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Math Help - Proof on Differentiation and Limit.

  1. #1
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    Unhappy Proof on Differentiation and Limit.

    Someone please help me ASAP on these problems in Real Analysis.


    1.> Find Lim supXn (upper limit) and Lim infXn (lower limit)
    n-->infinity n-->infinity

    where Xn is a recurrent sequence, X1= 1/2 and Xn+1= 1-(Xn)^2

    2.> Let f(x) be twice continously differentiable on (0,infinity). Prove that if
    Lim f(x) exists and f''(x) (i.e. 2nd derivative of f) is bounded on
    x-->infinity

    (0, infinity), then Lim f'(x) = 0 (i.e. 1st derivative of f=0)
    x-->infinity

    Please somene help.
    Last edited by mr fantastic; November 9th 2009 at 02:17 AM. Reason: Edited post title, removed shouts of urgency etc.
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  2. #2
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    Quote Originally Posted by concern View Post
    Someone please help me ASAP on these problems in Real Analysis.


    1.> Find Lim supXn (upper limit) and Lim infXn (lower limit)
    n-->infinity n-->infinity

    where Xn is a recurrent sequence, X1= 1/2 and Xn+1= 1-(Xn)^2

    2.> Let f(x) be twice continously differentiable on (0,infinity). Prove that if
    Lim f(x) exists and f''(x) (i.e. 2nd derivative of f) is bounded on
    x-->infinity

    (0, infinity), then Lim f'(x) = 0 (i.e. 1st derivative of f=0)
    x-->infinity

    PLEASE SOMEONE HELP ASAP. I URGENTLY NEED IT!!!

    Well, if it is that urgent I bet you already did some work on them: show us and tell us where are you stuck.

    Tonio
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  3. #3
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    I'm stuck here that's why I need help. The first question is all in my head. I've been writing out some numbers in the recurrent sequence and am trying to find out the pattern but all in vain. And the second question, I don't even know where to start. I mean, I can start with definition of given limit and then, I get blank. Well, we have been given this question because it's very difficult and the instructor thinks none can answer it so easily and that's what has happened. And now the deadline for this is turning in this week, and I really need someone to help me, or at least give me hint, if they don't want to fully share their knowledge (well, I would definitely share if I knew answers to posts in this section).

    So, if anyone can help, or at least give me hint (correct one though) that would be highly appreciated.

    Quote Originally Posted by tonio View Post
    Well, if it is that urgent I bet you already did some work on them: show us and tell us where are you stuck.

    Tonio
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  4. #4
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    Well, Tonio, do you have any answer to it yet! I'm getting to nowhere as i can't even get started ,especially in the second one.

    Quote Originally Posted by tonio View Post
    Well, if it is that urgent I bet you already did some work on them: show us and tell us where are you stuck.

    Tonio
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  5. #5
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    Quote Originally Posted by concern View Post
    Well, Tonio, do you have any answer to it yet! I'm getting to nowhere as i can't even get started ,especially in the second one.

    Oh, I have some good hints for you: x_{2n}\xrightarrow [n\to \infty] {}1\,,\,\,x_{2n-1}\xrightarrow [n\to \infty] {}0.

    You have to show, I'd say by induction, that the sequences with even indexes is monotone increasing and bounded above, and the seq. with odd indexes is monotone decreasing and bounded below, as for both sequences we get the same relation:

    x_{2n}=1-x_{2n-1}^2=1-(1-x_{2(n-1)}^2)^2=2x_{2(n-1)}+x_{2(n-1)}^4

    Since both sequences have limit just use the above relation to get the corresponding limits.

    About the second question: nothing... yet.

    Tonio
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  6. #6
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    thanks for this. but i'm really not getting much out here. and the second one, not a single clue. anyway, thanks for your help. lets see what other members say on this. i'm counting on drexel28 a lot.

    Quote Originally Posted by tonio View Post
    Oh, I have some good hints for you: x_{2n}\xrightarrow [n\to \infty] {}1\,,\,\,x_{2n-1}\xrightarrow [n\to \infty] {}0.

    You have to show, I'd say by induction, that the sequences with even indexes is monotone increasing and bounded above, and the seq. with odd indexes is monotone decreasing and bounded below, as for both sequences we get the same relation:

    x_{2n}=1-x_{2n-1}^2=1-(1-x_{2(n-1)}^2)^2=2x_{2(n-1)}+x_{2(n-1)}^4

    Since both sequences have limit just use the above relation to get the corresponding limits.

    About the second question: nothing... yet.

    Tonio
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  7. #7
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    Quote Originally Posted by concern View Post
    2.> Let f(x) be twice continously differentiable on (0,infinity). Prove that if
    Lim f(x) exists and f''(x) (i.e. 2nd derivative of f) is bounded on
    x-->infinity

    (0, infinity), then Lim f'(x) = 0 (i.e. 1st derivative of f=0)
    x-->infinity
    By Taylor's theorem, f(x+h) = f(x) + hf'(x) + \tfrac12h^2f''(\xi) for some \xi between x and x+h. The triangle inequality then tells you that |f'(x)| \leqslant\frac{|f(x+h)-f(x)|}h + \tfrac12h|f''(\xi)|.

    For x large enough (and assuming h>0) we can make |f(x+h)| and |f(x)| less than \varepsilon, for any given \varepsilon>0. Let M be a bound for |f"(x)|. Then |f'(x)| \leqslant\frac{2\varepsilon}h + \frac{hM}2.

    Now choose h = 2\sqrt{\varepsilon/M}. That gives |f'(x)| \leqslant 2\sqrt{\varepsilon M}.

    If you write that out a bit more carefully, you'll see that it ensures that |f'(x)| can be made arbitrarily small for all sufficiently large x, as required.
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  8. #8
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    Hey, thanks. But, can you explain the last line "if you write..." a bit more.

    Further, is it possible to do this other way, without using Taylor's Theorem? And, Drexel28, I'm still waiting on to hear what you have on this.
    Thanks Opalg for this. Can you help with the 1st one also in your way?

    Quote Originally Posted by Opalg View Post
    By Taylor's theorem, f(x+h) = f(x) + hf'(x) + \tfrac12h^2f''(\xi) for some \xi between x and x+h. The triangle inequality then tells you that |f'(x)| \leqslant\frac{|f(x+h)-f(x)|}h + \tfrac12h|f''(\xi)|.

    For x large enough (and assuming h>0) we can make |f(x+h)| and |f(x)| less than \varepsilon, for any given \varepsilon>0. Let M be a bound for |f"(x)|. Then |f'(x)| \leqslant\frac{2\varepsilon}h + \frac{hM}2.

    Now choose h = 2\sqrt{\varepsilon/M}. That gives |f'(x)| \leqslant 2\sqrt{\varepsilon M}.

    If you write that out a bit more carefully, you'll see that it ensures that |f'(x)| can be made arbitrarily small for all sufficiently large x, as required.
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