I was looking at hgfred([A+5/6,A+7/3],[2*A+23/3],4*z*(1-z)).
Currently, this fails because there is a call to hyp-cos (in hyp.lisp)
with arguments a = A+11/6, b = A+7/3, c = 2*(A+11/6)+1.
There is a test (alike1 (add 1 (mul 2 a1)) c) that fails. (a1 =
(a+b-1/2)/2 = (2*A+11/3)/2)
We want this test to succeed because 2*a1+1 actually does equal c,
when everything is expanded out.
What is the best way to achieve this? I could call $expand or
$ratsimp in hyp-cos, but I'm not sure what is right, or if something
else should be used.
(Even if this is fixed, the result, it's hard to tell if the answer is
correct or not. It differs from the answer given in Avgoustis'
thesis.)
Ray