将函数f(x)=1/xˇ2+5x+4展开成x+3的幂级数，并求展开式成立的区间
展开全部
思路：
1）Partial fraction:
f(x) = (1/3)[1/(x+1) -1/(x+4)]
2) Expansion:
1/(x+1) = 1/[(x+3)-2] =-(1/2)/[1 - (x+3)/2] = -(1/2)[1 + ((x+3)/2)^2 + ... + ((x+3)/2)^n + ...] n from 0 to oo
interval: -5 < x < -1
1/(x+4) = 1/[1+(x+3)] = 1 - (x+3) + (x+3)^2 + ... + (-1)^(n)(x+3)^n + ..., n from 0 to oo
interval: -4 < x < -2
3) Combination:
上面两展开式相加后乘1/3可得。
收敛区间：-4 < x < -2
1）Partial fraction:
f(x) = (1/3)[1/(x+1) -1/(x+4)]
2) Expansion:
1/(x+1) = 1/[(x+3)-2] =-(1/2)/[1 - (x+3)/2] = -(1/2)[1 + ((x+3)/2)^2 + ... + ((x+3)/2)^n + ...] n from 0 to oo
interval: -5 < x < -1
1/(x+4) = 1/[1+(x+3)] = 1 - (x+3) + (x+3)^2 + ... + (-1)^(n)(x+3)^n + ..., n from 0 to oo
interval: -4 < x < -2
3) Combination:
上面两展开式相加后乘1/3可得。
收敛区间：-4 < x < -2
本回答被提问者和网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
