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MATHEMATICS OF COMPUTATION Volume 74, Number 252, Pages 1923–1935 S 0025-5718(05)01736-9 Article electronically published on March 15, 2005
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS CHRISTOPHE DOCHE
Abstract. We know from Littlewood (1968) that the moments of order 4 of the classical Rudin–Shapiro polynomials Pn (z) satisfy a linear recurrence of degree 2. In a previous article, we developed a new approach, which enables us to compute exactly all the moments Mq (Pn ) of even order q for q 32. We were also able to check a conjecture on the asymptotic behavior of Mq (Pn ), namely Mq (Pn ) ∼ Cq 2nq/2 , where Cq = 2q/2 /(q/2 + 1), for q even and q 52. Now for every integer 2 there exists a sequence of generalized () Rudin–Shapiro polynomials, denoted by P0,n (z). In this paper, we extend our () earlier method to these polynomials. In particular, the moments Mq (P0,n ) have been completely determined for = 3 and q = 4, 6, 8, 10, for = 4 and q = 4, 6 and for = 5, 6, 7, 8 and q = 4. For higher values of and q, () we formulate a natural conjecture, which implies that Mq (P0,n ) ∼ C,q nq/2 , where C,q is an explicit constant.
1. Introduction Let T = {z ∈ C | |z| = 1} be the complex torus and let f be an Lq (T)–function. The moment of order q ∈ N of f satisfies 1 2iπt q dt. f e (1) Mq (f ) = 0
The Rudin–Shapiro polynomials [8] are defined by the recurrence relations (2)
n
n
Qn+1 (z) = Pn (z) − z 2 Qn (z)
Pn+1 (z) = Pn (z) + z 2 Qn (z),
and the first values P0 (z) = Q0 (z) = 1. Obviously M2 (Pn ) = 2n . In 1968, Littlewood [5] evaluated M4 (Pn ) and established that M4 (Pn ) ∼ 4n+1 /3. In 1980, Saffari [7] conjectured that Mq (Pn ) ∼
2(n+1)q/2 · q/2 + 1
In [2] we were able to prove this result for q even less than or equal to 52.
Received by the editor March 6, 2004 and, in revised form, May 31, 2004. 2000 Mathematics Subject Classification. Primary 11B83, 11B37, 42C05. Key words and phrases. Rudin–Shapiro polynomials, signal theory, Krawtchouk polynomials. c 2005 American Mathematical Society
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In this article, we investigate the moments of generalized Rudin–Shapiro polynomials as defined in [4]. More precisely let 2, put θ = e2iπ/ and ⎡
(3)
1 ⎢1 ⎢ ⎢ M (z) = ⎢1 ⎢. ⎣ ..
z θz θ2 z .. .
1
θ −1 z
z2 2 2 θ z θ4 z 2
··· ··· ··· .. .
θ (−1)2 z 2
···
.. .
⎤ z (−1) θ −1 z (−1) ⎥ ⎥ 2(−1) (−1) ⎥ θ z ⎥. ⎥ .. ⎦ . θ (−1)(−1) z (−1)
Then the generalized Rudin–Shapiro polynomials are defined by ()
()
P0,0 (z) = · · · = P−1,0 (z) = 1, ⎡ () ⎤ ⎤ () P0,n+1 (z) P0,n (z) ⎢ ⎥ ⎥ ⎢ .. .. ⎢ ⎥ = M z n × ⎢ ⎥ for n 0. . . ⎣ ⎣ ⎦ ⎦ () () P−1,n+1 (z) P−1,n (z) ⎡
(4)
The classical Rudin–Shapiro polynomials correspond to the choice = 2. () Since all the sequences P (z) play a quite symmetric role, we shall focus i,n () n∈N our attention on P0,n (z) n∈N in the remainder. In this paper we extend the methods () used in [2] for = 2 to larger . The results show that the moments Mq P0,n satisfy a linear recurrence and support the following conjecture. Conjecture 1. Let be an integer greater than 1, let q be an even integer and let () P0,n (z) n∈N be defined as in (4). We then have () (n+1)q/2 · Mq P0,n ∼ q/2 + − 1 −1 In the next part, we describe an efficient algorithm to compute the moments. In the third part, we show that these moments must satisfy a linear recurrence and we describe how to obtain it. In the fourth part, we check Conjecture 1 for some small values of and q; see Theorem 1. In the last part, we introduce a generalization of Krawtchouk polynomials in several variables to prove Conjecture 1 under a quite natural assumption; see Conjecture 2 and Theorem 2. Rudin–Shapiro polynomials are widely used in signal theory, and the corresponding moments are useful to approximate their maximum modulus on T. Furthermore, they are related to deep problems in harmonic analysis.
2. Moments computation algorithm ()
Since the number of terms of P0,n (z) grows exponentially with n, it becomes quickly impossible to compute the moments in a na¨ıve way using (1). Instead, let
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
us introduce Sn (a, a , z)
= =
−1
() ai Pi,n (z)
−1
i=0
1925
q/2 () aiPi,n (1/z)
i=0
ck,n (a, a )z k ,
k
where Pi,n is the complex conjugate of Pi,n and where a and a are two vectors of formal parameters, namely a = (a0 , . . . , a−1 ) and a = (a0 , . . . , a−1 ). The quantity 1 () 2iπt q () P (e ) dt Mq P0,n = 0,n ()
()
0
is then obviously equal to the constant term of the Laurent polynomial Sn (α0 , α0 , z) () where α0 = (1, 0, . . . , 0). More generally, the moment of Pi,n (z) is simply equal to c0,n αi , αi where αi is the vector whose coefficients are 0 except the i–th one, which is equal to 1. Now for n 0, let 1 Tn (a, a , z) = n Sn (a, a , y) = ckn ,n (a, a )z k . n k y =z The constant terms of Sn a, a , z and of Tn a, a , z are equal for all n but, unlike Sn , the number of terms of Tn is bounded above independently of n. In addition the Tn ’s can be obtained in a very simple way. Lemma 1. The polynomial Tn+1 (a, a , z) can be deduced from Tn (a, a , z) by the following operations: Let A(z) = tM (z) a and A (z) = tM (z) a , where tM is the complex conjugate of t M . Set dk (a, a )z k = Tn A(z), A (1/z), z . k
Then we have
Tn+1 (a, a , z) =
dk (a, a )z k .
k
Proof. By definition Tn+1 (a, a , z)
1
=
n+1 y
n+1
y =z
From (4), we have also () ai Pi,n+1 (z)
=
i=0
and
−1 i=0
() ai Pi,n+1 (1/z)
−1 −1 j=0
=
w
n
θ ij z j ai
()
Pj,n (z)
i=0
−1 −1 j=0
=y
i=0
θ
−ij
n n So Sn+1 (a, a , z) = Sn A z , A 1/z , z .
=z
1 1 Sn+1 (a, a , w). n n
=
−1
Sn+1 (a, a , y)
n /z j ai
()
Pj,n (1/z).
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We deduce that Tn+1 (a, a , z)
=
n n 1 1 Sn A w , A 1/w , w n n
=
1 Tn A(y), A (1/y), y .
y =z
w
=y
y =z
As Tn A(y), A (1/y), y = dk (a, a )y k this shows that k
Tn+1 (a, a , z) =
dk (a, a )z k ,
k
as claimed.
