Review 1-1

Review 1-1 - CE 30125 - Review 1 REVIEW NO. 1 TAYLOR SERIES...

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CE 30125 - Review 1 p. R1.1 REVIEW NO. 1 TAYLOR SERIES • Find away from given and the derivatives of evaluated at •No t e s and the derivatives of evaluated at are constant and not x -dependent. • When we use Taylor Series we do not carry all terms! • Our derivations typically carry enough terms to allow us to establish the error in our formula. fx  xa = fa = df dx ----- = 2 2 ! ------------------ d 2 f 2 -------- = 3 3 ! d 3 f 3 = 4 4 ! d 4 f 4 = + +++ = n n ! ------------------- d n f n = 1 n 1 + ! n 1 + d n 1 + n 1 + --------------- ++ + a x  f =
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CE 30125 - Review 1 p. R1.2 • Depending on what our purposes are, we may: • Truncate the series and only carry the term • The error term is • Carry enough terms so that we have a detailed form of the largest portion of the error term • The error term is The term is carried to ensure that we know that this next term is where we systematically truncate all terms Ox a  n fx fa xa df dx ----- = 2 2! ------------------ d 2 f 2 -------- = 3 ++ + = EO 3 = 2 2! d 2 f 2 = = 3 3! d 3 f 3 = 4 E 3! ---------------- 3 d 3 f 3 = = 4
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CE 30125 - Review 1 p. R1.3 • Carry even more terms so that we have information about the leading error terms as well as subsequent error terms. • The error term is fx  fa xa df dx ----- = 2 2! ------------------ d 2 f 2 -------- = 3 3! d 3 f 3 = 4 4! d 4 f 4 = ++ + + + = 5 5! d 5 f 5 = Ox a 6 E 3 3! d 3 f 3 = 4 4! d 4 f 4 = 5 5! d 5 f 5 = 6 +++ =
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CE 30125 - Review 1 p. R1.4 • Carry only the remainder term which represents all terms in the truncated series Error term is • We note that therefore depends on where you’re evaluating . • Typically we just estimate as a starting or a mid point in the interval as a constant! • However when you differentiate or integrate the error terms which involve , you must be very careful. It is best to consider a sequence of terms evaluated at fx  fa xa df dx ----- = 2 2! ------------------ d 2 f 2 -------- = 3 3! d 3 f 3 x = ++ + = a x  E 3 3! d 3 f 3 x = =  x = =
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CE 30125 - Review 1 p. R1.5 NUMERICAL SOLUTION TO LINEAR SYSTEMS OF ALGEBRAIC EQUA- TIONS • Solve the system of linear algebraic equations Direct Methods • All direct methods are based on some type of triangulation Gauss elimination • Develop an upper triangular matrix by manipulating and operations • Perform backward solution sweep operations A XB = A B On  3 2
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CE 30125 - Review 1 p. R1.6 LU decomposition - Cholesky decomposition (a factor method) • Decompose where is a lower triangular matrix and is an upper triangular matrix •So lv e •L e t Solve using a backward substitution
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Review 1-1 - CE 30125 - Review 1 REVIEW NO. 1 TAYLOR SERIES...

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