Science:Math Exam Resources/Courses/MATH307/December 2012/Question 01 (e)
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Question 01 (e)
In this question we will work with polynomials of degree 3 written
(e) Write down the MATLAB/OCTAVE code that plots the points (0,1), (1,2) and (2,2) and the polynomial p(x) that
(i) has zero slopes at x = 0 and x = 2,
(ii) comes closest in the least square sense to passing through the points (0,1), (1,2) and (2,2).
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In part (d) we found that, in order for p(x) to have zero slope at x = 0 and x = 2 and pass through the given points, the coefficients would need to satisfy
How can we find the least square solution of the system Cs = c above?
The least square solution s satisfies
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First, define the matrix C and the vector c
C=[-4, -2; -2, 1; 0, 1]; c=[2; 2; 1];
Then solve the least squares equation Cs = c for s
s=(C'*C)\(C'*c); % or s=inv(C'*C)*C'*c;
Finally, from s we retrieve the coefficient vector a
a_1 = [1; -3; 0; 0]; a_2 = [0; 0; 0; 1]; a = s(1)*a_1 + s(2)*a_2;
This determines the function p(x)
p = @(x) a(1)*x^3 + a(2)*x^2 + a(3)*x + a(4);
All that is left now is to plot the points (0, 1), (1, 2) and (2,2) as well as the polynomial p(x) on the interval [0, 2]:
hold on plot(0, 1, 'o') plot(1, 2, 'o') plot(2, 2, 'o') fplot(p, [0, 2]) % or eg plot(linspace(0, 2), p(linspace(0, 2)))