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527solns8 - 642:527 SOLUTIONS ASSIGNMENT 8 FALL 2009...

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642:527 SOLUTIONS: ASSIGNMENT 8 FALL 2009 Section 9.9: 4 (a) See text solution. (e) The formula is (23) of the text or (22) of the posted lecture notes. We have bardbl e 1 bardbl 2 = ( e 1 , e 1 ) (= e 1 · e 1 ) = 30, bardbl e 2 bardbl 2 = 6, bardbl e 3 bardbl 2 = 1, and bardbl e 4 bardbl 2 = 5; further, ( e 1 , u ) = 23, ( e 2 , u ) = 7, ( e 3 , u ) = 2, and ( e 4 , u ) = 1, so u = ( 23 / 30) e 1 + (7 / 6) e 2 + 2 e 3 + (1 / 5) e 4 . Section 9.10: 2 (a) See text solution. (c) With u = (3 , 0 , 1 , 4 , 1) we have ( ˆ e 1 , u ) = 5 / 5, ( ˆ e 2 , u ) = 6 / 6, ( ˆ e 3 , u ) = 4, so the best approximation is 3 summationdisplay i =1 ( ˆ e i , u ) ˆ e i = (1 , 0 , 2 , 0 , 0) + (2 , 0 , 1 , 0 , 1) + 4(0 , 0 , 0 , 1 , 0) = (3 , 0 , 1 , 4 , 1) = u . The best approximation is u itself, which just means that u is itself a linear combination of the vectors { ˆ e 1 , ˆ e 2 , ˆ e 3 } . 3. See text solution. Section 17.2: 5 To test whether a function f ( x ) is even, odd, or neither, one just looks at f ( x ); if f ( x ) = f ( x ) then f is even, and if f ( x ) = f ( x ) then f is odd. Of the functions given in this problem, (f), (h), and (k) are even and (j) is odd; the rest are neither. 12. It is not so easy to give a simple rule for testing for periodicity, but as a guideline, remember that the only elementary functions that are periodic are the trig functions and constant functions, so if f ( x ) does not involve trig functions it is probably not periodic. In this problem the periodic functions, with their elementary periods, are: (d) 2 π/ω ; (e) π/ 3; (f) π ; (g) π ; (j) π ; (k) π ; (l) 2 π ; (m) 2 π/ 3; (n) π ; (o) 2 π ; (p) π ; (q) 2 π ; (r) π/ 2. These are
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