lecture 18

# lecture 18 - CE 30125 Lecture 18 LECTURE 18 NUMERICAL...

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CE 30125 - Lecture 18 p. 18.1 LECTURE 18 NUMERICAL INTEGRATION CONTINUED Simpson’s 1/3 Rule • Simpson’s 1/3 rule assumes 3 equispaced data/interpolation/integration points • The integration rule is based on approximating using Lagrange quadratic (second degree) interpolation. • The sub-interval is defined as [ x o ,x 2 ] and the integration point to integration point spacing equals 79 fx  h x 2 x o 2 --------------- f 0 f 1 x 0 x 1 f (x) g (x) f 2 x 2 hh x 0 =0 x 1 =h x 2 =2h coordinate shift

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CE 30125 - Lecture 18 p. 18.2 • Lagrange quadratic interpolation over the sub-interval: where gx  f o V o x f 1 V 1 x f 2 V 2 x ++ = V o x xx 1 2 x o x 1 x o x 2 ------------------------------------------ = V o x x 2 3 hx –2 h 2 + 2 h 2 ----------------------------------- = V 1 x o 2 x 1 x o x 1 x 2 = V 1 x 4 2 x 2 2 h 2 ----------------------- = V 2 x o 1 x 2 x o x 2 x 1 = V 2 x x 2 2 h 2 ---------------- =
CE 30125 - Lecture 18 p. 18.3 • Integration rule is obtained by integrating • Simpson’s 1/3 Rule gx  If x x o x 2 = Ig x x o x 2 dx E + = o x 2 3 hx 2 h 2 + 2 h 2 ----------------------------------- f 1 4 2 x 2 2 h 2 ----------------------- f 2 x 2 2 h 2 ---------------- ++    xE + d x o =0 x 2 =2 h = I 1 2 h 2 -------- f o x 3 3 ---- 3 2 2 ----------- –2 h 2 x +   f 1 4 2 2 2 x 3 3 f 2 x 3 3 h x 2 2 2 h 0 E + = I 1 2 h 2 f o 8 h 3 3 ----- 12 h 3 2 –4 h 3 0 + f 1 8 h 3 16 3 h 3 f 2 8 h 3 3 h 4 h 2 2 E + = I h 3 -- f o 4 f 1 f 2  E + =

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CE 30125 - Lecture 18 p. 18.4 Evaluation of the Error for Simpson’s 1/3 Rule • Error is defined as: • We develop Taylor series expansions for , , and about EI h 3 -- f o 4 f 1 f 2 ++  = Ef x x h 3 f o 4 f 1 f 2 d 0 2 h = fx f o f 1 f 2 x 1 h = f 1 xh f 1 1 1 2 2 f 2 1 1 6 3 f 3 1 1 24 ----- + +++ 4 f 4 1 Ox h 5 + = f o f 1 h f 1 1 1 2 h 2 f 1 2 h 3 6 ----- f 3 1 1 24 h 4 f 4 1 Oh 5 + = f 1 f 1 = f 2 f 1 h f 1 1 1 2 h 2 f 2 1 h 3 6 f 3 1 1 24 h 4 f 4 1 5 + + + =
CE 30125 - Lecture 18 p. 18.5 • Substituting into the expression for E Ef 1 xh  f 1 1 1 2 -- 2 f 2 1 1 6 3 f 3 1 1 24 ----- 4 f 4 1 Ox h 5 ++ + + + x d 0 2 h = h 3 -- f 1 h f 1 1 1 2 h 2 f 2 1 h 3 6 ----- f 3 1 + 1 24 h 4 f 1 4 Oh 5 4 f 1 f 1 h f 1 1 1 2 h 2 f 1 2 + + + + h 3 6 f 3 1 1 24 h 4 f 4 1 5 + E 2 h f 1 1 2 h 2 h 2 f 1 1 1 6 h 3 h 3 + f 2 1 1 24 h 4 h 4 f 3 1 + = 1 120 -------- h 5 h 5 + f 4 1 6 h 3 6 f 1 h 2 f 2 1 1 12 h 4 f 4 1 5 +

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CE 30125 - Lecture 18 p. 18.6 • Error for Simpson’s 1/3 Rule
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lecture 18 - CE 30125 Lecture 18 LECTURE 18 NUMERICAL...

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