FinalSol - accuracy is not enough to oﬀset the superior...

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Unformatted text preview: accuracy is not enough to oﬀset the superior accuracy of the ﬁve-point NewtonCotes rule on the original interval. (c ) For an n-point interpolatory quadrature rule Qn , we have the general error bound 1 E ≤ hn+1 ￿f (n) ￿∞ . 4 The comparable bound for the rule Q2n−1 on an interval of twice the length is COT 4501 E≤ 1 (2h)2n ￿f (2n−1) ￿∞ . 4 Solutions for Exam Review Problems Again, provided the relevant derivatives of f are well behaved, we see that the (Chapter 8 and 9) higher-degree rule is more accurate than the lower-degree rule applied twice on intervals of half the size. 8.12. Averaging the ﬁrst-order accurate forward and backward diﬀerence formulas, we obtain ￿ ￿ f (x + h) − f (x − h) 1 f (x + h) − f (x) f (x) − f (x − h) + = , 2 h h 2h which is the centered diﬀerence formula we already know to be second-order accurate from the analysis in Section 8.6.1. 8.13. In the Taylor series expansion f ￿￿ (x) 2 f ￿￿￿ (x) 3 f (x + h) = f (x) + f (x)h + h+ h + ···, 2 6 ￿ we solve for f ￿ (x) to obtain f ￿ (x) = f ￿￿￿ (x) 2 f (x + h) − f (x) f ￿￿ (x) − h− h + ···. h 2 6 Similarly, in the Taylor series expansion f ￿￿￿ (x) f ￿￿ (x) (2h)2 + (2h)3 + · · · 2 6 4f ￿￿￿ (x) 3 = f (x) + 2f ￿ (x)h + 2f ￿￿ (x)h2 + h + ···, Chapter 8: Numerical Integration and Diﬀerentiation 3 f (x + 2h) = f (x) + f ￿ (x)(2h) + 104 we solve for f ￿ (x) to obtain f ￿ (x) = 2f ￿￿￿ (x) 2 f (x + 2h) − f (x) − f ￿￿ (x)h − h + ···. 2h 3 If we now subtract the second of these series for f ￿ (x) from twice the ﬁrst, we obtain f ￿ (x) = −3f (x) + 4f (x + h) − f (x + 2h) f ￿￿￿ (x) 2 + h + ···, 2h 3 so that −3f (x) + 4f (x + h) − f (x + 2h) 2h is a second-order accurate, one-sided diﬀerence approximation to f ￿ (x). f ￿ (x) ≈ 8.14. F (0) = F (h) + F (h) − F (h/2) −0.8333 + 0.9091 = −0.8333 + = −0.9849. −1 − 1 2 0.5 − 1 8.15. (a ) The power series expansion for the sine function is 2h 3 so that −3f (x) + 4f (x + h) − f (x + 2h) 2h is a second-order accurate, one-sided diﬀerence approximation to f ￿ (x). f ￿ (x) ≈ 8.14. F (0) = F (h) + F (h) − F (h/2) −0.8333 + 0.9091 = −0.8333 + = −0.9849. −1 − 1 2 0.5 − 1 8.15. (a ) The power series expansion for the sine function is sin(x) = x − so pn x5 x7 x3 + − + ···, 3! 5! 7! ￿ ￿ π π3 π5 π7 π3 π5 π7 =n − 3 + 5 − 7 + ··· = π − 2 + 4 − 6 + ··· n n 3! n 5! n 7! n 3! n 5! n 7! 5 7 3 π2π4π6 h+ h− h + ···, = π− 3! 5! 