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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 3 FALL 2007 Some of these solutions were written by Professor Dan Ocone. 4.5.8 The solutions in the text show how to do these. 9. In the definition (14) of the Gamma function we make the change of variable t = 2 u 2 , dt = 2 u du , which yields (1 / 2) = integraldisplay t 1 / 2 e t/ 2 dt = 2 integraldisplay e u 2 d, . and so, since u is just a dummy integration variable, ((1 / 2)) 2 = bracketleftbigg 2 integraldisplay e u 2 du bracketrightbiggbracketleftbigg 2 integraldisplay e v 2 dv bracketrightbigg = 4 integraldisplay integraldisplay e ( u 2 + v 2 ) du dv. Now we change an integral from Cartesian coordinates ( u, v ) to polar coordinates ( r, ); the integration region u, v 0, which is jusst the first quadrant, becomes 0 / 2, r , and we have change du dv to r dr d . Thus ((1 / 2)) 2 = 4 integraldisplay / 2 integraldisplay e r 2 r dr d = 2 parenleftBig 2 parenrightBig bracketleftBig e r 2 bracketrightBig = . So (1 / 2) = . The change of variables that we did here should really be done somewhat more carefully because the integal is improper, but we will omit these details. 4.6.1 Let y = a n x r + n . Then y = ( r + n ) a n x r + n 1 and y = ( r + n )( r + n 1) x r + n 2 . Plugging these into Bessels equation x 2 y + xy + ( x 2 2 ) y = 0 gives 0 = summationdisplay ( r + n )( r + n 1) x r + n + summationdisplay ( r + n ) a n x r + n summationdisplay 2 a n x r + n + summationdisplay a n x r + n +2 = summationdisplay bracketleftbig ( r + n )( r + n 1) + ( r + n ) 2 bracketrightbig x r + n + summationdisplay a n x r + n +2 Notice that a n x r + n +2 = 2 a n 2 x r + n , and that ( r + n )( r + n 1)+( r + n ) = ( r + n ) 2 . Thus 0 = summationdisplay bracketleftbig ( r + n ) 2 2 bracketrightbig x r + n + summationdisplay 2 a n 2 x r + n = bracketleftbig r 2 2 bracketrightbig a x r + bracketleftbig ( r + 1) 2 2 bracketrightbig a 1 x r +1 + summationdisplay 2 bracketleftbig( ( r + n ) 2 2 ) a n + a n 2 bracketrightbig x r + n ....
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- Fall '07