This paper is from 1993! I had thought it was much older on my first read through. Near $\alpha = 0$, $ \sqrt{-2\alpha} <\beta <0.450$. Since $\alpha_i=\frac{-w_i^2}{w_0^2}$, $ \sqrt{2\frac{w_i^2}{w_0^2}} <\beta <0.450$ Also the larger the value of $N$, the smaller the value of $l$ because $L$ is a constant, which means that it will push $w_0$ and $\epsilon$ out of the stability zone. To prove that the natural frequencies are proportional to $g$ we start again with the double pendulum and with the matrix: $$ \left( \begin{array}{ccc} \frac{l_1}{g} & \frac{m_2l_2}{(m_1+m_2)g} \\ \frac{l_1}{g} & \frac{l_2}{g} \\ \end{array} \right)\ \left( \begin{array}{ccc} \ddot \theta_1 \\ \ddot \theta_2 \\ \end{array} \right)\ = - \left( \begin{array}{ccc} \theta_1 \\ \theta_2 \\ \end{array} \right)\ $$ Since we expect oscillatory solutions let's start with the following ansatz: $$ \theta_1 = A_1e^{iwt} \\ \theta_2 = A_2e^{iwt} $$ If we now substitute $\theta_1,\theta_2$ with the expressions $$ \left( \begin{array}{ccc} \frac{l_1w^2}{g}+1 & \frac{m_2l_2w^2}{(m_1+m_2)g} \\ \frac{l_1w^2}{g} & \frac{l_2w^2}{g}+1 \\ \end{array} \right)\ \left( \begin{array}{ccc} A_1 \\ A_2 \\ \end{array} \right)\ = 0 $$ For this to be true for any amplitude, the matrix on the left must be singular i.e. must not be invertible, such that one cannot multiply both sides of the equation by the inverse, leaving the right matrix equal to zero. It follows that the determinant of the matrix must be equal to 0, so: $$ \left(\frac{l_1w^2}{g}+1 \right)\left(\frac{l_2w^2}{g}+1 \right)-\frac{l_1w^2}{g}\frac{m_2l_2w^2}{(m_1+m_2)g}=0 $$ And finally: $$ w_1=\frac{\sqrt{g/l}}{\sqrt{1-\sqrt m}} \\ w_2=\frac{\sqrt{g/l}}{\sqrt{1+\sqrt m}} $$ which means that $w_i^2 \propto g$ To prove this we are going to use the Lagrangian formulation and will start with the $N=2$ case (two coupled pendulums).  The Lagrangian $L$ of the system is given by $L=T-V$, where $T$ and $V$ are the kinetic energy and the potential energy of the system. $$ T=\frac{1}{2}m_1(x_1^2+y_1^2)+\frac{1}{2}m_2(x_2^2+y_2^2) \\ V= m_1gy_1+m_2gy_2 $$ For small angles ($\theta_i\approx 0 \rightarrow \cos(\theta_i)\approx 1$ and $\sin(\theta_i)\approx \theta_i$ ) we can write $T$ and $V$ as a function of the angles: $$ T\approx \frac{1}{2}m_1l_1^2\dot \theta_1^2+\frac{1}{2}m_2l_1^2\dot \theta_1^2+\frac{1}{2}m_2l_2^2\dot \theta_2^2+m_2l_1l_2\dot \theta_1\dot \theta_2 \\ V\approx \frac{1}{2}(m_1gl_1\theta_1^2+m_2gl_1\theta_1^2+m_2gl_2\theta_2^2) $$ If we calculate the Lagrange equations $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot \theta_i}\right)=\frac{\partial L}{\partial \theta_i}$ we end up with a system of coupled differential equations of the form: $$ \left( \begin{array}{ccc} \frac{l_1}{g} & \frac{m_2l_2}{(m_1+m_2)g} \\ \frac{l_1}{g} & \frac{l_2}{g} \\ \end{array} \right)\ \left( \begin{array}{ccc} \ddot \theta_1 \\ \ddot \theta_2 \\ \end{array} \right)\ = - \left( \begin{array}{ccc} \theta_1 \\ \theta_2 \\ \end{array} \right)\ $$ It is helpful to find a new basis in which the matrix $M$ is diagonal. The basis that diagonalizes $M$ is what we call the normal coordinates. In this basis we end up with a set of homogeneous second order linear differential equations, each of which corresponds to the motion of a simple harmonic oscillator (SHO). $$ \left( \begin{array}{ccc} \frac{1}{w_1^2} & 0 \\ 0 & \frac{1}{w_2^2} \\ \end{array} \right)\ \left( \begin{array}{ccc} \ddot X_1 \\ \ddot X_2 \\ \end{array} \right)\ = - \left( \begin{array}{ccc} X_1 \\ X_2 \\ \end{array} \right)\ $$ where $X_1$,$X_2$ are the normal coordinates and $w_1$,$w_2$ are the natural frequencies. Finally we get $$ \ddot X_i+w_i^2 X_i = 0 \\ i=1,2 $$ The generalization for $N$ pendulums is a similar process that just involves calculating a more complex lagrangian and dealing with a bigger matrix, but in the end we always end up with a diagonalized matrix if we use the normal coordinates. As we can see in this video by Steve Mould the chain of pendulums can be balanced if the pivot is vibrated up and down by a small enough amount and at a high enough frequency. [](https://youtu.be/gnn21smGVrQ?t=56 "Inverted Pendulum") This paper was written by D.J. Acheson while he was trying to give a new twist to an old problem in dynamics, first studied by Daniel Bernoulli in 1738. He had considered a hanging chain of several linked pendulums, all suspended from one another, and discovered various different modes of oscillation. Acheson proved that it was possible to take all these linked pendulums, turn them upside-down, so that they are all precariously balanced on top of one another, and then stabilise them in that position by vibrating the pivot up and down. He started calling this gravity-defying experiment "Not quite the Indian rope trick" because of a famous magic trick performed in India during the 19th century. During this trick, a magician would hurl a rope into the air. The rope would stand erect, with no external support. His boy assistant would climb the rope and then descend. Although is theorem doesn't hold for the case $N\rightarrow \infty$ (a perfectly flexible string can be approximated by a series of infinite pendulums all linked together), thus not explaining the Indian rope trick it is still a surprising mathematical result that has a certain magic on its own!  In 1908 Stephenson found that the upper vertical position of the pendulum might be stable when the driving frequency is fast. Yet no one could scientifically answer this unusual and counterintuitive phenomenon until almost fifty years later, a Russian physicist Pyotr Kapitza successfully analyzed the stability of this system by separating the motions into fast and slow, and introduced a new concept called Effective Potential (to read more about Kapitza's demonstration click [here](https://en.wikipedia.org/wiki/Kapitza%27s_pendulum)) Once Kapitza found the effective potential, he then use it to analyze the motion and stability of the system. Systems always tend to move towards positions of minimum potential energy, so by finding the minimum values of the potential energy expression, we can find the stability points. In particular Kapitza found that the vertical position was stable.