Is there an online Tylor series calculator that can handle this expression?
When I compute the the eigenvalues of the matrix $\begin{bmatrix}x^3& 2 + x^3& 3 - 2 x^3\2 + x^3& 4 - 5 x^3& 5 + 3 x^3\3 - 2 x^3& 5 + 3 x^3& x^3\end{bmatrix}$ using WolframAlpha, I get the following first eigenvalue (copied from WolframAlpha):
λ_1 = 1/3 (4 - 3 x^3) + 1/3 (-432 x^9 - 1449 x^6 + 810 x^3 + 3 sqrt(3) sqrt(-10664 x^18 + 34200 x^15 - 38845 x^12 - 162020 x^9 - 241908 x^6 + 30520 x^<PHONE_NUMBER>) + 1072)^(1/3) - (-78 x^6 - 18 x^3 - 130)/(3 (-432 x^9 - 1449 x^6 + 810 x^3 + 3 sqrt(3) sqrt(-10664 x^18 + 34200 x^15 - 38845 x^12 - 162020 x^9 - 241908 x^6 + 30520 x^<PHONE_NUMBER>) + 1072)^(1/3))
I want to see the first few terms of the Taylor expansion around x=0 of that expression, however, it seems WolframAlpha cannot handle it. I have been looking for an online Tylor series calculator that can do this but it seems the expression is too hard to crack.
Question: Can anyone share with me a website where I can compute the first few terms (until $x^3$) of the desired Tylor expansion about $x=0$? I do not know how to use programs like Matlab to do this. Do you have any recommendations?
In case it is of any help, here is the expression in latex code:
$$ \frac{1}{3} (4 - 3 x^3) + \frac{1}{3} (-432 x^9 - 1449 x^6 + 810 x^3 + 3 \sqrt{3} \sqrt{-10664 x^{18} + 34200 x^{15} - 38845 x^{12} - 162020 x^9 - 241908 x^6 + 30520 x^<PHONE_NUMBER>} + 1072)^{1/3} - \frac{(-78 x^6 - 18 x^3 - 130)}{(3 (-432 x^9 - 1449 x^6 + 810 x^3 + 3 \sqrt{3} \sqrt{-10664 x^{18} + 34200 x^{15} - 38845 x^{12} - 162020 x^9 - 241908 x^6 + 30520 x^<PHONE_NUMBER>} + 1072)^{1/3})}$$
which looks like this: $$ \frac{1}{3} (4 - 3 x^3) + \frac{1}{3} (-432 x^9 - 1449 x^6 + 810 x^3 + 3 \sqrt{3} \sqrt{-10664 x^{18} + 34200 x^{15} - 38845 x^{12} - 162020 x^9 - 241908 x^6 + 30520 x^<PHONE_NUMBER>} + 1072)^{1/3} - \frac{(-78 x^6 - 18 x^3 - 130)}{(3 (-432 x^9 - 1449 x^6 + 810 x^3 + 3 \sqrt{3} \sqrt{-10664 x^{18} + 34200 x^{15} - 38845 x^{12} - 162020 x^9 - 241908 x^6 + 30520 x^<PHONE_NUMBER>} + 1072)^{1/3})}$$ 565