s03 - Ma 450: Mathematics for Multimedia Solution: to...

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Unformatted text preview: Ma 450: Mathematics for Multimedia Solution: to Homework Assignment 3 Prof. Wickerhauser Due Friday, March 5th, 2010 1. Suppose that f ( t ) = 0 if | t | 1, and | f ( t ) | < 1 for all t (- 1 , 1). Prove that | f 1 ( t ) | < 2 for all t R , where f 1 is the 1-periodization of f . Solution: Since f ( t ) = 0 unless t (- 1 , 1) and { t + k : k Z } (- 1 , 1) contains exactly two elements for every t R , the set S t def = { k Z : f ( t + k ) 6 = 0 } can have at most two elements. Thus the sum defining f 1 ( t ) = k Z f ( t + k ) can have at most two nonzero terms. Call these f ( t 1 ) and f ( t 2 ), where t 1 , t 2 (- 1 , 1) satisfy | t 1- t 2 | = 1. But then the triangle inequality implies | f 1 ( t ) | = | f ( t 1 ) + f ( t 2 ) | | f ( t 1 ) | + | f ( t 2 ) | < 1 + 1 = 2 , according to the hypothesis that | f ( t ) | < for all t (- 1 , 1). 2 2. Define the reflection R to be the transformation Ru ( t ) def = u (- t ) acting on the vector space of functions of one real variable. Show that R is a linear transformation, and find a formula for the compositions RF , F R , and RF R , where F is the fraying operator of Equation 3.14. Solution: R is a linear transformation, since R [ au + bv ]( t ) = au (- t ) + bv (- t ) = aRu ( t ) + bRv ( t ). The compositions RF , F R , and RF R are therefore linear transformations as well. Using Equation 3.14 gives the following formulas: RF u ( t ) = r (- t ) u (- t ) + r ( t ) u ( t ) , if- t > 0, r ( t ) u (- t )- r (- t ) u ( t ) , if- t < 0, = r ( t ) u (- t )- r (- t ) u ( t ) , if t > 0, r (- t ) u (- t ) + r ( t ) u ( t ) , if t < 0; F Ru ( t ) = r ( t ) Ru ( t ) + r (- t ) Ru (- t ) , if t > 0, r (- t ) Ru ( t )- r ( t ) Ru (- t ) , if t < 0, = r ( t ) u (- t ) + r (- t ) u ( t ) , if t > 0, r (- t ) u (- t )- r ( t ) u ( t ) , if t < 0; RF Ru ( t ) = r (- t ) Ru (- t ) + r ( t ) Ru ( t ) , if- t > 0, r ( t ) Ru (- t )- r (- t ) Ru ( t ) , if- t < 0, = r ( t ) u ( t )- r (- t ) u (- t ) , if t > 0, r (- t ) u ( t ) + r ( t ) u (- t ) , if t < 0; In all cases, RF u (0) = F Ru (0) = RF Ru (0) = u (0). 2 3. Find an explicit formula by applying the loop operator of Equation 3.23 to the function u ( t ) = cos( t ), using the interval I = [0 , ], the rising cutoff function r ( t ) = r ( t ) defined in Equation 3.12, and the reach = / 2. (Note: the result should not be a continuous function.) 1 Solution: Use Equation 3.24 with = 0, = , = / 2 to compute L I def = L ( r , [0 , ] , 2 ) applied to u , noting that r = r : L ( r , [0 , ] , 2 ) u ( t ) = r ( 2 t ) u ( t )- r (- 2 t ) u (0- t ) , t (0 , 2 ), r (2- 2 t ) u ( t ) + r ( 2 t- 2) u (2 - t ) , t ( 2 , ), u ( t ) , otherwise....
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s03 - Ma 450: Mathematics for Multimedia Solution: to...

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