Equation involving continued fractions. I'm not sure if my solution is right, but I'm wondering about the equation:
$$
x = 1 + \frac 1{x + \frac 1{x+ \frac 1{x+\frac 1{x+ \frac 1{x + \ldots}}}}}
$$
I rewrite this equation in form:
$$
x = 1+\frac 1{x}
$$
And this is the golden ratio.
But I'm confused about transition from the continued fraction to this one. I'm not sure that when I replace the first $x$ in denominator I'm able to sum continued fractions. And maybe I should get something like
$$
x = 1 + \frac 1{1 + \frac 2{x+ \frac 1{x+\frac 1{x+ \frac 1{x + \ldots}}}}}
$$
 A: Your 'rewrite' is incorrect; $[x;x,x,x,\ldots]$ (to use the standard notation in which $[a;b,c,\ldots] = a+\dfrac{1}{b+\frac{1}{c+\ldots}}$) is not the same as $[1;x,x,x,\ldots]$, so you can't simply say that $x=1+\frac1x$. Instead, since $[x;x,x,\ldots]$ $= [1;x,x,\ldots]+x-1$, the equation that you get is that $x=1+\frac1{x+(x-1)}$ $=1+\frac1{2x-1}$. Can you take it from here?
ETA: I should note that even after you solve the equation for $x$ given by 'collapsing' the fraction, the problem isn't entirely done; since the equation here is improper, we're not immediately
guaranteed that an $x$ that solves the collapsed equation solves the original one. Fortunately, in this case we can look at the value of $x$ we get and see that the continued fraction that results is convergent.
A: Yout equation $x=1+\frac{1}{x}$ is wrong.
If $y=x-1$ then $$
y=\cfrac{1}{(y+1)+\cfrac{1}{(y+1)+\cfrac{1}{(y+1)+\cfrac{1}{\ddots}}}}$$
Or $y=\frac{1}{(y+1)+y}.$
You get $2y^2+y-1=0$ or $y=\frac{-1\pm 3}{4}=-1,\frac{1}{2}.$
So $x=0,\frac{3}{2}.$ But $x=0$ is absurd, so $x=3/2.$
