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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.

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    $\begingroup$ 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. $\endgroup$ – Nasser Apr 28 '18 at 6:47
  • $\begingroup$ @Nasser I mean the true part of the result. $\endgroup$ – m. bubu Apr 28 '18 at 6:57
  • $\begingroup$ 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. $\endgroup$ – Henrik Schumacher Apr 28 '18 at 7:13
  • $\begingroup$ The guys have already answered, what I want to ask is what is the point of the replacement rule /. y -> x ? $\endgroup$ – Konstantinos Apr 28 '18 at 7:21
  • $\begingroup$ @Konstantinos. Oh It has no relation to the question. I have used it for my own calculation purpose. $\endgroup$ – m. bubu Apr 28 '18 at 8:33

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