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16a - Lecture 16 1 Legendre Polynomials First we note that...

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Unformatted text preview: Lecture 16 1 Legendre Polynomials First, we note that n ! ( n- 2 j )! j ! j ! = ( n 2 j )( 2 j j ) , and so P n ( t ) = b n/ 2 c X j =0- 1 4 j n j 2 j j (1- t 2 ) j t n- 2 j . h Interlude: Recall that f n ( x,y,z ) = b n/ 2 c X j =0- 1 4 j n j 2 j j ( x 2 + y 2 ) j z n- 2 j . We can determine homogenous degree n polynomials that are invariant under rotation about any axis: say the axis determined by the unit vector u . Then, using the geometric properties of the cross product and the dot product, we find that these polynomials are all scalar multiples of f n ( r ) = b n/ 2 c X j =0- 1 4 j n j 2 j j || u × r || j ( u · r ) n- 2 j . i Recall that ∂ z f n = nf n and (2 n + 1) zf n ( x,y,z )- n ( x 2 + y 2 + z 2 ) f n- 1 ( x,y,z ) = ( n + 1) f n +1 ( x,y,z ) . From these properties of f n come the following properties of P n : nt n P n ( t ) + (1- t 2 ) P n ( t ) = nP n- 1 ( t ) (1) and ( n + 1) P n +1 ( t )- (2 n + 1) tP n ( t ) + nP n- 1 ( t ) = 0 . (2) Next, we show that P n satisfies the Legendre equation d dt [(1- t 2 ) dT dt ] + n ( n + 1) T = 0 . 1 From (1) it follows that (1- t 2 ) dP n dt = n [ P- n- 1( t )- tP n ( t )], and so d dt [(1- t 2 ) dP n dt ] = n [ P n- 1 ( t )- tP n ( t )]- nP n ( t ) . Now all that remains is to show that tP n ( t )- P n- 1 ( t ) = nP n ( t ). Multiply the LHS of this equation by (1- t 2 ), giving (1- t 2 )[ tP n ( t...
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16a - Lecture 16 1 Legendre Polynomials First we note that...

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