Re: PolyLogs
- To: mathgroup at smc.vnet.net
- Subject: [mg77429] Re: [mg77339] PolyLogs
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Fri, 8 Jun 2007 05:36:53 -0400 (EDT)
- References: <200706070742.DAA09727@smc.vnet.net>
I believe the problem of showing this symbolically is (probably)
impossible to solve and perhaps even "meaningless". The reason is
that the expression involves values of PlyLog and Log at points where
these functions have branch discontinuities, which means that the
usual symbolic identities these functions satisfy do not hold, and
their values are decided by a convention that makes them "continuous
form one side". This means identities involving such expressions will
usually not be decidable by pure symbolic manipulation. To illustrate
what I mean consider a "simpler" problem of deciding if the number
"o" is real. We get:
v = ComplexExpand[Im[-(Pi^2/12) - ArcSinh[1]*Log[68 - 48*Sqrt[2]] +
I*Pi*Log[3 - 2*Sqrt[2]] - (1/2)*Log[3 - 2*Sqrt[2]]^2 +
4*PolyLog[2, 2 - Sqrt[2]] - PolyLog[2, 3 + 2*Sqrt[2]] - Log[1 +
Sqrt[2]]^2]]
-Im[PolyLog[2, 3 + 2*Sqrt[2]]] + Pi*Log[3 - 2*Sqrt[2]]
Is this number 0? Well,
N[-Im[PolyLog[2, 3 + 2*Sqrt[2]]] + Pi*Log[3 - 2*Sqrt[2]], 100]
N::meprec:Internal precision limit $MaxExtraPrecision = 50. reached
while evaluating =B9 log(3-2 )-Im(Li2(3+2 )). >>
0``147.75234164756424
The answer seems to be yes. But let's perturb it in the complex plane
and take limits in different directions:
Limit[ComplexExpand[-Im[PolyLog[2, 3 + 2*Sqrt[2] + z]] + Pi*Log[3 -
2*Sqrt[2] + z]],
z -> 0, Direction -> -I]
Im[PolyLog[2, 3 + 2*Sqrt[2]]] + Pi*Log[3 - 2*Sqrt[2]]
N[%, 10]
-11.0756671441947228707`10.000000000000004
That's clearly non-zero. While:
Limit[ComplexExpand[-Im[PolyLog[2, 3 + 2*Sqrt[2] + z]] + Pi*Log[3 -
2*Sqrt[2] + z]],
z -> 0, Direction -> I]
Limit::ztest:Unable to decide whether numeric quantities {2 =B9 log
(3-2 )-2 Im(Li2(2 Power(<<2>>)+3))} are equal to zero. Assuming they
are.
0
I believe that (at least in this case) the numerical values of the
functions at points where the functions are discontinuous are defined
by a numerical procedure (as one sided limits) and therefore it is
pointless to ask for a purely symbolic proof of an identity involving
them (by a symbolic proof I mean here one that a CAS can produce
without human help and using only general symbolic identities).
Andrzej Kozlowski
On 7 Jun 2007, at 16:42, dimitris wrote:
> Hello.
>
> The following is part from my answer in another forum.
>
> In[602]:=
> o = -(Pi^2/12) - ArcSinh[1]*Log[68 - 48*Sqrt[2]] + I*Pi*Log[3 -
> 2*Sqrt[2]] - (1/2)*Log[3 - 2*Sqrt[2]]^2 +
> 4*PolyLog[2, 2 - Sqrt[2]] - PolyLog[2, 3 + 2*Sqrt[2]] - Log[1 +
> Sqrt[2]]^2;
>
> In[603]:=
> Developer`ZeroQ[o] (*version 5.2*)
> Out[603]=
> True
>
> However ZeroQ uses internally numerical evaluations so that its
> application can be considered
> as "...an plausibility argument but not a rigorous proof..." (I
> adopted this quote from Daniel Lichtblau).
>
> Can anyone really show within Mathematica that above expression
> simplifies really to 0?
>
> Dimitris
>
>
- References:
- PolyLogs
- From: dimitris <dimmechan@yahoo.com>
- PolyLogs