Chem Differential Eq HW Solutions Fall 2011 107

Chem Differential Eq HW Solutions Fall 2011 107 - Section...

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Section 7.2 The Fourier Transform 107 Solutions to Exercises 7.2 1. In computing ± f , the integral depends on the values of f on the interval ( - 1 , 1). Since on this interval f is odd, it follows that f ( x ) cos wx is odd and f ( x ) sin is even on the interval ( - 1 , 1). Thus ± f ( w )= 1 2 π ² -∞ f ( x ) e - iwx dx = 1 2 π =0 ³ ´µ ² 1 - 1 f ( x ) cos wxdx - i 2 π ² 1 - 1 f ( x ) sin = - 2 i 2 π ² 1 0 sin = 2 i 2 π cos w · · · 1 0 = i ¸ 2 π cos w - 1 w . 5. Use integration by parts to evaluate the integrals: ± f ( w 1 2 π ² -∞ f ( x ) e - iwx dx = 1 2 π ² 1 - 1 (1 -| x | )(cos - i sin ) dx = 1 2 π ² 1 - 1 (1 x | ) cos - i 2 π =0 ³ ´µ ² 1 - 1 (1 x | ) sin = 2 2 π ² 1 0 u ³ ´µ (1 - x ) dv ³ ´µ cos = 2 2 π ¹ (1 - x ) sin w º · · · · 1 0 + 2 2 πw ² 1 0 sin = - ¸ 2 π cos w 2 · · · · · 1 0 = ¸ 2 π 1 - cos w w 2 . 9. In Exercise 1, ± f (0) = 1 2 π × (area between graph of f
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