Re: [math-fun] Fred Lunnon, RWG, and whoever's arctrig confusion
Now I look again, I discover that a typo (or fly on my specs) swallowed the 2pi from the RHS from my notes, and it now looks as though WDS was right after all --- the sunset/offset is algebraic! [I think now that I actually realised that while in bed last night, but had forgotten about it again by the time I got near a computer again ... Apologies! WFL On 8/6/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Warren,
You don't appear to have seen later messages from myself and Goucher which addressed this issue. The relevant equation is not << FINITESUM_j A_j * arccos(B_j(u)) = C*pi >> but << ... = C >>> , which alters the result drastically.
Fred
On 8/5/15, Warren D Smith <warren.wds@gmail.com> wrote:
FWL: 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!
--WDS: I'd never claimed these 2 statements equivalent. Since FWL seems confused, let me clarify. Or repeat what I'd said before. Hopefully this also will unconfuse others such as RWG and "tender young minds." (Although, frankly, I think the problem lies more in cranky old minds.)
FACT: Suppose you have an equation of the form FINITESUM_j A_j * arccos(B_j(u)) = C*pi where C and the A_j are integers, and the functions B_j(u) are algebraic functions of u. Then: u is algebraic.
BECAUSE: You can rewrite the equation as realpart PRODUCT_j Y_j(u)^(A_j) = 0 where the Y_j(u) are complex numbers with unit norm that are algebraic functions of u, having realpart Y_j(u) = B_j(u).
Here by "algebraic function" F(u) I mean "if u algebraic then F(u) is, and if y algebraic with F(u)=y, then u is."
In gear applications, the fact the A_j are integers is related to the fact gear tooth counts are integers.
Are we clear now? This same trick is also how one can prove (or disprove) any of my so-called "arccos miracles" in an automated fashion, and note, this proof NOWHERE INVOLVES trig or arctrig, which was the whole point of the trick -- the arctrig intentionally gets removed from the scene, allowing simpler pure-algebraic simplifiers to do their thing.
Be happy.
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In view of the recent resolution of our earlier disagreement, it causes me huge glee --- er, modest regret --- to have to reveal that I may not be the only correspondent to experience difficulty in distinguishing sigmoid from olecranon. Consider the gear train comprising ring, sun, two planets with respective radii 12, 6, 3, 3 and sun offset = gamma = 0.57721566... . By Somsky's lemma that kicked everything off, this is mechanically realisable (a practical construction might benefit from embellishment with rods connecting sun to planets), with sunset transcendental. Tender young minds, particularly any anatomically challenged, are encouraged to reflect upon how gamma manages to be a root of a polynomial equation with rational coefficients, such as that resulting from A arccos( B Z^2 + C ) + ... = 2pi --- which I am hopeful of having eventually rendered correctly. Fred Lunnon On 8/6/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Now I look again, I discover that a typo (or fly on my specs) swallowed the 2pi from the RHS from my notes, and it now looks as though WDS was right after all --- the sunset/offset is algebraic!
[I think now that I actually realised that while in bed last night, but had forgotten about it again by the time I got near a computer again ... Apologies!
WFL
On 8/6/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Warren,
You don't appear to have seen later messages from myself and Goucher which addressed this issue. The relevant equation is not << FINITESUM_j A_j * arccos(B_j(u)) = C*pi >> but << ... = C >>> , which alters the result drastically.
Fred
On 8/5/15, Warren D Smith <warren.wds@gmail.com> wrote:
FWL: 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!
--WDS: I'd never claimed these 2 statements equivalent. Since FWL seems confused, let me clarify. Or repeat what I'd said before. Hopefully this also will unconfuse others such as RWG and "tender young minds." (Although, frankly, I think the problem lies more in cranky old minds.)
FACT: Suppose you have an equation of the form FINITESUM_j A_j * arccos(B_j(u)) = C*pi where C and the A_j are integers, and the functions B_j(u) are algebraic functions of u. Then: u is algebraic.
BECAUSE: You can rewrite the equation as realpart PRODUCT_j Y_j(u)^(A_j) = 0 where the Y_j(u) are complex numbers with unit norm that are algebraic functions of u, having realpart Y_j(u) = B_j(u).
Here by "algebraic function" F(u) I mean "if u algebraic then F(u) is, and if y algebraic with F(u)=y, then u is."
In gear applications, the fact the A_j are integers is related to the fact gear tooth counts are integers.
Are we clear now? This same trick is also how one can prove (or disprove) any of my so-called "arccos miracles" in an automated fashion, and note, this proof NOWHERE INVOLVES trig or arctrig, which was the whole point of the trick -- the arctrig intentionally gets removed from the scene, allowing simpler pure-algebraic simplifiers to do their thing.
Be happy.
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Fred Lunnon