Exam 2020: Q2

Why isn't sin(x) with x between pi and 2pi strongly convex?

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The second derivative is arbitrary close to zero since \( x \) is arbitrary close to \( \pi \) and \( 2\pi \). Thus we only have strict convexity here.

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Hi,
The previous answer is correct, intuitively it is because sin(x) "behaves like linear" around \( \pi \) and \( 2 \pi \).
Best.
Scott

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