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

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

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

and

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

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 $$j this integral is zero and for $$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.

• isn't is sufficient to write $1/p_n(z) =z^{-n} \sum_{l\geq 0} b_{nl}z^{-l}$ and multiply by $z^{-n-1}\sum_{j\geq 0} c_{nj}z^{-j}$ where $c_{nj}=\int_a^b t^{j+n}p_n(t)w(t)dt$? Jul 8 at 12:20
• Thanks for your comment and help @Giulio .I am not sure I can follow your thoughts: Firstly, can you be a little bit more detailed on how to reach the expression $1/p_n(z) = z^{-n} \sum_{l \geqslant 0} b_{nl} z^{-l}$ ?
– xyz
Jul 8 at 13:17
• @Giulio I made a few updates based on your comment to the post, if you wish to have a look.
– xyz
Jul 9 at 10:14
• wait, I think $k_n^j$ is just a notation for a quantity that depends on two indices. You can check in examples (take your favorite family of OPs) that these coefficients are not $j$th powers of numbers $k_n$ (in general)... Jul 10 at 17:59
• @Giulio That is a possibility indeed. Initially I thought that these must be powers. Either way, can you be more detailed on how to come with the expression for $1/p_n(z)?$ That would help me a lot.
– xyz
Jul 10 at 18:02

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 $$z\to\infty$$.

• Thanks for the answer. I appreciate every part of it (including the last explanation). Just one further question: Instead of writing $1 + \sum_{l \geqslant 1} b_{nl} \zeta^l$ one could write directly $\sum_{l \geqslant 0} b_{nl} \zeta^l$, correct? (I feel like this way is easier to continue with calculations).
– xyz
Jul 10 at 19:22
• yes it is easier to work with this notation to then continue the computation Jul 10 at 19:29
• Thanks. One further (hopefully the last one) question: don't we need to worry about the domains of convergence of the series defined in the answer?
– xyz
Jul 10 at 23:13
• A fundamental (and very useful!) fact in complex analysis is that a Taylor series of an analytic function $f$ near a point $\zeta_0$ converges absolutely in the largest open disk centered at $\zeta_0$ and which avoids singularities of $f$. This guarantees that if a polynomial $P(\zeta)$ does not vanish at $\zeta=0$, then the Taylor series $1/P(\zeta)=\sum_{l≥0}b_l\zeta^l$ converges absolutely for all $|\zeta|<R$ where $R$ is the minimum of the modulus of the roots of $P(\zeta)$ Jul 11 at 9:36