Moments of the Rudin-Shapiro Polynomials

The Journal of Fourier Analysis and Applications Moments of the Rudin-Shapiro Polynomials Christophe Doche, and Laurent Habsieger Communicated by Hans G. Feichtinger ABSTRACT. We develop a new approach of the Rudin-Shapiro polynomials. This enables us to compute their moments of even order q for q 3, and to chec a conjecture on the asymptotic behavior of these moments for q even and q 5. 1. Introduction The Rudin-Shapiro polynomials [6, 8] are polynomials of ±1 coefficients. They are defined by the recurrence relation P n+1 z) =P n z)+z n Q n z), 1.1) Q n+1 z) =P n z) z n Q n z). and the first values P 0 z) = Q 0 z) = 1. One easily prove by induction that the degree of P n and Q n is n 1. The moment of order q>0ofa polynomial P is given by the formula M q P )= 1 0 P e iπθ ) q dθ. Since P n and Q n have ±1 coefficients, we obviously get M P n )=M Q n )= n. In 1968 Littlewood [3] evaluated M 4 P n )=M 4 Q n ), a result which was Math Subject Classifications. 11B83, 4A05. Keywords and Phrases. Rudin-Shapiro polynomials, Golay polynomials, Signal theory. Partially supported by the European Community IHRP Program, within the Research Training Networ Algebraic Combinatorics in Europe, grant HPRN-CT- 001-007. c 00 Birhäuser Boston. All rights reserved ISSN 1069-5869

Christophe Doche, and Laurent Habsieger found again by many authors since. In 1980 Saffari [7] showed that the moments of order 4Q + may be computed from the moments of order 4q with q Q, and he proposed the following conjecture about the asymptotic behavior of the moments of the Rudin-Shapiro polynomials. Conjecture 1. For any even integer q we have M q P n ) q/n+1) q/+1 In this paper we show that the moments of order q satisfy a linear recurrence relation in n, with constant coefficients. For q 3, q 0 mod, we compute the minimal recurrence relation and the first values of the moments, which enables us to chec the conjecture for these q s. Our approach leads to a conjecture which implies Conjecture 1. We chec this new conjecture for q 5, q 0 mod 4, thus proving Conjecture 1 for even q 5. In section we show the existence of a linear recurrence relation. In section 3, we use linear algebra tools to reduce Conjecture 1 to an eigenvalue problem, and we state another stronger conjecture. Section 4 is devoted to the description of our experimental results.. Existence of a Recurrence Relation Let q denote a fixed even integer. For any set of variables a, a,b,b ), let us introduce R n z,a,a,b,b )= apn z)+bq n z) ) a P n z 1 )+b Q n z 1 ) )) q/. The expansion of R n may be written as R n z,a,a,b,b )= q n 1)/ = q n 1)/ We define the fundamental polynomial S n z,a,a,b,b )= q/ 1 = q/+1 c,n a, a,b,b )z. c n,na, a,b,b )z = 1 n y n =z R n y,a,a,b,b ). Let us note that, for any pair of complex numbers a, b), the moment of order q of the polynomial ap n z)+bq n z) equalsc 0,n a, ā, b, b), i.e. the constant term in z) of the polynomial S n z,a,ā, b, b). Thus any linear recurrence relation satisfied by the polynomial S n z,a,a,b,b ) will also be

