# Bessel function ratio approximation

Can we say anything about the ratio:

$$\frac{K_1(z)}{K_0(z)}?$$ In particular, can we describe its behaviour for small or large $z\in\mathbb{R}$.

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You should tag this question as e.g. limit , special-functions rather than differential-equations . –  doraemonpaul Oct 5 '13 at 17:23
We can analyse the asymptotic behaviour of the ratio:$$K_{1}\left(z\right)/K_{0}\left(z\right)$$ as $z$ becomes small or large by observing the leading order term of its series expansion. We then have: $$\frac{K_{1}\left(z\right)}{K_{0}\left(z\right)}\sim\left\{ \begin{array}{cc} -\frac{1}{z\ln\left(z\right)}, & \mathrm{as\;}z\to0\;\\ 1, & \mathrm{as\;}z\to\infty \end{array}\right.$$ –  David Simmons Oct 11 '13 at 12:51

Using the "smoothed" integral representation (i.e., just integrate in [0,L] and then we will take the limit L tending to infinity to conclude), we get that the leading order (in epsilon) is $$\frac{K_0}{K_1}(\epsilon)\approx\frac{\int_0^L dt}{\int_0^L \cosh(t)dt}=\frac{L}{\sinh(L)}\rightarrow 0\;\text{ if }L\rightarrow\infty.$$ In other words, your quotient diverges for small numbers, I think. For big numbers seems harder for me. Maybe using $K_0'=-K_1$ you can figure out something. I don't know.
We can analyse the asymptotic behaviour of the ratio: $K_1(z)/K_0(z)$ as z becomes small or large by observing the leading order term of its series expansion. We then have: $$\frac{K_{1}\left(z\right)}{K_{0}\left(z\right)}\sim\left\{ \begin{array}{cc} -\frac{1}{z\log z}, & z>0\;\mathrm{small}\\ 1, & z\;\mathrm{large} \end{array}\right..$$