p1038F07

# p1038F07 - EE103 Fall 2007 Lecture Notes(SEJ Section 8...

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EE103 Fall 2007 Lecture Notes (SEJ) Section 8 SECTION 8: INTRODUCTION TO ORTHOGONAL POLYNOMIALS ............. 122 Orthogonal Polynomials ........................................................................................... 124 Trigonometric Polynomials ...................................................................................... 125 Background: .......................................................................................................... 125 Trigonometric Interpolation ................................................................................ 127 Examples of Trigonometric Interpolation: ......................................................... 129

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EE103 Fall 2007 Lecture Notes (SEJ) Section 8 122 SECTION 8: INTRODUCTION TO ORTHOGONAL POLYNOMIALS Ex: Consider the following data for which we wish to compute a least-squares approximating polynomial. The X matrix, for the approximating quadratic, is The condition number of T X X is given by MatLab to be cond(X'*X) = 1.5598e+010. Keep in mind that the "basis polynomials" that we've been using are, for a quadratic least- squares approximating polynomial, 2 2 1 0 1 x x x x x ϕ = = () However, suppose we choose other "basis polynomials"; e.g., 2 2 1 0 10 5 0 1166667 10 5 1 xx x =− () ( .) . . Do not, at this point, be concerned with the derivation of these polynomials; however, as before, the X matrix becomes [x y] 10.0000 0 10.2000 0.0040 10.4000 0.0160 10.6000 0.0360 10.8000 0.0640 11.0000 0.1000 X = 100.0000 10.0000 1.0000 104.0400 10.2000 1.0000 108.1600 10.4000 1.0000 112.3600 10.6000 1.0000 116.6400 10.8000 1.0000 121.0000 11.0000 1.0000
EE103 Fall 2007 Lecture Notes (SEJ) Section 8 123 21 11 22 12 26 16 1 1 1 xx X ϕϕ = ⎡⎤ ⎢⎥ ⎣⎦ () () """""" and we wish to compute 210 ccc ,, so that 2 2 10 0 Px c x c x ϕ = ++ () is the least-squares approximating polynomial. In this case, the X matrix is and the condition number is cond(X'*X) = 100.4464, a remarkable reduction. However, this is not all: The matrix X'X = a diagonal matrix! Therefore, the system of equations, TT X Xc X y = can be trivially solved. Of course, we need to see why this has occurred. The functions 1 1 mm x , , " comprise a set of "basis polynomials" for the vector space of th m degree polynomials. That is, any th m degree polynomial is a linear combination of these "basis polynomials"; e.g., 1 0 ax a x ax a + " 0.1333 -0.5000 1.0000 -0.0267 -0.3000 1.0000 -0.1067 -0.1000 1.0000 -0.1067 0.1000 1.0000 -0.0267 0.3000 1.0000 0.1333 0.5000 1.0000 0.0597 0 0.0000 0 0.7000 0 0.0000 0 6.0000

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EE103 Fall 2007 Lecture Notes (SEJ) Section 8 124 Orthogonal Polynomials Suppose we can find, or construct, another set of "basis polynomials" for the vector space of th m degree polynomials; e.g., 11 0 mm x xx x ϕ ϕϕ ( ), ( ), , ( ), ( ) " and, suppose further, that 1 0 n ik jk k x x for all i j with i j = = ()() , Definition: When this latter condition holds, we say the "basis polynomials" are orthogonal . [Note: The definition depends upon the x values.] HW: Show that the "basis polynomials" 2 1 x x ,, are not orthogonal with respect to the data of the above example; then show that the functions 2 2 1 0 10 5 0 1166667 10 5 1 x =− () ( .) . () . are orthogonal . Recall the following about "orthogonal" vectors. The vectors 12 k vv v {, , , } " are said to be orthogonal if for all iji j ,( ) 0 T ij = Therefore, the matrix T VV is a diagonal matrix. To see this, consider the th ij i j (, ) , , element of the matrix T . That element is the inner product of the th i row of T V with the th j column of V . But the th i row of T V is the th i column of V . But we've assumed orthogonality and, therefore, the th , element is zero.
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p1038F07 - EE103 Fall 2007 Lecture Notes(SEJ Section 8...

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