Consider the following data points.
(1, 8)
(2, 4)
(4, 3)
Use partial derivatives to obtain the formula for the best least-squares fit to the data points.


To find the best least-squares fit to the given data points using partial derivatives, we will fit the data to a linear model of the form:

\[
y = ax + b
\]

where \( a \) and \( b \) are constants to be determined.

### Step 1: Define the Error Function
The error (or residual) for each data point is the difference between the actual \( y \)-value and the predicted value from the linear model. For a set of data points \( (x_1, y_1), (x_2, y_2), \dots, (x_n, y_n) \), the sum of squared errors (SSE) is:

\[
S(a, b) = \sum_{i=1}^{n} \left( y_i - (ax_i + b) \right)^2
\]

For the given data points \( (1, 8), (2, 4), (4, 3) \), we have \( n = 3 \), and the sum of squared errors becomes:

\[
S(a, b) = (8 - (a \cdot 1 + b))^2 + (4 - (a \cdot 2 + b))^2 + (3 - (a \cdot 4 + b))^2
\]

### Step 2: Minimize the Error Function
To minimize the error function, we take partial derivatives of \( S(a, b) \) with respect to \( a \) and \( b \), and set them equal to zero.

#### Partial Derivative with respect to \( a \):

\[
\frac{\partial S}{\partial a} = -2(8 - (a \cdot 1 + b)) \cdot 1 - 2(4 - (a \cdot 2 + b)) \cdot 2 - 2(3 - (a \cdot 4 + b)) \cdot 4
\]

Simplifying:

\[
\frac{\partial S}{\partial a} = -2 \left( (8 - a - b) + 2(4 - 2a - b) + 4(3 - 4a - b) \right)
\]
\[
\frac{\partial S}{\partial a} = -2 \left( 8 - a - b + 8 - 4a - 2b + 12 - 16a - 4b \right)
\]
\[
\frac{\partial S}{\partial a} = -2 \left( 28 - 21a - 7b \right) = 42a + 14b - 56
\]

#### Partial Derivative with respect to \( b \):

\[
\frac{\partial S}{\partial b} = -2(8 - (a \cdot 1 + b)) - 2(4 - (a \cdot 2 + b)) - 2(3 - (a \cdot 4 + b))
\]

Simplifying:

\[
\frac{\partial S}{\partial b} = -2 \left( (8 - a - b) + (4 - 2a - b) + (3 - 4a - b) \right)
\]
\[
\frac{\partial S}{\partial b} = -2 \left( 8 - a - b + 4 - 2a - b + 3 - 4a - b \right)
\]
\[
\frac{\partial S}{\partial b} = -2 \left( 15 - 7a - 3b \right) = 14a + 6b - 30
\]

### Step 3: Solve the System of Equations
Now, we solve the system of equations obtained from the partial derivatives:

1. \( 42a + 14b = 56 \)
2. \( 14a + 6b = 30 \)

We can solve this system to find \( a \) and \( b \).

#### Solving the system:

From the second equation, solve for \( b \) in terms of \( a \):

\[
14a + 6b = 30 \quad \Rightarrow \quad 6b = 30 - 14a \quad \Rightarrow \quad b = \frac{30 - 14a}{6} = 5 - \frac{7a}{3}
\]

Substitute this into the first equation:

\[
42a + 14\left( 5 - \frac{7a}{3} \right) = 56
\]
\[
42a + 70 - \frac{98a}{3} = 56
\]
Multiply through by 3 to eliminate the fraction:

\[
3(42a + 70) - 98a = 3 \cdot 56
\]
\[
126a + 210 - 98a = 168
\]
\[
28a = -42
\]
\[
a = -\frac{42}{28} = -1.5
\]

Now substitute \( a = -1.5 \) into the expression for \( b \):

\[
b = 5 - \frac{7(-1.5)}{3} = 5 + \frac{10.5}{3} = 5 + 3.5 = 8.5
\]

### Final Equation
The best least-squares fit for the data points is:

\[
y = -1.5x + 8.5
\]

Would you like more details on any of the steps? 

Here are some questions to consider:
1. How would the least-squares fit change with more data points?
2. What are the assumptions behind using a linear model for the fit?
3. Could we use a non-linear model, and how would we compute that?
4. How do we assess the goodness of the fit?
5. How do we generalize this method for fitting higher-order polynomials?

**Tip**: The least-squares method minimizes the sum of squared differences between the observed and predicted values, ensuring the best overall fit.

Consider the following data points.
(1, 8)
(2, 4)
(4, 3)
Use partial derivatives to obtain the formula for the best least-squares fit to the data points.

Learn how to find the least-squares fit for the data points (1, 8), (2, 4), (4, 3) using partial derivatives. This solution involves minimizing the sum of squared errors by solving a system of equations for the best linear fit. Suitable for university-level students studying calculus and linear algebra.

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