You could also use the coeff() method, like

s = series(...)
coeffs = [s.coeff(x, n) for n in range(10)]

I think in SymPy 0.7.1 or earlier, you have to use just coeff(x**n),
which won't work for n == 0.

Aaron Meurer

On Wed, Sep 12, 2012 at 8:32 PM, Nikolas Tezak <[email protected]> wrote:
> Hmm, I might be missing something, but it seems to me that that is not 
> addressing the particular point I asked about, namely that the current 
> behavior of the n=None lazy term generation does not yield 0 whenever there 
> is no contribution at a particular order.
>
> On Sep 13, 2012, at 11:15 AM, Chris Smith wrote:
>
>> On Thu, Sep 13, 2012 at 6:52 AM, nikolas <[email protected]> wrote:
>>> 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!
>>
>> I wrote about this in the docstring of Expr.series, I believe.  Try
>> "help(Expr.series)".
>>
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