sols_wk7 - Homework 7 due Thursday November 25 before 13:30...

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Unformatted text preview: Homework 7 due Thursday November 25 before 13:30 in corre-sponding group folder on door of oce CM110 1. Show that the table x 1 . 1 . 2 1 . 4 1 . 6 1 . 8 2 . f ( x ) 2 . 25 1 . 73 0 . 97-. 07-1 . 43-3 . 15 is consistent with the assumption that f ( x ) is a cubic polynomial in x , and nd that polynomial. Note that the x values are uniformly distributed. Hence if we calculate the dierence table 1 , then the fourth column should consist of a repeated number for the data to be consistent with the assumption that f is a cubic polynomial. The dierence table is: i f i f i 2 f i 3 f i 1 2 3 4 5 2 . 25 1 . 73 . 97-. 07-1 . 43-3 . 15-. 52-. 76-1 . 04-1 . 36-1 . 72-. 24-. 28-. 32-. 36-. 04-. 04-. 04 . Since x = 1 + 0 . 2 s , i.e. s = 5( x-1), then p 3 ( x ) = f + s f + s ( s-1) 2! 2 f + s ( s-1)( s-2) 3! 3 f = 2 . 25-. 52 s-. 24 s ( s-1) 2-. 04 s ( s-1)( s-2) 6 = 53 20 + 13 30 x-5 6 x 3 . Alternatively, for the benet of those who went down the path of a divided dierence table (or didnt) i x i f [ x i...
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sols_wk7 - Homework 7 due Thursday November 25 before 13:30...

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