You halve the step size h when solving an ODE. How do the global errors of Euler's method and RK4 compare after this change?
ABoth errors reduce by roughly half
BEuler's error reduces by half; RK4's error reduces by a factor of about 16
CEuler's error reduces by a factor of 4; RK4's reduces by a factor of 16
DRK4's error reduces by half; Euler's reduces by a factor of 16
Global error for Euler's method is O(h) — halving h halves the error. Global error for RK4 is O(h⁴) — halving h reduces error by 2⁴ = 16. This is the core advantage of high-order methods: the same reduction in step size yields dramatically more accuracy. A method's 'order' tells you the exponent: 4th-order means error scales as h⁴.
Question 2 Multiple Choice
Why do engineers often prefer implicit Runge-Kutta methods over explicit methods for certain ODE problems?
AImplicit methods require fewer function evaluations per step, making them faster
BImplicit methods achieve higher-order accuracy with fewer stages than explicit methods
CImplicit methods remain stable for stiff equations where explicit methods require prohibitively tiny step sizes
DImplicit methods do not require initial conditions, simplifying setup
For stiff equations — those with components that vary on very different timescales — explicit methods like RK4 must use extremely small step sizes to remain numerically stable, making them computationally impractical. Implicit methods allow stages to depend on each other (requiring a small linear system to be solved at each step), which gives them much better stability properties. The extra cost per step is outweighed by the ability to take much larger steps.
Question 3 True / False
In RK4, each subsequent stage uses results from earlier stages to refine the slope estimate, with the final update being a weighted average of all four slopes.
TTrue
FFalse
Answer: True
This is exactly how RK4 works. k₁ is the slope at the start; k₂ uses k₁ to estimate the slope at the midpoint; k₃ refines the midpoint estimate using k₂; k₄ estimates the slope at the endpoint using k₃. The update yₙ₊₁ = yₙ + (h/6)(k₁ + 2k₂ + 2k₃ + k₄) is the weighted average, analogous to Simpson's rule, which achieves O(h⁵) local and O(h⁴) global accuracy.
Question 4 True / False
The weights (1, 2, 2, 1)/6 in the RK4 update formula are arbitrary design choices that happen to produce good accuracy in practice.
TTrue
FFalse
Answer: False
The weights are engineered, not arbitrary. RK4 is designed so that a Taylor expansion of yₙ₊₁ matches the true solution's Taylor series through the h⁴ term. The weights (1, 2, 2, 1)/6 are identical to those in Simpson's rule, reflecting that both are solving the same approximation problem — how to best combine samples at specified points to achieve a high-order estimate. The design ensures no coincidence: the weights are derived analytically to cancel error terms through fourth order.
Question 5 Short Answer
Explain why RK4 achieves much higher accuracy than Euler's method without simply taking smaller steps.
Think about your answer, then reveal below.
Model answer: RK4 evaluates f at four intermediate points within each step interval and combines them with carefully chosen weights, analogous to how Simpson's rule improves on the rectangle rule in numerical integration. This sampling of f at multiple stages within [tₙ, tₙ + h] allows the method to approximate the true solution's behavior through the h⁴ term in its Taylor expansion, achieving O(h⁴) global error versus O(h) for Euler's method — so RK4 extracts much more accuracy from each step.
The analogy to numerical integration is precise: Euler's method is the rectangle rule (one sample at the left endpoint); RK4 is Simpson's rule (three samples with the correct weights). The key insight is that accuracy comes from how cleverly you sample f within the step interval, not just from how small the interval is.