If you call simplify, it simplifies a little bit. I guess it's not necessarily a good idea to auto-expand things like binomial(n, 5). Go ahead and open another issue for that if you want. Probably it should be changed to remain unevaluated, and to expand with expand(func=True).
Aaron Meurer On Fri, Nov 30, 2012 at 8:47 PM, Shriramana Sharma <[email protected]> wrote: > On Sat, Dec 1, 2012 at 9:13 AM, Shriramana Sharma <[email protected]> wrote: >>> easy way to fix it is to use summation(t**i*Ra(i),(i,0,5)).doit(). >> >> Wow nice, that works! > > ... but the problem is that it expands the binomial function too and > the output is not so readable as a result: > > t**5*(-n*(n - 4)*(n - 3)*(n - 2)*(n - 1)*ival(0)/120 + n*(n - 4)*(n - > 3)*(n - 2)*(n - 1)*ival(1)/24 - n*(n - 4)*(n - 3)*(n - 2)*(n - > 1)*ival(2)/12 + n*(n - 4)*(n - 3)*(n - 2)*(n - 1)*ival(3)/12 - n*(n - > 4)*(n - 3)*(n - 2)*(n - 1)*ival(4)/24 + n*(n - 4)*(n - 3)*(n - 2)*(n - > 1)*ival(5)/120) + t**4*(n*(n - 3)*(n - 2)*(n - 1)*ival(0)/24 - n*(n - > 3)*(n - 2)*(n - 1)*ival(1)/6 + n*(n - 3)*(n - 2)*(n - 1)*ival(2)/4 - > n*(n - 3)*(n - 2)*(n - 1)*ival(3)/6 + n*(n - 3)*(n - 2)*(n - > 1)*ival(4)/24) + t**3*(-n*(n - 2)*(n - 1)*ival(0)/6 + n*(n - 2)*(n - > 1)*ival(1)/2 - n*(n - 2)*(n - 1)*ival(2)/2 + n*(n - 2)*(n - > 1)*ival(3)/6) + t**2*(n*(n - 1)*ival(0)/2 - n*(n - 1)*ival(1) + n*(n - > 1)*ival(2)/2) + t*(-n*ival(0) + n*ival(1)) + ival(0) > > -- > Shriramana Sharma > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/sympy?hl=en. > -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
