Multi-Resolution Methods and B-Splines - Interpolation by Spline Functions

2.4 Interpolation by Spline Functions

2.4.6 Multi-Resolution Methods and B-Splines





B₁(ξ₁) . . . B_N(ξ₁)

... ...

B₁(ξ_N) . . . B_N(ξ_N)





has a special structure:A is a band matrix, because by (2.4.4.5) the func- tionsB_i(x) =B_i,r,t(x) have support [t_i, t_i+r], so that within thejth row of Aall elementsB_i(ξ_j) witht_i+r< ξ_jort_i> ξ_j are zero. Therefore, each row ofA contains at most relements diﬀerent from 0, and these elements are in consecutive positions. The componentsB_i(ξ_j) ofAcan be computed by the recursion described previously. The system (2.4.5.6), and thereby the interpolation problem, is uniquely solvable if A is nonsingular. The non- singularity ofAcan be checked by means of the following simple criterion due to Schoenberg and Whitney (1953), which is quoted without proof.

(2.4.5.7) Theorem.The matrix A= (B_i(ξ_j))of (2.4.5.6) is nonsingular if and only if all its diagonal elementsB_i(ξ_i)= 0are nonzero.

It is possible to show [see Karlin (1968)] that the matrix A is totally positive in the following sense: allr×rsubmatricesB ofAof the form

B= (a_i_p_,j_q)^r_p,q=1 with r≥1, i₁< i₂<· · ·< i_r, j₁< j₂<· · ·< j_r, have a nonnegative determinant, det(B) ≥ 0. As a consequence, solving (2.4.5.6) for nonsingular A by Gaussian elimination without pivoting is numerically stable [see de Boor and Pinkus (1977)]. Also the band structure ofAcan be exploited for additional savings.

For further properties of B-Splines, their applications, and algorithms the reader is referred to the literature, in particular to de Boor (1978), where one can also ﬁnd numerousfortranprograms.

(S1) V_j⊂V_j+1, j∈ZZ,

(S2) f(x)∈V_j ⇔f(2x)∈V_j+1, j∈ZZ, (S3) f(x)∈V_j⇔f(x+ 2^−j)∈V_j, j∈ZZ,

(S4) 6

j∈ZZ

V_j =L₂(IR),

(S5) 7

j∈ZZ

V_j={0}.

Usually, such subspaces are generated by a function Φ(x) ∈ L₂(IR) as follows: By translation and dilatation of Φ, we ﬁrst deﬁne additional functions (we call them “derivates” ofΦ)

Φ_j,k(x) := 2^j/2Φ(2^jx−k), j, k∈ZZ.

The spaceV_jis then deﬁned as the closed linear subspace ofL₂(IR) that is generated by the functionsΦ_j,k,k∈ZZ,

V_j := span{Φ_j,k|k∈ZZ}=(

|k|≤n

c_kΦ_j,k|c_k∈C, n≥0) . Then (S2) and (S3) are clearly satisﬁed.

(2.4.6.1) Deﬁnition. The functionΦ∈L₂(IR)is called scaling function, if the associated spacesV_jsatisfy conditions(S1)–(S5)and if, in addition, the functions Φ_0,k, k ∈ ZZ, form a Riesz-basis of V₀, that is, if there are positive constants 0< A≤B with

k∈ZZ

|c_k|²≤

k∈ZZ

c_kΦ_0,k²≤B

k∈ZZ

|c_k|²

for all sequences(c_k)_k∈ZZ ∈₂(then also for eachj∈ZZ, the functionsΦ_j,k, k∈ZZ, will form a Riesz-basis ofV_j).

IfΦis a scaling function then, for allj ∈ZZ, each functionf ∈V_j can be written as a convergent (inL₂(IR)) series

(2.4.6.2) f =

k∈ZZ

c_j,kΦ_j,k

with uniquely determined coeﬃcientsc_j,k,k∈ZZ, with (c_j,k)_k∈ZZ ∈₂. Also for a scaling functionΦ, condition (S1) is equivalent to the condition thatΦsatisﬁes a so-called two-scale-relation

(2.4.6.3) Φ(x) =

k∈ZZ

h_kΦ_1,k(x) =√

k∈ZZ

h_kΦ(2x−k)

with coeﬃcients (h_k)∈₂. For, (2.4.6.3) is equivalent toΦ∈V₀⊂V₁. But then alsoV_j⊂V_j+1 for allj∈ZZbecause of

Φ_j,l(x) =2^j/2Φ(2^jx−l) = 2^j/2√

k∈ZZ

h_kΦ(2^j+1x−2l−k)

k∈ZZ

h_kΦ_j+1,k+2l(x).

Example.A particularly simple scaling function is theHaar-function (2.4.6.4) Φ(x) := 1, for 0≤x <1,

0, otherwise.

The associated two-scale-relation is

Φ(x) =Φ(2x) +Φ(2x−1).

See Figure 3 for an illustration of some of its derivatesΦj,k.

...... .......................

...

Φj,k(x)

... x

...

