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# Worse than a continued fraction

Compute the result of the following:

$1 + \cfrac{2 + \cfrac{4 + \cfrac{8 + \ldots}{12 + \ldots}}{6 + \cfrac{12 + \ldots}{18 + \ldots}}}{3 + \cfrac{6 + \cfrac{12 + \ldots}{18 + \ldots}}{9 + \cfrac{18 + \ldots}{27 + \ldots}}}$

Informally, you have something like $$f(x) = x + \frac{f(2x)}{f(3x)}$$ for all $$x$$; expand $$f(1)$$ to infinity and compute its result.

(Note that I don't know the result; if I knew, I would have made this a problem. Source: friend's tweet)

Note by Ivan Koswara
1 year, 8 months ago

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Intuitively it would seem like $$f(x)$$ is an increasing function. Rearranging the recursive function gives $$f(3x)(f(x)-x)=f(2x)$$ so $$|f(x)-x| < 1$$. That would imply that $$f(x)$$ increases approximately linearly, but then again plugging in $$f(x)=x+c$$ gives no solution. · 1 year, 8 months ago

The solution to $$c$$ in $f(3x)(f(x)-x)=f(2x)$ where $$f(x)=\sqrt[n]{x^n+c}$$ as $$n\to \infty$$ and $$x=1$$ appears to be unbounded, so this path seems to be a dead end. Of course, all the math I'm doing is pretty hand-wavy to begin with. · 1 year, 8 months ago

An amazing idea !! · 1 year, 8 months ago

The answer is definitely $$\frac{5}{3}$$. Here's as far as I've gotten in proving it:

If you follow the fraction backward far enough you'll begin to realize that the terms inside the biggest parentheses tend towards $$1$$ as the fraction expands to infinity. This gives $$1+\frac{2}{3}=\frac{5}{3}$$, but this isn't exactly a proof, more a heuristic argument. · 1 year, 8 months ago

that's what I thought initially. However, if you track the fractions, you are not correctly removing the factors of 2 and 3. Staff · 1 year, 8 months ago

This problem really intrigues me, it seems so simple to state yet so hard to prove! · 1 year, 8 months ago

This is definetly a mind bender · 1 year, 8 months ago

Well...

The answer is definitely $$\frac{5}{3}$$.

this isn't exactly a proof, more a heuristic argument.

I think they are quite contradictory.

I'm also pretty confident the answer is much closer to the one found by iteration (slightly less than $$\sqrt{3}$$) than $$\frac{5}{3}$$. Intuitively (not yet proven!), any of the fractions has value lying in $$(0, 1)$$. You can thus bound the result by replacing fractions to deep in the iteration with one of the two extreme values appropriately:

\begin{align*} &1 + \cfrac{2 + \cfrac{4 + 0}{6 + 1}}{3 + \cfrac{6 + 1}{9 + 0}} = \frac{200}{119} \approx 1.6806722 \\ \le &1 + \cfrac{2 + \cfrac{4 + \ldots}{6 + \ldots}}{3 + \cfrac{6 + \ldots}{9 + \ldots}} \\ \le &1 + \cfrac{2 + \cfrac{4 + 1}{6 + 0}}{3 + \cfrac{6 + 0}{9 + 1}} = \frac{193}{108} \approx 1.7870370 \end{align*} · 1 year, 8 months ago

Some code to compute with different starting values and iteration depth: here

It's interesting to see that things converge quickly and that it is indeed approximately linear. · 1 year, 8 months ago

In fact, for large $$x$$, we have $$f(x) \to x + \frac{2}{3}$$ (as expected?). · 1 year, 8 months ago