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Let $n$ be any complex number and $q$ be a real number such that $q\in\Big(-\frac{1}{4};\frac{1}{4}\Big)$,$n\in\Big(-\infty;\infty\Big)$ ,it is then conjectured that the following is true

$$G(q,n)=\cfrac{1}{1-q+\cfrac{q(1-q)^n}{1-q^3+\cfrac{q(1-q^3)^n}{1-q^5+\cfrac{q(1-q^5)^n}{1-q^7+\cfrac{q(1-q^7)^n}{1-q^9+\ddots}}}}}\\=1+(n+1)\frac{q^2}{1!}-(n^2+n+4)\frac{q^3}{2!}+(n^3+6n^2+23n+30)\frac{q^4}{3!}-(n^4+22n^3+71n^2+218n+384)\frac{q^5}{4!}+(n^5+65n^4+265n^3+1315n^2+3634n+6000)\frac{q^6}{5!}-(n^6+171n^5+1225n^4+6765n^3+24334n^2+70464n+118080)\frac{q^7}{6!}+(n^7+420n^6+5908n^5+33810n^4+159859n^3+595770n^2+1638552n+2766960)\frac{q^8}{7!}-(n^8+988n^7+27146n^6+179200n^5+1057049n^4+4580212n^3+16233244n^2+44611440n+75519360)\frac{q^9}{8!}+ \dots$$

Special cases

($n=-1$)

$$\cfrac{1}{1-q+\cfrac{\frac{q}{\Big(1-q\Big)}}{1-q^3+\cfrac{\frac{q}{\Big(1-q^3\Big)}}{1-q^5+\cfrac{\frac{q}{\Big(1-q^5\Big)}}{1-q^7+\ddots}}}}=1-2q^3+2q^4-9q^5+29q^6-92q^7+316q^8-\dots$$

($n=i$)

$$\cfrac{1}{1-q+\cfrac{q(1-q)^i}{1-q^3+\cfrac{q(1-q^3)^i}{1-q^5+\cfrac{q(1-q^5)^i}{1-q^7+ \dots}}}}=1+\Big(1+i\Big)q^2-\Big(\frac{3}{2}+\frac{i}{2}\Big)q^3+\Big(4+\frac{11i}{3}\Big)q^4-\Big(\frac{157}{12}+\frac{49i}{6}\Big)q^5+\Big(\frac{475}{12}+\frac{337i}{12}\Big)q^6- \dots$$

where $i=\sqrt{-1}$

Note that by analysing the conjectured form it is evident that the coefficients of the series expansion are polynomial sequences. The form of the RHS suggests that the continued fraction is the exponential generating function $G(q)=\sum_{n=1}^{\infty}\frac{a_{n}}{n!}q^{n}$ for the given polynomials $a_{n}$(assuming we ignore matters of convergence).

How do we prove that the conjecture is true? and if true,can it be related to some well established theory?

PS:The radius of convergence had to be revised based on @mercio's argument supported by @Einar Rødland

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  • $\begingroup$ Do you have an explicit form for the polynomials in n? $\endgroup$ – marty cohen Jul 12 '17 at 2:55
  • $\begingroup$ @marty cohen :Not yet but i'm working on it.I have yet observed the relationship empirically.But it should be remembered that the polynomials have a combinatorial meaning $\endgroup$ – Nicco Jul 12 '17 at 8:30
  • $\begingroup$ what does "Thus the continued fraction is the exponential generating function for the given polynomials" mean ? $\endgroup$ – mercio Jul 13 '17 at 20:31
  • $\begingroup$ @mercio : the RHS is of the form $G(q)=\sum_{n=1}^{\infty}\frac{a_{n}}{n!}q^{n}$ where $a_{n}$ are the given polynomials $\endgroup$ – Nicco Jul 13 '17 at 20:58
  • $\begingroup$ So the conjecture is that the continued fraction has a power series expansion in $q$ with coefficients polynomials in $n$? I just want to make sure that this is the complete conjecture. $\endgroup$ – Somos Jul 13 '17 at 22:07
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Here is the general setup: We let $$A(x) = 1 + a_1 x^1 + a_2 x^2 + a_3 x^3 + \dots$$ where the $a_n$ are elements from any ring such as a polynomial ring. We seek the power series $$F(x, q) = \sum_{n,k\ge 0} b_{n,k} x^n q^k$$ that satisfies the functional equation $$F(x, q) = 1 / (1 - x + q A(x) F(x q^2, q)).$$ The theorem is that for all $n$ and $k$, $b_{n,k}$ is a multivariate polynomial in the variables $\{a_1,a_2,\dots\}.$

More precisely, $b_{0,k}=(-1)^k C_k$ where $C_k>0$ is Catalan numbers, $\; b_{1,k} = (-1)^k(u_k + v_k a_1)$, and so on involving more of the $a_i$s. Because of $b_{0,k}$, the convergence of $F(x,q)$ requires $|q|<1/4$.

