# a second order difference equation related to a real polynomials which seems to have only real roots

I am seeking solutions to the following difference equation: $$2c_k-c_{k-1}-c_{k+1}=\ln(k+A)-\ln(k+B)$$ where $$A>B>0$$.

This equation is related to a real polynomial (see here) which I want to prove that it has only real roots.

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

So $$2{c_k}-{c_{k-1}}-{c_{k+1}}=2\ln{b_k}-\ln{b_{k-1}}-\ln{b_{k+1}}=\ln(k+A)-\ln(k+B)$$

These polynomials showed up when we tried to find a polynomial approximation to Jensen's polynomials associated with Riemann $$\xi(z)$$ function.

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

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

Thanks- Mike

• What is the relation between this equation and those polynomials ? Feb 8 '14 at 16:28

$$c_k-2c_{k-1}+c_{k-2}=\ln(\frac{k-1+b}{k-1+a})$$

Let $b_k=c_k-c_{k-1}$. Then we have the relation $$b_k-b_{k-1}=\ln(\frac{k-1+b}{k-1+a}).$$ So we obtain
$$b_k=b_1+\ln(\prod_{i=1}^{k-1}\frac{i+b}{i+a})$$

Then we should solve the relaton $$c_k-c_{k-1}=b_1+\ln(\prod_{i=1}^{k-1}\frac{i+b}{i+a})$$ So $$c_k=c_1+kb_1+\ln(\prod_{i=1}^{k-1}\prod_{j=1}^{i}\frac{j+b}{j+a})$$

Do you want to find solution like this? And I don't know the relation between it and the question you have asked.

• Thanks a lot for the solution. This is exactly what I am looking for. Could you please check the index $j$ in your solution? It does not appeared in the term inside the product. Maybe you can simplify your expression using Pochhammer symbol.
– mike
Feb 9 '14 at 15:34
• @mike Thank you. I think you are right. Feb 10 '14 at 2:44