On Tue, Apr 24, 2012 at 5:07 PM, Kjetil brinchmann Halvorsen <kjetil1...@gmail.com> wrote: > I found the usefull summation function, which seems to work well > (examples from todays version from git) > > In [30]: summation(1/k**2,(k,1,oo)) > Out[30]: > 2 > π > ── > 6 > > In [31]: summation( (-1)**k * binomial(n,k), (k,0,n)) > Out[31]: 0 > > But: > > In [32]: summation( binomial(n,k), (k,0,n)) > Out[32]: > n > ____ > ╲ > ╲ ⎛n⎞ > ╲ ⎜ ⎟ > ╱ ⎝k⎠ > ╱ > ╱ > ‾‾‾‾ > k = 0 > > Which should be possible to do! > (I think sage can do that one) > > Kjetil > > -- > "If you want a picture of the future - imagine a boot stamping on the human > face - forever." > > George Orwell (1984)
I agree. Can you open an issue for this in our issue tracker (http://code.google.com/p/sympy/issues/list)? We can do hypergeometric series, which is why the summation() function is so powerful in the git master, but I think it's not recognizing this one as such because it's a finite sum (Tom, please correct me if I'm wrong). If you use the fact that binomial(n, k) == 0 for k > n, and replace n with oo in the summation limits, you get: In [150]: summation( binomial(n,k), (k,0,oo)) Out[150]: ⎧ n ⎪ 2 for (-n ≥ 0 ∧ -n < 1) ∨ -n < 0 ⎪ ⎪ ∞ ⎪ ____ ⎪ ╲ ⎨ ╲ ⎛n⎞ ⎪ ╲ ⎜ ⎟ ⎪ ╱ ⎝k⎠ otherwise ⎪ ╱ ⎪ ╱ ⎪ ‾‾‾‾ ⎩k = 0 And if you assume that n is positive (n = Symbol('n', integer=True, positive=True)), then the piecewise will evaluate to just 2**n. Aaron Meurer -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to sympy@googlegroups.com. To unsubscribe from this group, send email to sympy+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
