Quadrature rules for periodic integrands

32(2005) pp. 5–44.

Quadrature rules for periodic integrands.

Bi-orthogonality and para-orthogonality ^∗

Ruymán Cruz-Barroso

, Leyla Daruis

, Pablo González-Vera

, Olav Njåstad

aLa Laguna University ldaruis@ull.es, pglez@ull.es

bNorwegian University of Science and Technology

Abstract

In this paper, the algebraic construction of quadrature formulas for weigh- ted periodic integrals is revised. For this purpose, two classical papers ([10]

and [14]) in the literature are revisited and certain relations and connections are brought to light. In this respect, the concepts of “bi-orthogonality” and

“para-orthogonality” are shown to play a fundamental role.

Key Words: Trigonometric polynomials, Szegő polynomials, quadratures, bi-orthogonality, para-orthogonality.

AMS Classification Number: 41A55, 33C45

1. Introduction

Let the integral In(f) = R

Γf(z)dµ(z) be given with Γ a certain curve in the complex plane and dµ a positive measure on Γ. By an n-point quadrature rule for this integral we mean an expression likeIn(f) =P_n

j=1Ajf(zj)with zj 6=zk

if j 6=k and {zj}ⁿ_j=1 ⊂ Γ so that the weights or coefficients {Aj}ⁿ_j=1 and nodes {zj}ⁿ_j=1 are to be determined by imposing thatIn(f)exactly integrates i.e. In(f) coincides withIµ(f)for as many basis functions as possible in an appropriate func- tion space S where the above integral exists. Two situations have been most widely considered in the literature. Namely, on the one hand, the case when Γcoincides with a subinterval of the real line, that is, Γ = [a, b], −∞6 a < b 6∞ and on the other hand whenΓ is the unit circle, i.e. Γ = T ={z ∈C : |z|= 1}. Ob- serve that the second case is equivalent to dealing with real integrals of the form R_π

−πf(θ)dµ(θ),f being a 2π-periodic function (here by a slight abuse of notation

∗This work is partially supported by the research project PI-2002-023 of Gobierno de Canarias.

5

we writef(z) =f(θ), dµ(z) =dµ(θ) forz =e^iθ). As for the first case, it is well known that the construction of quadrature formulas to approximate integrals like R_b

af(x)dµ(x)represents an interesting research topic which has been exhaustively considered in the last decades and where orthogonal polynomials find one of their most direct and natural applications. Indeed, if {Qk}^∞_k=0 denotes the sequence of orthonormal polynomials for the measure µ, then In(f) =P_n

j=1Ajf(xj) with {xj}ⁿ_j=1the zeros ofQn(x)andλj =¡P_n

k=0Q²_k(xj)¢₋₁

forj = 1, . . . , n(Christoffel numbers) satisfiesIn(P) =R_b

aP(x)dµ(x) for any polynomialP of degree 2n−1.

In this case,{In(f) : n= 1,2, . . .}represent the well known sequence of Gaussian or Gauss–Christoffel quadrature formulas (see e.g. [8] for a survey). On the other hand, although quadratures on the unit circle and other related topics such as Szegő polynomials and the trigonometric moment problem have been receiving much re- cent attention because of their applications in rapidly growing fields of pure and applied mathematics (Digital Signal Processing, Operator Theory, Probability The- ory, . . .), there do not exist so many results about quadratures on the unit circle as in the real case. In this respect, the main aim of this paper is to emphasize the role played by certain sequences of orthogonal trigonometric polynomials in- troduced by Szegő [14] in the construction of quadrature rules on the unit circle by carrying out a comparision with the approach given by Jones et. al in [10]. In both approaches, a fundamental tool will be the so-called Szegő polynomials or polyno- mials which are orthogonal on the unit circle in the following sense: given n>1, it is known (see e.g. [13]) that a unique monic polynomial ρn(z)exists such that R_π

−πρn(e^iθ)e^−ikθdµ(θ) = 0for k= 0,1, . . . , n−1. Furthermore, if we assume that the support ofµ has infinitely many points, then R_π

−πρ²_n(e^iθ)dµ(θ) =k ρn k²_µ>0.

Setting ρ₀ ≡ 1, then {ρ_n}^∞_n=0 is called the orthogonal sequence of monic Szegő polynomials. On the other hand, the sequence {ϕn}^∞_n=0 with ϕn(z) = _kρ^ρn(z)

nkµ rep- resents an orthonormal sequence of Szegő polynomials (observe that such a se- quence is uniquely determined by assuming that the leading coefficient ofϕn(z)for n= 0,1, . . . is positive). Setting D={z ∈ C : |z|<1} (sometimes we will use E={z∈ C : |z|>1}, C=T∪D∪E) a fundamental property concerning the zeros of ρn(z)for n>1 (and apparently rather negative for our purposes) is the following (see e.g. [1]): “For eachn>1, all the zeros ofρn(z)lie inD”. Thus, un- like the Gauss–Christoffel formulas, now the zeros of Szegő polynomials can not be directly used as nodes in our quadratures. Following two initially different paths, throughout the paper we will see how this drawback can be overcome. The paper is organized as follows. In Section 2, some preliminary results concerning trigono- metric polynomials, Laurent polynomials and algebraic polynomials are presented.

