How to obtain the asymptotics of Legendre polynomials directly from their generating function

Question

I'm reading about Legendre polynomials for additional information since it is interesting to know! Moreover it would help me with a task I am working on. See https://math.stackexchange.com/questions/3945490

The generating function of Legendre polynomials $P_n(x)$ is defined as $$\frac{1}{\sqrt{1-2tx+t^2}}=\sum\limits_{n=0}^{\infty}P_n(x)t^n.$$

It is known that for large $n$ the asymptotic expansion of the Legendre polynomials is

$$P_n(x) \sim \frac{1}{\sqrt{2\pi n}}\frac{(x+(x^2-1)^{1/2})^{n+1/2}}{(x^2-1)^{1/4}}.$$

My question is how to prove this, without using integrals (if there exists such a proof). I searched a lot, but all the proofs I found regarding this start from the integral representation of the Legendre polynomials. Is it possible to prove it by starting directly from the generating function?

Thanks!

Answer 1

You may write $2x=a+1/a$ for certain $a$, $|a|>1$ (I guess you mean $|x|>1$), then $$ \frac1{\sqrt{1-2tx+t^2}}= \frac1{\sqrt{(1-at)(1-a^{-1}t)}}\\= \sum (-1)^n{-1/2\choose n}a^nt^n\cdot \sum (-1)^n{-1/2\choose n}a^{-n}t^n\\ :=\sum c_na^nt^n\cdot \sum c_na^{-n}t^n,\quad c_n=(-1)^n{-1/2\choose n}=\frac{1\cdot 3\cdot\ldots \cdot(2n-1)}{2\cdot 4\cdot \ldots \cdot (2n)}\sim \frac1{\sqrt{\pi n}} $$ (see the last relation in "Additional identities" section here, or deduce from the Wallis formula). Thus $$P_n(x)=\sum_{k=0}^n c_{n-k}c_k a^{n-2k}.$$ This gives (only small values of $k$ matter) the asymptotics $$P_n(x)a^{-n}\sqrt{\pi n}\to_{(*)} \sum_{k=0}^\infty c_k a^{-2k}=\frac1{\sqrt{1-a^{-2}}}$$ that is equivalent to what you ask for.

More details for $(*)$. We have $$ P_n(x)a^{-n}\sqrt{\pi n}-\sum_{k=0}^\infty c_k a^{-2k}=\sum \alpha_kc_ka^{-2k}, $$ where $$\alpha_k=\begin{cases}\sqrt{\pi n}c_{n-k}-1,& k\leqslant n\\ -1,&k>n.\end{cases}$$ For any fixed $k$ we have $\alpha_k\to 0$ when $n\to \infty$. On the other hand, $-1\leqslant \alpha_k\leqslant 100k$, say (for $k\geqslant n/2$ use the bound $c_{n-k}\leqslant 1$; for $k<n/2$ use the bound $c_{n-k}\leqslant c_{\lceil n/2\rceil}\leqslant 1/\sqrt{\pi n/2}$), thus the series $\sum \alpha_k c_k a^{-2k}$ is dominated by an absolutely convergent series $\sum 100 kc_ka^{-2k}$ and its sum goes to 0 by Dominated Convergence Theorem.)

Answer 2

As described in Analytic Combinatorics by Flajolet and Sedgewick, page 4, the pole $t_0$ of the generating function $F(t)$ of smallest absolute value governs the exponential asymptotics $P_n\sim (1/t_0)^n$. In this case $t_0=x-\sqrt{x^2-1}$, hence $P_n\sim (x+\sqrt{x^2-1})^n$.

To obtain the subexponential factor one expands $F(t)$ around $t_0$, $$F(t)\simeq 2^{-1/2}(x^2-1)^{-1/4}t_0^{-1/2}(1-t/t_0)^{-1/2}$$ $$=2^{-1/2}(x^2-1)^{-1/4}t_0^{-1/2}\sum_{n = 0}^{\infty}\frac{(2n - 1)!!}{2^n n!}(t/t_0)^n$$ $$\simeq (2\pi )^{-1/2}(x^2-1)^{-1/4}\sum_{n}n^{-1/2}(1/t_0)^{n+1/2}\,t^n,$$ which gives precisely the large-$n$ asymptotics for $P_n$ quoted in the OP.

Answer 3

One of the possibiliteis is a Liouville–Steklov method (see P. K. Suetin, “Classical Orthogonal Polynomials,” Nauka, Moscow, 1974 or 2005). It gives (Theorem 4.5) $$ (\sin \theta)^{1 / 2} P_{n}(\cos \theta)=\lambda_{n} \cos \left[\left(n+\frac{1}{2}\right) \theta-\frac{\pi}{4}\right]+R_{n}(\theta), $$ where $$ \left|R_{n}(\theta)\right| \leqslant \frac{c_{4}}{\theta n^{3 / 2}}\left(\frac{\pi}{2}-\theta\right), \quad 0<\theta \leqslant \frac{\pi}{2}, \\ |R_{n}(\theta) |\leqslant \frac{c_{5}}{(\pi-\theta) n^{3 / 2}}\left(\theta-\frac{\pi}{2}\right), \quad \frac{\pi}{2} \leqslant \theta<\pi. $$ But this approach need not only integrals but also ODE.