THE MONIC INTEGER TRANSFINITE DIAMETER

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1 MATHEMATICS OF COMPUTATION Volume 00, Number 0, Pages S (XX) THE MONIC INTEGER TRANSFINITE DIAMETER K. G. HARE AND C. J. SMYTH ABSTRACT. We stuy the problem of fining nonconstant monic integer polynomials, normalize by their egree, with small supremum on an interval I. The monic integer transfinite iameter t M I is efine as the infimum of all such supremums. We show that if I has length then t M I. We make three general conjectures relating to the value of t M I for intervals I of length less that 4. We also conjecture a value for t M 0 b where 0 b. We give some partial results, as well as computational evience, to support these conjectures. We efine functions L t an L t, which measure properties of the lengths of intervals I with t M I on either sie of t. Upper an lower bouns are given for these functions. We also consier the problem of etermining t M I when I is a Farey interval. We prove that a conjecture of Borwein, Pinner an Pritsker concerning this value is true for an infinite family of Farey intervals.. INTRODUCTION AND RESULTS In this paper we continue a stuy, recently initiate by Borwein, Pinner an Pritsker [], of the monic integer transfinite iameter of a real interval. We write the normalize supremum on an interval I as P I : sup P x egp x I Note that this is not a norm. Then the monic integer transfinite iameter t M I is efine as t M I : inf P I P where the infimum is taken over all non-constant monic polynomials with integer coefficients. We call t M I the monic integer transfinite iameter of I (also calle the monic integer Chebyshev constant [, ]). Clearly t M I t I, where t I enotes the integer transfinite iameter, efine using the 000 Mathematics Subject Classification. Primary C08; Seconary 30C0. Key wors an phrases. Chebyshev polynomials, monic integer transfinite iameter. Research of K.G. Hare was supporte in part by NSERC of Canaa an a Seggie Brown Fellowship, University of Einburgh. c005 American Mathematical Society

2 K. G. HARE AND C. J. SMYTH same infimum, but taken over the larger set of all non-constant polynomials with integer coefficients [3, 4, 5]. Further t I cap I, the capacity or transfinite iameter of I [6, 4], which can be efine again using the same infimum, but this time taken over all non-constant monic polynomials with real coefficients. It is well known that cap I I 4 for an interval I of length I. Further, if I 4 then t I t M I cap I by [] so that the challenge for evaluating t M I, as for t I, lies in intervals with I 4. For these intervals we know from [, Prop..] that t M I. However, in contrast to the stuy of t I, in the monic case it is possible to evaluate t M I exactly over some such intervals. Our first result is the following. Theorem.. All intervals I of length have t M I. In fact, slightly more is true: if I then t M I. Furthermore for any b there is an interval I with I b an t M I, while for b there is an interval I with I b an t M I. The proof, which is essentially a corollary of Theorem. (a) below, is iscusse in Section 5. The numbers, an in Theorem., like most numerical values given in this paper, are approximations to some exact algebraic number. These numbers are roune in the correct irection, if necessary, to ensure an inequality still hols. The polynomial equations that they satisfy is given within the text. We have trie to o this for all numerical values. To measure the range of lengths of intervals having a particular monic integer transfinite iameter t, we introuce the following two functions: L t : inf I L t : sup I I : t M I t ; I : t M I t It follows from [, Prop..3] that both L t an L t are nonecreasing functions of t. Also L t L t see Lemma 3.(a) below. We give (Proposition 3.) general metho for fining upper an lower bouns for L t an L t, an apply these methos to get such bouns for t. They are constructive, using both the LLL basis-reuction algorithm an the Simplex metho. These techniques were first applie in this area by Borwein an Erélyi [3], an then by Habsieger an Salvy [7]. These bouns are given in Theorem 4. an Proposition 4. see also Figures an. At t, we pushe this metho further, an were able to say more. Theorem.. We have

3 MONIC INTEGER TRANSFINITE DIAMETER 3 (a) L an (b) 44 L 475. Further properties of L an L are given in Lemma 3... DEFINITIONS, CONJECTURES AND FURTHER RESULTS In this section, we state some ol an some more new results, an (perhaps a little recklessly) make four conjectures. The following result is simple but funamental. It is useful for etermining lower bouns for t M I. Lemma BPP (Borwein, Pinner an Pritsker [, p.905]). Let Q x a x a 0 be a nonmonic irreucible polynomial with integer coefficients, all of whose roots lie in the interval I. Then P I a for every monic integer polynomial P, so that t M I a. Furthermore, if P I a then t M I a an P β egp a for every root β of Q, an Res P Q. The proof follows straight from the classical fact that, for the conjugates β i of β () Res P Q a egp is a nonzero integer, giving P β i i () P I i P β i egp a Res P Q egp a This result is a variant of a similar one in the theory of t I see Lemma 7.. We call such a value a in Lemma BPP an obstruction for I, with obstruction polynomial Q x. From Lemma BPP we see that t M I is boune below by the supremum of all such obstructions. If this supremum is attaine by some value a coming from Q x a x a 0, then we say a is a maximal obstruction, an Q x is a maximal obstruction polynomial. It is not known whether such a polynomial exists for all intervals I of length less than 4 (see Conjecture.3).

