# $f(ax)=f(x)^2-1$, what is $f$?

Suppose $$f(ax)=(f(x))^2-1$$ and suppose that $$f$$ is analytic in some neighborhood of $$x=0$$. Expanding in power series, we get $$a=1+\sqrt{5}$$ or $$1-\sqrt{5}$$. We take positive $$a$$. If $$f\neq{\rm const}$$ then $$f'(0)\neq0$$ - it can be any non-zero number. After that, we can uniquely define coefficients in power series $$f^{(n)}(0)/n!$$ step by step using differentiation of the functional equation $$f(0)=f(0)^2-1\ \Rightarrow\ f(0)=\frac{1\pm\sqrt{5}}2;\ 2f'(0)f(0)=af'(0)\ \Rightarrow\ f(0)=a/2;\ \ 2f''(0)f(0)+2(f'(0))^2=a^2f''(0)\ \Rightarrow\ f''(0)=\frac{2(f'(0))^2}{a^2-a};\ \ ....$$ Can $$f$$ be expressed in terms of some known functions?

It seems that $$f$$ is entire function of order $$1$$. Indeed, due to the Leibnitz formula, we have $$a^nf^{(n)}(0)=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}(0)f^{(k)}(0),$$ which leads to $$$$\label{l1} f^{(n)}(0)=\frac{\sum_{k=1}^{n-1}\binom{n}{k}f^{(n-k)}(0)f^{(k)}(0)}{a^n-a}.$$$$ It is true that $$|f''(0)|\leq|f'(0)|^2=:C^2$$, see above. Suppose that we already proved $$|f^{(k)}(0)|\leq C^k$$, $$1\leq k\leq n-1$$. Then $$|f^{(n)}(0)|\leq \frac{C^n2^n}{(\sqrt{5}+1)^{n-1}\sqrt{5}}\leq C^n.$$ Hence, the power series converges everywhere and $$|f(x)|\leq|f(0)|+e^{C|x|}-1$$, $$x\in\mathbb{C}$$.

(The functional equation is somewhat similar to $$\cos 2x=2\cos^2x-1$$.

If we denote $$g(x)=f(x)/f(0)$$ then $$g'(x)=g'(0)g(a^{-1}x)g(a^{-2}x)g(a^{-3}x)....$$

Also, the polynomial $$P_n(x)=f(a^nf^{-1}(x))=P\circ...\circ P(x),\ \ \ P(x)=x^2-1$$ is some analog of Chebyshev polynomial $$T_{2^n}(x)$$. The polynomials $$P_n(x)$$ are related to the representations of some class of infinite Lie algebras, but I forgot the story behind that...)

Remark. It is seen that $$f(\lambda x)$$ satisfies also the functional equation, for any $$\lambda$$. So, we can choose $$f'(0)$$ freely. Instead of this, let us choose the closest to $$0$$ zero as $$x_0:=1$$ (if zeroes exist), i.e. $$f(1)=0$$. Then $$f(a^{-1})=1$$ (not $$-1$$, since $$x_0$$ is the first zero and $$f(0)>0$$). Applying again the functional equation, we get $$f(1)=0,\ f(a^{-1})=1,\ f(a^{-2})=\sqrt{2},\ ...,\ f(a^{-n})=\sqrt{1+f(a^{-n+1})},\ ....$$ Theoretically, we can recover $$f(x)$$ from infinite number of values $$f(a^{-n})$$ covergent to $$f(0)=a/2$$.

