Empty
  • Empty
What's the general form of f(x)?
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Empty
  • Empty
$$\underbrace{f(f(...f}_{\text{n times}}(x)))= x$$
Michele_Laino
  • Michele_Laino
I think that we have the subsequent equation: \[\Large {f^n} \equiv I\] where I is the identity function
Empty
  • 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

Looking for something else?

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

More answers

Michele_Laino
  • 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
  • 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
  • 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
  • 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
  • dan815
|dw:1436810632074:dw|
ganeshie8
  • ganeshie8
I think any function that is symmetric about \(y=x\) works for even \(n\) : |dw:1436810458644:dw|
ganeshie8
  • ganeshie8
Because then the function is its own inverse and consequently \[f^2 = x \implies f^{2k} = x\]
dan815
  • dan815
im thinking its gotta be how to build up transformations to get to that 90 degree tangle rotation in n-1 transformations
dan815
  • dan815
this solujtion has to be some function of x and n
Empty
  • Empty
Ahhh How about adding the caveat that for instance: $$f(f(f(f(x))))=x$$ BUT $$f(f(x)) \ne x $$
dan815
  • dan815
nth complex root of -1/x?
dan815
  • dan815
oh how about f(x) is a nth root ration of x
Empty
  • 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
  • dan815
|dw:1436811053979:dw|
dan815
  • dan815
in n transformations we gotta end up with x again then
ganeshie8
  • ganeshie8
that is clever!
Empty
  • Empty
Ahhh ok, so what \(n\) do you plug in to get $$g(x)=\frac{-1}{x}$$
dan815
  • dan815
well i dont know that i thought wed just work with x straight away
Empty
  • Empty
You're on the right path haha :P
ganeshie8
  • ganeshie8
|dw:1436811396051:dw|
dan815
  • dan815
ohh ok lemme think
Empty
  • 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
  • dan815
page 52 of generating functionolgy
Empty
  • Empty
I'll look right now... ?
dan815
  • dan815
lol noo thats a joke
Empty
  • Empty
lol I was like wow he read further than me already rofl
dan815
  • dan815
ok so b^n has to go to 1 and a^{coefficient 1 power series = 1}
Empty
  • Empty
Are you sure?
dan815
  • dan815
oh 0 its gotta be 0
dan815
  • dan815
1+b+b^2+..+b^(n-1) = 0 b^n=1
dan815
  • dan815
1+b+b^2+..+b^(n-1)+ b^n =1
anonymous
  • anonymous
An example\[f(x)=-\frac{x}{x+1}\]?
anonymous
  • 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
  • anonymous
so \(f^{(n)}(x)=x\) where \(f^{(m)}(x)\ne x\) for \(m

Looking for something else?

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