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Unformatted text preview: Frobenius method example for Sept. 20 lecture Preliminary remarks; using the Gamma function. The Gamma function often appears in the solution to recursion equations for series solutions to o.d.e. The Gamma function ( x ) is defined for all reals numbers x except x = 0, x =- 1, x =- 2, etc., where it has vertical asymptotes. For x > 0 it is defined by the formula ( x ) = Z t x- 1 e- tdt. For x = n , n a positive integer, ( n ) = ( n- 1)!. Integration by parts shows that ( x +1) = x ( x ) , x > , and ( x ) is extended to x < 0 (except 0 and the negative integers) by imposing this identity everywhere. The usefulness of the Gamma function for recursions is due to the following cal- culation. Let k be a positive integer and let p be a real number. Then ( p +1) = p ( p ) = p ( p- 1)( p- 1) = p ( p- 1)( p- 2)( p- 2) = = p ( p- 1)( p- 2) ( p- k )( p- k ) . This is valid as long as none of p,p- 1 ,...,p- k are non-negative integers or zero. Dividing both sides of this identity by ( p- k ), leads to the identity: p ( p- 1)( p- 2) ( p- k ) = ( p +1) ( p- k ) . (1) 2. We solve x 2 y + ( x/ 2) y + (1 / 2) x 2 y = 0 in a series solution about the regular singular point x = 0. We look for two linearly independent solutions. The method of Frobenius says to look for solutions of the form | x | r X a n x n , where a = 1 and r is to be determined from the indicial equation. In doing the calculations, it is convenient to assume x > 0, so that | x | r = x r , and to write y = X a n x r + n . To do this, calculate x 2 y and xy by term by term differentiation and plug into the equation. The result is 0 = X ( r + n )( r + n- 1) a n x r + n + X (1 / 2)( r + n ) a n x r + n + X ( a n / 2) x r + n +2 . (2) 1 To combine the series, they should all be expressed in powers of r + n . Only the last series fails to conform, so we fix it by changing the index of summation to k = n + 2, and then relabeling the new index k using n again: X n =0 ( a n / 2) x r + n +2 = X k =2 ( a k- 2 / 2) x r + k...
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- Fall '06