The next statement is wrong because, perhaps, $$f$$ can have complex zeros along with real zeros. I leave it just for possible improvements. Due to $$f'(ax)=2a^{-1}f(x)f'(x)=(2a^{-1})^2f(x)f(a^{-1}x)f'(a^{-1}x)=...$$ (see above infinite product for $$g$$), we can conclude that $$f'(x)\neq0$$ for $$x\in[0,a)$$ and, hence $$f$$ is monotonic on this interval. $$f$$ is also monotonic for $$x\in(a,a^2)$$, since $$f'(ax)=2a^{-1}f(x)f'(x)$$. Due to $$f(a^2x)=f(x)^2(f(x)^2-2)$$, we see that $$x_1=a^2$$ is a double zero, $$x_2=a^4$$ is a zero of fourth order, etc. Following the same arguments as above, there are no other zeroes. Then $$f(x)=e^{h(x)}\prod_{n=0}^{\infty}\left(1-\frac{x}{a^{2n}}\right)^{2^n}.$$ Is $$h(x)=\ln a-\ln 2$$? Perhaps, something wrong can be here... but if this is true then $$\ln f(x)=\ln a-\ln 2-\sum_{k=1}^{\infty}\frac{a^{2k}x^k}{k(a^{2k}-2)}$$ or something like that.

While the previous section was wrong, the true Hadamard expansion perhaps exist. This is still rough. Let $$\{x_r\}$$ be the smallest primitive zeros of $$f$$, such that all are other zeroes are $$a^{2n}x_r$$. Dentote $$H(x)=\prod_r(1-x/x_r)$$ - I do not know about the convergence of the product, I am trying to explain some ideas. We can follow the arguments from the wrong section above. Then, perhaps, it is true that $$f(x)=\frac{a e^{dx}}{2}\prod_{n=0}^{\infty}H\left(\frac x{a^{2n}}\right)^{2^n}$$ with some constant $$d$$ (I hope it is $$0$$), since the order of $$f$$ does not exceed $$1$$. Substituting this identity into $$f(a^2x)=f(x)^2(f(x)^2-2)$$ we obtain also $$f(x)=\sqrt{2+e^{d(a^2-2)x}(2/a)H(a^2x)}.$$ Using $$f'(ax)=f'(0)\prod_{n=1}^{\infty}(2f(x/a^n)/a)$$ (see above $$g$$), we obtain $$f'(x)=f'(0)e^{\frac{dx}{a-1}}\prod_{n=1}^{\infty}H\left(\frac{x}{a^{2n}}\right)^{2^n-1}\prod_{n=1}^{\infty}H\left(\frac{x}{a^{2n-1}}\right)^{2^n-1}.$$ It is possible to obtain other formulas.

There is also another motivation to study $$f$$: $$f(a^nx)=P\circ..\circ P(f(x)),\ \ P(x)=x^2-1.$$ Thus, we can try to analyze the stability of values $$f(z)$$ under the action of the group of polynomials. This question is related to dynamical systems, fractals... Maybe the dynamical systems community already studied such analytic functions?

There is an exact relation to fractals and holomorphic dynamics. Using $$f(z)=P\circ...\circ P(f(a^{-n}z))$$, $$P(z)=z^2-1$$, we obtain that if $$z_0$$ (located in the strip $$S_n:=\{z:\ a^n\leq|z_0|) is a primitive zero of $$f$$, i.e. $$f(a^{-k}z)\neq0$$, $$k\geq1$$, then $$f(a^{-n}z_0)$$ is a non-trivial root of the polynomial $$P_n=P^{\circ n}$$. All such non-trivia roots have the form $$q= s_0\sqrt{1+s_1\sqrt{1+...+s_{n-1}\sqrt{1+s_n\sqrt{2}}}},$$ where $$s_j\in\{-1,1\}$$. Without loss of generality, consider the case $$f'(0)=1$$. Then, for sufficiently large $$n_0$$, we have $$\frac{a}2+a^{-n-n_0}z_0\approx f(a^{-n-n_0}z_0)= s_0\sqrt{1+s_1\sqrt{1+...+s_{n-1}\sqrt{1+s_n\sqrt{2}}}}.$$ Hence $$a^{-n-n_0}z_0\approx s_0\sqrt{1+s_1\sqrt{1+...+s_{n-1}\sqrt{1+s_n\sqrt{2}}}}-\frac{a}2(=q)$$ and all such $$q$$ lying in $$a^{-n_0}\leq|q| correspond to primitive roots of $$f$$ lying in $$S_n$$. I have computed them for $$n=20$$ (and $$n_0=4$$):

It looks like a Julia set. So, zeros of $$f$$ (even primitive zeros) are very complex. I hope I did not make large mistakes somewhere...

