I noticed that for the (real) function f ( x ) = 1 1 &#x2212;<!--

vestpn

vestpn

Answered question

2022-05-20

I noticed that for the (real) function f ( x ) = 1 1 x , f ( f ( f ( x ) = f 3 ( x ) = x for all real x.
This surprised me, and I was naturally curious about rational functions that elicit the identity after 4 or 5 or, in general, n iterations.
I looked at rational functions of the form f ( x ) = 1 z x for z R and wondered for which z did f n ( x ) := f ( f ( . . . f ( f ( x ) ) . . . ) ) = x.
After playing around for a bit on Wolfram and Desmos, I came up with the following conjecture: If n N and n 2 and f ( x ) = 1 2 cos ( π n ) x , then f n ( x ) = x.
Note that I do not claim that f is the only rational function of the form f ( x ) = 1 z x where z R .
So I have a few questions:
1) Is this conjecture true?
2) How would someone go about proving it, if it is true?
3) Is this a well-known theorem or a somewhat-immediate corollary or special case of a well-known theorem? If so, what is that theorem?
4) How would I find all z R such that if f ( x ) = 1 z x ,then f n ( x ) = x? Is such a thing easy to do?
5) What field of math would ask questions like this?

Answer & Explanation

Alessandra Weeks

Alessandra Weeks

Beginner2022-05-21Added 5 answers

The conjecture seems to be true and if it is, it could be proved by mathematical induction.I checked it for up to n = 4 and it worked. There are other values for z for example for n = 3, we also have z = 1 and for n = 4 we have z = ± 2 and z = 0
Note that with f 3 = f we have also f 5 = f and f 7 = f,...
You find these type of questions in discrete dynamical systems.
Quintacj

Quintacj

Beginner2022-05-22Added 2 answers

Thanks

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