Is variant of $\arccos$ analytic?

Question

If you look at the plot of the function $f(x):=\arccos(1-x^2)$ on $[0,1]$ then it almost looks as if this function would be perfectly smooth and could be extended smoothly through $0.$

But if you plot the function on $[-1,1]$ you immediately see that this function cannot be differentiable at $0.$

On the other hand, if you then define $g(x):=\arccos(1-x^2)$ for $x \ge 0$ and $g(x)=-\arccos(1-x^2)$ then it looks as if the function has become smooth.

Now obviously $\arccos(x)$ does not have a Taylor expansion at $x=-1,$ but $\arccos(1-x^2)$ is a perfect function around $x=0.$

My question is now: Is this function $g$ even analytic, i.e. can it be expanded in a power series around $x=0$?

Answer 1

$$ y=\arccos(1-x^2)\\ \cos(y)=1-x^2\\ x^2=1-\cos(y)=2\sin^2(y/2)\\ y=2\arcsin(|x|/\sqrt2) $$ In this more simplified form you can see where the kink comes from and how to continue the part over the positive half axis into a smooth function around $x=0$.

Answer 2

For $x\ne0$, the derivative of the initial function is

$$(\arccos(1-x^2))'=\frac{2x}{\sqrt{1-(1-x^2)^2}}=\frac{2x}{\sqrt{2x^2-x^4}}=\frac{2x}{|x|\sqrt{2-x^2}}=\frac{2\text{ sgn}(x)}{\sqrt{2-x^2}},$$ which is discontinuous at $x=0$.

You indeed restore continuity at $0$ by multiplying by $\text{sgn}(x)$.

The Taylor development can be established using the generalized binomial formula,

$$\frac1{\sqrt2}f'(x)=\left(1-\frac{x^2}2\right)^{-1/2}=1+\frac{x^2}4-\frac{3x^4}8+\frac{5x^6}{16}-\cdots$$ followed by term-wise integration.