Deriving Simpson's RuleDate: 06/07/98 at 16:11:29 From: Carol Subject: Calculus - Simpson's Rule I understand that using Simpson's rule is easier than using the Trapezoidal formula in order to find the area under a curve. I wanted to know how you derive the rule. I'm taking calculus AB and although my AB textbook mentions Simpson and his formula, it doesn't go into detail about how the rule was derived. Date: 06/07/98 at 18:11:06 From: Doctor Pat Subject: Re: Calculus - Simpson's Rule Carol, I don't know that Simpson's rule is "easier," but it is a quadratic rather than a linear approach, and therefore should be a little more accurate. Anyway, here is the derivation for the "rule." The Simpson method makes a parabola through three consecutive points, so we begin with the general equation of a parabola: y = Ax^2 + Bx + C and the area from x = -h to x = h is given by: Area = INT(-h to h) [Ax^2 + Bx + C] dx which is: Ax^3 Bx^2 ------ + ------ + Cx from -h to h 3 2 This simplifies to (2Ah^3)/3 + 2Ch and by some really tricky factoring becomes: (h/3)(2Ah^2 +6C) Now using the fact that the curve goes through (-h,y0), (0,y1), and (h,y2), we can substitute and get: y0 = Ah^2 - Bh + C y1 = C y2 = Ah^2 + Bh + C giving: C = y1 and: Ah^2 - Bh = y0 - y1 Ah^2 + Bh = y2 - y1 ---------------------- 2 Ah^2 = y0 + y2 - 2y1 Now we plug these into the Area integral above and get: Area = (h/3)[2Ah^2 + 6C] = (h/3)[y0 + y2 - 2y1 + 6y1] = (h/3)[y0 + 4y1 + y2] Simpson's rule just applies this formula to successive pieces of the curve like this: (h/3)[ y0 + 4y1 + y2] + (h/3)[ y2 + 4y3 + y4] + (h/3)[ y4 + 4y5 + y6] Now when we factor h/3 and combine the terms under addition we get: (h/3)[y0 + 4y1 + 2y2 + 4y3 + 2y4 + 4y5 + y6] and the deed is done. Hope that was clear. Maybe a good book in the library would have some pictures to help. Thanks for writing. -Doctor Pat, The Math Forum http://mathforum.org/dr.math/ |
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