Hermite Interpolation on the Unit Circle Considering up to the Second Derivative

Abstract

The paper is devoted to study the Hermite interpolation problem on the unit circle. The interpolation conditions prefix the values of the polynomial and its first two derivatives at the nodal points and the nodal system is constituted by complex numbers equally spaced on the unit circle. We solve the problem in the space of Laurent polynomials by giving two different expressions for the interpolation polynomial. The first one is given in terms of the natural basis of Laurent polynomials and the remarkable fact is that the coefficients can be computed in an easy and efficient way by means of the Fast Fourier Transform (FFT). The second expression is a barycentric formula, which is very suitable for computational purposes.

1. Introduction

One of the pioneering papers concerning Hermite interpolation on the unit circle is [1]. There a Fejér’s type theorem is proved (see [2, 3]), for nodal systems constituted by the 𝑛 roots of a complex number with modulus one. The main result asserts that the Hermite-Fejér interpolants uniformly converge for continuous functions on the unit circle. Some improvements to this result, considering nonvanishing derivatives and more smooth functions, were given in [4]. More recently, in [5], the order of convergence of Hermite-Fejér interpolants for analytic functions on a disk and on an annulus containing the unit circle was obtained.

The classical Hermite interpolation on the circle with nodal points equally spaced was studied in [6]. There it was constructed an orthogonal basis for the space of polynomials in order to obtain the expression of the interpolation polynomials. The coefficients of the interpolation polynomials in this basis can be computed by using the FFT. In [7], the same problem was studied and the corresponding expressions for the Laurent polynomials of interpolation were obtained in a more simple way. Another basis was constructed and again the coefficients can be computed by using the FFT. From these formulas, suitable expressions for the fundamental polynomials were obtained and the barycentric formulas for Hermite interpolation on the unit circle were deduced for the first time. The barycentric formulas were known for Hermite interpolation on the bounded interval (see [8]), but [7] was a new contribution on the circle.

A study about Hermite interpolation on two disjoint sets of nodes on the unit circle has been developed in [9] and problems considering more than one derivative were also considered. Indeed, lacunary Hermite interpolation problems have been also studied on some nonuniformly distributed nodes on the unit circle (see [10]).

In the present paper we study generalized Hermite interpolation problems on the unit circle considering nodal points equally spaced and using the values for the first two derivatives. First we obtain suitable basis for subspaces of the space of Laurent polynomials by considering appropriate interpolation conditions. This enables us to express the interpolation polynomials in such a way that the coefficients can be computed by using the FFT.

In the second part of the paper we deduce the barycentric formulas which constitute a new contribution of the paper. Like in the Lagrange interpolation (see [11]), the barycentric expressions are very useful for doing evaluations and calculus due to their stability (see [12]).

2. Laurent Hermite Interpolation Polynomials

We study the generalized Hermite interpolation problem on the unit circle 𝕋∶={𝑧∶|𝑧|=1} considering the first two derivatives. The nodal system {𝛼_𝑗}^𝑛−1_𝑗=0 is constituted by the 𝑛-roots of a complex number 𝜆, with |𝜆|=1; that is, it consists of complex numbers equally spaced on the unit circle. The problem to solve can be posed as follows.

If 𝑝(𝑛) and 𝑞(𝑛) are two nondecreasing sequences of nonnegative integers such that 𝑝(𝑛)+𝑞(𝑛)=3𝑛−1,𝑛=1,2,…, find the unique Laurent polynomial ℋ_{−𝑝(𝑛),𝑞(𝑛)}(𝑧)∈Λ_{−𝑝(𝑛),𝑞(𝑛)}[𝑧]=span{𝑧^𝑘∶−𝑝(𝑛)≤𝑘≤𝑞(𝑛)} satisfying the interpolation conditions ℋ_{−𝑝(𝑛),𝑞(𝑛)}𝛼_𝑗=𝑢_𝑗,ℋ^′_{(𝑛)(𝑛)−𝑝,𝑞}𝛼_𝑗=𝑣_𝑗,ℋ_(𝑛)(𝑛)^′′_−𝑝,𝑞𝛼_𝑗=𝑤_𝑗𝑗=0,⋯,𝑛−1, (1) where {𝑢_𝑗}^𝑛−1_𝑗=0,{𝑣_𝑗}^𝑛−1_𝑗=0 and {𝑤_𝑗}^𝑛−1_𝑗=0 are prefixed complex numbers.

For simplicity and without loss of generality we consider the case when 𝑝(𝑛)=𝑛+𝐸[𝑛/2] and 𝑞(𝑛)=𝑛+𝐸[(𝑛−1)/2] and we denote the polynomial ℋ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}(𝑧) by ℋ(𝑧).

To obtain an expression of the interpolation polynomial, first we introduce some auxiliary polynomials belonging to the Laurent space Λ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}[𝑧] that we denote by ℒ_0,𝑘, ℒ_1,𝑘, and ℒ_2,𝑘, with −𝐸(𝑛/2)⩽𝑘⩽𝐸[(𝑛−1)/2], and which are characterized by the following interpolation conditions: ℒ_0,𝑘𝛼_𝑗=𝛼^𝑘_𝑗,ℒ^′_0,𝑘𝛼_𝑗ℒ=0,_′′0,𝑘𝛼_𝑗=0for𝑗=0,…,𝑛−1, (2)ℒ_1,𝑘𝛼_𝑗=0,ℒ^′_1,𝑘𝛼_𝑗=𝛼_𝑗⁻¹if𝑘=0,𝑘𝛼_𝑗^𝑘−1ℒif𝑘≠0,_′′1,𝑘𝛼_𝑗ℒ=0for𝑗=0,…,𝑛−1,_2,𝑘𝛼_𝑗=0,ℒ^′_2,𝑘𝛼_𝑗ℒ=0,_′′2,𝑘𝛼_𝑗=𝛼_𝑗⁻²𝛼if𝑘=0,_𝑗−1𝑘()𝛼if𝑘=1,𝑘−1_𝑗^𝑘−2if𝑘∉{0,1}, (3)(4) for 𝑗=0,…,𝑛−1.

Proposition 1. For −𝐸(𝑛/2)⩽𝑘⩽𝐸[(𝑛−1)/2], the polynomials ℒ_0,𝑘(𝑧), ℒ_1,𝑘(𝑧), and ℒ_2,𝑘(𝑧) satisfying (2), (3), and (4), respectively, have the following expressions: ℒ_0,𝑘(𝑧)=1𝑛²𝜆2𝑘(𝑘+𝑛)𝑧^𝑘−𝑛+𝑛²−𝑘²𝑧^𝑘+𝑘()𝑧2𝜆𝑘−𝑛^𝑛+𝑘,ℒ_1,𝑘(𝑧)=1𝑛²𝜆(1−𝑛)2𝑧^−𝑛−1+𝑛+1𝑧2𝜆^𝑛1𝑖𝑓𝑘=0,𝑛²−𝜆2𝑘(𝑛+2𝑘−1)𝑧^𝑘−𝑛+𝑘(2𝑘−1)𝑧^𝑘+𝑘()𝑧2𝜆𝑛−2𝑘+1^𝑛+𝑘ℒ𝑖𝑓𝑘≠0,_2,𝑘(𝑧)=1𝑛²𝜆2𝑧^−𝑛1−1+𝑧2𝜆^𝑛1𝑖𝑓𝑘=0,𝑛²𝜆2𝑧^1−𝑛1−𝑧+𝑧2𝜆^𝑛+11𝑖𝑓𝑘=1,𝑛²𝜆2𝑘(𝑘−1)𝑧^𝑘−𝑛+𝑘(1−𝑘)𝑧^𝑘+1𝑘()𝑧2𝜆𝑘−1^𝑛+𝑘𝑖𝑓𝑘∉{0,1}. (5)

