aa r X i v : . [ m a t h . NA ] J un On Wiener – Hopf factorization of scalar polynomials
Victor M. Adukov
Faculty of Mathematics, Mechanics and Computer Science,Institute of Natural Sciences,South Ural State University (national research university), Chelyabinsk, Russia
Abstract
In the work we propose an algorithm for a Wiener – Hopf factorization of scalarpolynomials based on notions of indices and essential polynomials. The algo-rithm uses computations with finite Toeplitz matrices and permits to obtaincoefficients of both factorization factors simultaneously. Computation aspectsof the algorithm are considered. An a priory estimate for the condition numberof the used Toeplitz matrices is obtained. Upper bounds for the accuracy of thefactorization factors are established. All estimates are effective.
Keywords:
Wiener – Hopf factorization, polynomial factorization, Toeplitzmatrices
1. Introduction
The Wiener – Hopf technique is a powerful method used in various areasof mathematics, mechanics and mathematical physics (see, e.g., [11], [14], [16]).The core of the method is the Wiener – Hopf factorization problem for matrixfunctions (or the Riemann boundary-value problem) [10, 14]. In the scalarcase there is an explicit formula for a solution of the problem [12], but its usefor numerical computations is quite difficult. Therefore to factorize a scalarfunction, it is usually approximated by a rational function (see, e.g., [15]). Thusthe scalar factorization reduces to the polynomial factorization.The matrix case is more complicated because there are no explicit formulasfor the factorization factors and for important integer invariants of the problemsuch as partial indices. Moreover, it is difficult to develop approximate methodsof the factorization since the matrix factorization is unstable. For this reasonit is very important to find cases when the problem can be effectively or ex-plicitly solved. Almost all cases of constructive factorization known to date areconsidered in the review [17]. In [2] an explicit method for the factorizationof meromorphic matrix functions was proposed. In this paper the factorization
Email address: [email protected] (Victor M. Adukov)
Preprint submitted to Linear Algebra and its Applications October 7, 2018 roblem is called explicitly solved if it is reduced to the factorization of givenscalar functions and to solving of finitely many finite systems of linear algebraicequations. In [4] some classes of matrix functions are listed for which the factor-ization problem can be explicitly reduced to the factorization of analytic matrixfunctions.In this paper, we start a project to develop algorithms of the factorizationand their implementations for these classes of matrix functions. We proposeto consider the factorization of matrix polynomials, analytic and meromorphicmatrix functions, triangular matrix functions and matrix functions with onenon-meromorphic row. Some preliminary results in this direction were obtainedin [5].The first stage in solving of the factorization problem for the above men-tioned classes is the factorization of scalar functions that can be reduced to thepolynomial factorization.We consider coefficients of a polynomial p ( z ) as initial data and coefficientsof its factors as output data of the problem. The naive method of the polynomialfactorization is the following way. First we compute roots of a polynomial p ( z )and perform their separation. Second we find the coefficients of the factors of p ( z ) by their roots. It is well known that the roots of a polynomial are in generalnot well-condition functions of its coefficients (see, e.g., [13]), and coefficients ofa polynomial are also not well-condition functions of its roots [18].The latter means that, in general, we can not solve numerically the polyno-mial factorization problem by the naive way.Nevertheless there exist numerical methods for solving of this problem. Thebasic works in this direction are cited in [8]. It should be especially pointed outthe articles [8]–[7], where finding of the polynomial factorization was based oncomputations with Toeplitz matrices.Our approach resembles the method proposed by D.A. Bini and A. B¨ott-cher [8] in algorithm 3 . As the authors we solve systems with finite Toeplitzmatrices consisting of Laurent coefficients of the function 1 /p ( z ). The maindifference between the method of D.A. Bini and A. B¨ottcher and ours is that algorithm 3 permits to obtain only coefficients of the factor p − ( z ) for p ( z ),whereas we find coefficients of both factors simultaneously. This is importantbecause the polynomial division is, in general, not well-condition operation innumerical computations. Moreover, we use a different technique that can beextended to the factorization problem for analytic functions. In this case wecan obtain all coefficients of the polynomial factor and a required number ofTaylor coefficients of the analytic factor.The paper is organized as follows. In section 2 we consider the setting ofthe polynomial factorization problem and formulate some results on used normsof polynomials and their estimates. Section 3 contains basic tools for solvingof the problem. Here we introduce notions of indices and essential polynomialsin terms of which the problem will be solved. In Section 4 we prove the basicresults on the factorization of a polynomial p ( z ). In Section 5 some computationaspects of the factorization problem are considered. Here we obtain an a prioryestimate for the condition number of the used Toeplitz matrices and establish2pper bounds for the accuracy of the factorization factors. All estimates areeffective. Section 6 contains an algorithm and numerical examples.
2. Preliminaries
Let p ( z ) = p + p z + · · · + p ν z ν be a complex polynomial of degree ν > p = 0 , p ν = 1. We suppose that p ( z ) = 0 on the unit circle T , hence p ( z ) = 0 on a closed circular annulus K := { z ∈ C : r ≤ | z | ≤ R } for some0 < r < < R < ∞ . Denote by κ = ind T p ( z ) the index of p ( z ) with respectto T , i.e. the number of zeros of the polynomial inside the unit circle. Let ξ j , j = 1 , . . . , ν , be the zeros of p ( z ) and0 < | ξ | ≤ . . . ≤ | ξ κ | < r < < R < | ξ κ +1 | ≤ . . . ≤ | ξ ν | . Here the zeros are counted according to their multiplicity.In the work we will consider the following factorization of p ( z ) p ( z ) = p ( z ) p ( z ) , (1)where p ( z ) := ( z − ξ ) · · · ( z − ξ κ ) , p ( z ) := ( z − ξ κ +1 ) · · · ( z − ξ ν ) . Denote p − ( z ) = p ( z ) z κ , p + ( z ) = p ( z ). Then the representation p ( z ) = p − ( z ) z κ p + ( z ) , | z | = 1 , (2)is the Wiener – Hopf factorization of p ( z ) normalized by the condition p − ( ∞ ) =1. Throughout this paper, k x k means the H¨older 1-norm k x k = | x | + · · · + | x k | ,x = ( x , . . . , x k ) T ∈ C k . A norm k A k of a matrix A ∈ C ℓ × k is always theinduced norm k A k = max ≤ j ≤ k ℓ X i =1 | A ij | . Respectively, the norm of a polynomial p ( z ) = p + p z + · · · + p ν z ν is thenorm of the vector ( p , p , . . . , p ν ) T . For p ( z ) we will also use the maximumnorm k p k C = max z ∈ T | p ( z ) | on the unit circle T .The norms k · k and k · k C are equivalent. Clearly, k p k C ≤ k p k . Since for p ( z ) it is fulfilled the equality ν X k =0 | p k | = 12 π Z π | p ( e iϕ ) | dϕ,
3e have k p k ≤ √ ν + 1 k p k ≤ √ ν + 1 k p k C . Thus, k p k C ≤ k p k ≤ √ ν + 1 k p k C . (3)In order to study stability of the factorization problem, we will need esti-mates for the norm of inverses of some Toeplitz matrices. Such estimates willbe obtain in terms of k p k k p k , where p ( z ), p ( z ) are the factorization factorsof p ( z ). To get effective estimates, it will be required to estimate k p k k p k via k p k .Let q ( z ) = q ( z ) q ( z ), where q ( z ), q ( z ) are arbitrary monic complex poly-nomials and ν = deg q . It is obvious that k q k ≤ k q k k q k . In the work [9], D.W. Boyd proved the following inequality k q k C k q k C ≤ δ ν k q k C , where δ = e G/π = 1 . . . . and G is Catalan’s constant.The inequality is asymptoticaly sharp as ν → ∞ .Taking into account (3), in our case we obtain k p k ≤ k p k k p k ≤ δ ν p ( κ + 1)( ν − κ + 1) k p k . (4)However, the exponential factor δ ν can overestimate the upper bound. Insome special cases we can obtain more precise estimates. For example, this canbe done for a so-called spectral factorization of polynomials. Proposition 2.1.