....................................... ......................................................................................................................................................... ........................ .........................................................................................................

√2

−3/2 −1 0 1/2 1 4/2 2.5

Φ₁,−3 Φ₁,0

Φ₀,0=Φ

Φ₁,4

Fig. 3.Some derivatesΦj,k of the Haar-function

Scaling functions, and the associated spaces V_j they generate, play a central role in modern signal- and image processing methods. For example, any function f ∈ V_j ="

kc_j,kΦ_j,k can be considered as a function with a ﬁnite resolution (e.g. a signal or, in two dimensions, a picture obtained by scanning) where only details of size 2^−j are resolved (this is illustrated

most clearly by the spaceV_j belonging to the Haar-function). A common application is to approximate a high resolution functionf, that is a function f ∈ V_j with j large, by a coarser function ˆf (say a ˆf ∈ V_k with k < j) without losing to much information. This is the purpose ofmulti-resolution methods, whose concepts we wish to explain in this section.

We ﬁrst describe additional scaling functions based on splines. The Haar-function Φ is the simplest B-spline of order r = 1 with respect to the special sequence t:= (k)_k∈ZZ of all integers: Deﬁnition (2.4.4.3) shows immediately

B_0,1,t(x) =f_x⁰[0,1] = (1−x)⁰₊−(0−x)⁰₊=Φ(x).

Moreover, for allk∈ZZ, their translates are

B_k,1,t(x) =f_x⁰[k, k+ 1] =B_0,1,t(x−k) =Φ(x−k) =Φ_0,k. This suggests to consider general B-splines

(2.4.6.5) M_r(x) :=B_0,r,t(x) =rf_x^r−1[0,1, . . . , r]

of arbitrary order r≥1 as candidates for scaling functions, because they also generate all B-splines

B_k,r,t(x) =rf_x^r−1[k, k+ 1, . . . , k+r], k∈ZZ,

by translation,B_k,r,t(x) =B_0,r,t(x−k) =M_r(x−k). Advantages are that their smoothness increases withrsince, by definition,M_r(x) =B_0,r,t(x) is r−2 times continuously differentiable, and that they have compact support [see Theorem (2.4.4.5)]. It can be shown that all M_r, r ≥ 1, are in fact scaling functions in the sense of Definition (2.4.6.1); for example, their two-scale-relation reads

M_r(x) = 2^−r+1 r

l=0

r l

M_r(2x−l), x∈IR.

For proofs we refer the reader to the literature [see Chui (1992)]. In the interest of simplicity, we only consider low order instances of these functions and their properties.

Examples.For low orderrone ﬁnds the following B-splinesMr :=B₀,r,t:M₁(x) is the Haar-function (2.4.6.4), and forr= 2, 3 one obtains:

M₂(x) =

_x _{for 0}_≤_x_≤_1, 2−x for 1≤x≤2, 0 otherwise.

M₃(x) = 1 2





x² for 0≤x≤1,

−2x²+ 6x−3 for 1≤x≤2, (3−x)² for 2≤x≤3,

0 otherwise.

The translateH(x) :=M₂(x+ 1) ofM₂(x),

(2.4.6.6) H(x) = 1− |x| for−1≤x≤1 0 otherwise, is known ashat-function.

These formulas are special cases of the following representation ofMr(x) for generalr≥1:

(2.4.6.7) Mr(x) = 1 (r−1)!

r l=0

(−1)^l r

(x−l)^r₊⁻¹.

This is readily veriﬁed by induction overr, taking the recursion (k = 1, 2,. . ., r) (2.1.3.5) for divided diﬀerences

fx^r⁻¹[i, i+ 1, . . . , i+k] = 1 k

fx^r⁻¹[i+ 1, . . . , i+k]−fx^r⁻¹[i, . . . , i+k−1]

, andfx^r⁻¹(t) = (t−x)^r₊⁻¹ into account.

We now describe multi-resolution methods in more detail. As already stated, their aim is to approximate, as well as possible, a given “high resolution” function v_j+1 ∈V_j+1 by a function of lower resolution, say by a functionv_j∈V_j with

v_j+1−v_j= min

v∈Vjv_j+1−v.

Sincev_j+1−v_jis then orthogonal toV_j, andv_j ∈V_j⊂V_j+1, the orthogonal subspaceW_j ofV_j inV_j+1comes into play,

W_j :={w∈V_j+1| w, v!= 0 for allv∈V_j}, j∈ZZ.

We writeV_j+1=V_j⊕W_j to indicate that everyv_j+1∈V_j+1has a unique representation as a sumv_j+1=v_j+w_j of two orthogonal functionsv_j∈V_j andw_j∈W_j.

The spaces W_j are mutually orthogonal, W_j ⊥ W_k for j = k (i.e.

v, w! = 0 for v ∈ W_j, w ∈ W_k). If, say j < k, this follows from W_j ⊂ V_j+1⊂V_k andW_k ⊥V_k. More generally, (S4) and (S5) then imply

L₂(IR) =8

j∈ZZ

W_j =· · · ⊕W₋₁⊕W₀⊕W₁⊕ · · · .

