Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 15G / a / Solution

For the Fourier sine series on $(0,1)$, orthogonality givesThereforeThe series is that of the odd, two-periodic extension. At $x=0$ its value is zero, agreeing with $f(0)$; at $x=1$ every summand is zero and the series converges to zero, the average of the two one-sided extension limits. Thus it represents the function in the interior and in the square-integrable sense, not the prescribed value one at the endpoint $x=1$. This endpoint issue matters when using sine expansions with homogeneous boundary conditions.