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Math Help - Sort this numbers

  1. #1
    Super Member dhiab's Avatar
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    Sort this numbers

    Sort the following numbers of smallest to largest
     <br />
\left( {\sqrt 2 } \right)^{\frac{1}{{\sqrt 2 }}} ,\left( {\sqrt e } \right)^{\frac{1}{{\sqrt e }}} ,\left( {\sqrt 3 } \right)^{\frac{1}{{\sqrt 3 }}} ,\left( {\sqrt \pi } \right)^{\frac{1}{{\sqrt \pi }}} ,\left( 2 \right)^{\frac{1}{2}} ,\left( {\sqrt 5 } \right)^{\frac{1}{{\sqrt 5 }}} ,\left( 3 \right)^{\frac{1}{3}} ,\left( 4 \right)^{\frac{1}{4}} ,\left( 5 \right)^{\frac{1}{5}} <br /> <br />
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  2. #2
    MHF Contributor
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    Quote Originally Posted by dhiab View Post
    Sort the following numbers of smallest to largest
     <br />
\left( {\sqrt 2 } \right)^{\frac{1}{{\sqrt 2 }}} ,\left( {\sqrt e } \right)^{\frac{1}{{\sqrt e }}} ,\left( {\sqrt 3 } \right)^{\frac{1}{{\sqrt 3 }}} ,\left( {\sqrt \pi } \right)^{\frac{1}{{\sqrt \pi }}} ,\left( 2 \right)^{\frac{1}{2}} ,\left( {\sqrt 5 } \right)^{\frac{1}{{\sqrt 5 }}} ,\left( 3 \right)^{\frac{1}{3}} ,\left( 4 \right)^{\frac{1}{4}} ,\left( 5 \right)^{\frac{1}{5}} <br /> <br />
    Hi

    The function f(x) = x^{\frac{1}{x}} = e^{\frac{\ln x}{x}} is differentiable over ]0,+\infty[ and f'(x) = \frac{1-\ln x}{x^2}\:e^{\frac{\ln x}{x}}

    Therefore f is increasing over ]0,e] and decreasing over [e,+oo[ and
    \left( {\sqrt 2 } \right)^{\frac{1}{{\sqrt 2 }}} < \left( {\sqrt e } \right)^{\frac{1}{{\sqrt e }}} < \left( {\sqrt 3 } \right)^{\frac{1}{{\sqrt 3 }}} < \left( {\sqrt \pi } \right)^{\frac{1}{{\sqrt \pi }}} < \left( 2 \right)^{\frac{1}{2}} < \left( {\sqrt 5 } \right)^{\frac{1}{{\sqrt 5 }}}

    and \left( 5 \right)^{\frac{1}{5}} < \left( 4 \right)^{\frac{1}{4}} < \left( 3 \right)^{\frac{1}{3}}

    Then you can use the fact that \left( 4 \right)^{\frac{1}{4}} = \left(\left( 4 \right)^{\frac{1}{2}}\right)^{\frac{1}{2}} = \left( 2 \right)^{\frac{1}{2}}
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by running-gag View Post
    Hi

    The function f(x) = x^{\frac{1}{x}} = e^{\frac{\ln x}{x}} is differentiable over ]0,+\infty[ and f'(x) = \frac{1-\ln x}{x^2}\:e^{\frac{\ln x}{x}}

    Therefore f is increasing over ]0,e] and decreasing over [e,+oo[ and
    \left( {\sqrt 2 } \right)^{\frac{1}{{\sqrt 2 }}} < \left( {\sqrt e } \right)^{\frac{1}{{\sqrt e }}} < \left( {\sqrt 3 } \right)^{\frac{1}{{\sqrt 3 }}} < \left( {\sqrt \pi } \right)^{\frac{1}{{\sqrt \pi }}} < \left( 2 \right)^{\frac{1}{2}} < \left( {\sqrt 5 } \right)^{\frac{1}{{\sqrt 5 }}}

    and \left( 5 \right)^{\frac{1}{5}} < \left( 4 \right)^{\frac{1}{4}} < \left( 3 \right)^{\frac{1}{3}}

    Then you can use the fact that \left( 4 \right)^{\frac{1}{4}} = \left(\left( 4 \right)^{\frac{1}{2}}\right)^{\frac{1}{2}} = \left( 2 \right)^{\frac{1}{2}}
    Hello thankyou : How do you compare two numbers:


    \left( {\sqrt 2 } \right)^{\frac{1}{{\sqrt 2 }}} ,\left( {\sqrt 5 } \right)^{\frac{1}{{\sqrt 5 }}} Thank you
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