Then, in Section 3 the problem of the interpolation by trigonometric polynomials is analyzed whereas in Section 4 the so-called bi-orthogonal systems of trigonometric polynomials are introduced and their most relevant properties studied. The con- struction of quadrature rules exactly integrating trigonometric polynomials with degree as large as possible is considered in Section 5 and a connection with the unit circle presented in Section 6. Finally some illustrative numerical experiments

are shown in Section 7.

2. Preliminary results

We will start by fixing some definitions and notations. Thus, for a nonnegative integer n,Πn will denote the space of (in general complex) algebraic polynomials of degreenat most andΠthe space of all polynomials. On the other hand, a real trigonometric polynomial of degreenis a function of the form

Tn(θ) =a0+ Xn

k=1

(akcoskθ+bksinkθ), ak, bk ∈R, |an|+|bn|>0.

Clearly, whena0, ak and bk are in general complex numbers for k= 1, . . . , n, we shall be dealing with trigonometric polynomials with complex coefficients. Thus, when we refer to a trigonometric polynomial we are implicitly meaning with real coefficients. We also denote byTnthe space of trigonometric polynomials of degree nat most, i.e.

Tn=span{1,cosθ,sinθ, . . . ,cosnθ,sinnθ}

and hence, dim(T_n) = 2n+ 1. We occasionally deal with complex trigonomet- ric polynomials, where a0, ak and bk are arbitrary complex numbers. By using the transformation z = e^iθ and Euler’s formulas, for any complex trigonometric polynomial one can writeTn(θ) =Ln(e^iθ)where

Ln(z) = Xn

k=−n

ckz^k. (2.1)

Then

c0=a0, ck =1

2(ak−ibk), k= 1, . . . , n,

and when the trigonometric polynomialTn is real,a0,ak,bk are real andc−k =ck. FunctionsLn(z)as given above are called Laurent polynomials, or more generally, givenpand qintegers such that p6q, a Laurent polynomial is a function of the form

Ln(z) = Xq

j=p

αjz^j, αj∈C. (2.2)

We also denote byΛp,q the space of Laurent polynomials (2.2). Observe that Λp,q=span©

z^k :p6k6qª .

Hence,dim(Λp,q) =q−p+ 1. Thus,Ln(z)given by (2.1) belongs to Λ−n,n. Now, by recalling that a double sequence{µk}^∞_k=−∞of complex numbers is said to be “Hermitian” ifµ−k =µk, a Laurent polynomialL∈Λ−n,nis called Hermitian if the sequence of its coefficients is Hermitian. That is, withLn(z)in (2.1) we have ck =ck fork= 0,1, . . . , nand the following trivially holds,

Theorem 2.1. LetTn(θ)be a complex trigonometric polynomial, and setLn(e^iθ) = Tn(θ). ThenTn is real if and only ifLn is Hermitian.

Remark 2.2. If we define Λ^H_n ={L∈Λ−n,n : L Hermitian} then Λ^H_n is a real vector space of dimension2n+ 1and one can write

Tn =©

T(θ) : T(θ) =L(e^iθ)with L∈Λ^H_nª .

Let us next consider the connection between trigonometric polynomials and certain algebraic polynomials. For this purpose, letP(z)be an algebraic polynomial of degreen, i.e.,

P(z) = Xn

j=0

ajz^j, aj∈C, an6= 0.

Then, the reciprocal P^∗(z) of P(z) is a polynomial defined by P^∗(z) = zⁿP∗(z) where P∗(z) represents the “sub-star” conjugate of P(z), i.e., P∗(z) = P(1/¯z).

Thus,

P^∗(z) =zⁿP(1/z) =¯ zⁿP(1/z) =¯ Xn

j=0

an−jz^j

whereP(z) =P_n

j=0ajz^j. Now, a usefull property of the polynomials that we shall work with is the following: fork∈C\{0}, a polynomial P(z)is called “invariant”

or more precisely, “k-invariant” if

P^∗(z) =kP(z) ∀z∈C.

Some direct consequences of this definition are:

1. IfP(z)is invariant, thenP(0)6= 0.

2. Letαbe a zero of the invariant polynomialP(z). Then,1/α¯ is also a zero of P(z).

3. LetP(z) be an invariant polynomial of odd degreen. Then, the number of zeros ofP(z)onT (counting multiplicities) is also odd. On the other hand, ifP(z)is an invariant polynomial with even degreen, it has an even number of zeros onT.

