# Series default assumptions?

I have a question about the "Series" command. Specifically, the following input

Series[Sqrt[x^2], {x, 0, 2}]


gives me the following output

x+O[x]^3.


However, we know that this is not true for some $x$, e.g. for $x<0$. Because of the singular derivative at $x=0$, I would have expected Mathematica to give an error of some kind here. Moreover, when adding an assumption like this

Series[Sqrt[x^2], {x, 0, 2}, Assumptions -> (x < 0)],


I still get the same output

x+O[x]^3,


whereas I know that the leading order should be $-x$ rather than $x$.

I am trying to understand why Mathematica gives me this result. Does "Series" somehow make some general assumptions about the variable of the expansion?

EDIT: I am using Mathematica version 10.2.

• OK in Mma 11.1.0.0: the output under the assumption $x<0$ is $-x+O\left(x^3\right)$. – user64494 May 2 '17 at 16:02
• What version are you using? Recent versions appear to give sensible answers. – mikado May 2 '17 at 22:19
• Edited the question, I am using version 10.2. Is this is a bug in this version? – ScroogeMcDuck May 3 '17 at 10:24

## 2 Answers

This is an extended comment to demonstrate the effect of Assumptions in version 11.1.1 While a problem exists in some earlier versions, it has been corrected.

\$Version

"11.1.1 for Mac OS X x86 (64-bit) (April 18, 2017)"

Series[Sqrt[x^2], {x, 0, 2}] Series[Sqrt[x^2], {x, 0, 2}, Assumptions -> (x > 0)] Series[Sqrt[x^2], {x, 0, 2}, Assumptions -> (x < 0)] The last two are special cases of

Series[Sqrt[x^2], {x, 0, 2}, Assumptions -> Element[x, Reals]] • This seems more like an output I would have expected. I am using version 10.2, was this a bug in that version? – ScroogeMcDuck May 3 '17 at 10:27
• Yes, this appears to be a bug in (some?) earlier versions (e.g., it gives the result shown for x > 0 in all four cases using v10.4.1 on my Mac). – Bob Hanlon May 3 '17 at 15:00
• Cool, that's clear then. Could you add to your answer that there's this bug in some earlier versions? I'll accept the answer then since that answers my question – ScroogeMcDuck May 8 '17 at 13:04

Bob Hanlon has shown which solution Mathematica 11.1.1 has determined. We can check it with the Taylor formula.

s = Sum[1/j! Nest[(x - x0)*# &, D[Sqrt[x^2], {x, j}] /. x -> x0, j], {j, 0, 2}]


((x - x0) x0)/Sqrt[x0^2] + Sqrt[x0^2]

Limit[s, x0 -> 0, Direction -> -1]
x

Limit[s, x0 -> 0, Direction -> 1]
-x


Mathematica is right.

• I am using version 10.2, which gave me the output in my question. Guess that version was bugged then? The command "Limit" does give the right result in 10.2 as well though. – ScroogeMcDuck May 3 '17 at 10:28