We immediately derive from this lemma an efficient algorithm to compute the () moments of order q of the polynomials P0,n (z) for n 0. () Algorithm 1 (Computation of Mq P0,n ). Input: An integer , the order q of ()
the moments and a bound N 0. Output: The moments of order q of P0,n (z) for n N. q/2 Step 1. T (a, a , z) ← (a0 + · · · + a−1 )(a0 + · · · + a−1 ) A(z) ← tM (z) a A (z) ← tM (z) a () Mq P0,0 ← 1 n←1 Step 2. while n N do 1 T (a, a , z) ← T A(y), A (1/y), y () y =z Mq P0,n ← the constant term of T (α0 , α0 , z) n←n+1 endwhile () () Step 3. return Mq P0,0 , . . . , Mq P0,N Example. Suppose that = 3. To obtain the moments of the polynomials (3) P0,n (z) of order q = 4, first set T (a0 , a1 , a2 , a0 , a1 , a2 , z)
← [(a0 + a1 + a2 )(a0 + a1 + a2 )]2 .
Then in T (a0 , a1 , a2 , a0 , a1 , a2 , z) substitute • a0 by a0 + a1 + a2 • a1 by (a0 + a1 e2iπ/3 + a2 e4iπ/3 )z • a2 by (a0 + a1 e4iπ/3 + a2 e2iπ/3 )z 2 • a0 by a0 + a1 + a2 • a1 by (a0 + a1 e4iπ/3 + a2 e2iπ/3 )/z • a2 by (a0 + a1 e2iπ/3 + a2 e4iπ/3 )/z 2 . After that, the polynomial T (a0 , a1 , a2 , a0 , a1 , a2 , z) has 324 terms. The transformation 1 T (a0 , a1 , a2 , a0 , a1 , a2 , y) (5) T (a0 , a1 , a2 , a0 , a1 , a2 , z) ← 3 3 y =z
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
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discards the coefficients of degree k in z such that k ≡ 0 mod 3 and divides the degree of the other terms by 3. So T (a0 , a1 , a2 , a0 , a1 , a2 , z) has now only 108 terms (3) and T (1, 0, 0, 1, 0, 0, z) = 19 + 4(z + 1/z). So the moment of order 4 of P0,1 (z), that is the constant term of T (1, 0, 0, 1, 0, 0, z) is equal to 19. Another iteration gives a polynomial with 382 terms. This number reduces to 100 after applying (5), and (3) T (1, 0, 0, 1, 0, 0, z) = 93−6(z +1/z). Thus the moment of P0,2 (z) of order 4 is equal to 93. If we keep iterating these transformations, we successively get the moments (3) of order q = 4 of all the P0,n (z)’s. Remark. It is possible to extend Lemma 1 and Algorithm 1 to a wider class of polynomial sequences with very little change. In particular, we can choose any () () polynomials in C[z] for P0,0 (z), . . . , P−1,0 (z) and any matrix with coefficients () in C[z] for M (z). In this case, replace the assignment of T (a, a , z) and Mq P0,0 in the first step of Algorithm 1 by
−1 −1 q/2 () () T (a, a , z) ← ai Pi,0 (z) ai Pi,0 (1/z) i=0
and
i=0
q/2 () () () . Mq P0,0 ← the constant term of P0,0 (z)P0,0 (1/z) 3. Applications
Let E,q be the vector space over C generated by the basis B,q whose elements are n−1 n−1 n0 z m an0 0 · · · a−1 a0 · · · a−1 , where n0 , . . . , n−1 , n0 , . . . , n−1 are nonnegative integers such that n0 + · · · + n−1 = n0 + · · · + n−1 = q/2 and m belongs to [−q/2 + 1, q/2 − 1]. Obviously Tn (a, a , z) ∈ E,q for every n. Now let n = (n0 , . . . , n−1 ), n = (n0 , . . . , n−1 ) and n−1 n0 n−1 cm,n,n z m an0 0 · · · a−1 a0 · · · a−1 V (a, a , z) = m,n,n
be an element of E,q and define the map T,q by 1 V A(y), A (1/y), y . T,q V (a, a , z) = y =z
Clearly T,q is linear and the image of Tn (a, a , z) by T,q is Tn+1 (a, a , z). This im() plies that the polynomials Tn (a, a , z), hence the moments of order q of the P0,n (z)’s, satisfy a linear recurrence R,q (x) which must divide the minimal polynomial of the linear map T,q . In [2] we were able to determine these recurrences and therefore to compute exactly all the even moments q 32 of the classical Rudin–Shapiro polynomials. In the following, we establish some recurrences satisfied by the moments of the generalized Rudin–Shapiro polynomials for = 3, 4, 5, 6, 7, 8 and some values of q from 4 to 10. Computations have been performed with GP-PARI [1] and Maple.
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In theory, it is sufficient to know the minimal polynomial of the matrix of T,q in the basis B,q to compute the linear recurrence satisfied by the moments. When the dimension of the matrix is small, one computes directly its characteristic polynomial to deduce its minimal polynomial. But quickly the size of the basis and of the coefficients make such a computation impossible. In particular, 2 q/2 + − 1 (q − 1). dim E,q = −1 However, it is possible to get rid of a great number of vectors since many elements of B,q belong to the kernel of T,q . More precisely, it is easy to see that
n−1 −1 n0 a0 · · · a−1 ) = 0 if and only if T,q (z m an0 0 · · · a−1 n
−1
i(ni − ni ) ≡ m mod .
i=0
Even if, in general, this reduction is not sufficient to find a recurrence satisfied by the moments, the number of terms in the new basis cleared from unnecessary vectors gives a smaller upper bound D,q on the degree of the recurrence. Now given a sequence u0 , u1 , . . . , uN , . . . generated by an unknown linear recurrence, we are left with the problem of finding the minimal polynomial R of the recurrence. For this, consider the matrix of maximal rank r: ⎤ ⎡ u1 · · · ur−1 u0 ⎢ u1 u2 · · · ur ⎥ ⎥ ⎢ ⎢ .. .. .. ⎥ . . .. ⎣ . . . ⎦ ur−1
ur
···
u2r−2
Inverting it, we immediately deduce a conjectural minimal polynomial for the recurrence. When r is large, say r > 100, the computation of the inverse is difficult, especially since the coefficients grow exponentially. So it is better to perform the computations modulo small primes, then to apply the Chinese remainder theorem. In addition, for = 5, 7 and 8 the moments do not belong to Z. In this case, we determine the minimal equation over Z[e2iπ/ ] at the cost of longer computations. Once we have the conjectural recurrence polynomial we check that this relation is the right one for all n and not only for the very first values. To do this we check that all the terms of order up to D,q satisfy the equation. This is sufficient since the degree of the recurrence is known to be less than or equal to D,q .