Proof. The polynomial that interpolates 𝑧^𝑘 satisfying (2) has the following form: ℒ_0,𝑘(𝑧)=₀𝑎_𝑘𝑧^𝑘+₀𝑏_𝑘𝑧^𝑛+𝑘+₀𝑐_𝑘𝑧^𝑘−𝑛. (6) By applying the first interpolation condition we get that ₀𝑎_𝑘+₀𝑏_𝑘𝜆+₀𝑐_𝑘(1/𝜆)=1 and by using the second interpolation condition we have ₀𝑎_𝑘𝑘+₀𝑏_𝑘(𝑛+𝑘)𝜆+₀𝑐_𝑘(𝑘−𝑛)(1/𝜆)=0. Taking again derivatives and applying the third interpolation condition we obtain ₀𝑎_𝑘𝑘(𝑘−1)+₀𝑏_𝑘(𝑛+𝑘)(𝑛+𝑘−1)𝜆+₀𝑐_𝑘(𝑘−𝑛)(𝑘−𝑛−1)(1/𝜆)=0. Thus we have a linear system in the unknowns ₀𝑎_𝑘,₀𝑏_𝑘, and ₀𝑐_𝑘, which has the following solutions ₀𝑎_𝑘=(1/𝑛²)(𝑛²−𝑘²), ₀𝑏_𝑘=(𝑘/2𝑛²𝜆)(𝑘−𝑛), and ₀𝑐_𝑘=(𝑘𝜆/2𝑛²)(𝑘+𝑛), from which we obtain the expression of ℒ_0,𝑘(𝑧).
The polynomial ℒ_1,𝑘(𝑧) satisfying (3) has the form ℒ_1,𝑘(𝑧)=₁𝑎_𝑘𝑧^𝑘+₁𝑏_𝑘𝑧^𝑛+𝑘+₁𝑐_𝑘𝑧^𝑘−𝑛. (7) By imposing the first interpolation condition we get ₁𝑎_𝑘+₁𝑏_𝑘𝜆+₁𝑐_𝑘(1/𝜆)=0.
If we assume that 𝑘≠0, we take derivatives and use the second interpolation condition then we obtain ₁𝑎_𝑘𝑘+₁𝑏_𝑘(𝑛+𝑘)𝜆+₁𝑐_𝑘(𝑘−𝑛)(1/𝜆)=𝑘. By taking again derivatives and using the third interpolation condition we have ₁𝑎_𝑘𝑘(𝑘−1)+₁𝑏_𝑘(𝑛+𝑘)(𝑛+𝑘−1)𝜆+₁𝑐_𝑘(𝑘−𝑛)(𝑘−𝑛−1)(1/𝜆)=0. Hence we have a system with unknowns ₁𝑎_𝑘, ₁𝑏_𝑘, and ₁𝑐_𝑘, which has the following solutions: ₁𝑎_𝑘=(𝑘/𝑛²)(2𝑘−1),₁𝑏_𝑘=(𝑘/2𝑛²𝜆)(𝑛−2𝑘+1), and ₁𝑐_𝑘=−(𝑘𝜆/2𝑛²)(𝑛+2𝑘−1).
If 𝑘=0 then ℒ^′_1,0(𝑧)=₁𝑏₀𝑛𝑧^𝑛−1−₁𝑐₀𝑛𝑧^−𝑛−1. Thus if we evaluate at 𝛼_𝑗 and apply the second interpolation condition we get ₁𝑏₀𝑛𝜆−₁𝑐₀𝑛(1/𝜆)=1. By taking again derivatives and using the third interpolation condition we have ₁𝑏₀𝑛(𝑛−1)𝜆+₁𝑐₀𝑛(𝑛+1)(1/𝜆)=0. Then, if we solve the corresponding system we have that ₁𝑎₀=−1/𝑛²,₁𝑏₀=(1/2𝑛²𝜆)(𝑛+1) and ₁𝑐₀=(𝜆/2𝑛²)(1−𝑛). Hence we obtain the expression for ℒ_1,𝑘(𝑧) given in our statement.

Finally, the polynomial ℒ_2,𝑘(𝑧) satisfying (4) can be written as ℒ_2,𝑘(𝑧)=₂𝑎_𝑘𝑧^𝑘+₂𝑏_𝑘𝑧^𝑛+𝑘+₂𝑐_𝑘𝑧^𝑘−𝑛. (8) By applying the first interpolation condition we obtain ₂𝑎_𝑘+₂𝑏_𝑘𝜆+₂𝑐_𝑘(1/𝜆)=0 and if we use the second interpolation condition we get ₂𝑎_𝑘𝑘+₂𝑏_𝑘(𝑛+𝑘)𝜆+₂𝑐_𝑘(𝑘−𝑛)(1/𝜆)=0.

Now if we assume that 𝑘∉{0,1} and use the third interpolation condition we obtain ₂𝑎_𝑘𝑘(𝑘−1)+₂𝑏_𝑘(𝑛+𝑘)(𝑛+𝑘−1)𝜆+₂𝑐_𝑘(𝑘−𝑛)(𝑘−𝑛−1)(1/𝜆)=𝑘(𝑘−1). Then, if we solve the corresponding system we have ₂𝑎_𝑘=(𝑘/𝑛²)(1−𝑘), ₂𝑏_𝑘=(𝑘/2𝑛²𝜆)(𝑘−1), and ₂𝑐_𝑘=(𝜆𝑘/2𝑛²)(𝑘−1).

If 𝑘=0 and we use the third interpolation condition we obtain ₂𝑏₀𝑛(𝑛−1)𝜆+₂𝑐₀𝑛(𝑛+1)(1/𝜆)=1. By solving the system we have the following solution: ₂𝑎₀=−1/𝑛²,₂𝑏₀=1/2𝑛²𝜆, and ₂𝑐₀=𝜆/2𝑛².

Finally, if 𝑘=1 then ₂𝑏₁𝑛(𝑛+1)𝜆+₂𝑐₁𝑛(𝑛−1)(1/𝜆)=1. By solving the corresponding system in the unknowns ₂𝑎₁,₂𝑏₁, and ₂𝑐₁ we obtain that ₂𝑎₁=−1/𝑛²,₂𝑏₁=1/2𝑛²𝜆, and ₂𝑐₁=𝜆/2𝑛².

Next we prove that these auxiliary polynomials constitute a suitable basis of the space Λ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}[𝑧].

Proposition 2. The system ℒ_0,𝑘(𝑧)^{𝐸[(𝑛−1)/2]}_{𝑘=−𝐸[𝑛/2]}∪ℒ_1,𝑘(𝑧)^{𝐸[(𝑛−1)/2]}_{𝑘=−𝐸[𝑛/2]}∪ℒ_2,𝑘(𝑧)^{𝐸[(𝑛−1)/2]}_{𝑘=−𝐸[𝑛/2]} (9) is an orthogonal basis of the Laurent space Λ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}[𝑧] with respect to the inner product defined by ⟨⟩=1𝑃,𝑄𝑛^𝑛−1_𝑗=0𝑃𝛼_𝑗𝑄𝛼_𝑗+𝑃^′𝛼_𝑗𝑄^′𝛼_𝑗+𝑃^′′𝛼_𝑗𝑄^′′𝛼_𝑗 (10) for 𝑃,𝑄∈Λ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}[𝑧].
Moreover, it holds ℒ_0,𝑘(𝑧)²=1,ℒ_1,𝑘(𝑧)²=𝑘1,𝑖𝑓𝑘=0,₂ℒ,𝑖𝑓𝑘≠0,_2,𝑘(𝑧)²={},𝑘1,𝑖𝑓𝑘∈0,1₂(𝑘−1)²,𝑖𝑓𝑘∉{0,1}. (11)

