I want to solve the recursion relation given in equation 2.7(a/b) on page $6$ of this paper. (..the initial seed is $F_1 = G_1 = 1$ and the functions $\alpha$ and $\beta$ are defined on page $5$ in equation $2.4$ on page 5..) Eventually one defines the function $F_n^{(s)}$ and $G_n ^{(s)}$ as given in $2.8a$ and $2.8b$ on page 6..)

I am writing down the recursion relation again here,

Given $n$ vectors, $q_1,q_2,..,q_n$ we have for the set of functions $F_n$ and $G_n$,

$F_n(q_1,q_2,...,q_n) = \sum_{m=1}^{n-1} \frac{G_m(q_1,..,q_m)}{(2n+3)(n-1)}[ (2n+1)\alpha(k,k_1)F_{n-m}(q_{m+1},..,q_n)+2\beta(k,k_1,k_2)G_{n-m}(q_{m+1},..,q_n)] $

$G_n(q_1,q_2,...,q_n) = \sum_{m=1}^{n-1} \frac{G_m(q_1,..,q_m)}{(2n+3)(n-1)}[ 3\alpha(k,k_1)F_{n-m}(q_{m+1},..,q_n)+2n\beta(k,k_1,k_2)G_{n-m}(q_{m+1},..,q_n)] $

where $\alpha(k,k_1) = \frac{k.k_1}{k_1^2}$ and $\beta(k,k_1,k_2)=\frac{k^2(k_1.k_2)}{2k_1^2 k_2^2}$ and $k_1 = \sum_{i=1}^m q_i$ , $k_2 =\sum_{i=m+1}^{n} q_i$ and $k = k_1 + k_2$ (..and the seed of the recursion is $F_1 = G_1 = 1$..)

If $\pi$ is the set of permutations of a set of $n$ distinct elements then one defines,

$F_n(s) = \frac{1}{n!} \sum_{\pi} F_n (q_{\pi(1)},..,q_{\pi(n)})$

$G_n(s) = \frac{1}{n!} \sum_{\pi} G_n (q_{\pi(1)},..,q_{\pi(n)})$

One can see some example solutions for specific patterns of arguments for say $F_2^{(s)}$ and $F_3^{(s)}$ here in equation A.3 and A.4 on page 19.

I can in principle solve this by hand but its quite laborious and its not producing the comparatively more simple looking solutions as in A.3 and A.4 above. I would love to know if there is a way by which this can be automated to produce arbitrarily high order solutions.

  • 2
    $\begingroup$ It would be reasonable to include those equations if you expect someone to support. $\endgroup$
    – Artes
    Commented Apr 17, 2013 at 19:26
  • 1
    $\begingroup$ You have tried RSolve[] already? $\endgroup$ Commented Apr 17, 2013 at 20:25
  • $\begingroup$ Mathematica does not like expressions like RSolve[G[q1, q2] == 1/2 (G[q1, q2] + G[q2, q1]), G[q1, q2], {q1, q2}]. It says: the arguments should be ordered consistently. This will be hard I think. $\endgroup$ Commented Apr 17, 2013 at 22:42
  • $\begingroup$ @Artes I have typed up the recursion relations now explicitly in the question. $\endgroup$
    – user6818
    Commented Apr 18, 2013 at 17:37
  • $\begingroup$ And in general I don't see how I can get Mathematica to do any kind of vector manipulations without having to say its size and without breaking it down it its components. Like given any vector $q$ I am not sure it can such calculations like $\frac{q.q}{q^2} = 1$. $\endgroup$
    – user6818
    Commented Apr 18, 2013 at 18:05


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