[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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