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Re: Bug in analytical sum
*To*: mathgroup at smc.vnet.net
*Subject*: [mg100406] Re: Bug in analytical sum
*From*: Sebastian <meznaric at gmail.com>
*Date*: Tue, 2 Jun 2009 06:49:41 -0400 (EDT)
*References*: <200905300057.UAA23355@smc.vnet.net> <gvtmg5$gf8$1@smc.vnet.net>
On May 31, 11:34 am, d... at wolfram.com wrote:
> > Consider the following sum
>
> > f[k_] := Sum[n! (-I \[Alpha])^n Cos[\[Phi]/2]^n \[Alpha]^n Sin[\[Phi]=
/
> > 2]^n, {n,
> > 0, k}]/Sum[n! \[Alpha]^(2 n) Cos[\[Phi]/2]^(2 n), {n, 0, k}]
>
> > I am interested in taking alpha and k to infinity. Now clearly for
> > finite k this is just a rational function in alpha. So if we want to
> > take alpha to infinity we should get
>
> > (-i Tan[\[Phi]/2])^k.
>
> > But try this in Mathematica Limit[f[k], \[Alpha] -> \[Infinity]] and
> > you will get I Cot[\[Phi]/2]. This is Mathematica 7.0.0.
>
> > The reason seems to be that Mathematica evaluates the sum first and
> > obtains a fraction consisting of the incomplete gamma functions and Ei
> > integrals. It seems that the limits of those functions are not taken
> > properly.
>
> Offhand I do not know what is the correct result. But I can say that Limi=
t
> will have trouble managing branch cuts, if not given suitable assumptions
> (and it may have trouble anyway...). So you might instead do:
>
> In[18]:= Limit[f[k], \[Alpha] -> Infinity, Assumptions ->
> {k > 0, Element[\[Phi], Reals]}]
>
> Out[18]= (-(1/2))^k/(Cos[\[Phi]/2]^(2*k)*((-I)*Csc[\[Phi]])^k)
>
> Daniel Lichtblau
> Wolfram Research
Yes, this seems to be the problem. Looks like Mathematica is taking
the limit with k<0. Thanks, this solves the problem.
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