how to solve this equation $z=yb\cot(bx/2)$?

Question

how to solve this equation $z=yb\cot(bx/2)$? $b$ is unknown, the $x,y,z$ are known numbers, and $x\ne0,b>0$, we want to have the solution for $b$. Until now I have no idea about this.

Answer 1

You have to solve $$f(b)=y~ b\ cot\Big(\frac {b~x}{2}\Big)-z=0$$ where $x,y,z$ are given constants. I suppose you noticed that $f(b)=f(-b)$; so we can just focus on positive solutions. You also know that there are discontinuities if $b x = 2 n \pi$; so we shall just try for the first solution, assuming there is one between $b=0$ and $b=\frac{2 \pi}{x}$.

A simple and classical procedure for solving nonlinear equations $f(b)=0$ is Newton, provided a "reasonable" starting value $b_0$. From this point, Newton scheme is $$b_{n+1}=b_n-\frac{f(b_n)}{f'(b_n)}$$ For your problem, after some simplifications, $$f'(b)=\frac{y (b x-\sin (b x))}{\cos (b x)-1}$$ Fot he case you gave ($x=y=z=1$), we can notice (by simple inspection) that $f(0)=1$ and $f(\pi)=-1$. So, there is a root between $0$ and $\pi$ and we shall start with $b_0=\frac{\pi}{2}$. The successive iterates of Newton method are then : $b_1=2.57080$, $b_2=2.34796$, $b_3=2.33121$,$b_4=2.33112$ which is the solution for six significant figures.