# New notation : $f[0,0.1,0.2,0.3,0.4]$

I came across this question:

Suppose $f(x)=x^3-x+\frac{1}{4}$, What's the value of $f [0,0.1,0.2,0.3,0.4]$?

The problem is, I have no idea what is being asked, I'm unfamiliar with the notation "$f[0,0.1,0.2,0.3,0.4]$".

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I'd agree with user12477's suggestion about context. Sensible people don't just start using weird notation without first explaining it. – Gerry Myerson Feb 14 '12 at 11:11
@GerryMyerson: I have a bunch of questions without context, otherwise I wouldn't ask here. – Gigili Feb 14 '12 at 11:50
@Gigili: what Gerry is telling you here is that you should have mentioned the book/paper/whatever where you saw this notation. This is explicitly requested in the popup for the notation tag, among other things. – J. M. Feb 14 '12 at 13:05

@Gigili: Interpolation polynomials in Newton form go something like $d_0 + (x-x_1)(d_1 +(x-x_2)(d_2+\cdots))$. An $n$-th order polynomial is supposed to only have $n+1$ of those $d_k$'s (in much the same way that it can only have $n+1$ coefficients), so... – J. M. Feb 14 '12 at 12:14
As user12477 pointed out, it's Newton's divided differences interpolation polynomial and as J.M. pointed out, while $f(x)$ is a cubic function (polynomial of degree $n=3$), the divided difference on $n+1=4$ points (more than $n$ points generally) is equal to zero.