# Connection between Legendre polynomial and Bessel function

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)?

• – 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
• This comment comes is long after the asking of the question, but geometrically the relation occurs because near if you are at the north pole and want the eigenfunctions solutions of Laplace's equation you can pretend that the earth is flat and use $e^{im\phi} J_m(\kappa r)$. Alternatively you can remember the earth is a sphere and use the $Y^m_l(\theta, \phi)$ spherical harmonics. For wavelengths much shorter than the circumference of the earth, these two set of eigenfunctions must be proprtional to each other. – mike stone Jul 12 '18 at 14:53
• @mikestone: great insight (I wonder why I did not think about this myself). If you would expand a bit, I would be even a great additional answer... (do not get me wrong, I do understand your argument but it might be useful for future interested people to have a bit more meat to the bone) – Fabian Jul 12 '18 at 15:13
• @fabian After I wrote my comment I looked up the proof in Watson's Bessel Function treatise and I saw that he used this very argument to motivate the limit. The he gets down to the snarly proof which is probably the same as the one in W&W. My own (now forgotten) "proof" was to make this argument, and then fix up the factor of proportionality by some method. – mike stone Jul 13 '18 at 1:23

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 $$\lim_{\nu\to\infty} \left[\nu^\mu P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)\right] = J_\mu(x).$$