4. Experimental results Given , the moments of order 2 of the generalized Rudin–Shapiro polynomials () P0,n (z) are trivially equal to n , so that they satisfy the recurrence un+1 = un with u0 = 1. The corresponding polynomial is therefore R,2 (x) = x − . Now it is not hard to ensure that R,q (x) splits into R,q−2 (x/) (up to a suitable power of to make it monic) times a new factor F,q (x). This relies on the fact that E = a0 a0 + · · · + a−1 a−1
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
1929
is an eigenvector of T,2 with eigenvalue . Indeed, −1 −1 −1 1 ij −ik aj θ ak θ T,2 (E ) = i=0 j=0 k=0
y =z
=
−1 −1
θ i(j−k) aj ak
i=0 j,k=0
=
−1
ai ai +
i=0
−1 −1
θ i(j−k) aj ak
j,k=0 i=0 j=k
= E . So if Vλ is an eigenvector of T,q with eigenvalue λ, then E Vλ will be an eigenvector q/2 of T,q+2 with eigenvalue λ. As a consequence, E is an eigenvector of T,q with eigenvalue q/2 . For = 2, we found the recurrences R2,q (x) for q even and less than or equal to 32 [2]. We noticed that F2,q = 1 for q ≡ 2 mod 4; that is why only the values corresponding to q ≡ 0 (mod 4) are reported for = 2. Further information on R2,q (x) can be found in Table 1 where ρ(F2,q ) is the maximal modulus of the roots of F2,q (x). For larger , namely for • = 3, q = 4, 6, 8 and 10, • = 4, q = 4 and 6, • = 5, 6, 7 and 8, q = 4, we compute R,q (x) and verify that ρ(F,q ) is less than q/2 . So q/2 is simple and is the root R,q (x) of largest modulus. This implies that the asymptotic behavior of () of Mq Pj,n is C,q nq/2 , where C,q can be deduced from the minimal recurrence polynomial. Indeed, ⎞⎛ ⎞ ⎛ d,q −1 d,q −1 () ⎝ C,q = ⎝ rn,,q Mq Pj,n ⎠ rn,,q nq/2 ⎠ , n=0
n=0
Table 1. Properties of the polynomials R2,q (x)
q
deg R2,q
deg F2,q
ρ(F2,q )/2q/2
C2,q
4
2
1
0.50
4 3 16 5 64 7 256 9 1024 11 4096 13 16384 15 65536 17
8
12
10
0.69
12
36
24
0.74
16
78
42
0.76
20
144
66
0.75
24
240
96
0.72
28
369
129
0.73
32
536
167
0.75
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CHRISTOPHE DOCHE
where d,q −1
rn,,q xn = R,q (x)/ x − q/2 .
n=0
Example. Take = 3. The recurrence R3,4 (x) is equal to (x − 9)F3,4 (x) with F3,4 (x) = x19 + 3x18 − 48x17 − 54x16 + 270x15 − 2970x14 + 21303x13 +66096x12 − 75087x11 + 2991816x10 − 14545737x9 − 65800269x8 +215587899x7 − 1021961043x6 − 3231692721x5 + 30032262351x4 −43907655420x3 − 505583738145x2 + 2353579470675
and
(3) M4 P0,n ∼
1441593785878509060 961062523919006040
9n =
3 n 9 . 2
All the results are gathered in the following theorem. Theorem 1. For = 3, q = 4, 6, 8, 10, = 4, q = 4, 6 and = 5, 6, 7, 8, q = 4 the asymptotic behavior of the moments of the generalized Rudin–Shapiro polynomials () P0,n (z) defined as in (4) satisfy (6)
() q/2 · Mq P0,n (z) ∼ C,q nq/2 with C,q = q/2 + − 1 −1
See Tables 2, 3 and 4 for additional information. In the next section we formulate a natural conjecture which is true for all the cases we have investigated and implies (6) for all and all q. Table 2. Properties of the polynomials R3,q (x)
q
deg R3,q
deg F3,q
ρ(F3,q )/3q/2
D3,q
C3,q
4
20
19
0.74
36
6
83
63
0.73
166
3 2 27 10 27 5 81 7
8
240
157
0.64
525
10
574
334
0.65
1323
Table 3. Properties of the polynomials R4,q (x)
q
deg R4,q
deg F4,q
ρ(F4,q )/4q/2
D4,q
C4,q
4
35
34
0.76
74
6
208
173
0.62
500
8 5 16 5
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
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Table 4. Comparison of the properties of R,4 (x) for = 2, . . . , 8
deg R,4
deg F,4
ρ(F,4 )/2
D,4
C,4
2
2
1
0.50
6
3
20
19
0.74
36
4
35
34
0.76
74
5
57
56
0.80
135
6
80
79
0.84
219
7
113
112
0.86
336
8
145
144
0.88
484
4 3 3 2 8 5 5 3 12 7 7 4 16 9
5. Determination of the constant C,q This work was first carried out with Laurent Habsieger in the case = 2. In particular, he used classical properties of Krawtchouk polynomials to prove that 2q/2 · In the following we introduce a generalization of Krawtchouk polyC2,q = q/2+1 nomials for larger and we get an explicit expression for C,q mimicking his proof when = 2. Let n0 , . . . , n−1 and N be nonnegative integers. We shall use the following notation for the multinomial coefficient:
N n0 · · · n−1
⎧ ⎨
N! = n0 ! · · · n−1 ! ⎩ 0
if n0 + · · · + n−1 = N, if n0 + · · · + n−1 = N.
be the basis of E,q built from B,q , where we replace all the vectors of the Let B,q form (a0 a0 )n0 · · · (a−1 a−1 )n−1 by
(7)
(a0 a0 + · · · + a−1 a−1 )q/2 (a0 a0 )n0 · · · (a−1 a−1 )n−1 − q/2 q/2 + − 1 n0 · · · n−1 −1
except (a−1 a−1 )q/2 , which is replaced by (a0 a0 + · · · + a−1 a−1 )q/2 . Let F,q be the vector space generated by (a0 a0 + · · · + a−1 a−1 )q/2 and G,q the vector space generated by B,q \ (a0 a0 + · · · + a−1 a−1 )q/2 . Lemma 2. In the new basis B,q the projection of (a0 a0 )n0 · · · (a−1 a−1 )n−1 on F,q has coordinate
q/2 n0 · · · n−1
1 · q/2 + − 1 −1
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Proof. These equalities are trivial except for (a−1 a−1 )q/2 . In this case we see that the coordinate of (a−1 a−1 )q/2 along (a0 a0 + · · · + a−1 a−1 )q/2 is equal to q/2 q/2 + − 1 −1 n0 · · · n−1 −1 1− = 1− q/2 q/2 + − 1 q/2 + − 1 n n−1 =q/2 n · · · n −1 −1 0 −1 1 , = q/2 + − 1 −1 which is equal to 1 , q/2 q/2 + − 1 0 · · · 0 q/2 −1 as expected. Lemma 3. We have E,q = F,q ⊕ G,q with T,q (F,q ) = F,q and T,q (G,q ) ⊂ G,q . The first two statements are evident but to prove the last result we need to introduce additional material. Recall that N is a positive integer, and take k0 and k1 such that k0 + k1 = N . Then the classical Krawtchouk polynomials Kn (k0 , k1 , N ), see for example [3], are defined by ∞ Kn (k0 , k1 , N )z n = (1 − z)k0 (1 + z)k1 . n=0
At present, let us introduce the quantities Kn (k, N ), where as usual n stands for (n0 , . . . , n−1 ) and where k0 + · · · + k−1 = N . The Kn (k, N )’s can be viewed as a generalization of Krawtchouk polynomials in several variables. Indeed, they are defined by the generating sequence (8)
n
−1 Kn (k, N )an0 0 an1 1 · · · a−1 =
n
−1
k a0 + θ j a1 + · · · + θ j(−1) a−1 j .
j=0
Note that they are explicitly defined by the formula −1 kj Kn (k, N ) = θ ijni,j . n0,j · · · n−1,j j=0 0i,j−1 ni,j 0, j ni,j =ni
The symmetry and orthogonality properties of the classical Krawtchouk polynomials, which are the core of the proof of Lemma 3 for = 2, can also be generalized to Kn (k, N ) for larger . Lemma 4. Let k, n and n be vectors of integers. Then we have N N (9) Kn (k, N ) = Kk (n, N ) k0 · · · k−1 n0 · · · n−1 and ⎧ ⎨0 N (10) Kn (k, N )Kn (k, N ) = N N ⎩ k0 · · · k−1 k n0 · · · n−1
if n = n , otherwise.
