麻省理工學院6.00 Python中的牛頓法3

Question

這是麻省理工學院OCW 6.00使用Python進行計算和編程的第一個問題集的一部分。首先，我創建了一個函數來評估給定x值的多項式。然後是計算給定多項式的導數的函數。使用這些函數，我創建了一個函數，用於評估給定多項式和x值的一階導數。

然後，我試圖創建一個函數來估計任何給定的多項式在容差（ε）內的根。

測試用例處於預期輸出的底部。

我是新來的編程和新的Python，所以我在代碼中包含了一些註釋來解釋我認爲代碼應該做的事情。

def evaluate_poly(poly, x): 
""" Computes the polynomial function for a given value x. Returns that value.""" 
answer = poly[0] 
for i in range (1, len(poly)): 
    answer = answer + poly[i] * x**i 
return answer 


def compute_deriv(poly): 
""" 
#Computes and returns the derivative of a polynomial function. If the 
#derivative is 0, returns (0.0,).""" 
dpoly =() 
for i in range(1,len(poly)): 
    dpoly = dpoly + (poly[i]*i,) 

return dpoly 

def df(poly, x): 
"""Computes and returns the solution as a float to the derivative of a polynomial function 
""" 
dx = evaluate_poly(compute_deriv(poly), x) 
#dpoly = compute_deriv(poly) 
#dx = evaluate_poly(dpoly, x) 
return dx 




def compute_root(poly, x_0, epsilon): 
""" 
Uses Newton's method to find and return a root of a polynomial function. 
Returns a float containing the root""" 
iteration = 0 
fguess = evaluate_poly(poly, x_0) #evaluates poly for first guess 
print(fguess) 
x_guess = x_0 #initialize x_guess 
if fguess > 0 and fguess < epsilon: #if solution for first guess is close enough to root return first guess 
    return x_guess 
else: 
    while fguess > 0 and fguess > epsilon: 
     iteration+=1 
     x_guess = x_0 - (evaluate_poly(poly,x_0)/df(poly, x_0)) 
     fguess = evaluate_poly(poly, x_guess) 
     if fguess > 0 and fguess < epsilon: 
      break #fguess where guess is close enough to root, breaks while loop, skips else, return x_guess 
     else: 
      x_0 = x_guess #guess again with most recent guess as x_0 next time through while loop 
print(iteration) 
return x_guess 




#Example: 
poly = (-13.39, 0.0, 17.5, 3.0, 1.0) #x^4 + 3x^3 + 17.5x^2 - 13.39 
x_0 = 0.1 
epsilon = .0001 
print (compute_root(poly, x_0, epsilon)) 
#answer should be 0.80679075379635201 

第3個函數返回正確的答案，但compute_root（牛頓法）似乎並沒有進入while循環，因爲當我運行單元print(iteration)打印0。我會想，既然if fguess > 0 and fguess < epsilon:應爲返回false測試用例（聲明print(fguess)打印-13.2119），解釋器會去else並進入while循環，直到找到一個解決方案，是0

小量內我試圖消除第一ifelse康迪所以我只有一個return聲明，我也遇到了同樣的問題。

什麼可能導致函數跳過else case/while循環？我很難過！

感謝您的期待和/或幫助！

Answer 1

這似乎只是一個小小的疏忽。請注意，fguess的打印值爲-13.2119。在您的while條件（在else從compute_root）您需要fguess > 0 and fguess < epsilon，這是不符合，所以沒有進一步完成，你沒有迭代退出。

相反：

while fguess < 0 or fguess > epsilon: 

會給你你需要的東西：

-13.2119 
7 
0.806790753796352