The Reverse Ultra Log-Concavity of the Boros-Moll Polynomials

William Y.C. Chen^{1}, Cindy C.Y. Gu^{2}

Center for Combinatorics, LPMC-TJKLC Nankai University, Tianjin 300071, P. R. China 1chen@nankai.edu.cn, ^{2}guchunyan@cfc.nankai.edu.cn

Abstract. Based on the recurrence relations on the coefficients of the Boros-Moll polynomials P_{m}(a) = ∑_{i} d_{i}(m)a^{i} derived independently by Kauers and Paule, and Moll, we are led to the discovery of the reverse ultra log-concavity of the sequence {d_{i}(m)}. We also show that the sequence {i!d_{i}(m)} is log-concave for m ≥ 1. Two conjectures are proposed. Keywords: log-concavity, reverse ultra log-concavity, upper bound, imaginary roots, Boros-Moll polynomials. AMS Classification: 05A20; 33F10

1 Introduction

The main objective of this paper is to prove the reverse ultra log-concavity of the Boros-Moll polynomials. Boros and Moll [1, 2, 3, 7] studied a class of Jacobi polynomials in connection with the following integral:

Theorem 1.1

∫_{0}^{∞} \frac{1}{(x^{4} + 2ax^{2} + 1)^{m+1}} dx = \frac{π}{2^{m+3/2}(a + 1)^{m+1/2}} P_{m}(a).

where P_{m}(a) can be represented by

P_{m}(a) = 2^{−2m} ∑_{k} 2^{k}\binom{2m − 2k}{m − k}\binom{m + k}{k}(a + 1)^{k}.

The polynomials P_{m}(a) are called Boros-Moll polynomials. Let

P_{m}(a) = ∑_{i=0}^{m} d_{i}(m)a^{i}.

The coefficients d_{i}(m) are positive. They are also log-concave, as conjectured by Moll [7], and recently proved by Kauers and Paule [5].

Theorem 1.2 For 0 < i < m, we have

d_{i}^{2}(m) ≥ d_{i−1}(m)d_{i+1}(m).  (1.1)

Recall that a sequence (a_{n})_{n≥0} of real numbers is said to be log-concave if

a_{n}^{2} ≥ a_{n+1}a_{n−1},  (1.2)

for all n ≥ 1. If the sequence satisfies (1.2) with strict inequality, then it is said to be strictly log-concave. A polynomial is said to be (strictly) log-concave if its sequence of coefficients is (strictly) log-concave. Log-concave sequences and polynomials often arise in combinatorics, algebra and geometry, see, for example, Brenti [4] and Stanley [9].

A sequence {a_{k}}_{0≤k≤n} is called ultra log-concave if

\{a_{k}/\binom{n}{k}\}_{0≤k≤n}  (1.3)

is log-concave, see Liggett [6]. Note that this condition can be rewritten as

k(n − k)a_{k}^{2} − (n − k + 1)(k + 1)a_{k−1}a_{k+1} ≥ 0.  (1.4)

It is well known that if a polynomial has only real roots, then its coefficients form an ultra log-concave sequence. The above relation (1.4) implies the following inequality

ka_{k}^{2} − (k + 1)a_{k−1}a_{k+1} ≥ 0,

from which we can deduce that the sequence {k!a_{k}} is log-concave. This further implies that {a_{k}} is strictly log-concave.

The first result of this paper is to show that i!{d_{i}(m)} is log-concave, as stated below in an equivalent form.

Theorem 1.3 For 1 ≤ i ≤ m − 1, we have

i d_{i}^{2}(m) > (i + 1)d_{i−1}(m)d_{i+1}(m).  (1.5)

Despite the above property of d_{i}(m), we will show that the reverse ultra log-concavity holds, as described in the following theorem. This is the main result of this paper.

Theorem 1.4 For 1 ≤ i ≤ m − 1, we have

\frac{d_{i−1}(m)}{\binom{m}{i−1}} \frac{d_{i+1}(m)}{\binom{m}{i+1}} > (\frac{d_{i}(m)}{\binom{m}{i}})^{2} ,  (1.6)

or, equivalently,

(m − i + 1)(i + 1)d_{i−1}(m)d_{i+1}(m) − (m − i)id_{i}(m)^{2} > 0.  (1.7)

We conclude this paper with two conjectures. Roughly speaking, the first conjecture says that in spite of the reverse ultra log-concavity, the ultra log-concavity almost holds in the asymptotic sense. The second conjecture is concerned with the log-concavity of the sequence d_{i−1}(m)d_{i+1}(m)/d_{i}^{2}(m) for m ≥ 2.

