 Solution

Let $x_0$ be the equilibrium, so $V^{\prime }(x_0)=0$. For $\xi =x-x_0$, the Taylor expansion of the potential givesAn ordinary small-amplitude simple harmonic motion therefore requires a nondegenerate local minimum, $V^{\prime \prime }(x_0)>0$, with the quadratic term dominating the higher terms over the excursion. ThenA local minimum with $V^{\prime \prime }(x_0)=0$, such as a quartic minimum, need not give small-amplitude harmonic motion with a nonzero limiting angular frequency. If $V^{\prime \prime }(x_0)<0$, the equilibrium is unstable rather than oscillatory.