OK, looks sensible, although I said either abs() OR squared
differences. I don't think it makes sense to use both the L1 and the
L2 metric at the same time.
--
David.
On Aug 12, 2010, at 4:58 PM, TGS wrote:
# just to clean it up for my own understanding, the "difference"
approach as you had suggested would be
x <- seq(.2, .3, by = .00001)
f1 <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f1(x), type = "l")
abline(h = -.1)
abline(v = x[which.min(abs(diff((f1(x) - (-.1))**2)))], lty =
'dotted')
points(x = x[which.min(abs(diff((f1(x) - (-.1))**2)))], y = -.1)
# and the uniroot approach is:
x <- seq(.2, .3, by = .01)
f1 <- function(x){
x*cos(x)-2*x**2+3*x-1
}
f2 <- function(x){
-.1
}
f3 <- function(x){
f1(x) - f2(x)
}
plot(x,f1(x), type = "l")
abline(h = -.1)
abline(v = uniroot(f = f3, interval = c(.2, .3))$root, lty = 'dotted')
points(x = uniroot(f = f3, interval = c(.2, .3))$root, y = -.1)
# Thanks David!
On Aug 12, 2010, at 1:33 PM, David Winsemius wrote:
On Aug 12, 2010, at 4:15 PM, TGS wrote:
David, I was expecting this to work but how would I specify the
vector in "diff()" in order for the following to work?
x <- seq(.2, .3, by = .01)
f <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f(x), type = "l")
abline(h = -.1)
abline(v = x[which.min(abs(diff(c(-.1, f(x)))))], lty = 'dotted')
f2 <- function(x) -0.1
f3 <- function(x) f(x) -f2(x)
abline(v=uniroot(f3, c(0.2, 0.3) )$root)
points(x=uniroot(f3, c(0.2, 0.3) )$root, y= -0.1)
If you are going to use the differences, then you probably want to
minimize either the abs() or the square of the differences.
--
David.
On Aug 12, 2010, at 1:00 PM, David Winsemius wrote:
On Aug 12, 2010, at 3:54 PM, TGS wrote:
Actually I spoke too soon David.
I'm looking for a function that will either tell me which point is
the intersection so that I'd be able to plot a point there.
Or, if I have to solve for the roots in the ways which were
demonstrated yesterday, then would I be able to specify what the
horizontal line is, for instance in the case where y (is-not) 0?
Isn't the abline h=0 represented mathematically by the equation y=0
and therefore you are solving just for the zeros of "f" (whaich are
the same as for (f-0)? If it were something more interesting, like
solving the intersection of two polynomials, you would be solving
for the zeros of the difference of the equations. Or maybe I have
not understood what you were requesting?
On Aug 12, 2010, at 12:47 PM, David Winsemius wrote:
On Aug 12, 2010, at 3:43 PM, TGS wrote:
I'd like to plot a point at the intersection of these two curves.
Thanks
x <- seq(.2, .3, by = .01)
f <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f(x), type = "l")
abline(h = 0)
Would this just be the uniroot strategy applied to "f"? You then
plot the x and y values with points()
David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
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