Let p ( z ) = P mj =0 p j z j be a complex polynomial of degree m such that p m − j = ¯ p j for j = 0 , . . . , m , p = 1 . Suppose that p ( z ) = 0 on T .If p ( z ) = p ( z ) p ( z ) is factorization (1), then k p k k p k ≤ ( m + 1) k p k . Proof.
It is clear that if ξ k is a root of p ( z ), then so is 1 / ¯ ξ k . Hence m is theindex of p ( z ).Let | ξ k | < k = 1 , . . . , m and p ( z ) = ( z − ξ ) · · · ( z − ξ m ) , p ( z ) = ( z − / ¯ ξ ) · · · ( z − / ¯ ξ m )are the factorization factors of p ( z ). On the unit circle we have p ( t ) = ( − m ( ¯ ξ · · · ¯ ξ m ) t − m p ( t ) = p (0) t − m p ( t ) , | t | = 1 . It follows from this that k p k C = | p (0) | k p k C and p ( t ) = t m p (0) p ( t ) p ( t ) . Thus we have k p k C = 1 | p (0) | k p k C = k p k C k p k C . Applying inequality (3), we arrive at the assertion. (cid:3)
4t is possible that the equality k p k = k p k k p k holds. Proposition 2.2.
Let p ( z ) = P mj =0 p j z j be a real polynomial of degree m and p m − j = p j for j = 0 , . . . , m , p = 1 . Suppose that p ( z ) = 0 on T and all rootsof p ( z ) have negative real parts.If p ( z ) = p ( z ) p ( z ) is factorization (1), then k p k = k p k k p k . Proof.
In this case, if ξ k is a root of p ( z ), then so are ¯ ξ k and 1 /ξ k .Let | ξ k | < k = 1 , . . . , m . Then p ( z ) = ( z − ξ ) · · · ( z − ξ m ) , p ( z ) = ( z − /ξ ) · · · ( z − /ξ m ) . If z = ξ is a complex root of p ( z ), then ( z − ξ )( z − ¯ ξ ) has positive coefficients.Obviously, z − ξ has also positive coefficients for a real root ξ . Then p ( z ) haspositive coefficients as a product of such polynomials. Similar statement holdsfor p ( z ) and p ( z ).Thus we have k p k = p (1) = p (1) p (1) = k p k k p k . (cid:3) Since for a given polynomial p ( z ) more precise estimates can exist, we willused the inequalities k p k ≤ k p k k p k ≤ δ k p k , ≤ δ ≤ δ ν p ( κ + 1)( ν − κ + 1) , (5)instead of (4).
3. Basic tools
Let
M, N be integers,
M < N , and c NM = ( c M , c M +1 , . . . , c N ) a nonzerosequence of complex numbers. In this section we introduce notions of indicesand essential polynomials for the sequence c NM . These notions were given inmore general setting in the paper [3]. Here we will consider the scalar case only.The proofs of all statements of this section can be found in [3].Let us form the family of all Toeplitz matrices T k ( c NM ) = c k c k − . . . c M c k +1 c k . . . c M +1 ... ... ... c N c N − . . . c N + M − k , M ≤ k ≤ N, (6)which can be constructed with the help of the sequence c NM . We will usedthe short designation T k in place of T k ( c NM ) if there is not the possibility ofmisinterpretation.Our nearest aim is to describe a structure of the kernels ker T k . It is moreconvenient to deal not with vectors Q = ( q , q , . . . , q k − M ) t ∈ ker T k but withtheir generating polynomials Q ( z ) = q + q z + · · · + q k − M z k − M . We will use5he spaces N k of the generating polynomials instead of the spaces ker T k . Thegenerating function P Nj = M c k z k of the sequence c NM will be denoted by c NM ( z ).Let us introduce a linear functional σ by the formula: σ { z j } = c − j , − N ≤ j ≤ − M. The functional is defined on the space of rational functions of the form Q ( z ) = − M X j = − N q j z j . Besides this algebraic definition of σ we will use the following analytic definition σ { Q ( z ) } = 12 πi Z Γ t − c NM ( t ) Q ( t ) dt. (7)Here Γ is any closed contour around the point z = 0.Denote by N k ( M ≤ k ≤ N ) the space of polynomials Q ( z ) with the formaldegree k − M satisfying the orthogonality conditions: σ (cid:8) z − i Q ( z ) (cid:9) = 0 , i = k, k + 1 , . . . , N. (8)It is easily seen that N k is the space of generating polynomials of vectors inker T k . For convenience, we put N M − = 0 and denote by N N +1 the ( N − M +2)-dimensional space of all polynomials with the formal degree N − M + 1. Ifnecessary, the more detailed notation N k ( c NM ) instead of N k is used.Let d k be the dimension of the space N k and ∆ k = d k − d k − ( M ≤ k ≤ N + 1). The following proposition is crucial for the further considerations. Proposition 3.1.
For any non-zero sequence c NM the following inequalities M ≤ ∆ M +1 ≤ . . . ≤ ∆ N ≤ ∆ N +1 = 2 (9) are fulfilled. (cid:3) It follows from the inequalities (9) that there exist integers µ ≤ µ suchthat ∆ M = . . . = ∆ µ = 0 , ∆ µ +1 = . . . = ∆ µ = 1 , ∆ µ +1 = . . . = ∆ N +1 = 2 . (10)If the second row in these relations is absent, we assume µ = µ . Definition 3.1.
