R fixed point of a function

I am looking for a fixed point x when f(x)=x of a function, ofcourse numerically, but I have no idea how to solve it with R, I am trying with fsolve with following code, but possibly its not the right way to write this...I am not getting anything...Many thanks in advance

library(pracma)  
p<- c(x=0.1, y=0.1)
f1 <- function(p) {
                expPfLeft<- (160*p[1]) + ((1-p[1])*200)
                expPfRight<- (10*p[1])+ ((1-p[1])*370)
                expPfTop <- (200*p[2]) + ((1-p[2])*160)
                expPfBottom <- (370*p[2]) + ((1-p[2])*10) 

                return(c (exp(2*expPfLeft)/((exp(2*expPfLeft)+exp(2*expPfRight))) , 
                          exp(2*expPfTop)/((exp(2*expPfTop))+(exp(2*expPfBottom))) ) )


        }


fsolve(f1,p)

Assuming your function is defined correctly. You are looking for f(p[1], p[2]) = c(p[1], p[2])

You can create fix your answer by changing the return statement to:

return(c(exp(2*expPfLeft)/((exp(2*expPfLeft)+exp(2*expPfRight))), exp(2*expPfTop)/((exp(2*expPfTop))+(exp(2*expPfBottom)))) - p)

We can observe that your optimization function are actually two independent functions. p[1] is used in Left and Right. p[2] is used in Top and Bottom.

We can separate your functions.

f.x <- function(p) {
  expPfLeft<- (160*p) + ((1-p)*200)
  expPfRight<- (10*p)+ ((1-p)*370)
  return(exp(2*expPfLeft)/((exp(2*expPfLeft)+exp(2*expPfRight))) - p)
}

f.y <- function(p) {
  expPfTop <- (200*p) + ((1-p)*160)
  expPfBottom <- (370*p) + ((1-p)*10) 
  return(exp(2*expPfTop)/((exp(2*expPfTop))+(exp(2*expPfBottom))) - p)
}

Simplifying your expression so we can cheat a little for the starting value

We can see that there is only one approximate approximate solution at x = 1.

fsolve(f.x, 1)

$x
[1] 1

$fval
[1] 0

And similarly for the second function, there is a solution at 0.4689.

fsolve(f.y, 0.1)

$x
[1] 0.4689443

$fval
[1] 4.266803e-09

Doing all this defeats the purpose of optimization, and leads me to believe that your optimization function is mis-defined.