Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 12D (Methods)

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.

Writing $y=e^{-x}v$ gives $-v\prime \prime +v=\lambda v$, hence ${\lambda }_n=n^2{\pi }^2+1$ and $y_n=e^{-x}\sin (n\pi x)$. The Sturm–Liouville form is $-(e^{2x}y\prime )\prime =\lambda e^{2x}y$, so ${\int }_0^1{e}^{2x}{y}_n{y}_m=0$ for $n\ne m$. Expanding the source gives coefficients $c_n=2{\int }_0^{1}f\sin (n\pi x)dx$, hence