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:


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 $\gamma_0=1$.


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

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-


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

  • 3
    $\begingroup$ 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. $\endgroup$ – Per Alexandersson Feb 2 '14 at 11:02
  • 1
    $\begingroup$ @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 $\endgroup$ – mike Feb 2 '14 at 11:56
  • $\begingroup$ You haven't defined $b_k$ uniquely, is that your intention? We can assume $b_0=1$, but can $b_1$ be arbitrary? $\endgroup$ – Brendan McKay Feb 2 '14 at 12:37
  • $\begingroup$ @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 $\endgroup$ – mike Feb 2 '14 at 13:03
  • 1
    $\begingroup$ 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) $\endgroup$ – reuns Mar 10 '16 at 2:37

Just two pointers for this problem (sorry, no solution).

  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

J. Gélinas

| cite | improve this answer | |
  • $\begingroup$ Thanks for the heads up! For your comment 2, I heard that Hutchinson found out similar results for polynomials and functions in 1923. (Hutchinson_1923_On a Remarkable Class of Entire Functions) $\endgroup$ – mike Dec 16 '15 at 22:39

Just wanted to mention the new paper of Griffin, Ono, Rolen and Zagier https://www.pnas.org/content/early/2019/05/20/1902572116 which deals with exactly this kind of problem (if I understand you and them correctly).

| cite | improve this answer | |

It is unfortunate that this problem has not yet received a solution after four years, specially since it is related to the Riemann Hypothesis.

Now I have looked at it again, comparing the OP polynomials to the Jensen polynomials of the Bessel series (see my Dec 15 2015), and found a way forward ending up to the evaluation of a Hurwitz determinant with rational functions having integer coefficients. It is left for me to prove that this determinant is nonzero for a non negative parameter "d".

Here is some support for my conjecture, to be evaluated in your browser at http://pari.math.u-bordeaux.fr/gp.html

\\   Polynomial with negative roots for d > 0 ?
P(n,d='d) = sum(k=0,n, binomial(2*n,2*k)* a(k,d)* x^k)
a(k,d='d) = prod(j=1,k-1, prod(i=1,j, (2*i+1)/(d+2*i+1) ));
print( "# real roots ",[polsturm(P(n,2*Pi/31)) | n <- [1..16]])

\\   Hurwitz form of Sylvester matrix for factor of discriminant
Hm(n,m=n)  = matrix(2*m,2*m,k,j, binomial(2*n+1,2*j-k) * a(j-k\2) );
dHm(n,m=n) = matdet(Hm(n,m));  \\ m is minor order
for(n=1,3,print("n = ",n," det = ",dHm(n)))

If there is any interest, I will post the details of my transformation from the OP question to my new conjecture, a simple application of resultant theory, but I am interested in the Hurwitz determinant itself which seems to be a rational function of "d" with positive integer coefficients. The OP question is a particular case with $d = 2 \pi/31$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.