Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is the following result about Hankel determinants known or a simple consequence of some known results? Let

$f(x) = \frac{\displaystyle 1}{{\displaystyle 1 - \frac{{a x^{m + 2}}}{\displaystyle {1 - \frac{{b{x^{m + 2}}}}{\displaystyle {1 - {x^2}g(x)}}}}}} = \sum\limits_{n \ge 0} {{f_n}{x^n}} $

and

$g(x) = \sum\limits_{n \ge 0} {{g_n}{x^n}} $

be formal power series. Then the Hankel determinants

${d_f}(n) = \det \left( {{f_{i + j}}} \right)_{i,j = 0}^n$ and

${d_g}(n) = \det \left( {{g_{i + j}}} \right)_{i,j = 0}^n$

are closely related. To simplify notation let

${d_g}(n) = 1$ for $n < 0.$

Then for $n \ge 0$

$${d_f}(n + m + 1) = (-1)^{m+1\choose 2} {a^{m+n+1}}{b^n}{d_g}(n - 2)$$

Remark: For $m = 0$ this result is known and follows from the fact that the moment generating function of monic orthogonal polynomials can be represented as a continued fraction.

share|improve this question
    
Since nobody has given an answer let me state as a comment what I know now. The question can be answered by the method developed in the paper “Some q-orthogonal polynomials and related Hankel determinants” by George Andrews and Jet Wimp, Rocky Mountain J. Math 32 (2002), 429-442, and also by another method of Victor I. Buslaev, “On Hankel determinants of functions given by their expansions in P-fractions”, Ukrainian J. 62 (2010), 358-372. More details can be found in arXiv:1302.4235. –  Johann Cigler Nov 21 '13 at 12:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.