What's the general form of f(x)?

- Empty

What's the general form of f(x)?

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- Empty

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

- Michele_Laino

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

- 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

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## More answers

- 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

- 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!

- Michele_Laino

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

- 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.

- dan815

|dw:1436810632074:dw|

- ganeshie8

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

- ganeshie8

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

- dan815

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

- dan815

this solujtion has to be some function of x and n

- Empty

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

- dan815

nth complex root of -1/x?

- dan815

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

- Empty

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

- dan815

|dw:1436811053979:dw|

- dan815

in n transformations we gotta end up with x again then

- ganeshie8

that is clever!

- Empty

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

- dan815

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

- Empty

You're on the right path haha :P

- ganeshie8

|dw:1436811396051:dw|

- dan815

ohh ok lemme think

- Empty

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

- dan815

page 52 of generating functionolgy

- Empty

I'll look right now... ?

- dan815

lol noo thats a joke

- Empty

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

- dan815

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

- Empty

Are you sure?

- dan815

oh 0 its gotta be 0

- dan815

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

- dan815

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

- anonymous

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

- 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 \(a^{\sum b^k} x^{b^n}=a^0 x^1=x\)

- anonymous

so \(f^{(n)}(x)=x\) where \(f^{(m)}(x)\ne x\) for \(m

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