2 The Reverse Ultra Log-Concavity

In this section, we give the proofs of and Theorem 1.3 Theorem 1.4. We will need the following three recurrence formulas derived independently by Kauers and Paule [5] and Moll [8]:

d_{i}(m + 1) =\frac{m + i}{m + 1}d_{i−1}(m) + \frac{4m + 2i + 3}{2(m + 1)}d_{i}(m), 0 ≤ i ≤ m + 1,  (2.1)

d_{i}(m + 1) =\frac{(4m − 2i + 3)(m + i + 1)}{2(m + 1)(m + 1 − i)}d_{i}(m)

− \frac{i(i + 1)}{(m + 1)(m + 1 − i)}d_{i+1}(m), 0 ≤ i ≤ m,  (2.2)

d_{i}(m + 2) =\frac{−4i^{2} + 8m^{2} + 24m + 19}{2(m + 2 − i)(m + 2)}d_{i}(m + 1)

− \frac{(m + i + 1)(4m + 3)(4m + 5)}{4(m + 2 − i)(m + 1)(m + 2)}d_{i}(m), 0 ≤ i ≤ m + 1,  (2.3)

Recall that Kauers and Paule [5] obtained the following lower bound on d_{i}(m + 1)/d_{i}(m) for 0 ≤ i ≤ m,

\frac{d_{i}(m + 1)}{d_{i}(m)} ≥ Q(m, i),  (2.4)

where

Q(m, i) = \frac{4m^{2} + 7m + i + 3}{2(m + 1 − i)(m + 1)}.  (2.5)

Proof of Theorem 1.3. As the first step, by the recurrence relations (2.1) and (2.2) we may transform Theorem 1.3 to the following equivalent form

4(m + 1)^{2}(m − i + 1)d_{i}(m + 1)^{2} − 4(m + 1)(4m^{2} + 7m − 2i^{2} + 3)d_{i}(m)d_{i}(m + 1)

+ (4m + 4i + 3)(4m^{2} + 7m − i + 3)d_{i}(m)^{2} > 0,

which can be recast as

4(m + 1)^{2}(m − i + 1) (\frac{d_{i}(m + 1)}{d_{i}(m)})^{2} − 4(m + 1)(4m^{2} + 7m − 2i^{2} + 3)\frac{d_{i}(m + 1)}{d_{i}(m)}

+ (4m + 4i + 3)(4m^{2} + 7m − i + 3) > 0.  (2.6)

Notice that the discriminant of the above quadratic form is positive, since

△ = 16i^{2}(2i − 1)^{2}(m + 1)^{2} > 0.

Thus the quadratic function on the left hand side of (2.6) has two real roots,

x_{1} = \frac{4m^{2} + 7m + 3 − i}{2(m − i + 1)(m + 1)}, x_{2} = \frac{4m + 4i + 3}{2(m + 1)}.

Using the lower bound Q(m, i) for d_{i}(m + 1)/d_{i}(m), we deduce that for 1 ≤ i ≤ m − 1,

\frac{d_{i}(m + 1)}{d_{i}(m)} ≥ Q(m, i) > x_{1} > x_{2}.

Thus we obtain (2.6), and the proof is complete.

In order to prove Theorem 1.4, we need an upper bound for the ratio d_{i}(m+1)/d_{i}(m).

Theorem 2.1 We have for all m ≥ 2, 1 ≤ i ≤ m − 1,

\frac{d_{i}(m + 1)}{d_{i}(m)} < T (m, i),  (2.7)

where

T (m, i) = \frac{4m^{2} + 7m + 3 + i\sqrt{4m + 4i^{2} + 1} − 2i^{2}}{2(m − i + 1)(m + 1)} ,  (2.8)

and for m ≥ 1,

\frac{d_{0}(m + 1)}{d_{0}(m)} = T (m, 0), \frac{d_{m}(m + 1)}{d_{m}(m)} = T (m, m).  (2.9)

Proof. First, we consider (2.9). Setting i = 0 in (2.1) gives that for any m ≥ 1,

\frac{d_{0}(m + 1)}{d_{0}(m)} = \frac{4m + 3}{2(m + 1)},

which agrees with T (m, 0). While i = m, (2.1) implies

\frac{d_{m}(m + 1)}{d_{m}(m)} = \frac{(2m + 3)(2m + 1)}{2(m + 1)} = T (m, m).

Thus (2.9) holds for m ≥ 1.

We now proceed to conduct induction on m to show that (2.7) is valid for i ≥ 1. When m = 2 and i = 1, we have

\frac{d_{1}(3)}{d_{1}(2)} = \frac{43}{15} < T (2, 1) = \frac{31 + \sqrt{13}}{12}.

