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Math Help - "Recursive definition"

  1. #16
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by robocop_911 View Post
    Seems like my topic (questions) is/are really "hard" and odd. No wonder I didn't get any replies up till now!...
    *sigh*. Ok do you understand what it means when they recurrsive?
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  2. #17
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    The bit I'm not totally sure on, is are you allowed to include n in the recursive definition or does it have to be solely in terms of an?
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  3. #18
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    Quote Originally Posted by sean.1986 View Post
    The bit I'm not totally sure on, is are you allowed to include n in the recursive definition or does it have to be solely in terms of an?
    Hello sean.1986...
    Here is a link to LaTeX tutorial http://www.mathhelpforum.com/math-he...-tutorial.html

    here n = 1,2,3...what do you mean by "are you allowed to include n in the recursive definition"?
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  4. #19
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    For example...

    an = n(n+1)

    an+1 = (n+1)(n+2) = n^2 + 3n + 2 = n(n+1) + 2n + 2 = an + 2n + 2

    But is an + 2n + 2 an acceptable recursive definition, or do you need it totally in terms of an and constants?
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  5. #20
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    Quote Originally Posted by sean.1986 View Post
    For example...

    an = n(n+1)

    an+1 = (n+1)(n+2) = n^2 + 3n + 2 = n(n+1) + 2n + 2 = an + 2n + 2

    But is an + 2n + 2 an acceptable recursive definition, or do you need it totally in terms of an and constants?
     a_n and constants only!
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  6. #21
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    Quote Originally Posted by robocop_911 View Post
     a_n and constants only!
    d)

    a1 = 1

    a2 = 4

    an = 2(an-1) - (an-2) + 2

    And where I've put an-1 and an-2, I mean the previous 2 terms, not the current term with 1 or 2 subtracted from it.
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  7. #22
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    Quote Originally Posted by sean.1986 View Post
    d)

    a1 = 1

    a2 = 4

    an = 2(an-1) - (an-2) + 2

    And where I've put an-1 and an-2, I mean the previous 2 terms, not the current term with 1 or 2 subtracted from it.
    I think for d)  a_n = n^2 it should be following answer...

     a_{n+1} = a_{n} + 2n - 1

    Now, is this correct or not???
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  8. #23
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    Quote Originally Posted by robocop_911 View Post
    I think for d)  a_n = n^2 it should be following answer...

     a_{n+1} = a_{n} + 2n - 1

    Now, is this correct or not???
    You said that the answer cannot contain n values, only an and constants.
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  9. #24
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    Thumbs down

    Quote Originally Posted by sean.1986 View Post
    You said that the answer cannot contain n values, only an and constants.

    Is it possible to get a recursive definition without any "variable" like "n"?

    How to remove "n" from the given sequence?
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  10. #25
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    The definition I gave for d will compute the square of n, but I got that but looking at the definition of a square number on wiki.

    "The nth square number can be calculated from the previous two by doubling the (n − 1)-th square, subtracting the (n − 2)-th square number, and adding 2, because n^2 = 2(n − 1)^2 − (n − 2)^2 + 2. For example, 252 − 42 + 2 = 225 − 16 + 2 = 50 − 16 + 2 = 36 = 62."
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  11. #26
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    Quote Originally Posted by sean.1986 View Post
    The definition I gave for d will compute the square of n, but I got that but looking at the definition of a square number on wiki.

    "The nth square number can be calculated from the previous two by doubling the (n − 1)-th square, subtracting the (n − 2)-th square number, and adding 2, because n^2 = 2(n − 1)^2 − (n − 2)^2 + 2. For example, 252 − 42 + 2 = 225 − 16 + 2 = 50 − 16 + 2 = 36 = 62."
    I think there should be all a_n, n & constants
    I don't think its possible to get a recursive definition without "n" coming in it.
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  12. #27
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    It is, though. Look at my answer for d...
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  13. #28
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    In fact, the answers for c & d are the same.

    Edit - except of course the base cases.

    For c....

    a1 = 2
    a2 = 6
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  14. #29
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    Quote Originally Posted by robocop_911 View Post
    Does anybody know recursive definition?

    What is recursive definition of the sequence a_n:
    n = 1,2,3..... if

    a) a_n = 4n-2
    b) a_n = 1+(-1)^n
    c) a_n = n(n+1)
    d) a_n = n^2

    Can anyone also explain me how he/she got the above answers!

    Moo did the first one, I will try the remaining. Hopefully they match with sean's answers.

    b) a_{n+1} = 1+(-1)^{n+1} = 1-(-1)^n \Rightarrow a_n + a_{n+1} = 2 \Rightarrow a_{n+1} = 2 - a_n , a_1 = 0

    c) a_{n-1} = n(n-1) , a_{n+1} = n(n+1) \Rightarrow a_n = a_{n-1} + 2n , a_1 = 2

    d) a_{n+1} = n^2 + 2n + 1 \Rightarrow a_{n+1} = a_n + 2n + 1 , a_1 = 1
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  15. #30
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    Quote Originally Posted by robocop_911 View Post
    Does anybody know recursive definition?

    What is recursive definition of the sequence a_n:
    n = 1,2,3..... if

    a) a_n = 4n-2
    b) a_n = 1+(-1)^n
    c) a_n = n(n+1)
    d) a_n = n^2

    Can anyone also explain me how he/she got the above answers!
    A recursive definition of a sequence \{a_n, n \in \mathbb{N} \} is a rule that tells you the n th term in terms of one or more proceeding terms and possibly n, and a set of initial condition sufficient to allow you to apply the recursion to calculate the subsequent terms. Such a definition is not necessarilly unique.

    Lets look at (a)

    a_n = 4n-2

    now a_{n-1}=4n-6, so:

    a_n=a_{n-1}+4

    and the initial condition we need to start is that a_1=2.

    But we could also use two preceeding terms like:

    a_n=\frac{(a_{n-1}+a_{n-2})}{2}+6

    and the initial conditions we need to start is that a_1=2, and a_2=6 .

    RonL
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