Around Tait-Kneser Theorem. Peter Guthrie Tait ( ) and Adolf Kneser ( )

Transcription

1 Around Tait-Kneser Theorem Peter Guthrie Tait ( ) and Adolf Kneser ( ) 1

2 Osculating circles What do they look like? 2

3 Like this? 3

4 No, like this: 4

5 Theorem: The osculating circles of an arc with monotonic positive curvature are nested. 5

6 To quote from Tait s half a page long paper (1895): The proof is excessively simple. For if A, B, be any two points of the evolute, the chord AB is the distance between the centers of two of the circles, and is necessarily less than the arc AB, the difference of their radii. Let us elaborate... 6

7 Evolute Γ of a curve γ is the envelope of its normals: Γ is also the locus of centers of osculating circles of γ. 7

8 32 normals to an ellipse: The cusps are the centers of hyper-osculating circles, corresponding to the vertices of γ. 8

9 Evolute to involute, from Γ to γ: string construction. One obtains a 1-parameter family of equidistant involutes (integration, of sorts). 9

10 Corollary 1: an arc of the evolute Γ equals the increment of the radii of curvature of the involute γ. Corollary 2: the total signed length of the evolute is zero. (R 1 r 1 ) (R 2 r 1 ) + (R 2 r 2 ) (R 1 r 2 ) = 0. 10

11 Zero length, and absence of inflections, are sufficient for Γ to be an evolute. Proof of Tait-Kneser Theorem: z 1 z 2 < arc (z 1 z 2 ) = r 1 r 2. 11

12 Q: How can a curve be tangent to the leaves of a foliation and not be confined to one of them?? A: The foliation is not differentiable! Namely, a differentiable function, constant on the circles, is constant everywhere. Proof: if F is constant on the leaves then df is zero on the leaves. Then df is zero on the curve, hence F is constant on the curve, and therefore constant in the annulus. 12

13 Osculating Taylor polynomials f(x) g t (x) = n i=0 f (i) (t) (x t) i. i! Theorem: Let n be even and f (n+1) (x) 0 on interval I. Then, for a, b I, the graphs of g a (x) and g b (x) are disjoint. Proof. Assume f (n+1) (t) > 0 for t I, and a < b. Then g t (x) t hence g b (x) > g a (x). = f (n+1) (t) (x t) n > 0, n! 13

14 Osculating quadratic Taylor polynomials for y = x 3 :

15 Other examples Osculating trigonometric polynomials of 2π-periodic functions g t (x) = c + n i=1 (a i cos ix + b i sin ix). Osculating conics (through 5 consecutive points). Hyperosculation corresponds to sextactic points. 15

16 For f : RP 1 RP 1, one has the osculating fractionallinear transformations. The graphs are hyperbolas. Hyperosculation corresponds to zeros of the Schwarzian derivative S(f)(x) = f (x) f (x) 3 2 ( f ) 2 (x) = 0. f (x) Osculating cubics (through 9 consecutive points). Need to assume that it is the oval that osculates. Hyper-osculation corresponds to 3-extactic points. In general, the space of algebraic curves of degree d has dimension d(d + 3)/2. 16

17 Two types of cubic curves: In all cases, if no hyper-osculations occurs, then the osculating curves are disjoint. They form non-differentiable foliations. These are results of E. Ghys, V. Timorin and S. T. 17

18 Osculating ovals of cubic curves:

19 A related topic, 4-vertex theorem and its modifications: A plane oval has at least 4 vertices (curvature extrema, or hyper-osculating circles); A plane oval has at least 6 sextactic points (affine curvature extrema, or hyper-osculating conics); Given a diffeomorphism f : RP 1 RP 1, the Schwarzian derivative S(f) has at least 4 zeros (hyper-osculating fractionallinear transformations); A small perturbation of the oval of a cubic curve has at least ten 3-extactic points (hyper-osculating cubics). 19

20 These results are due to S. Mukhopadhyaya (1909), E. Ghys (1995), and V. Arnold (1996): 20

21 4-vertex theorem for immersed curves (conjectured by V. Arnold, proved by Yu. Chekanov and P. Pushkar, 2005) Regular homotopy, from an oval: 21

22 Another related (?) topic, Sylvester s problem and its variants: Let S be a finite set of points in the plane such that the line through every pair of points of S contains another point of S. Then S lies on a line (Educational Times, 1893). James Joseph Sylvester ( ) 22

23 First proof, by T. Gallai (1933, asked by P. Erdös). Proof from the Book (L. Kelly): Consider the closest point of S to a line through a pair of points of S. Two points, B and C, are on one side of the perpendicular from A to BC. Then point B is closer to the line AC. A B C 23

24 A surprise Sylvester-Gallai theorem fails over C: the 9 inflection points of a cubic curve in C 2 satisfy the condition but are not collinear (at most 3 are real). 24

25 Generalization: from linear functions to polynomials. Theorem (P. Borwein, 1983): Let S be a finite set of points in the plane, with different x-coordinates, such that the graph of a polynomial of degree n through every n + 1-tuple of them passes through another point of S. Then S lies on the graph of a polynomial of degree n. Another generalization: from lines to circles. Still another one, for conics (J. Wiseman, P. Wilson, 1988): Given a finite set S, not entirely on a conic, there exists a conic through exactly 5 points of the set S. Problem: what about cubic curves? 25