Results 1 to 6 of 6

Math Help - recurrence relation

  1. #1
    Junior Member
    Joined
    Jun 2010
    Posts
    69

    recurrence relation

    Hi
    Am I close with this or way off?

    First I have to decide if this is a linear constant coefficient recurrence relation
    a_{k+1} + a^2 _{k} + k^2

    I think a. The equation is not linear because the base a^2_{k} this a is squared.

    b. The equation is not constant coefficient recurrence relation because one of the coefficients k^2 is not constant.


    but, I also feel the equation could be linear because (excuse my simplicity, but I'm a beginner here), because the equation moves down in steps a_(k+1) --> a_(k)


    may I ask you advice please?

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dunsta View Post
    Hi
    Am I close with this or way off?

    First I have to decide if this is a linear constant coefficient recurrence relation
    a_{k+1} + a^2 _{k} + k^2

    I think a. The equation is not linear because the base a^2_{k} this a is squared.

    b. The equation is not constant coefficient recurrence relation because one of the coefficients k^2 is not constant.


    but, I also feel the equation could be linear because (excuse my simplicity, but I'm a beginner here), because the equation moves down in steps a_(k+1) --> a_(k)


    may I ask you advice please?

    thanks
    It is non-linear as you say.

    The k^2 term is not a coefficient

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2010
    Posts
    69
    Thanks Capn,
    I want to find a_2 and a_4
    when a_0 = 1
    Am I on the right path

    when a_0 = 1
    a_{k+1} + a^2 _{k} + k^2 for a_1

    we know a_0 = 1 so
    a_{0+1} + a^2 _{0} +0^2 = 1
    a_{1} + 1^2 + 0 = 1
    a_{1} + 1 + 0 = 1
    therefore a_{1}  = 0

    so now we know a_{1}  = 0
    a_{1+1} + 0^2 + 1^2  = 0
    a_2 = -1

    so now we know a_{2}  = -1
    a_{2+1} + -1^2  + 2^2 = -1
    a_3 + 1  +4 = -1
    a_3  = -6

    so now we know a_{3}  = -6
    a_{3+1} + -6^2 +3^2 =-6
    a_{4} + 36 +9 =-6
    a_{4} =-51

    have I worked through this correctly?

    thanks for helping.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771
    Is it a_{k+1} + a^2 _{k} + k^2 or a_{k+1} = a^2 _{k} + k^2? The former is not an equation. I strongly suspect that the first + is a typo.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2010
    Posts
    69
    sorry, it's
        a_{k+1} + a^2 _{k} + k^2 = 0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771
    In this case, a_{k+1}=-a_k^2-k^2. Therefore, a_1 = -1 - 0 = -1; a_2 = -1 - 1 = -2; a_3 = -4 - 4 = -8; a_4 = -64 - 9 = -73.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Recurrence relation
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: October 15th 2011, 11:27 PM
  2. Recurrence Relation Q
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 30th 2009, 11:57 PM
  3. Recurrence Relation HELP
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: May 3rd 2009, 01:18 PM
  4. recurrence relation
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 15th 2009, 06:20 PM
  5. Recurrence relation
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 16th 2008, 08:02 AM

Search Tags


/mathhelpforum @mathhelpforum