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Given $$\sqrt{\frac{B (-1 + q)}2}\cot{\sqrt{\frac{B (-1 + q)}2}}-1=0$$

what would be the solution for $q$, where $B$ is a parameter.

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  • $\begingroup$ This is an expression, not an equation. Put an equals sign somewhere then we can try and make q the subject of the equation. $\endgroup$ Commented Aug 27, 2019 at 14:41
  • $\begingroup$ @battletwink69 sorry, that was my mistake when I've edited the original post. Fixed it. $\endgroup$
    – Andrei
    Commented Aug 27, 2019 at 14:43

2 Answers 2

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You can write the equation as $y\cot y=1$. One of the solutions is $y=0$. The rest can only be found numerically. For $y=0$ you get $q=1$

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As Andrei answered, you want to find the zeros of $$f(y)=y\cot(y)-1$$ and you need numerical methods for that. If you plan to do it, discarding the trivial $y=0$, I would suggest to consider instead $$g(y)=y\cos(y)-\sin(y)$$ which does not show any discontinuity.

We can generate rather good approximations taking into account the fact the $n^{th}$ solution will be closer and closer to $(2n+1)\frac \pi 2$. Building the Taylor series around this point, this will give $$g(y)=-1-\frac{1}{2} (\pi (2 n+1)) \left(y-\pi \left(n+\frac{1}{2}\right)\right)-\frac{1}{2} \left(y-\pi \left(n+\frac{1}{2}\right)\right)^2+\frac{1}{12} \pi (2 n+1) \left(y-\pi \left(n+\frac{1}{2}\right)\right)^3+\frac{1}{8} \left(y-\pi \left(n+\frac{1}{2}\right)\right)^4+O\left(\left(y-\pi \left(n+\frac{1}{2}\right)\right)^{4}\right)$$

Now, using series reversion, $$\color{blue}{y_{n}=q-\frac{1}{q}-\frac{2}{3 q^3}-\frac{19}{24 q^5}-\frac{5}{8 q^7}+O\left(\frac{1}{q^9}\right)}\qquad \text{with}\qquad\color{blue}{q=(2n+1)\frac \pi 2}$$ which will be better and better when $n$ will increase.

Considering the first root, the above would give $y_1=4.49346$ while the numerical solution, obtained using Newton method, would be $4.49341$.

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