For y′=y with y(0)=1, explicit Euler follows the current slope for one step: yₙ₊₁=yₙ+h yₙ. With h=1/2, the values at t=0,1/2,1 are 1,3/2,9/4. The exact solution at t=1 is e, so the two-step result underestimates it. The update is exact for the chosen tangent-line model, not exact for the curved solution.
Predict
Predict whether making one step twice as small halves or quarters the immediate defect. Starting from the exact solution, Taylor’s theorem gives a defect proportional to h², so halving h quarters that local defect. But reaching the same final time then requires twice as many steps. Global error therefore has a different order, and propagation through the dynamics must also be controlled.
Experiment
Compare final-time approximations for the same problem.
h
Steps to t=1
Euler value
Exact recurrence
1/2
2
9/4
(1+1/2)²
1/4
4
625/256
(1+1/4)⁴
Compute the errors against e using a supplied high-precision reference. The recurrence validates the implementation independently of the continuous exact solution, which measures discretization error.
Understand
For y′=f(t,y), Euler uses yₙ₊₁=yₙ+h f(tₙ,yₙ). If the exact solution is C², Taylor expansion gives y(tₙ+h)=y(tₙ)+h f(tₙ,y(tₙ))+dₙ, with |dₙ|≤Mh²/2 for a bound M on |y″|. Here dₙ is the unscaled one-step defect; some texts divide it by h and call that quantity local truncation error, so state the convention. If f is Lipschitz in y with constant L on a region containing exact and numerical trajectories, the global error obeys |eₙ₊₁|≤(1+hL)|eₙ|+Mh²/2. A geometric sum over a fixed interval bounds it by a constant times h when the initial error is zero.
Apply
The recurrence also explains why consistent local slopes are not enough without stability. For y′=λy, Euler multiplies by 1+hλ each step. If λ<0, the true solution decays, but a large h can give amplification magnitude greater than one. Even a magnitude below one can alternate signs, violating positivity while remaining stable in norm. For nonlinear problems, ensure the iterates remain in the region where derivative and Lipschitz bounds hold. A theorem about a small neighborhood cannot justify a trace that has already left it.
For the linear decay equation, the exact one-step factor is e^(hλ), while Euler uses its first-degree approximation 1+hλ. Expanding their difference gives (hλ)²/2+O(h³) for fixed λ. This directly confirms the local-defect order and reveals why the constant grows with |λ|². A step small relative to the final time may still be large relative to a fast decay timescale. Step size should therefore be judged against the dynamics, not merely the number of digits in the time variable.
Continue
Implementation should bound the number of steps, use the old state for all components of one Euler update, and handle a final shortened step explicitly. A vector update performed component by component with overwritten values is generally a different method. Report local-defect convention, final-time error, step size and evaluation count. Next, Runge–Kutta methods combine several slope evaluations to cancel leading errors, improving order without automatically solving stiffness or event-detection problems.
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For y′=−2y, y(0)=3 and h=1/4, compute one explicit Euler step.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For y′=y, y₀=1 and h=1/2, compute y₂.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Derive the one-step defect for exact starting data when the solution is C².
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Why does an O(h²) defect per step usually lead to O(h) global error over a fixed interval?
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For y′=−5y with positive y₀, find conditions on h for explicit Euler to preserve nonnegativity and to be strictly stable.
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Specify an Euler implementation test that checks indexing and actual evaluation counts.
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Reflect and continue
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