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Math Help - inductive reasoning

  1. #1
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    inductive reasoning

    for the function f1(x)= 2x-1
    ______
    x+1

    f
    n+1(x)= f1(fn(x)), n>=1
    it can be shown that f35 = f5.
    Find f28.
    how can i get this?
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  2. #2
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    Quote Originally Posted by nh149 View Post
    for the function f1(x)= 2x-1
    ______
    x+1

    f
    n+1(x)= f1(fn(x)), n>=1
    it can be shown that f35 = f5.
    Find f28.
    how can i get this?
    Since 35= 5(7) and 28= 4(7), I think you need to calculate f_7 directly.

    Do you know how to do that? Since f_1(x)= \frac{2x-1}{x+1}, f_2(x)= f_1(f_1(x))= \frac{2\frac{2x-1}{x+1}- 1}{\frac{2x-1}{x+1}+1} and multiplying both numerator and denominator by x+ 1 gives f_2(x)= \frac{4x-2-x-1}{2x-1+x+1}= \frac{3x-3}{3x} = \frac{x-1}{x}.

    Once you get to f_7, I think you will find the rest is easy.
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  3. #3
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    i've been thinking of..

    f28=f4 since f35=f5
    because 35=5(7) and 28=4(7)..

    is it right?
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  4. #4
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    Quote Originally Posted by nh149 View Post
    for the function f_1(x)= \frac{2x-1}{x+1}
                       f_{n+1}(x)= f_1(f_n(x)), n>=1
    it can be shown that f_{35} = f_5.
    Find f_{28}.
    If y = \frac{2x-1}{x+1} then x = \frac{1+y}{2-y}. So the inverse function to f_1 is given by f_{-1}(x) = \frac{1+x}{2-x}.

    But if f_{35} = f_5 then the maps f_n repeat with period 30. Therefore f_{28} = f_{-2}, where f_{-2}(x) = f_{-1}(f_{-1}(x)). I think that's the least painful way to calculate f_{28}.

    You can check this by using HallsofIvy's calculation for f_2(x) = \frac{x-1}x to see that f_4(x) = f_2(f_2(x)) = {\frac{x-1}x - 1 \above1pt \frac{x-1}x}. This turns out to be the same as f_{-2}. So in fact the f_ns repeat with period 6.
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