On Jun 4, 2013, at 10:15 PM, Hans W Borchers wrote:
Bert Gunter gunter.berton at gene.com writes:
1. This looks like a homework question. We should not do homework
here.
2. optim() will only approximate the max.
3. optim() is not the right numerical tool for this anyway.
optimize() is.
David Winsemius dwinsemius at comcast.net writes:
[...]
On Jun 4, 2013, at 10:15 PM, Hans W Borchers wrote:
In the case of polynomials, elementary math ... methods can
actually be
executed with R:
library(polynomial) # -6 + 11*x - 6*x^2 + x^3
p0 -
Thank you all! This approach, using the 'polynom' library, did the trick.
library(polynom) # -6 + 11*x - 6*x^2 + x^3
p0 - polynomial(c(-6, 11, -6, 1)) # has zeros at 1, 2, and 3
p1 - deriv(p0); p2 - deriv(p1) # first and second derivative
xm - solve(p1) # maxima and minima of p0
xmax =
My script fits a third-order polynomial to my data with something like this:
model - lm( y ~ poly(x, 3) )
What I'd like to do is find the theoretical maximum of the polynomial (i.e. the
x at which model predicts the highest y). Specifically, I'd like to predict
the maximum between 0 = x = 1.
Hello,
As for the first question, you can use ?optim to compute the maximum of
a function. Note that by default optim minimizes, to maximize you must
set the parameter control$fnscale to a negative value.
fit - lm(y ~ poly(x, 3))
fn - function(x, coefs) as.numeric(c(1, x, x^2, x^3) %*%
1. This looks like a homework question. We should not do homework here.
2. optim() will only approximate the max.
3. optim() is not the right numerical tool for this anyway. optimize() is.
4. There is never a guarantee numerical methods will find the max.
5. This can (and should?) be done
Bert Gunter gunter.berton at gene.com writes:
1. This looks like a homework question. We should not do homework here.
2. optim() will only approximate the max.
3. optim() is not the right numerical tool for this anyway. optimize() is.
4. There is never a guarantee numerical methods will find
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