To derive: The equation ( 1 1 ′ ) from Rodrigues’s formula. Result used: The 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: The Rodrigues’s formula is given by, P n ( x ) = 1 2 n n ! d n d x n [ ( x 2 − 1 ) n ] . First substitute n = 0 in P n ( x ) to obtain the Legendre polynomial of degree zero as shown below. P 0 ( x ) = 1 2 0 ( 0 ) ! d 0 d x 0 [ ( x 2 − 1 ) 0 ] = 1 2 0 ( 0 ) ! d 0 d x 0 [ ( x 2 − 1 ) 0 ] = 1 ( 1 ) ( 1 ) ( 1 ) = 1 Thus, it is clear that P 0 ( x ) = 1 . Substitute n = 1 in P n ( x ) to obtain the Legendre polynomial of degree one as shown below. P 1 ( x ) = 1 2 1 ( 1 ) ! d d x [ ( x 2 − 1 ) 1 ] = 1 2 ( 1 ) ( 2 x ) = 2 x 2 = x Thus, it is clear that, P 1 ( x ) = x . Substitute n = 2 in P n ( x ) to obtain the Legendre polynomial of degree two as shown below. P 2 ( x ) = 1 2 2 ( 2 ) ! d 2 d x 2 [ ( x 2 − 1 ) 2 ] = 1 4 ( 2 ) d 2 d x 2 ( x 4 − 2 x 2 + 1 ) = 1 8 d d x ( 4 x 3 − 4 x ) = 1 8 ( 12 x 2 − 4 ) = 1 2 ( 3 x 2 − 1 ) Thus, it is clear that, P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) . Substitute n = 3 in P n ( x ) to obtain the Legendre polynomial of degree three as shown below. P 3 ( x ) = 1 2 3 ( 3 ) ! d 3 d x 3 [ ( x 2 − 1 ) 3 ] = 1 8 ( 6 ) d 3 d x 3 ( x 6 − 3 x 4 + 3 x 2 − 1 ) = 1 48 d 2 d x 2 ( 6 x 5 − 12 x 3 + 6 x ) = 1 48 d 2 d x 2 ( 6 x 5 − 12 x 3 + 6 x ) On further simplification, P 3 ( x ) = 1 48 d d x ( 30 x 4 − 36 x 2 + 6 ) = 1 48 ( 120 x 3 − 72 x ) = 1 2 ( 5 x 3 − 3 x ) Thus, it is clear that, P 3 ( x ) = 1 2 ( 5 x 3 − 2 x ) . Substitute n = 4 in P n ( x ) to obtain the Legendre polynomial of degree four as shown below. P 4 ( x ) = 1 2 4 ( 4 ) ! d 4 d x 4 [ ( x 2 − 1 ) 4 ] = 1 16 ( 24 ) d 4 d x 4 ( x 8 + 6 x 4 − 4 x 6 − 4 x 2 + 1 ) = 1 384 d 3 d x 3 ( 8 x 7 + 24 x 3 − 24 x 5 − 8 x ) = 1 384 d 2 d x 2 ( 56 x 6 + 72 x 2 − 120 x 4 − 8 ) It is further simplified as shown below. P 4 ( x ) = 1 384 d d x ( 336 x 5 + 144 x − 480 x 3 ) = 1 384 ( 1680 x 4 + 144 − 1440 x 2 ) = 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 P n ( x ) to obtain the Legendre polynomial of degree five as shown below. P 5 ( x ) = 1 2 5 ( 5 ) ! d 5 d x 5 [ ( x 2 − 1 ) 5 ] = 1 32 ( 120 ) d 5 d x 5 ( ( x 8 + 6 x 4 − 4 x 6 − 4 x 2 + 1 ) ( x 2 − 1 ) ) = 1 3840 d 5 d x 5 { x 10 + 6 x 6 − 4 x 8 − 4 x 4 + x 2 − x 8 − 6 x 4 + 4 x 6 + 4 x 2 − 1 } = 1 3840 d 5 d x 5 ( x 10 + 10 x 6 − 5 x 8 − 10 x 4 + 5 x 2 − 1 ) On further simplification, P 5 ( x ) = 1 3840 d 4 d x 4 ( 10 x 9 + 60 x 5 − 40 x 7 − 40 x 3 + 10 x ) = 1 3840 d 3 d x 3 ( 90 x 8 + 300 x 4 − 280 x 6 − 120 x 2 + 10 ) = 1 3840 d 2 d x 2 ( 720 x 7 + 1200 x 3 − 1680 x 5 − 240 x ) = 1 3840 d d x ( 5040 x 6 + 3600 x 2 − 8400 x 4 − 240 ) It is further simplified as shown below. P 5 ( x ) = 1 3840 ( 30240 x 5 + 7200 x − 33600 x 3 ) = 1 8 ( 63 x 5 − 70 x 3 + 15 x ) Thus, it is clear that, P 5 ( x ) = 1 8 ( 63 x 5 − 70 x 3 + 15 ) .

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