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solutionhw3 - Solutions to Homework Set 3 Test functions...

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Unformatted text preview: Solutions to Homework Set 3 Test functions and distributions : For part a) we take any test function ( x ) and look at ( , f + f ) Z - ( x ) { f ( x ) ( x ) + f ( x ) ( x ) } dx = Z - { [- ( x ) f ( x )- ( x ) f ( x )] ( x ) + ( x ) f ( x ) ( x ) } dx =- (0) f (0) , and compare it with ( , f (0) ) Z - ( x ) { f (0) ( x ) } dx =- (0) f (0) . The results are the same, and so the distributions f + f and f (0) are equal. For b) part i) we have t P Z - ( x ) ( x- t ) dx = t Z t-- + Z t + ( x ) ( x- t ) dx = ( t- )-- ( t + ) + Z t-- + Z t + ( x ) ( x- t ) 2 dx =- 2 ( t ) + Z t-- + Z t + ( x ) ( x- t ) 2 dx + O ( ) = Z t-- + Z t + ( x )- ( t ) ( x- t ) 2 dx + O ( ) P Z - ( x )- ( t ) ( x- t ) 2 dx as . The tricky part was recognizing that ( t + ) + ( t- ) = 2 ( t ) + O ( 2 ) , and that you can rewrite the resultant divergent bit 2 ( t ) / as 2 ( t ) = ( t ) Z t-- + Z t + 1 ( x- t ) 2 dx = Z t-- + Z t + ( t ) ( x- t ) 2 dx. For b) part ii) we have P Z - ( x ) ( x- t ) dx = P Z - ( y + t ) y dy. Taking the derivative, we get t P Z - ( x ) ( x- t ) dx = P Z - ( y + t ) y dy 1 = P Z - y ( y + t ) y !...
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solutionhw3 - Solutions to Homework Set 3 Test functions...

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