Physics - Eulers Method

Thanks to danske for this:

First off we need to expand the explanation of the Taylor series. 
There are more terms to the Taylor series than in: (x(t0 + h) = x(t0) + h x'(t0) + h^2/2 x''(t0) + O(h^3)). It continues off in an infinite series: 
x(t0 + h) = x(t0) + x'(t0)h + x''(t0)/2 h^2 + x'''(t0)/6 h^3 + x''''(t0)/24 h^4 + ... + x('n)(t0)/n! h^n + error 
Err...I'm going to write this quickly, so it may not be comprehensive or appropriately split-up for those two pages. 
If we're integrating over time, as in a simulation, then we need to be able to calculate the derivative over time for each variable we're attempting to integrate for. For position this derivative is velocity, for velocity this derivative is acceleration. If we could directly calculate the derivative for acceleration than we could use this as well to predict future acceleration. In a physical simulation acceleration (from forces) typically depend on the positions and velocities of all the objects in the system, so we don't have an equation for how acceleration (forces) are changing over time. If we did that would be like having an equation which would tell us that two objects will collide in the simulation before we actually get to that time. This is, however, what we will attempt to numerically emulate. 
Rewriting the Taylor series for physical quantities (and in the order usually used in physics texts) we have: 
x(t0 + h) = 0.5a(t0)h^2 + v(t0)h + x(t0) 
(the equation for position given constant acceleration) 
v(t0 + h) = a(t0)h + v(t0) 
(the equation for velocity given constant acceleration) 
if we had the derivative of acceleration then we could instead have: 
x(t0 + h) = 1/6 * a'(t0)h^3 + 0.5a(t0)h^2 + v(t0)h + x(t0) 
v(t0 + h) = 0.5a'(t0)h^2 + a(t0)h + v(t0) 
a(t0 + h) = a'(t0)h + a(t0) 
We can estimate a' using our simulation, however. We can calculate the forces in the simulation given positions and velocities off the objects. If only we knew the positions and velocities of the objects in the future we could know the forces in the future. Wait a second, computing positions and velocities of the future is what this simulation is doing. We can use it! 
How does this work? We can use our current forces and accelerations to provide us with a prediction of the positions and velocities of all the objects (h) in the future, calculate the forces and accelerations in our predictions, and then use those future accelerations to estimate (a') at (t0). We can then use that estimated (a') to add another term to our Taylor series, improving our estimation of future positions and accelerations, repeating the process until the difference between our prediction and corrected prediction is small. 
Here's what it looks like mathematically: 
a(t0 + h) = a'(t0)h + a(t0) 
a'(t0) = (a(t0 + h) - a(t0))/h 
(you'll recognize that as the definition of the derivative if h -> 0) 
Instead of using the "actual" value of (a(t0 + h)) we use our predicted value of the future. 
a'(t0) = (ap(t0 + h) - a(t0))/h 
If we put that back into our Taylor series (omitting some of the algebra) and calling our new values "corrected", we have, : 
xc(t0 + h) = 0.25*(ap(t0 + h) + a(t0))h^2 + v(t0)h + x(t0) 
vc(t0 + h) = 0.5*(ap(t0 + h) + a(t0))h + v(t0) 
So instead of using simply (a(t0)) in the Taylor series without (a'(t0)), we replace it with the average of the current acceleration and our predicted accelerations in the future ((a(t0) + ap(t0))/2). 
In terms of the simulation what we're doing is generating the world one time step in the future as if all the forces were constant during that time step, evaluating the forces in the future, and then using those future values to compute new average forces to use over the timestep. If the forces aren't constant over time then our new average forces will result in a more accurate simulation than pretending the forces are constant during each timestep. 
As a concrete example let us consider a spring being compressed over time. At (t0) it is producing some force, at (t0 + h) it is producing a greater force. If we integrate position over time with the initial force then we've underestimated the force of the spring during that time step. By "peeking" ahead we see that its force is increasing and that we should use a greater value as an average force over the time step. Doing so gives us a more accurate simulation of the forces from the spring. 
In another example which may be more intuitive, consider a car driving off a cliff sometime in the middle of the time step. At (t0) it is still on the ground. At (t0 + h) we find it hanging in mid air. Somewhere in the middle of the timestep the force from the ground when from the weight of the car to zero. Obviously the car has been experiencing an average acceleration somewhere between zero and -9.8m/s^2 over the timestep and our use of a zero acceleration is inaccurate. So we recalculate the future position and velocity of the car at (t0 + h) using an average acceleration of -4.9m/s^2 instead of zero. This simulates the car having driven off the cliff exactly half-way through the time step. Unless we want to instead decrease the time step to more accurately determine when the car went off the cliff then this result of simulating it going off half-way between when we last saw it on the ground and when we first found it hanging in mid-air, this is better than simulating it having just driven off the cliff an infinitesimal amount of time before we first found it hanging in the air. On average it will have driven off in the middle of the timestep, which is what using the average acceleration of -4.9m/s^2 over the timestep gives us. 
In coding this Modifier Euler method we do need to maintain at least two whole world states, the initial state at (t0) and our predicted state at (t0 + h), in order to compute the average forces to use in our corrected computations. The corrected values (using the computed averages) can then be stored in the memory allocated for the initial state (since we're done with it unless we want to re-correct with another estimate of average forces using the forces we find in our corrected world), so memory usage only doubles (or maybe as bad as triples depending on how we're forced to compute averages of the data). Since we're computing a future world at least twice for each time step the simulation rate will drop, but the accuracy improves more than if we halved the time step. If we're having problems with accuracy, or just want more of it, and can afford the memory usage and the added complexity of computing the average forces, then this may be faster than decreasing the timestep to achieve the same accuracy. 
In summary, what this method does is guess and check the world state in the future. We take our initial guess at the future assuming all the forces are constant over the time step, use the future forces in our first guess of the future to provide us with a better guess at average forces over the time step, and use those averages to make a better guess at the future. 
If code is your thing, here's a snippet computing a one-dimension mass on a spring: 
double x = 10.0; 
double v = 0.0; 
double k = 1000.0; 
double t = 0; 
double t_step = 0.001; 
double t_max = 10.0; 
while ( t < t_max ) { 
double a = ((x * -k) / m); 
double xp = (x + v * t_step + 0.5 * a * t_step * t_step); 
double vp = (v + a * t_step); 
double ap = ((xp * -k) / m); 
x = x + v * t_step + 0.25*(a + ap) * t_step * t_step; 
v = v + 0.5*(a + ap) * t_step; 
t += t_step; 


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