Results 1 to 2 of 2

Math Help - TA’s Challenge Problem #5

  1. #1
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1

    TA’s Challenge Problem #5

    This will be a very easy problem indeed.

    Find the continued-fraction expansion of n+\sqrt{n^2+k} where n,\,k are positive integers and k divides 2n.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by TheAbstractionist View Post
    This will be a very easy problem indeed.

    Find the continued-fraction expansion of n+\sqrt{n^2+k} where n,\,k are positive integers and k divides 2n.
    considering the condition d \mid 2n, i guess you're looking for a "simple" continued fraction of n + \sqrt{n^2+k}. let 2n=kd and x=n+\sqrt{n^2 +k}. then x=2n + \frac{1}{d + \frac{1}{x}}. therefore: n+\sqrt{n^2+k}= 2n \ + \ \frac{1}{d \ + \ \frac{1}{2n \ + \ \frac{1}{d \ + \ \frac{1}{2n \ + \ \cdots}}}}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Challenge Problem
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: July 12th 2010, 03:07 PM
  2. Challenge problem.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 21st 2010, 06:55 AM
  3. TA’s Challenge Problem #7
    Posted in the Math Challenge Problems Forum
    Replies: 5
    Last Post: August 7th 2009, 09:39 AM
  4. TA’s Challenge Problem #3
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: June 10th 2009, 12:54 AM
  5. TA’s Challenge Problem #2
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: June 9th 2009, 03:35 AM

Search Tags


/mathhelpforum @mathhelpforum