如何用python求差商？

Question

假设有一个点集[x0, x1, x2, ..., xn]，对应的函数值为[y0, y1, y2, ..., yn]，怎样求差商（1至n阶）。

k阶差商计算公式
f(x0, x1, ..., xk) = (f(x1, x2, ..., xk) - f(x0, x1, ... x(k-1))) / (xk - x0)
举个例子，一阶差商：
f(x0, x1) = (f(x1) - f(x0)) / (x1 - x0)
f(x1, x2) = (f(x2) - f(x1)) / (x2 - x1)
二阶差商：
f(x0, x1, x2) = (f(x1, x2) - f(x0, x1)) / (x2 - x1)

Answer 1

大概是这样:

fmap = {1:1, 2:2, 3:3}

def f(*x):
    if len(x)==1:
        rc = fmap[x[0]]
        print('f({})={}'.format(x[0], rc))
        return rc
    rc = (f(*x[1:])-f(*x[:-1]))/(x[-1]-x[0])
    template = 'f({})=(f({})-f({}))/({}-{})={}'
    print(template.format(', '.join([str(i) for i in x]),
                          ', '.join([str(i) for i in x[1:]]),
                          ', '.join([str(i) for i in x[:-1]]),
                          x[-1], x[0],
                          rc))
    return rc
    
f(1, 2, 3)

结果:

f(3)=3
f(2)=2
f(2, 3)=(f(3)-f(2))/(3-2)=1.0
f(2)=2
f(1)=1
f(1, 2)=(f(2)-f(1))/(2-1)=1.0
f(1, 2, 3)=(f(2, 3)-f(1, 2))/(3-1)=0.0

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我回答过的问题: Python-QA