For a givenv_j+1 ∈V_j+1, multi-resolution methods compute for m≤j the orthogonal decompositions v_m+1 = v_m+w_m, v_m ∈ V_m, w_m ∈ W_m, according to the following scheme:

(2.4.6.8)

v_j+1 → v_j → v_j−1 → · · ·

w_j w_j−1 · · ·

Then for m ≤ j, v_m ∈ V_m is also the best approximation of v_j+1 by an element ofV_m, that is,

v_j+1−v_m= min

v∈Vmv_j+1−v.

This is because the vectors w_k are mutually orthogonal and w_k ⊥V_m for k≥m. Hence for arbitraryv∈V_m, v_m−v∈V_m and

v_j+1−v=w_j+w_j−1+· · ·+w_m+ (v_m−v), so that

v_j+1−v²=w_j²+· · ·+w_m²+v_m−v²≥ v_j+1−v_m². In order to compute the orthogonal decompositionv_j+1=v_j+w_jof an arbitraryv_j+1∈V_j+1, we need thedual function Φ˜of the scaling function Φ. As will be shown later, ˜Φis uniquely determined by the property (2.4.6.9) Φ_0,k,Φ˜!=

# _∞

−∞

Φ(x−k) ˜Φ(x)dx=δ_k,0, k∈ZZ.

With the help of this dual function ˜Φ, we are able to compute the coeﬃ- cientsc_kof the representation (2.4.6.2) of functionf ∈V₀,f ="

l∈ZZc_lΦ_0,l, as scalar products:

f(x),Φ(x˜ −k)!=

c_l

# _∞

−∞

Φ(x−l) ˜Φ(x−k)dx

c_l

# _∞

−∞

Φ(x−l+k) ˜Φ(x)dx=c_k.

Also, for anyj ∈ZZ, the functions ˜Φ_j,k(x) := 2^j/2Φ(2˜ ^jx−k),k ∈ ZZ, satisfy

(2.4.6.10)

Φ_j,k,Φ˜_j,l!= 2^j

# _∞

−∞

Φ(2^jx−k) ˜Φ(2^jx−l)dx

# _∞

−∞

Φ(x−k+l) ˜Φ(x)dx

=δ_k−l,0, k, l∈ZZ.

The dual function ˜Φmay thus be used to compute also the coeﬃcientsc_j,l of the seriesf ="

k∈ZZc_j,kΦ_j,k representing a functionf ∈V_j: (2.4.6.11) f,Φ˜_j,l!=

c_j,k Φ_j,k,Φ˜_j,l!=c_j,l.

The following theorem ensures the existence of dual functions:

(2.4.6.12) Theorem. To each scaling functionΦthere exists exactly one dual functionΦ˜∈V₀.

Proof. Since the Φ_0,k, k ∈ ZZ, form a Riesz-basis of V₀, the function Φ_0,0=Φis not contained in the subspace

V := span{Φ_0,k| |k| ≥1}

ofV₀. Therefore, there exists exactly one elementg=Φ−u= 0 withu∈V, so that

g= min{ Φ−v |v∈V }.

uis the (unique!) orthogonal projection ofΦtoV, henceg⊥V, i.e. v, g!= 0 for allv∈V, and in particular,

Φ_0,k, g!= 0 for all|k| ≥1,

and 0 <g² = Φ−u, Φ−u!= Φ−u, g! = Φ, g!. The function ˜Φ:=

g/ Φ, g!has therefore the properties of a dual function.

Example.The Haar-function (2.4.6.4) is selfdual, ˜Φ=Φ. This follows from the orthonormality propertyΦ, Φ₀,k=δk,0,k∈ZZ.

It is, furthermore, possible to compute the best approximation v_j ∈ V_j of a given v_j+1 ∈ V_j+1 using the dual function ˜Φ. We assume that v_j+1∈V_j+1is given by its series representationv_j+1="

k∈ZZc_j+1,kΦ_j+1,k. Because of v_j ∈ V_j, the functionv_j we are looking for has the form v_j =

k∈ZZc_j,kΦ_j,k. Its coeﬃcientsc_j,khave to be determined so thatv_j+1−v_j ⊥ V_j,

v_j, v!= v_j+1, v! for allv∈V_j.

Now, since ˜Φ ∈ V₀, all functions ˜Φ_j,k, k ∈ ZZ, belong to V_j so that by (2.4.6.10)

c_j,l = v_j,Φ˜_j,l!= v_j+1,Φ˜_j,l!

k∈ZZ

c_j+1,k Φ_j+1,k,Φ˜_j,l!

k∈ZZ

c_j+1,k2^j+1/2

# _∞

−∞

Φ(2^j+1x−k) ˜Φ(2^jx−l)dx

k∈ZZ

c_j+1,k√ 2

# _∞

−∞

Φ(2x+ 2l−k) ˜Φ(x)dx.

