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Unformatted text preview: Solutions to Problem Set 2 Physics 341 by: Callum Quigley 1 The Dirac function To evaluate, ( ax ) consider the cases a > 0 and a < 0 separately. Define y = ax . Then for a > , 1 = Z  ( y )d y = a Z  ( ax )d x (1.1) and for a < , 1 = Z  ( y )d y = a Z + ( ax )d x = a Z  ( ax )d x. (1.2) Combining these, we find ( ax ) = ( x )  a  as required. Also, it should be clear that ( x ) = lim a 1 a x 2 + a 2 = , x = 0 , x 6 = 0 (1.3) as we expect for a function. With this definition, we can compute its integral over R : Z ( x )d x = lim a a Z  1 x 2 + a 2 d x = lim a a a = 1 , (1.4) as required. There are numerous ways to perform the above integral. The simplest probably being direct integration in terms of arctan. Another particularly nice way is to use the method of residues: that is, we consider x C and extend the integration contour, C , to enclose the upperhalf plane. Because the integrand decays sufficiently rapid as  x  , the contribution from this extension vanishes. This contour encloses a pole in the integrand at x = ia , and so we just pick up the residue Z  1 x 2 + a 2 d x = I C 1 ( x + ia )( x ia ) d x = 2 i Res x = ia 1 ( x + ia )( x ia ) = a . (1.5) 2 GramSchmidt Procedure The only thing youll need to solve this problem is to recall how the GramSchmidt procedure works. Starting from a given set of basis vectors { e i i} , we construct a new orthogonal basis { u i i} , where the kth vector is determined by the previous...
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 Spring '10
 Sghy
 Physics, mechanics

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