Re: [Fwd: [math-fun] triangoops
Nick, you chided
bill, i don't think your claim is accurate. it seems you've found the smallest c such that one edge goes through the origin. but i think one can always find a smaller triangle.
by experimentation, i find that c is suspiciously close to sqrt(3), but i have n o math to support this.
[...]
"R. William Gosper" wrote:
Nick Baxter points out that in answering
what is the smallest constant c so that the graph of the function f(x) = x^3 - c x contains the vertices of an equilateral triangle?
[...]
Claim: The midpoint of one triangle side is the origin. Otherwise, centering some side and rotating its endpoints back onto the curve will intersect the ot her two sides with the curve, indicating that c is non-minimal.
NONSENSE! The derivatives at both endpoints are equal, so you can slide them slightly in the same direction so that the other two sides obviously cross the curve, which is impossible when c is minimal. Grinding out dc/dx1=0, dc/dx2=0, the x coordinates in the minimal case are the roots near -.54557 and .83587 of 6 4 2 27 x - 54 sqrt(3) x + 72 x - 8 sqrt(3). Trying for the univariate c polynomial led toward degree 588, but numerically the answer is just c = sqrt(3) ! Erich, is there some way to see this? --rwg
what is the smallest constant c so that the graph of the function f(x) = x^3 - c x contains the vertices of an equilateral triangle?
<My braindamaged symmetry argument.>
NONSENSE! (Mine, not Nick's.)
Grinding out dc/dx1=0, dc/dx2=0,
Maybe better would have been to guess that for the critical c, the triangle would be unique (but for its negative twin). This would show up as a zero discriminant.
the x coordinates in the minimal case are the roots near -.54557 and .83587 of
6 4 2 27 x - 54 sqrt(3) x + 72 x - 8 sqrt(3).
The triangle abscissae are the three roots %pi 2 %pi %pi 2 %pi 2 sin(---) sin(-----) 2 sin(---) cos(-----) 18 9 9 9 [- 2 sqrt(---------------------), 2 sqrt(---------------------), 3 3 %pi %pi 2 cos(---) cos(---) 18 9 2 sqrt(-------------------)] 3 The area is 2, and the center is (sqrt(2)/3^(3/4),sqrt(2)/3^(5/4)). The slope of the side joining the first and third vertices is tan(pi/18). Macsyma users can see the picture by defining xy(x):=[x,x*(x^2-sqrt(3.))] and then executing block([equalscale:true,x1:-0.54557,x2:0.83587,x3], x3:-sqrt(3.)*(x2^3-x1^3)/2+2*x2-x1, paramplot(''([xy(-x3*(1-t)+x3*t),xy(x1)*(1-t)+t*xy(x2),xy(x2)*(1-t)+t*xy(x3), xy(x3)*(1-t)+t*xy(x1)]),t,0,1))
Trying for the univariate c polynomial led toward degree 588, but numerically the answer is just c = sqrt(3) !
Which can then be verified symbolically. Much harder might be to symbolically verify the centroid expression, found with Rich's old number-relator. The elegant results c=sqrt(3), area=2 raise hope that some dazzling insight will obviate my strenuous calculations, but those nontrivial abscissa values lower my anticipation. This looks like a classic case of intermediate expression swell. Erich, did you invent this bruiser? ---------- prod(uct )i(nte)g(r)al $.02 : P They're not terribly fruitful, since you can "do" I f(x) iff you can / / "do" | ln f(x), but sometimes the product limit seems the only way / to get, e.g., the Barnes-Alexeiewsky recurrence x / | ln t! dt = ln sqrt(2 pi) - (1 - ln x) x . / x-1 --rwg
=R. William Gosper prod(uct )i(nte)g(r)al $.02 : P They're not terribly fruitful, since you can "do" I f(x) iff you can / / "do" | ln f(x), but sometimes the product limit seems the only way [...] /
Agreed. Yet in part, with apologies to Hamming, "The purpose of notation is insight, not answers." Couldn't one as well argue for eliminating big Pi in favor of big Sigma log? Prodigals are indicated, in moderation, just when they are more natural, simple, or promote interesting ideas (eg generalizing factoring polynomials by their roots to arbitrary functions). Side-bar: Bill, are you following the discussion of negative binomials on seqfan? I was thinking you might have interesting comments.
participants (2)
-
Marc LeBrun -
R. William Gosper