Fit A Quadratic Polynomial To The Data Points (0,27), (1,0), (2,0), (3,0), Using Least Squares. Sketch
Introduction to Quadratic Polynomial Fitting
Fitting a quadratic polynomial to a set of data points is a fundamental task in data analysis and mathematical modeling. When data points are available, and the relationship between variables is believed to be quadratic in nature, the goal is to find a quadratic function that best approximates the data. This process is especially useful in various fields such as physics, engineering, economics, and biological sciences, where modeling the underlying trend of data points helps in predictions, understanding phenomena, and decision-making.
In this article, we will explore how to fit a quadratic polynomial to the specific data points (0, 27), (1, 0), (2, 0), and (3, 0) using the least squares method. We will also discuss how to sketch the resulting quadratic curve and interpret its significance.
Understanding the Data Points
The data points provided are:
- (0, 27)
- (1, 0)
- (2, 0)
- (3, 0)
Visually, these points suggest that the quadratic curve may have a peak or some specific shape. Observing the points, the value at x=0 is quite high (27), while at x=1, 2, and 3, the y-values are zero. This pattern hints at a quadratic function that reaches a maximum somewhere near x=0 and then decreases to zero for larger x-values.
The Quadratic Polynomial Model
A quadratic polynomial generally takes the form:
\[ y = ax^2 + bx + c \]
where:
- \( a \), \( b \), and \( c \) are real coefficients to be determined.
Our goal is to find the values of \( a \), \( b \), and \( c \) such that the polynomial best fits the given data points in the least squares sense.
Least Squares Method for Quadratic Fitting
The least squares method minimizes the sum of the squared residuals between the observed data points and the values predicted by the model. Given the data points \((xi, yi)\), the residual for each point is:
\[ ri = yi - (axi^2 + bxi + c) \]
The total error function \( S \) to minimize is:
\[ S(a, b, c) = \sum{i=1}^{n} ri^2 = \sum{i=1}^{n} [ yi - (axi^2 + bxi + c) ]^2 \]
For our four data points, the sum becomes:
\[ S(a, b, c) = (27 - c)^2 + (0 - (a + b + c))^2 + (0 - (4a + 2b + c))^2 + (0 - (9a + 3b + c))^2 \]
Our task is to find \( a \), \( b \), and \( c \) that minimize \( S \).
Formulating the Normal Equations
To minimize \( S \), we take partial derivatives with respect to \( a \), \( b \), and \( c \) and set them to zero:
\[
\frac{\partial S}{\partial a} = 0, \quad \frac{\partial S}{\partial b} = 0, \quad \frac{\partial S}{\partial c} = 0
\]
This results in a system of three equations known as the normal equations:
\[
\begin{cases}
\sum xi^4 \cdot a + \sum xi^3 \cdot b + \sum xi^2 \cdot c = \sum yi x_i^2 \\
\sum xi^3 \cdot a + \sum xi^2 \cdot b + \sum xi \cdot c = \sum yi x_i \\
\sum xi^2 \cdot a + \sum xi \cdot b + n \cdot c = \sum y_i
\end{cases}
\]
where \( n \) is the number of data points.
Let's compute these sums explicitly.
Calculating Sums for the Data Points
| \( xi \) | \( yi \) | \( xi^2 \) | \( xi^3 \) | \( xi^4 \) | \( yi xi \) | \( yi x_i^2 \) |
|------------|--------------|--------------|--------------|--------------|--------------|----------------|
| 0 | 27 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 | 0 |
| 2 | 0 | 4 | 8 | 16 | 0 | 0 |
| 3 | 0 | 9 | 27 | 81 | 0 | 0 |
Calculate the sums:
- \( \sum x_i^2 = 0 + 1 + 4 + 9 = 14 \)
- \( \sum x_i^3 = 0 + 1 + 8 + 27 = 36 \)
- \( \sum x_i^4 = 0 + 1 + 16 + 81 = 98 \)
- \( \sum y_i = 27 + 0 + 0 + 0 = 27 \)
- \( \sum yi xi = 27 \times 0 + 0 \times 1 + 0 \times 2 + 0 \times 3 = 0 \)
- \( \sum yi xi^2 = 27 \times 0 + 0 \times 1 + 0 \times 4 + 0 \times 9 = 0 \)
Now, the normal equations are:
\[
\begin{cases}
98a + 36b + 14c = 0 \\
36a + 14b + 4c = 0 \\
14a + 4b + 4c = 27
\end{cases}
\]
Solving the Normal Equations
Let's proceed to solve the system:
- Equation 1: \( 98a + 36b + 14c = 0 \)
- Equation 2: \( 36a + 14b + 4c = 0 \)
- Equation 3: \( 14a + 4b + 4c = 27 \)
---
Step 1: Express \( c \) from Equation 3:
\[
14a + 4b + 4c = 27 \implies 4c = 27 - 14a - 4b \implies c = \frac{27 - 14a - 4b}{4}
\]
Step 2: Substitute \( c \) into Equations 1 and 2.
- Substituting into Equation 2:
\[
36a + 14b + 4 \times \frac{27 - 14a - 4b}{4} = 0
\]
Simplify:
\[
36a + 14b + 27 - 14a - 4b = 0
\]
Combine like terms:
\[
(36a - 14a) + (14b - 4b) + 27 = 0 \implies 22a + 10b + 27 = 0
\]
This simplifies to:
\[
22a + 10b = -27 \quad \text{(Equation 4)}
\]
- Substituting \( c \) into Equation 1:
\[
98a + 36b + 14 \times \frac{27 - 14a - 4b}{4} = 0
\]
Simplify:
\[
98a + 36b + \frac{14 \times 27 - 14 \times 14a - 14 \times 4b}{4} = 0
\]
Calculate numerator:
\[
14 \times 27 = 378, \quad 14 \times 14a = 196a, \quad 14 \times 4b = 56b
\]
So:
\[
98a + 36b + \frac{378 - 196a - 56b}{4} = 0
\]
Express as:
\[
98a + 36b + \frac{378}{4} - \frac{196a}{4} - \frac{56b}{4} = 0
\]
Simplify fractions:
\[
98a + 36b + 94.5 - 49a - 14b = 0
\]
Combine like terms:
\[
(98a - 49a) + (36b - 14b) + 94.5 = 0 \implies
