Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 4 / 17D (Methods)

The complex Fourier series of a $2L$-periodic function isThe coefficient formula follows from orthogonality of the exponential functions. For an arbitrary square-integrable function, this representation means convergence in the $L^2$ norm. With appropriate additional regularity it also gives pointwise convergence, to the mean of the one-sided limits at a jump. No pointwise representation is asserted for an unrestricted function merely from periodicity.

The Parseval identity for a complex Fourier series isIt holds for every square-integrable function, with complex conjugation expressed by the absolute squares. It is the Pythagorean theorem extended to the complete orthonormal system of complex exponentials.

Integrate the two exponentials in $\cos (\alpha x)$ against $e^{-inx}$. For $\alpha \notin \mathbb{Z}$ there are no zero denominators, andThus the complex Fourier series isThis is the periodic extension, not a globally nonperiodic cosine formula. For each fixed complex $\alpha$, the coefficients are $O(n^{-2})$, giving absolute and uniform convergence. The extension is continuous at the endpoints because the two endpoint values agree; piecewise smooth pointwise convergence of a piecewise smooth Fourier series identifies the sum with it, including those endpoints.

Evaluate the absolutely convergent complex Fourier series at $x=\pi$. The factors $(-1{)}^n{e}^{in\pi }$ equal one, and the $n=0$ term is $1/{\alpha }^2$. Pairing positive and negative integers givesSince $\sin (\pi \alpha )\ne 0$ for nonintegral complex $\alpha$, rearranging provesThe last term means $\pi \cot (\pi \alpha )/(2\alpha )$, so half-integer arguments are included by the finite cotangent value zero.

For real nonintegral $\alpha$, the Parseval identity and the coefficient formula giveSolving for the sum yieldsAt $\alpha =5/2$, the sine square is one and the cotangent is zero, giving $2{\pi }^2/25$.