7! where h = 1/n, and hence we have a0 = π . Similarly, the power series expansion Exercises for the tangent function is x3 2x5 17x7 + + + ···, 3 15 315 u￿ = u2 , so 1 ￿ u￿ π 7= u3 ,￿ 2 π3 2π 5 17 π3 2π 5 17π 7 π +3+5 +7 + ··· = π + 2 + 4 + 6 + ··· qn = n ￿ u= n n 3 n 15 n3 315 −u1 u3 . n 3 n 15 n 315 3 5 7 (c ) = π + π h2 + 2π n4 + 17π n6 + · · · , 3 15 ￿ 315 u1 = u2 , and hence we have b0 = π￿. (b ) Using Richardson 2extrapolation with p = 2 and u2 = −GM u1 /(u2 + u3 )3/2 , 1 q = 2, we have ￿ u3 = u4 , F (1/6) − F (1/12) 3.0000 − 3.1058 ￿ u−2 = −GM u3 /0000 + 2 )3/2 . a0 = F (1/6) + = 3. (u2 + u3 = 3.1411. 4 1 2 −1 0.25 − 1 (b ) tan(x) = x + 9.3. Yes, since thisF (1/6) − F (1/12) homogeneous4641 − 3.with constant coeﬃﬁrst-order, linear, 3. system 2154 b0 = F (1written /6) + = 3.4641 + = 3.1325. cients can be 2−2 − ￿ ￿ 1 0.25 − 1 ￿￿ ￿￿ ￿ y1 −1 1 y1 ￿ y= ￿ = = Ay , y2 0 −2 y2 and the eigenvalues of A, −1 and −2, are both negative. 9.4. (a ) Yes, because the eigenvalue, −5, is negative, and hence the solutions decay exponentially. (b ) No, since |1 + hλ| = |1 + 0.5 (−5)| = 1.5 > 1. (c ) y1 = 1 + 0.5(−5) = −1.5. (d ) Yes, since backward Euler is unconditionally stable. (e ) y1 = 1 + 0.5 (−5y1 ) ⇒ y1 = 1/3.5 = 0.2857. 9.5. (a ) y1 = y0 + hf (y0 ) = 1 + 1 · (−1) = 0. (b ) y1 = y0 + hf (y1 ) = 1 + 1 · (−y1 ), so y = 0.5. 1 9.6. The ﬁxed-point iteration procedure is 3 yn+1 = 1 − yn /2, with y0 = 1 (or y0 = 0.5 if forward Euler is used to obtain the starting guess). Note that for the ﬁxed-point iteration function g (y ) = 1 − y 3 /2, we have g ￿ (y ) = 117 Exercises 9.1. (a ) = u2 , = t + u1 + u2 . u￿ 1 u￿ 2 (b ) u￿ 1 u￿ 2 u￿ 3 = u2 , = u3 , = u3 + tu1 . (c ) u￿ 1 u￿ 2 u￿ 3 = u2 , = u3 , = u3 − 2u2 + u1 − t + 1. 9.2. (a ) u￿ 1 u￿ 2 Exercises 116 = u2 , = u2 (1 − u2 ) − u1 . 1 117 (b ) u￿ 1 u￿ 2 u￿ 3 = = = u2 , u3 , −u1 u3 . (c ) u￿ 1 u￿ 2 u￿ 3 u￿ 4 = = = = u2 , −GM u1 /(u2 + u2 )3/2 , 1 3 u4 , −GM u3 /(u2 + u2 )3/2 . 1 3 9.3. Yes, since this ﬁrst-order, linear, homogeneous system with constant coeﬃcients can be written ￿ ￿￿ ￿ ￿￿ ￿ y1 −1 1 y1 ￿ y= = = Ay , ￿ y2 0 −2 y2 and the eigenvalues of A, −1 and −2, are both negative. 9.4. (a ) Yes, because the eigenvalue, −5, is negative, and hence the solutions decay exponentially. (b ) No, since |1 + hλ| = |1 + 0.5 (−5)| = 1.5 > 1. (c ) y1 = 1 + 0.5(−5) = −1.5. (d ) Yes, since backward Euler is unconditionally stable. (e ) y1 = 1 + 0.5 (−5y1 ) ⇒ y1 = 1/3.5 = 0.2857. 9.5. (a ) y1 = y0 + hf (y0 ) = 1 + 1 · (−1) = 0. (b ) y1 = y0 + hf (y1 ) = 1 + 1 · (−y1 ), so y1 = 0.5. ...
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This note was uploaded on 01/22/2012 for the course COT 4501 taught by Professor Davis during the Spring '08 term at University of Florida.

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