Maclaurin expansion of $\arccos(1-2x^2)$

Question

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?

Answer 1

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 $x$: $$\cos^{-1}(1-2x^2)=\sum_{n=0}^\infty\frac1{2^{2n-1}(2n+1)}\binom{2n}n{|x|^{2n+1}}\qquad{|x|\le1}$$

Answer 2

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 $

Answer 3

$arcos(1-2x^2) = \frac{\pi}{2} \sum\limits_{k=0}^{\infty} \frac{(1-2x^2)^{1+2k}(\frac{1}{2})k}{k!+2k\times k!}$

$\forall |1-2x^2|<1$

Answer 4

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 $$\lim_{x\to0^+}f'(x)=2\quad\hbox{but}\quad \lim_{x\to0^-}f'(x)=-2\ .$$