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
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Proof. The symmetry property (9) is immediate using (8). To show the orthogonality property (10), consider N n−1 n−1 (11) a−1 . Kn (k, N )Kn (k, N )an0 0 a0n0 · · · a−1 k · · · k 0 −1 n,n
k
Using (8) and swapping the sums, we see that (11) can be written as −1 kj N a0 +θ j a1 +· · ·+θ j(−1) a−1 a0 +θ −j a1 +· · ·+θ −j(−1) a−1 . k0 · · · k−1 j=0 k
This sum is equal to ⎤N ⎡ −1 ⎣ a0 + θ j a1 + · · · + θ j(−1) a−1 a0 + θ −j a1 + · · · + θ −j(−1) a−1 ⎦ , j=0
i.e., N (a0 a0 + a1 a1 + · · · + a−1 a−1 )N .
(12)
Expanding (12), one remarks that for each monomial involved in this expression the degrees of ai and ai are equal. In particular, if n = n , then
n
−1 −1 an0 0 a0n0 · · · a−1 a−1
n
does not appear in (12); hence the coefficient before this term in (11) is zero. If n = n , it is clear that the coefficient of (a0 a0 )n0 · · · (a−1 a−1 )n−1 in (11) is
N
N . n0 · · · n−1
This proves the result.
Proof of Lemma 3. Given m, n and n , let us compute the image of the monomial n−1 n−1 a−1 by T,q . The result is a sum of monomials having all the z m an0 0 a0n0 · · · a−1 same degree in z, which is (13)
m+
−1
i(ni − ni ).
i=0
n
−1 −1 Therefore, if (13) is nonzero, T,q (z m an0 0 a0n0 · · · a−1 a−1 ) belongs to G,q . This also easily implies that the monomial we are considering is an element of G,q . If (13) is zero, then nj nj −1 −1 −1 n−1 n−1 m n0 n0 ij −ij T,q (z a0 a0 · · · a−1 a−1 ) = ai θ ai θ .
j=0
n
i=0
i=0
Using (8) and Lemma 2 it is easy to ensure that the coefficient of the projection of n−1 n−1 T,q (z m an0 0 a0n0 · · · a−1 a−1 ) on F,q is equal to Cn,n =
k
Kk (n, q/2)Kk (n , q/2) · q/2 q/2 + − 1 k0 · · · k−1 −1
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With the help of (9), we get q/2 Kn (k, q/2)Kn (k, q/2) · Cn,n = q/2 q/2 q/2 + − 1 k0 · · · k−1 k n0 · · · n−1 n0 · · · n−1 −1 From (10) we know that Cn,n is zero if n = n . If n = n we have q/2 Kn (k, q/2)Kn (k, q/2) Cn = 2 k0 · · · k−1 q/2 q/2 + − 1 k
−1
n0 · · · n−1 q/2
· q/2 q/2 + − 1 n0 · · · n−1 −1
=
−1 n−1 But if n = n with n−1 = q/2, the monomial z m an0 0 a0n0 · · · a−1 a−1 has been replaced by the vector of the form (7), which can be written
n
(a0 a0 + · · · + a−1 a−1 )q/2 · q/2 Since T,q is a linear map such that (a0 a0 + · · · + a−1 a−1 )q/2 is one of its eigenvectors for the eigenvalue q/2 , this shows that the image of (14) by T,q has a nil coordinate on F,q . So the image by T,q of any vector basis of G,q has a zero component on F,q , and this ensures that T,q (G,q ) ⊂ G,q . (14)
n
−1 −1 a−1 − Cn z m an0 0 a0n0 · · · a−1
n
The computations we have done for small values of and q suggest that all the eigenvalues of T,q are less than or equal to q/2 in modulus. It seems also that (a0 a0 + · · · + a−1 a−1 )q/2 is the only eigenvector of E,q associated to q/2 . We formulate this result in the following conjecture. Conjecture 2. The eigenvalues of the restriction of T,q to G,q have a modulus smaller than q/2 . Proving this result would imply Conjecture 1. More precisely, Theorem 2. Assume Conjecture 2 is true. We then have () Mq P0,n ∼ C,q qn/2
with
q/2 · C,q = q/2 + − 1 −1
Proof. We know that q/2 is the largest eigenvalue of T,q and that its eigenspace is F,q of dimension 1. So () Mq P0,n ∼ C,q nq/2 , where C,q is the coordinate of the projection of the initial vector q/2 a0 + · · · + a−1 a0 + · · · + a−1
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
on F,q . Namely, C,q
=
=
=
q/2 n0 · · · n−1
1935
2
1 q/2 q/2 + − 1 n n0 · · · n−1 −1 q/2 1 q/2 + − 1 n n0 · · · n−1 −1
q/2 · q/2 + − 1 −1
A collection of GP-PARI [1] functions to compute the moments of the generalized Rudin–Shapiro polynomials for various values of and q is available at [6]. For () instance, the procedure findmoment gives the exact moment of order q of P0,n (z), for any n and for the values of and q discussed in this article. The files include all the factors F,q (x) and the requested initial moments. Acknowledgments I would like to thank Bahman Saffari, who suggested to me to investigate the moments of generalized Rudin–Shapiro polynomials and Laurent Habsieger, with whom I had many discussions on the subject. I am also grateful for the many helpful comments and suggestions of the referee. References 1. C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier, User’s guide to PARI / GP, ver. 2.1.3, 2002. http://www.parigp-home.de/ 2. C. Doche and L. Habsieger, Moments of the Rudin–Shapiro polynomials. Journal of Fourier Analysis and Applications 10 (2004). 3. I. Krasikov and S. Litsyn, On integral zeros of Krawtchouk polynomials. Journal of Combinatorial Theory, Ser. A 150 (1996), 441–447. http://www.eng.tau.ac.il/~ litsyn/publications.html MR1383506 (97i:33005) 4. Z. X. Lei, Some properties of generalized Rudin–Shapiro polynomials, Chinese Ann. Math. Ser. A 12 No 2 (1991), 145–153. MR1112426 (92k:11027) 5. J. E. Littlewood, Some problems in real and complex analysis, D. C. Heath and Co. Raytheon Education Co., Lexington, Mass., 1968. MR0244463 (39:5777) 6. A GP–PARI program to compute the moments of some generalized Rudin–Shapiro polynomials. http://www.math.u-bordeaux.fr/~cdoche/moments.gp 7. B. Saffari, personal communication, (2001). 8. H. S. Shapiro, Extremal problems for polynomials and power series, Ph.D. thesis, MIT, 1951. Division of ICS, Building E6A, Macquarie University, New South Wales 2109 Australia E-mail address: [email protected]
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS CHRISTOPHE DOCHE
Abstract. We know from Littlewood (1968) that the moments of order 4 of the classical Rudin–Shapiro polynomials Pn (z) satisfy a linear recurrence of degree 2. In a previous article, we developed a new approach, which enables us to compute exactly all the moments Mq (Pn ) of even order q for q 32. We were also able to check a conjecture on the asymptotic behavior of Mq (Pn ), namely Mq (Pn ) ∼ Cq 2nq/2 , where Cq = 2q/2 /(q/2 + 1), for q even and q 52. Now for every integer 2 there exists a sequence of generalized () Rudin–Shapiro polynomials, denoted by P0,n (z). In this paper, we extend our () earlier method to these polynomials. In particular, the moments Mq (P0,n ) have been completely determined for = 3 and q = 4, 6, 8, 10, for = 4 and q = 4, 6 and for = 5, 6, 7, 8 and q = 4. For higher values of and q, () we formulate a natural conjecture, which implies that Mq (P0,n ) ∼ C,q nq/2 , where C,q is an explicit constant.