Perhaps this expansion is of interest: $F(x,q)=+(1-(1)q+(2)q^2-(5)q^3+(14)q^4-(42)q^5+(132)q^6+\dots)x^0$ $+(1-(a_1+2)q+(3a_1+5)q^2-(9a_1+15)q^3+(29a_1+46)q^4-(95a_1+146)q^5+\dots)x^1$ $+(1-(2a_1+a_2+3)q+(a_1^2+8a_1+3a_2+9)q^2-(5a_1^2+29a_1+9a_2+30)q^3+\dots)x^2$ $+(1-(3a_1+2a_2+a_3+4)q+(3a_3+8a_2+15a_1+2a_2a_1+3a_1^2+14)q^2+\dots)x^3+\dots$

A proof could produce $F(x,q)$ as the limit of an iteration where $F_0(x,q)=1+O(q)$ and $F_{n+1}(x,q)=1 / (1 - x + q A(x) F_n(x q^2, q)).$

To apply the theorem to the present situation, let $A(x) = (1-x)^n$, then $G(q) = F(q,q)$.

I wrote code to implement this in $\texttt{PARI/GP}$ and here is a test run:

$\texttt{F = 1 + O(q^1); A = (1 - x + O(x^5))^n;}$ $\texttt{for(k=1, 4, F = 1 / (1 - x + q * A * subst(F, x, x*q^2)));}$ $\texttt{print(serlaplace(subst((F-1)/x, q x))*x+1)}$ $\texttt{1 + (n + 1)*x^2 + (-n^2 - n - 4)*x^3 + (n^3 + 6*n^2 + 23*n + 30)*x^4 + O(x^5)}$

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  • $\begingroup$ Very interesting.What's the name of the theorem? $\endgroup$ – Nicco Jul 15 '17 at 7:25
  • $\begingroup$ I don't have a name for it. It is one of several variations which have similar results. I suppose someone could get a common generalization and give that a name. The proofs are all similar and simple using the iteration I mention. $\endgroup$ – Somos Jul 15 '17 at 10:48
  • $\begingroup$ Somos :it would be nice if you could provide the complete proof in your answer $\endgroup$ – Nicco Jul 18 '17 at 16:50
  • $\begingroup$ What specifically requires proof? $\endgroup$ – Somos Jul 18 '17 at 18:33
  • $\begingroup$ I guess I meant to say deriving the given polynomials using that particular theorem $\endgroup$ – Nicco Jul 19 '17 at 2:40
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If $(f_n)$ is a sequence of Mobius transforms that converges (we identify Mobius transforms with a subset of $P^3(\Bbb C)$) to a Mobius transform $f$ that has an attractive fixpoint $a$ and a repulsive fixpoint $b$, then every sequence $(z_n)$ satisfying $z_n = f_n(z_{n+1})$ converges to $b$ except one, who converges to $a$.

Furthermore, if $\lim_{n \to \infty} f^{\circ n} (0) = a$ then $z_n = \lim_{ k\to \infty} f_n \circ \ldots \circ f_{n+k} (0)$ gives that sequence converging to $a$, which means that the corresponding continued fraction converges to $a$

The proof I have is a bit technical and boring :

We can assume that the attractive and repulsive fixpoints of $f$ are $0$ and $\infty$ respectively by conjugating everything by a suitable Mobius transform if needed, so that $f(z) = \lambda z$ with $|\lambda| < 1$. Also, for convenience, note $F_a^b = f_a \circ f_{a+1} \circ \ldots \circ f_b$

Since for any $R > 0$, the applications $Aut(P^1) \to \Bbb R_{> 0}$ given by $\tau \mapsto \sup_{|z| < R} |\tau(z)|$ and $\tau \mapsto \sup_{|z| < R} |\tau'(z)|$ are continuous, for every $R>0$, $S_R = \{\tau \in Aut(P^1) \mid \exists R' > R \exists \mu < (1+|\lambda|/2), \tau(B(0,R)) \subset B(0,R')$ and $\tau |_{B(0,R)}$ is $\mu$-Lipschitz $\}$ is open.

Since $f$ is in all of those sets, for any $R>0$ there is an $n_R$ such that $f_n \in S_R$ for $n \ge n_R$.

