CALCULATIONS FOR NEWMAN POLYNOMIALS

Transcription

1 iauliai Math. Semin., 4 (12), 2009, CALCULATIONS FOR NEWMAN POLYNOMIALS Tomas PLANKIS Vilnius University, Naugarduko 24, LT Vilnius, Lithuania; topl@hypernet.lt Abstract. In this paper, we will search for an eective algorithm to calculate the quantity min (deg(p ) + 1) H(P 2 ) P P (1) for Newman polynomials P 2 of degree at most 36. We will also give a few computational results and answer an open question mentioned in one of the earlier papers. Key words and phrases: algorithm, Newman polynomials Mathematics Subject Classication: 11C08, 11Y16, 68Q Introduction Let P (x) = a 0 + a 1 x a n x n C[x], where a n 0. We say that n = deg P is the degree of P, and H(P ) = max a i is its height. Then P (x) 0 i n is called a Newman polynomial (after [6]) if its every coecient belongs to the set {0, 1}. Sidon sets were studied by Martin and O'Bryant in [5]. As a result, some problems could be solved with the help of Newman polynomials. In [7], Yu considered the quantity lim inf k deg(p k )H(P 2 k ) P k (1) 2, where P k, k = 1, 2,..., is a sequence of Newman polynomials with deg(p 1 ) < deg(p 2 ) <... He conjectured (see Conjecture 2 in [7]) that this limit is always at least 1 if 0 as k. Berenhaut and Saidak [1] proved P k (1) deg P k

2 158 T. Plankis P k (1) deg(p k ) that the condition lim = 0 is indeed necessary in this conjecture. k More precisely, they showed that there is a sequence of Newman polynomials for which the above limit is 8 9. Martin and O'Bryant in their paper gave a lower bound for Sidon sets (see Theorem 1.4) which was obtained using probabilistic methods. This result was also obtained in [2] by Cilleruelo for Newman polynomials. It was shown that a corresponding limit can be as small as π 4 and so the conjecture of [7] was disproved. Similar result was obtained by Kolountzakis in [4]. Dubickas [3] proved that it is sucient to consider the quantity: Q 2 (P ) = (deg(p ) + 1)H(P 2 ) P (1) 2, because there is always a sequence of Newman polynomials P k, k = 1, 2,..., with increasing degrees such that lim inf Q 2(P k ) = Q 2 (P ) (see Theorem 1 in k [3]). In this paper, we will give concrete values of Q 2 (P ) for Newman polynomials of degrees 3 to 36 with the help of computer calculations and will search for an eective algorithm to calculate the minimum of Q 2 (P ) for polynomials of a given degree. We will answer one of the open questions in [1] as well. 2. Main result Let P n be the set of Newman polynomials of degree n, and let P = n=1 P n. For every P = a 0 + a 1 x a n x n, let P = a n + a n 1 x a 0 x n be its reciprocal polynomial. We say that P P n is self-symmetrical if P = P. Note that reciprocal polynomials are symmetric with respect to the height function H. Symmetry leads to the following simple conclusions. The set P n can be divided into three disjoint subsets of self-symmetrical polynomials, a set of polynomials S and a set of their reciprocals S. According to symmetry, it is enough to check two of them: self-symmetrical polynomials and the set S. Also, we can only consider Newman polynomials with the coecient a 0 = 1. Our main result is the following. Theorem 1. For each Newman polynomial P of degree at most 36, we have Q 2 (P ) Q 2 (P 0 ) = = ,

