Questions: Euler's Method for Numerical Solutions

Question 1

For the IVP y' = 2x, y(0) = 1, using Euler's method with step size h = 0.5, what is the approximate value of y(0.5)?

• A. 1.0 — Euler's approximation using the slope at x = 0
• B. 1.25 — the exact value of y(0.5) = x² + 1
• C. 1.5 — one full step of slope 1 from y₀ = 1
• D. 2.0 — the slope at x = 0.5 times the step size

Answer: (A)

Euler's method: y₁ = y₀ + h·f(x₀, y₀) = 1 + 0.5·f(0, 1) = 1 + 0.5·(2·0) = 1 + 0 = 1.0. The slope at x = 0 is f(0, 1) = 2(0) = 0, so the method predicts no change. The exact solution is y = x² + 1, giving y(0.5) = 1.25. The error is 0.25 — Euler's method misses the curvature entirely in this step because the slope is changing (it's 0 at x=0 but grows), yet we only use the slope at the left endpoint.

Question 2

If you reduce the step size h by half in Euler's method while approximating over the same total interval, what happens to the total (global) error?

• A. Error is reduced by a factor of 4 — halving h gives quadratic improvement
• B. Error is roughly halved — Euler's method is first-order accurate
• C. Error is unchanged — more steps accumulate more error, canceling the per-step improvement
• D. Error is reduced by a factor of √2 — each step's error decreases as √h

Answer: (B)

Euler's method is first-order accurate: global error is proportional to h. Halving h doubles the number of steps (from N to 2N) but each step's local error is proportional to h², so local error is quartered. Combined: 2N steps × (h/2)² local error per step ≈ 2N × h²/4 ∝ h. Net effect: halving h halves the global error. Option A would describe a second-order method like the improved Euler (Heun's method) or RK2.

Question 3

Halving the step size in Euler's method reduces the total accumulated error quadratically — that is, by a factor of 4.

• T. True
• F. False

Answer: False

Euler's method is first-order accurate, meaning global error is proportional to h (not h²). Halving h only halves the global error, not quarters it. A factor-of-4 improvement per halving would describe a second-order method. This distinction matters practically: to gain one extra decimal digit of accuracy with Euler's method, you must use ten times as many steps — much more expensive than higher-order methods like Runge-Kutta 4, which achieves the same digit with only about 1.8× more steps.

Question 4

Euler's method uses only the slope at the left endpoint of each step interval to predict the next value, without incorporating any slope information from within the interval.

• T. True
• F. False

Answer: True

The update rule y_{n+1} = y_n + h·f(x_n, y_n) uses f(x_n, y_n) — the slope at the start of the interval — and then walks a straight line to the next point. It ignores the fact that the slope changes across the interval. This is the source of local truncation error: the true curve curves, but Euler follows a tangent line. Higher-order methods like RK4 evaluate the slope at multiple points within each interval (the midpoint, endpoint, etc.) to get a more representative average slope, achieving much smaller error for the same step size.

Question 5

Why does Euler's method accumulate error over many steps, even though the linear approximation formula underlying each step is mathematically correct?

Model answer: Each step of Euler's method is a linear approximation: it assumes the slope is constant across the interval [x_n, x_{n+1}], equal to f(x_n, y_n). But the true solution curves — the slope changes — so each step drifts slightly from the true curve. This per-step drift (local truncation error, proportional to h²) is small for small h. However, to cover a fixed interval [x₀, X], you need 1/h steps, so errors accumulate over many steps. The total global error is (1/h) × h² = h — small but nonzero. Crucially, each step starts from the wrong point (because previous steps already introduced error), so errors compound rather than cancel.

The intuition is geometric: Euler walks along tangent lines. Each tangent line is slightly off from the curve, and you start the next step from the end of the wrong tangent line. The cumulative effect is that you wander progressively further from the true solution. Making h smaller keeps each tangent line closer to the curve and reduces compounding — but no matter how small h is, there is always some positive error unless the true solution is itself linear.