# How can this expression be calculated? $\dfrac{1 + \dfrac{3\cdots}{4\cdots}}{2 + \dfrac{5\cdots}{6\cdots}}$

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?

• This might help en.wikipedia.org/wiki/Continued_fraction
– user404484
Commented Jan 12, 2017 at 6:31
• 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.
– user856
Commented Jan 12, 2017 at 6:33
• @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. Commented Jan 12, 2017 at 6:33
• @Rahul $$0 + \dfrac{1 + \dfrac{3\cdots}{4\cdots}}{2 + \dfrac{5\cdots}{6\cdots}}$$ Commented Jan 12, 2017 at 6:35
• 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 $n \le f(n) \le n+1$.
– user856
Commented Jan 12, 2017 at 6:44

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.

• 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 $[\underbrace{1.73022677823852}\!12699, \underbrace{1.73022677823852}\!22390]$.
– user856
Commented Jan 12, 2017 at 20:21
• +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. Commented Jan 12, 2017 at 23:48