A class of compactly supported orthogonal symmetric complex wavelets with dilation factor 3

Acta Mathematica Scientia 2012,32B(4):1415–1425 http://actams.wipm.ac.cn

A CLASS OF COMPACTLY SUPPORTED ORTHOGONAL SYMMETRIC COMPLEX WAVELETS WITH DILATION FACTOR 3∗

)

Yang Shouzhi (


)

Shen Yanfeng (

Department of Mathematics, Shantou University, Shantou 515063, China E-mail: [email protected]; [email protected]



Li Youfa (

)

Department of Mathematics, Guangxi University, Nanning 530004, China E-mail: [email protected]

Abstract When approximation order is an odd positive integer, a simple method is given to construct compactly supported orthogonal symmetric complex scaling function with dilation factor 3. Two corresponding orthogonal wavelets, one is symmetric and the other is antisymmetric about origin, are constructed explicitly. Additionally, when approximation order is an even integer 2, we also give a method to construct compactly supported orthogonal symmetric complex wavelets. In the end, there are several examples that illustrate the corresponding results. Key words orthogonal complex wavelets; approximation order; symmetry; scaling function 2010 MR Subject Classification

1

42C15; 94A12

Introduction

It is well known that symmetry and orthogonality are two important properties. However, there is no other compactly supported orthogonal real-valued dyadic scaling function except Haar scaling function. Many valuable works were done when other dilation or complex-valued function was considered [1–9, 11–15]. Here we just state a few of them. Chui and Lian in [1] obtained orthogonal symmetric real-valued wavelets of dilation factor 3 by the method of undetermined coefficient and some special treatments. Several symmetric dyadic orthogonal complex wavelets were constructed in [2]. Shen et al. in [3] obtained complex pseudo splines with dilation 2 from the mask of pseudo splines and constructed symmetric or antisymmetric complex wavelets in L2 (R). In [4], Han provided an effective algorithm to construct a family of compactly supported symmetric (antisymmetric) orthogonal complex wavelets with dilation ∗ Received

November 9, 2010; revised August 31, 2011. This work was supported by the National Natural Science Foundation of China (11071152, 11126343), the Natural Science Foundation of Guangdong Province (10151503101000025, S2011010004511).

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factor 4. Moreover, Han also discussed a more general situation of dilation M ≥ 2 in [7]. In addition, complex wavelets are also very important in some applications, such as those dealing with complex value signals of radar, sonar and QAM (Quadrature Amplitude Modulation). In this article, we shall discuss the construction of compactly supported orthogonal symmetric complex wavelets with dilation factor 3. When approximation order m is an odd positive integer, a simple method is given to construct orthogonal symmetric complex scaling function of dilation factor 3. Then an algorithm is proposed to construct orthogonal wavelets, which are symmetric and antisymmetric about 0, respectively. Additionally, when approximation order m = 2, we also give a method for constructing compactly supported orthogonal symmetric complex wavelets. There are several examples supporting our results. Compared with [7], on one hand, by using a conclusion in [3], it is easy to get that the polynomial (related to the orthogonal scaling symbol) is positive on R, and inspired by [4] the paraunitary matrices used in the algorithm are simpler; on the other hand, a special situation of approximation of order m = 2 is also discussed. Now let us introduce some definitions and notations which will be used in the sequel. Let (a) ((a)) denote the real (imaginary) part and a the conjugate of a ∈ C. We say that complex function f (x) is symmetry (antisymmetry) about a, if f (a + x) = f (a − x) (f (a+x) = −f (a−x)), where a denotes a real constant. The Fourier translation of f (x) ∈ L1 (R)  ˆ is defined by f(ω) := R f (x)e−ixω dx. It is naturally to extend the definition to f (x) ∈ L2 (R). Let φ(x) be a scaling function with dilation factor 3 such that the following refinement equation  an φ(3x − n) (1.1) φ(x) = n∈Z

holds, where {an } is a finitely supported sequence on Z. The wavelet functions are often defined from the scaling function φ(x) by ψ j (x) =



bjn φ(3x − n), j = 1, 2,

n∈Z

where {bjn } for j = 1, 2 are also finitely supported sequences on Z. The refinement equation (1.1) has a corresponding form in frequency domain ˆ ˆ φ(ω) = a(ω/3)φ(ω/3), ω ∈ R, where  a(ω) :=

1 3

 n∈Z

(1.2)

an e−inω . If an am = 0 and ak = 0 for all k ∈ / [n, m] ∩ Z, then the coefficient

supports of  a is defined by coeffsupp( a) := [n, m]. The trigonometric polynomial  a(ω) is often −ikω −ikω called the scaling symbol. If  a(ω) = e  a(−ω) (or  a(ω) = −e  a(−ω)) for all k ∈ Z, we say  a(ω) is symmetric (or antisymmetric) about k/2.  We say φ(x) has orthogonal integer shifts if φ, φ(· − k) = R φ(x)φ(x − k)dx = δk , where {δk } is the Dirac sequence. Let φ(x) be an orthogonal scaling function with (1.2). Then  a(ω) satisfies the CQF (Conjugate Quadrature Filters) condition, i.e., | a(ω)|2 + | a(ω + 2π/3)|2 + | a(ω + 4π/3)|2 = 1,

ω ∈ R.

(1.3)

A scaling function φ(x) has approximation of order m, if all polynomials up to degree m−1

No.4

S.Z. Yang et al: A CLASS OF COMPLEX WAVELETS DILATION 3

can be represented as xj =



1417

ajk φ(x − k), j = 0, · · · , m − 1

k∈Z

for some coefficients {ajn }, j = 0, · · · , m − 1.

2

Scaling Function and Wavelets With Dilation Factor 3

In this section, we shall give a theorem for construction of complex scaling function with desired properties and propose an algorithm for obtaining symbols of the corresponding complex wavelets with dilation 3. First, we discuss several results and lemmas that are useful in the following. It is well known that φ(x) has approximation of order m if and only if the corresponding scaling symbol  a(ω) has the following form  a(ω) =

 1 + e−iω + e−i2ω m 3

Q(ω),

(2.1)

where (1 + e−iω + e−i2ω )  Q(ω). By (2.1), we have

   1 + e−iω + e−i2ω 2m  Pm (ω), | a(ω)|2 =   3

(2.2)

where Pm (ω) = |Q(ω)|2 is a cosine polynomial. Since both 2π/3 and 4π/3 are mth-order zeros of  a(ω), then | a(ω)|2 = 1 + O(ω 2m ) as ω → 0. By direct calculation, 4 π −2m  Pm (x) =  x − sin2 ( )  + O(xm ), 3 3

x → 0,

(2.3)

where x = sin2 (ω/2). Denote the (m−1)th-degree Taylor polynomial of Pm (x) by Pm (x). Then Pm (x) =

m−1  k=0

4 k 3



2m + k − 1 k

xk .

(2.4)

 −iω −i2ω 2m  2 ω 2  Pm sin Moreover, let |a(ω)|2 :=  1+e 3+e 2 , then |a(ω)| satisfies the CQF condition (1.3). For details, one can refer to [9] or [1]. Let

l  N −1+j PN,l (x) = xj , j j=0 where N and l are nonnegative integers with l < N . In [3], Shen et al. showed that PN,l (x) > 0 for all x ∈ R if and only if l is an even number. Therefore, when m is an odd positive integer, Pm (x) = P2m,m−1 ( 3x 4 ) > 0 for all x ∈ R. Riesz lemma is vital in constructing many compactly supported real scaling functions. For convenience, we state it here. Lemma 1 (Riesz lemma [10]) Let a0 , · · · , aN ∈ R with aN = 0 such that a0  + ak cos kω ≥ 0, 2 N

A(ω) :=

k=1

ω ∈ R.

(2.5)

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Then there exists a polynomial B(z) =

N 

bk z k

(2.6)

k=0

with real coefficients and exact degree N such that |B(z)|2 = A(ω),

z = e−iω .

By changing the condition of Riesz lemma, the following Lemma 2 is established. This conclusion can be found in [11, Proposition 5]. N  Lemma 2 Let N be an even positive integer. Suppose that A(x) = ak xk > 0 for all x ∈ R with aN > 0. Define A(ω) = b(z) =

N 

N  k=0



k ak sin2 ω2 .

k=0

Then there exits a symmetric polynomial

bk z k with complex coefficients such that

k=0

|b(z)|2 = A(ω),

z = e−iω .

