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.