Results 1 to 5 of 5

Math Help - simplify recurrence relation

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    3

    simplify recurrence relation

    I'm trying to solve this for b:

    b*2^n = 3*b*2^(n-1) + 2^n

    the answer is:

    b = -2^(n + 1)

    but i don't see how to get there.

    can someone enlighten me?

    thanks

    dan mcleran
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by danmcleran View Post
    I'm trying to solve this for b:

    b*2^n = 3*b*2^(n-1) + 2^n

    the answer is:

    b = -2^(n + 1)

    but i don't see how to get there.

    can someone enlighten me?

    thanks

    dan mcleran
     b2^n = 3b2^{n-1} + 2^n is not a recurrence relation. It is an equation in b and n.
    Solving it gives b = -2.

    Can you post the complete question? Maybe then, we can see why b = -2^{n + 1} is the correct answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2010
    Posts
    3

    Full problem

    an = 3an-1 + 2^nwith a1 = 5

    I was able to solve the homogeneous part to get:

    f(n) = a3^n

    I then substituted the particular solution into the above recurrence:

    b2^n = 3 b2^(n-1) + 2^n

    But I think I am too rusty at the algebra to solve for b. I was really looking for some algebra help I guess.

    Thanks

    Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615
    Hello, Dan!

    a_n \:=\: 3a_{n-1} + 2^n,\quad a_1 = 5

    I was able to solve the homogeneous part to get: . f(n) \:=\: a\!\cdot\!3^n

    I then substituted the particular solution into the above recurrence:
    . . b\cdot 2^n \:=\: 3b \cdot 2^{n-1} + 2^n

    But I think I am too rusty at the algebra to solve for b.
    As we say in the Marines, "Ya done good!"


    We have: . b\!\cdot\!2^n \:=\: 3b\!\cdot\!2^{n-1} + 2^n

    Divide by 2^{n-1}\!:\;\;2b \:=\:3b + 2 \quad\rightarrow\quad b \:=\:-2

    Hence: . f(n) \;=\;a\!\cdot\!3^n - 2\!\cdot\!2^n


    Since a_1 = 5, we have: . a\!\cdot\!3 - 2^2 \:=\:5 \quad\Rightarrow\quad a \,=\,3

    . . Hence: . f(n) \;=\;3\!\cdot\!3^n - 2\!\cdot\!2^n


    Therefore: . f(n) \;=\;3^{n+1} - 2^{n+1}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2010
    Posts
    3
    I see now. Thanks a lot.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Recurrence relation
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: December 10th 2011, 06:55 AM
  2. big o recurrence relation
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: April 4th 2011, 03:49 AM
  3. Recurrence Relation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: May 24th 2010, 03:04 AM
  4. Recurrence Relation Help
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 7th 2008, 08:49 AM
  5. Recurrence Relation
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: July 8th 2008, 09:47 AM

Search Tags


/mathhelpforum @mathhelpforum