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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. $1-1/r^{3/2} = \frac{3 (r-1)}{2}-\frac{15}{8} (r-1)^2+\frac{35}{16} (r-1)^3-\frac{315}{128} (r-1)^4+\frac{693}{256} (r-1)^5-\frac{3003 (r-1)^6}{1024}+\frac{6435 (r-1)^7}{2048}+O\left((r-1)^8\right).$
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  • $\begingroup$ @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. $\endgroup$
    – irchans
    Jul 8 '19 at 14:24
  • $\begingroup$ @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. $\endgroup$
    – irchans
    Jul 8 '19 at 14:34
  • $\begingroup$ 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. $\endgroup$
    – irchans
    Jul 8 '19 at 14:48
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The answer is yes.

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 $z=0$.

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