R help - Dec 2011 - uniroot function question

I have one equation, two unknowns, so I am trying to build the solution set
by running through possible values for one unknown, and then using uniroot
to solve for the accompanying second solution, then graphing the two
vectors.

p0 = .36
f = function(x) 0.29 * exp(5.66*(x - p0))
f.integral = integrate(f, p0, 1)
p1 = p0 + .01
i = 1
n = (1 - p0)/.01
p1.vector = rep(0,n)
p2.vector = rep(0,n)
for (i in 1:n) {
	p1.vector[i] = p1
	fcn = function(p2) p1*f(p1) + (.20/5.66)*(exp(5.66*(p2 - p0)) -
exp(5.66*(p1 - p0))) + (1 - p2)*f(p2) - as.numeric(f.integral$value)
	sol = uniroot(try, lower = p1, upper = 1)
	p2.vector[i] = p2
	i = i+1
	p1 = p1 + .01
}
plot(p1.vector,p2.vector)

p1, p2 both have to be between p0 and 1, p1 < p2.  Is there a better way to
do this?  I keep getting the error that my lower and upper bounds are not of
opposite sign, but I don't know how to find the correct interval values in
that case.  This may not even be a uniroot question (although I don't know
how to find those values in a general sense), if there is a better way to do
the same thing.  

Any ideas?

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kchkchkch wrote
> 
> I have one equation, two unknowns, so I am trying to build the solution
> set by running through possible values for one unknown, and then using
> uniroot to solve for the accompanying second solution, then graphing the
> two vectors.
> 
> p0 = .36
> f = function(x) 0.29 * exp(5.66*(x - p0))
> f.integral = integrate(f, p0, 1)
> p1 = p0 + .01
> i = 1
> n = (1 - p0)/.01
> p1.vector = rep(0,n)
> p2.vector = rep(0,n)
> for (i in 1:n) {
> 	p1.vector[i] = p1
> 	fcn = function(p2) p1*f(p1) + (.20/5.66)*(exp(5.66*(p2 - p0)) -
> exp(5.66*(p1 - p0))) + (1 - p2)*f(p2) - as.numeric(f.integral$value)
> 	sol = uniroot(try, lower = p1, upper = 1)
> 	p2.vector[i] = p2
> 	i = i+1
> 	p1 = p1 + .01
> }
> plot(p1.vector,p2.vector)
> 
> p1, p2 both have to be between p0 and 1, p1 < p2.  Is there a better way
> to do this?  I keep getting the error that my lower and upper bounds are
> not of opposite sign, but I don't know how to find the correct interval
> values in that case.  This may not even be a uniroot question (although I
> don't know how to find those values in a general sense), if there is a
> better way to do the same thing.  
> 
Shouldn't the line with uniroot be (why try as function?)

sol = uniroot(fcn, lower = p1, upper = 1)

Before the line with  for(i in 1:n) insert the following

fcn = function(p2) p1*f(p1) + (.20/5.66)*(exp(5.66*(p2 - p0)) - exp(5.66*(p1
- p0))) + (1 - p2)*f(p2) - as.numeric(f.integral$value) 
curve(fcn, from=0, to=1)

And it should be obvious what the problem is.

Berend

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