4. Let P(z) be invariant and set P(z) = P_n

j=0cjz^j = cn

Q_n

k=1(z−zk), then

|P(0)|=|c0|=|cn|Q_n

k=1|zk| and taking into account thatQ_n

k=1|zk|= 1 it follows that |c0| =|cn|. Consequently,cn =kc0 with |k|= 1. Setk =e^iω, ω ∈R, and define Q(z) =λP(z), λ6= 0. Then, Q^∗(z) =λkP(z) = ^λ_λkQ(z), that is,Q(z)is ^λ_λk-invariant. Set nowλ=Re^iγ, then ^λ_λk=e^i(ω−2γ). Thus, by takingγ∈Rsuch that γ= ^ω₂ +mπ, withm∈Z, then ^λ_λk= 1and Q(z) is1-invariant.

Remark 2.3. The term “k-invariant” was introduced by Jones et. al. in [10], whereas Szegő in [14] says that a polynomial P(z) is “autoreciprocal” if P^∗(z) = P(z)(1-invariant). Hence, we see that “invariant” polynomials are essentially “au- toreciprocal”.

LetP2n(z)be an invariant polynomial of degree2n. Then, there existsλ2n ∈ C\{0} such thatQ2n(z) =λ2nP2n(z)is1-invariant and we can write:

Ln(z) = Q2n(z) zⁿ =

Xn

j=−n

cjz^j, c−j =cj, j= 0,1, . . . , n

that is, Ln ∈ Λ^H_n and by Theorem 2.1, Ln(e^iθ) =Tn(θ) withTn ∈ Tn. Thus, we have

e^−inθP2n(e^iθ) =λ⁻¹_2nTn(θ).

Conversely, letTn∈ Tn. Then

Tn(θ) =Ln(e^iθ), Ln∈Λ^H_n.

Again, Ln(z) = ^P2n_zn^(z), where P2n(z) ∈ Π2n and 1-invariant. Indeed, P2n(z) = zⁿLn(z). Hence,

P_2n^∗ (z) =z²ⁿP2n(1/z) =¯ z²ⁿz⁻ⁿLn(1/¯z)

=zⁿP_n

j=−ncjz^−j=zⁿP_n

j=−nc−jz^−j =zⁿLn(z) =P2n(z).

Next, we will see how the connection between trigonometric polynomials and invariant algebraic polynomials allows us to recover some classical results about zeros of trigonometric polynomials. Thus, letαandβ be arbitrary constants, then sin¡_θ−α

2

¢sin³

θ−β 2

´

represents a trigonometric polynomial of degree one. Further- more, it can be easily proved by induction that the function

T(θ) =C Yn

j=1

sin

µθ−θ2j−1

2

¶ sin

µθ−θ2j

2

¶

, C6= 0 (2.3)

where{θ_j}²ⁿ_j=1are given constants, represents a trigonometric polynomial of degree n. We will now show that a converse result also holds, i.e. any trigonometric polynomial can be factorized as (2.3). Indeed, letTn ∈ Tn, thenTn(θ) =Ln(e^iθ), Ln∈∆^H_n and one can writeLn(z) = ^P2n_zn^(z) withP2n(z)an1-invariant polynomial of degree2n. Therefore,P2n(z) =cn

Q_2n

k=1(z−zk),cn6= 0(counting multiplicities) withzj 6= 0 and ifzj 6∈T, then1/z¯j is also a root ofP2n(z). Let 2m denote the number of zeros ofP_2n(z)onT(06m6n). Then

P2n(z) =cn

Y2m

j=1

(z−zj)

n−mY

k=1

(z−z˜k) µ

z− 1

˜ zk

¶

, cn 6= 0 (2.4)

where zj = e^iθj, θj ∈ R for 1 6 j 6 2m, are the zeros of P2m(z) on T and z˜k

and 1/z˜k for 1 6 k 6 n−m are the zeros not on T, so that z˜k = e^iωk with ωk ∈C, wich implies that1/˜zk =e^iωk. Furthermore, it can be easily checked that e^iθ−e^iω= 2isin¡_θ−ω

2

¢eⁱ(^θ+ω2 ). Therefore, P2n(e^iθ) = cn

Q_2m

j=1

¡e^iθ−e^iθj¢ Q_n−m

k=1

¡e^iθ−e^iωk¢ ¡

e^iθ−e^iωk¢

= c_n(−1)ⁿ2²ⁿQ_2m

j=1sin³

θ−θj

2

´ eⁱ

h_θ−

θj 2

i

×

× Q_n−m

k=1 sin¡_θ−ω

k

2

¢sin¡_θ−ω

k

2

¢eⁱ

h θ+^ωk+₂^ωk

i

. Then, it follows that,

P2n(e^iθ) =λne^inθ Y2m

j=1

sin

µθ−θj

2

¶n−mY

k=1

sin

µθ−ωk

2

¶ sin

µθ−ωk

2

¶

, λn 6= 0.

