Taylor Series

The expression for the order of error given above is found by comparison of the procedure with a Taylor series. Example m-file: Taylor Series Approximations to $1/(1-x)$ (http://siber.cankaya.edu.tr/ozdogan/NumericalComputations//mfiles/chapter0/demoTaylor.m demoTaylor.m)
Consider the function

\begin{displaymath}
f(x)=\frac{1}{1-x}
\end{displaymath}

Make the Taylor series expansion of this function up to third order.
demoTaylor(1.6,0.8)
All of the Taylor polynomials agree with $f(x)$ near $x_0=1.6$. The higher order polynomials agree over a larger range of $x$.
Cem Ozdogan 2010-10-09