The integers µ , µ defined in (10) will be called the essentialindices (briefly, indices) of the sequence c NM .The following proposition gives formulas for the indices.6 roposition 3.2. Let π = rank T [ N + M ] , where (cid:2) N + M (cid:3) is the integral part of N + M . Then the indices µ , µ are found by the formulas: µ = M + π − , µ = N − π + 1 . (cid:3) It follows from the definition of N k +1 that N k and z N k are subspaces of N k +1 , M − ≤ k ≤ N . Let h k +1 be the dimension of any complement H k +1 ofthe subspace N k + z N k in the whole space N k +1 .From (10) we see that h k +1 = 0 iff k = µ j ( j = 1 , h k +1 = 1 if µ < µ ,and h k +1 = 2 for µ = µ . Therefore, for k = µ j N k +1 = N k + z N k , (11)and for k = µ j N k +1 = (cid:0) N k + z N k (cid:1) ⊕ H k +1 . (12) Definition 3.2.
Let µ = µ . Any polynomials Q ( z ), Q ( z ) that form a basisfor the two-dimensional space N µ +1 are called the essential polynomials of thesequence c NM corresponding to the index µ = µ .If µ < µ , then any polynomial Q j ( z ) that is a basis for an one-dimensi-onal complement H µ j +1 is said to be the essential polynomial of the sequencecorresponding to the index µ j , j = 1 , Proposition 3.3.
Integers µ , µ , µ + µ = M + N , are the indices and poly-nomials Q ( z ) ∈ N µ +1 , Q ( z ) ∈ N µ +1 are the essential polynomials of thesequence c NM iff σ := σ { z − µ Q ,µ − M +1 Q ( z ) − z − µ Q ,µ − M +1 Q ( z ) } 6 = 0 . (13) Here Q j,µ j − M +1 is the coefficient of z µ j − M +1 in the polynomial Q j ( z ) . (cid:3) Now we can describe the structure of the kernels of the matrices T k in termsof the indices and the essential polynomials. Proposition 3.4.
Let µ , µ be the indices and Q ( z ) , Q ( z ) the essential poly-nomials of the sequence c NM .Then N k = { } , { q ( z ) Q ( z ) } , { q ( z ) Q ( z ) + q ( z ) Q ( z ) } , M ≤ k ≤ µ ,µ + 1 ≤ k ≤ µ ,µ + 1 ≤ k ≤ N, where q j ( z ) is an arbitrary polynomial of the formal degree k − µ j − , j = 1 , . (cid:3) . Construction of Wiener – Hopf factorization of scalar polynomials In this section we propose a method for solving the problem of the polynomialfactorization in terms of indices and essential polynomials of some sequence.Let p ( z ) = p + p z + · · · + p ν z ν be a polynomial of degree ν > p = 0. Wesuppose that p ( z ) = 0 on the unit circle T , hence p ( z ) = 0 on a closed annulus K := { z ∈ C : r ≤ | z | ≤ R } for some 0 < r < < R < ∞ . Let κ = ind T p ( z )be the index of p ( z ), i.e. the number of zeros of the polynomial inside the unitcircle. We can assume that 0 < κ < ν , otherwise the factorizations are trivial.Put n = max { κ , ν − κ } .We will find the factorization of p ( z ) in the form p ( z ) = p ( z ) p ( z ) , (14)where deg p = κ , deg p = ν − κ , the polynomial p ( z ) is monic, and all zerosof p ( z ) (respectively p ( z )) lie in the domain { z ∈ C : | z | < r } (respectively in { z ∈ C : | z | > R } ). This means that p ( z ) = p − ( z ) z κ p + ( z ) , z ∈ T , is the Wiener – Hopf factorization of p ( t ) normalized by the condition p − ( ∞ ) =1. Here p − ( z ) = p ( z ) z κ , p + ( z ) = p ( z ). Let p − ( z ) = ∞ X k = −∞ c k z k be the Laurent series for analytic function p − ( z ) in the annulus K . Here c k = 12 πi Z | z | = ρ t − k − p − ( t ) dt, k ∈ Z , r ≤ ρ ≤ R. (15)Note that c k are the Fourier coefficients of the function p − ( t ), t ∈ T .Form the sequence c n − κ − n − κ = ( c − n − κ , . . . , c − κ , . . . , c n − κ ) for the given n ≥ n . The generating function c n − κ − n − κ ( z ) of the sequence is a partial sum of theLaurent series of p − ( z ) and in formula (7) we can replace c n − κ − n − κ ( z ) by p − ( z ). Theorem 4.1.
For any n ≥ n the sequence c n − κ − n − κ is non-zero; the integers − κ , − κ are the indices; and Q ( z ) = z n − κ +1 p ( z ) , Q ( z ) = p ( z ) are essentialpolynomials of the sequence. Proof.
Let us prove the first statement of the theorem. Since Q ( z ) = z n +1 + α n z n + · · · + α n − κ +1 z n − κ +1 , then σ { z − n + κ − Q ( z ) } = c − κ + c − κ +1 α n + · · · + c α n − κ +1 . Otherwise, by analytic definition (7) of σ we have σ { z − n + κ − Q ( z ) } =12 πi Z T t − p − ( t ) p ( t ) dt = 12 πi Z T t − p − ( t ) dt = p − (0) = 0 . c − κ + c − κ +1 p − + · · · + c p − κ = 0 , and not all of the numbers c − κ , c − κ +1 , . . . , c are zero. For n ≥ n the sequence { c − κ , c − κ +1 , . . . , c } is a subsequence of c n − κ − n − κ . Hence, c n − κ − n − κ is also non-zero.Now we prove that Q ( z ) , Q ( z ) belong to the space N − κ +1 . A formaldegree of polynomials from this space should be n + 1. In our case we havedeg Q ( z ) = n + 1 and deg Q ( z ) = ν − κ < n + 1. Verify orthogonalitycondition (8). We have σ { z − j Q ( z ) } = 12 πi Z T t n − κ − j p − ( t ) dt = 0for n − κ − j ≥
0. Here we use analytic definition (7) of σ and the Wiener –Hopf factorization (14). In particular, σ (cid:8) z − j Q ( z ) (cid:9) = 0 , j = − κ + 1 , . . . , n − κ . Hence Q ( z ) ∈ N − κ +1 .Similarly, σ { z − j Q ( z ) } = 12 πi Z T t − j − p − ( t ) dt = 12 πi Z T t − κ − j − p − − ( t ) dt = 0for j = − κ + 1 , . . . , n − κ , and Q ( z ) ∈ N − κ +1 since p − ( z ) = z − κ p ( z ) isanalytic in the domain { z ∈ C : | z | > } .Moreover, σ { z κ Q ( z ) } = p − − ( ∞ ) = 1 , and Q (0) = p (0) = 0. Now we apply Propositiom 3.3. Let us find the test-number σ from formula (13): σ = σ { z κ Q ,n +1 Q ( z ) − z κ Q ,n +1 Q ( z ) } = − σ { z κ Q ( z ) } = − = 0 . Hence by Propositiom 3.3, the integers − κ , − κ are the indices; and Q ( z ), Q ( z ) are the essential polynomials of the sequence c n − κ − n − κ . (cid:3) In this theorem we have proved that there exist the essential polynomials Q ( z ) , Q ( z ) of the sequence c n − κ − n − κ such that the following additional propertiesare fulfilled:(i) deg Q ( z ) = n + 1, Q (0) = 0, Q ,n +1 = 1.(ii) deg Q ( z ) < n + 1, Q (0) = 0, σ { z κ Q ( z ) } = 1.Vice versa, if Q ( z ) , Q ( z ) ∈ N − κ +1 and satisfy conditions (i)–(ii), then σ = − Q ( z ) , Q ( z ) are the essential polynomials of the sequence c n − κ − n − κ . Definition 4.1.
