对于多项式,得到它的所有极值并通过突出显示所有单调碎片来绘制它

Question

有人问我这个有趣的问题,我认为值得在这里发布,因为Stack Overflow上没有任何相关的线程.

假设我在长度n向量中有多项式系数pc,其中n - 1变量的多项式x可以用其原始形式表示:

pc[1] + pc[2] * x + pc[3] * x ^ 2 + ... + pc[n] * x ^ (n - 1)

R核函数polyroot可以在复数域中找到该多项式的所有根.但是我们常常对极值感兴趣,对于单变量函数,局部最小值和最大值交替出现,将函数分解成单调的片段.

我的问题是:

1. 如何在多项式的实域中获得所有极值(实际上所有的鞍点)？
2. 如何使用双色方案绘制这个多项式:红色表示上升部分,绿色表示下降部分？

将它写成函数会很好,这样我们就可以轻松地探索/可视化多项式.

例如,考虑5度的多项式:

pc <- c(1, -2.2, -13.4, -5.1, 1.9, 0.52)

Answer 1

获得多项式的所有鞍点

事实上,通过使用polyroot多项式的一阶导数可以找到鞍点.这是一个功能.

SaddlePoly <- function (pc) {
  ## a polynomial needs be at least quadratic to have saddle points
  if (length(pc) < 3L) {
    message("A polynomial needs be at least quadratic to have saddle points!")
    return(numeric(0))
    }
  ## polynomial coefficient of the 1st derivative
  pc1 <- pc[-1] * seq_len(length(pc) - 1)
  ## roots in complex domain
  croots <- polyroot(pc1)
  ## retain roots in real domain
  ## be careful when testing 0 for floating point numbers
  rroots <- Re(croots)[abs(Im(croots)) < 1e-14]
  ## note that `polyroot` returns multiple root with multiplicies
  ## return unique real roots (in ascending order)
  sort(unique(rroots))
  }

xs <- SaddlePoly(pc)
#[1] -3.77435640 -1.20748286 -0.08654384  2.14530617

评估多项式

我们需要能够评估多项式以绘制它.我的这个答案定义了一个g可以评估多项式及其任意导数的函数.在这里,我复制此功能并将其重命名为PolyVal.

PolyVal <- function (x, pc, nderiv = 0L) {
  ## check missing aruments
  if (missing(x) || missing(pc)) stop ("arguments missing with no default!")
  ## polynomial order p
  p <- length(pc) - 1L
  ## number of derivatives
  n <- nderiv
  ## earlier return?
  if (n > p) return(rep.int(0, length(x)))
  ## polynomial basis from degree 0 to degree `(p - n)`
  X <- outer(x, 0:(p - n), FUN = "^")
  ## initial coefficients
  ## the additional `+ 1L` is because R vector starts from index 1 not 0
  beta <- pc[n:p + 1L]
  ## factorial multiplier
  beta <- beta * factorial(n:p) / factorial(0:(p - n))
  ## matrix vector multiplication
  base::c(X %*% beta)
  }

例如,我们可以在其所有鞍点处评估多项式:

PolyVal(xs, pc)
#[1]  79.912753  -4.197986   1.093443 -51.871351

为单调碎片绘制具有双色方案的多项式

这是一个查看/探索多项式的函数.

ViewPoly <- function (pc, extend = 0.1) {
  ## get saddle points
  xs <- SaddlePoly(pc)
  ## number of saddle points (if 0 the whole polynomial is monotonic)
  n_saddles <- length(xs)
  if (n_saddles == 0L) {
    message("the polynomial is monotonic; program exits!")
    return(NULL)
    }
  ## set a reasonable xlim to include all saddle points
  if (n_saddles == 1L) xlim <- c(xs - 1, xs + 1)
  else xlim <- extendrange(xs, range(xs), extend)
  x <- c(xlim[1], xs, xlim[2])
  ## number of monotonic pieces
  k <- length(xs) + 1L 
  ## monotonicity (positive for ascending and negative for descending)
  y <- PolyVal(x, pc)
  mono <- diff(y)
  ylim <- range(y)
  ## colour setting (red for ascending and green for descending)
  colour <- rep.int(3, k)
  colour[mono > 0] <- 2
  ## loop through pieces and plot the polynomial
  plot(x, y, type = "n", xlim = xlim, ylim = ylim)
  i <- 1L
  while (i <= k) {
    ## an evaluation grid between x[i] and x[i + 1]
    xg <- seq.int(x[i], x[i + 1L], length.out = 20)
    yg <- PolyVal(xg, pc)
    lines(xg, yg, col = colour[i])
    i <- i + 1L
    }
  ## add saddle points
  points(xs, y[2:k], pch = 19)
  ## return (x, y)
  list(x = x, y = y)
  }

我们可以通过以下方式可视化问题中的示例多项式:

ViewPoly(pc)
#$x
#[1] -4.07033952 -3.77435640 -1.20748286 -0.08654384  2.14530617  2.44128930
#
#$y
#[1]  72.424185  79.912753  -4.197986   1.093443 -51.871351 -45.856876