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 ?
from Hot Weekly Questions - Mathematics Stack Exchange
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