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hw5solns

# hw5solns - 151A HW 5 Solutions Will Feldman 1 See book for...

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151A HW 5 Solutions Will Feldman 1. See book for similar problems. 2. Using the Newton forward difference formula given in the problem statement P n ( x 0 + sh ) = f ( x 0 ) + n X k =1 s k 4 k f ( x 0 ) we want to calculate ( 4 2 P n )(10) given values for ( 4 k P n )(0) for k = 2 , 3 , 4. A derivation of the above formula can be found in the textbook. You can just think of it as a simplification of the formula for the Newton polynomial P n ( x ) = f [ x 0 ] + n k =1 f [ x 0 , ..., x k k - 1 i =0 ( x - x i ) in the case that the x j are evenly spaced. We can derive a formula for ( 4 2 P n )( x 0 + sh ) similar to the Newton forward difference formula above. Note that ( 4 P n )( x 0 + sh ) = P n (( x 0 + h ) + sh ) - P n ( x 0 + sh ) = P n ( x 0 + h ) + n X k =1 s k ( 4 k P n )( x 0 + h ) - " P n ( x 0 + h ) + n X k =1 s k ( 4 k P n ) # = P n ( x 0 + h ) - P n ( x 0 + h ) + n X k =1 s k ( 4 k P n )( x 0 + h ) - ( 4 k P n )( x 0 ) = 4 P n ( x 0 ) + n X k =1 s k ( 4 k +1 P n )( x 0 ) . By applying the same reasoning again we get that, ( 4 2 P n )( x 0 + sh ) = ( 4 2 P n )( x 0 ) + n X k =1 s k ( 4 k +2 P n )( x 0 ) . This is almost what we want since it is a formula for 4 2 P n which depends only on 4 k P n (0) for k 2. To finish the problem you just need to note that 4 P n ( x ) = Q n - 1 ( x ) where Q n - 1 is a polynomial of degree n - 1. This will imply that Δ

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