
Re: Ask for help: how to let mathematica output Fortran code with fewest float operations?
Posted:
Jan 13, 2013 1:13 PM


On Monday, January 14, 2013 12:13:06 AM UTC+8, Richard Fateman wrote: > On 1/13/2013 12:32 AM, Tang Laoya wrote: > > > When the test.f90 is outputted, the first solution of x is: > > >> > > >>x = a/3. + (6 + 2*a**2)/(3.*2**0.6666666666666666*(36*a + 20*a**3 + > > > > Sqrt(4*(6 + 2*a**2)**3 + (36*a + 20*a**3  216*b)**2)  > > 216*b)**0.3333333333333333)  > > > > (36*a + 20*a**3 + Sqrt(4*(6 + 2*a**2)**3 + (36*a + 20*a**3  > > 216*b)**2)  > > > > 216*b)**0.3333333333333333/(6.*2**0.3333333333333333) > > > > .............. > > in maxima, optimize(%) > > > > block([%1,%2,%3,%4,%5], > > %1:2*a^26, > > %2:36*a, > > %3:20*a^3, > > %4:216*b, > > %5:(%4+sqrt((%4+%3+%2)^2+4*%1^3)+%3+%2)^0.33333333333333, > > > > x = 0.13228342099735*%5+0.20998684164915*%1/%5+a/3)$ > > > > There is also a (separate) program to convert Maxima expressions > > to something like looks like Fortran, with ** instead of ^, > > and continuation lines. > > > > Maxima can be downloaded from sourceforge, free.
Dear Prof. Fateman,
Thank you very very much for your so kindly suggestions and help.
Thanks, Tang Laoya

