Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $z\in\mathbb{R}$.

share|improve this question
    
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: $$\frac{K_{1}\left(z\right)}{K_{0}\left(z\right)}\sim\left\{ \begin{array}{cc} -\frac{1}{z\ln\left(z\right)}, & \mathrm{as\;}z\to0\;\\ 1, & \mathrm{as\;}z\to\infty \end{array}\right.$$ –  Dave S Oct 11 '13 at 12:51

2 Answers 2

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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