find all positive solutions of $\frac{\ln^2(x^2)}{x^2} = \pi^2$

Question

This is a riddle I have to solve. What is surprising is the "all positive solutions" . Obviously this equation involves the Lambert function, but "positive' is meaningless for complex numbers... I have currently 4 solutions, 0.4745409994, -0.4745409994, i and -i, but one single positive (real) solution. However, the riddle solution must contain more that one value.

The problem is stated exactly like in the title of my question, not $\frac{\ln(x^2)}{x^2}^2 = \pi^2$

Could this be the trick ? $\ln(\ln(x)) = \ln^2(x)$ ?

Answer 1

The positive real solution you got seems to be unique.

We can evaluate: $$\ln^2(x^2)=(x\pi)^2\to \ln(x^2)=x\pi\to x^2=e^{\pi x}\to e^{x\pi}-x^2=0$$

Yes, I ignored negative roots here, but it doesn't matter here because the real roots will have the same absolute value.

What matters here is that $f(x)=e^{x\pi}-x^2\implies f'(x)=\pi e^{\pi x}-2x>0\forall x\in\Bbb R^+$ (and you can check the same for $e^{-\pi x}-x^2$ and $e^{\pi x}-x^{-2}$) so the function is monotone and will only have one solution.

Answer 2

We can use the Lambert W function to solve for $x$. $$\begin{align*} \left(\dfrac{\ln (x)}{x}\right)^2=\pi^2\end{align*}$$ \begin{array}{lll}% &\ln(x) = x\pi \quad & \ln(x) = -x\pi \\ \implies& x=\exp(-W(-\pi)) \qquad \implies &= \exp(-W(\pi)) \end{array}

Hence the two solutions are $\exp(-W(-\pi))\approx-0.159-0.585i$ and $\exp(-W(\pi))\approx 0.342$

Answer 3

Without Lambert function.

Consider that you look for the zero's of function $$f(x)=\log ^2\left(x^2\right)-\pi ^2 x^2$$ Let $x^2=t$ to make it $$g(t)=\log ^2\left(t\right)-\pi ^2 t\implies g'(t)=\frac{2 \log (t)}{t}-\pi ^2$$ Since $\forall t \,\,\, \log(t)<t$, then $ g'(t)< 0 \,\,\,\forall t$. If the is a root, it is unique and since $g(t)$ veries from $+\infty$ to $-\infty$, the root does exist.

Answer 4

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

Assuming that there is a typo, this is a minimal correction, with which the phrase "all positive solutions" is justified:

\begin{align} \tfrac1{x^2}\,\ln(x^2)^2 &= \tfrac1{\pi^2} ,\\ \tfrac1x\,\ln(x^2) &= \pm\tfrac1{\pi} ,\\ \tfrac1x\,\ln(x) &=\pm\tfrac1{2\,\pi} ,\\ -\tfrac1x\,\ln(x) &= \pm\tfrac1{2\,\pi} ,\\ \tfrac1x\,\ln(\tfrac1x) &= \pm\tfrac1{2\,\pi} ,\\ \ln(\tfrac1x)&=\W(\pm\tfrac1{2\,\pi}) ,\\ \tfrac1x&=\exp( \W(\pm\tfrac1{2\,\pi})) ,\\ x&=\exp(-\W(\pm\tfrac1{2\,\pi})) . \end{align}

And there are three distinct real solutions:

\begin{align} x&=\exp(-\W(\tfrac1{2\,\pi})) \approx 0.8706 ,\\ x&=\exp(-\Wp(-\tfrac1{2\,\pi})) \approx 1.2129 ,\\ x&=\exp(-\Wm(-\tfrac1{2\,\pi})) \approx 18.2460 . \end{align}

$\endgroup$