Let n ≥ n . Polynomials Q ( z ) , Q ( z ) ∈ N − κ +1 ( c n − κ − n − κ ) sat-isfying conditions (i)–(ii) will be called the factorization essential polynomialsof c n − κ − n − κ . 9 heorem 4.2. The factorization essential polynomials are uniquely determinedby conditions (i)–(ii). Let n ≥ n + 1 and R ( z ) , R ( z ) are any essential poly-nomials of the sequence c n − κ − n − κ . Then the factorization essential polynomials of c n − κ − n − κ can be found by the formulas Q ( z ) = − σ (cid:0) R , R ( z ) − R , R ( z ) (cid:1) ,Q ( z ) = 1 σ (cid:0) R ,n +1 R ( z ) − R ,n +1 R ( z ) (cid:1) . (16) Here R j, = R j (0) , R j,n +1 is the coefficient of z n +1 in the polynomial R j ( z ) , j = 1 , , and σ = (cid:12)(cid:12)(cid:12)(cid:12) R , R , R ,n +1 R ,n +1 (cid:12)(cid:12)(cid:12)(cid:12) = 0 , σ = σ { z κ (cid:0) R ,n +1 R ( z ) − R ,n +1 R ( z ) (cid:1) } 6 = 0 . Proof.
Let Q ( z ) , Q ( z ) and e Q ( z ) , e Q ( z ) be any couples of the factorizationessential polynomials. Since { Q ( z ) , Q ( z ) } is a basis of the space N − κ +1 , wehave e Q ( z ) = α Q ( z ) + α Q ( z ) , e Q ( z ) = β Q ( z ) + β Q ( z ) . Then 0 = e Q (0) = α Q (0) + α Q (0) = α Q (0). Hence α = 0. It followsfrom the conditions Q ,n +1 = Q ,n +1 = 1 that α = 1, and e Q ( z ) = Q ( z ). Insimilar manner we can prove e Q ( z ) = Q ( z ).Let R ( z ) , R ( z ) be any essential polynomials of the sequence c n − κ − n − κ for n ≥ n + 1. Then σ is the test-number of the essential polynomials R ( z ), R ( z ) and σ = 0.Suppose that σ = (cid:12)(cid:12)(cid:12)(cid:12) R , R , R ,n +1 R ,n +1 (cid:12)(cid:12)(cid:12)(cid:12) = 0 . Then λ (cid:18) R , R ,n +1 (cid:19) + λ (cid:18) R , R ,n +1 (cid:19) = 0 . Define Q ( z ) = λ R ( z ) + λ R ( z ). By definition, Q (0) = 0 and deg Q ≤ n .Put Q ( z ) = z e Q ( z ), where deg e Q ( z ) ≤ n −
1. Since Q ( z ) ∈ N − κ +1 ( c n − κ − n − κ ),we have e Q ( z ) ∈ N − κ ( c n − κ − − n − κ +1 ). However, if n ≥ n + 1, then the sequence c n − κ − − n − κ +1 has indices − κ , − κ , and, by Proposition 3.4, N − κ ( c n − κ − − n − κ +1 ) = { } .Thus, Q ( z ) ≡
0, and the polynomials R ( z ) , R ( z ) are linearly dependent. Butit is impossible and the contradiction proves the inequality σ = 0.The polynomials Q ( z ), Q ( z ), defined by (16), belong to N − κ +1 ( c n − κ − n − κ )and satisfy conditions (i)–(ii). Hence Q ( z ), Q ( z ) are the factorization essentialpolynomials. The theorem is proved. (cid:3) Now we can construct the Wiener – Hopf factorization of a polynomial withthe help of the factorization essential polynomials. Note that existence anduniqueness of the factorization essential polynomials were proved under condi-tion n ≥ n . 10 heorem 4.3. Let n ≥ n and let Q ( z ) , Q ( z ) be the factorization essentialpolynomials of the sequence c n − κ − n − κ . Then the Wiener – Hopf factorization ofthe polynomial p ( z ) can be constructed by the formula p ( z ) = p − ( z ) z κ p + ( z ) , where p − ( z ) = z − n − Q ( z ) , p + ( z ) = Q ( z ) . (17) Proof.
In Theorem 4.1 we prove that Q ( z ) = z n − κ +1 p ( z ), Q ( z ) = p ( z )are the factorization essential polynomials. By Theorem 4.2, the factorizationessential polynomials are determined uniquely. Thus formulas (17) hold. (cid:3) Now we can obtain the main result of the section about an explicit formulasfor the factors of the Wiener – Hopf factorization of a polynomial p ( z ). Theorem 4.4.
The matrices T − κ (cid:0) c n − κ − n − κ (cid:1) are invertible for all n ≥ n .Let n ≥ n + 1 . Denote by α = ( α , . . . , α n ) T and β = ( β , . . . , β n ) T thesolutions of the systems T − κ (cid:0) c n − κ − − n − κ +1 (cid:1) α = − ( c − κ − − n − κ ) T , T − κ (cid:0) c n − κ − n − κ (cid:1) β = e , (18) respectively. Here e = (1 , , . . . , T .Then α = . . . = α n − κ = 0 , β = 0 , β ν − κ +1 = . . . = β n = 0 , and the factorsfrom the Wiener – Hopf factorization of p ( z ) are found by the formulas p − ( z ) = z − κ ( α n − κ +1 + · · · + α n z κ − + z κ ) , p + ( z ) = β + β z + · · · + β ν − κ z ν − κ . Proof.