This leads to the following method for computing the coeﬃcientsc_j,k from the coeﬃcients c_j+1,k: First, we compute the numbers (they do not depend onj)

(2.4.6.13) γ_i:=√ 2

# _∞

−∞

Φ(2x+i) ˜Φ(x)dx, i∈ZZ. The coeﬃcientsc_j,l are then obtained by means of the formula

(2.4.6.14) c_j,l =

k∈ZZ

c_j+1,kγ_2l−k, l∈ZZ.

Example.Because of ˜Φ=Φfor the Haar-function, the numbersγiare given by γi:= 1/2 fori= 0,−1,

0 otherwise.

SinceΦhas compact support, only ﬁnitely manyγiare nonzero, so that also the sums (2.4.6.14) are ﬁnite in this case:

cj,l= 1

√2(cj+1,2l+cj+1,2l+1), l∈ZZ.

In general, the sums in (2.4.6.14) are infinite and thus have to be approximated by finite sums (truncation after finitely many terms). The ef- ficiency of the method thus depends on how fast the numbersγ_i converge to 0 as|i| → ∞.

It is possible to iterate the method following the scheme (2.4.6.8). Note that the numbersγ_ihave to be computed only once since they are indepen- dent ofj. In this way we obtain the followingmulti-resolution algorithmfor computing the coeﬃcientsc_m,k, k∈ZZ, of the seriesv_m="

k∈ZZc_m,kΦ_m,k for allm≤j:

(2.4.6.15)Given: c_j+1,k,k∈ZZ, andγ_i,i∈ZZ,s. (2.4.6.13).

Form=j,j−1,. . . forl∈ZZ

c_m,l:="

k∈ZZc_m+1,kγ_2l−k.

We have seen that this algorithm is particularly simple for the Haar- function as scaling function.

We now consider scaling functions that are given by higher order B- splines, Φ(x) = Φ_r(x) :=M_r(x), r >1. For reasons of simplicity we treat only caser= 2, which is already typical. Since the scaling functionsM₂(x) and the hat-functionH(x) =M₂(x+1) generate the same spacesV_j,j ∈ZZ, the investigation ofΦ(x) :=H(x) is suﬃcient.

First we show the property (S1), V_j⊂V_j+1, for the linear spaces generated byΦ(x) =H(x) and its derivatesΦ_j,k(x) = 2^j/2Φ(2^jx−k),j, k∈ZZ. This follows from the two-scale-relation ofH(x),

H(x) =1 2

H(2x+ 1) + 2H(2x) +H(2x−1) ,

that is immediately veriﬁed using the deﬁnitionH(x) (2.4.6.6). We leave it to the reader to prove thatH(x) has also the remaining properties required for a scaling function [see Exercise 32].

The functionsf ∈V_j have a simple structure: they are continuous and piecewise linear functions on IR with respect to the partition∆^j of ﬁneness 2^−j given by the knots

(2.4.6.16) ∆^j :={k·2^−j|k∈ZZ}.

Up to a common factor, the coeﬃcients c_j,k of the series representation f ="

k∈ZZc_j,kΦ_j,k are given by the values off on∆^j: (2.4.6.17) c_j,k= 2^−j/2f(k·2^−j), k∈ZZ.

Unfortunately, the functionsΦ_0,k(x) = H(x−k),k ∈ZZ, do not form an orthogonal system of functions: A short direct calculation shows

(2.4.6.18)

# _∞

−∞

H(x−j)H(x)dx=





2/3 forj= 0, 1/6 for|j|= 1, 0 otherwise, so that

Φ_0,k, Φ_0,l!=

# _∞

−∞

H(x−k+l)H(x)dx=

92/3 fork=l, 1/6 for|k−l|= 1, 0 otherwise.

As a consequence, the dual function ˜H will be diﬀerent fromH, but it can be calculated explicitly. According to the proof of Theorem (2.4.6.12), H˜ is a scalar multiple of the function

g(x) =H(x)−

|k|≥1

a_kH(x−k),

where the sequence (a_k)_|k|≥1 is the unique solution of the following equations:

g, Φ_0,k!= H(x)−

|l|≥1

a_lH(x−l), H(x−k)!= 0 for all|k| ≥1

that satisﬁes "

k|a_k|² < ∞. By (2.4.6.18), this leads for k ≥ 1 to the equations

(2.4.6.19) 4a₁+a₂ = 1, (k= 1), a_k−1+ 4a_k+a_k+1= 0, (k≥2), and fork≤ −1 to

4a₋₁+a₋₂ = 1, (k=−1), a_k−1+ 4a_k+a_k+1= 0, (k≤ −2).

It is easily seen that with (a_k) also (a_−k) is a solution of these equations, so that by the uniqueness of solutionsa_k =a_−k for all|k| ≥1. Therefore it is suﬃcient to ﬁnd the solutionsa_k,k≥1, of (2.4.6.19).

According to these equations, the sequence (a_k)_k≥1 is a solution of the homogeneous linear diﬀerence equation

c_k−1+ 4c_k+c_k+1= 0, k= 2,3, . . . .

The sequencec_k :=θ^k,k≥1, is clearly a solution of this diﬀerence equation, if θ is a zero of the polynomial x²+ 4x+ 1 = 0. This polynomial, however, has the two diﬀerent zeros

λ:=−2 +√

3 = −1 2 +√

3, µ:=−2−√

3 = 1/λ,

with |λ| < 1 < |µ|. The general solution of the diﬀerence equation is an arbitrary linear combination of these special special solutions (λ^k)_k≥1 and (µ^k)_k≥1, that is, the desired solutiona_khas the forma_k =Cλ^k+Dµ^k,k≥ 1, with appropriately chosen constantsC,D. The condition"

k|a_k|²<∞ impliesD = 0 because of |µ| >1. The constantC has to be chosen such that also the ﬁrst equation (2.4.6.19), 4a₁+a₂= 1, is satisﬁed, that is,

1 =Cλ(4 +λ) =Cλ(2 +√

3) =−C.

This impliesa_k=a_−k =−λ^k=−(−1)^k(2 +√

3)^−k fork≥1 so that g(x) =H(x) +

k≥1

(−1)^k(2 +√

3)^−k(H(x+k) +H(x−k)).

The dual function ˜H(x) =γg(x) we are looking for is a scalar multiple of g, whereγis determined by the condition H,H˜!= 1. Using (2.4.6.18) this leads to

γ = H, H! −(2 +√

3)⁻¹(H(x), H(x−1)!+ H(x), H(x+ 1)!)

3 − 1

3(2 +√ 3) = 1

√3.

Therefore, the dual function ˜H(x) is given by the inﬁnite series H˜(x) =

k∈ZZ

b_kH(x−k) with coeﬃcients

b_k:= (−1)^k√ 3 (2 +√

3)^|k|, k∈ZZ.

The computation of the numbers γ_i (2.4.6.13) for the multi-resolution method (2.4.6.15) leads to simple ﬁnite sums with known terms,

(2.4.6.20)

γ_i=√

2 H(2x+i),H˜(x)!

=√

k∈ZZ

b_k H(2x+i), H(x−k)!

=√

k∈ZZ

b_k H(2x+i+ 2k), H(x)!

=√

k:|2k+i|≤2

b_k H(2x+i+ 2k), H(x)!, because an elementary calculation shows

H(2x+j), H(x)!=







5/12 forj= 0, 1/4 for|j|= 1, 1/24 for|j|= 2, 0 otherwise.

Since 2 +√

3 = 3.73. . ., the numbers b_k = O((2 +√

3)^−|k|) and, conse- quently, the numbers γ_k converge rapidly enough to zero as |k| → ∞, so that the sums (2.4.6.14) of the multi-resolution method are well approximated by relatively short ﬁnite sums.

Example.For smalli, one ﬁnds the following values ofγi:

i γi

0 0.96592 58262 1 0.44828 77361 2 −0.16408 46996 3 −0.12011 83369 4 0.04396 63628 5 0.03218 56114 6 −0.01178 07514 7 −0.00862 41086 8 0.00315 66428 9 0.00231 08229 10 −0.00084 58199

Incidentally, the symmetry relationγ₋i=γiholds for alli∈ZZ[see Exercise 33].

There is a simple interpretation of this multi-resolution method: A functionf ="

k∈ZZc_j+1,kΦ_j+1,k∈V_j+1is a continuous function which is piecewise linear with respect to the partition∆^j+1 ={k·2^−(j+1) |k∈ZZ} of IR with [see (2.4.6.17)]

c_j+1,k= 2^−(j+1)/2f(k·2^−(j+1)), k∈ZZ.

It is approximated optimally by a continuous function ˆf = "

kZZc_j,kΦ_j,k which is piecewise linear with respect to the coarser partition∆^j ={l·2^−j| l∈ZZ}if we set forl∈ZZ

fˆ(l·2^−j) :=2^j/2c_j,l= 2^j/2

k∈ZZ

c_j+1,kγ_2l−k

= 1

√2

k∈ZZ

γ_2l−kf(k·2^−(j+1)).

The numbersγi thus have the following nice interpretation: The continuous piecewise linear (with respect to ∆⁰ ={l|l∈ ZZ}) function ˆf(x) with ˆf(l) :=

γ₂l/√

2, l∈ZZ, is the best approximation in V₀ of the compressed hat-function f(x) := H(2x) ∈V₁. The translated function f₁(x) :=H(2x−1) ∈V₁ is best approximated by ˆf₁ ∈V₀ where ˆf₁(l) :=γ₂l−1/√

2,l∈ZZ, [see Figure 4].

...... .......................

...

f,fˆ

...

... ... ... ... ... ...

−3 −2 −1

3 2

.........................................................................................................

fˆ f

...

