newton_forward_interpolation 牛頓向前插值法

參考： https://zhuanlan.zhihu.com/p/66793653

https://www.geeksforgeeks.org/newton-forward-backward-interpolation/

插值是對自變數的任意中間值估計函式值的技術，而計算給定範圍外的函式值的過程稱為外推。

正向差值:$y_1-y_0、y_2-y_1、y_3-y_2、\cdots、y_n-y_{n - 1} $ 分別用$ d_{y0}、d_{y1}、d_{y2}、\cdots、d_{y{(n-1)}}$ 表示，稱為第一次正向差值。 因此，第一個正向差異是:

\[\Delta Y_r = Y_{r + 1} -Y_{r} \]

前向插值表如圖所示：

公式：

NEWTON’S GREGORY FORWARD INTERPOLATION FORMULA :

\[f(a+hu)=f(a)+u\Delta f(a)+\frac{u\left ( u-1 \right )}{2!}\Delta ^{2}f(a)+...+\frac{u\left ( u-1 \right )\left ( u-2 \right )...\left ( u-n+1 \right )}{n!}\Delta ^{n}f(a) \]

這個公式對於插值f(x)在給定值集合的開始附近的值特別有用。 h是$ u=\frac{(x - a)} h $ 的區間，這裡a是第一項

程式碼

[newton_forward_interpolation.py]{..\src\arithmetic_analysis\newton_forward_interpolation.py}

"""
Prepare
   1. sys.path 中增加 TheAlgorithms\src 子模組
"""
import sys
sys.path.append('E:\dev\AI\TheAlgorithms\src')

安例一

NEWTON’S GREGORY FORWARD INTERPOLATION FORMULA :

\[f(a+hu)=f(a)+u\Delta f(a)+\frac{u\left ( u-1 \right )}{2!}\Delta ^{2}f(a)+...+\frac{u\left ( u-1 \right )\left ( u-2 \right )...\left ( u-n+1 \right )}{n!}\Delta ^{n}f(a) \]

這個公式對於插值f(x)在給定值集合的開始附近的值特別有用。 h是$ u=\frac{(x - a)} h $ 的區間，這裡a是第一項

for calculating u value

def ucal(u: float, p: int) -> float:

from arithmetic_analysis.newton_forward_interpolation import  ucal
"""
"""
'''
ucal(1, 2)    #    0
ucal(1.1, 2)  #    0.11000000000000011
ucal(1.2, 2)  #    0.23999999999999994
'''
print (ucal(1, 2))
print (ucal(1.1, 2) )
print (ucal(1.2, 2) )

0
0.11000000000000011
0.23999999999999994

已知：

sin45 = 0.7071

sin50 = 0.7660

sin55 = 0.8192

sin60 = 0.8660

求 sin52

from arithmetic_analysis.newton_forward_interpolation import  ucal

# calculating factorial of given number n
def fact(n):
    f = 1
    for i in range(2, n + 1):
        f *= i
    return f
 
# Driver Code
# Number of values given
n = 4
x = [ 45, 50, 55, 60 ]
     
# y[][] is used for difference table
# with y[][0] used for input
y = [[0 for i in range(n)]
        for j in range(n)]
y[0][0] = 0.7071
y[1][0] = 0.7660
y[2][0] = 0.8192
y[3][0] = 0.8660
 
# Calculating the forward difference table
for i in range(1, n):
    for j in range(n - i):
        y[j][i] = y[j + 1][i - 1] - y[j][i - 1]
 
# Displaying the forward difference table
for i in range(n):
    print(x[i], end = "\t")
    for j in range(n - i):
        print(y[i][j], end = "\t")
    print("")
 
# Value to interpolate at
value = 52
 
# initializing u and sum
sum = y[0][0]
u = (value - x[0]) / (x[1] - x[0])
for i in range(1,n):
    sum = sum + (ucal(u, i) * y[0][i]) / fact(i)
 
print("\nValue at", value,
      "is", round(sum, 6))

45	0.7071	0.05890000000000006	-0.005700000000000038	-0.0007000000000000339	
50	0.766	0.053200000000000025	-0.006400000000000072	
55	0.8192	0.04679999999999995	
60	0.866	

Value at 52 is 0.788003