1 | initial version |
sage: n, k = var('n, k')
sage: f(x,k) = sum((2/n)*(sin(n*x)*(-1)^(n+1)), n, 1, k)
#where n = 1,2,3 ... k
sage: f
(x, k) |--> -2*sum((-1)^n*sin(n*x)/n, n, 1, k)
I'm not sure what you think is wrong here. The 2 and a factor of -1 were both factored out, that's all.
However, I do agree that this doesn't expand. What is happening is that we are sending the sum to Maxima
if algorithm == 'maxima':
return maxima.sr_sum(expression,v,a,b)
and then ordinarily when it returns, it is still a Maxima object (which may be a bug?). But when we put it in the function, it becomes a Sage object - but we don't have a Sage "sum" object. So I think that is what would have to be fixed.
2 | No.2 Revision |
sage: n, k = var('n, k')
sage: f(x,k) = sum((2/n)*(sin(n*x)*(-1)^(n+1)), n, 1, k)
#where n = 1,2,3 ... k
sage: f
(x, k) |--> -2*sum((-1)^n*sin(n*x)/n, n, 1, k)
I'm not sure what you think is wrong here. The 2 and a factor of -1 were both factored out, that's all.
However, I do agree that this doesn't expand. What is happening is that we are sending the sum to Maxima
if algorithm == 'maxima':
return maxima.sr_sum(expression,v,a,b)
and then ordinarily when it returns, it is still a Maxima object (which may be a bug?). But when we put it in the function, it becomes a Sage object - but we don't have a Sage "sum" object. So I think that is what would have to be fixed.
That this is possible is shown by the following Maxima example (which I put on the ticket):
(%i1) f: -2*'sum((-1)^n*sin(n*x)/n,n,1,2);
2
==== n
\ (- 1) sin(n x)
(%o1) - 2 > ---------------
/ n
====
n = 1
(%i8) f, nouns;
sin(2 x)
(%o8) - 2 (-------- - sin(x))
2