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Unformatted text preview: 640:527: Homework 5: Solutions 1. Greenberg 4.6, # 1. Let y = a n x r + n . Then y = ( r + n ) a n x r + n 1 and y 00 = ( r + n )( r + n 1) x r + n 2 . Plugging these into Bessels equation x 2 y 00 + xy + ( x 2 2 ) y = 0 gives 0 = X ( r + n )( r + n 1) x r + n + X ( r + n ) a n x r + n X 2 a n x r + n + X a n x r + n +2 = X ( r + n )( r + n 1) + ( r + n ) 2 x r + n + X 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 = X ( r + n ) 2 2 x r + n + X 2 a n 2 x r + n = r 2 2 a x r + ( r + 1) 2 2 a 1 x r +1 + X 2 ( ( r + n ) 2 2 ) a n + a n 2 x r + n . This is essentially equation (3) of the text. I have just written out the first two terms of the series. Greenberg 4.6, # 2. Before starting this problem we make a general remark. Suppose 1 , 2 ,... is a given sequence. Suppose now we want to find the solution b ,b 1 ,b 2 ,... to the recursion equation b k = k b k 1 , k 1 , and b = B. (1) The solution is given by the formula b k = k k 1 k 2 1 b , b = B. (2) It is easy to see why this is true. Start with (1) and then apply the formula with k 1 instead of k to express b k 1 in terms of b k 2 : b k = k b k 1 = k ( k 1 b k 2 ). Repeat this process k times: b k = k k 1 k 2 b k 3 = = k k 1 k 2 1 b . The problem is to solve the Bessel equation of order = 1 / 2. This is the equation x 2 y 00 + xy + ( x 2 (1 / 4)) y = 0. We look for a solution of the form y = a n x r + n . The indicial equation recursion equation may be obtained by substituting = 1 / 2 into equation (3a) of section 4.6 in the text, and the result is r 2 (1 / 4) a = 0. Since a 6 = 0, it follows that r 2 1 / 4 = 0. This has the roots r 1 = 1 / 2 and r 2 = 1 / 2. These roots differ by an integer and there might be a problem, but we know there is always a solution of the form y = a n x r + n when r equals the larger root. So first we work out this solution and take r = r 2 = 1 / 2. The recursion equations for this choice of r can be obtained by substituting r = 1 / 2 and = 1 / 2 in equations (3b) and (3c) of the text. (I am using n as an index, rather than k .) We get (1 + 1 / 2) 2 1 2 a 1 = (5 / 4) a 1 = 0 (3) ( n + 1 / 2) 2 (1 / 4) a n + a n 2 = ( n +1) na n + a n 2 = 0 , n 2 . (4) The first equation implies that a 1 = 0. The second equation gives the recursion a n = a n 2 ( n + 1) n . (5) 1 By setting n = 3, a 3 = a 1 / 12 = 0, a 5 = a 3 / 30 = 0. Continuing, it is clear that a n = 0 for all odd n . For even n , we derive a recursion by replacing n in (5) by 2 k ....
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This note was uploaded on 09/29/2009 for the course 650 527 taught by Professor Danocone during the Fall '06 term at Rutgers.
 Fall '06
 DanOcone

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