Consequently,

Tn(θ) =Ln(e^iθ) = ^P2n_einθ^(eiθ)

=λn

Q_2m

j=1sin³

θ−θj

2

´ Q_n−m

k=1 sin¡_θ−ω

2k

¢sin¡_θ−ω

2k

¢, (2.5) whereλn 6= 0,θj∈Randωk∈Csuch that<(ωk) =ψk+ 2tπ,ψk ∈(−π, π],t∈Z andk= 1, . . . , n−m. Then, we have proved the following

Theorem 2.4. A real trigonometric polynomialTn(θ)of the precise degree nhas exactly 2n real or complex zeros provided that we count them as usual with their multiplicity and we restrict ourselves to the strip −π < <(θ)6 π. Furthermore, the non-real zeros appear in conjugate pairs.

Remark 2.5. The representation (2.3) is of course not unique.

Furthermore, from (2.3) and (2.5) it can be also proved

Theorem 2.6(L.Fejér and F.Riesz). A real trigonometric polynomial T(θ) is nonnegative for all real θ, if and only if, it can be written in the form

T(θ) =|g(z)|², z=e^iθ

whereg(z)is an algebraic polynomial of the same degree as T(θ).

Proof. Assume T(θ) a trigonometric polynomial of degree n such that T(θ) =

P(e^iθ)

e^inθ with P(z) a polynomial of degree 2n. Since T(θ)> 0 for all θ ∈ R, then possible real zeros ofT(θ)must have even multiplicity. Furthermore, if θ=αis a real zero ofT(θ)thenz=e^iα is a zero of P(z)onT. Hence, from (2.4),P(z)can be expressed as:

P(z) =λnp²_m(z)qn−m(z)q^∗_n−m(z), λn6= 0

wherepm(z)∈Πm for06m6nand qn−m(z)∈Πn−m. SinceT(θ)>0, for any θ∈R,

T(θ) = |T(θ)|=

¯¯

¯^P(e_einθ^iθ)

¯¯

¯=|λn|¯

¯p²_m(e^iθ)¯

¯¯

¯qn−m(e^iθ)¯

¯¯

¯¯qn−m(e^iθ)

¯¯

¯

= |λn|¯

¯p²_m(e^iθ)¯

¯¯

¯qn−m(e^iθ)¯

¯²=¯

¯g(e^iθ)¯

¯²

whereg(z) =p

|λn|pm(z)qn−m(z)∈Πn.

Conversely, let g(z) be an algebraic polynomial of degree n, then by setting z=e^iθ it follows that

|g(z)|²=g(z)g(z) =g(z)g∗(z) = g(z)g^∗(z)

zⁿ =P2n(z) zⁿ

whereP2n(z) =g(z)g^∗(z)is clearly an1-invariant polynomial of degree2nso that

|g(z)|² = Ln(z) ∈ Λ^H_n, and by Theorem 2.1, |g(z)|² represents a trigonometric polynomial of degreenwhich is clearly nonnegative for anyθ∈R. ¤

3. Interpolation by Trigonometric Polynomials

As it is well known, polynomial interpolation finds in the construction of quadra- ture formulas one of its most immediate applications. On the other hand, when considering quadrature rules based on trigonometric polynomials, similar results on interpolation will be needed. In this respect, some of the already known results will now be proved by means of the close connection between trigonometric poly- nomials and Hermitian Laurent polynomials shown in the preceding section. First we have,

Theorem 3.1. Given(2n+ 1)distinct nodes{θj} ⊂(−π, π], there exists a unique Tn∈ Tn such that

Tn(θj) =yj, j= 1, . . . ,2n+ 1, (3.1) {yj}²ⁿ⁺¹_j=1 being a given set of real numbers.

Proof. SetT(θ) =a0+P_n

k=1akcoskθ+bksinkθ. We first show that the constants {ak}ⁿ_k=0∪ {bk}ⁿ_k=1 are uniquely determined from conditions (3.1). Now, T(θ) = L(e^iθ)withL∈Λ−n,n so that (3.1) is equivalent to

L(zj) =yj, zj =e^iθk, j= 1, . . . ,2n+ 1. (3.2) NowL(z)∈Λ−n,n implies thatL(z) = ^P(z)_zn ,P(z)∈Π2n so that (3.2) yields

P(zj) =zⁿ_jyj, j= 1, . . . ,2n+ 1. (3.3) Sincezj 6=zk,P(z)is uniquely determined by (3.3) and henceT(θ)has the desired interpolation properties. It remains to show thatT(θ) has real coefficients. This

will be proved by showing thatP(z)is1-invariant. To see this we will show that alsoP^∗(z)satisfies the interpolation conditions (3.3). Indeed,

P^∗(zj) =z_j²ⁿP(1/z¯j) =z_j²ⁿP(zj) =z_j²ⁿz_jⁿyj=z_jⁿyj, yj∈R.

