Interpolation by Polynomials with Symmetries
- n
the Imaginary Axis
Izchak Lewkowicz ECE dept, Ben-Gurion University, Israel Ben-Gurion University, 24 May 2012 Joint work with D. Alpay, Math. dept BGU.
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Interpolation by Polynomials with Symmetries on the Imaginary - - PowerPoint PPT Presentation
Interpolation by Polynomials with Symmetries on the Imaginary Axis Izchak Lewkowicz ECE dept, Ben-Gurion University, Israel Ben-Gurion University, 24 May 2012 Joint work with D. Alpay, Math. dept BGU. p. 1 Outline The function
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Yj=Fa(xj) j=1 , ... , m Yj=Fb(xj) j=1 , ... , m
a
b
a
b
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s → ∞ F (s)
0 I
C D
C D
0 I
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a
α , β , γ , δ ≥ 0
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q−1
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q−1
Il x1Il x2
1Il
... xm−1
1
Il
Il xnIl x2
nIl
... xm−1
n
Il
Co
Cn−1
Y1
Yn
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x1 x2 x3 ˜ F1(s) Y1 ˜ F2(s) Y2 ˜ F3(s) Y3.
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F „ 1 2 3 « → „ 4 1 −4 « F (s)= ˜ F1(s)+ ˜ F2(s)+ ˜ F3(s)
˜ F1(s) = (4−s2)(9−s2)( ˜ α(1−s2)+ 1
6
) ˜ α≥0, ˜ F2(s) = (1−s2)(9−s2)( ˜ β(4−s2)− 1
60s2 )
˜ β≥0, ˜ F3(s) = (1−s2)(4−s2)( ˜ γ(9−s2)− 1
90s2 )
˜ γ≥0.
1 36
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xj−xk=0 xj+x∗
k=0
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q−1
P #(s) = P (s) P (xj) = Yj j=1, ... , m.
xj−xk=0 = ⇒ Yj = Yk xj+x∗
k=0
= ⇒ Yj = Y ∗
k
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q−1
Il x1Il x2
1Il
... x2n−1
1
Il
Il xnIl x2
nIl
... x2n−1
n
Il Il −x∗
1Il (−x∗ 1)2Il ... (−x∗ 1)2n−1Il
Il −x∗
nIl (−x∗ n)2Il ... (−xn ∗)2n−1Il
Co
Cn−1 Cn
C2n−1
Y1
Yn Y ∗
1
Y ∗
n
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n
j + s)
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n
j + s)
ω∈R ω=ixj
p=1 , ... , l λp
M −1
2n−1
P
k=0
ikCkωk
n
Q
j=1
|xj−iω|2
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F @ 1 2 3 1 A → @ 18 75 50 1 A
F (s) = P (s) +β Ψ(s) = −3s4+34s2−13 +β (1−s2)(4−s2)(9−s2)
2
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q−1
P #(s) = AP (s)B A,B∈Cl×l non−singular P (xj) = Yj j=1, ... , m.
xj−xk=0 = ⇒ Yj = Yk xj+x∗
k=0
= ⇒ Yj = (AYkB)∗
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q−1
Il x1Il x2
1Il
... x2n−1
1
Il
Il xnIl x2
nIl
... x2n−1
n
Il Il −x∗
1Il (−x∗ 1)2Il ... (−x∗ 1)2n−1Il
Il −x∗
nIl (−x∗ n)2Il ... (−xn ∗)2n−1Il
Co
Cn−1 Cn
C2n−1
Y1
Yn (AY1B)∗
(AYnB)∗
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q−1
P #(s) = AP (s)(A∗)−1 A∈Cl×l non−singular P (xj) = Yj j=1, ... , m. Ψ(s)=
n
Q
j=1
(xj−s)M(x∗
j +s)
AM=(AM)∗ M parameter
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2
0 −1
0 1 )
π eigenvalues in C+ l−π eigenvalues in C−
Ψ =
π eigenvalues in C+ l−π eigenvalues in C−
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Il−ν
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n
Il−ν
j + s)
ω∈R
P
k=0
R−1Ck(R∗)−1(iω)k
n
Q
j=1
|xj−iω|2
ω∈R )
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s
s
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s
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2n
R(F (s)) := snF( 1
s)(sn)#
= (−1)n
2n
P
k=0
Cks2n−k = (−1)n(Cos2n+C1s2n−1+...+C2n−1s+C2n)
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Yj=Fa(xj) j=1 , ... , m Yj=Fb(xj) j=1 , ... , m
a
b
a
b
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Yj=Fa(xj) j=1 , ... , m Yj=Fb(xj) j=1 , ... , m
2n
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Systems, Vol. 10, pp. 265-285, 1997.
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