ELEC 400M / EECE 571M Homework 3: Nonlinear Transform
Homework is not handed in. Please solve for your own benefit.
• Exercise 3.11, solution attached • Exercise 3.13, solution attached
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• Exercise 3.11
z = Φ(x) = [1,×21,x2]T: decision boundary is w ̃Tz = w ̃0 +w ̃1z1 +w ̃2z2 = 0 in Z and
w ̃ T z = w ̃0 + w ̃1×21 + w ̃2×2 = 0 in X (a)w ̃1>0,w ̃2<0:x2 −x21 =−w ̃0
|w ̃1 |
|w ̃2 |
|w ̃1 w ̃2 |
tilde(w) <0 0
tilde(w)0 >0
This is a hyperbola.
5 4 3 2 1 0
-1 -2 -3 -4 -5
(b) w ̃1 > 0, w ̃2 = 0: w ̃1×21 + w ̃0 = 0 → x1 = ±−w ̃0 w ̃1
→ two vertical lines in the (x2, x1)-plane (and w ̃0 > 0)
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
1
(c) w ̃1 > 0, w ̃2 > 0, w ̃0 < 0: x2 |w ̃1 |
This is an ellipse.
5
4
3
2
1
0
-1
-2
-3
-4
+ x21 |w ̃2 |
= − w ̃0 |w ̃1 w ̃2 |
> 0
xx 22
-5
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
1
(d)w ̃1 >0,w ̃2 >0,w ̃0 >0: |w ̃0|+|w ̃1|x21 +|w ̃2|x2 =0 This equation has not solution x1, x2 ∈ R.
• Exercise 3.13
(a) z = [1, x1, x2, x21]T
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(b) z = [1, x1, x2, x21, x2]T
(c) z = [1, x1, x2, x21, x2]T
(d) z = [1, x1, x2, x21, x2]T
(e) z = [1, x1, x2, x21, x1x2, x2]T (f) z = [1,×1,x2]T = x
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