Practice question · Multiple choice
Truncation error shrinks as the step size h decreases, while round-off error grows. Why does that leave an optimal h rather than "as small as possible"?
Hints
- Sketch the two error terms against h, one rising and one falling. Where is their sum smallest?
- In a difference quotient, what is being divided by h - and what does that do to a small rounding error?
Show the answer
A. Because the total error is the sum of two terms moving oppositely.
Why
Truncation error falls with h and round-off grows like 1/h, because subtracting two nearly equal floats and dividing by a tiny h amplifies the representation error. The sum has a minimum — for a forward difference it sits near √(machine epsilon), about 10⁻⁸ — and pushing h to 10⁻¹⁵ makes the answer dramatically worse.
Practise Numerical Errors and Stability
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