The following curious-looking fraction, with numerical value approximately $1.7302267782385217$, appears in this Reddit question:


This led me to consider the following functional equation; note that the fraction above is equal to $f(1)$.

Proposition: There is a unique meromorphic function $f$ on $\mathbb C\setminus[-1,0]$, satisfying $$f(z)=z+\frac{f(2z)}{f(2z+1)}$$ for all $z\in\mathbb C\setminus[-1,0]$.

Proof sketch: Let $X$ be the function space of all bounded holomorphic functions on $|z|>10$, equipped with the sup norm. One checks that the map $\mathcal F$ defined by $$(\mathcal Fg)(z)=\frac{g(2z)+2z}{g(2z+1)+2z+1}$$ is a contraction on the ball of radius $2$ in $X$, so it has a unique fixed point $g_0$; then we can take $f(z)=g_0(z)+z$. Now if we define $$S_n=\{z\in\mathbb C\,:\,d(z,[-1,0])>10\cdot2^{-n}\},$$ then the functional equation gives the meromorphic extension of $f$ from $S_n$ to $S_{n+1}$. $\square$

I am interested in what we can say about $f$ in general, and the value $f(1)$ in particular. Here are a couple of my observations:

  • Note that $f(1)\approx\sqrt3$ comes from the exceptionally good approximation $$f(z)=\sqrt{(z+1)^2-1}+O((z+1)^{-3}).$$
  • Also, I suspect that near $z=0$ we have $f(z)\sim Az^\alpha$, where $\alpha=\frac{\log f(1)}{\log 2}$. This also controls the singularities of $f$ at the dyadic rationals in $[-1,0]$.


  1. Is $f(1)$ irrational? transcendental? Of course, a closed-form expression for $f(1)$ would be great, but I'm not hopeful that there is one.
  2. How do we compute $f(1)$ to high accuracy, say 1000 digits?
  3. Have there been similar functional equations studied in the literature? The closest result I know of is the Woods−Robbin product identity, which can be written as $$2\times\cfrac{3\times\cfrac{5\times\cfrac{9\times\cdots}{10\times\cdots}}{6\times\cfrac{11\times\cdots}{12\times\cdots}}}{4\times\cfrac{7\times\cfrac{13\times\cdots}{14\times\cdots}}{8\times\cfrac{15\times\cdots}{16\times\cdots}}}=\sqrt2,$$ which comes from the nice closed-form solution $h(z)=\sqrt{z(z-1)}$ of the corresponding functional equation $h(z)=z\frac{h(2z-1)}{h(2z)}$. What happens when no closed-form is available?
  • 1
    $\begingroup$ I've seen a similar expression in a Mathologer video from 2016 but unfortunately I don't think he gives any references to the literature about it. $\endgroup$ Nov 19 at 22:56
  • 2
    $\begingroup$ Same question, in fractured English, at wlord.org/… $\endgroup$ Nov 19 at 23:36
  • 2
    $\begingroup$ I edited to use \cfrac to make this easier to read. And notice that \backslash differs from \setminus, thus: $$\begin{align} & \mathbb C\backslash[0,1] \\ {} \\ & \mathbb C\setminus[0,1] \end{align}$$ $\endgroup$ Nov 21 at 19:23

To find $f(1)$ to high precision, we will expand $f$ as a Laurent series in $(z+1)^{-1}$, and solve for the coefficients. Setting $f(z)=g(z+1)$, we want to find $$g(z)=z+0+a_1z^{-1}+a_2z^{-2}+\cdots$$ satisfying $$g(z)=z-1+\frac{g(2z-1)}{g(2z)},$$ or $(g(z)-z+1)g(2z)=g(2z-1)$. Hence $$\begin{array}{rcrcrcrcrl} \bigg(1&+&a_1z^{-1}&+&a_2z^{-2}&+&a_3z^{-3}&+&a_4z^{-4}&+\cdots\bigg)\\ {}\times\bigg(2z&+&0&+&\cfrac{a_1}2z^{-1}&+&\cfrac{a_2}4z^{-2}&+&\cfrac{a_3}8z^{-3}&+\cdots\bigg)\\ =\bigg(2z&-&1&+&\cfrac{a_1}2z^{-1}&+&\cfrac{a_1+a_2}4z^{-2}&+&\cfrac{a_1+2a_2+a_3}4z^{-3}&+\cdots\bigg), \end{array}$$ and we can recursively solve for $(a_n)$ as $(-\frac12,0,-\frac18,-\frac1{32},-\frac3{64},-\frac{53}{2048},\ldots)$.

Now we can recursively approximate $f$ using $$\tilde f(z)=\begin{cases}z+1+\frac{a_1}{z+1}+\frac{a_2}{(z+1)^2}+\cdots+\frac{a_m}{(z+1)^m},&|z|\geq C,\\z+\frac{\tilde f(2z)}{\tilde f(2z+1)},&|z|<C.\end{cases}$$

Error analysis: Consider the plot of $f$ below. (The wild behaviour near the interval $[-1,0]$ suggests that $f$ should have no meromorphic continuation there.) Plot of f around -1.

We see that $f$ appears to have no poles in $|z+1|>1$. Hence $a_n$ grows sub-exponentially, so the relative error $\left|\frac{\tilde f}f-1\right|$ on $|z|>C$ is at most $\frac{K_\varepsilon}{((1-\varepsilon)|z+1|)^{m+2}}$ for any $\varepsilon>0$.

Hence for 1000-digit accuracy, it should be sufficient to take eg. $m=500$, $C=100$. Python takes care of this computation in a couple of seconds, giving the value of $f(1)$ as:


Repeating the same computation with larger values of $m$ and $C$ and more digits yields the same result, so we can be fairly confident that the digits given here are correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.