Fig. 4.The functionsf and ˆf

We now return to general scaling functions. Within the framework of multi-resolution methods it is essential that the functions Φ_j,k, k ∈ ZZ, form a Riesz-basis of the spaces V_j. It is of equal importance that also the orthogonal complementsW_j ofV_j inV_j+1have similarly simple Riesz- bases: One can show [proofs can be found in e.g. Daubechies (1992), Louis et al. (1994)], that to any scaling functionΦ generating the linear spaces V_j there is a functionΨ ∈W₀ with the following properties:

a)The functions Ψ_0,k(x) :=Ψ(x−k),k∈ZZ, form a Riesz-basis [(see (2.4.6.1))]ofW₀, and for eachj∈ZZits derivatesΨ_j,k(x) := 2^j/2Ψ(2^jx−k), k∈ZZ, form a Riesz-basis ofW_j.

b)The functionsΨ_j,k,j, k∈ZZ, form a Riesz-basis ofL₂(IR).

Each function Ψ with these properties is called a wavelet belonging to the scaling function Φ(wavelets are not uniquely determined by Φ).Ψ is called an orthonormal wavelet,if in addition the Ψ_0,k, k ∈ ZZ, form an orthonormal basis of W₀, Ψ_0,k, Ψ_0,l! = δ_k,l. In this case an orthonormal basis ofL₂(IR) is given by the derivatesΨ_j,k, j, k∈ZZ, ofΨ,

Ψ_j,k, Ψ_l,m!=δ_j,lδ_k,m.

Example.To the Haar-functionΦ(2.4.6.4) belongs theHaar-waveletdeﬁned by Ψ(x) :=

91, for 0≤x <1/2,

−1, for 1/2≤x <1, 0, otherwise.

Indeed, the relationΨ(x) =Φ(2x)−Φ(2x−1) ﬁrst impliesΨ₀,k∈V₁for allk∈ZZ; then the orthogonality

Ψ₀,k, Φ₀,l= 0 for allk, l∈ZZ givesΨ₀,k∈W₀ for allk∈ZZ, and ﬁnally

Φ₁,k(x) = 1

√2(Ψ₀,k(x) +Φ₀,k(x)), k∈ZZ,

provesV₁=V₀⊕W₀. Also, sinceΨ, Ψ₀,k=δk,0, the Haar-wavelet is orthonormal.

One can show that every orthonormal scaling function Φ has also an orthonormal wavelet, which even can be given explicitly by means of the two-scale-relation ofΦ[see e.g. the literature cited above]. It is very diﬃcult however, to determine for a non-orthonormalΦan associated wavelet. The situation is better for the scaling functions Φ = Φ_r := M_r given by B- splines. Here, explicit formulas for the special waveletsΨ_r are known that have a minimal compact support (namely the interval [0,2r−1]) among all wavelets belongingM_r: they are given by ﬁnite sums of the form

Ψ_r(x) =

3r−2

k=0

q_kM_r(2x−k).

However, these wavelets are not orthonormal forr >1, but one still knows explicitly their associated dual functions ˜Ψ_r∈W₀with their usual properties Ψ_r(x−k),Ψ˜_r(x)!=δ_k,0fork∈ZZ[see Chui (1992) for a comprehensive account].

The example of the Haar-function shows, how simple the situation be- comes if both the scaling functionΦand the waveletΨ are orthonormal and

if both functions have compact support. A drawback of the Haar-function is that it is not even continuous, whereas already M₂ is continuous and the smoothness of the B-Splines M_r with r≥3 and its wavelets Ψ_r even increases with r. Also for allr ≥ 1,Φ_r =M_r has compact support, and has a waveletΨ_rwith compact support. But unfortunately neitherΦ_r nor Ψ_r are orthonormal forr >1.

Therefore the following deep result of Daubechies is very remarkable:

She was the ﬁrst to give explicit examples of scaling functions and associated wavelets that have an arbitrarily high order of diﬀerentiability, have compact support, and are orthonormal. For a detailed exposition of these important results and their applications, we refer the reader to the rele- vant special literature [see e.g. Louis et al. (1994), Mallat (1998), and, in particular, Daubechies (1992)].

Exercises for Chapter 2

1. LetLi(x) be the Lagrange polynomials (2.1.1.3) for pairwise diﬀerent support abscissasx₀, . . . , xn, and letci:=Li(0). Show that

n i=0

cix^j_i =

91 forj= 0,

0 forj= 1,2, . . . , n, (−1)ⁿx₀x₁. . . xn forj=n+ 1, (a)

n i=0

Li(x)≡1.

(b)

2. Interpolate the function lnxby a quadratic polynomial atx= 10, 11, 12.

(a) Estimate the error committed forx= 11.1 when approximating lnxby the interpolating polynomial.

(b) How does the sign of the error depend onx?

3. Consider a functionf which is twice continuously diﬀerentiable on the interval I= [−1,1]. Interpolate the function by a linear polynomial through the support points (xi, f(xi)),i= 0, 1,x₀,x₁∈I. Verify that

α= ¹₂max

ξ∈I |f(ξ)|max

x∈I |(x−x₀)(x−x₁)|

is an upper bound for the maximal absolute interpolation error on the interval I. Which values x₀,x₁ minimize α? What is the connection between (x− x₀)(x−x₁) and cos(2 arccosx)?