1. Introduction Let T = {z ∈ C | |z| = 1} be the complex torus and let f be an Lq (T)–function. The moment of order q ∈ N of f satisfies 1 2iπt q dt. f e (1) Mq (f ) = 0
The Rudin–Shapiro polynomials [8] are defined by the recurrence relations (2)
n
n
Qn+1 (z) = Pn (z) − z 2 Qn (z)
Pn+1 (z) = Pn (z) + z 2 Qn (z),
and the first values P0 (z) = Q0 (z) = 1. Obviously M2 (Pn ) = 2n . In 1968, Littlewood [5] evaluated M4 (Pn ) and established that M4 (Pn ) ∼ 4n+1 /3. In 1980, Saffari [7] conjectured that Mq (Pn ) ∼
2(n+1)q/2 · q/2 + 1
In [2] we were able to prove this result for q even less than or equal to 52.
Received by the editor March 6, 2004 and, in revised form, May 31, 2004. 2000 Mathematics Subject Classification. Primary 11B83, 11B37, 42C05. Key words and phrases. Rudin–Shapiro polynomials, signal theory, Krawtchouk polynomials. c 2005 American Mathematical Society
1923
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CHRISTOPHE DOCHE
In this article, we investigate the moments of generalized Rudin–Shapiro polynomials as defined in [4]. More precisely let 2, put θ = e2iπ/ and ⎡
(3)
1 ⎢1 ⎢ ⎢ M (z) = ⎢1 ⎢. ⎣ ..
z θz θ2 z .. .
1
θ −1 z
z2 2 2 θ z θ4 z 2
··· ··· ··· .. .
θ (−1)2 z 2
···
.. .
⎤ z (−1) θ −1 z (−1) ⎥ ⎥ 2(−1) (−1) ⎥ θ z ⎥. ⎥ .. ⎦ . θ (−1)(−1) z (−1)
Then the generalized Rudin–Shapiro polynomials are defined by ()
()
P0,0 (z) = · · · = P−1,0 (z) = 1, ⎡ () ⎤ ⎤ () P0,n+1 (z) P0,n (z) ⎢ ⎥ ⎥ ⎢ .. .. ⎢ ⎥ = M z n × ⎢ ⎥ for n 0. . . ⎣ ⎣ ⎦ ⎦ () () P−1,n+1 (z) P−1,n (z) ⎡
(4)
The classical Rudin–Shapiro polynomials correspond to the choice = 2. () Since all the sequences P (z) play a quite symmetric role, we shall focus i,n () n∈N our attention on P0,n (z) n∈N in the remainder. In this paper we extend the methods () used in [2] for = 2 to larger . The results show that the moments Mq P0,n satisfy a linear recurrence and support the following conjecture. Conjecture 1. Let be an integer greater than 1, let q be an even integer and let () P0,n (z) n∈N be defined as in (4). We then have () (n+1)q/2 · Mq P0,n ∼ q/2 + − 1 −1 In the next part, we describe an efficient algorithm to compute the moments. In the third part, we show that these moments must satisfy a linear recurrence and we describe how to obtain it. In the fourth part, we check Conjecture 1 for some small values of and q; see Theorem 1. In the last part, we introduce a generalization of Krawtchouk polynomials in several variables to prove Conjecture 1 under a quite natural assumption; see Conjecture 2 and Theorem 2. Rudin–Shapiro polynomials are widely used in signal theory, and the corresponding moments are useful to approximate their maximum modulus on T. Furthermore, they are related to deep problems in harmonic analysis.
2. Moments computation algorithm ()
Since the number of terms of P0,n (z) grows exponentially with n, it becomes quickly impossible to compute the moments in a na¨ıve way using (1). Instead, let
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
us introduce Sn (a, a , z)
= =
−1
() ai Pi,n (z)
−1
i=0
1925
q/2 () aiPi,n (1/z)
i=0
ck,n (a, a )z k ,
k
where Pi,n is the complex conjugate of Pi,n and where a and a are two vectors of formal parameters, namely a = (a0 , . . . , a−1 ) and a = (a0 , . . . , a−1 ). The quantity 1 () 2iπt q () P (e ) dt Mq P0,n = 0,n ()
()
0
is then obviously equal to the constant term of the Laurent polynomial Sn (α0 , α0 , z) () where α0 = (1, 0, . . . , 0). More generally, the moment of Pi,n (z) is simply equal to c0,n αi , αi where αi is the vector whose coefficients are 0 except the i–th one, which is equal to 1. Now for n 0, let 1 Tn (a, a , z) = n Sn (a, a , y) = ckn ,n (a, a )z k . n k y =z The constant terms of Sn a, a , z and of Tn a, a , z are equal for all n but, unlike Sn , the number of terms of Tn is bounded above independently of n. In addition the Tn ’s can be obtained in a very simple way. Lemma 1. The polynomial Tn+1 (a, a , z) can be deduced from Tn (a, a , z) by the following operations: Let A(z) = tM (z) a and A (z) = tM (z) a , where tM is the complex conjugate of t M . Set dk (a, a )z k = Tn A(z), A (1/z), z . k
Then we have
Tn+1 (a, a , z) =
dk (a, a )z k .
k
Proof. By definition Tn+1 (a, a , z)
1
=
n+1 y
n+1
y =z
From (4), we have also () ai Pi,n+1 (z)
=
i=0
and
−1 i=0
() ai Pi,n+1 (1/z)
−1 −1 j=0
=
w
n
θ ij z j ai
()
Pj,n (z)
i=0
−1 −1 j=0
=y
i=0
θ
−ij
n n So Sn+1 (a, a , z) = Sn A z , A 1/z , z .
=z
1 1 Sn+1 (a, a , w). n n
=
−1
Sn+1 (a, a , y)
n /z j ai
()
Pj,n (1/z).
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CHRISTOPHE DOCHE
We deduce that Tn+1 (a, a , z)
=
n n 1 1 Sn A w , A 1/w , w n n
=
1 Tn A(y), A (1/y), y .
y =z
w
=y
y =z
As Tn A(y), A (1/y), y = dk (a, a )y k this shows that k
Tn+1 (a, a , z) =
dk (a, a )z k ,
k
as claimed.
