具有R的复平面中的多个根 [英] Multiple roots in the complex plane with R
问题描述
我一直试图找到一个函数,该函数返回方程的所有复杂解,例如:
I've been trying to find a function that returns all complex solutions of an equation such as:
16^(1/4) = 2+i0, -2+i0, 0+i2, 0-i2
就目前而言,如果我在控制台中输入 16 ^(1/4)
,它只会返回2。我可以为此编写一个函数,但是我想知道是否有
As it stands, if I enter 16^(1/4)
into the console, it only returns 2. I can write a function for this but I was wondering if there is a simple way to do this in R.
推荐答案
您需要 polyroot()
:
polyroot(z = c(-16,0,0,0,1))
# [1] 0+2i -2-0i 0-2i 2+0i
其中 z
是多项式系数的升序向量。
Where z
is a "vector of polynomial coefficients in increasing order".
在上面的示例中,我传递给 z
的向量是该方程的紧凑表示:
The vector I passed to z
in the example above is a compact representation of this equation:
-16x^0 + 0x^1 + 0x^2 + 0x^3 + 1x^4 = 0
x^4 - 16 = 0
x^4 = 16
x = 16^(1/4)
编辑:
如果 polyroot
的语法使您感到困扰,您只需要编写一个包装函数即可为您提供更好的(如果通用性较差)接口:
If polyroot
's syntax bothers you, you just could write a wrapper function that presents you with a nicer (if less versatile) interface:
nRoot <- function(x, root) {
polyroot(c(-x, rep(0, root-1), 1))
}
nRoot(16, 4)
# [1] 0+2i -2-0i 0-2i 2+0i
nRoot(16, 8)
# [1] 1.000000+1.000000i -1.000000+1.000000i -1.000000-1.000000i
# [4] 1.000000-1.000000i 0.000000+1.414214i -1.414214-0.000000i
# [7] 0.000000-1.414214i 1.414214+0.000000i
这篇关于具有R的复平面中的多个根的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!