Proof. It is easy to check that ℒ_0,𝑘(𝑧),ℒ_0,𝑙(𝑧=1)𝑛^𝑛−1_𝑗=0ℒ_0,𝑘𝛼_𝑗ℒ_0,𝑙𝛼_𝑗+ℒ^′_0,𝑘𝛼_𝑗ℒ^′_0,𝑙𝛼_𝑗+ℒ^′′_0,𝑘𝛼_𝑗ℒ^′′_0,𝑙𝛼_𝑗=1𝑛^𝑛−1_𝑗=0𝛼^𝑘_𝑗𝛼^𝑙_𝑗=ℒ1,if𝑘=𝑙,0,otherwise,_0,𝑘(𝑧),ℒ_1,𝑙(𝑧=1)𝑛^𝑛−1_𝑗=0ℒ_0,𝑘𝛼_𝑗ℒ_1,𝑙𝛼_𝑗+ℒ^′_0,𝑘𝛼_𝑗ℒ^′_1,𝑙𝛼_𝑗+ℒ^′′_0,𝑘𝛼_𝑗ℒ_1,𝑙^′′𝛼_𝑗ℒ=0,_0,𝑘(𝑧),ℒ_2,𝑙(𝑧=1)𝑛^𝑛−1_𝑗=0ℒ_0,𝑘𝛼_𝑗ℒ_2,𝑙𝛼_𝑗+ℒ^′_0,𝑘𝛼_𝑗ℒ^′_2,𝑙𝛼_𝑗+ℒ^′′_0,𝑘𝛼_𝑗ℒ^′′_2,𝑙𝛼_𝑗ℒ=0,_1,𝑘(𝑧),ℒ_1,𝑙(𝑧=1)𝑛^𝑛−1_𝑗=0ℒ_1,𝑘𝛼_𝑗ℒ_1,𝑙𝛼_𝑗+ℒ_1,𝑘𝛼_𝑗ℒ^′_1,𝑙𝛼_𝑗+ℒ^′′_1,𝑘𝛼_𝑗ℒ^′′_1,𝑙𝛼_𝑗=𝑘1,if𝑘=𝑙=0,₂ℒ,if𝑘=𝑙≠0,0,otherwise,_1,𝑘(𝑧),ℒ_2,𝑙(𝑧=1)𝑛^𝑛−1_𝑗=0ℒ_1,𝑘𝛼_𝑗ℒ_2,𝑙𝛼_𝑗+ℒ^′_1,𝑘𝛼_𝑗ℒ^′_2,𝑙𝛼_𝑗+ℒ^′′_1,𝑘𝛼_𝑗ℒ^′′_2,𝑙𝛼_𝑗ℒ=0,_2,𝑘(𝑧),ℒ_2,𝑙(𝑧=1)𝑛^𝑛−1_𝑗=0ℒ_2,𝑘𝛼_𝑗ℒ_2,𝑙𝛼_𝑗+ℒ^′_2,𝑘𝛼_𝑗ℒ^′_2,𝑙𝛼_𝑗+ℒ^′′_1,𝑘𝛼_𝑗ℒ^′′_2,𝑙𝛼_𝑗=1𝑛^𝑛−1_𝑗=0𝑘(𝑘−1)𝛼_𝑗^𝑘−2𝑙(𝑙−1)𝛼_𝑗^𝑙−2=1𝑛^𝑛−1_𝑗=0𝑘(𝑘−1)𝑙(𝑙−1)𝛼_𝑗^𝑘−𝑙={},𝑘1,if𝑘=𝑙,with𝑙∈0,1₂(𝑘−1)²,if𝑘=𝑙,with𝑙∉{0,1},0,otherwise. (12)

Now we are in conditions to obtain the expression of the polynomial ℋ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}(𝑧) satisfying (1) that we denote for simplicity ℋ(𝑧). It is clear that it can be written as ℋ=ℋ₀+ℋ₁+ℋ₂, (13) where ℋ₀,ℋ₁, and ℋ₂ are the solutions of the following problems: ℋ₀∈Λ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}[𝑧] and it satisfies ℋ₀(𝛼_𝑗)=𝑢_𝑗, ℋ^′₀(𝛼_𝑗)=0, and ℋ₀^′′(𝛼_𝑗)=0 for 𝑗=0,…,𝑛−1. ℋ₁∈Λ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}[𝑧] and it satisfies ℋ₁(𝛼_𝑗)=0, ℋ^′₁(𝛼_𝑗)=𝑣_𝑗, and ℋ₁^′′(𝛼_𝑗)=0for𝑗=0,…,𝑛−1.  ℋ₂∈Λ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}[𝑧] and it satisfies ℋ₂(𝛼_𝑗)=0, ℋ^′₂(𝛼_𝑗)=0, and ℋ₂^′′(𝛼_𝑗)=𝑤_𝑗for𝑗=0,…,𝑛−1.

Proposition 3. The polynomials ℋ₀, ℋ₁, ℋ₂, and ℋ defined before have the following expressions: (i)ℋ₀(𝑧)=1𝑛^{2𝐸[(𝑛−1)/2]}_{[]𝑘=−𝐸𝑛/2}1𝑛^𝑛−1_𝑗=0𝑢_𝑗𝛼_𝑗^𝑘×𝜆2𝑘(𝑘+𝑛)𝑧^𝑘−𝑛+𝑛²−𝑘²𝑧^𝑘+𝑘()𝑧2𝜆𝑘−𝑛^𝑛+𝑘, (14)(ii)ℋ₁(𝑧)=1𝑛^{2𝐸[(𝑛−1)/2]}_{𝑘=−𝐸[𝑛/2]}1𝑛^𝑛−1_𝑗=0𝑣_𝑗𝛼_𝑗^𝑘−1×−𝜆2(𝑛+2𝑘−1)𝑧^𝑘−𝑛+(2𝑘−1)𝑧^𝑘+1()𝑧2𝜆𝑛−2𝑘+1^𝑛+𝑘, (15)(iii)ℋ₂(𝑧)=1𝑛^{2𝐸[(𝑛−1)/2]}_{[]𝑘=−𝐸𝑛/2}1𝑛^𝑛−1_𝑗=0𝑤_𝑗𝛼_𝑗^𝑘−2×𝜆2𝑧^𝑘−𝑛−𝑧^𝑘+1𝑧2𝜆^𝑛+𝑘, (16)(iv)𝐻(𝑧)=1𝑛^{3𝐸[(𝑛−1)/2]}_{[]𝑘=−𝐸𝑛/2}𝜆2^𝑛−1_𝑗=0𝑢_𝑗𝛼_𝑗^𝑘𝑘()−𝑘+𝑛_{𝑛−1𝑗=0}𝑣_𝑗𝛼_𝑗^𝑘−1()+𝑛+2𝑘−1_{𝑛−1𝑗=0}𝑤_𝑗𝛼_𝑗^𝑘−2𝑧^𝑘−𝑛+^𝑛−1_𝑗=0𝑢_𝑗𝛼_𝑗^𝑘𝑛²−𝑘²+^𝑛−1_𝑗=0𝑣_𝑗𝛼_𝑗^𝑘−1×(2𝑘−1)−^𝑛−1_𝑗=0𝑤_𝑗𝛼_𝑗^𝑘−2𝑧^𝑘+12𝜆^𝑛−1_𝑗=0𝑢_𝑗𝛼_𝑗^𝑘𝑘()+𝑘−𝑛_{𝑛−1𝑗=0}𝑣_𝑗𝛼_𝑗^𝑘−1()+𝑛−2𝑘+1_{𝑛−1𝑗=0}𝑤_𝑗𝛼_𝑗^𝑘−2𝑧^𝑛+𝑘. (17)

