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9 Nov 2019
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Least-Squares: Here we return to polynomial interpolation, let's assume that we are given 'nodes' x0,x1 ,x2 = {-1 ,0,2}, and function values f0 ,f1, f2 = {1 ,2 ,1 } corresponding to the three points (-1,1), (0,2) and (2,1). We will try and build a polynomial interpolant through these three points. First, let's try and pass a parabolic interpolant p(x) = p0 + p1x + p2x2 through these three points. Show that such a parabolic interpolant must satisfy p(x0) = f0, P (x1) = f1 and p (x2) = Â2, which corresponds to p (-1) = 1, p(0) = 2, and p (2) = 1, which corresponds to P0 + ( -1 ) p1 + (-1)2p2 = 1 P0 + (+0)p1 + (+ 0)2p2 = 2 p0 + (+2) p1 + (+ 2)2 p2 = 1 which corresponds to or equivalently Solve this equation using row-rcduction and find the vector p- (i.e.,, find P0,P1,P2)· Plot p (x ) and show that p (x ) indeed passed through the poinits (-1,1), (0,2) and (2,1). Find the QR decomposition of A. Can you solve the equation Ap- = b- using the QR decomposition? Hopefully you get the
Help with these questions
Least-Squares: Here we return to polynomial interpolation, let's assume that we are given 'nodes' x0,x1 ,x2 = {-1 ,0,2}, and function values f0 ,f1, f2 = {1 ,2 ,1 } corresponding to the three points (-1,1), (0,2) and (2,1). We will try and build a polynomial interpolant through these three points. First, let's try and pass a parabolic interpolant p(x) = p0 + p1x + p2x2 through these three points. Show that such a parabolic interpolant must satisfy p(x0) = f0, P (x1) = f1 and p (x2) = Â2, which corresponds to p (-1) = 1, p(0) = 2, and p (2) = 1, which corresponds to P0 + ( -1 ) p1 + (-1)2p2 = 1 P0 + (+0)p1 + (+ 0)2p2 = 2 p0 + (+2) p1 + (+ 2)2 p2 = 1 which corresponds to or equivalently Solve this equation using row-rcduction and find the vector p- (i.e.,, find P0,P1,P2)· Plot p (x ) and show that p (x ) indeed passed through the poinits (-1,1), (0,2) and (2,1). Find the QR decomposition of A. Can you solve the equation Ap- = b- using the QR decomposition? Hopefully you get the
Elin HesselLv2
14 Oct 2019