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Asked by: gvr
Created at: 2025-07-30 07:20:17
Final Answer:
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Let's solve part (b) **step by step**:
### **Given**
We want to minimize the error function:
\[
E(m, b) = \sum_{i=1}^n (y_i - mx_i - b)^2
\]
with respect to \(m\) and \(b\).
We need to find the critical points, i.e., where:
\[
\frac{\partial E}{\partial m} = 0 \quad \text{and} \quad \frac{\partial E}{\partial b} = 0
\]
---
## **Step 1: Compute Partial Derivatives**
### **Partial with respect to \(m\):**
\[
\frac{\partial E}{\partial m} = \frac{\partial}{\partial m} \sum_{i=1}^n (y_i - mx_i - b)^2
\]
Using the chain rule:
\[
= \sum_{i=1}^n 2(y_i - mx_i - b)(-x_i)
\]
\[
= -2 \sum_{i=1}^n x_i (y_i - mx_i - b)
\]
Set this to zero:
\[
-2 \sum_{i=1}^n x_i (y_i - mx_i - b) = 0
\]
\[
\sum_{i=1}^n x_i (y_i - mx_i - b) = 0
\]
---
### **Partial with respect to \(b\):**
\[
\frac{\partial E}{\partial b} = \frac{\partial}{\partial b} \sum_{i=1}^n (y_i - mx_i - b)^2
\]
Using the chain rule:
\[
= \sum_{i=1}^n 2(y_i - mx_i - b)(-1)
\]
\[
= -2 \sum_{i=1}^n (y_i - mx_i - b)
\]
Set this to zero:
\[
-2 \sum_{i=1}^n (y_i - mx_i - b) = 0
\]
\[
\sum_{i=1}^n (y_i - mx_i - b) = 0
\]
---
## **Step 2: Expand and Rearrange Equations**
### **From \(\frac{\partial E}{\partial m} = 0\):**
\[
\sum_{i=1}^n x_i y_i - m \sum_{i=1}^n x_i^2 - b \sum_{i=1}^n x_i = 0
\]
\[
m \sum_{i=1}^n x_i^2 + b \sum_{i=1}^n x_i = \sum_{i=1}^n x_i y_i \tag{1}
\]
---
### **From \(\frac{\partial E}{\partial b} = 0\):**
\[
\sum_{i=1}^n y_i - m \sum_{i=1}^n x_i - b \sum_{i=1}^n 1 = 0
\]
But \(\sum_{i=1}^n 1 = n\):
\[
m \sum_{i=1}^n x_i + b n = \sum_{i=1}^n y_i \tag{2}
\]
---
## **Final System of Equations (as required):**
\[
\boxed{
\begin{aligned}
& m \left(\sum_{i=1}^n x_i^2\right) + b \left( \sum_{i=1}^n x_i \right) = \sum_{i=1}^n x_i y_i \\
& m \left(\sum_{i=1}^n x_i \right) + b n = \sum_{i=1}^n y_i
\end{aligned}
}
\]
---
### **Explanation**
- These are two linear equations in the two unknowns, \(m\) and \(b\). They can be solved using algebra (substitution, elimination, or matrix methods).
- The solution to this system gives the slope and intercept of the **best-fit line** (in the least squares sense).
### **Final Answer**
\[
\boxed{
\begin{aligned}
& m \left(\sum_{i=1}^n x_i^2\right) + b \left( \sum_{i=1}^n x_i \right) = \sum_{i=1}^n x_i y_i \\
& m \left(\sum_{i=1}^n x_i \right) + b n = \sum_{i=1}^n y_i
\end{aligned}
}
\]
These are the normal equations for least squares linear regression.
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