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.