Question

问题一道高数题？

Answer 1

p(x)=nx/(1-x^2)
e^[∫p(x) dx] = (1-x^2)^(-n/2)
(1-x^2)fn'(x) +nx.fn(x) =0
fn'(x) +[nx/(1-x^2)].fn(x) =0
两边乘以 (1-x^2)^(-n/2)
(1-x^2)^(-n/2).[fn'(x) +[nx/(1-x^2)].fn(x)] =0
d/dx [(1-x^2)^(-n/2).fn(x)] =0
(1-x^2)^(-n/2).fn(x) =C
fn(0) =1, =>C=1
fn(x) = (1-x^2)^(n/2)
an
=∫(0->1) fn(x) dx
=∫(0->1) (1-x^2)^(n/2) dx
=[x.(1-x^2)^(n/2)]|(0->1) +n∫(0->1) x^2.(1-x^2)^[(n-2)/2] dx
=0 -n∫(0->1) (1-x^2).(1-x^2)^[(n-2)/2] dx +n∫(0->1) (1-x^2)^[(n-2)/2] dx
= -nan +na(n-2)
(1+n)an = na(n-2)
an/a(n-2) = n/(1+n)
a(n+2)/an = (n+2)/(n+3) = 1 - 1/(n+3)
lim(n->∞) [a(n+2)/an]^n
=lim(n->∞) [1 - 1/(n+3)]^n
=e^(-1)