# How can I select the true expansion?

I have a series expansion which gives two results.

Series[-(1/  Sqrt[-1 + x + x z1^2 \[Chi]^2]) , {x, 0, 1}] /. y -> x


I want only the series which is true. I know I can do by using Assumptions or Assuming by discarding the other. But I want the other way i.e. Mathematica should pick the true one.

• why can't you do res=Series[....] then type Last[res] this return the True part? I do not know what you actually mean by i.e. Mathematica should pick the true one. Apr 28 '18 at 6:47
• @Nasser I mean the true part of the result. Apr 28 '18 at 6:57
• series = Series[-(1/Sqrt[-1 + x + x z1^2 \[Chi]^2]), {x, 0, 1}]; Simplify[series, ! Im[x (1 + z1^2 \[Chi]^2)] < 0]. Note also that True in Piecewise means that this is the default value, that means the values of the expression when all other conditions are violated. Apr 28 '18 at 7:13
• The guys have already answered, what I want to ask is what is the point of the replacement rule /. y -> x ? Apr 28 '18 at 7:21
• @Konstantinos. Oh It has no relation to the question. I have used it for my own calculation purpose. Apr 28 '18 at 8:33