Raymond Toy wrote:
> ....
> But when invertbigfloat calls fpquotient, there is a
> problem. fpquotient is called:
>
> (FPQUOTIENT (36028797018963968 1) (535045584704600947821 2))
>
>
> The first argument is 1b0, using fpprec=56 bits. The second arg is
> gamma(1/4), using 69 bits. But I'm pretty sure these fp functions
> assume all args are supposed to have the same precision.
>
yes, this would be a real problem if the two numbers are different
precisions and the headers are missing.
If you put the headers back, and did the division, e.g.
( ...divide.. ((bigfloat simp 56) 360..... 1) (bigfloat simp 69)
53..... 2))
then that would work. I don't know what the function needed for
..divide.. is. is there an mdiv or simpdiv or something like that?
Otherwise (defun ..divide.. (a b)(simplifya (list '(mquotient) a b) t))
> So, perhaps, the correct solution is to make simpgamma round the result
> from bffac to fpprec bits before returning.
>
> Ray
>
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