Proof. (i) It is clear that ℋ₀(𝑧) can be written as ℋ₀(𝑧)=^{𝐸[(𝑛−1)/2]}_{𝑘=−𝐸[𝑛/2]}𝑎_𝑘ℒ_0,𝑘(𝑧). (18) For computing the coefficients we take into account that ⟨ℋ₀(𝑧),ℒ_0,𝑘(𝑧)⟩=𝑎_𝑘‖ℒ_0,𝑘(𝑧)‖²=𝑎_𝑘. On the other hand, it holds that ⟨ℋ₀(𝑧),ℒ_0,𝑘(𝑧)⟩=(1/𝑛)∑^𝑛−1_𝑗=0𝑢_𝑗𝛼_𝑗^𝑘 and therefore 𝑎_𝑘=(1/𝑛)∑^𝑛−1_𝑗=0𝑢_𝑗𝛼_𝑗^𝑘. Hence, using the expression of ℒ_0,𝑘(𝑧), we have (i).
(ii) Proceeding in the same way, ℋ₁(𝑧) can be written as ℋ₁(𝑧)=^{𝐸[(𝑛−1)/2]}_{𝑘=−𝐸[𝑛/2]}𝑏_𝑘ℒ_1,𝑘(𝑧). (19) On one hand, if we calculate the inner product we have ℋ₁(𝑧),ℒ_1,𝑘(𝑧)=𝑏_𝑘ℒ_1,𝑘(𝑧)²=𝑏₀𝑏,if𝑘=0,_𝑘𝑘²,otherwise. (20) On the other hand, it holds that 1𝑛^𝑛−1_𝑗=0ℋ^′₁𝛼_𝑗ℒ^′_1,𝑘𝛼_𝑗=1𝑛^𝑛−1_𝑗∑₌₀𝑣_𝑗𝛼_𝑗⁻¹1,if𝑘=0,𝑛^𝑛−1_𝑗∑₌₀𝑘𝑣_𝑗𝛼_𝑗^𝑘−1,otherwise (21) and therefore 𝑏_𝑘=1𝑛^𝑛−1_𝑗∑₌₀𝑣_𝑗𝛼_𝑗⁻¹1,if𝑘=0,𝑛𝑘^𝑛−1_𝑗∑₌₀𝑣_𝑗𝛼_𝑗^𝑘−1,otherwise. (22) Hence, taking into account the expression of ℒ_1,𝑘(𝑧), we obtain (ii).
(iii) We write ℋ₂(𝑧) as follows ℋ₂(𝑧)=^{𝐸[(𝑛−1)/2]}_{𝑘=−𝐸[𝑛/2]}𝑐_𝑘ℒ_2,𝑘(𝑧). (23) By computing the inner product, we obtain ℋ₂(𝑧),ℒ_2,𝑘(𝑧)=𝑐_𝑘ℒ_2,𝑘(𝑧)²=𝑐_𝑘{},𝑐,if𝑘∈0,1_𝑘𝑘²(𝑘−1)²1,otherwise,𝑛^𝑛−1_𝑗=0ℋ₂^′′𝛼_𝑗ℒ^′′_2,𝑘𝛼_𝑗=1𝑛^𝑛−1_𝑗∑₌₀𝑤_𝑗𝛼_𝑗⁻²1,if𝑘=0,𝑛^𝑛−1_𝑗∑₌₀𝑤_𝑗𝛼_𝑗⁻¹1,if𝑘=1,𝑛^𝑛−1_𝑗∑₌₀𝑤_𝑗𝑘(𝑘−1)𝛼_𝑗^𝑘−2,otherwise. (24) Therefore 𝑐_𝑘=1𝑛^𝑛−1_𝑗∑₌₀𝑤_𝑗𝛼_𝑗⁻²1,if𝑘=0,𝑛^𝑛−1_𝑗∑₌₀𝑤_𝑗𝛼_𝑗⁻¹1,if𝑘=1,𝑘(𝑘−1)𝑛^𝑛−1_𝑗∑₌₀𝑤_𝑗𝛼_𝑗^𝑘−2,otherwise (25) and we obtain the expression of ℒ_2,𝑘(𝑧) given in (iii).
(iv) It is straightforward from ℋ(𝑧)=ℋ₀(𝑧)+ℋ₁(𝑧)+ℋ₂(𝑧). (26)

Next we consider the particular cases in which the nodes are 𝑛 roots of 1 and −1, obtaining the following results.

Corollary 4. (i) If {𝑦_𝑗}^2𝑛−1_𝑗=0 are the 2𝑛 roots of −1, then the Hermite interpolation polynomial satisfying (1) is ℋ_{−3𝑛,3𝑛−1}(𝑧)=18𝑛^3𝑛−1_𝑘=−𝑛−12^2𝑛−1_𝑗=0𝑢_𝑗𝑦_𝑗^𝑘𝑘()−𝑘+2𝑛_{2𝑛−1𝑗=0}𝑣_𝑗𝑦_𝑗^𝑘−1()+2𝑛+2𝑘−1_{2𝑛−1𝑗=0}𝑤_𝑗𝑦_𝑗^𝑘−2𝑧^𝑘−2𝑛+^2𝑛−1_𝑗=0𝑢_𝑗𝑦_𝑗^𝑘4𝑛²−𝑘²+^2𝑛−1_𝑗=0𝑣_𝑗𝑦_𝑗^𝑘−1()−2𝑘−1_{2𝑛−1𝑗=0}𝑤_𝑗𝑦_𝑗^𝑘−2𝑧^𝑘−12^2𝑛−1_𝑗=0𝑢_𝑗𝑦_𝑗^𝑘𝑘()+𝑘−2𝑛_{2𝑛−1𝑗=0}𝑣_𝑗𝑦_𝑗^𝑘−1()+2𝑛−2𝑘+1_{2𝑛−1𝑗=0}𝑤_𝑗𝑦_𝑗^𝑘−2𝑧^2𝑛+𝑘. (27)
(ii) If {𝑦_𝑗}^2𝑛_𝑗=0 are the 2𝑛+1 roots of −1, then the Hermite interpolation polynomial satisfying (1) is ℋ_{−3𝑛−1,3𝑛+1}(𝑧)=1(2𝑛+1)³×^𝑛_𝑘=−𝑛−12^2𝑛_𝑗=0𝑢_𝑗𝑦_𝑗^𝑘𝑘()−𝑘+2𝑛+1_2𝑛𝑗=0𝑣_𝑗𝑦_𝑗^𝑘−1()+2𝑛+2𝑘_2𝑛𝑗=0𝑤_𝑗𝑦_𝑗^𝑘−2𝑧^{𝑘−2𝑛−1}+^2𝑛_𝑗=0𝑢_𝑗𝑦_𝑗^𝑘(2𝑛+1)²−𝑘²+^2𝑛_𝑗=0𝑣_𝑗𝑦_𝑗^𝑘−1()−2𝑘−1_2𝑛𝑗=0𝑤_𝑗𝑦_𝑗^𝑘−2𝑧^𝑘+−12^2𝑛_𝑗=0𝑢_𝑗𝑦_𝑗^𝑘𝑘()+𝑘−2𝑛−1_2𝑛𝑗=0𝑣_𝑗𝑦_𝑗^𝑘−1()+2𝑛−2𝑘+2_2𝑛𝑗=0𝑤_𝑗𝑦_𝑗^𝑘−2𝑧^2𝑛+1+𝑘. (28)

