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Chem Differential Eq HW Solutions Fall 2011 11

# Chem Differential Eq HW Solutions Fall 2011 11 - π 2 6 =...

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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 k =0 1 (2 k + 1) sin(2 k + 1) πx = 1 2 + 2 π 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 PiSum 1 2k 1 Sin 2k 1 Pix , 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 n =1 ( - 1) n - 1 n 2 π 2 12 = 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 n =1 ( - 1) n - 1 ( - 1) n n 2 π 2 6 = n =1 1 n 2 . Summing over the even and odd integers separately, we get
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Unformatted text preview: π 2 6 = ∞ ² n =1 1 n 2 = ∞ ² k =0 1 (2 k + 1) 2 + ∞ ² k =1 1 (2 k ) 2 . But ∑ ∞ k =1 1 (2 k ) 2 = 1 4 ∑ ∞ k =1 1 k 2 = 1 4 π 2 6 . So π 2 6 = ∞ ² k =0 1 (2 k + 1) 2 + π 2 24 ⇒ ∞ ² k =0 1 (2 k + 1) 2 = π 2 6-π 2 24 = π 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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