# Thread: Calculating Phi (golden ratio) - continued fraction.

1. ## Calculating Phi (golden ratio) - continued fraction.

My teacher told our class that the $\displaystyle \varphi$ can be calculated by:

$\displaystyle 1 + \frac{1}{1 + \frac{1}{1+ \frac{1}{...}}}$

I don't understand how or why this works. I understand that as the denominators' continue on forever, the 1 + (fraction) changes ever so slightly. How do I go about solving this?

But I'm curious... What if I wanted to include a different set of numbers in a similar fashion?

$\displaystyle 2*\frac{3x+1}{2*\frac{3x+1}{2*3x+1 ...}}$

My gut feeling is, eventually, this number will reach infinity, if x > -1.

*note* I'm not used to the math programming language, so I know what is appearing is slightly off. If need be, I'll transcribe my intentions without the math programming language.

2. Originally Posted by EMyk01
My teacher told our class that the $\displaystyle \varphi$ can be calculated by:

$\displaystyle 1 + \frac{1}{1 + \frac{1}{1+ \frac{1}{...}}}$

I don't understand how or why this works. I understand that as the denominators' continue on forever, the 1 + (fraction) changes ever so slightly. How do I go about solving this?

[snip]
This type of question is an example of a continued fraction and has been asked many times. If you are patient you will find the threads using the Search tool.
In brief:

$\displaystyle \displaystyle x = 1 + \frac{1}{1 + \frac{1}{1+ \frac{1}{...}}} \Rightarrow x = 1 + \frac{1}{x}$ and you solve for x.

3. If You set...

$\displaystyle \displaystyle \varphi = 1+ \frac{1}{1+\frac{1}{1 + ...}}}$ (1)

... is...

$\displaystyle \displaystyle \varphi = 1+\frac{1}{\varphi}$ (2)

Now You can find $\displaystyle \varphi$ solving the equation (2)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Look at this