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opre - Numerical Analysis Time 3:003:50 Location LT-1...

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Numerical Analysis Time: 3:00–3:50 Location: LT-1 Instructor: Subramania Pillai. I office: CC 103, phone 0832-2580442 e-mail: [email protected] 6-Feb-2009 (1) I.S.Pillai, CC 103 1
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The Basic Problem Given a set of tabular values ( available from an experiment) of a function f . X x 0 x 1 x 2 ... x n f ( X ) f ( x 0 ) f ( x 1 ) f ( x 3 ) ... f ( x n ) Where the explicit nature of the function f is not known . 6-Feb-2009 (1) I.S.Pillai, CC 103 2
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The problem is .. To find a function which fits the given data. To find a function φ such that φ ( x i ) = f ( x i ) for all 0 i n. Such a function φ is called an interpolating function . 6-Feb-2009 (1) I.S.Pillai, CC 103 3
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Geometrically The problem is : To find a function / polynomial whose graph passes through the given set of ( n + 1) - points, ( x 0 , f ( x 0 )) , ( x 1 , f ( x 1 )) , · · · , ( x n , f ( x n )) . 6-Feb-2009 (1) I.S.Pillai, CC 103 4
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Polynomial interpolation If φ is a polynomial then the process is called polynomial interpolation and φ is called an interpolating polynomial. 6-Feb-2009 (1) I.S.Pillai, CC 103 5
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Some types of interpolation Polynomial interpolation Piecewise Polynomial( Spline ) interpolation Rational interpolation Trigonometric interpolation Exponential interpolation We study : Polynomial and Piecewise polynomial interpolations. 6-Feb-2009 (1) I.S.Pillai, CC 103 6
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Example Find a polynomial which fits the following data. X x 0 x 1 x 2 f ( X ) f ( x 0 ) f ( x 1 ) f ( x 2 ) 6-Feb-2009 (1) I.S.Pillai, CC 103 7
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Example Solution: P 3 ( x ) = ( x - x 1 )( x - x 2 ) ( x 0 - x 1 )( x 0 - x 2 ) f ( x 0 ) + ( x - x 0 )( x - x 2 ) ( x 1 - x 0 )( x 1 - x 2 ) f ( x 1 ) + ( x - x 0 )( x - x 1 ) ( x 2 - x 0 )( x 2 - x 1 ) f ( x 2 ) . This is a polynomial of degree 2 . Called Lagrange’s interpolation formula. 6-Feb-2009 (1) I.S.Pillai, CC 103 8
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Lagrange’s Interpolation formula Similarly, p n ( x ) = ( x - x 1 )( x - x 2 ) · · · ( x - x n ) ( x 0 - x 1 )( x 0 - x 2 ) · · · ( x 0 - x n ) f ( x 0 ) + ( x - x 2 )( x - x 3 ) · · · ( x - x n ) ( x 1 - x 0 )( x 1 - x 2 ) · · · ( x 1 - x n ) f ( x 1 ) + · · · + ( x - x 0 )( x - x 1 ) · · · ( x - x n - 1 ) ( x n - x 0 )( x n - x 1 ) · · · ( x 1 - x n - 1 ) f ( x n ) is the “ unique polynomial of degree n ” which interpolates f at the n + 1 points x 0 , x 1 , · · · , x n . 6-Feb-2009 (1) I.S.Pillai, CC 103 9
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Lagrange’s Interpolation formula If we write l i ( x ) = ( x - x 0 )( x - x 1 ) · · · ( x - x i - 1 )( x - x i +1 ) · · · ( x - x n ) ( x i - x 0 )( x i - x 1 ) · · · ( x i - x i - 1 )( x i - x i +1 ) · · · ( x i - x n ) then p n ( x ) = n i =0 l i ( x ) f ( x i ) is the “ unique(how?) polynomial of degree n ” which interpolates f at the n + 1 points x 0 , x 1 , · · · , x n . 6-Feb-2009 (1) I.S.Pillai, CC 103 10
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We proved: Theorem 1. (Existence and Uniqueness theorem for interpolating polynomials.) If x 0 , x 1 , x 2 , · · · , x n are n + 1 distinct numbers and f is a function whose values are given at these numbers, then a unique polynomial p ( x ) of degree n with p ( x i ) = f ( x i ) for all i = 1 , 2 , · · · , n. This polynomial is given by p ( x ) = n i =0 l i ( x ) f ( x i ) 6-Feb-2009 (1) I.S.Pillai, CC 103 11
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NOTATION Suppose f : [ a, b ] R is given at the n + 1 points x 0 , x 1 , · · · , x n .
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