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.
 A: hint
Put $t=x^4-6x^2$.
we check that when $x\to 0,  \;\; t $ goes also to zero.
so we can use the expansion of $$\sqrt {1+t}=1+t/2+.... $$
where we replace $t $ by $x^4-6x^2$.
and use $$t^n=x^{2n}\sum_{k=0}^n\binom {n}{k}x^{2k}(-6)^{n-k} $$
Substitute in the sums and you get
$$\sqrt{x^4-6x^2+1}=\sum_{n=0}^\infty a_{2n}x^{2n}$$
where
$$a_{2n}=\sum_{k=0}^n\binom{1/2}{n-k}\binom{n-k}k(-6)^{n-k}$$
A: Oh yes, substituting $t=-6x^2+x^4$ into this and "simplifying" looks like a lot of fun, indeed. =D Just kidding. Let's remember the generating function of Legendre polynomials: $$\frac1{\sqrt{1-2xt+t^2}}=\sum^\infty_{n=0}P_n(x)\,t^n.$$ Multiplying with $t-x$ gives $$\frac{t-x}{\sqrt{1-2xt+t^2}}=-x+\sum^\infty_{n=1}[P_{n-1}(x)-x\,P_n(x)]\,t^n, \tag{1}$$ and the recurrence relation (http://mathworld.wolfram.com/LegendrePolynomial.html, (43)) gives
$$P_{n-1}(x)-x\,P_n(x)=(n+1)\frac{x\,P_n(x)-P_{n-1}(x)}n,$$ so integrating this from $0$ to $t$, we obtain
$$\sqrt{1-2xt+t^2}=1-xt+\sum^\infty_{n=2}\frac{x\,P_{n-1}(x)-P_n(x)}{n-1}t^n \tag{2}.$$ Substituting $x=3$ in (2), we arrive at
$$\sqrt{1-6t+t^2}=1-3t+\sum^\infty_{n=2}\frac{3\,P_{n-1}(3)-P_n(3)}{n-1}t^n \tag{3}.$$ For the final result, we substitute $t=z^2$ to obtain
$$\sqrt{1-6z^2+z^4}=1-3z^2+\sum^\infty_{n=2}\frac{3\,P_{n-1}(3)-P_n(3)}{n-1}z^{2n}.$$ BTW, $P_n(3)$ is always an integer, because (http://mathworld.wolfram.com/LegendrePolynomial.html, (33))
$$P_n(3)=\sum^n_{k=0}\binom{n}{k}^2 2^k.$$ The coefficients in (3) seem to be integers, too, at least that's what numerical results suggest (http://swift.sandbox.bluemix.net/#/repl/597d0432c3917c7502ddfa97).
A: 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
$$-\sqrt{3+\sqrt{10}}<x<-1-\sqrt{2}\lor 1-\sqrt{2}<x<\sqrt{2}-1\lor 1+\sqrt{2}<x<\sqrt{3+\sqrt{10}}$$
