高等数学的证明题关于n阶导数的
d[(x^2-1)^n]/dx=2nx(x^2-1)^(n-1)
==>
(x^2-1)d[(x^2-1)^n]/dx=2nx(x^2-1)^n
==>
d^(n+1){(x^2-1)d[(x^2-1)^n]/dx}/dx^(n+1)=
=d^(n+1){2nx(x^2-1)^n}/dx^(n+1)
==>
(x^2-1)d^(n+2)[(x^2-1)^n]/dx^(n+2)+
+2(n+1)xd^(n+1)[(x^2-1)^n]/dx^(n+1)+
+(n+1)nd^n[(x^2-1)^n]/dx^n=
=2nxd^(n+1){(x^2-1)^n}/dx^(n+1)+
+2n(n+1)d...全部
d[(x^2-1)^n]/dx=2nx(x^2-1)^(n-1)
==>
(x^2-1)d[(x^2-1)^n]/dx=2nx(x^2-1)^n
==>
d^(n+1){(x^2-1)d[(x^2-1)^n]/dx}/dx^(n+1)=
=d^(n+1){2nx(x^2-1)^n}/dx^(n+1)
==>
(x^2-1)d^(n+2)[(x^2-1)^n]/dx^(n+2)+
+2(n+1)xd^(n+1)[(x^2-1)^n]/dx^(n+1)+
+(n+1)nd^n[(x^2-1)^n]/dx^n=
=2nxd^(n+1){(x^2-1)^n}/dx^(n+1)+
+2n(n+1)d^n{(x^2-1)^n}/dx^n
==>
(1-x^2)d^(n+2)[(x^2-1)^n]/dx^(n+2)-
-2xd^(n+1){(x^2-1)^n}/dx^(n+1)+
+n(n+1)d^n{(x^2-1)^n}/dx^n=0
==>
(1-x^2)Pn"(x)-2xPn'(x)+n(n+1)Pn(x)=0。
。收起