Hence, by virtue of the uniqueness of polynomialP(z), it follows thatP^∗(z) =P(z)

and the proof is completed. ¤

As for an explicit representation ofTn∈ Tn satisfying (3.1), because of unique- ness, one can write

T_n(θ) =

2n+1X

j=1

l_j(θ)y_j (3.4)

where lj(θ) = lj,n(θ) ∈ Tn such that lj(θk) = δj,k =

½ 1 if j=k

0 if j6=k . Since lj(θk) = 0fork= 1, . . . ,2n+ 1,k6=j, clearly by (2.5),

lj(θ) =λj 2n+1Y

k=1,k6=j

sin

µθ−θk

2

¶

, λj 6= 0,

λj being a normalization constant such thatlj(θj) = 1. More precisely, setting Wn(θ) =

2n+1Y

k=1

sin

µθ−θk

2

¶

then, it follows that

lj(θ) =λj Wn(θ) sin

³θ−θj

2

´, j= 1, . . . ,2n+ 1.

Thus,

lj(θj) =λj lim

θ→θj

Wn(θ) sin

³θ−θj

2

´ =λj lim

θ→θj

Wn(θ)

θ−θj

2

= 2λjW_n⁰(θj).

Hence, takingλj =_2W0¹

n(θj) one haslj(θj) = 1and we can write lj(θ) = Wn(θ)

2W_n⁰(θj) sin

³θ−θj

2

´ , j= 1, . . . ,2n+ 1.

Furthermore, when dealing with the construction of certain quadrature formulas exactly integrating trigonometric polynomials of degree as high as possible, the following result will be required:

Theorem 3.2. Let θ1. . . θn+1 be (n+ 1) distinct nodes on (−π, π]. Then there exists a unique trigonometric polynomial Hn∈ Tn satisfying

Hn(θj) =Hn^(k)(θj) =yj j= 1, . . . , n+ 1 H_n⁰(θj) =Hn^(k)0(θj) =y_j⁰ j= 1, . . . , n+ 1, j6=k





 (3.5)

wherek ∈ {1, . . . , n+ 1} is previously fixed and {yj}ⁿ⁺¹_j=1 ∪ {y_j⁰}ⁿ⁺¹_j=1,j6=k is a set of (2n+ 1) real numbers.

Proof. SetHn(θ) =Ln(e^iθ)∈Λ−n,n. Then (3.5) becomes Hn(θj) =Ln(e^iθj) = Ln(zj) =yj with zj =e^iθj ∈Tfor all j = 1, . . . , n+ 1and zj 6=zk ifj 6=k. On the other hand, H_n⁰(θ) =L_n⁰(e^iθ)ie^iθ. Hence, L_n⁰(zj) = −izjH_n⁰(θj) =−izjy_j⁰ for j= 1, . . . , n+ 1andj6=k. SinceLn ∈Λ−n,n, thenLn(z) = ^P2n_zn^(z) withP2n(z)∈ Π2n such that P2n(zj) = z_jⁿLn(zj) = z_jⁿyj, yj ∈ R and zj ∈ T. Furthermore, P_2n⁰ (z) =nzⁿ⁻¹L_n(z) +zⁿL_n⁰(z), hence

P_2n⁰(z_j) =nz_jⁿ⁻¹L_n(z_j) +zⁿ_jL_n⁰(z_j) =z_jⁿ⁻¹³

ny_j−iy_j⁰´

, j= 1, . . . , n+ 1, j6=k.

Thus our Hermite-type trigonometric interpolation problem reduces to finding P_2n(z)∈Π_2n such that

P2n(zj) =z_jⁿyj j= 1, . . . , n+ 1 P_2n⁰ (zj) =zⁿ⁻¹_j

³

nyj−iy_j⁰

´

j= 1, . . . , n+ 1, j6=k





 (3.6)

Now, sincezj6=zlforj6=l, it is known that the interpolation problem (3.6) has a unique solutionP2n(z)andTn(θ) =Ln(e^iθ) = ^P2n_einθ^(eiθ) will be the unique solution to (3.5). As in Theorem 3.1, it remains to prove thatTn(θ)is a real trigonometric polynomial. To do this, we will show thatP_n^∗(z)is also a solution to (3.6), hence because of uniqueness we haveP2n(z) =P_2n^∗ (z)and the conclusion follows. Indeed,

P_2n^∗ (z_j) =z_j²ⁿP_2n(1/z_j) =z_j²ⁿP_2n(z_j)

=z_j²ⁿz_jⁿyj =z_jⁿyj =P2n(zj), j= 1, . . . , n+ 1.

