On 2D Inverse Problems/Stieltjes continued fractions: Difference between revisions

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Let {ak} be a sequence of n positive numbers. The Stieltjes continued fraction is an expression of the form, see [KK] & also [JT],
βa(z)=anz+1an1z+1+1a1z
or its reciprocal βaβa*(z)=1.

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

βa,1/βa:+n1+,

since

{Re(z1),Re(z2)>0Re(z1+z2)>0,Re(z)>0Re(1/z)>0,Re(z)>0,a>0Re(az)>0.
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 βa(z)=z(ξ+kξkz2+θk2), where ξ,ξk and θk,k, 're non-negative real numbers, and the fractions 're characterized by it.
The function βa is determined by the pre-image of unity (i.e. n points, counting multiplicities), since
βa(z)=p(z2)zq(z2)=1p(z2)zq(z2)=0,
and a complex polynomial is determined by its roots up to a multiplicative constant by the fundamental theorem of algebra.
Let σl be the elementary symmetric functions of the set M. That is,
k(zμk)=kσnkzk.
Then, the coefficients ak of the continued fraction are the pivots in the Gauss-Jordan elimination algorithm of the following n×n square Hurwitz matrix:
HM=(σ1σ3σ5σ701σ2σ4σ600σ1σ3σ5001σ2σ4000σ1σ300000σn)

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

Exercise (**). Prove that

</math>

a1=1/σ1,a2=σ12det(σ1σ31σ2),a3=det(σ1σ31σ2)2σ1det(σ1σ301σ2σ40σ1σ3),
Exercise (*). Use the previous exercise to prove that
kak=1kμk=1/σn.

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