hmwk_1_solutions

hmwk_1_solutions - UNIVERSITY OF NOTRE DAME Department of...

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Unformatted text preview: UNIVERSITY OF NOTRE DAME Department of Civil Engineering and Geological Sciences CE 30125 J.J. Westerink Homework Set #1 Solutions Problem 1 f ( x ) =- . 78 x 4- . 25 x 3- . 45 x 2- . 25 x + 1 . 2 For clarity we can begin by creating a table of derivatives of f ( x ). We will be developing the Taylor Polynomial about the point x = 0, or from our notes we understand this to be a = 0, thus: General x = a = 0 f ( x )- . 78 x 4- . 25 x 3- . 45 x 2- . 25 x + 1 . 2 1 . 2 df dx- 3 . 12 x 3- . 75 x 2- . 9 x- . 25- . 25 d 2 f dx 2- 9 . 36 x 2- 1 . 5 x- . 9- . 9 d 3 f dx 3- 18 . 72 x- 1 . 5- 1 . 5 d 4 f dx 4- 18 . 72- 18 . 72 d 5 f dx 5 a) Develop 2, 4 and 6 term Taylor expansions 1 2 2 Term: g TS ( x ) = f ( a ) + ( x- a ) f ( a ) Using the above table: g TS ( x ) = 1 . 2 + ( x- 0)(- . 25) g TS ( x ) = 1 . 2- . 25 x 4 Term: g TS ( x ) = f ( a ) + ( x- a ) f ( a ) + 1 2 ( x- a ) 2 f 00 ( a ) + 1 3! ( x- a ) 3 f 000 ( a ) Using the above table: g TS ( x ) = 1 . 2 + ( x- 0)(- . 25) + 1 2 ( x- 0) 2 (- . 9) + 1 6 ( x- 0) 3 (- 1 . 5) g TS ( x ) = 1 . 2- . 25 x- . 45 x 2- . 25 x 3 6 Term: g TS = f ( a )+( x- a ) f ( a )+ 1 2 ( x- a ) 2 f 00 ( a )+ 1 3! ( x- a ) 3 f 000 ( a )+ 1 4! ( x- a ) 4 f ( iv ) ( a )+ 1 5! ( x- a ) 5 f ( v ) ( a ) Using the table above: g TS ( x ) = 1 . 2 + ( x- 0)(- . 25) + 1 2 ( x- 0) 2 (- . 9) + 1 6 ( x- 0) 3 (- 1 . 5) + 1 24 ( x- 0) 4 (- 18 . 72) + 1 120 ( x- 0) 5 (0) g TS ( x ) = 1 . 2- . 25 x- . 45 x 2- . 25 x 3- . 78 x 4 3 b) Plot the exact function f ( x ) , as well as all three Taylor series on the interval [0 , 1] . 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 1.2 x y g T S ( x ) 2-Term g T S ( x ) 4-Term f ( x ) g T S ( x ) 6-Term Figure 1 c) Plot the estimated error and actual error for all three Taylor se- ries on a log log plot, where x is the distance from 0. Note: There is no error between the 6-term expansion and the original function f ( x ), since we can only plot positive nonzero values in a log log plot only the errors for the 4-term and 2-term expansion are shown in the following figure. The estimated error is determined from the leading order error term in the Taylor expansion. 1 ( n + 1)! ( x- a ) n +1 d n +1 f dx n +1 | x = ξ where ξ ∈ [ a,x ]. There are many ways to estimate ξ , since it is difficult to determine which value for ξ is the most reasonable. The following 4 figure will use ξ = a = 0, however the midpoint ξ = 1 2 ( a + x ) is also a good choice. In this problem these estimates are: 1 2 ( x- a ) 2 f 00 ( a ) = 1 2 x 2 · (- . 9) =- . 45 x 2 for 2-term and 1 4! ( x- a ) 4 f ( iv ) ( a ) = 1 4! x 4 · (- 18 . 72) =- . 78 x 4 for 4-term d) How do the errors generally behave?...
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This note was uploaded on 03/02/2012 for the course CE 30125 taught by Professor Westerink,j during the Fall '08 term at Notre Dame.

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hmwk_1_solutions - UNIVERSITY OF NOTRE DAME Department of...

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