Orthogonal Polynomial
- 1. Orthogonal Polynomial Dr. Varun Kumar Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 1 / 9
- 2. Outlines 1 Introduction 2 Inner product and orthogonal polynomial 3 Orthogonal properties of Legendre’s polynomial Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 2 / 9
- 3. Introduction: ⇒ An orthogonal polynomial sequence is a family of polynomials such that any two different polynomials in the sequence are orthogonal to each other under some inner product. ⇒ The most widely used orthogonal polynomials are the classical orthogonal polynomials, consisting of the Hermite polynomials, Laguerre polynomials Jacobi polynomials Chebyshev polynomials Legendre polynomials ⇒ Developed in the late 19th century from a study of continued fractions by P. L. Chebyshev and was pursued by A. A. Markov and T. J. Stieltjes. Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 3 / 9
- 4. Inner product and orthogonal polynomial Property of Inner product 1 (a, b) = (b, a) 2 (a, b + c) = (a, b) + (a, c) 3 (a, a) > 0 ∀ a 6= 0 4 (~ a, ~ b) = |~ a|.|~ b| cos θ Orthogonal polynomial (p, q) = Z 1 −1 p(x)q(x)dx (1) Example: (A) Let p(x) = 1 and q(x) = 1 (1, 1) = Z 1 −1 1 × 1dx = 2 → Non-orthogonal (2) Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 4 / 9
- 5. Continued– (B) Let p(x) = 1 and q(x) = x (1, x) = Z 1 −1 1 × xdx = 0 → Orthogonal (3) (C) Let p(x) = x and q(x) = x (x, x) = Z 1 −1 x2 dx = 2 3 → Non-Orthogonal (4) (D) Let p(x) = x and q(x) = x2 (x, x) = Z 1 −1 x3 dx = 0 → Orthogonal (5) Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 5 / 9
- 6. Orthogonal properties of Legendre’s polynomial Prove that Z 1 −1 pm(x)pn(x)dx = 0, m 6= n = 2 2n + 1 m = n (6) Proof: Case 1: when m 6= n By Legendre’s DE (1 − x2)d2y dx2 − 2x dy dx + n(n + 1)y = 0 or d dx (1 − x2 ) dy dx + n(n + 1)y = 0 (7) Let pm(x) and pn(x) are solution of (7) Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 6 / 9
- 7. Continued– Here, y = pm(x) or pn(x) d dx (1 − x2 ) dpn dx + n(n + 1)pn = 0 (8) and d dx (1 − x2 ) dpm dx + m(m + 1)pm = 0 (9) Multiply (8) by pm(x) and (9) by pn(x) and subtracting them pm d dx (1−x2 ) dpn dx −pn d dx (1−x2 ) dpm dx + n(n+1)−m(m+1) pmpn = 0 (10) Integrating (10) between -1 to 1 Z 1 −1 pm d dx (1 − x2 ) dpn dx dx − Z 1 −1 pn d dx (1 − x2 ) dpm dx dx + n(n + 1) − m(m + 1) Z 1 −1 pm(x)pn(x)dx = 0 (11) Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 7 / 9
- 8. Applying integration by parts in integral (11) pm (1 − x2 ) dpn dx
- 11. 1 −1 − Z 1 −1 dpm dx (1 − x2 ) dpn dx dx − h pn (1 − x2 ) dpm dx
- 14. 1 −1 − Z 1 −1 dpn dx (1 − x2 ) dpx dx dx i + n(n + 1) − m(m + 1) Z 1 −1 pm(x)pn(x)dx = 0 (12) ⇒ n(n + 1) − m(m + 1) R 1 −1 pm(x)pn(x)dx = 0 ⇒ Since, m 6= n Z 1 −1 pm(x)pn(x)dx = 0 m 6= n (13) Case 2: As per Legendre’s relation (1 − 2xh + h2 )−1/2 = ∞ X n=0 hn pn(x) = p0(x) + hp1(x) + ... (14) Squaring on both side (1 − 2xh + h2 )−1 = ∞ X n=0 h2n p2 n(x) + ∞ X m,n=0,m6=n 2hm+n pm(x)pn(x) (15) Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 8 / 9
- 15. Integrating (15) between -1 to 1 Z 1 −1 1 (1 − 2xh + h2) dx = ∞ X n=0 h2n Z 1 −1 p2 n(x)dx + ∞ X m,n=0,m6=n 2hm+n Z 1 −1 pm(x)pn(x)dx (16) or ∞ X n=0 h2n Z 1 −1 p2 n(x)dx = − log(1 − 2xh + h2 )
- 18. 1 −1 = 1 h h log(1 + h) − log(1 − h) i (17) Here, log(1 + h) = h − h2 2 + h3 3 − .... and log(1 − h) = −h − h2 2 − h3 3 − ..... From (17) ∞ X n=0 h2n Z 1 −1 p2 n(x)dx = 2 h h h + h3 3 + .... i = ∞ X n=0 h2n 2n + 1 (18) Hence, Z 1 −1 p2 n(x)dx = 1 2n + 1 (19) Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 9 / 9