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Math Help - Big O, Little O and Logarithm Integral

  1. #1
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    Big O, Little O and Logarithm Integral

    Hello everyone,

    I've spent the last hours trying to understand Big O and little O notations in depth. Usually the limit of the quotient method works but here, I'm dealing with integrals which seems to make things harder.

    integral(dt/ln(t),t=2..sqrt(x))=O(sqrt(x)) when x->+∞
    integral(dt/(ln(t))^2,t=sqrt(x)..x)=o(integral(dt/ln(t),t=sqrt(x)..x) when x->+∞

    I'd really appreciate some tips or the general method on proving these two assertions. Sorry for not knowing Latex, I'll get to it soon ^^

    Thanks for any help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by DwProd View Post

    integral(dt/ln(t),t=2..sqrt(x))=O(sqrt(x)) when x->+∞
    Put:

    I(x)=\displaystyle \int_{t=2}^{\sqrt{x}}\dfrac{1}{\ln(t)}\;dt

    Now \ln (t) is a positive increasing function on (1,\infty) so 1/\ln(t) is decreasing so:

    I(x)<\displaystyle \int_{t=2}^{\sqrt{x}}\dfrac{1}{\ln(2)}\;dt=\dfrac{  1}{\ln(2)} \times (\sqrt{x}-2)

    and you should be able to complete the proof from there.

    CB
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  3. #3
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    Yes that is what I've been able to do so far but I'm not sure this allows me to verify the first assertion. I think it has to do with the
    ∃x0>0 ∃M>0 ∀x>x0 abs(f(x))<M*abs(g(x)) => f(x)=O(g(x)) but I'm not even sure it is a proper definition and why the condition x->+∞ is useful here. Thanks for your answer by the way!
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by DwProd View Post
    integral(dt/(ln(t))^2,t=sqrt(x)..x)=o(integral(dt/ln(t),t=sqrt(x)..x) when x->+∞
    Let:

    \displaystyle I(x)=\int_{\sqrt{x}}^x \dfrac{1}{(\ln(t))^2}\; dt

    again we observe that on (1,\infty) 1/\ln(t) is decreasing so for t>\sqrt{x}:

    \dfrac{1}{(\ln(t))^2}<\dfrac{1}{\ln(\sqrt{x})}\tim  es \dfrac{1}{\ln(t)}

    CB
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  5. #5
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    Ok I think I get it, I can just create the quotients with those inequalities (since everything is positive and ≠ 0) and then just use the Squeeze theorem.
    If this is all I need to do, please just confirm so that I can put the little [SOLVED] Thank you very much for your help CaptainBlack!
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by DwProd View Post
    Ok I think I get it, I can just create the quotients with those inequalities (since everything is positive and ≠ 0) and then just use the Squeeze theorem.
    If this is all I need to do, please just confirm so that I can put the little [SOLVED] Thank you very much for your help CaptainBlack!
    I don't think you need the squeeze theorem just the definition of big-O and little-o notation

    CB
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