Two Bessel Functions

Question

Let $I_k$ denote the modified Bessel function of the first kind of index $k$. Show that, if $y > x > 0$, then $$-I_0(x)I_1(y) + I_0(y)I_1(x) <0$$

I am looking for a source that provides this information.

You are surely aware that the derivative of $I_0$ is $I_1$. This property has hopefuly a role to play... — Jean Marie, Aug 17 '16 at 20:21

Martin Nicholson · Accepted Answer · 2016-08-18T14:36:29.960

2

From the integral representation $$ I_0(x)I_0(y)=\frac{1}{\pi}\int_0^\pi I_0(\sqrt{x^2+y^2-2xy\cos\theta})d\theta, $$ using the formula $I_1(x)=I_0'(x)$ we immediately see that $$-I_0(x)I_1(y)+I_0(y)I_1(x)=\frac{1}{\pi}\int_0^\pi \frac{I_1(\sqrt{x^2+y^2-2xy\cos\theta})}{\sqrt{x^2+y^2-2xy\cos\theta}}(x-y)(1+\cos\theta)d\theta. $$ For $0<x<y$ this is clearly less than $0$, thus proving $$-I_0(x)I_1(y)+I_0(y)I_1(x)<0,~0<x<y.$$

Since $I_1(x)/x$ is monotonically increasing for $x>0$, it is also possible to prove a stronger inequality

$$ -I_0(x)I_1(y)+I_0(y)I_1(x)<-I_1(y-x),~0<x<y. $$

edited Aug 18 '16 at 14:36

answered Aug 18 '16 at 09:41

Martin Nicholson

3,883

This is great, thanks so much. Would you happen to have a source where the integral representation comes from? – JBSHaldane Aug 18 '16 at 19:31
This integral representation is obtained by integrating the addition theorem for the function $I_0$, $I_0(\sqrt{x^2+y^2-2xy\cos\theta})=\sum_{n=-\infty}^\infty I_n(x)I_n(y)\cos n\theta$, (see equation 11.41.(6) on page 365, Treatise on the theory of Bessel funcions, by G.N. Watson) with respect to $\theta$ – Martin Nicholson Aug 18 '16 at 21:44

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

There is a useful lemma that settles the question very nicely here.

If $f(z)$ and $g(z)$ are two real analytic functions in a neighbourhood $U$ of the origin, $$ f(z)=\sum_{n\geq 0}a_n z^n,\qquad g(z)=\sum_{n\geq > 0}b_n z^n $$ with $a_n,b_n>0$ and the sequence $\left\{\frac{a_n}{b_n}\right\}_{n\geq 0}$ being an increasing/decreasing sequence,
then the ratio $r(x)=\frac{f(x)}{g(x)}$ is an increasing/decreasing function over $U$.

The proof is not difficult: it is enough to consider $f'g-g\, f'$ in terms of power series to deduce that $\frac{dr}{dx}$ has constant sign. In our case, the coefficients of $I_0(x)$ and $I_1(x)$ have a simple structure, and the above lemma leads to:

$\displaystyle\frac{I_0(x)}{I_1(x)}$ is a decreasing function over $\mathbb{R}^+$;
$\displaystyle\frac{x\cdot I_0(x)}{I_1(x)}$ is an increasing function over $\mathbb{R}^+$.

Additionally, $\frac{I_0}{I_1}$ has a very nice representation in terms of continued fractions, that makes the above statements almost trivial.

Two Bessel Functions

2 Answers2