f[x_, a_] := Simplify[Sqrt[a*x + x^2], Assumptions -> x \[Element] Reals]

Then if I run:

 Series[f[x, 0], {x, Infinity, 1}]

 Series[f[x, 1], {x, Infinity, 1}]

The first results in Abs[x] as expected but the leading term of the second Series is x. How do I force it to be Abs[x]? The background is that this is a simplified version of a more complicated function and I am interested in the asymptotic behavior at both +Infinity and -Infinity.

  • $\begingroup$ Welcome to stackexchange, you can format code with the {} button, or starting a new paragraph with four spaces. $\endgroup$ – Feyre Aug 31 '16 at 14:50
  • $\begingroup$ It is not clear, why are you thinking about the absolute value? You have in mind a real x and expand it in the vicinity of the infinity. It is positive. The result of your first expression Series[f1[x, 0], {x, Infinity, 1}] was for me (MMa11, Win7) as follows: SeriesData[x, DirectedInfinity[1], {1}, -1, 2, 1], Not Abs[x]. $\endgroup$ – Alexei Boulbitch Aug 31 '16 at 14:56
  • $\begingroup$ I would also like to know the asymptotic behavior at -Infinity. Running $\endgroup$ – user42740 Aug 31 '16 at 14:58
  • $\begingroup$ The expansion at -infinity does show a long-standing problem with Series. About all I can say is I am hoping to see it improved in the not-too-distant future, possibly version 11.1. (As a matter of general policy we do not make comments like that. I'm making an exception mostly as a way to hold myself to it.) $\endgroup$ – Daniel Lichtblau Aug 31 '16 at 15:04
  • $\begingroup$ Welcome! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – user9660 Aug 31 '16 at 16:01

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