Chem Differential Eq HW Solutions Fall 2011 18

Chem Differential Eq HW Solutions Fall 2011 18 - (-1) n ²...

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18 Chapter 2 Fourier Series Solutions to Exercises 2.6 1. From Example 1, for a ± =0 , ² i, ² 2 i, ² 3 i,. .. , e ax = sinh πa π ± n = -∞ ( - 1) n a - in e inx ( - π<x<π ); consequently, e - ax = sinh πa π ± n = -∞ ( - 1) n a + in e inx ( - π<x<π ) , and so, for - π<x<π , cosh ax = e ax + e - ax 2 = sinh πa 2 π ± n = -∞ ( - 1) n ² 1 a + in + 1 a - in ³ e inx = a sinh πa π ± n = -∞ ( - 1) n n 2 + a 2 e inx . 2. From Example 1, for a ± =0 , ² i, ² 2 i, ² 3 i,. .. , e ax = sinh πa π ± n = -∞ ( - 1) n a - in e inx ( - π<x<π ); consequently, e - ax = sinh πa π ± n = -∞ ( - 1) n a + in e inx ( - π<x<π ) , and so, for - π<x<π , sinh ax = e ax - e - ax 2 = sinh πa 2 π ± n = -∞
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Unformatted text preview: (-1) n ² 1 a-in-1 a + in ³ e inx = i sinh πa π ∞ ± n =-∞ (-1) n n n 2 + a 2 e inx . 5. Use identities (1); then cos 2 x + 2 sin 3 x = e 2 ix + e-2 ix 2 + 2 e 3 ix-e-3 ix 2 i = ie-3 ix + e-2 ix 2 + e 2 ix 2-ie 3 ix . 9. If m = n then 1 2 p ´ p-p e i mπ p x e-i nπ p x dx = 1 2 p ´ p-p e i mπ p x e-i mπ p x dx = 1 2 p ´ p-p dx = 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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