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.

In Abramovitz and Stegun (Eq. 9.1.71) I found this curious relation $$\lim_{\nu\to\infty} \left[ \nu^\mu P_\nu^{-\mu}\left(\cos \frac{x}{\nu} \right) \right]= J_\mu(x) \qquad(1)$$ valid for $x>0$. In fact it can be used to obtain a rather good approximation $$ P_\nu^{-\mu}(\cos\theta) \approx \frac{1}{\nu^\mu} J_\mu(\nu \theta)$$ of the Legendre polynomial in terms of a Bessel function for small $\theta$ (but $\nu\theta$ potentially large). This relation is a way to understand the eikonal approximation of wave scattering (which is the reason I noted it in the first place).

As I am looking into the eikonal approximation, I would appreciate if somebody could help me proving equation (1)?

share|improve this question
2  
See Whittaker & Watson p367 –  Raymond Manzoni Apr 8 '12 at 0:16
    
@RaymondManzoni: thank you. They express the left side in terms of a confluent hypergeometric function and then show that it converges to the left hand side; in fact, they don't directly do it for $P_\nu^{-\mu}\left(\cos \frac{x}{\nu} \right)$ but rather for $P_\nu^{-\mu}\left(1- \frac{x^2}{2\nu^2} \right)$. This kind of proof is more complicated than I expected (I thought one could use some integral representation and then perform the limit). Does somebody know an more compact proof without referring to hypergeometric functions? –  Fabian Apr 8 '12 at 5:07
add comment

1 Answer 1

up vote 7 down vote accepted

Consider the differential equation for the associated Legendre polynomials, $$(1-z^2)w''(z) - 2z w(z) + \left(\nu(\nu+1) - \frac{\mu^2}{1-z^2}\right)w(z) = 0.$$ Change variables. Let $z = \cos \frac{x}{\nu}$. (Notice, for example, that $\frac{d}{dz} = -\frac{\nu}{\sin\frac{x}{\nu}} \frac{d}{dx}$.) In the limit $\nu\to\infty$ the DE takes the form $$x^2 w''(x) + x w'(x) + (x^2-\mu^2)w(x) = 0$$ which is, of course, Bessel's equation. Therefore,
$$\lim_{\nu\to\infty} P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right) \propto J_\mu(x).$$ Since it is getting late, I leave it as an exercise to find the constant.

Addendum: The argument above tells us that in the limit $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)$ is some combination of solutions to Bessel's equation. The singular solution $Y_\mu(x)$ is ruled out since $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)$ is not singular at $x=0$.

Using the integral representation for $-1<z<1$ and $\mathrm{Re}\,\mu > 0$, $$P_\nu^{-\mu}(z) = \frac{(1-z^2)^{-\mu/2}}{\Gamma(\mu)} \int_z^1 d t\, P_\nu(t)(t-z)^{\mu-1},$$ we find for $x\ll 1 \ll \nu$ that $$P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right) \sim \frac{1}{\Gamma(\mu+1)} \left(\frac{x}{2\nu}\right)^\mu.$$ (Here we exploit the fact that for $x\ll 1$, $P_\nu\left(\cos \frac{x}{\nu}\right) = 1+O(x^2)$.) But for small $x$ we have $$J_\mu(x) \sim \frac{1}{\Gamma(\mu+1)} \left(\frac{x}{2}\right)^\mu$$ and so $$\lim_{\nu\to\infty} \left[\nu^\mu P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)\right] = J_\mu(x).$$

share|improve this answer
add comment

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.