Dual basis:
f1(x)=a1+a2*x+a3*x^2
f2(x)=b1+b2*x+b3*x^2
f3(x)=c1+c2*x+c3*x^2
y1=1+x
y2=1+2x
y3=1+x+x^2
fi(yi)=1; fi(yj)=0 (i "\\ne" j)
1 = a1+a2
0 = b1+b2
0 = c1+c2
1 = b1+2b2
0 = a1+2a2
0 = c1+2c2
1 = c1+c2+c3
0 = a1+a2+a3
0 = b1+b2+b3
Solving this system we obtain
a1=2, a2=-1, a3=-1, b1=-1, b2=1, b3=0, c1=0, c2=0, c3=1
Answer:
f1=2-x-x^2
f2=-1+x
f3=x^2
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