For the first index of the sequence c n − κ − n − κ by Proposition 3.2 we have µ = − n − κ + π − , where π = rank T − κ (cid:0) c n − κ − n − κ (cid:1) . By Theorem 4.1, we have µ = − κ for n ≥ n .Hence rank of the ( n + 1) × ( n + 1) matrix T − κ (cid:0) c n − κ − n − κ (cid:1) is equal to n + 1.If α = ( α , . . . , α n ) T satisfies the equation T − κ (cid:0) c n − κ − − n − κ +1 (cid:1) α = − ( c − κ − − n − κ ) T ,then the vector (0 , α , . . . , α n , T belongs to the space ker T − κ +1 (cid:0) c n − κ − n − κ (cid:1) , i.e. Q ( t ) = α z + · · · + α n z n + z n +1 ∈ N − κ +1 .Similarly, ( β , . . . , β n , T ∈ ker T − κ +1 (cid:0) c n − κ − n − κ (cid:1) , and Q ( z ) = β + β z + . . . + β n z n ∈ N − κ +1 . Moreover, σ { z κ Q ( z ) } = c − κ β + · · · + c − n − κ β n = 1.Let us prove β = 0. If we suppose β = 0, then the non-zero vector( β , . . . , β n , T belongs to ker T − κ (cid:0) c n − κ − − n − κ +1 (cid:1) . This is impossible since thematrix T − κ (cid:0) c n − κ − − n − κ +1 (cid:1) is invertible for n ≥ n + 1.Therefore the polynomials Q ( z ), Q ( z ) satisfy conditions (i)–(ii) and theyare the factorization essential polynomials of c n − κ − n − κ . By Theorem 4.3, Q ( z ) = z n +1 p − ( z ) , Q ( z ) = p + ( z ) . Hence, α = . . . = α n − κ = 0 , β ν − κ +1 = . . . = β n = 0 . The theorem is proved. (cid:3) . Some computational aspects of the factorization problem It is well known that the index κ of a scalar function is stable under smallperturbations of the function. In particular, the following statement holds. Proposition 5.1. If p ( z ) = 0 on the unit circle T and a polynomial e p ( z ) satis-fies the inequality k p − e p k < min | z | =1 | p ( z ) | , then e p ( z ) = 0 on T and ind T p ( z ) = ind T e p ( z ) . The proof is carried out using standard arguments. (cid:3)
There are many ways to calculate the index. Numerical experiments showthat it is more convenient to use formula (12.6) from [12]. Let p ( e iϕ ) = ξ ( ϕ ) + i η ( ϕ ), ϕ ∈ [0 , π ], where ξ ( ϕ ) and η ( ϕ ) are real continuously differentiablefunctions on [0 , π ]. Then κ = 12 π Z π ξ ( ϕ ) η ′ ( ϕ ) − ξ ′ ( ϕ ) η ( ϕ ) ξ ( ϕ ) + η ( ϕ ) dϕ. (19)The integral can be computed numerically by the Gaussian quadraturemethod. Since κ is integer, the result is rounded up to the nearest integer. Theorem 4.4 shows that solving of the factorization problem is equivalent tosolving of linear systems with the invertible matrix T − κ ( c n − κ − n − κ ), n ≥ n .Here we obtain an upper bound for the condition number k ( T − κ ( c n − κ − n − κ )) = k T − κ ( c n − κ − n − κ ) k k T − − κ ( c n − κ − n − κ ) k in terms of the given polynomial p ( z ).Recall that n = max { κ , ν − κ } , m K = min z ∈ K | p ( z ) | , ρ = max { r, /R } . Proposition 5.2.
For n ≥ n k T − κ ( c n − κ − n − κ ) k < m K ρ (1 − ρ ) . (20) Proof.
From the structure of the matrix it is obvious that k T − κ ( c n − κ − n − κ ) k ≤ k c n − κ − n − κ k = − X j = − n − κ | c j | + n − κ X j =0 | c j | , where c j = 12 πi Z | z | =¯ ρ t − j − p − ( t ) dt, r ≤ ¯ ρ ≤ R, j ∈ Z . j ≥ ρ = R . Then | c j | ≤ π Z π | dt | R j +1 | p ( t ) | ≤ R j min | z | = R | p ( z ) | ≤ ρ j m K . In the same way, if j < | c j | ≤ r | j | min | z | = r | p ( z ) | ≤ ρ | j | m K . Therefore, k T − κ ( c n − κ − n − κ ) k ≤ k c n − κ − n − κ k ≤ m K n + κ X j =1 ρ j + n − κ X j =0 ρ j ≤ m K ρ − ρ . (cid:3) Remark 5.1.
It is clear that | c j | ≤ m , where m = min | z | =1 | p ( z ) | . Hence k T − κ ( c n − κ − n − κ ) k ≤ k c n − κ − n − κ k ≤ n + 1 m . (21)This rough estimate can be more precise for a polynomial of small degree.Further, we will also use the estimates k c − κ − − n − κ k ≤ k c n − κ − n − κ k < m K ρ (1 − ρ ) (22)or k c − κ − − n − κ k ≤ nm .To obtain bounds for k T − − κ ( c n − κ − n − κ ) k we will get a formula for the inverse T − − κ ( c n − κ − n − κ ) in terms of the factorization p ( z ) = p ( z ) p ( z ).If p ( z ) = z κ + p , κ − z κ − + · · · + p , and p ( z ) = z ν − κ + p ,ν − κ − z ν − κ − + · · · + p , , then we denote p ( k )1 ( z ) = z k + p , κ − z k − + · · · + p , κ − k , ≤ k ≤ κ , and p ( k )2 ( z ) = p ,k z k + p ,k − z k − + · · · + p , , ≤ k ≤ ν − κ . Let T − − κ ( c n − κ − n − κ ) = ( b ij ) ni,j =0 and B j ( z ) = P ni =0 b ij z i be the generatingpolynomial of the j -th column B j of the inverse. Theorem 5.1.
