Chem Differential Eq HW Solutions Fall 2011 19

Chem Differential Eq HW Solutions Fall 2011 19 - Section...

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Section 2.6 Complex Form of Fourier Series 19 If m ± = n , then 1 2 p ± p - p e i p x e - i p x dx = 1 2 p ± p - p e i ( m - n ) π p x dx = - i 2( m - n ) π e i ( m - n ) π p x ² ² ² p - p = - i 2( m - n ) π ³ e i ( m - n ) π - e - i ( m - n ) π ´ = - i 2( m - n ) π (cos[( m - n ) π ] - cos[ - ( m - n ) π ]) = 0 . 13. (a) At points of discontinuity, the Fourier series in Example 1 converges to the average of the function. Consequently, at x = π the Fourier series converges to e + e - 2 = cosh( ). Thus, plugging x = π into the Fourier series, we get cosh( )= sinh( πa ) π µ n = -∞ ( - 1) n a 2 + n 2 ( a + in ) =( - 1) n ¶·¸¹ e inπ = sinh( πa ) π µ n = -∞ ( a + in ) a 2 + n 2 . The sum n = -∞ in a 2 + n 2 is the limit of the symmetric partial sums i N µ n = - N n a 2 + n 2 =0 . Hence n = -∞ in a 2 + n 2 = 0 and so cosh( )= sinh( πa ) π µ n = -∞ a a 2 + n 2 coth( )= a π µ n = -∞ 1 a 2 + n 2 , upon dividing both sides by sinh( ). Setting t = ,weget coth t = t π 2 µ n = -∞ 1 ( t π ) 2 + n 2 = µ n = -∞ t t 2 +( πn ) 2 , which is (b). Note that since a is not an integer, it follows that t is not of the form kπi , where
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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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