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sol03 - MIT OpenCourseWare http/ocw.mit.edu 6.055J 2.038J...

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MIT OpenCourseWare http://ocw.mit.edu 6.055J / 2.038J The Art of Approximation in Science and Engineering Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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6.055J/2.038J (Spring 2008) Solution set 3 Do the following warmups and problems. Due in class on Friday, 14 Mar 2008 . Open universe: Collaboration, notes, and other sources of information are encouraged. However, avoid looking up answers until you solve the problem (or have tried hard). That policy helps you learn the most from the problems. Bring a photocopy to class on the due date , trade it for a solution set, and figure out or ask me about any confusing points. Your work will be graded lightly: P (made a reasonable effort), D (did not make a reasonable effort), or F (did not turn in). Warmups 1. Explain a Unix pipeline What does this pipeline do? ls -t | head | tac [Hint: If you are not familiar with Unix commands, use the man command on Athena or on any nearby Unix or GNU/Linux system.] The ls -t lists the files and subdirectories in a directory ordered by modification time with the most recently modified files at the beginning. The head selects the first ten lines, which means the first ten names. The tac reverses the order of the lines, so the 10th-most-recently-modified file (or subdirectory) comes first, then the 9th-most-recently-modified file, etc. with the most-recently- modified file at the end of the list. 2. Symmetry for algebra Use symmetry to find ( a b ) 3 . The original expression is antisymmetric in a and b : The result changes sign if you swap a and b . The expansion has third-degree terms such as a 3 or a 2 b . One category of third-degree terms is like a 3 and includes a 3 and b 3 . The antisymmetric combination is a 3 b 3 . The other category of third-degree terms is like a 2 b and includes a 2 b and ab 2 . The antisymmetric combination is a 2 b ab 2 . The expansion therefore has the antisymmetric form ( a b ) 3 = A ( a 3 b 3 ) + B ( a 2 b ab 2 ) where A and B are constants to be determined. Setting b = 0 shows that A = 1 , because ( a 0) 3 = A ( a 3 0) + B (0 0) or a 3 = Aa 3 . To find B , think about the naive expansion of ( a b ) 3 . The basic expression a b has two terms, so ( a b ) 3 has eight terms (before collecting like terms). So the absolute values of the coefficients of each term in the form
2 Solution set 3 / 6.055J/2.038J: Art of approximation in science and engineering (Spring 2008) A ( a 3 b 3 ) + B ( a 2 b ab 2 ) have to add to eight. With A = 1 , this requirement shows that B = ± 3 . The choice B = 3 gives the correct sign for the a 2 b term (which has one negative factor from the b ). So ( a b ) 3 = ( a 3 b 3 ) 3( a 2 b ab 2 ) . Problems 3. Highway vs city driving In lecture we derived a measure of how important drag is for a car moving at speed v for a distance d : E drag ρ v 2 Ad .

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