I'm writing an advanced interface, but I don't yet have a concept of derivatives or integrals, and I don't have an easy way to construct infinite many functions (which could effectively delay or tween their frame's contributing distance [difference between B and A] over the next few frames).
I can store values for a frame, and I can also consume them or previous values and map into A.
For example, each frame could calculate the distance between B and A, then add that distance to A, but they would be perfectly in sync.
I can keep track of the last N frames' distances and constantly shift old distances off, but this would create a delay, not an elastic effect. Somehow, the function that's popping off past-frames' distances needs to adjust for how long it's been for those 10 frames.
Is there any function I can rewrite each frame, which picks up the progress from it's predecessor, and contributes it's correct delta amount, adjusting for the new total distance between B and A?
Does this question make sense? How can I achieve behavior where A is constantly catching up to B in a non-linear, exponential way?
You can truly simulate the physics of a dampened spring, which leads to a differential equation.
$$m\ddot x_A+d\dot x_A+k(x_A-x_B)=0$$
where $x$ is the position, $m$ the mass, $d$ a damping coefficient and $k$ the stiffness constant of the spring.
As you probably have enough with a qualitatively realistic solution, there is no need for a sophisticated ODE solver and the Euler method should be good enough.
Turn the equation in a system of two first order equations:
$$\begin{cases}\dot v_A=\dfrac{kx_B-kx_A-dv_A}m,\\\dot x_A=v_A.\end{cases}$$
Hence you will iterate
$$\begin{cases}v_{A,n+1}=v_{A,n}+\dfrac{kx_{B,n}-kx_{A,n}-dv_{A,n}}m\Delta t,\\x_{A,n+1}=x_{A,n}+ v_{A,n}\Delta t\end{cases}$$
for some time increment. This way you can compute a position from the previous, but you also need to remember the speed.
By varying the damping and stiffness coefficients, you can achieve more or less tight following of $B$, with an oscillatory behavior or not.
If you want a 2D simulation, $x$ will be a vector, and the shape of the equations remains.