Interpolation and Design
Marie-Laurence Mazure
Laboratoire Jean Kuntzmann
Université Joseph Fourier
Grenoble, France
Oslo, December 7th, 2009
WHY POLYNOMIALS?
1
Why polynomials? I- For Design
• Given a < b and P0, . . . , Pn ∈ Rd (control points) draw a
parametric curve in Rd, defined on [a, b], such that
– it mimics the control polygon [P0 . . . Pn]
– changes on the control polygon imply predictible changes
on the curve
– shape preservation properties
2
Why polynomials? I- For Design
• Given a < b and P0, . . . , Pn ∈ Rd (control points) draw a
parametric curve in Rd, defined on [a, b], such that
– it mimics the control polygon [P0 . . . Pn]
– changes on the control polygon imply predictible changes
on the curve
– shape preservation properties
• Use the degree n Bernstein basis relative to (a, b):
x−b
n
n
Bi (x) := ( i )
a−b
!n−i
x−a
b−a
!i
0≤i≤n
3
Why polynomials? I- For Design
• Given a < b and P0, . . . , Pn ∈ Rd (control points) draw a
parametric curve in Rd, defined on [a, b], such that
– it mimics the control polygon [P0 . . . Pn]
– changes on the control polygon imply predictible changes
on the curve
– shape preservation properties
• Use the degree n Bernstein basis relative to (a, b) :
x−b
n
n
Bi (x) := ( i )
a−b
!n−i
x−a
b−a
!i
0≤i≤n
and define the curve via
F (x) =
n
X
Bin(x)Pi,
x ∈ [a, b]
i=0
4
Why polynomials? I- For Design
• Given a < b and P0, . . . , Pn ∈ Rd (control points) draw a
parametric curve in Rd, defined on [a, b], such that
– it mimics the control polygon [P0 . . . Pn]
– changes on the control polygon imply predictible changes
on the curve
– shape preservation properties
• Use the degree n Bernstein basis relative to (a, b):
x−b
a−b
and define the curve via
Bin(x) := ( ni )
F (x) =
n
X
!n−i
x−a
b−a
Bin(x)Pi,
!i
0≤i≤n
x ∈ [a, b]
i=0
“Bézier curves” (∼ 60)
5
Polynomial Design
P3
P3
P3
P2
P2
P4
P4
P5
P1
P6
P5
P1
P6
P6
P0
P0
Parametric polynomial curves
F (x) =
6
X
Bi6(x)Pi
i=0
6
Underlying reason: Blossoms
Theorem. (L.Ramshaw, 87) For any F ∈ Pnd, there exists a unique
function f : Rn → Rd, called its blossom, such that
• symmetry: f is symmetric on Rn
• n-affinity: f is affine in each variable, i.e., for any x1 , . . . , xn−1 , γ, δ ∈ I
with γ < δ, for all x ∈ I,
f (x1 , . . . , xn−1 ,x) = 1 − α(x) f (x1 , . . . , xn−1 ,γ) + α(x)f (x1 , . . . , xn−1 ,δ)
where α(x) := (x − γ)/(δ − γ)
• diagonal property: f (x, . . . , x) = F (x) for all x ∈ R.
7
Underlying reason: Blossoms
Theorem. (L.Ramshaw, 87) For any F ∈ Pnd, there exists a unique
function f : Rn → Rd, called its blossom, such that
• symmetry: f is symmetric on Rn
• n-affinity: f is affine in each variable, i.e., for any x1 , . . . , xn−1 , γ, δ ∈ I
with γ < δ, for all x ∈ I,
f (x1 , . . . , xn−1 ,x) = 1 − α(x) f (x1 , . . . , xn−1 ,γ) + α(x)f (x1 , . . . , xn−1 ,δ)
where α(x) := (x − γ)/(δ − γ)
• diagonal property: f (x, . . . , x) = F (x) for all x ∈ R.