We immediately derive from this lemma an efficient algorithm to compute the () moments of order q of the polynomials P0,n (z) for n 0. () Algorithm 1 (Computation of Mq P0,n ). Input: An integer , the order q of ()
the moments and a bound N 0. Output: The moments of order q of P0,n (z) for n N. q/2 Step 1. T (a, a , z) ← (a0 + · · · + a−1 )(a0 + · · · + a−1 ) A(z) ← tM (z) a A (z) ← tM (z) a () Mq P0,0 ← 1 n←1 Step 2. while n N do 1 T (a, a , z) ← T A(y), A (1/y), y () y =z Mq P0,n ← the constant term of T (α0 , α0 , z) n←n+1 endwhile () () Step 3. return Mq P0,0 , . . . , Mq P0,N Example. Suppose that = 3. To obtain the moments of the polynomials (3) P0,n (z) of order q = 4, first set T (a0 , a1 , a2 , a0 , a1 , a2 , z)
← [(a0 + a1 + a2 )(a0 + a1 + a2 )]2 .
Then in T (a0 , a1 , a2 , a0 , a1 , a2 , z) substitute • a0 by a0 + a1 + a2 • a1 by (a0 + a1 e2iπ/3 + a2 e4iπ/3 )z • a2 by (a0 + a1 e4iπ/3 + a2 e2iπ/3 )z 2 • a0 by a0 + a1 + a2 • a1 by (a0 + a1 e4iπ/3 + a2 e2iπ/3 )/z • a2 by (a0 + a1 e2iπ/3 + a2 e4iπ/3 )/z 2 . After that, the polynomial T (a0 , a1 , a2 , a0 , a1 , a2 , z) has 324 terms. The transformation 1 T (a0 , a1 , a2 , a0 , a1 , a2 , y) (5) T (a0 , a1 , a2 , a0 , a1 , a2 , z) ← 3 3 y =z
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
1927
discards the coefficients of degree k in z such that k ≡ 0 mod 3 and divides the degree of the other terms by 3. So T (a0 , a1 , a2 , a0 , a1 , a2 , z) has now only 108 terms (3) and T (1, 0, 0, 1, 0, 0, z) = 19 + 4(z + 1/z). So the moment of order 4 of P0,1 (z), that is the constant term of T (1, 0, 0, 1, 0, 0, z) is equal to 19. Another iteration gives a polynomial with 382 terms. This number reduces to 100 after applying (5), and (3) T (1, 0, 0, 1, 0, 0, z) = 93−6(z +1/z). Thus the moment of P0,2 (z) of order 4 is equal to 93. If we keep iterating these transformations, we successively get the moments (3) of order q = 4 of all the P0,n (z)’s. Remark. It is possible to extend Lemma 1 and Algorithm 1 to a wider class of polynomial sequences with very little change. In particular, we can choose any () () polynomials in C[z] for P0,0 (z), . . . , P−1,0 (z) and any matrix with coefficients () in C[z] for M (z). In this case, replace the assignment of T (a, a , z) and Mq P0,0 in the first step of Algorithm 1 by
−1 −1 q/2 () () T (a, a , z) ← ai Pi,0 (z) ai Pi,0 (1/z) i=0
and
i=0
q/2 () () () . Mq P0,0 ← the constant term of P0,0 (z)P0,0 (1/z) 3. Applications
Let E,q be the vector space over C generated by the basis B,q whose elements are n−1 n−1 n0 z m an0 0 · · · a−1 a0 · · · a−1 , where n0 , . . . , n−1 , n0 , . . . , n−1 are nonnegative integers such that n0 + · · · + n−1 = n0 + · · · + n−1 = q/2 and m belongs to [−q/2 + 1, q/2 − 1]. Obviously Tn (a, a , z) ∈ E,q for every n. Now let n = (n0 , . . . , n−1 ), n = (n0 , . . . , n−1 ) and n−1 n0 n−1 cm,n,n z m an0 0 · · · a−1 a0 · · · a−1 V (a, a , z) = m,n,n
be an element of E,q and define the map T,q by 1 V A(y), A (1/y), y . T,q V (a, a , z) = y =z
Clearly T,q is linear and the image of Tn (a, a , z) by T,q is Tn+1 (a, a , z). This im() plies that the polynomials Tn (a, a , z), hence the moments of order q of the P0,n (z)’s, satisfy a linear recurrence R,q (x) which must divide the minimal polynomial of the linear map T,q . In [2] we were able to determine these recurrences and therefore to compute exactly all the even moments q 32 of the classical Rudin–Shapiro polynomials. In the following, we establish some recurrences satisfied by the moments of the generalized Rudin–Shapiro polynomials for = 3, 4, 5, 6, 7, 8 and some values of q from 4 to 10. Computations have been performed with GP-PARI [1] and Maple.
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CHRISTOPHE DOCHE
In theory, it is sufficient to know the minimal polynomial of the matrix of T,q in the basis B,q to compute the linear recurrence satisfied by the moments. When the dimension of the matrix is small, one computes directly its characteristic polynomial to deduce its minimal polynomial. But quickly the size of the basis and of the coefficients make such a computation impossible. In particular, 2 q/2 + − 1 (q − 1). dim E,q = −1 However, it is possible to get rid of a great number of vectors since many elements of B,q belong to the kernel of T,q . More precisely, it is easy to see that
n−1 −1 n0 a0 · · · a−1 ) = 0 if and only if T,q (z m an0 0 · · · a−1 n
−1
i(ni − ni ) ≡ m mod .
i=0
Even if, in general, this reduction is not sufficient to find a recurrence satisfied by the moments, the number of terms in the new basis cleared from unnecessary vectors gives a smaller upper bound D,q on the degree of the recurrence. Now given a sequence u0 , u1 , . . . , uN , . . . generated by an unknown linear recurrence, we are left with the problem of finding the minimal polynomial R of the recurrence. For this, consider the matrix of maximal rank r: ⎤ ⎡ u1 · · · ur−1 u0 ⎢ u1 u2 · · · ur ⎥ ⎥ ⎢ ⎢ .. .. .. ⎥ . . .. ⎣ . . . ⎦ ur−1
ur
···
u2r−2
Inverting it, we immediately deduce a conjectural minimal polynomial for the recurrence. When r is large, say r > 100, the computation of the inverse is difficult, especially since the coefficients grow exponentially. So it is better to perform the computations modulo small primes, then to apply the Chinese remainder theorem. In addition, for = 5, 7 and 8 the moments do not belong to Z. In this case, we determine the minimal equation over Z[e2iπ/ ] at the cost of longer computations. Once we have the conjectural recurrence polynomial we check that this relation is the right one for all n and not only for the very first values. To do this we check that all the terms of order up to D,q satisfy the equation. This is sufficient since the degree of the recurrence is known to be less than or equal to D,q .