4. Suppose a functionf(x) is interpolated on the interval [a, b] by a polynomial Pn(x) whose degree does not exceedn. Suppose further thatfis arbitrarily often diﬀerentiable on [a, b] and that there existsM such that|f⁽ⁱ⁾(x)| ≤M for i = 0, 1, . . . and any x ∈ [a, b]. Can it be shown, without additional hypotheses about the location of the support abscissasxi∈[a, b], thatPn(x) converges uniformly on [a, b] tof(x) asn→ ∞?

5. (a) The Bessel function of order zero, J₀(x) = 1

# π 0

cos (xsint)dt,

is to be tabulated at equidistant argumentsxi=x₀+ih,i= 0, 1, 2,. . . . How small must the incrementh be chosen so that the interpolation error remains below 10⁻⁶ if linear interpolation is used?

(b) What is the behavior of the maximal interpolation error

0≤maxx≤1|Pn(x)−J₀(x)|

asn→ ∞, ifPn(x) interpolatesJ₀(x) atx=x⁽_iⁿ⁾:=i/n,i= 0, 1,. . ., n?

Hint: It suﬃces to show that|J₀⁽^k⁾(x)| ≤1 fork= 0, 1,. . . . (c) Compare the above result with the behavior of the error

0≤maxx≤1|S∆_n(x)−J₀(x)|

asn→ ∞, whereS∆_nis the interpolating spline function with knot set

∆n={x⁽_iⁿ⁾}andS_∆ _n(x) =J₀(x) forx= 0, 1.

6. Interpolation on product spaces: Suppose every linear interpolation problem stated in terms of functionsϕ₀,ϕ₁,. . .,ϕnhas a unique solution

Φ(x) = n

i=0

αiϕi(x)

with Φ(xk) =fk,k = 0, . . ., n, for prescribed support argumentsx₀, . . ., xn with xi =xj, i=j. Show the following: Ifψ₀,. . ., ψm is also a set of functions for which every linear interpolation problem has a unique solution, then for every choice of abscissas

x₀, x₁, . . . , xn, xi=xjfori=j, y₀, y₁, . . . , ym, yi=yjfori=j, and support ordinates

fik, i= 0, . . . , n, k= 0, . . . , m, there exists a unique function of the form

Φ(x, y) = n ν=0

m µ=0

ανµϕν(x)ψµ(y) withΦ(xi, yk) =fik,i= 0,. . .,n,k= 0,. . .,m.

7. Specialize the general result of Exercise 6 to interpolation by polynomials.

Give the explicit form of the functionΦ(x, y) in this case.

8. Given the abscissas

y₀, y₁, . . . , ym, yi=yjfori=j,

and, for eachk= 0,. . .,m, the values

x⁽₀^k⁾, x⁽₁^k⁾, . . . , x⁽n^k_k⁾, x⁽_i^k⁾=x⁽_j^k⁾ fori=j, and support ordinates

fik, i= 0, . . . , nk, k= 0, . . . , m,

suppose without loss of generality that theykare numbered in such a fashion that

n₀≥n₁≥ · · · ≥nm. Prove by induction overmthat exactly one polynomial

P(x, y)≡ m µ=0

n_µ

ν=0

ανµx^νy^µ exists with

P(x⁽_i^k⁾, yk) =fik, i= 0, . . . , nk, k= 0, . . . , m .

9. Is it possible to solve the interpolation problem of Exercise 8 by other polynomials

P(x, y) = M µ=0

N_µ

ν=0

ανµx^νy^µ,

requiring only that the number of parametersανµagree with the number of support points, that is,

m µ=0

(nµ+ 1) = M µ=0

(Nµ+ 1) ? Hint: Study a few simple examples.

10. Calculate the inverse and reciprocal diﬀerences for the support points xi: 0 1 −1 2 −2

fi: 1 3 3/5 3 3/5

and use them to determine the rational espressionΦ²^,²(x) whose numerator and denominator are quadratic polynomials and for whichΦ²^,²(xi) =fi, ﬁrst in continued-fraction form and then as the ratio of polynomials.

11. LetΦ^m,n be the rational function which solves the system S^m,n for given support points (xk, fk),k= 0, 1,. . .,m+n:

(a₀+a₁xk+· · ·+amx^mk)−fk(b₀+b₁xk+· · ·+bnxⁿk) = 0, k= 0,1, . . . , m+n.

Show thatΦ^m,n(x) can be represented as follows by determinants:

Φ^m,n(x) =|fk, xk−x, . . . ,(xk−x)^m,(xk−x)fk, . . . ,(xk−x)ⁿfk|^m_k₌₀⁺ⁿ

|1, xk−x, . . . ,(xk−x)^m,(xk−x)fk, . . . ,(xk−x)ⁿfk|^m_k₌₀⁺ⁿ

Here the following abbreviation has been used:

_α_k, . . . , ζk^m⁺ⁿ

k=0 = det







α₀ · · · ζ₀ α₁ · · · ζ₁

· ·

αm+n · · · ζm+n





^.

