Skew log-concavity of the Boros-Moll sequences

Abstract

Let ${\{T(n,k)\}}_{0\le n<\mathrm{\infty },0\le k\le n}$ be a triangular array of numbers. We say that $T(n,k)$ is skew log-concave if for any fixed n, the sequence ${\{T(n+k,k)\}}_{0\le k<\mathrm{\infty }}$ is log-concave. In this paper, we show that the Boros-Moll sequences are almost skew log-concave.

Introduction and main result

Boros and Moll [1, 2] explored a special class of Jacobi polynomials in their study of a quartic integral. They have shown that for any $a>-1$ and any nonnegative integer m,

$${\int }_0^{\mathrm{\infty }}\frac{1}{{(x^4+2a{x}^2+1)}^{m+1}}dx=\frac{\pi }{2^{m+3/2}{(a+1)}^{m+1/2}}{P}_m(a),$$

where

$$P_m(a)=\sum _{j,k}\left(\begin{array}{c}2m+1\\ 2j\end{array}\right)\left(\begin{array}{c}m-j\\ k\end{array}\right)\left(\begin{array}{c}2k+2j\\ k+j\end{array}\right)\frac{{(a+1)}^j{(a-1)}^k}{2^{3(k+j)}}.$$ 1.1

Using Ramanujan’s master theorem, Boros and Moll [2] derived the following formula for $P_m(a)$:

$$P_m(a)=2^{-2m}\sum _k{2}^k\left(\begin{array}{c}2m-2k\\ m-k\end{array}\right)\left(\begin{array}{c}m+k\\ k\end{array}\right){(a+1)}^k,$$ 1.2

which implies that the coefficient of $a^i$ in $P_m(a)$ is positive for $0\le i\le m$. Let $d_i(m)$ be given by

$$P_m(a)=\sum _{i=0}^m{d}_i(m)a^i.$$ 1.3

The polynomial $P_m(a)$ is called the Boros-Moll polynomial, and the sequence ${\{d_i(m)\}}_{0\le i\le m}$ of the coefficients is called a Boros-Moll sequence. From (1.3), we know that $d_i(m)$ can be given by

$$d_i(m)=2^{-2m}\sum _{k=i}^m{2}^k\left(\begin{array}{c}2m-2k\\ m-k\end{array}\right)\left(\begin{array}{c}m+k\\ k\end{array}\right)\left(\begin{array}{c}k\\ i\end{array}\right).$$ 1.4

Some combinatorial properties of ${\{d_i(m)\}}_{0\le i\le m}$ have been proved. Boros and Moll [1] proved that the sequence ${\{d_i(m)\}}_{0\le i\le m}$ is unimodal, and the maximum element appears in the middle. Recall that a sequence ${\{a_i\}}_{0\le i\le m}$ of real numbers is said to be unimodal if there exists an index $0\le j\le m$ such that

$$a_0\le a_1\le \cdots \le a_{j-1}\le a_j\ge a_{j+1}\ge \cdots \ge a_m$$

and ${\{a_i\}}_{0\le i\le m}$ is said to be log-concave if

$$a_i^2-a_{i+1}{a}_{i-1}\ge 0,1\le i\le m,$$ 1.5

where $a_{-1}=a_{m+1}=0$. Moll [2] conjectured that the sequence ${\{d_i(m)\}}_{0\le i\le m}$ is log-concave. Kauers and Paule [3] proved this conjecture based on recurrence relations found using a computer algebra approach. Recently, Chen and Xia [4] showed that the sequence ${\{d_i(m)\}}_{0\le i\le m}$ satisfies the strongly ratio monotone property which implies the log-concavity and the spiral property. They [5] also confirmed a conjecture of Moll which says that ${\{i(i+1)(d_i^2(m)-d_{i-1}(m)d_{i+1}(m))\}}_{1\le i\le m}$ attains its minimum at $i=m$. Chen et al. [6] proved that the Boros-Moll sequences are interlacing log-concave. Chen and Gu [7] showed that the sequence ${\{d_i(m)\}}_{0\le i\le m}$ satisfies the reverse ultra log-concavity. Chen and Xia [8] proved that the Boros-Moll sequences are 2-log-concave, and Xia [9] studied the concavity and convexity of the Boros-Moll sequences.

In this paper, we give a new definition, i.e., skew log-concavity. Let ${\{T(n,k)\}}_{0\le n<\mathrm{\infty },0\le k\le n}$ be a triangular array of numbers. We say that $T(n,k)$ is skew log-concave if for any fixed n, the sequence ${\{T(n+k,k)\}}_{0\le k<\mathrm{\infty }}$ is log-concave. We will show that the Boros-Moll sequences are almost skew log-concave.

