Mfrhtyj

Let $k>1$ be an integer.

Consider equations of type

$$f_k(x+y) = f_k(x) + f_k(y) + f_k^k(x) f_k(y) + f_k^k(y) f_k(x) , f(-z) = f(z), $$

where $f_k$ is the $k$ th function and $f_k^k $ is the $k$ th power of $f_k$.

Valid for all real $x,y,z$ and $f$ is real-analytic.

Consider the nonconstant solutions.

When considering $f_2$ I came to the conclusion that $f_2(z) = 0 $ is the only solution.
( this was achieved by contradictions while investigating---and equating---$f(-1),f(0),f(1),f(2),f(3), f(2) = f(1+1) = f(3-1) $ by using algebra )

As a side note I considered the julia sets of $f_kleft(2 operatorname{arc}f_k(z)right) = 2 z + 2 z^{k+1} $ , more specific if they are connected or not. For $k>4$ they are totally disconnected and otherwise they are totally connected. For $k = 2$ it is a circle shape. Not sure if it relates or not?

Notice $k$ needs to be even to have $f(-z) = - f(z) $.

Although odd $k$ might be intresting too if we drop that condition.

So I wonder about general case; for what even $k$ do we get a solution and when do we fail like The case $f_2$?

How does $f_k$ behave?
Probably like a double exponential ... but a bit more precise understanding would be nice.

Is there uniqueness?

To be a bit more concrete I particularly wonder about

$$f(x+y) = f(x) + f(y) + f^4(x) f(y) + f^4(y)f(x) $$

and

$$f(x+y) = f(x) + f(y) + f^6(x) f(y) + f^6(y)f(x). $$

Do any of those 2 equations have a nonconstant solution?

Note that those are around The switch from connected to disconnected associated Julia sets.

Another remark :

$$g(x + y) = g(x) + g(y) + g(x)g(y) $$

has the following solution $g(z)=exp(C z) - 1$.

Maybe this implies we should study instead

$$f(x+y) = f(x) + f(y) + frac12left( f^k(x) f(y) + f^k(y)f(x) right)?? $$

The related Julia sets are similar btw.

Are there representations known for these functions or their functional inverses , like integral representations , differential equations , infinite sums , analogues to eulers addition theorem or gauss integral for The agm , continued fractions , ...
have “ very “similar things been studied before ?
( not Somos )

Fix a non-negative integer $k$ and $tinBbb R$. Let $f:Bbb Rto Bbb R$ be a function such that$$(1) f(x+y)=f(x)+f(y)-frac{t}{2}left(f^k(x)f(y)+f(x)f^k(y)right)$$
for all $x,yin Bbb R$. (Here we interpret $0^0$ as $1$.) Plugging in $x,y=0$ in $(1)$, we have
$$f(0)big(1-tf^k(0)big)=0.$$
This shows that
$(i)$ $f(0)=0$,
$(ii)$ $tneq 0$, $k$ is odd, and $f(0)=frac{1}{sqrt[k]{t}}$,
$(iii)$ $t> 0$, $k>0$ is even, and $f(0)=frac{1}{sqrt[k]{t}}$, or
$(iv)$ $t=1$ and $k=0$.

We first deal with $(ii)$ and $(iii)$. Plugging in $y=0$ in $(1)$ yields
$$big(sqrt[k]{t}f(x)big)^k+big(sqrt[k]{t}f(x)big)-2=0$$
for all $xin Bbb R$. By Descartes' rule of signs, the polynomial $u^k+u-2=0$ has exactly one real root $u=1$ for odd $k$, and this polynomial has two real roots $u=1$ and $u=-epsilon_k$ for even $k$ (where $epsilon_k>1$ is the only positive root of $u^k-u-2=0$). For even $k>0$ and for $t>0$, suppose that there exists $cin Bbb R$ s.t. $f(a)=-frac{epsilon_k}{sqrt[k]{t}}$. Plugging in $c$ for both $x$ and $y$ gives
$$1<epsilon_k^2=epsilon_k(epsilon_k^k-2)in{1,-epsilon_k}.$$
This is absurd, so $c$ does not exist. Therefore, in both cases $(ii)$ and $(iii)$, the only solution is
$$(2) f(x)=frac1{sqrt[k]{t}} forall x$$

We now deal with the case $k=0$ separately (and this will complete the case $(iv)$ automatically). From $(1)$, we see that
$$f(x+y)=left(1-frac{t}{2}right)big(f(x)+f(y)big).$$
If $t=0$, we have Cauchy's functional equation, whose solutions are well-known and the only analytic solutions are the functions $f$ of the form $$(3) f(x)=m x$$ for some constant $m$. If $t=1$, taking $y=0$ we have the constant solution $$(4) f(x)=f(0) forall x.$$ If $tnotin{0,1}$, then $f(0)=0$. By taking $y=0$, $$(5) f(x)=0 forall x.$$

For the remaining part of this answer, we assume that $k>0$ and $f(0)=0$. If $t=0$, we get once again Cauchy's functional equation, and we already know the solutions. Suppose from now on that $tne0 $.

