最佳平方逼近-python

分享一下最近学习的数值分析课后作业

一次逼近

import numpy as np
from sympy import *
import math
import matplotlib.pyplot as plt

#n代表n-1次逼近
n = 2

#建立hilbert矩阵
H = np.zeros((n,n))

for i in range(n):
    for j in range(n):
        H[i][j] = 1/(i+j+1)


f = np.zeros((n,1))

for i in range(n):
    x = symbols("x")
    f[i][0] = integrate((math.e**x)*(x**i),(x,0,1))

#a为系数向量
a = np.linalg.inv(H).dot(f)
print(a)

x = np.linspace(0,1,20)
y = math.e**(x)

def fun(a,x):
    return a[0]+x*a[1]
y_pred = fun(a,x)

plt.plot(x,y,'b')
plt.plot(x,y_pred,'r')

二次逼近

import numpy as np
from sympy import *
import math
import matplotlib.pyplot as plt

#n代表n-1次逼近
n = 3

#建立hilbert矩阵
H = np.zeros((n,n))

for i in range(n):
    for j in range(n):
        H[i][j] = 1/(i+j+1)


f = np.zeros((n,1))

for i in range(n):
    x = symbols("x")
    f[i][0] = integrate((math.e**x)*(x**i),(x,0,1))

#a为系数向量
a = np.linalg.inv(H).dot(f)
print(a)

x = np.linspace(0,1,20)
y = math.e**(x)

def fun(a,x):
    return a[0]+x*a[1]+x**2*a[2]
y_pred = fun(a,x)

plt.plot(x,y,'b')
plt.plot(x,y_pred,'r')

三次逼近

import numpy as np
from sympy import *
import math
import matplotlib.pyplot as plt

#n代表n-1次逼近
n = 4

#建立hilbert矩阵
H = np.zeros((n,n))

for i in range(n):
    for j in range(n):
        H[i][j] = 1/(i+j+1)


f = np.zeros((n,1))

for i in range(n):
    x = symbols("x")
    f[i][0] = integrate((math.e**x)*(x**i),(x,0,1))

#a为系数向量
a = np.linalg.inv(H).dot(f)
print(a)

x = np.linspace(0,1,20)
y = math.e**(x)

def fun(a,x):
    return a[0]+x*a[1]+x**2*a[2]+x**3*a[3]
y_pred = fun(a,x)

plt.plot(x,y,'b')
plt.plot(x,y_pred,'r')