Question

def fx(x):
    v = float(x*x*x - 5*x - 7)
    return v


err = int(input("Enter zeroes in approxiamtion in solution: "))
b = float(input("Enter upper limit for root: "))
a = float(input("Enter lower limit for root: "))
c = 0.0
while fx(c) > (1 * 10 ** -err):
    c = b - ((b - a)/(fx(b) - fx(a))) * fx(b)
    print(a, b, c, sep=', ')
    a = b
    b = c
    # c = 0
    print()
print("The root is approximately = ", b)

即使循环给出了正确的值，为什么在执行此代码时仍会出现零除错误？

编辑：我将^运算符更改为**，并删除了c = 0，但现在答案甚至与实际答案并不接近。 ^接线员奇怪地给了我正确的答案。

带有**运算符的输出：

Enter zeroes in approxiamtion in solution: 3
Enter upper limit for root: 2
Enter lower limit for root: 1
The root is approximately =  2.0

带有^运算符的输出：

Enter zeroes in approxiamtion in solution: 3
Enter upper limit for root: 2
Enter lower limit for root: 1
1.0, 2.0, 6.5
Traceback (most recent call last):

2.0, 6.5, 2.1658986175115205
  File "C:/Users/anves/PycharmProjects/nt/nt.py", line 46, in <module>

    c = b - (((b - a)/(fx(b) - fx(a))) * fx(b))
6.5, 2.1658986175115205, 2.302797651275858

ZeroDivisionError: float division by zero
2.1658986175115205, 2.302797651275858, 2.934211734811194

2.302797651275858, 2.934211734811194, 2.705017710718186

2.934211734811194, 2.705017710718186, 2.7438299655247143

2.705017710718186, 2.7438299655247143, 2.7474171932954787

2.7438299655247143, 2.7474171932954787, 2.7473464241055843

2.7474171932954787, 2.7473464241055843, 2.7473465403033757

2.7473464241055843, 2.7473465403033757, 2.747346540307211

2.7473465403033757, 2.747346540307211, 2.747346540307211

最后一个值是未打印的正确值吗？

Answer 1

我刚刚运行了您的代码，发现fx(b)和fx(a)是相同的值。因此，当您计算出fx(b)减去fx(a)时，您会得到0 希望这会有所帮助。

Terminal -----------------------------------
Enter zeroes in approxiamtion in solution: 5
Enter upper limit for root: 5
Enter lower limit for root: 5
93.0
93.0
Traceback (most recent call last):
  File "d:/PythonScripts/first_python_project/curso_python.py", line 13, in <module>
    c = b - ((b - a)/(fx(b) - fx(a))) * fx(b)
ZeroDivisionError: float division by zero

添加更多内容：当您达到较低的值时，fx(b)和fx(a)会变得非常相似。像这样：

Enter zeroes in approxiamtion in solution: 3
Enter upper limit for root: 5
Enter lower limit for root: 3
93.0
5.0
3.0, 5.0, 2.8863636363636362

2.614751596543952
93.0
5.0, 2.8863636363636362, 2.825218326106125

1.424401534793665
2.614751596543952
2.8863636363636362, 2.825218326106125, 2.7520503761266304

0.08317571970829363
1.424401534793665
2.825218326106125, 2.7520503761266304, 2.747512887487003

0.002935214303143141
0.08317571970829363
2.7520503761266304, 2.747512887487003, 2.747346905212986

6.43830337132556e-06
0.002935214303143141
2.747512887487003, 2.747346905212986, 2.747346540335565

5.002736003234531e-10
6.43830337132556e-06
2.747346905212986, 2.747346540335565, 2.747346540307211

1.7763568394002505e-15
5.002736003234531e-10
2.747346540335565, 2.747346540307211, 2.747346540307211

1.7763568394002505e-15
1.7763568394002505e-15
Traceback (most recent call last):
  File "d:/PythonScripts/first_python_project/curso_python.py", line 13, in <module>
    c = b - ((b - a)/(fx(b) - fx(a))) * fx(b)
ZeroDivisionError: float division by zero

Answer 2

我看到两个问题。一种是表达式（1 * 10 ^ -err）。 Python中的运算符 ^ 表示按位XOR，10 ^ -err是负整数。因此，在找到根之后，循环不会结束。它应该是（10 ** -err）。另外，您必须将abs（f（c））与它进行比较。

第二个问题是c = 0作为根的初始值。它应该在（a，b）间隔之内，而不是零。

Answer 3

当我添加使用abs()和**的建议，然后捕获异常时，我得到如下代码：

def fx(x):
    v = float(x*x*x - 5*x - 7)
    return v


err = int(input("Enter zeroes in approxiamtion in solution: "))
b = float(input("Enter upper limit for root: "))
a = float(input("Enter lower limit for root: "))
c = 0.0

while abs(fx(c)) > ( 10 ** -err):
    try:
        c = b - ((b - a)/(fx(b) - fx(a))) * fx(b)
    except ZeroDivisionError:
        break
    print(a, b, c, sep=', ')
    a = b
    b = c
    c = 0
    print()
print("The root is approximately = ", b)

当我运行它时，我得到了：

Enter zeroes in approxiamtion in solution: 3
Enter upper limit for root: 2
Enter lower limit for root: 1
1.0, 2.0, 6.5

2.0, 6.5, 2.1658986175115205

6.5, 2.1658986175115205, 2.302797651275858

( steps omitted )

2.7473464241055843, 2.7473465403033757, 2.747346540307211

2.7473465403033757, 2.747346540307211, 2.747346540307211

The root is approximately =  2.747346540307211

这会产生正确的答案吗？

从输出的倒数第二行可以看到，最后两个数字相同（b和c），下一次循环a和b将具有这些值，因此fx(b) - fx(a)将得出{{ 1}}和一个ZeroDivisionError。

Answer 4

您需要重新设计逻辑，但是我可以告诉您(1 * 10 ** -err)并没有您认为的那样。 ^是按位互斥或运算符。对于指数，您需要**。因此您的代码应为

(1 * 10 ** -err)

我尝试运行它，由于fx（0）为负数，它立即停止运行，该值小于上面括号中生成的非常小的正值。

这是一个工作版本，尽管我不确定代码实际上完成了什么...

def fx(x):
    v = float(x*x*x - 5*x - 7)
    return v


err = int(input("Enter zeroes in approximation in solution: "))
b = float(input("Enter upper limit for root: "))
a = float(input("Enter lower limit for root: "))
c = 0.0
ctr = 0 #                           ^^^^^^^^^^^^^^^^^^
while abs(fx(c)) > (1 * 10 ** -err) and fx(b) != fx(a): # ^ is binary xor!
    ctr += 1
    c = b - ((b - a)/(fx(b) - fx(a))) * fx(b)
    print(ctr, a, b, c, sep=', ')
    a = b
    b = c
    c = 0
    print()
print("The root is approximately = ", b, ' after ', ctr, ' iterations')
input()

我尝试运行了3次，范围为1到10和1/3/8零。所有三倍的结果以b完全相同。到达那里只是花了不同的迭代次数。

Answer 5

出现此错误是因为在代码中的某个时候

让我们说fx(a)=1.0000000000000001和fy(b)=1.0

而不是fx(a)-fy(b)=0.0000000000000001，您将获得0.0。

因此零除法错误即将来临

ZeroDivisionError：浮点数被零除？

5 个答案: