Lecture 6A

Lecture 6A - Approximation, and Interpolation Consider the...

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EE103 SLIDES 6A (SEJ) 1 Approximation, and Interpolation Consider the system of linear equations A xb = where A is mxn, m > n, and the columns of 12 n Aaa a ,, , , " , are linearly independent (i.e., rank A = n ). Such a system of equations usually has no solution. We define, for a given n x R , the error vector ex A x b =− () The minimum norm problem is the problem of choosing an n x R that minimizes the norm of the error, ||() || . For the purposes of these notes, the only norms that are of concern are the following: {} 1 1 2 2 1 1, , 1. 2. 3. max n i i n t i i i in zz z z = = = = == = " The minimum error problem is, therefore, n xR min ||() || EE103 SLIDES 6A (SEJ) 2 Minimum Norm Least-squares problems are those for which the norm is the Euclidean norm 2 T z = || || In this case we may write 22 1 nn n n m T i i ex ex e x ∈∈ = ⇔= = min || ( ) || min || ( ) || min ( ) ( ) min ( ) That is, if the Euclidean norm is used, we choose an x that minimizes the sum of the squared errors; hence the term “least squares”. Let 2 2 T f xe x e x e x () () ; we wish to minimize f x and, of course, if x is a minimizer, then x must satisfy the vector equation 0 fx ∇=
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EE103 SLIDES 6A (SEJ) 3 Least Squares 2 22 TT T T t T T fx ex ex A x b A x b xAA x bA x bb fx xAA bA == = + ⇒∇ = () () () ( )( ) 0 T f xx A A b A ∇= = ( ) A Ax Ab = LLx Ab = 1 xA A A b = Choleski Factorization EE103 SLIDES 6A (SEJ) 4 Least Squares 1.0 1 1.5 2 32 . 0 1.0 4 55 . 0 65 . 5 y x ⎡⎤ ⎢⎥ ⎣⎦ To determine a second degree approx polynomial, via least-squares.
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Lecture 6A - Approximation, and Interpolation Consider the...

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