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Unformatted text preview: Solutions to Homework Set 3 Test functions and distributions : For part a) we take any test function Ï• ( x ) and look at ( Ï•, fÎ´ + f Î´ ) â‰¡ Z âˆžâˆž Ï• ( x ) { f ( x ) Î´ ( x ) + f ( x ) Î´ ( x ) } dx = Z âˆžâˆž { [ Ï• ( x ) f ( x ) Ï• ( x ) f ( x )] Î´ ( x ) + Ï• ( x ) f ( x ) Î´ ( x ) } dx = Ï• (0) f (0) , and compare it with ( Ï•, f (0) Î´ ) â‰¡ Z âˆžâˆž Ï• ( x ) { f (0) Î´ ( x ) } dx = Ï• (0) f (0) . The results are the same, and so the distributions fÎ´ + f Î´ and f (0) Î´ are equal. For b) part i) we have âˆ‚ âˆ‚t P Z âˆžâˆž Ï• ( x ) ( x t ) dx = âˆ‚ âˆ‚t Z tâˆž + Z âˆž t + Ï• ( x ) ( x t ) dx = Ï• ( t ) Ï• ( t + ) + Z tâˆž + Z âˆž t + Ï• ( x ) ( x t ) 2 dx = 2 Ï• ( t ) + Z tâˆž + Z âˆž t + Ï• ( x ) ( x t ) 2 dx + O ( ) = Z tâˆž + Z âˆž t + Ï• ( x ) Ï• ( t ) ( x t ) 2 dx + O ( ) â†’ P Z âˆžâˆž Ï• ( x ) Ï• ( t ) ( x t ) 2 dx as â†’ . The tricky part was recognizing that Ï• ( t + ) + Ï• ( t ) = 2 Ï• ( t ) + O ( 2 ) , and that you can rewrite the resultant divergent bit 2 Ï• ( t ) / as 2 Ï• ( t ) = Ï• ( t ) Z tâˆž + Z âˆž t + 1 ( x t ) 2 dx = Z tâˆž + Z âˆž t + Ï• ( t ) ( x t ) 2 dx. For b) part ii) we have P Z âˆžâˆž Ï• ( x ) ( x t ) dx = P Z âˆžâˆž Ï• ( y + t ) y dy. Taking the derivative, we get âˆ‚ âˆ‚t P Z âˆžâˆž Ï• ( x ) ( x t ) dx = P Z âˆžâˆž Ï• ( y + t ) y dy 1 = P Z âˆžâˆž âˆ‚ âˆ‚y Ï• ( y + t ) y !...
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 Fall '07
 Stone
 Derivative, Work, Trigraph, dx, P t, Y1 Y1

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