12. Generalize Theorem (2.3.1.12):

(a) For 2n+ 1 support abscissasxkwith

a≤x₀< x₁<· · ·< x₂n< a+ 2π

and support ordinates y₀, . . . , y₂n, there exists a unique trigonometric polynomial

T(x) = ¹₂a₀+ n j=1

(ajcosjx+bjsinjx) with

T(xk) =yk for k= 0,1, . . . ,2n.

(b) Ify₀,. . .,y₂nare real numbers, then so are the coeﬃcientsaj,bj. Hint: Reduce the interpolation by trigonometric polynomials in (a) to (complex) interpolation by polynomials using the transformationT(x) =

j=−ncje^ijx. Then showc₋j=cj to establish (b).

13. (a) Show that, for realx₁, . . . , x₂n, the function t(x) =

k=1

sinx−xk

2 is a trigonometric polynomial

12a₀+ n

j=1

(ajcosjx+bjsinjx) with real coeﬃcientsaj,bj.

Hint: Substitute sinϕ= (1/2i)(e^iϕ−e⁻^iϕ).

(b) Prove, using (a) that the interpolating trigonometric polynomial with support abscissasxk,

a≤x₀< x₁· · ·< x₂n< a+ 2π, and support ordinatesy₀, . . . , y₂n is identical with

T(x) =

j=0

yjtj(x),

where

tj(x) :=

k=0k=j

sinx−xk

:!2n k=0k=j

sinxj−xk

2 .

14. Show that forn+ 1 support abscissasxkwith a≤x₀ < x₁<· · ·< xn< a+π

and support ordinatesy₀,. . .,yn, a unique “cosine polynomial”

C(x) = n

j=0

ajcosjx exists withC(xk) =yk,k= 0, 1,. . .,n.

Hint: See Exercise 12.

15. (a) Show that for any integerj

k=0

cosjxk= (2m+ 1)h(j),

k=0

sinjxk= 0, with

xk:= 2πk

2m+ 1, k= 0,1, . . . ,2m, and

h(j) := 1 forj= 0 mod 2m+ 1, 0 otherwise.

(b) Use (a) to derive for integersj,kthe following orthogonality relations:

i=0

sinjxisinkxi= 2m+ 1

2 [h(j−k)−h(j+k)],

i=0

cosjxicoskxi= 2m+ 1

2 [h(j−k) +h(j+k)],

i=0

cosjxisinkxi= 0.

16. Suppose the 2π-periodic function f: IR → IR has an absolutely convergent Fourier series

f(x) = ¹₂a₀+ ∞ j=1

(ajcosjx+bjsinjx).

Let

Ψ(x) =¹₂A₀+ m j=1

(Ajcosjx+Bjsinjx)

be trigonometric polynomials with

Ψ(xk) =f(xk), xk:= 2πk 2m+ 1, fork= 0, 1,. . ., 2m.

Show that

Ak=ak+ ∞ p=1

[ap(2m+1)+k+ap(2m+1)−k], 0≤k≤m,

Bk=bk+ ∞ p=1

[bp(2m+1)+k−bp(2m+1)−k], 1≤k≤m .

17. Formulate a Cooley-Tukey method in which the array ˜β[ ] is initialized di- rectly_˜

β[j] =fj

rather than in bit-reversed fashion.

Hint: Deﬁne and determine explicitly a mapσ =σ(m, r, j) with the same replacement properties as (2.3.2.6) but σ(0, r,0) =rinstead of (2.3.2.7).

18. LetN = 2ⁿ: Consider theN-vectors

f:= [f₀, . . . , fN−1]^T, β= [β₀, . . . , βN−1]^T.

(2.3.2.1) expresses a linear transformation between these two vectors, β = (1/N)T f, whereT = [tjk],tjk:=e⁻²^πijk/N.

(a) Show thatT can be factored as follows:

T =QSP(Dn−1SP)· · ·(D₁SP), whereS is theN×N matrix







1 1

1 −1 . ..

1 1

1 −1





^.

The matricesDl= diag(1, δ₁⁽^l⁾,1, δ₃⁽^l⁾,· · ·,1, δ_N⁽^l⁾₋₁),l= 1,. . .,n−1, are diagonal matrices with

δ^lr= exp(−2πi˜r/2ⁿ⁻^l⁻¹), r˜= _r

2^l

, rodd.

Q is the matrix of the bit-reversal permutation (2.3.2.8), andP is the matrix of the followingbit-cycling permutationζ:

ζ(α₀+α₁·2 +· · ·+αn−12ⁿ⁻¹) :=αn−1+α₀·2 +α₁·4 +· · ·+αn−22ⁿ⁻¹. (b) Show that the Sande-Tukey method for fast Fourier transforms corre- sponds to multiplying the vectorffrom the left by the (sparse) matrices in the above factorization.

Dalam dokumen 4 Systems of Linear Equations 190 (Halaman 131-153)