see also, formes à pôles, Paul de Faget de Casteljau (59)
8
Underlying reason: Blossoms
Theorem. (L.Ramshaw, 87) For any F ∈ Pnd, there exists a unique
function f : Rn → Rd, called its blossom, such that
• symmetry: f is symmetric on Rn
• n-affinity: f is affine in each variable, i.e., for any x1 , . . . , xn−1 , γ, δ ∈ I
with γ < δ, for all x ∈ I,
f (x1 , . . . , xn−1 ,x) = 1 − α(x) f (x1 , . . . , xn−1 ,γ) + α(x)f (x1 , . . . , xn−1 ,δ)
where α(x) := (x − γ)/(δ − γ)
• diagonal property: f (x, . . . , x) = F (x) for all x ∈ R.
see also, formes à pôles, Paul de Faget de Casteljau (59)
Ex: n = 3, d = 1, if F (x) = a0 + a1 x + a2 x2 + a3 x3 ,
f (x1 , x2 , x3 ) = a0 + a1
x1 + x2 + x3
x1 x2 + x2 x3 + x3 x1
+ a2
+ a3 x1 x2 x3
3
3
10
Polynomial design
D
K
M
C
E
A
I
J
L
B
Polynomial function of degree 3: De Casteljau algorithm for the
evaluation of E := F (x) (here x = (a + b)/2), from the Bézier points of F
w.r. to (a, b):
A := f (a, a, a) = F (a), B := f (a, a, b), C := f (a, b, b), D := f (b, b, b) = F (b)
11
Polynomial design
D
K
M
C
E
A
I
J
L
B
Polynomial function of degree 3: De Casteljau algorithm for the
evaluation of E := F (x) (here x = (a + b)/2), from the Bézier points of F
w.r. to (a, b):
A := f (a, a, a) = F (a), B := f (a, a, b), C := f (a, b, b), D := f (b, b, b) = F (b)
{B} = Osc1 F (a) ∩ Osc2 F (b)
{C} = Osc2 F (a) ∩ Osc1 F (b)
12
Why polynomials? II- for Interpolation
• Given
– E, N -dimensional space of sufficiently differentiable functions on I,
– x1 < x2 < · · · < xr−1 < xr in I, interpolation abscissae
P
– µ1, . . . , µr > 0 with ri=1 µi = N , interp. multiplicities
– αi,j ∈ R, 1 ≤ i ≤ r, 0 ≤ j ≤ µi − 1, data
• Hermite interpolation problem: Find F ∈ E such that
F (j)(xi) = αi,j , 1 ≤ i ≤ r, 0 ≤ j ≤ µi − 1.
(1)
• Taylor interpolation: r = 1, µ1 = N
• Lagrange interpolation: r = N , µ1 = · · · = µN = 1
13
Why polynomials? II- for Interpolation
Theorem. Unique solution in E := PN −1
Due to Z(F ) ≤ n for any nonzero F ∈ Pn.
14
Why polynomials? II- for Interpolation
Theorem. Unique solution in E := PN −1
Due to Z(F ) ≤ n for any nonzero F ∈ Pn.
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
-0.1
-0.1
-0.2
-0.2
-0.3
-0.3
0
0.2
0.4
0.6
0.8
1
0
0.2
0.4
0.6
0.8
1
Interpolation in P3.
15
PROBLEMS AS DEGREE GROWS
16
“High” degree ... I- Design problems
Trebble clef with parametric polynomial curves
of degree 18!!!!!
17
“High” degree ... I- Design problems
Trebble clef with parametric polynomial curves
of degree 18!!!!!
18
“High” degree ... I- Design problems
• The mimicking becomes not so good.
• Change of one control point implies global change of the curve.
• Difficult to predict the changes.
19
“High” degree ... II- Interpolation
problems
50
40
30
20
10
0
-10
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
Regularly spaced Lagrange interpolation by parametric polynomial
curves with, from left to right, 5; ; points.
20
“High” degree ... II- Interpolation
problems
50
50
40
40
30
30
20
20
10
10
0
0
-10
-10
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
Regularly spaced Lagrange interpolation by parametric polynomial
curves with, from left to right, 5;7;
points.
21
“High” degree ... II- Interpolation
problems
50
50
50
40
40
40
30
30
30
20
20
20
10
10
10
0
0
0
-10
-10
-10
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
Regularly spaced Lagrange interpolation by parametric polynomial
curves with, from left to right, 5;7;9 points.
