问题
I want to find the solution of:
-x^3+6*x^2+51*x+44=0
but with R. Is it possible?
I found the package Ryacas, but nobody seems to be able to make it work.
May sound trivial, but I'm not able to find an easy way to do this...
Do you have an alternative?
Thanks guys!
回答1:
You can use polynom package:
library(polynom)
p <- polynomial(c(44,51,6,-1))
# 44 + 51*x + 6*x^2 - x^3
solve(p)
# [1] -4 -1 11
But you simply can use the function polyroot from base package:
polyroot(c(44,51,6,-1))
# [1] -1+0i -4+0i 11+0i
If you keep the real part with Re:
Re(polyroot(c(44,51,6,-1)))
# [1] -1 -4 11
回答2:
Here we solve for the roots using the relationship between a matrix and its characteristic polynomial.
Given the polynomial a0 + a1*x^1 + a2*x^2 + x^3, define the matrix:
0 0 -a0
1 0 -a1
0 1 -a2
The eigenvalues of this matrix are the roots of the polynomial.
Substituting y = -x in your polynomial equation gives this
y^3 + 6*y^2 - 51*y + 44=0
And gives this example
> z <- matrix(c(0,1,0,0,0,1,-44,51,-6),3,3)
> z
[,1] [,2] [,3]
[1,] 0 0 -44
[2,] 1 0 51
[3,] 0 1 -6
> eigen(z)
$values
[1] -11 4 1
$vectors
[,1] [,2] [,3]
[1,] 0.6172134 0.73827166 0.98733164
[2,] -0.7715167 -0.67115606 -0.15707549
[3,] 0.1543033 -0.06711561 -0.02243936
Or, since we've substituted -y for x:
> eigen(-z)$values
[1] 11 -4 -1
See: http://www-math.mit.edu/~edelman/publications/polynomial_roots.pdf
回答3:
I just stumbled upon this question and I am not sure if anything inherently changed around the Ryacas package, but it seems to work great in 2020, here is a helpful vignette to get started: https://cran.r-project.org/web/packages/Ryacas/vignettes/getting-started.html
Following the vignette, things seem to work as expected when I run the code:
library(Ryacas)
# initialize equation:
eq <- "-x^3+6*x^2+51*x+44"
# simplify the equation:
library(glue)
yac_str(glue("Simplify({eq})"))
[1] "6*x^2-x^3+51*x+44"
# factor:
yac_str(glue("Factor({eq})"))
[1] "(-1)*(x-11)*(x+4)*(x+1)"
You can evaluate the expression like this, plugging in whatever values for x:
# evaluate
evaluate(eq,list(x=c(0,1,10,100,-100)))
[[1]]
$src
[1] "-x^3+6*x^2+51*x+44"
attr(,"class")
[1] "source"
[[2]]
[1] "[1] 44 100 154 -934856 1054944\n"
Here you can see the results where x=0 produced an answer of 44, x=1 produced an answer of 100, etc...
If you evaluated the new simplified or factored versions and evaluated those, you would of course end up with the same exact results:
evaluate(yac_str(glue("Simplify({eq})")),list(x=c(0,1,10,100,-100)))
[[1]]
$src
[1] "6*x^2-x^3+51*x+44"
attr(,"class")
[1] "source"
[[2]]
[1] "[1] 44 100 154 -934856 1054944\n"
Notice the formula changed in the $src output, but we get the same results.
Here's the factored one too:
evaluate(yac_str(glue("Factor({eq})")),list(x=c(0,1,10,100,-100)))
[[1]]
$src
[1] "(-1)*(x-11)*(x+4)*(x+1)"
attr(,"class")
[1] "source"
[[2]]
[1] "[1] 44 100 154 -934856 1054944\n"
The only real difference between what I outlined here and what's outlined in the vignette is the actual formula, and the fact that I used library(glue) instead of paste0, which is also a fair option.
来源：https://stackoverflow.com/questions/32599215/is-it-possible-to-solve-an-algebraic-equation-in-r