It is also possible to introduce more general functions satisfying $$f(ax)=bf(x)^2+cf(x)+d\ \ \ {\rm or\ even}\ \ \ f(ax)=Q(f(x))$$ with some polynomial or rational $$Q$$. Such class of functions contain $$\sin,\cos,\exp,...$$

One of the main questions is still open: are there any relations between $$f$$ and some known functions?

Of course, any information about asymptotics, expansions, numerical results, differential equations, etc. is very welcome.

A relation with the approximation of the golden ratio. Finally, I found one relation with more or less known functions. Consider $$g(z)=\lim_{n\to\infty}a^n\biggl(\underbrace{\sqrt{1+\sqrt{1+...\sqrt{1+z}}}}_n-\frac{a}2\biggr).$$ The function $$g$$ is analytic (not entire), it is considered in Paris, R. B. "An Asymptotic Approximation Connected with the Golden Number." Amer. Math. Monthly 94, 272-278, 1987. It satisfies the functional equation $$g(z)=ag(\sqrt{1+z}).$$ Hence, $$f$$ is exactly the inverse function to $$g$$. It is useful to note that $$g'$$ admits an explicit representation $$g'(z)=\frac{a}{2\sqrt{1+z}}\cdot\frac{a}{2\sqrt{1+\sqrt{1+z}}}....$$

Edit - 04 June 2019. There are few remarks: due to the product expansion for $$g'(x)$$, all primitive zeros are simple and each simple zero of $$f$$ is primitive. Due to the comments and answers below (many thanks to the authors), the order of $$f$$ is $$\rho=\ln 2/\ln a<1$$. This value can be obtained by substituting $$e^{A|z|^{\rho}}$$ into $$f(az)=f(z)^2-1$$, which gives $$a^{\rho}=2$$. Because the order $$\rho<1$$, we have $$d=0$$ in the Weierstrass-Hadamard expansion. Something like this... For me, it was a bit strange to see the more or less explicit entire function $$f$$ whose zeros form fractals.

Edit - 05 June 2019 As mentioned above, the function $$f$$ has a lot of multiple roots. There is a linear transform which leaves simple roots of only. Without loss of generality, consider the case $$f'(0)=1$$. Recall that, see above, if $$(x_r)$$ are simple roots of $$f$$ then $$f$$ can be expressed in terms of $$H(x)=\prod(1-x/x_r)$$ as $$f(x)=\sqrt{2+(2/a)H(a^2x)}$$ or $$f(x)^2-1=1+(2/a)H(a^2x),$$ which leads to $$f(x)=1+(2/a)H(ax).$$ The function $$H$$ has simple zeros only forming a fractal structure.

Edit - 06 June 2019 People from IMRN said that $$f(az)=P(f(z))$$ is a Poincare equation. They provided also somereference:

[main] P. Fatou, "Memoire sur les equations fonctionnelles", Bull. Soc. Math. Fr., 47, 161-271; 48, 33-94, 208-314 (1919).

[2] A. Eremenko and G. Levin, "Periodic points of polynomials", Ukrain. Mat. Zh., 41 (1989), 1467--1471

[3] A. Eremenko, M. Sodin, "Iterations of rational functions and the distribution of the values of Poincare functions", Teor. Funktsii Funktsional. Anal. i Prilozhen., No. 53 (1990), 18--25; translation in J. Soviet Math. 58 (1992), no. 6, 504–509

At the moment, I did not find a detailed analysis of Hadamard expansion for Poincare functions (especially for the case $$P(z)=z^2-1$$), but there is something in, e.g.,

[4] G. Derfel, P. Grabner, F. Vogl, "Complex asymptotics of Poincare functions and properties of Julia sets", Math. Proc. Cambridge Philos. Soc., 145 (2008), 699-718

