Here's one solution based on one of Warren's miracles. Plot (39, 21, 3, 3) (ring, sun, planet, planet). This gives an offset of exactly 15. This is due to the odd equation 2 * acos(-11/16) - 3 acos(7/8) = pi Integral offsets such as these are rare except in certain degenerate cases. On Tue, Aug 4, 2015 at 4:22 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Unfortunately, my previous carefully-honed missile was directed at an obsolete target (arc-lengths of belts), rather than the current one (offset of sun).
For offset, we want to solve the equation in Z with rational A,B,C,D A arccos( B Z^2 + C) + ... = D , so that cos D is a polynomial with rational coefficients in cos(Z^2) and sin(Z^2) , which can be further rationalised to a polynomial in cos^2(Z^2) alone.
Solving this yields Z^2 = arccos(algebraic) ; and APG reminds us that this never has properly algebraic solutions, only rational or (more usually) transcendental!
Fred Lunnon
On 8/4/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, the only solutions to:
cos(algebraic * pi) = algebraic
are of the form:
cos(rational * pi) = algebraic
Proof: --------
It is clear that cos(algebraic * pi) and exp(algebraic * (pi i)) are algebraically dependent, so we can consider the equivalent problem to find solutions to:
exp(algebraic * (pi i)) = different algebraic
which rearranges to:
(-1)^algebraic = different algebraic
By Gelfond-Schneider, this occurs only when the exponent is rational.
--------
Clearly the converse is also true, so the algebraic numbers x for which cos(x * pi) is also algebraic are precisely the rationals.
Sincerely,
Adam P. Goucher
Sent: Tuesday, August 04, 2015 at 8:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Offset Somsky Gears, coaxial offset
Before this intrusion of ineptitude risks the moral endangerment of any further tender young minds, I feel obliged to point out the logical inequivalence between the statements cos(rational * 2pi) = algebraic ! and arccos(rational) = algebraic * 2pi ? --- the second of which is in general (to put it mildly) problematic.
Those remaining in doubt about the matter should reflect upon the kindergarten fact arccos(1/3) = dihedral angle of regular tetrahedron, the expression of which as an algebraic angle ought to be worth a Fields at least!
Fred Lunnon
On 8/3/15, Warren D Smith <warren.wds@gmail.com> wrote:
Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill Gosper
--Answer (WDS): I showed how the non-algebraic equations involving arctrig, could be restated in a way involving integer powers of complex numbers (with coordinates inside the complexes involving square roots). So then really, they are algebraic equations in disguise. And then yes, you do not need any decimals at all to know algebraic.
Furthermore in an early post I explored general bipartite-planar gear-circle-contact graphs, and the equations there are just about distances, which are of course algebraic functions.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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