The autoregressive filter problem for multivariable degree one symmetric polynomials
aa r X i v : . [ m a t h . C A ] J a n The autoregressive filter problem for multivariable degree onesymmetric polynomials
Jeffrey S. Geronimo ∗ , Hugo J. Woerdeman † , and Chung Y. Wong ‡ Abstract
The multivariable autoregressive filter problem asks for a polynomial p ( z ) = p ( z , . . . , z d )without roots in the closed d -disk based on prescribed Fourier coefficients of its spectral densityfunction 1 / | p ( z ) | . The conditions derived in this paper for the construction of a degree onesymmetric polynomial reveal a major divide between the case of at most two variables vs. thethe case of three or more variables. The latter involves multivariable elliptic functions, whilethe former (due to [J. S. Geronimo and H. J. Woerdeman, Ann. of Math. (2) , 160(3):839–906, 2004]) only involve polynomials. The three variable case is treated with more detail, andentails hypergeometric functions. Along the way, we identify a seemingly new relation between F (cid:16) , ; z (cid:17) and F (cid:16) , ; e z (cid:17) . Keywords:
Stable polynomial, spectral density function, Fourier coefficients, hypergeometricfunctions, elliptic functions, autoregressive filter, multivariable Toeplitz matrices.
AMS subject classifications:
The identification problem for wide sense stationary autoregressive stochastic processes is a clas-sical signal processing problem. We consider (wide sense) stationary processes X m = X ( m ,...,m d ) depending on d discrete variables defined on a fixed probability space (Ω , A , P ). We shall assumethat the random variables X m are centered , i.e., their means E ( X m ) equal zero. Recall that thespace L (Ω , A , P ) of square integrable random variables endowed with the inner product of centeredrandom variables h X, Y i := E ( Y ∗ X )is a Hilbert space. A sequence X = ( X m ) m ∈ Z d is called a stationary process on Z d if for m, k ∈ Z d we have that E ( X ∗ m X k ) = E ( X ∗ m + p X k + p ) =: R X ( m − k ) for all p ∈ Z d . It is known that the function R X , termed the covariance function of X , defines a positive semidef-inite function, that is, k X i,j =1 α i α j R X ( r i − r j ) ≥ ∗ School of Mathematics, Georgia Institute of Technology, 225 North Ave, Atlanta, GA 30332. † Department of Mathematics, Drexel University, 3141 Chestnut Street, Philadelphia, PA 19104. Research sup-ported by Simons Foundation grant 355645 and National Science Foundation grant DMS 2000037. ‡ Department of Mathematics, County College of Morris, 214 Center Grove Rd., Randolph, NJ 07869. k ∈ N , α , . . . , α k ∈ C , r , . . . , r k ∈ Z d . Bochner’s theorem [3, 4] on positive semidefinitefunctions states that for such a function R X there is a positive measure µ X defined for Borel setson the torus [0 , π ] d such that R X ( r ) = Z e − i h r,u i dµ X ( u )for all d -tuples of integers r ∈ Z d . The measure µ X is referred to as the spectral distribution measure of the process X .For n = ( n , . . . , n d ) ∈ N d we let n = Q dj =1 { , . . . , n j } . A centered stationary stochastic process X is said to be AR( n ) if there exist complex numbers a k , k ∈ n \ { } , such that for every t , x t + X k ∈ nk =0 a k x t − k = e t , t ∈ Z d , (1)where { e k ; k ∈ Z d } is a white noise zero mean process with variance σ . Here AR stands forauto-regressive. Let H be the standard half-space in Z d ; that is H = { ( k , . . . , k d ) ∈ Z d : there is j ∈ { , . . . , d } with k = · · · = k j − = 0 and k j > } . The AR( n ) process is said to be causal if there is a solution to (1) of the form x t = X k ∈ H ∪{ } φ k e t − k , t ∈ Z d , (2)with P k ∈ H ∪{ } | φ k | < ∞ . Causality based on halfspaces and multivariable generalizations of theone variable case go back to the influential papers by Helson and Lowdenslager [12, 13]. It is notdifficult to see that the AR ( n ) process X is causal if and only if the polynomial˜ p ( z ) = 1 + X k ∈ nk =0 a k z k has no roots in the closed d -disk; we call such a polynomial stable . A causal AR( n ) process is infact positive orthant causal , which by definition means that there is a solution to (1) of the form x t = X k ≥ k =0 φ k e t − k , t ∈ Z d , (3)where k = ( k , . . . , k d ) ≥ k j ≥ j = 1 , . . . , d .The multivariate autoregressive filter design problem is the following. “Given are covariances c k = E ( X ∗ X k ) , k ∈ n. What conditions must the covariances satisfy in order that these are the covariances of a causalAR( n ) process? And in that case, how does one compute the filter coefficients a k , k ∈ n \ { } and σ ?” The papers [16], [19], [17] are useful sources for an explanation how the autoregressive filtersare used in signal processing.The following characterization for the two variable autoregressive filter design problem appearedin [7]. 2 heorem 1.1. [7] Let n, m ∈ N and c kl , ( k, l ) ∈ { , . . . , n }×{ , . . . , m } , be given complex numbers.There exists a causal autoregressive process with the given covariances c kl if and only if there existcomplex numbers c kl , ( k, l ) ∈ { , . . . , n } × {− m, . . . , } , such that1. the ( n + 1)( m + 1) × ( n + 1)( m + 1) doubly indexed Toeplitz matrix Γ = ( c t − s ) s,t ∈{ ,...,n }×{ ,...,m } is positive definite;2. the matrix ( c s − t ) s ∈{ ,...,n }×{ ,...,m } ,t ∈{ ,...,n }×{ ,...,m } has rank equal to nm .In this case one finds the vector σ [ a nm · · · a n · · · a m · · · a as the last row of the inverse of Γ . If we consider the polynomial p ( z ) = σ ˜ p ( z ), then the Fourier coefficients of | p | coincide exactlywith the covariances c k . In other words, d | p | ( k ) = c k , k ∈ n, where b f ( k ) denotes the k th Fourier coefficient of the function f . In the remainder of the paper wewill formulate the problems and results in terms of this direct connection.In this paper we will focus on the case where the polynomial p ( z ) is a degree one symmetricpolynomials in d variables, i.e., p ( z , . . . , z d ) = p + p ( z + · · · + z d ) . In general, a symmetric polynomial is a polynomial where a permutation of the variables does notchange the polynomial. It is easy to see that p is stable if and only if d | p | < | p | . The correspondingautoregressive filter problem is as follows. Problem.
