Chem Differential Eq HW Solutions Fall 2011 128

Chem Differential Eq HW Solutions Fall 2011 128 - f ( x ) =...

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128 Chapter 7 The Fourier Transform and its Applications Solutions to Exercises 7.8 1. As we move from left to right at a point x 0 , if the graph jumps by c units, then we must add the scaled Dirac delta function by x 0 ( x ). If the jump is upward, c is positive; and if the jump is downward, c is negative. With this in mind, by looking at the graph, we see that f ± ( x )= 1 2 δ - 2 ( x )+ 1 2 δ - 1 ( x ) - 1 2 δ 1 ( x ) - 1 2 δ 2 ( x )= 1 2 ( δ - 2 ( x )+ δ - 1 ( x ) - δ 1 ( x ) - δ 2 ( x )) . 13. We do this problem by reversing the steps in the solutions of the previous exercises. Since f ( x ) has zero derivative for x< - 2o r x> 3, it is therefore constant on these intervals. But since f ( x ) tends to zero as x →±∞ , we conclude that f ( x ) = 0 for x< - 2or x> 3. At x = - 2, we have a jump upward by one unit, then the function stays constant for - 2 <x< - 1. At x = - 1, we have another jump upward by one unit, then the function stays constant for - 1 <x< - 1. At x = 1, we have another jump upward by one unit, then the function stays constant for 1 <x< 3. At x = 3, we have a jump downward by three units, then the function stays constant for x>
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Unformatted text preview: f ( x ) = 0 if x &lt;-2 , 1 if-2 &lt; x &lt;-1 , 2 if-1 &lt; x &lt; 1 , 3 if 1 &lt; x &lt; 3 , 0 if 3 &lt; x. 17. We use the deFnition (7) of the derivative of a generalized function and the fact that the integral against a delta function a picks up the value of the function at a . Thus ( x ) , f ( x ) = ( x ) ,-f ( x ) =- ( x ) , f ( x ) =- ( x )- 1 ( x ) , f ( x ) =-f (0) + f (1) . 21. rom Exercise 7, we have ( x ) = 1 a ( U-2 a ( x )- U-a ( x ) )-1 a ( U a ( x )- U 2 a ( x ) ) . Using (9) (or arguing using jumps on the graph), we Fnd ( x ) = 1 a ( -2 a ( x )--a ( x ) )-1 a ( a ( x )- 2 a ( x ) ) = 1 a ( -2 a ( x )--a ( x )- a ( x )+ 2 a ( x ) ) . 25. Using the deFnition of and the deFnition of a derivative of a generalized...
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