4 4 K. G. HARE AND C. J. SMYTH We say that the monic integer polynomial P x is an optimal monic integer Chebyshev polynomial for I if P I t M I. If I has a maximal obstruction a with t M I a an an optimal monic integer Chebyshev polynomial P then we say that P attains the maximal obstruction a. Throughout this paper, P x will enote a monic integer polynomial, Q x a nonmonic integer polynomial an R x any integer polynomial. One very nice property of the monic integer transfinite iameter problem, not share by its nonmonic cousin, is that often exact values can be compute for t M I. In all cases where this has been one, incluing Theorem., it was achieve by fining a maximal obstruction, an a corresponing optimal monic integer Chebyshev polynomial. Simple examples of this are given ([, Theorem.5]) by the intervals I 0 n for n, where Q x nx is a maximal obstruction polynomial, an P x x is an optimal monic integer Chebyshev polynomial. For n, t M 0, with Q x x an P x x x. This was the case too in [, Section 5] in the proof of the Farey Interval conjecture for small-enominator intervals. A much less obvious example is the interval I , of length 073. Here, we have t M I P I , with maximal obstruction polynomial 7x 3 7x an where P is the optimal monic integer Chebyshev polynomial P x x x 9858 x x 499 x 5 7x 4 4x 3 8x x 8848 x 7 7x 6 94x 5 70x 4 3x 3 8x x 7935 x 8 4x 7 97x 6 7x 5 78x 4 0x 3 8x 9795 x 8 34x 7 64x 6 08x 5 65x 4 33x 3 8x x 5846 x 8 7x 7 x 6 x 5 0x 4 8x 3 5x x 48 of egree (Tighter enpoints for this interval, an its length, can be egp compute by solving the equation P x 3 7.) The iscovery of this polynomial require the use of Lemma 6. below. For the nonmonic transfinite iameter t, Pritsker [3, Theorem.7] has recently prove that no integer polynomial R x can attain R x I t I, this value being achieve only by a normalize prouct of infinitely many polynomials. An immeiate consequence of his result is the following. Proposition.. If an interval I has an optimal monic integer Chebyshev polynomial then t M I t I. A funamental question for both the monic an nonmonic integer transfinite iameter of an interval is whether its value can be compute exactly. In [, Conjecture 5.], Borwein et al make a conjecture for Farey intervals

5 b b MONIC INTEGER TRANSFINITE DIAMETER 5 (intervals c c where b b c c an b c b c ) concerning the exact value of their monic transfinite iameter. Conjecture BPP (Farey Interval Conjecture [, p. 8], [, Conjecture 5.]). Suppose that integer. Then b c b c is a Farey interval, neither of whose enpoints is an t M b c b c min c c Borwein et al verify their conjecture for all Farey intervals having the enominators c c less than. In Section 8 we exten the verification to some infinite families of Farey intervals (Theorems 8. an 8.3). We next investigate what happens to t M 0 b when b is close to n. For these intervals, some surprising things happen. Using the polynomial P x x, we know that t M 0 b b n if b n. In fact it appears likely that t M 0 b, clearly a non-ecreasing function of b, has a left iscontinuity at t n n. On the other han, we show in Theorem 9. that t M is locally constant on an interval of positive length δ n to the right of n. Further, Theorem 9. gives much larger values for δ n for n 3 an 4, as well as an upper boun for δ. In fact, more may be true. Conjecture. (Zero-enpoint Interval Conjecture). If I 0 b is an interval with b, then t M I n, where n max b is the smallest integer n for which n b. What little we know about t M 0 b for b is given in Theorem 9. (c), (). Both Conjecture BPP an Conjecture. are a consequence of the following conjecture. Conjecture. (Maximal obstruction implies t M I Conjecture). If an interval I of length less than 4 has a maximal obstruction m, then t M I m. We were at first tempte to conjecture here that t M I, as well as equaling its maximal obstruction, is always attaine by some monic integer polynomial. However, the following counterexample eliminates this possibility in general. Counterexample.. The polynomial 7x 3 4x x is a maximal obstruction polynomial for the interval I However, there is no monic integer polynomial P with P I equal to the maximal obstruction 7 3 for I.

6 6 K. G. HARE AND C. J. SMYTH This result is prove in Section 0. Our next result proves the existence of maximal obstructions for many intervals. Theorem.. Every interval not containing an integer in its interior has a maximal obstruction. Base on Conjecture. an Theorem. we make the following conjecture. Conjecture.3 (Maximal Obstruction Conjecture). Every interval of length less than 4 has a maximal obstruction. We o not have much irect evience for this conjecture. However, our next conjecture, Conjecture.4, implies it. To escribe this implication, we nee the following notion, taken from Flammang, Rhin an Smyth [5]. An irreucible polynomial Q x a x a 0 x with a 0, all of whose roots lie in an interval I, an for which a is greater than the (nonmonic) transfinite iameter t I is calle a critical polynomial for I. Here we are intereste only in nonmonic critical polynomials. It may be that every interval of length less than 4 has infinitely many nonmonic critical polynomials see Proposition. below. We make the following weaker conjecture. Conjecture.4 (Critical Polynomial Conjecture). Every interval of length less than 4 has at least one nonmonic critical polynomial. From Theorem. below, this conjecture is true for intervals not containing an integer. For intervals I of length less than 4 that o contain an integer (say 0), then, since t I, the polynomial x is a critical polynomial for I. Thus nonmonic is an important wor in this conjecture. In Theorem 7. we prove that Conjecture.4 implies Conjecture.3. More interestingly, we also prove in Corollary 7. that Conjecture. an Conjecture.3 together imply Conjecture.4. We observe in passing the following conitional result for the integer transfinite iameter t. Proposition.. Suppose that an interval I has infinitely many critical polynomials Q i x a i ix i a 0 i. Then t I i a i i i This result is prove in Section 7. Montgomery [, p.8] conjecture this result unconitionally for the interval I 0. inf

