Ralf Hemmecke <[email protected]> writes:

>>> [[x^n]f(x) : -xf'(x) + f(x)^3 - f(x)^2 = 0, f(0) = 1, f'(0) = 1].
>>>
>>> ???
>>>
>>> Nowhere is actually explained what that notation means. :-(
>>
>> OK, thanks for the hint, and for discovering this inconsistency in the
>> article.
>>
>>> Is this a common notation? Can you please give a definition for that
>>> notation?
>>
>> Hm, I thought so, but possibly it's only notation common in
>> combinatorics.
>
> Oh, I knew the [x^n]f(x) notation, but ...
>
>>> Let me guess... The sequence that is written after guessADE is obtained by
>>>
>>> [coefficient(f(x), n) for f in F |
>>>      -xf'(x) + f(x)^3 - f(x)^2 = 0 and f(0) = 1 and f'(0) = 1]
>>
>> well, this rather looks like we would be iterating over all f in F...
>
> ... you iterate over all n. n lives where exactly? Unspecified. Right?

No, I do not iterate over all n.  n is a non negative integer, we are
dealing with formal power series only, i.e., no Puiseux or Laurent
expansions are considered.

> Do you always assume that you have full information of f, i.e. f is a
> given (analytic) function? (Well, analytic is too much, because you
> actually deal with formal power series.)

as I wrote:

>> I have to admit that sometimes the equation doesn't actually determine
>> the sequence, even with initial values given.  (see Section 3.1,
>> description of guessRec)

unfortunately, rather strange things can happen, see Section 6 of the
article.

>>> where F is the space of all (differentiable) functions? Oh, no, F is
>>> K[[x]], right?
>
>> Yes, for guessADE, guessHolo, guessAlg, guessPade, guessFE we consider
>> the given sequence as the first few terms of a formal power series.
>
> But your sequence is fully given, i.e. you could compute the
> coefficient up to any order n.

no, although that's one would hope for.  But it may happen that the
equation does not determine a sequence completely.

Does this answer your question?

Martin

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