(2.7)

¯1 , · · · , xN/2 , x ¯N/2 } ⊂ C\R be the set of all complex roots of A(x) = Furthermore, let {x1 , x N 2 

sin2 ω2 − xk which is 0. Then the trigonometric polynomial b(z) can be chosen as C0 k=1

symmetric about zero, where C0 ∈ C is a constant. Remark 1

N

N/2

If A(0) = 1, C0 can be selected as C0 = (−1)

2

k=1

1 xk .

Corollary 1 Suppose that all the real roots of A(x) = 0 defined in Lemma 2 are of even multiplicities. Exactly, let xj for j = 1, · · · , m be all 2-order real roots of A(x) = 0 (xi = xj , possibly for some i = j, i = 1, · · · , m); let xj , x ¯j , j = m + 1, · · · , N/2 be all the non-real roots of A(x) = 0. Then Lemma 2 still holds. Now, let us introduce a sufficient condition of the stability of compactly supported scaling function. We say that a(ω) (a compactly supported scaling symbol with dilation factor 3) satisfies Cohen condition, if there exists a compact set κ such that (1) the origin is an interior point of κ, (2) |κ| = 2π, and for all ω ∈ [−π, π], there is k ∈ Z, satisfies ω + 2πk ∈ κ,   (3) inf inf a( 3ωj ) > 0. j≥1 ω∈κ

Theorem 1 Assume that m is a positive odd integer, and Pm (x) is the polynomial defined in (2.4). Then there exists a symmetric polynomial pm (ω) with complex coefficients such that

ω | pm (ω)|2 = Pm sin2 . 2 Furthermore, let am (ω) := e

imω



1 + e−iω + e−i2ω 3

m pm (ω),

(2.8)

then am (ω) can generate an orthogonal symmetric complex scaling function φ(x) with symmetric center 0.

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Proof We have already known that Pm (x) > 0 for all x ∈ R. And it is easy to see that Pm (0) = 1. By Lemma 2, we have m−1 2

pm (ω) = b(z) = C0



sin2

k=1

(m−1)/2

where C0 = (−1)

m−1 2

k=1

1 xk ,

 ω − xk , 2

and {x1 , x ¯1 , · · · , x(m−1)/2 , x ¯(m−1)/2 } is all the non-real roots

of Pm (x) = 0. It is straightforward to show that am (ω) = am (−ω) and am (ω) satisfies the CQF condition (1.3). We remark that φ(x) ∈ L2 (R) by applying the result of Wang [12, Theorem 2.5]. In order to prove that am (ω) generates an orthogonal scaling function, we just need to check that am (ω) satisfies Cohen condition. We select the compact set κ = [−π, π], then for all j ≥ 1, there exists −π/3 ≤ ω/3j ≤ π/3. So |am (ω/3j )|2 ≥

 2 2j 3

Pm (0) =

 4 j 9

> 0.

Hence am (ω) satisfies Cohen condition. Then φ(x) is an orthogonal scaling function. Moreover, by the symmetry of am (ω), we have φ(x) = φ(−x). So φ(x) is symmetric about 0. 2 In [4], Han provided an effective algorithm to construct a family of compactly supported symmetric orthogonal complex wavelets with dilation factor 4. Inspired by this idea, we shall provide an algorithm to construct a class of compactly supported orthogonal complex wavelets symmetric about 0 with dilation factor 3. First, we need the following theorem. Theorem 2 Suppose that A(ω) = (A1 (ω), A2 (ω), A3 (ω))T be a 2π periodic trigonometric polynomial vector with complex coefficients such that A1 (ω) = A1 (−ω), A2 (ω) = A2 (−ω), A3 (ω) = −A3 (−ω), and |A1 (ω)|2 + |A2 (ω)|2 + |A3 (ω)|3 = 1,

ω ∈ R.

