IND ENG 160 Lecture Notes - Lecture 5: Stationary Point, Symmetric Matrix

• School: Berkeley
• Department: Industrial Engineering And Operations Research
• Course: IND ENG 160
• Professor: Javad Lavaei

Document Summary

Suppose we have and we want to nd x2(cid:21) x r2 x = (cid:20)x1 min/maxxf (x) 2 such that the coordinates are local minima. If x is a min for f (x), then x = [x of x2 and x . 2 is the minimum of the new function. Stationary point: the gradient at that point is zero. But remember, a stationary point can be a min, max, or saddle. The 1- dimensional equivalent of the gradient is just the derivative, so this is consistent. Example min x r2 (x1 1)2 + (x2 2)2. A stationary point generalizes to the gradient of a function, f (x). The second derivative equivalent in the multidimensional case is the hessian, which is a symmetric matrix of second partial derivatives. If f (x ) = 0, if: 2f (x ) 0, then it is a local minima, 2f (x ) 0, then it is a local maxima.