If $k=1$, we let $g(x)=1-tf(x)$. Then $$g(x+y)=1-tf(x+y)=1-tf(x)-tf(y)+t^2f(x)f(y)=g(x)g(y).$$
This is another well-known functional equation. Its solution is $g(x)=0$ for all $x$, or $g(x)=expbig(gamma(x)big)$ where $gamma$ is a solution to Cauchy's functional equation. In particular, this means that all analytic solutions are
$$(6) f(x)=frac1t forall x wedge f(x)=frac{1-exp(lambda x)}{t} forall x.$$

Suppose now that $k>1$. At this point I gave up on finding general solutions to $(1)$, but it is possible to show that the only analytic solution to $(1)$ is the zero function $f(x)=0$. To see this, take derivative of $(1)$ wrt $y$ to get
$$(7) f'(x+y)=f'(y)-frac{t}{2}f(x)f'(y)left(f^{k-1}(x)+kf^{k-1}(y)right).$$
Putting $y=0$ we get
$$(8) f'(x)=f'(0)left(1-frac{t}{2}f^{k}(x)right).$$
Using $(8)$, $(7)$ becomes
$$f'(0)left(1-frac{t}{2}f^k(x+y)right)=f'(0)left(1-frac{t}{2}f^{k}(y)right)left(1-frac{t}{2}f(x)left(f^{k-1}(x)+kf^{k-1}(y)right)right).$$
By symmetry,
$$f'(0)left(1-frac{t}{2}f^k(x+y)right)=f'(0)left(1-frac{t}{2}f^{k}(x)right)left(1-frac{t}{2}f(y)left(f^{k-1}(y)+kf^{k-1}(x)right)right).$$
If $f'(0)ne 0$, then
begin{align}left(1-frac{t}{2}f^{k}(y)right)&left(1-frac{t}{2}f(x)left(f^{k-1}(x)+kf^{k-1}(y)right)right)\&=left(1-frac{t}{2}f^{k}(x)right)left(1-frac{t}{2}f(y)left(f^{k-1}(y)+kf^{k-1}(x)right)right)end{align}
and so
$$f(x)f(y)Big(2f^{k-2}(x)-2f^{k-2}(y)-tf^{2k-2}(x)+tf^{2k-2}(y)Big)=0.$$
If $f$ is non-zero, then
$$2f^{k-2}(x)-2f^{k-2}(y)-tf^{2k-2}(x)+tf^{2k-2}(y)=0$$
for all $x,y$. Differentiating the previous equation wrt $y$ yields
$$2(k-2)f^{k-3}(x)f'(x)-(2k-2)tf^{2k-3}(x)f'(x)=0.$$
Observe that $f'$ is also non-zero. Therefore
$$2(k-2)-(2k-2)tf^{k}(x)=0$$
or
$$f^k(x)=frac{k-2}{(k-1)t}$$
so $f$ is constant, but this contradicts the fact that $f'$ is non-zero.

In conclusion, an analytic solution to $(1)$ is $f(x)=0$. There are exceptional solutions, depending on the parameters $t$ and $k$. These (analytic) solutions are listed below.
$(a)$ $t=0$: $f(x)=m x$.
$(b)$ $tneq 0$ and $k$ is odd: $f(x)=frac{1}{sqrt[k]{t}}$.
$(c)$ $t>0$ and $k>0$ is even: $f(x)=frac{1}{sqrt[k]{t}}$.
$(d)$ $t=1$ and $k=0$: $f(x)=kappa$.
$(e)$ $tne 0$ and $k=1$: $f(x)=frac1t$.
$(f)$ $tne 0$ and $k=1$: $f(x)=frac{1-exp(lambda x)}{t}$.

There are non-analytic solutions at least in the case $t=0$ and the case $tne 0$ with $k=1$. There may be non-analytic solutions in the case $tne 0$, $k>1$, and $f(0)=0$, but I have no idea.