Edit - 10 June 2019 There is a closed-form expression for $$f(z)$$ based on the explicit formula for inverse $$g(w)=f^{-1}(w)$$, see above, $$g(w)=(w-a/2)\frac{2a}{a+2\sqrt{1+w}}\cdot\frac{2a}{a+2\sqrt{1+\sqrt{1+w}}}\dot....$$ We can obtain explicit Vi`ete-type formulas, involving nested radicals, for all zeros of $$f$$. Then, Weierstrass-Hadamard factorization gives us $$f(z)=\frac{a}{2}\prod_{\sigma\in\Sigma}\left(1+\frac{2z}{a}\prod_{n=1}^{\infty}\frac{a+2\sigma_n\sqrt{1+\sigma_{n-1}\sqrt{1+...+\sigma_1\sqrt{1}}}}{2a}\right),$$ where $$\Sigma=\{\sigma:\mathbb{N}\to\{\pm1\},\ \ \lim_{n\to\infty}\sigma_n=1\}.$$ This factorization is one of those I was looking for.

• Yes, but I really need an analytic $f$. – AAK May 30 '19 at 11:16
• @AAK, how do you know that? – Paul May 30 '19 at 11:57
• If $f'(0)=0$ then $f''(0)=f'''(0)=...=0$, see above. – AAK May 30 '19 at 12:03
• Note that if $f(ax) = f(x^2) + 1$, then $g(a^{1/n}x) = g(x)^2 -1$, where $g(x) = f(x^n)$. So $1\pm \sqrt{5}$ is not the only possible choice for $a$ and there are nonconstant solutions with $f'(0) = 0$. That said, I think all such solutions can expressed in terms of the basic solution with $f'(0) = 1$ and $a = 1\pm\sqrt{5}$. – eyeballfrog May 30 '19 at 19:53
• the constant functions $f(z)=\frac{1+\sqrt{5}}{2}$ an $g(z)=\frac{1-\sqrt{5}}{2}$ work. In fact, if you look into the set of all analytic functions around $0$ that satisfies the functional equation, you will that they are in fact the only solutions. – Oliver Diaz May 30 '19 at 20:38

This is more of a comment than a solution, but it is too long for a comment.

Without loss of generality you can concentrate on $$a=2$$. Here is a chain of substitutions starting from $$f(ax)=f(x)^2-1\rm :$$

$$y=\log x$$, $$x=e^y$$, $$g(v)=f(e^v)$$

This transforms $$f(ax)=f(x)^2-1$$ to $$g(y+\log a)=g(y)^2-1\rm .$$

$$z=\frac{\log 2}{\log a}y$$, $$y=\frac{\log a}{\log 2}z$$, $$h(v)=g(\frac{\log a}{\log 2}v)$$

This transforms $$g(y+\log a)=g(y)^2-1$$ to $$h(z+\log 2)=h(z)^2-1\rm .$$

Finally, undo the first transform:

$$w=e^z$$, $$z=\log w$$, $$i(v)=h(\log v)$$

This transforms $$h(z+\log 2)=h(z)^2-1$$ to $$i(2w)=i(w)^2-1\rm .$$

So if you find a solution to $$i(2x)=i(x)^2-1$$, your solution to $$f(ax)=f(x)^2-1$$ will be $$f(x)=i(x)^\frac{\log a}{\log 2}\rm .$$

• Gives an easier to solve problem! Definitely an answer for me! – MathQED May 30 '19 at 21:10
• Should be $i(x^{\log_2(a)})$ if I'm not mistaken. This is not in general analytic at $x = 0$. – eyeballfrog May 30 '19 at 21:25
• The problem determines the value of $a$ as one of $1\pm\sqrt{5}$, provided $f'(0) \neq 0$. What is the rationale for assuming $a = 2$? – Sangchul Lee May 30 '19 at 22:00

Some partial results

If $$f$$ satisfies $$f(a x) = f(x)^2 - 1$$, then $$g = f(\lambda x^n)$$ satisfies $$g(a^{1/n}x) = g(x^2)-1$$ for every $$n$$ and $$\lambda$$. This means any solution $$f$$ creates a family of possible solutions. Analysis of the power series suggests that all analytic functions that satisfy $$f(a x) = f(x)^2 -1$$ for some $$a$$ are of the form $$f_*(\lambda x^n)$$ with $$n$$ an integer and have $$a = a_*^{1/n}$$, where $${f_*}'(0) = 1$$ and $$f_*(a_* x) = f_*(x)^2 - 1$$. So it is reasonable to restrict to the case $$f'(0) \ne 0$$ provided we keep this in mind.

