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Recurrence Relations

  1. Mar 14, 2008 #1
    [SOLVED] Recurrence Relations

    1. The problem statement, all variables and given/known data
    I need to express this recursive statement as a nonrecursive formula, using the technique of itteration.

    [tex]a_n = (n+1)a_{n-1}[/tex]
    [tex]a_0 = 2[/tex]

    3. The attempt at a solution
    [tex]a_n = (n+1)a_{n-1}[/tex]
    [tex]a_n = (n+1)(n+1)a_{n-2} = (n+1)^{2}a_{n-2}[/tex]
    [tex]a_n = (n+1)(n+1)(n+1)a_{n-3} = (n+1)^{3}a_{n-3}[/tex]
    [tex]a_n = (n+1)^{n}a_{n-n}[/tex]
    [tex]a_n = 2(n+1)^{n}[/tex]

    I plugged both the recursive formula, and my answer, into MS-Excel, and they don't match up. They work for n = 0,1, but then my answer start getting larger than the recursive one.
  2. jcsd
  3. Mar 14, 2008 #2
    Here's the spreadsheet I'm using to check my answer.

    First column is n
    Second column is the recursive
    Third column is my proposed answer

    Attached Files:

  4. Mar 14, 2008 #3


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    Hi Goldenwind! :smile:

    With these recurrence questions, always start by working out what a_n+2 is in terms of a_n.

    Then a_n+3.

    By then, you should have a good idea of what is happening.

    So … what is happening? :smile:
  5. Mar 14, 2008 #4
    [tex]a_n = (n+1)a_{n-1}[/tex]

    Which is the same as
    [tex]\frac{a_{n+1}}{n+1} = a_{n}[/tex]

    [tex]\frac{a_{n+2}}{{(n+1)}^2} = a_{n}[/tex]

    [tex]\frac{a_{n+3}}{{(n+1)}^3} = a_{n}[/tex]

    I still don't see my mistake :(
    This is pretty much what I did above, only the other way around >.>
  6. Mar 14, 2008 #5


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    No no no … :frown:

    First, don't rearrange the formula - it's simplest as it is!

    Then … be careful!

    You should have [tex]a_{n+2}\,=\,(n+2)a_{n+1}\,=\,(n+2)(n+1)a_n\,.[/tex]

    So … the nonrecursive formula is … ? :smile:
  7. Mar 14, 2008 #6


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    I always start a problem involving any kind of sequence by working out a few values!

    [itex]a_1= (1+1)a_0= 2a_0[/itex]
    [itex]a_2= (2+1)a_1= 3a_1= 3(2)a_0[/itex]
    [itex]a_3= (3+1)a_2= 4a_2= 4(3)(2)a_0[/itex]
    [itex]a_4= (4+1)a_3= 5a_3= 5(4)(3)(2)a_0[/itex]

    Anything come to mind?
  8. Mar 14, 2008 #7
    [itex]a_n = 2(n+1)![/itex]
    ...is what it looks like to me.

    However, even if I know the formula, I need to demonstrate that this *is* the formula, via iteration.

    As in, I start out with [itex]a_n = (n+1)a_{n-1}[/itex], then replace [itex]a_{n-1}[/itex] with [itex]a_{n-1} = (n+1)a_{n-2}[/itex], then replace [itex]a_{n-2}[/itex] with [itex]a_{n-2} = (n+1)a_{n-3}[/itex], and so on, until I get to [itex]a_{n-n} = a_0[/itex], where I can simply sub in the given base value, as shown above.
  9. Mar 14, 2008 #8
    Yeah, I see my mistake... when I replaced n with n-1, or n-1 with n-2, I didn't change the (n+1) at all. Doing so would lead to the factorial, instead of [itex](n+1)^n[/itex].

    Thank you for your help :)
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