If the tabular points are evenly spaced, the calculation of y(x) can be simplified.
For convenience, choose n odd, and let the spacing be h. I have n x-values
Let be the interval midpoint:
I will measure from . Let the kth point be given by
Perhaps an example will make the simplification clear. Let n = 5. Shifting the polynomial so that falls at x=0, I have the linear system:
Each column is multiplied by a distinct power of h. Pulling out the h's:
Or, giving the matrices names: . The unknowns are the coefficients c. The solution is then
C is a column vector of length n. The approximation y(x) is then
,
where to remove the offset , we choose
Now. Let the product The multiplication simply divides the kth entry of u by . Let
, and we can write the general case for arbitrary n as

,

Where the approximation is calculated at an arbitrary position x in the interval spanned by the tabular points.
At this point, there is one more simplification which interests me...