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.

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see en.wikipedia.org/wiki/Continued_fraction

– hyportnex

2 days ago

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1 Answer

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This is what’s called a continued fraction. Let’s denote your continued fraction as f(x)f(x). Notice that if we subtract xx from f(x)f(x) and take its reciprocal, we get f(x)f(x) back. Therefore, 1f(x)−x=f(x)\frac{1}{f(x)-x} = f(x) or f(x)2−xf(x)−1=0f(x)^2-xf(x)-1 = 0 Solving for f(x)f(x) using the quadratic formula, we get f(x)=x±√x2+42f(x) = \frac{x\pm\sqrt{x^2+4}}{2} It’s not hard to see that if xx is positive, then f(x)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>0x > 0, we have f(x)=x+√x2+42f(x) = \frac{x+\sqrt{x^2+4}}{2} This is the same as x2+√(x2)2+1\frac{x}{2}+\sqrt{\left(\frac{x}{2}\right)^2+1} (where we have just distributed the 12\frac{1}{2}). Notice that f(1)=1+√52=ϕf(1) = \frac{1+\sqrt{5}}{2} = \phi.