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Math Help - Continued Fractions

  1. #1
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    Continued Fractions

    Hello, I can't seem to understand how to use continued fractions to find the value of x in: x^2 + x - 72 = 0

    Can someone please explain how to solve this problem? Basically, I need to prove that x = -9 or 8.
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  2. #2
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    Try this.

    x^{2}+x-72

    x^{2}+x=72

    x(x+1)=72

    x+1=\frac{72}{x}

    x=-1+\frac{72}{x}

    Now, perform the iterations.

    Hopefully, it convegres to the solutions. It may not do it too fast, though.
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  3. #3
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    Actually, I'm totally new to the continued fractions thing. Could you explain what these iterations are?
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  4. #4
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    See what x equals?. See the x under the 72?. Keep subbing it in, ad infinitum.

    Do it enough and it should converge to the solution(s)
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  5. #5
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    Keep subbing what into x?
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  6. #6
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    Oh, so sub -1 + 72/x into x, right?

    And what do you mean keep subbing to infinity?
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  7. #7
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    Quote Originally Posted by galactus View Post
    Hopefully, it convegres to the solutions. It may not do it too fast, though.
    Continued fractions converges always and very quickly.

    Let [a_0;a_1,a_2,..] be a continued fraction then, if \frac{p_k}{q_k} represents that k-th approximation k=0,1,2,... in reduced terms we have:
    \left| x - \frac{p_k}{q_k} \right| \leq \frac{1}{q_kq_{k+1}}
    But, q_k are and increasing sequence of integers so q_k \geq k (this is a theorem).
    This means,
    \left| x - \frac{p_k}{q_k} \right| \leq \frac{1}{k^2}

    EDIT: Mistake fixed.
    Last edited by ThePerfectHacker; October 21st 2007 at 06:17 PM.
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  8. #8
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    can you explain that to me in lay man's terms?
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  9. #9
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    Quote Originally Posted by vperera View Post
    can you explain that to me in lay man's terms?
    (I made a pathetic mistake instead of \frac{1}{4^k} it should be \frac{1}{k^2} for k\geq 1).

    Consider x = 1+\frac{1}{x} the positive solution is \phi \approx 1.618.

    Now it means x = \frac{1}{1+\frac{1}{x}} = \frac{1}{1+\frac{1}{1+\frac{1}{x}}}

    You form the continued fraction or simply 1's so [1;1,1,1,1,....

    Now let C_k represent the fraction (approximation) after the k terms.

    So C_1 = [1;1] = 1+\frac{1}{1} = 2 and the difference between the real value is |1.618 - 2| = .618 \leq \frac{1}{1^2}.

    Now C_2 = [1;1,1] = 1+\frac{1}{1+\frac{1}{1}} = \frac{3}{2} and the difference between the real value is |1.618 - 1.5| = .118\leq \frac{1}{2^2} = .25.

    This tells us the difference between the approximation and the actual value gets very close.
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  10. #10
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    I still don't understand.. could you use my problem as an example? x^2 +x - 72 = 0
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