Taylor Polynomial

I always forget this basic series. So recall:
$$P(x)=\frac{f(x_0)(x-x_0)^0}{0!} + \frac{f'(x_0)(x-x_0)^1}{1!} + \frac{f''(x_0)(x-x_0)^2}{2!}+ \cdots$$
Sample Application: Please see numericalmethods.xls
$$\begin{split} ln(1.10) &= ln(1+0.10) \\ & \Rightarrow ln(1+x), \,\,\  x=0.10 \end{split}$$
$$\begin{split}  ln(1+x) &= \int {\frac {1}{1+x}dx} \\ &= \int{1-x+x^2-x^3+x^4-\cdots} \\ &= x - \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots \end{split}$$
If $x=0.10$, then
$$\begin{split} g(x)&=ln(1+x)\\ &= x - \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots \\ &= 0.10 - \frac{0.10^2}{2}+\frac{0.10^3}{3}-\frac{0.10^4}{4}+\frac{0.10^5}{5}-\cdots &= 0.09531 \end{split}$$
And as you involve higher polynomials, you will arrive at higher or exact answers.