Furthermore,(P_2n^∗ )⁰(z) = 2nz²ⁿ⁻¹P_2n(1/z) +z²ⁿ¡ P_2n¢⁰

(1/z)¡₋₁

z²

¢, yielding:

(P_2n^∗ )⁰(zj) =z²ⁿ⁻²_j h

2nzjP2n(zj)−P_2n⁰(zj) i

. (Here we are making use of the fact ¡

P¢⁰

(z) = (P⁰)(z)). Therefore, for j = 1, . . . , n+ 1, j6=k:

(P_2n^∗ )⁰(zj) = z_j²ⁿ⁻² h

2nzjz_jⁿyj−z_j⁻⁽ⁿ⁻¹⁾(nyj+iy_j⁰) i

= z_jⁿ⁻¹ h

2nyj−nyj−iy_j⁰ i

=z_jⁿ⁻¹[nyj−iy_j⁰] =P_2n⁰(zj).

¤

As for an explicit representation of the interpolating trigonometric polynomial Hn(θ)satisfying (3.5), by virtue of uniqueness we can write for anyk∈ {1, . . . , n+ 1},

Hn(θ) =H_n^(k)(θ) =t^(k)_k (θ)yk+

n+1X

j=1,j6=k

h

t^(k)_j (θ)yj+s^(k)_j (θ)y⁰_ji

(3.7) wheret^(k)_j (θ)ands^(k)_j (θ)are trigonometric polynomials inTn, such that

t^(k)_j (θr) =δj,r 16j, r6n+ 1

³ t^(k)_j ´⁰

(θ_r) = 0 16j, r6n+ 1, r6=k s^(k)_j (θr) = 0 16r6n+ 1, j6=k

³ s^(k)_j

´⁰

(θr) =δj,r 16j, r6n+ 1, r6=k, j6=k.

(3.8)

Define now W_n(θ) = Q_n+1

j=1sin³

θ−θj

2

´

. If we proceed as in the previous case, after some elementary calculations we deduce the following expressions for such trigonometric polynomials for16j6n+ 1, j6=k:

s^(k)_j (θ) = W_n²(θ) sin

³θj−θk

2

´

2 sin

³θ−θj

2

´

sin¡_θ−θ

k

2

¢[W_n⁰(θj)]² ∈ Tn, (3.9)

t^(k)_j (θ) = ^Wn2(θ)

sin²_θ−

θj 2

sin

θ−θk 2

[^2Wn⁰(θj)]²×

× h

sin

³θj−θk

2

´ + cos

³θj−θk

2

´ sin

³θ−θj

2

´i

∈ Tn

(3.10) and

t^(k)_k (θ) =

"

Wn(θ) 2W_n⁰(θk) sin¡_θ−θ

k

2

¢

#₂

∈ Tn. (3.11)

In the rest of the section we shall be concerned with certain interpolation prob- lems using an even number of nodes, say 2n, in subspaces T˜n ofTn of dimension 2n. For instance,T˜n=Tn\span{cosnθ} orT˜n=Tn\span{sinnθ}. In this respect, it should be recalled that a system of continuous functions{f0, . . . , fm}on an in- terval[a, b]represents a Haar system on [a, b]if and only if for any k,16k6m, {f0, . . . , fk} is a Chebyshev system on[a, b]. Clearly,

{1,cosθ,sinθ, . . . ,cosnθ,sinnθ},

can not be a Haar system on[−π, π](check simply that{1,cosθ}is not a Chebyshev system). Hence, we can not initially assume that given2nnodes{θj}²ⁿ_j=1on(−π, π]

there existsTn∈ Tn\span{cosnθ}or inTn\span{sinnθ}such thatTn(θj) =yj for allj= 1, . . . ,2n. However, we can prove the following

Theorem 3.3. Let{θj}²ⁿ_j=1⊂(−π, π]be2ndistinct nodes, let{yj}²ⁿ_j=1be arbitrary real numbers, and consider the interpolation problem:

T˜n(θj) =yj, j= 1, . . . ,2n. (3.12) Then the following hold:

1. If P_2n

j=1θj 6=kπ for all k∈ Z, then there is a unique solution of (3.12) in Tn\span{cosnθ} and a unique solution of (3.12) inTn\span{sinnθ}.

2. IfP_2n

j=1θj=kπfor an odd integerk, then there is a unique solution of (3.12) in Tn\span{cosnθ}.

3. If P_2n

j=1θj = kπ for an even integer k, then there is a unique solution of (3.12) in Tn\span{sinnθ}.

Proof. Assume first that we are trying to findT˜n(θ)∈ Tn\span{sinnθ}satisfying (3.12). Thus, we can write:

T˜n(θ) =a0+

n−1X

j=1

(ajcosjθ+bjsinjθ) +ancosnθ=Ln(e^iθ)∈Λ−n,n

withLn(z) =P_n

j=−ncjz^j, where cj= aj−ibj

2 , c−j=cj, 16j6n−1, c0=a0.

