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Math Help - Calculating Phi (golden ratio) - continued fraction.

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

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

    <br /> <br />
1 + \frac{1}{1 + \frac{1}{1+ \frac{1}{...}}}<br /> <br />

    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?

    <br /> <br />
2*\frac{3x+1}{2*\frac{3x+1}{2*3x+1 ...}}<br /> <br />

    Thank you for your help.

    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.
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  2. #2
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    Quote Originally Posted by EMyk01 View Post
    My teacher told our class that the \varphi can be calculated by:

    <br /> <br />
1 + \frac{1}{1 + \frac{1}{1+ \frac{1}{...}}}<br /> <br />

    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 x = 1 + \frac{1}{1 + \frac{1}{1+ \frac{1}{...}}} \Rightarrow x = 1 + \frac{1}{x} and you solve for x.
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  3. #3
    MHF Contributor chisigma's Avatar
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    If You set...

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

    ... is...

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

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

    Kind regards

    \chi \sigma
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  4. #4
    Member rtblue's Avatar
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    Look at this
    Last edited by rtblue; February 2nd 2011 at 04:36 PM.
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