On 2D Inverse Problems/Stieltjes continued fractions

< On 2D Inverse Problems
Let \{a_k\} be a sequence of n positive numbers. The Stieltjes continued fraction is an expression of the form, see [KK] & also [JT],
\beta_a(z) = a_nz + \cfrac{1}{a_{n-1}z + \cfrac{1}{ \ddots + \cfrac{1}{a_1 z} }}
or its reciprocal \beta_a\beta^*_a(z)=1.

The function defines a rational n-to-1 map of the right half of the complex plane onto itself,

\beta_a,1/\beta_a:\mathbb{C^+}\xrightarrow[]{n\leftrightarrow 1}\mathbb{C^+},

since


\begin{cases}
Re(z_1), Re(z_2) > 0 \implies Re(z_1+z_2) > 0, \\
Re(z)>0 \implies Re(1/z) > 0, \\
Re(z) > 0, a > 0 \implies Re(az) > 0.
\end{cases}
Exercise(***). Use the mapping properties of Stieltjes continued fractions to prove that their interlacing, simple and symmetric zeros and poles lie at the origin and the imaginary axes and that the properties and rationality characterize the continued fractions.
Exercise(**). Prove that the continued fractions 've the representation \beta_a(z) = z(\xi_{\infty} + \sum_k\frac{\xi_k}{z^2+\theta_k^2}),\mbox{ where }\xi_{\infty}, \xi_k \mbox{ and } \theta_k, k\in\mathbb{N}, 're non-negative real numbers, and the fractions 're characterized by it.
The function \beta_a is determined by the pre-image of unity (i.e. n points, counting multiplicities), since

\beta_a(z) = \frac{p(z^2)}{zq(z^2)}=1 \iff p(z^2)-zq(z^2) = 0,
and a complex polynomial is determined by its roots up to a multiplicative constant by the fundamental theorem of algebra.
Let \sigma_l be the elementary symmetric functions of the set \Mu. That is,

\prod_k (z-\mu_k) = \sum_k \sigma_{n-k} z^k.
Then, the coefficients a_k of the continued fraction are the pivots in the Gauss-Jordan elimination algorithm of the following n\times n square Hurwitz matrix:
 H_\Mu = \begin{pmatrix}
\sigma_1 & \sigma_3 & \sigma_5 & \sigma_7 & \ldots & 0\\
1 & \sigma_2 & \sigma_4 & \sigma_6& \ldots & 0\\
0 & \sigma_1 & \sigma_3 & \sigma_5& \ldots & 0\\
0 & 1 & \sigma_2 & \sigma_4& \ldots & 0\\
0 & 0 & \sigma_1 & \sigma_3& \ldots & 0\\
\vdots & \vdots & \vdots & \vdots& \ddots& \vdots\\
0 & 0 & 0 & 0& \ldots& \sigma_n\\
\end{pmatrix}

and, therefore, can be expressed as the ratios of monomials of the determinants of the blocks of \Mu.

Exercise (**). Prove that

</math>


a_1 = 1/ \sigma_1, 

a_2 = \frac{\sigma_1^2}{\det\begin{pmatrix}\sigma_1 & \sigma_3
\\ 1 & \sigma_2
\end{pmatrix}},

a_3 = \frac{\det
\begin{pmatrix}\sigma_1 & \sigma_3 
\\ 1 & \sigma_2 \end{pmatrix}^2}{\sigma_1\det
\begin{pmatrix}\sigma_1 & \sigma_3 & 0 
\\ 1 & \sigma_2 & \sigma_4 
\\ 0 & \sigma_1 & \sigma_3
\end{pmatrix}}, \ldots
Exercise (*). Use the previous exercise to prove that

\prod_k a_k = \frac{1}{\prod_k \mu_k} = 1/\sigma_n.