Question

给定一些 f 和微分方程 x ＆＃39;（ t ）= f （< em> x （ t ）），我如何计算 x ^{（ n ）}（< em> t ） x （ t ）？

例如，给定 f （ x （ t ））= sin（ x （吨）），我想获得 x ^（3）（ t ）=（cos（ x （ t < / em>））² - sin（ x （ t ））²）sin（ x （吨））。

到目前为止，我已经尝试了

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

给了我

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

但我不确定如何告诉SymPy Derivative(x(t), t)是什么，让它自动找出Derivative(x(t), t, t)等。

答案：

根据我在下面收到的答案，这是我的最终解决方案：

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

示例：

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

Answer 1

声明f并使用替换：

>>> f = diff(x(t))
>>> diff(sin(x(t)), t, 2).subs(f, sin(x(t)))
-sin(x(t))**3 + cos(x(t))*Derivative(sin(x(t)), t)

Answer 2

这是一种方法，它返回所有导数的列表，直到n - 顺序

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]

使用f=sp.sin返回

 [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 - 衍生物的递归函数：

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))

-sin(x(t))**3 + sin(x(t))*cos(x(t))**2

如何找到SymPy给出一阶导数的n阶导数？

答案：

2 个答案: