Functional Square Root

[JmsNxn]

What about that (x) = f(sqrt(x))? Is f(x) = x^4 the only answer to it?

Tommy is correct in his deduction, as far as I can tell. No reason to doubt him. But \(\infty\) is also a solution if you ask for meromorphic solutions. So there are two constant solutions \(0\) and \(\infty\), but other than that, the only other solution is \(x^4\), and Tommy's proof is good enough.

[Catullus]

What about that (x) = f(sqrt(x))? Is f(x) = x^4 the only answer to it?

Tommy is correct in his deduction, as far as I can tell. No reason to doubt him. But \(\infty\) is also a solution if you ask for meromorphic solutions. So there are two constant solutions \(0\) and \(\infty\), but other than that, the only other solution is \(x^4\), and Tommy's proof is good enough.

Hmm, I saw some weird formatting on your reply.

[tommy1729]

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

Let f(x) = v(x) + x 

then f(x) + 1 = f(f(x)) = f( v(x) + x ) = v( v(x) + x ) + v(x) + x = v(x) + x + 1.

so v( v(x) + x ) = 1. since this implies that v is independant from x , we get that v(x) is a constant function. 

Hence v(x) = 1. 
And thus f(x) = x + 1.

 ***

notice that equation

g(x+1) = g(x) + 1 has solution g(x) = x + 1-periodic(x).

this is cruxial for superfunctions !

***

also fun :

h(x+1) = h(h(x))

let t(x) be the inverse of h :

t h(x+1) = t h(h(x))

x + 1 = h(x).


I could give more similar looking equations and solve them ... but why ?


regards

tommy1729

[Leo.W]

What about that (x) = f(sqrt(x))? Is f(x) = x^4 the only answer to it?

Hi Catullus

The answer can be different seen from varied viewpoints.

If you're talking about the functional equation \[\sqrt{f}(z)=f(\sqrt{z})\] where f can be piecewise analytic, then f(x)=x^4 is not the only answer, or even not an answer. This is because every function have at least 2 different functional square root, for this f, it can be x^2 and -x^2, for a simple identity function, it has infinitely many functional square root, like f(x)=c-x, f(x)=c/x, f(x)=x, for any constant c, for all these f(x), f(f(x))=x.
And for most cases, some of the functional square roots can even be not analutic in the whole complex plane, which means your equation could be a little bit badly-defined if only considering singlevalued functions.

For multivalued functions, we can say f(x)=x^4 is a nice solution.
Or if we consider a properly defined or regularized functional square root, then f(x)=x^4 can be a nice singlevalued analytic solution, and the only one.
In this way, you're actually asking for an analytic function g which satisfies,
\[g(x)=\sqrt{f}(x)\Leftrightarrow g(g(x))=f(x),g(x)=g(g(\sqrt{x}))\]
Since g is now analytic, it has its maybe-multivalued inverse function g^-1
Let \[g^{-1}(g(x))=x\text{ takes true for some x, but for all x: }g^{-1}(g(x))=\sigma(x)\]
Then we take g^-1 for both side and substitute sqrt(x)=t, this gives \[\sigma(t^2)=\sigma(g(t))\]
This functional equation has infinitely many solutions, each combines with such a symmetry, thus we can't get any information about g(z).

But there's ways to find some other solutions which may give rise to multivalued or piecewise analytic solutions.
For example, we derive a new sigma(z) from our old solution f(x)=x^4, we use its functional square root g(x)=x^2 and its multivalued inverse, gives \[\sigma(z)=e^{\frac{ln(z^2)}{2}},f(z)=\sqrt{z^8}\]
which is also a solution, only piecewise analytic, can be proved easily using singlevalued complex analysis.

regards
leo

[Leo.W]

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

The equation shows f is one Abel function of itself, and thus also the superfunction of itself.

If you assume that f has a singlevalued inverse function, then by the definition f must satisfy:
\[f^t(x)=f^{-1}(f(x)+t)\Leftrightarrow f^{t+1}(x)=f(x)+t\]
by taking t=-1, we simply get
\[f^0(x)=x=f(x)-1\]
this shows f(x)=x+1 is the only solution.

If not, we can still infer that for any natural number n, \[f^{n+1}(x)=f(x)+n\]
ps. This can be extended by the definition of iteration as
\[f^n(x)=f^{m+(n-m)}(x)=f^m(f^{n-m}(x))=f^m(f(x)+n-m-1), n>m\]
Consider n=3\[f(f(x))+1=f^2(f(x))=f^3(x)=f(f^2(x))=f(f(x)+1)\]
Let f(x)=t, \[f(t)+1=f(t+1)\]
Then this gives f(t+1)-f(t)=1, a recurrence, pretty easy to solve and get f(t)=t+C for some C, and easily prove C=1
So, this proves again that f is the only solution, if it's provided that f's domain and range are the same, which is the whole plane.

