数值分析，方便计算工程数值模 program newton c...Polynomial reconstruction in Newton form. parameter(nmax=300,n1max=20000) double precision a(nmax),dd(nmax,nmax-1),sum,rmax double precision x(nmax),y(nmax),yy,t1,x2(n1max),y2(n1max) open(unit=9,file='newton.out') o
具体代码如下所示：
import numpy as np
from matplotlib import pyplot as plt
from scipy.interpolate import interp1d
x=np.linspace(0,10*np.pi,num=20)
y=np.sin(x)
f1=interp1d(x,y,kind='linear')#线性插值
f2=interp1d(x,y,kind='cubic')#三次样条插值
x_pred=np.linspace(0,10*np
