# Rules of inverse function?

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.

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

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 $$q=1$$
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 $$4.49341$$.