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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...

Xetrov
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3 Answers3

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

Dando18
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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: $$ \varphi =2\cos {\frac {\pi }{5}}=2\sin {\frac {3\pi }{10}} $$

Erik Pillon
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  • I hope you read that no trigonometric functions are allowed... Also, do you have proof that it is the fastest way ? – Xetrov Jun 06 '17 at 15:43
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    Sorry, I have to bag you're pardon... But actually what do you mean by fastest ? – Erik Pillon Jun 06 '17 at 15:44
  • More decimal places for more iterations of a function – Xetrov Jun 06 '17 at 15:52
  • Then I thin the fastest way (but I don't have any proof about it) it is exactly the coninuous fraction! Lol. But if you want, also considering the fact that the ration of tho successive numbers of the Fibonacci numbers tends to $\phi$... Only sums and ratios are needed – Erik Pillon Jun 06 '17 at 16:01
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    A fast way to walk on convergents while skipping most of them https://math.stackexchange.com/questions/1733530/newtons-method-for-square-roots-jumps-through-the-continued-fraction-converge – Jaume Oliver Lafont Jun 06 '17 at 20:00
  • @Goodra That request is not well-defined, because you also need the function to be nicely computable so that you can iterate it in the first place. – Ian Jun 06 '17 at 20:03
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$\varphi=\frac{1+\sqrt{5}}{2}$ 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.

Jaume Oliver Lafont
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