在 Sympy 中解决不等式 [英] Solving Inequalities in Sympy

问题描述

我试图在 sympy 中解决以下不等式:

(10000/x) - 1 <0

于是我发出命令:

solve_poly_inequality( Poly((10000/x) - 1 ), '<')

但是，我得到了:

[Interval.open(-oo, 1/10000)]

然而，我的手动计算给出了 x <0 或 x >10000.

我错过了什么?由于 -1，我无法将其表示为有理函数.

提前致谢！

解决方案

您正在使用低级求解例程.我建议使用更高级别的例程 solve 或 solveset，例如

<预><代码>>>>solveset((10000/x) - 1 <0, x, S.Reals)(−∞, 0) ∪ (10000, ∞)

您的尝试是正确但看起来错误的原因是您没有指定要使用的生成器，因此 Poly 使用 1/x 作为其变量(我们称之为 g>) 所以它解决了问题 1000*g - 1 <0...当 g 小于您发现的 1/1000 时为真.

你可以通过书写看到这个生成器标识

<预><代码>>>>聚(1000/x - 1)Poly(1000*1/x - 1, 1/x, domain='ZZ')

I tried to solve the following inequality in sympy:

(10000 / x) - 1 < 0

So I issued the command:

solve_poly_inequality( Poly((10000 / x) - 1 ), '<')

However, I got:

[Interval.open(-oo, 1/10000)]

However, my manual computations give either x < 0 or x > 10000.

What am I missing? Due to the -1, I cannot represent it as a rational function.

Thanks in advance!

解决方案

You are using a low-level solving routine. I would recommend using the higher-level routines solve or solveset, e.g.

>>> solveset((10000 / x) - 1 < 0, x, S.Reals)
(−∞, 0) ∪ (10000, ∞)

The reason that your attempt is right but looks wrong is that you did not specify the generator to use so Poly used 1/x as its variable (let's call it g) so it solved the problem 1000*g - 1 < 0...which is true when g is less than 1/1000 as you found.

You can see this generator identification by writing

>>> Poly(1000/x - 1)
Poly(1000*1/x - 1, 1/x, domain='ZZ')

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