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Math Help - prove or desprove by example

  1. #1
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    prove or desprove by example

    1.prove or desprove by example:
    a)
    sup\{\frac{[x]}{[x]+1}|x\geq0\}=1
    b)prove or desprove that x^{3} 'uniformly' continuous in R
    c)if f(x) monotonicly decreasing in [0,\infty) and lim_{x->\infty}=0then f(x)>0 for x\in[0,\infty)
    d)f,g are continues in [a,b] if x\in[a,b] \int_{a}^{x}f(t)dt=\int_{a}^{x}g(t)dt then f=g in [a,b]


    regarding the first one i couldnt find a value for which the supremum is 1
    for part a) i cant say if its correct or not

    ?
    Last edited by transgalactic; September 3rd 2011 at 03:37 AM.
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  2. #2
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    Re: prove or desprove by example

    Quote Originally Posted by transgalactic View Post
    1.prove or desprove by example:
    a)
    sup\{\frac{[x]}{[x]+1}|x\geq0\}=1
    b)prove that x^{3} 'uniformly' continuous in R
    c)if f(x) monotonicly decreasing in [0,\infty) and lim_{x->\infty}=0then f(x)>0 for x\in[0,\infty)
    d)f,g are continues in [a,b] if x\in[a,b] \int_{a}^{x}f(t)dt=\int_{a}^{x}g(t)dt then f=g in [a,b]


    regarding the first one i couldnt find a value for which the supremum is 1
    That makes no sense. The supremum does not depend upon a "value". It should be obvious that 1 is an upper bound for this. Is there a smaller upper bound? That is, is there a number \alpha< 1 such that \frac{x}{x+1}< \alpha for all x> 1? What do you get if you solve that inequality for x?

    for part a) i cant say if its correct or not

    ?
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  3. #3
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    Re: prove or desprove by example

    ok so 1 is the upper bound and in order to prove that its supremum ,we need to show
    that its the smallest upper bound.

    suupose 1 is not the smallest and sup\{\frac{[x]}{[x]+1}|x\geq0\}=t<1.
    now i think we should use lim_{x->\infty}\frac{[x]}{[x]+1}
    but i cant use lhopital law here like normal variable because we have to get rid of [x].
    fo x>0
    x-1<[x]<x+1
    so lim_{x->\infty}\frac{[x]}{[x]+1} > lim_{x->\infty}\frac{x-1}{x+1+1}
    so i have the result of a smaller limit.
    how i know the result of the bigger one
    ?
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    Re: prove or desprove by example

    Instead of using the L'Hopital's rule, I would follow the suggestion above and solve [x]/([x] + 1) < t for x. If [x] means the integer part of x, I would solve it when x is an integer.
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    Re: prove or desprove by example

    so you want me to replace [x] with x,and prove that t is its supremum
    ?
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    Re: prove or desprove by example

    Quote Originally Posted by transgalactic View Post
    so you want me to replace [x] with x,and prove that t is its supremum
    ?
    No,
    Quote Originally Posted by emakarov
    I would follow the suggestion above and solve [x]/([x] + 1) < t for x
    I have not said anything about proving that t is a supremum. The suggestion is to solve the inequality on integers, so you can remove the square brackets.
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  7. #7
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    Re: prove or desprove by example

    ahhhh instead of R i solve it on N
    but how to make the transition back to R
    ?
    Last edited by transgalactic; September 2nd 2011 at 10:46 AM.
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  8. #8
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    Re: prove or desprove by example

    but how to make the transition back to R
    ?
    Each integer is a real number.

