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I'm trying to find a approximation solution for the following equation: ${e^{ - x}}\left[ {{I_o}\left( x \right) + {I_1}\left( x \right)} \right] = C$ where $I_0$ and $I_1$ is the modified Bessel functions of the first kind of order 0 and 1. C is a constant. Do you have any suggestion? Thank you.

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1 Answer 1

In the limit of large $x$, the left hand side looks like $\sqrt{2/(\pi x)}$, which results in equation that is easily soluble for $x$, so long as $C$ is sufficiently small.

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Thank for your respond. –  Bình Văn Nguyễn Jan 25 '13 at 2:33
For small $x$, I used the relation ${I_1}\left( x \right) = {I_0}\left( x \right)\frac{x}{2}{e^{\frac{{ - {x^2}}}{8}}}$. And when x is small, $I_0(x)=1$. We obtain: ${e^{ - x}}\left[ {1 + \frac{x}{2}{e^{\frac{{ - {x^2}}}{8}}}} \right] \simeq {e^{ - x}}\left[ {1 + \frac{x}{2}} \right] \simeq \left( {1 - x + \frac{{{x^2}}}{2}} \right)\left( {1 + \frac{x}{2}} \right) = C$. The solution for the above cubic equation can be found in wikipedia. –  Bình Văn Nguyễn Jan 25 '13 at 8:05
The relation between $I_0(x)$ and $I_1(x)$ can be found in: "Exponential Approximation of the Modified Bessel Function". –  Bình Văn Nguyễn Jan 25 '13 at 8:08

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