Corollary 5. (i) If {𝑧_𝑗}^2𝑛−1_𝑗=0 are the 2𝑛 roots of 1, then the Hermite interpolation polynomial satisfying (1) is ℋ_{−3𝑛,3𝑛−1}(𝑧)=18𝑛³×^𝑛−1_𝑘=−𝑛12^2𝑛−1_𝑗=0𝑢_𝑗𝑧_𝑗^𝑘𝑘()−𝑘+2𝑛_2n−1𝑗=0𝑣_𝑗𝑧_𝑗^𝑘−1()+2𝑛+2𝑘−1_{2𝑛−1𝑗=0}𝑤_𝑗𝑧_𝑗^𝑘−2𝑧^𝑘−2𝑛+^2𝑛−1_𝑗=0𝑢_𝑗𝑧_𝑗^𝑘4𝑛²−𝑘²+^2𝑛−1_𝑗=0𝑣_𝑗𝑧_𝑗^𝑘−1()−2𝑘−1_{2𝑛−1𝑗=0}𝑤_𝑗𝑧_𝑗^𝑘−2𝑧^𝑘+12^2𝑛−1_𝑗=0𝑢_𝑗𝑧_𝑗^𝑘𝑘()+𝑘−2𝑛_{2𝑛−1𝑗=0}𝑣_𝑗𝑧_𝑗^𝑘−1()+2𝑛−2𝑘+1_{2𝑛−1𝑗=0}𝑤_𝑗𝑧_𝑗^𝑘−2𝑧^2𝑛+𝑘. (29)
(ii) If {𝑧_𝑗}^2𝑛_𝑗=0 are the 2𝑛+1 roots of 1, then the Hermite interpolation polynomial satisfying (1) is ℋ_{−3𝑛−1,3𝑛+1}(𝑧)=1(2𝑛+1)³×^𝑛_𝑘=−𝑛12^2𝑛_𝑗=0𝑢_𝑗𝑧_𝑗^𝑘𝑘()−𝑘+2𝑛+1_2𝑛𝑗=0𝑣_𝑗𝑧_𝑗^𝑘−1()+2𝑛+2𝑘_2𝑛𝑗=0𝑤_𝑗𝑧_𝑗^𝑘−2𝑧^{𝑘−2𝑛−1}+^2𝑛_𝑗=0𝑢_𝑗𝑧_𝑗^𝑘(2𝑛+1)²−𝑘²+^2𝑛_𝑗=0𝑣_𝑗𝑧_𝑗^𝑘−1()−2𝑘−1_2𝑛𝑗=0𝑤_𝑗𝑧_𝑗^𝑘−2𝑧^𝑘+12^2𝑛_𝑗=0𝑢_𝑗𝑧_𝑗^𝑘𝑘()+𝑘−2𝑛−1_2𝑛𝑗=0𝑣_𝑗𝑧_𝑗^𝑘−1()+2𝑛−2𝑘+2_2𝑛𝑗=0𝑤_𝑗𝑧_𝑗^𝑘−2𝑧^2𝑛+1+𝑘. (30)

Remark 6. (a)  From (i),(ii), and (iii) in Proposition 3 it is immediate to obtain an expression for the fundamental polynomials of Hermite interpolation.
(b)  Notice that the coefficients of the expressions given before can be computed in an easy and efficient way by using the FFT.

3. Barycentric Expression

In this section, our aim is to obtain a barycentric expression for the interpolation polynomial ℋ_{−𝑛−𝐸[𝑛/2],𝑛+𝐸[(𝑛−1)/2]}(𝑧). We distinguish two cases according to the nodal system having an even or odd number of points.

3.1. Nodal System with an Even Number of Points

First we assume that the nodal system has an even number of points that we denote by 2𝑛 and we try to obtain the expression of ℋ_{−3𝑛,3𝑛−1}(𝑧) that we denote for simplicity, ℋ(𝑧). Since ℋ(𝑧) can be written in terms of the fundamental polynomials of Hermite interpolation first we obtain suitable expressions for these polynomials.

Lemma 7. The polynomials ℰ_𝑗(𝑧)=(𝑧^2𝑛−𝜆)³/𝑧^3𝑛(𝑧−𝛼_𝑗), ℱ_𝑗(𝑧)=(𝑧^2𝑛−𝜆)³/𝑧^3𝑛(𝑧−𝛼_𝑗)², and 𝒢_𝑗(𝑧)=(𝑧^2𝑛−𝜆)³/𝑧^3𝑛(𝑧−𝛼_𝑗)³ for 𝑗=0,…,2𝑛−1 satisfy (i)ℰ_𝑗(𝛼_𝑖)=ℰ^′_𝑗(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖=0,…,2𝑛−1,ℰ_𝑗^′′(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖≠𝑗, and ℰ_𝑗^′′(𝛼_𝑗)=16𝜆𝛼^𝑛_𝑗𝑛³/𝛼³_𝑗.(ii)ℱ_𝑗(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖=0,…,2𝑛−1,ℱ^′_𝑗(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖≠𝑗,  ℱ^′_𝑗(𝛼_𝑗)=8𝜆𝛼^𝑛_𝑗𝑛³/𝛼³_𝑗,ℱ_𝑗^′′(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖≠𝑗, and ℱ_𝑗^′′(𝛼_𝑗)=−(24𝜆𝛼^𝑛_𝑗𝑛³/𝛼⁴_𝑗),(iii)𝒢_𝑗(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖≠𝑗, 𝒢_𝑗(𝛼_𝑗)=8𝜆𝛼^𝑛_𝑗𝑛³/𝛼³_𝑗, 𝒢^′_𝑗(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖≠𝑗, 𝒢^′_𝑗(𝛼_𝑗)=−(12𝜆𝛼^𝑛_𝑗𝑛³/𝛼⁴_𝑗),𝒢_𝑗^′′(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖≠𝑗, and 𝒢_𝑗^′′(𝛼_𝑗)=4𝜆𝛼^𝑛_𝑗𝑛³(2𝑛²+7)/𝛼⁵_𝑗.

Proof. Taking into account that (𝑧^2𝑛−𝜆)³=Π^2𝑛−1_𝑗=0(𝑧−𝛼_𝑗)³, then it is clear that ℰ_𝑗(𝛼_𝑗)=ℰ^′_𝑗(𝛼_𝑗)=0,   for all  𝑖=0,…,2𝑛−1, ℰ_𝑗^′′(𝛼_𝑖)=0,   for all  𝑖≠𝑗,andℰ_𝑗^′′(𝛼_𝑗)≠0. In the same way, it is immediate to see that ℱ_𝑗(𝛼_𝑖)=0,  for all  𝑖=0,…,2𝑛−1,ℱ^′_𝑗(𝛼_𝑖)=0 for 𝑖≠𝑗, ℱ^′_𝑗(𝛼_𝑗)≠0, ℱ_𝑗^′′(𝛼_𝑖)=0 for 𝑖≠𝑗,  and  ℱ_𝑗^′′(𝛼_𝑗)≠0. Furthermore, 𝒢_𝑗(𝛼_𝑖)=0 for 𝑖≠𝑗, 𝒢_𝑗(𝛼_𝑗)≠0  and 𝒢^′_𝑗(𝛼_𝑖)=0 for 𝑖≠𝑗,𝒢^′_𝑗(𝛼_𝑗)≠0 and 𝒢_𝑗^′′(𝛼_𝑖)=0 for 𝑖≠𝑗 and 𝒢_𝑗^′′(𝛼_𝑗)≠0.

