5
$\begingroup$

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.

Edit

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 $\pm\sqrt{\frac{60-5 \pi ^2}{20-2 \pi ^2}}\approx \pm 6.39100$.

$\endgroup$
1
  • $\begingroup$ 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

0
$\begingroup$

$$\Phi(a,b)=\int_0^{\frac \pi 2}\Big[(2\pi-x)(2\pi+x)\left(\frac{\pi }{2}-x\right) \left(x+\frac{\pi }{2}\right) x\tan (x)-(ax^2+bx^4) \Big]^2\,dx$$ 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, $\Phi_{\text{opt}}(a,b)=9.182\times 10^{-6}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.