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)](https://mathoverflow.net/questions/156444/zeros-of-polynomials-defined-by-the-recursive-relations-for-the-coefficients) 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



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

> 
> $$c\_k-2c\_{k-1}+c\_{k-2}=\ln(\frac{k-1+b}{k-1+a})$$
> 
> 
> 


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.
> 
> 
> 


