Question

将函数f(x)=1/(x^2-2x-3) 展成x的幂级数,并写出收敛域

Answer 1

=1/(x-3)(x+1)=-1/4[1/(x+1)-1/(x-3)]=-0.25/(1+x)-1/12(1-x/3)，然后利用1/(1-x)=1+x+x^2+x^3+...，得
-0.25（1-x+x^2-x^3+x^4-...)-1/12(1+x/3+x^2/9+x^3/27+...)，合并同类项即可。收敛范围：第一个级数是|x|<1，第二个是|x/3|<1，于是要求|x|<1

Answer 2

f(x)=1/(x²-2x-3)
=1/[(x+1)(x-3)]
=1/4 [1/(x-3)-1/(x+1)]
=1/4 [-1/3·1/(1-x/3) - 1/(1+x)]
因为1/(1-x/3)=∑<n=0到∞> (x/3)^n (-3<x<3)
1/(1+x)=∑<n=0到∞> (-x)^n (-1<x<1)
则f(x)=1/4 [-1/3 ·∑<n=0到∞> (x/3)^n + ∑<n=0到∞> (-x)^n ] (-1<x<1)
=-1/4 ∑<n=0到∞>[1/3^(n+1) + (-1)^(n+1)]x^n (-1<x<1)

Answer 3

你好！

f(x) = 1/(x² -2x+3) = 1/[(x-3)(x+1)] = -1/4 [1/(x+1) + 1/(3-x) ]

1/(x+1) = ∑ (-1)^(n-1) x^(n-1) ，-1<x<1

1/(3-x) = 1/3 * 1/[ 1+ (-x/3)]
= 1/3 ∑ (-1)^(n-1) (-x/3)^(n-1)
= ∑ x^(n-1) / 3^n
-1< -x/3 <1 即 -3 < x < 3

f(x) = - 1/4 ∑ [ (-1)^(n-1) + 1/3^n ] x^(n-1) ，-1 < x < 1