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Thread: supremum proof question

  1. #1
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    supremum proof question

    prove that $\displaystyle sup\{\frac{xsinx}{x+1}|x>0\}=1$
    here is the solution:
    for $\displaystyle x>0$ $\displaystyle x+1>x$ so $\displaystyle 0<\frac{x}{x+1}<1$ so 1 is upper bound so
    there is a supremum suppose 1 is not the smallest upper bound
    suppose $\displaystyle sup\{f(R)\}=1-\epsilon$. $\displaystyle lim_{x->\infty}\frac{x}{x+1}=1$
    so by the limit definition for every epsilon there is M>0 so for every
    x for which x>M $\displaystyle |\frac{x}{x+1}-1|<\epsilon$
    so $\displaystyle \frac{x}{x+1}>1-\epsilon$
    and it works for every x for which x>M
    we take $\displaystyle x_{0}=2\pi([M]+45)+\frac{\pi}{2}$ $\displaystyle sin(x_{0})=1$
    for this $\displaystyle x_{0}$ it works too
    $\displaystyle \frac{x_{0}}{x_{0}+1}=\frac{x_{0}sin(x_{0})}{x_{0} +1}>1-\epsilon$
    where is the contradiction here??
    we were talking about the supremum of $\displaystyle \frac{x}{x+1}$ not
    $\displaystyle \frac{xsin(x)}{x+1}$
    ??
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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 $\displaystyle 1-\varepsilon$ then using the fact that $\displaystyle \lim_{x\to+\infty} \frac x{x+1}=1$ we find $\displaystyle x_0$ such that $\displaystyle \frac{x_0\sin x_0}{x_0+1}>1-\varepsilon$ which is a contradiction.
    But here, we don't have to prove the result by contradiction. Since $\displaystyle \frac{x\sin x}{x+1}\leq 1$ we know that the supremum is $\displaystyle \leq 1$. Then consider for $\displaystyle n\in\mathbb{N}$: $\displaystyle x_n :=2n\pi+\frac{\pi}2$ (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 $\displaystyle \sup \left\{\frac{x\sin x}{x+1}\right\}\neq 1$ and since we know that this supremum is $\displaystyle \leq 1$ we should have $\displaystyle \sup \left\{\frac{x\sin x}{x+1}\right\}< 1$. Hence we can find $\displaystyle \varepsilon>0$ such that $\displaystyle \sup \left\{\frac{x\sin x}{x+1}\right\}= 1-\varepsilon$. Do you agree that we contradict that if we can find $\displaystyle x_0$ such that $\displaystyle \frac{x_0\sin x_0}{1+x_0}$ ?
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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 $\displaystyle f(x):=\frac{x\sin x}{x+1}$. We have $\displaystyle f(x_n) =\frac{2\pi n+\frac{\pi}2}{2\pi n+\frac{\pi}2+1}$, and if you take the limit when $\displaystyle n\to \infty$ you will find $\displaystyle 1$. 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 $\displaystyle \sup\left\{f(x);x>0\right\}\geq \sup \left\{f(x_n),n\in\mathbb{N}\right\} $ and compute $\displaystyle \sup \left\{f(x_n),n\in\mathbb{N}\right\}$.
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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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