Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 18

Assume $f\in L^2[-L,L]$, so the squared-error integral is finite. WriteAt a minimum, differentiating with respect to each coefficient givesBy Fourier orthogonality, the constant has squared norm $2L$, each nonconstant cosine has squared norm $L$, and all distinct modes are orthogonal. ThusThese are exactly the Fourier coefficients of the Fourier cosine series of the even function. To check that the stationary choice is a unique global minimum, perturb the coefficients by $\delta a_n$. The residual is orthogonal to every fitted mode, so the cross terms vanish andwith equality only when every perturbation is zero. The truncated series is the orthogonal projection onto a finite-dimensional subspace spanned by the constant and the first $N$ cosine modes.

Evenness makes every sine coefficient zero. For the constant coefficient,An integration by parts gives, for $n\ge 1$,The even coefficients vanish and the odd coefficients are $-4L/(n^2{\pi }^2)$. ThereforeThe periodic function is continuous and piecewise smooth, so the Dirichlet-Jordan convergence theorem identifies the Fourier series with $f$ at every point, including the endpoints. The absolute summability of the coefficients also gives uniform convergence.

Set $x=0$ in the convergent Fourier series from part (b). Since $f(0)=0$ and every cosine equals one,Rearranging givesConvergence at the evaluation point is essential when extracting a numerical series identity from a Fourier series. Here the function is continuous there and the coefficients are absolutely summable.

For a real $2L$-periodic square-integrable function with the convention $f\sim a_0/2+{\sum }_{n\ge 1}(a_n\cos (n\pi x/L)+b_n\sin (n\pi x/L))$, Parseval's identity isIt states that the squared L2 norm equals the sum of the squared components in the orthogonal Fourier basis, with the normalization factors shown. In the present even-function case every $b_n$ vanishes.

Here $f(x{)}^2=x^2$, soSubstituting the coefficients from part (b) into Parseval's identity givesHence the sum over all positive odd integers is ${\pi }^4/96$. The requested series starts at $n=1$, whose first denominator is $3^4$, so it excludes the contribution $1^{-4}$: