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Question on functions

  1. Apr 25, 2012 #1
    Consider the function defined by
    [tex] f(x) = 0.6x + 2100 [/tex]

    Suppose this function is iterated [itex]n[/itex] times.

    Express [itex] f^{n}(x) [/itex] as an elementary function of [itex]x[/itex].

    The problem is actually a simplified version of a finance problem I am trying to solve. I am not sure there is a solution so if a solution does not exist, please help me prove it.

    BiP
     
  2. jcsd
  3. Apr 25, 2012 #2
    This is straightforward if you iterate a few times.
    f2(x) = 0.6(0.6x + 2100) + 2100 = (0.6^2)x + 0.6*2100 + 2100
    f3(x) = 0.6((0.6^2)x + 0.6*2100 + 2100) + 2100 = (0.6^3)x + (0.6^2)2100 + 0.6*2100 + 2100
    This should be enough to see the general pattern for fn(x).
     
  4. Apr 25, 2012 #3
    Would this be correct?

    [tex] f^{n}(x) = .6^{n}x + 3500(.6^{n-1} + .6^{n-2} + .6^{n-3} + ... + .6^{0}) [/tex] for [itex] n>1 [/itex]

    BiP
     
  5. Apr 25, 2012 #4
    Yes, assuming you meant 2100 instead of 3500. It is a linear function of x, which satisfies the elementary function requirement. The constants are just rather large polynomials of 0.6.
     
  6. Apr 25, 2012 #5
    Interesting.
    What if we make [itex] n → ∞ [/itex] ?

    Can we simplify in that case?

    My guess is that the answer will be [itex] 3500*(\frac{.6}{1-.6}) = 5250 [/itex].

    BiP
     
  7. Apr 25, 2012 #6
    It should be 3500*(1/(1-0.6)) = 35000/4 = 8750 for the sum of the geometric series. Is 3500 the correct value? You have 2100 in its place in the first post.
     
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