Next, numeric calculations. For $$a = 1+\sqrt{5}$$ case, the power series for $$f_*$$ has infinite radius of convergence and converges quickly, meaning we can calculate it numerically easily. For $$x>0$$, it scales roughly as $$\exp(0.941x^r)$$, where $$r = \ln(2)/\ln(1+\sqrt{5})$$ (thanks DinosaurEgg for that exponent) . For $$x < 0$$, it oscillates between 0 and 1 like this:

Each zero is $$a^2$$ times the previous and has double the order, as you found.

For $$a = 1-\sqrt{5}$$, the power series converges extremely slowly. Its behavior is very erf-like:

Lastly, for your infinite product representation, $$h(x)$$ scales as something like $$0.072x^r\ln(x)$$ at large positive $$x$$. Overall it looks like this

so I don't think it will have a simple form.

EDIT: I calculated a power series representation to 80 terms and totally forgot to check for zeros in $$\mathbb C\setminus \mathbb R$$. Bad news: it has them. A lot of them. For every solution to $$f(z) = \pm \sqrt{2}$$ and $$n \ge 1$$, $$a^{2n}z$$ is a zero. And there are infinitely many solutions to that, which means the infinite product plan probably won't work.

EDIT2: OK, so it seems the power series for the $$a = 1-\sqrt{5}$$ case actually does converge, just incredibly slowly--the coefficients go something like $$\exp(-0.036n^{1.6})$$, which grows faster than any exponential function but takes a hell of a lot of time to get there. Not sure if that exponent is $$1+r$$ or $$(1+\sqrt{5})/2$$.

• actually, the asymptotic behavior as $x\rightarrow\infty$ is of order $e^{\delta x^r}$, $r=\frac{\ln 2}{\ln a}$, as determined by performing dominant balance analysis on the equation. ($\delta$ is undetermined by the method I used, and also i couldn't produce subleading terms) – DinosaurEgg May 31 '19 at 0:14
• @DinosaurEgg Well that's certainly consistent with my result, as $\ln(2)/\ln(1+\sqrt{5}) \approx 0.590234$. – eyeballfrog May 31 '19 at 0:22
• I think $\delta$ must depend on the value of $f'(0)$ but I have no idea how to determine it. Care to do a graph of the exponent varying the derivative ? – DinosaurEgg May 31 '19 at 0:47
• @DinosaurEgg It's a perfect power law: $\delta = 0.941f'(0)^r$. Of course that still doesn't explain the $0.941$. – eyeballfrog May 31 '19 at 3:18
• Very interesting. Complex zeros are very important. So, I am going to rewrite the Hadamard decomposition... – AAK May 31 '19 at 5:01

$$f(ax) = f(x)^2-1$$

now assuming $$a x > 0$$ we have

$$f(a^{\log_a(ax)})=f(a^{\log_a(x)})^2-1\Rightarrow F(u+1)=F(u)^2-1$$

with $$u = \log_a x$$. Now, regarding $$F(u)$$, considering $$F(0,y) = y$$ we have using the recursion formula

$$F(5,y) = \left(\left(\left(\left(\left(y^2-1\right)^2-1\right)^2-1\right)^2-1\right)^2-1\right)^2-1$$

This recursion has some interesting properties.

$$\frac{F(2,y)}{F(1,y)} = y^2 \left(y^2-2\right)\\ \frac{F(3,y)}{F(2,y)} =\left(y^2 \left(y^2-2\right)-1\right)\left(y^2 \left(y^2-2\right)+1\right)\\ \frac{F(4,y)}{F(3,y)} =y^4 \left(y^2-2\right)^2 \left(y^4 \left(y^2-2\right)^2-2\right)\\ \frac{F(5,y)}{F(4,y)} =\left(y^2-1\right)^4 \left(y^4-2 y^2-1\right)^2 \left(y^4 \left(y^2-2\right)^2 \left(y^4 \left(y^2-2\right)^2-2\right)-1\right)\\ \vdots$$

All the remainders are $$-1$$ and the quotients are pretty factorizable.