Thus,c−j =cj for all 06j 6n. Setting as usual zj =e^iθj for all j = 1, . . . ,2n, (zj6=zk ifj6=k), (3.12) becomes

T˜n(θj) =Ln(e^iθj) =Ln(zj) =yj, j= 1, . . . ,2n giving rise to the linear system

n−1X

k=−(n−1)

ckz_j^k+cn(z_jⁿ+z_j⁻ⁿ) =yj, j= 1, . . . ,2n. (3.13) Now, the system (3.13) has a unique solution if and only if∆n 6= 0, where

∆n =

¯¯

¯¯

¯¯

¯¯

¯¯

¯¯

¯¯

¯¯

z⁻⁽ⁿ⁻¹⁾₁ z₁⁻⁽ⁿ⁻²⁾ · · · 1 · · · z₁ⁿ⁻¹ ¡

zⁿ₁ +z⁻ⁿ₁ ¢ z⁻⁽ⁿ⁻¹⁾₂ z₂⁻⁽ⁿ⁻²⁾ · · · 1 · · · z₂ⁿ⁻¹ ¡

zⁿ₂ +z⁻ⁿ₂ ¢

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

z⁻⁽ⁿ⁻¹⁾_2n z_2n⁻⁽ⁿ⁻²⁾ · · · 1 · · · z_2nⁿ⁻¹ ¡

z_2nⁿ +z_2n⁻ⁿ¢

¯¯

¯¯

¯¯

¯¯

¯¯

¯¯

¯¯

¯¯ .

By introducing the Vandermonde determinant associated withz1, . . . , z2n, i.e.,

γn=

¯¯

¯¯

¯¯

¯¯

¯

1 z1 · · · z²ⁿ⁻¹₁ 1 z2 · · · z²ⁿ⁻¹₂

... ... ... 1 z_2n · · · z²ⁿ⁻¹_2n

¯¯

¯¯

¯¯

¯¯

¯

it can be easily checked that

∆n= (z1· · ·z2n)ⁿ⁻¹(1−z1· · ·z2n)γn. (3.14) On the other hand, if we consider our interpolation problem in

T˜n(θ) ∈ Tn\span{cosnθ}, the associated determinant ∆˜n of the corresponding system satisfies

∆˜n= (z1· · ·z2n)ⁿ⁻¹(1 +z1· · ·z2n)γn. (3.15) Since zj = e^iθj, then z1· · ·z2n = e^iλn with λn = P_2n

j=1θj. If λn 6= kπ for any integer k, then clearly z1· · ·z2n 6= ±1 and from (3.14) and (3.15), both determinants ∆n and ∆˜n are nonzero since γn 6= 0, which means that the in- terpolation problem (3.12) has a unique solution both in Tn\span{sinnθ} and Tn\span{cosnθ}. Next, assume that λn = kπ for some integer k. Thus, if k is even, then e^iλn = 1 and (3.15) is different from zero, whereas if k is odd, then e^iλn = −1 and (3.14) does not vanish. Thus, for instance, if ∆n 6= 0, we have found a uniqueLn ∈Λ−n,n, Ln(z) =P_n

j=−ncjz^j such thatc−n =cn and satisfy- ingLn(zj) =yj forj = 1, . . . ,2n. Therefore, T˜n(θ) =Ln(e^iθ)∈ Tn\span{sinnθ}

andT˜n(θj) =yj for j= 1, . . . ,2n. To check that T˜n(θ)is actually a real trigono-

metric polynomial we proceed as in Theorem 3.1. ¤

Next, a Lagrange-type representation for the trigonometric polynomial T˜n(θ) satisfying the conditions of Theorem 3.3 will be given. Indeed, set

ηn= 1 2

X2n

j=1

θj= 1 2λn

and assume that ηn 6= kπ for any integer k so that ∆n 6= 0. Thus, T˜n(θ) ∈ Tn\span{sinnθ}and by virtue of uniqueness, one hasT˜n(θ) =P_2n

j=1t˜j(θ)yj where

˜tj ∈ Tn\span{sinnθ}and ˜tj(θk) =δj,k for16j, k62n. Fix j ∈ {1, . . . ,2n} and define αj = P_2n

k=1,k6=jθj. Now, we can write s˜j(θ) = ^˜lj_e^(einθ^iθ) where ˜lj(z) ∈ Π2n

such that ˜lj(zk) = zⁿ_jδj,k where, as usual, zk = e^iθk for k = 1, . . . ,2n. Since

˜tj ∈ Tn\span{sinnθ}, the leading coefficient of˜lj(z)must coincide with˜lj(0), and one has ˜lj(z) = cj(z−wj)Q_2n

k=1,k6=j(z−zj) = cjz²ⁿ+· · ·+ ˜lj(0). But ˜lj(0) = cjwj

Q_2n

j=1,j6=kzj, hence

wj = 1

Q_2n

j=1,j6=kzj

= Y2n

j=1

zj =e^{−P2nj=1,j6=kθj} =e^−iαj.

