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

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) = ∑(-1)^n(x-1)^n,

g(x)= ∫dt/t = ∑(-1)^n(x-1)^(n+1)/(n+1),

h'(x) = 1/(1+x-1) = ∑(-1)^n(x-1)^n/2^n,

h(x)= -ln2+∫dt/(1+t) = -ln2+∑(-1)^n(x-1)^(n+1)/[(n+1)2^n],

得 f(x)=∑(-1)^n(x-1)^(n+1)/(n+1) - ∑(-1)^n(x-1)^(n+1)/[(n+1)2^n]

= ∑(-1)^n(1+1/2^n)(x-1)^(n+1)/(n+1).

收敛域 -1