Approximate solution problem of rank-one modification matrix secular equation

Question

In Golub's paper , page 327,the eigenvalues of a rank-one modification of a $n\times n$ symmetric matrix can be computed by findng the zeros of the secular equation \begin{equation*} w(\lambda_j)=1-\sum_{i}^n \frac{u_i^2}{d_i-\lambda_j}. \end{equation*} where $u(n\times 1)$ is the modified column vector; $d_i$ $(i=1,2,3...n)$ is the eigenvalues of the matrix before modification,which are arranged from smallest to largest. My question is that when $n$ is a large number,only first $k$ terms of the vector $u$ and $d_i$ are known, and the new eigenvalues $\lambda_j$ $(j=1,2,3...n)$are arranged from smallest to largest,how can I get the first $k$ terms of new eigenvalues $\lambda_j$? Added known conditions are that we have different pairs of $uh$ and $dh$ $(h=1,2,3...)$, and which make \begin{equation*} w(\lambda_j)≈1-\sum_{i}^k \frac{u1_i^2}{d1_i-\lambda_j}≈1-\sum_{i}^k \frac{u2_i^2}{d2_i-\lambda_j}≈1-\sum_{i}^k \frac{u3_i^2}{d3_i-\lambda_j}... \end{equation*} Is there any way that we can calculate the new eigenvalues $\lambda$? and how to choose which sets of $uj$ and $dj$ to solve for the eigenvalues? The method I tried was to choose two sets of $d1$、$u1$ and $d2$、$u2$, square their $w(\lambda)$ differences, and then find the extreme point of the new formula to find the approximation of the $\lambda$. It seems that the larger the difference between the chosen $d1_1$ and $d2_1$, the closer the approximate solution of the $\lambda$ is to the true value.