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

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!

• what's wrong with integrals? see math.stackexchange.com/q/1536211/87355 Dec 16 '20 at 18:04
• Nothing but I prefer a standard proof. Dec 16 '20 at 18:16
• I seems to me you need to assume that $\,x^2-1>0\,$ for this result. Is that correct? If so, you need to add that in the statement of the result. Dec 16 '20 at 22:54
• You should link or mention that I've answered several questions for you on this problem on MSE 3947721. The techniques in Bender and Orszag on saddle point analysis will yield the answer. Likewise, Olver is good book, which is the source the DLMF references. Dec 17 '20 at 0:19
• We could be facetious and note that, according to the original problem statement, both infinite sums and limits are evidently acceptable mathematical techniques. One could take any proof involving an integral and convert the integral into the limit of a sum, thereby obtaining a "standard" proof. Dec 17 '20 at 6:30

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 $$k use the bound $$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.)

• Excellent approach!! Dec 17 '20 at 11:14
• Rigorously speaking, I use Wallis formula which is hard to get itself without integrals. But I guess this should be possible (say, from certain discrete versions of the product expansion of $\sin x$). Dec 17 '20 at 11:58
• Hi @Fedor Petrov thanks for your great answer! I would ask why $c_n \sim 1/\sqrt{\pi n}$? , and how to conclude the last equation where the sum equals $1/ \sqrt{1-a^2}$ Dec 17 '20 at 19:19
• @user726608 I added Wallis' references (on $c_n$ asymptotics). You mean, how to conclude that the sum equals $1/\sqrt{1-a^{-2}}$, or how to get the asymptotical relation in this line? Dec 17 '20 at 20:28
• @user726608 the first one follows from Stirling's formula (write as quotient of two factorials), the second one by the above definition of $c_n$ writing $a^{-2}$ instead of $a$ and $t=1$. Dec 17 '20 at 20:39

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.

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.