I am trying to calculate a asymptotic series expansion of a list.

Series[{0,-((R^2 ν (2+(-1+ν (2+ν)) Cos[2 θ]))/((-1+ν) (1+ν^2) r^2)) + 
    (R^4 ν (-7+5 ν (2+ν)+16 Cos[2 θ]+(-1+3 ν (2+ν)) Cos[4 θ]))/(4 (-1+ν) (1+ν^2) r^4) + 
    O[1/r]^6,-((R^2 ν (3+ν) Cos[θ] Sin[θ])/((1+ν^2) r^3))+O[1/r]^4,0}, {r,Infinity,3}]

I don't understand why the 1/r^4 term shows up in the output. Even more, I can copy & paste this input

Series[-((R^2 ν (2+(-1+ν (2+ν)) Cos[2 θ]))/((-1+ν) (1+ν^2) r^2)) + 
    (R^4 ν (-7+5 ν (2+ν)+16 Cos[2 θ]+(-1+3 ν (2+ν)) Cos[4 θ]))/(4 (-1+ν) (1+ν^2) r^4) + 

but evaluating the following line gives an error message about the argument


Any input is highly appreciated.


1 Answer 1


You have to remove the O[_]^_ from your input:

Series[{0, -((R^2 ν (2 + (-1 + ν (2 + ν)) Cos[2 θ]))/((-1 + ν) (1 + ν^2) r^2)) + (R^4 ν (-7 + 5 ν (2 + ν) + 16 Cos[2 θ] + (-1 + 3 ν (2 + ν)) Cos[4 θ]))/(4 (-1 + ν) (1 + ν^2) r^4), -((R^2 ν (3 + ν) Cos[θ] Sin[θ])/((1 + ν^2) r^3)), 0}, {r, Infinity, 3}]
  • $\begingroup$ Yes, but why? I mean why can't Mathematica interpret it properly? $\endgroup$
    – user37022
    Mar 23, 2013 at 2:01
  • 4
    $\begingroup$ Because O[1/r]^6 is just a display form for a more complicated thing. For example, if you compute Series[x, {x, \[Infinity], 5}], the frontend displays a nice x + O[1/x]^6, but that's not the internal representation, which is SeriesData[x, DirectedInfinity[1], {1}, -1, 6, 1], as you can discover with FullForm. For this reason, you can either accurately copy and paste SeriesDatas without corrupting them or (better) leave out the O[_] placeholder (possibly with Normal). $\endgroup$
    – Federico
    Mar 23, 2013 at 2:09

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