On 2/22/07, Daniel Asimov <dasimov@earthlink.net> wrote:
I plotted this with equal scale using the Mac utility "Grapher", and from the plot it's clear that this closed curve is strictly convex everywhere.
It does have arcs (near the maximum of curvature) where the curvature comes very close to 0.
Correct --- I hadn't meant to imply that the curvature actually went negative --- just that it looked as if it was thinking about it! [Also I cheated, and tried varying Bill's constant coefficients ...]
NOTE: There is a converse to the four-vertex theorem (which says that the curvature function k(s) (s=arclength) of a C^2 simple closed planar curve must have a least two local maxima and two local minima).
The converse says that for any continuous real-valued function on the circle with at least two local maxima and two local minima is the curvature function of some simple closed curve in the plane. (Cf. < http://www.ams.org/notices/200702/fea-gluck.pdf >.)
What an interesting and well-presented account! I wish I claim to have understood it all, but it would easily bear re-reading.
We want our egg to be strictly convex, bilaterally symmetric, and as simple as possible. Hence we'd like the curvature function to be everywhere positive, satisfy k(t) = k(-t) (0 <= t <= 2pi), and have the fewest possible critical points: two local maxes and two local mins.
The important point here is that t is not arclength, gradient, or any other intrinsic feature of the curve. Quoting from loco. cit. p195 [S^1 = circle, |R^2 = plane] : "Full Converse to the Four Vertex Theorem. Let κ : S^1 → |R be a continuous function that is either a nonzero constant or else has at least two local maxima and two local minima. Then there is an embedding α : S^1 → |R^2 whose curvature at the point α(t) is κ(t) for all t ∈ S^1."
... I'd like to see simple closed curves with curvature functions like these.
This would surely make an interesting project, but a highly nontrivial one. Any plotting algorithm must recreate the embedding referred to above, the construction of which constitutes the major portion of the proof! Furthermore, it's not at all clear to me at this stage whether such an embedding, and hence the resulting curve, would necessarily be unique --- indeed, I strongly suspect it may not be.
RWG wrote: ...
Or maybe it just wanted some action with an American whoopee crane.
P.S. Fred, they're called *whooping* cranes, not whoopee cranes.
Um, not guilty, I'm afraid. And I'm quite sure that Bill's ornithological nomenclature is impeccable, at any rate when he can restrain himself from punning in dubious taste! Fred Lunnon