# Re: [R] using integrate with optimize nested in the integration

From: Sundar Dorai-Raj <sundar.dorai-raj_at_pdf.com>
Date: Fri 29 Jul 2005 - 06:48:53 EST

Again, not a good example, since f is linear in n so the max will always be at 15.

Try this:

f <- function(x, n) -(x - 2.5 * n)^2 # max is at 2.5*n g <- function(n) {

o <- vector("numeric", length(n))
for(i in seq(along = n))
o[i] <- optimize(f, c(0, 15), maximum = TRUE, n = n[i])\$maximum     n * o
}
integrate(g, lower = 0, upper = 5)

# int_0^5 (2.5 * n^2) dn = 2.5/3 * 5^3 = 104.1667

--sundar

Gregory Gentlemen wrote:
> Thanks for the prompt reply.
> Your right, that was a weak example.
> Consider this one though:
>
> f <- function(n,x) (x-2.5)^2*n
> g <- function(y) optimize(f,c(0,15), maximum=TRUE,x=y)\$maximum*y
>
> then if you try:
> integrate(g,lower=0,upper=5)
> it produces:
> Error in optimize(f, c(0, 15), maximum = TRUE, x = y) :
> invalid function value in 'optimize'
>
> Any ideas?
> My problem is a similar more complex function in which optimizaiton depends on the value of the integrator.
>
> "Huntsinger, Reid" <reid_huntsinger@merck.com> wrote:
> In your example, f is a function, but
> optimize(f,c(0,15),maximum=TRUE)\$maximum is just a number (the point at
> which f reaches its maximum value). I'm not sure what you want, but if you
>
> f <- function(x,y) x^3 + yx + 1
>
> and defined
>
> g <- function(y) optimize(f,c(0,5),maximum=TRUE,y)\$maximum
>
> then g(t) is the x-value at which the function f(x,t) (over x in (0,5))
> reaches its maximum for this fixed t. That could then be integrated.
>
> Reid Huntsinger
>
> -----Original Message-----
> From: r-help-bounces@stat.math.ethz.ch
> [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Gregory Gentlemen
> Sent: Thursday, July 28, 2005 3:58 PM
> To: r-help@stat.math.ethz.ch
> Subject: [R] using integrate with optimize nested in the integration
>
>
> Hi guys
> im having a problem getting R to numerically integrate for some function,
> say f(bhat)*optimize(G(bhat)), over bhat. Where id like to integrate this
> over some finite range, so that here as we integrate over bhat optimize
> would return a different optimum.
>
> For instance consider this simple example for which I cannot get R to return
> the desired result:
>
> f <- function(bhat) exp(bhat)
> g <- function(bhat) optimize(f,c(0,15),maximum=TRUE)\$maximum*bhat
> integrate(g,lower=0,upper=5)
> which returns:
> 187.499393759 with absolute error < 2.1e-12
>
> However this is an approximation of : 15*(5^2/2 - 0)=187.5, not what I
> intended on getting. Its not identifying that f is a function of bhat ...
> any advice or ways I can get integrate to not treat this as a constant?
>
> Any help is appreciated.
>
> Gregoy Gentlemen
>
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