Therefore, by (2.5) it follows that

˜

sj(θ) = ˜cjsin

µθ+αj

2

¶ Y2n

j=1,j6=k

sin

µθ−θj

2

¶

where˜cj is to be determined such that˜sj(θj) = 1. Setting Wn(θ) =

Y2n

j=1

sin

µθ−θj

2

¶

∈ Tn,

we have

˜

sj(θ) = ˜cjsin

µθ+αj

2

¶ Wn(θ) sin³

θ−θ_j 2

´.

Now,

1 = lim

θ→θj

˜ cjsin

µθ+αj

2

¶ Wn(θ) sin

³θ−θj

2

´ = 2˜cjsin

µθj+αj

2

¶

W_n⁰(θj).

Observe that ¹₂(θj +αj) = ¹₂P_2n

j=1θj = ηn 6= kπ for any integer k, so that sin

³θj+αj

2

´

= sinηn6= 0 and hence

˜

sj(θ) = 1

2W_n⁰(θj) sinηn sin

µθ+αj

2

¶ Wn(θ) sin

³θ−θj

2

´, j= 1, . . . ,2n. (3.16)

When dealing with the interpolantT˜n(θ)∈ Tn\span{cosnθ}it can be easily verified that the fundamental Lagrange-type trigonometric polynomials˜sj(θ)are now given by

˜

sj(θ) = 1

2W_n⁰(θj) cosηn cos

µθ+αj

2

¶ Wn(θ) sin

³θ−θj

2

´, j= 1, . . . ,2n. (3.17)

withαj andηn as previously given.

4. Bi-orthogonal systems

Let ω(θ) be a weight function on (−π, π], i.e., ω(θ) > 0 on (−π, π] and 0 <

R_π

−πω(θ)dθ < ∞. The main aim of this section is briefly collecting some results by Szegő (see [14]) concerning properties of an orthogonal basis for the spaceT of real trigonometric polynomials with respect to the inner product onT induced by ω(θ), namely,

hf, giω= Z _π

−π

f(θ)g(θ)ω(θ)dθ, ∀f, g∈ T (4.1)

As indicated in [14], we might consider an arbitrary measure dµ(θ) on the unit circle; in what follows, however we restrict ourselves for the sake of simplicity, to the previously defined case, i.e. to the case whenµ(θ)is absolutely continuous.

Furthermore, when only real functions are considered, complex conjugation in (4.1) can be omitted. For this purpose, let us first consider the basis of Tn given by {1,cosθ,sinθ, . . . ,cosnθ,sinnθ}which is clearly orthogonal forω(θ)≡1on[−π, π]

and let us see how this property can be extended to an arbitrary weight function ω(θ). Certainly, this can be done by orthogonalizing the elementary functions

1,cosθ,sinθ, . . . ,cosnθ,sinnθ

arranged in a linear order, according to Gram-Schmidt process. Thus, a set {f0, f1, g1, . . . , fn, gn}

of trigonometric polynomials is generated such thatf₀is a nonzero constant, f₁∈span{1,cosθ}, g₁∈span{1,cosθ,sinθ}, f₂∈span{1,cosθ,sinθ,cos 2θ}

g2∈span{1,cosθ,sinθ,cos 2θ,sin 2θ} . . . fn∈ Tn\span{sinnθ}, gn ∈ Tn

and it holds that

hfj, fkiω=κjδj,k , κj >0 hgj, gkiω=κ_j⁰δj,k , κ_j⁰ >0 hfj, gkiω= 0, j= 0,1, . . . , n , k= 1, . . . , n.

(4.2) When the process is repeated for eachn∈N, thenf0∪ {fk, gk}^∞_k=1 represents an orthogonal basis forT with respect to ω(θ). Now, if we set

f0=a0,06= 0 fj =aj,0+P_j

k=1(aj,kcoskθ+bj,ksinkθ) gj =cj,0+P_j

k=1(cj,kcoskθ+dj,ksinkθ)

(4.3)

then, because of the linear independence it clearly follows that

¯¯

¯¯ an,n bn,n

cn,n dn,n

¯¯

¯¯6= 0, n>1.

Conversely, we also have (see [14])

Theorem 4.1. Letf0∪ {fk, gk}^∞_k=1 be a system of trigonometric polynomials such thatf₀(θ)≡c6= 0 and forn>1:

fn(θ) =an,0+P_n

k=1(an,kcoskθ+bn,ksinkθ), gn(θ) =cn,0+P_n

k=1(cn,kcoskθ+dn,ksinkθ). Assume that forn>1, ¯

¯¯

¯ an,n bn,n

cn,n dn,n

¯¯

¯¯6= 0.

Then,f0∪ {fk, gk}^∞_k=1 is a basis forT.