Finding the fixed points of a function

Question

I am trying to find the fixed point of a logistic distribution function and determine how the fixed point changes for different parameter values. The code looks like:

nfxp.reps <- 0
err <- 10
p <- seq(0, 1, by = 0.0001)
pold <- p
gamma <- 6
k <- 3
while (err > 1E-12) {
  nfxp.reps <- nfxp.reps + 1
  if (nfxp.reps > 999) { 
    stop("The number of NFXP reps needs to be increased. \n")
  } 
  pnew <- plogis(-k + gamma * pold) 
  err <- max(abs(pnew - pold))
  pold <- pnew
}

The above code works very well in the above parameter choices: gamma and k - find 3 fixed points, 2 stable and 1 unstable (where p=0.5). However if I change the above parameter non-proportionally, where the middle fixed point is either above or below 0.5, say for:

gamma<-7
k<-3

The loop is unable to locate the middle fixed point which is p=0.3225 (if gamma=7, k=3)

UtopiaLtd · Accepted Answer · 2013-04-18 21:38:44Z

1

Fixed point iteration by construction cannot find the unstable equilibria in your setup since it is repelling. In other words, unless you start right at the unstable equilibria the nfxp algorithm will always move away from it.

An alternative approach is to use a root solving approach. Of course, there are no guarantees that all fixed points will be found. Here is a simple example:

library(rootSolve) # for the uniroot.all function
pfind<-function(k=3,gamma=7) 
{
pdiff <-function(p0) p0-plogis(-k + gamma * p0) 
uniroot.all(p.diff,c(0,1))
}
> fps= pfind()
> fps
[1] 0.08036917 0.32257992 0.97925817

We can verify this:

pseq =seq(0,1,length=100)
plot(x=pseq ,y= plogis(-k + gamma *pseq),type= 'l')
abline(0,1,col='grey')
points(matrix(rep(fps,each=2), ncol=2, byrow=TRUE),pch=19,col='red')

Hope this helps.

edited Apr 18, 2013 at 21:38

UtopiaLtd

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answered Apr 18, 2013 at 21:03

acryd

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Comments

agstudy · Accepted Answer · 2013-03-06 23:02:29Z

1

I rearrange your code in a new function.

p.fixed <- function(p0,tol = 1E-9,max.iter = 100,k=3,gamma=7,verbose=F){
  pold <- p0
  pnew <-  plogis(-k + gamma * pold) 
  iter <- 1
    while ((abs(pnew - pold) > tol) && (iter < max.iter)){
      pold <- pnew
      pnew <- plogis(-k + gamma * pold) 
      iter <- iter + 1
      if(verbose)
         cat("At iteration", iter, "value of p is:", pnew, "\n")
    }
    if (abs(pnew - pold) > tol) {
      cat("Algorithm failed to converge")
      return(NULL)
    }
    else {
      cat("Algorithm converged, in :" ,iter,"iterations \n")
      return(pnew)
    }
}

some tests:

p.fixed(0.2,k=3,gamma=7)
Algorithm converged, in : 30 iterations 
[1] 0.08035782
> p.fixed(0.2,k=5,gamma=5)
Algorithm converged, in : 7 iterations 
[1] 0.006927088
> p.fixed(0.2,k=5,gamma=5,verbose=T)
At iteration 2 value of p is: 0.007318032 
At iteration 3 value of p is: 0.006940548 
At iteration 4 value of p is: 0.006927551 
At iteration 5 value of p is: 0.006927104 
At iteration 6 value of p is: 0.006927089 
At iteration 7 value of p is: 0.006927088 
Algorithm converged, in : 7 iterations 
[1] 0.006927088

edited Mar 6, 2013 at 23:02

answered Mar 6, 2013 at 22:55

agstudy

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3 Comments

user1682980 Over a year ago

Thanks for the above. However as you can see the above code also is unable to locate the unstable equilibria under any parameter specifications. Under k=3 and gamma=7 the fixed points should be (0.08035, 0.3225, 097926) it seems to always miss out the middle one.

agstudy Over a year ago

@user1682980 how do you know the result? how can I check this?

user1682980 Over a year ago

just through a graphical test, the final code requires addition of a large number of covariates to the structure so this is just a simple loop to start with so that I can be certain that all fixed points are found, when there are multiple fixed points for certain parameter values.

great_minds_think_alike · Accepted Answer · 2018-11-08 18:46:35Z

0

I don't really understand which distribution you are using; this is my standard code for the fixed point method which I always use and change if needed (you have to fill in your function f(x) in ftn;

fixed_point <- function(x0, eps = 1e-6, max_iter = 100){
  x.old <- x0
  x.new <- ftn(x.old)
  iter <- 1
  while((abs(x.new-x.old) > eps) && (iter < max_iter){
    x.old <- x.new
    x.new <- ftn(x.old)
    iter <- iter + 1
  }
 if (abs(x.new-x.old) > eps){
  cat("failed to converge\n")
  return(NULL)
 } else {
  return(x.new)
 }
}

answered Nov 8, 2018 at 18:46

great_minds_think_alike

314 bronze badges

Comments

peterweethetbeter · Accepted Answer · 2019-01-23 08:54:06Z

Not sure what exactly you did wrong but I will give you my code which always works for finding the fixed point. The last function below can be used to compute function g, which is defined as g(x) = c*ftn(x) + x.

fixpt_own <- function(x0, tol = 1e-6, max.iter = 100) {
  xold <- x0
  xnew <- ftn_g(xold)
  iter <- 1
  cat("At iteration 1 value of x is:", xnew, "\n")
  while ((abs(xnew-xold) > tol) && (iter < max.iter)) {
    xold <- xnew;
    xnew <- ftn_g(xold);
    iter <- iter + 1
    cat("At iteration", iter, "value of x is:", xnew, "\n")
  }
  if (abs(xnew-xold) > tol) {
    cat("Algorithm failed to converge\n")
    return(NULL)
  } else {
    cat("Algorithm converged\n")
    return(xnew)
  }
}
fixpt_own(3,1e-6,150)


ftn_g <- function(x){
  c <- 4;
  g <- c*(((1+x)/x - log(x))/(1+x)^2) + x;
  return(g)
}

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