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MIT 18 01 - 18 01 Lecture 18

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18.01 Calculus Jason Starr Fall 2005 Lecture 18. October 25, 2005 Homework. Problem Set 5 Part I: (c). Practice Problems. Course Reader: 3G-1, 3G-2, 3G-4, 3G-5. 1. Approximating Riemann integrals. Often, there is no simpler expression for the antideriva-tive than the expression given by the Fundamental Theorem of Calculus. In such cases, the simplest method to compute a Riemann integral is to use the definition. However, this is not necessarily the most efficient method. Often trapezoids or segments under a parabola give a better approximation to the Riemann integral than do vertical strips.� � � 18.01 Calculus Jason Starr Fall 2005 2. The trapezoid rule. The problem is to find an approximation of the Riemann integral, � b I = ydx a for a function y(x) defined on the interval [a, b]. Choose a partition of the interval [a, b] into n equal subintervals. The points of this partition are, b − a xk = a +(b − a)k, Δxk = . n n The values of these points are, yk = f(xk ). The Riemann sum using always the left endpoint is , nIl = yk−1Δxk. k=1 The Riemann sum using always the right endpoint is, nIr = yk Δxk . k=1 The average of the two is, nItrap = yk−1 + yk Δxk . 2 k=1 This is usually a better approximation than either of the two approximations individually. Part of the reason is that the term (yk−1 + yk )Δxk /2 is the area of the trapezoid containing the points (xk−1, 0), (xk−1, yk−1), (xk , 0) and (xk , yk ). In particular, if the graph of y = f(x) is a line, this trapezoid is precisely the region between the graph and the x-axis over the interval [xk−1, xk ]. Thus, the approximation above gives the exact integral for linear integrands. Writing out the sum gives, Itrap = b − a ((y0 + y1) + (y1 + y2) + (y2 + y3) + ··· + (yn−2 + yn−1) + (yn−1 + yn)). 2n Gathering like terms, this reduces to, I = (b − a)(y0 y1 y2 + yn−1 + yn)/2n.trap + 2 + 2 ··· + 23. Simpson’s rule. Again partition the interval [a, b] into n equal subintervals. For reasons that will become apparent, n must be even. So let n = 2m where m is a positive integer. Again define, (b − a)k (b − a)k b − a b −a xk = a + = a + , Δxk = = . n 2m n 2m� � � �� � 18.01 Calculus Jason Starr Fall 2005 Pair off the intervals as ([x0, x1], [x1, x2]), ([x2, x3], [x3, x4]), etc. Thus the lth pair of intervals is, ([x2l−2, x2l−1], [x2l−1, x2l]). The idea is to approximate the area of the graph over the pair of intervals by the area under the unique parabola containing the 3 points (x2l−2, y2l−2), (x2l−1, y2l−1), (x2l, y2l). For notation’s sake, denote 2l − 1 by k. Thus the 3 points are (xk−1, yk−1), (xk , yk ), and (xk+1, yk+1) (this is slightly more symmetric). The first problem is to find the equation of this parabola. Since the parabola contains the point (xk , yk ), it has the equation, y = A(x − xk )2 + B(x − xk ) + yk , Plugging in x = xk−1 and x = xk+1, and using that xk+1 − xk = xk − xk−1 equals Δx, yk+1 = A(Δx)2 + B(Δx) + yk , yk−1 = A(Δx)2 − B(Δx) + yk. Summing the two sides gives, yk+1 + yk−1 = 2A(Δx)2 + 2yk . Solving for A gives, 1 A = 2(Δx)2 (yk−1 − 2yk + yk+1). Similarly, taking the difference of the two sides gives, yk+1 − yk−1 = 2B(Δx). Solving for B gives, 1 B = 2(Δx)(yk+1 − yk−1). Thus, the equation of the parabola passing through (xk−1, yk−1), (xk , yk ) and (xk+1, yk+1) is, y = A(x − xk )2 + B(x − xk )2 + yk , A yk−1 − 2yk + yk+1)/ x)2 , B yk+1 − yk−1)/ x). = ( 2(Δ= ( 2(ΔThe next problem is to compute the area under the parabola from x = xk−1 to x = xk+1. This is a straightforward application of the Fundamental Theorem of Calculus, xk+1 A B A(x − xk )2 + B(x − xk ) + yk dx = (x − xk )3 + (x − xk )2 + yk (x − xk )3 2 xk+1 . xk−1 xk−118.01 Calculus Jason Starr Fall 2005 Plugging in and using that xk+1 − xk = xk − xk−1 equals Δx, this is, 2A (Δx)3 + 2yk (Δx). 3 Substituting in the formula for A and simplifying, this is, Δx Δx Δx (yk−1 − 2yk + yk+1) + (6yk ) = (yk−1 + 4yk + yk+1). 3 3 3 Back-substituting 2l −1 for k and (b −a)/2m for Δx, the approximate area for the pair of intervals [x2l−2, x2l−2] and [x2l−1, x2l] is, ΔIl = b − a (y2l−2 + 4y2l−1 + y2l). 6m Finally, summing this contribution over each choice of l gives the Simpson’s rule approximation, mb − a � ISimpson = (y2l−2 + 4y2l−1 + y2l). 6m l=1 Writing out the sum gives, b−aISimpson = ((y0 + 4y1 + y2) + (y2 + 4y3 + y4) + (y4 + 4y5 + y6)+6m ··· + (y2m−4 + 4y2m−3 + y2m−2) + (y2m−2 + 4y2m−1 + y2m)). Gathering like terms, ISimpson reduces to, (b − a)(y0 y1 y2 y3 y4 y5 y6 + y2m−3 y2m−2 y2m−1 + y2m)/6m.+ 4 + 2 + 4 + 2 + 4 + 2 ··· + 4 + 2 + 4Example. Approximate ln(2) using a partition into 4 equal subintervals with the Trapezoid Rule and with Simpson’s Rule. The value ln(2) equals the Riemann integral, � 2 1 dx. 1 x The points of the partition are x0 = 4/4, x1 = 5/4, x2 = 6/4, x3 = 7/4 and x4 = 8/4. The corresponding values are y0 = 4/4, y1 = 4/5, y2 = 4/6, y3 = 4/7, y4 = 4/8. Thus the Trapezoid Rule gives, 1 4 4 4 4 4 1171 Itrap = b − a (y0 + 2y1 + 2y2 + 2y3 + y4) = ( + 2 + 2 + 2 + ) =1680 ≈2n 8 4 5 6 7 8 For Simpson’s Rule, because n equals 4, m equals 2. Thus, 1 4 4 4 4 4 1747 0.6970 ISimpson = b − a (y0 + 4y1 + 2y2 + 4y3 + y4) = ( + 4 + 2 + 4 + ) =2520 ≈6m 12 4 5 6 7 8 0.693318.01 Calculus Jason Starr Fall 2005 According to a calculator, the true value is, ln(2) = Note that trapezoids overestimate the area, because 1/x is concave up. The approximating parabo-las cross the graph of y = 1/x, thus the underestimation to the left of (xk , yk ) somewhat cancels the overestimation to the right of (xk , yk ), explaining the better approximation. 0. ± 10−46931 4. One review problem. This is a related rates review problem for Exam …


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MIT 18 01 - 18 01 Lecture 18

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