Proving that quotient of orthogonal polynomials is a Padé approximant of Stieltjes transform.

Question

Conext and notation. In the question below, $R_n(z) / S_n(z)$ denotes the n-th convergent of the continued fraction

$$ \frac{1|}{|z-b_1} + \frac{-a_2|}{|z-b_2} + \dots + \frac{-a_n|}{|z-b_n} + \dots,$$

where $a_n,b_n$ are the coefficients of a generic orthogonal sequence of monic polynomials $\{p_n(z)\}$ that satisfies the three term recurrence relation

$$ p_n(z) = (z-b_n)p_{n-1}(z) - a_np_{n-2}(z), \quad \text{ for } \, \, n= 1,2,\dots$$

The problem. Define the function

$$ \hat w(z) = \int_a^b \frac{1}{z-t}w(t) dt,$$ which is normally known as the Stieltjes transformation. I wish to prove that

\begin{equation} \tag{1} \hat w(z) - \frac{R_n(z)}{S_n(z)} = \frac{1}{p_n(z)}\int_a^b \frac{p_n(t)}{z-t}w(t) \, dt.\end{equation}

and

\begin{equation} \tag{2} \hat w(z) - \frac{R_n(z)}{S_n(z)} = \frac{k_n^1}{z^{2n+1}} + \frac{k_n^2}{z^{2n+2}} + \dots, \quad |z| > R,\end{equation}

where $R$ is big enough to guarantee uniform convergence of the series $\displaystyle{\sum_{j=0}^\infty \frac{t^j}{z^{j+1}}}.$

My attempt. I was able to prove $(1)$ with ease. Identity $(2)$ gave me quite some more problems. Follows my attempt:

We have that

\begin{equation*} \begin{split} \hat w(z) - \frac{R_n(z)}{S_n(z)} &= \frac{1}{p_n(z)} \int_a^b \frac{p_n(t)}{z-t}w(t) \, dt \\[.25cm] &= \frac{1}{p_n(z)} \int_a^b \sum_{j=0}^\infty \frac{t^j}{z^{j+1}}p_n(t) w(t) \, dt \\[.25cm] &= \frac{1}{p_n(z)} \sum_{j=0}^\infty \frac{1}{z^{j+1}} \boxed{\int_a^b t^j p_n(t) w(t) \, dt}. \end{split} \end{equation*}

So, all we have to do is study the boxed integral above. From the theory of orthogonal polynomials, we know that for $j<n$ this integral is zero and for $j=n$ we have that

$$ \int_a^b t^np_n(t) w(t) \, dt = \gamma_n h_n, $$

where $\displaystyle{h_n = \int_a^b p_n^2(t)w(t) \, dt}$ and $\gamma_n$ is such that $\displaystyle{t^n = \sum_{i=0}^n \gamma_i p_i(t)}$ (recall that $\{p_0(t),\dots,p_n(t)\}$ forms a basis for the vectorial space of the polynomials in one variable of degree equal or smaller than $n$). So everything we have to do is to study the boxed integral for values of $j$ such that $j > n.$ For this cases, it is clear that $\{ p_0(t),\dots,p_j(t)\}$ forms a basis for the vectorial space of the polynomials in one variable of degree equal or smaller than $j$. Therefore, we can find scalars $\delta_i$ such that $$ t^j = \sum_{i=0}^j \delta_i p_i(t). $$ Then, $$ \int_a^b t^j p_n(t) w(t) \, dt = \sum_{i=0}^j \delta_i \int_a^b p_i(t)p_n(t) w(t) \, dt = \delta_n h_n. $$ Therefore, we have that

$$ \hat w(z) - \frac{R_n(z)}{S_n(z)} = \frac{1}{p_n(z)}\left[ \frac{\gamma_n h_n}{z^{n+1}} + \sum_{j=n+1}^\infty \frac{\delta_n h_n}{z^{j+1}} \right].$$

Thanks for any help in advance.

Answer 1

Let $c_{nl}$ be the coefficient of $z^{n-l}$ in $p_n(z)$, then $p_n(z)=z^n (1+\sum_{l=1}^n c_{nl} z^{-l})$ and so $$1/p_n(z) =z^{-n}\frac1{1+\sum_{l=1}^nc_{nl} z^{-l}}.$$

The function $\frac1{1+\sum_{l=1}^nc_{nl}\zeta^l}$ is holomorphic in a neighborhood of $\zeta=0$, hence it admits a Taylor series $$ \frac1{1+\sum_{l=1}^nc_{nl}\zeta^l} = 1+\sum_{l\geq 1}b_{nl}\zeta^l. $$

Set $\zeta=z^{-1}$ and combine with what you already know.

The values $k_n^j$ are just quantities depending on two indices $n,j$ and not $j$th powers of constants $k_n$ depending on $n$ only. E.g., for the monic Legendre polynomials, orthogonal on $(-1,1)$ wrt Lebesgue, you have $$ \frac 1{p_3(z)}\sum_{j\geq 0}\frac 1{z^{j+1}}\int_{-1}^1 p_3(t)t^j dt = \frac {8}{175z^7}+ \frac {88}{1125z^9}+ \frac {656}{6875z^{11}} +\frac {4144}{40625z^{13}}+\dots $$ as $z\to\infty$.