(2.9)

Let k be the highest degree of {Aj (ω), j = 1, 2, 3}. Then trigonometric polynomial A(ω) can be written as ⎡ ⎡ ⎡ ⎤ ⎡ ⎤ ⎤ ⎤ f1 f3 f3 f1 ⎢ ⎢ ⎥ −i(k−1)ω ⎢ ⎥ −ikω ⎥ ikω ⎢ ⎥ i(k−1)ω ⎢ ⎢ ⎢ ⎢ ⎥ ⎥ ⎥ A(ω) = ⎣ f2 ⎦ e + ⎣ f4 ⎦ e + · · · + ⎣ f4 ⎦ e + ⎣ f2 ⎥ . (2.10) ⎦e −g1 −g3 g3 g1 Denote ⎤  h h c0 f2 cos ω − i ic0 g1 sin ω ⎥ ⎢ c0 f1 cos ω − i c c ⎥ ⎢   ⎥ ⎢ ⎥ ⎢ 2 2 UA (ω) := ⎢ ⎥, f f − 0 2 1 ⎥ ⎢ c c ⎥ ⎢   ⎣ h ⎦ ic0 f1 sin ω ic0 f2 sin ω c0 g1 cos ω + i c ⎡



(2.11)

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 where c = |f1 |2 + |f2 |2 + |g1 |2 , h = (f1 f3 + f2 f4 − g1 g3 ) and c0 = c2 2c +h2 . Then B(ω) =: UA (ω)A(ω) satisfies (2.9) and (2.10). Moreover, deg(B) < deg(A) and UA (ω) is a paraunitary matrix. Proof First let us prove that deg(B) < deg(A) and B(ω) satisfies (2.10), B(ω) = (B1 (ω), B2 (ω), B3 (ω))T = UA (ω)A(ω) ⎛   ⎞ h c0 (f1 A1 (ω) + f2 A2 (ω)) cos ω − i + ig1 A3 (ω) sin ω ⎟ ⎜ c ⎟ ⎜  ⎟ ⎜ =⎜ 2/c(f1 A2 (ω) − f2 A1 (ω)) ⎟. ⎟ ⎜ ⎠ ⎝    h c0 (f1 A1 (ω) + f2 A2 (ω)) sin ω + ig1 A3 (ω) cos ω − i c It is easy to see that B(ω) satisfies (2.10) and deg(B2 ) < k. Here, we only prove that deg(B1 ) < k, as the proof of deg(B3 ) < k is similar. Suppose that B1 (ω) = c1 ei(k+1)ω + c2 eikω + · · · + c2 e−ikω + c1 e−i(k+1)ω , where c1 = c0 (|f1 |2 + |f2 |2 − |g1 |2 )/2 and c2 = c0 ((f1 f3 + f2 f4 − g1 g3 ) − i2(|f1 |2 + |f2 |2 h/c))/2. Form (2.9), some tedious manipulation yields that (reference [13]) |f1 |2 + |f2 |2 = |g1 |2 , (f1 f3 + f2 f4 ) = (g1 g3 ).

(2.12)

Then we have c1 = c2 = 0. So deg(B) < deg(A). Our task now is to show that UA (ω) is a paraunitary matrix. By (2.12) and the definition of UA we find ⎛ ⎞ U1,1 0 U1,3 ⎜ ⎟ T UA (ω)UA (ω) = ⎜ 1 0 ⎟ ⎝ 0 ⎠, U3,1 0 U3,3 where U1,1 = c20 ((|f1 |2 + |f2 |2 )(cos2 ω + h2 /c2 ) + |g1 |2 sin2 ω), U1,3 = ic20 (|g1 |2 − |f1 |2 − |f2 |2 )(cos ω − ih/c) sin2 ω, U3,1 = ic20 (||f1 |2 + |f2 |2 − g1 |2 )(cos ω − ih/c) sin2 ω, U3,3 = c20 ((|f1 |2 + |f2 |2 ) sin2 ω + |g1 |2 (cos2 ω + h2 /c2 )). T

By direct calculation we have UA (ω)UA (ω) = I3 . Consequently, the trigonometric polynomial B(ω) also satisfies (2.9). 2 In Theorem 2, if A(ω) has four entries (A1 (ω), A2 (ω), A3 (ω), A4 (ω))T that satisfy Aj (ω) = Aj (−ω) for j = 1, 2 and Aj (ω) = −Aj (−ω) for j = 3, 4, the corresponding conclusion can be found in [3, Lemma 2]. If A(ω) is a polynomial with real coefficients, then the corresponding situation was also studied systematically by Petukhov in [13]. Recall the trigonometric polynomial am (ω) defined √ in (2.8), we have that the support of  3m−1 3 k am (ω) is coeffsupp(am ) = [− 3m−1 , ]. Let a (ω) := ak+3j e−ijω . Then m 2 2 3 j∈[−[ 3m−1 ],[ 3m−1 ]] 6 6

am (ω) has polyphase decomposition 1 1  k am (3ω)z k , am (ω) = √ 3 k=−1

z = e−iω .

