Question

一道关于幂级数展开的问题

f(x)=(1/4)*ln[(1+x)/(1-x)]+(1/2)*arctanx-x
展开成x的幂级数
请高手给出详细步骤，不胜感激!

Answer 1

f(x)=(1/4)*ln[(1+x)-ln(1-x)]+(1/2)*arctanx-x
已知当|x|<1时,
ln(1+x)=x-x^2/2+x^3/3-x^4/4+……=∑(-1)^(n+1) * x^n/n n from 1 to ∞
同理ln(1-x)=-x-x^2/2-x^3/3-x^4/4-……=∑(-1) * x^n/n n from 1 to ∞
ln(1+x)-ln(1-x)=2[x+x^3/3+x^5/5+……]=2∑x^(2n+1) / (2n+1) n from 0 to ∞
arctanx=∫[0->x] 1/(1+x^2) dx
当|x|<1时,它=∫[0->x] 1-x^2+x^4-x^6+…… dx= x-x^3/3+x^5/5-x^7/7……
=∑(-1)^n*x^(2n+1)/(2n+1) n from 0 to ∞

所以f(x)=(1/4)*ln[(1+x)/(1-x)]+(1/2)*arctanx-x
=(1/2)∑x^(2n+1) / (2n+1) + (1/2)∑(-1)^n*x^(2n+1)/(2n+1) - x
=[x+x^5/5+x^9/9+x^13/13+……]-x=∑x^(4n+1)/(4n+1) n from 1 to ∞ |x|<1

Answer 2

解答如下：