User:Daviddaved/Stieltjes continued fractions

Let $a_{k}$$a_{k}$ be a set of n positive numbers. The Stieltjes continued fraction is an expression of the form

$\beta (z)=a_{n}z+{\cfrac {1}{a_{n-1}z+{\cfrac {1}{\ddots +{\cfrac {1}{a_{1}z}}}}}}.$$\beta (z)=a_{n}z+{\cfrac {1}{a_{n-1}z+{\cfrac {1}{\ddots +{\cfrac {1}{a_{1}z}}}}}}.$

The function and its reciprocal define all rational n-to-1 maps of the right half of the complex plane onto itself,

$\beta ,1/\beta :\mathbb {C^{+}} {\xrightarrow[{}]{n\leftrightarrow 1}}\mathbb {C^{+}} ,$$\beta ,1/\beta :\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)\implies Re(1/z)>0,\\Re(z)>0,a>0\implies Re(az)>0.\end{cases}}$${\begin{cases}Re(z_{1}),Re(z_{2})>0\implies Re(z_{1}+z_{2})>0,\\Re(z)\implies Re(1/z)>0,\\Re(z)>0,a>0\implies Re(az)>0.\end{cases}}$

The function $\beta$$\beta$ is determined by the pre-image set $\mathrm {M} =\{\mu _{k}\}$$\mathrm {M} =\{\mu _{k}\}$ of the point {z = 1}, since

$\beta (z)={\frac {p(z^{2})}{zq(z^{2})}}=1\iff p(z^{2})-zq(z^{2})=0,$$\beta (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}$$\sigma _{l}$ be the elementary symmetric functions of the set $\mathrm {M}$$\mathrm {M}$. That is,

$\prod _{k}(z-\mu _{k})=\sum _{k}\sigma _{n-k}z^{k}.$$\prod _{k}(z-\mu _{k})=\sum _{k}\sigma _{n-k}z^{k}.$

Then the coefficients $a_{k}$$a_{k}$ of the continued fraction are the pivots in the Gauss-Jordan elimination algorithm of the following n by n square Hurwitz matrix:

$S(\mathrm {M} ):={\begin{bmatrix}\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{bmatrix}}$$S(\mathrm {M} ):={\begin{bmatrix}\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{bmatrix}}$

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

Exercise 1 : Prove that

$a_{1}=1/\sigma _{1}.$$a_{1}=1/\sigma _{1}.$

Exercise 2 :

Let A be a diagonal matrix with the alternating in sign diagonal entries:

$A={\begin{bmatrix}1/a_{1}&0&0&\ldots &0\\0&-1/a_{2}&0&\ldots &0\\0&0&\ddots &\ddots &\vdots \\\vdots &\vdots &\ddots &-1/a_{n-1}&0\\0&0&\ldots &0&1/a_{n}\\\end{bmatrix}}$$A={\begin{bmatrix}1/a_{1}&0&0&\ldots &0\\0&-1/a_{2}&0&\ldots &0\\0&0&\ddots &\ddots &\vdots \\\vdots &\vdots &\ddots &-1/a_{n-1}&0\\0&0&\ldots &0&1/a_{n}\\\end{bmatrix}}$

and D the (0,1)-matrix

$D={\begin{bmatrix}0&1&0&\ldots &0\\1&0&1&\ldots &0\\0&1&\ddots &\ddots &\vdots \\\vdots &\vdots &\ddots &0&1\\0&0&\ldots &1&1\\\end{bmatrix}}.$$D={\begin{bmatrix}0&1&0&\ldots &0\\1&0&1&\ldots &0\\0&1&\ddots &\ddots &\vdots \\\vdots &\vdots &\ddots &0&1\\0&0&\ldots &1&1\\\end{bmatrix}}.$

Prove that the continued fraction $\beta (z)$$\beta (z)$ evaluated at a point $z=\mu$$z=\mu$ equals to 1 if and only if $\mu$$\mu$ is an eigenvalue of the matrix AD.

Exercise 3 :

Use Exercise 1 to prove that

$\prod _{k}a_{k}={\frac {1}{\prod _{k}\mu _{k}}}=1/\sigma _{n}.$$\prod _{k}a_{k}={\frac {1}{\prod _{k}\mu _{k}}}=1/\sigma _{n}.$