Question

--Red In( - ) () -2 21 (+1) 2+1(n + m)} ml(m + 2n + 1) (1 - 1) de 22+1(n!) (2n +1) 8. By evaluating ac 2h +G ah where G(h) is the generating function for Legendre polynomials, show that 1 - 2 Σ (2n +1) Po (1 - 2ch + ha) Hence, or otherwise, prove that Pn(x) dx 2h 9. Given that {(2, 2) = ( 12h the hm-dh m>1. prove that 2am+ | 112,0)P.a)dt (m + n) (2n + 1) 20

Answer 1

2011 G=(1-24h + h²) ² Generating Function. Now we have the following relation: 100 (i-zahtu? & hh Pn(x) no ah a žh Pala) 2h 26 dh 2h d dh dh h=o (1-2 4h+62) 21 [ 2 -21052 (3406). an 00 n-l => ah En Pale). Sh nzo 00 = -h/2h-22) (1-2 4h th 2, 3 2 I hh Pulad (1=3e h=0 2h da & G dh ) an mh Pula) + . + h=o = h (2h-2x) ( 1–2 xht h?) (1-28hth?) 2 h Pula). h=o D] n -2h²+2th & 1-2xhth2 (anti) h Pula) no (1-24h +h) ? – 12 & Guti) in Pale). hao (1-2 xh th 2

Now Let 2 { (2mti) hm Pm la) 1-h? (1-2xh+h?] mo we werit Pulx) and Integrating Multiply both sides by get (1-->) Pn(a) da (1-2 4h+22) 32 § 6 mon) hin S Pen los Prasida 2 o men Now Pan (2) Pulx) = man 2 2 ntl Honce (1-2²) Palas de (ma chanzysh (anti) h 2 antl 1 n. 25 = ) 2h Pn(x) dx (1-2 wh the sh 1-42

a 9 hmidh fle.a)= (1-24h thee a u ni Koo K pik(-). Now (0-24241) a m-1 So flaia) = h hh Pkle) dh. ko - Z K=0 mika h Pk (e) dhe mtk -1 fla, a) P (n) da= h . Pic (e) Pn(x) dx dh. Kad h mtk-1 dh Py(x) Poladda N AT -1 K=0 0 Forkan s Pelad litos de Ple) Puledda - 2 ant! Zero For ktn it is a mtn-1 h dh X 2 2ntl so s tlma) Paladde. So [mta 19 ) mtn аа" (2nty (min)