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Limit of a recursively defined sequence

  1. Sep 10, 2009 #1
    The problem is:

    a(n+1) = ( a(n)^2 +16) / ( 2a(n) + 6) converges to the limit:...?
    for n>=1 and a(1) = -10

    Also note (n+1) and (n) are subscripts


    Also another question..:

    If f(x) is a function that is 8 times continuously differentiable such that the coefficient of x^5 in its 8th MacLaurin polynomial is 0.2, then f^5(0) =

    If someone could solve these and explain what they are doing it would be extremely helpful. My textbook has no questions like this so I'm completely lost.
     
  2. jcsd
  3. Sep 10, 2009 #2

    HallsofIvy

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    Assuming it has a limit, a, then
    [tex]\lim_{n\to \infty} a_{n+1}= \lim_{n\to\infty}\frac{a_n^2+ 16}{2a_n+ 6}[/tex]
    [tex]a= \frac{a^2+ 16}{2a+ 6}[/tex]
    so [itex]a(2a+ 6)= 2a^2+ 6a= a^2+ 16[/itex] or 2a^2+ 6a= a^2+ 16 so [itex]a^2+ 6a- 16= 0[/itex]. That gives two possible values for a.

    Do you know what a MacLaurin polynomial is and how its coefficients are calculated?
     
  4. Sep 10, 2009 #3
    Following up on HallsofIvy.

    The two values of a in the first question are potential attractor values for the sequence. Depending on your initial sequence term only one will be valid. Test the first few terms to see which one works.

    As to the second question. The temrs of a function's Maclaurin series is defined in terms of the various derivatives of the function. It is possible to reverse engineer the values of those derivatives from the series' coefficients.

    --Elucidus
     
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