2
$\begingroup$

I'm working on a notebook, trying to expand the root of a cubic polynomial in Taylor series. When I type:

Series[(Sqrt[46656 a^2 - 864 (3 2^(1/3) a^(2/3) + 2 a B)^3] + 216 a)^(1/3) , {a, 0, 2}]

Mathematica takes an indefinite amount of time and I am forced to halt execution. After this occurs, even simple functions like Exp[x] will not compute and I have to restart the kernel.

Am I doing something wrong here? My computer is a month old, so I know the problem isn't old hardware.

Improve this question
$\endgroup$
10
  • $\begingroup$ Works for me, but takes a bit more than two minutes on an i7-2820QM. $\endgroup$
    – Yves Klett
    Commented Nov 4, 2013 at 17:56
  • 1
    $\begingroup$ Do you really need an exact, symbolic result? It is likely to be huge so it may not be useful to you. You could convert the input to inexact numbers to a certain precision and work with that. $\endgroup$
    – Szabolcs
    Commented Nov 4, 2013 at 18:00
  • 1
    $\begingroup$ Series[N[(Sqrt[46656 a^2 - 864 (3 2^(1/3) a^(2/3) + 2 a bb)^3] + 216 a)^(1/3), 30], {a, 0, 2}] $\endgroup$
    – Szabolcs
    Commented Nov 4, 2013 at 18:00
  • 3
    $\begingroup$ You're not doing anything incorrect here. It seems that the Series code is using a fairly high order in some internal computations. I need to check whether there is solid reason for that, or whether it needs to be tamed to some extent. $\endgroup$
    – Daniel Lichtblau
    Commented Nov 4, 2013 at 18:33
  • 1
    $\begingroup$ If you take @Szabolcs advice: Series[(Sqrt[46656 a^2 - 864 (3 2^(1/3) a^(2/3) + 2 a B)^3] + 216 a)^(1/3) // N, {a, 0, 2}] works very quickly $\endgroup$
    – Yves Klett
    Commented Nov 4, 2013 at 18:47

1 Answer 1

Reset to default
1
$\begingroup$

Assuming you're interested in the series expansion for positive values of the parameter a, you can use:

FullSimplify @ Series[
    (Sqrt[46656 a^2 - 864 (3 2^(1/3) a^(2/3) + 2 a B)^3] + 216 a)^(1/3),
    {a, 0, 2},
    Assumptions -> a>0
] //TeXForm

$6 \sqrt[3]{a}+2 \sqrt[3]{2} \sqrt{a} \sqrt{-B}+\frac{2}{3} 2^{2/3} a^{2/3} B+\frac{2}{27} a^{5/6} (-B)^{3/2}-\frac{4}{81} a \left(\sqrt[3]{2} B^2\right)+\frac{5 a^{7/6} (-B)^{5/2}}{243 \sqrt[3]{2}}+\frac{8 a^{4/3} B^3}{2187}+\frac{35 a^{3/2} (-B)^{7/2}}{6561\ 2^{2/3}}+\frac{8\ 2^{2/3} a^{5/3} B^4}{19683}+\frac{2555 a^{11/6} (-B)^{9/2}}{4251528}-\frac{496 a^2 \left(\sqrt[3]{2} B^5\right)}{1594323}+O\left(a^{13/6}\right)$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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