Results 1 to 10 of 10

Math Help - Continued Fractions

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    11

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2007
    Posts
    11
    Actually, I'm totally new to the continued fractions thing. Could you explain what these iterations are?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    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)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2007
    Posts
    11
    Keep subbing what into x?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2007
    Posts
    11
    Oh, so sub -1 + 72/x into x, right?

    And what do you mean keep subbing to infinity?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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 07:17 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2007
    Posts
    11
    can you explain that to me in lay man's terms?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Sep 2007
    Posts
    11
    I still don't understand.. could you use my problem as an example? x^2 +x - 72 = 0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. continued fractions
    Posted in the Algebra Forum
    Replies: 16
    Last Post: March 21st 2010, 02:39 AM
  2. Convergence of infinite continued fractions
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: June 13th 2009, 08:51 AM
  3. Value of a continued fraction.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 11th 2009, 11:12 AM
  4. finite continued fractions ?
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 5th 2008, 11:02 PM
  5. 2, 5, 8, 11, 14... (continued)
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 13th 2006, 02:48 PM

Search Tags


/mathhelpforum @mathhelpforum