(2.13)

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Then (1.3) is equalent to 2 |a0m (ω)|2 + |a1m (ω)|2 + |a−1 m (ω)| = 1,

ω ∈ R.

(2.14)

Noticing that am (ω) = am (−ω), we have a0m (ω) = a0m (−ω). Moreover, √  3 1 −1 am (ω) + am (ω) = (a1+3j + a−1+3j )e−ijω . 3 3m−1 3m−1 j∈[−[

6

],[

6

]]

It is easy to see that a1m (ω) + a−1 m (ω) is symmetric about 0. Analogously we can obtain that 1 a−1 (ω) − a (ω) is antisymmetrical about 0. In fact, we have the following algorithm. m m Algorithm 1 Step 1 Let am (ω) be the scaling symbol obtained in Theorem 1. Then am (0) = 1 and am (−ω) = am (ω). Moreover, let a0m (ω), a1m (ω) and a−1 m (ω) be polyphase decomposition of am (ω). Step 2 Denote ⎤ ⎡ ⎡ ⎤ 1 0 0 ⎥ a0m (ω) ⎢ ⎥ ⎢ 1 1 ⎢ ⎥ ⎢ √ √ ⎥ A0m (ω) = ⎢ 0 (2.15) ⎥⎢ a1m (ω) ⎥ ⎣ ⎦. 2 2 ⎥ ⎢ ⎦ ⎣ 1 1 a−1 m (ω) √ 0 −√ 2 2 Then A0m (ω) satisfies conditions (2.9) and (2.10). Step 3 Recursively applying (2.11) until (ω)Aj−1 Ajm (ω) = UAj−1 m (ω) m and deg(Ajm (ω)) = 0. Thus Ajm (ω) = (h1 , h2 , 0)T , where h1 , h2 |h1 |2 + |h2 |2 = 1. Define a paraunitary matrix ⎡ h1 − h2 T T ⎢ U (ω) = UA0m (ω) · · · UAj−1 (ω) ⎢ h1 ⎣ h2 m 0 0 Step 4

The wavelet symbols bjm (ω) =

1 3

 k∈Z

1

⎢ √  ⎢ 3 ⎢ 1 e−iω eiω ⎢ 0 am (ω) b1m (ω) b2m (ω) = ⎢ 3 ⎣ 0 

are complex values satisfy

0

⎤

⎥ 0⎥ ⎦. 1

bjk e−ikω , j = 1, 2, can be derived by ⎡



(2.16)

0 1 √ 2 1 −√ 2

⎤−1 0 ⎥ 1 ⎥ √ ⎥ U (3ω). 2⎥ ⎥ ⎦ 1 √ 2

(2.17)

We remark that bjm (ω) defined in Algorithm 1 satisfies bjm (ω) = (−1)j+1 bjm (−ω) for j = 1, 2. From (2.11), we deduce that the entries of U (ω) satisfy that U11 , U12 , U21 , U22 and U33 are symmetric and others are antisymmetric about 0. So b1m and b2m in (2.17) can be rewritten as √ √ √

3 2 2 Uj1 + Uj2 (e−iω + eiω ) − Uj3 (e−iω − eiω ) , j = 1, 2. bjm (ω) = (2.18) 3 2 2

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Then the symmetry of b1m (ω) and antisymmetry of b2m (ω) are obtained. By calculation, we can see that ⎡ ⎤ am (ω) am (ω + 2π/3) am (ω + 4π/3) ⎢ ⎥ 1 1 1 ⎥ H(ω) := ⎢ b (ω) b (ω + 2π/3) b (ω + 4π/3) m m m ⎣ ⎦ b2m (ω) b2m (ω + 2π/3) b2m (ω + 4π/3)  j is a paraunitary matrix. Define ψ j (x) = 13 bk φ(3 · −k) for j = 1, 2, then ψ j ∈ L2 (R) and k∈Z