For n ≥ ν we have B j ( z ) = p ( j )1 ( z ) p ( z ) , ≤ j ≤ κ ,z j − κ p ( z ) , κ + 1 ≤ j ≤ n − ν + κ ,z j − κ p ( z ) p ( n − j )2 ( z ) , n − ν + κ + 1 ≤ j ≤ n. (23) Here the second row is absent if n = ν . roof. If n ≥ ν , then T − κ ( c n − κ − n − κ ) is invertible, and Q ( z ) = z n − κ +1 p ( z ), Q ( z ) = p ( z ) are the factorization essential polynomials of the sequence c n − κ − n − κ with the test-number σ = −
1. Let B ( t, s ) = P ni,j =0 b ij t i s − j be the generatingfunction of the inverse matrix T − − κ ( c n − κ − n − κ ) = ( b ij ) ni,j =0 . By Theorem 2.2 from[1], we have B ( t, s ) = − s − n − Q ( t ) Q ( s ) − Q ( s ) Q ( t )1 − ts − . Hence ( t − s ) n X j =0 B j ( t ) s n − j = t n − κ +1 p ( t ) p ( s ) − s n − κ +1 p ( s ) p ( t ) . Equating the coefficients of s j , we arrive to relation (23). (cid:3) Corollary 5.1.
For n ≥ ν , we have k p k ≤ k T − − κ ( c n − κ − n − κ ) k ≤ k p k k p k ≤ δ k p k . (24) Proof.
It follows from (23) that k B j k = k p ( j )1 p k , ≤ j ≤ κ , k p k , κ + 1 ≤ j ≤ n − ν + κ , k p p ( n − j )2 k , n − ν + κ + 1 ≤ j ≤ n. Hence, k T − − κ ( c n − κ − n − κ ) k = max ≤ j ≤ n k B j k ≥ k p k . Moreover, k p ( j )1 p k ≤k p ( j )1 k k p k ≤ k p k k p k , and k p p ( n − j )2 k ≤ k p k k p ( n − j )2 k ≤ k p k k p k . Thusthe second inequality also holds. (cid:3) Both inequality (24) are sharp for p ( z ) = p ( z ) p ( z ), where p ( z ), p ( z )are polynomials with real nonnegative coefficients. In this case k p k = p (1) = k p k k p k (see also Proposition 2.1).Now, using (20), (21), (24), and (5) we get the final result Corollary 5.2.
For n ≥ νk ( T − κ ( c n − κ − n − κ )) ≤ δ m K ρ (1 − ρ ) k p k , (25) and k ( T − κ ( c n − κ − n − κ )) ≤ (2 n + 1) δ m k p k . (26) To realize the factorization method proposed in Section 4 we must calculatethe Laurent coefficients c − n − κ , . . . , c − κ , . . . , c n − κ , of the function p − ( z ) for n ≥ n = max { κ , ν − κ } . 14n general, the coefficients can be found only approximately. In order todo this, we will apply the method suggested by D.A. Bini and A. B¨ottcher(see Theorem 3.3 in [8]). For future applications we will consider more generalsituation than in this work. Moreover, our proof of inequality (27) differs fromthe proof in the above mentioned theorem.Let f ( z ) be a function that analytic in the annulus K = { z ∈ C : r ≤ | z | ≤ R } , 0 < r < < R < ∞ . By f k denote the Laurent coefficients of f ( z ): f k = 12 πi Z | t | = ρ t − k − f ( t ) dt, r ≤ ρ ≤ R. For ℓ, k ∈ Z , ℓ ≥
2, define e f k ( ℓ ) = 1 ℓ ℓ − X j =0 f ( ω j ) ω kj , where ω j = e πiℓ j , j = 0 , . . . , ℓ −
1, are the zeros of the polynomial z ℓ − Theorem 5.2.
Let M K = max z ∈ K | f ( z ) | , ρ = max { r, /R } , and ℓ be an evenpositive integer. Then | f k − e f k ( ℓ ) | < M K (1 − ρ ℓ ) ρ ℓ/ (27) for k = − ℓ/ , . . . , , . . . , ℓ/ . Proof.
Define I = 12 πi Z ∂K f ( t ) t ℓ − k − t ℓ − dt. Calculate the integral by residue theorem. The integrand is analytic in theannulus K except simple poles at the points ω , . . . , ω ℓ − . Sinceres z = ω j f ( z ) z ℓ − k − z ℓ − f ( ω j ) ℓω kj , we have I = ℓ P ℓ − j =0 f ( ω j ) ω kj = e f k ( ℓ ).Therefore, f k − e f k ( ℓ ) = 12 πi Z | t | = R f ( t ) t k +1 (cid:20) − t ℓ t ℓ − (cid:21) dt + 12 πi Z | t | = r f ( t ) t ℓ − k − t ℓ − dt = − πi Z | t | = R f ( t ) t k +1 ( t ℓ − dt + 12 πi Z | t | = r f ( t ) t ℓ − k − t ℓ − dt. Then | f k − e f k ( ℓ ) | ≤ π Z | t | = R | f ( t ) | R k +1 | t ℓ − | | dt | + 12 π Z | t | = r | f ( t ) | r ℓ − k − | t ℓ − | | dt | ≤ R R k min | t | = R | t ℓ − | + M r r ℓ − k min | t | = r | t ℓ − | . Here M R = max | t | = R | f ( t ) | , M r = max | t | = r | f ( t ) | . It is easily seen thatmin | t | = R | t ℓ − | = R ℓ − , min | t | = r | t ℓ − | = 1 − r ℓ . Hence, | f k − e f k ( ℓ ) | ≤ M R R k ( R ℓ −
1) + M r r ℓ − k (1 − r ℓ ) . Now, from the definitions of M K and ρ , it follows that | f k − e f k ( ℓ ) | ≤ M K (1 − ρ ℓ ) (cid:2) ρ ℓ + k + ρ ℓ − k (cid:3) . If k = − ℓ/ , . . . , ℓ/
2, then ρ ℓ + k ≤ ρ ℓ/ and ρ ℓ − k < ρ ℓ/ . Estimate (27) has beenobtained. (cid:3) By the theorem, in order to compute every element of the sequence f M , f M +1 , . . . , f N with the given accuracy, we have to select an appropriate number ℓ . p ( z ) , p ( z )Recall that we consider the coefficients of the polynomial p ( z ) as the initialdata of the factorization problem. First of all we must study the sensitivity ofthe factors p ( z ), p ( z ) with respect to variations in these data. Moreover, wecompute approximately the Laurent coefficients c k of the function f ( z ) = p − ( z )by the formula c k ≈ e c k = 1 ℓ ℓ − X j =0 p ( ω j ) ω kj . (28)We can not consider these variations as a perturbation of p ( z ). Hence we muststudy the sensitivity of the factors p ( z ), p ( z ) with respect to change in theLaurent coefficients separately.For this reason, we first study the behavior of the factorization essentialpolynomials Q ( z ), Q ( z ) of the sequence c n − κ − n − κ under small perturbations.Recall that the sequence consists of the Laurent coefficients of p − ( z ) and hasindices − κ , − κ , where κ = ind T p ( z ) and n ≥ n .We will need some modification of a well known result on the absolute errorfor the solutions of linear systems. The statement can be proved by the standardmethod. Lemma 5.1.
