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)

#Algebra #Fractions #InfiniteSum

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 5 years, 11 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.

Daniel Liu - 5 years, 11 months ago

An amazing idea !!

Sriram Venkatesan - 5 years, 11 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.

Ivan Koswara - 5 years, 11 months ago

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

Ivan Koswara - 5 years, 11 months ago

Indeed. If we assume that $f(x)\sim x+c$ as $x\to\infty$ , then $f(2x)/f(3x)\sim(2x+c)/(3x+c)\to2/3$ in the limit. Consistency then implies that $c=2/3$ .

Ricardo Moritz Cavalcanti - 5 months, 3 weeks 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.

Garrett Clarke - 5 years, 11 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:

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

Ivan Koswara - 5 years, 11 months ago

This is definetly a mind bender

Agnishom Chattopadhyay - 5 years, 11 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.

Calvin Lin Staff - 5 years, 11 months ago

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

Garrett Clarke - 5 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$