数值微分：错误的行为不符合预期

Question

I want to differentiate a function f(x, y) with respect to x and y numerically using the difference quotient

def dfdx_v2(x, y, h):
    return 0.5 * (f(x+h, y) - f(x-h, y)) / h


def dfdy_v2(x, y, h):
    return 0.5 * (f(x, y+h) - f(x, y-h)) / h

where h is the length of the interval. In my assumption, the implementation above should be more and more accurate the smaller h gets. However, I noticed that this is not the case. When I plot the error for h = [1e-12, 1e-4] I get the following graph:

I would have expected, that the error gets smaller and smaller if I let h go to zero. However, for very small h the error increases substantially. Another interesting observation is that area where the error oscillates very strongly. Why does that happen? I would have expected, that the error is a monotonic increasing function without any fluctuations. Are there any ideas why the error behaves this way?

Here is the code that I used to produce the graph above:

import numpy as np
import matplotlib.pyplot as plt


def f(x, y):
    return x * x + y * x + x * y**2 + np.exp(x) + np.sin(x) * np.cos(x) + np.sin(y)


def dfdx_v1(x, y):
    return 2 * x + y + y**2 + np.exp(x) + np.cos(x)**2 - np.sin(x)**2


def dfdy_v1(x, y):
    return x + 2 * x * y + np.cos(y)


def dfdx_v2(x, y, h):
    return 0.5 * (f(x+h, y) - f(x-h, y)) / h


def dfdy_v2(x, y, h):
    return 0.5 * (f(x, y+h) - f(x, y-h)) / h


def main():
    x = 2.1415
    y = -1.1415

    h_min = 1e-12
    h_max = 1e-4
    n_steps = 10000

    h = np.linspace(h_min, h_max, n_steps)

    error = list()

    for i in range(n_steps):
        error.append(np.sqrt((dfdx_v1(x, y) - dfdx_v2(x, y, h[i]))**2 + (dfdx_v1(x, y) - dfdx_v2(x, y, h[i]))**2))

    plt.plot(h, error, linewidth=0.2, label="Error")
    plt.yscale("log")
    plt.legend()
    plt.show()


if __name__ == "__main__":
    main()

---

Answer 1

The fact that the error isn't monotonically increasing is a known problem of numerical differentiation. See this link .

Basically, when step h becomes too small, expressions such as f(x+h, y) - f(xh, y) are not computed correctly due to the presence of numerical errors. This happens because computers work in finite-precision arithmetic and, therefore, compute f(x+h, y) and f(xh, y) with a small error. Subtracting two very close numbers, which f(x+h, y) and f(xh, y) are for small enough h , usually gives a result dominated by error. Dividing by h won't correct the existing error.

Answer 2

This has to do with the precision of float s in Python (or most any other programming language). What happens when h tends to zero, is that the difference between f(x, y+h) and f(x, yh) also tends to zero - however, the latter will not be equaly smooth. Depending on the exact value of the parameters, the difference will sometimes be (a tiny bit) larger and sometimes a smaller.

The error in dfdy_v2 will be bigger than in h due to the numbers being bigger. Thus the exponent of the error will be bigger. This will both lead to the rippling, due to the quantification of f and the strictly increasing error, when h is approaching zero.