Now I have studied Andy's generalisation and proof of the extrinsic criterion, I take off my hat to a model of clarity and concision, making everything look easier than it is (as earlier discovered to my cost). However, I do not understand the necessity for C^2 --- why would not C^1 suffice? In contrast my attempts to cast light by comparing "extrinsic" with "intrinsic" side-by-side have evidently backfired in spades, resulting in dismissive refutations of conjectures which I remain unaware of having proposed in the first place. Anyway moving on, in the spirit of his neat one-liner, there follows a proposed more general intrinsic criterion for real 2-space zero-detection: Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0 ***. Proof (offered tentatively): By the implicit function theorem, the constraint f(x, y) = 0 defines a function x(y) single-valued in any interval of y where df/dx <> 0 . Such an interval meeting R is finite since R is compact, and its endpoints lie in the interior of R since f <> 0 on the boundary; therefore df/dx = 0 at those endpoints. The converse is trivial. QED. The result may well generalise to an m-dimensional curve in n-space, but I prefer to take one step at a time. It is stronger than the earlier extrinsic criterion because nec. and suff., so should be applicable to situations such as solution of equations and curve-tracing, besides computational theorem-proving. The driver behind its practical utility is that for f polynomial there exists a constructive algorithm to detect critical points: compute discriminant g(y) of f(x, y) with respect to x , then solve g = 0 to yield their (real) y-components. Locating corresponding x-components is rather more tedious [ Gröbner bases prove helpful here in practice, despite theoretical complications ] though for some applications these may not be required. In particular if y_0 denotes the maximum critical y-component, we may rest assured --- pace compactness and boundary condition --- that there are no zeroes in the region y > y_0 ; and similarly (exchanging variables) for x . Fred Lunnon On 6/1/14, Andy Latto <andy.latto@pobox.com> wrote:
I don't understand what you mean by "the critical points for the intrinsic criterion"; given a function f(x,y), what is your definition of what you're calling the critical points for the intrinsic criterion? Points where the tangent line is horizontal? What about point where the tangent line does not exist? is the origin a critical point in your sense of y^2 = x^3? Of x^2 = y^3?
Andy
On Sat, May 31, 2014 at 11:13 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Some confusion here, I'm afraid. The origin is not a critical point for the intrinsic criterion: the critical points lie at (x,y) = (0,+1), (0,-1) . WFL
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Sat, May 31, 2014 at 8:05 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote: .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R contains the origin
I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in Spane?) remaining firmly in 2-space:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R intersects the unit circle
if and only if
The boundary of R intersects the unit circle
or
R contains the origin.
One direction of this is true:
If R intersects the unit circle, but the boundary of R does not, then since the unit circle is connected, R contains the entire unit circle, and since it is simply connected, it contains the entire unit disc.
But for the converse, a small disc around the origin is an obvious counterexample.
This is the second conjecture you've posted about all plane algebraic curves where the answer is "well, it's trivially false for the circle". Since you're presumably not just spewing out random conjectures, presumably you have some intuition about plane algebraic curves that you are trying to capture and failing. Maybe it would work better for you to post what this intuition is informally, rather than making failed attempts to formalize it and having math-fun play the "well, this is obviously false, let's try to guess what Fred meant to say" game.
Andy
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