1
$\begingroup$

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) + 
    O[1/r]^6,{r,Infinity,3}]

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

Series[a/r^2+b/r^4+O[1/r]^6,{r,Infinity,3}]

Any input is highly appreciated.

$\endgroup$
5
$\begingroup$

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}]
| improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, but why? I mean why can't Mathematica interpret it properly? $\endgroup$ – user37022 Mar 23 '13 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 '13 at 2:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.