二維逆問題/斯蒂爾傑斯連分數

Let $\{a_{k}\}$ be a sequence of n positive numbers. The Stieltjes continued fraction is an expression of the form, see [KK] & also [JT],

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

or its reciprocal $\beta _{a}\beta _{a}^{*}(z)=1.$

該函式定義了複平面右半部分到其自身的n對1的有理對映，

$\beta _{a},1/\beta _{a}:\mathbb {C^{+}} {\xrightarrow[{}]{n\leftrightarrow 1}}\mathbb {C^{+}} ,$

因為

${\begin{cases}Re(z_{1}),Re(z_{2})>0\implies Re(z_{1}+z_{2})>0,\\Re(z)>0\implies Re(1/z)>0,\\Re(z)>0,a>0\implies Re(az)>0.\end{cases}}$

練習(***)。利用斯蒂爾傑斯連分數的對映性質證明其交錯、簡單且對稱的零點和極點位於原點和虛軸上，並且這些性質和有理性表徵了連分數。

練習(**)。證明連分數具有表示形式$\beta _{a}(z)=z(\xi _{\infty }+\sum _{k}{\frac {\xi _{k}}{z^{2}+\theta _{k}^{2}}}),{\mbox{ where }}\xi _{\infty },\xi _{k}{\mbox{ and }}\theta _{k},k\in \mathbb {N}$，是非負實數，並且連分數由此表徵。

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 $\mathrm {M}$. 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:

$H_{\mathrm {M} }={\begin{pmatrix}\sigma _{1}&\sigma _{3}&\sigma _{5}&\sigma _{7}&\ldots &0\\1&\sigma _{2}&\sigma _{4}&\sigma _{6}&\ldots &0\\0&\sigma _{1}&\sigma _{3}&\sigma _{5}&\ldots &0\\0&1&\sigma _{2}&\sigma _{4}&\ldots &0\\0&0&\sigma _{1}&\sigma _{3}&\ldots &0\\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&0&\ldots &\sigma _{n}\\\end{pmatrix}}$

因此，可以表示為$\mathrm {M}$塊的行列式的單項式比率。

練習 (**)。證明

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$a_{1}=1/\sigma _{1},a_{2}={\frac {\sigma _{1}^{2}}{\det {\begin{pmatrix}\sigma _{1}&\sigma _{3}\\1&\sigma _{2}\end{pmatrix}}}},a_{3}={\frac {\det {\begin{pmatrix}\sigma _{1}&\sigma _{3}\\1&\sigma _{2}\end{pmatrix}}^{2}}{\sigma _{1}\det {\begin{pmatrix}\sigma _{1}&\sigma _{3}&0\\1&\sigma _{2}&\sigma _{4}\\0&\sigma _{1}&\sigma _{3}\end{pmatrix}}}},\ldots$

練習 (*)。利用前面的練習證明

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