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}


\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.

  • $\begingroup$ 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 $c_{nj}=\int_a^b t^{j+n}p_n(t)w(t)dt$? $\endgroup$
    – Giulio R
    Jul 8 at 12:20
  • $\begingroup$ 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 $1/p_n(z) = z^{-n} \sum_{l \geqslant 0} b_{nl} z^{-l}$ ? $\endgroup$
    – xyz
    Jul 8 at 13:17
  • $\begingroup$ @Giulio I made a few updates based on your comment to the post, if you wish to have a look. $\endgroup$
    – xyz
    Jul 9 at 10:14
  • $\begingroup$ 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)... $\endgroup$
    – Giulio R
    Jul 10 at 17:59
  • $\begingroup$ @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. $\endgroup$
    – xyz
    Jul 10 at 18:02

1 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 $z\to\infty$.

  • $\begingroup$ 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). $\endgroup$
    – xyz
    Jul 10 at 19:22
  • $\begingroup$ yes it is easier to work with this notation to then continue the computation $\endgroup$
    – Giulio R
    Jul 10 at 19:29
  • $\begingroup$ 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? $\endgroup$
    – xyz
    Jul 10 at 23:13
  • $\begingroup$ 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 $P(\zeta)$ $\endgroup$
    – Giulio R
    Jul 11 at 9:36

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