Is there a proof that there is no sequence for root 2 like there is for the golden ratio?

Question

The golden ratio has can be approximated as a ratio of 2 successive terms of the Fibonacci sequence. 1 1 2 3 5 8 etc... The rule is pretty simple. Is there a proof that there are no rules where you can generate the square root of 2?

I looked at the Pell equation, but you can't get a rule out of it, is there anything else or is the Golden Ratio special? For example I tried picking random numbers to start with and then if you keep doing the same rule eventually you get the golden ratio again if you do the ratio between successive terms!

Answer 1

As people say in the comments, you need to define what sort of rules are allowed. The classic Babylonian sequence $a_1=1, a_n=\frac 12(a_{n-1}+\frac 2{a_{n-1}})$ converges rapidly to $\sqrt 2$. If you want a homogeneous linear recurrence you need the characteristic polynomial to have a root of $\sqrt 2$ and all other roots smaller than $\sqrt 2$ in absolute value. You can't do that with integer coefficients, but you can have the characteristic polynomial $(x-\sqrt 2)(x+\sqrt 2-1)=x^2+x-2-\sqrt 2$ so the recurrence would be $x_n=x_{n-1}+(2+\sqrt 2)x_{n-2}$

Answer 2

The series you mentioned for the golden ratio is actually the series of convergents of its continued fraction. $$\frac{1+\sqrt{5}}{2}=1+\frac{1}{1+\frac{1}{1+\dotsb}}$$

Every number has a continued fraction, but numbers which are the solution to a second degree equation with rational coefficients (quadratic irrationals) have periodic ones.

$$\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dotsb}}}$$

You can prove this equality by noticing that

$$x=1+\frac{1}{1+\color{red}{1+\frac{1}{2+\frac{1}{2+\dotsb}}}}=1+\frac{1}{1+x}$$

So if you stop this sequence at the first points you get increasingly good approximations.