We assume that the theorem holds for m, where m > 2. Then we aim to show that it also holds for m + 1, namely, for 1 ≤ i ≤ m,

d_{i}(m + 2) < T (m + 1, i)d_{i}(m + 1).  (2.10)

Using the recurrence (2.3), we may rewrite (2.10) in the following form

\frac{−4i^{2} + 8m^{2} + 24m + 19}{2(m − i + 2)(m + 2)}d_{i}(m + 1)

− \frac{(m + i + 1)(4m + 3)(4m + 5)}{4(m + 1)(m + 2)(m − i + 2)}d_{i}(m) < T (m + 1, i)d_{i}(m + 1).  (2.11)

In order to derive an upper bound for d_{i}(m + 1)/d_{i}(m), it is necessary to show that

R(m, i) = \frac{−4i^{2} + 8m^{2} + 24m + 19}{2(m − i + 2)(m + 2)} − T (m + 1, i)  (2.12)

is positive. Since m ≥ i, we have 4m + 4i^{2} + 5 < 12m + 4m^{2} + 9. It follows that

R(m, i) = \frac{4m^{2} + 9m + 5 − 2i^{2} − i\sqrt{4m + 4i^{2} + 5}}{2(m − i + 2)(m + 2)}

≥ \frac{4m^{2} + 9m + 5 − 2i^{2} − i(2m + 3)}{2(m − i + 2)(m + 2)}

= \frac{(4m^{2} − 2i^{2} − 2mi) + (9m − 3i) + 5}{2(m − i + 2)(m + 2)},

which is positive for 1 ≤ i ≤ m. Hence (2.11) is equivalent to the following inequality

\frac{d_{i}(m + 1)}{d_{i}(m)} < \frac{(m + i + 1)(4m + 3)(4m + 5)}{4(m + 1)(m + 2)(m − i + 2)R(m, i)}.  (2.13)

Note that the right hand side of (2.13) can be expressed as

F (m, i) = \frac{(m + i + 1)(4m + 3)(4m + 5)}{2(m + 1)(4m^{2} − 2i^{2} + 9m + 5 − i\sqrt{4m + 4i^{2} + 5})}.  (2.14)

By the inductive hypothesis, it suffices to show that for 1 ≤ i ≤ m − 1,

T (m, i) ≤ F (m, i).  (2.15)

Let A = \sqrt{4m + 4i^{2} + 1} and B = \sqrt{4m + 4i^{2} + 5}. It is easy to check that F (m, i)−T (m, i) equals

\frac{(i^{2} − 4i^{4}) − i(5 + 4m^{2} + 9m − 2i^{2})A + i(3 + 4m^{2} + 7m − 2i^{2})B + i^{2}AB}{2(m + 1)(m − i + 1)(4m^{2} + 9m + 5 − 2i^{2} − iB)} .  (2.16)

Since

(4m^{2} + 9m + 5 − 2i^{2})^{2} − (iB)^{2} = (4m + 5)^{2}(m + i + 1)(m − i + 1) > 0,

the denominator of (2.16) is positive. It remains to show that the numerator of (2.16) is also positive. Observe that every term in the numerator contains a factor i. So we may divide this factor. Put

X = (i − 4i^{3}) + iAB
Y = (5 + 4m^{2} + 9m − 2i^{2})A − (3 + 4m^{2} + 7m − 2i^{2})B.

We claim that X > 0 and X > Y. Since m ≥ i, we have A ≥ 2i + 1 and B ≥ 2i + 1. Moreover, since i ≥ 1, we find

X = (i − 4i^{3}) + iAB ≥ i − 4i^{3} + i(2i + 1)^{2} = 4i^{2} + 2i > 0.

To show that X − Y > 0, we will consider X^{2} − Y^{2}. Let us introduce G(m, i) and H(m, i) as given by

G(m, i) = (32m^{4} − 32m^{2}i^{2} + 128m^{3} − 64mi^{2} + 190m^{2} − 30i^{2} + 124m + 30)AB,
H(m, i) = 128m^{5} + 608m^{4} + 1128m^{3} + 1014m^{2} + 436m + 128m^{4}i^{2} + 384m^{3}i^{2}
+ 408m^{2}i^{2} − 128m^{2}i^{4} + 200mi^{2} − 256mi^{4} − 120i^{4} + 50i^{2} + 70.

It can be checked that

X^{2} − Y^{2} = G(m, i) − H(m, i).  (2.17)

Since i ≤ m − 1, it is easily seen that G(m, i) > 0. To verify G(m, i) > H(m, i), we will show that G(m, i)^{2} − H(m, i)^{2} > 0. In fact, for 1 ≤ i ≤ m − 1

G(m, i)^{2} − H(m, i)^{2} = 16(4m + 5)^{2}(16mi^{2} + 12i^{2} − 1)(m + i + 1)^{2}(m − i + 1)^{2} > 0.

This implies that X > Y. Hence the numerator of (2.16) is positive. Consequently, (2.15) holds for 1 ≤ i ≤ m − 1.

Up to now, we still need to consider the case i = m. It remains to show that

\frac{d_{m}(m + 2)}{d_{m}(m + 1)} < T (m + 1, m).