Let A be an invertible matrix and x = A − b . If k A − e A k ≤ q k A − k for some < q < , then e A is invertible and for e x = e A − e b we have k x − e x k ≤ k A − k − q h k A − k k A − e A k k b k + k b − e b k i . (cid:3) (29)16 heorem 5.3. Let n ≥ n and let e c n − κ − n − κ be a sequence such that k c n − κ − n − κ − e c n − κ − n − κ k ≤ qδ k p k (30) for some < q < . Then (i) The indices of the sequence e c n − κ − n − κ are − κ , − κ . (ii) The sequence has the factorization essential polynomials e Q ( z ) , e Q ( z ) . (iii) For n ≥ n + 1 the following estimates are fulfilled: k Q − e Q k ≤ δ k p k − q (cid:20) δ k p k m K ρ − ρ + 1 (cid:21) k c n − κ − − n − κ +1 − e c n − κ − − n − κ +1 k , k Q − e Q k ≤ δ k p k − q k c n − κ − n − κ − e c n − κ − n − κ k . Proof. (i). Form the matrix T − κ ( e c n − κ − n − κ ). If k T − κ ( c n − κ − n − κ ) − T − κ ( e c n − κ − n − κ ) k ≤ q k T − − κ ( c n − κ − n − κ ) k , (31)then T − κ ( e c n − κ − n − κ ) is invertible. This means that the sequence e c n − κ − n − κ has theindices − κ , − κ . Let us find the condition that guarantees the fulfillment ofinequality (31).From inequalities (24) we have qδ k p k ≤ q k p k k p k ≤ q k T − − κ ( c n − κ − n − κ ) k . Since k T − κ ( c n − κ − n − κ ) − T − κ ( e c n − κ − n − κ ) k ≤ k c n − κ − n − κ − e c n − κ − n − κ k , then from inequal-ity (30) it follows (31). Hence the indices of e c n − κ − n − κ are − κ , − κ .(ii). Let e α = ( e α , . . . , e α n ), e β = ( e β , . . . , e β n ) be the solutions of systems (18)for the perturbed sequence e c n − κ − n − κ . Then e Q ( z ) = e α z + · · · + e α n z n + z n +1 and e Q ( z ) = e β + · · · + e β n z n belong to N − κ +1 ( e c n − κ − n − κ ) and satisfy the conditions ofDefinition 4.1. Thus e Q ( z ), e Q ( z ) are the factorization essential polynomials of e c n − κ − n − κ .(iii). For n ≥ n + 1 the inequality k c n − κ − − n − κ +1 − e c n − κ − − n − κ +1 k ≤ qδ k p k is valid,condition (31) is fulfilled, and, by Lemma 5.1, we have k Q − e Q k = k α − e α k ≤ k T − − κ ( c n − κ − − n − κ +1 ) k − q × (cid:2) k T − − κ ( c n − κ − − n − κ +1 ) k k T − κ ( c n − κ − − n − κ +1 ) − T − κ ( e c n − κ − − n − κ +1 ) k k c − κ − − n − κ k + k c n − κ − − n − κ +1 − e c n − κ − − n − κ +1 k (cid:3) . Taking into account estimates (24) and (22) we arrive at the desired inequalityfor k Q − e Q k . The estimate for k Q − e Q k can be proved analogously. (cid:3) p , p under small perturba-tions of p ( z ). Let m = min | z | =1 | p ( z ) | . By Proposition 5.1, if k p − e p k < m ,then e p ( z ) = 0 on T and ind T p ( z ) = ind T e p ( z ). Let e p ( z ) = e p ( z ) e p ( z ) be thefactorization of e p ( z ). By e c j we denote the Laurent coefficients of e p − ( z ).Now we estimate k p − e p k , k p − e p k via k p − e p k . Theorem 5.4.
Let n ≥ n + 1 . If k p − e p k ≤ min n qm , q (1 − q ) m (2 n +1) δ k p k o , then k p − e p k < (2 n + 1) δ k p k (1 − q ) m (cid:20) δ k p k m K ρ − ρ + 1 (cid:21) k p − e p k , (32) and k p − e p k < (2 n + 1) δ k p k (1 − q ) m k p − e p k . (33) Proof.
Let us apply Theorem 5.3. To do this we must estimate k c n − κ − n − κ − e c n − κ − n − κ k via k p − e p k . We have k c n − κ − n − κ − e c n − κ − n − κ k = n − κ X j = − n − κ | c j − e c j | , and | c j − e c j | ≤ π Z | t | =1 | p − ( t ) − e p − ( t ) | | dt | ≤ k p − − e p − k C . Hence, k c n − κ − n − κ − c n − κ − n − κ k ≤ (2 n + 1) k p − − e p − k C . For any Banach algebra A the following inequality k a − − e a − k A ≤ k a − k A − q k a − e a k A holds if k a − e a k A ≤ q k a − k A for some 0 < q < A = C ( T ), k p − k C = m , and k p − e p k C ≤ k p − e p k . Hence, if k p − e p k ≤ qm , then k p − − e p − k C ≤ − q ) m k p − e p k C ≤ − q ) m k p − e p k and k c n − κ − n − κ − e c n − κ − n − κ k ≤ (2 n + 1)(1 − q ) m k p − e p k . If k p − e p k ≤ min { qm , q (1 − q ) m (2 n +1) δ k p k } , then condition (30) of Theorem 5.3 isfulfilled. Applying this theorem we arrive desired statement. (cid:3) Now we consider perturbations of the polynomials p ( z ), p ( z ) caused by theapproximation of the sequence c n − κ − n − κ by e c n − κ − n − κ , where e c k = ℓ P ℓ − j =0 1 p ( ω j ) ω kj .Let e p ( z ), e p ( z ) be polynomials that define by Eq. (18) for the sequence e c n − κ − n − κ .18 heorem 5.5. Let n ≥ n + 1 , ℓ is an even integer such that ℓ ≥ n + κ ) , and ρ ℓ/ − ρ ℓ < qm K (4 n + 2) δ k p k . (34) Then k p − e p k < (4 n − δ k p k (1 − q ) m K (cid:20) δ (1 + ρ ) k p k (1 − q )(1 − ρ ) m K + 1 (cid:21) ρ ℓ/ − ρ ℓ , k p − e p k < (4 n + 2) δ k p k (1 − q ) m K ρ ℓ/ − ρ ℓ . Proof.
