Page 2 of 2 FirstFirst 12
Results 16 to 23 of 23

Math Help - The Battle Of Hastings

  1. #16
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,675
    Thanks
    302
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    Let l be the length of the square root fraction.

    If l is even then C_{l-1} is the first solution.

    If l is odd then C_{2l-1} is the first solution.

    Once you compute the convergent in reduced terms you will get p/q. Then choose x=p\mbox{ and }y=q to obtain a solution.
    Let me see if I've got this right then:

    x^2 - 2y^2 = 1

    \sqrt{2} = [1; 2, 2, 2, ~ ... ~ ]

    Thus l = 1 and l is odd.

    So I have
    C_{2l - 1} = C_{2 \cdot 1 - 1} = C_1 = 1 + \frac{1}{2} = \frac{3}{2} = \frac{x}{y}

    Thus y = \frac{2}{3}x.

    So
    x^2 - 2\left ( \frac{2}{3} x \right )^2 = 1 ==> x = 3. Thus y = 2. Which is a correct solution.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by topsquark View Post
    Let me see if I've got this right then:

    x^2 - 2y^2 = 1

    \sqrt{2} = [1; 2, 2, 2, ~ ... ~ ]

    Thus l = 1 and l is odd.

    So I have
    C_{2l - 1} = C_{2 \cdot 1 - 1} = C_1 = 1 + \frac{1}{2} = \frac{3}{2} = \frac{x}{y}

    Thus y = \frac{2}{3}x.

    So
    x^2 - 2\left ( \frac{2}{3} x \right )^2 = 1 ==> x = 3. Thus y = 2. Which is a correct solution.

    -Dan
    You solution is correct but the procedure is not.


    \frac{3}{2} means chose x=\mbox{ numerator }=3 and y=\mbox{ deminator }=2.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Harold must have been quite a leader!

    I reckon not. He got shot in the eye at Hastings and the Normans won.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,552
    Thanks
    542
    Hello, Galactus!

    I reckon not. He got shot in the eye at Hastings and the Normans won.
    I was referring to the size of his army.

    Since y = 226,153,980, the first formation had: . {\color{blue}3.18 \times10^{18}}\,\text{ soldiers.}


    And the Normans won?
    They must have drafted Klingons, Vogans,and Trafalmadorians.

    Follow Math Help Forum on Facebook and Google+

  5. #20
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I know what you meant, Soroban. Despite the astronomical size of his army, he still lost. Though, I am pretty sure, his army was not actually that large. Over 3 quintillion in the first formation. You would think such an army would be invincible.

    I believe the realsitic size of the Saxon army is thought to be have been about 8,000 or so.

    Just relating history.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Newbie
    Joined
    Jul 2007
    Posts
    8
    Continued Fraction Calculator

    <br />
x^2 - 61y^2 = 1<br />

    <br />
\sqrt{61} = [7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 ~ ... ~ ]<br />

    Thus l = 11and l is odd

    So I have
    [tex] C_{2l - 1} = C_{2 \cdot 11 - 1} = C_{21} = 7 + \frac{1}{1+\frac{1}{4+\frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{2+\frac{1}{1+\frac{1}{3+\frac{1}{4+\frac{1}{1+\frac{1}{14+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{3+ \frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2+\frac{ 1}{1+\frac{1}{2+\frac{1}{2+\frac{1}{1+\frac{1}{1+\ frac{1}{3+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1 }{1}}}}}}}}}}}}}}}}}}}}}}}}}}}[/math]
     = \frac{2713847760}{183241189}
    so
    <br />
x = 2713847760<br />
    <br />
y = 183241189<br />
    there is something wrong with the result
    all the maths was done by the site above
    can u point out what i did wrong

    edit: i have corrected 1 mistake are there any others
    edit: whats making it show the code
    Last edited by henryzz; July 27th 2007 at 05:48 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,675
    Thanks
    302
    Awards
    1
    Quote Originally Posted by henryzz View Post
    Continued Fraction Calculator

    <br />
x^2 - 61y^2 = 1<br />

    <br />
\sqrt{61} = [7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 ~ ... ~ ]<br />

    Thus l = 7and l is odd
    How are you getting l = 7? I'm counting 11 digits in the repeating part.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Newbie
    Joined
    Jul 2007
    Posts
    8
    thanks will edit
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Japanese Samurai and his soldiers after a battle
    Posted in the Math Puzzles Forum
    Replies: 4
    Last Post: May 1st 2011, 10:05 PM
  2. Metropolis-Hastings/Objective Prior
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 25th 2009, 02:48 PM
  3. RPG Battle
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: September 29th 2008, 01:53 PM

Search Tags


/mathhelpforum @mathhelpforum