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.


PROOF BY CONTRADICTION : infinite radius is not possible :

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.


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 !!!

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.

  • 1
    $\begingroup$ @GEdgar I guess it is a typo and $x$ should be replaced by $z$. $\endgroup$
    – Abezhiko
    Nov 29, 2023 at 13:46
  • $\begingroup$ typo fixed guys $\endgroup$
    – mick
    Nov 29, 2023 at 18:44
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    $\begingroup$ an idea to try would be to start with an entire function $g(z)=1+\sum_{n \ge 1} q_n z^n$ for which $g(n), q_n$ are smallish and then look for coefficients $b_n$ st $f(z)=g(z)+h(z)\sin \pi z$ where $h(z)=\sum b_n z^n$; since the $n$th derivative of $h \sin \pi z$ at zero is a linear combination of $b_k$ of the same parity up to $n-1$ you should be able to determine them inductively (more or less ) uniquely so $a_n=g(n)^2=f(n)^2$, eg $q_1+b_0\pi=g(1)^2$ and then the question is to show that they are small enough to determine an entire function which may be doable with good initial choice of $g$ $\endgroup$
    – Conrad
    Nov 29, 2023 at 19:11
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    $\begingroup$ that's a good point - note though that any power series with non negative coefficients and finite radius of convergence has singularity at the positive value there so the function cannot be extended and in particular defining $f(n)$ may not be possible - the idea I mentioned should give you solutions with $f(n)$ complex, while I do not think that the problem really makes sense for $f(n)$ real $\endgroup$
    – Conrad
    Nov 29, 2023 at 19:31
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    $\begingroup$ Note (maybe this helps): If $z = 0$ then $f( 0 ) = f( 0 )^{2} + ( \dots ) \cdot 0 = f( 0 )^{2}$, wich implies $f( 0 ) \in \{ 0,\, 1 \}$. Furthermore, the only existing constant solution is given by $f( x ) = 0$ since $\sum\limits_{n = 0}^{\infty}\left[ ( \text{constant} )^{2} \cdot z^{n} \right] = ( \text{constant} )^{2} \cdot \sum\limits_{n = 0}^{\infty}\left[ z^{n} \right] = ( \text{constant} )^{2} \cdot \frac{1}{1 - x}$ for $| x | < 1$ and $\text{constant} = ( \text{constant} )^{2} \cdot \frac{1}{1 - x} \Longleftrightarrow \text{constant} \in \{ 0,\, 1 - x \}$. $\endgroup$ Dec 2, 2023 at 16:19

1 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...

  • $\begingroup$ Seems ok at first sight. I will look at the details later and then probably accept. Thank you. $\endgroup$
    – mick
    Dec 5, 2023 at 11:48

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