On Aug 12, 2010, at 12:49 AM, Dennis Murphy wrote:
Hi:
Try the following:
f <- function(x) 5*cos(2*x)-2*x*sin(2*x)
curve(f, -5, 5)
abline(0, 0, lty = 'dotted')
This shows rather clearly that your function has multiple roots,
which isn't
surprising given that it's a linear combination of sines and
cosines. To
find a specific root numerically, use function uniroot on f, as
follows:
uniroot(f, c(0, 2))
Except he was asking for the root of the derivative. If the classroom
assignment allows use of R's limited symbolic differentiation you
could try:
> df.dx <- D(expression(5*cos(2*x)-2*x*sin(2*x)), "x")
> df.dx
-(5 * (sin(2 * x) * 2) + (2 * sin(2 * x) + 2 * x * (cos(2 * x) *
2)))
(Which as one of the examples in the deriv help page notes is not the
most simple form.)
I was assuming that the OP wanted a solution to:
d( abs(f(x)) )/dt = 0 in the domain [1,2]
So:
f.df.dx <- function (x) {
eval(parse(text=D(expression(5*cos(2*x)-2*x*sin(2*x)), "x") ) )
}
# no abs() but we should be satisfied with either a minimum or a
maximum
uniroot(f.df.dx, c(1,2) )
$root
[1] 1.958218267
$f.root
[1] 1.138013788e-05
$iter
[1] 4
$estim.prec
[1] 6.103515625e-05
It doesn't agree with my earlier method and I think this one has a
greater probablity of being correct. I don't think I needed to take
second differences.
--
David.
$root
[1] 0.6569286
$f.root
[1] -0.0001196119
$iter
[1] 6
$estim.prec
[1] 6.103516e-05
This catches the root that lies between x = 0 and x = 2. If you want
to find
a set of roots, you can try a loop. Fortunately, since the function
is even,
you really only need to find the roots on one side of zero, since
the ones
on the other side are the same with opposite sign.
lo <- seq(0, 4.5, by = 1.5)
hi <- seq(1.5, 6, by = 1.5)
roots <- numeric(length(lo))
for(i in seq_along(lo)) roots[i] <- uniroot(f, c(lo[i], hi[i]))$root
roots
See ?uniroot for other options and tolerance settings.
HTH,
Dennis
On Wed, Aug 11, 2010 at 6:21 PM, TGS <cran.questi...@gmail.com> wrote:
How would I numerically find the x value where the derivative of the
function below is zero?
x <- seq(1,2, by = .01)
y <- 5*cos(2*x)-2*x*sin(2*x)
plot(x,abs(y), type = "l", ylab = "|y|")
--
David Winsemius, MD
West Hartford, CT
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