to the probability of crossing a horizontal line and not crossing a vertical line (since their sum must be 2/pi). Equivalently, the conditional probabilities, of crossing or not crossing a vertical line, given that a horizontal line is crossed, are equal. If you believe there are no accidents in mathematics, then surely there is some bijection that explains this fact ...

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Re: [math-fun] why can a line of slope 10^n determine n decimal digits of π?
by Dan Asimov 25 Nov '10

25 Nov '10

Gary, it doesn't. I'm not sure what led you to think that the numbers are 31, 314, etc., but rather the actual numbers are only approximations to these integers. --Dan << Thanks, Joshua. But I still don't get it. Why does a line of slope -10 carve out exactly 31 arcs and not, say, 30? >> Those who sleep faster get more rest.

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[math-fun] Inhomogeneous contiguity
by Bill Gosper 25 Nov '10

25 Nov '10

For the past 30 yrs I should have been nagging the computer algebra community to add a huge-bang-for-the-buck feature to each of those many algorithms (Risch integration, indefinite summation, Berlekamp factoring, various ODE techniques) which internally generate and solve linear systems. Namely, provide a way of designating some of the input symbols as constrainable, in order to get a closed integral, sum, factorization, etc.) I think from the outset Macsyma's NUSUM had an undocumented list named %CONSTRAINABLES, and it smashes the list %CONSTRAINTS with the results of its internal call to SOLVE. A fairly powerful application is my Macsyma function CONTIGUATE, which takes an arbitrary list of contiguous hypergeometric functions in sum or pFq form, and finds a contiguity relation if there is one, simply by lumping all the summands times undetermined coefficients. Surprisingly(?), this can produce inhomogeneous relations, e.g., HypergeometricPFQRegularized[{1}, {a, 1 + b}, z] + (-2 + a - b) HypergeometricPFQRegularized[{1}, {a, 2 + b}, z] - z HypergeometricPFQRegularized[{1}, {a, 3 + b}, z] == ((2 + b)^(-(3/2) - b) E^(2 + b))/(Sqrt[2 \[Pi]] Gamma[-1 + a]) FunctionExpand and FullSimplify shrug at this, but FindInstance claims to refute it with a crazy triad of Gaussian rationals and some choice of exponential branch cuts. But when I stipulate Reals it resumes shrugging. Conjoining just Element[a,Reals] runs forever(?). --rwg PVC! Yanqui no!

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Re: [math-fun] Coincidentally, 109
by N. J. A. Sloane 24 Nov '10

24 Nov '10

There were several postings on this subject around Nov 10. The main sequence underlying the discussion is A000371 in the OEIS, and I have just finished editing the entry to bring it up to date. It arises in several different contexts. The sequence Marc was looking at was A003465, which is one-half of A000371. Neil

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[math-fun] Marden's/Siebeck's Theorem
by Henry Baker 22 Nov '10

22 Nov '10

I just ran across this "Most Marvelous Theorem in Mathematics": http://www.maa.org/joma/Volume8/Kalman/index.html It is well-known/well-taught that the roots of p'(x) lie within the convex hull of the roots of p(x). However, in the case of a cubic, we can say a lot more: the roots of p'(x) are the _foci_ of the inscribed _ellipse_ that passes through the midpoints of the sides of the triangle formed by the roots of p(x). This is Marden's/Siebeck's theorem. So in the case of (my) standardized cubic p(x)=x^3-(3/4)*x-cos(3*a)/4 [roots are cos(a), cos(a+2*pi/3), cos(a-2*pi/3), even when a is complex], the roots of p'(x) are +-1/2. So even as the triangle becomes enormous (a is complex), the ellipse becomes closer & closer to a circle, and the triangle becomes closer & closer to an equilateral triangle. Marden's/Siebeck's Theorem is one property of ellipses that Newton was almost certainly not aware of. Are there any consequences of this theorem that show up in Newton's 2-body problem? Given an ellipse, is the triangle that the ellipse is inscribed within unique? If so, are there any interesting physical/mechanical consequences of those triangle vertices? My conjecture is that this inscribed ellipse may describe an maximum speed trajectory that lies within the triangle, subject to certain acceleration constraints, but I don't know enough calculus of variations to be able to get any further.

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