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Math Help - Proving logarithmic inequality

  1. #1
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    Apr 2010
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    Proving logarithmic inequality

    Hello, everyone!
    I'm stuck at the following problem.

    Let us have an equation that I intend to solve with some numerical method:

    |x (e^{1/x} - 1) - 1| = \epsilon, where \epsilon is a small number and x is positive.

    First off, I want to get rid of the modulus sign to avoid solving a system of equations. To do so, I need to show that
    \forall x > 0: x (e^{1/x} - 1) > 1
    (This fact I established empirically and want to prove. The plot can be seen here: 1 is actually a limit).

    The inequality is equivalent to the following one:

    e^{1/x} > \frac{1}{x} + 1
    {\frac{1}{x}} > \ln{\left(\frac{1}{x} + 1\right)}

    Denote 1 + \frac{1}{x} = u ( u > 1), then

    u > 1 + \ln{u}

    And here's where I'm out of ideas. How do I analytically prove the last inequality for all u > 1?

    Any thoughts would be very appreciated!

    PS I probably should have posted this in Calculus subforum, my bad.
    Last edited by fiontie; April 12th 2010 at 11:05 AM.
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  2. #2
    Junior Member
    Joined
    Dec 2008
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    Quote Originally Posted by fiontie View Post
    Hello, everyone!
    I'm stuck at the following problem.

    Let us have an equation that I intend to solve with some numerical method:

    |x (e^{1/x} - 1) - 1| = \epsilon, where \epsilon is a small number and x is positive.

    First off, I want to get rid of the modulus sign to avoid solving a system of equations. To do so, I need to show that
    \forall x > 0: x (e^{1/x} - 1) > 1
    (This fact I established empirically and want to prove. The plot can be seen here: 1 is actually a limit).

    The inequality is equivalent to the following one:

    e^{1/x} > \frac{1}{x} + 1
    {\frac{1}{x}} > \ln{\left(\frac{1}{x} + 1\right)}

    Denote 1 + \frac{1}{x} = u ( u > 1), then

    u > 1 + \ln{u}

    And here's where I'm out of ideas. How do I analytically prove the last inequality for all u > 1?

    Any thoughts would be very appreciated!

    PS I probably should have posted this in Calculus subforum, my bad.
    Are familiar with Taylor's theroem? Because if so it's a direct application of it to show e^{1/x} > \frac{1}{x} + 1.
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  3. #3
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    Oct 2009
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    Quote Originally Posted by fiontie View Post
    Hello, everyone!
    I'm stuck at the following problem.

    Let us have an equation that I intend to solve with some numerical method:

    |x (e^{1/x} - 1) - 1| = \epsilon, where \epsilon is a small number and x is positive.

    First off, I want to get rid of the modulus sign to avoid solving a system of equations. To do so, I need to show that
    \forall x > 0: x (e^{1/x} - 1) > 1
    (This fact I established empirically and want to prove. The plot can be seen here: 1 is actually a limit).

    The inequality is equivalent to the following one:

    e^{1/x} > \frac{1}{x} + 1

    {\frac{1}{x}} > \ln{\left(\frac{1}{x} + 1\right)}

    Denote 1 + \frac{1}{x} = u ( u > 1), then

    u > 1 + \ln{u}

    And here's where I'm out of ideas. How do I analytically prove the last inequality for all u > 1?

    Any thoughts would be very appreciated!

    PS I probably should have posted this in Calculus subforum, my bad.

    Define f(u)=\ln u-u+1 \Longrightarrow f'(u)=\frac{1}{u}-1<0\,\,\,\forall u>1\Longrightarrow the function is monotone decreasing , so 0=f(1)> f(u)\,\,\forall u>1 and we're done

    Tonio

    Ps. Very nice and good idea to define u as you did.
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  4. #4
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    Apr 2010
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    @tonio: elegance is easy

    Thank you guys for the quick responses, they saved my day!
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