Why do we use Taylor expansion expressing potential energy

[Ahmed Abdalla]

My textbook doesn’t go into it, can someone tell me why Taylor expansion is used to express spring potential energy? A lot of the questions I do I think I can just use F=-Kx and relate it to U(x) being F=-Gradiant U(x) but I see most answers using the Taylor expansion instead to get 1/2 kx^2. Usually it has to do with x being a small value but I never understood why the Taylor expansion is used

 

[Lord Jestocost]

Have a look at [PDF]Hooke's Law - UCSB Physics

 

[Chandra Prayaga]

The Hooke's law force, F = -Kx, defines an ideal spring. In a real spring, Hooke's law is valid only for small stretches (small x) of the spring. The Taylor expansion of the potential energy about the equilibrium position (in this case, x = 0), is used to express more general potential energies, not necessaily of an ideal spring. Since any function (inifinitely differentiable) can be expanded in a Taylor series, even the potential energy can be expanded in a Taylor series. If you are expanding the potential energy about its minimum (at x = 0), then
U(x) = U(0) + x (dU/dx)₀ + (x² / 2!) (d²U/dx²)₀ + ...(higher order terms)
where the subscript 0 on the derivatives means that they are evaluated at x = 0. If x = 0 is a minimum, then the first derivative in the above expansion is zero, and that leaves the quadratic term as the dominant term in the Taylor expansion of the potential energy, and is identified as the "spring" potential energy.

 

[Ahmed Abdalla]

The Hooke's law force, F = -Kx, defines an ideal spring. In a real spring, Hooke's law is valid only for small stretches (small x) of the spring. The Taylor expansion of the potential energy about the equilibrium position (in this case, x = 0), is used to express more general potential energies, not necessaily of an ideal spring. Since any function (inifinitely differentiable) can be expanded in a Taylor series, even the potential energy can be expanded in a Taylor series. If you are expanding the potential energy about its minimum (at x = 0), then
U(x) = U(0) + x (dU/dx)₀ + (x² / 2!) (d²U/dx²)₀ + ...(higher order terms)
where the subscript 0 on the derivatives means that they are evaluated at x = 0. If x = 0 is a minimum, then the first derivative in the above expansion is zero, and that leaves the quadratic term as the dominant term in the Taylor expansion of the potential energy, and is identified as the "spring" potential energy.

Thank you that explained it in great detail!