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Need help in using Resolve to prove equality over specified domain

Posted 9 years ago

I'm working on a problem from an old textbook that involves composite functions, domains, and equalities that are valid within a given domain. After being given f(x)=1/(1-x), I'm asked to find the domain of a composite function, as shown here:

f[x_]:=1/(1-x);FunctionDomain[f@*f@*f@x, x]

which gave

x < 0 || 0 < x < 1 || x > 1

Then, I'm asked to prove that Composite[f,f,f][x] is equal to x for all x in the domain of Composite[f,f,f][x]. This is how I chose to prove it.

Resolve[ForAll[x, Element[x, FunctionDomain[f@*f@*f@x, x]], 
  f@*f@*f@x == x]]

Instead of Mathematica returning a "True" or some kind of thumbs-up, it returned

Resolve::elemc: Unable to resolve the domain or region membership condition x\[Element](x<0||0<x<1||x>1). >>

So, I tried using Reals for the domain in Element, and Mathematica returned "True".

In[17]:= f[x_] := 1/(1 - x); Resolve[
 ForAll[x, Element[x, Reals], f@*f@*f@x == x]]

Out[17]= True

But how can that be? The composite function goes to infinity for x=0 and x=1.

It's likely that I don't completely understand the data-types that Resolve, ForAll, Element, and FunctionDomain are using and returning. Or, maybe I completely missed something simple. That's likely, too. How can I make this yield what I expect to see? Thanks! (Notebook attached, too. Problem at the very bottom.)

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POSTED BY: Glen Deering
4 Replies
Posted 9 years ago

Thanks again!

POSTED BY: Glen Deering

That composite certainly does not "go to infinity" at the points 0 and 1. To do so would imply that the limiting values are infinite. Since the composite is the identity nearby, the limiting values are 0 and 1 respectively.

Limit[f[f[f[x]]], x -> 0, Direction -> -1]
Limit[f[f[f[x]]], x -> 0, Direction -> 1]

(*Out[8]= 0
Out[9]= 0 *)

Limit[f[f[f[x]]], x -> 1, Direction -> -1]
Limit[f[f[f[x]]], x -> 1, Direction -> 1]

(* Out[10]= 1
Out[11]= 1 *)
POSTED BY: Daniel Lichtblau
Posted 9 years ago

Okay. Quite right about the Limit of the composite going to zero for x->0 (and going to 1 for x->1, for that matter). And, a simple plot of the the composite shows a straight line up through zero and 1 with not even a hiccup. I should have looked a little more carefully before I spoke.

However, what I noticed initially was that Mathematica encountered an infinite expression when I tried

Composition[f,f,f][x]/.x->0

It still gave me zero as the output (which I didn't notice at first), but I was focused on the infinite expression message as the reason that FunctionDomain gave me a discontinuous domain.

Well, obviously (perhaps), that's why "Reals" works as a domain quantifier in the above code. I think you gave me the answer to the problem I was working on. But why doesn't FunctionDomain work as a quantifier in this problem?

POSTED BY: Glen Deering

It just isn't the way Element works. But specification of the desired predicate is easier than you might have thought: just give the inequalities

Resolve[
 ForAll[x, FunctionDomain[f@*f@*f@x, x], f@*f@*f@x == x]]

(* Out[72]= True *)
POSTED BY: Daniel Lichtblau
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