[R] what is the purpose of an error message in uniroot?

[Chris Andrews]

Matt,

Some time back I didn't like the uniroot restriction either so I wrote a short function "manyroots" that breaks an interval into many shorter intervals and looks for a single root in each of them.  This function is NOT guaranteed to find all roots in an interval even if you specify many subintervals.  Graphing is always a good idea.  There is always a chance the function dips to or below the axis and back up in an arbitrarily short interval.  For example, f(x) = x^2.  If one of your subintervals doesn't happen to end at 0, you're out of luck.  (This is in contrast to uniroot working on a continuous function that is positive at one end of an interval and negative at the other end:  at least one root is guaranteed and uniroot will find it given enough iterations.)  Furthermore, if you are working with a polynomial, just use polyroot.

Chris

(Sorry for the dearth of code comments in the following)

manyroots <- function(f, interval, ints=1, maxlen=NULL,
  lower = min(interval), upper = max(interval),
  tol = .Machine$double.eps^0.25, maxiter = 1000, ...)
{
    if (!is.numeric(lower) || !is.numeric(upper) || lower >=
        upper)
        stop("lower < upper  is not fulfilled")
    if (is.infinite(lower) || is.infinite(upper))
        stop("Interval must have finite length")
    if (!is.null(maxlen))
        ints <- ceiling((upper-lower)/maxlen)
    if (!is.numeric(ints) || length(ints)>1 || floor(ints)!=ints || ints<1)
        stop("ints must be positive integer")

    ends <- seq(lower, upper, length=ints+1)
    fends <- numeric(length(ends))
    for (i in seq(along=ends)) fends[i] <- f(ends[i], ...)

    zeros <- iters <- prec <- rep(NA, ints)

    for (i in seq(ints)) {
        cat(i, ends[i], ends[i+1], fends[i], fends[i+1], "\n")
        if (fends[i] * fends[i+1] > 0) {
#            cat("f() values at end points not of opposite sign\n")
            next;
        }
        if (fends[i] == 0 & i>1) {
#            cat("this was found in previous iteration\n")
            next;
        }
        
        val <- .Internal(zeroin(function(arg) f(arg, ...), ends[i],
            ends[i+1], tol, as.integer(maxiter)))
        if (as.integer(val[2]) == maxiter) {
            warning("Iteration limit (", maxiter, ") reached in interval (",
              ends[i], ",", ends[i+1], ").")
        }
        zeros[i] <- val[1]
        iters[i] <- val[2]
        prec[i] <- val[3]
    }
    zeros <- as.vector(na.omit(zeros))
    fzeros <- numeric(length(zeros))
    for (i in seq(along=zeros)) fzeros[i] <- f(zeros[i], ...)

    list(root = zeros, f.root = fzeros,
        iter = as.vector(na.omit(iters)),
        estim.prec = as.vector(na.omit(prec)))
}

gg <- function(x) x*(x-1)*(x+1)
manyroots(gg, c(-4,4), 13, maxiter=200, tol=10^-10)

hh <- function(x,x2) x^2-x2
manyroots(hh, c(-10, 10), maxlen=.178, x2=9)

manyroots(sin, c(-4,20), maxlen=.01)
#but
ss <- function(x) sin(x)^2
manyroots(ss, c(-4,20), maxlen=.01)
plot(ss, -4,20)
abline(h=0)



From: rolf at math.unb.ca

mckellercran at gmail.com wrote:

> > This is probably a blindingly obvious question:

        Yes, it is.

> > Why does it matter in the uniroot function whether the f() values at
> > the end points that you supply are of the same sign?

        Plot some graphs.

        Think about the *name* of the function --- *uni*root.

        Does that ring any bells?

        And how do you know there *is* a root in the interval
        in question?  Try your ``uniroot2'' on f(x) = 1+x2
        and the interval [-5,5].

        To belabour the point --- if the f() values are of the
        same sign, then there are 0, or 2, or 4, or ....
        roots in the interval in question.

Rolf,
Only if f is continuous.... (of course finding roots of discontinuous functions is a greater challenge)

        The ***only chance*** you have of there being a unique
        root is if the f() values are of opposite sign.

        The algorithm used and the precision estimates returned
        presumably depend on the change of sign.  You can get
        answers --- sometimes --- if the change of sign is not
        present, but the results could be seriously misleading.

        Without the opposite sign requirement the user will often
        wind up trying to do something impossible or getting
        results about which he/she is deluded.

                                cheers,

                                        Rolf Turner
                                        rolf at math.unb.ca

P. S.  If the f() values are of the same sign, uniroot() DOES
NOT give a warning!  It gives an error.

                                        R. T.




-- 
Christopher Andrews, PhD
SUNY Buffalo, Department of Biostatistics
242 Farber Hall, candrews at buffalo.edu, 716 829 2756