Functional equation: $f(z) = \sum_{n=0}^{\infty} f(n)^2 z^n$?

Question

Im looking for functions $f(z)$ such that

$f(z) = \sum_{n=0}^{\infty} f(n)^2 z^n = f(0)^2 + f(1)^2 z + f(2)^2 z^2 + f(3)^2 z^3 + ...$

and $f(n)$ are all real.

And I wonder how fast $f(n)$ grows.

I had this question in my head for a long time but I got reminded by these

Function $f(x)$, such that $\sum_{n=0}^{\infty} f(n) x^n = f(x)$

Does there exist a function that generates itself?

and I am grateful to them.

edit

PROOF BY CONTRADICTION : infinite radius is not possible :

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Note $f(z)$ can not be a nonconstant polynomial. This also implies that :

$f(n) = 0 $ for all $n > N$ for some $N$, cannot be true unless $f(z)=0$.

Also notice that the radius must be $\infty$ since $f(n)$ goes to infinity.

So apart from the trivial constant functions, we are dealing with real entire transcendental functions.

That implies that $f(n)^2$ converges to $0$ faster than exponential. So it might make more sense to speak about how fast $\frac{1}{f(n)}$ grows.

EVEN STRONGER :

One can show that most $f(n)$ are not $0$ ; the tail of $f(n)$ must have no zero values, since all taylor coefficients are positive.
And finally realize that therefore $f(1),f(2)$ must be nonzero, we can show that the start also can contain no zero's.

So we end up with concluding that only $f(0)$ can be zero, what makes perfect sense since the function is strictly increasing on the positive reals.

But now I run into a problem.

Since $f(z)$ is strictly increasing , so is $f(n)$.

But strictly increasing $f(n)$ implies $0 < C < f(n)^2$ for some constant $C$.

Hence the radius is not infinite unless $f(z)$ is a constant !!!

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So we end up studying functions with finite radius, than can somehow get the real values $f(n)$. Analytic continuation around a pole or such might work. Even lacunary series than can be extended might work.

So the situation is not so clear and simple.

Answer 1

I'm not sure if it's correct, but these are my thoughts (I hope they help):

• According to the convergence criteria $\lim\limits_{z \to \infty}\left[ f\left( x \right) \right] = \text{constant}$ must apply (if $\lim\limits_{n \to \infty}\left[ a_{n} \right] = \pm \infty$ then $\sum\limits^{\infty} a_{n}$ would not converge) and $f\left( n \right) \in \mathbb{R}$.
• Lets define $a_{n} := f\left( n \right)^{2}$. If $f\left( n \right) \in \mathbb{R}$ then $f\left( n \right)^{2} = a_{n} \in \mathbb{R}^{+}_{0}$, since the square of a real number is always positive or zero.
• Scince $f\left( m \right) = \sum\limits_{n = 0}^{\infty}\left[ a_{n} \cdot m^{n} \right]$ and $a_{n} \in \mathbb{R}^{+}_{0}$ apply, $0 \leq f\left( n \right) \leq f(n + 1) = f\left( n \right) + c$ where $c \in \mathbb{R}^{+}_{0}$ has also to apply aka $f{:}~ \mathbb{N} \cup \left\{ 0 \right\} \to \mathbb{N} \cup \left\{ 0 \right\}$ must be monotonically increasing. If there is an $m \in \mathbb{N}$ such that $a_{m} \geq 1$, then $f\left( x \right)$ would always diverge for $\left| x \right| \geq 1$, but if it diverges for any $a_{n}$, then no $a_{m}$ , then there is also no $a_{m + 1}$, $a_{m + 2}$, ..., therefore no real $f\left( m \right)$, $f\left( m + 1 \right)$, ..., which causes $f\left( x \right)$ to diverge. Therefore $0 \leq f\left( n \right) < 1$ must apply.

Let's assume that a $g \in \mathbb{N}$ exists such that $f\left( g - 1 \right) = 0 > f\left( g \right)$ aka $a_{g - 1} = 0 > a_{g}$ would hold. Now let's assume $f\left( g \right) = f\left( g + 1 \right) = f\left( g + 2 \right) = \dots$ aka the simplest non-trivial case. Then we could say: $$ \begin{align*} f\left( z \right) &= \sum\limits_{n = 0}^{\infty}\left[ a_{n} \cdot z^{n} \right]\\ f\left( z \right) &= \sum\limits_{n = 0}^{g - 1}\left[ a_{n} \cdot z^{n} \right] + \sum\limits_{n = g}^{\infty}\left[ a_{n} \cdot z^{n} \right]\\ f\left( z \right) &= 0 + \sum\limits_{n = g}^{\infty}\left[ a_{n} \cdot z^{n} \right]\\ f\left( z \right) &= \sum\limits_{n = g}^{\infty}\left[ a_{n} \cdot z^{n} \right]\\ f\left( z \right) &= \sum\limits_{n = g}^{\infty}\left[ a_{g} \cdot z^{n} \right]\\ f\left( z \right) &= a_{g} \cdot \sum\limits_{n = g}^{\infty}\left[ z^{n} \right]\\ \end{align*} $$

The problem is obvious: $\sum\limits_{n = g}^{\infty}\left[ z^{n} \right] = \frac{z^{g}}{1 - z}$ diverges ($\lim\limits_{x \to g}\left[ f\left( x \right) \right] \to \infty$), for all $\left| z \right| \geq 1$ wich contradicts $f\left( m \right) \in \mathbb{R}$. Any case other than $f\left( g \right) = f\left( g + 1 \right) = \dots$ would imply faster growth of $f\left( x \right)$ and thus also imply $\lim\limits_{x \to m}\left[ f\left( x \right) \right] \to \infty$, which also contradicts $f\left( m \right) \in \mathbb{R}$. Accordingly, there should be no non-constant function $f\left( x \right)$ that satisfies this equation. The only solution would be $f\left( x \right) = 0$.

But I may be wrong...