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

Find a (presumably, generalized hypergeometric-based function $f(n,a,k)$), yielding for $n=1, a=\frac{1}{2}$,the rational function (ratio of fifth-degree polynomials) \begin{equation} f(1,\frac{1}{2},k)=\frac{64 k^5 +128 k^4 -340 k^3 -1032 k^2 -1099 k -384}{k (2 k-1) (2 k+5) (4 k-1) (4 k+1)} \end{equation} and for $n=1, a=1$, \begin{equation} f(1,1,k)=\frac{8 k^5+36 k^4 -82 k^3 -681 k^2 -1366 k -885}{128 (k+2) (k+3) (k+4) (4 k+5) (4 k+7)} \end{equation} These are, respectively, eqs. (25) and (8) in arXiv:1207.1297v2, "Bures and Hilbert-Schmidt $2 \times 2$ Determinantal Moments". Presumably, for general $n$, we have ratios of $5 n$-degree polynomials in $k$. (These ratios pertain to the Bures case. In the Hilbert-Schmidt counterpart, general formulas--incorporating a $_5F_4$ hypergeometric function--yielding ratios of $3 n$-degree polynomials have previously been found, though not yet rigorously demonstrated arXiv:1109.2560, sec. D.6.)

share|improve this question
add comment

1 Answer 1

Are there any other conditions this function must satisfy? Otherwise, there is a trivial solution: $$\frac{(2-2a) \left(64 k^5+128 k^4-340 k^3-1032 k^2-1099 k-384\right)}{k (2 k-1) (2 k+5) (4 k-1) (4 k+1)}+\frac{(2 a-1) \left(8 k^5+36 k^4-82 k^3-681 k^2-1366 k-885\right)}{128 (k+2) (k+3) (k+4) (4 k+5) (4 k+7)}$$

share|improve this answer
    
Yes, there are other conditions that must be satisfied. In general, $f(n,a,k)$ should yield ratios of $5 n$-degree polynomials in $k$ (as indicated, but not prominently enough perhaps, in the original posting). Our objective is to develop a "Bures"--degree $5 n$--counterpart to the detailed "Hilbert-Schmidt" (HS)--degree $3 n$--results featured in the two arXiv preprints indicated in the original posting. Computational limitations prevent us from directly finding the $n \geq 2$ counterparts, but the (HS) line of reasoning employed in sec. D.4 of arXiv:1109.2560 may be transferable. –  Paul Slater May 13 '13 at 21:00
add comment

Your Answer

 
discard

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

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