Legendre多项式的生成函数 √1-2+=P 士√x2-1 1-2t+t2√1-2t+t2-2(x-1)t
Properties of Legendre Polynomials (cont.) Applications of Legendre Polynomials Generating Function Recurrence Formulas for Legendre Polynomials Legendreõª)¤¼ê 1 √ 1−2xt+t 2 = X∞ l=0 Pl(x)t l |t| < x± √ x 2−1 Proof 1 √ 1 − 2xt + t 2 = 1 p 1 − 2t + t 2 − 2(x − 1)t = 1 1 − t 1 − 2(x − 1)t (1 − t) 2 −1/2 C. S. Wu 1Ôù ¥¼ê()
Legendre多项式的生成函数 √1-2n+=P() 士√x2-1 2 1-2xt+t21-t (1-t)2 1>((2)-(2-)(
Properties of Legendre Polynomials (cont.) Applications of Legendre Polynomials Generating Function Recurrence Formulas for Legendre Polynomials Legendreõª)¤¼ê 1 √ 1−2xt+t 2 = X∞ l=0 Pl(x)t l |t| < x± √ x 2−1 Proof 1 √ 1 − 2xt + t 2 = 1 1 − t 1 − 2(x − 1)t (1 − t) 2 −1/2 = 1 1−t X∞ k=0 1 k! − 1 2 − 3 2 · · · 1 2 −k − 2(x−1)t (1−t) 2 k C. S. Wu 1Ôù ¥¼ê()
Legendre多项式的生成函数 √1-2n+=P() 士√x2-1 1-2xt+t2 y11 2 (1-t)2 ((3)-(2-)ax =k1(x-1)(1-~(2k+1)
Properties of Legendre Polynomials (cont.) Applications of Legendre Polynomials Generating Function Recurrence Formulas for Legendre Polynomials Legendreõª)¤¼ê 1 √ 1−2xt+t 2 = X∞ l=0 Pl(x)t l |t| < x± √ x 2−1 Proof 1 √ 1 − 2xt + t 2 = 1 1 − t 1 − 2(x − 1)t (1 − t) 2 −1/2 = 1 1−t X∞ k=0 1 k! − 1 2 − 3 2 · · · 1 2 −k − 2(x−1)t (1−t) 2 k = X∞ k=0 (2k − 1)!! k! (x − 1)k t k (1 − t) −(2k+1) C. S. Wu 1Ôù ¥¼ê()
Legendre多项式的生成函数 √1-2rt+t2 ∑P(x)<士 2 1-2rt+ (1-t)2 k-1 2k-1)!! (2k+m)! n!(2k) k=0
Properties of Legendre Polynomials (cont.) Applications of Legendre Polynomials Generating Function Recurrence Formulas for Legendre Polynomials Legendreõª)¤¼ê 1 √ 1−2xt+t 2 = X∞ l=0 Pl(x)t l |t| < x± √ x 2−1 Proof 1 √ 1 − 2xt + t 2 = 1 1 − t 1 − 2(x − 1)t (1 − t) 2 −1/2 = X∞ k=0 (2k − 1)!! k! (x − 1)k t k (1 − t) −(2k+1) = X∞ k=0 (2k − 1)!! k! (x − 1)k t kX∞ n=0 (2k + n)! n!(2k)! t n C. S. Wu 1Ôù ¥¼ê()
Legendre多项式的生成函数 √1-2rt+t2 ∑P(x)<士 2 1-2rt+ (1-t)2 k-1 2k-1)!! (2k+m)! k=0 n=o n! (+k)! =0Lk=0 k!k!(-k)!(2
Properties of Legendre Polynomials (cont.) Applications of Legendre Polynomials Generating Function Recurrence Formulas for Legendre Polynomials Legendreõª)¤¼ê 1 √ 1−2xt+t 2 = X∞ l=0 Pl(x)t l |t| < x± √ x 2−1 Proof 1 √ 1 − 2xt + t 2 = 1 1 − t 1 − 2(x − 1)t (1 − t) 2 −1/2 = X∞ k=0 (2k − 1)!! k! (x − 1)k t k (1 − t) −(2k+1) = X∞ k=0 (2k − 1)!! k! (x − 1)k t kX∞ n=0 (2k + n)! n!(2k)! t n = X∞ l=0 "X l k=0 (l + k)! k!k!(l − k)! x − 1 2 k # t l C. S. Wu 1Ôù ¥¼ê()