Then, for any $n \ge N_R$, the sequence $F_n^{n+k} (B (0,R)))$ is a decreasing sequence of non-empty closed subsets of $B(0,R)$ whose diameters decrease exponentially to $0$, so their intersection is a point
$z_n^{(R)} \in B(0,R)$. By construction, $z_n^{(R)} = f_n(z_{n+1}^{(R)})$ and $(z_n^{(R)},z_{n+1}^{(R)},\ldots)$ is the only such sequence that stays in $B(0,R)$.

If you do this for a $R' < R$ you obtain another sequence $(z_n^{(R')})$ defined from an index $n_{R'} > n_R$ that is the only sequence that stays in $B(0,R')$. But since it also stays forever in $B(0,R)$, we must have $z_n^{(R)} = z_n^{(R')}$ forall $n \ge n_{R'}$.

This proves that the sequence(s) we obtain converge to $0$, and by extending it the other way, we obtain the unique sequence $(z_0,z_1,\ldots )$ that converges to $0$ .

Furthermore, if $z_0 \in \Bbb C = P^1 \setminus \{ \infty \}$ then by taking $R > |z_0|$, we get that for $n \ge n_R$, $F_n^{n+k} (z_0) \in F_n^{n+k}(B(0,R))$ and so $F_n^{n+k} (z_0)$ converges to $z_n^{(R)} = z_n$. Then by continuity of the first $n_R$ functions, the sequence $(F_0^n)(z_0)$ converges to $z_0$. This is better than what I announced since this works with any $z_0$ that is not the repulsive fixpoint, not just $0$.

Finally, say we have a recurrent sequence not converging to $0$. Then there is an $R$ such that it gets out of $B(0,R)$ infinitely many times. But then, conjugating the $f_n^{-1}$ with $z \mapsto 1/z$ and applying the same reasoning, for $n$ large enough, $f_n^{-1}$ is close enough to $f^{-1}$ so that $P^1 \setminus B(0,R)$ is stable by all the $f_n^{-1}$ and in fact it multiplies its argument's modulus by at least $2/(1+|\lambda|) > 1$. Then as soon as the sequence enters this zone for $n$ large enough it will have to converge to $\infty$ exponentially. And it will enter it because it goes there an infinite amount of times.

Thus, every sequence converge to the repulsive fixpoint except one which converges to the attractive one and is given by the value of the continued fraction as long as $0$ is not the repulsive fixpoint itself (I am not sure what happens then)


In your case, your Mobius transforms converge to $z\mapsto q/(1+z)$, whose fixpoints are $\frac {-1 \pm \sqrt {1+4q}}2$. The attractive one being the one on the right side of the line $\Re(z) = -1/2$.

And so if the sequence $q, q/(1+q), q/(1+q/(1+q)) \ldots$ converges the attractive fixpoint (and I think it does for every $z \notin (\infty ; -1/4])$, then your continued fraction converges to some $G_n(q) \in \Bbb C$, and the sequence $(z_{n,q,k})$ given by $z_{n,q,0} = G_n(q)$, $z_{n,q,0} = 1/(1-q+z_{n,q,1})$ and $z_{n,q,k} = q(1-q^{2k-1})^n/(1-q^{2k+1}+z_{n,q,k+1})$ for $k \ge 1$ is the only such recursive sequence that converges to $\frac{-1+\sqrt{1+4q}}2$


A very similar argument can be made on the ring of formal power series except it is a lot easier because ultrametric spaces are very nice :

If $(a_k,b_k) \in K[[q]]$ and if you have a sequence of formal power series $z_k \in K((q))$ such that $z_k(1 + qb_k + z_{k+1}) = qa_k$, then there is a unique one that stays forever in $qK[[q]]$, and all the other sequences eventually get stuck in $-1+qK[[q]]$

Here, since the maps $z_{k+1} \to z_k$ are contracting when restricted to $qK[[q]]$, in order to know $z_k$ modulo $q^n$ you only need to know $z_{k+1}$ modulo $q^{n-1}$.

Therefore in order to compute this unique sequence, to compute $z_k$ modulo $q^n$ you simply unfold the recurrence about $n$ times and then the contribution of $z_{k+n}$ has a $q^n$ coefficient and so it disappears modulo $q^n$.

And you can say this much only just by looking at the very first coefficients of $a_k$ and $b_k$.

If you use that your coefficients actually converge to something then you can deduce that $z_k$ converges one coefficient at a time to the expansion of $\frac{-1+\sqrt{1+4q}}2 = q-2q^2 + \ldots$ and that all the other sequences will converge to the other.

