High order parametric polynomial approximation of conic sections Ga - PowerPoint PPT Presentation

High order parametric polynomial approximation of conic sections Gaˇ sper Jakliˇ c (joint work with J. Kozak, M. Krajnc, V. Vitrih and E. ˇ Zagar) FMF and IMFM, University of Ljubljana PINT, University of Primorska DWCAA09, Alba di Canazei September 9th 2009 1 / 28

Outline 1 Parametric approximation of implicit curves 2 Conic sections 3 Best solution 4 Examples 5 H¨ ollig-Koch conjecture 6 Sphere approximation 2 / 28

1.5 1.0 0.5 � 1.5 � 1.0 � 0.5 0.5 1.0 � 0.5 � 1.0 � 1.5 Figure: The unit circle (blue dashed), quintic polynomial approximation given by Lyche and Mørken in 1995 (brown) and the new quintic approximant (red). 3 / 28

Parametric approximation • Let ( x , y ) ∈ D ⊂ R 2 , f ( x , y ) = 0 , be a segment of a regular smooth planar curve f f f f f f f f f . • Suppose � x r ( t ) � r : I ⊂ R → R 2 : t �→ r r r r r r r r y r ( t ) is a parametric approximation of the curve segment f f and f f f f f f f f ( x r ( t ) , y r ( t )) = ε ( t ) , t ∈ [ a , b ] . 4 / 28

• Consider the regular parameterization of r r r r r r r r r with respect to the normal of f f , i.e., every point ( x , y ) on a curve f f f f f f f f defines a unique parameter t := t ( x , y ) on a curve r r r : r r r r r r f ( x , y ) r r r r r r r r r r ( t ) r r r r r r r r • This provides an upper bound on Hausdorff and parametric distance. 5 / 28

Theorem If the curve r r r r r r r r r can be regularly reparameterized by the normal to f , and ε is small enough, the distance between curves is bounded by | ε ( t ( x , y )) | + O ( ε ( t ( x , y )) 2 ) . max � ( x , y ) ∈D f 2 x ( x , y ) + f 2 y ( x , y ) 6 / 28

Conic sections • Only ellipse and hyperbola are interesting to consider (no polynomial parameterization). • A particular conic section is given as f ( x , y ) := x 2 ± y 2 − 1 = 0 . • The main problem: find two nonconstant polynomials x n and y n of degree at most n , such that x 2 n ( t ) ± y 2 n ( t ) = 1 + ε ( t ) , where ε is a polynomial of degree at most 2 n . 7 / 28

• Assume that at least one point is interpolated, i.e., ε (0) = 0. In order to get ε as small as possible, it is natural to choose ε ( t ) := t 2 n . • If also a tangent direction is prescribed at the interpolation point, we have ( x ′ n (0) , y ′ ( x n (0) , y n (0)) = (1 , 0) , n (0)) = (0 , 1) . • Thus n � a ℓ t ℓ , x n ( t ) = 1 + ℓ =1 n � b ℓ t ℓ , y n ( t ) = t + ℓ =2 and 8 / 28

• x 2 n ( t ) ± y 2 a 2 n ± b 2 t 2 n . � � n ( t ) = 1 + n � • A reparameterization t �→ | a 2 n ± b 2 2 n n | t gives n n � � α ℓ t ℓ , β ℓ t ℓ , x n ( t ) := 1 + y n ( t ) := β 1 > 0 , ℓ =1 ℓ =1 where x 2 n ( t ) ± y 2 n ( t ) = 1 + sign ( a 2 n ± b 2 n ) t 2 n . Many acceptable solutions exist. 9 / 28

Table: The number of appropriate solutions in all three cases for n = 1 , 2 , . . . , 10. 1 2 3 4 5 6 7 8 9 10 n elliptic 1 1 3 6 15 27 63 120 246 495 hyp. a 2 n < b 2 1 0 1 2 5 8 19 32 68 120 n hyp. a 2 n > b 2 0 1 0 2 0 9 0 32 0 125 n 10 / 28

Solutions • Solving the equation x 2 n ( t ) ± y 2 n ( t ) = 1 ± t 2 n is equivalent to solving x 2 n ( t ) ± y 2 n ( t ) = 1 in the factorial ring R [ t ] / t 2 n . • There are additional restrictions, classic algebraic tools can not be applied. 11 / 28

Idea • Rewrite 2 n − 1 � t − ei i i i i i 2 k +1 i i i 2 n π � � ( x n ( t ) + i i i i i i i i i y n ( t )) ( x n ( t ) − i i i i i i i y n ( t )) = i i , k =0 where the right-hand side is the factorization of 1 + t 2 n over C . • Thus n − 1 � t − ei i i i i i i i i σ k 2 k +1 2 n π � � x n ( t ) + i i i i i i i i y n ( t ) = γ i , γ ∈ C , | γ | = 1 , k =0 where σ k = ± 1. 12 / 28

Best solution • The best solution should have the minimum error term ε . • It turns out that this happens if 1 β 1 = . sin π 2 n • Surprisingly, any solution for the elliptic case, for which x n is even and y n is odd, can be transformed to the hyperbolic solution by the map x n ( t ) �→ x n ( i i i i i i i i i t ) , y n ( t ) �→ − i i y n ( i i t ) . i i i i i i i i i i i i i i • In particular, this is true for the best solution too, thus it is enough to consider the elliptic case only. 13 / 28

