# Hilbert Matrix and Approximation Theory

$\det \left[ \frac{1}{i+j-1} \right]_{i,j}$

By guessing where this determinant is $0$ or $\infty$ we can guess the right formula. In Wikipedia, I found this problem:

"Assume that $I = [a, b]$ is a real interval. Is it then possible to find a non-zero polynomial $P$ with integral coefficients, such that the integral $\int_a^b P(x)^2\, dx$ is smaller than any given bound $\varepsilon > 0$, taken arbitrarily small?"

To answer this question, Hilbert derives an exact formula for the determinant of the Hilbert matrices and investigates their asymptotics. He concludes that the answer to his question is positive if the length $b − a$ of the interval is smaller than $4$.

I'm asking for a reference / proof to this exercise. I think you can expand $P$ in Legendre polynomials, or use the Gram determinant.

In general, why is this matrix related to approximation theory?

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This matrix (with $i,j$ ranging over $\lbrace 0, 1, 2, \ldots, d \rbrace$) arises naturally in such problems because it is the Gram matrix of the basis $(1,x,x^2,\ldots,x^d)$ of the space of polynomials of degree at most $d$ with respect to the inner product $\langle P,Q \rangle = \int_0^1 P(x) Q(x) dx$. – Noam D. Elkies Mar 14 '13 at 2:00
I believe that the answer to your question lies in the paper by Hilbert which you can find on the wikipedia page on Hilber matrix. – Vít Tuček Mar 14 '13 at 13:40

In the generically titled, Ein Beitrag zur Theorie des Legendre'schen Polynoms Hilbert says integral $\int_a^b P(x)^2 dx$ defines quaratic form over the space of polynomials of degree $\leq n$.

Over $[0,1]$ the determinant in the basis $\{ 1, x, x^2, \dots, x^n \}$ is:

$D_{[a,b]} = \int_{[a,b]^n} x_1^{n-1}x_2^{n-2}\dots x_{n-2}^2 x_{n-1} \prod_{i < j}(x_i - x_j)^2 d\mathbf{x} = \left( \frac{b-a}{2} \right)^{n^2} D$

Usually Legendre polynomials are defined over [0,1], but you can do a change of variables. In that basis the quadratic form is diagonalized:

$\int_a^b P(x)^2 \, dx = \frac{2}{2n-1}b_1^2 + \frac{2}{2n-3} b_2^2 + \dots ,$

where we expand in Legendre polynomials $P(x) = b_1 p_1(x) + b_2 p_2(x) + \dots .$ You are finding the determinant of this change of variables:

$\left| \begin{array}{cccc} 1 & \frac{1}{2} & \dots & \frac{1}{n} \\\\ \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{n+1} \\\\ \vdots & & & \vdots \\\\ \frac{1}{n} & \frac{1}{n+1} & \dots & \frac{1}{2n+1} \end{array}\right| = \frac{\left[1^{n-1}2^{n-2}\dots (n-2)^2(n-1)\right]^4}{1^{2n-1}2^{2n-2}\dots (2n-2)^2(2n-1)} .$

The discriminant of this quadratic form to be proportional to $\left(\frac{b-a}{4}\right)^n$ which tends to 0 for $b-a < 4$.

Polynomials with integer coefficients form an lattice in the space of polynomials. For this discriminant to tend to 0, there must be polynomials with arbitrarily small norm.

I still find this result counterintuitive... We can find explicit examples, when $|a|,|b|<1$

$\int_{a}^b x^{2n} \, dx = \frac{b^{2n+1}-a^{2n+1}}{2n+1}$

but not sure about $a=0,b=2$, for example.

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Trying to make the result more intuitive: Recall that the Chebyshev polynomials $T_n$ defined by $T_n(2 \cos x) = 2 \cos(nx)$ take $[-2,2]$ to $[-2,2]$ and have integer coefficients, so $\int_{x=-2}^2 T_n(x)^2 dx \leq \int_{x=-2}^2 4 dx = 16$. So, for $[a,b]=[-2,2]$, we can make the integral constant independent of $n$, and it isn't surprising (but still needs proof) that for $b-a <4$ we can do better. – David Speyer Jul 17 '14 at 19:16
@DavidSpeyer Do you think Chebyshev polynoimals are a better choice than Legendre polynomials that Hilbert picked? In Legendre basis this is a rational quadratic form so it can take arbitrarily small values as $n \to \infty$, I guess. – john mangual Jul 17 '14 at 21:03
Legnedre's give the optimal answer; Chebyshev makes it obvious to me that $b-a=4$ should be the cut off. – David Speyer Jul 17 '14 at 21:12
@DavidSpeyer Legendre polynomials also arise as the matrix elements of $SO(3)$ representations. I got interested because of Hilbert's use of Minkowski theorem and geometry of numbers, but never collected my thoughts on this. – john mangual Jul 17 '14 at 21:26
@DavidSpeyer I got started reading on this since I didn't realize that just because $P(x) \in \mathbb{Z}[x]$ doesn't mean $\int_0^x P(s)^2 ds \in \mathbb{Z}[x]$. So this is a rational quadratic form in the coefficients of $P(x)$. – john mangual Jul 31 '14 at 19:08