𝑔
(
𝑤
)
=
−
𝑐
𝑜
𝑠
(
2
𝜋
𝑤
𝑇
𝑤
)
+
𝑤
𝑇
𝑤
, where
𝑤
∈
𝑅
𝑁
. Is
𝑔
(
𝑤
)
convex?

Question

𝑔
(
𝑤
)
=
−
𝑐
𝑜
𝑠
(
2
𝜋
𝑤
𝑇
𝑤
)
+
𝑤
𝑇
𝑤
, where 
𝑤
∈
𝑅
𝑁
. Is 
𝑔
(
𝑤
)
 convex?

BlackTom AI · Accepted Answer

Let me restate the problem in my own words and list the options clearly so we can analyze them one by one.
Question: g(w) = − c cos(2π w^T w) + w^T w, where w ∈ R^N. Is g(w) convex?
Answer options:
1) No
2) Yes
3) Not enough information is provided to answer the question.

Now, evaluate each option with careful reasoning.
Option 1: No
This option claims that g is not a convex function for all admissible values of the parameters. The convexity of g depends on the interplay between the quadratic term w^T w (which is convex in w) and the oscillatory term − c cos(2π w^T w). If we denote t = w^T w, then g(w) = f(t) with f(t) = t − c cos(2π t). For convexity in t, f''(t) = c (2π)^2 cos(2π t). This second derivative oscillates in sign as t varies (cos(2π t) ranges between −1 and 1). Therefore f(t) is not convex for all t whenever c ≠ 0, because f''(t) can be negative for some t. Since t = w^T w is a nonnegative scalar that can take a continuum of values as w varies, the composed function g(w) inherits this nonconvex behavior in w. Thus, unless c = 0 (a trivial, purely quadratic case), g(w) fails to be convex over w. This reasoning supports the claim that g is not convex in general. However, if c were exactly zero, g would reduce to w^T w, which is convex, but the statement does not specify c’s value. Still, with the usual assumption c > 0 in many contexts, the nonconvex cosine term disrupts convexity, making this option plausible.
Option 2: Yes
This option asserts convexity. The presence of − c cos(2π w^T w) introduces an oscillatory component that depends on w only through w^T w. The cosine term can create local curvature changes (both convex and concave regions) as t = w^T w varies, which is incompatible with global convexity in w. The second-derivative analysis in t shows that f''(t) = c (2π)^2 cos(2π t) changes sign with t, so f is not globally convex in t, hence not convex when composed with t = w^T w to yield g(w). Therefore, this option is inconsistent with the typical mathematical behavior of g unless c = 0. It is generally incorrect under the common assumption c ≠ 0.
Option 3: Not enough information is provided to answer the question.
This cautionary option would be chosen if one believed that the sign or value of c is unspecified, making it impossible to determine convexity. Indeed, the convexity conclusion can depend on the value of c (e.g., if c = 0, g is convex; if c ≠ 0, g is not convex due to the oscillatory term). If the problem statement truly leaves c completely unspecified, one might argue that we cannot categorically declare convexity or nonconvexity for all possible c. However, in many contexts, questions assume a nonzero c and seek the general behavior; under that typical interpretation, the conclusion about nonconvexity holds. If the intention is to allow any c, then this option would be ethically appropriate to consider, but you would still need to specify the range or sign of c to make a precise claim.

In summary, the reasoning above shows that the convexity status hinges on the role of the oscillatory cosine term through t = w^T w. The presence of a nonconstant oscillatory component typically prevents g from being globally convex for nonzero c, while the purely quadratic case (c = 0) would be convex. The most commonly accepted stance for c > 0 is that g is not convex; if c is exactly zero, the convexity would hold. Therefore, among the presented options, the claim that g is not convex aligns with the standard interpretation when c ≠ 0, while the “Not enough information” option is only strictly correct if one insists on confirming the exact value of c is unspecified and must be universal for all c. Nevertheless, the mathematical intuition leans toward the nonconvex conclusion for typical nonzero c values.

25S-STATS-102B-LEC-3 S25 Midterm Exam- Requires Respondus LockDown Browser

𝑔 ( 𝑤 ) = − 𝑐 𝑜 𝑠 ( 2 𝜋 𝑤 𝑇 𝑤 ) + 𝑤 𝑇 𝑤 , where 𝑤 ∈ 𝑅 𝑁 . Is 𝑔 ( 𝑤 ) convex?

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类似问题

Suppose 𝑔 ( 𝑤 | 𝑦 𝑖 , 𝑥 𝑖 ) = ( 𝑦 𝑖 − 𝑥 𝑖 𝑇 𝑤 ) 2 , where 𝑤 ∈ 𝑅 𝑝 , 𝑦 𝑖 ∈ 𝑅 , and 𝑥 𝑖 ∈ 𝑅 𝑝 , for 𝑖 = 1 … 𝑛 . Is ∑ 𝑖 = 1 𝑛 𝑔 ( 𝑤 | 𝑦 𝑖 , 𝑥 𝑖 ) convex?

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