7 MONIC INTEGER TRANSFINITE DIAMETER 7 3. UPPER AND LOWER BOUNDS FOR L t AND L t FOR FIXED t The following lemma contains some simple properties, as well as alternative efinitions, of L an L. Lemma 3.. We have (a) L t L t for t 0; (b) L t 0 for 0 t ; (c) L t t for 0 t ; () L t sup I : t M I t for all I with I for t ; (e) L t inf I : t M I t for all I with I for t 0; (f) L t L t 4t for t. Proof. First note that, by [, equation (.)], t M I interval, from which (b) follows. Part (c) follows from the fact that x t t t. To prove (), take t. Then the set S : : t M I t for all I with I for the zero-length contains 0 (by (b)), so is nonempty. Put s sup S, an take S. Since I I implies that t M I t M I ([, Prop..3]), any with 0 also lies in S, so that S 0 s or 0 s. Hence L t s. On the other han, for each s there is an interval I with I an t M I t. Hence L t, giving L t s. Now (a) follows straight from (b) an (). The proof of (e), similar to that of (), is left as an exercise for the reaer. Finally, part (f) follows from the fact that for I 4 we have t M I t I cap I I 4 (see for instance []). Next, we give a simple lemma, neee for applying Proposition 3. below. Lemma 3.. Suppose that I i a i b i i n are intervals with a a a n a, an put M : max n i b i a i, m : min n i b i a i. Then (a) Any interval of length at least M contains an integer translate of some I i. (b) Any interval of length at most m is containe in an integer translate of some I i. Proof. Given an interval I of length, we can, after translation by an integer, assume that I a b, where a j a a j, for some j n. (a) Suppose that M. Then b j a j M a, so that a j b j a a.

8 8 K. G. HARE AND C. J. SMYTH (b) Suppose that m. Then b j a j m a, so that a a a j b j. The following proposition will be use to obtain explicit upper an lower bouns for L t an L t for particular values of t. Proposition 3.. (a) If Q x a x a 0, with integer coefficients an a, has roots spanning an interval of length, then for any t a we have L t Proof. (b) Suppose that we have a finite set of polynomials Q i x a i ix i a 0 i with all i a i i t with the property that every interval of length contains an integer translate of the roots of at least one of the polynomials Q i. Then L t (c) Suppose that we have a finite set of intervals I i such that for each I i there is a monic integer polynomial P i with P i I i t. Suppose too that every interval of length is containe in an integer translate of some I i. Then L t () If P I t for some monic integer polynomial P an interval I of length, then L t (a) Given such a Q x an interval I of length, an t a, then from Lemma BPP we have t M I a t so that, from the efinition of L t, we have L t. (b) Suppose that every interval I of length contains some integer translate of the set of roots of some Q i. Then, by Lemma BPP, t M I i t. Hence t M I t for any interval of length I, an so L t. (c) Here, for every interval I of length with I r I i say, (with r ), we have t P i I i P i I r P i x r I a i i t M I so that any I with t M I t has I. Hence L t. () If P I t an I then t M I t, so that L t.

9 MONIC INTEGER TRANSFINITE DIAMETER 9 i Polynomials Q i Intervals a i b i 7x 3 7x x 6 8x 5 6x 4 3x 3 9x 3x x 3 4x x x 6 8x 5 43x 4 5x 3 x x x x 6 8x 5 43x 4 5x 3 x x x 3 4x x x 6 8x 5 6x 4 3x 3 9x 3x x 3 7x x 6 36x 5 7x 4 6x 3 7x x 6 46x 5 9x 4 7x 3 8x x x 6 39x 5 90x 4 6x 3 8x x x 6 47x 5 05x 4 3x 3 8x x x 6 59x 5 5x 4 4x 3 9x x x 4 9x 3 3x x x 6 7x 5 53x 4 x 3 x 3x x 4 3x 3 6x x 6 9x 5 65x 4 6x 3 4x 5x x 6 07x 5 55x 4 7x 3 8x 4x x 6 09x 5 65x 4 36x 3 0x 4x x 6 3x 5 306x 4 7x 3 34x x x 6 4x 5 337x 4 04x 3 48x TABLE. Obstruction polynomials use for Theorem. to prove that L 475.

10 0 K. G. HARE AND C. J. SMYTH i Polynomials P i Intervals I i x 600 x 3 4x 36 x 4 4x 3 4x x 55 x 8 36x 7 96x 6 67x 5 64x 4 39x 3 4x 3x x 8 37x 7 96x 6 49x 5 9x 4 55x 3 8x 4x 0 x x 3 4x 77 x 4 0x 3 5x x 84 x 7 43x 6 x 5 44x 4 x 3 x 60 3 x 446 x x 99 x 4 7x 3 5x x 909 x 6 53x 5 46x 4 0x 3 4x x 394 x 4 7x 3 5x x 453 x 4 8x 3 8x 7 x 4 9x 3 7x x x 7556 x 5 6x 4 x 3 5x 3x 56 x 4 8x 3 8x x x x x 7038 x 4 8x 3 8x x 5 7x 4 4x 3 8x x x 5 6x 4 x 3 5x 3x x 4080 x 934 x 4 8x 3 8x 59 x 4 9x 3 9x 85 x 8 7x 7 440x 6 377x 5 8x 4 47x 3 x x 884 x 8 4x 7 53x 6 440x 5 90x 4 54x 3 3x x x 500 x 59 x 4 9x 3 9x 9 x 8 7x 7 440x 6 377x 5 8x 4 47x 3 x x x 336 x 768 x 6 3x 5 6x 4 8x 3 9x x 3 x 8 7x 7 440x 6 377x 5 8x 4 47x 3 x x x x x x 6369 x 6 3x 5 6x 4 8x 3 9x x 768 TABLE. Optimal monic integer Chebyshev polynomials use for Theorem. to prove that L