Moments of the Rudin-Shapiro Polynomials 3 satisfied by the moments of order q of the polynomial ap n z) +bq n z), for any choice of a, b). The Rudin-Shapiro polynomials correspond to the special case a, b) =1, 0). We start with a basic formula. Lemma. For every nonnegative integer n, we have S n+1 z,a,a,b,b )= 1 S n y,a + b, a + b, a b)y,a b )y 1). y =z Proof. By definition, the polynomial S n+1 z,a,a,b,b )equals 1 n+1 y n+1 =z By 1.1), we have R n+1 y,a,a,b,b )= 1 y 1 =z 1 n y n =y 1 R n+1 y,a,a,b,b ). ap n+1 z)+bq n+1 z) = a + b)p n z)+a b)z n Q n z), a P n+1 z 1 )+b Q n+1 z 1 ) = a + b )P n z 1 )+a b )z n Q n z 1 ), and therefore R n+1 z,a,a,b,b )=R n z,a + b, a + b, a b)z n, a b )z n ). We then deduce that S n+1 z,a,a,b,b ) = 1 = 1 1 n y1 =z y 1 =z S n and the lemma follows. y n =y 1 R n y,a+ b, a + b, a b)y n, a b ) )y n y1,a+ b, a + b, a b)y 1, a b )y 1 1 ), We now loo for a finite-dimensional vector space which contains the polynomials S n z,a,a,b,b ) and which is invariant under the transformation described in Lemma. The polynomials S n z,a,a,b,b ) are bihomogeneous in a, b), a,b ) ) of bidegree q/,q/), and their degree in z belongs to the set q/ + 1,...,q/ 1}. Let us introduce the new set of variables u = aa + bb, v = ab + a b, w = aa bb, x = ab a b

4 Christophe Doche, and Laurent Habsieger so that u v = w x and S 0 z,a,a,b,b )=u + v) q/. Let E be the complex vector space of the polynomials in z,z 1,u,v,w,x) which are homogeneous in u, v, w, x) ofdegreeq/, and whose degree in z belongs to q/+1,...,q/ 1}. LetE 0 be the subspace of E consisting of the elements of E which are invariant under the conjugacy action x, z) x, z 1 ). Let F be the quotient space of E 0 by the principal ideal generated by the polynomial u v w + x. We shall now consider S 0 z,a,a,b,b ) as an element of F. Let us define on F a linear map T by T P z,z 1,u,v,w,x) ) = 1 P y,y 1, u, c y w+s y x), v, s y w+c y x) ) y =z where c y =y + y 1 )/ ands y =y y 1 )/. One easily checs that T is well-defined on F, and corresponds to the transformation given in Lemma : S n z,a,a,b,b )=T n S 0 z,a,a,b,b ) ). Since F is finite-dimensional, the characteristic polynomial of T induces a linear recurrence relation satisfied by S n z,a,a,b,b ), with constant coefficients. It appears that the minimal recurrence relation comes from the minimal polynomial of T, whose degree is smaller than the dimension of F. 3. Another Conjecture It is obvious that u q/ is an eigenvector of T, associated to the eigenvalue q/. It would be nice to prove that this eigenspace is of dimension one, to compute the projection of S 0 z,a,a,b,b ) and to show that the other eigenvalues of T have a smaller modulus. We shall give partial results in that direction, using smaller vector spaces. An easy computation gives T S 0 z,a,a,b,b ) ) = T u + v) q/) = q/ We define 0 q/ q/ ) u q/ c z w + s z x). S 0 z,a,a,b,b )= S 0z,a,a,b,b )+S 0 z,a,a, b, b ) = ) q/ u q/ v, 0 q/

Moments of the Rudin-Shapiro Polynomials 5 so that T S 0 z,a,a,b,b ) ) = T S 0 z,a,a,b,b ) ). Let us study the subspace G F spanned by B 1 B where and B 1 = B = q/ 1 m=0 q/ 1 m=1 j,,l 0 j++l=q/ j,,l 0 j++l=q/ 1 z m + z m )u j v w l} } z m z m )u j v w l x. Then S 0 z,a,a,b,b ) belongs to G and T G) G. Indeed, for every element z m + z m )u j v w l of B 1,wehaveT z m + z m )u j v w l) =0ifm is odd and T z m + z m )u j v w l) = q/ z m/ + z m/ )u j cz +1 w + c ) z 1 x + s z wx v l if m is even. Similarly, for every element z m z m )u j v w l x of B,wehave T z m z m )u j v w l x ) =0ifm is even and ) T z m z m )u j v w l x = q/ z m 1)/ + z m 1)/) u j cz +1 q/ z m+1)/ + z m+1)/) u j cz +1 q/ z m 1)/ z m 1)/) u j cz +1 q/ z m+1)/ z m+1)/) u j cz +1 w + c z 1 x + s z wx) v l w w + c z 1 x + s z wx) v l w w + c z 1 x + s z wx) v l x w + c z 1 x + s z wx) v l x if m is odd. We then use hyperbolic trigonometry formulas to express these images as linear combination of elements of B 1 B. Let } B 0 = u q/ v : 0 q/, B 0 = u q/ } u q/ l v w l B 1 = j,,l 0 j++l+1=q/,l 0 0<+l q/4 u j v w l+1 }, )! + l)!l)!!l! +l)! } u q/ +l +1

