https://piazza.com/class_profile/get_resource/ln18bjs43q41tr/loi47owuije4pj

$$ \sum w * x^i + \frac{1}{2}c||w||^2 $$

a)

$$ ∇L(w) = \sum x^i + \frac{2*1}{2}cw\\ = \sum x^i + cw\\ $$

b)

$$ \text{set =0|}\\ 0 = \sum x^i + cw\\ -cw = \sum x^i\\ w = -\sum cx^i $$

a)

$$ L(w) = w_1^2 + 2w_2^2 + w_3^2 - 2w_3w_4 + w_4^2 + 2w_1 - 4w_2 + 4 \\

L(w) = (2w_1 + 2, 4w_2 - 4, 2w_3 -2w_4, -2w_3 +2w_4) $$

b)

$$ ∇L(w) = (0 + 2, 0- 4, 0 -0, -0 +0)\\ = (2, -4, 0, 0)\\

w_{i+1} = (0, 0, 0, 0) - \mu (2, -4, 0, 0)\\

w_{i+1} = (-2\mu, -4\mu, 0, 0) $$

c)

$$ 2w_1+2=0 \\ 4w_2−4=0\\ 2w_3−2w_4=0\\ 2w_4−2w_3=0\\

\text{Solving this system, we get:}\\

w_1=−1\\ w_2=1\\ w_3=w_4\\

\text{Plugging in these values}\\ L(w) = 1^2 + 2(1)^2 + w_4^2 - 2w_3w_3 + w_3^2 + 2(1)- 4(1) + 4 \\ = 1 + 2 + w_3^2 - 2w_3^2 + w_3^2 + 2(1)- 4(1) + 4 \\ = 1\\

\text{Therefore any vector of the form where x is a real number will be a minimizer}\\ (-1, 1, x, x) $$

d) There is not a unique solution for w since there are infinitely many solutions for w_3 and w_4 despite w_1, w_2 being well defined.

a)

$$ L(w) = \sum (y^i - w * x^i)^2 + \frac{1}{2}\lambda||w||^2 \\

∇L(w) = -2x_i \sum (y^i - w * x^i)^2 + \lambda w \\

$$