Exercise · ZW-INTERPOLATION-ERROR-008 · ★★☆ Intermediate
If \(|f'''(x)|\le 6\) and nodes are \(0,1,2\) (\(n=2\)), what is the bound at \(x=0.5\)?
Choose one option.
Show answer
- D. \(\frac{3}{8}\)
Explanation
Using \(\frac{6}{3!}\cdot|0.5\cdot(-0.5)\cdot(-1.5)|=1\cdot0.375=\frac{3}{8}\).
Review the concept: Interpolation error
The interpolation error is the difference between \(f(x)\) and the polynomial \(P(x)\) that approximates it at the nodes \(x_0,\dots,x_n\); it depends on \(f^{(n+1)}\) and on \(\prod(x-x_i)\).
\[f(x)-P_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^{n}(x-x_i)\]
More practice problems: interpolation error
- ZW-INTERPOLATION-ERROR-001 What is the error factor prod_i=0^n(x-x_i) for nodes x_0=0,\ x_1=1?
- ZW-INTERPOLATION-ERROR-002 What is the value of (n+1)! when n=3?
- ZW-INTERPOLATION-ERROR-003 In the error formula for the degree-n interpolant, which derivative of f appears?
- ZW-INTERPOLATION-ERROR-004 At which points does the interpolation error vanish?
- ZW-INTERPOLATION-ERROR-005 Which theorem is used in the proof of the error formula?
- ZW-INTERPOLATION-ERROR-006 For f(x)=e^x with nodes 0,1 (n=1), what is the error bound at x=0.5 if |f''|≤ e on [0,1]?
- ZW-INTERPOLATION-ERROR-007 With nodes x_0=0,\ x_1=2 and f(x)=x^3 (n=1), what is prod(x-x_i) at x=1?
- ZW-INTERPOLATION-ERROR-009 For f(x)=cos (x) with nodes 0,π (n=1), what is the bound at x=π/2 using |f''|≤ 1?