Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 18 / b

Take the inner product of the Fourier series with $\sin (n\pi x/L)$. Fourier orthogonality eliminates the constant, all cosine terms, and all sine terms except the $n$th. The remaining squared norm issoFor a square-integrable function, the calculation is justified by taking inner products of finite Fourier partial sums and passing to their mean-square limit.

The analogous orthogonal projection onto the cosine terms givesFor $n=0$, this means $a_0=L^{-1}{\int }_{-L}^{L}f(x)dx$; the actual constant in the Fourier series is $a_0/2$ because the constant function has squared norm $2L$.

Let $S_N$ be the Fourier partial sum through frequency $N$. By orthogonality and the coefficient formulas,Expanding the nonnegative squared L2 norm of the error gives the exact finite identityFor $f$ in the L2 space, completeness of the Fourier basis makes the left-hand side tend to zero. Dividing the resulting equality by $2L$ proves Parseval's identity:This argument uses convergence in L2 rather than an unqualified exchange of an infinite sum with the integral of its square. The function in part (c) is continuous and piecewise polynomial, so these hypotheses hold.