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Help me with this legendre equation. THE QUESTION IS IN THE LAST PIC. QUESTION IS NOT INCOMPLETE JUST FOLLOW THE STEPS IN THE REFERENCE PICS.
Help me with this legendre equation. THE QUESTION IS IN THE LAST PIC. QUESTION IS NOT INCOMPLETE JUST FOLLOW THE STEPS IN THE REFERENCE PICS.
\fRecurrence formulas Differentiate the generating function g(x, t) witt ag(x, t) 00 x-t ot (1 - 2xt + +2) 3/2 nen(x)tn-1 (15.15). n=0 Rearrange x - t V (1 - 2xt + t2) 1/2 = (1 - 2xt + t2) nen(x)tn-1 n=0 00 00 (x - t ) > Pn(x)th = (1-2xt + +2) > nen ( x ) tn - 1 n=0 n=0 (t - x) > Pn(x ) th + (1 - 2xt + +2 ) nen (x) th-1 = 0 n=0 n=0 11 Arfken, G., Weber, H., and Harris, F. (2013). Mathematical Methods for Physicists (7th ed.). Elsevier. Recurrence formulas 00 (1 - 2xt + t2) > nin(x)tn-1 +(t-x) Pn(x ) tn = 0 n= 0 n=0 (15.16) Expanding [ nPn (x)t-1 - 2 _ nx Pn(x ) tn + V nen ( x ) tn + 1 n=0 n=0 n=0 + Pn (x ) tn+ 1 xin(x)th = 0 (15.17) n=0 n=0 12 Arfken, G., Weber, H., and Harris, F. (2013). Mathematical Methods for Physicists (7th ed.). Elsevier. Recurrence formulas Collecting th 00 00 2 ) nxPn ( x ) tn + n=0 00 n=0 )nin ( * ) tn - 1 + nen (x ) tn+ 1 + n=0 n=0 n=0 [ (2n + 1 ) x Pr (x ) th n=0 00 = > (n+ 1 ) Pr(x )tn+ 1 + nen(x)tn-1 n=0 n=0\fRecurrence formulas (2n + 1)xPn(x) = (n+ 1)Pn+1(x) +nPn-1(x) (15.18) where n = 1, 2, 3, ... Eq. (15.18) will allow the generation of successive Pn(x) from starting values of Po and P1. For example n = 1, (2 + 1)xP1(x) = (1+1)P1+1(x) + 1P1-1(x) 3xP1 (x) = 2P2(x) + Po(x) 2P2 (x) = 3xP1 (x) - Po(x) - P2(x) = = (3x2 -1) (15.19) See table 15.1 17 Arfken, G., Weber, H., and Harris, F. (2013). Mathematical Methods for Physicists (7th ed.). Elsevier. Legendre Polynomials Table 15.1 Legendre Polynomials Po(x) = 1 PI(x) =x P2(x) = (3x2 - 1) P3(x) = (5x3 - 3x) PA(x) = (35x4 - 30x2 + 3) P5(x) = (63x5 - 70x3 + 15x) P6(x) = 16 (231x6 - 315x4 + 105x2 - 5) P7(x) = 16 (429x7 - 693x5 + 315x3 - 35x) Pg(x) = 128 (6435x8 - 12012x6 + 6930x4 - 1260x2 + 35) 18 Arfken, G., Weber, H., and Harris, F. (2013). Mathematical Methods for Physicists (7th ed.). Elsevier. Recurrence formulas Recurrence formula involving Pn Differentiate the generating function g (x, t) wrt x ag(x, t) t (1 - 2xt + +2) 3/2 - E Pi (x)tn n=0 00 t (1 - 2xt + t2) 1/2 = (1 - 2xt + +2) > PR(x)tn n= 0 Pn(x) th = (1 - 2xt + +2 ) Pr ( x ) tn n=0 n=0 00 (1 -2xt + t2) > PR(x) th - t ) Pn(x)th = 0 (15.20) n=0 n=0 Step by Step Solution
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