I would like to solve the equation: $$x = \cot(x/2).$$ So far, I was able to find only an approximate solution by Taylor expanding $\cot(x/2)$ up to a 3rd term and solving for $x$. But I was hoping there is a clever way of finding $x$.

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    $\begingroup$ What is the motivation for thinking that you can find the exact form for the roots? $\endgroup$
    – A. P.
    Nov 5, 2022 at 3:53
  • $\begingroup$ Well, I'm sure given your writing: " But I was hoping there is a clever way of finding $x$" . So my first question makes perfect sense. There is an extremely big difference between "finding $x$" and "finding an approximate value of $x$". However, I will stop commenting in this space as you seem to have suggested. $\endgroup$
    – A. P.
    Nov 5, 2022 at 20:49

1 Answer 1


If you look for the zeros of function $$f(x)=x-\cot \left(\frac{x}{2}\right)$$ because of the discontinuities, it is better to consider $$g(x)=x \sin \left(\frac{x}{2}\right)-\cos \left(\frac{x}{2}\right)$$

For the first root, use Taylor series $$g(x)=\sum_{n=0 }^\infty(-1)^{n+1}\,\frac{(4 n+1) }{(2 n)!}\,\left(\frac{x}{2}\right)^{2n}$$ which is just a polynomial in $x^2$.

If you do not require a very precise result, using $$g(x)=-1+\frac{5 x^2}{8}-\frac{3 x^4}{128}+O\left(x^6\right)$$ gives $$x=2 \sqrt{\frac{2}{3} \left(5-\sqrt{19}\right)}=1.30752$$ while the solution, as given by Newton method, is $1.30654$.

Adding one more term and solving the cubic in $x^2$ (more tedious) would give $x=1.30653$.

For the other roots, consider that, for $x>0$ $$x \sin \left(\frac{x}{2}\right)-1 \leq g(x) \leq x \sin \left(\frac{x}{2}\right)+1$$ So, in your opinion, where will be located the next roots ?

Show some working and I could explain how you could get almost the exact solutions.


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