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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
2 years, 4 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.

Daniel Liu - 2 years, 4 months ago

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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.

Daniel Liu - 2 years, 4 months ago

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An amazing idea !!

Sriram Venkatesan - 2 years, 4 months 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, 4 months ago

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

Calvin Lin Staff - 2 years, 4 months ago

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

Garrett Clarke - 2 years, 4 months ago

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This is definetly a mind bender

Agnishom Chattopadhyay - 2 years, 4 months ago

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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:

\[\begin{align*} &1 + \cfrac{2 + \cfrac{4 + 0}{6 + 1}}{3 + \cfrac{6 + 1}{9 + 0}} = \frac{200}{119} \approx 1.6806722 \\ \le &1 + \cfrac{2 + \cfrac{4 + \ldots}{6 + \ldots}}{3 + \cfrac{6 + \ldots}{9 + \ldots}} \\ \le &1 + \cfrac{2 + \cfrac{4 + 1}{6 + 0}}{3 + \cfrac{6 + 0}{9 + 1}} = \frac{193}{108} \approx 1.7870370 \end{align*}\]

Ivan Koswara - 2 years, 4 months 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, 4 months ago

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

Ivan Koswara - 2 years, 4 months ago

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