Math Problem Statement

In this question, you will find the least-squares line, y = Ax + B through the points (0, −4), (1, 4), (4, 5) and (5, 10).

The error (sum of squared errors) is a function of A and B. State this function. (Give an answer in terms of A and B.) E =
Correct: Your answer is correct.

We want to find the values of A and B that minimize this function. As usual, we do this by calculating partial derivatives and then solving the simultaneous equations that result from setting both derivatives equal to zero.

Calculate the partial derivatives. (Give answers in terms of A and B.) ∂E ∂A

Correct: Your answer is correct.

∂E ∂B

Correct: Your answer is correct.

Set both partial derivatives equal to zero. Rearrange the equation ∂E ∂A = 0 to give A in terms of B. A =
Incorrect: Your answer is incorrect.

Now, substitute the above into the equation ∂E ∂B = 0 and solve for B. (Give an exact numerical answer with no variables. Exact answer means give your answer as a fraction or a terminating decimal, DO NOT round.) B =
Incorrect: Your answer is incorrect.

Finally, find the value of A. (Give an exact numerical answer with no variables. Exact answer means give your answer as a fraction or a terminating decimal, DO NOT round.)

Solution

Step 1: Define the Error Function

We want to find the least-squares line y=Ax+By = Ax + B through the points (0,−4)(0, -4), (1,4)(1, 4), (4,5)(4, 5), and (5,10)(5, 10).

The error function is the sum of squared differences between the actual yy-values and the predicted yy-values from the line y=Ax+By = Ax + B:

E(A,B)=∑(yi−(Axi+B))2E(A, B) = \sum \left( y_i - (Ax_i + B) \right)^2 Given the points, this becomes: E(A,B)=(−4−(A(0)+B))2+(4−(A(1)+B))2+(5−(A(4)+B))2+(10−(A(5)+B))2E(A, B) = \left( -4 - (A(0) + B) \right)^2 + \left( 4 - (A(1) + B) \right)^2 + \left( 5 - (A(4) + B) \right)^2 + \left( 10 - (A(5) + B) \right)^2 Simplifying: E(A,B)=(−4−B)2+(4−(A+B))2+(5−(4A+B))2+(10−(5A+B))2E(A, B) = (-4 - B)^2 + (4 - (A + B))^2 + (5 - (4A + B))^2 + (10 - (5A + B))^2

Step 2: Calculate the Partial Derivatives

To minimize the error, we calculate the partial derivatives of E(A,B)E(A, B) with respect to AA and BB.

  1. Partial derivative with respect to AA: ∂E∂A=2⋅(4−(A+B))⋅(−1)+2⋅(5−(4A+B))⋅(−4)+2⋅(10−(5A+B))⋅(−5)\frac{\partial E}{\partial A} = 2 \cdot (4 - (A + B)) \cdot (-1) + 2 \cdot (5 - (4A + B)) \cdot (-4) + 2 \cdot (10 - (5A + B)) \cdot (-5) Simplify: ∂E∂A=−2(4−A−B)−8(5−4A−B)−10(10−5A−B)\frac{\partial E}{\partial A} = -2(4 - A - B) - 8(5 - 4A - B) - 10(10 - 5A - B) Expanding and combining: ∂E∂A=−2(4−A−B)−8(5−4A−B)−10(10−5A−B)\frac{\partial E}{\partial A} = -2(4 - A - B) - 8(5 - 4A - B) - 10(10 - 5A - B)

  2. Partial

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Math Problem Analysis

Mathematical Concepts

Statistics
Linear Regression
Calculus

Formulas

Least-squares line equation: y = Ax + B
Sum of squared errors: E(A, B) = Σ(y_i - (Ax_i + B))^2
Partial derivatives of the error function

Theorems

Least Squares Method

Suitable Grade Level

Undergraduate level

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