By formula (27), we have | c k − e c k | < m K ρ ℓ/ − ρ ℓ . Therefore, k c n − κ − − n − κ +1 − e c n − κ − − n − κ +1 k < n − m K ρ ℓ/ − ρ ℓ , k c n − κ − n − κ − e c n − κ − n − κ k < n + 2 m K ρ ℓ/ − ρ ℓ . If condition (34) is fulfilled, then k c n − κ − n − κ − e c n − κ − n − κ k < qδ k p k and in order toobtain the required estimates it is sufficient to apply Theorem 5.3. (cid:3) From this theorem it is easy to obtain
Corollary 5.3.
Let ε < q − q and α = ε (1 − q ) m K δ k p k min (cid:26) (4 n − (cid:18) δ k p k (1 + ρ ) m K (1 − ρ ) (cid:19) , (4 n + 2) δ k p k (cid:27) . If ℓ is an even integer such that ℓ > n + κ , log (cid:16)q α + α (cid:17) | log ρ | , (35) then k p − e p k < ε , k p − e p k < ε . (cid:3)
6. Algorithm and numerical examples
The above results can be summarized in the form of the following algorithm.
Algorithm.
Wiener – Hopf factorization of a scalar polynomialINPUT. The coefficients of the polynomial p ( z ), the parameter ρ of the annulus K , m = min | z | =1 | p ( z ) | , m K = min z ∈ K | p ( z ) | , the given accuracy ∆ for thecoefficients of p ( z ): k p − e p k < ∆.COMPUTATION. 19. Compute the index κ of p ( z ) by formula (19). The result is rounded upto the nearest integer.2. Compute k p k , choose n > ν = deg p . For the sake of simplicity, put q = 1 / δ = 1, or δ = κ + 1, or δ = δ ν p ( κ + 1)( ν − κ + 1). Here δ = e G/π is Boyd’s constant.4. Find accuracy ε , ε for p , p by formulas (32), (33). Compute thetheoretically guaranteed accuracy ε := 10 − d < max { ε , ε } .5. Estimate the condition number k ≤ ˜ d by formula (25) or (26). Put e ε := 10 − d − ˜ d .6. Find an even integer ℓ satisfying inequality (35), where ε := e ε .7. Form the sequence e c n − κ − n − κ by formula (28).8. Form the Toeplitz matrix T − κ +1 ( e c n − κ − n − κ ) and find a basis { R , R } of itskernel. The last operation can be done with the help of SVD.9. Find the factorization essential polynomials Q ( z ) = α z + · · · + α n z n + z n +1 , Q ( z ) = β + β z + · · · + β n z n by (16).10. Verify that the absolute values of the coefficients α , . . . , α n − κ , β ν − κ +1 ,. . . , β n are less than ε and delete these coefficients (see Theorem 4.4).11. ˜ p ( z ) := z − n + κ − Q ( z ), ˜ p ( z ) := Q ( z ).12. endOUTPUT. The coefficients ˜ p k , ˜ p k of the factors ˜ p ( z ), ˜ p ( z ) with the guaranteedaccuracy ε .In the following examples we use the Maple computer algebra system. Allcomputations were performed on a desktop.The polynomial p ( z ) in Example 6.1 satisfies the conditions of Proposi-tion 2.2 and its Wiener – Hopf factorization is actually the spectral factorization. Example 6.1.
Let p ( z ) = ( z +1 / z +1 / · · · ( z +1 / z +2)( z +3) · · · ( z +12) .Taking into account the values of the coefficients of p ( z ) , we choose the precision Digits := . Assume that the accuracy of the input data ∆ is equal to − .We may take ρ := 0 . . Then m = 3 . × , m K = 30 . . We have ν = 22, κ = 11, k p k = 20237817600. Put n = ν + 1 = 23. By Propo-sition 2.2, δ = 1. The computation of the theoretically guaranteed accuracy ε gives the following result ε = 0 . × − . By formula (26), we obtain thefollowing estimate k ( T − κ ( c n − κ − n − κ )) ≤ . × . It follows from this that e ε = 10 − and we get ℓ = 136.In this example the exact output is known p ( z ) = ( z + 1 / z + 1 / · · · ( z + 1 / , p ( z ) = ( z + 2)( z + 3) · · · ( z + 12) . Table 1 shows the results of computations of the factors ˜ p ( z ), ˜ p ( z ). Itcontains coefficients ˜ p k , ˜ p k , absolute errors | ˜ p k − p k | , | ˜ p k − p k | for the coefficients20 able 1: Coefficients ˜ p k , ˜ p k k ˜ p k | ˜ p k − p k | ˜ p k | ˜ p k − p k | k ˜ p − p k k ˜ p − p k p k , p k , and k ˜ p − p k , k ˜ p − p k . For ˜ p k , ˜ p k the number of decimal placesobtained accurately is shown.Thus k ˜ p − p k = 0 . × − < . × − = ε , and k ˜ p − p k =2 . × − < . × − = ε . We obtain p ( z ), p ( z ) with the desiredaccuracy.The following example was taken from [8]. Since p ( z ) has real coefficients p j and p ν − j = p j , we can use Proposition 2.1 and the factorization of p ( z ) is alsothe spectral factorization. Example 6.2.
Let p ( z ) = P i =0 z i + 4 z , Digits := , ∆ = 10 − . We maytake ρ = 0 . . Now m = 1 . , m K = 0 . . In this example ν = 10, κ = 5, k p k = 15, n = ν + 1 = 11. By Proposition 2.1, δ = κ + 1 = 6. For the accuracy ε we obtain t ε = 0 . × − . Fromformula (26) it follows the following estimate k ( T − κ ( c n − κ − n − κ )) ≤ . e ε = 10 − and ℓ = 418.The computed coefficients of the factors ˜ p ( z ), ˜ p ( z ) are given by Table 2.We only indicate 5 decimal places here. Table 2: Coefficients ˜ p k , ˜ p k k p k p k p ( z ), we can use the fol-lowing relation between the factors p ( z ) and p ( z ) in the spectral factorization:21 ( z ) = z κ p (1 /z ) /p (0). For our example we have k ˜ p − z κ ˜ p (1 /z ) / ˜ p (0) k =5 . × − . Moreover, the residual error is k ˜ p ˜ p − p k = 8 . × − .In the next example the random polynomial p ( z ) was generated with thehelp of package Random Tools . Example 6.3.
Let p = z − z + 1310 z + (cid:18) i (cid:19) z + (cid:18) − i (cid:19) z + (cid:18) − i (cid:19) z + (cid:18) i (cid:19) z + (cid:18) −