Given a and b . Find, if possible, a degree one stable symmetric polynomial in d variables so that d | p | (0 , , . . . ,
0) = a, d | p | (1 , , . . . ,
0) = b. Clearly, due to the symmetry, we have that d | p | (1 , , . . . ,
0) = d | p | (0 , , . . . ,
0) = · · · = d | p | (0 , . . . , , , so that it suffices to just require c | p | (1 , , . . . ,
0) = b . Notice that a > Theorem 1.2.
The above problem has a solution in d = 2 variables if and only if | b | < a . In thatcase, the polynomial p ( z ) = p + p ( z + z ) is given via a ¯ b ¯ bb a | b | a b | b | a a | p | p ¯ p p ¯ p = . c , − is the only unknown in the matrix Γ, and item 2 in Theorem 1.1requires (cid:20) c , − c , − c , c , (cid:21) = (cid:20) c , − ¯ bb a (cid:21) to be of rank 1, which leads to c , − = | b | a .The main result in this paper addresses the case of d variables, which we will state in the nextsection. Recall that the hypergeometric function F is defined for | z | < F (cid:18) a, bc ; z (cid:19) = ∞ X n =0 ( a ) n ( b ) n ( c ) n z n n ! . Here the Pochhammer function ( q ) n is defined by( q ) n = ( , n = 0; q ( q + 1) · · · ( q + n − , otherwise . When we specify the result for d = 3 variables we obtain the following. Theorem 1.3.
The above problem has a solution in d = 3 variables if and only if | b | < a . In thatcase, one finds the polynomial p ( z ) = p + p ( z + z + z ) by determining c ≥ so that a ( a + 2 c ) a + 2 ac − | b | = ( a + 2 c ) ( a + 2 c ) − | b | F (cid:18) , | b | (( a + 2 c ) − | b | )(( a + 2 c ) − | b | ) (cid:19) , (4) and a ¯ b ¯ b ¯ bb a c cb c a cb c c a is positive definite. Next a solution p ( z ) is found via the equation a ¯ b ¯ b ¯ bb a c cb c a cb c c a | p | p ¯ p p ¯ p p ¯ p = . As one can see there is a significant difference between two and three variables. In two variablesthe unknown in the matrix is easily found by setting c = | b | a , while in three variables one needs tosolve the highly nontrivial equation (4) to find the unknown c in the matrix. The number c playsthe role of c = d | p | (1 , − , , . . . ,
0) = d | p | ( − , , , . . . ,
0) = · · · = d | p | (0 , . . . , , , − , where again we used the symmetry of the polynomial. We will see that c is required to be nonneg-ative (see Proposition 2.2).The paper is organized as follows. In Section 2 we present our main result giving necessary andsufficient condition for the existence of an autoregressive filter with a stable symmetric degree onepolynomial in d variables, as well as a method how to find the polynomial. In Section 3 we furtherspecify the results for the case of three variables and present a new relation between F (cid:16) , ; z (cid:17) and F (cid:16) , ; e z (cid:17) . Finally, in Section 4 we explore finding formulas for other Fourier coefficients inthe three variable case. 4 The main result
We will begin by determining some of the Fourier coefficients of | p ( z ) | , where p ( z ) = p + p ( z + · · · + z d ), z = ( z , . . . z d ). It will be convenient to do a simple scaling and assume that p = 1. Nextwe will write p = − s . We will use the notation D = { z ∈ C : | z | < } , T = { z ∈ C : | z | = 1 } , D = D ∪ T , N = { , , , . . . } . Lemma 2.1.
The polynomial p ( z ) = 1 − s ( z + · · · + z d ) is stable if and only if | s | < d .Proof. Let | s | < d and ( z , . . . z d ) ∈ D d . Then we have that | s ( z + · · · + z d ) | <
1, and thus p ( z ) = 0. This gives that p ( z ) is stable.When | s | ≥ d , then z = · · · = z d = sd yields a root of p ( z ) inside D d . Thus p ( z ) is not stable. (cid:3) For q ∈ Z we let q + = max { , q } and q − = max { , − q } . Proposition 2.2.
