Mehrdad - 8 months ago 43

Python Question

Given some *f* and the differential equation *x*'(*t*) = *f*(*x*(*t*)), how do I compute *x*^{(n)}(*t*) in terms of *x*(*t*)?

For example, given *f*(*x*(*t*)) = sin(*x*(*t*)),

I want to obtain *x*^{(3)}(*t*) = (cos(*x*(*t*))^{2} − sin(*x*(*t*))^{2}) sin(*x*(*t*)).

So far I've tried

`>>> from sympy import diff, sin`

>>> from sympy.abc import x, t

>>> diff(sin(x(t)), t, 2)

which gives me

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

but I'm not sure how to tell SymPy what

`Derivative(x(t), t)`

`Derivative(x(t), t, t)`

Here's my final solution based on the answers I received below:

`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

Example:

`>>> diff((x(t), sympy.sin(x(t))), t, 3, True)`

sin(x(t))*cos(2*x(t))

Answer

Here is one approach that returns a list of all derivatives up to `n`

-th order

```
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]`

With `f=sp.sin`

it returns

`[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]`

EDIT: A recursive function for the computation of the `n`

-th derivative:

```
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`

Source (Stackoverflow)