Newton Forward Divided Difference Formula
C
The Newton polynomial can be expressed in a simplified form when $ x_0,x_1,...,x_k $ are arranged consecutively with equal spacing. Introducing the notation $h=x_{i+1} - x_i$ for each $ i = 0, 1,...,k-1$ and $ x = x_0+sh$, the difference $x-x_i$ can be written as $(s-i)h$. So the Newton polynomial becomes $$\sum_{i=0}^k = {s \choose i}i!h^i[y_0,...,y_i]. $$This is called the Newton forward divided difference formula.
Let's check this method for the next function: $$f(x*3^x)$$ with $\varepsilon = 0.0001$
import numpy as np
import math
x = np.array([-1.0, -0.5, 0, 0.5, 1.0], float)
y = np.array([-0.3333333333333333,-0.28867513459481287, 0.0, 0.8660254037844386, 3.0], float)
eps = 0.0001
n = len(x)
for i in range(n):
a = float(x[i]*(3**x[i]))
print("f(x",i,") =",a)
def coef(x, y):
x.astype(float)
y.astype(float)
n = len(x)
a = []
for i in range(n):
a.append(y[i])
for j in range(1, n):
for i in range(n-1, j-1, -1):
a[i] = float(a[i]-a[i-1])
return np.array(a)
print(coef(x, y))
r = float(-0.25)
t = float((r - x[0])/(x[1]-x[0]))
def tsum(t, n):
sum = float(1.0)
for i in range(0, n, 1):
sum = float(sum*(t-i))
return float(sum)
def Eval(a, x, t):
x.astype(float)
n = len(a)
temp = a[0]
count = 1
for i in range(1, n, 1):
temp += (tsum(t, i-1)*a[i])/math.factorial(i)
count = count + 1
print("Count of parts in the interpolation formula:",count)
return temp
result = Eval(coef(x,y),x,t)
print("Value of functions f(x) in",r,"=",result)
f(x 0 ) = -0.3333333333333333
f(x 1 ) = -0.28867513459481287
f(x 2 ) = 0.0
f(x 3 ) = 0.8660254037844386
f(x 4 ) = 3.0
[-0.33333333 0.0446582 0.24401694 0.33333333 0.35726559]
Count of parts in the interpolation formula: 5
Value of functions f(x) in -0.25 = -0.06957804087824197
