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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 1periodization 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 )  < π 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 , π ], the rising cutoff function r ( t ) = r ( t ) defined in Equation 3.12, and the reach = π/ 2. (Note: the result should not be a continuous function.) 1 Solution: Use Equation 3.24 with α = 0, β = π , = π/ 2 to compute L I def = L ( r , [0 , π ] , π 2 ) applied to u , noting that r = ¯ r : L ( r , [0 , π ] , π 2 ) u ( t ) = r ( 2 π t ) u ( t ) r ( 2 π t ) u (0 t ) , t ∈ (0 , π 2 ), r (2 2 π t ) u ( t ) + r ( 2 π t 2) u (2 π t ) , t ∈ ( π 2 , π ), u ( t ) , otherwise....
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This note was uploaded on 05/16/2010 for the course MATH 450 taught by Professor Wickerhauser during the Spring '10 term at Washington University in St. Louis.
 Spring '10
 Wickerhauser
 Math

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