6 Christophe Doche, and Laurent Habsieger B 1 = q/ 1 m=1 j,,l 0 j++l=q/ z m + z m )u j v w l}. Put B = B 0 B 1 B 1 B ; it is still a basis of G. LetG 0 denote the subspace spanned by u q/ and let G 1 be the subspace of G spanned by the elements of B distinct from u q/. This choice of basis is motivated by the following lemma. Lemma 3. T G 1 ) G 1. Proof. Let π denote the projection on the subspace spanned by B 0. every element z m + z m )u j v w l of B 1, we find π T z m + z m )u j v w l)) 0 if m>0, = 0 if m =0andl 1 mod. For When m =0andl 0 mod, we obtain π T u j v w l)) = q/ ) u j u v ) v l. Moreover, for every element z m z m )u j v w l x of B,wehave π T z m z m )u j v w x) ) l =0. Let us tae a vector u q/ l v w l,with, l 0and + l q/4, and let us decompose its image by π T in the basis B 0.Weobtain π T u q/ l v w l)) ) = q/ 1) i i) ) u q/ i l v i+l u q/ + C,l u q/, i +l +1 with i=0 ) ) C,l = q/ 1) i 1 i i +l +1 i=0 ) ) 1 = q/ 1) i t i+l dt i i=0 0 ) 1 = q/ t l 1 t ) dt 0

Moments of the Rudin-Shapiro Polynomials 7 = )! + l)!l)!!l! +l)! q/ +l +1 Therefore we always get π T u j v w l )! + l)!l)! u q/ ) ) G 1.!l! +l)! +l +1 Let us now compute the projection of S 0 z,a,a,b,b )ong 0. Lemma 4. S 0 z,a,a,b,b ) u) q/ /q/+1) G 1. Proof. We have S 0z,a,a,b,b ) = = 0 q/ 0 q/ ) q/ u q/ v q/ ) u q/ v uq/ +1 ) + Cu q/. This shows that S 0 z,a,a,b,b ) Cu q/ G 1 with ) q/ 1 C = +1 = 1 ) q/ 1 t dt 1 = 1 0 q/ 1 1 and the lemma is proved. 1 + t) q/ dt = q/ q/+1, 0 q/ We can now state a conjecture which implies Conjecture 1. Conjecture 5. The eigenvalues of the restriction of T to G 1 have a modulus smaller than q/. Let us now give a few applications of this conjecture. By Lemmas 3 and 4, we have the following result. Theorem 6. Assume Conjecture 5 is true. For a, a,b,b ) C 4,wethen have S n z,a,a,b,b ) q/ ) n q/ aa + bb ) q/+1 when n goes to infinity. Considering the constant term of S n a, ā, b, b) leads to the following corollary.