11 MONIC INTEGER TRANSFINITE DIAMETER Proof of Theorem.. Applying Proposition 3.(a) with Q x 7x 3 7x, we have L Here, a more precise value coul be etermine by calculating the span of the roots of Q x to a higher precision. We apply Proposition 3.(b) an Lemma 3.(a) using the polynomials Q i of Table, with the intervals a i b i containing their roots. (Here, the enpoints liste in Table are approximations of the minimal an maximal root of the obstruction polynomial in question. A higher precision was use for the computation of the upper boun of L 475.) We put Q 3 x Q x, whose roots are containe in a 3 b 3 : a b, an apply the Proposition to the 3 polynomials Q Q 3. Each has a. Then because max i b i a i b 6 a 5 475, any interval I of length I 475 must, by Lemma 3.(a), contain some integer translate of some interval a i b i, an so all the roots of the corresponing polynomial Q i. Hence L 475. We apply Proposition 3.(c) by starting with the 0 intervals I i i 0 in Table, an putting I i I i an P i x P i x for i 0, with I I an P x P x. (Here again, the enpoints liste in Table are approximations only. To fin a more accurate egp values, we woul solve for the roots of P x. Higher precision values were use to compute the lower boun L ) Each polynomial P i liste has a critical point at (an also at in the case of egp the last polynomial), with P i i. The value of P i x at all other critical points, as well as at the interval enpoints, is strictly less than egp i. This shows in each case that P i I i. Then all intervals I i have t M I i an, writing I i a i b i i we have (3) 0 min i b i a i b 5 a From this it follows by Lemma 3.(b) that every interval I of length less than is a subinterval of an integer translate of some I i, so that t M I P i I P i I i. This proves part (a) of the Theorem. Part (b) of the Theorem follows on applying Proposition 3. () with P x x x. We then have, for I L t I., that P I, so that

12 K. G. HARE AND C. J. SMYTH i Polynomial Q i t i i 7x 3 7x x x 3 3x x x x 4 x 3 4x x x 3 4x x 4 8x 3 8x x 5 5x 4 39x 3 40x x x 6 x 5 x 4 8x 3 0x 4x TABLE 3. Upper bouns for L t. Here L t i for t t i, where i is the span of the roots of the ith polynomial (see Theorem 4.). 4. GENERAL BOUNDS FOR L t AND L t In this section we fin upper an lower bouns for L t an L t, vali for t from 0 5 to close to 0 9. Our first result gives the upper bouns. Theorem 4.. (a) For all t i an i in Table 3 an for all t t i we have L t (b) For all t i an i in Table 4 an for all t t i we have L t The Theorem is prove by applying Proposition 3. (a) an (b) for a range of values in 0 5. Here again, the iameter given in Table 3 can be compute more exactly by consiering the ifference between the maximal an minimal roots of the obstruction polynomial. For Table 4, a calculation similar to that one for Table was one for each t i. The rouning proceure was that use for Table. Then the monotonicity of L t an L t gives the result for all t in this range. For the lower bouns, we first efine the normalize polynomial P α (4) P α x x x α x x of egree, an let α be the root in 0 of the equation (5) 4α α α α 5 α The following result gives the lower bouns. Proposition 4.. For 0 α ln4 ln5 we have α i. i.

13 MONIC INTEGER TRANSFINITE DIAMETER 3 5 (a) L α α, where α is the root of P α α 5 if α α ; 5 if α α (b) L 5 α max α α in t FIGURE. Upper an lower bouns for L t (Theorem 4. an Proposition 4.). grey line upper boun; black line lower boun. For the proof, we nee the following simple observation. Lemma 4.. If L t then L t. This follows straight from the fact that, given an interval I of length, every interval of length has an integer translate that is a subinterval of I.

14 4 K. G. HARE AND C. J. SMYTH t FIGURE. Upper an lower bouns for L t (Theorem 4. an Proposition 4.). grey line upper boun; black line lower boun. Proof of Proposition 4.. It shoul first be pointe out that this proposition is in fact true for all α, an not just those in the range specifie. That being ln4 sai, for α ln5 we woul have 5α, in which case we coul appeal to Lemma 3. (f) for the exact answer. (a) We will procee to analyze P α x I, picking α an such that, at the enpoints of the interval I, Pα x equals the largest local maximum of Pα x in the interior of I. (Notice that P α is alreay normalize, so P α I P α I.) (See Figures 3 an 4.) Notice first that P α x 5 α x α x 5 α

15 MONIC INTEGER TRANSFINITE DIAMETER x FIGURE 3. The normalize polynomial P α x (see (4)) with α 0 35 α, α 559 an t M I α P α x I α which has a local maximum of 5α at x 0, an a local maximum of m α α α α α at x 5 4α. Now the equation 5 m α α 5 has a root α efine by (5), with m α α for α α an m α 5 α for α α. Hence if α α then P α x for x 5, so that 5 P α I α α, where I α α α with α the root α 5 of P 5 α α α. However, if α α then we have the same result, but only for α the root in 5 of P α 5 α α. This gives the lower boun L 5 α α, but with a left iscontinuity in α (as a function of α) at α α. A plot of this lower boun, along with the upper bouns from Theorem 4. an Table 4, is given in Figure. 5 α

