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python - Python 3 错误“溢出错误：（34，'数值结果超出范围'）

问题描述

我认为类似的问题已经在这里和这里得到了回答。执行梯形和辛普森 1by3 规则的代码也有类似的问题。我已经在运行良好的桌面上运行了代码，但是使用不同的机器，代码失败了。我正在共享代码和错误消息。

代码

import numpy as np

exp = 5.33333
ul = 10
ll = -10
n = 50

def f(x):                           # The function
    return np.e**x**2

def trapezoidal(ll, ul, n):           #Setting up Trapezoidal rule
    traint = (f(ll) + f(ul)) / 2
    h = (ul - ll) / 2
    for i in range(1,n):
        k = ll + i * h
        traint = traint + f(k)
    traint = h * traint
    return traint

def simpson1by3(ll, ul, n):           #Setting up Simpson 1by3 rule
    simint = (f(ll) + f(ul)) / 2
    h = (ul - ll) / 2
    for i in range(1,n,2):
        simint = simint + (2 * f(ll + i * h)) + (4 * f(ll + ((i + 1) * h)))
    simint = simint * h / 3
    return simint

res_sim = simpson1by3(ll, ul, n)
res_trap = trapezoidal(ll, ul, n)

e_rel_trap = (res_trap - exp) / exp
e_rel_simp = (res_simp - exp) / exp

f1 = open("trap.txt", "w+")
f2 = open("simp.txt", "w+")

for n in range(1,50):
    h = (ul - ll) / 2
    res_trap = trapezoidal(ll, ul, n)
    res_simp = simpson1by3(ll, ul, n)
    f1.write(str(n) + " " + str(res_trap) + '\n')
    f2.write(str(n) + " " + str(res_simp) + '\n')

f1.close()
f2.close()

错误信息是，

Traceback (most recent call last):
  File "simpson1by3+trapezoidal.py", line 30, in <module>
    res_sim = simpson1by3(ll, ul, n)
  File "simpson1by3+trapezoidal.py", line 26, in simpson1by3
    simint = simint + (2 * f(ll + i * h)) + (4 * f(ll + ((i + 1) * h)))
  File "simpson1by3+trapezoidal.py", line 11, in f
    return np.e**x**2
OverflowError: (34, 'Numerical result out of range')

标签： pythoninteger-overflow

解决方案

似乎您的积分间隔设置错误。如果我没记错，间隔应该是（ba）/2n，即，ul-ll/n或者ul-ll/2n取决于你的设置，但不是ul-ll/2。将间隔设置为ul-ll/2使您的实际积分上限太大，并且对于像 e**x**2 这样的函数，它会溢出最大 Pythonic 浮点范围（在我的机器上大约为 1.8e308 == 2^1024 == FP256 MAX )

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