<s>[INST] Finding the general Taylor Series of a function I have to find the general Taylor Series Expansion, about $0$, of the following function.     
$$ \sqrt{x^4 -6x^2+1} $$
I have tried to use the identity     
$$ \left(1+t\right)^{1/2} = \sum_{n\ge 0}\frac{(-1)^{n+1}}{4^n (2n-1)} \binom{2n}{n}t^n $$.   
and then substitute for $t$ accordingly, but to no avail, since the resulting expression becomes messy.   
Any help will be appreciated.   
Thanks.
 [/INST] Keep it simple, for $|x|<1$
$\sqrt{1+x}=1+\dfrac{x}{2}-\dfrac{x^2}{8}+\dfrac{x^3}{16}+O(x^4)$
Plug $x\to x^4-6 x^2$ and get
$$\sqrt{1+ x^4-6 x^2}= \dfrac{1}{16} \left(x^4-6 x^2\right)^3-\dfrac{1}{8} \left(x^4-6 x^2\right)^2+\dfrac{1}{2} \left(x^4-6 x^2\right)+1=\\=1 - 3 x^2 - 4 x^4 - 12 x^6+O(x^7)$$
The initial series had $4$ terms and so will the second one. The additional terms are wrong, I let you imagine the reason why. If you need more terms for the second one you must use more terms in the original series. 
All this provided that $|x^4-6x^2|<1$ that is
$$-\sqrt{3+\sqrt{10}}<x<-1-\sqrt{2}\lor 1-\sqrt{2}<x<\sqrt{2}-1\lor 1+\sqrt{2}<x<\sqrt{3+\sqrt{10}}$$
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