Chem Differential Eq HW Solutions Fall 2011 114

# Chem Differential Eq HW Solutions Fall 2011 114 - 114...

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114 Chapter 7 The Fourier Transform and its Applications To prove (1), note that cos cwt = e icwt + e - icwt 2 , so 1 2 π ± -∞ ² f ( w ) cos cwte iwx dw = 1 2 π ± -∞ ² f ( w )( e icwt + e - icwt 2 ) e iwx dw = 1 2 ³ 1 2 π ± -∞ ² f ( w ) e iw ( x + ct ) dw 1 2 π ± -∞ ² f ( w ) e iw ( x - ct ) dw ´ = 1 2 [ f ( x + ct )+ f ( x - ct )]; because the frst integral is simply the inverse Fourier trans±orm o± ² f evaluated at x + ct , and the second integral is the inverse Fourier trans±orm o± ² f evaluated at x - ct . This proves (1). To prove (2), we note that the le±t side o± (2) is an inverse Fourier trans±orm. So (2) will ±ollow i± we can show that (3) F µ± x + ct x - ct g ( s ) ds = 2 w ² g ( w ) sin cwt. Let G denote an antiderivative o± g . Then (3) is equivalent to F ( G ( x + ct ) - G ( x - ct )) ( w )= 2 w · G ± ( w ) sin cwt. Since · G ± = iw ² G , the last equation is equivalent to (4) F ( G ( x + ct )) ( w ) -F ( G ( x - ct )) ( w )=2 i ² G ( w ) sin cwt. Using Exercise 19, Sec. 7.2, we have
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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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