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I tried solving $x\log x=1$ where $x \in (1,e) \subset \mathbb{R} $ but couldn't find a nice way to solve it.

  1. Given that $x \in (1,e)$, I tried $x:=e^\alpha$ for $\alpha \in [0,1]$ so that I obtain $\alpha e^\alpha=1$. This resembles the familiar exponential function $f(t)=e^{\alpha t}$.

  2. I tried to recover some useful relations and I found the following:

a) $\forall t \in [0,1]\quad f^n(t)f^{n+1}(t)=\alpha^{2n}$

b) Using the mean-value theorem, I can show that:

$\int_{0}^{1} f^n(t)dt=\alpha^{n-1}(\frac{1}{\alpha}-1)$ which isn't too difficult to show given that $\exists c_n \in (0,1) \quad f^{n+1}(c_n)=f^n(1)-f^n(0)=\alpha^n f'(c_0)$

I found that $c_0 = \frac{1}{\alpha}\log(\frac{1}{\alpha^2}-\frac{1}{\alpha})$

But, after all this work I still couldn't figure out how to find $\alpha$. Perhaps there's an additional relation which I simply failed to discover?

  1. Anyway, at this point I tried Newton's method:

Using $\large t_{n+1}=t_n-\frac{f(t_n)}{f'(t_{n+1)}}=t_n-\frac{t_n\ln(t_n)+t_n^2}{1+t_n}$

I found that $\alpha \approx 0.567143$. But, I don't consider this a nice solution as I haven't found the exact value of $\alpha$. Might there be a better way that doesn't use numerical methods?

Parcly Taxel
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    https://en.wikipedia.org/wiki/Omega_constant –  Sep 05 '16 at 09:59
  • https://en.wikipedia.org/wiki/Lambert_W_function – Vincent Sep 05 '16 at 10:00
  • See http://math.stackexchange.com/questions/54210/is-xx-y-solvable-for-x as well. – user357980 Sep 05 '16 at 10:09
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    You may have spotted that not only $x=e^\alpha$ but also $x=\dfrac1{\alpha}$, so $x \approx 1.76322$ – Henry Sep 05 '16 at 10:10
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    You can't solve that without numerical methods, there is no closed-form formula. Of course there's the Lambert function, but saying $x=\exp(W(1))$ is not very different from "$x$ is the solution of $x\ln x=1$". For evaluation, you will need numerical methods anyway. –  Sep 05 '16 at 10:27

2 Answers2

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The exact solution requires the use of the Lambert W function. $$x\log x=1$$ $$(\log x)e^{\log x}=1$$ Now use the relation of the Lambert W function, $W(z)e^{W(z)}=z$: $$\log x=W(1)$$ $$x=e^{W(1)}=\frac1{W(1)}$$ $W(1)$ is sometimes called the omega constant, denoted as $\Omega$. This happens to be equivalent to the $\alpha$ you found with Newton's method.

Parcly Taxel
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With e.g. the Lambert-W-Function you get $\displaystyle \alpha=\frac{1}{W(1)}$ with $\displaystyle W(1)=\sum\limits_{n=1}^\infty \frac{(-n)^{n-1}}{n!}$.

But the numerical method is easier:

You have $z^\frac{1}{z}=\frac{1}{e}$ and therefore $x=\frac{1}{z}$.

$z=\lim\limits_{n\to\infty}z_n$ for $z_{n+1}=(\frac{1}{e})^{z_n}$ with $z_0:=1$.

user90369
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