使用 Ryacas 包或替代方案在 R 中符号求解非线性方程 [英] Solve nonlinear equation symbolically in R with `Ryacas` package or an alternative

问题描述

出于纯粹的好奇，我对是否有一个函数/包可以解决 R 中的简单非线性方程感兴趣?

Out of pure curiosity I'm interested if there is a function/package that allows to solve simple nonlinear equations in R?

假设我想(象征性地)解决0 = C + 1/x^2.上面例子的预期结果是 x = sqrt(-1/-C)

Let's say I want to (symbolically) solve 0 = C + 1/x^2. The expected result for the above example is x = sqrt(-1/-C)

我尝试了 Ryacas 包:

library("Ryacas")
Solve(yacas("C+1/x^2"))

返回一个错误:

Sym("Solve(", x, ",", y, ")") 中的错误:缺少参数y"，没有默认

Error in Sym("Solve(", x, ",", y, ")") : argument "y" is missing, with no default

所以我做到了:

Solve(yacas("C+1/x^2"), 0)

它没有返回任何有用的东西:

which returns nothing useful:

Yacas vector:
character(0)

我按照 ?yacas 上的说明安装 yacas.看起来好像 yacas 有效，demo(Ryacas) 生成输出.这是第一部分:

I followed the instructions on ?yacas to install yacas. It seems as if yacas works, demo(Ryacas) generates output. Here is the first portion:

    demo(Ryacas)
    ---- ~~~~~~

Type  <Return>   to start : 

>   x <- -3 + (0:600)/300

>   exp0 <- expression(x ^ 3)

>   exp1 <- expression(x^2 + 2 * x^2)

>   exp2 <- expression(2 * exp0)

>   exp3 <- expression(6 * pi * x)

>   exp4 <- expression((exp1 * (1 - sin(exp3))) / exp2)

>   res1 <- yacas(exp4); print(res1)
expression(3 * (x^2 * (1 - sin(6 * (x * pi))))/(2 * x^3))

>   exp5 <- expression(Simplify(exp4))

>   res2 <- yacas(exp5); print(res2)
expression(3 * (1 - sin(6 * (x * pi)))/(2 * x))

>   plot(x, eval(res2[[1]]), type="l", col="red")

<小时>

有什么提示吗?

---

Any hints?

推荐答案

我们可以使用包Ryacas(感谢@Bhas 的提示)一个接口到库yacas对于符号方程求解:

We can use package Ryacas (thanks for the hint @Bhas) an interface to the library yacas for symbolic equation solving:

library(Ryacas)

expr <- yacas("C+1/x^2 == 0")  #Generate yacas expression | note the double equals!

solv <- Solve(expr,"x") # Solve the expression for x
[1] x == root(abs(1/C), 2) * complex_cartesian(cos(argument(-1/C)/2), sin(argument(-1/C)/2))                      
[2] x == root(abs(1/C), 2) * complex_cartesian(cos((argument(-1/C) + 2 * pi)/2), sin((argument(-1/C) + 2 * pi)/2))

Yacas 显然会生成一个复数解，因为对于 C 的正值，这个方程只有复数根(负数的平方根).由于我们有一个二次方程，因此也需要两个解.complex_cartesian 部分是指复平面内的旋转，即依赖于 C 的值(基本上是 z=a*i + b).

Yacas obviously generates a complex solution because for positive values of C this equation only has complex roots (square-root of a negative-number). Two solutions is also expected, since we have a quadratic equation. The complex_cartesian part refers to the rotation in the complex plane, that is dependend on the value of C (basically the value of a in a complex number of type z=a*i + b).

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