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

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Empty
 one year ago
Best ResponseYou've already chosen the best response.0$$\underbrace{f(f(...f}_{\text{n times}}(x)))= x$$

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I think that we have the subsequent equation: \[\Large {f^n} \equiv I\] where I is the identity function

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please keep in mind that my formula above is a functional writing, keep in mind if we substitute f(x)=sin(x), for example

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh 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
 one year ago
Best ResponseYou've already chosen the best response.0if 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
 one year ago
Best ResponseYou've already chosen the best response.0Here'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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I think any function that is symmetric about \(y=x\) works for even \(n\) : dw:1436810458644:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Because then the function is its own inverse and consequently \[f^2 = x \implies f^{2k} = x\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2im thinking its gotta be how to build up transformations to get to that 90 degree tangle rotation in n1 transformations

dan815
 one year ago
Best ResponseYou've already chosen the best response.2this solujtion has to be some function of x and n

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh How about adding the caveat that for instance: $$f(f(f(f(x))))=x$$ BUT $$f(f(x)) \ne x $$

dan815
 one year ago
Best ResponseYou've already chosen the best response.2nth complex root of 1/x?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh how about f(x) is a nth root ration of x

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2in n transformations we gotta end up with x again then

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh ok, so what \(n\) do you plug in to get $$g(x)=\frac{1}{x}$$

dan815
 one year ago
Best ResponseYou've already chosen the best response.2well i dont know that i thought wed just work with x straight away

Empty
 one year ago
Best ResponseYou've already chosen the best response.0You're on the right path haha :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436811396051:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Here's a fun hint: $$f(x)=ax^b$$ $$f(f(x))=a^{1+b}x^{b^2}$$ $$f(f(f(x)))=a^{1+b+b^2}x^{b^3}$$ $$\cdots$$ $$f^n(x)=a^{\sum_{k=0}^{n1}b^k} x^{b^n}$$

dan815
 one year ago
Best ResponseYou've already chosen the best response.2page 52 of generating functionolgy

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I'll look right now... ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0lol I was like wow he read further than me already rofl

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ok so b^n has to go to 1 and a^{coefficient 1 power series = 1}

dan815
 one year ago
Best ResponseYou've already chosen the best response.21+b+b^2+..+b^(n1) = 0 b^n=1

dan815
 one year ago
Best ResponseYou've already chosen the best response.21+b+b^2+..+b^(n1)+ b^n =1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0An example\[f(x)=\frac{x}{x+1}\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the hint he's giving is that we need \(b\) to be the primitive \(n\)th root of unity, since $$b^n=1\\b^n1=0\\(b1)\sum_{k=0}^{n1} b^k=0\implies \sum_{k=0}^{n1} b^k=0$$ so this reduces \(a^{\sum b^k} x^{b^n}=a^0 x^1=x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \(f^{(n)}(x)=x\) 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 \(\omega_n=\exp(2\pi i/n)\)
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