User:Andrewdavinci

$$varphi = $$ 1.61803398874989484820458683436563811772 0309179805762862135448622705260462818902 44970720720418939113748475





$$\varphi = \sqrt[3](\ 2+\sqrt{5})$$

$$\varphi - 1 = \tfrac{1}{\varphi}$$.

$$\tfrac{1}{\sqrt[3](\ 2+\sqrt{5}} = \sqrt[3](\ 2+\sqrt{5}) - 4$$

$$\varphi$$2 = $$\varphi + 1$$

$$\varphi = \frac{1+\sqrt{5}}{2}$$


 * $$\varphi = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}\,.$$


 * $$-\frac{\varphi}{2}=\sin666^\circ=\cos(6\cdot 6 \cdot 6^\circ).{}$$


 * $$\varphi = 1+2\sin(\pi/10) = 1 + 2\sin 18^\circ$$
 * $$\varphi = {1 \over 2}\csc(\pi/10) = {1 \over 2}\csc 18^\circ$$
 * $$\varphi = 2\cos(\pi/5)=2\cos 36^\circ.\,$$


 * $$\sum_{n=1}^{\infty}|F(n)\varphi-F(n+1)|

= \varphi\,.$$


 * $$\varphi^{n+1} = \varphi^n + \varphi^{n-1}\,.$$



\begin{align} 3\varphi^3 - 5\varphi^2 + 4 & = 3(\varphi^2 + \varphi) - 5\varphi^2 + 4 \\ & = 3[(\varphi + 1) + \varphi] - 5(\varphi + 1) + 4 \\ & = \varphi + 2 \approx 3.618. \end{align} $$

The golden ratio's decimal expansion can be calculated directly from the expression
 * $$\varphi = {1+\sqrt{5} \over 2},$$

√5 ≈ 2.2360679774997896964.


 * $$x_{n+1} = \frac{(x_n + 5/x_n)}{2}$$

xn and xn−1.


 * $$x_{n+1} = \frac{x_n^2 + 1}{2x_n - 1},$$

x − 1 − 1/x = 0,
 * $$x_{n+1} = \frac{x_n^2 + 2x_n}{x_n^2 + 1}.$$

F25001 and F25000,