For obtaining the exact nonvanishing values we proceed as follows.(i)If we define 𝑒_𝑗(𝑦)=ℰ_𝑗(𝛼_𝑗𝑦), we obtain 𝑒_𝑗(𝑦)=𝜆𝛼^𝑛_𝑗(𝑦^2𝑛−1)³/𝛼_𝑗𝑦^3𝑛(𝑦−1). By taking derivatives and evaluating at 𝑦=1 we get 𝑒_𝑗^′′(1)=16𝜆𝛼^𝑛_𝑗𝑛³/𝛼_𝑗, from which we deduce ℰ_𝑗^′′(𝛼_𝑗)=16𝜆𝛼^𝑛_𝑗𝑛³/𝛼³_𝑗.(ii)In the same way, if we define 𝑓_𝑗(𝑦)=ℱ_𝑗(𝛼_𝑗𝑦) we get that 𝑓_𝑗(𝑦)=𝜆𝛼^𝑛_𝑗(𝑦^2𝑛−1)³/𝛼²_𝑗𝑦^3𝑛(𝑦−1)². By taking derivatives and evaluating at 𝑦=1 we obtain 𝑓^′_𝑗(1)=8𝑛³𝜆𝛼^𝑛_𝑗/𝛼²_𝑗 and 𝑓_𝑗^′′(1)=−(24𝜆𝛼^𝑛_𝑗𝑛³/𝛼²_𝑗), from which we deduce the values of ℱ^′_𝑗(𝛼_𝑗) and ℱ_𝑗^′′(𝛼_𝑗).(iii)Finally, if we define 𝑔_𝑗(𝑦)=𝒢_𝑗(𝛼_𝑗𝑦) we get 𝑔_𝑗(𝑦)=𝜆𝛼^𝑛_𝑗(𝑦^2𝑛−1)³/𝛼³_𝑗𝑦^3𝑛(𝑦−1)³. By evaluating 𝑔_𝑗 and its derivatives at 𝑦=1 we have that 𝑔_𝑗(1)=8𝑛³𝜆𝛼^𝑛_𝑗/𝛼³_𝑗,𝑔^′_𝑗(1)=−(12𝑛³𝜆𝛼^𝑛_𝑗/𝛼³_𝑗), and 𝑔_𝑗^′′(1)=4𝑛³(2𝑛²+7)𝜆𝛼^𝑛_𝑗/𝛼³_𝑗, from which we obtain the values of 𝒢_𝑗(𝛼_𝑗), 𝒢^′_𝑗(𝛼_𝑗), and 𝒢_𝑗^′′(𝛼_𝑗).

Proposition 8. The fundamental polynomials of the Hermite interpolation in the Laurent space Λ_{−3𝑛,3𝑛−1}[𝑧], 𝒜_𝑗(𝑧), ℬ_𝑗(𝑧), and 𝒞_𝑗(𝑧), for 𝑗=0,…,2𝑛−1, characterized by 𝒜_𝑗𝛼_𝑖=𝛿_𝑖,𝑗,𝒜^′_𝑗𝛼_𝑖=0,𝒜_𝑗^′′𝛼_𝑖ℬ=0,∀𝑖=0,…,2𝑛−1,_𝑗𝛼_𝑖=0,ℬ^′_𝑗𝛼_𝑖=𝛿_𝑖,𝑗,ℬ_𝑗^′′𝛼_𝑖𝒞=0,∀𝑖=0,…,2𝑛−1,_𝑗𝛼_𝑖=0,𝒞^′_𝑗𝛼_𝑖=0,𝒞_𝑗^′′𝛼_𝑖=𝛿_𝑖,𝑗,∀𝑖=0,…,2𝑛−1, (31) have the following expressions 𝒜_𝑗(𝑧)=𝛼_𝑗1−𝑛²16𝜆𝛼^𝑛_𝑗𝑛³𝑧^2𝑛−𝜆³𝑧^3𝑛𝑧−𝛼_𝑗+3𝛼²_𝑗16𝜆𝛼^𝑛_𝑗𝑛³𝑧^2𝑛−𝜆³𝑧^3𝑛𝑧−𝛼_𝑗²+𝛼³_𝑗8𝜆𝛼^𝑛_𝑗𝑛³𝑧^2𝑛−𝜆³𝑧^3𝑛𝑧−𝛼_𝑗³,ℬ_𝑗(𝑧)=3𝛼²_𝑗16𝜆𝛼^𝑛_𝑗𝑛³𝑧^2𝑛−𝜆𝑎³𝑧^3𝑛𝑧−𝛼_𝑗+𝛼³_𝑗8𝜆𝛼^𝑛_𝑗𝑛³𝑧^2𝑛−𝜆³𝑧^3𝑛𝑧−𝛼_𝑗²,𝒞_𝑗(𝑧)=𝛼³_𝑗16𝜆𝛼^𝑛_𝑗𝑛³𝑧^2𝑛−𝜆³𝑧^3𝑛𝑧−𝛼_𝑗. (32)

Proof. It is clear that 𝒜_𝑗(𝑧), ℬ_𝑗(𝑧), and 𝒞_𝑗(𝑧) can be written in the following form: 𝒜_𝑗(𝑧)=𝑎_𝑗,0ℰ_𝑗(𝑧)+𝑏_𝑗,0ℱ_𝑗(𝑧)+𝑐_𝑗,0𝒢_𝑗(𝑧),ℬ_𝑗(𝑧)=𝑎_𝑗,1ℰ_𝑗(𝑧)+𝑏_𝑗,1ℱ_𝑗(𝑧),𝒞_𝑗(𝑧)=𝑎_𝑗,2ℰ_𝑗(𝑧), (33) with ℰ_𝑗(𝑧), ℱ_𝑗(𝑧), and 𝒢_𝑗(𝑧) given in Lemma 7.
To compute 𝑎_𝑗,2 take into account that it must be 𝒞_𝑗^′′(𝛼_𝑗)=1. Then applying the preceding lemma we get that 1=𝑎_𝑗,2(16𝜆𝛼^𝑛_𝑗𝑛³/𝛼³_𝑗), from which it follows that 𝑎_𝑗,2=𝛼³_𝑗/16𝜆𝛼^𝑛_𝑗𝑛³.
For computing 𝑎_𝑗,1 and 𝑏_𝑗,1 we use that ℬ^′_𝑗(𝛼_𝑗)=1 and ℬ_𝑗^′′(𝛼_𝑗)=0 and we obtain the following system: 1=𝑎_𝑗,1ℰ^′_𝑗𝛼_𝑗+𝑏_𝑗,1ℱ^′_𝑗𝛼_𝑗,0=𝑎_𝑗,1ℰ_𝑗^′′𝛼_𝑗+𝑏_𝑗,1ℱ_𝑗^′′𝛼_𝑗. (34) By applying Lemma 7 and solving the system we get the result.
Finally, to obtain the coefficients 𝑎_𝑗,0,𝑏_𝑗,0, and 𝑐_𝑗,0 in the expression of 𝒜_𝑗(𝑧), we proceed in the same way. By applying the interpolation conditions we have the system 0=𝑎_𝑗,0ℰ_𝑗^′′𝛼_𝑗+𝑏_𝑗,0ℱ_𝑗^′′𝛼_𝑗+𝑐_𝑗,0𝒢_𝑗^′′𝛼_𝑗,0=𝑏_𝑗,0ℱ^′_𝑗𝛼_𝑗+𝑐_𝑗,0𝒢^′_𝑗𝛼_𝑗,1=𝑐_𝑗,0𝒢_𝑗𝛼_𝑗. (35) Then, by using Lemma 7 and solving the system we conclude our result.

It is straightforward to deduce, from the preceding Proposition 8, the so-called barycentric expression for ℋ(𝑧).

