#16813: symbolic Legendre / associated Legendre functions / polynomials
-------------------------------------+-------------------------------------
Reporter: rws | Owner:
Type: enhancement | Status: new
Priority: major | Milestone: sage-6.4
Component: symbolics | Resolution:
Keywords: | Merged in:
Authors: | Reviewers:
Report Upstream: N/A | Work issues:
Branch: | Commit:
u/rws/symbolic_legendre___associated_legendre_functions___polynomials|
0f86b77a9aa818add6848cacf9a3f371d49c7d3a
Dependencies: | Stopgaps:
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Comment (by rws):
I think there must be another different formula, because Wolfram has this
for `Q(2,x)`:
{{{
sage: ((3*x^2-1)/2*(log(x+1)-log(1-x))/2-3*x/2).subs(x=3)
-13/2*I*pi + 13/2*log(4) - 13/2*log(2) - 9/2
sage: ((3*x^2-1)/2*(log(x+1)-log(1-x))/2-3*x/2).subs(x=3).n()
0.00545667363964419 - 20.4203522483337*I
}}}
which yields the correct value without use of `conjugate`.
The first few `Q(n,x)` from Wolfram are:
{{{
Q(0,x) = 1/2 log(x+1)-1/2 log(1-x)
Q(1,x) = x (1/2 log(x+1)-1/2 log(1-x))-1
Q(2,x) = 1/2 (3 x^2-1) (1/2 log(x+1)-1/2 log(1-x))-(3 x)/2
Q(3,x) = -(5 x^2)/2-1/2 (3-5 x^2) x (1/2 log(x+1)-1/2 log(1-x))+2/3
}}}
which makes clear that instead of `log((1+x)/(1-x)).conjugate()` we should
just use `log(1+x)-log(1-x)`, of course. Oh well.
--
Ticket URL: <http://trac.sagemath.org/ticket/16813#comment:23>
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