16 6 K. G. HARE AND C. J. SMYTH x FIGURE 4. The normalize polynomial P α x (see (4)) with α 0 5 α, α 449 an t M I α P α x I α (b) We know that L is a non-ecreasing function, an that L Combining these facts with Lemma 4. we get that L α max α This is isplaye numerically, along with the upper bouns from Theorem 4. an Table 3, in Figure. 5. INTERVALS OF LENGTH : PROOF OF THEOREM. Proof of Theorem.. From Theorem. (a) we know that every interval I of length has t M I. Now since every interval of length has some integer translate that contains, we have t M t M I for all such intervals, so that t M I for all I with I

17 MONIC INTEGER TRANSFINITE DIAMETER 7 If b then again from Theorem. (a), with the polynomial Q x 7x 3 7x, there is an interval I of length b with t M I To complete the proof, note that for b. t M b b t M 0 b 4 on applying [, Prop.4 with the polynomial x ], an then t M 0 b 4 b 4 using the polynomial P x x on 0 b COMPUTATIONAL METHODS 6.. Fining optimal monic integer Chebyshev polynomials P. We now escribe how the polynomials of Table were foun. These are optimal monic integer polynomials P having P I on various intervals of length just greater than. For these intervals, the maximal obstruction polynomial is Q x x, an the maximal obstruction is m. The metho applies more generally, however, to any interval I having a maximal obstruction polynomial Q, so we shall escribe the metho for this more general situation. We suppose that the maximal obstruction is m a, where Q x a x a 0, so that we seek a monic integer polynomial P with P I m. Firstly, potential factors of P of small egree k were ientifie using LLL [, 8, 9]. The basis use was x x k, with the inner prouct R R R x R x x I b k c k Here R x b k x k b 0 an R x c k x k c 0. The b k c k component of the inner prouct was inserte to iscourage nonmonic polynomials from appearing in the basis returne by LLL. Now, at least one element in the basis will contain an x k term an, because of the b k c k penalty, such an element is almost always monic. (In fact always in the examples we compute.) So we obtaine a monic polynomial of egree with small L norm, which usually also ha a small supremum norm. These monic polynomials with small L norm are not necessarily irreucible. At this point we examine each of their irreucible factors f i, again monic polynomials, an applie Lemma 6.(a) below to eliminate some of them. We then use the metho of Borwein an Erélyi [3] to search for exponents α i such that P eg P : i f α i eg f i i has the esire property P I m. To o this,

18 8 K. G. HARE AND C. J. SMYTH we neee to minimize t subject to the constraint i α i eg f i log for all x I with i α i, 0 α i. Some aitional constraints on the α i that we mae use of are given by Lemma 6. (b), (c). The main ifference between our application an the original one is that here the polynomials f i are all monic. By choosing a large number of points x I to iscretize the problem, we get a system of linear equations, on which the Simplex metho can be use to get a goo estimate of min t [3, 7, 6]. In practice, with a high enough precision an a large enough number of sample points, we obtain min t m exactly, an the corresponing α i then give the require P. We then check that P is inee an optimal monic integer Chebyshev polynomial for I by checking algebraically that P egp m at all roots of the maximal obstruction polynomial Q, an furthermore that all other local maxima of P in this interval are strictly smaller than m. The following lemma, use to help construct these polynomials P, specifies extra properties that their factors f i an normalize exponents α i must have. Lemma 6.. Let I be an interval that has a maximal obstruction polynomial Q x a x a 0. Suppose further that P x attains the maximal obstruction, an that P x egp i f α i eg f i i, with i α i. Then (a) The resultant Res f i Q is equal to for every factor f i x of P. (b) For every root β of Q we have f i x α i f i β 0 i eg f i f i β (c) Fix a root β of Q, an put ˆf i f i β eg f i. Let F be the multiplicative subgroup of 0 generate by a an the ˆf i with b a an b b k an inepenent generating set for F, with say ˆf eg f i i j b c j i j for some integers c j i. Then if j ; c j iα i i 0 if j Proof. We have i P β i a egp, where the prouct is taken over the roots β i of Q, so that, from (), Res P Q. Then (a) follows from the fact that the resultant of a prouct with Q is the prouct of the resultants with Q. t

19 MONIC INTEGER TRANSFINITE DIAMETER 9 The secon part follows from the fact that all the roots β of Q must be critical points of P x. Further, since P x attains the maximal obstruction, we have from Lemma BPP that for all such β we have P β egp a, giving the thir part. Note that Lemma 6. simplifies consierably when the maximal obstruction polynomial is linear, say a x a 0. Then it says that f i 0 eg f a a i a a an with P 0 a 0. The inepenent generating set b b k for F was foun using the integer relation-fining program PSLQ, which we use to search for linear integer relations between loga an the log ˆf i. As we have seen, the metho for fining an optimal monic integer Chebyshev polynomial P epens on first fining the (in practice there was only one) maximal obstruction polynomial for the interval. We now escribe how to o this. 6.. Fining obstruction polynomials Q. The obstruction polynomial 7x 3 7x, as well as those liste in Table an 3, were foun using the technique of Robinson [5] (see also [0, 7]). In this metho, the aim is to search for all egree polynomials Q x a x a 0 having all their roots in an interval I 0, for fixe egree, an fixe lea coefficient, a, with a. We escribe below how I 0 is chosen. Robinson s metho uses the fact that for k the span of the roots of the kth erivative of Q is containe in the span of the roots of the k th erivative of Q. In particular, these erivatives have all their roots in I 0. Starting with the st erivative of Q, we get a range of possible vali values for a. Consier then the n erivative to fin vali ranges for a. Continuing in this fashion, we obtain a list of polynomials, each one having all its roots in I 0. We now sieve this list, first by eliminating all polynomials that are reucible, or have integer content greater than. Having obtaine a list of irreucible polynomials, we can then prune it further, as follows. If Q x an R x are both irreucible polynomials, with the same egree an lea coefficient, an the span of the roots of R x contain the roots of Q x, then for any interval I where R x is an obstruction polynomial, Q x is also an obstruction polynomial, an hence R x is not neee. After construction of these polynomials, we can, for fixe a, an t a fin an upper boun for L t by fining the polynomial Q whose roots have the smallest span, an then appealing to Proposition 3. (a). This was one in Table 3, formalize in Theorem 4. (a), an isplaye in Figure.