3 Calculations for Newman polynomials 159 where the coecients of the polynomial P 0 of degree 35 in ascing order are given by According to Table 2 and Table 3, we can answer one question of [1]. In the paper mentioned above, author asked the following question: is it true that the same value of Q 2 occurs only for families of polynomials for which P (1) deg P is the same? As we can see, for Q 2 = , there are at least two dierent Newman polynomials of degrees 23 and 29. The following table will provide such Q 2. This answers the second open question in [1]. Q 2 8 Degrees P (1) deg P 3, 7, 11, 15 1, 6, , 9 1, , 27, 34, , 4 5, 25 27, , , 29, Table 1: Values of Q 2 (P ) for some P with dierent degree Now we will analyze some data. The following two tables represent our calculations. Here are given the values min (Q 2 ) of the given polynomial set P P n. Degree min (Q 2 ) P(1) H(P 2 ) Coecients of polynomial: a 0, a 1,..., a Degree Table 2: Calculations (PC)

4 160 T. Plankis How many polynomials in P n give us min (Q 2 )? The answer is not so P simple. It looks obvious that this number deps on degree n, which is wrong. With the slight modication of the algorithm given below, we can get all such polynomials for a given degree n. We have calculated them for polynomials from degree 2 to 20 and got some interesting results. Degree min (Q 2) P(1) H(P 2 ) Coecients of polynomial: a 0, a 1,..., a Degree Table 3: Calculations (Supercomputer) The number of such polynomials t to increase when the degree is increasing. However, the uctuation is very high, especially for prime numbers. Table 4 shows this. The next question would be: does these polynomials have something in common? In fact, almost nothing, because for a given degree such polynomials can be: all prime; all composite; mixed. Degree n Solutions Table 4: Number of polynomials Unfortunately, this characteristic does not dep on primality of the degree n. The only thing in common is that P (1) is the same for the solutions mentioned above. These coecients form the linear regression y = a+bx (see Figure 1). Using Maple, we calculated the coecients and got the equation y = x

5 Calculations for Newman polynomials 161 with dispersion d = Calculated correlation coecients are very close to 1, and that shows that linear model ts almost perfect. The main question would be: is the dispersion correct, or does it t to increase? We do not know the answer Figure 1: Graph for P (1) Next we will look at asymptotic behaviour of the value of min P P n Q 2 (P ). For this purpose we will use the regression analysis. As we can see from Figure 1, data t to decrease according to a particular function. We tried a few most common nonlinear models for this case: logarithmic; power; asymptotic regression (Metcherlich Law of Diminishing Returns). The last one proved itself to be useful. Other models have limit going to innity. The model mentioned above has the form a + be cx. With the help of SPSS 16.0 (Statistical Package for the Social Sciences) we calculated these coecients. Results can be found in Table 5. Parameter Estimate Std. Error 95% Condence Interval Lover Bound Upper Bound a b c Table 5: Coecients for model

6 162 T. Plankis We get an estimate for asymptotic behavior because lim a + x be cx = a (see Figure 2). We must notice that this limit is very close to π 4. In fact, if we take standard error into consideration, we will see that π can be a good candidate for the lower bound (see Theorem 2 in [2]). We will notice that these data look almost periodic with the period T = 3. To be precise, from 23rd degree it looks that data is periodic with period T = 3. If that is true, then we need to calculate not all Newman polynomials, but only those with deg P = 2 (mod 3) in order to nd the next possible minimum. 0,96 0,94 0,92 0,90 0,88 0,86 0,84 0, Figure 2: Graph for Q 2 Now we will give the basic algorithm to calculate Q(P ). 3. Basic algorithm We need to calculate Q 2 (P ) for every P P n. So we need to go trough every polynomial and calculate two additional parameters: height H(P 2 ); length P (1) 2. Before giving the algorithm, let us consider how we will describe the polynomial. It is natural to describe polynomial using its coecients. For this purpose, programming languages contain few convenient data structures. We will use vector associating its indexes with the indexes of coecients of a polynomial. Let A be a vector. Then A[0] = a 0, A[1] = a 1,... Because of conclusions mentioned above it is sucient to do essential work with coecients a 1,..., a deg P 1.