they are orthogonal complex wavelets with ψ 1 (x) = ψ 1 (−x) and ψ 2 (x) = −ψ 2 (−x). Since 3m−1 3m−1 3m−1 coeffsupp(am ) = [− 3m−1 2 , 2 ], supp(φ) = [− 4 , 4 ] can be got from Theorem 1. Then 3m−1 3m−1 j we can deduce that supp(ψ ) ⊆ [− 4 , 4 ], j = 1, 2 from Algorithm 1. Thus we have constructed the desired wavelet basis of L2 (R): {ψ 1 (3j · −k), ψ 2 (3j · −k), j, k ∈ Z}. In (2.4), if m is an even positive integer, then Pm (x) cannot satisfy Pm (x) > 0 for all x ∈ R. Hence Theorem 1 is invalid. Karoui [14] already studied constructing orthogonal wavelets with better regularity properties by adding extra zeros in the interval (π/2, π). Here we use his idea  to make Pm (y) ≥ 0 for all y ∈ R and with real zeros only in (−1, 0). Let T2 be an operator on R[x], as in [14],  2    1 + e−i(ω+2πj/3) + e−i2(ω+2πj/3) 4   P (cos(ω + 2πj/3)). T2 (P )(ω) = (2.19)   3 j=0 Karoui showed that span{y, y 2 − 1/2} ⊆ ker T2 with y = cos ω. Define P2 (y) := P2 ( 1−y 2 ) + (1 − y)2 (ay + b(y 2 − 1/2)), where a, b are undetermined constants and P2 (x) = 1 + 16 x. By direct 3 calculation, we can show that    1 + e−iω + e−i2ω 4   P (cos(ω)) (2.20) |a2 (ω)|2 =   2 3 satisfies (1.3). Suppose that the only real root y0 ∈ (−1, 0) of P2 (y) = 0, is a double root, i.e., P2 (y) = (y − y0 )2 Q(y) with Q(y) > 0 for all y ∈ R, then P2 (y) ≥ 0 with real zeros only in (−1, 0). Using Corollary 1, we can still get an orthogonal complex scaling function with symmetry when m = 2, see the example in Section 3.

3

Examples

In this section, we shall give three examples. First, we discuss two case: m = 3, 5. By Algorithm 1 give two examples for compactly supported orthogonal symmetric complex scaling functions and wavelets. Then we give a special example when m = 2. Example 1 Let m = 3 in (2.4). Then P3 (x) = 1 + 8x + (112/3)x2 . By Theorem 1, we √ 1 can select p3 (x) = x − 28 (−3 − i2 3). Hence

3 −iω + e−i2ω i3ω 1 + e a3 (ω) = e p3 (sin2 (ω/2)). 3 We calculate the Sobolev regularity exponent ν2 (φ) = ν2 (a3 ) = 1.1599 by [15]. The wavelet symbols can be derived from Algorithm 1:    

2 2 4 26 4 29 2 13 + i+ − i (e−iω + eiω ) b1 (ω) = − 3 27 3 27 13 27 39

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2 2 2 −i3ω 1 1 1 −i2ω i2ω + i (e i(e + +e )+ + ei3ω ) 27 13 27 39 27 39 

2 1 1 √ − i (e−i4ω + ei4ω ), − 27 26 27 39  

 2 −iω 1 2 1 2 iω i(e + i (e−i2ω − ei2ω ) b2 (ω) = − −e )+ 13 3 39 9 13  

1 2 −i3ω 1 2 1 i3ω i(e i (e−i4ω − ei4ω ). − −e )+ − √ + 9 13 3 78 9 13

See Fig. 1 for the graphs of the scaling function and two wavelets in this example. A

C

1.5

1.5

1

1

1

0.5

0.5

0.5

0

0

0

−0.5

−0.5

−0.5

−1

−1

−1

−1.5 −2

Fig.1

B

1.5

0

2

−1.5 −2

0

2

−1.5 −2

0

2

A is the scaling function φ and B, C are wavelets ψ 1 , ψ 2 in Example 1; solid line

denotes the real part and dashed line the imaginary part. 880 2 14080 3 183040 4 Example 2 Let m = 5 in (2.4). Then P5 (x) = 1 + 40 3 x + 9 x + 27 x + 81 x . By Theorem 1, select  p5 (x) = (−23.0698 + 41.5636i) 0.422483 − 0.130613i + (−0.278846 + 0.0561102i)/x  +(−0.278846 + 0.0561102i)x + 1/16x−2 + 1/16x2 .

