To derive: The first six Legendre polynomial functions from Legendre polynomial of degree n . Result used: Equation ( 11 ) is, P n ( x ) = ∑ m = 0 M ( − 1 ) m ( 2 n − 2 m ) ! 2 n m ! ( n − m ) ! ( n − 2 m ) ! x n − 2 m where M = n 2 or n − 1 2 whichever is an integer. And Equation ( 1 1 ′ ) is, P 0 ( x ) = 1 , P 1 ( x ) = x , P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) , P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) , P 5 ( x ) = 1 8 ( 63 x 5 − 70 x 3 + 15 x ) Calculation: First substitute n = 0 in equation (11) to obtain the Legendre polynomial of degree zero as shown below. P 0 ( x ) = ( 2 ( 0 ) ) ! 2 0 ( ( 0 ) ! ) 2 x 0 ( Here M is 0 ) = 0 ! 2 0 ( ( 0 ) ! ) 2 x 0 = 1 ( 1 ) ( 1 ) 2 ( 1 ) = 1 Substitute n = 1 in equation (11) to obtain the Legendre polynomial of degree zero as shown below. P 1 ( x ) = ( − 1 ) 0 ( 2 ( 1 ) − 2 ( 0 ) ) ! 2 1 ( 0 ) ! ( ( 1 ) − ( 0 ) ) ! ( ( 1 ) − 2 ( 0 ) ) ! x ( 1 ) − 2 ( 0 ) = 2 ! ( 2 ) ( 1 ) ! ( 1 ) ! ( 1 ) ! x − 0 = 2 2 ( x ) = x Thus, it is clear that, P 1 ( x ) = x . Substitute n = 2 in equation (11) to obtain the Legendre polynomial of degree zero as shown below. P 2 ( x ) = ( − 1 ) 0 ( 2 ( 2 ) − 2 ( 0 ) ) ! 2 2 ( 0 ) ! ( 2 − ( 0 ) ) ! ( 2 − 2 ( 0 ) ) ! x ( 2 ) − 2 ( 0 ) + ( − 1 ) 1 ( 2 ( 2 ) − 2 ( 1 ) ) ! 2 2 ( 1 ) ! ( 2 − ( 1 ) ) ! ( 2 − 2 ( 1 ) ) ! x ( 2 ) − 2 ( 1 ) = 4 ! 4 ( 1 ) ( 2 ) ! ( 2 ) ! x 2 − 2 ! 4 ( 1 ) ! ( 1 ) ! ( 0 ) ! x 0 = 3 2 x 2 − 1 2 = 1 2 ( 3 x 2 − 1 ) Thus, it is clear that, P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) . Substitute n = 3 in equation (11) to obtain the Legendre polynomial of degree zero as shown below. P 3 ( x ) = ( − 1 ) 0 ( 2 ( 3 ) − 2 ( 0 ) ) ! 2 3 ( 0 ) ! ( 3 − ( 0 ) ) ! ( 3 − 2 ( 0 ) ) ! x ( 3 ) − 2 ( 0 ) + ( − 1 ) 1 ( 2 ( 3 ) − 2 ( 1 ) ) ! 2 3 ( 1 ) ! ( 3 − ( 1 ) ) ! ( 3 − 2 ( 1 ) ) ! x ( 3 ) − 2 ( 1 ) = 6 ! 8 ( 3 ) ! ( 3 ) ! x 3 + ( − 1 ) 1 4 ! 8 ( 1 ) ! ( 2 ) ! ( 1 ) ! x 1 = 5 2 x 3 − 3 2 x = 1 2 ( 5 x 3 − 2 x ) Thus, it is clear that, P 3 ( x ) = 1 2 ( 5 x 3 − 2 x ) . Substitute n = 4 in equation (11) to obtain the Legendre polynomial of degree zero as shown below. P 4 ( x ) = { ( − 1 ) 0 ( 2 ( 4 ) − 2 ( 0 ) ) ! 2 4 ( 0 ) ! ( 4 − ( 0 ) ) ! ( 4 − 2 ( 0 ) ) ! x ( 4 ) − 2 ( 0 ) + ( − 1 ) 1 ( 2 ( 4 ) − 2 ( 1 ) ) ! 2 4 ( 1 ) ! ( 4 − ( 1 ) ) ! ( 4 − 2 ( 1 ) ) ! x ( 4 ) − 2 ( 1 ) + ( − 1 ) 2 ( 2 ( 4 ) − 2 ( 2 ) ) ! 2 4 ( 2 ) ! ( 4 − ( 2 ) ) ! ( 4 − 2 ( 2 ) ) ! x ( 4 ) − 2 ( 2 ) } = { 8 ! 16 ( 4 ) ! ( 4 ) ! x 4 − 6 ! 16 ( 1 ) ! ( 3 ) ! ( 2 ) ! x 2 + ( 4 ) ! 16 ( 2 ) ! ( 2 ) ! ( 0 ) ! x 0 } = 35 8 x 4 − 30 8 x 2 + 3 8 = 1 8 ( 35 x 4 − 30 x 2 + 3 ) Thus, it is clear that, P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) . Substitute n = 5 in equation (11) to obtain the Legendre polynomial of degree zero as shown below. P 5 ( x ) = { ( − 1 ) 0 ( 2 ( 5 ) − 2 ( 0 ) ) ! 2 5 ( 0 ) ! ( 5 − ( 0 ) ) ! ( 5 − 2 ( 0 ) ) ! x ( 5 ) − 2 ( 0 ) + ( − 1 ) 1 ( 2 ( 5 ) − 2 ( 1 ) ) ! 2 5 ( 1 ) ! ( 5 − ( 1 ) ) ! ( 5 − 2 ( 1 ) ) ! x ( 5 ) − 2 ( 1 ) + ( − 1 ) 2 ( 2 ( 5 ) − 2 ( 2 ) ) ! 2 5 ( 2 ) ! ( 5 − ( 2 ) ) ! ( 5 − 2 ( 2 ) ) ! x 5 − 2 ( 2 ) } = { 10 ! 32 ( 5 ) ! ( 5 ) ! x 5 − 8 ! 32 ( 1 ) ! ( 4 ) ! ( 3 ) ! x 3 + 6 ! 32 ( 2 ) ! ( 3 ) ! ( 1 ) ! x 1 } = 63 8 x 5 − 70 8 x 3 + 15 8 x = 1 8 ( 63 x 5 − 70 x 3 + 15 ) Thus, it is clear that, P 5 ( x ) = 1 8 ( 63 x 5 − 70 x 3 + 15 ) .

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