1
$\begingroup$

Fixed in 11.2


I have an expression that I want to expand around a given point, and when I do it without simplifying it gives a different result than when I simplify it beforehand.

Full expression

Tttfinal=1/8 (-1 + x^2) Sqrt[1/((1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (
  r0^2 (3 - 3 x^2 + x^4))/L^2)] Sqrt[(
 r0^4 (1 - y^2)^2)/((1 - x^2)^4 (2 - y^2))] ((
   16 x ((1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (
      r0^2 (3 - 3 x^2 + x^4))/L^2))/(L (-1 + x^2)^2) + (
   1/((-1 + x^2)^3))
   Sqrt[((-1 + x^2)^2 ((1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (
       r0^2 (3 - 3 x^2 + x^4))/L^2))/
    r0^2] (-x (-1 + x^2) (-4 x (1 - x^2) + (6 Q^2 x (1 - x^2)^2)/
         r0^2 + (r0^2 (-6 x + 4 x^3))/L^2) + 
      2 (1 + x^2) ((1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (
         r0^2 (3 - 3 x^2 + x^4))/L^2)) - (1/((-1 + x^2)^2))
   x ((1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (r0^2 (3 - 3 x^2 + x^4))/
      L^2) ((4 x Sqrt[(1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (
        r0^2 (3 - 3 x^2 + x^4))/L^2])/
      r0 + (Sqrt[((-1 + x^2)^2 ((1 - x^2)^2 - (Q^2 (1 - x^2)^3)/
            r0^2 + (r0^2 (3 - 3 x^2 + x^4))/L^2))/
         r0^2] (x (-1 + x^2) (-4 x (1 - x^2) + (6 Q^2 x (1 - x^2)^2)/
              r0^2 + (r0^2 (-6 x + 4 x^3))/L^2) - 
           2 (1 + x^2) ((1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (
              r0^2 (3 - 3 x^2 + x^4))/L^2)))/(x (-1 + 
           x^2) ((1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (
           r0^2 (3 - 3 x^2 + x^4))/L^2)) - (
      4 x Sqrt[(1 - x^2)^2 - (Q^2 (1 - x^2)^3)/r0^2 + (
        r0^2 (3 - 3 x^2 + x^4))/L^2] (2 - y^2))/(r0 (-2 + y^2))))    

with the assumptions:

$Assumptions = 
 And[x >= 0, x <= 1, y >= -1, y <= 1, Q > 0, L > 0, r0 > 0]

Result:

Series[Tttfinal, {x, 1, 2}] // Simplify
Series[Tttfinal // Simplify, {x, 1, 2}] // Simplify

SeriesData[x, 1, {
 Rational[-1, 4] L^(-2) r0^3 (2 - y^2)^Rational[-1, 2] (-1 + y^2), 
  Rational[3, 8] L^(-2) r0^3 (2 - y^2)^Rational[-1, 2] (-1 + y^2), 
  Rational[-3, 8] L^(-2) r0^3 (2 - y^2)^Rational[-1, 2] (-1 + y^2), 
  Rational[1, 16] L^(-2) r0^(-1) (
   13 r0^4 + 8 L^2 (Q^2 + r0^2)) (2 - y^2)^Rational[-1, 2] (-1 + y^2),
   Rational[1, 64] L^(-2) r0^(-1) (
   32 L^4 + 128 L^2 Q^2 - 15 r0^4) (
    2 - y^2)^Rational[-1, 2] (-1 + y^2), 
  Rational[1, 128] L^(-2) r0^(-3) (
   21 r0^6 + 128 L^2 r0^2 (Q^2 + 2 r0^2) + 32 L^4 (8 Q^2 + 9 r0^2)) (
    2 - y^2)^Rational[-1, 2] (-1 + y^2)}, -3, 3, 1]
SeriesData[x, 1, {
 Rational[-1, 2] L^(-2) r0^(-1) (
   r0^4 + L^2 (Q^2 + r0^2)) (2 - y^2)^Rational[-1, 2] (-1 + y^2), 
  Rational[1, 2] (L^2 - 4 Q^2)
    r0^(-1) (2 - y^2)^Rational[-1, 2] (-1 + y^2), 
  Rational[1, 4]
    r0^(-3) ((-4) Q^2 r0^2 + 8 r0^4 + L^2 (8 Q^2 + 9 r0^2)) (
    2 - y^2)^Rational[-1, 2] (-1 + y^2)}, 0, 3, 1]

To be clear, the result I expect to be correct is the one where I simplify the expression.

$\endgroup$
7
  • 3
    $\begingroup$ Your expression for Tttfinal is missing some brackets. $\endgroup$
    – TimRias
    Sep 5, 2017 at 15:00
  • $\begingroup$ How do you know the two results are not equivalent? $\endgroup$
    – m_goldberg
    Sep 6, 2017 at 0:14
  • $\begingroup$ how can they be equivalent? they are series with different coefficients. sorry i will edit with the correct expression $\endgroup$ Sep 6, 2017 at 15:41
  • $\begingroup$ I am not able to replicate the result indicated above. $\endgroup$ Sep 6, 2017 at 16:31
  • 1
    $\begingroup$ Okay, I see it now. Adding the "bugs" tag. It's been fixed for a future release though. $\endgroup$ Sep 6, 2017 at 20:37

1 Answer 1

1
$\begingroup$

fyi, Fixed in 11.2

Mathematica graphics

Mathematica graphics

code using data in post:

res1=Series[Tttfinal,{x,1,2}]//Normal//Simplify
res2=Series[Tttfinal//Simplify,{x,1,2}]//Normal//Simplify
res1-res2
(* 0 *)
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1
  • $\begingroup$ Thank you! I guess I'l have to wait until my institution upgrades to the newest version. $\endgroup$ Sep 16, 2017 at 9:18

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