Follow-up: I think I have found a way that works.
I guess if one is interested in expanding about 0, one can simply do
>>> f = x + 5 * x**3
>>> coeffs = reversed(series(f, x, 0, n = 4).as_poly(x).all_coeffs())
>>> list(coeffs)
[0, 1, 0, 5]
So, to expand about x0!= 0, I can substitute x->x0 + x1 and expand in
orders of x1.
If this is a bad way to do it, please let me know!
Thanks,
Nikolas
On Thursday, September 13, 2012 9:42:30 AM UTC+9, nikolas wrote:
>
> Hey guys,
>
> for a series expansion of some function f(x) about some point x0 up to
> order n+1, I would like to easily generate the sequence of all coefficients.
> I.e. if we have
>
> f(x) = a0 + a1 * (x-x0) + a2 * (x-x0)**2 + ... + an (x-x0)**n +
> O((x-x0)**n)
>
> is there a straightforward way to obtain a generator for the sequence a0,
> a1, a2, ... an that includes the non-zero terms?
> With the current implementation of series, I get:
>
> >>> f = x + 5 * x**3
> >>> coeffs = series(f, x, 0, n = None)
> >>> for c in coeffs:
> >>> print str(c) + ", ",
> x, 5 * x**3,
>
> It would be nice to add a flag, e.g. include_nozero=True such that
> >>> f = x + 5 * x**3
> >>> coeffs = series(f, x, 0, n = None)
> >>> for c in coeffs:
> >>> print str(c) + " + ",
> 0, x, 0, 5 * x**3,
>
> Then I could simply divide each term by (x-x0)**k to get the actual
> coefficient...
> Is there some other way to achieve this functionality easily? It would
> seem that under the hood the machinery to do all this would already be in
> place...
> Thanks!
>
> Nikolas
>
>
>
>
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