Sympy:简化平方的平方根 [英] Sympy: Simplifying square roots of squares

问题描述

Sympy 似乎无法简化涉及变量平方根的表达式:

在[28]中:a = x**2在[29]中:b = a**(1/2)在 [30] 中:b出[30]:0.5⎛ 2⎞⎝x ⎠在 [31] 中:b.simplify()出[31]:0.5⎛ 2⎞⎝x ⎠

我不认为它适用于 simplify 的其他变体，特别是我认为 b.powsimp() 应该可以工作.

在[32]中:b.powsimp()出[32]:0.5⎛ 2⎞⎝x ⎠

有谁知道为什么这不起作用，或者我做错了什么?

解决方案

你的例子有两个问题.

第一个 sqrt(x**2)==x 仅用于 正 实数.

其次，对于 SymPy 1/2 和 0.5 不是一回事.第一个是 Rational 实例，第二个是 float 实例.

最后一个例子:

<预><代码>>>>x = Symbol('x', real=True)>>>(x**2)**(1./2)∣x∣**1.0>>>(x**2)**(S(1)/2) # S() 是 sympify() 的缩写∣x∣

sympify 将 python 对象转换为更合适的 SymPy 对象.

Sympy does not seem to be able to simplify an expression where the square root of a square of a variable is involved:

In [28]: a = x**2
In [29]: b = a**(1/2)
In [30]: b
Out[30]: 
    0.5
⎛ 2⎞   
⎝x ⎠   
In [31]: b.simplify()
Out[31]: 
    0.5
⎛ 2⎞   
⎝x ⎠   

I do not get this to work with other variants of simplify, in particular I would have thought that b.powsimp() should work.

In [32]: b.powsimp()
Out[32]: 
    0.5
⎛ 2⎞   
⎝x ⎠   

Does anyone know why this does not work, or what I am doing wrong?

解决方案

There are two problems with your example.

First sqrt(x**2)==x only for positive real numbers.

Second, for SymPy 1/2 and 0.5 are not the same things. The first is a Rational instance, the second is a float instance.

Finally, an example:

>>> x = Symbol('x', real=True)
>>> (x**2)**(1./2)
∣x∣**1.0
>>> (x**2)**(S(1)/2) # S() is short for sympify()
∣x∣

sympify transforms python objects to more adequate SymPy objects.

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