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Math Help - supremum proof question

  1. #1
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    supremum proof question

    prove that sup\{\frac{xsinx}{x+1}|x>0\}=1
    here is the solution:
    for x>0 x+1>x so 0<\frac{x}{x+1}<1 so 1 is upper bound so
    there is a supremum suppose 1 is not the smallest upper bound
    suppose sup\{f(R)\}=1-\epsilon. lim_{x->\infty}\frac{x}{x+1}=1
    so by the limit definition for every epsilon there is M>0 so for every
    x for which x>M |\frac{x}{x+1}-1|<\epsilon
    so \frac{x}{x+1}>1-\epsilon
    and it works for every x for which x>M
    we take x_{0}=2\pi([M]+45)+\frac{\pi}{2} sin(x_{0})=1
    for this x_{0} it works too
    \frac{x_{0}}{x_{0}+1}=\frac{x_{0}sin(x_{0})}{x_{0}  +1}>1-\epsilon
    where is the contradiction here??
    we were talking about the supremum of \frac{x}{x+1} not
    \frac{xsin(x)}{x+1}
    ??
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  2. #2
    Super Member girdav's Avatar
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    Re: supremum proof question

    No, the solution shows that if the supremum is 1-\varepsilon then using the fact that \lim_{x\to+\infty} \frac x{x+1}=1 we find x_0 such that \frac{x_0\sin x_0}{x_0+1}>1-\varepsilon which is a contradiction.
    But here, we don't have to prove the result by contradiction. Since \frac{x\sin x}{x+1}\leq 1 we know that the supremum is \leq 1. Then consider for n\in\mathbb{N}: x_n :=2n\pi+\frac{\pi}2 (the idea is quite the same as the solution gives, but I used a "direct way").
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  3. #3
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    Re: supremum proof question

    why its a contradiction
    what it contradicts
    ?
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  4. #4
    Super Member girdav's Avatar
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    Re: supremum proof question

    In the solution, we assumed that \sup \left\{\frac{x\sin x}{x+1}\right\}\neq 1 and since we know that this supremum is \leq 1 we should have \sup \left\{\frac{x\sin x}{x+1}\right\}< 1. Hence we can find \varepsilon>0 such that \sup \left\{\frac{x\sin x}{x+1}\right\}= 1-\varepsilon. Do you agree that we contradict that if we can find x_0 such that \frac{x_0\sin x_0}{1+x_0} ?
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  5. #5
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    Re: supremum proof question

    ok
    we assumed that the supremum is 1-epsilon

    but we fund a certain x in which the function is bigger
    so its not a supremum



    regarding your solution
    you picked a certain x
    how it proves that 1 is a supremum
    ?
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  6. #6
    Super Member girdav's Avatar
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    Re: supremum proof question

    We define f(x):=\frac{x\sin x}{x+1}. We have f(x_n) =\frac{2\pi n+\frac{\pi}2}{2\pi n+\frac{\pi}2+1}, and if you take the limit when n\to \infty you will find 1. Now apply the definition of supremum and you are done.
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  7. #7
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    Re: supremum proof question

    supremum is the smallest upper bound
    i dont how this definition will help here

    ?
    and you found a limit to a certain group of x
    not all of them
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  8. #8
    Super Member girdav's Avatar
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    Re: supremum proof question

    To conclude, use \sup\left\{f(x);x>0\right\}\geq \sup \left\{f(x_n),n\in\mathbb{N}\right\} and compute \sup \left\{f(x_n),n\in\mathbb{N}\right\}.
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  9. #9
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    Re: supremum proof question

    why the supremum of this subgroup is >= of the bigger group
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