Newton method for interpolation of polynomials I have a function $f(x) = \frac1{1+25x^2}$, and 7 equally spaced nodes on the interval $[-1, 1]$
My points are $(-1, 1/27), (-0.6, 11), (-0.3, 4.25), (0, 2), (0.3, 4.25), (0.6, 11), (1, 1/27)$. 
I'm asked to find the interpolating polynomial $p(x)$ of degree 6 for $f(x)$ by the Newton approach, but how would the interpolating polynomial $p(x)$ be any different from $f(x)$?
 A: Are you sure you have the correct data points? These should lie on the curve 
\begin{equation}
f(x) = \frac{1}{1+25x^2}.
\end{equation}
But if you look at them overlayed with $f(x)$ they clearly do not...
The interpolating polynomial, $p_n(x)$ is a polynomial of order n, so in this case it will be of order 6, because you have $n+1 = 7$ data points. So this is clearly different from your $f(x)$, because $f(x)$ is not even a polynomial for starters.
Also, $p_6(x)$ will pass through your 7 data points that (should) lie on $f(x)$. But, it will not be the same as $f(x)$ between each of your data points; this is because it interpolates between them (hence the whole idea of this exercise).
The interpolating polynomial $p_6(x)$ will take the form
\begin{split}
p_6(x) = a_0 & + a_1(x - x_0) + a_2(x - x_0)(x - x_1) + ... \\
& + a_6(x - x_0)(x - x_1)(x - x_2)(x - x_3)(x - x_4)(x - x_5)
\end{split}
Now to evaluate this polynomial you need the data points $x_n$; you already know these. And you need to calculate the coefficients $a_n$ using recursive divided difference formulae.
