Newton form vs. Lagrange form for interpolating polynomials I'm just wondering, what are the advantages of using either the Newton form of polynomial interpolation or the Lagrange form over the other?
It seems to me, that the computational cost of the two are equal, and seeing as the interpolated polynomial is unique, why ever use one over the other?
I get that they give different forms of the polynomial, but when is one form superior to the other?
 A: Here are two differences:


*

*Lagrange's form is more efficient when you have to interpolate several data sets on the same data points.

*Newton's form is more efficient when you have to interpolate data incrementally.
A: Lagrange method is mostly a theoretical tool used for proving theorems. Not only it is not very efficient when a new point is added (which requires computing the polynomial again, from scratch), it is also numerically unstable.
Therefore, Newton's method is usually used. However, there is a variation of the Lagrange interpolation, which is numerically stable and computationally efficient! Unfortunately, this method is not very known...
I am attaching a link to a paper from SIAM Review called "Barycentric Lagrange interpolation", which is not difficult to read. I hope you will find it interesting.
http://epubs.siam.org/doi/abs/10.1137/S0036144502417715
(A typo for that article is noted at https://people.sc.fsu.edu/~jburkardt/py_src/barycentric_interp_1d/barycentric_interp_1d.html)
A: Frankly, Lagrange interpolation is mostly just useful for theory. Actually computing with it requires huge numbers and catastrophic cancellations. In floating point arithmetic this is very bad. It does have some small advantages: for instance, the Lagrange approach amounts to diagonalizing the problem of finding the coefficients, so it takes only linear time to find the coefficients. This is good if you need to use the same set of points repeatedly. But all of these advantages do not make up for the problems associated with trying to actually evaluate a Lagrange interpolating polynomial.
With Newton interpolation, you get the coefficients reasonably fast (quadratic time), the evaluation is much more stable (roughly because there is usually a single dominant term for a given $x$), the evaluation can be done quickly and straightforwardly using Horner's method, and adding an additional node just amounts to adding a single additional term. It is also fairly easy to see how to interpolate derivatives using the Newton framework.
A: Here is an example of a problem that is much easier using Newton interpolation than Lagrange interpolation. Let $p(x)$ be the unique polynomial of degree $n$ such that
$$p(k) = 3^k, 0 \le k \le n.$$
What is $p(n + 1)$? 
