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Unformatted text preview: Solutions for Homework 5 Foundations of Computational Math 2 Spring 2011 Problem 5.1 Recall that we have derived different sets of linear equations for the coefficients of an inter polating cubic spline. Assume that f ( x ) = x 3 and analyze the equations and boundary conditions that define s ( x ) in the forms below and determine what can be said about the relationship between s ( x ) and f ( x ). 5.1.a . s ( x ) is determined by Ts ′′ = d where s ′′ is a vector containing s ′′ i 1 ≤ i ≤ n 1 and boundary conditions s ′′ = f ′′ ( x ) and s ′′ n = f ′′ ( x n ). 5.1.b . s ( x ) is determined by ˜ Ts ′ = ˜ d where s ′ is a vector containing s ′ i 1 ≤ i ≤ n 1 and boundary conditions s ′ = f ′ ( x ) and s ′ n = f ′ ( x n ). Solution: Consider first when s ( x ) is determined by Ts ′′ = d where s ′′ is a vector containing s ′′ i 1 ≤ i ≤ n 1 and boundary conditions s ′′ = f ′′ ( x ) and s ′′ n = f ′′ ( x n ). We have s ′′ i = 6 x i , x i +1 = x i + h i +1 , x i − 1 = x i h i , and f i = x 3 i . Given that we have set the boundary conditions to the acutal values of f ′′ we need only show that the basic equation is satisfied by s ( x ) = x 3 . We have μ i s ′′ i − 1 + 2 s ′′ i + λ i s ′′ i +1 = d i μ i = h i h i + h i +1 , λ i = h i +1 h i + h i +1 d i = 6 h i + h i +1 bracketleftbig ( f i +1 f i ) h i +1 ( f i f i − 1 ) h i bracketrightbig If we substitute in for s ′′ i , s ′′ i +1 , s ′′ i − 1 and multiply both sides by ( h i + h i +1 ) / 6 we get h i ( x i h i ) + 2 x i ( h i + h i +1 ) + h i +1 ( x i + h i +1 ) = bracketleftbig ( f i +1 f i ) h i +1 ( f i f i − 1 ) h i bracketrightbig = bracketleftbig ( x i + h i +1 ) 3 x 3 i h i +1 x 3 i ( x i h i ) 3 h i bracketrightbig = 1 h i +1 (3 h i +1 x 2 i + 3 h 2 i +1 x i + h 3 i +1 ) 1 h i (3 h i x 2 i 3 h 2 i x i + h 3 i ) = 3 x 2 i + 3 h i +1 x i + h 2 i +1 3 x 2 i + 3 h i x i h 2 i = 3 h i +1 x i + h 2 i +1 + 3 h i x i h 2 i So now simplifying the left side yields the desired result h i ( x i h i ) + 2 x i ( h i + h i +1 ) + h i +1 ( x i + h i +1 ) = 3 h i +1 x i + h 2 i +1 + 3 h i x i h 2 i 3 h i +1 x i + h 2 i +1 + 3 h i x i h 2 i = 3 h i +1 x i + h 2 i +1 + 3 h i x i h 2 i 1 So s ( x ) = x 3 is the spline produced by this choice of equations and boundary conditions given f ( x ) = x 3 . We can repeat the exercise for when s ( x ) is determined by ˜ Ts ′ = ˜ d where s ′ is a vector containing s ′ i 1 ≤ i ≤ n 1 and boundary conditions s ′ = f ′ ( x ) and s ′ n = f ′ ( x n ). We have λ i s ′ i − 1 + 2 s ′ i + μ i s ′ i +1 = ˜ d i μ i = h i h i + h i +1 , λ i = h i +1 h i + h i +1 ˜ d i = 3 bracketleftbig λ i ( f i f i − 1 ) h i + μ i ( f i +1 f i ) h i +1 bracketrightbig If we substitute in for s ′ i , s ′ i +1 , s ′ i − 1 and multiply both sides by ( h i + h i +1 ) / 3 we get h i +1 ( x i h i ) 2 + 2( h...
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 Spring '11
 gallivan
 µi, λI, Bspline

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