Proposition 9. (i) The polynomial ℋ₀(𝑧)∈Λ_{−3𝑛,3𝑛−1}[𝑧] satisfying the conditions ℋ₀(𝛼_𝑗)=𝑢_𝑗, ℋ^′₀(𝛼_𝑗)=0, and ℋ₀^′′(𝛼_𝑗)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑗=0,…,2𝑛−1, has the following barycentric expression: ℋ₀(𝑧)=^2𝑛−1_𝑗=0𝑢_𝑗𝛼^𝑛_𝑗𝛼³_𝑗𝑧−𝛼_𝑗³+3𝛼²_𝑗2𝑧−𝛼_𝑗²+𝛼_𝑗1−𝑛²2𝑧−𝛼_𝑗×^2𝑛−1_𝑗=01𝛼^𝑛_𝑗𝛼³_𝑗𝑧−𝛼_𝑗³+3𝛼²_𝑗2𝑧−𝛼_𝑗²+𝛼_𝑗1−𝑛²2𝑧−𝛼_𝑗⁻¹. (36)
(ii) The polynomial ℋ₁(𝑧)∈Λ_{−3𝑛,3𝑛−1}[𝑧] satisfying the conditions ℋ₁(𝛼_𝑗)=0, ℋ^′₁(𝛼_𝑗)=𝑣_𝑗, and ℋ₁^′′(𝛼_𝑗)=0, 𝑓𝑜𝑟𝑎𝑙𝑙𝑗=0,…,2𝑛−1, has the following barycentric expression: ℋ₁(𝑧)=^2𝑛−1_𝑗=0𝑣_𝑗𝛼^𝑛_𝑗𝛼³_𝑗𝑧−𝛼_𝑗²+3𝛼²_𝑗2𝑧−𝛼_𝑗×^2𝑛−1_𝑗=01𝛼^𝑛_𝑗𝛼³_𝑗𝑧−𝛼_𝑗³+3𝛼²_𝑗2𝑧−𝛼_𝑗²+𝛼_𝑗1−𝑛²2𝑧−𝛼_𝑗⁻¹. (37)
(iii) The polynomial ℋ₂(𝑧)∈Λ_{−3𝑛,3𝑛−1}[𝑧] satisfying the conditions ℋ₂(𝛼_𝑗)=0, ℋ^′₂(𝛼_𝑗)=0, and ℋ₂^′′(𝛼_𝑗)=𝑤_𝑗, 𝑓𝑜𝑟𝑎𝑙𝑙𝑗=0,…,2𝑛−1, has the following barycentric expression: ℋ₂(𝑧)=^2𝑛−1_𝑗=0𝑤_𝑗𝛼^𝑛_𝑗𝛼³_𝑗2𝑧−𝛼_𝑗×^2𝑛−1_𝑗=01𝛼^𝑛_𝑗𝛼³_𝑗(𝑧−𝛼_𝑗)³+3𝛼²_𝑗2(𝑧−𝛼_𝑗)²+𝛼_𝑗1−𝑛²2𝑧−𝛼_𝑗⁻¹. (38)
(iv) The polynomial ℋ(𝑧)∈Λ_{−3𝑛,3𝑛−1}[𝑧] satisfying (1) has the following barycentric expression: ℋ(𝑧)=^2𝑛−1_𝑗=01𝛼^𝑛_𝑗𝑢_𝑗𝛼³_𝑗𝑧−𝛼_𝑗³+3𝛼²_𝑗2𝑧−𝛼_𝑗²+𝛼_𝑗1−𝑛²2𝑧−𝛼_𝑗+𝑣_𝑗𝛼³_𝑗𝑧−𝛼_𝑗²+3𝛼²_𝑗2𝑧−𝛼_𝑗+𝑤_𝑗𝛼³_𝑗2𝑧−𝛼_𝑗×^2𝑛−1_𝑗=01𝛼^𝑛_𝑗𝛼³_𝑗𝑧−𝛼_𝑗³+3𝛼²_𝑗2𝑧−𝛼_𝑗²+𝛼_𝑗1−𝑛²2𝑧−𝛼_𝑗⁻¹. (39)

Proof. (i) It is immediate if we take into account that ℋ₀(𝑧)=∑^2𝑛−1_𝑗=0𝑢_𝑗𝒜_𝑗(𝑧), with 𝒜_𝑗(𝑧) given in Proposition 8. Thus, if we divide the polynomial by 1=∑^2𝑛−1_𝑗=0𝒜_𝑗(𝑧), that is, ℋ₀(𝑧)=∑^2𝑛−1_𝑗=0𝑢_𝑗𝒜_𝑗(𝑧)∑^2𝑛−1_𝑗=0𝒜_𝑗(𝑧), (40) and we use the expressions of 𝒜_𝑗(𝑧), after doing some simplifications, we obtain the result.
(ii) Take into account that ℋ₁(𝑧)=∑^2𝑛−1_𝑗=0𝑣_𝑗ℬ_𝑗(𝑧), with ℬ_𝑗(𝑧) given in Proposition 8 and proceed in the same way as in (i).
(iii) Take into account that ℋ₂(𝑧)=∑^2𝑛−1_𝑗=0𝑤_𝑗𝒞_𝑗(𝑧), with 𝒞_𝑗(𝑧) given in Proposition 8 and proceed in the same way as in (i).
(iv) Take into account that ℋ(𝑧)=^2𝑛−1_𝑗=0𝑢_𝑗𝒜_𝑗(𝑧)+𝑣_𝑗ℬ_𝑗(𝑧)+𝑤_𝑗𝒞_𝑗(𝑧), (41) with 𝒜_𝑗(𝑧),ℬ_𝑗(𝑧), and 𝒞_𝑗(𝑧) given in Proposition 8 and proceed in the same way as in (i).

3.2. Nodal System with an Odd Number of Points

Now we assume that the nodal system {𝛼_𝑗}^2𝑛_𝑗=0 is constituted by the 2𝑛+1 roots of 𝜆, with |𝜆|=1. In this case we obtain the barycentric expression for the Laurent polynomial of Hermite interpolation ℋ_{−3𝑛,3𝑛+1}(𝑧) that we denote by ℋ(𝑧), characterized by ℋ𝛼_𝑗=𝑢_𝑗,ℋ^′𝛼_𝑗=𝑣_𝑗,ℋ^′′𝛼_𝑗=𝑤_𝑗,for𝑗=0,…,2𝑛, (42) where {𝑢_𝑗}^2𝑛_𝑗=0,{𝑣_𝑗}^2𝑛_𝑗=0  and  {𝑤_𝑗}^2𝑛_𝑗=0 are fixed complex numbers.

Proceeding like in the previous case, first we obtain the following auxiliary results.

Lemma 10. The Laurent polynomials, ℰ_𝑗(𝑧)=(𝑧^2𝑛+1−𝜆)³/𝑧^3𝑛+1(𝑧−𝛼_𝑗),ℱ_𝑗(𝑧)=(𝑧^2𝑛+1−𝜆)³/𝑧^3𝑛+1(𝑧−𝛼_𝑗)²,  and  𝒢_𝑗(𝑧)=(𝑧^2𝑛+1−𝜆)³/𝑧^3𝑛+1(𝑧−𝛼_𝑗)³,  for  𝑗=0,⋯,2𝑛, satisfy(i)ℰ_𝑗(𝛼_𝑖)=0,ℰ^′_𝑗(𝛼_𝑖)=0,   for  all  𝑖=0,…,2𝑛,ℰ_𝑗^′′(𝛼_𝑖)=0,   for  all  𝑖≠𝑗, and ℰ_𝑗^′′(𝛼_𝑗)=2𝜆𝛼^𝑛_𝑗(2𝑛+1)³/𝛼²_𝑗,(ii)ℱ_𝑗(𝛼_𝑖)=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖=0,…,2𝑛,ℱ^′_𝑗(𝛼_𝑖)=0,  for  all  𝑖≠𝑗,ℱ^′_𝑗(𝛼_𝑗)=𝜆𝛼^𝑛_𝑗(2𝑛+1)³/𝛼²_𝑗,ℱ_𝑗^′′(𝛼_𝑖)=0,  for  all  𝑖≠𝑗,𝑎𝑛𝑑ℱ_𝑗^′′(𝛼_𝑗)=−(2𝜆𝛼^𝑛_𝑗(2𝑛+1)³/𝛼³_𝑗),(iii)𝒢_𝑗(𝛼_𝑖)=0,  for  all  𝑖≠𝑗,𝒢_𝑗(𝛼_𝑗)=𝜆𝛼^𝑛_𝑗(2𝑛+1)³/𝛼²_𝑗,𝒢^′_𝑗(𝛼_𝑖)=0,  for  all   𝑖≠𝑗,𝒢^′_𝑗(𝛼_𝑗)=−(𝜆𝛼^𝑛_𝑗(2𝑛+1)³/𝛼³_𝑗),𝒢_𝑗^′′(𝛼_𝑖)=0,  for  all  𝑖≠𝑗,𝑎𝑛𝑑𝒢_𝑗^′′(𝛼_𝑗)=(𝜆𝛼^𝑛_𝑗(2𝑛+1)³(𝑛²+𝑛+2))/𝛼⁴_𝑗.

