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I am looking for the approximation of the following function:

$$\rho(a,b)=1-e^{-(a+b)}\sum_{m=1}^{\infty}\left(\sqrt{\frac{b}{a}}\right)^m I_m(2\sqrt{ab})$$ where $I_m(x)$ is the modified Bessel function. Since $I_m(x)$ is again an infinite series, up to how many terms I need to do summation in both the cases above in order to get some better approximation? Is there any rule of thumb?

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Actually, you can use the trapezoidal rule to compute the modified Bessel function of the first kind. That seems to me simpler than using infinite series. But don't do forward recursion over the Bessel function, as that is unstable. – J. M. Sep 9 '11 at 2:20
Bessel function $I_m(2 z) \sim \frac{z^m}{\Gamma(m+1)}$ for large $m \gg \vert z \vert$. This should give a pessimistic bound on how many terms you need to sum. – Sasha Sep 9 '11 at 2:24
Here's an idea: for large $m$, $I_m(z)$ is well approximated by $\frac1{\sqrt{2\pi m}}\left(\frac{e z}{2m}\right)^m$. Find the value of $m$ that makes this approximation tiny (perhaps via Newton-Raphson), compute the modified Bessel function there, and recurse back. On the other hand, you might also want to take into account the other stuff multiplying the modified Bessel function when you're estimating $m$. – J. M. Sep 9 '11 at 2:29
Thanks for your comments. But up to what value of $m$ shall I perform summation? – shaikh Sep 9 '11 at 2:43
@shaikh, what typical $(a,b)$-regime are you interested in? – Did Sep 10 '11 at 12:27
up vote 5 down vote accepted

If one is interested in expressing $\rho(a,b)$ with a single summation/integration operation, one can use, for every positive $a$ and $b$ such that $b<a$, $$ \rho(a,b)=1+e^{-(a+b)}\int\limits_0^\pi e^{2\sqrt{ab}\cos\theta}\frac{b-\sqrt{ab}\cos\theta}{a+b-2\sqrt{ab}\cos\theta}\frac{\mathrm{d}\theta}\pi. $$ See below for some explicit formulas when $a=b$ or $a<b$.

Case $b< a:$ This uses only the integral formula for modified Bessel functions with integer parameter, which says that for every integer $m$ and every $x$ with nonnegative real part, $$ I_m(x)=\int\limits_0^\pi e^{x\cos\theta}\cos(m\theta)\frac{\mathrm{d}\theta}\pi. $$ Writing the cosine as the real part of a complex exponential and summing this over $m\ge1$ yields, for every real number $s$ such that $|s|<1$, $$ \sum\limits_{m\ge1}s^mI_m(x)=\text{Re}\int\limits_0^\pi e^{x\cos\theta}\frac{se^{i\theta}}{1-se^{i\theta}}\frac{\mathrm{d}\theta}\pi. $$ Multiplying the numerator and the denumerator by the conjugate quantity $1-se^{-i\theta}$ and using the relation $|1-se^{i\theta}|^2=1+s^2-2s\cos\theta$, one sees that $$ \sum\limits_{m\ge1}s^mI_m(x)=\int\limits_0^\pi e^{x\cos\theta}\frac{s\cos\theta-s^2}{1+s^2-2s\cos\theta}\frac{\mathrm{d}\theta}\pi. $$ Plugging in $x=2\sqrt{ab}$ and $s=\sqrt{b/a}$ (hence the restriction $b< a$) yields $\rho(a,b)$ as above.

Case $a=b$: One sees that $$ \rho(a,a)=1-e^{-2a}\sum\limits_{m\ge1}I_m(2a). $$ Using the fact that $I_m(x)=I_{-m}(x)$ for every integer $m$ and the formula (written somewhere in the WP page already mentioned) for the generating function $$ \sum\limits_{m=-\infty}^{+\infty}I_m(x)\cos(m\theta)=e^{x\cos\theta}, $$ at $\theta=0$, one sees that $$ \rho(a,a)=\frac12+\frac12e^{-2a}I_0(2a)=\frac12+\frac12e^{-2a}\int\limits_0^\pi e^{2a\cos\theta}\frac{\mathrm{d}\theta}\pi. $$ Case $a<b$: Comparing $\rho(a,b)+\rho(b,a)$ with the generating function of the family $(I_m(x))$ that we recalled above, one gets $$ \rho(a,b)+\rho(b,a)=1+e^{-(a+b)}I_0(2\sqrt{ab}), $$ for every $a$ and $b$. Hence, when $a< b$, $$ \rho(a,b)=e^{-(a+b)}\int\limits_0^\pi e^{2\sqrt{ab}\cos\theta}\frac{b-\sqrt{ab}\cos\theta}{a+b-2\sqrt{ab}\cos\theta}\frac{\mathrm{d}\theta}\pi. $$ Remark: The result is discontinuous around $a=b$ since, for every fixed $a$, $\rho(a,b)\to\rho(a,a)+\frac12$ when $b\to a^-$ and $\rho(a,b)\to\rho(a,a)-\frac12$ when $b\to a^+$.

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I'd like to see how this formula was derived. – Matt Groff Sep 10 '11 at 8:55
@Matt, here you are. (But a "please" would have been nice.) – Did Sep 10 '11 at 9:26
@J. M. ...Except that, to sum the series, I assumed that $|s|<1$, that is, that $b<a$. Thanks for drawing my attention to this point. I modified my post. – Did Sep 10 '11 at 12:00
@Didier Piau: Thank you so much! I tried not to put pressure on you. I'm sorry I didn't say please. Please forgive me! :-) – Matt Groff Sep 10 '11 at 16:19
Yes, it is. See alternative formula for $\rho(a,a)$ in the revised version. – Did Sep 13 '11 at 5:29

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