[arbitrandomuser]

> > It's basically taking the first order perturbation of a Taylor expansion around a local minima. > > I'm not sure that's less complicated, at least for my level understanding.

You know Taylor expansions ? So every complicated function of x can be written as soe expansion of infinite powers of x, a0 +a1 ( x )+a2 ( x^2 ) ... And so on Every complicated force can then be written as a function of displacement x like this , now if x is small ignore the sqaure and higher order terms what you get is the linear part , which looks something like F = a1x + a0 , the a0 part only shifts the default position (you can rewrite x as x-c so that a0 goes to zero,) lo and behold you have F=Kx everything is like your familiar hooks law spring as long as x is small enough.

[ridiculous_fish]

This is on the right track but not entirely right. In your F = a1x + a0, here a0 is a force, not a position, so you cannot neglect it on the basis of "default position." Instead you set it to zero because the system is at a stable equilibrium.

Here's the presentation I've seen. Usually we like to work in potentials, not forces, because potentials are nice scalar functions, while force is an ugly vector function. So say you have a potential V which is at a local minimum at position x.

Expand the potential at x around a small displacement dx: V(x + dx). This gives us the Taylor series V(x+dx) = V(x) + a1 V'(x) dx + a2 V''(x) dx^2 + a3 V'''(x) dx^3...

We can neglect V(x) since it's just a constant, and adding a constant to the potential does not affect the physics. And (the crux) we can neglect V'(x) because the potential is at a minimum, so the derivative is zero.

That leaves the quadratic and higher-order terms. Neglecting the higher order terms on the basis that dx is small, we get the harmonic potential, or Hooke's Law in the language of forces.