# CS 61C

From univariate to multivariate:

• Second Derivative -> Hessian
• Taylor Series…
$f(x + \Delta) = f(x) + \grad f(x) \Delta + 1/2 \Delta^T \gradient^2f(x) \Delta + O(||\Delta||^3)$

To cut the tail without error, use the mid-value theorem. Given some $$x$$ and $$\delta$$:

1. There exists $$z$$ on the segment $$[x, x+\delta]$$ such that $$f(x + \delta) = f(x) + \grad f(z) \delta$$.
2. There exists some $$z'$$ on the segment $$[x, x+ \delta]$$ such that $$f(x + \delta) = f(x) + \grad f(x) \delta + 1/2 \delta^T \grad f(z') \delta$$.

The segment connecting $$x$$ to $$x + \delta$$ has dimension 1. The neighborhood around a given point forms a set. A sphere is ‘hollow’, a ball is solid. With an open ball, there is no way to reach the boundary; those points are excluded from the set. A closed set contains the boundary. A neighborhood is an open set that contains the point and an open ball. Equivalently, a set of a neighborhood of a point if the point is interior.

A local minimum will have $$\grad = 0$$ because otherwise, we would have $$f(x_* + \Delta) < f_(x_*)$$. To generalize minimization:

n=1 n>1
$$f' = 0$$ $$\grad f = 0$$
$$f'' > 0$$ $$\grad^2 f \succcurlyeq 0$$

After computing the gradient and setting it equal to 0, there are any number of potential solutions. A symmetric (Hessian) matrix has $$n$$ eigenvalues $$\lambda 1, \lambda_2, \lambda_3, ...$$ such that: $$Hy = \lambda y$$ which are computed by setting $det(\lambda I - H) = 0$. The matrix $$H$$ is positive semidefinite (PSD, $$H \succcurlyeq 0$$) if all the eigenvalues are positive. One property is that $$H \succcurlyeq 0 \equiv \forall \Delta: \Delta^T H \Delta > 0$$. To prove that $$H$$ is sign indefinite, it suffices to show that: $$\exist y, z \in \mathbb{R}^n \text{ s.t. } y^THy.> 0, z^T H z < 0$$ To show that a Hessian is PSD or NSD, compute the determininat or compute the eigenvalues. To show that it is sign indefinite, show that the determinant is zero,

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