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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.

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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.

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\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

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