Past exam of the numerical linear algebra and continuous optimisation course of the University of Oxford / 2010 / B / 6 / v / Solution

At the current exact path point $v(\mu )$, stationarity is already zero and $Cy=\mu e$. Therefore $F_{\bar{\mu }}(v(\mu ))=(0,(\mu -\bar{\mu })e{)}^T$. Its Newton correction solvesThe matrix has no explicit dependence on the barrier parameter, so comparison with part (iv) giveswhen $J$ is nonsingular. If it is singular, the displayed vector is still a solution of the correction equations, but uniqueness of the correction cannot be inferred from differentiability of the path alone.

This step is exactly the tangent predictor in the stated first-order Taylor expansion. For nearby positive $\bar{\mu }$, it approximates the new path point with error $o(|\bar{\mu }-\mu |)$; a bounded second path derivative improves this to $O((\bar{\mu }-\mu {)}^2)$. Continuity also preserves strict positivity for sufficiently small parameter changes. Larger changes may require damping to keep both the constraint values and multiplier estimates positive.