17
$\begingroup$

I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...

$\dfrac{1 + \dfrac{3}{4}}{2 + \dfrac{5}{6}} \approx 0.6176 $

$\dfrac{1 + \dfrac{3 + \dfrac{7}{8}}{4 + \dfrac{9}{10}}}{2 + \dfrac{5 + \dfrac{11}{12}}{6 + \dfrac{13}{14}}} \approx 0.6175 $

Going all the way up to 62 gives a result of 0.627841944566, so it seems to converge.

Is it possible to find a value for this? Will it have a closed form solution?

$\endgroup$
9
  • $\begingroup$ This might help en.wikipedia.org/wiki/Continued_fraction $\endgroup$
    – user404484
    Commented Jan 12, 2017 at 6:31
  • 2
    $\begingroup$ Not to detract from your question, but I find the following different expression more natural: $$1+\frac{2+\frac{4+\frac{8+\cdots}{9+\cdots}}{5+\frac{10+\cdots}{11+\cdots}}}{3+\frac{6+\frac{12+\cdots}{13+\cdots}}{7+\frac{14+\cdots}{15+\cdots}}}$$ That's a proper binary tree, that is. And here the powers of two are apparent. $\endgroup$
    – user856
    Commented Jan 12, 2017 at 6:33
  • $\begingroup$ @anonymous I considered if it would be an example of a continued fraction when I was tagging the post, but it doesn't seem to me like it is since each fraction has a numerator. That being said I'm honestly not sure. $\endgroup$
    – the6p4c
    Commented Jan 12, 2017 at 6:33
  • 2
    $\begingroup$ @Rahul $$0 + \dfrac{1 + \dfrac{3\cdots}{4\cdots}}{2 + \dfrac{5\cdots}{6\cdots}}$$ $\endgroup$
    – Axoren
    Commented Jan 12, 2017 at 6:35
  • 1
    $\begingroup$ More compactly (thanks @Axoren), your expression is $f(0)$, where $f$ is the (unique?) function that satisfies $f(n) = n + \frac{f(2n+1)}{f(2n+2)}$ and $n \le f(n) \le n+1$. $\endgroup$
    – user856
    Commented Jan 12, 2017 at 6:44

1 Answer 1

6
$\begingroup$

Define $$f_m(n) = \begin{cases} n+\cfrac{f_m(2n+1)}{f_m(2n+2)} & \text{if $n<m$,} \\ n & \text{otherwise.} \end{cases}$$ Then $$f_0(0) = 0, \quad f_1(0) = 0 + \frac12, \quad f_2(0) = 0 + \frac{1+\frac34}2, \quad f_3(0) = 0 + \frac{1+\frac34}{2+\frac56}, \quad \dots$$ and you're looking for the value of $\lim_{m\to\infty}f_m(0)$.

One can show via reverse induction over $n=m,\ldots,1,0$ that $f_m(n) \in [n, n+1]$. So in the limit, defining $f(n)=\lim_{m\to\infty}f_m(n)$, we have $f(n) \in [n, n+1]$.

Using interval arithmetic we can then obtain rigorous bounds on $f(n)$. Define the interval-valued function $$[f]_m(n) = \begin{cases} n+\cfrac{[f]_m(2n+1)}{[f]_m(2n+2)} & \text{if $n<m$,} \\ [n, n+1] & \text{otherwise,} \end{cases}$$ where the usual interval arithmetic rules apply, $$x+[a,b] = [x+a,x+b], \qquad \frac{[a_1,b_1]}{[a_2,b_2]} = \left[\frac{a_1}{b_2}, \frac{a_2}{b_1}\right]$$ (because all our intervals are positive, except $[f]_0(0)$ which never appears in a denominator). It should be possible to show that $f_{k}(n) \in [f]_m(n)$ for all $k\ge m$, and so $f(n) \in [f]_m(n)$. Assuming that's true, $$[f]_{1023}(0) = [\underbrace{0.62784196682396}\!542323, \underbrace{0.62784196682396}\!734620]$$ narrows down the desired number $f(0)$ to $14$ significant digits.

$\endgroup$
2
  • $\begingroup$ P.S. My preferred number $1+\frac{2+\frac{4+\frac{8+\cdots}{9+\cdots}}{5+\frac{10+\cdots}{11+\cdots}}}{3+\frac{6+\frac{12+\cdots}{13+\cdots}}{7+\frac{14+\cdots}{15+\cdots}}}$ lies in $[\underbrace{1.73022677823852}\!12699, \underbrace{1.73022677823852}\!22390]$. $\endgroup$
    – user856
    Commented Jan 12, 2017 at 20:21
  • $\begingroup$ +1 This interval arithmetic approach is awesome and provides a good basis for testing exact results. That last statement just needs to be proved and you have bounds around the solution which are arbitrarily tight. $\endgroup$
    – Axoren
    Commented Jan 12, 2017 at 23:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .