# Is $\frac{\arccos\left((\sqrt{r}+1)/(r+1/\sqrt{r})\right)}{\pi \left|1-r^{-3/2}\right|}$ analytic at $r=1$?

Is the function $$f:(0,\infty)\rightarrow(0, 1)$$, defined below, analytic at $$r=1$$?

$$f(r) := \frac{\arccos\left(\frac{\sqrt{r}+1}{r+\frac{1}{\sqrt{r}}}\right)}{\pi \left|1-\frac{1}{r^{3/2}}\right|}\quad \mathrm{if\ } r>0\mathrm{\ and\ } r\neq1,$$ and $$f(1) :=\frac{\sqrt{2}}{3 \pi }$$ where $$\arccos(x)\in [0,\pi].$$

If you are interested, this function arises from retrograde motion.

The following statements seem to be true:

1. $$\lim_{r\rightarrow\infty} f(r) = 1/2$$,
2. $$\lim_{r\rightarrow 0} \frac{f(r)}{r^{3/2}}=1/2$$,
3. $$f(r) = f(1/r)r^{(3/2)}$$,
4. $$g(r) = \frac{\sqrt{r}+1}{r+\frac{1}{\sqrt{r}}}$$ is analytic at $$r=1$$,
5. $$g(r) = 1-\frac{1}{4} (r-1)^2+\frac{1}{4} (r-1)^3-\frac{11}{64} (r-1)^4+ \frac{3}{32} (r-1)^5 -\frac{21}{512} (r-1)^6+\frac{7}{512} (r-1)^7+ O((r-1)^8),$$
6. $$1-g(r) = \frac{(r-1)(1-1/\sqrt{r})}{r+1/\sqrt{r}}\geq 0$$,
7. $$\mathrm{sgn}(x)\arccos(1-x^2) = \sqrt{2} x + \frac{x^3}{6\sqrt{2}} +\frac{3 x^5}{80 \sqrt{2}} + \frac{5 x^7}{448 \sqrt{2}}+O(x^9)$$, and
8. $$1-1/r^{3/2} = \frac{3 (r-1)}{2}-\frac{15}{8} (r-1)^2+\frac{35}{16} (r-1)^3-\frac{315}{128} (r-1)^4+\frac{693}{256} (r-1)^5-\frac{3003 (r-1)^6}{1024}+\frac{6435 (r-1)^7}{2048}+O\left((r-1)^8\right).$$
• @BGreen Numerically, I get that $\lim_{h-> 0^+} (\frac{f(1+h)- f(1)}h) = \frac1{\pi\; 2 \sqrt{2}}$ where $f(1)= \sqrt{2}/(3 \pi)$. Proving that seems a bit tricky. Commented Jul 8, 2019 at 14:24
• @BGreen I spent a little time trying to prove the right side derivative using the approximations $$\frac{\sqrt{r}+1}{r+\frac{1}{\sqrt{r}}} \approx 1-\frac{1}{4} (r-1)^2$$ and $\arccos(1-x^2) \approx \sqrt{2} x$, but I have not succeeded yet. Commented Jul 8, 2019 at 14:34
• One possible approach is to find converging power series expression for $\frac{\sqrt{r}+1}{r+\frac{1}{\sqrt{r}}}$, $\arccos(x)$, and $1-r^{(-3/2)}$. I think this may work. Commented Jul 8, 2019 at 14:48

1) Consider $$f$$ as a function of a complex variable $$z$$. If $$g(z)$$ is analytic at $$z=1$$ then $$g(\sqrt{z})$$ is also analytic as a composition of holomorphic functions. Thus it's enough to establish analyticity of the function $$g(z)=f(z^2)=\frac{\cos ^{-1}\left(\frac{z^2+z}{z^3+1}\right)}{\pi\sqrt{\left(\frac{1}{z^3}-1\right)^2}}.$$
2) Denote for convenience $$h(z)=g(1+z)$$ so the point of interest is $$z=0$$ now. Using the equality $$\cos ^{-1}(z)=2 \tan ^{-1}\left(\frac{\sqrt{1-z^2}}{z+1}\right)$$ gives $$h(z)= \frac{2 \tan ^{-1}\left(\frac{\left(z^2+z+1\right) \sqrt{1-\frac{(z+1)^2}{\left(z^2+z+1\right)^2}}}{z^2+2z+2}\right)}{\pi \sqrt{\left(\frac{1}{(z+1)^3}-1\right)^2}}= \frac{2 \tan ^{-1}\left(\sqrt{z^2}\frac{\left(z^2+z+1\right) \sqrt{\frac{\left(z^2+2 z+2\right)}{\left(z^2+z+1\right)^2}}}{z^2+2z+2}\right)}{\pi \sqrt{z^2} \sqrt{\frac{ \left(z^2+3 z+3\right)^2}{(z+1)^6}}}.$$ All functions in the rhs are analytic at $$z=0$$ except $$\sqrt{z^2}$$. Since $$\tan^{-1}(z)$$ is odd, function $$\tan^{-1}(\sqrt{z^2}u(z))/\sqrt{z^2}$$ is analytic for $$u$$ analytic at $$z=0$$.