       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
• Sender: owner-wri-mathgroup at wolfram.com

```Is this what you want?

(timings on a 1GHz Linux box):

Mathematica 4.2 for Linux
-- Motif graphics initialized --

In:= !!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:= <<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= {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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