20 0 K. G. HARE AND C. J. SMYTH Similarly, given this list of polynomials, we can compute the least such that any interval of length will contain an integer translate of at least one of the polynomials in our list. Then with Proposition 3. (b) we get an upper boun for L t. for given, we must choose I 0 carefully. If I 0 is too short, we might miss an important obstruction polynomial. On the other han, if I 0 is long, we will fin, along with the obstruction polynomials we seek, also (possibly multiple) integer translates of these polynomials. This is inefficient, as we en up oing more calculations than we nee to. So we wish to pick I 0 so that it is long enough to ensure that we have all important obstruction polynomials, an yet small enough that we are not oing more work than necessary. We o this by ensuring that I 0, the interval which contains the roots of the polynomials we have foun, has the property that I 0 is just greater. This ensures that there are no other useful obstruction polynomials that we might have misse, since any obstruction polynomial having a span of length will then have some integer translate lying in I 0. (We might have to re-run the calculation if I 0 is too small base on the current value of.) We can achieve tighter upper bouns for L t by consiering the list of all obstruction polynomials we foun such that a t. This computation was one for t (Table an Theorem.) an also for 0 other values of t (Table 4, Theorem 4. (b) an Figure ). To save space, the list of relevant polynomials for each t is not given in the table. (This information is available upon request from the authors.) 7. CRITICAL POLYNOMIALS: RESULTS AND PROOFS We first establish a relationship between critical polynomials an maximal obstructions. We efine a maximal nonmonic critical polynomial of an interval I to be a critical polynomial Q x a x a 0 such that the value a is maximal for Q within the set of nonmonic critical polynomials for I. Such a polynomial is well efine, as a result of the following Theorem. Theorem 7.. Suppose that an interval I has a nonmonic critical polynomial. Then I has a maximal nonmonic critical polynomial, Q x a x a 0 say, an furthermore Q is also a maximal obstruction polynomial, so that a is the maximal obstruction. To prove this result, we will apply the following version of a classical lemma.

21 MONIC INTEGER TRANSFINITE DIAMETER Lemma 7. ([, p. 77]). Let Q x an R x be two (not necessarily monic) integer polynomials. Further suppose that Q x a x a 0 is a critical polynomial for the interval I, an that the integer polynomial R x satisfies R I a. Then Q ivies R. Proof. From equations () an (), with R x replacing P x, we must have Res Q R 0. This result, essentially known to early workers on integer transfinite iameter (Gorškov, Sanov, Trigub, Aparicio Bernaro,...), has appeare in the literature in various forms see for instance Chunovsky [4, Lemma.3], Montgomery [, Chapter 0], Borwein an Erélyi [3], Flammang, Rhin an Smyth [5]. Proof of Theorem 7.. We first observe that nonmonic critical polynomials are obstruction polynomials. Conversely, if an obstruction is greater than t I then its associate polynomial is also a critical polynomial. Assume that I has a nonmonic critical polynomial, an consier the nonempty set A a of obstructions coming from the nonmonic critical polynomials of I. Any integer polynomial R x (not necessarily monic), must, by Lemma 7., contain as factors all critical polynomials Q whose obstructions a are strictly greater than R x I. Therefore R x I for any limit point of A, an hence t I. So if A has a limit point, then it must be inf A. Thus sup A is attaine, an there is a maximal nonmonic critical polynomial Q say. Then Q is also a maximal obstruction polynomial. Corollary 7.. Conjecture.3 an Conjecture. together imply Conjecture.4. Proof. From the proof above, we see that an obstruction that is greater than t I is associate to a critical polynomial. The existence of such an obstruction is a consequence of Conjecture.3 an Conjecture.. Proof of Proposition.. Now t I inf i i a i i, by the efinition of a critical polynomial. But if this inequality were strict, then we coul fin an integer polynomial R with t I R I inf i i a i i. But then, from Lemma 7., R woul have to be ivisible by all the Q i, which is impossible. 8. FAREY INTERVALS AND THE PROOF OF THEOREM. Every close interval I has a least positive integer q such that some rational p q with p q lies in the interior of I. If q then I belongs to a unique Farey interval b c b c whose enpoints are consecutive fractions in

