Results 1 to 4 of 4

Math Help - recurrence relation and initial condition

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    15

    recurrence relation and initial condition

    Could someone make sure my answers are correct. Not sure about the first one but after the S0 s1 = 0? Here are the questions:

    6.1 In the following sequences determine s5 if s0, s1, ... sn, ... is a sequence satisfying the given recurrence relation and initial condition.
    a. sn= -sn-1 - n2 for n >= 1, s0 = 3

    -3(3) - 33 = -9 + 9 - 0

    S1 = 0
    The rest of the answers up to S5 would be zero correct?

    b. sn = 5sn-1 - 3sn-2 for n >= 2, s0 = -1, s1 = -2

    S2 = 5(-2) - 3(-1) = -10 + 3 = -7
    S3 = 5(-7) - 3(-2) = -35 + 6 = -29
    S4 = 5(-29) - 3(-7) = -145 + 21 = 124
    S5 = 5(-124) - 3(29) = -620 + 87 = -533

    Any help would be much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by papa_chango123 View Post
    The rest of the answers up to S5 would be zero correct?
    It seems thus.
    In the second problem do you want me to solve the recurrence relation or just compute it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2006
    Posts
    15
    Both if you don't mind. So is problem one right? Is the answer zero?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by papa_chango123 View Post

    b. sn = 5sn-1 - 3sn-2 for n >= 2, s0 = -1, s1 = -2
    As I already said the first one is correct.

    The charachteristic equation is,
    k^2=5k-3
    Thus,
    k^2-5k+3=0
    Thus,
    k=\frac{5\pm \sqrt{13}}{2}
    Thus the general solution to the recurrence relation is,
    C_1\left( \frac{5+\sqrt{13}}{2} \right)^n+C_2 \left( \frac{5-\sqrt{13}}{2} \right)^n
    When,
    n=0
    We have,
    s_0=C_1+C_2=-1
    When,
    n=1
    We have,
    s_1=C_1 \left( \frac{5+ \sqrt{13}}{2} \right)+C_2\left( \frac{5-\sqrt{13}}{2} \right)=-2
    Multiply by 2,
    C_1(5+\sqrt{13})+C_2(5-\sqrt{13})=-4
    5(C_1+C_2)+\sqrt{13}(C_1-C_2)=-4
    From first equation,
    -5+\sqrt{13}(C_1-C_2)=-4
    \sqrt{13}(C_1-C_2)=1
    Thus,
    \left\{ \begin{array}{c}C_1+C_2=-1\\ C_1-C_2=\frac{\sqrt{13}}{13} \end{array} \right\}
    Add then subtract, respectively,
    2C_1=\frac{-13+\sqrt{13}}{13}
    2C_2=\frac{-13-\sqrt{13}}{13}
    Thus,
    (C_1,C_2)=\left( \frac{-13+\sqrt{13}}{26},\frac{-13-\sqrt{13}}{26} \right)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] PDE Initial Condition Question
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 30th 2010, 02:57 AM
  2. recurrence relation with initial conditions
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: April 13th 2010, 08:53 PM
  3. Initial condition DE
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 12th 2009, 09:17 PM
  4. Integrating with initial condition
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 22nd 2009, 11:04 AM
  5. Initial condition problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 26th 2008, 10:55 AM

Search Tags


/mathhelpforum @mathhelpforum