Please see the bottom of the post for an updated version of the question.
Original Question. I’m looking for an analytical proof of an upper bound for $-r’’_\nu(x)/r’_\nu(x)$ for $\nu \ge 0$ and $x \ge 0$, where $r_\nu(x)=I_{\nu+1}(x)/I_\nu(x)$ and $I_\nu(x)$ is the modified Bessel function of the first kind with order $\nu$ and argument $x$. The derivatives are taken with respect to $x$.
The background is I’m looking at Newton’s method to find the root of $r_\nu(x)=\rho$ given $\nu \ge 0$ and some $0 < \rho < 1$. It is known that $0 < r_\nu(x) < 1$, $r_\nu’(x) > 0$, $r’’_\nu(x) < 0$, so Newton’s method works well. The upper bound in the question is used to bound the (initial) convergence rate.
Numerical experiment shows $-r’’/r’$ is bounded by 1 (not tight) when $\nu = 0$, and the bound is smaller for larger $\nu$. However, I’m unable to find any analytical result on this.
Any help would be very much appreciated!
Update 1. What I have got so far is the following. It is known that (Amos, 1974, Eq. 12)
$$ r’ = 1-\frac{2\nu+1}{x}r-r^2 $$
and that (Amos, 1974, Eq. 15)
$$ r' < \frac{r}{x} $$
It follows that
$$ r’’=\frac{2\nu+1}{x}\left(\frac{r}{x}-r’\right)-2rr’ $$
and
$$ \frac{r’’}{r’}=\frac{2\nu+1}{x}\left(\frac{r-xr'}{xr’}\right)-2r > -2r > -2 $$
But the bound $2$ looks a bit coarse for the problem at hand (ideally I’d like a bound less than $4/3$), and it’s pretty far from the numerically found bound (about 0.833) so I feel like there should be some room for improvement.
Reference
Amos, D. E. (1974). "Computation of Modified Bessel Functions and Their Ratios." Mathematics of Computation, 28(125):239-251.
Update 2. It turns out the condition $\nu \ge 1$ may not be more interesting than say $\nu \ge 1.2345$. In fact, for any given $\nu \ge -1/2$, $r_\nu(x)$ (as a function of $x$ for $x > 0$) is (1) monotonic increasing, (2) concave, and (3) taking value in $(0,1)$. So the range of $\nu$ may be relaxed to $\nu \ge -1/2$.
Once we relax the range of $\nu$, the best upper bound that works for all $\nu$ is indeed $2$. That $2$ is an upper bound is shown in "Update 1" above. To see the bound cannot be better, I plot $r''_\nu/r'_\nu$ versus $r_\nu$ for different choices of $\nu$:
$r''_\nu/r'_\nu$ versus $r_\nu$" />
It is apparent from the plot that as $\nu \downarrow -1/2$, the minimum value of $r''_\nu/r'_\nu$ tends to $-2$. When $\nu=-1/2$, it is known that $r_\nu(x) \equiv \tanh (x)$, and it can be shown that $\lim_{x\rightarrow +\infty} \tanh''(x)/\tanh'(x) = -2$$\tanh''(x)/\tanh'(x) =-2\tanh(x) \rightarrow -2$ as $x\rightarrow +\infty$.
Therefore, a better posed form of the question is:
Find an upper bound $U(\nu)$ such that $-r''_\nu(x)/r'_\nu(x) \le U(\nu)$ for all $\nu \ge -1/2$ and all $x > 0$.
$U(\nu)\equiv 2$ is an answer, but I’m looking for a more interesting bound which, for example, satisfies $\lim_{\nu\rightarrow +\infty}U(\nu)=0$.
