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### hmwk_5_2010_solutions

Course: CE 30125, Fall 2010
School: Notre Dame
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Word Count: 1462

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30125 CE Computational Methods: Assignment 5 December 8, 2010 Problem 1 A fourth order accurate central approximation to the second derivative (p=2, n=4) requires p + n - 1 = 2 + 4 - 1 = 6 - 1 = 5 nodes. The second central derivative can therefore be approximate to O(h4 ) as: (2) fi E = 1 fi2 + 2 fi1 + 3 fi + 4 fi+1 + 5 fi+2 h2 T.S. expansions about xi are: (1) fi2 = fi 2hfi (2) + 2h2 fi (1) + (1) +...

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30125 CE Computational Methods: Assignment 5 December 8, 2010 Problem 1 A fourth order accurate central approximation to the second derivative (p=2, n=4) requires p + n - 1 = 2 + 4 - 1 = 6 - 1 = 5 nodes. The second central derivative can therefore be approximate to O(h4 ) as: (2) fi E = 1 fi2 + 2 fi1 + 3 fi + 4 fi+1 + 5 fi+2 h2 T.S. expansions about xi are: (1) fi2 = fi 2hfi (2) + 2h2 fi (1) + (1) + fi1 = fi hfi fi+1 = fi + hfi (1) fi+2 = fi + 2hfi 4h3 (3) 2h4 (4) 4h5 (5) 4h6 (6) f+ f f+ f + O(h7 ) 3i 3i 15 i 45 i h2 (2) h3 (3) h4 (4) h5 (5) h6 (6) fi fi + fi fi + f + O(h7 ) 2 6 24 120 720 i fi = fi h2 (2) h3 (3) h4 (4) h5 (5) h6 (6) fi + fi + fi + fi + f + O(h7 ) 2 6 24 120 720 i (2) + 2h2 fi + 4h3 (3) 2h4 (4) 4h5 (5) 4h6 (6) f+ f+ f+ f + O(h7 ) 3i 3i 15 i 45 i Substituting into our assumed form of fi (2) and re-arranging: (2) fi E = 1 fi2 + 2 fi1 3 fi + 4 fi+1 + 5 fi+2 h2 (1 + 2 + 3 + 4 + 5 ) (21 2 + 4 + 25 ) (1) 2 4 (2) fi + fi + (21 + + + 25 )fi 2 h h 2 2 41 2 4 45 21 2 4 25 2 (4) 41 2 4 45 3 (5) (3) +( + + )hfi + ( + + + )h fi + ( + + )h fi 3 6 6 3 3 24 24 3 15 120 120 15 41 2 4 45 4 (6) +( + + + ) h fi 45 720 720 45 = We want fi (2) and 4th order accuracy, so we note that the coecients must satisfy the following equalities: 1 + 2 + 3 + 4 + 5 = 0 21 2 + 4 + 25 = 0 2 4 21 + + + 25 = 1 2 2 41 2 4 45 + + =0 3 6 6 3 21 2 4 25 + + + =0 3 24 24 3 1 We can then write this system of equations in matrix form AX = B 0 1 1 1 1 1 1 2 1 0 1 2 2 0 2 0.5 0 0.5 2 3 = 1 1.33333 0.16667 0 0.16667 1.33333 4 0 0 5 0.66667 0.041667 0 0.041667 0.66667 1 Solving this equation in Matlab (see asgn5 prob1.m) yields 1 = - 12 , 2 = 4 , 3 = - 5 , 4 = 4 , and 3 2 3 1 5 = - 12 . The equation for our fourth order accurate central approximation to the second derivative now becomes 1 (6) (1) (2) (3) (4) (5) )f + O(h5 ) = (0)fi + (0)fi + (1)fi + (0)fi + (0)fi + (0)fi + ( 90 i The fourth order accurate central approximation to the second derivative is as follows: fi2 + 16fi1 30fi + 16fi+1 fi+2 +E 12h2 where 1 4 (6) E h fi = 90 2 Problem 2 For the third derivative p = 3 and we must use our N+1 = 4 nodes to establish an interpolating polynomial of degree N 3. We let the four points be x0 = 0, x1 = h, x2 = 2h, and x3 = 3h. Using Newton forward interpolation, we then establish the following interpolating polynomial: g ( x) = f0 + 3 f0 2 f0 f0 (x x0 )(x x1 ) + (x x0 )(x x1 )(x x2 ) (x x0 ) + 2 h 2h 6h3 with the following associated error term: e(x) = f (3) ( ) 4 f0 (x x0 )(x x1 )(x x2 )(x x3 ) (x x0 )(x x1 )(x x2 )(x x3 ) = 24 24h4 