Let p ( z ) = 1 − s ( z + · · · + z d ) , | s | < d . Then for k = ( k , . . . , k d ) ∈ Z d , d | p | ( k ) = ∞ X n =0 X P n i = n (cid:18) n + k +1 + · · · + k + d n + k +1 , . . . , n d + k + d (cid:19) (cid:18) n + k − + · · · + k − d n + k − , . . . , n d + k − d (cid:19) | s | n s P j k + j ¯ s P j k − j . Here n , . . . , n d ≥ range over all nonnegative numbers that sum up to n . In particular, d | p | (0 , . . . , > , d | p | (1 , − , , . . . , ≥ . (5) Proof.
For ( z , . . . , z d ) ∈ T d we have1 p ( z ) = ∞ X n =0 s n ( z + · · · + z d ) n = ∞ X n =0 X P n i = n (cid:18) nn , . . . , n d (cid:19) s n z n · · · z n d d , p ( z ) = ∞ X n =0 ¯ s n ( z − + · · · + z − d ) n = ∞ X n =0 X P n i = n (cid:18) nn , . . . , n d (cid:19) ¯ s n z − n · · · z − n d d . Multiplying the two and extracting the coefficient of z k · · · z k d d gives the stated formula for c | p | ( k ).Finally, when k = (0 , . . . ,
0) the number s only appears in | s | n which is always ≥
0, and > n = 0, and when k = (1 , − , , . . . ,
0) the number s only appears in | s | n +2 which is always ≥
0. Clearly, all the multinomial coefficients are nonnegative, and thus (5) follows. (cid:3)
Proposition 2.3.
Let p ( z ) = 1 − s ( z + · · · + z d ) , | s | < d . Put a = d | p | (0 , , . . . , , b = d | p | (1 , , . . . , , c = d | p | (1 , − , , . . . , . Then a > , c ≥ , and the matrix A = a ¯ b ¯ b · · · ¯ bb a c · · · cb c a · · · c ... ... . . . ... b c c · · · a , (6)5 s positive definite. Furthermore A − s ... − s = ... . (7) Proof.
Let 1 | p ( z ) | = X k ∈ Z d c k z k denote its Fourier series. Thus \ | p ( z ) | = c k , k ∈ Z d . Since | p ( z ) | is positive, the multiplicationoperator on L ( T d ) with symbol | p ( z ) | is positive definite. Its matrix representation with respectto the standard monomial basis is ( c k − ℓ ) k,ℓ ∈ Z d . Consequently, any principal submatrix ( c k − ℓ ) k,ℓ ∈ Λ ,Λ ⊆ Z d , is positive definite. If we let Λ = { , e , . . . , e d } , where e j is the j th standard basis vectorof C d , we obtain ( c k − ℓ ) k,ℓ ∈ Λ = a ¯ b ¯ b · · · ¯ bb a c · · · cb c a · · · c ... ... . . . ... b c c · · · a , (8)where a = d | p | (0 , , . . . , , b = d | p | (1 , , . . . , , c = d | p | (1 , − , , . . . , . Thus (8) is positive definite.Next, we have that 1 | p ( z , . . . , z d ) | p ( z , . . . , z d ) = 1 p ( z , . . . , z d ) = X k ∈ N d φ k z − k , z ∈ T d , where φ = 1. Comparing the coefficients of 1 , z , . . . , z d on both sides we get the equality (7). (cid:3) Proposition 2.4.
Let p s ( z ) = 1 − s ( z + · · · + z d ) , | s | < d . Put a ( s ) = d | p s | (0 , , . . . , , b ( s ) = d | p s | (1 , , . . . , . Then a ( s ) is a function of | s | and strictly increasing function for | s | ∈ [0 , d ) , and { a ( s ) : | s | ∈ [0 , d ) } = [1 , γ d ) , where γ d = ∞ X n =0 X P n i = n (cid:18) nn , . . . , n d (cid:19) d − n . (9) We have γ = γ = γ = ∞ and γ d < ∞ for d ≥ . Finally, { | b ( s ) | a ( s ) : | s | ∈ [0 , d ) } = [0 , − γ d ) , where ∞ = 0 . roof. By the established asymptotic that was first ascertained in [21] and later generalized by[22, Theorem 4] and [5, Theorem 5.1], we have X P n i = n (cid:18) nn , . . . , n d (cid:19) d − n ≈ d d/ (4 πn ) (1 − d ) / = Θ( n (1 − d ) / ) as n → ∞ . Thus γ d = ∞ for d ≤
3, and γ d < ∞ for d > a ( s ) = ∞ X n =0 X P n i = n (cid:18) nn , . . . , n d (cid:19) | s | n , thus a ( s ) is a continuous function and is increasing as | s | increases. Further, a (0) = 1 andlim | s |→ d − a ( s ) = γ d , yielding that the range of a ( s ) is the interval [1 , γ d ). Similarly, | b ( s ) | = ∞ X n =0 X P n i = n (cid:18) n + 1 n + 1 , n , . . . , n d (cid:19) (cid:18) nn , . . . , n d (cid:19) | s | n +1 is a continuous function and is increasing as | s | increases. Also, note that b (0) = 0. By (7) we havethat a ( s ) − dsb ( s ) = 1 , and thus | b ( s ) | a ( s ) = 1 d | s | a ( s ) − a ( s ) = 1 d | s | (1 − a ( s ) ) . Since | b (0) | a (0) = 0 and lim | s |→ d − | b ( s ) | a ( s ) = 1 − γ d , the last statement follows. (cid:3) The main result is the following.
