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

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.

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

## Exam 2020: Q2

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

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.

Hi,

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

Best.

Scott

## Add comment