22
“High” degree ... II- Interpolation
problems
50
40
30
20
10
0
-10
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
Regularly spaced Lagrange interpolation by parametric polynomial
curves with, from left to right,11;
;
points.
23
“High” degree ... II- Interpolation
problems
50
50
40
40
30
30
20
20
10
10
0
0
-10
-10
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
Regularly spaced Lagrange interpolation by parametric polynomial
curves with, from left to right,11;12;
points.
24
“High” degree ... II- Interpolation
problems
50
50
50
40
40
40
30
30
30
20
20
20
10
10
10
0
0
0
-10
-10
-10
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
Regularly spaced Lagrange interpolation by parametric polynomial
curves with, from left to right,11;12;13 points.
25
HOW TO MAKE UP FOR THE LATTER FLAWS?
USE POLYNOMIAL SPLINES!
26
Polynomial splines
• Given a < b, choose
– t1 < · · · < tq : interior knots in ]a, b[,
– positive integers m1, . . . , mq : knot multiplicities, with
1 ≤ mk ≤ n for 1 ≤ k ≤ q
• Spline space S:= set of all functions S : [t0, tq+1] := [a, b] →
R s.t.
– S is C n−mk at tk , 1 ≤ k ≤ q;
– for 0 ≤ k ≤ q, there exists Fk ∈ Pn such that S(x) = Fk (x)
for all x ∈ [tk , tk+1].
• dimS = n + 1 + m, with m :=
Pq
k=1 mk
• cubic splines = sections in P3 and simple knots
27
Polynomial splines and Design
Design with parametric cubic spline curves and 19 control points
28
Polynomial splines and Design
Design with parametric cubic spline curves and 19 control points
29
Polynomial splines and Design
• Excellent mimicking!
• Change of one control point implies local change of the curve!
• Easy to predict the changes!
30
Underlying reason: B-spline bases!
S(x) =
m
X
Nin(x)Pi,
x ∈ [a, b]
i=−n
31
Underlying reason: B-spline bases!
S(x) =
m
X
Nin(x)Pi,
x ∈ [a, b]
i=−n
(N−n, . . . , Nm):=B-spline basis.
32
Underlying reason: B-spline bases!
S(x) =
m
X
Nin(x)Pi,
x ∈ [a, b]
i=−n
(N−n, . . . , Nm):=B-spline basis. For simple knots it satisfies :
• Support property: Supp(Nin) = [ti, ti+n+1]
• Positivity property: Nin(x) > 0 for x ∈]ti, ti+n+1[
• Normalisation property:
Pq
n = 1I
N
i=−n i
with
t−n := t−n+1 := · · · = t−1 := a
tq+2 := · · · = tq+n := tq+n+1 := b
33
Due to existence of Blossoms in S!
• (y1, . . . , yn) ∈ [a, b]n is admissible if, for all i = 1, . . . , q s.t.
min(y1, . . . , yn) < ti < max(y1, . . . , yn), ti appears at least mi times
among (y1, . . . , yn).
A:=set of all admissible (y1, . . . , yn) ∈ [a, b]n
• Theorem. For any S ∈ Snd, there exists a unique function
s : A → Rd, called its blossom, such that
• symmetry: s is symmetric on A
• n-affinity on A
• diagonal property: s(x, . . . , x) = S(x) for all x ∈ [a, b].
34
Polynomial splines and Interpolation
N =n+1+m
Hermite interpolation problem: Find S ∈ S such that
S (j)(xi) = αi,j , 1 ≤ i ≤ r, 0 ≤ j ≤ µi − 1.
with the consistency condition:
if xi = tk for some k, 1 ≤ k ≤ q, then µi − 1 ≤ n − mk .
35
Polynomial splines and Interpolation
Hermite interpolation problem: Find S ∈ S such that
S (j)(xi) = αi,j , 1 ≤ i ≤ r, 0 ≤ j ≤ µi − 1.
(2)
with the consistency condition:
if xi = tk for some k, 1 ≤ k ≤ q, then µi − 1 ≤ n − mk .
