Jan24-12

# Jan24-12 - Math 260, Spring 2012 Jerry L. Kazdan Class...

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Unformatted text preview: Math 260, Spring 2012 Jerry L. Kazdan Class Outline: Jan. 24, 2012 1. Example: quadratic polynomials p ( x ) with p (1) = 0. 2. Polynomial Interpolation. Find a quadratic polynomial with p (1) = 1 , p (2) =- 1 , p (4) = 3 . Methods: Naive basis: p 1 ( x ) := 1 , p 2 ( x ) = x, p 3 ( x ) := x 2 . Seek p ( x ) = A 1 p 1 ( x ) + A 2 p 2 ( x ) + A 3 p 3 ( x ) = A 1 + A 2 x + A 3 x 2 Lagrange’s basis: p 1 ( x ) := ( x- 2)( x- 4) (1- 2)(1- 4) , p 2 ( x ) := ( x- 1)( x- 4) (2- 1)(2- 4) , p 3 ( x ) := ( x- 1)( x- 2) (4- 1)(4- 2) . Note p 1 (1) = 1 while p 1 (2) = p 1 (4) = 0, etc. As above, seek p ( x ) = A 1 p 1 ( x ) + A 2 p 2 ( x ) + A 3 p 3 ( x ) . Since p 1 (1) = 1 while p 2 (1 = p 3 (1) = 0, by setting x = 1 we immediately find that A 1 = 1. What are A 2 and A 3 ? Newton’s basis: p 1 ( x ) := 1 , p 2 ( x ) = ( x- 1) , p 2 ( x ) := ( x- 1)( x- 2) and seek p ( x ) = A 1 p 1 ( x ) + A 2 p 2 ( x ) + A 3 p 3 ( x ) ....
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## This note was uploaded on 03/06/2012 for the course MATH 260 taught by Professor Staff during the Spring '12 term at UPenn.

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Jan24-12 - Math 260, Spring 2012 Jerry L. Kazdan Class...

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