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.