Bound on angle implicitly defined via trigonometric equation

Question

Introduction / setup: In the article Spectral Gap for a Class of Random Billiards by Renato Feres and Hong-Kun Zhang (link: https://www.math.wustl.edu/~feres/spectralgapnew.pdf) a trigonometric equation is established (in proposition 8, equation (3.2)) between two angles $\theta$ and $\Theta = \Theta_{\theta,1}$, namely \begin{equation} K\sin\Theta = 1 - \cos\left( \frac{\theta + \Theta}{2} \right). \end{equation} Here $K \geq 2$ is the curvature / inverse of the radius of the circles $\Gamma_1$ and $\Gamma_2$, as seen in the following figure. In proposition 9 (inequality (3.7), here you have to note that they define $\sigma(\theta) := \frac{\theta^2}{8K}$) they go on to say that from this equation we can deduce that if $0 \leq \theta \leq \sin^{-1}\left(\frac{1}{K}\right) := \theta_{0}$, then $$\Theta \leq (1+\theta)\frac{\theta^2}{8K}.$$ Here it is also useful to note that we always have $0 \leq \Theta \leq \theta$.

My question: How can I prove that this bound indeed holds?

What I have tried / what may be useful to know: I do not believe that it is important how they established this equation and how the dynamics works (it comes from a billiard dynamics system). What is useful to know is the above and that they say one can derive this inequality by using estimates from Taylor approximations on the above trigonometric equation. You should for example think of: if $\theta \in \left[0,\frac{\pi}{2}\right]$, then $\sin(\theta) \leq \theta$ and $\sin(\theta) \geq \theta - \frac{\theta^3}{6}$, etc (see for example their proof of the lower bound in the inequality (3.7) from proposition 9). I have tried to do this along with trying some other trigonometric identities, but I just cannot get the right expression.

Many thanks in advance.

Answer 1

This is not an answer

I do not clearly understand the steps leading to $$\Theta \leq (1+\theta)\frac{\theta^2}{8K}$$ but I think that other could be proposed for lower bounds and approximate solutions.

Consider that we look for the zero of function

$$\begin{equation} f(\Theta)=K\sin(\Theta) - 1 + \cos\left( \frac{\theta + \Theta}{2} \right). \end{equation}$$ that we shall solve using Newton method starting at $\Theta_0=0$.

Whe have $$f(0)=\cos \left(\frac{\theta }{2}\right)-1 \qquad f'(0)=K-\frac{1}{2} \sin \left(\frac{\theta }{2}\right)\qquad f''(0)=\frac{1}{8} \sin \left(\frac{\theta }{2}\right)-K$$ So, the first iterate of Newton method is $$\Theta_{(2)}=\frac{2 \left(1-\cos \left(\frac{\theta }{2}\right)\right)}{2 K-\sin \left(\frac{\theta }{2}\right)}=\frac{4 \sin ^2\left(\frac{\theta }{4}\right)}{2 K-\sin \left(\frac{\theta}{2}\right)}$$

Now, the key point is that $$f(0)\,f''(0)=\left(1-\cos \left(\frac{\theta }{2}\right)\right) \left(K-\frac{1}{8} \sin\left(\frac{\theta }{2}\right)\right)$$ is always positive and since $f''(0)<0$, by Darboux theorem we have $$\Theta >\Theta_{(2)}$$

Doing the same using Halley method, we can generate a still better lower bound $$\Theta_{(3)}=\frac{8 \left(1-\cos \left(\frac{\theta }{2}\right)\right) \left(2 K-\sin \left(\frac{\theta }{2}\right)\right)}{16 K^2-16 K \sin \left(\frac{\theta }{2}\right)+3-\left(2 \cos \left(\frac{\theta }{2}\right)+\cos (\theta )\right)}$$ and $$\Theta>\Theta_{(3)}$$

A few examples $$\left( \begin{array}{cccccc} K & \theta & \Theta_{(2)}& \Theta_{(3)} & \text{solution} & (1+\theta)\frac{\theta^2}{8K} \\ 3 & 0.25 & 0.00265597 & 0.00265626 & 0.00265597 & 0.00325521 \\ 3 & 0.50 & 0.01080819 & 0.01081311 & 0.01081333 & 0.01562500 \\ 3 & 0.75 & 0.02467012 & 0.02469528 & 0.02469795 & 0.04101563 \\ 3 & 1.00 & 0.04434953 & 0.04442783 & 0.04444361 & 0.08333333 \\ 3 & 1.25 & 0.06982097 & 0.07000397 & 0.07006672 & 0.14648438 \\ 3 & 1.50 & 0.10089993 & 0.10125131 & 0.10144440 & 0.23437500 \\ & & & & & \\ 4 & 0.25 & 0.001981463 & 0.001981587 & 0.001981463 & 0.00244141 \\ 4 & 0.50 & 0.008019914 & 0.008021925 & 0.008022013 & 0.01171875 \\ 4 & 0.75 & 0.018206670 & 0.018216777 & 0.018217828 & 0.03076172 \\ 4 & 1.00 & 0.032555342 & 0.032586290 & 0.032592377 & 0.06250000 \\ 4 & 1.25 & 0.050988364 & 0.051059548 & 0.051083238 & 0.10986328 \\ 4 & 1.50 & 0.073325468 & 0.073460103 & 0.073531467 & 0.17578125 \end{array} \right)$$

What is sure is that $$f\left(\frac{\theta ^2 (\theta +1)}{8 K}\right)=K \sin \left(\frac{\theta ^2 (\theta +1)}{8 K}\right)+\cos \left(\frac{\theta \left(\theta ^2+\theta +8 K\right)}{16 K}\right)-1$$ is always positive.

