Sort this numbers

Printable View

• Aug 22nd 2009, 12:49 AM
dhiab
Sort this numbers
Sort the following numbers of smallest to largest
$\displaystyle \left( {\sqrt 2 } \right)^{\frac{1}{{\sqrt 2 }}} ,\left( {\sqrt e } \right)^{\frac{1}{{\sqrt e }}} ,\left( {\sqrt 3 } \right)^{\frac{1}{{\sqrt 3 }}} ,\left( {\sqrt \pi } \right)^{\frac{1}{{\sqrt \pi }}} ,\left( 2 \right)^{\frac{1}{2}} ,\left( {\sqrt 5 } \right)^{\frac{1}{{\sqrt 5 }}} ,\left( 3 \right)^{\frac{1}{3}} ,\left( 4 \right)^{\frac{1}{4}} ,\left( 5 \right)^{\frac{1}{5}}$
• Aug 22nd 2009, 08:59 AM
running-gag
Quote:

Originally Posted by dhiab
Sort the following numbers of smallest to largest
$\displaystyle \left( {\sqrt 2 } \right)^{\frac{1}{{\sqrt 2 }}} ,\left( {\sqrt e } \right)^{\frac{1}{{\sqrt e }}} ,\left( {\sqrt 3 } \right)^{\frac{1}{{\sqrt 3 }}} ,\left( {\sqrt \pi } \right)^{\frac{1}{{\sqrt \pi }}} ,\left( 2 \right)^{\frac{1}{2}} ,\left( {\sqrt 5 } \right)^{\frac{1}{{\sqrt 5 }}} ,\left( 3 \right)^{\frac{1}{3}} ,\left( 4 \right)^{\frac{1}{4}} ,\left( 5 \right)^{\frac{1}{5}}$

Hi

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

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

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

Then you can use the fact that $\displaystyle \left( 4 \right)^{\frac{1}{4}} = \left(\left( 4 \right)^{\frac{1}{2}}\right)^{\frac{1}{2}} = \left( 2 \right)^{\frac{1}{2}}$
• Aug 23rd 2009, 04:36 AM
dhiab
Quote:

Originally Posted by running-gag
Hi

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

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

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

Then you can use the fact that $\displaystyle \left( 4 \right)^{\frac{1}{4}} = \left(\left( 4 \right)^{\frac{1}{2}}\right)^{\frac{1}{2}} = \left( 2 \right)^{\frac{1}{2}}$

Hello thankyou : How do you compare two numbers:

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