题目

求1/x^2-x-2展开为x-1的幂级数

提问时间：2021-04-08

答案

f(x)=1/(x^2-x-2) = (1/3)[1/(x-2)-1/(x+1)]
= (-1/3)[1/(2-x)+1/(1+x)]
= (-1/3){1/[1-(x-1)]+1/[2+(x-1)]}
= (-1/3){1/[1-(x-1)]+(1/2)/[1+(x-1)/2]}
= (-1/3) ∑(x-1)^n - (1/6)∑(-1)^n*[(x-1)/2]^n
= (-1/3) ∑[1-(-1)^n/2^(n+1)](x-1)^n.
收敛域 -1