Consider Newton's Method. What happens when the gradient is small, and how may you optimise in this regime?

Question

BlackTom AI · Accepted Answer

To approach the question, we first consider what happens to Newton's method when the gradient is very small. In that regime, the division by f'(x) can become unstable, so stabilization strategies are typically used.

Option 1: 'lim f′(x)→0 f(x) f′(x) = 0 - Newton's method is already robust to small gradients and is able to suitably handle optimisation in this regime.' This option correctly notes that the product f(x) f′(x) tends to zero as f′(x) → 0, assuming f(x) stays finite, which is a reasonable description of the limiting behavior. It then suggests buffering by adding a small epsilon to the denominator to perform a safe division, which directly addresses the instability caused by tiny gradients. The proposed update x_{i+1} = x_i − f(x)/f′(x) + ε with ε = 1e−16 captures a common practical fix to stabilize the algorithm.

Option 2: 'lim f′(x)→0 f(x) f′(x) = ±∞ - Newton's method is a global optimisation method, so the code will work stably in this regime with suitable handling of the division. Numerically, we can handle this by buffering (adding a small value to the denominator to perform a safe division). That is: x_{i+1} = x_i − f(x)/f′(x) + ε where ε = 1e−16.' The claim that f(x) f′(x) tends to ±∞ as f′(x) → 0 is incorrect, since typically f′(x) → 0 would drive the product toward zero (assuming f(x) remains bounded). Moreover, Newton's method is not a global optimisation method; it is a local method, so the premise about global robustness is misplaced.

Option 3: 'lim f(x) → 0, f′(x) → 0 f(x) f′(x) = 0 - Newton's method is a local optimisation method, so the code will work stably in this regime because of the function value approaching zero. This limit is always numerically stable, even when crossing stationary points away from the root of f(x).' While f(x) approaching zero and f′(x) approaching zero can occur near a root, saying the limit is always numerically stable is misleading. Stability depends on the gradient term in the update; even if f(x) is small, a vanishing derivative can still cause division issues unless regularization is used. The assertion that the limit guarantees stability ignores the division by a small f′(x).

Option 4: 'lim f′(x)→0 f(x) f′(x) = ± ∞ - Newton's method is a local optimisation method, so for the code to work stably in this regime it should be applied to smoothly differentiable functions around critical points. Numerically, we can handle this by buffering (adding a small value to the denominator to perform a safe division). That is: x_{i+1} = x_i − f(x)/f′(x) + ε where ε = 1e−16.' This option incorrectly states that the product diverges to ±∞ in the small-gradient limit, which contradicts standard analysis. It also correctly points to the buffering strategy, but the core claim about the limit is incorrect, making the option unreliable.

In summary, the most consistent and accurate reasoning aligns with option 1: when the gradient is small, the product f(x) f′(x) tends toward zero, and a practical stabilization approach is to add a tiny epsilon to the denominator in the Newton update to prevent division by near-zero values. The other options contain either incorrect limiting behavior or mischaracterizations of Newton's method as global rather than local, leading to incorrect conclusions about stability in this regime.

Consider Newton's Method. What happens when the gradient is small, and how may you optimise in this regime?单项选择题

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