Theorem. (SW 53) Unique solution to problem (2) iff knots and
interpolation abscissae satisfy some interlacing property known
as the Schoenberg-Whitney conditions
Due to existence of B-spline bases and recurrence relations
between them under knot insertion ( e.g., C.de Boor, R.DeVore,
85)
36
Schoenberg-Whitney conditions
Schoenberg-Whitney conditions for simple knots and Lagrange interpolation:
xk < tk < xk+n+1
for 1 ≤ k ≤ q.
37
Schoenberg-Whitney conditions
Schoenberg-Whitney conditions for simple knots and Lagrange interpolation:
xk < tk < xk+n+1
for 1 ≤ k ≤ q.
Given interpolation abscissae (x1, . . . , xN ) marked by stars,
choose n = 3 and simple knots as marked by circles, i.e.,
(t0, t1 . . . , tq , tq+1) := (x1, x3, x4, . . . , xN −3, xN −2, xN )
38
Schoenberg-Whitney conditions
Schoenberg-Whitney conditions for simple knots and Lagrange interpolation:
xk < tk < xk+n+1
for 1 ≤ k ≤ q.
Given interpolation abscissae (x1, . . . , xN ) marked by stars,
choose n = 3 and simple knots as marked by circles, i.e.,
(t0, t1 . . . , tq , tq+1) := (x1, x3, x4, . . . , xN −3, xN −2, xN )
Not-a-knot!
39
Schoenberg-Whitney conditions
Inadequate choices for the knots
40
Not-a-knot
50
40
30
20
10
0
-10
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
Lagrange interpolation by parametric cubic splines curves with
not-a-knot procedure.
Left: regularly spaced interpolation abscissae
41
Not-a-knot
50
50
40
40
30
30
20
20
10
10
0
0
-10
-10
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
-20
-20
-15
-10
-5
0
5
10
15
20
25
30
Lagrange interpolation by parametric cubic splines curves with
not-a-knot procedure.
Left: regularly spaced interpolation abscissae
Right: distance between consecutive interpolation abscissae =
distance between the corresponding points to interpolate
42
Non desired oscillations with polynomial
splines
1.2
1
0.8
0.6
0.4
0.2
0
-0.2
-2
-1
0
1
2
3
4
5
6
Regularly spaced Lagrange interpolation by
parametric cubic splines.
43
HOW TO MAKE UP FOR THE FLAWS OF
POLYNOMIAL SPLINE INTERPOLATION?
*****
INTRODUCE SHAPE PARAMETERS
44
HOW TO MAKE UP FOR THE FLAWS OF
POLYNOMIAL SPLINE INTERPOLATION?
*****
INTRODUCE SHAPE PARAMETERS
Solution I: INSERT CONNECTION MATRICES!
45
Polyn. splines with connection matrices
• Choose
– Interior knots t1 < · · · < tq and their multiplicities
m1, . . . , mq as before, with now 0 ≤ mk ≤ n for k = 1, . . . , q.
– for k = 1, . . . , q, a lower-triangular matrix of order (n − mk )
Mk with positive diagonal elements : connection matrix
at tk ,
• Spline space S :=space of all continuous functions S : I → R
such that
– for 0 ≤ k ≤ q, there exists Fk ∈ Pn such that S(x) = Fk (x)
for all x ∈ [tk , tk+1];
– for k = 1, . . . , q: the following connection condition is
fulfilled:
T
T
+
+
−
−
0
(n−m
)
0
(n−m
)
k (t )
k (t )
S (tk ), . . . , S
= Mk · S (tk ), . . . , S
.
k
k
46
Connection matrices against oscillations
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
-0.2
-0.2
-2
-1
0
1
2
3
4
5
6
-2
-1
0
1
2
3
4
5
6
Regularly spaced Lagrange interpolation by parametric cubic splines with
connection matrices; not-a-knot;
M
1
k := I2 everywhere
except at the two
0
8 0
8
and
M
.
central knots where M5 =
=
6
1
0 64
0 64
47
INTRODUCE SHAPE PARAMETERS
*****
Solution II: REPLACE POLYNOMIALS BY .....!
48
Interpolation/Design with shape effects
E (n + 1)-dimensional space contained in C n(I) or C n−1(I).
