在下面的代码中，当我选择例如“0”等于1时，列表“近似”在打印时显示两个元素，它只显示一个元素（初始x）。
这是什么原因？
#This exercise shows an immediate way to find the root of a real valued funciton, using successive better approximations
#This method is known as Newton Raphson method
print 'Find the root of a given function - NEWTON RAPHSONS METHOD'
print 'The function is the following: ...'
x=input('Choose an initial estimate:') #An initial estimate is necessary to be choosen to start the iteration process
x=float(x) #The number inserted is transformed into a floating point number
max_n_iterations=input('Choose max n. iterations:') #The user decides the maximum number of iterations to run
approximations = [] #Vector collecting all the intermediate solutions; i.e. the approximated roots evaluated before reaching the final solution
iterations= []
def f(x): #The given function has to be inserted manually in the code
return 2*x**3+45*x+1/x**2+16
def D(f): #Evaluates the first derivative of the given function, using the definition of derivative
def df(x, h): #x is the initial estimate, h is the increment
return (f(x+h) - f(x))/h #Difference quotient
return df #First derivative
def newtons_method(f, x, h=0.000001, epsilon=0.000001,): #This is the main process: f is the given function, x and h are the same as above, epsilon is the tolerance level. Epsilon and h have to be choosen sufficiently small
df = D(f) #df is just the first derivative, as before
for i in range(max_n_iterations): #The code runs until the maximum number of iterations is reached
x1 = x - f(x)/df(x, h) #Essence of Newton Raphson method: the iteration process
approximations.append(x) #Every intermediate solution is collected into a vector, as defined above
iterations.append(1)
if abs(x1 - x) < epsilon: #When the absolute difference between two successive roots is less than the tolerance level (i.e. the sequence of solutions strongly converges), the program exists the cycle
break
x = x1 #The next solution becomes the starting solution in the new cycle
return x #Final solution
def final_solution(): #The final solution from the Newton Raphson method
return newtons_method(f,x)
df=D(f) #These values have to be inserted again to allow the execution of the final step
h=0.000001
epsilon=0.000001
x=newtons_method(f,x)
if abs((x-f(x)/df(x,h))-x) < epsilon: #If (strong) convergence has been reached
print 'Solution is:', final_solution() #Prints the final solution
print 'Approximations before reaching convergence:', approximations #Prints the vector of intermediate solutions
print 'Convergence has been reached after', len(iterations), 'iterations'
print 'Newton Raphson method was successful!'
elif abs((x-f(x)/df(x,h))-x) >= epsilon: #If (strong) convergence has not been reached
print 'Approximated solution is:', final_solution()
print 'Approximations evaluated:', approximations
print 'Convergence has not been reached after', max_n_iterations, 'iterations'
print 'Newton Raphson method was not successful'
最佳答案
在newtons_method()返回值之前，它必须再次调用自己。例如：
def newtons_method(f, x, h=0.000001, epsilon=0.000001,):
...
if abs((x - f(x)/df(x, h))-x)< epsilon:
print 'Solution is:', round(newtons_method(f,x),6) # function called again here
...
return x
else:
print 'Approximated solution is:', round(newtons_method(f,x),4) # and again here
...
return x
所以对
newtons_method()的第一个调用永远不会返回，因为它必须在return之前调用自己，然后那个函数调用必须在return之前调用自己，然后…您能修改代码使
newtons_method()不会以这种方式递归调用吗？关于python - 牛顿拉夫森法方程求解器算法，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/25388213/