<s>[INST] Approximate solution to a transcendental equation I'm working on a physics problem and stumbled upon the following equation:
$$h=\frac{2n\pi+\arctan\left(\frac{c}{b}\right)}{b}$$
where $n \in \mathbb{Z}$, $c \in [0,20]$ and $h \in \mathbb{R}^+$. This equation has to be solved for $b$, which can be done numerically ofcourse, but if possible, I would love to have a quite accurate expression in the given interval. Ive already did some work and obtained the following fair approximation for $b$:
$$b\approx\frac{(1+2n)\pi c}{2+ch}$$ However, this equation breaks down at low $c$. Thus, I was wondering if more accurate (analytical) expressions are possible.
 [/INST] Let $\frac c b=x$ to make the equation
$$2 n\pi   x+x \tan ^{-1}(x)=k \qquad \text{with} \qquad k=ch$$ Expand the lhs as a Taylor series around $x=0$
$$2 n\pi   x+x \tan ^{-1}(x)=2 n\pi   x+x^2-\frac{x^4}{3}+\frac{x^6}{5}+O\left(x^8\right)$$ Use series reversion
$$x=\frac{k}{2 \pi  n}-\frac{k^2}{8 \pi ^3 n^3}+\frac{k^3}{16 \pi ^5
   n^5}+\frac{k^4 \left(4 \pi ^2 n^2-15\right)}{384 \pi ^7 n^7}+\frac{k^5 \left(7-4
   \pi ^2 n^2\right)}{256 \pi ^9 n^9}+O\left(k^6\right)$$
In view of this results, we can also perform one single iteration iteration of Newton method
$$x_0=\frac k {2 n\pi}\implies x_1=\frac k {2 n\pi}-\frac{k \cot ^{-1}\left(\frac{2 \pi  n}{k}\right)}{2 \pi  n \left(2 \pi  n
   \left(\frac{k}{k^2+4 \pi ^2 n^2}+1\right)+\cot ^{-1}\left(\frac{2 \pi 
   n}{k}\right)\right)}$$
Edit (after @River Li's answer)
We can expand the function
$$f(b)=b-\frac{\tan ^{-1}\left(\frac{c}{b}\right)+2 \pi  n}{h}$$ as series solution around $b=\frac {2n\pi}h$,  use series reversion and have, as an approximation
$$b=\frac {2n\pi}h+t+\sum_{k=2}^p a_k \,t^k+ O(t^{p+1})$$ where $$t=\frac{c^2 h^2+4 \pi ^2 n^2}{h \left(c^2 h^2+c h+4 \pi ^2 n^2\right)}\,\tan ^{-1}\left(\frac{c h}{2 \pi  n}\right)$$ The first coefficient is
$$a_2=\frac{2 \pi  c h^2 n}{\left(c^2 h^2+4 \pi ^2 n^2\right) \left(c^2 h^2+c h+4 \pi ^2   n^2\right)}$$
Repeating @River Li's test cases, the second term is perfectly negligible.
What we can also do is one iteration of Halley method using $b_0=\frac {2n\pi}h$ and have as estimate
$$b=\frac {2n\pi}h+\frac{\left(c^2 h^2+4 \pi ^2 n^2\right) \left(c h (c h+1)+4 \pi ^2 n^2\right) \cot
   ^{-1}\left(\frac{2 \pi  n}{c h}\right)}{h \left(\left(c h (c h+1)+4 \pi ^2
   n^2\right)^2-2 \pi  c h n \cot ^{-1}\left(\frac{2 \pi  n}{c h}\right)\right)}$$
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