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I have been trying to solve the following goat grazing problem:

A goat is tied to the outside of a circular fence. If the length of the rope is the same as the circumference of the fence, what is the maximum area upon which the goat can feed?

I'm using integration to calculate the area but to find the limits for the integral, I need to solve the following transcendental equation: $$(2 \pi - \theta)\cos \theta + \sin \theta = 0$$

Is it possible to compute the exact solution of this equation or do I just have to solve it numerically?

russell.price
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    $2\pi$ is a solution. I think all others can only be computed numerically – Andrei Jun 16 '23 at 19:59
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    Related goat problem post with contour integral solution. Also, $2\pi-\theta=x$ gives $\tan(x)=x$ for which there is a BesselJZero Mathematica and a series solution – Тyma Gaidash Jun 16 '23 at 20:23
  • @Andrei I'm aware of $2\pi$ being a solution. I'm interested in the other solution that lies in the range $[0, 2\pi]$. I suspected that I might have to compute it numerically. – russell.price Jun 16 '23 at 22:23
  • @TymaGaidash Learning about Ingo Ullisch's result which you link to is how I found the problem I mentioned in my question. Apparently, it's the first known goat grazing problem that has been published. – russell.price Jun 16 '23 at 22:26

2 Answers2

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To solve the equation for $\theta=1.7897758\dots$, use $2\pi-\theta=x$:

$$(2\pi-\theta)\cos(\theta)+\sin(\theta)=0\iff \tan(x)=x$$

Now use the series solution in

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

Bessel J zero $j_{v,x}$, and Stirling S1 $S_n^{(m)}$ to get:

$$\bbox[5px,border: 5px groove black]{\theta=\frac\pi2-\sum_{m=1}^\infty\sum_{n=0}^m\frac{S_m^{(n)}\Gamma(n+m-1)(-1)^ni^{m+n}2^{m-n}}{\left(\frac{3\pi}2-i\right)^{n+m-1} m!\Gamma(m)}=\pi-\tan^{-1}(j_{\frac32,1})}$$

shown here is the series and the closed form also works

Тyma Gaidash
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If, as @Tyma Gaidash did, you let $2\pi-\theta=x$, you nedd to find the zero's of function $$f(x)=x-\tan(x)$$ whic is better to write as $$g(x)=x\cos(x)-\sin(x)$$ in order to remove the discontinuities.

Because of the multiplying $x$, the solutions are closer and closer to $(2n+1)\frac \pi 2$.

Expanding as series around this value an simplifying the trigonometric functions since $n$ is an integer. This gives $$g(x)=\sum_{k=0}^\infty \frac{(-1)^n \left(2 (k-1) \cos \left(\frac{\pi k}{2}\right)-\pi (2 n+1) \sin \left(\frac{\pi k}{2}\right)\right)}{2 \Gamma (k+1)}\, \left(x- (2 n+1)\frac{\pi}{2} \right)^k$$

Truncate to some order and use power series reversion to obtain $$x_{(n)}=q-\frac{1}{q}-\frac{2}{3 q^3}-\frac{13}{15 q^5}-\frac{146}{105q^7}-\frac{781}{315q^9}+O\left(\frac{1}{q^{11}}\right)$$ where $q=(2 n+1)\frac{\pi}{2}$.

For the first root, the above truncated series would give $x_{(1)}=\color{red}{4.493409}66$ while the "exact" solution is $4.49340946$.

Claude Leibovici
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