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| -rw-r--r-- | 2-point.tex | 12 |
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diff --git a/2-point.tex b/2-point.tex index 3bacf89..af0af87 100644 --- a/2-point.tex +++ b/2-point.tex @@ -1173,7 +1173,17 @@ assuming the last equation is satisfied. The trivial solution, which gives the b \[ \lambda_\mathrm{min}=\mu-\sqrt{4f''(1)}=\mu-\mu_\mathrm m \] -as expected. We need to first the nontrivial solutions with nonzero $X$, but because the coefficients are so nasty this will be a numeric problem... Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is zero. +as expected. We need to first the nontrivial solutions with nonzero $X$, but +because the coefficients are so nasty this will be a numeric problem... +Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is +zero. In this case, if the associated normalized eigenvector is $\hat X$, its magnitude is set by +\begin{equation} + \|X\|^2=\frac1{\hat X^TC\hat X}\left(1-\frac{f''(1)}{y^2}\right) +\end{equation} +In practice, we find that $\hat X^TC\hat X$ is positive. Therefore, for the +solution to make sense we must have $y^2>f''(1)$. In practice, there is at most +\emph{one} $y$ which produces a zero eigenvalue of $B-yC$ and satisfies this +inequality, so the solution seems to be unique. \section{Conclusion} |