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)

## Comments

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TopNewestIntuitively 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. – Daniel Liu · 2 years, 1 month ago

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– Daniel Liu · 2 years, 1 month 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.Log in to reply

An amazing idea !! – Sriram Venkatesan · 2 years, 1 month ago

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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. – Garrett Clarke · 2 years, 1 month ago

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– Calvin Lin Staff · 2 years, 1 month ago

that's what I thought initially. However, if you track the fractions, you are not correctly removing the factors of 2 and 3.Log in to reply

– Garrett Clarke · 2 years, 1 month ago

This problem really intrigues me, it seems so simple to state yet so hard to prove!Log in to reply

– Agnishom Chattopadhyay · 2 years, 1 month ago

This is definetly a mind benderLog in to reply

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:

\[\begin{align*} &1 + \cfrac{2 + \cfrac{4 + 0}{6 + 1}}{3 + \cfrac{6 + 1}{9 + 0}} = \frac{200}{119} \approx 1.6806722 \\ \le &1 + \cfrac{2 + \cfrac{4 + \ldots}{6 + \ldots}}{3 + \cfrac{6 + \ldots}{9 + \ldots}} \\ \le &1 + \cfrac{2 + \cfrac{4 + 1}{6 + 0}}{3 + \cfrac{6 + 0}{9 + 1}} = \frac{193}{108} \approx 1.7870370 \end{align*}\] – Ivan Koswara · 2 years, 1 month ago

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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. – Ivan Koswara · 2 years, 1 month ago

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– Ivan Koswara · 2 years, 1 month ago

In fact, for large \(x\), we have \(f(x) \to x + \frac{2}{3}\) (as expected?).Log in to reply