# When $x=2$ in the infinitely continued fraction $x+\frac{1}{x^2+\frac{1}{x^3+\ldots}}$, what algebraic value does it converge to?

Say you have the infinitely continued fraction: $$x+\cfrac{1}{x^2+\cfrac{1}{x^3+\cfrac{1}{x^4+\ddots}}}$$

When $$x=1$$, you can see that it's $$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ddots}}}$$ which converges upon the golden ratio $$\phi = 1.61803398875...$$.

What if I were to plug in $$x=2$$? $$2+\cfrac{1}{4+\cfrac{1}{8+\cfrac{1}{16+\ddots}}}$$ Using a caluclator, it seems to converge upon the irrational number: $$2.24248109286...$$

Is there any way I can represent this irrational number algebraically?

• The chances are that you can't represent this number algebraically. The chances are that it is transcendental, like $\pi$, not algebraic, like $\root3\of2$. Commented Aug 15, 2015 at 0:38
• @GerryMyerson Are you sure?
– Sam
Commented Aug 15, 2015 at 0:40
• It has very good rational approximations. The theorem of Thue-Siegel-Roth says algebraic irrationals can't have rational approximations that are too good. That's where I'd recommend starting, if you want to try to prove the number is transcendental. Commented Aug 15, 2015 at 0:46

Let $F(x)=x^0+\cfrac1{x^1+\cfrac1\cdots}~.$ Then $F(2)$ is OEIS A$214070$, for which no closed form is currently known.