The main results of this paper can be stated as follows.

Theorem 1.1

Let $d_i(m)$ be defined by (1.4). We have, for any fixed $m\ge 1$,

$$d_i^2(m+i)>d_{i-1}(m+i-1)d_{i+1}(m+i+1),i\ge 1,$$ 1.6

and

$$d_i^2(i)<d_{i-1}(i-1)d_{i+1}(i+1),i\ge 1.$$ 1.7

Proof of Theorem 1.1

From (1.4), we see that $d_m(m)=2^{-m}\left(\begin{array}{c}2m\\ m\end{array}\right)$, which implies that (1.7) holds.

By (1.4),

$$d_m(m+1)=\frac{(2m+3)(2m+1)}{2(m+1)}{2}^{-m}\left(\begin{array}{c}2m\\ m\end{array}\right),$$

which yields

$$d_i^2(i+1)>d_{i-1}(i)d_{i+1}(i+2).$$

Therefore, (1.6) holds when $m=1$.

Hence, in the following, we always assume that $m\ge 2$ and $i\ge 1$. We first recall the following three recurrence relations derived by Kauers and Paule [3]:

$$d_i(m+1)=\frac{m+i}{m+1}{d}_{i-1}(m)+\frac{(4m+2i+3)}{2(m+1)}{d}_i(m),0\le i\le m+1,$$ 2.1

$$\begin{array}{c}{d}_i(m+1)=\frac{(4m-2i+3)(m+i+1)}{2(m+1)(m+1-i)}{d}_i(m)\hfill \\ \phantom{d_i(m+1)=}-\frac{i(i+1)}{(m+1)(m+1-i)}{d}_{i+1}(m),0\le i\le m,\hfill \end{array}$$ 2.2

and

$$\begin{array}{rcl}{d}_i(m+2)& =& \frac{-4i^2+8m^2+24m+19}{2(m+2-i)(m+2)}{d}_i(m+1)\\ & & -\frac{(m+i+1)(4m+3)(4m+5)}{4(m+2-i)(m+1)(m+2)}{d}_i(m),0\le i\le m+1.\end{array}$$ 2.3

Now we represent the difference $d_i^2(m+i)-d_{i-1}(m+i-1)d_{i+1}(m+i+1)$ in terms of $d_i(m+i)$ and $d_i(m+i+1)$. Thanks to (2.1), (2.2) and (2.3),

$$\begin{array}{c}{d}_i^2(m+i)-d_{i-1}(m+i-1)d_{i+1}(m+i+1)\hfill \\ =A{d}_i^2(m+i+1)+B{d}_i(m+i+1)d_i(m+i)+C{d}_i^2(m+i),\hfill \end{array}$$ 2.4

where

$$A=\frac{(4m+6i+5)(m+1+i)(m+i){(m+1)}^2(4m+6i-1)}{(i+1)i(4m+4i+1)(4m+4i-1)(m+2i)(m+2i-1)},$$ 2.5

$$B=-\frac{(m+1)(m+i)D}{(i+1)i(4m+4i+1)(4m+4i-1)(m+2i)(m+2i-1)},$$ 2.6

$$C=\frac{E}{4(m+2i-1)(m+2i)(4m+4i-1)(4m+4i+1)(m+1+i)i(i+1)}$$ 2.7

with

$$\begin{array}{c}D=-15+400mi+35i+13m+140m^2+292i^2+864m{i}^2+688m^{2}i\hfill \\ \phantom{D=}+176m^3+336i^3+64m^4+72i^4+320m^{3}i+560m^2{i}^2+384m{i}^3,\hfill \end{array}$$ 2.8

$$\begin{array}{c}E=-68mi-45i-45m-66m^2+2i^2+2\text{,}614m{i}^2+1\text{,}901m^{2}i+451m^3+1\text{,}164i^3\hfill \\ \phantom{E=}+1\text{,}560m^4+3\text{,}320i^4+7\text{,}732m^{3}i+14\text{,}176m^2{i}^2+11\text{,}328m{i}^3+1\text{,}152i^6+1\text{,}984m^5\hfill \\ \phantom{E=}+3\text{,}392i^5+11\text{,}888m^{4}i+16\text{,}856i^{4}m+27\text{,}772m^3{i}^2+31\text{,}332m^2{i}^3+8\text{,}128m^{5}i\hfill \\ \phantom{E=}+23\text{,}040m^4{i}^2+9\text{,}216i^{5}m+33\text{,}216m^3{i}^3+25\text{,}216m^2{i}^4+6\text{,}720m^5{i}^2+11\text{,}584m^4{i}^3\hfill \\ \phantom{E=}+1\text{,}152i^{6}m+11\text{,}072m^3{i}^4+5\text{,}568m^2{i}^5+2\text{,}048m^{6}i+1\text{,}152m^6+256m^7.\hfill \end{array}$$ 2.9