Theorem Coefficients of the best solution for the elliptic case are  k ( n − k ) � k 2 π � 2 n + π �  P ( j , k , n − k ) cos n j ; k is even ,  α k = j =0   0; k is odd , 0; k is even ,    k ( n − k ) � k 2 π 2 n + π � β k = � P ( j , k , n − k ) sin ; k is odd , n j   j =0 where P ( j , k , r ) denotes the number of integer partitions of j ∈ N with ≤ k parts, all between 1 and r, where k , r ∈ N , and P (0 , k , r ) := 1 . 14 / 28

Examples Table: Polynomial approximation of the unit circle and maximal normal (radial) error. n x n and y n error x 3 ( t )=1 − 2 t 2 3 2 y 3 ( t )=2 t − t 3 √ 2) t 2 + t 4 x 4 ( t )=1+( − 2 − y 4 ( t )=( √ 2+ √ 4 0 . 414213 √ √ 2)( t − t 3 ) 2+ 2 − √ √ 5) t 2 +(1+ 5) t 4 x 5 ( t )=1+( − 3 − 5 0 . 089987 √ √ 5) t 3 + t 5 y 5 ( t )=(1+ 5) t +( − 3 − √ √ 3) t 2 +2(2+ 3) t 4 − t 6 x 6 ( t )=1 − 2(2+ 6 0 . 013886 √ √ √ √ √ √ 3) t 3 +( 6) t 5 y 6 ( t )=( 2+ 6) t − 2(3+2 2+ . . . . . . . . . 1 . 07280 · 10 − 15 15 . . . It can be shown that the error is O ( n − 2 n ). 15 / 28

1.5 1.0 0.5 � 1.5 � 1.0 � 0.5 0.5 1.0 � 0.5 � 1.0 � 1.5 Figure: The unit circle. 16 / 28

1.5 1.0 0.5 � 1.5 � 1.0 � 0.5 0.5 1.0 � 0.5 � 1.0 � 1.5 Figure: The unit circle and its polynomial approximant for n = 2. 17 / 28

1.5 1.0 0.5 � 1.5 � 1.0 � 0.5 0.5 1.0 � 0.5 � 1.0 � 1.5 Figure: The unit circle and its polynomial approximant for n = 3. 18 / 28

1.5 1.0 0.5 � 1.5 � 1.0 � 0.5 0.5 1.0 � 0.5 � 1.0 � 1.5 Figure: The unit circle and its polynomial approximant for n = 4. 19 / 28

1.5 1.0 0.5 � 1.5 � 1.0 � 0.5 0.5 1.0 � 0.5 � 1.0 � 1.5 Figure: The unit circle and its polynomial approximant for n = 5. 20 / 28

1.5 1.0 0.5 � 1.5 � 1.0 � 0.5 0.5 1.0 � 0.5 � 1.0 � 1.5 Figure: The unit circle and its polynomial approximant for n = 6. 21 / 28

1.5 1.0 0.5 � 1.5 � 1.0 � 0.5 0.5 1.0 � 0.5 � 1.0 � 1.5 Figure: The unit circle and its polynomial approximant for n = 7. 22 / 28

Approximants and curvatures 1.0 0.8 0.6 0.4 0.2 � 1.0 � 0.5 0.0 0.5 1.0 23 / 28

Cycling Figure: Unit circle together with the cycles of the approximant for n = 20 and t ∈ [ − 1 , 1]. 24 / 28

H¨ ollig-Koch conjecture Conjecture A polynomial planar parametric curve of degree n can interpolate 2 n given points with an approximation order 2 n. Theorem H¨ ollig-Koch conjecture holds true for conic sections. Idea of a proof: • asymptotic approach, • a particular nonlinear system has to be studied, • an existence of a solution guarantees the optimal approximation order, • solution is obtained by canonical form and optimal solutions for ellipse and hyperbola. 25 / 28

Ellipse 0.5 1.0 1.5 � 0.5 0.5 1.0 � 0.2 � 0.2 � 0.4 � 0.4 � 0.6 � 0.6 � 0.8 � 0.8 2 x 2 + xy + 5 3 y 2 + y = 0 with Figure: Approximation of the ellipse 1 the best approximant of degree n = 5 , 7. 26 / 28

Hyperbola 1 1 2 4 6 8 10 12 2 4 6 8 10 12 � 1 � 1 � 2 � 2 � 3 � 3 5 x 2 + xy + 1 8 y 2 + y = 0 Figure: Approximation of the hyperbola 1 with the best approximant of degree n = 3 , 4. 27 / 28

Sphere approximation Particular polynomials of degree 5 in u and v yield: 5) u 2 + (1 + 5) v 2 + (1 + √ √ 5) u 4 )(1 + ( − 3 − √ √ 5) v 4 ) x ( u , v ) = (1 + ( − 3 − 5) u 3 + u 5 )(1 + ( − 3 − 5) v 2 + (1 + √ √ √ √ 5) v 4 ) y ( u , v ) = ((1 + 5) u + ( − 3 − √ √ 5) v 3 + v 5 z ( u , v ) = (1 + 5) v + ( − 3 − 28 / 28