# Rearrange $y = x\tan y$ to solve for $y$ given $x$

I have an equation: $$x = \frac{y}{\tan y}$$ or rewritten as: $$y = x\tan y$$

How can I rearrange this so that I can calculate $y$ given $x$?

Doing so would involve calculating the inverse of the tanc function,

$$y = \operatorname{tanc}^{-1}\frac{1}{x}$$

There is no expression for this function in terms of other elementary functions, so you can't find an analytic solution. You could find the solutions of your equation numerically, though, once you're given an explicit numerical value for $x$.

• For those who are interested, a lot of information about the situation when $x = 1$ (i.e. numerical solutions to the equation $\tan y = y$) is linked at the following URL, including a .pdf file of the LaTeX slides I used for an Iowa MAA Section talk I gave on this equation (8 April 2006). mathforum.org/kb/message.jspa?messageID=7014308 Commented Jan 11, 2012 at 15:37

$$\def\tanc{\operatorname{tanc}}$$ Using Lagrange reversion

$$\tanc(z)=a\iff z=\left(k+\frac12\right)\pi+\frac i2\ln\left(\frac{2 i}{az-i}+1\right)= \left(k+\frac12\right)\pi+\sum_{n=1}^\infty\left(\frac i2\right)^n\left.\frac{d^{n-1}}{dz^{n-1}}\frac{\ln^n\left(\frac{2 i}{az-i}+1\right)}{n!}\right|_{\left(k+\frac12\right)\pi}$$

Applying a Stirling S1 series:

$$\frac{d^{n-1}}{dz^{n-1}}\frac{\ln^n\left(\frac{2 i}{az-i}+1\right)}{n!}=\sum_{m=0}^\infty \frac{(2i)^{n+m}S_{n+m}^{(n)}}{(n+m)!}\frac{d^{n-1}(a z-i)^{-(n+m)}}{dz^{n-1}}\mathop=^{\left|\frac{2i}{az-i}\right|<1}\sum_{m=0}^\infty\frac{S_{n+m}^{(n)}\Gamma(2n+m-1)(-1)^{n+1}a^{n-1}(2 i)^{n+m}}{(m+n)!\Gamma(m+n)(az-i)^{2n+m-1}}$$

and finally use $$\sum\limits_{n=1}^\infty \sum \limits_{m=n}^\infty a_{m,n}=\sum_ \limits{m=1}^\infty\sum \limits_{n=0}^ma_{m,n}$$ and the Gauss hypergeometric function:

\bbox[3px,border: 1px solid blue]{\begin{align}\tanc_k^{-1}(z)\mathop=^{\left|\frac1{\pi i k z-1}+1\right|<\frac12}\pi k+\sum_{m=1}^\infty\sum_{n=0}^m\frac{S_m^{(n)}\Gamma(n+m-1)i^n 2^{m-n} (\pi k)^m}{\left(\frac i z-\pi k\right)^{n+m-1}\Gamma(m)m!}\,_2\operatorname F_1\left(1-n,-m;2-m-n;1-\frac i{\pi k z}\right)\mathop=^{2<\left|\left(k+\frac12\right)\pi z-i\right|}\left(k+\frac12\right)\pi-\sum_{m=1}^\infty\sum_{n=0}^m\frac{S_m^{(n)}\Gamma(n+m-1)(-1)^ni^{m+n}2^{m-n}z^{n-1}}{\left(z\left(k+\frac12\right)\pi-i\right)^{n+m-1} m!\Gamma(m)}\end{align}}

Both cases are shown here and, if infinite, sums are interchangeable. The other series was derived similarly from a slightly modified Lagrange reversion setup. The $$k=0$$ branch may have convergence problems and the sum gradually converges more slowly as $$k\to\infty$$ for the simpler series.

• This also solves $\tan(x)=x$ Commented May 28, 2023 at 15:45