22 K. G. HARE AND C. J. SMYTH the Farey sequence of orer q. We efine this interval to be the minimal Farey interval containing I. Theorem. follows irectly from our next result. Theorem 8.. Let I be an interval not containing an integer in its interior, b an b c c be the minimal Farey interval containing I. Then c c x b b is a critical polynomial for I. Moreover, the maximal obstruction for I is b c if c b c I c I; b c if c b c I c I; min c c if c b c an I b c c ; c c otherwise. Proof. Now the polynomial Q x c x b c c x b c has a local c maximum of c c c at x b b c c. Thus, by continuity, there exist integers r an r such that R x : Q x r c c x b b r has normalize supremum less than c c. Hence c c x b b is an obstruction polynomial. Now b b c c I, as otherwise I woul be containe b in one of the Farey intervals or. c b b c c b b c c Since the polynomials c c x b b, c x b an c x b are critical only if their roots are in I, an are, as factors of R, by Lemma 7. the only three possible maximal critical polynomials in this Farey interval, we get the final result. Theorem 8.. Let b c b c with c b c that b mo c an b B mo c where c t M b c b c be a Farey interval, an suppose B c. Then Proof. From [, p. 905] we have that there exists a monic quaratic integer b polynomial P x which has the property that P b c an P c c. Since its critical point is at a half integer, it is strictly monotonic on the B c Farey interval. Hence it attains its maximum at one of its enpoints, an b P b P. c c Theorem 8.3. Let P x x a x a 0 be an irreucible integer polynomial with real roots. Then there exist infinitely many Farey intervals for which P x attains the maximal obstruction. c

23 MONIC INTEGER TRANSFINITE DIAMETER 3 Proof. We know (Pell s Equation) that the equation x a xy a 0 y has an infinite number of solutions x y b i c i. These solutions have the b property that P i c i. Further, by choosing a suitable subsequence c i we may assume that both the c i an the b i c i are monotonically increasing. Thus for any interval I : b i c i b i c i not containing a half-integer, we see that P x attains the maximal obstruction c i with Q x c i x b i, so that t M I c i. This happens infinitely often as the b i c i ten to a root of P x. We can fin a b b i b i b c c i c i such that i b c i c is a Farey interval, an hence P x attains its maximal obstruction c i on this interval. It shoul be note that this metho of proof will not work for polynomials of egree 3 or higher, as the resulting Thue equation x n a n x n y a 0 y n has only a finite number of integer solutions [8]. 9. STUDY OF t M b In this section we consier intervals 0 b, with t M b enoting t M 0 b. Our first result for such intervals is a consequence of Theorem 8.. b n. Then n is the maximal obstruc- Corollary 9.. Let n tion of 0 b. an n Theorem 9.. For all n there exists δ n 0 ε δ n t M 0 n ε n n n 4 n such that for all Proof. Consier the polynomial P n x x n x nx. It has the following properties: n ; P n n n P n x has a local maximum (with respect to x) at x n ; P n x is strictly increasing (with respect to x) on 0 n ; P n x has a root β n strictly greater than n n 4 n ; P n x is strictly ecreasing on n β n. Let α n be the minimal root, strictly greater than β n, of the equation P n x. Thus P n x emonstrates that t M α n n n n, where α n β n n n 4 n. Theorem 9.. We have that (a) t M b 4 for b ;

24 4 K. G. HARE AND C. J. SMYTH (b) t M b 3 for b ; (c) t M b for b 6 ; () t M 38. Hence, in the notation of Theorem 9., 0 76 δ δ , δ an Proof. The optimal monic polynomials neee for Parts (a) an (b) are given in Table 5. In each case they attain the maximal obstruction 4 an 3 respectively. As before, a slightly larger interval can be compute exactly, by solving P x 4 or P x egp egp 3 respectively. The values of an have been roune own to ensure that the inequality still hols. Part (c) follows from the first part of Table, using the map x x, with the same comments to the exact values as above. Part () is prove using Lemma BPP using the obstruction polynomial 7x 3 4x 7x. Here.38 is an approximation to its largest root, roune up to ensure that () hols. The factors use for the construction of the polynomials in Table 5 were foun using the techniques iscusse in Section 6, making use of the constraints given by Lemma 6.. Bouns have been given on the exponents of certain factors for large integer Chebyshev polynomials use for estimating t I. For example, for the interval I 0, Pritsker [] shows that x x γ, where 0 96 γ , must appear as a factor in any polynomial R (normalize to have egree ), for which R I is sufficiently close to t I. Following [5], we now etermine a lower boun for γ b such that x γ b must ivie any normalize monic integer polynomial P such that P 0 b approximates t M b sufficiently closely. Suppose that the function m b is an upper boun for t M b. Then by Proposition 5.3 an Lemma 5. of [5] we have that γ b is boune below by the least positive root of x x x x x x b x m b So in particular, if t M b b for b 0 as in Conjecture., then our lower boun for γ b woul have infinitely many iscontinuities in this range (Figure 5 black lines). However, we know, by using the polynomial x, that we have a provable, albeit weaker, upper boun m b b. This gives us a proven lower boun for γ b (Figure 5 grey line). Proposition 9.. We have lim b 0 γ b.

25 MONIC INTEGER TRANSFINITE DIAMETER FIGURE 5. Lower bouns for γ b. grey line lower boun using m b b (known), soli line lower boun assuming m b b (Conjecture.). Proof. Define T x b x x x x x x b x b Now T x b has a positive local maximum at x 4b as b 0,while T b b 0 for 0 b 0 04, so that T x b 0 has a root in b 4b. Further, since T x b is increasing for x 0 4b this root is the least positive root of T x b 0. Hence γ b b, giving the result. 0. PROOF OF COUNTEREXAMPLE. For the proof of Counterexample. we nee the following p-aic result.

