Re: [R] uniroot function question

kchkchkch wrote > > Heh. Yes, > Berend Hasselman wrote >> >> 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

Re: [R] uniroot function question

Heh. Yes, Berend Hasselman wrote > > 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)))

Re: [R] uniroot function question

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

[R] 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 = integ