Golden Ratio Approximation

Question

$$\sqrt{1000}-30.0047 \approx \varphi $$ $$[(\sqrt{1000}-30.0047)^2-(\sqrt{1000}-30.0047)]^{5050.3535}\approx \varphi $$ Simplifying Above expression we get
$$1.0000952872327798^{5050.3535}=1.1618033..... $$ Is this really true that $$[\varphi^2-\varphi]^{5050.3535}=\varphi $$

Answer 1

This is not an answer but it is too long for a comment.

Working with illimited precision, let $$a=\sqrt{1000}-\frac{300047}{10000}\approx 1.6180766016837933200$$ $$(a^2-a)^{\frac {50503535}{10000}}\approx 1.6180331121536741389$$ $$\phi\approx 1.6180339887498948482$$ Using as exponent $5050.3592$ (same number of digits as in your post) instead (and doing the same), you would get $$(a^2-a)^{\frac {50503592}{10000}}\approx 1.6180339909260630347$$ which is closer but still not exact (Vincenzo Oliva clearly explained the problem).

Answer 2

No, not at all. Factoring and then using a well known property of $\varphi$: $$\left(\varphi(\varphi-1)\right)^{5050.3535}=\left(\frac{\varphi}{\varphi}\right)^{5050.3535}=1\ne\varphi. $$

The argument needs to be a bit larger than $1$. You're dealing with an approximation.

Answer 3

Of course, this is an approximation. No need for more proof than those already given by Vincenzo Oliva and Claude Leibovici - kindest regards !

I would add that it is easy to find a lot of such approximations.

Some amazing ones, to be compared to :

$1.6180339887...\simeq\frac{1+\sqrt5}{2}=\varphi$

$1.6180339886...\simeq\cos (\sqrt2\: e^{-2})-\cos(\sqrt[3]{2}\:e^\pi )$

$1.6180339884...\simeq\sqrt{\cosh(\gamma)+\cos(\gamma)}-\frac{\gamma^3}{\sin(5)}$ where $\gamma$ is the Euler constant.

$1.6180339881...\simeq\cosh(G^2\sinh(1))+\cos\left(\frac{\sqrt[3]\pi}{\cos(3)} \right)$ where $G$ is the Catalan constant.

They comes from : https://fr.scribd.com/doc/14161596/Mathematiques-experimentales (page 4)

This paper roughly explains the method to compute a lot of approximations of this kind (In French, presently no translation avalable).