4. Experimental results Given , the moments of order 2 of the generalized Rudin–Shapiro polynomials () P0,n (z) are trivially equal to n , so that they satisfy the recurrence un+1 = un with u0 = 1. The corresponding polynomial is therefore R,2 (x) = x − . Now it is not hard to ensure that R,q (x) splits into R,q−2 (x/) (up to a suitable power of to make it monic) times a new factor F,q (x). This relies on the fact that E = a0 a0 + · · · + a−1 a−1
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
1929
is an eigenvector of T,2 with eigenvalue . Indeed, −1 −1 −1 1 ij −ik aj θ ak θ T,2 (E ) = i=0 j=0 k=0
y =z
=
−1 −1
θ i(j−k) aj ak
i=0 j,k=0
=
−1
ai ai +
i=0
−1 −1
θ i(j−k) aj ak
j,k=0 i=0 j=k
= E . So if Vλ is an eigenvector of T,q with eigenvalue λ, then E Vλ will be an eigenvector q/2 of T,q+2 with eigenvalue λ. As a consequence, E is an eigenvector of T,q with eigenvalue q/2 . For = 2, we found the recurrences R2,q (x) for q even and less than or equal to 32 [2]. We noticed that F2,q = 1 for q ≡ 2 mod 4; that is why only the values corresponding to q ≡ 0 (mod 4) are reported for = 2. Further information on R2,q (x) can be found in Table 1 where ρ(F2,q ) is the maximal modulus of the roots of F2,q (x). For larger , namely for • = 3, q = 4, 6, 8 and 10, • = 4, q = 4 and 6, • = 5, 6, 7 and 8, q = 4, we compute R,q (x) and verify that ρ(F,q ) is less than q/2 . So q/2 is simple and is the root R,q (x) of largest modulus. This implies that the asymptotic behavior of () of Mq Pj,n is C,q nq/2 , where C,q can be deduced from the minimal recurrence polynomial. Indeed, ⎞⎛ ⎞ ⎛ d,q −1 d,q −1 () ⎝ C,q = ⎝ rn,,q Mq Pj,n ⎠ rn,,q nq/2 ⎠ , n=0
n=0
Table 1. Properties of the polynomials R2,q (x)
q
deg R2,q
deg F2,q
ρ(F2,q )/2q/2
C2,q
4
2
1
0.50
4 3 16 5 64 7 256 9 1024 11 4096 13 16384 15 65536 17
8
12
10
0.69
12
36
24
0.74
16
78
42
0.76
20
144
66
0.75
24
240
96
0.72
28
369
129
0.73
32
536
167
0.75
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CHRISTOPHE DOCHE
where d,q −1
rn,,q xn = R,q (x)/ x − q/2 .
n=0
Example. Take = 3. The recurrence R3,4 (x) is equal to (x − 9)F3,4 (x) with F3,4 (x) = x19 + 3x18 − 48x17 − 54x16 + 270x15 − 2970x14 + 21303x13 +66096x12 − 75087x11 + 2991816x10 − 14545737x9 − 65800269x8 +215587899x7 − 1021961043x6 − 3231692721x5 + 30032262351x4 −43907655420x3 − 505583738145x2 + 2353579470675
and
(3) M4 P0,n ∼
1441593785878509060 961062523919006040
9n =
3 n 9 . 2
All the results are gathered in the following theorem. Theorem 1. For = 3, q = 4, 6, 8, 10, = 4, q = 4, 6 and = 5, 6, 7, 8, q = 4 the asymptotic behavior of the moments of the generalized Rudin–Shapiro polynomials () P0,n (z) defined as in (4) satisfy (6)
() q/2 · Mq P0,n (z) ∼ C,q nq/2 with C,q = q/2 + − 1 −1
See Tables 2, 3 and 4 for additional information. In the next section we formulate a natural conjecture which is true for all the cases we have investigated and implies (6) for all and all q. Table 2. Properties of the polynomials R3,q (x)
q
deg R3,q
deg F3,q
ρ(F3,q )/3q/2
D3,q
C3,q
4
20
19
0.74
36
6
83
63
0.73
166
3 2 27 10 27 5 81 7
8
240
157
0.64
525
10
574
334
0.65
1323
Table 3. Properties of the polynomials R4,q (x)
q
deg R4,q
deg F4,q
ρ(F4,q )/4q/2
D4,q
C4,q
4
35
34
0.76
74
6
208
173
0.62
500
8 5 16 5
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
1931
Table 4. Comparison of the properties of R,4 (x) for = 2, . . . , 8
deg R,4
deg F,4
ρ(F,4 )/2
D,4
C,4
2
2
1
0.50
6
3
20
19
0.74
36
4
35
34
0.76
74
5
57
56
0.80
135
6
80
79
0.84
219
7
113
112
0.86
336
8
145
144
0.88
484
4 3 3 2 8 5 5 3 12 7 7 4 16 9
5. Determination of the constant C,q This work was first carried out with Laurent Habsieger in the case = 2. In particular, he used classical properties of Krawtchouk polynomials to prove that 2q/2 · In the following we introduce a generalization of Krawtchouk polyC2,q = q/2+1 nomials for larger and we get an explicit expression for C,q mimicking his proof when = 2. Let n0 , . . . , n−1 and N be nonnegative integers. We shall use the following notation for the multinomial coefficient:
N n0 · · · n−1
⎧ ⎨
N! = n0 ! · · · n−1 ! ⎩ 0
if n0 + · · · + n−1 = N, if n0 + · · · + n−1 = N.
be the basis of E,q built from B,q , where we replace all the vectors of the Let B,q form (a0 a0 )n0 · · · (a−1 a−1 )n−1 by
(7)
(a0 a0 + · · · + a−1 a−1 )q/2 (a0 a0 )n0 · · · (a−1 a−1 )n−1 − q/2 q/2 + − 1 n0 · · · n−1 −1
except (a−1 a−1 )q/2 , which is replaced by (a0 a0 + · · · + a−1 a−1 )q/2 . Let F,q be the vector space generated by (a0 a0 + · · · + a−1 a−1 )q/2 and G,q the vector space generated by B,q \ (a0 a0 + · · · + a−1 a−1 )q/2 . Lemma 2. In the new basis B,q the projection of (a0 a0 )n0 · · · (a−1 a−1 )n−1 on F,q has coordinate
q/2 n0 · · · n−1
1 · q/2 + − 1 −1
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CHRISTOPHE DOCHE
Proof. These equalities are trivial except for (a−1 a−1 )q/2 . In this case we see that the coordinate of (a−1 a−1 )q/2 along (a0 a0 + · · · + a−1 a−1 )q/2 is equal to q/2 q/2 + − 1 −1 n0 · · · n−1 −1 1− = 1− q/2 q/2 + − 1 q/2 + − 1 n n−1 =q/2 n · · · n −1 −1 0 −1 1 , = q/2 + − 1 −1 which is equal to 1 , q/2 q/2 + − 1 0 · · · 0 q/2 −1 as expected. Lemma 3. We have E,q = F,q ⊕ G,q with T,q (F,q ) = F,q and T,q (G,q ) ⊂ G,q . The first two statements are evident but to prove the last result we need to introduce additional material. Recall that N is a positive integer, and take k0 and k1 such that k0 + k1 = N . Then the classical Krawtchouk polynomials Kn (k0 , k1 , N ), see for example [3], are defined by ∞ Kn (k0 , k1 , N )z n = (1 − z)k0 (1 + z)k1 . n=0
At present, let us introduce the quantities Kn (k, N ), where as usual n stands for (n0 , . . . , n−1 ) and where k0 + · · · + k−1 = N . The Kn (k, N )’s can be viewed as a generalization of Krawtchouk polynomials in several variables. Indeed, they are defined by the generating sequence (8)
n
−1 Kn (k, N )an0 0 an1 1 · · · a−1 =
n
−1
k a0 + θ j a1 + · · · + θ j(−1) a−1 j .
j=0
Note that they are explicitly defined by the formula −1 kj Kn (k, N ) = θ ijni,j . n0,j · · · n−1,j j=0 0i,j−1 ni,j 0, j ni,j =ni
The symmetry and orthogonality properties of the classical Krawtchouk polynomials, which are the core of the proof of Lemma 3 for = 2, can also be generalized to Kn (k, N ) for larger . Lemma 4. Let k, n and n be vectors of integers. Then we have N N (9) Kn (k, N ) = Kk (n, N ) k0 · · · k−1 n0 · · · n−1 and ⎧ ⎨0 N (10) Kn (k, N )Kn (k, N ) = N N ⎩ k0 · · · k−1 k n0 · · · n−1
if n = n , otherwise.