By direct computation, we find that

d_{m}(m + 2) = \frac{(m + 1)(4m^{2} + 18m + 21)}{2^{m+4}(2m + 3)}\binom{2m + 4}{m + 2}

and

d_{m}(m + 1) = \frac{2m + 3}{2^{m+2}} \binom{2m + 2}{m + 1}.

We get

\frac{d_{m}(m + 2)}{d_{m}(m + 1)} = \frac{(m + 1)(4m^{2} + 18m + 21)}{2(2m + 3)(m + 2)}.

On the other hand,

T (m + 1, m) = \frac{2m^{2} + 15m + 14 + m\sqrt{4m^{2} + 4m + 5}}{4(m + 2)}.

We see that for m ≥ 2,

T (m + 1, m) > \frac{d_{m}(m + 2)}{d_{m}(m + 1)}

⟺(2m^{2} + 15m + 14 + m\sqrt{4m^{2} + 4m + 5})(2m + 3) − 2(m + 1)(4m^{2} + 18m + 21) > 0
⟺(2m^{2} + 3m)\sqrt{4m^{2} + 4m + 5} > 4m^{3} + 8m^{2} + 5m
⟺ ((2m^{2} + 3m)\sqrt{4m^{2} + 4m + 5})^{2} − (4m^{3} + 8m^{2} + 5m)^{2} > 0
⟺4m^{2}(4m + 5) > 0,

which is evident. This completes the proof of the theorem.

We are now ready to prove Theorem 1.4. Like the first step in the proof of Theorem 1.3, we use the recurrences (2.1) and (2.2) to restate (1.7) as follows

4(m − i + 1)^{2}(m + 1)^{2} (\frac{d_{i}(m + 1)}{d_{i}(m)})^{2}

− 4(m − i + 1)(m + 1)(4m^{2} − 2i^{2} + 7m + 3)\frac{d_{i}(m + 1)}{d_{i}(m)}

− (32mi^{2} − 56m^{3} − 73m^{2} − 42m + 13i^{2} − 9 − 16m^{4} + 16i^{2}m^{2}) < 0.  (2.18)

Observe that the discriminant of the above quadratic form is positive for i ≥ 1, since

△ = 16i^{2}(m + 1)^{2}(4i^{2} + 4m + 1)(m − i + 1)^{2} > 0.

It follows that the quadratic function on the left hand side of (2.18) has two real roots

x_{1} = \frac{4m^{2} − 2i^{2} + 7m + 3 − i\sqrt{4m + 4i^{2} + 1}}{2(m − i + 1)(m + 1)},

x_{2} = \frac{4m^{2} − 2i^{2} + 7m + 3 + i\sqrt{4m + 4i^{2} + 1}}{2(m − i + 1)(m + 1)}.

By the definition of Q(m, i) in (2.5), we see that x_{1} < Q(m, i). Note that x_{2} = T (m, i) as given in Theorem 2.1. In view of Theorem 2.1, we deduce that

x_{1} < \frac{d_{i}(m + 1)}{d_{i}(m)} < x_{2},

for 1 ≤ i ≤ m − 1. Hence we conclude that (2.18) holds. This completes the proof of Theorem 1.4.

We conclude this paper with two conjectures. Let

c_{i}(m) = \frac{d_{i}^{2}(m)}{d_{i−1}(m)d_{i+1}(m)}, 1 ≤ i ≤ m − 1.

Then Theorem 1.3 and Theorem 1.4 lead to the following bounds on c_{i}(m) for 1 ≤ i ≤ m − 1,

1 + \frac{1}{i} ≤ c_{i}(m) ≤ (1 + \frac{1}{i}) (1 + \frac{1}{m − i}) .  (2.19)

Numerical evidence indicates that the upper bound in (2.19) is very close to c_{i}(m) even when m is small. Let u_{i}(m) = (1 + \frac{1}{i}) (1 + \frac{1}{m−i}). For example, when m = 6, the values of c_{i}(m)/u_{i}(m) are given below

0.9462708849, 0.9642110408, 0.9752109510, 0.9821688283, 0.9867303609.

Conjecture 2.2 For 1 ≤ i ≤ m − 1, we have

lim_{m→∞} \frac{c_{i}(m)}{u_{i}(m)} = 1.  (2.20)

Conjecture 2.3 For m ≥ 2, the sequence {1/c_{i}(m)}_{i=2}^{m−2} is log-concave.

Conjecture 2.2 implies that the Boros-Moll polynomials are almost ultra log-concave. Further conjectures can be made based on Conjecture 2.3 in the spirit of Moll’s conjectures on the k-log-concavity [7].

Acknowledgments. This work was supported by the 973 Project, the PCSIRT Project of the Ministry of Education, the Ministry of Science and Technology, and the National Science Foundation of China.

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