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MathGroup Archive 2003

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Re: Re: Time-consuming series coefficients

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40281] Re: [mg40253] Re: Time-consuming series coefficients
  • From: Dr Bob <majort at cox-internet.com>
  • Date: Sat, 29 Mar 2003 05:19:58 -0500 (EST)
  • References: <200303280929.EAA28015@smc.vnet.net>
  • Reply-to: majort at cox-internet.com
  • Sender: owner-wri-mathgroup at wolfram.com

It's much faster if you can settle for Real coefficients:

Timing[r1 = t[50, 0.5, 0.5];]
Timing[r2 = t[50, 1/2, 1/2];]

{26.359*Second, Null}
{117.094*Second, Null}

Here's the 50th term:

r1[[50]]
r2[[50]]

1.537289031168204647417604240534`15.1095*^-384*u^50

u^50/650495762166524942620327\
52046025534616345389045416905\
03324445219002255037295850186\
25764965108270261552651350872\
14256298696147289052021655675\
87441628548325476827580943368\
26271610823959353848971904559\
96378423061320073085482172177\
99291851563365273912682332718\
05331947551783651036564265975\
29163905199318242511544176388\
76962825641686522180458098768\
04878179358511681294670044409\
731448045568

Bobby

On Fri, 28 Mar 2003 04:29:56 -0500 (EST), Rolf Mertig <rolf at mertig.com> 
wrote:

> 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
>
>
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



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