Thinking more about this question where I proposed some approximate solution of the first positive root of equation $\color{blue}{x\tan(x)=k}$ for any $k >0$, I notice that, for the $[3,4]$ Padé approximant built at $x=0$ $$\tan(x)=\frac{5 x \left(21-2 x^2\right)}{105-45 x^2+x^4} $$ the denominator cancels at $$x=\pm\sqrt{\frac{1}{2} \left(45-\sqrt{1605}\right)}\approx 1.57123\qquad x=\pm\sqrt{\frac{1}{2} \left(45+\sqrt{1605}\right)}\approx 6.52160$$ that is to say close to $\frac \pi 2$ (which seems "normal") and "rather close" to $2\pi$ (more surprising). This is the only case for all explored $[2n-1,2n]$ Padé approximants.

This gave me the idea of working the series expansion $$(2\pi-x)(2\pi+x)\left(\frac{\pi }{2}-x\right) \left(x+\frac{\pi }{2}\right) x\tan (x)=\pi ^4 x^2+\left(\frac{\pi ^4}{3}-\frac{17 \pi ^2}{4}\right) x^4+O\left(x^6\right)\tag 1$$ reducing the original problem to $$\frac{\pi^2 x^2\left(\left(\frac{\pi ^2}{3}-\frac{17}{4}\right) x^2 +\pi^2\right) }{x^4-\frac{17 \pi ^2 }{4}x^2+\pi ^4 }=k$$ which is just a quadratic equation in $x^2$ the retained solution of which being $$x= \pi\sqrt{\frac{51 k+12 \pi ^2-\sqrt{2025 k^2+\pi ^2 \left(192 \pi ^2-1224\right) k+144 \pi ^4}}{2 \left(12 k+\pi ^2 \left(51-4 \pi ^2\right)\right)}}$$ which seems to lead to quite good approximations (in the table below $k=10^n$) $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ -3 & 0.03161750711 & 0.03161750711 \\ -2 & 0.09983363885 & 0.09983363855 \\ -1 & 0.31105293142 & 0.31105284820 \\ 0 & 0.86034131667 & 0.86033358902 \\ 1 & 1.42887264708 & 1.42887001121 \\ 2 & 1.55524418982 & 1.55524512931 \\ 3 & 1.56922698357 & 1.56922710099 \\ 4 & 1.57063925088 & 1.57063926287 \end{array} \right)$$ For large values of $k$, the asymptotics is $$x=\frac{\pi }{2}-\frac{ \pi ^3(4 \pi ^2-3)}{720\,k}+\frac{\pi ^5 \left(4 \pi ^2-3\right) \left(436 \pi ^2-3207\right)}{7776000\, k^2}+O\left(\frac{1}{k^3}\right)$$ where we can notice that each coefficient is again very close to $\pm\frac \pi 2$

Is there a way to justify the appearance of this value close to $\pm 2\pi$ or is it just an happy coincidence ?

By the way, it is possible to improve the coefficients appearing in $(1)$ minimizing $$\Phi(a,b)=\int_0^{\frac \pi 2}\left((2\pi-x)(2\pi+x)\left(\frac{\pi }{2}-x\right) \left(x+\frac{\pi }{2}\right) x\tan (x)-(ax^2+bx^4) \right)^2\,dx$$ and get analytical formulae for $a,b$ in terms of $\zeta (p)$ $(p=3,5,7,9)$ and powers of $\pi$. Compared to the initial values, the value of $\Phi(a,b)$ is reduced by a factor of $2.22$. The expressions for the optimal $a,b$ are not given here but they are available to any one who would like to get them.


About he "coincidence", consider the $[4,2]$ Padé approximant $$\left(\frac{\pi }{2}-x\right) \left(x+\frac{\pi }{2}\right) x\tan (x)=\frac{\frac{\pi ^2}{4}x^2+\frac{\left(720-60 \pi ^2-\pi ^4\right) }{60 \left(\pi ^2-12\right)}x^4 } {1-\frac{2 \left(\pi ^2-10\right) }{5 \left(\pi ^2-12\right)}x^2 }$$ and the roots of the denominator are $\pm\sqrt{\frac{60-5 \pi ^2}{20-2 \pi ^2}}\approx \pm 6.39100$.

  • $\begingroup$ A great approximation, especially considering the agreement with the exact values for such a large range of parameters. No idea why this works though. $\endgroup$
    – Yuriy S
    Oct 8, 2018 at 21:44

1 Answer 1


$$\Phi(a,b)=\int_0^{\frac \pi 2}\Big[(2\pi-x)(2\pi+x)\left(\frac{\pi }{2}-x\right) \left(x+\frac{\pi }{2}\right) x\tan (x)-(ax^2+bx^4) \Big]^2\,dx$$ is minimal for $$a=-\frac{35}{4 \pi ^6}\Big[371 \pi ^6 \zeta (3)-18906 \pi ^4 \zeta (5)+65520 \pi ^2 \zeta (7)+816480 \zeta (9) \Big]$$ $$b=\frac{63}{ \pi ^8}\Big[311 \pi ^6 \zeta (3)-14838 \pi ^4 \zeta (5)+50400 \pi ^2 \zeta (7)+635040 \zeta (9) \Big]$$ and, numerically, $\Phi_{\text{opt}}(a,b)=9.182\times 10^{-6}$.


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