Local maxima of Legendre polynomials

When I plotted the (normalized) Legendre polynoials, I couldn't help noticing that all the local maxima lay on a really nice curve:

What is the equation of the curve (and how can we arrive to that equation)?

-
Where "normalized" refers to $\int_{-1}^1 P_n^2=1$ (rather than having leading coefficient $1$ or other kinds of normalization). –  ˈjuː.zɚ79365 Jun 12 '13 at 2:09

I claim that that curve is $$y=\pm \frac{\sqrt{2/\pi}}{ \sqrt[4]{1-x^2}}.$$ This argument will not be rigorous, and will cite a source I haven't fully understood.

Take a look at Whittaker and Watson, A course in Modern Analysis, p. 316. They write: $$P_n(\cos \theta) = \frac{4}{\pi} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \left( \frac{\cos[(n+1/2) \theta - \pi/4]}{(2 \sin \theta)^{1/2}} + \frac{\cos[(n+3/2) \theta - 3\pi/4]}{2(2n+3) (2 \sin \theta)^{3/2}}+ \cdots \right)$$

... Shew also that the first few terms of this series give an approximate value of $P_n(\cos \theta)$ for all values of $\theta$ between $0$ and $\pi$ which are no nearly equal to either $0$ or $\pi$.

This approximation seems to be very good in practice. In the picture below, the blue curve is $P_5(x) \sqrt{\sin(\cos^{-1} x)}$ and the red curve is $512/(63 \sqrt{11} \pi)*\cos((11/2) \cos^{-1}(x) - \pi/4)$. (These are the same normalizations user79365 is using in his post.)

I assume that, in more modern language, the authors are saying that this is an asymptotic series for $P_n(\cos \theta)$, valid on $(0, \pi)$. I have taken the liberties of applying some algebraic rearrangements, replacing their parameter $\phi$ by its definition and stopping the sum after two terms rather than the four they give. Also, they are using the normalization where $\int_{-1}^1 P_n^2 = 2/(2n+1)$, so you'd want to multiply by $\sqrt{(2n+1)/2}$.

For fixed $\theta$, that second term is bounded by $c/n$. The later terms (not displayed) die off even faster as $n$ grows. So $$\tilde{P}_n(\cos \theta)\approx \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \frac{\cos[(n+1/2) \theta - \pi/4]}{(2 \sin \theta)^{1/2}}$$ as $n \to \infty$. The $\sim$ over the $P$ indicates that I am now using your normalization. As long as $\theta/\pi$ is irrational, that $\cos$ term will swing between $1$ and $-1$, coming arbitrarily close to both ends. So $$\lim \sup_{n \to \infty} \tilde{P}_n(\cos \theta) = \lim_{n \to \infty} \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \frac{1}{(2 \sin \theta)^{1/2}}$$

Taking $1/\sqrt{\sin \theta} =1/\sqrt[4]{1-x^2}$ out of everything, we need to compute $$\lim_{n \to \infty} \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \frac{1}{\sqrt{2}}$$ This limit is similar to Wallis's product, and I think the second and third proofs on the Wikipedia page should be adaptable to evaluate it. I used the third proof, by Stirling's approximation: $$\sqrt{2n+1} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} = \sqrt{2n+1} \frac{2^{2n} (n!)^2}{(2n+1)!} \approx \sqrt{2n+1} \frac{2^{2n} (n/e)^{2n} (2 \pi n)}{((2n+1)/e)^{2n+1} \sqrt{2 \pi (2n+1)}}$$ $$=\frac{2 \pi n \sqrt{2n+1}}{(1+1/2n)^{2n} \cdot (2n+1)/e \cdot \sqrt{2 \pi (2n+1)}}=\frac{e \sqrt{2 \pi}}{(2+1/n) (1+1/2n)^{2n}} \approx \frac{\sqrt{2 \pi}}{2}.$$

Putting back in the other constants gave the result I state above.

Remark It's fun to note that $$\int_{-1}^1 \left(\frac{\sqrt{2/\pi}}{\sqrt[4]{1-x^2}} \right)^2 dx = 2.$$ So $P_n^2$ is, on average, half the envelope above it.

Thanks to user79365 for providing the following image: The Legendre polynomials are in blue and the above curve is in red. It's also interesting to note the bunching up at values like $x=\pm 1/2$ and $x = 0$, when $\cos^{-1}(x)$ is a rational multiple of $\pi$.

