[Newbie] expansion of a function as a power series

search for "truncated power series"  and you'll learn how to compute series very fast. There is an interesting paper by Kung and Traub on the topic.
RJF

----- Original Message -----
From: Daniel Lakeland <dlakelan at street-artists.org>
Date: Wednesday, July 25, 2007 7:50 pm
Subject: Re: [Maxima] [Newbie] expansion of a function as a power series

> On Wed, Jul 25, 2007 at 07:00:41PM -0700, Richard Fateman wrote:
> > 
> > 
> > ----- Original Message -----
> > From: Daniel Lakeland <dlakelan at street-artists.org>
> > Date: Wednesday, July 25, 2007 2:36 pm
> > .....
> > > You compute the nth derivative of your function and evaluate it 
> at the
> > > expansion point, then divide by n!
> > 
> > 
> > This is extremely unlikely to be more computationally 
> efficiently.  More likely it will be exponentially more expensive 
> than the method used by taylor   (computing in a truncated power 
> series domain).   Computing implicit  power series by the 
> powerseries command uses quite different tools.
> > RJF
> 
> You're absolutely right, I timed computing the 6th order coefficient
> of the taylor series for his example problem by direct methods, and by
> using coeff on the output of taylor. The direct method took 10 times
> the time of taylor. Still both were less than a clock tick with
> "showtime:true" I had to run them in a loop 1000 times to see the
> aggregated time difference. I imagine most of the time goes to
> computing the 6th derivative.
> 
> I'm not quite sure how taylor gets around computing the derivatives
> the long way, but I'm impressed at the difference.
> 
> That being said, my first inclination would have been to call taylor
> and then coeff or ratcoef. That looks to be the best way as well.
> 
> -- 
> Daniel Lakeland
> dlakelan at street-artists.org
> http://www.street-artists.org/~dlakelan
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