Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 4

For a nonzero integer $k$, direct integration giveswhereas ${\int }_{-\pi }^{\pi }1dx=2\pi$. The product-to-sum formulastherefore establish Fourier orthogonality: for positive integers $m,n$,Here ${\delta }_{mn}$ is the Kronecker delta. In particular the constant function is orthogonal to every nonconstant sine and cosine in this system. Integrating the given Fourier series, then multiplying it by $\cos (mx)$ or $\sin (mx)$ and integrating, isolates its Fourier coefficients. Using the permitted interchange of the sum and integral givesIf $f$ is an even function, then $f(x)\sin (nx)$ is an odd function, while $f(x)\cos (nx)$ is an even function. ThusThe Fourier series consequently reduces to a Fourier cosine series.

The function $x\sin (px)$ is an even function, so every Fourier sine coefficient is zero. Introduce the auxiliary integralFor a nonzero integer $k$, integration by parts givesConsequentlyFor $n\ge 1$, a product-to-sum formula yieldsIf $n\ne p$, this becomesThe exceptional Fourier coefficient at $n=p$ must be evaluated before dividing by $p-n$:Thus the full Fourier series isThis equality holds on $[-\pi ,\pi ]$ and by periodicity everywhere. Indeed the periodic extension is continuous at the joined endpoints, since both endpoint values are zero, and is piecewise smooth. The theorem on pointwise convergence of a piecewise smooth Fourier series therefore applies, including at the endpoints. The Fourier coefficients also decay as $n^{-2}$, making the resulting series absolutely and uniformly convergent.

Set $p=3$ and $x=0$ in the Fourier series from b. Since the function is continuous there, pointwise convergence of a piecewise smooth Fourier series givesIt follows thatThe $n=1$ and $n=2$ terms are $1/8$ and $-1/5$, respectively. Subtracting them gives the required absolutely convergent series: