浮点运算错误 [英] Floating Point Arithmetic error

问题描述

我正在使用以下函数近似一个函数的派生：

I'm using the following function to approximate the derivative of a function at a point:

def prime_x(f, x, h):

    if not f(x+h) == f(x) and not h == 0.0: 
        return (f(x+h) - f(x)) / h
    else:
        raise PrecisionError

作为一个测试，作为 f 作为 f ， x 其中 fx 是：

As a test I'm passing f as fx and x as 3.0. Where fx is:

def fx(x):

    import math
    return math.exp(x)*math.sin(x)

$ b $其中 exp（x）*（sin（x）+ cos（x））作为导数。现在，根据Google和我的计算器

Which has exp(x)*(sin(x)+cos(x)) as derivative. Now, according to Google and to my calculator

exp（3）*（sin（3）+ cos（3））= -17.050059 。

到目前为止这么好。但是当我决定使用 h 的小值测试功能时，我得到以下内容：

So far so good. But when I decided to test the function with small values for h I got the following:

print prime_x(fx, 3.0, 10**-5)
-17.0502585578
print prime_x(fx, 3.0, 10**-10)
-17.0500591423
 print prime_x(fx, 3.0, 10**-12)
-17.0512493014
print prime_x(fx, 3.0, 10**-13)
-17.0352620898
print prime_x(fx, 3.0, 10**-16)
__main__.PrecisionError: Mantissa is 16 digits

为什么当h降低（某一点之后）误差增加？我期待相反，直到 f（x + h）等于 f（x）。

Why does the error increase when h decreases (after a certain point)? I was expecting the contrary until f(x+h) was equal to f(x).

推荐答案

浮点算术（和整数运算和定点运算）具有一定的粒度：值只能以某一步长变化。对于IEEE-754 64位二进制格式，该步长大约为值（约2.22•10 ^-16 ）的大约2 ^{-52。对于物理测量来说，这非常小。}

Floating-point arithmetic (and integer arithmetic and fixed-point arithmetic) has a certain granularity: Values can only change by a certain step size. For IEEE-754 64-bit binary format, that step size is about 2^–52 times the value (about 2.22•10^–16). That is very small for physical measurements.

然而，当您使 h 非常小时，f（ x ）和f（x + h））与步长相比不是很大。差异只能是步长的整数倍。

However, when you make h very small, the difference between f(x) and f(x+h) is not very large compared to the step size. The difference can only be an integral multiple of the step size.

当导数为 d 时，f（x） 是关于 h • d 。即使您以浮点格式计算f（ x ）和f（ x + h ），测量值他们的差异必须是步长大小的倍数，所以它必须是圆的（ h ？ d / em>）• s ，其中round（ y ）是 y 四舍五入到最接近的整数。显然，当你使 h 更小时， h ？ d / s 更小，所以舍入它的效果一个整数相对较大。

When the derivative is d, the change in f(x) is about h•d. Even if you calculate f(x) and f(x+h) as well as possible in the floating-point format, the measured value of their difference must be a multiple of the step size s, so it must be round(h•d/s)•s, where round(y) is y rounded to the nearest integer. Clearly, as you make h smaller, h•d/s is smaller, so the effect of rounding it to an integer is relatively larger.

另一种观察方式是，对于给定的f（x ），有一定的计算f（x ）周围值之间的误差量。当您使 h 更小时，f（ x + h ） - f（x ）变小，错误保持不变。因此，错误相对于 h 增加。

Another way of looking at this is that, for a given f(x), there is a certain amount of error in calculating values around f(x). As you make h smaller, f(x+h)–f(x) gets smaller, but the error stays the same. Therefore, the error increases relative to h.

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