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Homework Help: Need help with a recurion formula

  1. May 15, 2005 #1
    Hi everyone,

    Right now my class is studying recursion formulas - er, when the next answer is based off the answer before it, if that makes any sense? I am having a hard time understanding this.

    A picture of the attached problem is included...Thanks for any and all help for this problem.

    skrying
     

    Attached Files:

  2. jcsd
  3. May 15, 2005 #2

    dextercioby

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    That is a geometrical progression.The "n+1" term is gotten from the "n"-th by dividing through 2.

    It converges to 0.What else...?

    Daniel.
     
  4. May 15, 2005 #3

    saltydog

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    You have:
    [tex]a_n=\frac{1}{2^n}[/tex]

    That's just an expression for the n'th term. Something like:

    [tex]a_n=2a_{n-1}+3a_{n-2}[/tex]

    would be a recursive formula defining the n-th term as a function of the n-1 term and the n-2 term.
     
    Last edited: May 15, 2005
  5. May 15, 2005 #4
    Thanks Dexter and Salty

    :rofl: Thank you dexter and salty for your input. It does make more sense when you put it that way. I think I am just making all this harder than it really is perhaps?

    Thanks, Skrying
     
  6. May 15, 2005 #5
    How far can this recursion formula go?

    I had another quick question (and probably a dumb one) since math of any kind is not my forte'....how far does or can this formula continue..?

    Thanks!
     
  7. May 15, 2005 #6

    OlderDan

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    What Dexter was saying is

    [tex]a_n = \frac{1}{2^n}[/tex]

    can be rewritten. Since

    [tex]a_{n-1} = \frac{1}{2^{n-1}}[/tex]

    [tex]a_n = \frac{1}{2^n} = \frac{1}{2 \bullet 2^{n-1}}=\frac{1}{2} a_{n-1} [/tex]

    That is the recursion formula, and it goes on forever. There is no limit on the value of n.
     
  8. May 15, 2005 #7
    Thank you OlderDan

    Thanks OlderDan...that looks like the same way some of my classmates were working on the problem. I really appreciate your help!

    Thanks Skrying!
     
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