A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Empty

  • one year ago

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

  • This Question is Closed
  1. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1436810632074:dw|

  9. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  12. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this solujtion has to be some function of x and n

  13. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    nth complex root of -1/x?

  15. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  16. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1436811053979:dw|

  18. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    in n transformations we gotta end up with x again then

  19. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is clever!

  20. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  22. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You're on the right path haha :P

  23. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1436811396051:dw|

  24. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ohh ok lemme think

  25. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    page 52 of generating functionolgy

  27. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'll look right now... ?

  28. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    lol noo thats a joke

  29. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  30. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  31. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Are you sure?

  32. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh 0 its gotta be 0

  33. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  34. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  35. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \(\omega_n=\exp(2\pi i/n)\)

  38. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.