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I'm trying to find the roots $\omega_n$ of, \begin{equation} \tan(\omega) = \frac{2b\omega}{b^2\omega^2 -1} \end{equation}

I know it must be done numerically, so I would use the bisection method. In order to use that I must know the interval to search for each root for any $b \in (0,1)$. The problem is that there is different behaviour of the singularities at $b>1/2$. Does anyone know a way to solve the roots of this?

Josh Albert
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    Have you plotted the functions? Perhaps you can gain insights as to their respective characteristics. Have you heard of Newton's method or the secant method? – Mark Viola Feb 11 '17 at 16:41
  • Yes, I plotted and it can be seen here: https://postimg.org/image/f85kkxe8n/ The dashed lines are $b \in (1/2,1), solids are $b \in (0,1/2]$. Note where the singularities lie for them. – Josh Albert Feb 12 '17 at 00:11
  • Which roots are you interested in? There are infinitely-many real roots. Could you also clarify what you mean by "there is different behavior of the singularities at $b > 1/2$"? I don't notice a difference between $b<1/2$ and $b>1/2$ in the plot you linked. – Antonio Vargas Feb 12 '17 at 10:42
  • I would like to be able to find all up to an arbitrary $n$. The plot shows the first four roots. I'll clarify, @antonio-vargas. Except for the first three roots ($n=0,1,2$), the dashed lines ($b>1/2$) begin and end at $\pi/2 + n\pi$, sequential singularities of $\tan\omega$. For the solid ($b<1/2$) some lines begin and end at tan singularities and others asymptote to other intermediate points. Knowing the interval to search for a particular root is what I need for applying the bisection method. – Josh Albert Feb 13 '17 at 00:54

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It would be simpler to look for the zeros of function $$f(\omega)=\left(b^2 \omega ^2-1\right) \sin (\omega )-2 b \,\omega\, \cos (\omega )$$ which does not show any discontinuity.

Focusing on the first postive root, series expansion followed by power series reversion gives, as an approximation, $$\omega=t+\frac {20 b^2+10 b+1}{40(6 b^2+6 b+1 ) } t^3+$$ $$\frac {48720 b^4+45360 b^3+15300 b^2+2140 b+107}{67200(6 b^2+6 b+1 )^2}t^5+O(t^7)$$where $$t=\sqrt{\frac{6 (2 b+1)}{6 b^2+6 b+1} }$$

As shown below, the approximation is decent (it could be improved at the price of more terms in the initial series expansion). $$\left( \begin{array}{ccc} b & \text{estimate} & \text{solution} \\ 0.05 & 2.69706 & 2.85774 \\ 0.10 & 2.49284 & 2.62768 \\ 0.15 & 2.32792 & 2.43990 \\ 0.20 & 2.19140 & 2.28445 \\ 0.25 & 2.07603 & 2.15375 \\ 0.30 & 1.97688 & 2.04223 \\ 0.35 & 1.89049 & 1.94582 \\ 0.40 & 1.81433 & 1.86151 \\ 0.45 & 1.74654 & 1.78704 \\ 0.50 & 1.68568 & 1.72067 \\ 0.55 & 1.63066 & 1.66106 \\ 0.60 & 1.58059 & 1.60715 \\ 0.65 & 1.53479 & 1.55811 \\ 0.70 & 1.49267 & 1.51325 \\ 0.75 & 1.45378 & 1.47201 \\ 0.80 & 1.41772 & 1.43394 \\ 0.85 & 1.38417 & 1.39866 \\ 0.90 & 1.35285 & 1.36584 \\ 0.95 & 1.32353 & 1.33521 \\ 1.00 & 1.29600 & 1.30654 \\ \end{array} \right)$$

In any manner, with these estimates, Newton method converges in a couple of iterattions.

Claude Leibovici
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