Exercise · ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-007 · ★★☆ Intermediate
If the step \(h\) is halved in the centered formula, how does the truncation error change approximately?
Choose one option.
Show answer
- D. It is divided by 4
Explanation
The error is \(O(h^2)\), so halving \(h\) divides the error by \(2^2=4\).
Review the concept: Finite differences for derivatives
Finite differences are approximations of the derivatives of a function \(f\) from its values at discrete points separated by a step \(h\). They turn differential problems into algebraic systems that a computer can solve.
\[f'(x)\approx \frac{f(x+h)-f(x)}{h}\]
More practice problems: finite differences for derivatives
- ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-001 What is the forward finite difference formula for f'(x)?
- ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-002 What is the backward finite difference formula for f'(x)?
- ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-003 What is the centered finite difference formula for f'(x)?
- ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-004 In finite differences, what does h represent?
- ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-005 You have f(1)=2 and f(1.1)=2.5. What is the forward approximation of f'(1) with h=0.1?
- ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-006 Let f(x)=x^2. Compute the centered approximation of f'(2) with h=0.5.
- ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-008 Expanding f(x+h)=f(x)+hf'(x)+(h^2)/(2)f''(x)+cdots, what is the leading term of the error in (f(x+h…
- ZW-FINITE-DIFFERENCES-FOR-DERIVATIVES-009 When a domain is discretized, what replaces the derivative in the finite difference method?
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