Question

将函数f（x）=ln（x/1+x）展开成x-1的幂级数，并指出其收敛域

Answer 1

简单计算一下即可，答案如图所示

Answer 2

f(x) = ln[x/(1+x)] = ln[(1+x-1)/(2+x-1)] = ln(1+x-1) - ln(2+x-1)
= ln(1+x-1) - ln2 - ln{1+(x-1)/2] = g(x) - ln2 -h(x)
g'(x) = 1/(1+x-1) = ∑<n=0,∞>(-1)^n(x-1)^n,
g(x)= ∫<1,x>dt/t = ∑<n=0,∞>(-1)^n(x-1)^(n+1)/(n+1),
h'(x) = 1/(1+x-1) = ∑<n=0,∞>(-1)^n(x-1)^n/2^n,
h(x)= -ln2+∫<1,x>dt/(1+t) = -ln2+∑<n=0,∞>(-1)^n(x-1)^(n+1)/[(n+1)2^n],
得 f(x)=∑<n=0,∞>(-1)^n(x-1)^(n+1)/(n+1) - ∑<n=0,∞>(-1)^n(x-1)^(n+1)/[(n+1)2^n]
= ∑<n=0,∞>(-1)^n(1+1/2^n)(x-1)^(n+1)/(n+1).
收敛域 -1<x-1≤1, -1<(x-1)/2≤1, 联立解得 0<x≤2,