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Let $f:\Bbb R^{\geqslant-1}\to\Bbb R$ be $$f(x)=\begin{cases} \dfrac{x\sqrt{1-x^2}}{\arccos(x)}, &-1\leqslant x<1\\ 1, &x=1\\ \dfrac{x\sqrt{x^2-1}}{\ln(x+\sqrt{x^2-1})}, &x>1 \end{cases}\tag 1$$ The question is whether this function is analytic at $x=1$.

For $x<1$ I used that $\arccos$ around 1 is basically a square root$^1$, more specifically that

$$\arccos (1-x) = \sqrt{2x}\,h(x)\tag 2$$ where $h$ is analytic at $x=0$ with $h(x)\neq0$ and radius of convergence of 2. $h$ can be represented as $$h(x) = {\arcsin(\sqrt{x/2}) \over \sqrt{x/2}} = \sum_{j=0}^\infty \binom{2j}{j} \frac{x^j}{(2j+1)8^j} \tag 3$$

Using this yields the following representation of $f$:

$$\begin{align} f(1-x) &= \frac{(1-x)\sqrt{1-(1-x)^2}}{\arccos(1-x)}\\ &= \frac{(1-x)\sqrt{2x-x^2}}{\sqrt{2x}\,h(x)}\\ &= \frac{(1-x)\sqrt{1-x/2}}{h(x)} \tag 4\\ \end{align}$$

As $h\neq 0$, the function on the right side is analytic for $|x| < 2$ with $f(1) = 1$ because $h(0)=1$.

If it can be shown that this representation of $f$ is also valid for $x\in(-2,0)$ then $f$ is analytic at $1$.

Now I am stuck here because a similar representation should exist for $\ln(x+\sqrt{x^2-1})$, i.e. that expression is basically a square root around $x=1$. This is clear from the power series expansion of $\ln$, but it does not yield an explicit representation. Obviouly, $h$ does not only work over $\Bbb R$, but due to $(3)$ it works also over $\Bbb C$, but noidea how to apply that. The final step would be to express

$$\operatorname{arcosh} x = \ln(x+\sqrt{x^2-1})$$

in terms of $\arcsin$ and square-root like in $(2)$ and $(3)$, maybe using $\cosh x=\cos (ix)$?

Note: The function is from this answer.

$^1$What I didn't even try is to find $n$-th derivative of either functions and equating them. This is likely to run into too much complications.

emacs drives me nuts
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1 Answers1

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The double-angle formula $\cosh2\theta=1+2\sinh^2\theta$ gives

$$2\,\text{arsinh}(u)=\text{arcosh}(1+2u^2)$$

for $u\geq0$. Letting $1+2u^2=x$, we get

$$2\,\text{arsinh}\sqrt{\frac{x-1}{2}}=\text{arcosh}(x),$$

so your function for $x>1$ is

$$\frac{x\sqrt{x^2-1}}{\text{arcosh}(x)}=x\sqrt{\frac{x+1}{2}}\cdot\frac{\sqrt\frac{x-1}{2}}{\text{arsinh}\sqrt\frac{x-1}{2}}.$$

Similarly, you've found that the function for $x<1$ is

$$\frac{x\sqrt{1-x^2}}{\arccos(x)}=x\sqrt\frac{x+1}{2}\cdot\frac{\sqrt\frac{1-x}{2}}{\arcsin\sqrt\frac{1-x}{2}}.$$

Now just compare the power series around $u=0$ of $\arcsin(u)/u$ and $\text{arsinh}(u)/u$, and you're done.

See Power series expansion of $\text{arsinh}(x)$ at $0$ or Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor

mr_e_man
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  • What confuses me here is which branch of either square-root is the right one. Expressions like $\arcsin(\sqrt x) /\sqrt{x}$ can be made sense of by the power-series expansion of $\arcsin$: As $\arcsin$ is an odd function, the $\sqrt{\ }$ can be shortened out. Is this enough to settle the question about which branches to pick? – emacs drives me nuts Jun 14 '22 at 17:11
  • Use the positive square root. – mr_e_man Jun 14 '22 at 17:23
  • The function was defined for real numbers $x$. To define it for complex $x$ near $1$, take $f(x)=x\sqrt{(x+1)/2}/g(x)$, where the square root is analytic with $\sqrt{(1+1)/2}=+1$, and $g(x)$ is analytic with power series $$g(x)=\sum_k\frac{(-1)^k(2k-1)!!}{(2k)!!}\frac{((x-1)/2)^k}{2k+1}=\sum_k\frac{(2k-1)!!}{(2k)!!}\frac{((1-x)/2)^k}{2k+1}.$$ – mr_e_man Jun 14 '22 at 17:36