Question

高数求函数f(x)=1/(x^2-5x+6)展开成(x-5)的幂级数,

Answer 1

记t=x-5,展开成t的幂级数即可

x=t+5

f(x)=1/(x-2)(x-3)=1/(x-3)-1/(x-2)=1/(t+5-3)-1/(t+5-2)=1/(t+2)-1/(t+3)

=1/[2(1+t/2)]-1/[3(1+t/3)]

=1/2[1-t/2+t^2/4-t^3/8+...]-1/3[1-t/3+t^2/9-t^3/27+...]

=1/6-(1/4+1/9)t+(1/8-1/27)t^3-.