Question

求幂级数∑(∞,n=1)2nx^(2n-1)/(2n-1)收敛域及和函数

Answer 1

∑(∞,n=1)2nx^(2n-1)/(2n-1)收敛域及和函数
1.收敛域
显然收敛区间为(-1,1)
2nx^(2n-1)/(2n-1)=(2n-1+1)x^(2n-1)/(2n-1)=x^(2n-1)+x^(2n-1)/(2n-1)
∑(∞,n=1)x^(2n-1)在x=±1时发散,所以
收敛域为(-1,1)
2.和函数
2nx^(2n-1)/(2n-1)=(2n-1+1)x^(2n-1)/(2n-1)=x^(2n-1)+x^(2n-1)/(2n-1)
∑(∞,n=1)2nx^(2n-1)/(2n-1)=
∑(∞,n=1)x^(2n-1)+∑(∞,n=1)x^(2n-1)/(2n-1)=s1+s2
s1=x/(1-x^2)
∑(∞,n=1)x^(2n-1)/(2n-1)=s2
s2'=[∑(∞,n=1)x^(2n-1)/(2n-1)]'
=∑(∞,n=1)[x^(2n-1)/(2n-1)]'
=∑(∞,n=1)x^(2n-2)
=1/(1-x^2)
s1=∫(0到x)s1'dx=∫(0到x)1/(1-x^2)dx=1/2ln|(1+x)/(1-x)|
所以幂级数∑(∞,n=1)2nx^(2n-1)/(2n-1)收敛域及和函数为
s(x)=x/(1-x^2)+1/2ln|(1+x)/(1-x)|.

Answer 2

-1<x<1 （x=-1或者x=1都不行）所以没有等号，
设s(x)为和函数，则 积分x*s'(x)=Σ(∞,n=1)x^(2*n)=x^2/(1-x^2);
即：x*s'(x)=x^2/(1-x^2)
s'(x)=x/(1-x^2)
s(x)=-1/2*ln(1-x^2)+c;因为s(0)=0 c=0;
s(x)=-1/2*ln(1-x^2)