What is the quickest way to calculate $\phi$?

Question

What is the quickest way to calculate $\phi$? I am looking for methods that don't include complex calculations such as cube root, or $\sin$ etc. but $\sqrt {\ \ }$ is okay. I think it is $\dfrac{1+\sqrt5}{2}$ but I do not have any proof. I will calculate square roots by using $$\sqrt x = x_\infty ; x_{n+1} = \dfrac{x_n+\dfrac{x}{x_n}}{2}$$

I am calculating by hand

Edit: If anybody hasn't read the comments below, then it says: "which produces $\phi$ the fastest for more iterations of that formula/function. a single fraction is obviously one function, but how many decimal places does it produce? I'd rather not have continued fractions...

Answer 1

I think you can try with the very quick formula $$ \phi =1+{\frac {1}{1+{\frac {1}{1+{\frac {1}{1+{\frac {1}{1+...}}}}}}}}$$ By the way, without posting stupid solutions, I would suggest the trigonometric interpetation: $$ \varphi =2\cos {\frac {\pi }{5}}=2\sin {\frac {3\pi }{10}} $$

Answer 2

There are several ways:

• Limit of ratio between consecutive fibonacci numbers: $\phi = \frac{F_{n+1}}{F_n}$ as $n \to \infty$

• Continued square root: $\varphi = \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}$

• Continued fraction: $\varphi = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}}$

• Convergence sum: $\varphi = \frac{13}{8} + \sum_{n=0}^\infty \frac{(-1)^{n+1} (2n+1)!}{(n+2)!n!4^{2n+3}}$

• Trig. Identity (included simply for information's sake): $\varphi = 2 \sin \frac{3\pi}{10} = 2 \sin 54^\circ$

• Fractional Approximations: $\varphi \approx \frac{46\, 368}{28\, 657}$ (accurate to $10$ digits)

• Definition: $\varphi$ is defined as the solutions of $x^2 - x - 1$, which gives $\varphi = \frac{1 + \sqrt{5}}{2}$

Answer 3

$\varphi=\frac{1+\sqrt{5}}{2}$ is the largest root of the polynomial

$$f(x)=x^2-x-1$$

which has derivative

$$f'(x)=2x-1$$

You can apply Newton-Raphson method to approximate roots with quadratic convergence

$$x_{n+1}=x_{n}-\frac{f(x_n)}{f'(x_n)}$$

which here becomes

$$\begin{align} x_{n+1}&=\frac{x_n^2+1}{2x_n-1} \end{align}$$

and start from ${x_0=2}$, for instance.

The resulting sequence of approximations are the convergents

$$\frac{5}{3}, \frac{34}{21}, \frac{1597}{987}, \frac{3524578}{2178309}, \frac{17167680177565}{10610209857723},...$$

These are the truncations of the Egyptian fraction

$$\begin{align} \varphi&=2-\sum_{k=0}^\infty \frac{1}{F(2^{k+2})}\\ &=2-\frac{1}{3}-\frac{1}{21}-\frac{1}{987}-\frac{1}{2178309}-\frac{1}{10610209857723}-... \end{align}$$

where F(n) are the Fibonacci numbers starting from $F(0)=0, F(1)=1$.

Here are more examples of similar Egyptian fractions that correspond to different initial values for the algorithm.