# 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 $\root3\of2$. – Gerry Myerson Aug 15 '15 at 0:38
• @GerryMyerson Are you sure? – Sam Aug 15 '15 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. – Gerry Myerson Aug 15 '15 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.