[Newbie] expansion of a function as a power series

On Wed, 25 Jul 2007 14:00:22 -0400
"Stavros Macrakis" <macrakis at alum.mit.edu> wrote:

> ratcoeff(ser,x,2) gives you the coefficient of x^2.  This is more
> reliable than "part", which simply counts terms from the beginning.
> If there is more than one term in the constant part, or terms with 0
> coefficients, "part" won't get the right answer.  For example,
> 
>       part(taylor(cos(x),x,0,10),3)/x^3 => x/24 (wrong)
> but
>       ratcoeff(taylor(cos(x),x,10),x,3) => 0 (correct)
> 
Ah, thanks.

The other question still stands: is there a general way to get a higher
taylor coefficient of a function that's more computationally efficient
than computing all the lower ones?

And can one do it in maxima?

Rupert

P.S. I haven't really thought about this, so it might be mathematically
trivial or impossible...
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