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.

$\endgroup$
  • $\begingroup$ Works for me, but takes a bit more than two minutes on an i7-2820QM. $\endgroup$ – Yves Klett Nov 4 '13 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 Nov 4 '13 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 Nov 4 '13 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 Nov 4 '13 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 Nov 4 '13 at 18:47
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

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

$\endgroup$

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.