Where does the original function for divided differences come from? I understand the mechanics of dealing with divided difference functions but I am trying to understand what is actually happening.
We start with an equation
$P_n(x)= a_0+a_1(x-x_0)+a_2(x-x_0)(x-x_1)+...+ a_n(x-x_0)...(x-x_n)$.
From there we take a number of points, $x_0,..,x_n$. Then we solve using divided differences to find the nth degree interpolating polynomial. My question is why are we using
$P_n(x)= a_0+a_1(x-x_0)+a_2(x-x_0)(x-x_1)+...+ a_n(x-x_0)...(x-x_n)$ as a starting equation? Why not $P_n(x)= a_0+a_1x+a_2+x+...+ a_nx^n$?
 A: The formulas for divided differences appear quite naturally when you think of computing the interpolating polynomial by gradually adding new interpolation points. Let's say that you have computed $p_k(x)$, the interpolating polynomial of some function $f$ on the points $x_0, \cdots, x_k$. How do you use this polynomial to compute $p_{k+1}(x)$, the interpolating polynomial on the points $x_0,\cdots, x_k, x_{k+1}$? In general, you can it as write
$$
p_{k+1}(x) = p_k(x) + Q_{k+1}(x),
$$
where $Q_{k+1}$ is a polynomial with degree $\leq k+1$. Since both polynomial satisfy $p(x_i) = f(x_i), i =0, \cdots, k$, you conclude that
$$
Q_{k+1}(x) = c_{k+1} (x-x_0)\cdots (x-x_k).
$$
Addicionally, by requiring that $Q_{k+1}(x_{k+1})=f(x_{k+1})$, you can fully compute $Q_{k+1}$ and therefore $p_{k+1}$.
This constant $c_{k+1}$ is precisely what we usually define a $f[x_0, \cdots, x_{k+1}]$. If you start with $x_0$ and keep adding points, you'll get the usual recursive formula for divided differences.
A posteriori, like it was mentioned in another answer, you can identify the divided differences with derivatives. For instance,
$$
f[x_0,x_1]:=\dfrac{f(x_1)-f(x_0)}{x_1-x_0} = f'(\xi), \quad \xi \in (x_0,x_1)
$$
$$
f[x_0, x_1, x_2]= \frac{f''(\xi)}{2!}, \quad \xi \in (x_0, x_2)
$$
$$ \vdots$$
$$
f[x_0, \cdots, x_n]=\frac{f^{(n)}(\xi)}{n!},\quad \xi \in (x_0,x_n)
$$
A: It helps to be aware of the beautiful analogy between calculus and the calculus of finite differences.
In the calculus of finite differences, the analog of the derivative operator is the forward difference operator $D$ defined by $Df(x) = f(x+1) - f(x)$. In calculus we learn that the derivative of $x^n$ is $n x^{n-1}$. In the difference calculus, the expression $x^n$ is replaced by the “falling factorial” $x^{\underline{n}} = x(x-1)\cdots(x-n+1)$. For example, $x^{\underline{3}} = x(x-1)(x-2)$. If $f(x) = x^{\underline{n}}$ then $Df(x) = n x^{\underline{n-1}}$. In calculus, we can take derivatives repeatedly to find a formula for the coefficients of a polynomial. In the difference calculus, if $f(x) = \sum_{n=0}^N a_n x^{\underline{n}}$ then we can apply the difference operator $D$ repeatedly to figure out the coefficients $a_n$. This is a very illuminating way to understand Newton’s divided difference method.
