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Math Help - Contienued fractions

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    Contienued fractions

    how do they work, does anyone know how to explain it cause i need to do a portfolio and give it in tomorow, if not: death
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  2. #2
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    Quote Originally Posted by Cilia View Post
    how do they work, does anyone know how to explain it cause i need to do a portfolio and give it in tomorow, if not: death
    It just an alternative way of expressing a number (rational or irrational). This have a fabolous property that they from the best possible convergent of a number with such a denominator.
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  3. #3
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    look, i kinda know it sounds weird but i've asked ppl aound me 2 but, their not of the friendly type or do not have tha mathematical genius to understand it....so PLEASE help me!
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    haha, i seriusly do know what to do. if u got the basic continued fraction (not the algebraic one, but the 1 one) and its something like:
    t1=1+1
    t2=1+(1/(1+1))
    t3=1+(1/1+(1+1)
    ...
    and ur suposed to give a formula for tn+1 in terms of tn
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  5. #5
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    Quote Originally Posted by Cilia View Post
    haha, i seriusly do know what to do. if u got the basic continued fraction (not the algebraic one, but the 1 one) and its something like:
    t1=1+1
    t2=1+(1/(1+1))
    t3=1+(1/1+(1+1)
    ...
    and ur suposed to give a formula for tn+1 in terms of tn
    t_n=[1:1,1,1,....,1]
    t_{n+1}=[1;t_n]
    Thus,
    t_{n+1}=1+\frac{1}{t_n}
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  6. #6
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    how that, for large values of n, tn satisfies an equation which can be expressed in form tn^2-tn-1.

    obtain an equation in tn for a continous fraction where ther is a 2+ instead of a 1+.
    ...then solve the equation to fin the EXACT value
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  7. #7
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    Quote Originally Posted by Cilia View Post
    how that, for large values of n, tn satisfies an equation which can be expressed in form tn^2-tn-1.

    obtain an equation in tn for a continous fraction where ther is a 2+ instead of a 1+.
    ...then solve the equation to fin the EXACT value
    It is relying on a theorem which is not so easy to show that all continued fractions converge (the ones with integers).

    So we know that x=[1;1,1,...,]
    Converges to some number.
    But,
    x=[1;x]
    For it repeats.
    Thus,
    x=1+\frac{1}{x}
    Thus,
    x^2=x+1
    Thus,
    x^2-x-1=0
    Thus,
    x=\frac{1\pm \sqrt{5}}{2}
    This tells us that this is one of these two possible values.
    It cannot be negative thus,
    x=\frac{1+\sqrt{5}}{2}=\psi
    The Divine Proportion (My body is shaped in Divine Proportion).
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  8. #8
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    Hello, Cilia!

    This is a vast and intricate topic which could take months to explain.
    If you know the very basics, I can show you a tiny sliver of the whole sprectrum.


    Your example:

    t_1 \:=\:1 + 1\qquad t_2 \:=\:1 + \frac{1}{1+1}\qquad t_3 \:=\:1 + \frac{1}{1 + \frac{1}{1+1}}


    By inspection ("eyeballing" it),
    . . we see that each term is: one plue one over the preceding term.

    That is: . t_{n+1} \;= \; 1 + \frac{1}{t_n}


    Assuming that the fraction goes on forever: . 1 + \frac{1}{1 + \frac{1}{1+\frac{1}{1 + ...}}}
    . . can we determine its value?


    \text{Let }x\:=\:1 + \frac{1}{1 + \frac{1}{1+\frac{1}{1 + ...}}}<br />
\begin{array}{ccc} \\ \\ \bigg\}\text{ this is }x \end{array}<br />

    Hence, we have: . x \:=\:1 + \frac{1}{x}\quad\Rightarrow\quad  x^2 - x - 1 \:=\:0

    Quadratic Formula: . x \:=\:\frac{\text{-}(\text{-}1) \pm \sqrt{(\text{-}1)^2 - 4(1)(\text{-}1)}}{2(1)} \;=\;\frac{1 \pm \sqrt{5}}{2}

    Since x is positive, we have: . x \:=\:\frac{1 + \sqrt{5}}{2} \:= \:1.618033989...
    . . which happens to be the Golden Mean, \phi.


    \text{Therefore: }\;1 + \frac{1}{1 + \frac{1}{1+\frac{1}{1 + ...}}} \;=\;\phi

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