Theorem 2.5.
Let d ≥ and define γ d via (9) . Given are a > and b ∈ C Then there exists astable degree one symmetric polynomial p ( z , . . . , z d ) so that d | p | (0 , , . . . ,
0) = a, d | p | (1 , , . . . ,
0) = b, if and only if | b | < (1 − γ d ) a . In that case, the polynomial p ( z ) may be found by finding c ≥ sothat a ( a + ( d − c ) a + ( d − ac + d | b | = 1(2 π ) d − Z [0 , π ] d − p g ( t , . . . , t d ) dt · · · dt d , (10) where g ( t , . . . , t d ) = − | b | a + ( d − c X ≤ j ≤ d cos t j + | b | ( a + ( d − c ) X ≤ j,k ≤ d cos( t j − t k ) × − | b | a + ( d − c X ≤ j ≤ d cos t j + | b | ( a + ( d − c ) − X ≤ j,k ≤ d cos( t j − t k ) , nd the matrix a ¯ b ¯ b · · · ¯ bb a c · · · cb c a · · · c ... ... . . . ... b c c · · · a is positive definite. Subsequently, p ( z ) = p + p ( z + · · · + z d ) is found via the equation a ¯ b ¯ b · · · ¯ bb a c · · · cb c a · · · c ... ... . . . ... b c c · · · a | p | p ¯ p ... p ¯ p = ... . Remark 2.6.
When we put s = ba +( d − ac − d | b | , the right hand side of (10) may be rewritten as πi ) d − Z T d − | − s ( z + · · · + z d ) | p | − s ( z + · · · + z d ) | − | s | dz z · · · dz d z d . In determining the Fourier coefficients of | p ( z ) | , where p ( z ) = 1 − s ( z + · · · + z d ) , | s | < d , we let w = z + · · · + z d , which we will treat as a parameter, and write p ( z ) = p ( z , z , w ) = p ( w ) − s ( z + z ) , where p ( w ) = 1 − sw . We write f ( z ) = | p ( z ) | in Fourier series with w as a parameter f ( z ) = X k,l ∈ Z c kl ( w ) z k z l . Proposition 2.7.
Let p ( z ) = p ( z , z , w ) = p ( w ) − s ( z + z ) , p ( w ) = 1 − sw , | s | < d , and write f ( z ) = | p ( z ) | in Fourier series as f ( z ) = X k,l ∈ Z c kl ( w ) z k z l . Then c ( w ) c , − ( w ) c − , ( w ) c ( w ) c ( w ) c − , ( w ) c ( w ) c , − ( w ) c ( w ) − = (11) | − sw | − ¯ s (1 − sw ) − ¯ s (1 − sw ) − s (1 − ¯ s ¯ w ) ( | − sw | + p | − sw | − | s | | − sw | ) 0 − s (1 − ¯ s ¯ w ) 0 ( | − sw | + p | − sw | − | s | | − sw | ) nd c ( w ) c , − ( w ) c − , ( w ) c − , − ( w ) c ( w ) c ( w ) c − , ( w ) c − , ( w ) c ( w ) c , − ( w ) c ( w ) c , − ( w ) c ( w ) c ( w ) c ( w ) c ( w ) − = " | − sw | − ¯ s (1 − sw ) − ¯ s (1 − sw ) 0 − s (1 − ¯ s ¯ w ) s + ( | − sw | + p | − sw | − | s | | − sw | ) s − ¯ s (1 − sw ) − s (1 − ¯ s ¯ w ) s s + ( | − sw | + p | − sw | − | s | | − sw | ) − ¯ s (1 − sw )0 − s (1 − ¯ s ¯ w ) − s (1 − ¯ s ¯ w ) | − sw | . Proof.
The first inverse follows from [18, Theorem 1.1]. With p ( z , z ) = p + p z + p z + p z z and using the notation from [18, Theorem 1.1] we have A = p p p p p , B = p p p p
00 0 p ,C = p p − p p , C = p p − p p
00 0 0... ... ... ,D = | p | + | p | − | p | p p · · · p p | p | + | p | − | p | p p · · · p p | p | + | p | − | p | . . . . . .... . . . . . . . . . . . . ,D = | p | + | p | − | p | p p · · · p p | p | + | p | − | p | p p · · · p p | p | + | p | − | p | . . . . . .... . . . . . . . . . . . . . To invert D we write D = K K ∗ , where K is an upper triangular bidiagonal Toeplitz operatorwith α on the main diagonal and β on the superdiagonal, where α > β are so that α + | β | = ( | p | + | p | − | p | ) + | p p | , ab = ( p p )( | p | + | p | − | p | ) . Similarly for D . Now we use the formula c ( w ) c , − ( w ) c − , ( w ) c ( w ) c ( w ) c − , ( w ) c ( w ) c , − ( w ) c ( w ) − = AA ∗ − B ∗ B − C ∗ D − C − C ∗ D − C to obtain (11).For the second inverse, we use that the (4,1) entry in the inverse is 0 as p ( z ) does not have a p z z term. It now follows from the inverse block matrix formula P H H H ∗ Q H H ∗ H ∗ R − = (cid:20) P H H ∗ Q (cid:21) −
000 0 0 + (cid:20) Q H H ∗ R (cid:21) − − Q −
00 0 0 , (12)which holds if there is a zero in the (3,1) block of the inverse. (cid:3) roposition 2.8. For p ( z ) = 1 − s ( z + · · · + z d ) , | s | < d , we have d | p | (0 , . . . ,
0) = 1(2 πi ) d − Z T d − | − s ( z + · · · + z d ) | p | − s ( z + · · · + z d ) | − | s | dz z · · · dz d z d . Proof.
