
 [No title] (Site not responding. Last check: 20071107) 
  The spectral density of $J$ is formally defined as $\rho^J(x)={(\d \Omega_x e_0,e_0)}/{\d x}$, where $\Omega_x$ is the resolution of the identity of $J$. 
  It is easily seen that the spectral functions of $E_t$ corresponding to the first $t$ basis vectors are given by the formula \begin{equation} \rho_{(0)\,jk}(x)=\delta_{jk}{1\over 4}\rho_{\rm ch}({{x+2}\over{4}}),\qquad j,k=0,1,\dots,t1, \end{equation} where $\rho_{\rm ch}(\la)=\frac 8\pi\sqrt{\la(1\la)}$ if $\la\in [0,1]$ and $\rho_{\rm ch}(\la)=0$ otherwise, is the spectral density of the Jmatrix $J_{\rm ch}$ with the elements $a_i=1/2$, $b_i=1/4$, $i=0,1,\dots$. 
  To find the spectral density of $L_t$, let us apply an arbitrary singlevalued integrable function $f(x)$ to both sides of (\ref{B}) and rewrite the result in the form \begin{equation} \int_{\infty}^{\infty}f(\la)\left\{\sum_{i=0}^{t1} \frac{\d \Omega_{\varepsilon_i}}{\d\varepsilon_i} \left\frac{\d\varepsilon_i}{\d\la}\right\right\}\d\la= \int_{\infty}^{\infty}f(\la)\frac{\d\widetilde \Omega_{\la}} {\d\la}\d\la,\label{id} \end{equation} where $\la=S_t(\varepsilon_i)$, $i=0,1,\dots,t1$, and $\Omega_x$, $\widetilde \Omega_x$ are the resolutions of the identity of $L_t$ and $E_t+V$, respectively. 
 www.ma.utexas.edu /mp_arc/html/papers/96686 (1316 words) 