Proof. It is easy to prove the result following the same steps of the proof of the preceding Lemma 7.

Proposition 11. The fundamental polynomials of Hermite interpolation in the Laurent space Λ_{−3𝑛−1,3𝑛+1}[𝑧], 𝒜_𝑗(𝑧)ℬ_𝑗(𝑧) y 𝒞_𝑗(𝑧), for 𝑗=0,…,2𝑛, characterized by 𝒜_𝑗𝛼_𝑖=𝛿_𝑖,𝑗,𝒜^′_𝑗𝛼_𝑖=0,𝒜_𝑗^′′𝛼_𝑖ℬ=0,∀𝑖=0,…,2𝑛,_𝑗𝛼_𝑖=0,ℬ^′_𝑗𝛼_𝑖=𝛿_𝑖,𝑗,ℬ_𝑗^′′𝛼_𝑖𝒞=0,∀𝑖=0,…,2𝑛,_𝑗𝛼_𝑖=0,𝒞^′_𝑗𝛼_𝑖=0,𝒞_𝑗^′′𝛼_𝑖=𝛿_𝑖,𝑗,∀𝑖=0,…,2𝑛, (43) have the following expressions: 𝒜_𝑗(𝑧)=𝛼²_𝑗𝜆𝛼^𝑛_𝑗(2𝑛+1)³𝑧^2𝑛+1−𝜆³𝑧^3𝑛+1𝑧−𝛼_𝑗³+𝛼_𝑗𝜆𝛼^𝑛_𝑗(2𝑛+1)³𝑧^2𝑛+1−𝜆³𝑧^3𝑛+1𝑧−𝛼_𝑗²−𝑛²+𝑛2𝜆𝛼^𝑛_𝑗(2𝑛+1)³𝑧^2𝑛+1−𝜆³𝑧^3𝑛+1𝑧−𝛼_𝑗,ℬ_𝑗(𝑧)=𝛼²_𝑗𝜆𝛼^𝑛_𝑗(2𝑛+1)³𝑧^2𝑛+1−𝜆³𝑧^3𝑛+1𝑧−𝛼_𝑗²+𝛼_𝑗𝜆𝛼^𝑛_𝑗(2𝑛+1)³𝑧^2𝑛+1−𝜆³𝑧^3𝑛+1𝑧−𝛼_𝑗,𝒞_𝑗(𝑧)=𝛼²_𝑗2𝜆𝛼^𝑛_𝑗(2𝑛+1)³𝑧^2𝑛+1−𝜆³𝑧^3𝑛+1𝑧−𝛼_𝑗. (44)

Proof. It is similar to the proof of Proposition 8.

It is straightforward to deduce, from the preceding Proposition, the so-called barycentric expression for ℋ(𝑧).

Proposition 12. (i)  The polynomial ℋ₀(𝑧)∈Λ_{−3𝑛−1,3𝑛+1}[𝑧] satisfying the conditions ℋ₀(𝛼_𝑗)=𝑢_𝑗, ℋ^′₀(𝛼_𝑗)=0, and ℋ₀^′′(𝛼_𝑗)=0, 𝑓𝑜𝑟𝑎𝑙𝑙𝑗=0,…,2𝑛, has the barycentric expression ℋ₀(𝑧)=^2𝑛_𝑗=0𝑢_𝑗𝛼^𝑛_𝑗𝛼²_𝑗𝑧−𝛼_𝑗³+𝛼_𝑗𝑧−𝛼_𝑗²−𝑛²+𝑛2𝑧−𝛼_𝑗×^2𝑛_𝑗=01𝛼^𝑛_𝑗𝛼²_𝑗𝑧−𝛼_𝑗³+𝛼_𝑗𝑧−𝛼_𝑗²−𝑛²+𝑛2𝑧−𝛼_𝑗⁻¹. (45)
(ii)  The polynomial ℋ₁(𝑧)∈Λ_{−3𝑛−1,3𝑛+1}[𝑧] satisfying the conditions ℋ₁(𝛼_𝑗)=0, ℋ^′₁(𝛼_𝑗)=𝑣_𝑗, and ℋ₁^′′(𝛼_𝑗)=0, 𝑓𝑜𝑟𝑎𝑙𝑙𝑗=0,…,2𝑛, has the barycentric expression ℋ₁(𝑧)=^2𝑛_𝑗=0𝑣_𝑗𝛼^𝑛_𝑗𝛼²_𝑗𝑧−𝛼_𝑗²+𝛼_𝑗𝑧−𝛼_𝑗×^2𝑛_𝑗=01𝛼^𝑛_𝑗𝛼²_𝑗𝑧−𝛼_𝑗³+𝛼_𝑗𝑧−𝛼_𝑗²−𝑛²+𝑛2𝑧−𝛼_𝑗⁻¹. (46)
(iii)  The polynomial ℋ₂(𝑧)∈Λ_{−3𝑛−1,3𝑛+1}[𝑧] satisfying the conditions ℋ₂(𝛼_𝑗)=0, ℋ^′₂(𝛼_𝑗)=0, and ℋ₂^′′(𝛼_𝑗)=𝑤_𝑗, 𝑓𝑜𝑟𝑎𝑙𝑙𝑗=0,…,2𝑛, has the barycentric expression ℋ₂(𝑧)=^2𝑛_𝑗=0𝑤_𝑗𝛼^𝑛_𝑗𝛼²_𝑗2𝑧−𝛼_𝑗×^2𝑛_𝑗=01𝛼^𝑛_𝑗𝛼²_𝑗𝑧−𝛼_𝑗³+𝛼_𝑗𝑧−𝛼_𝑗²−𝑛²+𝑛2𝑧−𝛼_𝑗⁻¹. (47)
(iv)  The polynomial ℋ(𝑧)∈Λ_{−3𝑛−1,3𝑛+1}[𝑧] satisfying the conditions (42) has the barycentric expression ℋ(𝑧)=^2𝑛_𝑗=01𝛼^𝑛_𝑗𝑢_𝑗𝛼²_𝑗𝑧−𝛼_𝑗³+𝛼_𝑗𝑧−𝛼_𝑗²−𝑛²+𝑛2𝑧−𝛼_𝑗+𝑣_𝑗𝛼²_𝑗𝑧−𝛼_𝑗²+𝛼_𝑗𝑧−𝛼_𝑗+𝑤_𝑗𝛼²_𝑗2𝑧−𝛼_𝑗1𝛼^𝑛_𝑗×^2𝑛_𝑗=01𝛼^𝑛_𝑗𝛼²_𝑗(𝑧−𝛼_𝑗)³+𝛼_𝑗(𝑧−𝛼_𝑗)²−𝑛²+𝑛2(𝑧−𝛼_𝑗)⁻¹. (48)

Proof. It is similar to the proof of Proposition 9.

Remark 13. Notice the novelty of the barycentric expressions that we have obtained for Hermite interpolation on the unit circle using up to the second derivative. These formulas can be implemented for use in a simple way and they are very useful for computations due to their stability.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The research was supported by Ministerio de Educación y Ciencia of Spain under Grant no. MTM2011-22713.