Exercise 1: MSE Loss at m=2, b=1 — Possible Solution
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GIVEN
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Data: (1,3), (2,5), (3,7), (4,8), (5,11)
m=2, b=1

STEP 1: COMPUTE THE PREDICTION AND SQUARED ERROR FOR EACH POINT
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x=1, y=3: prediction = 2*1+1 = 3. Error^2 = (3-3)^2 = 0
x=2, y=5: prediction = 2*2+1 = 5. Error^2 = (5-5)^2 = 0
x=3, y=7: prediction = 2*3+1 = 7. Error^2 = (7-7)^2 = 0
x=4, y=8: prediction = 2*4+1 = 9. Error^2 = (9-8)^2 = 1
x=5, y=11: prediction = 2*5+1 = 11. Error^2 = (11-11)^2 = 0

STEP 2: AVERAGE THE SQUARED ERRORS
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Sum of squared errors: 0+0+0+1+0 = 1
Number of points: 5

L(2,1) = 1/5 = 0.2

RESULT
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L(m=2, b=1) = 0.2

Interestingly, m=2, b=1 fits four of the five points EXACTLY -
only the point (4,8) is off, predicted as 9 instead of 8 - a single
error of 1 accounts for the entire loss.

WHY THIS WORKS AS AN ANSWER
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Each data point's own prediction and squared error is computed
individually and shown explicitly, rather than only the final
averaged result, making clear exactly which point contributes to
the loss and which points are already fit perfectly.