f(x)=In(1+x)在x=0处的Taylor展开式为

y = ln (1 + x)
dy/dx = 1/ (1 + x)d2y/dx2 = -(1 + x)^-2d3y/dx3 = 2(1 + x)^-3d4y/dx4 = -6(1 + x)^-4d5y/dx5 = 24 (1 +x)^-5.....therefore, f(0) = ln(1+0) = ln1 = 0f'(0) = 1/(1 + 0) = 1f''(0) = -(1+0)^-2 = -1f'''(0) = 2(1 + 0)^-3 = 2f^4(0) = -6(1 + 0)^-4 = -6f^5(0) = 24 (1 + 0)^-5 = 24maclaurin's series for ln (1+x) = 0 + x - (1/2)x^2 + (2/3)!x^3 - (6/4)!x^4 + (24/5)!x^5+...= x - (1/2)x^2 + (1/3)x^3 - (1/4)x^4 + (1/5)x^5+...

令 g(x) = ln(1+x),g(0) = 0；
[ln(1+x)] ' = 1 / (1+x),g'(0) = 1；
[ln(1+x)] '' = -1 / (1+x)^2,g''(0) = -1；
[ln(1+x)] ''' = 2 / (1+x)^3,g''(0) = 2!；
一般有：[ln(1+x)] ^(k) = (-1)^(k-1) * (k-1)!/ (1+x)^k,g^(k)(0) = (-1)^(k-1) * (k-1)!；
根据泰勒展开式有：
∴ ln(1+x) = x - x^2 / 2 + x^3 / 3 + ......+ (-1)^(n-1) * x^n / n + .