Let
be a sequence of n positive numbers. The Stieltjes continued fraction is an expression of the form, see [KK] & also [JT],
![{\displaystyle \beta _{a}(z)=a_{n}z+{\cfrac {1}{a_{n-1}z+{\cfrac {1}{\ddots +{\cfrac {1}{a_{1}z}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16d309b9d2c87b6fa3e652aeee1b9ffacbac89ca)
or its reciprocal
The function defines a rational n-to-1 map of the right half of the complex plane onto itself,
![{\displaystyle \beta _{a},1/\beta _{a}:\mathbb {C^{+}} {\xrightarrow[{}]{n\leftrightarrow 1}}\mathbb {C^{+}} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfc1458c2d45bfeeac4a2bc83393ff6c2b4b3aa3)
since
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9e29320ea2f2df23f6a839d887ddc26cf850f99)
- 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
, 're non-negative real numbers, and the fractions 're characterized by it.
The function
is determined by the pre-image of unity (i.e. n points, counting multiplicities), since
![{\displaystyle \beta _{a}(z)={\frac {p(z^{2})}{zq(z^{2})}}=1\iff p(z^{2})-zq(z^{2})=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd99348f03b49cff550232eefcf388a0423d1096)
and a complex polynomial is determined by its roots up to a multiplicative constant by the fundamental theorem of algebra.
Let
be the elementary symmetric functions of the set
. That is,
![{\displaystyle \prod _{k}(z-\mu _{k})=\sum _{k}\sigma _{n-k}z^{k}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f38fe9ff9a6637be71f689ddc20341f8920b64e)
Then, the coefficients
of the continued fraction are the pivots in the Gauss-Jordan elimination algorithm of the following
square Hurwitz matrix:
![{\displaystyle 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59f6f0f50c445a963e55b4b74490a17ed702e81a)
and, therefore, can be expressed as the ratios of monomials of the determinants of the blocks of
.
- Exercise (**). Prove that
</math>
![{\displaystyle 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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/13f816a4618f7da92e656556ff7736803334028b)
- Exercise (*). Use the previous exercise to prove that
![{\displaystyle \prod _{k}a_{k}={\frac {1}{\prod _{k}\mu _{k}}}=1/\sigma _{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3195bdc84f49667182c3d1ed72e2f4f3dc29483e)