Questions
ENGG1810 ENGG9810 (ND) Week 13 Quiz
Single choice
Consider Newton's Method. What happens when the gradient is small, and how may you optimise in this regime?
Options
A.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.
B.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
=
1
e
−
16
.
C.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
)
.
D.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
=
1
e
−
16
.
View Explanation
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Step-by-Step Analysis
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 addre......Login to view full explanationLog in for full answers
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