Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2024 / Paper 2 / A3 / c / Solution

Use reproducible random starts and retain the best successful result:

rng = np.random.default_rng(2024)
starts = rng.uniform(-4, 4, size=100)
results = [opt.minimize(lambda z: f(z[0]), x0=[x0]) for x0 in starts]
successful = [result for result in results if result.success]
best = min(successful, key=lambda result: result.fun)
print(best.x[0], best.fun)

The estimate is $x_{m i n} ≃ - 0.309213$ , $f_{m i n} ≃ 0.0192409$ . This random-restart optimization improves coverage of the different attraction regions, but their number alone does not guarantee global optimality.

Here an analytic check is possible. For

∣ x ∣ \geq π /5

, the envelope gives

f (x) \geq 1 - e^{- π^{2} /125} ≃ 0.0759

, above the candidate. On

[0, π /5]

,

f \geq 1

. In the remaining interval

[- π /5, 0]

, the derivative condition is equivalent, with

z = - x

, to

cot (5 z) = 0.08 z

. The left side decreases strictly from

+ \infty

to

- \infty

while the right side increases, giving one minimum. This certifies the selected well as global rather than relying only on the restart experiment.

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