In general we have that x ¯ y ¯ yy v y v − = 1 xv − | y | v − ¯ y − ¯ y − y x − | y | v | y | v − y | y | v x − | y | v . Combining this with Proposition 2.7 we find c ( w ) = vxv − | y | , where v = 12 ( | − sw | + p | − sw | − | s | | − sw | ) , x = | − sw | , y = − s (1 − ¯ s ¯ w ) . We have xv − | y | = 12 (cid:16) | − sw | − | s | | − sw | + | − sw | p | − sw | − | s | | − sw | (cid:17) = v p | − sw | − | s | | − sw | = v | − sw | p | − sw | − | s | . Thus c ( w ) = 1 | − sw | p | − sw | − | s | = 1 | − s ( z + · · · + z d ) | p | − s ( z + · · · + z d ) | − | s | . To find the 0th Fourier coefficient of | p ( z ) | we need to compute1(2 πi ) d − Z T d − c ( z + · · · + z d ) dz z · · · dz d z d , which yields the stated formula. (cid:3) It is easy to check the following lemma.
Lemma 2.9.
Suppose that the ( d + 1) × ( d + 1) matrix A = a ¯ b ¯ b · · · ¯ bb a c · · · cb c a · · · c ... ... . . . ... b c c · · · a (13) is invertible. Then the first column of the inverse equals a + ( d − ac − d | b | a + ( d − c − b ... − b . roof. Simply multiply A by the vector to obtain the first standard basis vector. (cid:3) Proof of Theorem 2.5.
By the last statement in Proposition 2.4 we see that | b | a ∈ [0 , − γ d ) isnecessary and sufficient.Next, the polynomial p ( z ) after normalization so that p (0) = 1 will satisfy (7). Starting with A as in (13) we can, by Lemma 2.9, rescale the matrix as a +( d − ca +( d − ac − d | b | A so that the (1,1) entry ofits inverse is 1, which corresponds to the situation where p (0) = 1. Then, again using Lemma 2.9,we find that s = − dpdz | z =0 corresponds to the value s = ba +( d − ac − d | b | . Using this value for s aswell as c | p | (0 , . . . ,
0) = a a +( d − ca +( d − ac − d | b | , we find that Proposition 2.8 yields equality (10). (cid:3) In this section we provide further details when d = 3. To be consistent with earlier results in [8]and [25], we consider the polynomial p ( z , z , z ) = 1 − z + z + z r , r > . Comparing this with the previous section, we make the conversion s = r and require s >
0. Thisis not a significant restriction as a phase appearing in s can always be absorbed in the variablesvia ( z , z , z ) → e iθ ( z , z , z ).We will use the complete elliptic integral of the first kind, which is K ( m ) = Z π p − m sin ( t ) dt = Z √ − t √ − mt dt = π F (cid:18) , m (cid:19) . Theorem 3.1.
Let p ( z , z , z ) = 1 − z + z + z r , r > , and f ( z ) = | p ( z ) | , z = ( z , z , z ) . Write f ( z ) = X k,l,m ∈ Z c klm z k z l z m , ( z , z , z ) ∈ T . Then c = r π Z π √ r + 1 − r cos t √ r − − r cos t dt = (14)2 r π ( r − ( r + 3) K ( 16 r ( r − ( r + 3) ) = r ( r − ( r + 3) F (cid:18) , r ( r − ( r + 3) (cid:19) . Proof of Theorem 3.1.
From Proposition 2.8 with s = r and z = e it we obtain c = 12 π Z π | − e it r | q | − e it r | − r dt. Using that | − e it r | = (1 − cos tr ) + sin tr = r ( r − r cos t + 1), formula (14) follows.Next, use cos t = 2 cos t − ( π − t ) −
1, do a change of variable t → π − t , use thesymmetry of the integrand, and (14) becomes2 r π Z π p ( r + 1) − r sin t p ( r + 3)( r − − r sin t dt. (15)11ow we let p = r ( r +1) and q = r ( r +3)( r − , and use the first formula in Section 2.616 of [9], whichis the equality ∗ Z π dx p (1 − p sin x )(1 − q sin x ) = 1 p − p Z π dα q − q − p − p sin α . This transforms (15) into2 r π p ( r − ( r + 3) Z π q − r ( r − ( r +3) sin t dt = 2 r π ( r − ( r + 3) K ( 16 r ( r − ( r + 3) ) . (cid:3) The following result is inspired by a generating function entry by Paul D. Hanna [10] regardingsequence A002893 on the On-Line Encyclopedia of Integer Sequences (oeis.org). Hanna arrived atthis entry as a variation of the generating function for the triangle of cubed binomial coefficients(sequence A181543 on oeis.org) and numerically verified it for hundreds of terms [11].
Theorem 3.2.
