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Iteration, linear function. convergence and divergence

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    I need to understand and prove the following: That if a>1 the function diverges, except for a special case x_0= b/(1-a). Then if a=-1 diverges for some cases and converges if x_0 is b/2. Again, not to clear on this.


    2. Relevant equations

    lim n →∞ a^n(x_0+b/(a-1))-b/(a-1)

    3. The attempt at a solution

    For a>1, then we just need to look at x_0 and b/(a-1)? If x_0 > b/(a-1), then this is + and still diverges, then we rely on -b/(a-1)? I am kind of confused and any help would be appreciated.
     
    Last edited by a moderator: Oct 20, 2011
  2. jcsd
  3. Oct 20, 2011 #2

    Mark44

    Staff: Mentor

    MOD Note: Is this your limit? What you wrote is hard to understand.
    [tex]\lim_{n \to \infty} a^n \left(x_0 + \frac{b}{a - 1}\right) - \frac{b}{a - 1}[/tex]
     
  4. Oct 20, 2011 #3
    Yes.
     
  5. Oct 21, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ??? For a equal to 1 you have [itex]a^n= 1[/itex] so the two fractions cancel, but not a not equal to 1.
     
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