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MathGroup Archive 2003

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Re: Time-consuming series coefficients

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40253] Re: Time-consuming series coefficients
  • From: Rolf Mertig <rolf at mertig.com>
  • Date: Fri, 28 Mar 2003 04:29:56 -0500 (EST)
  • Organization: Mertig Consulting
  • Reply-to: rolf at mertig.com
  • Sender: owner-wri-mathgroup at wolfram.com

Is this what you want?

(timings on a 1GHz Linux box):

Mathematica 4.2 for Linux
Copyright 1988-2002 Wolfram Research, Inc.
 -- Motif graphics initialized --

In[1]:= !!tay
t[n_,p_,q_] := Block[{f, res = 1},
  f[i_] = (E^(z*p*(q)^(i - 1)) - 1)*u + 1 + O[z]^(n + 1);
    Do[res = Expand[Normal[res*f[j]]] + O[z]^(n + 1), {j, 1, n}];
     SeriesCoefficient[res, n]];

Table[Print["n = ",j, " time = ",
   ti[j]=First[Timing[r[j]=t[j,1/2,1/2]]]];ti[j]/Second,{j,10,50,10}]

In[1]:= <<tay
n = 10 time = 0.35 Second
n = 20 time = 2.82 Second
n = 30 time = 16.83 Second
n = 40 time = 74.97 Second
n = 50 time = 262.01 Second

Out[1]= {0.35, 2.82, 16.83, 74.97, 262.01}


Rolf Mertig
Mertig Consulting
http://www.mertig.com
http://www.mathxls.com


"marchibald" <marchibald at maths.wits.ac.za> wrote in message 
news:<b5uoel$lfe$1 at smc.vnet.net>...
> Hi,
> 
> I have the following function from which I would like to extract the 
> coefficient of z^{50}, and then expand the result as a series in u:
> 
> f[z_,u_] := Product[1 + u (Exp(zpq^{i - 1}) - 1), {i,50}]
> 
> where p=q=1/2.
> 
> The product from i=1 to 10 can be done in a reasonable amount of time, but 
> anything larger takes far too long.
> 
> Does anyone have an idea of how to expand f[z,u] as a series quickly, or 
even 
> to extract the coefficient of z^{50} some other way?
> 
> Thanks,
> 
> M. Archibald
> University of the Witwatersrand




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