## Empty one year ago What's the general form of f(x)?

1. Empty

$$\underbrace{f(f(...f}_{\text{n times}}(x)))= x$$

2. Michele_Laino

I think that we have the subsequent equation: $\Large {f^n} \equiv I$ where I is the identity function

3. Empty

I have found a function that isn't the identity function that satisfies this condition, but wasn't sure if it was something perhaps known and I was curious to see how people try to tackle this! :D

4. Michele_Laino

please keep in mind that my formula above is a functional writing, keep in mind if we substitute f(x)=sin(x), for example

5. Empty

Ahhh yes ok I see, I misread it and thought you were saying that $$f= I$$ but I see now that that's not what you're saying!

6. Michele_Laino

if f(x)= sin(x), i get: $\Large \left( {{{\sin }^n}} \right)\left( x \right) = x$ where: $\Large \left( {{{\sin }^n}} \right)\left( x \right) \ne {\left( {\sin x} \right)^n}$

7. Empty

Here's a hint and little story, while playing around with these things called Rational Tangles (part of Knot Theory), there's a certain function that represents rotating a tangle of two strings by 90 degrees and the fraction it represents gets altered by this function: $$g(x)=\frac{-1}{x}$$ Which is its own inverse. So this is my hint and also how I guessed at a general solution to the question I asked, but I feel like there is an even more general solution than the one I've found so far.

8. dan815

|dw:1436810632074:dw|

9. ganeshie8

I think any function that is symmetric about $$y=x$$ works for even $$n$$ : |dw:1436810458644:dw|

10. ganeshie8

Because then the function is its own inverse and consequently $f^2 = x \implies f^{2k} = x$

11. dan815

im thinking its gotta be how to build up transformations to get to that 90 degree tangle rotation in n-1 transformations

12. dan815

this solujtion has to be some function of x and n

13. Empty

Ahhh How about adding the caveat that for instance: $$f(f(f(f(x))))=x$$ BUT $$f(f(x)) \ne x$$

14. dan815

nth complex root of -1/x?

15. dan815

oh how about f(x) is a nth root ration of x

16. Empty

I dunno, hehehe, I guess a way to check would be to plug in specific values, maybe for this one: $$f(f(f(x)))x$$

17. dan815

|dw:1436811053979:dw|

18. dan815

in n transformations we gotta end up with x again then

19. ganeshie8

that is clever!

20. Empty

Ahhh ok, so what $$n$$ do you plug in to get $$g(x)=\frac{-1}{x}$$

21. dan815

well i dont know that i thought wed just work with x straight away

22. Empty

You're on the right path haha :P

23. ganeshie8

|dw:1436811396051:dw|

24. dan815

ohh ok lemme think

25. Empty

Here's a fun hint: $$f(x)=ax^b$$ $$f(f(x))=a^{1+b}x^{b^2}$$ $$f(f(f(x)))=a^{1+b+b^2}x^{b^3}$$ $$\cdots$$ $$f^n(x)=a^{\sum_{k=0}^{n-1}b^k} x^{b^n}$$

26. dan815

page 52 of generating functionolgy

27. Empty

I'll look right now... ?

28. dan815

lol noo thats a joke

29. Empty

lol I was like wow he read further than me already rofl

30. dan815

ok so b^n has to go to 1 and a^{coefficient 1 power series = 1}

31. Empty

Are you sure?

32. dan815

oh 0 its gotta be 0

33. dan815

1+b+b^2+..+b^(n-1) = 0 b^n=1

34. dan815

1+b+b^2+..+b^(n-1)+ b^n =1

35. anonymous

An example$f(x)=-\frac{x}{x+1}$?

36. anonymous

the hint he's giving is that we need $$b$$ to be the primitive $$n$$-th root of unity, since b^n=1\\b^n-1=0\$$b-1)\sum_{k=0}^{n-1} b^k=0\implies \sum_{k=0}^{n-1} b^k=0 so this reduces \(a^{\sum b^k} x^{b^n}=a^0 x^1=x$$

37. anonymous

so $$f^{(n)}(x)=x$$ where $$f^{(m)}(x)\ne x$$ for $$m<n$$ is solved by the family $$f(x)=ax^{\omega_n}$$ where $$a$$ is a nonzero real and $$\omega_n=\exp(2\pi i/n)$$