# Zeros of polynomials related to Jensen polynomial associated with Riemann xi function $\xi(x)$

We encountered polynomials defined by the recursive relations for the coefficients $b_k>0$ as defined below: $$p_{n}(x)=\sum_{k=0}^{n}\binom{2n}{2k}b_k x^k$$ $$\frac{b_k^2}{b_{k-1}b_{k+1}}=1+\frac{\pi}{31(k+1/2)}=\frac{k+A}{k+B}>1$$

These polynomials showed up when we tried to find a polynomial approximation to Jensen polynomial associated with Riemann $\xi(z)$ function. For detailed background information on jensen polynomial and its relation to entire function like Riemann $\xi(z)$ function, see ref. 1 and ref. 2 below.

Here we provide some basic information from ref. 1 and ref. 2. Riemann $\xi(z)$ function is defined as $$\xi (z/2)=8\int_0^{\infty}\Phi(t)\cos(zt)dt$$ where $$\Phi(x)=\sum_{n=1}^{\infty}(2n^4\pi^2e^{9t}-3n^2\pi e^{5t})exp(-n^2\pi e^{4t})$$

The Riemann Hypothesis is equivalent to that all the infinite zeros of $\xi(z)$ are real.

Taking Taylor expansion on $\cos(zt)$, we obtain

$$\frac{1}{8}\xi(z/2)=\sum_{m=0}^\infty (-1)^m a_m\frac{z^{2m}}{(2m)!}$$ where $$a_m=\int_0^{\infty}t^{2m} \Phi(t)dt$$

On setting $x=-z^2$ and $\xi_1(x)=\frac{1}{8}\xi(z)$, we obtain $$\xi_1(x)=\sum_{m=0}^\infty a_{m}\frac{x^{m}}{(2m)!}$$

The function $\xi_1(x)$ is then an entire function of order 1/2.

The Jensen polynomial $g_n(x)$ associated with $\xi_1(x)$ is defined as $$g_n(x)=\sum_{m=0}^n \binom{n}{m}a_{m}\frac{m!}{(2m)!}x^{m}$$

A Theorem due to Polya and Schur states that A real entire function $\phi(x)=\sum_{m=0}^\infty c_m \frac{x^m}{m!}$ to be in Laguerre-Plya class (i.e., all infinite zeros of $\phi(x)$ are real) if and only if the associated Jensen polynomials $g_n(x)=\sum_{m=0}^n \binom{n}{m}c^{m}x^{m}$ (n=1,2,3...) have only real zeros.

Therefore the Riemann Hypothesis is equivalent to that all the zeros of $g_n(x)$ are real.

Polya conjectured and Craven, Norfolk and Varga proved (cf ref. 1 and ref. 2) the following necessary condition (now also known as Turan inequality) for all the zeros of $g_n(x)$ to be real:

$$a_m^2\gt\frac{m-\frac{1}{2}}{m+\frac{1}{2}}a_{m-1}a_{m+1}$$

Since it is too hard to directly prove that all the zeros of $g_n(x)$ are real, we try to see if a polynomial (like $p_n(x)$ defined at the top) similar to $g_n(x)$ has all the real zeros.

Here by similar we mean that their coefficients obey similar recursive relations.

@gaoxinge found a closed-form solution of coefficients $b_k$ here

Let $\binom{2n}{2k}b_k=\gamma_k\binom{2n}{2k-2}b_{k-1}$. Then we have

$$\gamma_k=\frac{k-1+B}{k-1+A}\frac{(n-k+1)(2n-2k+1)}{k(2k-1)}\gamma_{k-1}$$ So we obtain $$\gamma_k=\frac{(k-1+B)_k}{(k-1+A)_k}\frac{(n-k+1)_k(2n-2k+1)_k}{(k)_k(2k-1)_k}\gamma_{0}$$ where $(A)_k$ is the Pochhammer symbol and $\gamma_0=1$.

$$\binom{2n}{2k}b_k=\prod_{j=0}^k\gamma_j$$

$$p_{n}(x)=\sum_{k=0}^{n}(\prod_{j=0}^k\gamma_j) x^k$$

Numerical results showed that the roots for $p_{n}(x)$ with $1\leq n\leq 150$ are all real.

We are looking for a proof (or a reference on such proof) that all the zeros of $p_n(x)$ are real.

Any references on similar proofs will be helpful to us.

Best regards-

Mike

ref.1 G. Csordas, T. S. Norfolk and R. S. Varga, The Riemann Hypothesis and the Turán Inequalities, Transactions of the American Mathematical Society, Vol. 296, No. 2 (Aug., 1986), pp.521-541

ref.2 T. Craven, G. Csordas; Jensen polynomials and the Turan and Laguerre inequalities. Pacific J. Math., 136 (2) (1989), pp. 241–260

• Hm, you want to look up log-concavity... It looks like the $b_i$ are log-concave, and the convolution with binomial coefficients should not be a problem. See the works of Petter Brändén. – Per Alexandersson Feb 2 '14 at 11:02
• @Per Thanks a lot for the comment. The coefficients $\binom{n}{i}b_i$ definitely satisfy Newton inequality so that the polynomials $p_n(x)$ satisfy the necessary condition to have real zeros. Could you please point me to a specific paper (or a specific theorem) by Petter Branden on arXiv? Thanks- Mike – mike Feb 2 '14 at 11:56
• You haven't defined $b_k$ uniquely, is that your intention? We can assume $b_0=1$, but can $b_1$ be arbitrary? – Brendan McKay Feb 2 '14 at 12:37
• @Brendan $b_0=b_1=1$ is OK. $b_0$ can be divided through, $b_1$ can be absorbed into x. So their values won't affect if the polynomials have real roots or not. This is true for all such polynomials. Thanks – mike Feb 2 '14 at 13:03
• there is also zeros of entire functions represented by Fourier transforms, Peter Hallum which was intended to list all the functions (and the methods of proofs) related to $\xi(z)$ such that we know how to prove that all its zeros are real (or tend to be) – reuns Mar 10 '16 at 2:37

1. Hurwitz used a version of Sturm's theorem on numerators of continued fractions to study the zeros of the Bessel functions, and Watson's treatise on Bessel functions (1944) has this in section 9.7. The series in question can be written as $$\sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!} \prod_{j=1}^k \frac{2j-1}{2j+2p},\quad(p>-1).$$ (Hurwitz 1889 Ueber die Nullstellen der Bessel'schen Function)
2. There is a converse to Newton's inequalities for real polynomials with positive coefficients: if $a_i>0$ and $a_i^2 > 4 a_{i-1} a_{i+1}$, then all roots are real and distinct: A Sufficient Condition for All the Roots of a Polynomial To Be Real David C. Kurtz, The American Mathematical Monthly Vol. 99, No. 3 (Mar., 1992), pp. 259-263