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I'm interested in the roots of the equation:

$I_1(bx) - x I_0(bx) = 0$

Where $I_n(x)$ is the modified Bessel function of the first kind and $b$ is real positive constant.

More specifically, I'm interested in the behaviour of the largest non-negative root for $x \geq 0$. I have solved the problem numerically in a crude fashion to see what the function looks like, but as far as I can tell there isn't a closed form.

If the value of the largest root is given by $x(b)$, then based on the numerical solution I think the following is true:

$x(b) < 0 \; $ for $\; b < 2$

$\lim_{b\to\infty} x(b) = 1$

Can anyone give me some pointers to the possible approaches I might take to approximating $x(b)$ for $b>2$ ?


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I assume you meant that $I_n(x)$ is the modified Bessel function of the $n$th kind, yes? – Cameron Buie Feb 19 '13 at 15:00
up vote 1 down vote accepted

Notice that the equation also can be rewritten as $$ \frac{I_1(b x)}{I_0(b x)} = x $$ Using the asymptotic series expansion for large $b$ and some fixed $x>0$ we get: $$ 1 - \frac{1}{2 x b} - \frac{3}{8 b^2 x^2} - \frac{1}{8 b^3 x^3} + \mathcal{o}\left(b^{-1}\right) = x $$ resulting in $$ x(b) = 1 - \frac{1}{2 b} - \frac{3}{8 b^2} - \frac{9}{16 b^3} + \mathcal{o}\left(b^{-3}\right) $$ Confirming $\lim_{b \to + \infty} x(b) = 1$.

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Thanks for your answer! I tried the expansion for $x(b)$ you suggested but it didn't look quite right, but I played with the coefficients and the following series did look correct: $x(b) = 1 - \frac{1}{2 b} - \frac{1}{4 b^2} - \frac{1}{8 b^3} + \mathcal{o}\left(b^{-3}\right)$ Could this be the correct one? Thanks – CBowman Feb 20 '13 at 14:49
@CBowman Yes, this was a typo. The series I gave was the series for $I_0(bx)/I_1(b x)$. Will correct in the moment. – Sasha Feb 20 '13 at 15:15
Thanks again, the updated series works perfectly. Do you mind If I ask how you got from the asymptotic expansion to the series for $x(b)$ ? It's not obvious to me. – CBowman Feb 21 '13 at 16:21
I used method of indeterminate coefficients, i.e. let $x(b) = 1 + \frac{c_1}{b} + \frac{c_2}{b^2} + \frac{c_3}{b^3} + \mathcal{o}\left(\frac{1}{b^3}\right)$. Then substitute to get a system of linear equations for $c_1$, $c_2$ and $c_3$ and solve. – Sasha Feb 21 '13 at 16:58

I'm a little confused by your notation. Here's what I see:

Assume $b>0$. I can tell you why there is a discontinuity at $b=2$, because

$$I_1(b x) - x I_0(b x) = \left (\frac{b}{2}-1 \right ) x + O(x^3)$$

so when $b=2$, this expression is very close to zero near $x=0$ anyway. Away from $b=2$, I get a plot for the zero like this:

enter image description here

The abcissa is $b$, while the ordinate is the zero for that value of $b$. I do see the limiting behavior you see, i.e., the zero tends to $1$ as $b \rightarrow \infty$.

For $b<2$, $I_1(b x) - x I_0(b x) < 0$ and there are no zeroes.

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Yes I had myself very confused over notation! I should have checked more carefully, I've now edited the question, hopefully it now makes more sense. – CBowman Feb 19 '13 at 16:09

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