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//Eg-7.9
//pg-355
clear
clc
y = [2 9 24 47 78];
t = [0 1 2 3 4];
//since P0 = 1 b0 will be as
P0 = [1 1 1 1 1];
b0 = sum(y)/5;
//P1 = t-G1
G1 = sum(t)/5;
//b1 = summation(P1*y)/summation(P1^2)
P1 = t - [G1 G1 G1 G1 G1];
b1 = sum(P1.*y)/sum(P1^2);
//P2 = (t-G2)P1 - d2
//G2 = summation(t*P1^2)/summation(P1^2)
G2 = sum(t.*P1^2)/sum(P1^2);
//d2 = summation(t*P1*P0)/summation(P0^2)
d2 = sum(t.*P1.*P0)/sum(P0^2);
P2 = (t-[G2 G2 G2 G2 G2]).*P1 - [d2 d2 d2 d2 d2]
b2 = sum(P2.*y)/sum(P2^2);
printf('Therefore the expression is V = (%f)P0 + (%f)P1 + (%f)P2,\n where P0 = 1, P1 = (t-2), P2 = (t-2)\n\n Finally V = 4*t^2 + 3*t + 2\n\n',b0,b1,b2)
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