E contains constants
Theorem. (MLM05, MLM08)
Interpolation in DE
⇐⇒
Design in E
⇓
Interpolation in E
⇓: comes from Rolle’s theorem
49
1. Interpolation in E . . .
• . . . either includes Taylor interpolation: means that Zn+1(F ) ≤ n
for any nonzero F ∈ E ⊂ C n (I).
This corresponds to Extended
Chebyshev spaces on I (S. Karlin W. Studden 66, L. Schumaker 81, T. Lyche
85, . . . ),
e.g., the spaces spanned by
– 1, x, . . . , xn−2, cosh x, sinh x (on I = R) “hyperbolic spaces”
– 1, x, . . . , xn−2, cos x, sin x (on |I| < 2π) “trigonometric spaces”
– 1, xk+1, . . . , xk+n with k > 0 (on I =]0, +∞[) ....
50
1. Interpolation in E . . .
• . . . either includes Taylor interpolation: means that Zn+1(F ) ≤ n
for any nonzero F ∈ E ⊂ C n (I).
This corresponds to Extended
Chebyshev spaces on I (S. Karlin W. Studden 66, L. Schumaker 81, T. Lyche
85, . . . ),
e.g., the spaces spanned by
– 1, x, . . . , xn−2, cosh x, sinh x (on I = R) “hyperbolic spaces”
– 1, x, . . . , xn−2, cos x, sin x (on |I| < 2π) “trigonometric spaces”
– 1, xk+1, . . . , xk+n with k > 0 (on I =]0, +∞[) ....
• . . . or does not include Taylor interpolation: means that Zn(F ) ≤
n for any nonzero F ∈ E ⊂ C n−1 (I). This corresponds to Quasi Extended
Chebyshev spaces on I, e.g., the spaces spanned by
– 1, x, . . . , xn−2, (1 − x)p, xq , with p, q > n − 1 (on I = [0, 1])
(P. Costantini 86+00, P. Kaklis 90, . . . ),
– 1, x, . . . , xn−2, xn−1, x2p+n−1 with p > 0 (on I = R)
– 1, x, . . . , xn−2, xn−1, x2p+n with p > 0 (on I = [0, +∞[) ....
51
1. Interpolation with shape effects
1.6
1.6
1.4
1.4
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
-0.2
-0.2
-0.4
-0.4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Regularly spaced Lagrange interpolation by parametric curves in
the EC-space spanned by
• Left: 1, xk+1 , xk+2 , xk+3 on [1, 2], with, from top to bottom on the left part
of the curve, k=0 (pol. space P3 ); 1;2;
• Right: on [0,1], from top to bottom on the left part of the curve
1, x, cos αx, sin αx for α = 6.2; 4.5; 1, x, x2 , x3 ; 1, x, cosh αx, sinh αx for α =
10; 20.
52
1. Interpolation with shape effects
2
2
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Regularly spaced Lagrange interpolation by parametric curves in the QECspace spanned by 1, x, (1 − x)p , xq on [0, 1], with, from top to bottom on the
left part of the curve, (left) p = q = 3 (pol. space P3 ); 7; 11; and (right)
q = 3, p = 3; 5; 8.
53
2. Design in E. . .
. . . means that blossoms exist in E. Geometric definition in
terms of intersections of osculating flats.
(H. Pottmann 93, MLM+HP 96, ....)
Theorem The blossom f : I n → Rd of any F ∈ Ed satisfies
• symmetry: f is symmetric on Rn
• n-pseudoaffinity: for any x1 , . . . , xn−1 , γ, δ ∈ I with γ < δ, for all x ∈ I,
f (x1 , . . . , xn−1 , x) = 1 − α(x) f (x1 , . . . , xn−1 , γ) + α(x)f (x1 , . . . , xn−1 , δ)
where α is a strictly increasing function which is independent of F
• diagonal property: f (x, . . . , x) = F (x) for all x ∈ I.
54
2. Design in E. . .
. . . means that blossoms exist in E. Geometric definition in
terms of intersections of osculating flats.
(H. Pottmann 93, MLM+HP 96, ....)
