性常徵分方程的幂级数解法 第9页 反复利用递推关系,就可以求得 n(n +p) (n-1)(n+p)(n+p-1)22n-4 (-)11 !(p+1)n n+1/2)(n+p+1/2)22n-1 (n+1/2)(n-1/2)(n+p+1/2(n+p-1/2)22n-3 3/2)n(p+3/2)n2m 0. 用1=v代入,即得 2r(u+1) 就有解 Jv(a (k+u+1) 用 代入,有 u2(2)=cz 当v≠(正)整数时,也可取co=2"/T(-+1),又得 (9.8) 补充讨论一下P=-1/2的情形·前面曾提解,这时仍然可以取c1=0·可为如果c1≠0 则 C2n+1 (3/2)n(1)n22n 注意解(1)n=n!,这在u2(2)中只不过是再增加一项 S++m+m()-V 即在u2(2)中只不过是再加本程一解Wu Chong-shi ￾✁✂ ✄☎✆✝✞✟✠✡☛☞✌✍✎✏✑ (✄ ) ✒ 9 ✓ ÏÐÑÒÓÔ♣Õ●Ö❁➬❘❒ c2n = − 1 n(n + ρ) 1 2 2 c2n−2 = (−) 2 1 n(n − 1)(n + ρ)(n + ρ − 1) 1 2 4 c2n−4 = · · · = (−) n n! 1 (ρ + 1)n 1 2 2n c0, (9.5) c2n+1 = − 1 (n + 1/2)(n + ρ + 1/2) 1 2 2 c2n−1 = (−) 2 (n + 1/2)(n − 1/2)(n + ρ + 1/2)(n + ρ − 1/2) 1 2 4 c2n−3 = · · · = (−) n 1 (3/2)n 1 (ρ + 3/2)n 1 2 2n c1 = 0. (9.6) Ò ρ1 = ν ✣✤●✂❒ w1(z) = c0z ν X∞ k=0 (−) k k!(ν + 1)k z 2 2k . ❝ c0 = 1 2 νΓ (ν + 1) ●Ö➚❇ Jν(z) = X∞ k=0 (−) k k!Γ (k + ν + 1) z 2 2k+ν . (9.7) Ò ρ2 = −ν ✣✤●➚ w2(z) = c0z −ν X∞ k=0 (−) k k!(−ν + 1)k z 2 2k , Ü ν 6=(▼ ) ✉t❄●❅❁❝ c0 = 2ν/Γ (−ν + 1) ●❊❒ J−ν(z) = X∞ k=0 (−) k k!Γ (k − ν + 1) z 2 2k−ν . (9.8) ➺ð✷✸❫õ ρ = −1/2 ✼❻❦✾✗ö ➻ ✽➼❇●➴❄➯Ý❁➬❝ c1 = 0 ✾❁❑ ❡❢ c1 6= 0 ● ◆ c2n+1 = (−) n (3/2)n(1)n 1 2 2n c1, ⑩❶❇ (1)n = n! ●➴✒▲ w2(z) r✶❈￾❆❯➽❲❫➘ z −1/2X∞ n=0 c2n+1z 2n+1 = c1 X∞ n=0 (−) n n!Γ (n + 3/2) z 2 2n+1/2 · r π 2 = c1 r π 2 J1/2(z). ✂▲ w2(z) r✶❈￾❆❯➾❲❋❀❫❇✾