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Re: Bug in analytical sum
*To*: mathgroup at smc.vnet.net
*Subject*: [mg100295] Re: [mg100244] Bug in analytical sum
*From*: danl at wolfram.com
*Date*: Sun, 31 May 2009 06:35:08 -0400 (EDT)
*References*: <200905300057.UAA23355@smc.vnet.net>
> 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 Limit
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
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