Taylor Series for $\log(x)$

Solution 1:

$$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots \qquad (|x|<1)$$

There is no expansion around $x=1$ because the log is singular at $0$.

Solution 2:

Abromowitz & Stegun gives a number of forms.

Among these, the bilinear expansion is known for its used in digital filter theory:

$$\log(z) = 2\left[\left({z-1\over z+1}\right)+ {1\over3} \left({z-1\over z+1}\right)^3 + {1\over5} \left({z-1\over z+1}\right)^5+ \cdots\right],$$

for $\Re z > 0.$

Series of log(x)

Solution 3:

For $x \in \mathbb{R}$ satisfying $0 < x < 2$,

$$f(x) = \ln(x) = \left(x-1\right)-\frac{1}{2}\left(x-1\right)^2 + \frac{1}{3} \left(x-1\right)^3-\frac{1}{4} \left(x-1\right)^4 + \cdots$$ $$ f(x) = \displaystyle\sum\limits_{n=1}^{\infty} \left[\frac{\left(-1\right)^{n+1}}{n}\left(x-1\right) ^n\right] $$