# [R] uniroot function question

Berend Hasselman bhh at xs4all.nl
Wed Dec 14 18:57:00 CET 2011

```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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