Functional Square Root

[Catullus]

Does anyone know of a function , such that ?

[tommy1729]

Yes x^4 obviously.

Regards

Tommy1729

[MphLee]

The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\)

If all the functions are granted to be bijective, Tommy's answer is evident since \({\rm id}=sfs\) thus \(s^{-2}=f\). Let \(s={\rm sqrt}\) you get Tommy's solution. Asking less means we transform the functional equation in easier f.eq. For example.

• if we look for just \(f\) surjective, then we are solving for \(\) \(x^2=f(\sqrt{x})\);
  
• if we ask \(fs=sf\) then we are solving \(f(x)^4=f(f(x))\);
  

[Catullus]

The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\)

If all the functions are granted to be bijective, Tommy's answer is evident since \({\rm id}=sfs\) thus \(s^{-2}=f\). Let \(s={\rm sqrt}\) you get Tommy's solution. Asking less means we transform the functional equation in easier f.eq. For example.

• if we look for just \(f\) surjective, then we are solving for \(\) \(x^2=f(\sqrt{x})\);
  
• if we ask \(fs=sf\) then we are solving \(f(x)^4=f(f(x))\);
  

Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and  is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If , then  would have to equal . But is not the answer.

[tommy1729]

The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\)

If all the functions are granted to be bijective, Tommy's answer is evident since \({\rm id}=sfs\) thus \(s^{-2}=f\). Let \(s={\rm sqrt}\) you get Tommy's solution. Asking less means we transform the functional equation in easier f.eq. For example.

• if we look for just \(f\) surjective, then we are solving for \(\) \(x^2=f(\sqrt{x})\);
  
• if we ask \(fs=sf\) then we are solving \(f(x)^4=f(f(x))\);
  

Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and (x) is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If f(x) = x^0+x+x^/2, (x) would have to equal sqrt(x^2)+sqrt(x)+sqrt(x^2/). But f(x)=x^0+x+x^2/2 is not the answer.

Well x^4 is analytic everywhere.

And cosh(x) also not a solution.

Let g(x) = c * x^a.



f(x) = g(g(x)) = c * ( c^a x^(a^2) ) = c^(a+1) x^(a^2)

f(sqrt(x)) = c^(a+1) x^(a^2 /2) = g(x)

so c^(a+1) = c , a^2 = 2 * a.


So a = 0 or a = 2.

So we get 

g(x) = C for any C.

or 

g(x) = c x^2 for c^3 = c.

c^3 = c implies c = -1 or 0 or 1.

regards

tommy1729

[tommy1729]

The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\)

If all the functions are granted to be bijective, Tommy's answer is evident since \({\rm id}=sfs\) thus \(s^{-2}=f\). Let \(s={\rm sqrt}\) you get Tommy's solution. Asking less means we transform the functional equation in easier f.eq. For example.

• if we look for just \(f\) surjective, then we are solving for \(\) \(x^2=f(\sqrt{x})\);
  
• if we ask \(fs=sf\) then we are solving \(f(x)^4=f(f(x))\);
  

Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and (x) is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If f(x) = x^0+x+x^/2, (x) would have to equal sqrt(x^2)+sqrt(x)+sqrt(x^2/). But f(x)=x^0+x+x^2/2 is not the answer.

Well x^4 is analytic everywhere.

And cosh(x) also not a solution.

Let g(x) = c * x^a.



f(x) = g(g(x)) = c * ( c^a x^(a^2) ) = c^(a+1) x^(a^2)

f(sqrt(x)) = c^(a+1) x^(a^2 /2) = g(x)

so c^(a+1) = c , a^2 = 2 * a.


So a = 0 or a = 2.

So we get 

g(x) = C for any C.

or 

g(x) = c x^2 for c^3 = c.

c^3 = c implies c = -1 or 0 or 1.

regards

tommy1729

also 

Let g(x) = c(x) x^a

f(x) = g(g(x)) = c(g(x)) * g(x)^a = c(g(x)) * c(x)^a * x^(a^2)

f(sqrt(x)) = c(g(sqrt(x)) * c(sqrt(x))^a * x^(a^2/2)

f(sqrt(x))/g(x) = 1 = c(g(sqrt(x)) * c(sqrt(x))^a / c(x) * x^(a^2 / 2 - a)

1 = c(g(x))* c(x)^a / c(x^2) * x^(a^2 - 2a)

WLOG we can take

a = 0 ( and consider c a function of x and a !! )

so we get :

c(x^2) = c(g(x))

c(x^2) = c(c(x))

d (x^2)^2 = d (d (x^2))^2 = d^3 x^4 = d x^4.

just as before.

So it seems we have uniqueness ; there are no other solutions.


regards

tommy1729

[Catullus]

On the topic of function composition. Is the successor function the only function such that ?

[JmsNxn]

On the topic of function composition. Is the successor function the only function f(x) such that f(f(x)) = f(x)+1?

This depends on how you want to phrase this question. I believe you could probably find solutions to this equation in tropical algebra, and crazy areas like that.

In Complex Function theory, the answer is yes, but with a caveat.

If you assume that \(x \in \widehat{\mathbb{C}}\), then \(f = \infty\) is also a solution.

But, there's much much much more from this equation.

\[
f(f(x)) = f(x) + 1\\
\]

Since \(\infty\) is a fixed point (assuming \(f(\infty) = \infty\)), then... let's do leibniz variable change: \(F(x) = 1/f(1/x)\). This function satisfies \(F(0) = 0\), assuming that \(f\) is meromorphic at infinity. Now take \(F'(0) = C\). We are asking that:

\[
F(F(x)) = F(x) + 1\\
\]

In a neighborhood of \(x \approx 0\). This is kind of like a golden ratio/fibonacci equation, which I've never seen before. I think It's unique though. 

\[
\begin{align*}
F'(0)\cdot F'(0) = F'(0) = 1\\
F'(F(x))\cdot F'(x) = F'(x)\\
\end{align*}
\]

Therefore \(F' =1\). Therefore \(F(x) = x+1\). Therefore there are two solutions in complex analysis.

There's the constant solution \(\infty\), and the successor function \(x \mapsto x+1\).

So you are right, but there's an exception in complex analysis. The "successorship" hyperoperator could be an essential singularity... it looks like how \(e^{1/z}\) looks like \(0\) and \(\infty\) near \(z\approx 0\). We can have a hyperoperator successorship that looks like \(\infty\) and \(x \mapsto x +1\). So, SINGULAR solutions to the above question can exist.

[Catullus]

Thank you for helping me.

[Catullus]

What about that ? Is the only answer to it?