Uniform Asymptotic Expansion of a ratio modified Bessel functions of the first kind

Question

Let $I_{\nu}(\cdot)$ be the modified Bessel function of the first kind of order $\nu$. Can someone please confirm if, when $\nu\rightarrow\infty$, the following approximation

$$\frac{I_{\frac{\nu}{2}}(\frac{\nu}{2}x)}{I_{\frac{\nu}{2}-1}(\frac{\nu}{2}x)}=\frac{x}{1+(1+x^{2})^{1/2}}+O\!\left(\frac{1}{\nu}\right)$$

holds uniformly with respect to $x$? How can I show this?

Thanks for any help.

Answer 1

If you look at this paper and, in particular at equation $(3.3a)$ $$\frac{I_{n}(x)}{I_{n-1}(x)}=\frac {x}{2n+x\frac{I_{n+1}(x)}{I_{n}(x)}} $$ you then end with $$\frac{I_{n}(nx)}{I_{n-1}(nx)}=\frac {x}{2+x\frac{I_{n+1}(nx)}{I_{n}(nx)}} $$ So, solving for $L$ equation $$L=\frac x{2+x L}\implies L=\frac{x}{1+\sqrt{1+x^{2}}}$$

Answer 2

Suppose that $x>0$ and $\nu$ is large and positive. First note that, by $(10.29.2)$, $$ \frac{{I_\nu (\nu x)}}{{I_{\nu - 1} (\nu x)}} = \frac{{I_\nu (\nu x)}}{{I'_\nu (\nu x) + (1/x)I_\nu (\nu x)}} = \cfrac{1}{{\displaystyle\frac{{I'_\nu (\nu x)}}{{ I_\nu (\nu x)}} + \frac{1}{x}}}. $$ If we denote $$ \eta = \sqrt {1 + x^2 } + \log \left( {\frac{x}{{1 + \sqrt {1 + x^2 } }}} \right), $$ then $$ I_\nu (\nu x) = \frac{{\mathrm{e}^{\nu \eta } }}{{\sqrt {2\pi \nu } (1 + x^2 )^{1/4} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\nu \sqrt {1 + x^2 } }}} \right)} \right) $$ and $$ I'_\nu (\nu x) = \frac{{(1 + x^2 )^{1/4} \mathrm{e}^{\nu \eta } }}{{\sqrt {2\pi \nu } x}}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\nu \sqrt {1 + x^2 } }}} \right)} \right), $$ as $\nu\to +\infty$, uniformly with respect to $x>0$. These follow from $\S10.41.\mathrm{(ii)}$ and the double asymptotic properties of these expansions (cf. $\S10.41.\mathrm{(iv)}$). Accordingly, $$ \frac{{I'_\nu (\nu x)}}{{I_\nu (\nu x)}} = \cfrac{{\displaystyle\frac{{(1 + x^2 )^{1/4} \mathrm{e}^{\nu \eta } }}{{\sqrt {2\pi \nu } x}}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\nu \sqrt {1 + x^2 } }}} \right)} \right)}}{{\displaystyle\frac{{\mathrm{e}^{\nu \eta } }}{{\sqrt {2\pi \nu } (1 + x^2 )^{1/4} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\nu \sqrt {1 + x^2 } }}} \right)} \right)}} = \frac{{\sqrt {1 + x^2 } }}{x} + \mathcal{O}\!\left( {\frac{1}{{\nu x}}} \right), $$ and therefore, \begin{align*} \frac{{I_\nu (\nu x)}}{{I_{\nu - 1} (\nu x)}} & = \cfrac{1}{{\displaystyle\frac{{\sqrt {1 + x^2 } }}{x} + \mathcal{O}\!\left( {\frac{1}{{\nu x}}} \right) + \frac{1}{x}}} = \cfrac{x}{{\displaystyle\sqrt {1 + x^2 } + 1 + \mathcal{O}\!\left( {\frac{1}{\nu }} \right)}} \\ & = \frac{x}{{\sqrt {1 + x^2 } + 1}} + \mathcal{O}\!\left( {\frac{x}{{\nu (\sqrt {1 + x^2 } + 1)^2 }}} \right) = \frac{x}{{\sqrt {1 + x^2 } + 1}} + \mathcal{O}\!\left( {\frac{1}{\nu }} \right). \end{align*} Throughout the whole derivation, the implied constants in the big-$\mathcal{O}$'s are independent of $x$ and $\nu$.

Answer 3

Kiefer and Weiss showed this more generally. Adapting their result in equation $7a$ to a similar parlance:

$$ \frac{I_{\nu+n}(\nu x)}{I_{\nu}(\nu x)} = \left(\frac{x}{1+\sqrt{x^2+1}}\right)^n\left(1 - \frac{1+n\sqrt{x^2+1}}{2(x^2+1)}\frac{n}{\nu}+ O\!\left(\frac{1}{\nu^2}\right)\right).$$

Similar relations are given for Hankel functions and additional Bessel functions.