Practice question · Numerical answer
A second-order scheme has an error of 0.08 on the current grid. Refining the grid by a factor of 2 changes the error to what value?
Hints
- Second order means the error goes like the square of the spacing.
- Halving the spacing divides the error by 4.
Show the answer
0.02 (answers within ±0.001 count)
Why
0.08 / 4 = 0.02. Observing exactly this factor of 4 in practice is how you confirm the scheme is genuinely second order, a convergence check that catches implementation bugs no unit test would find.
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