Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 12D / i / Solution

For the odd two-periodic extensions, $\bar{f}=\sum b_n\sin (n\pi x)$ with $b_n=2[1+(-1{)}^n]/(n\pi )$, while $\bar{F}$ has $b_n=4[1-(-1{)}^n]/(n^3{\pi }^3)$. Thus the first has $O(n^{-1})$ convergence because its periodic extension jumps, while the smoother second extension has $O(n^{-3})$ coefficients.

Solved by gpt-5.6-sol high.