[Catullus]

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

The equation shows f is one Abel function of itself, and thus also the superfunction of itself.

If you assume that f has a singlevalued inverse function, then by the definition f must satisfy:
\[f^t(x)=f^{-1}(f(x)+t)\Leftrightarrow f^{t+1}(x)=f(x)+t\]
by taking t=-1, we simply get
\[f^0(x)=x=f(x)-1\]
this shows f(x)=x+1 is the only solution.

is also a solution.

What about ? What is f?
What about ?

[Leo.W]

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

The equation shows f is one Abel function of itself, and thus also the superfunction of itself.
...
this shows f(x)=x+1 is the only solution.

F(x) = ∞ is also a solution.

What about Σ n = 0 to ∞ f^n(x)/n! = exp(f(x)). What is f?
What about Σ n = 0 to ∞ f^n(x)/n! = f(exp(x))?

Well if you insist, then there will be also a solution to the equation x=x+1, which is x=∞, that'll make sense to some extent, but only considering finite solutions as we usually do, then f(x)=x+1 is the only. 

For these 2 equations, if you're asking finite analytic solution, I guess you can solve it by the fixed point method. For example:

Assume L is one solution of f(L)=L, then f^n(L)=L, gives L*e=exp(L), thus we know all fixed points of f(x) has to solve L*e=exp(L), for example L=1, f(1)=1.
Then we take derivatives of both sides at x=1, gives e^(f'(1))=e^(f(1))f'(1), so f'(1) satisfies ex=e^x, let's again assume f'(1)=1, then we can solve by induction for f''(1), f'''(1) and on.
But if you really take attempts, you'll find that f'(1) cannot be 1, which would make the sum diverge (seen easily from the second derivative), to make it converge, you have to find a solution to ex=e^x where the absolute value of x mustnot exceed 1, which is impossible.
And this shows the fixed point of f cannot be 1.

If the fixed point of f is any L not 1, we'll have L*e=exp(L) and exp(f'(L))=exp(L)f'(L)=e*L*f'(L), again we have to find a solution of exp(x)=e*L*x whose absolute value remains smaller than 1. As I computed,
\[L=-W_1(-\tfrac{1}{e})\approx{3.08884-7.46149i}, s=f'(L)=-W_0\bigg(\frac{1}{eW_1(-\tfrac{1}{e})}\bigg)\approx{0.0158263+0.0434912i}\]
Then we can work out the series:
\[\begin{matrix}L\approx{3.088843015613044-7.461489285654255i},\\f(z+L)-L\approx(0.01582633039817652 +
    0.04349124710235049 i) z\\+ (0.0008417750386726426-
    0.0006753346652964609 i) z^2\\- (0.00002833079697273823 +
    0.00001790952001363610 i) z^3\\- (2.664193985539843*10^{-7} -
    1.1390607179533432*10^{-6} i) z^4\\+ (4.329246420079456*10^{-8} -
    4.56979799629351*10^{-9} i) z^5\\- (7.19187320031943*10^{-10} +
    1.526422873318864*10^{-9} i) z^6\\- (4.803646824452359*10^{-11} -
    4.709364757070538*10^{-11} i) z^7\\+ (2.453010990405934*10^{-12} +
    1.215725690054748*10^{-12} i) z^8\\+ (1.410524507975909*10^{-14} -
    1.1230584812952190*10^{-13} i) z^9+O(z^{10})\end{matrix}\]
for 16 digits precision in decimal

You can solve the other in the similar way, but I don't see f(x) has a closed form expression.

[Catullus]

If the fixed point of f is any L not 1, we'll have L*e=exp(L) and exp(f'(L))=exp(L)f'(L)=e*L*f'(L), again we have to find a solution of exp(x)=e*L*x whose absolute value remains smaller than 1. As I computed,
\[L=-W_1(-\tfrac{1}{e})\approx{3.08884-7.46149i}, s=f'(L)=-W_0\bigg(\frac{1}{eW_1(-\tfrac{1}{e})}\bigg)\approx{0.0158263+0.0434912i}\]

The real part of L is close to the Tetra-Euler Number. 

[tommy1729]

If the fixed point of f is any L not 1, we'll have L*e=exp(L) and exp(f'(L))=exp(L)f'(L)=e*L*f'(L), again we have to find a solution of exp(x)=e*L*x whose absolute value remains smaller than 1. As I computed,
\[L=-W_1(-\tfrac{1}{e})\approx{3.08884-7.46149i}, s=f'(L)=-W_0\bigg(\frac{1}{eW_1(-\tfrac{1}{e})}\bigg)\approx{0.0158263+0.0434912i}\]

The real part of L is close to the Tetra-Euler Number. 

my intuition says that is no coincidence ... but no detailed rational ideas yet.

[Catullus]

How about ?
How about the super function of ?