# -*- coding: utf-8 -*-
import numpy as np
import matplotlib.pyplot as plt
def f(x):
return np.sin(x)
def intf(x):
return -np.cos(x)
a = 2
b = 3
# use N draws
N= 10000
X = np.random.uniform(low=a, high=b, size=N) # N values uniformly drawn from a to b
Y =f(X) # CALCULATE THE f(x)
# 蒙特卡洛法计算定积分：面积=宽度*平均高度
Imc= (b-a) * np.sum(Y)/ N
exactval=intf(b)-intf(a)
print ("Monte Carlo estimation=",Imc, "Exact number=", intf(b)-intf(a))
# --How does the accuracy depends on the number of points(samples)? Lets try the same 1-D integral
# The Monte Carlo methods yield approximate answers whose accuracy depends on the number of draws.
Imc=np.zeros(1000)
Na = np.linspace(0,1000,1000)
exactval= intf(b)-intf(a)
for N in np.arange(0,1000):
X = np.random.uniform(low=a, high=b, size=N) # N values uniformly drawn from a to b
Y =f(X) # CALCULATE THE f(x)
Imc[N]= (b-a) * np.sum(Y)/ N
plt.plot(Na[10:],np.sqrt((Imc[10:]-exactval)**2), alpha=0.7)
plt.plot(Na[10:], 1/np.sqrt(Na[10:]), 'r')
plt.xlabel("N")
plt.ylabel("sqrt((Imc-ExactValue)$^2$)")
plt.show()
