Questions: x1 y1 1 -5 2 -5 3 9 4 49 5 125 6 255 7 445 y1 sim a x^3 + b x1^2 + c x1+d

x1  y1
1  -5
2  -5
3  9
4  49
5  125
6  255
7  445

y1 sim a x^3 + b x1^2 + c x1+d
Transcript text: \[ \begin{tabular}{c|c} $x_{1}$ & $y_{1}$ \\ \hline 1 & -5 \\ \hline 2 & -5 \\ \hline 3 & 9 \\ 4 & 49 \\ 5 & 125 \\ 6 & 255 \\ 7 & 445 \\ \end{tabular} \] \[ y_{1} \sim a x^{3} + b x_{1}^{2} + c x_{1}+d \]
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Solution

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Solution Steps

To solve this problem, we need to find the coefficients \(a\), \(b\), \(c\), and \(d\) for the polynomial equation \(y_1 = ax^3 + bx^2 + cx + d\) that best fits the given data points. This can be done using polynomial regression, which involves setting up a system of equations based on the data points and solving for the coefficients.

Step 1: Construct the Design Matrix

We start by constructing the design matrix \( A \) for a cubic polynomial based on the given data points. The matrix \( A \) is formed as follows:

\[ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \\ 125 & 25 & 5 & 1 \\ 216 & 36 & 6 & 1 \\ 343 & 49 & 7 & 1 \end{bmatrix} \]

Step 2: Solve for Coefficients

Next, we solve the system of equations \( A \cdot \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = y \) using least squares to find the coefficients \( a \), \( b \), \( c \), and \( d \). The resulting coefficients are:

\[ \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} \approx \begin{bmatrix} 5.0 \\ -20.0 \\ 15.0 \\ -5.0 \end{bmatrix} \]

Step 3: Formulate the Polynomial

The polynomial that fits the data points can be expressed as:

\[ y_1 = 5.0 x^3 - 20.0 x^2 + 15.0 x - 5.0 \]

Final Answer

The coefficients of the polynomial are:

  • \( a = 5.0 \)
  • \( b = -20.0 \)
  • \( c = 15.0 \)
  • \( d = -5.0 \)

Thus, the polynomial is:

\[ \boxed{y_1 = 5.0 x^3 - 20.0 x^2 + 15.0 x - 5.0} \]

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