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. $1-1/r^{3/2} = \frac{3 (r-1)}{2}-\frac{15}{8} (r-1)^2+\frac{35}{16} (r-1)^3-\frac{315}{128} (r-1)^4+\frac{693}{256} (r-1)^5-\frac{3003 (r-1)^6}{1024}+\frac{6435 (r-1)^7}{2048}+O\left((r-1)^8\right).$
  • $\begingroup$ @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

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 $z=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.