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I need the solutions $x$, such that $x\ln|x|+\frac{1}{4}=0 $ is true. Wolframalpha gives http://www.wolframalpha.com/input/?i=xln|x|%2B1%2F4%3D0, but I never heard of the Lambert W-Function before. Can you give me a hint how to find the solutions of this equation? Best.

  • http://en.wikipedia.org/wiki/Lambert_W_function – Claude Leibovici Feb 20 '14 at 14:50
  • After I saw this solutions I read this article about the Lambert W-Function on Wikipedia but nevertheless I have no idea to determine the solutions. –  Feb 20 '14 at 14:56
  • It is just a special function Lambert and Euler worked. It has a lot of applications. If you need more info, just post. – Claude Leibovici Feb 20 '14 at 14:59
  • Your problem is a special case $b=e, ; a=-1/4$ of the Wiki example http://en.wikipedia.org/wiki/Lambert_W_function#Example_4. Or do you want do no how you actually compute a function value? – gammatester Feb 20 '14 at 15:10
  • Thank you. I want to know how to derivate (I'm not sure if that is the suitable english word, I use http://www.dict.cc ) the solutions, how to start a possible computation to get this solutions without knowing explicitly the Lambert W-Function (sorry for my english :( ) –  Feb 20 '14 at 15:28
  • A very simple hint: Take a starting value $x_0 = 0.5$ and iterate $x_{n+1} = -0.25 / \ln x_n$. This process approaches one solution $x = 0.11610128\dots$ – gammatester Feb 20 '14 at 15:41
  • ok thank you. I will try to do it. Now I know that it is not solvable with a simple analysis 1,2-calculation. –  Feb 21 '14 at 08:14

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The Lambert-W function is the solution to the inverse of $f(x)=xe^x$, where $W(xe^x)=f^{-1}(xe^x)=x$. This is also true for $W(xlnx)$. So let's get your equation into that form: $$ xln|x|+\frac{1}{4} = 0 $$ $$ xln|x| = -\frac{1}{4} $$ Now, in order to take the $W$ of both sides, we need to resolve this absolute value by splitting our equation into two parts, one with $ln(x)$ and one with $ln(-x)$. Here's the positive branch: $$ W(xlnx) = W(-\frac{1}{4}) $$ $$ lnx = W(-\frac{1}{4}) $$ $$ x = e^{W(-\frac{1}{4})} $$ This branch gives us two of our answers, $e^{W_0(-1/4)} \approx .699491$ and $e^{W_{-1}(-1/4)} \approx .116101$. Then, our negative counterpart is as follows: $$ xln(-x) = -\frac{1}{4} $$ $$ -xln(-x) = \frac{1}{4} $$ $$ W\biggl(-xln(-x)\biggr) = W\left(\frac{1}{4}\right) $$ $$ ln(-x) = W\left(\frac{1}{4}\right) $$ $$ -x = e^{W\left(\frac{1}{4}\right)} $$ $$ x = -e^{W\left(\frac{1}{4}\right)} $$ Thus, our final value for $x$ is $-e^{W_0\left(\frac{1}{4}\right)} \approx -1.226161$. There exist, in fact, 3 more solutions in total from the $W_{-1}$ and $W_1$ branches for both $lnx$ and $ln(-x)$, however those solutions are complex answers, not real.

Cole Kesey
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