Also regarding the functions $$F(k,y)$$ they have a behavior shown in the following plot. Here are shown $$F(0,y),\cdots, F(5,y)$$. All curves intersect at

$$\left(\frac{1}{2} \left(\sqrt{5}-1\right),\frac{1}{2} \left(1-\sqrt{5}\right)\right),\ \ \left(\frac{1}{2} \left(1+\sqrt{5}\right),\frac{1}{2} \left(1+\sqrt{5}\right)\right)$$

I hope this helps.

• Yes, this was a motivation, see above about polynomials $P_n$. – AAK May 31 '19 at 10:12
• $a=\frac{1+\sqrt{5}}2$ is the fixed point for $P(x)=x^2-1$ and, hence, for any $P_n:=P\circ ... \circ P$ $(=F(n,y))$. But in any vicinity of $a$, there are many points $y$ such that $P_n(y)\to\infty$. Other $y$, such that $P_n(y)$ is bounded, form a fractal structure. But I am not a specialist in holomorphic dynamics. – AAK May 31 '19 at 10:56

As the thread isn't closed I assume there is still some demand for answers.

The function you are looking for (or constructed) is the (inverse of) a Koenigs linearisation map. The polynomial $$g = x^2 -1$$ has a fixed point at $$a/2$$ ($$=$$ the golden ratio as you pointed out) and the derivative of $$g$$ at $$a/2$$ is $$a$$. Usually one transfers everything to 0 so for sake of completeness define $$h = g(x+a) -a$$ so that $$h(0) = 0$$ and $$h'(0) = a$$. Then by Koenigs linearisation theorem (for example see 6.1 in here https://arxiv.org/pdf/math/9201272.pdf) there exists an analytic $$\varphi$$ such that $$\varphi(0) = 0, \varphi'(0) \neq 0$$ and $$\varphi^{-1} \circ h\circ \varphi(x) = ax$$ in a neighbourhood of 0. (In relation to the reference $$\varphi = \phi^{-1}$$.) Now set $$f = \varphi + a$$. Maybe it helps finding more references about various expansions of $$f$$ one might be interested in. Good luck!

• Yes, these are very nice lectures. Somebody already recommended me this book as an introduction to holomorphic dynamics. Maybe it does not contain some specific expansions for the particular solutions of Poincare equations as I am looking for, but the book is very useful in general. – AAK Apr 7 at 11:41
• Hi AAK. The point (although perhaps not a very helpful one) I was trying to make is that the function you constructed fits into a larger well-known theory. (It stroke me as quite the coincidence that your functional equation does have any solution.) For example your function also does extend to a meromorphic function on all of $\mathbb{C}$. Does this help? I am not sure what kind of expansion you are looking for. – Gari Apr 9 at 9:03
• Hi Gari. We already found that $f(z)$, satisfying the Poincare equation $f(az)=f(z)^2-1$ with the golden ratio $a$, is an entire function (analytic everywhere!). Moreover, the Taylor expansion and Weierstrass-Hadamard expansions are also obtained. At the moment, I am looking for some relations with known functions, some integrodifferential equations for $f$, etc. – AAK Apr 9 at 10:49
• If I had to bet I would think that $f$ does not satisfy any algebraic differential equation but I would not know how to prove it. So my bet is that $f,f’,f’’,....$ are alg independent over $\mathbb{C}(z)$ but perhaps I am wrong. Would that be the kind of question you would be interested in? – Gari Apr 9 at 20:57
• Hi AAK, I recently stumbled upon a result of Becker and Berweiler and remembered our back and forth here. Their paper is: "Hypertranscendency of conjugacies in complex dynamics" (in Math. Ann.) but it turns out that Ritt already showed that $f$ as above does not satisfy any algebraic differential equations. gdz.sub.uni-goettingen.de/id/PPN235181684_0095?tify={%22pages%22:[680],%22view%22:%22info%22} Hope that helps! – Gari Aug 28 at 8:29