University:
Princeton University
Course:
PHY 103 | General Physics I
Academic year:

2023
Views:

192

Pages:

1
Author:

Jack Alasco

1) Evaluate ∫ − 1 1 𝑃 0 ( 𝑥 ) 𝑃 2 ( 𝑥 ) 𝑑 𝑥 ∫ −1 1 P 0 (x)P 2 (x)dx to show that these functions are orthogonal on (-1,1). Answer: 𝑃 0 ( 𝑥 ) = 1 P 0 (x)=1 𝑃 2 ( 𝑥 ) = 1 2 ( 3 𝑥 2 − 1 ) P 2 (x)= 2 1 (3x 2 −1) So, ∫ − 1 1 1 2 ( 3 𝑥 2 − 1 ) 𝑑 𝑥 = 1 2 [ ( 3 𝑥 3 3 ) − 𝑥 ] − 1 1 ∫ −1 1 2 1 (3x 2 −1)dx= 2 1 [( 3 3x 3 )−x] −1 1 = 1 2 [ ( 1 3 − 1 ) − ( ( − 1 ) 3 − ( − 1 ) ) ] = 2 1 [(1 3 −1)−((−1) 3 −(−1))] = 1 2 [ ( 1 − 1 ) − ( − 1 + 1 ) ] = 2 1 [(1−1)−(−1+1)] = 1 2 [ 2 − 2 ] = 0 = 2 1 [2−2]=0 2) Show in two ways that 𝑃 𝑙 ( 𝑥 ) P l (x) and 𝑃 𝑙 ′ ( 𝑥 ) P l ′ (x) are orthogonal on (-1,1). (Hint: See problem 4 and problem 4.4) Answer: ∫ 𝑎 𝑏 𝑓 ( 𝑥 ) 𝑔 ( 𝑥 ) 𝑑 𝑥 = 0 ∫ a b f(x)g(x)dx=0 Variable: 𝑦 = − 𝑥 y=−x ∫ − 𝑎 𝑎 𝑓 ( 𝑦 ) 𝑔 ( 𝑥 ) 𝑑 𝑥 = − ∫ − 𝑎 𝑎 𝑓 ( − 𝑦 ) 𝑔 ( 𝑦 ) 𝑑 𝑦 ∫ −a a f(y)g(x)dx=−∫ −a a f(−y)g(y)dy = ∫ 𝑎 𝑏 𝑓 ( − 𝑦 ) 𝑔 ( 𝑦 ) 𝑑 𝑦 = − ∫ 𝑎 𝑏 𝑓 ( 𝑦 ) 𝑔 ( 𝑦 ) 𝑑 𝑦 =∫ a b f(−y)g(y)dy=−∫ a b f(y)g(y)dy ⇒ ∫ 𝑎 𝑏 𝑓 ( 𝑦 ) 𝑔 ( 𝑦 ) 𝑑 𝑦 = 0 ⇒∫ a b f(y)g(y)dy=0 𝑓 ( 𝑥 ) f(x) = even function 𝑥 x 𝑓 ( 𝑥 ) f(x) = odd function 𝑥 x So, ⇒ ∫ − 𝑎 𝑎 𝑃 𝑙 ( 𝑥 ) 𝑃 𝑙 ′ ( 𝑥 ) = 0 ⇒∫ −a a P l (x)P l ′ (x)=0 𝑃 𝑙 P l is a polynomial in order 1 and 𝑃 𝑙 ′ P l ′ acts as a polynomial in order (l-1). So, ⇒ ∫ − 𝑎 𝑎 𝑃 𝑙 ( 𝑥 ) 𝑃 𝑙 ′ ( 𝑥 ) 𝑑 𝑥 = 0 ⇒∫ −a a P l (x)P l ′ (x)dx=0