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Learning a technique to figuring out the Explicit Formula

  1. Mar 29, 2005 #1
    I have a lot of trouble trying to figure out the explcit forumla of a series of numbers.

    I can see the pattern in a recursion forumla, however.

    [tex] 2, -4, 8, -16, ...[/tex] Is a multiplication of [tex]-2[/tex] to the term before it. Which is cake to write in a recursion forumla. But what about an Explicit Formula?

    What techniques and ideas do you look for first? Are there any clues?
     
  2. jcsd
  3. Mar 29, 2005 #2

    dextercioby

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    That is a SEQUENCE...U need to find the expression for the general term...

    It's not difficult.

    [tex] a_{1}=2 [/tex]
    [tex] a_{n+1}=-2a_{n},\forall n\geq 1[/tex]

    Now find a_{n} as a function of "n"...

    Daniel.
     
  4. Mar 29, 2005 #3
    Whoa. Wait, what?

    Can you breakdown this forumla for me:
    [tex] a_{n+1}=-2a_{n},\forall n\geq 1[/tex]
    ?
     
  5. Mar 29, 2005 #4

    Whoa. Wait, What?

    Can you breakdown this formula for me:
    [tex] a_{n+1}=-2a_{n},\forall n\geq 1[/tex]
    ?
     
  6. Mar 29, 2005 #5
    The [itex]n+1[/itex]st term is the [itex]n[/itex]th term multiplied by [itex]-2[/itex]. That's all that equation says. You then need to solve the equation for [itex]a_{n}[/itex] in terms of [itex]n[/itex]. It is what is called a "first order linear homogeneous constant-coefficient difference equation."

    Here is a hint as to how to solve it: Guess the solution [itex]a_n = Ak^n[/itex] for some constants [itex]k[/itex] and [itex]A[/itex], and try to determined what [itex]k[/itex] is by substitution into the equation. Then solve for [itex]A[/itex] by using the initial condition [itex]a_1 = 2[/itex]
     
    Last edited: Mar 29, 2005
  7. Mar 29, 2005 #6

    dextercioby

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    That's called "reccurence relation".It defines a sequence of numbers...

    There's no guessing here.It's a simple geometric progression with the ratio "-2".

    Daniel.
     
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