where f0 = f1 f0 2 f0 = f2 2f1 + f0 3 f0 = f3 3f2 + 3f1 f0 Since we are looking for the third derivatives of the interpolating function and the error term, we note that these terms dierentiate to zero, and simply focus in on the third and fourth order terms. Dierentiating these terms yields: g (1) (x) = 3 f0 3 f0 3 f 0 (x x1 )(x x2 ) + (x x0 )(x x2 ) + (x x0 )(x x1 ) 3 3 6h 6h 6h3 with the following associated error term: e(1) (x) = f (3) ( ) f (3) ( ) (x x1 )(x x2 )(x x3 ) + (x x0 )(x x2 )(x x3 ) 24 24 f (3) ( ) f (3) ( ) (x x0 )(x x1 )(x x3 ) + (x x0 )(x x1 )(x x2 ) 24 24 Dierenting a second time yields: + g (2) (x) = 3 f 0 3 f0 3 f 0 (x x1 ) + ( x x2 ) + (x x0 ) 6h3 6h3 6h3 3 f0 3 f 0 3 f0 ( x x2 ) + (x x0 ) + ( x x1 ) 3 3 6h 6h 6h3 with the following associated error term: + e(2) (x) = f (3) ( ) f (3) ( ) f (3) ( ) (x x2 )(x x3 ) + (x x1 )(x x3 ) + (x x1 )(x x2 ) 24 24 24 + f (3) ( ) f (3) ( ) f (3) ( ) (x x2 )(x x3 ) + (x x0 )(x x3 ) + (x x0 )(x x2 ) 24 24 24 + f (3) ( ) f (3) ( ) f (3) ( ) (x x1 )(x x3 ) + (x x0 )(x x3 ) + (x x0 )(x x1 ) 24 24 24 f (3) ( ) f (3) ( ) f (3) ( ) (x x1 )(x x2 ) + (x x0 )(x x2 ) + (x x0 )(x x1 ) 24 24 24 Finally, dierentiating a third time and evaluating at x = x3 gives us the backward approximations to the third derivative and the error associated with the formula. + g (3) (x) = 6 3 f0 f3 3f2 + 3f1 f0 3 f0 = = 3 3 6h h h3 3 Dierentiating the error term one more time leads to e(3) (x) = f (3) ( ) f (3) ( ) f (3) ( ) f (3) ( ) (x x3 ) + (x x2 ) + ( x x3 ) + (x x1 ) 24 24 24 24 + f (3) ( ) f (3) ( ) f (3) ( ) f (3) ( ) (x x2 ) + (x x1 ) + ( x x3 ) + (x x2 ) 24 24 24 24 + f (3) ( ) f (3) ( ) f (3) ( ) f (3) ( ) (x x0 ) + (x x3 ) + ( x x2 ) + (x x0 ) 24 24 24 24 + f (3) ( ) f (3) ( ) f (3) ( ) f (3) ( ) (x x3 ) + (x x1 ) + ( x x3 ) + (x x0 ) 24 24 24 24 + f (3) ( ) f (3) ( ) f (3) ( ) f (3) ( ) (x x1 ) + (x x0 ) + ( x x2 ) + (x x1 ) 24 24 24 24 f (3) ( ) f (3) ( ) f (3) ( ) f (3) ( ) (x x2 ) + (x x0 ) + ( x x1 ) + (x x0 ) 24 24 24 24 The error term simplies to the following: + e(3) = f (x) (3) ( ) f (3) ( ) f (3) ( ) f (3) ( ) (x x0 ) + (x x1 ) + ( x x2 ) + (x x3 ) 4 4 4 4 f (3) ( ) [(x x3 ) + (x x2 ) + (x x1 ) + (x x0 )] 4 We then evaluate this at x3 in order to get the error associated with the backward approximation: = e(3) (x3 ) = f (3) ( ) f (3) ( ) 3f (3) ( ) [(0) + (h) + (2h) + (3h)] = [6h] = h 4 4 2 Finally, we re-index so that the node j corresponds with x3 d3 f dx3 = xj fj 3fj 1 + 3fj 2 fj 3 h3 3 (3) E fj h = 2 4 Problem 3 For this problem, we evaluate the approximations for f(4) and veried them using a Python script (see prob 3.py ). (a) 3 (fi ) 3 (fi+1 fi ) 2 (fi+1 fi ) 2 (fi+2 2fi+1 + fi )) (fi+2 2fi+1 + fi )) 4 f i = = = = = 4 4 4 4 h h h h h4 h4 = (fi+3 fi+2 2fi+2 + 2fi+1 + fi+1 fi ) (fi+3 3fi+2 + 3fi+1 fi ) fi+3 3fi+2 + 3fi+1 fi = = h4 h4 h4 fi+4 fi+3 3fi+3 + 3fi+2 + 3fi+2 3fi+1 fi+1 + fi fi+4 4fi+3 + 6fi+2 4fi+1 + fi = = h4 h4 (b) 3 f i 3 (fi fi1 ) 2 (fi fi1 ) 2 (fi+1 2fi + fi1 ) (fi+1 2fi + fi1 ) = = = = 4 4 4 4 h h h h h4 (fi+2 3fi+1 + 3fi fi1 ) (fi+2 3fi+1 + 3fi fi1 ) = 4 h h4 fi+3 fi+2 3fi+2 + 3fi+1 + 3fi+1 3fi fi + fi1 fi+3 4fi+2 + 6fi+1 4fi + fi1 = = h4 h4 = (c) 2 2 fi = h4 2 (fi+1 fi ) = h4 2 (fi+1 fi ) h4 ( fi+2 2 fi+1 + fi ) (fi+2 2fi+1 + fi ) = 4 h h4 fi+2 4fi+1 + 6fi 4fi1 + fi2 (fi+2 3fi+1 + 3fi fi1 ) = 4 h h4 = = 2 5 Problem 4 For this problem we are working with the following function: f (x) = 3x5 4x3 + x 6 which has the following derivatives: f (1) (x) = 15x4 12x2 + 1 f (2) (x) = 60x3 24x f (3) (x) = 180x2 24 (a) The matlab script computes all the values asked for in the homework assignment. The results are shown in Table 1. h 1.0 0.5 0.1 0.05 0.01 f2 60.000000 -6.000000 -5.831040 -5.903970 -5.980032 f1 -6.000000 -5.906250 -5.903970 -5.950499 -5.990004 f0 -6.000000 -6.000000 -6.000000 -6.000000 -6.000000 f0 (1) approx -33.000000 0.375000 1.075800 1.019739 1.000800 Eactual 34.000000 0.625000 -0.075800 -0.019738 -0.000800 E1 -8.000000 -2.000000 -0.080000 -0.020000 -0.00080 Table 1: Functional values, approximations, and errors for problem 4 (b) (i) As can be seen in Figure 1, the estimated error curve has a slope of 2 on the log-log plot, and the actual error curve generally has a slope of 2. The slope of the estimated error makes sense since taking logs of both sides and then rearranging yields the following: 1 1 (3) 1 (3) (3) log(E1 ) = log( h2 f0 ) = log( f0 ) + log(h2 ) = 2 log(h) + log( f0 ) 3 3 3 1 (3) log(E1 ) = 2 log(h) + log( f0 ) 3 Since log(E1 ) is on the y-axis and log(h) is on the x-axis in the plot, this equation is in the form y = mx + b. The slope (m) is 2 which is what we see in the plot. Simply put, we expect a second order accurate formula to result in a straight line with a slope of 2 on the log-log plot. (ii) When we have larger h, we see that the two curves may dier. This is because the leading order truncated error term may not dominate the rest of the Taylor series in this case since the higher order terms may be non-negligible. We would have to inspect the actual functional values, and their derivatives in relation to grid spacing, to understand why the actual error is worse than expected as h approaches 1, but is actually better than expected for some range. Regardless, we see that as h becomes suciently small, the actual error approaches the estimated error since the error term eventually dominates the rest of the series. 6 Figure 1: Error Curves. 7 Figure 2: Molecule for problem 5 part (a). Problem 5 (a) For this problem, we use the fourth-order accurate central dierence approximation to the rst derivative: 1 4 (6) fi+2 + 16fi+1 30fi + 16fi1 fi2 (2) + E, E h fi fi = = 2 12h 90 With this, we have the following partial derivatives with respect to x and y: fi+2,j + 16fi+1,j 30fi,j + 16fi1,j fi2,j 2f = x2 12x2 2f fi,j +2 + 16fi,j +1 30fi,j + 16fi,j 1 fi,j 2 = y 2 12y 2 We then consider Laplaces equation and x = y = h 2 f= 2f 2f +2 x2 y 1 (fi+2,j + 16fi+1,j + 16fi1,j fi2,j fi,j +2 + 16fi,j +1 + 16fi,j 1 fi,j 2 60fi,j ) 12h2 This equation can then be represented by the molecule shown in Figure 2. (b) By simply adjusting the i and j values in the above equation, we can derive the following two approximations: 2 2 fi,j = fi1,j = 1 (fi+1,j +16fi,j +16fi2,j fi3,j fi1,j +2 +16fi1,j +1 +16fi1,j 1 fi1,j 2 60fi1,j ) 12h2 1 (fi+4,j +1 +16fi+3,j +1 +16fi+1,j +1 fi,j +1 fi+2,j +3 +16fi+2,j +2 +16fi+2,j fi+2,j 1 60fi+2,j +1 ) 12h2 These approximations can be represented by the molecules shown in Figure 3. With this, we can see that no nodal values would be shared in both approximations. 2 fi+2,j +1 = 8 Figure 3: Molecule for problem 5 part (b). Red molecule is for 9 2 fi1,j . Blue molecule is for 2 fi+2,j +1 .
