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Math Help - Small Challenge

  1. #1
    Senior Member Dinkydoe's Avatar
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    Small Challenge

    This is a challenge question:

    Let f:[0,1]\to [0,1] with f(0)=0, f(1)=1 differentiable

    Show there exists distinct points x,y\in [0,1] such that f'(x)f'(y)=1.


    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; July 21st 2010 at 02:01 PM. Reason: Approved.
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    From Lagrange theorem it follows that there exist c\in[0,1] such that

    f'(c) = \frac{f(1)-f(0)}{1-0} = 1 so we can assume f'(x)>1 or f'(x)<1 for all x\in[0,1]\setminus\{c\}

    (otherwise the proof it trivial). Suppose we have f'(a)<1 and f'(b)>1 for some a,b\in[0,1].

    If f'(a)f'(b)=1 then we're done.

    If f'(a)\le0 then f'(a)<\frac{1}{f'(b)}<f'(c) and thus the existence of d such

    that f'(d)=\frac{1}{f'(b)} is guaranteed by the intermediate value theorem for derivatives and we're done.

    Otherwise 0<f'(a)<1 and f'(b)>1 so either f'(a)<\frac{1}{f'(b)} < f'(c) (=1) or f'(c)<\frac{1}{f'(a)}<f'(b). Either

    way, using the intermediate value theorem for derivatives we're done again.

    Now suppose the previous case is not true. That is, f'(x)<1 for all x\in[0,1]\setminus\{c\} or f'(x)>1 for all x\in[0,1]\setminus\{c\}

    Therefore the function g(x) = x-f(x) is either monotone increasing or monotone decreasing (by looking at the derivative).

    Either way, it means that g(0)\ne g(1). But from the conditions on f(x) it follows that g(0)=g(1)=0, a contradiction.
    Last edited by Unbeatable0; July 21st 2010 at 01:57 PM. Reason: Flaw in proof
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Well done.
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    Quote Originally Posted by Dinkydoe
    0< f'(a)< 1 doesn't guarantee \frac{1}{f'(a)}< f'(b)
    Not 0<f'(a)<1 alone.

    f'(a)<\frac{1}{f'(b)} < f'(c) corresponds to the option f'(a)f'(b) < 1.

    f'(c)<\frac{1}{f'(a)}<f'(b) corresponds to the option f'(a)f'(b) > 1

    I'm interested - what's your solution?
    Last edited by Unbeatable0; July 21st 2010 at 04:40 PM.
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  5. #5
    Senior Member Dinkydoe's Avatar
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    I allready corrected my statement ;p

    Didnt expect such fast reply xp
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  6. #6
    Senior Member Dinkydoe's Avatar
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    My solution was a little shorter:

    We may assume f(x)\neq x. (otherwise there's nothing to prove).
    By the fundamental theorem of calculus we have \int_{0}^1f'(x)dx= f(1)-f(0)= 1. Hence we can't have f'(x)\leq 1 or f'(x)\geq 1 for all x. (since we also assumed f(x)\neq x)

    But then there exist points a,b \in [0,1] such that f'(a)<1 and f'(b)>1 and the intermediate value theorem now gives that f'(x) reaches all points in (f'(a),f'(b)) and trivially we can find p,q in this interval such that pq= 1.
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  7. #7
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    Quote Originally Posted by Dinkydoe View Post
    My solution was a little shorter:

    We may assume f(x)\neq x. (otherwise there's nothing to prove).
    By the fundamental theorem of calculus we have \int_{0}^1f'(x)dx= f(1)-f(0)= 1. Hence we can't have f'(x)\leq 1 or f'(x)\geq 1 for all x. (since we also assumed f(x)\neq x)

    But then there exist points a,b \in [0,1] such that f'(a)<1 and f'(b)>1 and the intermediate value theorem now gives that f'(x) reaches all points in (f'(a),f'(b)) and trivially we can find p,q in this interval such that pq= 1.
    The fundumental theorem of calculus has the assumption that f'(x) is continuous,

    which is not given, and there are example of derivatives which are not continuous, so I think your argument is

    not perfectly justified.
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  8. #8
    Senior Member Dinkydoe's Avatar
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    The fundumental theorem of calculus has the assumption that is continuous,

    which is not given, and there are example of derivatives which are not continuous, so I think your argument is

    not perfectly justified.
    Check Fundamental theorem of calculus - Wikipedia, the free encyclopedia

    Second Fundamental Theorem of Calculus

    We don't have to assume f'(x) is continuous.
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    Quote Originally Posted by Dinkydoe View Post
    Check Fundamental theorem of calculus - Wikipedia, the free encyclopedia

    Second Fundamental Theorem of Calculus

    We don't have to assume f'(x) is continuous.
    It still does require f'(x) to be integrable, which is not always the case (you can google some examples).
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  10. #10
    Senior Member Dinkydoe's Avatar
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    It still does require to be integrable, which is not always the case (you can google some examples).
    f'(x) is by definition integrable, since we start out with the 'antiderivative' f(x) as a continuous function. ;p
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    Quote Originally Posted by Dinkydoe View Post
    f'(x) is by definition integrable, since we start out with the 'antiderivative' f(x) as a continuous function. ;p
    That's not the definition.

    See here: A differentiable function whose derivative is not integrable
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  12. #12
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    Quote Originally Posted by Unbeatable0 View Post

    I think what he meant is that since f(x) is defined and derivable in [0,1] and f(x) is clearly a primitive function of f'(x) , the fundamental

    theorem of integral calculus gives us that f'(x) is (Riemann) integrable and \int\limits^1_0f'(x)\,dx=f(1)-f(0)

    Tonio
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    Quote Originally Posted by tonio View Post
    I think what he meant is that since f(x) is defined and derivable in [0,1] and f(x) is clearly a primitive function of f'(x) , the fundamental

    theorem of integral calculus gives us that f'(x) is (Riemann) integrable and \int\limits^1_0f'(x)\,dx=f(1)-f(0)

    Tonio
    That's not true. A derivative need not in general be integrable over a closed interval.
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  14. #14
    Senior Member Dinkydoe's Avatar
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    The examples given in your link are of oscillating functions that are unbounded. (at least from what I read) From say, f:[-1,1]\to \mathbb{R}?

    I hope you didn't forget our function is f:[0,1]\to [0,1] and is by definition bounded.

    And I remember that a function can still be integrable, even though it may have a countable set of problem-points.

    The second last post gives a function, from where they agree it is integrable....(oscillating but still bounded)

    But, maybe we need help of an expert here.
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  15. #15
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    Quote Originally Posted by Dinkydoe View Post
    The examples given in your link are of oscillating functions that are unbounded. (at least from what I read) From say, f:[-1,1]\to \mathbb{R}?

    I hope you didn't forget our function is f:[0,1]\to [0,1] and is by definition bounded.

    Well, I think we need help of an expert here.
    It's f'(x) that's not bounded in the example there.

    f(x) = x^2 \sin\frac{1}{x^3} is bounded on every closed interval. [0,1], for example.

    (of course while defining f(0) = 0 )

    The above function indeed satisfies f:[0,1]\to [0,1].
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