Python中的梯形规则
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Trapezoidal rule in Python
提问by user3199345
I'm trying to implement the trapezoidal rule in Python 2.7.2. I've written the following function:
我正在尝试在 Python 2.7.2 中实现梯形规则。我编写了以下函数：
def trapezoidal(f, a, b, n):
h = float(b - a) / n
s = 0.0
s += h * f(a)
for i in range(1, n):
s += 2.0 * h * f(a + i*h)
s += h * f(b)
return s
However, f(lambda x:x**2, 5, 10, 100) returns 583.333 (it's supposed to return 291.667), so clearly there is something wrong with my script. I can't spot it though.
但是， f(lambda x:x**2, 5, 10, 100) 返回 583.333（它应该返回 291.667），所以很明显我的脚本有问题。不过我看不出来。
回答by unutbu
You are off by a factor of two. Indeed, the Trapezoidal Ruleas taught in math class would use an increment like
你错了两倍。实际上，数学课中教授的梯形规则将使用类似的增量
s += h * (f(a + i*h) + f(a + (i-1)*h))/2.0
(f(a + i*h) + f(a + (i-1)*h))/2.0is averaging the height of the function at two adjacent points on the grid.
(f(a + i*h) + f(a + (i-1)*h))/2.0是函数在网格上两个相邻点的高度的平均值。
Since every two adjacent trapezoids have a common edge, the formula above requires evaluating the function twice as often as necessary.
由于每两个相邻的梯形都有一个公共边，因此上面的公式需要将函数计算为必要的两倍。
A more efficient implementation (closer to what you posted), would combine common terms from adjacent iterations of the for-loop:
更有效的实现（更接近您发布的内容）将结合来自相邻迭代的公共术语for-loop：
f(a + i*h)/2.0 + f(a + i*h)/2.0 = f(a + i*h)
to arrive at:
到达：
def trapezoidal(f, a, b, n):
h = float(b - a) / n
s = 0.0
s += f(a)/2.0
for i in range(1, n):
s += f(a + i*h)
s += f(b)/2.0
return s * h
print( trapezoidal(lambda x:x**2, 5, 10, 100))
which yields
这产生
291.66875
回答by abarnert
The trapezoidal rulehas a big /2fraction (each term is (f(i) + f(i+1))/2, not f(i) + f(i+1)), which you've left out of your code.
该梯形规则有一个大的/2分数（每学期是(f(i) + f(i+1))/2，不是f(i) + f(i+1)），你已经离开你的代码了。
You've used the common optimization that treats the first and last pair specially so you can use 2 * f(i)instead of calculating f(i)twice (once as f(j+1)and once as f(i)), so you have to add the / 2to the loop step and to the special first and last steps:
您已经使用了特殊处理第一对和最后一对的通用优化，因此您可以使用2 * f(i)而不是计算f(i)两次（一次 asf(j+1)和一次 as f(i)），因此您必须将 添加/ 2到循环步骤以及特殊的第一个和最后一个步骤：
s += h * f(a) / 2.0
for i in range(1, n):
s += 2.0 * h * f(a + i*h) / 2.0
s += h * f(b) / 2.0
You can obviously simplify the loop step by replacing the 2.0 * … / 2.0with just ….
您显然可以通过将 替换为2.0 * … / 2.0just来简化循环步骤…。
However, even more simply, you can just divide the whole thing by 2 at the end, changing nothing but this line:
然而，更简单的是，你可以在最后将整个事物除以 2，只改变这一行：
return s / 2.0
