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Looking a good approximation of the $n^{th}$ positive root of the equation $$\color{blue}{\tan(x)=k x}$$ As already done many times, I expanded as Taylor series around $x=(2n+1)\frac \pi 2$ and used series reversion.

As a result, this write $$\color{blue}{x=q-\sum _{m=1}^{\infty }\frac{P_{m}(k) }{a_m } \frac 1 {(kq)^{2 m-1}}}\qquad \text{where}\qquad \color{blue}{q=(2n+1)\frac \pi 2}$$

Concerning the $a_m$, the first terms are $\{1,3,15,105,315,3465\}$; this is sequence $A088989$ in $OEIS$ and they nicely correspond to the denominators of the coefficients of odd powers of $\frac 1q$ in the solution series of the equation for $k=1$.

Concerning the polynomials $P_{m}(k)$, the first are listed below $$\left( \begin{array}{cc} m & P_{m}(k) \\ 1 & 1 \\ 2 & 3 k-1 \\ 3 & 30 k^2-20 k+3 \\ 4 & 525 k^3-525 k^2+161 k-15 \\ 5 & 4410 k^4-5880 k^3+2744 k^2-528 k+35 \\ 6 & 145530 k^5-242550 k^4+152460 k^3-44990 k^2+6193 k-315 \end{array} \right)$$

Even truncated to six terms, this provides quite good estimates even for the first solution $$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 0.5 & 4.27478222835695 & 4.27478227145813 \\ 1.0 & 4.49340947613302 & 4.49340945790907 \\ 1.5 & 4.56745216948297 & 4.56745216641733 \\ 2.0 & 4.60421677783123 & 4.60421677720058 \\ 2.5 & 4.62613829114934 & 4.62613829098161 \\ 3.0 & 4.64068363082997 & 4.64068363077555 \\ 3.5 & 4.65103583026074 & 4.65103583024022 \\ 4.0 & 4.65877826296638 & 4.65877826295768 \\ 4.5 & 4.66478672978692 & 4.66478672978287 \\ 5.0 & 4.66958478090799 & 4.66958478090596 \end{array} \right)$$

Could the above polynomials be identified ?

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  • $\begingroup$ @metamorphy. Thanks for pointing ! I shall edit. Cheers. $\endgroup$ – Claude Leibovici Sep 1 at 7:23
  • $\begingroup$ For those who could be interested, I have been able to generate the polynomials up to $P_8$. $\endgroup$ – Claude Leibovici Sep 2 at 8:15
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If instead of using $a_m = \mathrm{lcm} \{ 1, 3, \dotsc, 2m-1 \}$ you use $b_m = (2m-1)!!$, you can write $$x=q-\sum _{m=1}^{\infty }\frac{Q_{m}(k) }{b_m } \frac 1 {(kq)^{2 m-1}}$$ and then $$Q_{m+1}(k) = \sum_{i=0}^m (-1)^{m-i} C(m, i) k^i$$ where the coefficients $C(m, i)$ satisfy $C(0, i) = (2i-1)!!$ and $$C(m, i) = \frac {(2m-1)2m(2m+1)} {(2m-i)(2m-i+1)} [C(m-1, i-1) + C(m-1, i)]$$

For a reference, see L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 170.

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  • $\begingroup$ Thank you very much ! I did not know that this problem was already solved. I recovered the pdf of the book and I am working on. Cheers :-) $\endgroup$ – Claude Leibovici Sep 1 at 8:57

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