Exercise · ZW-INTERPOLATION-ERROR-009 · ★★☆ Intermediate
For \(f(x)=\cos(x)\) with nodes \(0,\pi\) (\(n=1\)), what is the bound at \(x=\pi/2\) using \(|f''|\le 1\)?
Choose one option.
Show answer
- C. \(\frac{\pi^2}{8}\)
Explanation
Using \(\frac{1}{2}\cdot\frac{\pi}{2}\cdot\frac{\pi}{2}=\frac{\pi^2}{8}\), which is the option with denominator 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-008 If |f'''(x)|≤ 6 and nodes are 0,1,2 (n=2), what is the bound at x=0.5?