Exercise 3.13

${\Phi }_2(x)=(1,x_1,x_2,x_1^2,x_1{x}_2,x_x^2)$, make the corresponding $\tilde{w}$ as ${\tilde{w}}_0,\dots ,{\tilde{w}}_5$, we have

(a)

${(x_1-3)}^2+x_2-1=-1-6x_1+x_2+x_1^2$, we choose ${\tilde{w}}_0=-1$, ${\tilde{w}}_1=-6$, ${\tilde{w}}_2=1$, ${\tilde{w}}_3=1$, ${\tilde{w}}_4=0$, ${\tilde{w}}_5=0$.

(b)

${(x_1-3)}^2+{(x_2-4)}^2-1=24-6x_1-8x_2+x_1^2+x_2^2$ , we choose ${\tilde{w}}_0=24$, ${\tilde{w}}_1=-6$, ${\tilde{w}}_2=-8$, ${\tilde{w}}_3=1$, ${\tilde{w}}_4=0$, ${\tilde{w}}_5=1$

(c)

$2{(x_1-3)}^2+{(x_2-4)}^2-1=33-12x_1-8x_2+2x_1^2+x_2^2$, we choose ${\tilde{w}}_0=33$, ${\tilde{w}}_1=-12$, ${\tilde{w}}_2=-8$, ${\tilde{w}}_3=2$, ${\tilde{w}}_4=0$, ${\tilde{w}}_5=1$

(d)

${(x_1-3)}^2-{(x_2-4)}^2-1=-8-6x_1+8x_2+x_1^2-x_2^2$, we choose ${\tilde{w}}_0=-8$, ${\tilde{w}}_1=-6$, ${\tilde{w}}_2=8$, ${\tilde{w}}_3=1$, ${\tilde{w}}_4=0$, ${\tilde{w}}_5=-1$

(e)

$2{(x_1+x_2-3)}^2+{(x_1-x_2-4)}^2-1=33-20x_1-4x_2+3x_1^2+2x_1{x}_2+3x_2^2$, we choose ${\tilde{w}}_0=33$, ${\tilde{w}}_1=-20$, ${\tilde{w}}_2=-4$, ${\tilde{w}}_3=1$, ${\tilde{w}}_4=0$, ${\tilde{w}}_5=3$

(f)

$2x_1+x_2-1$, we choose ${\tilde{w}}_0=-1$, ${\tilde{w}}_1=2$, ${\tilde{w}}_2=1$, ${\tilde{w}}_3=0$, ${\tilde{w}}_4=0$, ${\tilde{w}}_5=0$.