Results 1 to 5 of 5

Math Help - 2nd-order recurrence sequence (simultaneous equation mind block)

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    12

    2nd-order recurrence sequence (simultaneous equation mind block)

    Hi,

    I'm trying to find the closed form of a second-order sequence. I'm down to the 'general solution' which is ( Un = A6/5^n + B(-2)^n ), but I'm having a simultaneous equation mind block.

    Here's where I'm stuck:

    U0 = 2 gives A + B = 2
    U1 = 12 gives 6/5A - 2B = 12

    Please can someone help to jog my memory on simultaneous equations, I'm staring blankly at these and Mathcad isn't helping.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by mezhopking View Post
    U0 = 2 gives A + B = 2
    U1 = 12 gives 6/5A - 2B = 12

    Please can someone help to jog my memory on simultaneous equations....
    You can review this lesson, but a quick set-up might be:

    . . . . .A + B = 2

    . . . . .B = 2 - A

    . . . . .(6/5)A - 2(2 - A) = 12

    Then solve the linear equation.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    12
    Thanks for the jog, but I'm now not sure I factored out correctly .

    The original equation was:

    r^2 + 4/5r - 12/5 = 0; which is (r - 6/5)(r + 2/1)?

    Does this look correct to you?

    Thanks...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by mezhopking View Post
    The original equation was:

    r^2 + 4/5r - 12/5 = 0; which is (r - 6/5)(r + 2/1)?

    Does this look correct to you?
    If you're not sure of your factoring, then multiply the factors back out and make sure you get what you'd started with.

    (You should; the factorization is correct.)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    12
    Thanks! I really don't know why I forgot about such a simple test ... Looks like all is correct.

    Ta much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear second order recurrence sequence
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: March 28th 2011, 12:46 PM
  2. Recurrence and Fibonacci sequence
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 12th 2010, 07:27 PM
  3. Replies: 4
    Last Post: January 9th 2010, 04:02 PM
  4. Solving simultaneous recurrence relations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 14th 2008, 03:36 PM
  5. recurrence sequence
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: October 30th 2008, 03:02 AM

Search Tags


/mathhelpforum @mathhelpforum