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
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Proof. The symmetry property (9) is immediate using (8). To show the orthogonality property (10), consider N n−1 n−1 (11) a−1 . Kn (k, N )Kn (k, N )an0 0 a0n0 · · · a−1 k · · · k 0 −1 n,n
k
Using (8) and swapping the sums, we see that (11) can be written as −1 kj N a0 +θ j a1 +· · ·+θ j(−1) a−1 a0 +θ −j a1 +· · ·+θ −j(−1) a−1 . k0 · · · k−1 j=0 k
This sum is equal to ⎤N ⎡ −1 ⎣ a0 + θ j a1 + · · · + θ j(−1) a−1 a0 + θ −j a1 + · · · + θ −j(−1) a−1 ⎦ , j=0
i.e., N (a0 a0 + a1 a1 + · · · + a−1 a−1 )N .
(12)
Expanding (12), one remarks that for each monomial involved in this expression the degrees of ai and ai are equal. In particular, if n = n , then
n
−1 −1 an0 0 a0n0 · · · a−1 a−1
n
does not appear in (12); hence the coefficient before this term in (11) is zero. If n = n , it is clear that the coefficient of (a0 a0 )n0 · · · (a−1 a−1 )n−1 in (11) is
N
N . n0 · · · n−1
This proves the result.
Proof of Lemma 3. Given m, n and n , let us compute the image of the monomial n−1 n−1 a−1 by T,q . The result is a sum of monomials having all the z m an0 0 a0n0 · · · a−1 same degree in z, which is (13)
m+
−1
i(ni − ni ).
i=0
n
−1 −1 Therefore, if (13) is nonzero, T,q (z m an0 0 a0n0 · · · a−1 a−1 ) belongs to G,q . This also easily implies that the monomial we are considering is an element of G,q . If (13) is zero, then nj nj −1 −1 −1 n−1 n−1 m n0 n0 ij −ij T,q (z a0 a0 · · · a−1 a−1 ) = ai θ ai θ .
j=0
n
i=0
i=0
Using (8) and Lemma 2 it is easy to ensure that the coefficient of the projection of n−1 n−1 T,q (z m an0 0 a0n0 · · · a−1 a−1 ) on F,q is equal to Cn,n =
k
Kk (n, q/2)Kk (n , q/2) · q/2 q/2 + − 1 k0 · · · k−1 −1
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CHRISTOPHE DOCHE
With the help of (9), we get q/2 Kn (k, q/2)Kn (k, q/2) · Cn,n = q/2 q/2 q/2 + − 1 k0 · · · k−1 k n0 · · · n−1 n0 · · · n−1 −1 From (10) we know that Cn,n is zero if n = n . If n = n we have q/2 Kn (k, q/2)Kn (k, q/2) Cn = 2 k0 · · · k−1 q/2 q/2 + − 1 k
−1
n0 · · · n−1 q/2
· q/2 q/2 + − 1 n0 · · · n−1 −1
=
−1 n−1 But if n = n with n−1 = q/2, the monomial z m an0 0 a0n0 · · · a−1 a−1 has been replaced by the vector of the form (7), which can be written
n
(a0 a0 + · · · + a−1 a−1 )q/2 · q/2 Since T,q is a linear map such that (a0 a0 + · · · + a−1 a−1 )q/2 is one of its eigenvectors for the eigenvalue q/2 , this shows that the image of (14) by T,q has a nil coordinate on F,q . So the image by T,q of any vector basis of G,q has a zero component on F,q , and this ensures that T,q (G,q ) ⊂ G,q . (14)
n
−1 −1 a−1 − Cn z m an0 0 a0n0 · · · a−1
n
The computations we have done for small values of and q suggest that all the eigenvalues of T,q are less than or equal to q/2 in modulus. It seems also that (a0 a0 + · · · + a−1 a−1 )q/2 is the only eigenvector of E,q associated to q/2 . We formulate this result in the following conjecture. Conjecture 2. The eigenvalues of the restriction of T,q to G,q have a modulus smaller than q/2 . Proving this result would imply Conjecture 1. More precisely, Theorem 2. Assume Conjecture 2 is true. We then have () Mq P0,n ∼ C,q qn/2
with
q/2 · C,q = q/2 + − 1 −1
Proof. We know that q/2 is the largest eigenvalue of T,q and that its eigenspace is F,q of dimension 1. So () Mq P0,n ∼ C,q nq/2 , where C,q is the coordinate of the projection of the initial vector q/2 a0 + · · · + a−1 a0 + · · · + a−1
EVEN MOMENTS OF GENERALIZED RUDIN–SHAPIRO POLYNOMIALS
on F,q . Namely, C,q
=
=
=
q/2 n0 · · · n−1
1935
2
1 q/2 q/2 + − 1 n n0 · · · n−1 −1 q/2 1 q/2 + − 1 n n0 · · · n−1 −1
q/2 · q/2 + − 1 −1
A collection of GP-PARI [1] functions to compute the moments of the generalized Rudin–Shapiro polynomials for various values of and q is available at [6]. For () instance, the procedure findmoment gives the exact moment of order q of P0,n (z), for any n and for the values of and q discussed in this article. The files include all the factors F,q (x) and the requested initial moments. Acknowledgments I would like to thank Bahman Saffari, who suggested to me to investigate the moments of generalized Rudin–Shapiro polynomials and Laurent Habsieger, with whom I had many discussions on the subject. I am also grateful for the many helpful comments and suggestions of the referee. References 1. C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier, User’s guide to PARI / GP, ver. 2.1.3, 2002. http://www.parigp-home.de/ 2. C. Doche and L. Habsieger, Moments of the Rudin–Shapiro polynomials. Journal of Fourier Analysis and Applications 10 (2004). 3. I. Krasikov and S. Litsyn, On integral zeros of Krawtchouk polynomials. Journal of Combinatorial Theory, Ser. A 150 (1996), 441–447. http://www.eng.tau.ac.il/~ litsyn/publications.html MR1383506 (97i:33005) 4. Z. X. Lei, Some properties of generalized Rudin–Shapiro polynomials, Chinese Ann. Math. Ser. A 12 No 2 (1991), 145–153. MR1112426 (92k:11027) 5. J. E. Littlewood, Some problems in real and complex analysis, D. C. Heath and Co. Raytheon Education Co., Lexington, Mass., 1968. MR0244463 (39:5777) 6. A GP–PARI program to compute the moments of some generalized Rudin–Shapiro polynomials. http://www.math.u-bordeaux.fr/~cdoche/moments.gp 7. B. Saffari, personal communication, (2001). 8. H. S. Shapiro, Extremal problems for polynomials and power series, Ph.D. thesis, MIT, 1951. Division of ICS, Building E6A, Macquarie University, New South Wales 2109 Australia E-mail address: [email protected]
