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conjecture Right, so this was something I stumbled upon whilst looking at forms of the golden ratio (if you replace x with one in the diagram above then you get phi) and then I wondered about what if you messed with the value of x, did a massive amount of trial and error (lost in a notebook somewhere) then came up with the equation in the right, which worked exactly for every value I tested (about up to 100)

Now I have found stack exchange, in wondering if there is a way to prove this, I plugged in some values into wolfram alpha and geogebra, which seems to show it works, but I want a proof, not just trial and error answers. graphs!

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This is what's called a continued fraction. Let's denote your continued fraction as $f(x)$. Notice that if we subtract $x$ from $f(x)$ and take its reciprocal, we get $f(x)$ back. Therefore, $$\frac{1}{f(x)-x} = f(x)$$ or $$f(x)^2-xf(x)-1 = 0$$ Solving for $f(x)$ using the quadratic formula, we get $$f(x) = \frac{x\pm\sqrt{x^2+4}}{2}$$ It's not hard to see that if $x$ is positive, then $f(x)$ must be positive (as the reciprocal of a positive number is always positive, as is the sum of two positive numbers). Therefore, for $x > 0$, we have $$f(x) = \frac{x+\sqrt{x^2+4}}{2}$$ This is the same as $\frac{x}{2}+\sqrt{\left(\frac{x}{2}\right)^2+1}$ (where we have just distributed the $\frac{1}{2}$). Notice that $f(1) = \frac{1+\sqrt{5}}{2} = \phi$.

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