Exercise 2: The Single-Step-Optimal Learning Rate for (x-5)^2 — Possible Solution
====================================================================

STEP 1: SET UP THE THRESHOLD CONDITION
------------------------------
This chapter's own convergence-threshold formula, |1-2*alpha|,
applies directly here because f(x)=(x-5)^2 has the exact same
"2" coefficient in its own gradient (f'(x)=2(x-5)) as this
chapter's own worked example (which used (x-3)^2, also with a "2"
coefficient) - the threshold formula depends on that coefficient,
not on where the minimum happens to sit.

For the fastest possible (single-step) convergence, this chapter's
own finding was that |1-2*alpha|=0 gives immediate convergence.

STEP 2: SOLVE FOR alpha
------------------------------
1 - 2*alpha = 0
2*alpha = 1
alpha = 0.5

STEP 3: VERIFY BY RUNNING ONE STEP FROM x=0
------------------------------
f'(0) = 2*(0-5) = -10

x_new = 0 - 0.5*(-10) = 0 + 5 = 5

RESULT
------------------------------
alpha=0.5 is the single-step-optimal learning rate for this
function - and running gradient descent with it from x=0 lands
EXACTLY on x=5, the true minimum, confirmed directly, in exactly
one step.

WHY THIS WORKS AS AN ANSWER
------------------------------
The threshold condition is correctly recognized as applying to this
new function (since it shares the same coefficient structure as this
chapter's own worked example, not by coincidence but because it's
the same coefficient), alpha is solved for algebraically rather than
guessed, and the result is independently verified by actually
running the update step and confirming it lands exactly on the true
minimum.