8 Christophe Doche, and Laurent Habsieger Corrolary 7. Assume Conjecture 5 is true. For a, b) C, the moment of order q of ap n z)+bq n z) is equivalent to when n goes to infinity. q/ q/ a + b ) ) n, q/+1 The case a, b) =1, 0) gives the asymptotic behavior of the moments of order q of the Rudin-Shapiro polynomials. 4. Experimental Results We get an exact formula for the moments of order q of the Rudin- Shapiro polynomials as soon as we find a recurrence formula they satisfy, and the first values of the moments at least up to the order of the recurrence). We compute these first moments using the polynomials S n z,a,a,b,b ). We determine them by induction with Lemma. At each step we save the constant term of S n z,1, 1, 0, 0), which is the moment of order q of P n z). We compute the minimal recurrence relation R q studying the corresponding operator T q. For q 0 mod 4, q 3, we get the characteristic polynomial of T q using the command charpoly of PARI [5]. For q =8and 3, the computations are first made modulo large primes and the Chinese Remainder Theorem allows to find the requested polynomial. We then test all the divisors of this characteristic polynomial on the first moments to obtain the minimal recurrence relation. Obviously R X) =X and it is nown that R 4 X) =X X 8. We find R 8 X) =X 1 16X 11 1X 10 + 4416X 9 + 3904X 8 47411X 7 +133939X 6 + 3461888X 5 4718590X 4 46976048X 3 6811549696X 548578304X + 1803886643. Now it is not hard to ensure that R q X) splits into R q X/) up to a suitable power of to mae it monic) times a new factor F q X). This relies on the fact that T u) =u. So when λ is an eigenvalue of T q associated with the eigenvector V λ,thenλ is an eigenvalue of T q+ with the eigenvector uv λ. Each new recurrence relation is thus obtained by multiplying a new factor F q toanownone. Let σf q ) be the maximal modulus of the roots of F q. We observe on the computed examples that σf q ) < q/ for all even q less than 3 so that q/ is the largest root of R q in modulus and that it is simple. Further details on the recurrences can be found in Table 1.

Moments of the Rudin-Shapiro Polynomials 9 q deg R q deg F q σf q )/ q/ c q 4 1 0.50 4/3 8 1 10 0.69 16/5 1 36 4 0.74 64/7 16 78 4 0.76 56/9 0 144 66 0.75 104/11 4 40 96 0.7 4096/13 8 369 19 0.73 16384/15 3 536 167 0.75 65536/17 Table 1. This implies that the moment of order q of P n z) is equivalent to c q nq/. Considering the sequence of linear combinations of the moments associated to the polynomial R q X)/X q/ ), we obtain a geometric sequence with ratio q/, whose first term is nown. This enables us to compute c q and to chec Conjecture 1 directly for q 3. Nevertheless in order to prove Conjecture 5 it is enough to study the spectral radius of the restriction of T q to G 1. This is how we show Conjecture 1 for 36 q 5, q 0 mod 4. More precisely, let N 1 be the matrix of T q in the basis B\u q/ }.Let N denote the l norm of the matrix N and let ρn) be the spectral radius of N. It is nown [4] that ρn) N 1/ for every positive integer. We computed N1 1/ for small values of and checed Conjecture 5 in this way. Thus Conjecture 1 is also true for these values of q, even though we do not now the corresponding recurrence relations. Some of these results have been recently extended to generalized Rudin- Shapiro polynomials [1] and a fast program to compute the exact value of the moments of these polynomials is available at []. It includes all the corresponding minimal recurrences as well as the initial moments. References [1] Doche, C, Even Moments of Generalized Rudin-Shapiro Polynomials, Preprint. [] Doche, C, A GP PARI program to compute the moments of some generalized Rudin Shapiro polynomials. Available from http://www.math.u-bordeaux.fr/ ~ cdoche/. [3] Littlewood, J. E., Some Problems in Real and Complex Analysis, D. C. Heathand Co. Raytheon Education Co., Lexington, Mass., 1968. [4] Ljubič, Ju. I., An Algorithm for Calculating the Spectral Radius of an Arbitrary Matrix, Urain. Mat. Z. 17 1965) 18 135. [5] The PARI Group, Bordeaux, PARI/GP, Version.1.5 00). Available from http://www.parigp-home.de/. [6] Rudin, W., Some Theorems on Fourier Coefficients, Proc. Amer. Math. Soc. 10 1959) 855 859. [7] Saffari, B., personal communication 001).

10 Christophe Doche, and Laurent Habsieger [8] Shapiro, H. S., Extremal Problems for Polynomials and Power Series, Ph.D.thesis, MIT 1951). Received April 1, 1000 Revision received December 31, 000 Division of ICS, Building E6A Room 360, Macquarie University NSW 109, Australia e-mail: doche@ics.mq.edu.au IGD, CNRS UMR 508, Université Claude Bernard Lyon 1, 43, boulevard du 11 novembre 1918 696 Villeurbanne Cedex, France e-mail: Laurent.Habsieger@euler.univ-lyon1.fr