Question

The Algorithm of Newton's Divided-Difference. To obtain the divided-difference coefficients of the interpolating polynomial $P(x)$ on the $n + 1$ distinct numbers $x_0, x_1, \dots, x_n$ for the function $f(x)$: Input: numbers $x_0, x_1, \dots, x_n$; values $f(x_0), f(x_1), \dots, f(x_n)$. Output: the numbers $F_{0,0}, F_{1,1}, \dots, F_{n,n}$ where $$P(x) = F_{0,0} + \sum_{i=1}^{n} F_{i,i} \prod_{j=0}^{i-1} (x - x_j)$$ Step 1: For $i = 0, 1, 2, \dots, n$, set $F_{i,0} = f(x_i)$ Step 2: For $i = 1, 2, \dots, n$ For $j = 1, 2, \dots, i$ set $F_{i,j} = \frac{F_{i,j-1} - F_{i-1,j-1}}{x_i - x_{i-j}}$ Step 3: Output $(F_{0,0}, F_{1,1}, \dots, F_{n,n})$ (where $F_{i,i} = f[x_0, x_1, \dots, x_i])$

          The Algorithm of Newton's Divided-Difference.
To obtain the divided-difference coefficients of the interpolating polynomial $P(x)$
on the $n + 1$ distinct numbers $x_0, x_1, \dots, x_n$ for the function $f(x)$:
Input: numbers $x_0, x_1, \dots, x_n$; values $f(x_0), f(x_1), \dots, f(x_n)$.
Output: the numbers $F_{0,0}, F_{1,1}, \dots, F_{n,n}$ where
$$P(x) = F_{0,0} + \sum_{i=1}^{n} F_{i,i} \prod_{j=0}^{i-1} (x - x_j)$$
Step 1: For $i = 0, 1, 2, \dots, n$,
set $F_{i,0} = f(x_i)$
Step 2: For $i = 1, 2, \dots, n$
For $j = 1, 2, \dots, i$
set $F_{i,j} = \frac{F_{i,j-1} - F_{i-1,j-1}}{x_i - x_{i-j}}$
Step 3: Output $(F_{0,0}, F_{1,1}, \dots, F_{n,n})$ (where $F_{i,i} = f[x_0, x_1, \dots, x_i])$

The Algorithm of Newton's Divided-Difference.
To obtain the divided-difference coefficients of the interpolating polynomial P(x)
on the n + 1 distinct numbers x0, x1, …, xn for the function f(x):
Input: numbers x0, x1, …, xn; values f(x0), f(x1), …, f(xn).
Output: the numbers F0,0, F1,1, …, Fn,n where

P(x) = F0,0 + ∑i=1^n Fi,i∏j=0^i-1 (x - xj)

Step 1: For i = 0, 1, 2, …, n,
set Fi,0 = f(xi)
Step 2: For i = 1, 2, …, n
For j = 1, 2, …, i
set Fi,j = (Fi,j-1 - Fi-1,j-1)/(xi - xi-j)
Step 3: Output (F0,0, F1,1, …, Fn,n) (where Fi,i = f[x0, x1, …, xi])

Added by Gabriel C.

Question

Please give Ace some feedback