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Thread: Can't manage to solve this equation with Lambert W function

  1. #1
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    Unhappy Can't manage to solve this equation with Lambert W function

    Hi dear forumers !

    I have some trouble solving for n in this equation :
    $\displaystyle p=\frac{\left ( 1-a^2 \right )^{(n-1)/2}}{a\sqrt{2\pi n }}$

    I know the solution involves the Lambert W function (inverse of n*e^n). However I can't get n to be isolated on one side .

    $\displaystyle pa\sqrt{2\pi n }=\left ( 1-a^2 \right )^{(n-1)/2}$
    $\displaystyle p^2a^22\pi n=\left ( 1-a^2 \right )^{(n-1)}$
    $\displaystyle \ln \left ( p^2a^22\pi n \right )=(n-1)\ln\left ( 1-a^2 \right )$
    $\displaystyle \ln \left ( p^2a^22\pi \right )=(n-1)\ln\left ( 1-a^2 \right )-\ln(n)$
    $\displaystyle \ln \left ( p^2a^22\pi \right )+\ln\left ( 1-a^2 \right )=n\ln\left ( 1-a^2 \right )-\ln(n)$

    From there I am stuck. Thank you very much for your help !
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  2. #2
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    Re: Can't manage to solve this equation with Lambert W function

    $n = \Large -\frac{W\left(\frac{\log \left(1-a^2\right)}{2 \pi (a-1) a^2 (a+1) p^2}\right)}{\log \left(1-a^2\right)}$
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  3. #3
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    Re: Can't manage to solve this equation with Lambert W function

    Thank you romsek ; in fact I found the same answer by solving with wolfram alpha. Do you have the steps for this ?
    I am very interested in the steps to get there for personal satisfaction and advancing my math knowledge !
    Last edited by Jo37; Dec 11th 2018 at 01:21 PM.
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  4. #4
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    Re: Can't manage to solve this equation with Lambert W function

    Quote Originally Posted by Jo37 View Post
    Thank you romsek ; in fact I found the same answer by solving with wolfram alpha. Do you have the steps for this ?
    I am very interested in the steps to get there for personal satisfaction and advancing my math knowledge !
    sorry I don't feel like grinding through all that algebra.
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  5. #5
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    Re: Can't manage to solve this equation with Lambert W function

    The problem is of the form $$p = \frac {C^{\frac n 2}}{D n^\frac 1 2}$$ for constants $C$ and $D$. Rewrite it as
    $$Dp = n^{-\frac 1 2}C^{\frac n 2} $$ Now raise both sides to the $-2$ power:
    $$(Dp)^{-2} = nC^{-n} =n e^{-n \ln C}$$Multiply both sides by $-\ln C$:
    $$-(Dp)^{-2}\ln C = -n \ln Ce^{-n \ln C}$$The right side is now in the form $xe^x$ so you can write$$
    -n \ln C = W(-(Dp)^{-2}\ln C)$$This you can easily solve for $n$. Put in the constants and simplify. (And correct any mistakes you see).
    Thanks from topsquark and Jo37
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