Answer 2

I prefer to add a second answer.

In the previous answer, I did not take into account the fact that $$ 0 \leq \theta \leq \sin^{-1}\left(\frac{1}{K}\right)$$ So, here, I shall use $$\theta= \lambda \sin^{-1}\left(\frac{1}{K}\right)\qquad \text{with} \qquad 0 \leq \lambda \leq 1$$ and show that we obtain the solution at the price of a sixth order iterative scheme (the formalae are totally explicit).

$$\left( \begin{array}{cccccc} K & \lambda & \text{estimate} & \text{solution} \\ 2 & 0.2 & 0.0006943476059 & 0.0006943476059 \\ 2 & 0.4 & 0.0028130590587 & 0.0028130590587 \\ 2 & 0.6 & 0.0064090581895 & 0.0064090581890 \\ 2 & 0.8 & 0.0115341038081 & 0.0115341038026 \\ 2 & 1.0 & 0.0182382653140 & 0.0182382653141 \\ & & & & & \\ 3 & 0.2 & 0.0001935610212 & 0.0001935610212 \\ 3 & 0.4 & 0.0007784680401 & 0.0007784680401 \\ 3 & 0.6 & 0.0017608343448 & 0.0017608343448 \\ 3 & 0.8 & 0.0031464504755 & 0.0031464504754 \\ 3 & 1.0 & 0.0049407510147 & 0.0049407510147 \\ & & & & & \\ 4 & 0.2 & 0.0000800579296 & 0.0000800579296 \\ 4 & 0.4 & 0.0003212000055 & 0.0003212000055 \\ 4 & 0.6 & 0.0007248162266 & 0.0007248162266 \\ 4 & 0.8 & 0.0012922098228 & 0.0012922098227 \\ 4 & 1.0 & 0.0020245928060 & 0.0020245928060 \end{array} \right)$$

Expanded as a series for large values of $K$ $$\Theta=\frac{\lambda ^2}{8 K^3}\Bigg[1-\frac{\lambda ^2-12 \lambda -16}{48 K^2}+\frac{\lambda ^4-90 \lambda ^3+370 \lambda ^2+720 \lambda +1024}{5760 K^4}+O\left(\frac{1}{K^6}\right) \Bigg]$$

Answer 3

Another approach

Let $\Theta=4 \tan ^{-1}(x)$, $c=\cos \left(\frac{\theta }{2}\right)$ and $s=\sin \left(\frac{\theta }{2}\right)$ to make the equation

$$\left(1+c\right)x^4+2\left(2K+ s\right)x^3+2x^2-2\left(2K- s\right)x+\left(1-c\right)=0\tag 1$$ which can be solved with (nasty) radicals.

For the given conditions, the discriminant $\Delta$ is always negative and there are only two real roots in $x$.

Now, back to approximations : if, as it look, $\Theta$ is small, $x$ is small too.

Considering that $(1)$ is an expansion to $O(x^5)$, a series reversion gives $$x=\frac{(1-c)}{2 (2 K-s)}+\frac{(1-c)^2}{4 (2 K-s)^3}+\frac{(1-c)^3 }{8 (2 K-s)^5}\left(4 K^2+2-s^2\right)+$$ $$\frac{(1-c)^4 }{32 (2 K-s)^7}\left(4 (c+11) K^2-4 (c+1) sK+\left((c-9) s^2+10\right)\right)$$

For $K=3$ and $\lambda=0.8$, this gives $$\Theta=\color{red}{0.00314645047544}11$$ while the exact solution is $$\Theta= \color{red}{0.0031464504754461}$$

For $K=4$ and $\lambda=0.6$, this gives $$\Theta=\color{red}{0.00072481622655383}49$$ while the exact solution is $$\Theta=\color{red}{0.0007248162265538377}$$

We still could improve considering that $(1)$ is an expansion to $O(x^{5+p})$ with $p>1$ and have $p$ extra terms. For example, the next term is $$-\frac{(1-c)^5}{32 (2 K-s)^9}\left(48 K^4+K^2 \left(12 c-24 s^2+96\right)-12 K (c s+s)+\left(3 c s^2+3 s^4-18 s^2+14\right)\right)$$

Expanded as a series for large values of $K$, we have $$\Theta=\frac{1-c}{K}+\frac{(1-c) s}{2 K^2}+O\left(\frac{1}{K^3}\right)=\frac{2 \sin ^2\left(\frac{\theta }{4}\right)}{K}+\frac{\sin \left(\frac{\theta }{2}\right) \sin ^2\left(\frac{\theta }{4}\right)}{K^2}+O\left(\frac{1}{K^3}\right)$$