#16813: symbolic Legendre / associated Legendre functions / polynomials
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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 maldun):
Oh yeah it's again the non uniqueness of the representation of the complex
logarithm
{{{
sage: log((x+1)/(1-x)).subs(x=3)
I*pi + log(2)
sage: (log(x+1)-log(1-x)).subs(x=3).simplify_log()
-I*pi + log(2)
sage: log((x+1)/(1-x)).subs(x=3).conjugate()
-I*pi + log(2)
}}}
confusing as hell ...
I think Wolfram uses the log(1+x)-log(1-x) representation simply by the
fact that it is independent of the branch in the following sense:
Let log(x) = ln|x| + i*arg(x) + 2kπi and log(y) = ln|y| + i*arg(y) + 2kπi
then
{{{
log(x) - log(y) = ln|x| + i*arg(x) + 2kπi- ln|y| + i*arg(y) + 2kπi =
= ln|x/y| + i*(arg(x) - arg(y)) + 0
}}}
I.e. if we have the same branch on the logarithm the module of 2kπi
cancels out.
That means the formula isn't exactly wrong, it uses simply a different
branch of the logarithm. But the representation of log as difference saves
us indeed a lot of trouble, and as showed above is independent of the
branch we use.
Nevertheless I think we should stick to the recursion with W(n,x), because
from a computational view it is a lot better since:
1) The computational complexity is the same (solving a two term recursion)
2) we save computation time since we don't have to simplify expressions
containing logarithms but only polynomials which are much simpler to
handle and expand.
--
Ticket URL: <http://trac.sagemath.org/ticket/16813#comment:24>
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