同情，我如何声明Piecewise子函数中多个变量具有多个限制的函数？
这是我的背景和尝试：
from sympy import Piecewise, Symbol, exp
from sympy.abc import z
x1 = Symbol('x1')
x2 = Symbol('x2')
f = 2*pow(z,2)*exp(-z*(x1 + x2 + 2))
p = Piecewise((f, z > 0 and x1 > 0 and x2 > 0), (0, True))
我收到的错误是：
TypeError Traceback (most recent call last)
<ipython-input-47-5e3db02fe3dc> in <module>()
----> 1 p = Piecewise((f, z > 0 and x1 > 0 and x2 > 0), (0, True))
C:\Anaconda3\lib\site-packages\sympy\core\relational.py in __nonzero__(self)
193
194 def __nonzero__(self):
--> 195 raise TypeError("cannot determine truth value of Relational")
196
197 __bool__ = __nonzero__
TypeError: cannot determine truth value of Relational
啊，有一种同情And为此功能：
from sympy import Piecewise, Symbol, exp, And
from sympy.abc import z
x1 = Symbol('x1')
x2 = Symbol('x2')
f = 2*pow(z,2)*exp(-z*(x1 + x2 + 2))
p = Piecewise((f, And(z > 0, x1 > 0, x2 > 0)), (0, True))
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