Oh I didn't even notice that the summation is finite. So of course it converges. I think the algorithm computes finite sums by computing an infinite sum as an intermediary somehow (like summation(..., (k, 0, oo)) - summation(..., (k, n + 1, oo)); assuming my memory serves me correctly), but obviously the convergence conditions are not important if the summation is finite.
Aaron Meurer On Sat, Jan 18, 2014 at 4:33 AM, F. B. <franz.bona...@gmail.com> wrote: > > On Friday, January 17, 2014 11:27:49 PM UTC+1, Aaron Meurer wrote: >> >> >> We'll have to see if the conditions for this integral can be improved. >> Any idea what the full convergence conditions should be? >> > > That distribution is given by the polynomial expansion of 1 = ( p + (1-p) > )^n, the variable k is just the polynomial term index, so I guess it > converges to 1 for every value of p. > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to sympy+unsubscr...@googlegroups.com. > To post to this group, send email to sympy@googlegroups.com. > Visit this group at http://groups.google.com/group/sympy. > For more options, visit https://groups.google.com/groups/opt_out. -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. For more options, visit https://groups.google.com/groups/opt_out.
