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Possible error with Series expansion

Mathematica Asked on November 5, 2021

I’m expanding the following expression around x=1

Series[-((-(16/(x^2 (-2 + x^2))) + 32/(-1 + x^2)^2 - (
  16 x Sqrt[-(-2 + x^2) (-1 + x^2)^2])/(-1 + x^2)^3 + (
  4 (4 x Sqrt[-(-2 + x^2) (-1 + x^2)^2] - (
     4 Sqrt[-(-2 + x^2) (-1 + x^2)^2])/(x (-2 + x^2)) + (
     4 Sqrt[-(-2 + x^2) (-1 + x^2)^2] (2 - y^2))/(
     x (-2 + x^2) (-2 + y^2))))/(-1 + x^2)^3)/(128 [Pi])),{x,1,2}]

and i get the result

SeriesData[x, 1, {
 Rational[-1, 8]/Pi, Rational[1, 8]/Pi, Rational[-11, 32]/Pi, 
  Rational[1, 16]/Pi, Rational[-117, 128]/Pi}, -2, 3, 1]

while if I use Simplify on the expression first I get

Series[-((-1 - 2 x^2 + x^4 + 2 x Sqrt[2 - x^2])/(
 8 [Pi] x^2 (-2 + x^2) (-1 + x^2)^2)),{x,1,2}]

and the result is now different, despite the expression being the same

SeriesData[x, 1, {Rational[-1, 8]/Pi}, 2, 3, 1]

Can anyone think of an explanation for this? Is this a bug? What result should I consider to be correct?

Thank you

Edit: forgot to add the assumptions

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

2 Answers

Calling

fxy = -((-(16/(x^2 (-2 + x^2))) + 32/(-1 + x^2)^2 - (16 x Sqrt[-(-2 + x^2) (-1 + x^2)^2])/(-1 + x^2)^3 + (4 (4 x Sqrt[-(-2 + x^2) (-1 + x^2)^2] - (4 Sqrt[-(-2 + x^2) (-1 + x^2)^2])/(x (-2 + x^2)) + (4 Sqrt[-(-2 + x^2) (-1 + x^2)^2] (2 - y^2))/(x (-2 + x^2) (-2 + y^2))))/(-1 + x^2)^3)/(128 [Pi]));

we have

fx = Factor[fxy]

now fx depends only on x, with the variable dependency on y being eliminated.

Answered by Cesareo on November 5, 2021

A comparison between the Mathematica result and the of the question is:

Plot[{-(Sqrt[(-1 + x)^2]/(16 [Pi] (x - 1)^3)) + (-1 + 
    Sqrt[(-1 + x)^2])/(16 [Pi] (x - 1)^2) + (
   4 - 11 Sqrt[(-1 + x)^2])/(
   64 [Pi] (x - 1)) + (-11 + 2 Sqrt[(-1 + x)^2])/(
   64 [Pi]) + ((8 - 101 Sqrt[(-1 + x)^2]) (x - 1))/(
   256 [Pi]) + ((-133 - 93 Sqrt[(-1 + x)^2]) (x - 1)^2)/(
   256 [Pi]), -(1/(8 [Pi] (x - 1)^2)) + 1/(8 [Pi] (x - 1)) - 11/(
   32 [Pi]) + (x - 1)/(16 [Pi]) - (117 (x - 1)^2)/(128 [Pi])}, {x, 
  0, 1}]

Plot of the two functions

I got the Mathematica result with this version of the input

Series[-((-(16/(x^2 (-2 + x^2))) + 
      32/(-1 + x^2)^2 - (16 x Sqrt[-(-2 + x^2) (-1 + x^2)^2])/(-1 + 
          x^2)^3 + (4 (4 x Sqrt[-(-2 + x^2) (-1 + 
                 x^2)^2] - (4 Sqrt[-(-2 + x^2) (-1 + x^2)^2])/(x (-2 +
                 x^2)) + (4 Sqrt[-(-2 + x^2) (-1 + x^2)^2] (2 - 
                y^2))/(x (-2 + x^2) (-2 + y^2))))/(-1 + 
          x^2)^3)/(128 [Pi])), {x, 1, 2}, {y, 1, 2}]

SeriesData[x, 1, {
 Rational[-1, 16] Pi^(-1) ((-1 + x)^2)^Rational[1, 2], 
  Rational[1, 16] Pi^(-1) (-1 + ((-1 + x)^2)^Rational[1, 2]), 
  Rational[1, 64] Pi^(-1) (4 - 11 ((-1 + x)^2)^Rational[1, 2]), 
  Rational[1, 64] Pi^(-1) (-11 + 2 ((-1 + x)^2)^Rational[1, 2]), 
  Rational[1, 256] Pi^(-1) (8 - 101 ((-1 + x)^2)^Rational[1, 2]), 
  Rational[1, 256]
    Pi^(-1) (-133 - 93 ((-1 + x)^2)^Rational[1, 2])}, -3, 3, 1]

Since this is a rational function of x and y the version with the with the development in x and y is necessary for a meaningful result. This is special case in which the development into y does not result in a y dependent term of the Series. That does not matter, but it matters to forget the y development.

The input term looks this:

Plot3D[-((-(16/(x^2 (-2 + x^2))) + 
      32/(-1 + x^2)^2 - (16 x Sqrt[-(-2 + x^2) (-1 + x^2)^2])/(-1 + 
          x^2)^3 + (4 (4 x Sqrt[-(-2 + x^2) (-1 + 
                 x^2)^2] - (4 Sqrt[-(-2 + x^2) (-1 + x^2)^2])/(x (-2 +
                 x^2)) + (4 Sqrt[-(-2 + x^2) (-1 + x^2)^2] (2 - 
                y^2))/(x (-2 + x^2) (-2 + y^2))))/(-1 + 
          x^2)^3)/(128 [Pi])), {x, 0.1, 1}, {y, 0.1, 1}, 
 PlotRange -> Full, AxesLabel -> {"x", "y"}]

Plot3D of the given function

Answered by Steffen Jaeschke on November 5, 2021

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