=₁³-171. The vectors for Po. P₁, and p2 are 1 1 Fit a cubic trend function to the data (-2,6). (-1,8), (0.8). (1.7), (2,6). Let po(t) = 1. P₁ (t)=t and p₂ (t)=1²-2. The orthogonal cubic polynomial is p3(t)= -1 0 and -2, respectively. 1 2
=₁³-171. The vectors for Po. P₁, and p2 are 1 1 Fit a cubic trend function to the data (-2,6). (-1,8), (0.8). (1.7), (2,6). Let po(t) = 1. P₁ (t)=t and p₂ (t)=1²-2. The orthogonal cubic polynomial is p3(t)= -1 0 and -2, respectively. 1 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Fitting a Cubic Trend Function
#### Problem Statement:
Fit a cubic trend function to the given data points: \[ (-2, 6), (-1, 8), (0, 8), (1, 7), (2, 6) \]
#### Definitions:
Let the polynomial functions be defined as:
\[ p_0(t) = 1 \]
\[ p_1(t) = t \]
\[ p_2(t) = t^2 - 2 \]
The orthogonal cubic polynomial is defined as:
\[ p_3(t) = \frac{5}{6}t^3 - \frac{17}{6}t \]
#### Explanation:
To fit the cubic trend function, we use these orthogonal polynomials and their associated vectors:
For \( p_0, p_1, \) and \( p_2 \), the vectors are:
\[ p_0 \rightarrow
\begin{pmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{pmatrix}
\]
\[ p_1 \rightarrow
\begin{pmatrix}
-2 \\
-1 \\
0 \\
1 \\
2
\end{pmatrix}
\]
\[ p_2 \rightarrow
\begin{pmatrix}
-1 \\
-1 \\
0 \\
1 \\
2
\end{pmatrix}
\]
#### Objective:
To find and express the cubic trend function \(\hat{p}(t)\), incorporating the given data points and orthogonal polynomials.
\[ \hat{p}(t) = \boxed{} \]
This expression involves determining the coefficients that best fit the respective orthogonal polynomials to the given data set.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb34eb3b-53e5-4e6c-9638-9efc4a2e33af%2Fd401cb1c-280e-47c5-a058-bc670a4aa983%2Fdzgpymp_processed.png&w=3840&q=75)
Transcribed Image Text:### Fitting a Cubic Trend Function
#### Problem Statement:
Fit a cubic trend function to the given data points: \[ (-2, 6), (-1, 8), (0, 8), (1, 7), (2, 6) \]
#### Definitions:
Let the polynomial functions be defined as:
\[ p_0(t) = 1 \]
\[ p_1(t) = t \]
\[ p_2(t) = t^2 - 2 \]
The orthogonal cubic polynomial is defined as:
\[ p_3(t) = \frac{5}{6}t^3 - \frac{17}{6}t \]
#### Explanation:
To fit the cubic trend function, we use these orthogonal polynomials and their associated vectors:
For \( p_0, p_1, \) and \( p_2 \), the vectors are:
\[ p_0 \rightarrow
\begin{pmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{pmatrix}
\]
\[ p_1 \rightarrow
\begin{pmatrix}
-2 \\
-1 \\
0 \\
1 \\
2
\end{pmatrix}
\]
\[ p_2 \rightarrow
\begin{pmatrix}
-1 \\
-1 \\
0 \\
1 \\
2
\end{pmatrix}
\]
#### Objective:
To find and express the cubic trend function \(\hat{p}(t)\), incorporating the given data points and orthogonal polynomials.
\[ \hat{p}(t) = \boxed{} \]
This expression involves determining the coefficients that best fit the respective orthogonal polynomials to the given data set.
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