Let $\mu_j=\int_a^b x^jw(x)dx$ be the $j$th moment of the weight distribution $w(x)$. Show that the linear system of equations

$$\left[\begin{matrix} \mu_0 & \mu_1 & \cdots & \mu_{n-1} \\ \mu_1 & \mu_2 & \cdots & \mu_n \\ \vdots & \vdots & \ddots & \vdots \\ \mu_{n-1} & \mu_n & \cdots & \mu_{2n-2} \end{matrix}\right]\left[\begin{matrix}c_0\\c_1\\\vdots\\c_{n-1}\end{matrix}\right]=\left[\begin{matrix}\mu_n\\\mu_{n+1}\\\vdots\\\mu_{2n-1}\end{matrix}\right]$$

has as solution the coefficients of a polynomial $x^n-\sum_{j=0}^{n-1} c_jx^j$, which is a member of the family of orthogonal polynomials associated with the weight function $w$.

By "has as solution" does that just mean the $c_i$ here are the same as in the sum?

I've tried expanding this matrix out and solving it but that doesn't really get anywhere.

Is there a trick that I'm not spotting?

Thanks for the help!

  • $\begingroup$ To answer your first question, yes, it's what it means. $\endgroup$ – Jean-Claude Arbaut Mar 28 '16 at 21:45
  • $\begingroup$ Is there a trick to go about solving this question? $\endgroup$ – J. Bant Mar 28 '16 at 21:50

An important thing to notice is the matrix is a gramian matrix G:

$$G_{i,j}=\int_a^b x^ix^jw(x)dx=(x^i|x^j)$$

where the right hand side denotes the inner product $(f|g)=\int_a^b f(x)g(x)w(x)dx$, on the linear space of polynomials.

Then, reading one line $i$, your system simply states that:

$$\sum_{j=0}^{n-1} (x^i|x^j)c_j=(x^i|x^n)$$


$$(x^i|\sum_{j=0}^{n-1} c_jx^j)=(x^i|x^n)$$

That is, for all $i=0,\dots,n-1$, $(x^i|p_n)=0$, with


Therefore the $p_n$ form a family of orthogonal polynomials: $p_n$, of degree $n$, is orthogonal to all polynomials of degree $<n$, w.r.t. the inner product $(f|g)=\int_a^b f(x)g(x)w(x)dx$.

One also needs to check this system has actually a solution. For a given $n$, and two vectors $u=(d_0,\dots,d_{n-1})^T$ and $v=(c_0,\dots,c_{n-1})^T$, the product $Gv$ is a vector whose $i$-th row is $(x^i|\sum_{j=0}^{n-1} c_jx^j)$, as shown above. Thus

$$u^TGv=\sum_{i=0}^{n-1} d_i(x^i|\sum_{j=0}^{n-1} c_jx^j)=\left(\sum_{i=0}^{n-1}d_ix^i\right|\left. \sum_{j=0}^{n-1}c_jx^j\right)$$

That is, given two polynomials $P,Q$ of degree at most $n-1$, and their coefficients given bu $u,v$, then $u^TGv=(P|Q)$.

Thus $u^TGu=(P|P)=\int_a^b P^2(x)w(x)dx$, which is $>0$ when $P\neq0$, granted the density function is positive. Thus the matrix $G$ is symmetric definite positive, and it's invertible.

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  • $\begingroup$ Ah, that makes a lot of sense. Thank you! $\endgroup$ – J. Bant Mar 28 '16 at 22:20

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