Playing around with the assumptions more (in Maple), that is the answer when c is positive, and the answer is 1 when c is zero or negative.
Aaron Meurer On Oct 9, 2010, at 9:45 PM, Aaron S. Meurer wrote: > Ah! If I plug that important bit of information into Maple, I get that the > answer is 2^c*(1/(1+r))^c+(1/(1+r))^c*r-(1/(1+r))^c. > > No doubt SymPy would need to know it at some point, too. > > Aaron Meurer > > On Oct 9, 2010, at 9:12 PM, Ben Goodrich wrote: > >> On Oct 9, 11:07 pm, Ben Goodrich <[email protected]> wrote: >>>> I tried running the limit through Maple, and it couldn't do it (it just >>>> returned an unevaluated form), and Wolphram Alpha times out. Any idea >>>> what the answer should be? >>> >>> I conjecture the limit is 1 >> >> for c in the (0,1) interval I meant to add. > -- 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.
