如何使用SymPy查找给定一阶导数的n阶导数？

用户541686

鉴于一些˚F和微分方程X “（吨）= ˚F（X（吨）），我如何计算X ^（Ñ）（吨）在以下方面X（吨）？

例如，给定f（x（t））= sin（x（t）），我想获得x ^（3）（t）=（cos（x（t））² − sin（x（t））²）sin（x（t））。

到目前为止，我已经尝试过

>>> from sympy import diff, sin
>>> from sympy.abc import x, t
>>> diff(sin(x(t)), t, 2)

这给了我

-sin(x(t))*Derivative(x(t), t)**2 + cos(x(t))*Derivative(x(t), t, t)

但是我不确定如何告诉SymPy是什么Derivative(x(t), t)，并Derivative(x(t), t, t)自动找出答案，等等。

回答：

这是根据我在下面收到的答案得出的最终解决方案：

def diff(x_derivs_known, t, k, simplify=False):
    try: n = len(x_derivs_known)
    except TypeError: n = None
    if n is None:
        result = sympy.diff(x_derivs_known, t, k)
        if simplify: result = result.simplify()
    elif k < n:
        result = x_derivs_known[k]
    else:
        i = n - 1
        result = x_derivs_known[i]
        while i < k:
            result = result.diff(t)
            j = len(x_derivs_known)
            x0 = None
            while j > 1:
                j -= 1
                result = result.subs(sympy.Derivative(x_derivs_known[0], t, j), x_derivs_known[j])
            i += 1
            if simplify: result = result.simplify()
    return result

例：

>>> diff((x(t), sympy.sin(x(t))), t, 3, True)
sin(x(t))*cos(2*x(t))

Stelios

这是一种方法，它返回直到n-th阶的所有导数的列表

import sympy as sp

x = sp.Function('x')
t = sp.symbols('t')

f = lambda x: x**2 #sp.exp, sp.sin
n = 4 #3, 4, 5

deriv_list = [x(t), f(x(t))]  # list of derivatives [x(t), x'(t), x''(t),...]
for i in range(1,n):
    df_i = deriv_list[-1].diff(t).replace(sp.Derivative,lambda *args: f(x(t)))
    deriv_list.append(df_i)

print(deriv_list)

[x(t), x(t)**2, 2*x(t)**3, 6*x(t)**4, 24*x(t)**5]

随着f=sp.sin它的回报

 [x(t), sin(x(t)), sin(x(t))*cos(x(t)), -sin(x(t))**3 + sin(x(t))*cos(x(t))**2, -5*sin(x(t))**3*cos(x(t)) + sin(x(t))*cos(x(t))**3]

编辑：用于计算n-th导数的递归函数：

def der_xt(f, n):
    if n==1:
        return f(x(t))
    else:
        return der_xt(f,n-1).diff(t).replace(sp.Derivative,lambda *args: f(x(t)))

print(der_xt(sp.sin,3))