Theorem The blossom f : I n → Rd of any F ∈ Ed satisfies
• symmetry: f is symmetric on Rn
• n-pseudoaffinity: for any x1 , . . . , xn−1 , γ, δ ∈ I with γ < δ, for all x ∈ I,
f (x1 , . . . , xn−1 , x) = 1 − α(x) f (x1 , . . . , xn−1 , γ) + α(x)f (x1 , . . . , xn−1 , δ)
where α is a strictly increasing function which is independent of F
• diagonal property: f (x, . . . , x) = F (x) for all x ∈ I.
⇔
Bernstein-type bases in E
55
2. Design with shape effects
Continuous line: polynomial curve on any [a, b]
Above: the EC-space spanned by 1, x, cosh x, sinh x on [a, b] for b − a = 3; 6; 18
Below: the EC-space spanned by 1, x, cos x, sin x on [a, b] for b − a = 3; 4.5; 6
56
Tension effects with EC/QEC splines
spaces
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
-0.2
-0.2
-1
0
1
2
3
4
5
6
-1
0
1
2
3
4
5
6
Tension effects with Lagrange interpolation by C 2 spline curves and not-aknot. Interpol. abscissae xi = i for 1 ≤ i ≤ N = 14.
Left: sections in the hyperbolic space spanned by 1, x, cosh αx, sinh αx with,
from the most to the least oscillatory, α = 1; 8; 16 (tension splines, [D.G.
Schweikert, 66] [P. E. Koch and T. Lyche, 90])
Right:
up to changes of variables, sections in the space spanned by
1, x, (1 − x)p , xp , with, from the most to the least oscillatory, p = 3 (pol.
splines); 7; 15 (variable degree polynomial splines, P.Costantini 85).
57
INTRODUCE SHAPE PARAMETERS
******
Solution III: MIX THE TWO IDEAS WITH
DIFFERENT SECTION SPACES!
58
Cheb. splines with connection matrices P.J. Barry, 96
• Choose
– interior knots, their multiplicities, and associated connection matrices as before
– or k = 0, . . . , q, Ek : the section spaces. For each k, Ek
is (n + 1)-dimensional, it contains constants, and DEk is
an EC (or QEC) on [tk , tk+1]
• Chebyshevian spline space S :=space of all continuous
functions S : I → R such that
– for k = 0, . . . , q: the restriction of S to [tk , tk+1] is in Ek
– for k = 1, . . . , q: the following connection condition is
fulfilled:
T
T
+
+
−
−
0
(n−m
)
0
(n−m
)
k (t )
k (t )
S (tk ), . . . , S
= Mk . S (tk ), . . . , S
k
k
59
Interpolation and Design
Assume that n + 1 + m = N .
Proposition. In the Chebyshevian spline space S, if a given
Hermite interpolation problem (2) has a unique solution, then it
satisfies the Schoenberg-Whitney conditions
60
Interpolation and Design
Assume that n + 1 + m = N .
Proposition. In the Chebyshevian spline space S, if a given
Hermite interpolation problem (2) has a unique solution, then it
satisfies the Schoenberg-Whitney conditions
Theorem.
(MLM in preparation)
In the Chebyshevian spline space S
Interpolation under SW-conditions
⇐⇒
Design in S
in DS + in . . .
⇓
Interpolation under SW-conditions
in S + in . . .
A.Kayumov and MLM, CRAS, 344 (1), 65–70, 2007 (the EC case)
61
Proof
• “Design in means S” means existence of blossoms in S defined
on the set A
62
Proof
• “Design in means S” means existence of blossoms in S defined
on the set A
• Theorem
(MLM 02):
Equivalence between
(i) blossoms in S (defined on the set A of all admissible n-tuples);
(ii) B-spline bases exist in S and in all spline spaces obtained from S by
knot insertion
63
Proof
• “Design in means S” means existence of blossoms in S defined
on the set A
• Theorem
(MLM 02):
Equivalence between
(i) blossoms in S (defined on the set A of all admissible n-tuples);
(ii) B-spline bases exist in S and in all spline spaces obtained from S by
knot insertion
• (MLM09) also equivalence with existence of B-spline-like bases in
DS and all spline spaces obtained from it by knot insertion
64
Proof
• “Design in means S” means existence of blossoms in S defined
on the set A
• Theorem
(MLM 02):
Equivalence between
(i) blossoms in S (defined on the set A of all admissible n-tuples);
(ii) B-spline bases exist in S and in all spline spaces obtained from S by
knot insertion
• (MLM09) also equivalence with existence of B-spline-like bases in
DS and all spline spaces obtained from it by knot insertion
• Blossoms automatically imply recurrence relations under knot
insertion between them
65
Illustration
the case n = 3 with simple knots: explicit practical necessary
and sufficient conditions for existence of blossoms (MLM 01)
⇓
• design in S
• interpolation under SW-conditions in S
66
Shape effects with Chebyshevian splines
Hyperbolic/Trigonometric splines:
(dashes/dots): shape effects obtained
by varying the distance between 2 consecutive knots.
