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).
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).
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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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 \( x = 0.05 \):
\[
\frac{(0.05)(-0.05)(-0.15)}{6} \cdot 3 \cdot (5\pi)^3 \cdot \cos(5\pi \xi) \approx 0.726
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb460c0c-d029-4e90-a450-1d82490780a1%2Ffed7f384-94f1-40fb-9b31-125129ff989c%2Fqxks8aa_processed.jpeg&w=3840&q=75)
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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