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Recurrence Relation Problem

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]a_{n} = a_{n-1} + n[/itex]
    [itex]a_{0} = 0[/itex]

    3. The attempt at a solution
    [itex] h_{n} = h_{n-1}[/itex]
    [itex] t^{2} - t = 0[/itex]
    [itex] t=0 t=1 [/itex]
    [itex] h_{n} = B[/itex]

    [itex] p_{n} = bn + c [/itex]
    [itex] p_{n} = p_{n-1} + n [/itex]
    [itex] bn + c = b(n-1) + n [/itex]
    [itex] bn + c = (b+1)n -b [/itex]

    I'm sure I've gone wrong somewhere, I just can't figure out where!
     
  2. jcsd
  3. Apr 15, 2013 #2

    mfb

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    I think there is a "+c" missing on the right side.

    I have no idea how the first 2 and the following 4 lines are related to the other groups.
     
  4. Apr 15, 2013 #3
    Ok sorry I should have explained.

    Basically, how I solve a nonhomogenous RR is by ignoring the nonhomogenous term, and solving the homogenous term, 1st group.

    Then I solve the homogenous term by picking a particular function of the same order of the nonhomogenous part and call this Pn. This is the second group.

    Then, the General Solution to the RR is [itex]h_{n} + p_{n}[/itex]


    Yes you're right. So with the correction I get the following.

    [itex] p_{n} = bn + c [/itex]
    [itex] p_{n} = p_{n-1} + n [/itex]
    [itex] bn + c = b(n-1) + c + n [/itex]
    [itex] bn + c = (b+1)n -b + c [/itex]

    I don't know how to solve for b and c I'm afraid.
     
  5. Apr 15, 2013 #4
    Hi Darth Frodo, aren't you just trying to solve a simple arithmetic serie ?
    u(n)=n
    a(n)=1+2+...+n=n(n+1)/2 ?
     
  6. Apr 15, 2013 #5

    mfb

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    The solution for a_n cannot be expressed as sum of a constant (h_n) and bn+c with constant b and c.

    Right.
     
  7. Apr 15, 2013 #6

    SammyS

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    Please, give a complete statement of the problem.

    My mind-reading skills seem to be eroding lately.
     
  8. Apr 16, 2013 #7

    HallsofIvy

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    If the problem is to solve [itex]a_n= a_{n-1}+ n[/itex], [itex]a_0= 0[/itex], then most of what you are doing is unnecessary. Instead, first calculate a few values directly:
    [itex]a_1= a_0+ 1= 0+ 1= 1[/itex]
    [itex]a_2= a_1+ 2= 1+ 2= 3[/itex]
    [itex]a_3= a_2+ 3= 3+ 3= 6[/itex]
    [itex]a_4= a_3+ 4= 6+ 4= 10[/itex]
    [itex]a_5= a_4+ 5= 15[/itex]

    In other words, [itex]a_n= 1+ 2+ 3+ 4+ \cdot\cdot\cdot+ n[/tex], an arithmetic series as both oli4 and mfb said, for which there is a well known formula
     
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