-
In the final result (first equation) the numerator should be $\sqrt 2$. BTW beautiful answer! –  Dave Jun 12 '13 at 17:40
Thanks! Fixed now. –  David Speyer Jun 12 '13 at 19:11
I uploaded this plot of the envelope to SE imgur account, but wasn't sure if you wanted it as an illustration of your answer. BTW I posted something about keeping $n$ fixed here. –  ˈjuː.zɚ79365 Jun 13 '13 at 5:55
Just awesome. Thanks for this answer. –  Antonio Vargas Jun 20 '13 at 14:11

Estimates of the form $$(1-x^2)^{1/4}|P_n(x)|<Cn^{-1/2}\tag{A1}$$ (here and in the sequel $-1\le x\le 1$) have a long history. Stiltjes obtained (A1) with unspecified $C$ in 1890. Gronwall (1913) and Fejer (1925) had explicit constants in (A1). Bernstein (1937) proved (A1) with sharp constant $C=\sqrt{2/\pi}$; his proof is presented in Orthogonal Polynomials by Szegő, section 7.3.

In terms of the $L^2$-normalized polynomials $\widetilde P_n=\sqrt{(2n+1)/2}\,P_n$, the sharp (Bernstein) form of (A1) becomes
$$(1-x^2)^{1/4}|\widetilde P_n(x)|\le \sqrt{\frac{2n+1}{\pi n}} \tag{A2}$$ where the right-hand side obviously approaches $\sqrt{2/\pi}$.

But the stronger and simpler inequality $$(1-x^2)^{1/4}|\widetilde P_n(x)| \le \sqrt{\frac{2}{\pi}} \tag{1}$$ is also true and is very tight. On the plot below, (A2) is in green while (1) is in red; the blue curve is $\widetilde P_9$.

Proof of (1). Let's return to original Legendre polynomial $P_n$. In terms of $P_n$ (1) takes the form $$(1-x^2)^{1/4}|P_n(x)|\le \frac{2}{\sqrt{\pi (2n+1)}} \tag2$$

Theorem 7.3.3 in Orthogonal Polynomials by Szegő (originally proved by Bernstein) is more precise than (A1). It says that $$(1-x^2)^{1/4}|P_n(x)| \le |P_n(0)|,\quad \text{if n is even} \tag3$$ $$(1-x^2)^{1/4}|P_n(x)| < \frac{2}{\sqrt{(2n+1)^2+1}}|P_n'(0)|,\quad \text{if n is odd} \tag4$$ Sketch of proof of (3)-(4). Let $u(\theta)=\sqrt{\sin\theta}P_n(\cos\theta)$ for $\theta\in [0,\pi/2]$. Check that $u''(\theta)+\varphi(\theta)u(\theta)=0$ where $\varphi(\theta)=(n+1/2)^2+1/(2\sin\theta)^2$. Show that $u^2+(u')^2/\varphi$ has positive derivative on the interval $(0,\pi/2)$. At $\theta=\pi/2$, either $u$ or $u'$ vanishes, and (3)-(4) follow. $\Box$

From the coefficients of Legendre polynomials we see that $$|P_{2k}(0)|=\frac{(2k-1)!!}{(2k)!!} \tag5$$ $$|P_{2k+1}'(0)|=\frac{(2k+1)!!}{(2k)!!} \tag6$$ The case of $n=2k$ turns out to be easier. A refined version of Wallis' product formula, proved by D.K. Kazarinoff in On Wallis' formula, states $$\frac{(2k-1)!!}{(2k)!!} = \frac{1}{\sqrt{\pi(n+\theta)}},\quad \frac14<\theta<\frac12 \tag7$$ From (5) and (7) we have $$|P_{2k}(0)|<\frac{1}{\sqrt{\pi(k+1/4)}} = \frac{2}{\sqrt{\pi (4k+1)}}$$ which in view of (3) yields (2).

Instead of referring to Kazarinoff's paper one can prove directly that $$\left(\frac{(2k)!!}{(2k-1)!!}\right)^2\frac{4}{4k+1} \searrow \pi \tag8$$ Indeed, the limit is $\pi$ by Wallis' formula, and to show that the left hand side is decreasing, one has to check that $$\left(\frac{2k+2}{2k+1}\right)^2\frac{4k+1}{4k+5}< 1 \tag9$$ Clearing the denominators in (9), one obtains $(2k+2)^2 (4k+1)- (2k+1)^2(4k+5) = -1 <0$ which gives (9).

