Exercise · ZW-LOCAL-AND-GLOBAL-ERROR-006 · ★★☆ Intermediate
For \(y'=y\), \(y(0)=1\), and \(h=0.1\), what is approximately the local error of the first Euler step?
Choose one option.
Show answer
- D. 0.00517
Explanation
Euler gives \(y_1=1.1\) and the exact solution is \(e^{0.1}\approx1.10517\), so the difference is about \(0.00517\).
Review the concept: Local and global error
The local error is the difference between the exact ODE solution over one step and the method's approximation starting from the exact value.
\[e_{n+1}=y(t_{n+1})-y_{n+1}\]
More practice problems: local and global error
- ZW-LOCAL-AND-GLOBAL-ERROR-001 What does the local error measure in a numerical ODE method?
- ZW-LOCAL-AND-GLOBAL-ERROR-002 What does the global error measure in a numerical ODE method?
- ZW-LOCAL-AND-GLOBAL-ERROR-003 If a method's local error is O(h^2), what order does the global error typically have?
- ZW-LOCAL-AND-GLOBAL-ERROR-004 With Euler's method, what is the local error when computing y_n+1 from y(t_n)?
- ZW-LOCAL-AND-GLOBAL-ERROR-005 What best describes how local errors build up into the global error?
- ZW-LOCAL-AND-GLOBAL-ERROR-007 If halving h roughly halves the global error of Euler's method, what is the order of the global err…
- ZW-LOCAL-AND-GLOBAL-ERROR-008 If the local error of one step is C h^3, roughly how many steps accumulate over the interval [0,1]?
- ZW-LOCAL-AND-GLOBAL-ERROR-009 A method has local error O(h^2). If h goes from 0.1 to 0.05, how does the global error change appro…
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