Question

幂级数的展开！！求解答！！

幂级数的展开！！求解答！！如图，最好是手写！！

Answer 1

g(x) =1/(x+1) => g(1) =1/2

g'(x) =-1/(x+1)^2 => g'(1)/1! =-1/4
g''(x) =2/(x+1)^3 => g''(1)/2! =1/8
..
g^(n)(x) =(-1)^n . n!/(x+1)^(n+1) => g^(n)(1)/n! = (-1/2)^n
1/(x+1) = 1/2 -(1/4)(x-1)+....+(-1/2)^n.(x-1)^n +....
similary
1/(x+2) = 1/3 -(1/9)(x-1)+....+(-1/3)^n.(x-1)^n +....
/
1/(x^2+3x+2)
=1/[(x+2)(x+1)]
=1/(x+1) - 1/(x+2)
=[ 1/2 -(1/4)(x-1)+....+(-1/2)^n.(x-1)^n +....] -[1/3 -(1/9)(x-1)+....+(-1/3)^n.(x-1)^n +....]
= 1/6 - (5/36)(x-1)+....+[ (-1/2)^n + (-1/3)^n ] (x-1)^n +....
收敛区域: (-1,1)