hmwk_5_2010_solutions

# Polynomial of degree n 3 we let the four points be x

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polynomial of degree N 3. We let the four points be x 0 = 0, x 1 = h, x 2 = 2h, and x 3 = 3h. Using Newton forward interpolation, we then establish the following interpolating polynomial: g ( x ) = f 0 + Δ f 0 h ( x - x 0 ) + Δ 2 f 0 2 h 2 ( x - x 0 )( x - x 1 ) + Δ 3 f 0 6 h 3 ( x - x 0 )( x - x 1 )( x - x 2 ) with the following associated error term: e ( x ) = f (3) ( ξ ) 24 ( x - x 0 )( x - x 1 )( x - x 2 )( x - x 3 ) = Δ 4 f 0 24 h 4 ( x - x 0 )( x - x 1 )( x - x 2 )( x - x 3 ) where Δ f 0 = f 1 - f 0 Δ 2 f 0 = f 2 - 2 f 1 + f 0 Δ 3 f 0 = f 3 - 3 f 2 + 3 f 1 - f 0 Since we are looking for the third derivatives of the interpolating function and the error term, we note that these terms differentiate to zero, and simply focus in on the third and fourth order terms. Differentiating these terms yields: g (1) ( x ) = Δ 3 f 0 6 h 3 ( x - x 1 )( x - x 2 ) + Δ 3 f 0 6 h 3 ( x - x 0 )( x - x 2 ) + Δ 3 f 0 6 h 3 ( x - x 0 )( x - x 1 ) with the following associated error term: e (1) ( x ) = f (3) ( ξ ) 24 ( x - x 1 )( x - x 2 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 2 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 1 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 1 )( x - x 2 ) Differenting a second time yields: g (2) ( x ) = Δ 3 f 0 6 h 3 ( x - x 1 ) + Δ 3 f 0 6 h 3 ( x - x 2 ) + Δ 3 f 0 6 h 3 ( x - x 0 ) + Δ 3 f 0 6 h 3 ( x - x 2 ) + Δ 3 f 0 6 h 3 ( x - x 0 ) + Δ 3 f 0 6 h 3 ( x - x 1 ) with the following associated error term: e (2) ( x ) = f (3) ( ξ ) 24 ( x - x 2 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 1 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 1 )( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 2 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 1 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 1 ) + f (3) ( ξ ) 24 ( x - x 1 )( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 0 )( x - x 1 ) Finally, differentiating a third time and evaluating at x = x 3 gives us the backward approximations to the third derivative and the error associated with the formula. g (3) ( x ) = 6 Δ 3 f 0 6 h 3 = Δ 3 f 0 h 3 = f 3 - 3 f 2 + 3 f 1 - f 0 h 3 3

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Differentiating the error term one more time leads to e (3) ( x ) = f (3) ( ξ ) 24 ( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 1 ) + f (3) ( ξ ) 24 ( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 1 ) + f (3) ( ξ ) 24 ( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 0 ) + f (3) ( ξ ) 24 ( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 0 ) + f (3) ( ξ ) 24 ( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 1 ) + f (3) ( ξ ) 24 ( x - x 3 ) + f (3) ( ξ ) 24 ( x - x 0 ) + f (3) ( ξ ) 24 ( x - x 1 ) + f (3) ( ξ ) 24 ( x - x 0 ) + f (3) ( ξ ) 24 ( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 1 ) + f (3) ( ξ ) 24 ( x - x 2 ) + f (3) ( ξ ) 24 ( x - x 0 ) + f (3) ( ξ ) 24 ( x - x 1 ) + f (3) ( ξ ) 24 ( x - x 0 ) The error term simplifies to the following: e (3) ( x ) = f (3) ( ξ ) 4 ( x - x 0 ) + f (3) ( ξ ) 4 ( x - x 1 ) + f (3) ( ξ ) 4 ( x - x 2 ) + f (3) ( ξ ) 4 ( x - x 3 ) = f (3) ( ξ ) 4 [( x - x 3 ) + ( x - x 2 ) + ( x - x 1 ) + ( x - x 0 )] We then evaluate this at x 3 in order to get the error associated with the backward approximation: e (3) ( x 3 ) = f (3) ( ξ ) 4 [(0) + ( h ) + (2 h ) + (3 h )] = f (3) ( ξ ) 4 [6 h ] = 3 f
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