Question

f（x）=1/(x^2+x-2)展开成（x-1）的幂级数,并确定其收敛域

Answer 1

f（x）=1/(x^2+x-2)=1/(x-1)*1/(x+2)

下面只要考虑1/(x+2)的展开式即可：

1/(x+2)=1/(3+(x-1))=1/3*1/(1+(x-1)/3)=1/3*∑[-(x-1)/3]^k

求和对k进行,k从0变化到正无穷

幂级数就容易求了吧

接下来考虑其收敛域,显然|-(x-1)/3|