Using the same notation as in Theorem 3.1, we have c = r r − F (cid:18) , r − r − (cid:19) , r > . Proof.
By Proposition 2.2, we have c ( r ) = P ∞ n =0 P n + n + n = n (cid:18) nn , n , n (cid:19) r − n , r > . Letting x = r − , and g ( x ) = ∞ X n =0 X n + n + n = n (cid:18) nn , n , n (cid:19) x n , h ( x ) = 11 − x F (cid:18) , x (1 − x )(1 − x ) (cid:19) , the stated equality now comes down to proving that g ( x ) = h ( x ), | x | < . We will show that both g ( x ) and h ( x ) satisfy the Heun differential equation (see [14]) with initial values x (1 − x )(1 − x ) y ′′ + (1 − x + 27 x ) y ′ + (9 x − y = 0 , y (0) = 1 , y ′ (0) = 3 . (16)If we write g ( x ) = P ∞ n =0 g n x n , | x | < , then it follows from [23, Theorem 1; see also Table 1]that n g n − (10 n − n + 3) g n − + 9( n − g n − = 0 , n ≥ , g = 1 , g = 3 . (17)But then it is a straightforward computation that g ( x ) satisfies (16). Indeed, plugging y = g ( x ) = P ∞ n =0 g n x n in the left hand side of (16) and extracting the coefficient of x n − we obtain n ( n − g n − n − n − g n − +9( n − n − g n − + ng n − n − g n − +27( n − g n − +9 g n − − g n − == n g n + ( − n + 30 n − − n + 20 − g n − + (9 n − n + 54 + 27 n −
54 + 9) g n − = n g n − (10 n − n + 3) g n − + 9( n − g n − = 0 , where in the last step we use (17). ∗ due to a change of variables sin α = √ − p sin x √ − p sin x . h ( x ). Introduce z ( x ) = x (1 − x )(1 − x ) and w ( z ) = F (cid:16) , ; z (cid:17) . Then (see, forinstance, [9, Section 9.15]) (1 − z ) zw ′′ ( z ) + (1 − z ) w ′ ( z ) − w ( z ) = 0 . We have that h ( x ) = − x w ( z ( x )), h ′ ( x ) = − x ) w ( z ( x )) + x (1 − x ) w ′ ( z ( x )), and h ′′ ( x ) = 18(1 − x ) w ( z ( x )) + 54(15 x + 1)(1 − x ) w ′ ( z ( x )) + 4(27 x ) (1 − x ) w ′′ ( z ( x )) . Plugging y = h ( x ) in the left hand side of (16) yields x (1 − x )(1 − x ) h ′′ ( x ) + (1 − x + 27 x ) h ′ ( x ) + (9 x − h ( x ) =108 x (1 − x ) (cid:18) (1 − z ( x )) z ( x ) w ′′ ( z ( x )) + (1 − z ( x )) w ′ ( z ( x )) − w ( z ( x )) (cid:19) = 0 . In addition, it is easy to check that h (0) = 1 , h ′ (0) = 3.Thus both g ( x ) and h ( x ) satisfy (16), and thus by uniqueness we find that h ( x ) = g ( x ). (cid:3) Remark 3.3.
Using the Birkhoff-Trjitzinsky method (see [2], and [15] for complete proofs; see also[24] and [20]) one can obtain that the asymptotics of g n = h n is 0 . · n n (1 + O ( n − )). Fromthis one can deduce that g ( x ) is transcendental over Q ( x ); see [20, Corollary 2.1]. This implies thatthe autoregressive filter problem in three and more variables is significantly more involved fromthe case of one or two variables in the sense that one can no longer expect necessary and sufficientconditions via polynomial expressions with rational coefficients, such as the low rank requirementin two variables. Corollary 3.4. r − F (cid:18) , r − r − (cid:19) = 1( r − ( r + 3) F (cid:18) , r ( r − ( r + 3) (cid:19) . Proof.
Combine Theorems 3.1 and 3.2. (cid:3)
There are formulas that relate F (cid:16) , ; z (cid:17) and F (cid:16) , ; e z (cid:17) (see, for instance, [1, page 112]),but the above equality seems to be of a different nature than those already known.We end this section by providing a proof for Theorem 1.3. Proof of Theorem 1.3.
By Theorem 2.5 we see that | b | a < c fromTheorem 3.2. Let d = 3. As before, the polynomial p ( z ) after normalization so that p (0) = 1 willsatisfy (7). Starting with A as in (13) we can, by Lemma 2.9, rescale the matrix as a +( d − ca +( d − ac − d | b | A so that the (1,1) entry of its inverse is 1, which corresponds to the situation where p (0) = 1. Then,again using Lemma 2.9, we find that r = − dpdz | z =0 corresponds to the value r = ba +( d − ac − d | b | .Using this value for r as well as c = c | p | (0 , . . . ,
0) = a a +( d − ca +( d − ac − d | b | , we find that Proposition2.8 yields equality (4). (cid:3) The three variable case: other Fourier coefficients
In [8] the current authors considered the two variable analog, and obtained the following expressionfor the Fourier coefficients of f ( z , z ) = | − z + z r | − , r > Theorem 4.1. [8, Theorem 1] Let p ( z , z ) = 1 − z + z r with r > , and let c k ,k denote the Fouriercoefficients of its spectral density function f ( z , z ) = | − z + z r | − . Then we have c k ,k = 1 q − r r − r r − ! | k | + | k | , k k ≤ , and c k ,k = (cid:0) | k | + | k || k | (cid:1) r | k | + | k | F , | k | + | k | + 1 , | k | + | k | +12 | k | + 1 , | k | + 1 ; 4 r ! , k k > . In an attempt to obtain a three variable generalization of the above result, we have foundfollowing expressions for the Fourier coefficients c J , J ∈ {− , , } of f ( z , z ) = | − z + z + z r | − , r > Theorem 4.2.