In fact all the operations needed to compute the coefficients of $G$ are ring operations, so by taking $R = \Bbb Z[n,n(n-1)/2, n(n-1)(n-2)/6, \ldots]$ you get the result that your continued fraction may indeed correspond to a power series expansion whose coefficients are polynomials in $n$.

And you need this $R$ because you have to expand $(1-q)^n = 1-nq-n(n-1)q^2/2 \ldots$ to write the coefficients of the Mobius transforms as power series.


It is very likely that the power series you obtain has radius of convergence $1/4$ and coincides with the "pointwise" solution obtained earlier but I'm not sure how to prove it neatly.

And even if it is, it doesn't say anything away from $B(0,1/4)$, and the behaviour also may change for $|q| > 1$ because the limit transformation is not the same (if it converges at all)

For $n=1$ with this time I hope a correct code I get this : enter image description here

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    $\begingroup$ I agree that the radius of convergence should be $1/4$. The way I reached that conclusion was that if $G_k$ is the continued fraction starting from the $k$th level down, as $k$ increases the $q^{2k+1}$ terms vanish and you get $G_\infty(q)=[1+\sqrt{1-4q}]/2$. Liked the ring argument by the way. $\endgroup$ – Einar Rødland Jul 16 '17 at 10:40
  • $\begingroup$ wow I think I completely messed up my code, I put the coeffs in the wrong places lol (brb doing it again) $\endgroup$ – mercio Jul 16 '17 at 15:08
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Let's just treat the expression as a formal power series in $q$ and $n$ where we expand $(1+u)^n=1+nu/1!+n(n-1)u^2/2!+\cdots$ regardless of what kind of value $n$ takes.

We write such a power series $$F(q,n) = \sum_{k=0}^\infty f_k(n) q^k / k!$$ where $f_k(n)$ in turn are power series in $n$. I'll use $F$, $H$ to denote arbitrary power series, while $G$ will refer to the one defined in the problem.

Define $F(q,n)$ to be of Type $M$ for some non-negative integer $M$ if $f_k(n)$ is an integer coefficient polynomial of degree at most $k$ for all $k\le M$. We say that it is of Infinite Type or Type $\infty$ if it holds for all $k$. Refer to $f_0(n)$, or just $f_0$ since this should be an integer, as the Leading coefficient.

Lemma: The power series $(1-q^k)^n$ is of Type $\infty$ with leading coefficient 1.

Lemma: If $F,H$ are power series of type $M$, then so are $F+H$, $F-H$, $FH$, $F/H$ (provided $h_0\not=0$), and $F+qH$; the leading coefficients are $f_0+h_0$, $f_0-h_0$, $f_0h_0$, $f_0/h_0$, and $f_0$ respectively.

Lemma: If $F$ is of type $M$ and $H$ is of type $M-1$, then $F+qH$ is of type $M$ and has leading coefficient $f_0$.

Now, let's look at the continued fraction. We may rewrite this as $G_0(q,n)$ which we define recursively as $$ G_{k}(q,n) = \frac{1}{1-q^{2k+1}+q\left(1-q^{2k+1}\right)^n G_{k+1}(q,n)}. $$ All $G_k$ have leading coefficient 1, and so are of Type 0 (ie integer leading coefficient).

If $G_{k+1}$ is of type $M$, by the above lemmas, $G_k$ will be of type $M+1$.

Combining these, since $G_{k+M}$ is of type 0 for any $k,M\ge0$, $G_k$ must be of type $M$ for any $M$, which means all $G_k$ are of Type $\infty$.

This is almost the full result, except that we now have terms of type $g_k(n)q^n/n!$ where $g_k$ has degree at most $k$ while in the result we should have had similar coefficients on the form $g_k(n)q^{k+1}/k!$: ie an extra power of $q$. However, this actually follows from the fact that we can write $$ G_k(q,n) = \frac{1}{1-qH_k(q,n)} = 1 + qH_k + q^2 H_k^2 + \cdots $$ where $H_k$ is of type $\infty$ and therefore has terms of type $h_k(n)q^k/k!$.

Next, the degree 1 term of $G_0(q,n)$ vanishes because the denominator of $$ G_0(q,n) = \frac{1}{1-q+q(1-q)^n G_1(q,n)} $$ has the form $1+q^2H(q,n)$.

So this completes the proof that $G$ has leading coefficient 1 and terms on the form $h_k(n)q^{k+1}/k!$ where $h_k(n)$ are integer coefficient polynomials of degree at most $k$ and with $h_0=0$ (for the $q^1$ coefficient).

It should also follow, using the same reasoning, that for any integer $n$, the coefficients of $G(q)$ will also be integers.

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