Chem Differential Eq HW Solutions Fall 2011 12

Chem Differential Eq HW Solutions Fall 2011 12 - smooth...

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12 Chapter 2 Fourier Series and f o ( x )= f ( x ) - f ( - x ) 2 = ± - 1i f - 1 <x< 0 , 1i f 0 <x< 1 . Note that, f e ( x )= | x | for - 1 <x< 1. The Fourier series of f is the sum of the Fourier series of f e and f o . From Example 1 with p =1 , f e ( x )= 1 2 - 4 π 2 ² k =0 1 (2 k +1) 2 cos[(2 k +1) πx ] . From Exercise 1 with p =1 , f o ( x )= 4 π ² k =0 1 2 k +1 sin[(2 k +1) πx ] . Hence f ( x )= 1 2 + 4 π ² k =0 ³ - cos[(2 k +1) πx ] π (2 k +1) 2 + sin[(2 k +1) πx ] 2 k +1 ´ . 25. Since f is 2 p -periodic and continuous, we have f ( - p )= f ( - p +2 p )= f ( p ). Now a ± 0 = 1 2 p µ p - p f ± ( x ) dx = 1 2 p f ( x ) p - p = 1 2 p ( f ( p ) - f ( - p )) = 0 . Integrating by parts, we get a ± n = 1 p µ p - p f ± ( x ) cos nπx p dx = 1 p =0 · ¸¹ º f ( x ) cos nπx p p - p + p b n · ¸¹ º 1 p µ p - p f ( x ) sin nπx p dx = p b n . Similarly, b ± n = 1 p µ p - p f ± ( x ) sin nπx p dx = 1 p =0 · ¸¹ º f ( x ) sin nπx p p - p - p a n · ¸¹ º 1 p µ p - p f ( x ) cos nπx p dx = - p a n .
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Unformatted text preview: smooth derivative. We have f ( x ) = -c d if 0 &lt; x &lt; d, i f d &lt; | x | &lt; p, c d if-d &lt; x &lt; . The Fourier series of f is obtained by dierentiating term by term the Fourier series of f (by Exercise 26). Now the function in this exercise is obtained by multiplying f ( x ) by-d c . So the desired Fourier series is-d c f ( x ) =-d c 2 cp d 2 n =1 1-cos dn p n 2 -n p sin n p x = 2 n =1 1-cos dn p n sin n p x....
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