We can expect the case $n=2k+1$ to be harder, because (4) is not sharp, unlike (3). And indeed, the combination of (4), (6), and (7) does not yield (2). Instead, we can use a monotonicity fact similar to (8), though less elegant:
$$\left(\frac{(2k)!!}{(2k+1)!!}\right)^2\frac{(4k+3)^2+1}{4k+3} \searrow \pi \tag{10}$$ Indeed, the limit is $\pi$ by Wallis' formula, and to show that the left hand side is decreasing, one has to check that $$\left(\frac{2k+2}{2k+3}\right)^2 \frac{(4k+7)^2+1}{4k+7}\frac{4k+3}{(4k+3)^2+1} < 1 \tag{11}$$ Clearing the denominators in (11), one obtains $$(2k+2)^2((4k+7)^2+1)(4k+3) -(2k+3)^2((4k+3)^2+1)(4k+7) = -(4k+5)^2-5<0$$ which gives (11).

From (10) it follows that $$\frac{(2k+1)!!}{(2k)!!}\frac{2}{\sqrt{(4k+3)^2+1}} < \frac{2}{\sqrt{\pi(4k+3)}} \tag{12}$$ Combining (6) and (12) we obtain $$\frac{2}{\sqrt{(4k+3)^2+1}}|P_{2k+1}'(0)| < \frac{2}{\sqrt{\pi(4k+3)}}$$ which in view of (4) yields (2). $\Box$

-
I don't know how to prove the precise inequality you want, but I think the explanation for why the maxima are all so close to equal is simply that the aymptotic I use is very good. Check out the image I just added to my post. –  David Speyer Jun 13 '13 at 13:17
@DavidSpeyer I added the proof now: all polynomials $\widetilde{P_n}$ indeed stay between $\pm \sqrt{2/\pi}(1-x^2)^{-1/4}$. –  ˈjuː.zɚ79365 Jun 20 '13 at 14:31
Nice work! There are some slick manipulations here. –  David Speyer Jun 20 '13 at 15:40
I'm planning to give this a bounty for awesomeness once the system will let me. –  David Speyer Jun 20 '13 at 16:03

All the local maxima do not sit on the same "nice curve".

Under Wikipedia's normalisation with $\|P_n\|^2 = \frac{2}{2n+1}$ we can use Bonnet's recursion formula to get that

$$(n+1)P_{n+1}(0) = - n P_{n-1}(0)$$

Since your "normalised" $\sqrt{\frac{2}{2n+1}}\tilde{P}_n = P_n$ we get that under your normalization

$$\frac{n+1}{n} \sqrt{\frac{2n-1}{2n+3}} |\tilde{P}_{n+1}(0)| = |\tilde{P}_{n-1}(0)|$$

The multiplicative factor is

$$\frac{n+1}{n} \sqrt{\frac{2n-1}{2n+3}} = \sqrt{ \frac{2n^3 + 3n^2 - 1}{2n^3 + 3n^2}} < 1$$

Hence we have that the local maximum at $x = 0$ for $n = 0\pmod 4$ is strictly increasing, and hence cannot all belong to the same curve.

Perhaps the question you want to ask is: what is the smallest convex function defined on $(-1,1)$ that is greater than or equal to the normalised Legendre polynomials $\tilde{P}_{n}$?

-
I believe the technical term is envelope except this is a discrete family of curves... Equivalently, one could state the question in terms of the closure of the union of graphs of $\widetilde P_n$. –  ˈjuː.zɚ79365 Jun 12 '13 at 13:10
In any case, I am not convinced that this question is more likely to have a clean solution compared to say, considering $\overline{P}_{n = 2k}$ normalised so that they evaluate to 1 at the origin. –  Willie Wong Jun 12 '13 at 13:13
In pictures: $L^2$ normalization versus constant term normalization. I guess I like the first one better. The both look good, though. –  ˈjuː.zɚ79365 Jun 12 '13 at 13:23
You could also work with Legendre functions en.wikipedia.org/wiki/Legendre_function instead of Legendre polynomials, so that the parameter $n$ can be any real number. I think this is a nice question, and am looking forward to seeing an answer. –  David Speyer Jun 12 '13 at 15:26