Using the same notation as in Theorem 3.1, we have c = r π Z π r − e it r r − r cos t + 1 r − r cos t − − ! dt = − r r π Z π r − cos t √ r − r cos t + 1 √ r − r cos t − dt,c − , , = r π Z π r r − r cos t − r − r cos t + 1 − r r − r cos t + 1 r − r cos t − dt, and c = r π Z π e it ( r − e it )( p ( r + 1 − r cos t )( r − − r cos t ) + r − − r cos t ) dt = −
12 + r π Z π r cos t − cos 2 t √ r + 1 − r cos t √ r − − r cos t dt = r π Z π cos t r r − r cos t + 1 r − r cos t − dt + r π Z π r − cos t √ r − r cos t + 1 √ r − r cos t − dt −
12 = r π Z π cos t r r − r cos t + 1 r − r cos t − dt + c r . (18) Proof of Theorem 4.2.
From Proposition 2.7 we get c ( e it ) = 2 r r − e it (cid:16)p ( r + 1 − r cos t )( r − − r cos t ) + r − − r cos t (cid:17) − c = π R π c ( e it ) e − it dt we consequently obtain c = r π Z π e it ( e it − r ) √ r − − r cos t ( √ r + 1 − r cos t + √ r − − r cos t ) dt. Multiplying numerator and denominator in the integrand with √ r + 1 − r cos t −√ r − − r cos t ,we obtain r π Z π √ r + 1 − r cos te it ( r − e it ) √ r − − r cos t dt − r π Z π e it ( r − e it ) dt. The second term equals , and for the first term we can take its real part (since we know that c is real). This gives c = −
12 + r π Z π r cos t − cos 2 tr + 1 − r cos t √ r + 1 − r cos t √ r − − r cos t dt = −
12 + r π Z π r cos t − cos 2 t √ r + 1 − r cos t √ r − − r cos t dt. The last equality for c is obtained by using z ( r − z ) = r ( z + r − z ) and applying it to the firstexpression for c .Next, from Proposition 2.7 we find c − , ( e it ) = 4 r √ r + 1 − r cos t √ r − − r cos t ( √ r + 1 − r cos t + √ r − − r cos t ) . Multiplying numerator and denominator with ( √ r + 1 − r cos t − √ r − − r cos t ) we obtain c − , ( e it ) = r ( √ r + 1 − r cos t − √ r − − r cos t ) √ r + 1 − r cos t √ r − − r cos t ! = r r r − r cos t − r − r cos t + 1 − r r − r cos t + 1 r − r cos t − ! . Use now c − , , = π R π c − , ( e it ) dt to obtain the result.The proof for c is similar. (cid:3) In Theorem 3.1 we have expressed c in terms of the complete elliptic integral of the firstkind. We can express the other Fourier coefficients above in terms of the complete elliptic integralof the first, second and third kind, which are K ( m ), E ( m ) and Π( n, m ), respectively, where E ( m ) = Z π p − m sin t dt = π F (cid:18) − , m (cid:19) , and Π( n, m ) = Z π − n sin ( t )) p − m sin ( t ) dt. roposition 4.3. Using the same notation as in Theorem 3.1, we have c = r c − ,c = 13 ( c − r − r − K ( r ( r +3)( r − ) − ( r + 3)( r − E ( r ( r +3)( r − ) − r − r + 1)Π( r ( r +3)( r − , r ( r +3)( r − )4 π ( r − p ( r + 3)( r − c − −
12 + (19)( r + 3)( r − K ( r ( r +3)( r − ) − ( r + 3)( r − E ( r ( r +3)( r − ) + 4( r − r + 1)Π( r − , r ( r +3)( r − )4 π ( r − p ( r + 3)( r − ,c = 3 r c ,c , , − = 12 ( rc − c ) ,c , , − = rc − c . Other Fourier coefficients c J , J ∈ {− , , } , are obtained via c J = c σ ( J ) = c − J , where σ is apermutation.Proof. First observe that1 | p ( z , . . . , z d ) | p ( z , . . . , z d ) = 1 p ( z , . . . , z d ) = X k ∈ N d φ k z − k , z ∈ T d , (20)where φ = 1. If we extract the Fourier coefficients indexed by Λ = { , } on both sides, we obtain c c , − c , − , c , − , − c − c − , , − c − , − , c − , − , − c c c , − , c , − , c − c − , , c − , − , c − , − , c c , − c c , , − c − c − , , − c − , , c − , , − c c c c c − c − , , c − , , c − , , c c , , − c , − , c , − , − c c , , − c , − , c , − , − c c c , − , c , − , c c c , − , c , − , c c , , − c , , c , , − c c , , − c c , , − c c c c c c c c − r − r − r = . Since p is a symmetric polynomial with real coefficients we have that c J = c σ ( J ) = c − J , where σ isa permutation. Thus we obtain c − c r = 1 , (1 − r ) c − c r − c , − , r = 0 , c − c , − , r − c , − , − r = 0 ,c − c , , − + c + c , − , r = 0 , c − c r − c r = 0 . This yields the stated relations between the different Fourier coefficients (see also [25, Proposition3.1.1]). 