There are advantages to using either version of the formula. If you use the "first" then you do need to take care about overflow/underflow when the number of interpolation points is large, but that isn't an issue here. The "first" behaves better than the "second" for extrapolation, as discussed in the paper by Trefethen et al. And yes, the choice of interpolation points is important.

But none of that should make a difference here. What's actually causing the dubious results jlokier reports isn't exactly proximity to the interpolation points -- I don't think it could happen if one of them weren't exactly zero. What's happening is that the very smallest values of x in jlokier's code are denormal numbers; that is, ones whose floating-point representation has less precision than usual.

Usually a floating-point number looks like 1.<stuff> x 2^n; for IEEE double precision, the <stuff> is 52 bits long. So if you start with 1, 1/2, 1/4, 1/8, and keep dividing by 2, you always have stuff=0 and n decreases by 1 each time. But what should you do once you reach the smallest value of n that will fit in the floating-point format? One answer is just to say that after that you get zero. But you can get a little extra dynamic range if instead you say that when n is as small as it's allowed to be, you can use 0.<stuff> instead of 1.<stuff>. These numbers have reduced precision for the sake of extra dynamic range, and they are called "denormal". Lots of things go subtly (or not so subtly) wrong when dealing with denormal numbers.

One thing that goes wrong is the interpolation formula we're discussing here. The formula looks like f(x) ~= l(x) . sum of wj f(xj) / (x-xj) where xj are the interpolation points (in this case 0 and 1), l(x) is the product of (x-xj), so in this case x(x-1), and wj = product (xj-xi) where xi ranges over all the other interpolation points besides xj, so w0 = -1 and w1 = +1. So we have f(x) ~= x(x-1) . [-10/x + 20/(x-1)]. In fact this is exact provided x is neither 0 or 1, because the thing we're approximating is a polynomial of degree <= (# interpolation points) - 1.

So, what I claimed before was: suppose x = xj + h where h is small, and suppose you compute l(x) as the product of its factors. Then we get l(x) = h times product (xj+h-xi), and the sum is wj f(xj) / h + the sum of the other terms; and if h is tiny then the other terms in the sum are tiny in comparison, and the factors of h and 1/h cancel and everything's fine.

But that doesn't happen when xj=0 and x (and hence h) is denormal -- because you can't pull the denormal trick at the large end of the range, so the reciprocal of a denormal floating-point number is floating-point infinity!

This isn't a cancellation problem. It isn't a matter of the formula going unstable when x is very close to one of the interpolation points. But it is a problem, and it never occurred to me until jlokier demonstrated it, because denormals are kinda obscure and I didn't think to think about what might happen if they get involved. It would be avoided if you made your interpolator check for values very close to one of the xj. But -- I think, but this stuff is treacherous -- it would also be avoided if you checked for infinities as well as NaNs on the output side, which would be more efficient (provided your inputs are usually sensible). Or -- I think, but this stuff is treacherous -- you could just check explicitly for denormal values of x and maybe of the xj, because if none of those is denormal this pathology can't arise.