How do you prove that the divided differences of two sets are equal?

Question

My text book is saying that if two sets are equal $\{x_0, x_1,\dots , x_i\} = \{y_0, y_1,\dots ,y_j\}$, then their divided differences are equal. ( $f[y_0, y_1, \dots ,y_j] = f[x_0, x_1, \dots , x_i]$. ) Why?!

Well of course the polynomial interpolation is the same for both sets,....now I'm just not sure what the question wants.

Is this proposition implying that the order of the points isn't important? If so how do I prove that?

Answer 1

There is essentially two options here.

1. Your textbook defines $f[x_0,x_1,\dotsc,x_n]$ as the coefficient of $x^n$ in the polynomial of degree at most $n$ which interpolates $f$ at the nodes $x_i$. This polynomial is unique and so it cannot depend on the order of the nodes, hence $$f[x_0,x_1,\dotsc,x_n]=f[z_0,z_1,\dotsc,z_n]$$ whenever $\{z_0,z_1,\dotsc,z_n\}$ is a permutation of $\{x_1,x_2,\dotsc,x_n\}$. Your textbook will use an argument inspired by Neville's algorithm to derive the recurrence relation satisfied by the divided differences.
2. Your textbook defines the divided difference $f[x_0,x_1,\dotsc,x_n]$ recursively and eventually deduces that $f[x_0,x_1,\dotsc,x_n]$ is the coefficient of $x^n$ in the polynomial of degree at most $n$ which interpolates $f$ at the nodes $x_i$. The previous argument can then be applied.