Chem Differential Eq HW Solutions Fall 2011 11

Chem Differential Eq HW Solutions Fall 2011 11 - 2 6 = n =1...

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Section 2.3 Fourier Series of Functions with Arbitrary Periods 11 Thus the Fourier series of g is 1 2 ± 1+ 4 π ² k =0 1 (2 k +1) sin(2 k +1) πx ³ = 1 2 + 2 π ² k =0 1 (2 k +1) sin(2 k +1) πx. f x_ Which x 0, 0, 0 x 1, 1, x 1, 0 s n_, x_ 1 2 2 Pi Sum 1 2k 1 Sin 2k 1 Pi x , k, 0, n ; Plot Evaluate f x ,s 20, x , x, 1, 1 Which x 0, 0, 0 x 1, 1, x 1, 0 The 41st partial sum of the Fourier series and the function on the interval (-1, 1). 17. (a) Take x = 0 in the Fourier series of Exercise 4 and get 0= p 2 3 - 4 p 2 π 2 ² n =1 ( - 1) n - 1 n 2 π 2 12 = ² n =1 ( - 1) n - 1 n 2 . (b) Take x = p in the Fourier series of Exercise 4 and get p 2 = p 2 3 - 4 p 2 π 2 ² n =1 ( - 1) n - 1 ( - 1) n n 2 π 2 6 = ² n =1 1 n 2 . Summing over the even and odd integers separately, we get
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Unformatted text preview: 2 6 = n =1 1 n 2 = k =0 1 (2 k + 1) 2 + k =1 1 (2 k ) 2 . But k =1 1 (2 k ) 2 = 1 4 k =1 1 k 2 = 1 4 2 6 . So 2 6 = k =0 1 (2 k + 1) 2 + 2 24 k =0 1 (2 k + 1) 2 = 2 6- 2 24 = 2 8 . 21. From the graph, we have f ( x ) = -1-x if-1 < x < , 1 + x if 0 < x < 1 . So f (-x ) = 1-x if-1 < x < ,-1 + x if 0 < x < 1; hence f e ( x ) = f ( x ) + f (-x ) 2 = -x if-1 < x < , x if 0 < x < 1 ,...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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