Chem Differential Eq HW Solutions Fall 2011 120

# Chem Differential Eq HW Solutions Fall 2011 120 - 120...

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120 Chapter 7 The Fourier Transform and its Applications with f ( s )= 1 2 t e - ( x - s ) 2 t and F ± 1 2 t e - ( x - s ) 2 t ² ( w e - ( iwx + w 2 t ) (see the solution of Exercise 5). So u ( x, t )=( i ) n ³ d n dw n e - ( iwx + w 2 t ) ´ w =0 . To compute this last derivative, recall the Taylor series formula f ( x µ n =0 f ( n ) ( a ) n ! ( x - a ) n . So knowledge of the Taylor series gives immediately the values of the derivatives at a . Since e aw = µ n =0 ( aw ) n n ! , we get ³ d j dw j e aw ´ w =0 = a j . Similarly, ³ d k dw k e bw 2 ´ w =0 = 0i f k is odd , b j (2 j )! j ! if k =2 j. Returning to u ( x, t ), we compute the n th derivative of e - ( iwx + w 2 t ) using the Leibniz rule and use the what we just found and get u ( x, t i ) n ³ d n dw n e - w 2 t e - iwx ´ w =0 =( i ) n n µ j =0 ± n j ² d j dw j e - w 2 t · d n - j dw n - j e - iwx w =0 i ) n [ n 2 ] µ j
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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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