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  1. #1
    MHF Contributor Mathstud28's Avatar
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    Ok

    I will start a new thread here to ask this question...suppose that f(x)<g(x),\forall{x}\in\mathbb{R} and also suppose that f(x)\sim{g(x)} could you argue that the original inequality could be untrue?
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    It depends on where f(x) ~ g(x)

    And what does f(x) ~ g(x) mean ? It means that there exists \epsilon (x) such as f(x)=g(x)+\epsilon (x), where \epsilon (x) tends to be very small as approaching the domain where f(x) ~ g(x).

    This doesn't impeach f(x) to be inferior to g(x)...
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    That is not what I meant

    I have always known f(x)\sim{g(x)} to mean \lim_{x \to {\infty}}\frac{f(x)}{g(x)}=1 or \lim_{x \to {\infty}}\frac{g(x)}{f(x)}=1...or in other words they are asymptotically equivalent
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I have always known f(x)\sim{g(x)} to mean \lim_{x \to {\infty}}\frac{f(x)}{g(x)}=1 or \lim_{x \to {\infty}}\frac{g(x)}{f(x)}=1...or in other words they are asymptotically equivalent
    well, f(x) = g(x) certainly satisfies that definition, and f(x) > g(x) for all x could satisfy it as well. so yes, we could argue that the original inequality was false. but of course, to do that, we would need more information that you have provided.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    I dont know if this is a good example

    but what relating to the classic \pi(x)....I am pretty sure that Li(x)>\frac{x}{\ln(x)} \forall{x}\in\mathbb{R}...but assuming the limit exists that then Li(x)\sim\frac{x}{\ln(x)}...I am not sure that is the best example but Its the only one I can think of right now...so would you still agree with the original inequality?
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  6. #6
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    Another example is Stirling approximation.
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