7 Calculations for Newman polynomials 163 Data: deg(p) Result: min Q 2(P ) P P initialization; for i 1 to 2 deg(p 1) 1 do increment vector by 1; create polynomial of degree deg(p ) adding a 0 and a deg P ; calculate H(P 2 ); calculate P (1); calculate Q(P ); Algorithm 1: Basic algorithm The increase of the vector could be achieved by applying the sum modulo 2. The complexity of this part is deg P. Next we need to calculate P 2. We give the simple algorithm for this: for i 0 to deg P do for j 0 to deg P do calculating new polynomial P 2 [i + j] = P 2 [i + j] + P [i] P [j]; Algorithm 2: Squaring a polynomial The complexity of this part is (deg(p )) 2, and to nd H(P 2 ) the complexity is 2 deg(p ). To calculate P (1), the complexity is deg P. All this gives us the complexity for calculating Q(P ) for some P is 4 deg P + (deg P ) 2. This leads to overall complexity of (4n + n 2 ) 2 n 1 for nding the smallest Q(P ) for all P P n. Working code was written in C++ and implemented on PC (2.4 GHz Intel Core 2 Duo) and supercomputer (SGI Altix 4700). As we see, this algorithm is not very fast. If the power of polynomial increases by 1, the amount of time needed to compute Q 2 for all possible polynomials doubles. According to conclusions mentioned in the rst section, we can cut approximately half of the time, but that does not change the fact that the complexity of this algorithm is exponential. In the next section we will talk about possible improvements. 4. Improvements First of all we will mention that, for deg P = 22, supercomputer needs about 20 seconds to nish the algorithm above. It is quite obvious that, for deg(p ) = 36, we will need a few days to accomplish calculations. What improvements can be done?

8 164 T. Plankis Instead of searching through all polynomials of degree n, we could search only for polynomials with the same P (1) if the exact P (1) would be known (for further notice we call subset). That would lead to checking just a very small amount of polynomials compared to all P. More precisely, according to the combinatoric (see permutations with repetitions) we will need to check (deg(p ) 1)! n 1!n 2! polynomials, where n 1 = P (1) 2 and n 2 = deg(p ) 1 n 1. The last section gives us a hint that we could check the interval [[y d], [y + d]]. This means we need to check nine subsets. The algorithm looks as follows. Data: deg P Result: min Q2(P ) P P initialization; for i [y d] to [y + d] do calculate subset for i; for all j in subset do create polynomial of degree deg P adding a 0 and a deg P ; calculate H(P 2 ); calculate P (1); calculate Q(P ); Algorithm 3: Improved algorithm It is dicult to calculate the complexity of the algorithm. But it should be less than n! n 2! n!, where n = deg P 1. Because n! 2πn( n e )n, we get 2 the complexity deg(p ) 2 2deg P deg P. Let look at deg P = 35. Predicted N(P ) = We need to calculate polynomials for interval [21,29]. That means we need to check 1'971'173'936 polynomials, and by our estimation it would be 2'333'606'220. There is a possibility to use 'periodicity' (see residual analysis) in order to predict P (1) more precisely. Then we could reduce from 1'971'173'936 polynomials to 493'212'976 or even less. This could be the next research. References [1] K. S. Berenhaut, F. Saidak, A note on the maximal coecients of squares of Newman polynomials, J. Number Theory 125, (2007). [2] J. Cilleruelo, Maximal coecients of squares of Newman polynomials, preprint (2008).

9 Calculations for Newman polynomials 165 [3] A. Dubickas, Heights of powers of Newman and Littlewood polynomials, Acta Arith. 128(2), (2007). [4] M. N. Kolountzakis, Coecients of squares of Newman polynomials, preprint (2008), [5] G. Martin, K. O'Bryant, Continuous Ramsey theory and Sidon sets, preprint (2002), arxiv:math/ v1. [6] D. J. Newman, An L 1 extremal problem for polynomials, Proc. Amer. Math. Soc. 16, (1965). [7] G. Yu, An upper bound for B 2 [g] sets, J. Number Theory 122, (2007). Received 26 January 2009