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Math Help - Need solution to mixed equation.

  1. #1
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    Need solution to mixed equation.

    Who can solve ln x = 1/x? I can get an approximate solution to any degree of accuracy, but is there an exact solution?
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  2. #2
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    Let's just use Newton's method. It's always a good standby.

    x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    f(x)=ln(x)-\frac{1}{x}

    f'(x)=\frac{1}{x}+\frac{1}{x^{2}}

    Let's make an initial guess of 1.5:

    x_{n+1}=1.5-\frac{ln(1.5)-\frac{1}{1.5}}{\frac{1}{1.5}+\frac{1}{(1.5)^{2}}}\  approx 1.7350814027

    Iterate:

    1.7350814027-\frac{ln(1.7350814027)-\frac{1}{1.7350814027}}{\frac{1}{1.7350814027}+\fr  ac{1}{(1.7350814027)^{2}}}\approx 1.76291539065

    Keep going until you reach the desired accuracy.

    The solution is 1.76291539065....

    As far as an exact solution, I do not think so.

    \frac{9ln(\frac{2}{3})}{10}+\frac{21}{10}=\frac{3(  3ln(2/3)+7)}{10}

    is pretty close of an approximation.
    Last edited by galactus; December 31st 2008 at 02:27 PM. Reason: I'm sorry, I see you have an approximate solution.
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  3. #3
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    Quote Originally Posted by billyates View Post
    Who can solve ln x = 1/x? I can get an approximate solution to any degree of accuracy, but is there an exact solution?
    The exact solution can be written in terms of the Lambert W-function:

    \ln x = \frac{1}{x} \Rightarrow x = e^{1/x} \Rightarrow 1 = \frac{1}{x} \cdot e^{1/x}.

    Therefore \frac{1}{x} = W(1) \Rightarrow x = \frac{1}{W(1)} where W is the Lambert W-function.

    W(1) is the Omega constant: http://en.wikipedia.org/wiki/Omega_constant.
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