Exercise 9.14

Answers

(a)

sign (x)

is monotonically increasing function on

x

, to have

h (x)

is monotonically increasing in

x

, for

x \geq z

, we want to have

w_{0} + w_{1} x_{1} + w_{2} x_{2} + w_{3} x_{1}^{2} + w_{4} x_{2}^{2} + w_{5} x_{1} x_{2} \geq w_{0} + w_{1} z_{1} + w_{2} z_{2} + w_{3} z_{1}^{2} + w_{4} z_{2}^{2} + w_{5} z_{1} z_{2},

i.e.

w_{1} (x_{1} - z_{1}) + w_{2} (x_{2} - z_{2}) + w_{3} (x_{1}^{2} - z_{1}^{2}) + w_{4} (x_{2}^{2} - z_{2}^{2}) + w_{5} (x_{1} x_{2} - z_{1} z_{2}) \geq 0

So we want $w_{1}, w_{2}, w_{3}, w_{4}, w_{5} \geq 0$ for $h (x)$ to be monotonically increasing in $x$

(b)

For

h (x)

to be invariant under and arbitrary rotation of

x

, it needs to depend only on

∥ x ∥ = \sqrt{x_{1}^{2} + x_{2}^{2}}

So we constrain $w_{1} = w_{2} = w_{5} = 0$ and $w_{2} = w_{3}$ , thus $h (x) = sign (w_{0} + w ∥ x ∥^{2})$

(c)

Since the 2D quadratic function is less than 0 in the middle of the bowl, if we want the positive set to be convex, we want to have

w_{3} < 0, w_{4} < 0

, such that the bowl is now upside down, and the middle part is larger than 0. The set is convex and enclosed by the points where

h (x) = 0

niuers

2021-12-08 10:28