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Notre Dame - CE - 30125
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Notre Dame - CE - 30125
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Notre Dame - ENGL - 20118
Introduction II: Potted History of the Fairy TaleFairy Tales from Around the WorldThe Literary Fairy Tale in EuropeItaly:Giovan Straparola (1480-1558)Giambattista Basile (1575-1632)France:Madame dAulnoy (1650-1705)Charles Perrault (1628-1703)[Ris
Notre Dame - ENGL - 20118
OED: Feminism: 1. The qualities of females.2. [After F. fminisme.] Advocacy of the rights of women (based on the theory of equality of thesexes).(term coined in the 19c.)Proto-Feminism (pre-19c.): Concerned primarily with questions of womens morality
Notre Dame - PHIL - 10100
Philosophy 101: IntroductionTo PhilosophySpring 2005Professor RamseyCourse ObjectivesIntroduction to Central Themes Repository For Unanswered Questions Is There A God? What Makes An Action Right? Do We Have Free Will? How Do We Know What Is Real
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Philosophical MethodLogic: A Calculus For Good ReasonClarification, Not Obfuscation Distinctions and DisambiguationExamples and Counterexamples Revealing Our Deepest Convictions Testing Our Principles and DefinitionsLogic: Primary PhilosophicalToo
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PHILOSOPHY OFRELIGIONPreliminary Issues: Agreement vs. Tolerance Different Religions ARE Incompatible Religious Claims Arent True FOR Individuals Reason and Faith Unusual Standards For Belief Recent Trends Go Against Western Tradition Classic Tri
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Objections And RepliesGaunilos Objection: Argument TooStrong Proving The Existence Of The Perfect Island First Reply: Argument Concerns OnlyThings In General (Not Any Specific Thing) Second Reply: Fully Perfect Island NotPossible But What About A
Notre Dame - PHIL - 10100
The Cosmological ArgumentAquinas: 1225-1274; Clarke: 1675-1729Background Sources Of Explanation: Three Options: Explained By a) Other, b) Nothing, c) Self Principle of Sufficient Reason Individual Things &amp; Events Need AnExplanation Positive Facts
Notre Dame - PHIL - 10100
The Teleological Argument Aquinas, Paley (1743-1805)The Argument: Two Ways To View It:First Way: Argument By Analogy 1. Aspects Of Natural World Are LikeMachines 2. Machines Are Produced By IntelligentDesign 3. Therefore, Aspects Of Natural World
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Theodicy And The Problem OfEvilThe Argument Against Western Theism:Reason To Doubt That A Christian GodExists1. Christianity Assumes God Is Omniscient,Omnipotent, Perfectly Good, And Loves Us2. Massive Evil Exists Moral Evil (Suffering Caused By U
Notre Dame - PHIL - 10100
Responses To The ArgumentAnd RebuttalsFirst Response: Challenge (2, 5) Denies Suffering Is Real Rebuttals: Makes God A Deceiver Hard To Take SeriouslySecond Response: Challenge Premise (3)In Many WaysChallenging InconsistencyBetween God And Evil