Maclaurin expansion of $\arccos(1-2x^2)$ Maclaurin expansion of $\arccos(1-2x^2)$
This is what I tried.
$f'(x)=2(1-x^2)^{-1/2} \\
f''(x)=2(1-x^2)^{-3/2}+3 \cdot 2 x^2(1-x^2)^{-5/2} \\
f^{(3)}(x)=18x(1-x^2)^{-5/2}+2\cdot 3\cdot 5x^3(1-x^2)^{-7/2} \\
f^{(4)}(x)=18(1-x^2)^{-5/2}+180x^2(1-x^2)^{-7/2}+2\cdot 3\cdot 5\cdot 7x^4(1-x^2)^{-9/2}$
From this I get $f'(0)=2 $, $f''(0)=0 $, $f^{(3)}(0)=2 $, $f^{(4)}(0)=0 $
But I don't know how to find a general term. Can this be solved in easier steps? 
 A: If you check your differentiation very carefully, you will find that $f(x)=\arccos(1-2x^2)$ is not continuously differentiable at $x=0$, so the MacLaurin expansion does not exist.
Details: if $x\ne0$ we have
$$\frac{d}{dx}\arccos(1-2x^2)=\frac{4x}{\sqrt{1-(1-2x^2)^2}}
  =\frac{2x}{|x|\sqrt{1-x^2}}\ .$$
So
$$\lim_{x\to0^+}f'(x)=2\quad\hbox{but}\quad \lim_{x\to0^-}f'(x)=-2\ .$$
A: As David has pointed out, $\cos^{-1}(1-2x^2)$ is not differentiable at $x=0$, so it does not have a Maclaurin series in a strict sense.
But now let $x\ge0$ and $\cos^{-1}(1-2x^2)=u$. We have
$$1-2x^2=\cos u$$
$$\sqrt{\frac{1-\cos u}2}=x=\sin\frac u2$$
$$\sin^{-1}x=\frac u2$$
$$u=2\sin^{-1}x=\cos^{-1}(1-2x^2)$$
Now the Maclaurin series of $2\sin^{-1}x$ can be derived easily:
$$2\sin^{-1}x=\sum_{n=0}^\infty\frac1{2^{2n-1}(2n+1)}\binom{2n}nx^{2n+1}$$
Therefore $\cos^{-1}(1-2x^2)$ may be written as a "quasi-Maclaurin series", using $|x|$ and not just $x$:
$$\cos^{-1}(1-2x^2)=\sum_{n=0}^\infty\frac1{2^{2n-1}(2n+1)}\binom{2n}n{|x|^{2n+1}}\qquad{|x|\le1}$$
A: Hint:
Let $t=\arcsin x$
$\implies x=\sin t,-\dfrac\pi2\le t\le\dfrac\pi2$
$\arccos(1-2x^2)=\arccos(\sin2t)=\dfrac\pi2-\arcsin(\sin2t)$
Put $y=x$ in Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
A: $arcos(1-2x^2) = \frac{\pi}{2} \sum\limits_{k=0}^{\infty} \frac{(1-2x^2)^{1+2k}(\frac{1}{2})k}{k!+2k\times k!}$
$\forall |1-2x^2|<1$
