Solutions Manual to Quantum Chemistry.

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Equation……..

Taking (/ )m d dx of (4.84) gives ( ) 0 ( ) ( 1)( 2) ( 1)( ) m nm n n f x cnn n n m x a ∞ − = = − − −+ − ∑ ” . The factors n, ( 1),… n − make the terms with 0, n = 1, n = …, n m= −1 vanish, so ( )( ) ( 1)( 2) ( 1)( ) m nm n n m f x cnn n n m x a ∞ − = = − − −+ − ∑ ” (Eq. 1). (If this is too abstract for you, write the expansion as 2 01 2 ( ) k k f x c cx c x c x = + + ++ + ” ” and do the differentiation.) With x = a in Eq. 1, the ( )n m x a − − factor makes all terms equal to zero except the term with n m= , which is a constant. Equation (1) with x = a gives ( )( ) ( 1)( 2) ( 1) ! m m m f a c mm m m m c m = − − −+= ” and ( )( )/ ! m mc f am = ………..DOWNLOAD

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