16inally, we turn to c . To prove the first expression for c , by (18) it suffices to prove Z π cos t r r − r cos t + 1 r − r cos t − dt = (21) − ( r + 3)( r − E ( r ( r +3)( r − ) + ( r − r − K ( r ( r +3)( r − ) − r − r + 1)Π( r ( r +3)( r − , r ( r +3)( r − ) r ( r − p ( r + 3)( r − . The left hand side of (21) can be rewritten as Z π cos t r r − r cos t + 1 r − r cos t − dt = Z π r r − r cos t + 1 r − r cos t − dt − Z π (1 − cos t ) r r − r cos t + 1 r − r cos t − dt. (22)We will first show that Z π r r − r cos t + 1 r − r cos t − dt = 4 (cid:16) K ( r ( r +3)( r − ) + ( r − r + 1)Π( r ( r +3)( r − , r ( r +3)( r − ) (cid:17) ( r − p ( r + 3)( r − . (23)To prove (23) we need to show the equality Z π r r − r cos t + 1 r − r cos t − dt = Z π p ( r + 3)( r − − r sin t ( r + 3)( r − − r sin t dt. To prove the above equality we make some simplifications. In the second integral because everythingis in terms of sin t the integral from [0 , π ] is equal to 4 times the integral [0 , π/ t = 1 − t/ t → t/ , π/
2] and divide by 4 to obtain, r − p ( r − r + 1) Z π vuut r ( r − sin t r ( r − r +1) sin t dt = r r − r + 3 Z π q − r ( r +3)( r − sin t − r ( r +3)( r − sin t dt. Now let p = − r ( r − and q = − r ( r − r +1) . Then 1 − q = ( r +3)( r − r − r +1) and q − p − q = r ( r +3)( r − . Theintegrals become Z π s − p sin t − q sin t dt = 1 p − q Z π q − q − p − q sin t q − q sin t dt. (24)On the right hand integral make the change of variable sin t = √ − q sin x √ − q sin x † , then [0 , π/ → [0 , π/ tdt = p − q sin x p − q cos x − p − q sin x √ − q sin x ( − q sin x cos x )1 − q sin x dx. Using cos t = q − (1 − q ) sin x − q sin x = cos x √ − q sin x , we find dt = p − q − q sin x dx. † The reverse change of variables is sin x = sin t √ − q + q sin t and we get dx = √ − q − q + q sin t dt. Z π (1 − cos t ) r r − r cos t + 1 r − r cos t − dt = 2 Z π (1 − cos t ) vuut r − r − cos t r − r − cos t dt. (25)By using the change of variables u = cos t (and thus dt = − √ − u du ), we can rewrite this as2 Z − vuut ( r +12 r − u )(1 − u )( r − r − u )( u − ( − du. Let a = r + 12 r , b = r − r , c = 1 , y = 1 , d = − , and observe that a > b > c ≥ y > d . We can now use [6, Equations 252.17 and 362.16], which yield Z yd s ( c − u )( a − u )( b − u )( u − d ) du =( a − d )( c − d ) g α ( k − α ) (cid:0) α E ( k ) + ( k − α ) K ( k ) + (2 k α − α − k )Π( α , k ) (cid:1) , where g = 2 p ( a − c )( b − d ) , α = d − ca − c , k = ( a − b )( c − d )( a − c )( b − d ) . We obtain that (25) equals( r − ( r + 3) E ( r ( r +3)( r − ) − ( r − ( r + 1) K ( r ( r +3)( r − ) + 4( r + 1) Π( − r ( r − , r ( r +3)( r − ) r p ( r − ( r + 3) . (26)Next we observe that [6, Equation 117.03], after multiplying with ( r − ( r +3)4 r ( r +1) , gives( r + 1) Π( − r ( r − , r ( r + 3)( r − ) = (27)( r − r + 1)Π( 4 r ( r + 3)( r − , r ( r + 3)( r − ) + 4 K ( 16 r ( r + 3)( r − ) . Putting these together with (23), yields (21).To prove the second equality for c from the first, we use (see [6, Formula 117.02])Π( n, m ) = K ( m ) − Π( mn , m ) + π r n (1 − n )( n − m ) , with n = r ( r +3)( r − and m = r ( r +3)( r − . The constant here works out to equal π r +3) ( r − ( r − r +1) . Thus (19) follows. (cid:3)
Equation (20) yields the relations c klm − c k − ,l,m + c k,l − ,m + c k,l,m − r = 0 , ( k, l, m )
6∈ − N . | − z + z + z r | − , r > {− , , } , we will need to be able to findmanageable expressions for (part of) the inverse of more involved infinite (block) Toeplitz matrices,which is a challenge. Acknowledgment.
We would like to thank Professor Bruce C. Berndt for his feedback onCorollary 3.4. Also we thank Paul D. Hanna for giving us background on his entries on the On-Line Encyclopedia of Integer Sequences. Finally, we thank Professor Stephen Melczer for his helpwith Remark 3.3.
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