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Math Help - Problem of a function

  1. #1
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    Problem of a function

    Suppose that f:\mathbb{R}\mapsto\mathbb{R} is a continuous, bounded, strictly increasing function.

    Questions:

    (a) Show that there is a point a_{1}\epsilon\mathbb{R} such that f(a_{1})>a_{1}
    (b) For each n\epsilon\mathbb{N}, define a_{n+1}=f(a_{n}). Explain why a_{2}>a_{1}. Then explain why a_{n+1}>a_{n} for all n\epsilon\mathbb{N}
    (c) Explain why the sequence {a_{n}} is bounded above.
    (d) Explain why the sequence {a_{n}} converges to some number, L.
    (e) Explain why f(L) = L.

    My work so far:

    (a) f is bounded and strictly increasing, therefore f'(x) > 0, so there is a point x\epsilon\mathbb{R} such that x < f(x). I know this reason is not enough, still working on the proper proof.

    (b) Now a_{2}=a_{1+1}=f(a_{1})>a_{1} as defined by part (a).
    Let S = {n \epsilon \mathbb{N}:a_{n+1}>a_{n}}
    1 \epsilon S as a_{1+1}=a_{2} > a_{1}
    let k \epsilon S, then a_{k+1} > a_{k}
    Now a_{k+1+1} = f(a_{k+1}) > a_{k+1} , thus proved k+1 \epsilon S
    Therefore S is inductive, and proved a_{n+1}>a_{n}

    (c) Because f is continuous and bounded.

    (d) Because f is monotonic, implies {a_{n}} is monotonic, thus {a_{n}} is bounded. All bounded monotonic sequence converges, thus the sequence converges to L.

    (e) {a_{n}}converges to L, then f({a_{n}}) converges to L because f is continuous.

    I know I made some mistakes up there, please check, thank you.

    KK
    Last edited by tttcomrader; January 26th 2007 at 11:04 AM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    (a) Show that there is a point a_{1}\epsilon\mathbb{R} such that f(a_{1})>a_{1}
    Let us assume that f(x)\geq x for all x\in \mathbb{R}. Then f(x) is not bounded because the function x is not bounded. Which is a contradiction.

    Now follow the logic....
    FOR ALL x\in \mathbb{R} we have f(x)\geq x.

    The negation of that (which must be because by contradiction) is,
    FOR SOME x\in \mathbb{R} we have f(x)<x.
    Because the negation of a universal quantifier is an existencial quantifier. And the negation of \geq is <.
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  3. #3
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    Quote Originally Posted by tttcomrader View Post

    (a) f is bounded and strictly increasing, therefore f'(x) > 0, so there is a point x\epsilon\mathbb{R} such that x < f(x). I know this reason is not enough, still working on the proper proof.
    The problem with that is we do not know whether f is differenciable or not.

    (b) Now a_{2}=a_{1+1}=f(a_{1})>a_{1} as defined by part (a).
    Let S = {n \epsilon \mathbb{N}:a_{n+1}>a_{n}}
    1 \epsilon S as a_{1+1}=a_{2} > a_{1}
    let k \epsilon S, then a_{k+1} > a_{k}
    Now a_{k+1+1} = f(a_{k+1}) > a_{k+1} , thus proved k+1 \epsilon S
    Therefore S is inductive, and proved a_{n+1}>a_{n}
    I got the same thing. It is inductive because a_{n}>a_{n-1} then a_{n+1}=f(a_n)>f(a_{n-1})=a_n because it is strictly increasing.
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  4. #4
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    I'm sorry, I don't really understand what you were doing. Did you proved f(x) < x by contradiction? But I'm trying to prove f(x) > x.

    What is negation?

    Sorry about my lack of understanding.

    KK
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    I'm sorry, I don't really understand what you were doing. Did you proved f(x) < x by contradiction? But I'm trying to prove f(x) > x.

    What is negation?

    Sorry about my lack of understanding.

    KK
    Sorry I proved a statement that you are not looking for.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    The real number f(a_1) is an upper bound.
    Because,
    f(a_1)>a_1>a_2>a_3>....
    Isn't a_{n+1}>a_n? So shouldn't a_1<a_2<a_3...?

    So would you say f is bounded above by f(a_n)?

    Thanks.

    KK
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  7. #7
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    Quote Originally Posted by tttcomrader View Post
    Isn't a_{n+1}>a_n? So shouldn't a_1<a_2<a_3...?

    So would you say f is bounded above by f(a_n)?

    Thanks.

    KK
    Sorry, again. I am doing this entire problem with reverse inequalities, igonore those posts.

    No you cannot say f(a_n) an an upper bounded because a_n is not a real number.
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