1. a. A table of values for f(x) is: I f(r) 0.0 0.0 0.1 3.0 0.2 0.0 Use quadratic interpolation to estimate f(0.05).

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**1. Quadratic Interpolation and Error Estimation for \( f(x) \)**

a. **A table of values for \( f(x) \):**

\[
\begin{array}{c|c}
x & f(x) \\
\hline
0.0 & 0.0 \\
0.1 & 3.0 \\
0.2 & 0.0 \\
\end{array}
\]

**Use quadratic interpolation to estimate \( f(0.05) \):**

The quadratic interpolation formula is:

\[
L_2(x) = \frac{(x-0)(x-0.2)}{(0.1-0)(0.1-0.2)} \times 3
\]

Substituting \( x = 0.05 \):

\[
L_2(0.05) = \frac{(0.05)(-0.15)}{0.1(-0.1)} \times 3 = 1.25
\]

b. **If \( f(x) = 3 \sin(5\pi x) \), obtain a reasonable bound on the error in your estimate of \( f(0.05) \):**

Error bound formula:

\[
\frac{(x-0)(x-0.1)(x-0.2)}{6} \cdot f'''(\xi)
\]

Estimate with \( x = 0.05 \):

\[
\frac{(0.05)(-0.05)(-0.15)}{6} \cdot 3 \cdot (5\pi)^3 \cdot \cos(5\pi \xi) \approx 0.726
\]
Transcribed Image Text:**1. Quadratic Interpolation and Error Estimation for \( f(x) \)** a. **A table of values for \( f(x) \):** \[ \begin{array}{c|c} x & f(x) \\ \hline 0.0 & 0.0 \\ 0.1 & 3.0 \\ 0.2 & 0.0 \\ \end{array} \] **Use quadratic interpolation to estimate \( f(0.05) \):** The quadratic interpolation formula is: \[ L_2(x) = \frac{(x-0)(x-0.2)}{(0.1-0)(0.1-0.2)} \times 3 \] Substituting \( x = 0.05 \): \[ L_2(0.05) = \frac{(0.05)(-0.15)}{0.1(-0.1)} \times 3 = 1.25 \] b. **If \( f(x) = 3 \sin(5\pi x) \), obtain a reasonable bound on the error in your estimate of \( f(0.05) \):** Error bound formula: \[ \frac{(x-0)(x-0.1)(x-0.2)}{6} \cdot f'''(\xi) \] Estimate with \( x = 0.05 \): \[ \frac{(0.05)(-0.05)(-0.15)}{6} \cdot 3 \cdot (5\pi)^3 \cdot \cos(5\pi \xi) \approx 0.726 \]
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