26 6 K. G. HARE AND C. J. SMYTH Proposition 0.. Suppose that Q x a x a 0 x is a maximal obstruction polynomial for the interval I, an that the maximal obstruction is attaine by some monic integer polynomial P x. Then gc a 0 a an, for every prime p iviing a we have a i a p a i p In particular, if a is square-free then a Q x Q 0 has integer coefficients. Here p is the usual p-aic valuation on. For the proof, it is extene to. Proof. Take β to be any root of Q x, an p any prime factor of a. Let P x be of egree m. Then, as P x attains the obstruction, P β a m, so that P β p m a p. If β p then P β p, a contraiction, as P x has integer coefficients. Hence β p an P β p β m p m a p, giving (6) β p a p Applying (6) for all roots β j of Q x we get j β j p a p. But also from a Q x j x β j we have that j β j p a 0 a p. Hence a 0 p. Doing this for all p a we obtain a 0 a. Furthermore, if i a for any i then the Newton polygon of P (see for instance [9, p. 73]) p tells us that β j p a p for some j, contraicting (6). i In the case of a square-free, a p for i a, so that i p a p, an hence, using all primes p iviing a, we see that a i a is an integer. Proof of Counterexample.. The fact that 7x 3 4x x is a maximal obstruction polynomial for the interval I can be verifie by showing that it is a critical polynomial. This follows from the fact that the polynomial i 0 a i a p R x x 878 5x 3 4x x x 3 4x x x 6 4x 5 0x 4 0x 3 9x x 3x 5 6x 4 3x 3 8x x

27 MONIC INTEGER TRANSFINITE DIAMETER 7 has R I 7 3, so that t I 7 3. As 7x 3 4x x has all its roots in I, it is therefore a critical polynomial. As always, the interval is an approximation only, an a tighter one can easily be compute. We now claim that 7x 3 4x x is the maximal nonmonic critical polynomial for I. For any critical polynomial a x a 0 for I with a R I must be a factor of R, by Lemma 7.. But among the four irreucible factors of R, 7x 3 4x x is the only one having all its roots within I. As it is nonmonic, it must inee be the maximal nonmonic critical polynomial for I. By Theorem 7., this polynomial is the maximal obstruction polynomial. However, 7 7x3 4x x oes not have integer coefficients so that, by Proposition 0., the interval has no optimal monic integer Chebyshev polynomial.. SOME FINAL COMMENTS ON THE COMPUTATIONS AND FIGURES Consier Figure. We see that L t 0 for for t, an further that L t 4t for t. So in fact the area of interest is for t between an. That being sai, the upper boun is only given up to approximately This is because the upper boun from Proposition 3.(a) is given by high egree polynomials with small lea coefficient. In our search, we compute only up to egree 6. As this is the limit to our knowlege of the upper boun. If we wishe to exten these calculations, we coul exten the knowlege of the upper boun, but the computation time becomes excessive. For example, even if we compute up to egree 0, which is probably beyon our computational range, we woul only get up to As it was, the computations up to egree 6 took over 3000 CPU hours, an the computation time approximately triples for each aitional egree. Similar comments apply to bouning L t (Figure ) for t close to. In this case, it actually turne out that none of the polynomials with lea coefficient an egree 6 were useful in the calculations for such t, an hence we only get an upper boun for L t for t up to t While we know from Lemma 3.(c) that L t t for t, we o not know L t exactly in this range. In orer to get an upper boun for L t in at least part of this range, it woul in principle be possible to exten the calculation ownwars from t. The lower boun of for t was chosen, as we compute obstruction polynomials of egree, with coefficients up to. If we were to compute up to 3 instea, we woul be able to exten this graph own to t 3. This woul, however, be a massive unertaking, because we woul have times as many possible lea coefficients. Furthermore, we observe that, for a given egree, the

28 8 K. G. HARE AND C. J. SMYTH computations took longer the higher the lea coefficient was, so this factor is probably an unerestimate. It may be possible to exten these calculations though in a more sophisticate manner, somehow oing a less extensive an more intelligent search for obstruction polynomials of higher egree or larger lea coefficients. This woul be a worthwhile project, an coul lea to some interesting new results. Lastly, consier Figure 5. This coul very easily have been extene all the way to 0. The reason that we chose not to o this is because the hypothetical lower boun (the black lines) starts to merge into itself, an the Figure becomes unreaable. (The lower boun jumps at every n which get more frequent as n 0.) Acknowlegement. We thank the referee for helpful comments.

29 MONIC INTEGER TRANSFINITE DIAMETER 9 i t i i i t i i TABLE 4. Upper bouns for L t. Here L t i for t t i, where i is the span of the roots of the ith polynomial (see Theorem 4.).

30 30 K. G. HARE AND C. J. SMYTH t M b 4 for b by P x x 640 x 5 43x 4 456x 3 79x 3x 47 x x 6 334x x 4 784x 3 54x 50x 35 t M b 3 for b by P x x x 5 3x 4 7x 3 x 6x x 7 33x 6 406x 5 93x 4 79x 3 79x x 0840 x 8 484x x x 5 083x 4 704x 3 405x 53x x 8 484x x x 5 537x 4 594x 3 78x 6x x 8 78x 7 807x 6 756x 5 97x 4 864x 3 56x 50x x x x x x x x x x x x x 3 x 5x TABLE 5. Optimal monic integer polynomials use for the proof of Theorem 9..

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32 3 K. G. HARE AND C. J. SMYTH DEPARTMENT OF PURE MATHEMATICS, UNIVERSITY OF WATERLOO, WATERLOO, ONTARIO, CANADA, NL 3G aress: SCHOOL OF MATHEMATICS, UNIVERSITY OF EDINBURGH, JAMES CLERK MAXWELL BUILDING, KING S BUILDINGS, MAYFIELD ROAD, EDINBURGH EH9 3JZ, UK. aress: c.smyth@e.ac.uk