67
Chebyshev splines against oscillations
1.2
1
0.8
0.6
0.4
0.2
0
-0.2
-2
-1
0
1
Lagrange interpol.
2
3
4
5
6
by Chebyshev spline curves with connection matrices.
Interpol. Absc. xi = i for 1 ≤ i ≤ N = 14, not-a-knot.
1
Everywhere
0
8
except at the two central knots where M5 =
and M6 =
1
0 64
Left: sections in the space spanned by 1, x, cosh αx, sinh αx, with,
Mk := I2
8 0
;
0 64
from the
most to the least oscillatory α = 0.1; 1; 4.
68
Chebyshev splines against oscillations
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
-0.2
-0.2
-2
-1
0
1
Lagrange interpol.
2
3
4
5
6
-2
-1
0
1
2
3
4
5
6
by Chebyshev spline curves with connection matrices.
Interpol. Absc. xi = i for 1 ≤ i ≤ N = 14, not-a-knot.
1
Everywhere
0
8
except at the two central knots where M5 =
and M6 =
1
0 64
Left: sections in the space spanned by 1, x, cosh αx, sinh αx, with,
Mk := I2
8 0
;
0 64
from the
most to the least oscillatory α = 0.1; 1; 4.
Right: up to changes of variables, sections in the space spanned by 1, x, (1 −
x)p , xp , with, from the most to the least oscillatory p = 3; 7; 15.
69
INTERPOLATING DESIGN ?
70
Interpolating design
Lagrange interpolation in S under SW conditions. Choose
n = 3, the convenient number of arbitrary knots (indicated by
circles), then the interpolation abscissae as indicated by stars
71
Interpolating design
(?) for C 2 trigonometric splines: tk+2 −tk < 2π for 0 ≤ k ≤ q−1
Shape effects with Lagrange interpolation by C 2 splines within existence of
blossoms. Regularly spaced knots.
Left: trigonometric fish with knot spacing=0.1 (dotted line: nearly polynomial spline); 3 (continuous line)
72
Interpolating design
(?) for C 2 trigonometric splines: tk+2 −tk < 2π for 0 ≤ k ≤ q−1
Shape effects with Lagrange interpolation by C 2 splines within existence of
blossoms. Regularly spaced knots.
Left: trigonometric fish with knot spacing=0.1 (dotted line: nearly polynomial spline); 3 (continuous line)
Right: hyperbolic fish with knot spacing=0.1 (dotted line: nearly polynomial
spline); 14 (continuous line)
73
Interpolating design
C 2 Hyperbolic/Trigonometric fish with regularly spaced knots within existence
of blossoms. From left to right:
Up: 4H 6T – 4T 6H – 2H 4T 4H; everywhere knot spacing=3.
Down: 3H 1T 3H 1T 2H – 3H 1T 4H 1T 1H – 1T 8H 1T; everywhere knot
spacing=4.5.
74
Interpolating design
A few more examples of Hyperbolic/Trigonometric
interpolating fish within existence of blossoms
75
Interpolating design
Lagrange interpolation by C 2 spline curves within existence of blossoms. Distance between two consecutive knots=α× distance between the corresponding
interpolation points
Up: Hyperbolic fish, with α = 1; 2; 4.
Down: Trigonometric fish, with α = 0.25; 0.5; 1.
76
The end....
Thanks for
your attention !
77