Using expand is a very bad idea, since it is so slow. ratsimp is
faster, but may be overkill.
I think testing ratsimp(c-(2*a1+1) )= 0 would be ok.
Raymond Toy wrote:
> 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
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