Construct a5 (ω) = e

i3ω

1 + e−iω + e−i2ω 3

5

p5 (sin2 (ω/2)).

We can also arrive at ν2 (φ) = 1.3665. The wavelet symbols can be derived from Algorithm 1: b1 (ω) = 0.334063 − 0.298353i − (0.207359 − 0.146753i)(e−iω + eiω ) +(0.0250731 + 0.0239254i)(e−i2ω + ei2ω ) +(0.0133969 − 0.00482949i)(e−i3ω + ei3ω ) +(0.00619337 − 0.0191475i)(e−i4ω + ei4ω ) − 0.00356885(e−i5ω + ei5ω ) −(0.00109314 − 0.00105363i)(e−i6ω + ei6ω ) +(0.000326024 + 0.00141317i)(e−i7ω + ei7ω ),

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b2 (ω) = (−0.275091 − 0.2032i)(e−iω − eiω ) − (0.04161 − 0.17124i)(e−i2ω − ei2ω ) +(0.0563849 + 0.0269705i)(e−i3ω − ei3ω ) +(0.105258 − 0.0492435i)(e−i4ω − ei4ω ) −(0.0225515 + 0.0202334i)(e−i5ω − ei5ω ) −(0.012882 − 0.000432322i)(e−i6ω − ei6ω ) −(0.0059752 − 0.0107652i)(e−i7ω − ei7ω ). See Fig. 2 for the graphs of the scaling function and two wavelets in this example. A

C

1.5

1.5

1

1

1

0.5

0.5

0.5

0

0

0

−0.5

−0.5

−0.5

−1

−1

−1

−1.5

Fig.2

B

1.5

−2

0

2

−1.5

−2

0

2

−1.5

−2

0

2

A is the scaling function φ and B, C are wavelets ψ 1 , ψ 2 in Example 2; solid line

denotes the real part and dashed line the imaginary part.

Example 3 We just give an example when arccos(−1+2e−2 ) (see [14]) is a double root of   Pm (x) = 0. We can get a = 2.69192 and b = 2.7219. Then Pm (y) = (y + 0.729329)2(4.33469 − 6.72221y + 2.7219y 2) = (y + 0.729329)2Q(y). Q(y) can be rewritten as

4.33469 −3.36111 Q(y) = (1, y) (1, y)T . −3.36111 2.7219 It is easy to see the matrix is positive definite. As a conclusion, Theorem 1 also can be established. The scaling symbol and wavelet symbols are a(ω) = 0.35866 − 0.0992493i + (0.283633 − 0.0680414i)(e−iω + eiω ) +(0.0804057 + 0.0340207i)(e−i2ω + ei2ω ) − (0.0126632 − 0.0496246i)(e−i3ω + ei3ω ) −(0.0307054 − 0.0340207i)(e−i4ω + ei4ω ), b1 (ω) = −0.112889 + 0.411502i + (0.0615651 − 0.262981i)(e−iω + eiω ) −(0.0284939 + 0.0290149i)(e−i2ω + ei2ω ) − (0.0143233 − 0.0190766i)(e−i3ω + ei3ω ) −(0.0192913 − 0.00913833i)(e−i4ω + ei4ω ), b2 (ω) = −0.389413i(e−iω − eiω ) + (0.0887002 + 0.0375303i)(e−i2ω − ei2ω ) −(0.0139696 − 0.0547438i)(e−i3ω − ei3ω ) − (0.0338729 − 0.0375303i)(e−i4ω − ei4ω ).

No.4

S.Z. Yang et al: A CLASS OF COMPLEX WAVELETS DILATION 3

1425

We can also calculate that ν2 (φ) = 2.0112. See Fig. 3 for the graphs of the scaling function and two wavelets in this example. A

C

1.5

1.5

1

1

1

0.5

0.5

0.5

0

0

0

−0.5

−0.5

−0.5

−1

−1

−1

−1.5 −2

Fig.3

B

1.5

0

2

−1.5 −2

0

2

−1.5 −2

0

2

A is the scaling function φ and B, C are wavelets ψ 1 , ψ 2 in Example 3; solid line

denotes the real part and dashed line the imaginary part.

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