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I have the expression

$$(1-x^4)^{-1/4}$$

where $0<x<1$, which gives me a real number. When I try to expand it around $x=1$, Mathematica gives me complex coefficients, which makes me think that he is picking some particular (and unwanted) branch.

$Assumptions = And[0 < x < 1];

Series[(1 - x^4)^(-1/4), {x, 1, 0}] // Simplify // Normal

(*(1/2-\[ImaginaryI]/2)/(-1+x)^(1/4)*)

The lowest order coefficient is

$$\frac{1-i}{2(x-1)^{1/4}}$$

which is complex (the numerator has a complex phase of $-\pi/4$ and the denominator of $\pi/4$, giving the coefficient a total phase of $-\pi/2$).

If instead I previously perform a change of variable $t=(1-x)^{1/4}$, which removes the multivalued behavior at the singularity, Mathematica will give me a real coefficient (as expected):

Normal[Series[1/(1 - x^4)^(1/4) /. x -> 1 - t^4, {t, 0, 0}]] /.t -> (1 - x)^(1/4)

The result is

$$\frac{1}{\sqrt{2}(1-x)^{1/4}}$$

It is not economical for me perform constantly changes of variables like that, so I want to know if there is a way to control the output in cases like that.

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The challenge is that Series[f[x], {x, 1, n}] creates an expansion in terms of x - 1. Since (1 - x^4)^(-1/4) is complex for x > 1, so are the expansion coefficients. If Series did its expansion for 0 < x < 1, as given in $Assumptions, in terms of 1 - x, real coefficients would result instead. A comment by J.M. in Question 112291 suggests how this might be accomplished. For generality, define

mySeries[f_, lst_] := 
    Quiet@Module[{z = First@lst, lst1 = ReplacePart[lst, 2 -> -(lst[[2]])]}, 
    (Series[f /. z -> -z, lst1] // Normal) /. z -> -z]

Then, for the expression in the question,

mySeries[(1 - x^4)^(-1/4), {x, 1, 2}]
(* 1/(Sqrt[2] (1 - x)^(1/4)) + (3 (1 - x)^(3/4))/(8 Sqrt[2]) +
   (13 (1 - x)^(7/4))/(128 Sqrt[2]) *)

as desired.

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  • $\begingroup$ Thank you, that was a very useful and elegant solution. $\endgroup$ – dpravos Apr 18 '16 at 8:02

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