# 用Python实现Newton插值法

## 1. n阶差商实现

```def diff(xi,yi,n):
"""
param xi:插值节点xi
param yi:插值节点yi
param n: 求几阶差商
return: n阶差商
"""
if len(xi) != len(yi):  #xi和yi必须保证长度一致
return
else:
diff_quot = [[] for i in range(n)]
for j in range(1,n+1):
if j == 1:
for i in range(n+1-j):
diff_quot[j-1].append((yi[i]-yi[i+1]) / (xi[i] - xi[i + 1]))
else:
for i in range(n+1-j):
diff_quot[j-1].append((diff_quot[j-2][i]-diff_quot[j-2][i+1]) / (xi[i] - xi[i + j]))
return diff_quot
```

```xi = [1.615,1.634,1.702,1.828]
yi = [2.41450,2.46259,2.65271,3.03035]
n = 3
print(diff(xi,yi,n))```

[[2.53105263157897, 2.7958823529411716, 2.997142857142854], [3.0440197857724347, 1.0374252793901158], [-9.420631485362996]]

## 2. 牛顿插值实现

```def Newton(x):
f = yi[0]
v = []
r = 1
for i in range(n):
r *= (x - xi[i])
v.append(r)
f += diff_quot[i][0] * v[i]
return f
```

```x = 1.682
print(Newton(x))```

2.5944760289639732

## 3.完整Python代码

```def Newton(xi,yi,n,x):
"""
param xi:插值节点xi
param yi:插值节点yi
param n: 求几阶差商
param x: 代求近似值
return: n阶差商
"""
if len(xi) != len(yi):  #xi和yi必须保证长度一致
return
else:
diff_quot = [[] for i in range(n)]
for j in range(1,n+1):
if j == 1:
for i in range(n+1-j):
diff_quot[j-1].append((yi[i]-yi[i+1]) / (xi[i] - xi[i + 1]))
else:
for i in range(n+1-j):
diff_quot[j-1].append((diff_quot[j-2][i]-diff_quot[j-2][i+1]) / (xi[i] - xi[i + j]))
print(diff_quot)

f = yi[0]
v = []
r = 1
for i in range(n):
r *= (x - xi[i])
v.append(r)
f += diff_quot[i][0] * v[i]
return f
```