It is easy to check that

$$\mathrm{\Delta }=B^2-4AC=\frac{{(m+1)}^2(m+i)F}{i{(i+1)}^2{(4i+4m+1)}^2{(4i+4m-1)}^2{(2i+m)}^2{(2i+m-1)}^2},$$

where

$$\begin{array}{rcl}F& =& 5\text{,}184i^8+19\text{,}008i^{7}m+27\text{,}648i^6{m}^2+19\text{,}968i^5{m}^3+7\text{,}168i^4{m}^4+1\text{,}024i^3{m}^5\\ & & +6\text{,}912i^7+16\text{,}128i^{6}m+768i^5{m}^2-33\text{,}024i^4{m}^3-44\text{,}288i^3{m}^4-26\text{,}880i^2{m}^5\\ & & -8\text{,}192i{m}^6-1\text{,}024m^7+5\text{,}184i^6+13\text{,}920i^{5}m+9\text{,}584i^4{m}^2-5\text{,}936i^3{m}^3\\ & & -11\text{,}648i^2{m}^4-5\text{,}888i{m}^5-1\text{,}024m^6+6\text{,}096i^5+23\text{,}488i^{4}m+35\text{,}600i^3{m}^2\\ & & +26\text{,}512i^2{m}^3+9\text{,}728i{m}^4+1\text{,}408m^5+2\text{,}000i^4+7\text{,}232i^{3}m+9\text{,}536i^2{m}^2\\ & & +5\text{,}360i{m}^3+1\text{,}088m^4-1\text{,}048i^3-2\text{,}336i^{2}m-1\text{,}728i{m}^2-404m^3\\ & & -143i^2-175im-64m^2+40i+20m.\end{array}$$

Note that A is positive. Hence, in order to prove that the right-hand side of (2.4) is positive, it suffices to prove that when Δ is nonnegative,

$$\frac{d_i(m+i+1)}{d_i(m+i)}>\frac{-B+\sqrt{\mathrm{\Delta }}}{2A}.$$ 2.10

Therefore, in the following, we assume that $\mathrm{\Delta }\ge 0$.

Recall that Kauers and Paule [3] proved the following inequality:

$$\frac{d_i(m+1)}{d_i(m)}\ge \frac{4m^2+7m+i+3}{2(m+1)(m+1-i)},0\le i\le m.$$

Replacing m by $m+i$, we see that

$$\frac{d_i(m+i+1)}{d_i(m+i)}\ge \frac{4i^2+8im+4m^2+8i+7m+3}{2(m+1+i)(m+1)},i\ge 0.$$ 2.11

It is a routine to verify that

$$\begin{array}{c}{(A\frac{4i^2+8im+4m^2+8i+7m+3}{(m+1+i)(m+1)}+B)}^2-\mathrm{\Delta }\hfill \\ =\frac{4(i+m){(m+1)}^2(6i+4m+5)(6i+4m-1)G}{i{(i+1)}^2{(4i+4m+1)}^2{(4i+4m-1)}^2{(2i+m)}^2{(2i+m-1)}^2},\hfill \end{array}$$ 2.12

where

$$\begin{array}{rcl}G& =& 28i^{4}m+108i^3{m}^2+144i^2{m}^3+80i{m}^4+16m^5-32i^4-66i^{3}m\\ & & -46i^2{m}^2-12i{m}^3-32i^3-78i^{2}m-64i{m}^2-17m^3+2i^2+2im+2i+m.\end{array}$$

Note that when $m\ge 2$ and $i\ge 1$, G is positive. Thus the right-hand side of (2.12) is positive. On the other hand,

$$\begin{array}{c}A\frac{4i^2+8im+4m^2+8i+7m+3}{(m+1+i)(m+1)}+B\hfill \\ =\frac{(i+m)(m+1)(-3-12i+28im+48i^2+72i^3+32i{m}^2+96i^{2}m)}{(i+1)(4i+4m+1)(4i+4m-1)(2i+m)(2i+m-1)},\hfill \end{array}$$

which is positive. Therefore, from (2.12), we have

$$A\frac{4i^2+8im+4m^2+8i+7m+3}{(m+1+i)(m+1)}+B>\mathrm{\Delta },$$

which can be rewritten as

$$\frac{4i^2+8im+4m^2+8i+7m+3}{2(m+1+i)(m+1)}>\frac{-B+\sqrt{\mathrm{\Delta }}}{2A}.$$ 2.13

From (2.11) and (2.13), we obtain (2.10) and this completes the proof.