The error of Euler's method at a fixed end time is the accumulation of every step's small error, and it shrinks in proportion to the step. Halve h and the error is about halved, which the lab shows as a ratio near 2 and names first order. On one equation the ratio is exactly 2; on the others it only approaches 2, and the lesson says which is which.
Core ideas
Three ideas: what is being measured, where the total comes from, and how a ratio of errors names the order.
To compare step sizes, fix an end time T and ask for E(h) = |y_N − y(T)|, where N = (T − t₀)/h steps are needed to arrive. A smaller h takes more steps to reach the same T, and that is what makes the three errors comparable. An error at the end of four steps and an error at the end of eight of the same size are errors at two different times.
Each step starts a little off and adds a little more. On y′ = 2t one step of size h from t misses by exactly h², and there are 1/h of them, so the error at T = 1 is h²·(1/h) = h. The last step contributes one of those pieces and no more.
If the error is about C·h, halving h halves it, and E(h)/E(h/2) is near 2. If it were about C·h² the ratio would be near 4. The order is the power: log₂ of the ratio. A ratio is read from two errors, and the lab forms two ratios from three, so one can check the other.
The material
The measurement, the one equation where Euler's error is exactly h, why the total is a sum of equal pieces, two equations where the ratio only approaches 2, and a rounded case.
The lab runs Euler's method three times on one equation, with step sizes h, h/2 and h/4, to the same end time T = 1. For each it prints the error E(h) against the known solution, then the two ratios E(h)/E(h/2) and E(h/2)/E(h/4), and the order those ratios imply.
On this equation the error is exactly h: E(1/4) = 1/4, E(1/8) = 1/8, E(1/16) = 1/16. Euler's method on y′ = 2t is a left sum, and the left sum of 2t over N pieces from t = 0 to t = 1 is h²·N·(N − 1) = 1 − h, which differs from 1 by exactly h. Both ratios are exactly 2, the order is 1, and here that is not an approximation.
Look at one step. The true solution t² moves from t² to (t + h)², a rise of 2th + h². Euler's step rises by h·2t. The difference is h², whatever t is, so every step falls short by the same h². With N = 1/h steps the total is (1/h)·h² = h. At h = 1/4 each of the four steps misses by 1/16, and the four add to 1/4.
For a right-hand side with continuous first derivatives, Euler's error at a fixed T is approximately C·h for a constant C that depends on the equation and T but not on h. Halving h therefore halves the error, and the ratio E(h)/E(h/2) approaches 2.
The lab demonstrates the claim on chosen equations by exact ratios, and does not prove it for every equation. On an equation where f also depends on y, an early error is carried forward by the later steps, which changes the constant C and not the power of h.
Take y′ = t², with exact solution t³/3. Euler's method is again a left sum, and the lab prints the errors 11/96, 23/384 and 47/1536. The ratios are 44/23 ≈ 1.91304 and 92/47 ≈ 1.95745, both rounded here, and the order tile reads ≈ 0.968973, also rounded. The ratios are close to 2 and not equal to it.
The reason can be read off the numbers. For this equation the error is exactly h/2 − h²/6: at h = 1/4 that is 1/8 − 1/96 = 11/96. The first term halves exactly when h does. The second term is four times smaller after each halving, so it falls faster, and its effect on the ratio fades: the ratio stays below 2 and climbs towards it as h falls.
On y′ = y with exact solution e^t, the true value e is not a fraction, so the lab cannot subtract exactly. Each error is printed rounded with ≈, the two ratios are rounded, and the order is rounded.
That is the exactness rule at work: Euler's values (1 + h)ⁿ are exact, the true value is not, and the lab says so in every tile where the two meet.
Interactive
Errors at h, h/2 and h/4, and their ratios
Method
From an equation with a known solution to a number you can state.
Pick T and h, h/2, h/4 so that (T − t₀)/h is a whole number of steps for each. Every error is then an error at the same T.
Compute y_N for each step size and the gap |y_N − y(T)|. Keep the gap exact when the true value at T is rational and mark it ≈ when it is not.
Divide each error by the next: E(h)/E(h/2) and E(h/2)/E(h/4). Two ratios that agree are better evidence than one.
A ratio of 2 is first order, 4 second, 16 fourth. A ratio near but not equal to a power of two is read as approaching it, and the order is log₂ of the ratio.
Find the error of one Euler step on y′ = 2t, add the steps, and check the total against the lab.
The three errors agree with the lab. The shortfall of a single step is h², a tiny number, and it is the number of steps that turns it into an error of size h: halving h quarters each piece and doubles the number of pieces, which halves the total.
For a rehearsal, run h = 1/10 by the same formula: ten steps of 1/100 give 1/10.
You should be able to run the method to a fixed T with h, h/2 and h/4, subtract the true value, divide successive errors, and say that a ratio near 2 means first order. You should also be able to explain, for y′ = 2t, why the total error is h and not the error of one step.
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