[GiNaC-list] series((x+x^2)^2,x,0) is broken
Vladimir V. Kisil
V.Kisil at leeds.ac.uk
Mon Jun 26 11:52:34 CEST 2023
>>>>> On Mon, 26 Jun 2023 10:26:55 +0200, "Richard B. Kreckel" <kreckel at in.terlu.de> said:
RK> Now, I may be missing something: Isn't
RK> Order(x^k*sin(1/x))==Order(x^k) at x==0? And isn't
RK> Order(x^-1)==Order(x)^-1 not finite either but still fine in a
RK> Laurent series?
I think a lot of confusion is coming because our function Order is
not clearly defined. First of all, if we speak on Order(x^4) do we mean
the asymptotic behaviour for x→0 or x→∞? For simple cases like
series(sin(x),x==0,4) = x-1/6*x^3+Order(x^4)
both orders are the same, but this is not true in general. Ginsh
answer
series(sin(x)+x^10,x==0,4) = 1*x+(-1/6)*x^3+Order(x^4)
suggests that we are speaking for x→0 only.
Next, either GiNaC::Order() is only meaningful in the context of series
expansions of analytic functions or it is a sort of big-O concept? For
the latter take f(x) = x^k * ( sin(1/x) +1) +x^m with k > m.
Then for x→0 we have f(x) = O(x^k) but 1/f(x) = O(x^{-m}).
Finally, if we only consider power expansion of analytic functions
then having a zero of an integer order n for f(x) at some point
implies that 1/f(x) has a pole of the same order n there. But I am not
sure that it will be safe to translate this into some properties of
GiNaC::Order(). Currently we have:
> series(x^(100),x==0,4);
Order(x^4)
> series(x^(-100),x==0,4);
1*x^(-100)
The root of the issue is that things like Order() are not about
identities, they are merely about inequalities—which are not
implemented very much in GiNaC presently..
--
Vladimir V. Kisil http://www1.maths.leeds.ac.uk/~kisilv/
Book: Geometry of Mobius Maps https://doi.org/10.1142/p835
Soft: Geometry of cycles http://moebinv.sourceforge.net/
Jupyter notebooks: https://github.com/vvkisil?tab=repositories
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