I would like to compute ∑∞k=0(1+(−1)k)xk. In Mathematica, one can do
In[1]:= Sum[(1+(-1)^k)*x^k,{k,0,Infinity}]
-2
Out[1]= -------
2
-1 + xHow does it work in Sage? I tried
sage: var('k')
k
sage: sum((1+(-1)^k)*x^k, k, 0, oo)
sum(((-1)^k + 1)*x^k, k, 0, +Infinity)which doesn't give me any new information.
For example
giac('normal(sum((1+(-1)^k)*x^k,k,0,inf))').sage()
-2/(x^2 - 1)Nice ! I didn't know this one.
Alternate way to express, possibly more "natural"' to Python/Sage :
sage: sum((1+(-1)^k)*x^k, k, 0, oo)._giac_().normal()._sage_()
-2/(x^2 - 1)Note a convergence condition given by Sympy's solution (currently not directly translatable in Sage) :
sage: sum((1+(-1)^k)*x^k, k, 0, oo)._sympy_().doit()
Piecewise((1/(1 - x), Abs(x) < 1), (Sum(x**k, (k, 0, oo)), True)) + Piecewise((1/(x + 1), Abs(x) < 1), (Sum((-1)**k*x**k, (k, 0, oo)), True))This convergence radius is important.
Thank you both!
Asked: 1 year ago
Seen: 194 times
Last updated: Jul 09 '23
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