    We know that 1 is an upper bound of \{[x]/([x]+1)\mid x\ge0\}. The idea is to prove that anything less than 1 is not an upper bound. Suppose t < 1. Then t is not an upper bound iff the inequality [x]/([x]+1) > t is true for some x\ge0. Well, the suspicion is that it is true for infinitely many x\ge 0, and even true for infinitely many positive integers. So, try to solve this inequality when x is an integer and see if this is true.
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  9. #9
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    Re: prove or desprove by example

    ahhh if i will prove for integer that \frac{x}{x+1}>t then t cannot be an upper bound and specially supremum.
    ok the integer group is not bounded from top so N> \frac{x}{x+1}
    but i dont know how to continue and i dont know where to put "t"
    i only remmember the part when N is not bounded and we put it to be larger then the fracture.

    another way i remmber is
    lim_{x->\infty}\frac{x}{x+1}=1
    so for every \epsilon >0 there is M >0 for which every x>M
    |\frac{x}{x+1}-1|<\epsilon
    i can open this
    to
    1-\epsilon<\frac{x}{x+1}<\epsilon+1
    so i can choose 1-\epsilon=t
    and
    t<\frac{x}{x+1}
    correct?
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  10. #10
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    Re: prove or desprove by example

    Quote Originally Posted by transgalactic View Post
    ahhh if i will prove for integer that \frac{x}{x+1}>t then t cannot be an upper bound and specially supremum.
    ok the integer group is not bounded from top so N> \frac{x}{x+1}
    but i dont know how to continue and i dont know where to put "t"
    You should review how to solve linear inequalities such as x / (x + 1) > t. Without it, frankly, talking about limits, continuity, or any other calculus topic looses meaning. (I am assuming this is the issue because this thread has four suggestions to solve this inequality, and my last post explains how this would help solve the problem.)

    Quote Originally Posted by transgalactic View Post
    another way i remmber is
    lim_{x->\infty}\frac{x}{x+1}=1
    so for every \epsilon >0 there is M >0 for which every x>M
    |\frac{x}{x+1}-1|<\epsilon
    i can open this
    to
    1-\epsilon<\frac{x}{x+1}<\epsilon+1
    so i can choose 1-\epsilon=t
    and
    t<\frac{x}{x+1}
    correct?
    There is sense in this, but, again, I think it's meaningless doing limits without a solid grasp of algebra.
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  11. #11
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    Re: prove or desprove by example

    i know how to solve innequalitties
    you didnt say which one.i have presented two ways of continuing this solution
    could you say which one to follow
    ?
    be more spesific
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  12. #12
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    Re: prove or desprove by example

    Quote Originally Posted by emakarov View Post
    We know that 1 is an upper bound of \{[x]/([x]+1)\mid x\ge0\}. The idea is to prove that anything less than 1 is not an upper bound. Suppose t < 1. Then t is not an upper bound iff the inequality [x]/([x]+1) > t is true for some x\ge0. Well, the suspicion is that it is true for infinitely many x\ge 0, and even true for infinitely many positive integers. So, try to solve this inequality when x is an integer and see if this is true.
    .
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  13. #13
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    Re: prove or desprove by example

    ahhhhhhhhhhhhh ok
    x/(x+1) >t
    x>t(x+1)
    x-tx>1
    x>1/(1-t)
    0<t<1 from the data in the question
    but how it proves that t is not upper bound
    ?
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  14. #14
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    Re: prove or desprove by example

    Since 0<t<1, 1/(1-t) is well-defined positive number. You can take x to be any integer greater than 1/(1-t), and then [x] / ([x] + 1) > t, disproving that t is the upper bound of {[x] / ([x] + 1) | x >= 0}.
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  15. #15
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    Re: prove or desprove by example

    ahh i understand now part a

    regarding part b:
    i know a guide line that any curve above f(x)=x is not uniformly continues

    so we must prove that there is \epsilon>0 so for all \delta>0
    there is x,y in R which |x-y|<\delta and |f(x)-f(y)|=>\epsilon

    |x^3-y^3|=|(x-y)(x^2+xy+y^2)|<|\delta (x^2+xy+y^2)|
    so i choose \epsilon=2
    x=\frac{2}{\sqrt{\delta}}
    y=\frac{3}{\sqrt{\delta}}
    but i know only that |x-y|<\delta
    i need to find some delta expression so |x-y| will be greater that it.

    so the result of the x,y value input will give use greater then 2.

    but here i have smaller then delta only expression
    what do?
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