# A strange infinite fraction, and a functional equation

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

$$1+\cfrac{2+\cfrac{4+\cfrac{8+\cdots}{9+\cdots}}{5+\cfrac{10+\cdots}{11+\cdots}}}{3+\cfrac{6+\cfrac{12+\cdots}{13+\cdots}}{7+\cfrac{14+\cdots}{15+\cdots}}}$$

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 $$[-1,0]$$.

Questions:

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?
• 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. Nov 19 at 22:56
• Same question, in fractured English, at wlord.org/… Nov 19 at 23:36
• 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} 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|

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.)

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:

1.7302267782385217544381698541152699925740183216029488688864885028332513907333693345646270488159502214950843363020350470213441253957340326352328561791171244190587818887372617471589668574204820635351503841244668894369715741246287234805951537498344216968670493698854114058250241488445330585729167971359616966922714014756959002552196973038237715813737228888533086007706008494191029702586009155155857178120229752847309351394036332450991676860619347322383467758347490263505456371425968875446610282501221354852043094164992448913961373100134068975739552107884094399667105504894539961757197393217850962993090814905127642047885264595559651678209082120226478492561364943417813452236281748368640276148641318099905936687132637746067779650063715657905148563047541537134149138041295813105265141506542718462128816629514689451477708923545221188314450643071922013515179061067834863656328477242382897885733818064215138871260520071841200572613786364112252919672535780909630608259772405491818052189306022828479334607094679

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.