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

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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 ) = 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. 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 , FR , and RFR , 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 , FR , and RFR are therefore linear transformations as well. Using Equation 3.14 gives the following formulas: RFu ( 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; FRu ( 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; RFRu ( 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, RFu (0) = FRu (0) = RFRu (0) = u (0). 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 0 ( t ) defined in Equation 3.12, and the reach = π/ 2. (Note: the result should not be a continuous function.) 1
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Solution: Use Equation 3.24 with α = 0, β = π , = π/ 2 to compute L I def = L ( r 0 , [0 , π ] , π 2 ) applied to u , noting that r 0 = ¯ r 0 : L ( r 0 , [0 , π ] , π 2 ) u ( t ) = r 0 ( 2 π t ) u ( t ) - r 0 ( - 2 π t ) u (0 - t ) , t (0 , π 2 ), r 0 (2 - 2 π t ) u ( t ) + r 0 ( 2 π t - 2) u (2 π - t ) , t ( π 2 , π ), u ( t ) , otherwise. First note that if t belongs to (0 , π ) then so does π - t , and thus u ( π - t ) = cos( π - t ) = - cos( t ), by an identity valid for all real t . The same identity shows that for 0 < t < π 2 the rising cutoff function satisfies r 0 ( 2 π t ) = sin π 4 (1 + 2 π t ) = sin π 4 + t 2 ; r 0 ( - 2 π t ) = cos π 4 (1 + 2 π t ) = cos π 4 + t 2 . Second, for π 2 < t < π , the rising cutoff function satisfies r 0 (2 - 2 π t ) = sin π 4 (3 - 2 π t ) = sin π 4 + t 2 , since sin( π - x ) = sin( x ) for all real x . But sin( - x ) = - sin( x ), so r 0 ( 2 π t - 2) = sin π 4 ( - 1+ 2 π t ) = - cos π 4 + t 2 . Combining these evaluations for both t (0 , π 2 ) and t ( π 2 , π ) yields the same formula: L I u ( t ) = sin π 4 + t 2 cos( t ) - cos π 4 + t 2 cos( t ) . But the identity sin x + cos x = 2 cos( x - π 4 ) may be used to simplify it: sin π 4 + t 2 - cos π 4 + t 2 = 2 sin t 2 , so L I u ( t ) = 2 cos( t/ 2) cos( t ) on (0 , π 2 ) ( π 2 , π ). But u ( t ) = cos( t ) is a factor in this expression, so it agrees with it at t = π 2 where u ( π 2 ) = cos( π 2 ) = 0. Thus in fact L ( r 0 , [0 , π ] , π 2 ) u ( t ) = 2 cos( t/ 2) cos( t ) for all t (0 , π ).
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